Grade 10 Math by infinity-learn

IL FOUNDATION SERIES

MATHEMATICS

A Reliable Companion for JEE | NEET | Olympiads

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Contents Module 1 1.

Real Numbers

2. Polynomials

3. Pair of Linear Equations in Two Variables

4. Quadratic Equations

5. Inequations

125

6. Arithmetic and Geometric Progressions

143

7. Triangles

168

8. Coordinate Geometry

201

9. Trigonometry

238

1.1

REAL NUMBERS

INTRODUCTION TO REAL NUMBERS

Natural Numbers: Natural numbers are counting numbers denoted by ’N’. N = {1, 2, 3, 4, 5...} Whole numbers: Whole numbers are a set of natural numbers, including zero, and denoted by ’W’. W = {0, 1, 2, 3...} Integers: Integers are sets of natural numbers, their negatives, and zero, denoted by ’I’ or ’Z’. Z = {…− 3,−2,−1, 0, 1, 2, 3…}

p , where p and q are both integers (having no q common factor) and q ≠ 0, which is called a rational number and is denoted by ’Q’. Rational Numbers: A number in the form of

∴Q =

p ; p, q z and q ≠ 0 q

p If p and q have no factor in common, then defines a rational number in its lowest terms. For q 12 3 example 4 is a rational number in its lowest terms, whereas 16 is not. Note: Every rational number, when expressed in decimal form, is either a terminating decimal or a non-terminating repeating (or recurring) decimal. The converse is also true. Example: 1 = 0.5 (Terminating decimal) 2 15 = 5 (Terminating decimal) 3 1 = 0.333... (Non-terminating repeating decimal) = 0.3. 3 Rule 1: A rational number has a terminating decimal representation if its denominator , when written in its standard form, has 2 or 5 or both as factors and has no other factors. Rule 2: If the denominator in standard form has some factor other than 2 and 5, then the rational number is a repeating decimal or non-terminating recurring decimal. Rule 3: Between any two different rational numbers, a and b, there exists another rational number a+b a+b 2 i.e., a < 2 < b

p Irrational numbers: A number which cannot be written in the form of , where p and q are q integers, and q ! 0 , is called an irrational number. 1

REAL NUMBERS

(or) A number whose decimal expansion is non-terminating and non-recurring is called an irrational number. 3

Example: 2, 3, 5, and 3 are irrational numbers. Real Numbers: The set of rational numbers and irrational numbers form a set of real numbers and it is denoted by ’R’. Structure of real number system: Real numbers (R)

Rational numbers (Q)

Irrational numbers (Q’)

p Fractions of the type q

Integers (I)

(Where q ≠ 0, q ≠ 1,G.C.D of p,q is 1)

Natural numbers (N)

1.2

Whole numbers (W)

THE FUNDAMENTAL THEOREM OF ARITHMETIC

"Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique apart from the order in which the prime factors occur". Example: For the composite number 250, 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 21 . 53 Note 1: The prime factorisation of 250 can be shown as a factor tree, 250 125

2 5

25 1

250 = 2 . 5

Note 2: Every natural number (>1), N, can be expressed as, n

1 2 3 k i) N = P1 × P2 × P3 f Pk where P1, P2, f, PK are distinct primes and n1, n2, ..., nk N. n n n n n n ii) N = 2 1 ⋅ 3 2 ⋅ 5 3 … where N = 2 1 ⋅ 3 2 ⋅ 5 3 …

IL Foundation Series Class 10

1.2.1 H.C.F. and L.C.M. by prime factorisation method Given two positive integers, M and N, with prime factorisations, m

M = 2 1 $ 3 2 $ 5 3f n n n N = 2 1 $ 3 2 $ 5 3f min # m1, n1 -

Then H.C.F. (M, N) = 2

max # m1, n1 -

and L.C.M. (M, N) = 2

min # m2, n2 -

max # m2, n2 -

min # m3, n3 -

max # m3, n3 -

Note : This concept can similarly be extended for three or more positive integers. For any two numbers a and b, we can write Product of two numbers = (H.C.F. of two numbers) × (L.C.M. of two numbers) Product (a,b) = H.C.F. (a,b) × L.C.M. (a,b) Example: Given numbers are 12, 15 and 21 Applying prime factorisation, we get 12 = 22 . 31 . 50 . 70 15 = 20 . 31 . 51 . 70 21 = 20 . 31 . 50 . 71 H.C.F. (12, 15, 21) = 20 . 31 . 50 . 70 = 3 and L.C.M. (12, 15, 21) = 22 . 31 . 51 . 71 = 420

SOLVED EXAMPLES Example 1: Find the H.C.F. of 96 and 404 by the prime factorisation method. Hence find their L.C.M. Solution: By prime factorisation: 96 = 25 . 31 404 = 22 . 1011 = 22 . 30 . 1011 H.C.F. = 22 . 30 . 1010 = 4 (96 × 404) H.C.F. (96, 404) (96 × 404) = 4 = 9696

Also, L.C.M. (96, 404) =

REAL NUMBERS

Example 2 : Find the H.C.F. and L.C.M. of 6, 72 and 120 using the prime factorisation method. Solution: By Prime factorisation 6 = 21 . 31 . 50 72 = 23 . 32 . 50 120 = 23 . 31 . 51 H.C.F. = 21 . 31 . 50 = 6 L.C.M. = 23 . 32 . 51 = 360 Example 3: Factorize 5005 using prime factorisation Solution: 5005 = 5 × 7 × 11 × 13 5 7 11 13

5005 1001 143 13 1

Example 4: Find H.C.F. and L.C.M. of 26 and 91 and hence verify that H.C.F. × L.C.M. = Product of two numbers. Solution: By prime factorisation, 26 = 21 . 70 . 131 91 = 20 . 71 . 131 H.C.F. = 20 . 70 . 131 = 13 L.C.M. = 21 . 71 . 131 = 182 Verification : H.C.F. × L.C.M. = Product of two numbers On left-hand side H.C.F. × L.C.M. = 13 × 182 = 2366 On right-hand side Product of two numbers = 26 × 91 = 2366

IL Foundation Series Class 10

Since the left-hand side is equal to the right-hand side H.C.F. × L.C.M. = Product of two numbers is true. Example 5: Check whether 6n can end with the digit ’0’ for any natural number ’n’. Solution: We know that any positive integer ending with the digit ’0’ is divisible by 10 and hence divisible by 2 and 5. So, prime factorisation must contain 2 and 5 as prime factors. But, given number = 6n = (2 × 3)n = 2n × 3n 2 and 3 are the only prime factors of 6n. Hence, 5 is not a prime factor of 6n due to the uniqueness of the fundamental theorem of arithmetic. Hence, 6n does not end with the digit ’0’ for any natural number n.

1.3

REVISITING IRRATIONAL NUMBERS

p Rational Number: A number which can be expressed in the form of where p, q are integers and q q ≠ 0 and (p, q are coprimes) are called rational numbers. 3 1 -4 1 Examples: , 4 7 , , 1 , - 0.12, 32.1737373..., etc. 5 2 2

p Irrational Number: A number which cannot be expressed in the form of where p and q are inteq gers, and q ≠ 0 is called an irrational number. (or) A real number which is not rational is called an irrational number. (or) A number whose decimal expansion is non-terminating and non-recurring is called irrational number. Examples: 2, 3, 5, 6, p, etc.

REAL NUMBERS

1.4

PROVING OF IRRATIONAL NUMBERS

Theorem: Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer. Proof: Let the prime factorisation of a be as follows: a = p1, p2,..., pn, where p1, p2,..., pn, are primes, not necessarily distinct. Therefore, a2 = (p1 p2 ... pn) (p1 p2 ... pn) = p12 p22 ... pn2. Now, we are given that p divides a2. Therefore, from the Fundamental theorem of arithmetic, it follows that p is one of the prime factors of a2. However, using the uniqueness part of the Fundamental theorem of arithmetic, we realise that the only prime factors of a2 are p1, p2 ,..., pn. So p is one of p1, p2 ,..., pn. Now, since a = p1, p2 ,..., pn, p divides a.

SOLVED EXAMPLES Example 1: Prove that 2 is irrational. Solution: Suppose 2 is rational. By definition of rational number, 2 = a where a, b are integers, b ≠ 0 and a, b are coprime. b a 2 2 2 = b 2 a 2= 2 b 2 a = 2 . b2 ...(1) 2 divides a2 2 divides a ( 2 is prime) ...(2) a = 2c where c is some integer a2 = (2c)2 2b2 = 4c2 ( by (1)) b2 = 2c2 2 divides b2 2 divides b ( 2 is prime) ...(3)

IL Foundation Series Class 10

from (2) & (3), we get, 2 is a common factor of a, b a and b are not coprimes. This is a contradiction. 2 is an irrational number. Example 2: Prove that 5 is an irrational number by contradiction method. Solution:

p Suppose 5 represents a rational number. Then 5 can be expressed in the form , q where p and q are integers, and have no common factor, q ≠ 0. p q Squaring both sides, we get 5=

p2 p2 = 5q2 q2 divides p2 5=

...(1)

5 divides p.

...(2)

Let p = 5m

p2 = 25m2

Putting the value of p2 in (1), we get 25m2 = 5q2 5m2 = q2 5 divides q2 5 divides q

...(3)

Thus, from (2), 5 divides p, and from (3), 5 also divides q. It means 5 is a common factor of p and q. This contradicts the supposition, so there is no common factor of p and q. Hence, 5 is an irrational number. Example 3: Prove that 3 + 2 5 is irrational. Solution: We know that 5 is irrational. If possible, let 3 + 2 5 is rational a By definition of rational number, 3 + 2 5 = b where a, b are integers, b ≠ 0 and a, b are coprime. a 2 5= b -3 a - 3b a - 3b 2 5= 5 = 2b b

an irrational number can't be expressed in the form of q where q ! 0. 7

REAL NUMBERS

Hence, our assumption was wrong. 3 + 2 5 is irrational. Example 4: Prove that

1 is irrational. 2

Solution: 1 is rational 2 By definition of rational number, 1 a = b where a, b are integers, b ≠ 0 and a, b are coprime. 2

If possible, let's assume

2 b 1 = a b 2= a

An irrational number cannot be expressed in the form of q where q ! 0. Hence, our assumption was wrong. 1 is an irrational number. 2 Example 5: Prove that 2 + 5 is irrational. Solution: By definition of rational number, a 2 + 5 = b where a,b are integers, b ! 0 and a, b are coprimes. a 5= b - 2 2 a a 2 a 2 - 2 5= -2. . 2+ 2 b b b 2a a2 2a 2 5= +2 b b b2 2 2a a2 - 3b2 a . 2= 2= -3 b b2 b2 2 2 a2 - 3b2 b . a - 3b 2= 2 = b2 2ab 2a 2

5 =

p An irrational number cannot be expressed in the form of q where q ! 0. Hence, our assumption was wrong.

2 + 5 is an irrational number.

IL Foundation Series Class 10

1.5

REAL NUMBERS AND THEIR DECIMAL EXPANSIONS

Every rational number has a terminating decimal expansion or non-terminating recurring decimal expansion. p Condition for terminating decimal expansion: Let x = q be a rational number where p, q are integers, q 0 and p, q are coprimes, Then x has terminating decimal expansion if and only if the prime factorisation of q is of the form 2m × 5n where m and n are non-negative integers. p Note : Let x = q is a rational number where p and q are coprime integers, and q 0. If x has terminating decimal expansion and q = 2m × 5n for some non-negative integers m, n, then the number of digits in decimal point is max {m, n}. 11 Example: For a rational number , 40 = 23 × 51 and 11, 40 are coprimes. 40 max {3, 1} = 3 11 Number of digits in the decimal part of is 3. 40 Alternately, 11 11 11.5 2 11 # 25 275 275 = 3 1 = 3 1 2 = = = 0.275 3 3 3 = 40 2 .5 1000 2 .5 .5 2 .5 10 There are 3 digits in decimal part. Period and periodicity of recurring decimals: A non-terminating recurring decimal such as 32.7183183183... can be expressed in short as 32.7183 . Here, 183 is called period and it has 3 digits representing its periodicity. In a non-terminating recurring decimal, the group of digits in the decimal part which is repeating is called a period. The number of digits in the period is called periodicity. The concept of period and periodicity is useful in expressing a non-terminating recurring decimal in p the form of q where p and q are integers, and q ≠ 0. Example: Let us take the real number 32.71831 Let x = 32.7183 x = 32.71831831831......(1) Here periodicity = 3 Multiply 103 (i.e. 1000) on both sides 1000x = 1000 × 32.71831831831...

REAL NUMBERS

1000x = 32718.31831831......(2) But x = 32.71831831... On subtracting (1) from (2), we get 999 x = 32685.6 326856 x = 9990

326856

32.7183 = 9990

SOLVED EXAMPLES Example 1: Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating recurring decimal expansion. 129 2 ·57 ·7 5 Solution: i)

77 ii) 210

15 iii) 1600

129 3 × 43 = p 7 5 = 2 2 ×5 ×7 2 × 57 × 7 5 q 2

` q = 2 ×5 ×7

It has non-terminating recurring decimal expansion. 77 7×11 11 ii) 210 = 7×30 = 30 =

p 11 1 1 = q 2 ×3 ×5 1

q = 2l × 3l × 51 It has non-terminating decimal expansion. 15 3×5 iii) 1600 = 320×5 =

3 = p 2 ×5 2 q 6

` q = 26 × 52 = 2m × 5n It has a terminating decimal expansion. Example 2: After how many places of decimal, will the decimal expansion of Solution: 43 = 43 5 = 215 = 215 = 0.0215 × 2 4 ·5 3 2 4 ·5 3 5 10 4 10000 The decimal expansion terminates after 4 places of decimal.

43 terminate ? 24 $ 53

IL Foundation Series Class 10

1.6

INTERVAL NOTATION

Interval notation serves as a means to express a range on a number line, providing a concise representation of subsets within the real number continuum. Essentially, it is a method of denoting the set of numbers that fall between two specified values. For instance, consider the set of numbers represented by the inequality 0 ≤ x ≤ 5; this forms an interval encompassing 0, 5, and all the numerical values positioned between 0 and 5. If we aim to represent the set of real numbers {x |-2 < x < 5} using interval notation, we can succinctly express it as the interval (-2, 5). -5

-4

-3

-2

-1

2 3 (-2, 5)

1.6.1 Types of interval notation 1.

Open interval

This category of interval excludes the boundary points specified in the inequality. For instance, in the set {x|-3 < x < 1}, the endpoints, -3 and 1, are not part of the interval. This characteristic is conveyed through open interval notation as (-3, 1). not included

-5

-4

-3

-2

-1

(-3,1) 2. Closed interval

This interval type incorporates the endpoints defined by the inequality. For instance, in the set {x | -3 ≤ x ≤ 1}, both endpoints, -3 and 1, are included. This inclusiveness is denoted by closed interval notation as [-3, 1]. included

-5

-4

-3

-2

-1

[-3,1]

REAL NUMBERS

Half-open interval

This category of interval includes just one of the inequality endpoints. For instance, in the set {x|-3 ≤ x < 1}, only the endpoint -3 is part of the interval. This characteristic is conveyed through half-open interval notation as [-3, 1). included

-5

not included

-4

-3

-2

-1

[-3,1)

The symbols employed in interval notation are as follows: •

[ ]: A square bracket denotes an interval where both endpoints are encompassed in the set.

•

( ): A round bracket signifies an interval excluding both endpoints in the set.

•

( ]: A semi-open bracket is utilised when the left endpoint is excluded and the right endpoint is included in the set.

•

[ ): Another semi-open bracket is used when the left endpoint is included and the right endpoint is excluded from the set.

Various forms of interval notation find visual representation on a number line. Examine the convenient table that provides a clear differentiation between these interval types, illustrating their placement on the number line.

Interval Notation

Inequality

Type of Interval

(a, b)

{x | a < x < b}

Open Interval

[a, b]

{x | a # x # b}

Closed interval

[a, ∞)

{x | x $ a}

Half-Open Interval

(a, ∞)

{x | x > a}

Half-Open Interval

(- ∞, a)

{x| x < a}

Half-Open Interval

(- ∞, a]

{x | x ≤ a}

Half-Open Interval

IL Foundation Series Class 10

1.7

ABSOLUTE VALUE

1.7.1 Absolute value of an integer The absolute value of an integer is the numerical value of the integer regardless of its sign. If 'a' is an integer then its absolute value is denoted by |a| and is defined as, |a| = a if a ≥ 0, = -a if a < 0 Example: |6| = 6 and |-6| = -(-6) = 6. Here i)|a| = a, if a is positive. ii)|a| = 0, if a is zero. iii)|a| = -a, if a is negative.

1.8

COMPLEX NUMBERS

A number which is in the form of a + ib (where a, b ! R ) is called a complex number, and a + ib is usually denoted by z. The set of complex numbers is denoted by C. In z = a + ib, a is called the real part and denoted by Re (z); b is called the imaginary part and denoted by Im (z). Complex Number (z) = a + ib Im(z)=0 Purely real

Re(z)=0 Purely imaginary

Im(z)≠0 Imaginary

Complex number 0 + i0 is purely real as well as purely imaginary but not imaginary. If z1 and z2 are two non-real complex numbers, then either z1 = z2 or z1 ≠ z2 only holds. i.e. between two complex numbers either =, ≠ symbols are used. There is no meaning for z1 > z2, z1 < z2. There is no comparison among complex numbers. Example: 1) If a + ib = 3 + i4, then a = 3, b = 4. 2) If 3 + 4i, 4 + 6i are two complex numbers, then 3 + 4i ≠ 4 + 6i is only the right way and we cannot compare which is greater. 1.8.1 Argand plane Every complex number z = a + ib can be represented by an ordered pair (a, b) and hence can be represented by a point with coordinates (a, b) in the complex plane (Argand plane or Argand diagram). 13

REAL NUMBERS

Modulus: Let z = x + iy be represented by point P in the Argand plane, then the distance of P from origin (O) is called the modulus of z and is denoted by |z|. Y (Imaginary Axis)

|z|

Argand Plane P(x,y) X (Real Axis)

If z = x + iy, then | z | = distance from (0, 0) to (x, y) = Example: If z = 3 + 4i , then |z|=

x 2 + y 2 = (Re (z)) 2 + (Im (z)) 2

3 2 + 4 2 = 5.

Note: The modulus of complex numbers never be negative. |z|= z (if z > 0 ). |z|= -z (if z < 0 ) is not correct if z is not non-real complex number. 1.8.2 Algebra of complex numbers Addition

If z1, z2 are two complex numbers such that z1 = a1 + ib1, z2 = a2 + ib2, then sum of z1, z2 is defined as (a1 + a2) + i (b1 + b2 ) and is denoted as z1 + z2 ` z1 + z2 = (a1 + a2) + i(b1 + b2)

= (Sum of real parts of z1, z2 ) + i(Sum of imaginary parts of z1, z2 )

Example: If z1 = 3 + 6i, z2 = 7 + 5i, then z1 + z2 = (3 + 7) + i(6 + 5) = 10 + 11i Subtraction

If z1 = a1 + ib1, z2 = a2 + ib2 then z1 - z2 is defined as (a1 - a2) + i(b1 - b2).

` z1 - z2 = ^a1 - a2h + i ^b1 - b2h = ^ Re ^ z1h - Re ^ z2hh + i ^ Im ^ z1h - Im ^ z2hh

Example : If z1 = 2 + i, z2 = 1 + 2i , then z1 - z2 = ^2 - 1h + ^1 - 2h i = 1- i

Multiplication

If z1 = a1 + ib1, z2 = a2 + ib2 , then product of z1, z2 defined as â1 a2 - b1 b2h + i â1 b2 +a2 b1) and denoted as z1z2. z1 z2 = â1 a2 - b1 b2h + i â1 b2 + a2 b1h

Example: If z1 = 2 + i, z2 = 1 + i then z1 z2 = ^2 + ih^1 + ih = ^2 - 1h + i ^2 + 1h = 1 + 3i 14

IL Foundation Series Class 10

Conjugate of a complex number

If the sum and product of two complex numbers are real, then two complex numbers are said to be conjugate to each other. If z = a + ib, then the conjugate of z is a - ib , and the conjugate of z is denoted by zr. Division

If z1 and z2(≠0) are two complex numbers, then z1 = z1 zrr2 = z1 zr22 (where zr2 is conjugate of z2 ) z2 z2 z2 z2 Example : 4 + 2i # 1 + 2i = (4 + 2i) (1 + 2i) = (4 - 4) + i (10) = 2i 1+4 5 1 - 2i 1 + 2i Integral powers of ‘i’

We have i = - 1 & i2 =- 1; i 3 = i2 i = (- 1) i =- i; i 4 = ^i2h2 = (- 1) 2 = 1 Note : 1) i 0 is defined as 1 2) in = ir, where r (0 # r # 3) be the remainder when n(n > 4) is divided by 4. Proof : Let m be the quotient and r be the remainder when n is divided by 4. ` i n = i 4m + r = i 4m # i r = ^i 4hm # i r = (1) m # i r = i r Example: Find the value of i127 . Solution: i127 = i "4^31h+ 3 , = i3 =- i Note: The values of the negative integral powers of i are found as given below: 1 1#i i i -1 = i = i # i = - 1 =- i 1 1 i -2 = 2 = - 1 =- 1 i i i 1 1#i i - 3 = 3 = 3 # = 4 = 1 = i. i i i i Cube root of unity

The cube root of unity is represented as 3 1 and it has three roots. The three cube roots of unity are 1, ~ , ~ 2, which, on multiplication, gives the answer of unity (1). Among the roots of the cube roots of unity, one root is a real root, and the other two roots are imaginary roots. The values of the imaginary cube roots of unity are as follows. • •

(-1 + i 3) 2

w2 =

(-1 - i 3) 2

REAL NUMBERS

Properties of cube root of unity

• • • •

The cube roots of unity has two imaginary roots ^~, ~2h and one real root (1). The sum of the roots of the cube root of unity is equal to zero. (1 + ~ + ~2 = 0) he square of one imaginary root (~) of the cube root of unity is equal to another imaginary T root ^~2h . The product of the imaginary roots of the cube roots of unity is equal to 1. ^~ # ~2 = ~3 = 1h

Example: Find the value of ~67 , using the values of the cube root of unity. Solution: To find the value of ~67 , we can use the property ~3 = 1 (since ~ is a cube root of unity). Therefore, we can rewrite ~67 as ~3 # 22 + 1 . Using the property ~3 = 1 , we can simplify this expression: ~67 = ~3 # 22 + 1 = ^~3h22 $ ~1 = 122 $ ~ =~ So, the value of ~67 is simply ~ .

QUICK REVIEW •

he fundamental theorem of arithmetic: Every composite number can be expressed T (factorised) as a product of primes, and this factorisation is unique except for the order in which the prime factors occur.

•

very composite number can be uniquely expressed as the product of powers of primes in E ascending or descending order.

•

If p is a positive prime, then p is an irrational number. For example, 11 etc. are irrational numbers.

•

et x be a rational number whose decimal expansion terminates. Then, x can be expressed in L p the form q , where p and q are coprime, and the prime factorisation of q is of the form 2 m # 5 n , where m and n are non-negative integers. p Let x = q be a rational number, such that the prime factorisation of q is of the form 2 m # 5 n where m and n are non-negative integers. Then, x has a terminating decimal expansion, which terminates after k places of decimals, where k is the larger of m and n. p Let x = q be a rational number, such that the prime factorisation of q is not of the form 2 m # 5 n 2, where m,n are non-negative integers. Then, x has non-terminating repeating decimal expansion.

•

5, 7,

IL Foundation Series Class 10

The symbols employed in interval notation are as follows: 1. [ ]: A square bracket denotes an interval where both endpoints are encompassed in the set. 2. ( ): A round bracket signifies an interval excluding both endpoints in the set. 3. ( ]: A semi-open bracket is utilised when the left endpoint is excluded and the right endpoint is included in the set. 4. [ ): Another semi-open bracket is used when the left endpoint is included and the right endpoint is excluded in the set. •

omplex Number: A number which is in the form of a + ib (where a, b ! R ) is called C complex number and a + ib is usually denoted by z. And the set of complex numbers is denoted by C.

•

lgebra of complex Numbers: If z1, z2 are two complex numbers such that A z1 = a1 + ib1, z2 = a2 + ib2 then 1. Addition: z1 + z2 = ^a1 + a2h + i ^b1 + b2h

2. Subtraction: z1 - z2 = ^a1 - a2h + i ^b1 - b2h

3. Multiplication: z1 z2 = ^a1 a2 - b1 b2h + i ^a1 b2 + a2 b1h z z zr z zr 4. Division: z12 = z12 zr22 = 1 22 (If z2 (! 0)) z2

WORKSHEET - 1 I.

THE FUNDAMENTAL THEOREM OF ARITHMETIC 1. Express the following as a product of prime factors:

i) 156

ii) 7429

2. Find the L.C.M. and H.C.F. of 6 and 20 by the prime factorisation method. 3. Find the L.C.M. and H.C.F. of 510 and 92 and hence verify that L.C.M. # H.C.F = product of two numbers. 4. Find the L.C.M. and H.C.F of 12,15 and 21 by prime factorisation. 5. Given that H.C.F (306, 657) = 9, Find L.C.M. (306, 657). 6. Check whether 4 n can end with the digit '0’ for any natural number n. 7. Write the H.C.F of the smallest composite number and the smallest prime number. 8. The H.C.F of two numbers is 145, and their L.C.M. is 2175. If one number is 725, then find the other. 9. Find the largest number, which divides 245 and 1029, leaving the remainder 5 in each case. 10. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.

REAL NUMBERS

II. REVISITING IRRATIONAL NUMBERS 1. Prove that the following are irrational numbers: i)

ii)

iii)

iv) 5 2

2. Show that 3 2 is an irrational number. 3. Show that 5 - 3 is irrational. 4. Prove that

5 + 3 is irrational.

5. If p and q are primes then prove that

p + q is irrational.

6. Without actually performing long division, state whether the following rational numbers will have a terminating or non-terminating repeating decimal expansion: 17 35 14588 129 i) 8 ii) 50 iii) 2 # 2 # 17 iv) 625 2 5 7 7. What can you say about the prime factorisation of the denominators of the following rationals: i) 43.123456789 ii) 43. 123456789 8. After how many places of decimal will the decimal expansion of the following rational numbers terminates: 3 13 7 i) 8 ii) 125 iii) 80 p 9. Express each of the following in the form of q : i) 43. 12

ii) 3.1 47

10. Express the following in the form of x+iy: i) ^2 - 3ih^3 + 4ih a - ib iv) a + ib vii)

ii) ^1 + 2ih3

2 + 5i 2 - 5i v) 3 - 2i + 3 + 2i

1 1 + cos 2i + i sin 2i

iii) 0.12 3 iii) ^1 - ih3 ]1 + ig

4 + 2i 3 + 4i vi) 1 - 2i + 2 + 3i

III. INTERVAL NOTATION 1. Write the interval notation for the set of all real numbers x such that - 1 # x # 3 . 2. Represent the solution set for x in the inequality x $- 5 using interval notation. 3. Express the interval x such that 2 # x < 8 in interval notation. 4. Write the interval notation for the set of all real numbers x such that x < - 3. 5. Represent the solution set for x in the inequality - 2 # x < 3 using interval notation. 6. Write the interval notation for the set of all real numbers x such that - 4 # x # 0 . IV. ABSOLUTE VALUE 1. Find the absolute value of - 7 .

IL Foundation Series Class 10

2. If x = 9 , what are the possible values of x ? 3. Evaluate 3 + - 5 . 4. If a - 3 = 6 , what are the possible values of a? V. COMPLEX NUMBERS

1. Express the result of ^- 2 + 3ih + ^4 - 5ih in the standard form a + bi. 2. If w = 5 - 2i and z =- 3 + 7i , calculate the sum w + z .

3. Determine the real and imaginary parts of the complex number c = 8 + 6i . 4. If m =- 4 + 3i and n = 2 - 5i , find the sum m + n in the standard form. 5. Determine the real and imaginary parts of the complex number c = 6 + 9i . 6. If w = 4 - 7i and z =- 1 + 5i , calculate the difference w - z .

7. Express the result of ^- 3 + 2ih - ^5 - 6ih in the standard form a + bi . 8. Explain the process of multiplying two complex numbers.

9. Express the result of (4 - i) # (3 + 2i) in the standard form a + bi . 10. If w = 3 - 2i and z =- 2 + 5i , calculate the product w # z . 11. If m = 1 + 4i and n =- 3 - i , find the product m # n in the standard form. 12. If r =- 1 - 4i , find r and r + r . 13. If z1 = 6 - 3i and z2 = 2 + i , find the quotient z1 ' z2 . 3 - 2i 14. Express the result of 1 + i in the standard form a + bi . w 15. If w =- 4 + 2i and z = 1 - 3i , calculate the quotient z . 16. Find the value of ~6 + ~7 + ~5 . 17. Find the value of ~3n + ~3n + 1 + ~3n + 2 , where ~ is cube roots of unity. 18. What is ~100 + ~200 + ~300 equal to, where ~ is the cube root of unity? 19. Find the value of ^1 - ~ + ~2h2 + ^1 - ~2 + ~h2 .

20. If ~ is a complex cube root of unity, then what is ~10 + ~ -10 equal to?

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. Which of the following numbers has terminating decimal expansion?

37 a) 45

21 23 56

17 c) 49

89 22 32

REAL NUMBERS

43 will terminate after 24 # 53 c) 5 places d) 1 place

2. The decimal expansion of the rational number a) 3 places

b) 4 places

3. If HCF (a, 8) = 4, LCM (a, 8) = 24, then 'a' is a) 8

b) 10

c) 12

d) 14

4. If p is a prime number, then LCM of p, p2 and p3 is b) p3

a) p

5. Decimal expansion of a) 0.115

c) p2

d) p6

c) 11.5

d) 0.0115

23 is 2 # 52 3

b) 1.15

6. The product of H.C.F. and L.C.M. of the smallest prime number and the smallest composite number is a) 4

b) 6

c) 8

d) 10

7. Which of the following is rational? a)

6+ 9

2+ 4

4 + 9

3+ 5

8. A pair of irrational numbers whose product is a rational number is a) 16 , 4

5, 2

3 , 27

36 , 2

9. Euclid's division lemma states that if a and b are any two positive integers, then there exists unique integers q and r such that: a) a = bq + r, 0 1 r 1 b

b) a = bq + r, 0 # r # b

c) a = bq + r, 0 # r < b

d) a = bq + r, 0 1 b 1 r

10. The value of x in the factor tree is: x 5 5 2

a) 30

b) 150

c) 100

d) 50

11. For any integer 'a' and 3, unique integers q and r exist such that a = 3q + r . Then, the possible values of r are a) 1, 2

b) 0, 1

c) 0, 1, 2, 3

d) 0, 1, 2

12. The H.C.F. of two numbers is 23, and their L.C.M. is 1449. If one of the numbers is 161, then the other number is a) 23 20

b) 207

c) 1449

d) none of these

IL Foundation Series Class 10

13. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is a) 10

b) 100

c) 504

d) 2520

14. If 'a' and 'b' are two positive integers such that a = 14b, the H.C.F of 'a' and 'b' is a) a

b) b

c) 1

d) 0

63 15. The decimal expansion of 72 # 175 is a) terminating b) non-terminating c) non-terminating repeating

d) an irrational number

16. The missing number in the following factor tree is 3 18

a) 5

b) 7

c) 12

d) 6

17. Let K be a positive integer. If 3 divides K2 , then 3 also divides a) K

b) K3

c) K 4

d) none of these

18. 2 - 5 is a) an integer

b) a rational number

c) an irrational number

d) none of these

19. Assertion (A): If the product of two numbers is 5780 and their HCF is 17, then their LCM is 340. Reason (R): HCF is always a factor of LCM.

a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

b) Both Assertion (A) and Reason (R) are true but Reason(R) is not the correct explanation of Assertion (A). c) Assertion (A) is true but Reason (R) is false. d) Assertion (A) is false but Reason (R) is true.

20. Assertion (A): 5 + 3 is an irrational number. Reason (R): The sum or difference between a rational and an irrational number is always irrational

REAL NUMBERS

a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). b) Both Assertion (A) and Reason (R) are true but Reason(R) is not the correct explanation of Assertion (A). c) Assertion (A) is true but Reason (R) is false. d) Assertion (A) is false but Reason (R) is true. II. SUBJECTIVE QUESTIONS 1. The product of three consecutive integers is divisible by 6. Is this statement true or false? Justify your answer. 2. Find the rational number which, in its decimal expansion is 327.7081. 3. What can you say about the prime factors of q when a number is expressed in the p form q ? 4. If p and q are two prime numbers, then what is their HCF? p 5. Write the condition to be satisfied by q so that a rational number q has a non-terminating decimal expansion. 6. Write the exponent of 2 in the prime factorisation of 144. 7. Find the H.C.F. of the smallest composite number and the smallest prime number. 8. If two positive numbers, a and b ,are written as a = x5 y2, b = x3 y3 , where x and y are prime numbers, then find the H.C.F. (a, b). 9. If H.C.F. (a, b) = 12 and a# b = 1800, then find LCM ^a, bh ?

6243 10. The decimal expansion of the rational number 3 # 4 will terminate after how many 2 5 p decimal places. If q is a rational number (q ! 0), what is a condition of q so that the p decimal representation of q is terminating? 11. Express the interval x such that - 3 # x < 5 in interval notation. 12. Write the interval notation for the set of all real numbers x such that x #- 2 . 13. Represent the solution set for x in the inequality - 1 < x # 4 using interval notation. 14. Express the interval x such that x 2 7 in interval notation. 15. If b = 11 , what are the possible values of b? 16. If c + 2 = 10 , what are the possible values of c? 17. If x = 6 - i and y =- 2 + 3i , calculate the sum x + y and express the result in the standard form. 18. If x = 6 - i and y =- 2 + 3i , calculate the difference x - y and express the result in the standard form. 22

IL Foundation Series Class 10

19. If x = 3 - 2i and y =- 1 + 4i , calculate the difference x - y and express the result in the standard form. 20. If m =- 6 + 2i and n = 3 - 4i , find the difference m - n in the standard form. 21. If w = 3 + 4i , find its complex conjugate w . 22. If z =- 2 - 7 i , calculate z z and express the result in the standard form. 23. If p = 5 - 2i , determine the imaginary part of p. x 24. If x = 5 - i and y =- 2 + 3i , calculate the quotient y and express the result in the standard form. 25. What is the value of the expression of the cube root of unity, (1 + ~) + (1 + ~) 2 + (1 + ~) 3 ? 26. Prove that(1 + ~) 3 - ^1 + ~2h3 = 0.

27. If ~ is a complex cube root of unity, then find the value of ^1 - ~ + ~2h6 + (1- ~2 + ~h6 .

2.1

POLYNOMIALS

REVISITING POLYNOMIALS

Polynomial: An algebraic expression in which the exponents of the variable(s) are non-negative integers or whole numbers is called a polynomial. Examples: i) 7x2 + 8x - 9 is a polynomial. ii) 8 + x + x2 is not a polynomial. Polynomials are of different types: 1. A polynomial with one term is called a monomial. Example: 5x 2. A polynomial with two terms is called a binomial. Example: 2x - 3 3. A polynomial with three terms is called a trinomial. Example: 2x2 + 4x - 7 The standard form of a polynomial: Let x be a variable, n be a positive integer and a0, a1, a2, f, an be constants (real numbers). Then, the standard form of a polynomial is: f (x) = an x n + an -1 x(n -1) + an -2 x(n -2) + .... + a1 x1 + a0 , an ¹ 0

Note: In this polynomial, n is called the degree of the polynomial. Degree of a polynomial: The highest exponent of a variable in a polynomial is called its degree. 1 Examples: i) For polynomial f (x) = 3x + 2 , the degree is 1. 7 3 ii) For polynomial g (x) = 2 x - 2 x2 + 8 - x3 , the degree is 3. Constant polynomial: A polynomial with zero degree is called a constant polynomial. Example: f ^ xh = 7 is a constant polynomial. Zero polynomial: A polynomial with the coefficients of all the terms as zeroes is called zero polynomial. f (x) = 0 $ x n + 0 $ x n - 1 + 0 $ x n - 2 + f0 $ x1 + 0 The degree of a zero polynomial is not defined.

IL Foundation Series Class 10

Linear polynomial: A polynomial with degree 1 is called a linear polynomial. Its general form is f (x) = ax + b, a ≠ 0. Example: f ( x ) = 2 x - 3 is a linear polynomial. Quadratic polynomial: A polynomial with degree 2 is called a quadratic polynomial. Its general form is f (x) = ax2 + bx + c, a ! 0 . Examples: i. f ^ xh = 2x2 + 3x - 1 ii. g ^xh = x2 - 5 Cubic polynomial: A polynomial with degree 3 is called a cubic polynomial. Its general form is f (x) = ax3 + bx2 + cx + d, a ≠ 0 Example: f ( x )= 2 x 3 − x 2 + x + 9 Bi-quadratic polynomial: A polynomial with degree 4 is called a bi-quadratic or quartic polynomial. Example: f ^ xh = 5x 4 + 4x3 + 3x2 + 2x1 + 1 Zero of a polynomial: A real number k is said to be the zero of a polynomial f(x) if f(k) = 0 Example: For polynomial p ^ xh = x2 - 3x - 4 , p(4) = 42 - 3 × 4 - 4 = 16 - 16 = 0 and p (- 1) = (- 1) 2 - 3 (- 1) - 4 = 1 + 3 - 4 = 0 ` 4 and - 1 are zeroes of p (x) . Note:

- b - ( constant term) 1. The zero of a linear polynomial f (x) = ax + b, a ≠ 0 is a = coefficient of x 2. An n th degree polynomial has at most n zeroes.

2.2

GEOMETRICAL MEANING OF THE ZEROES OF A POLYNOMIAL

2.2.1 Geometrical representation of a polynomial and its zero(es) For a polynomial, let y = f ( x ) . Then, for every real value of x, there is a unique real value of y, giving rise to the ordered pair (x, y). This ordered pair (x, y) represents a point on a coordinate plane. The set of all such points gives us the graph of the polynomial or geometrical representation of the polynomial. The zero(es) of a polynomial f(x) is/are the x-coordinate(s) of intersecting points of the graph of y = f ^ xh and x-axis. 25

POLYNOMIALS

Graph of a constant polynomial: Let us take the constant polynomial, f(x) = 4. x

y = f ^ xh

-1

Y 5 y = f(x) = 4

4 3 2 1 X’

-5 -4 -3 -2 -1 0 -1

Y’

The graph is a straight line parallel to the x-axis. This line does not meet the x-axis. So, its zeroes do not exist. Graph of a linear polynomial: Let us take the linear polynomial, f ^ xh = 2x + 3. x

y = f ^ xh

-1

Y 5 y = f(x) = 2x + 3

4 3 2 _3

(- 2, 0(

X’ -3

-2

-1

-1 Y’

-3 The graph is a straight line. This line meets the x-axis at point b- 32 , 0 l. So, it's zero = . 2 Graph of a quadratic polynomial: Let us take the quadratic polynomial, f ^ xh = x2 - 3x - 4. 26

IL Foundation Series Class 10

-2

-1

y = f(x)

-4

-6

-4

X’

7 6 5 4 3 2 (-1, 0) 1 -4 -3 -2 -1

-1

y = f(x) = x2 - 3x - 4 (4, 0) 1

2 3

4 5 6 7

-2

-3 -4 -5 -6 -7 -8 Y’

The graph of a curve is called a parabola. It meets the x-axis at two points (-1, 0) and (4, 0). ` The zeroes of this quadratic polynomial are -1 and 4. Note: 1. For a quadratic polynomial, f (x) = ax2 + bx + c, a ! 0, the graph is a parabola. It opens upwards, shaped like a

when a 2 0.

ii) It opens downwards, shaped like

when a < 0.

2. The number of zeroes of a polynomial is the number of times its graph intersects with the x-axis.

SOLVED EXAMPLES Example 1: The graphs of y = p(x) are given in the figure below for some polynomials p ^ xh . Find the number of zeroes of p ^ xh in each case.

POLYNOMIALS

X’

Y’

X’

Y’

(i)

(ii)

(iii)

Solution: The number of zeroes in each of the graphs above are: (i) 1, since the line intersects with the x-axis only once. (ii) 3, since the line intersects with the x-axis thrice. (iii) 3, since the line intersects with the x-axis thrice. Example 2: Which of the following graphs represents a cubic polynomial?

(a)

(b)

(c)

(d)

Solution: (b) is the graph of a cubic polynomial because it cuts the x-axis only three times.

IL Foundation Series Class 10

2.3

RELATIONSHIP BETWEEN ZEROES AND COEFFICIENTS OF A POLYNOMIAL FOR QUADRATIC AND CUBIC POLYNOMIAL

2.3.1 Relationship between zeroes and coefficients of a polynomial The zeroes and coefficients of a polynomial are interrelated as follows: i) Let a be a zero of a linear polynomial f (x) = ax + b, a ≠ 0, then: - b - ( constant term) a = a = coefficient of x ii) Let α and β be the zeroes of a quadratic polynomial f ( x ) = ax 2 + bx + c, a ≠ 0 , then: -b c a + b = a and ab = a , - ( coefficient of x) constant term i.e., the sum of zeroes = and the product of zeroes = coefficient of x2 coefficient of x2 cb, c be the zeroes of a cubic polynomial f (x) = ax3 + bx2 + cx + d, a ≠ 0, then: iii) Let a, ba, ,and -b -d c a + b + c = a , ab + bc + ca = a and abc = a , -^coefficient of x2h i.e., the sum of zeroes = , coefficient of x3 the sum of the product of zeroes taken two at a time = the product of zeroes =

coefficient of x coefficient of x3

- (constant term) . coefficient of x3

2.3.2 Finding polynomial, if all its zeroes are known i) If a is a zero of a linear polynomial f ^ xh , then f (x) = K (x - a) , where K is a non-zero constant. ii) If a, b are zeroes of a quadratic polynomial f ^ xh , then f ( x )= k  x 2 − (α + β ) x + αβ  , where K is a non-zero constant, i.e., f ( x ) = k  x 2 − ( sum of zeroes ) x + ( product of zeroes ) 

iii) If a, b, c are zeroes of a cubic polynomial f ^ xh , then

f ( x= ) k  x3 − (α + β + γ ) x 2 + (αβ + βγ + γα ) x − αβγ  , where k is a non-zero constant, i.e.,

ff= ((x )) =kK6xx3 3−- (sum of zeroes) x2+( sum of product of zeroes taken two at a time) x - (product of zeroes)]

POLYNOMIALS

SOLVED EXAMPLES Example 1: F ind the zeroes of the following quadratic polynomials and verify the relationships between the zeroes and coefficients: a) x2 - 2x - 8

b) 6x2 - 3 - 7x

Solution: 2

a) x2 - 2x - 8 = x - 4x + 2x - 8 = x ^ x - 4h + 2 ^ x - 4h = ^ x + 2h^ x - 4h

` The value of x2 - 2x - 8 is zero when x + 2 = 0 or x - 4 = 0 , i.e., x = -2 or x = 4. ` The zeroes are -2 and 4, i.e., a =- 2, b = 4 .

Verification: Here, x2 - 2x - 8 = ax2 + bx + c ` a = 1, b =- 2, c =- 8 -b i) a + b = a - (- 2) &-2+4 = 1 & 2 = 2 (True) c ii) ab = a -8 & (- 2) $ 4 = 1 ⇒ − 8 = −8 (True) b) 6x2 - 3 - 7x = 6x2 - 7x - 3 = 6x2 - 9x + 2x - 3 = 3x ^2x - 3h + 1 ^2x - 3h = ^3x + 1h^2x - 3h The value qf the equatiqn is 0 when (3x + 1) (2x - 3) = 0 & 3x + 1 = 0 or 2x - 3 = 0 -1 3 & x = 3 or x = 2 . -1 3 . . The zeroes are 3 , 2 .

IL Foundation Series Class 10

Verification: We have: -1 3 a = 3 ,b = 2 a = 6, b =- 7, c =- 3 -b i) a + b = a ⇒ - 1 + 3 = - (- 7) 2 6 3 ⇒ -2 + 9 = 7 & 7 = 7 6 6 6 6 - 2 + 9 7⇒ 7 7 (True) =6 &6 =6 6 c ii) αβ = a −1 3 −3 ⇒ ⋅ = 3 2 6 −1 −1 ⇒ = (True) 2 2 Example 2: Find a quadratic polynomial for which the sum of zeroes is 4, and the product of zeroes is 1. Solution: Given: The sum of zeroes = 4 The product of zeroes = 1 ` The quadratic polynomial is given by: f ( x) = k  x 2 − ( sum of zeroes ) x + ( product of zeroes )  where k is a non-zero constant. = k  x 2 − ( 4 ) x + (1) 

(

)

= k x2 − 4x + 1

Example 3: Find the quadratic polynomial whose zeroes are −2, Solution:

1 . 2

Given:

1 a =- 2, b = 2

1 -3 ` a + b =- 2 + 2 = 2 1 ab = (- 2) $ 2 =- 1 31

POLYNOMIALS

` The quadratic polynomial is given by: f ( x )= k  x 2 − (α + β ) x + αβ , where k is a non-zero constant.

   −3  = k  x 2 −   x + ( −1)   2    3   = k  x 2 + x − 1 2   k = 2x2 + 3x − 2 2

(

)

= l 2 x 2 + 3 x − 2 where =l

k is a non-zero constant. 2

Example 4: Find a quadratic polynomial whose zeroes are 3 + 5 and 3 - 5 . Solution: The sum of the zeroes = 3 + 5 + 3 - 5 = 6 The product of the zeroes = (3 + 5) × (3 − 5) = 9 − 5 = 4 ` The quadratic polynomial is given by:

(( x)) =kK6xx2 2−- (sum of zeroes) x + (product of zeroes)] where k is a non-zero constant. f= = k éë x 2 - 6 x + 4ùû

Example 5: Verify that 1, 4, and 7 are the zeroes of the cubic polynomial f ^ xh = x3 - 12x2 + 39x - 28 and then verify the relationship between the zeroes and coefficients. Solution: Comparing the given polynomial with ax3 + bx2 + cx + d, we get: a = 1, b = -12, c = 39, d = -28 Further, f (1) = (1) 3 - 12 (1) 2 + 39 (1) - 28 = 1 - 12 + 39 - 28 = 40 - 40 = 0 f (4) = (4) 3 - 12 (4) 2 + 39 (4) - 28 = 64 - 192 + 156 - 28 = 220 - 220 =0 f (7) = (7) 3 - 12 (7) 2 + 39 (7) - 28 = 343 - 588 + 273 - 28 = 616 - 616 = 0 Therefore, 1, 4, and 7 are the zeroes of x3 - 12x2 + 39x - 28. 32

IL Foundation Series Class 10

So, we take a = 1, b = 4,and c = 7 . ( −12 ) =−b α + β + γ =1 + 4 + 7 =12 =− a 1

αβ + βγ + γα = (1)( 4 ) + ( 4 )( 7 ) + ( 7 )(1) = 4 + 28 + 7 = 39 =

( −28 ) = −d

28 = − αβγ = (1) × ( 4 ) × ( 7 ) =

39 c = 1 a

Example 6: Find a cubic polynomial with the sum, the sum of the product of its zeroes taken two at a time and the product of its zeroes as 3,-1 and -3, respectively. Solution: Let a, b, and c be the zeroes of cubic polynomial f ^ xh . Given that a + b + c = 3 ab + bc + ca =- 1 abc =- 3 Cubic polynomial f ( x ) K  x 3 - (α + β + γ ) x 2 + (αβ + βγ + γα ) x - αβγ  = = K  x 3 - 3 x 2 + ( -1) x - ( -3)  K  x 3 - 3x 2 - x + 3 , where K is any non-zero real number. 2

Example 7: If a and b are the zeroes of the quadratic polynomial f (x) = 6x + x - 2, find findthe thevalue valueof of a + b . b a Solution: Since a and b are the zeroes of the polynomial f ^ xh = 6x2 + x - 2 , −1 −2 −1 , αβ = = 6 6 3 2 2 α β α +β + =

α +β = β

αβ

−1 1 2  −1  − 2× + α β (α + β )2 − 2αβ  6  3 = ⇒ + = = 36 3 1 1 β α αβ − − 3 3 1 + 24 α β 36 = −25 × 3= −25 ⇒ + = 1 β α 36 1 12 − 3 33

POLYNOMIALS

Example 8: If a and b are the zeroes of the polynomial ax2 + bx + c, then form the polynomial 1 1 whose zeroes are a and . b Solution: Since a and b are the zeroes of polynomial ax2 + bx + c . -b c a + b = a ; ab = a b -b 1 1 b+a -a = c = c The sum of the zeroes = a + = b ab 1 1 1 aa The product of the zeroes = a $ = c = c b a But the required polynomial is: x2 - (sum of the zeroes) x + product of the zeroes æ -b ö a b a Þ x 2 - çç ÷÷÷ x + or x 2 + x + çè c ø c c c Example 9: If a and b are the zeroes of the polynomial x2 + 4x + 3 , form the quadratic polynomial b a whose zeroes are 1 + a and 1+ . b Solution: Since a and b are the zeroes of the quadratic polynomial x2 + 4x + 3 . Then, α + β = −4, αβ = 3

Sum of the zeroes = 1 +

β α αβ + β2 + αβ + α2 α2 + β2 + 2αβ (α + β)2 (−4)2 16 +1+ = = = = = α β αβ αβ αβ 3 3

α β αβ 2αβ + α 2 + β2 (a + b) 2 (- 4) 2 16  β  α  = 3 = 3 Product of the zeroes = 1 +   1 +  =1 + + + = = ab β α αβ αβ  α  β  Therefore, the required polynomial is: k ×  x 2 − ( sum of the zeroes)x + ( product of zeroes)] 16 16   ⇒ k ×  x2 − x +  3 3  16 16   3) ⇒ 3 ×  x 2 − x + ( if k = 3 3   ⇒ 3 x 2 − 16 x + 16

IL Foundation Series Class 10

2.4

GRAPH OF QUADRATIC POLYNOMIAL, IDENTIFYING SIGN OF a, b, c IF GRAPH IS GIVEN

2.4.1 Quadratic equation Quadratic equation: An equation of the form ax2 + bx + c = 0, where a, b, c ! C and a ! 0 is called a quadratic equation in one variable. The numbers a, b and c are called coefficients. Note: 1. ax2 + bx + c = 0 (where a, b, c ! C and a ≠ 0) is called the general form of the quadratic equation in x. Example: 3x2 - 4x + 5 = 0 is a quadratic equation. 2. The graph of the quadratic expression in one variable forms a parabola. Roots of a quadratic equation: A complex number a is called a solution of a quadratic equation f(x) = ax2 + bx + c = 0 if f (a) = aa2 + ba + c = 0. A quadratic equation cannot have more than two roots. If a quadratic equation f(x) = ax2 + bx + c = 0 is satisfied by more than two distinct roots, then f ^ xh = 0 becomes as an 'identity', i.e., a = b = c = 0 . The roots of the quadratic equation: ax + bx + c = 0 are given by x = 2

- b ! b2 - 4ac . 2a

Here, b2 - 4ac is called the discriminant, and it is denoted by D or D. If a, b are the two roots of a quadratic equation ax2 + bx + c = 0, then: -b The sum of the roots = a + b = a c The product of the roots = ab = a Nature of the roots of a quadratic equation: Nature of the roots of the quadratic equation ax2 + bx + c = 0 depends on the discriminant ^D = b2 - 4ach. Case (i): If a, b, c are real numbers:

POLYNOMIALS

Nature of roots of ax2 +bx+c = 0, a ≠ 0

If > 0, then the roots are real and distinct

If = 0, then the roots are real and equal

Example:

x 2 - 3x + 2 = 0 Here, a = 1, b = -3, c = 2

x2 - 4x + 4 = 0 Here, a = 1, b = -4, c = 2

= b - 4ac = 1 > 0 the roots are real and distinct

If < 0, then the roots are non-realcomplex numbers and conjugate to each other Example:

= b 2 - 4ac = 0 the roots are real and equal

x2 + 2x + 3 = 0 Here, a = 1, b = 2, c = 3

= b 2 - 4ac = -8 < 0 the roots are non-real complex and conjugate to each other

Case (ii): If a, b, c are rational numbers. Nature of roots of ax2 +bx+c = 0, a ≠ 0

If > 0 and is a perfect square, then the roots are rational and distinct

If > 0 and is not a perfect square, then the roots are irrationals and conjugate to each other

If = 0, then the roots are rational and equal

If < 0, then the roots are non-real complex and conjugate to each other

Example:

x 2 - 5x + 4 = 0 Here, a = 1, b = -5, c = 4

x 2 + 3x - 2 = 0 Here, a = 1, b = 3, c = -2

= b2 - 4ac = 9 > 0, and 9 is a perfect square the roots are rational and distinct

= b2 - 4ac = 17 > 0, but not a perfect square the roots are irrational and conjugate to each other

x 2 - 2x + 1 = 0 Here, a = 1, b = -2, c = 1 = b - 4ac = 0 the roots are rational and equal 2

2x 2 - 3x + 2 = 0 Here, a = 2, b = -3, c = 2 = b2 - 4ac = -7 < 0 the roots are non-real complex and conjugate to each other

SOLVED EXAMPLES Example 1: Find the quadratic equation whose roots are 3 and 5. Solution: If a, b are the roots of the quadratic equation, then the quadratic equation is:

IL Foundation Series Class 10

x2 - (a + b) x + ab = 0 Here, a = 3, b = 5 The required quadratic equation is x2 - ^3 + 5h x + 3 ^5 h = 0 ` x2 - 8x + 15 = 0 Example 2: If a, b are the roots of the equation x2 + x - 20 = 0 , then find the values of | a - b | and

α 2 − β2

Solution: The sum of the roots a + b =- 1 The product of the roots ab =- 20

α − β=

(α + β)2 − 4αβ

α − β=

(−1)2 − 4 ( −20 )

α − β=

1 + 80

α −= β

= −9 | 9 81 9, hence= | 9 | 9 also | =

∴ α − β =9 α 2 − β2 =

( α + β )( α − β ) = ( −1) 9 = 9

Example 3: If the roots of ax2 + bx + c = 0 differ by unity, then show that b2 - a2 = 4ac . Solution: Let α , β be roots of ax2 + bx + c = 0. Given α − β = 1 We know that α += β

−b c . ,= αβ a a

| α − β |2 = 12 (α + β ) 2 − 4αβ = 1 2

c  −b  1   −4 = a  a  b 2 − 4ac =1 a2 b 2 − 4ac = a2 b2 − a 2 = 4ac Hence, proved.

POLYNOMIALS

2.4.2 Signs of a and ax2 + bx + c 1. If the quadratic equation ax2 + bx + c = 0 has complex roots (D < 0) , then a and ax2 + bx + c will have the same signs 6x ! R . Note: i) 6x ! R, if ax2 + bx + c = 0 and a 2 0 , then b2 - 4ac 1 0. ii) 6x ! R, if ax2 + bx + c = 0 and a 1 0 , then b2 - 4ac 1 0. 2. If the quadratic equation ax2 + bx + c = 0 has equal roots, then a and ax2 + bx + c will have the b same signs 6x ! R 2a 3. If the quadratic equation ax2 + bx + c = 0 has real roots a and b (D > 0, a < b) , then: i. a < x < b + a and ax2 + bx + c will have the opposite signs. ii. x < a or x > b + a and ax2 + bx + c will have the same signs.

SOLVED EXAMPLES Example 1: For what values of x is the expression - 7x2 + 8x - 9 = 0 negative? Solution: Let f ^ xh =- 7x2 + 8x - 9 = 0 Here a =- 7, b = 8, c =- 9 The coefficient of x2 = a =- 7 1 0 D = b2 - 4ac = 82 - 4 (- 7) (- 9) =- 188 < 0 ` 6x ! R, f(x) is negative. Example 2: For what values of x is the expression x2 - 5x + 14 positive? Solution: Let f ^ xh = x2 - 5x + 14 Here, a = 1, b =- 5, c = 14 The coefficient of x2 = a = 1 2 0 D = b2 - 4ac = (- 5) 2 - 4 # 1 # 14 =- 31 < 0 ``66xx!!RR, ,ff((xx))isispositive positive. .

IL Foundation Series Class 10

2.4.3 Graphs of different quadratic expressions Graph of y = ax2 + bx + c Case

Graphical representation

i. If a 2 0 and T 2 0 , then the graph of y = ax2 + bx + c intersects with the X-axis at two distinct points.

Y y = f(x)

(α, 0)

(β, 0)

(-_ ,-_( b

ii. If a 2 0 and T = 0 , then the graph of y = ax2 + bx + c touches the X-axis and lies entirely above the X-axis.

Y y = f(x) X

( ( _b,0 2a

iii. If a 2 0 and T 1 0 , then the graph of y = ax2 + bx + c lies entirely above the X-axis.

y = f(x) -b , _ (_ 2a 4a( X

iv. If a 1 0 and T 2 0 , then the graph of y = ax2 + bx + c intersects with the X-axis at two distinct points.

(α, 0)

,_ (_ 2a 4a ( -b

(β, 0)

y = f(x)

POLYNOMIALS

v. If a 1 0 and T = 0 , then the graph of y = ax2 + bx + c touches the X-axis and lies entirely below the X-axis.

Y b ,0 (-_ 2a (

y = f(x)

vi. If a 1 0 and T 1 0 , then the graph of y = ax2 + bx + c lies entirely below the X-axis.

Y X -_b , _ 2a 4a

(

y = f(x)

The graph of a quadratic function: ax2 + bx + c = 0 , a ≠ 0. Characteristics of the functions

b2 - 4ac 1 0

b2 - 4ac = 0

b2 - 4ac 2 0

When a 2 0

X Minima

When a 1 0

Minima X

O X Max

Max

Minima

IL Foundation Series Class 10

2.5 MAXIMUM AND MINIMUM VALUE OF QUADRATIC POLYNOMIAL The maximum and minimum values of a quadratic expression ^ax2 + bx + ch : b k2 4ac - b2 ax2 + bx + c a = + x a 2a + 4a2

-b 4ac - b2 If a 2 0 , then the minimum value of ax2 + bx + c at x = 2a is 4a .

--bb 4ac - b2 2 2 ++ ++ ii) If a 1 0 , then the maximum value of axax bxbx c catatxx== 22aa is 4a .

SOLVED EXAMPLES Example 1: Find the maximum value of the quadratic expression 2x - 7 - 5x2 . Solution: Here, a =- 5 , b = 2 and c = - 7. - 34 4ac - b2 4 (- 5) (- 7) - 2 = = 5 4a 4 (- 5) -2 -b 1 2a = - 10 = 5 2

1 34 Since a =- 5 < 0 , the given expression 2x - 7 - 5x2 has maximum value at x = 5 , and it is - 5 . Example 2: Find the maximum or minimum value of the quadratic expression 3x2 + 2x + 11. Solution: Here, a = 3, b = 2, c = 11 4ac − b 2 4 = 3= 11 − 4 32 = = 4a 4(3) 3 −b −2 −1 = = 2a 6 3 1 32 Since a = 3 > 0, the given expression 3x2 + 2x + 11 has minimum value at x =- 3 , and it is 3 .

QUICK REVIEW •

n algebraic expression in which the exponents of the variable(s) are non-negative integers is A called a polynomial.

•

Polynomials of degrees 1, 2, and 3 are called linear, quadratic and cubic polynomials.

•

he zero(es) of a polynomial f(x) is/are x-coordinate(s) of intersecting points of graph of T y = f ^ xh and x-axis. 41

POLYNOMIALS

•

he number of zeroes of a polynomial is the number of times its graph intersects with the T x-axis.

•

If a and b are the zeroes of the quadratic polynomial ax2 + bx + c , then:

•

-b a+b= a c ab = a b, c are the zeroes of a cubic polynomial ax3 + bx2 + cx + d , then: If a, b,a,and

a, bb+, c = - b a+ a c αβ + βγ + γα = a a, ab b, c = - d a • An expression of the form ax2 + bx + c , where a, b, c ! C and a ! 0 is called a quadratic expression in one variable. • •

- b ! b2 - 4ac 2a 2 The maximum and minimum values of a quadratic expression ^ax + bx + ch: The roots of the quadratic equation ax2 + bx + c = 0 are given by x = b k2 4ac - b2 ax2 + bx + c a = + x a 2a + 4a2

-b 4ac - b2 . i) If a 2 0 , then the minimum value of ax2 + bx + c at x = 2a is 4a -b 4ac - b2 . ii) If a 1 0 , then the maximum value of ax2 + bx + c at x = 2a is 4a

WORKSHEET - 1 I.

GEOMETRICAL MEANING OF THE ZEROES OF A POLYNOMIAL

1. Find the number of zeroes lying between -2 and 2 of the polynomial f ^ xh , whose graph is given below: Y y = f (x) X’

-4 -2

Y’

IL Foundation Series Class 10

2. The graph of a polynomial p ^ xh intersects with the x-axis three times at distinct points. Could 4 - 4x - x2 - x3 be an expression for p ^ xh ? 3.

Find the number of zeroes for the polynomial y = p ^ xh using the given graph. Y

X’

y = p(x) Y’

II. RELATIONSHIP BETWEEN ZEROES AND COEFFICIENTS OF A POLYNOMIAL 1.

If p ^ xh = x 4 + ax3 + bx2 + cx + d, p ^1 h = p ^2 h = p ^3 h = 0 , then find the value of p ^4 h + p ^0 h .

2. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and coefficients.

i) 4s2 - 4s + 1

ii) 3x2 - x - 4

iii) x2 + 2 2x - 6

3. Find the quadratic polynomial, whose sum and product of zeroes, respectively, are given below: 4.

1 i) 4 , - 1

ii)

1 2, 3

iii) 0, 5

Find the quadratic polynomial, whose zeroes are given below: i) 3, 4

2 ii) - 1, 3

iii)

-2 , 3 4 3

iv) 3 - 7 , 3 + 7

-1 5. Find a cubic polynomial whose zeroes are 3, -1, 3 . 6. Find a cubic polynomial with the sum, the sum of the products of its zeroes taken two at a time, and the product of its zeroes as 2, -7, and -14, respectively. 7. If a , b are zeroes of x2 + 5x + 5 then find the values of: 8.

i) a -1 + b -1

ii) a2 + b2

Find a quadratic polynomial, one of whose zeroes is

iii) a3 + b3 5 and the sum of zeroes is 4.

1 9. If a and a are zeroes of polynomial 4x2 - 2x + ^ k - 4h , then find the value of k. 10. If one zero of quadratic polynomial 2x2 + px + 4 is 2 , then find the other zero. Also, find the value of p. 43

POLYNOMIALS

11.

Find the zeroes of

3 x2 + 10x + 7 3 .

12. If a, b are zeroes of f ^ xh = x2 - 5x + k such that a - b = 1 , then find the value of k. 13. 14.

If p, q are zeroes of polynomial f ^ xh = 2x2 - 7x + 3 , find the value of p2 + q2 . Find the quadratic polynomial whose zeroes are

2 and 2 2 .

15.

4 1 Write the quadratic polynomial whose zeroes are - 5 and 3 .

16.

Find the zeroes of the polynomial 4t2 - 5 .

17.

Find a quadratic polynomial whose zeroes are 2 + 3 and 2 - 3 .

21 15 18. Find the quadratic polynomial whose sum and product of the zeroes are 8 and 6 , respectively. 19.

From a quadratic polynomial whose one zero is 8 and the product of the zeroes is -56 .

20. If one zero of the polynomial p (x) = ^a2 + 9h x2 + 45x + 6a is reciprocal of the other, find the value of a. 21. If the zeroes of the polynomial x2 + px + q are double in value to the zeroes of 2x2 - 5x - 3 , find the value of p and q. 22.

Find the zeroes of the quadratic polynomial f (x) = abx2 + ^b2 - ach x - bc .

m n 23. If m and n are the zeroes of the polynomial 3x2 + 11x - 4 , find the value of n + m .

24. For what value of k, the number -4 is a zero of the polynomial x2 - x - ^2k + 2h . Also, find the other zero. 25.

3 Find the quadratic polynomial whose zeroes are - 2 3 and - 2 .

26. If a and b are the zeroes of the polynomial 2x2 + 3x + 5 , then find the value of

a+b . ab

27. What should be added to the polynomial x2 - 5x + 4 , so that 3 is the zero of the polynomial? 1 28. If zeroes of the polynomial 6x2 + 4x + 2a are a and a , then find the value of a. III. GRAPH, IDENTIFYING SIGN OF a, b, c IF THE GRAPH IS GIVEN, AND THE MAXIMUM AND MINIMUM VALUE OF QUADRATIC POLYNOMIAL 1.

The graphs of y = ax2 + bx + c is given below. Identify the signs of a, b, and c.

IL Foundation Series Class 10

X’ P

O A

b ,-_ D _ ( ( 2a 4a Y’

The graphs of y = ax2 + bx + c is given below. Identify the signs of a, b, and c. Y

P A

X’

(-_2ab ,-_4aD (

Y’

The graphs of y = ax2 + bx + c is given below. Identify the signs of a, b and c. Y

(

X’

D O _ _b , 2a 4a

(

P Y’

For what real values of x will the expression - x2 + 4x - 8 be negative?

Find the minimum or maximum value of 2x - 8 + 7x2 .

Find the minimum or maximum value of 8 - 2x - 3x2 .

POLYNOMIALS

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1 1 1. The sum and the product of the zeroes of a quadratic polynomial are - 2 and 2 , respectively, then the polynomial is: a) 2x2 + x + 1

d) 2x2 + x - 1

b) 44

c) 8

d) 42

The polynomial whose zeroes are -5 and 4 is: a) x2 - 5x + 4

c) 2x2 - x - 1

If p ^ xh = x2 + 5x + 2 , then p ^3 h + p ^2 h + p ^0 h is: a) 40

b) 2x2 - x + 1

b) x2 + 5x - 4

c) x2 + x - 20

d) x2 - 9x - 20

4. The graph of y = p(x) given below. The number of zeroes of p(x) is: Y

X’

Y’

a) 0

b) 2

c) 4

d) 3

5. The product and sum of the zeroes of the quadratic polynomial ax2 + bx + c, respectively, are: b c c b c b c a) - a , a b) b , 1 c) a , a d) a , - a 6. If the sum of the zeroes of the quadratic polynomial 3x2 - kx + 6 is 3 then the value of k is: a) 9

b) 3

c) -3

d) 6

7. If the graph of the polynomial p(x) intersects with the x-axis at 3 distinct points, then, p(x) can be: a) Linear

b) Cubic

c) Quadratic

d) Constant polynomial

3 3 a) 2 , 2

3 3 b) - 2 , - 2

c) 3, 4

d) -3,-4

8. The zeroes of the polynomial p ^ xh = 4x2 - 12x + 9 are:

9. If a and b are the zeroes of the polynomial 2x2 + 5x + 1 , then the value of a + b + ab is: a) -2 46

b) -1

c) 1

d) 3

IL Foundation Series Class 10

10. If -1 is a zero of the polynomial f ^ xh = x2 - 7x - 8 , then the other zero is: a) 6

b) 8

c) -8

d) 1

a) 2

b) 4

c) -2

d) -4

11. If the sum of the zeroes of the polynomial p ^ xh = 2x3 - 3kx2 + 4x - 5 is 6, then the value of k is:

12. A quadratic polynomial whose sum and product of zeroes are -3 and 4, respectively, is: a) x2 - 3x + 12

b) x2 + 3x + 12

c) 2x2 + x - 24

d) None of these

13. The product of the zeroes of - 2x2 + kx + 6 is: a) 3 14.

b) -3

c) 1

d) 9

Which are the zeroes of p ^ xh = x2 + 3x - 10 ? a) 5, -2

b) -5, 2

c) -5, -2

d) None of these

15. The graph of x = p(y) is given, for a polynomial p(y). The number of zeroes of p(y) is: Y

X’

Y’

a) 4

b) 3

c) 2

d) 1

16. If the product of two of the three zeroes of the polynomial 2x3 - 9x2 + 13x - 6 is 2, the third zero of the polynomial is: -3 d) 2 17. If the two zeroes of the quadratic polynomial 7x2 - 15x - k are a reciprocal of each other, then the value of k is:

a) -1

b) -2

3 c) 2

1 a) 7

b) -7

c) 5

−7, b = −1 a) a =

b) a = 5, b = −1

18.

d) None of these

If the zeroes of the quadratic polynomial x2 + ^a + 1h x + b are 2 and -3, then: c) a = 2, b = −6

d) a = 0, b = −6

POLYNOMIALS

19. If p ^ xh = x3 - 3x2 + 2x + 5 and p ^ah = p ^ b h = p ^ c h = 0 Then the value of ^2 - ah^2 - bh^2 - ch is:

a) 3

b) 5

c) 7

d) 9

20. Given that one of the three zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of the other two zeroes is: c c b a) - a b) a c) 0 d) - a 21. If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is -1, then the product of the other two zeroes is: 22. 23.

a) b - a + 1

b) b - a - 1

c) a - b + 1

d) a - b - 1

The zeroes of the quadratic polynomial x2 + 99x + 127 are: a) Both positive

b) Both negative

c) One positive and one negative d) Both equal

The zeroes of the quadratic polynomial x 2 + kx + k, k ≠ 0,

a) Cannot both be positive

b) Cannot both be negative

c) Are always unequal

d) Are always equal

24.

If the zeroes of the quadratic polynomial ax2 + bx + c, c ! 0 are equal, then:

a) c and a have opposite signs

b) c and b have opposite signs

c) c and a have the same sign

d) c and b have the same sign

25. If one of the zeroes of a quadratic polynomial of the form x2 + ax + b is the negative of the other, then it:

a) Has no linear term and the constant term is negative

b) Has no linear term and the constant term is positive

c) Can have a linear term but the constant term is negative

d) Can have a linear term but the constant term is positive

26. 27. 28. 29. 48

The quadratic equation one of whose roots is 3 + 2 is: a) x2 + 6x + 7 = 0

b) x2 - 6x - 7 = 0

c) x2 - 6x + 7 = 0

d) x2 + 6x - 7 = 0

The quadratic equation one of whose roots is - 3 - 2i is: a) x2 + 6x + 13 = 0

b) x2 - 6x + 13 = 0

c) x2 + 6x - 13 = 0

d) x2 - 6x - 13 = 0

3 - 4i is a root of ax2 + bx + c = 0 then a + b + c = ? a) 20a

b) 2c

c) 32a

d) 20b

If a root of 8x2 - 6x + a = 0 is the square of the other then a =? a) 27, 1

b) 27, -1

c) -27, 1

d) -1, -27

IL Foundation Series Class 10

30.

Assertion (A): x2 + 4x + 5 has two zeroes.

Reason (R): A quadratic polynomial can have, at the most, two zeroes.

a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

b) Both Assertion (A) and Reason (R) are true, but Reason(R) is not the correct explanation of Assertion (A).

c) Assertion (A) is true, but Reason (R) is false.

d) Assertion (A) is false, but Reason (R) is true.

31.

Assertion (A): A quadratic polynomial whose zeroes are 5 + 2 and 5 − 2 is x 2 − 10 x + 23.

Reason (R): If a and b are zeroes of the quadratic polynomial p(x), then

p (x) = x2 - (a + b) x + ab.

a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

b) Both Assertion (A) and Reason (R) are true, but Reason(R) is not the correct explanation of Assertion (A).

c) Assertion (A) is true, but Reason (R) is false.

d) Assertion (A) is false, but Reason (R) is true.

II. FILL IN THE BLANKS 1 1 1. If a and b are the zeroes of the polynomial, f ^ xh = x2 + x + 1 , then a + = _________. b 2 2. The value of m, if the polynomial, p ^ xh = 4x - 4x - m is exactly divisible by ^ x - 3h , is __________. 3. If a zero of the polynomial x2 - 4x + 1 is (2 + 3 ) , then the other zero is __________. 4.

A cubic polynomial can have at most ________ zeroes.

5. If one of the zeroes of the quadratic polynomial ^ k - 1h x2 + kx + 1 is -3, then the value of k is ___________. 6. If the product of zeroes of the polynomial ax2 - 6x - 6 is 4, then the value of a is _________. 7.

The graph of a quadratic polynomial is a __________.

If (-4) is a zero of the polynomial x2 - 2x - ^7p + 3h , then the value of p is__________.

9. __________ should be added to the polynomial x2 + 7x - 35 so that 3 is the zero of the polynomial. 49

POLYNOMIALS

1 10. If a and a are the zeroes of the polynomial 4x2 - 2x + ^ k - 4h , then the value of k is __________. 11. If the sum of the zeroes of the quadratic polynomial 3x2 - kx + 6 is 3, then the value of k is __________. 12.

If p(x) = x 2 + 5 x + 2, then the value of p(3) + p(2) + p(0) is ________.

13. If one of the zeroes of the quadratic polynomial ax2 + bx + c is 0, then the other zero is ____________. 14. If a and b are the zeroes of the polynomial 2x2 + 5x + 1 , then the value of a + b + ab is _____________. 15. If the product of zeroes of the quadratic polynomial f ^ xh = x2 - 4x + k is 3, then the value of k is ___________. 16. The sum of the zeroes of a quadratic polynomial is 2 3 . The product of its zeroes is __________. 17. The zeroes of the polynomial x2 - x - 6 are _______________. 18. If 1 is a zero of the polynomial, P ^ xh = ax2 - 3 ^a - 1h x - 1 , then the value of a is ____________. 19. If x = 1 is a zero of the polynomial, f ^ xh = x3 - 2x2 + 4x + k , then the value of k is ___________. 20. If one of the zeroes of the polynomial is 3 + 2 , then the product of the zeroes is _________. III. SUBJECTIVE QUESTIONS 1.

The following graphs represent polynomials. Find the number of zeroes in each case. Y

X’

iii) X’

X’

ii) X’

Y’

iv) X’

Y’

vi) X’

iii) X’

iv) X’

Y’

IL Foundation Series Class 10 Y’

X’

vi) X’

Y’

Write a quadratic polynomial whose sum and product of zeroes are 3 and -2 .

If (x + a) is a factor of 2x2 + 2ax + 5x + 10 , find a.

Show that the polynomial x2 - 2x + 5 has no real zeroes.

5. If a , b are the zeroes of the polynomial, fp ^ xh = x2 - p ^ x + 1h - c such that (a + 1) (b + 1) = 0 . What is the value of c? 6. The graph of a polynomial p(x) does not intersect with the x-axis. However, it intersects with the y-axis at one point. Find the number of zeroes of p(x). 7. Find the quadratic polynomial whose zeroes are -5 and 4. 8.

If (x-2) is a factor of x2 - 3ax + 3a - 7, then find the value of a.

If -1 is a zero of the polynomial f ^ xh = x2 - 7x - 8 , then find the other zero.

1 1 10. The sum and the product of the zeroes of a quadratic polynomial are - 4 and 4 , respectively. Find the polynomial. 1 1 11. If a, b are zeroes of x2 - 4x + 1 , then find the value of + − αβ . α β

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

3.1 INTRODUCTION 3.1.1 Algebraic expression An algebraic expression is a mathematical phrase containing numbers, variables, and arithmetic operations, which can be simplified, evaluated, or solved. 3 Examples: 2 x − 1, x + 5 x − 2, 7x − y

Equation: An equation is a mathematical statement stating that two expressions are equal. Examples: x − 15 = 3 x + 1, 3 z − 1 = 17 3.1.2 Linear equation A linear equation is an algebraic equation of degree 1 involving variables raised to the power of 1. Examples: x + 11 = 2, 3 x − 2y = 5, x − y + z = 21 Linear equation in one variable

A linear equation in one variable is an algebraic equation of degree 1 that involves only one variable. Example: 5 x − 1 = 34 Note: ax + b = 0 is the general form of a linear equation in one variable. Solution of an equation

The replacement value of a variable in an equation that makes LHS = RHS is called the solution of the equation. Example: 2 x + 1 = 11 ⇒ 2 x = 10 ⇒ x = 5 ∴ 5 is the solution of an equation 2 x + 1 = 11.

IL Foundation Series Class 10

3.2 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 3.2.1 Linear equations in two variables An equation of the form ax + by + c = 0, where a ≠ 0 and b ≠ 0, ( a, b, c ∈ R ) is called a linear equation in two variables, x and y. Example: i) x + 3y − 3 = 0 ii) 2 x − 3y = 10 Note: If ax + by + c = 0 and a = 0, b ≠ 0 (or) a ≠ 0, b = 0, then ax + by + c = 0 becomes by + c = 0 or ax + c = 0. These equations are called linear equations in one variable. 3.2.2 Solutions of a linear equation in two variables Let the linear equation be ax + by + c = 0, a ≠ 0 and b ≠ 0. Then, any ordered pair of values of x and y that satisfies (i.e., makes LHS = RHS) the equation ax + by + c = 0 is called its solution. Example: The solution of 2 x + y = 7 is x = 2, y = 3. Because LHS = 2 ( 2 ) + 3 = 7 and RHS = 7. ∴ LHS = RHS ∴ Solution ( x, y ) = ( 2, 3 ). A pair of linear equations/simultaneous equations in two variables:

A pair of linear equations/simultaneous equations in two variables refers to a system of two equations, each of which is linear and involves two variables. The general form of such equations can be expressed as ax + by + c = 0, where a, b, and c are constants, and x and y are the variables. Solving this system involves finding values for x and y that satisfy both equations simultaneously. Example: x − 2y = 10 and 2 x + y = 5

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

3.3

G RAPHICAL METHOD OF SOLUTION OF A PAIR OF LINEAR EQUATIONS

3.3.1 Consistent pair of linear equations: A pair of linear equations in two variables, which has at least one solution, is called a consistent pair of linear equations. Example: x - y = 5 and 3x - 3y = 15 3.3.2 Inconsistent pair of linear equations: A pair of linear equations that have no solution is called an inconsistent pair of linear equations. Example: x + y = 2 and x + y = 5 The behaviour of lines representing a pair of linear equations in two variables and the existence of solutions are as follows: (i) The lines may intersect in a single point. In this case, the pair of equations has a unique solution (consistent pair of equations).

X’

a 1 x + b1 y + c1 = 0

’

a 2 x + b2 y + c2 = 0

IL Foundation Series Class 10

(ii) The lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations).

a 1 x + b1 y + c1 = 0 X’

a2 x + b2 y + c2 = 0

Y’ (iii) The lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations].

Y a1 x + b1 y + c1 = 0 a 2 x + b2 y + c2 = 0 X’

Y’ Examples: i) x − 2y = 0 and 3 x + 4y − 20 = 0

(The lines intersect)

ii) 2 x + 3y − 9 = 0 and 4 x + 6y − 18 = 0

(The lines coincide) 55

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

iii) x + 2y − 4 = 0 and 2 x + 4y − 12 = 0

(The lines are parallel)

3.3.3 Graphical representation of a pair of general equations: The lines represented by a pair of linear equations a1 x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0 are i) intersecting, then ii) coincident, then iii) parallel, then

a1 b1 ≠ . a2 b2

a1 b1 c1 = = . a2 b2 c2

a1 b1 c1 = ≠ . a2 b2 c2

a1 b1 c , and 1 , find out whether the given pair of linear equations a2 b2 c2 2 x + 3y − 9 = 0 and 4 x + 6y − 18 = 0 are consistent, or inconsistent. Example: On comparing the ratios

Solution: On comparing with the general pair of linear equations, we get: a1 = 2, b1 = 3, c1 = −9, a2 = 4, b2 = 6, c2 = −18 We observe that a1 b1 c1 1 = = = a2 b2 c2 2 So, the given pair of linear equations is coincident. Hence, it is consistent.

SOLVED EXAMPLES Example 1: Show that x = 2, y = 1 is solution of the system of equations 3 x − 2y = 4 and 6 x − 4y = 8. Solution: The given system of equations is: 3 x − 2y = 4

-----(i)

6 x − 4y = 8

-----(ii)

Putting x = 2 and y = 1 in equation (i) and (ii) respectively, we get: L HS 3 2 2 1 4 RHS L HS 6 2 4 1 8 RHS So, x = 2, y = 1 is a solution of the given system of equations. 56

IL Foundation Series Class 10

Example 2: On comparing the ratios

a1 b1 c , and 1 , find out whether the given pair of linear a2 b2 c2

equations 6 x − 3y + 10 = 0 and 2 x − y + 9 = 0 are consistent, or inconsistent. Solution: On comparing with the general pair of linear equations, we get: a1 = 6, b1 = −3, c1 = 10, a2 = 2, b2 = −1, c2 = 9 e observe that, W a 6 3 1 = = a2 2 1 b1 −3 3 = = b2 −1 1 c1 10 = c 9 2 a1 b1 c1 = ≠ a2 b2 c2 So, the given pair of linear equations is parallel. Hence, it is inconsistent. Example 3: On the same set of axes, graph 3 x + y = 4 and x + y = −2, then solve the equations simultaneously. Solution: Step 1: Construct tables of values. 3x + y = 4

x + y = −2

-2

-3

-4

Step 2: Graph the equations.

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Y 6

x + y = -2

-6

-4

-2

4 2

3x + y = 4

-2 -4 -6

The lines intersect at ( 3,−5 ). ∴ The solution of the simultaneous equations 3 x + y = 4 and x + y = −2 is x = 3, y = −5. Example 4: Solve the following system of equations graphically: x + 3y = 6, 2 x − 3y = 12 and, find the value of a, if 4 x + 3y = a . Solution: Graph of the equation x + 3y = 6: We have, x + 3y = 6 ⇒ x = 6 − 3y We have, x + 3y = 6 ⇒ x = 6 − 3y When y = 1, we have x = 6 − 3 = 3 When y = 2, we have x = 6 − 6 = 0 Thus, we have the following table: x

lotting the points A ( 3,1 ) and B ( 0, 2 ) and drawing a line joining them, we get the graph of the P equation x + 3y = 6. Graph of the equation 2 x − 3y = 12: 2 x − 12 3 2 × 3 − 12 When x = 3, we have y = = −2 3 We have, 2 x − 3y = 12 ⇒ y =

IL Foundation Series Class 10

When x = 0, we have y =

0 − 12 = −4 3

Thus, we have the following table: x

-2

-4

lotting the points C ( 3,−2 ) and D ( 0,−4 ) on the same graph paper and drawing a line joining them, P we obtain the graph of the equation 2 x − 3y = 12. Y 4

x + 3y = 6

B (0,2)

X’

A (3,1) 4

2x - 3y = 12

-2

P (6,0)

C (3,-2) -4

D (0,-4)

-6 Y’

Clearly, two lines intersect at P ( 6, 0 ). Hence, x = 6, y = 0 is the solution of the given system of equations. Putting x = 6, y = 0 in a = 4 x + 3y , we get a = ( 4 × 6 ) + ( 3 × 0 ) = 24 Hence, the value of the a is 24.

3.4

A LGEBRAIC METHODS OF SOLVING A PAIR LINEAR EQUATIONS

Using graphs to solve simultaneous equations can be time-consuming and sometimes inaccurate. Algebraic methods provide a better way of solving things. There are three algebraic methods: the substitution method, the elimination method, and the cross-multiplication method.

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

3.4.1 Substitution method Steps to solve simultaneous equations using the substitution method: Step (i): Express one variable in terms of the other variable from one of the two equations. Step (ii): Substitute that value of the variable in the other equation and solve it for the variable left. Step (iii): Substitute this value in any of the two original equations and solve it to find the value of the other variable. Example: Solve x + y − 5 = 0 and y − 2 = 2 x. Solution: Given equations: x + y − 5 = 0

-----(i)

y − 2 = 2x

-----(ii)

from (i), y = 5 − x substituting this in (ii), we get

( 5 − x ) − 2 = 2x ⇒ − x − 2 x = −3 ⇒ −3 x = −3 ⇒ x =1 Substituting this x = 1 in (i) or (ii), we get 'y' value. From (ii) y = 2x + 2 = 2 ( 1 ) + 2 = 4 ∴ Solution is x = 1, y = 4 3.4.2 Elimination method Steps to solve simultaneous equations using the elimination method (or) addition-subtraction method: Step (i): Multiply one or both equations (if necessary) to transform them so that either addition or subtraction will eliminate one variable. Step (ii): Solve the resulting single variable equation. Step (iii): Substitute this value in any of the two original equations and find the value of the second variable. Example: Solve 5 x − 3y + 3 = 0 and 6 x − 5y − 2 = 0. Solution:

Given equations: 5 x − 3y + 3 = 0

-----(i)

6 x − 5y − 2 = 0

-----(ii)

IL Foundation Series Class 10

(i) ⇒ 5 x − 3y = −3

-----(iii)

(ii) ⇒ 6 x − 5y = 2

-----(iv)

To eliminate the variable x, we try to get the coefficients of x as LCM of 5 and 6, i.e., 30. Multiplying on both sides of the equation (iii) by 6 and equation (iv) by 5, we get: 30 x 18y 18 30 x 25y 10

7 y 28 ⇒ y = −4 Substituting y = −4 in (i), we get: 5 x − 3 ( −4 ) + 3 = 0 ⇒ 5 x = −15 ⇒ x = −3 Hence, the solution is x = −3, y = −4. 3.4.3 Cross multiplication method Let the system of linear equations be a1 x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0, where

a1 b1 ≠ . a2 b2

∴ It has a unique solution. The coefficients of the above system should be written in the following order. x

∴

y 1 x = = b1c2 − b2 c1 c1a2 − c2a1 a1b2 − a2b1

⇒

x 1 y 1 = , = b1c2 − b2 c1 a1 b2 − a2 b1 c1a2 − c2a1 a1 b2 − a2 b1

∴x =

b1c2 − b2 c1 c a −c a ,y= 1 2 2 1 a1b2 − a2b1 a1b2 − a2b1

∴ This is the required solution. 61

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Example: Solve 5 x − 3y + 3 = 0 and 6 x − 5y − 2 = 0. Solution: Given equations are 5 x − 3y + 3 = 0 6 x − 5y − 2 = 0 The coefficients of the above system can be written in the following order. x

-3

-5

-2

-5

x y 1 3 2 5 3 3 6 2 5 5 5 6 3

x y 1 6 15 18 10 25 18

x y 1 21 28 7

⇒x=

21 28 ,y= −7 −7

Hence, the solution is x 3, y 4.

SOLVED EXAMPLES Example 1: Solve the equations 5 x + 3y = 9 and y = 7 − 3 x by substitution method. Solution: Let the given equations be: 5 x + 3y = 9

-----(i)

y = 7 − 3 x

-----(ii)

Substitute equation (ii) into equation (i) to give an equation using x only. 5 x + 3 ( 7 − 3 x ) = 9 ⇒ 5 x + 21 − 9 x = 9 ⇒ −4 x = −12 ⇒ x=3

IL Foundation Series Class 10

Now substitute x = 3 into equation (ii) to find y. y = 7 − 3 x y = 7 − 3 × 3 y = −2 ∴ The solution is x = 3 and y = −2. Example 2: Solve the following system of equations by using the method of elimination by equating the coefficients:

x y x y + + 1 = 15 and + = 15. 10 5 8 6

Solution: The given system of equations is: x y + = 14 10 5 x y + = 15 8 6 This system of equations can be re-written as: x + 2y = 140

-----(i)

3 x + 4y = 360

-----(ii)

Let us eliminate y from the equations (i) and (ii). The coefficients of y in the given equations are 2 and 4, respectively. The LCM of 2 and 4 is 4. Multiplying (i) by 2, we get: 2 x + 4y = 280

-----(iii)

3 x + 4y = 360

-----(iv)

Subtracting (iv) from (iii), we get: − x = −80 ⇒ x = 80 Putting x = 80 in equation (i), we get: 8 0 + 2y = 140 ⇒ 2y = 60 ⇒ y = 30 Hence, the solution of the given system of equations is x = 80, y = 30 . Example 3: One number is greater than thrice the other number by 2, and 4 times the smaller number exceeds the greater one by 5. Find the numbers. Solution: Let the greater number be x, and the smaller number be y. Then, according to the given conditions, we have

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

x − 3y = 2 4 y − x = 5 Thus, we have: x − 3y = 2

-----(i)

− x + 4y = 5

-----(ii)

Adding (i) and (ii), we get y = 7. Putting y = 7 in (i), we get x − 3 ( 7 ) = 2 ⇒ x = 2 + 21 = 23. ∴ Greater number = 23 and smaller number = 7. Example 4: 2 tables and 3 chairs together cost `2000, whereas 3 tables and 2 chairs together cost `2500. Find the total cost of 1 table and 5 chairs. Solution: Let the cost of a table be `x, and that of a chair be `y. Then, 2 x + 3y = 2000

-----(i)

3 x + 2y = 2500

-----(ii)

Multiplying equation (i) by 3 and (ii) by 2, we get: 6 x + 9y = 6000

-----(iii)

6 x + 4y = 5000

-----(iv)

Subtracting (iv) from (iii), we get: 5 y = 1000 y = 200 Substituting the value of y in equation (i), we get: 2 x + 3 ( 200 ) = 2000 ⇒ 2 x + 600 = 2000 ⇒ 2 x = 1400 ⇒ x = 700 So, the cost of a table = `700 and the cost of a chair = `200. Hence, cost of 1 table and 5 chairs ` x 5y

IL Foundation Series Class 10

` 700 5 200 = `1700 Example 5: A fraction is such that if the numerator is multiplied by 3 and the denominator is 18 reduced by 3, we get , but if the numerator is increased by 8 and the denominator is doubled, we 11 2 get . Find the fraction. 5 Solution: x Let the fraction be . y 3x 18 x+8 2 Then, according to the given conditions, we have = and = or y − 3 11 2y 5 1 1x = 6y − 18 and 5 x + 40 = 4y or 1 1x − 6y = −18

-----(i)

and 5 x − 4y = −40

-----(ii)

Multiplying equation (i) by 2 and (ii) by 3, we get: 2 2 x − 12y = −36

-----(iii)

1 5 x − 12y = −120

-----(iv)

Subtracting (iv) from (iii), we get: 7 x = 84 ⇒ x = 12 Substituting x = 12 in (i), we get: ⇒ 11( 12 ) − 6y = −18 ⇒ 132 − 6y = −18 ⇒ −6y = −150 ⇒ y = 25 Hence, the fraction is

12 . 25

3.5 EQUATIONS REDUCIBLE TO A PAIR OF LINEAR EQUATIONS Equations which contain the variables only in the denominators are called reciprocal equations. These types of equations need to be reducible to a pair of linear equations.

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

5 9 + = −10 is a reciprocal equation, where x ≠ 0, y ≠ 0. x y Such equations can be solved easily by assuming the reciprocals of the given variables as new variables. 8 5 3 2 Example: Solve − = 34, − = 13, where x ≠ 0, y ≠ 0 x y x y Solution: For example,

The given system of equations is 8 5 − = 34 -----(i) x y 3 2 − = 13 -----(ii) x y 1 1 = u and = v , then the given equations become x y 8u − 5v = 34 -----(iii) Let

3u − 2v = 13 -----(iv) Now solving (iii) and (iv), we get u = 3, v = −2. 1 1 Now u = 3 ⇒ = 3 ⇒ x = 3 x And v = −2 ⇒ Hence x =

1 −1 = −2 ⇒ y = 2 y

1 −1 and y = . 3 2

3.6 SIMULTANEOUS LINEAR EQUATION IN THREE VARIABLES With simultaneous equations in three variables, we have three equations to solve. Step 1: Take two of the equations and eliminate one of the variables. Step 2: Take another two of the equations and eliminate the same variable. Step 3: Solve the two new simultaneous equations from Steps 1 and 2. Step 4: Use substitution to find the values of the other two variables.

IL Foundation Series Class 10

Example: Solve simultaneously: a − b + c = 7, a + 2b − c = −4 and 3a − b − c = 3. Solution: Let, a − b + c = 7 -----(i) a + 2b − c = −4 -----(ii) 3a − b − c = 3 -----(iii) Eliminating the variable c by adding equations (i) and (ii), 2a + b = 3 -----(iv) Again, eliminating the variable c by adding equations (i) and (iii), 4a − 2b = 10 ⇒ 2a − b = 5-----(v) Now, further adding the equations (iv) and (v), we get: ⇒ 4a = 8 ⇒a=2 Substitute the value of a in equation (iv), we get: 2( 2 ) + b = 3 ⇒ 4+b = 3 ⇒ b = −1 Substitute the values of a and b in equation (i), we get: 2 − ( −1 ) + c = 7 ⇒ 2 +1+ c = 7 ⇒3+c = 7 ⇒c=4 ∴ The solution of the equations is a = 2, b = −1 and c = 4.

SOLVED EXAMPLES Example 1: Solve:

5 2 15 7 − = −1 and + = 10 where x + y ≠ 0 and x − y ≠ 0. x+y x−y x+y x−y

Solution: 1 1 = u and = v. Let x+y x−y 67

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Then, the given system of equations becomes: 5 u − 2v = −1

-----(i)

1 5u + 7 v = 10

-----(ii)

Multiplying equation (i) by 3, this system of equations becomes: 1 5u − 6v = −3

-----(iii)

1 5u + 7 v = 10

-----(iv)

Subtracting equation (iv) from equation (iii), we get: −13v = −13 ⇒ v = 1 Putting v = 1 in equation (i), we get: 1 5 u − 2 = −1 ⇒ u = 5 1 1 1 Now, u = ⇒ = ⇒ x+y =5 -----(v) 5 x+y 5 1 = 1⇒ x − y = 1 and v = 1 ⇒ -----(vi) x−y Adding equations (vi) and (v), we get 2 x = 6 ⇒ x = 3 . Putting x = 3 in equation (v), we get y = 2. Hence, x = 3, y = 2 is the solution of the given system of equations. Example 2: Solve the system of equations: 8v 3u 5uv and 6v 5u 2uv . Solution: learly, the given equations are not linear in the variables u and v but can be reduced into linear C equations by an appropriate substitution. If we put u = 0 in either of the two equations, we get v = 0. Thus, u = 0, v = 0 form one solution of the given system of equations. To find the other solutions, we assume that u ≠ 0, v ≠ 0 . Since u ≠ 0, v ≠ 0 . Therefore, uv ≠ 0. n dividing each of the given equations by uv, we get: O 8 3 − =5 -----(i) u v 6 5 − = −2 -----(ii) u v 1 1 Taking = x and = y , the above equations become: u v

IL Foundation Series Class 10

8 x − 3y = 5

-----(iii)

6 x − 5y = −2 -----(iv) Multiplying equation (iii) by 3 and equation (iv) by 4, we get: 2 4 x − 9y = 15 -----(v) 2 4 x − 20 y = −8 -----(vi) Subtracting equation (vi) from equation (v), we get: 23 1 1y = 23 ⇒ y = 11 Putting y =

23 in equation (iii), we get: 11

69 69 124 31 = 5 ⇒ 8x = + 5 ⇒ 8x = ⇒x= 11 11 11 22 31 1 31 22 Now, x = ⇒ = ⇒u= 22 31 u 22

8 x −

23 1 23 11 ⇒ = ⇒v= 11 23 v 11 Hence, the given system of equations has two solutions given by: and, y =

i) u = 0, v = 0 22 11 ii) u = , v = 31 23 Example 3: Solve: 217 x + 131y = 913 and 131x + 217 y = 827. Solution: We have, 2 17 x + 131y = 913

-----(i)

1 31x + 217 y = 827

-----(ii)

Adding equations (i) and (ii), we get: 3 48 x + 348y = 1740 ⇒ x + y = 5

-----(iii)

Subtracting equation (ii) from equation (i), we get: 8 6 x − 86y = 86 ⇒ x − y = 1

-----(iv)

Adding equations (iii) and (iv), we get: 2 x = 6 ⇒ x = 3 Putting x = 3 in equation (iii), we get

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

3 y 5 y 5 3 y 2 Hence, x = 3 and y = 2 is the solution of the given system of equations. Example 4: Find a solution to the following system: x − 2y + 3 z = 9, − x + 3y − z = −6 and 2 x − 5y + 5 z = 17 Solution: Let x − 2y + 3 z = 9

-----(i)

− x + 3y − z = −6

-----(ii)

2 x − 5y + 5 z = 17

-----(iii)

Eliminate x by adding equations (i) and (ii). y + 2 z = 3 -----(iv) he second step is multiplying equation (i) by -2 and adding the result to equation (iii). These two T steps will eliminate the variable x. − y − z = −1

-----(v)

I n equations (iv) and (v), we have created a new two-by-two system. We can solve for z by adding the two equations. z = 2 Next, we substitute z = 2 into equation (iv) and solve for y. y + 2 ( 2 ) = 3 ⇒ y+4=3 ⇒ y = −1 Finally, substitute z = 2 and y = −1 into equation (i) to get the value of x. x − 2 ( −1 ) + 3 ( 2 ) = 9 ⇒ x+2+6 = 9 ⇒ x =1 Hence, x = 1, y = −1 and z = 2 is the solution of the given system of equations.

IL Foundation Series Class 10

QUICK REVIEW •

pair of linear equations/simultaneous equations in two variables refers to a system of two A equations, each of which is linear and involves two variables. Solving this system involves finding values for x and y that satisfy both equations simultaneously.

•

A pair of linear equations in two variables can be represented and solved by the: i) Graphical method

•

ii) Algebraic method

Graphical method: The graph of a pair of linear equations in two variables is represented by two lines. i) If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent. ii) If the lines coincide, then there are infinitely many solutions - each point on the line being a solution. In this case, the pair of equations is dependent (consistent). iii) If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is inconsistent.

•

I f a pair of linear equations is given by a1 x + b1y + c1 = 0 and a2 x b2 y c2 0 , then the following situations can arise: System

No. of solutions

Condition

Graphical representation

Nature of lines

Consistent

Unique solution

a1 b1 ≠ a2 b2

Intersecting lines X

Consistent

a1 b1 c1 Infinite solution a = b = c 2 2 2

Coincident lines X

Y l1

Inconsistent No solution

a1 b1 c1 ≠ = a2 b2 c2

Parallel lines O

•

lgebraic methods: We have discussed the following methods for finding the solution(s) of a A pair of linear equations: i) Substitution method ii) Elimination method 71

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

iii) Cross-multiplication method •

here are several situations which can be mathematically represented by two equations that T are not linear to start with. But we alter them so that they are reduced to a pair of linear equations.

•

system of three equations in three variables can be solved by following a series of steps that A eliminate one variable. These steps involve rearranging the order of equations, multiplying both sides of an equation by a non-zero number, and adding or subtracting a non-zero multiple of one equation from another.

WORKSHEET - 1 I. INTRODUCTION, GRAPHICAL METHOD OF SOLUTION OF A PAIR OF LINEAR EQUATION IN TWO VARIABLES 1. Test whether the point ( −2, 3 ) lies on the line represented by each equation. i) y = 1 − x

ii) x + y = 3

iii) 2 x − y = 7

1 iv) x + y = 2 2

v) y = 3 x + 7

vi) 2y = 3 x

2. Show that the point ( 2,5 ) lies on both the lines y = 2 x + 1 and x + y = 7. 3. Complete the table of values for each equation. i) x + 2y = 0 x

-2

-1

-2

-1

y ii) y = x + 4 x y 4. Use the graph to write the solution to each pair of simultaneous equations. i) x − y = 4 and 2 x + y = 5 ii) 2 x + y = 5 and y = 2 x − 3 iii) x − y = 4 and y = 2 x − 3

IL Foundation Series Class 10

2x + y = 5 4

y = 2x - 3

X’ -4

-2

4 X

-2

x-y=4

-4

Y’

5. Graph each pair of equations on the same set of axes. Then, find the solution to the pair of simultaneous equations.

y = 2 x and y = 3 − x

ii)

y = 2 x + 1 and y = x − 4

iii)

x + y = 3 and 4 x + y = 6

iv)

y = − x + 2 and y = 3 x + 4

y = 2 x − 5 and y = 5 x + 1

vi)

2 x + y = 6 and y = 1 − x

vii)

y = 7 − x and y = 3 x + 5

viii)

x + 2y = 7 and 2 x − y = 4

ix)

3 x − 2y = 12 and x + 2y = 8

y = x + 3 and 2 x − y = 2

xi)

5 x − y = 5 and x + y = 4

xii)

15 x + 3y = 20 and y = x − 4

6. i) On the same set of axes, draw the graphs of y = 1 − 2 x and 2 x + y = 4. ii) Why isn’t there a solution to the simultaneous equations y = 1 − 2 x and 2 x + y = 4? a b c 7. On comparing the ratios 1 , 1 and 1 , find out whether the system of equations describe a2 b2 c2 consistent and independent, consistent and dependent, or inconsistent: i) y = x + 4 and y = x − 4

ii) y = x + 3 and y = 2 x + 6

iii) x + y = 4 and −4 x + y = 9

iv) 3 x + y = 3 and 6 x + 2y = 6

v) y − x = 5 and 2y − 2 x = 8

vi) 4 x − 2y = 6 and 6 x − 3y = 9

8. The sides of an angle are parts of two lines whose equations are 2y + 3 x = −7 and 3y − 2 x = 9. The angle’s vertex is the point where the two sides meet. Find the coordinates of the vertex of the angle. 9. The graphs of y − 2 x = 1, 4 x + y = 7, and 2y − x = −4 contain the sides of a triangle. Find the coordinates of the vertices of the triangle. 10. The solution of the system of equations Ax + y = 5 and Ax + By = 20 is ( 2,−3 ). What are the values of A and B? Justify your reasoning.

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

11. Write three equations such that they form a system of equations with y = 5 x − 3. The systems should have no, one, and infinitely many solutions, respectively. 12. Determine the values of m and n so that the following system of linear equations has an infinite number of solutions:

( 2m − 1 ) x + 3y − 5 = 0 3x + ( n − 1 ) y − 2 = 0 13. Determine the value of k so that the following linear equations have no solution:

( 3 k + 1 ) x + 3y − 2 = 0 ( k2 + 1 ) x + ( k − 2 ) y − 5 = 0 14. For which value(s) of λ , do the pair of linear equations λ x + y = λ 2 and x + λ y = 1 have i) no solution? ii) infinitely many solutions? iii) a unique solution? 15. For what value of α the system of equations α x + 3y = α − 3 and 12x + α y = α will have no solution? 16. Determine, graphically, the vertices of the triangle formed by the lines y = x, 3y = x, x + y = 8. 17. Draw the graphs of the pair of linear equations x − y + 2 = 0 and 4 x − y − 4 = 0. Calculate the area of the triangle formed by the lines so drawn and the x-axis. II. ALGEBRAIC METHODS OF SOLVING A PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 1. Solve using the substitution method. i) y = 3 − 2 x and 3 x + y = 5

ii) 3y + x = 4 and x = 2y − 1

iii) 3 x + 5y = 3 and x = 8 − 4y

iv) 9 x − 2y = 3 and 3 x − 6 = y

v) 3s − 4t = 14 and 5s + t = 8

vi) m − 2n = 16 and 4m + n = 1

2. Solve using the elimination method. i) x + 3y = 7 and − x + 4y = 7

ii) x − y = 3 and 2 x + y = 6

iii) x − 2y = 11 and 3 x + 2y = 17

iv) 5 x − 3y = 8 and −5 x + y = 4

v) 9 x + 3y = −3 and 2 x − 3y = −8

vi) 6 x − 3y = 18 and 6 x + 3y = −12

3. Solve using the cross-multiplication method. i) 5 x + 3y = 19 and x − 6y = 11

ii) 3 x + 2y = 3 and 9 x − 8y = −2

IL Foundation Series Class 10

iii) 5 r − 3s = 24 and 3 r + 5s = 28

iv) 5 x − 7 y = −16 and 2 x + 8y = 26

v) 6s + 9t = 12 and 4s + 6t = 5

vi) 10a + 6b = 8 and 5a + 3b = 2

4. If ( 1, 2 ) and ( −3, 4 ) are two solutions of f ( x ) = mx + b , find m and b. ⎛ 3 ⎞ 5. If ( 0,−3 ) and ⎜ − , 6 ⎟ are two solutions of px − qy = −1, find p and q. ⎝ 2 ⎠ 6. Determine a and b for which ( −4, −3 ) is a solution of the system: ax + by = −26 and bx − ay = 7 7. Solve for x and y in terms of a and b: 5 x + 2y = a and x − y = b 8. Weights of atoms and molecules are measured in atomic mass units (u). A molecule of C2 H6 (ethane) is made up of 2 carbon atoms and 6 hydrogen atoms and weighs 30.07 u. A molecule of C3 H8 (propane) is made up of 3 carbon atoms and 8 hydrogen atoms and weighs 44.097 u. Find the weights of a carbon atom and a hydrogen atom.

Ethane molecule

Propane molecule

9. You want to burn 380 calories during 40 minutes of exercise. You burn about 8 calories per minute inline skating and 12 calories per minute swimming. How long should you spend doing each activity? 10. At a school concert, there were 640 guests. There were 70 more women than men. How many of the audience were men? 11. At a circus, there were twice as many children as there were adults in attendance. Altogether, 1020 attended the circus. How many were children? 12. Tickets to a concert cost `5 for children and `14 for adults. Altogether, 650 people attended the concert and ticket sales totalled `5824. Find the number of children that attended the concert. 13. Tracey bought a total of 17 DVDs and CDs. Each DVD cost her `25 and each CD cost `18. Altogether, Tracey spent `390. How many DVDs did she buy? 14. Aaron is three times as old as Sejuti. The sum of their ages is 48. How old are Aaron and Sejuti?

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

15. The sum of the ages of Mrs. Mehta and her daughter Sweety is 70. The difference between their ages is 38 years. How old is Sweety? 16. Susan invested a certain amount of money in two schemes, A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received ` 1860 as annual interest. However, is she had interchanged the amount of investment in the two schemes, she would have received ` 20 more as annual interest. How much money did she invest in each scheme? 17. The coach of a cricket team buys 7 bats and 6 balls for `3800. Later, he buys 3 bats and 5 balls for `1750. Find the cost of each bat and each ball. 18. A two-digit number is 4 more than 6 times the sum of its digits. If 18 is subtracted from the number, the digits are reversed. Find the number. 19. The difference between two numbers is 26, and one number is three times the other. Find them. 20. The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction. III. EQUATIONS REDUCIBLE TO A PAIR OF LINEAR EQUATIONS, SIMULTANEOUS LINEAR EQUATIONS IN THREE VARIABLES 1. Solve the pair of equations: 2 1 5 2 i) + = 0 and + = −5 x y x y ii)

1 3 6 5 − = 2 and + = −34 x y x y

2. Ritu can row downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. 3. A motorboat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream. 4. Abdul travelled 300 km by train and 200 km by taxi. It took him 5 hours and 30 minutes. But if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi. 5. 2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroidery and that taken by 1 man alone. 6. Meena went to a bank to withdraw `2000. She asked the cashier to give her `50 and `100 notes only. Meena got 25 notes in all. Find how many notes `50 and `100 she received. 7. There are two examination rooms, A and B. If 10 candidates are sent from A to B, the number of students in each room is the same. If 20 candidates are sent from B to A, the number of 76

IL Foundation Series Class 10

students in A is double the number of students in B. Find the number of students in each room. 8. Solve the system of linear equations: i) x − 3y + 6 z = 21

3 x + 2y − 5 z = −30

2 x − 5y + 2 z = −6

ii) x + 2y + 5 z = −1

2x − y + z = 2

3 x + 4y − 4 z = 14

iii) x − 6y − 2 z = −8

− x + 5y + 3 z = 2

3 x − 2y − 4 z = 18

9. Your aunt receives an inheritance of `20,000. She wants to put some of the money into a savings account that earns 2% interest annually and invest the rest in certificates of deposit (CDs) and bonds. A broker tells her that CDs pay 5% interest annually and bonds pay 6% interest annually. She wants to earn `1000 interest per year, and she wants to put twice as much money in CDs as in bonds. How much should she put in each type of investment? 10. Model High prevailed in Saturday’s track meet with the help of 20 individual-event placers, earning a combined 68 points. A first-place finish earns 5 points, a second-place finish earns 3 points, and a third-place finish earns 1 point. Model High had a strong second-place showing as many second-place finishers as first and third-place finishers combined. Find how many athletes finished in each place. 11. The sum of three numbers is 85. The second is 7 more than the first. The third is 2 more than four times the second. Find the numbers. 12. In triangle ABC, the measure of angle B is three times that of angle A. The measure of angle C is 20 more than that of angle A. Find the angle measures.

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. Which is the correct expression for the statement "thrice a number is 10 less than twice the another number"?

a) 3 x = 2y − 10

b) 2 x = 3y − 10

c) 3 x − 10 = 2y

d) none of these

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

2. How many solutions are possible for the given set of equations? 2 x − 3y + 1 = 0 3 x + 2y + 2 = 0 a) One

b) More than one

c) Infinite

d) None

3. The number of solutions of the given set of equations is: x + 2y + 3 = 0 3 x + 6y + 9 = 0 a) Only one

b) Infinite

c) Exactly two

d) None

4. The number of solutions for the given set of equations is: 2 x − 3y + 2 = 0 6 x + 9y + 5 = 0 a) Only one

b) Infinite

c) No solution

d) Exactly two

5. x = 1 and y = 3 is a solution of the given set of equations. 2 x + ky = 2 x − ny = 2 What are the values of n and k? a) n = 3, k = 2

b) n =

−1 1 ,k = 3 3

1 −1 c) n = , k = 3 3

d) none of these

6. If x ∈ A and y ∈ A such that A = {1, 2, 3, 4 }, then find the solution set for x + y = 6. a) { ( 1, 5 ), ( 2, 4 ), ( 3, 3 ), ( 4, 2 ) }

b) { ( 1, 5 ), ( 2, 4 ), ( 3, 3 ) }

c) { ( 2, 4 ), ( 3, 3 ), ( 4, 2 ) }

d) { ( 3, 3 ), ( 4, 2 ) }

7. What is the solution of the given set of equations? 0.2u + 0.3v = 1.3 0.4u + 0.5v = 2.3 a) u = 2, v = 3

b) u = −2, v = 3

c) u = 2, v = 2

d) u = 3, v = 3

c) r = 1, s = 1

d) r = 2, s = 0

8. Find r and s for the following set of equations, 2 r + 3s = 0

3 r − 8s = 0

a) r = 0, s = 1

b) r = 0, s = 0

IL Foundation Series Class 10

9. Find the values of x and y for the following set of equations. x 2y + +1 = 0 2 3 y x − −3 = 0 3 a) x = 2, y = −2

b) x = 3, y = −3

c) x = −2, y = −3

d) x = 2, y = −3

10. For what values of p and q, the given set of equations will have infinite solutions? 2 x + 3y = 7

( p − q ) x + ( p + q )y = 3p + q − 2 a) p = 5, q = 1

b) p = −5, q = −1

c) p = 5, q = −1

d) p = −1, q = 5

11. By solving the following set of equations, find u and v (if u ≠ 0 and v ≠ 0 ). 6u + 3v = 6uv 2u + 4v = 5uv a) u = 1, v = 2

b) u = 2, v = 1

c) u = −1, v = 2

d) u = −1, v = −2

5 −1 b) x = , y = 2 2

1 −1 c) x = , y = 2 2

d) None of these

12. Solve for x and y. 2 3 + =2 x+y x−y 5 10 35 + = x+y x−y 6 a) x = 5, y = −1

13. Find the values of u and v for the following equations. 1 1 3 + = 3u + v 3u − v 4 1 1 −1 − = 2 ( 3u + v ) 2 ( 3u − v ) 8 a) u = 1, v = −1

b) u = −1, v = −1

c) u = 1, v = 1

d) u = −1, v = 1

14. Solve for x and y. ax + by = c bx + ay = 1 + c a) x =

bc − ac + b bc − ac − a ,y = 2 2 2 b −a b − a2

c) x = a, y = b

b a ,y = 2 2 b −a b − a2 bc − ac + b d) x = b, y = 2 b − a2 b) x =

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

15. Solve for x and y. ax + by = a − b bx − ay = a + b a) x = 1, y = 1

b) x = −1, y = 1

c) x = 1, y = −1

d) x = −1, y = −1

16. Solve for x and y. x − y = 0.9 11 =1 2( x + y ) a) x =

−16 23 ,y = 5 10

b) x =

16 23 ,y = 5 10

c) x =

23 16 ,y = 10 5

d) none of these

17. Solve for x and y. x + y = a + b ax − by = a 2 − b 2 a) x = b, y = −b

b) x = b, y = a

c) x = −a, y = −b

d) x = a, y = b

b) x = a 2 , y = b

c) x = a, y = b 2

d) x = a 2 , y = b 2

18. Find x and y. x y + = a+b a b x y + 2 =2 2 a b a) x = a, y = b

19. Find the value of p for which the given set of equations will have a unique solution. px + 2y = 5 3x + y = 1 a) p = 6

b) All values

c) Any real values except 6

d) None of these

20. Find the value of t for which no solution of a given set of equations is possible. 3x + y = 1

( 2t − 1 ) x + ( t − 1 ) y = 2t + 1 a) t = 2

b) t = −2

c) t =

1 2

d) t = −

1 2

IL Foundation Series Class 10

21. Find the value of k for which a given set of equations will have infinitely many solutions. kx + 3y = k − 3 12x + kx = k a) k = 6

b) k = −6

c) k ≠ 6

d) Not possible

22. Find the value of k for which the following two lines are coincident. 2 x + 3y = 4

( k + 2 ) x + 6y = 3 k + 2 a) k ≠ 2

b) k = 2

c) k = −2

d) k ≠ −2

23. Find the values of α and β for which the given set of equations will have infinite solutions. 2 x + 3y = 7 2α x + ( α + β ) y = 28 a) α = −4, β = −8

b) α = −4, β = 8

c) α = 4, β = −8

d) α = 4, β = 8

24. Find the value of m for which the given set of equations has no solution. 3x + y = 1

( 2m − 1 ) x + ( m − 1 ) y = 2m + 1 a) m = 1

b) m = −1

c) m = 2

d) m = −2

25. Find the values of x and y (if p ≠ 0 and q ≠ 0 ). p ( x + y ) + q ( x − y ) = p2 − pq + q 2 p ( x + y ) − q ( x − y ) = p2 + pq + q 2 a) x =

q2 q2 ,y = p − 2p 2p

b) x =

q2 p2 ,y = 2p 2q

c) x =

p p ,y = 2 q q

d) x =

q2 q2 ,y = p + 2p 2p

26. There is a certain two-digit number, the sum of whose digits is 11. If 45 is added to the number, the digits in the number are reversed. The original number is: a) 83

b) 38

c) 56

d) 65

27. ABCD (taken in order) is a cyclic quadrilateral in which ∠A = (x + y + 10) , ∠B = (y + 20) , ∠C = (x + y − 30), and ∠D = (x + y). Find the values of x and y. a) x = 40 , y = 60

b) x = 60 , y = 40

c) x = 20 , y = 80

d) None of these

28. The ratio of the present ages of a father and his son is 3:1. After 14 years, the ratio of their ages will be 2:1. What are the present ages of the father and his son, respectively? a) 42, 14

b) 30, 10

c) 45, 15

d) 60, 20

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

29. If the length of a rectangle is increased by 3 metres and the breadth is increased by 2 metres, its area will increase by 67sq .metres. If its length is decreased by 5 metres and its breadth is increased by 3 metres, its area will decrease by 9 sq.metres. Find the area of the rectangle (in sq. metres). a) 150

b) 153

c) 162

d) 154

30. In a triangle PQR,∠R is greater than ∠Q by 9. If ∠P = α  , ∠Q = (3α − 2), and ∠R = β , then find all the angles of the triangle. a) 30 , 60 , 90

b) 45 , 45 , 90

c) 25 , 73 , 82

d) None of these

31. Which of the following is correct? Column I

Column II

(i) 6u - 5v = 20, 12u - 10v = 40

(p) Unique solution

(ii) x - 3y = 3, 3x - 9y = 2

(q) Infinitely many solutions

(iii) 3x - 5y = 25, 7x - 2y = 15

(r) No solution

a) (i)-q, (ii)-r, (iii)-p

b) (i)-q, (ii)-p, (iii)-r

c) (i)-r, (ii)-q, (iii)-p

d) (i)-r, (ii)-p, (iii)-q

32. If x men can do a piece of work in y days, in how many days will z men do the same work? a)

xz y

xy z

33. Find the values of x and y if a) x = 4, y = 3

yz x

d) xyz

5 2 7 36 24 − = and 1, y x 6 x y

b) x = −4, y = +3

c) x = −4, y = −3

d) x = 4, y = −3

34. A fraction becomes 2 when 9 is added to its numerator and 1 when 2 is subtracted from its denominator. Then the fraction is: 6 7 5 4 a) b) c) d) 8 9 7 7 35. What should be the value of p if 3 x + 2y = 8 and 6 x + 4y = p have infinitely many solutions? a) 3

b) 16

c) 5

d) 6

IL Foundation Series Class 10

36. What should be the value of m in the pair of equations to have a unique solution? 4 x + my + 9 = 0 3 x + 4y + 8 = 0

16 15 d) m ≠ 3 4 37. If the sum of two numbers is 640 and their difference is 280, then the numbers are:

a) m ≠ 16

b) m ≠ 15

c) m ≠

a) 140, 500

b) 180, 460

c) 130, 510

d) 150, 490

38. A certain number of two digits is four times the sum of the digits. If 9 is added to the number, the digits in the number are reversed. Find the number. a) 13

b) 15

c) 12

d) 14

39. Six years ago, a man was three times as old as his son. In 6 years’ time, he will be twice as old as his son. Find their present ages. a) 30, 15

b) 40, 20

c) 42, 18

d) 41, 19

40. The total salary of 15 men and 8 women is `3050. The difference in salaries of 5 women and 3 men is `50. Find the sum of the salaries of 3 men and 3 women. a) ` 900

b) ` 750

c) ` 950

d) ` 1000

II. FILL IN THE BLANKS 1. If 99 x + 101y = 400 and 101x + 99y = 600, then x + y is __________. 2. The number of common solutions for the system of linear equations 5 x + 4y + 6 = 0 and 10 x + 8y = 12 is __________. 3. The sum of the heights of A and B is 320 cm, and the difference of heights of A and B is 20 cm. The height of B is __________. 4. If a : b = 7 : 3 and a + b = 20, then b = __________. 5. If

1 1 1 1 + = k and − = k, then the value of y is __________. (0 / does not exist). x y x y

6. If p + q = k, p − q = n and k > n , then q is __________. (positive/negative) 7. If a + b = x, a − b = y and x < y, then b is __________. (positive/negative) 8. If 3a + 2b + 4c = 26 and 6b + 4a + 2c = 48, then a b c __________. 9. The sum of the ages of X and Y,12 years ago was 48 years, and the sum of the ages of X and Y,12 years hence will be 96 years. The present age of X is __________. 10. The number of non-negative integral solutions for the equation 2 x + 3y = 12 is __________.

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

11. Two distinct natural numbers are such that the sum of one number and twice the other number is 6. The two numbers are __________. 12. If 2 x + 3y = 5 and 3 x + 2y = 10, then x − y = __________. 13. If a + b = p and ab = p, then the value of p is __________. (where a and b are positive integers). 14. If ( a + b ) : ( b + c ) : ( c + a ) = 6 : 7 : 8 and ( a + b + c ) = 14, then the value of c is ______. III. SUBJECTIVE QUESTIONS 1. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid `27 for a book kept for seven days, while Susy paid `21 for a book she kept for five days. Find the fixed charge and the charge for each extra day. 2. The cost of 4 pens and 4 pencil boxes is `100. Three times the cost of a pen is `15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box. 3. A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number. 4. A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number. 5. Solve: 331a + 247 b = 746 and 247 a + 331b = 410. 6. Six gallery seats and three balcony seats for a play were sold for `162. Four gallery seats and five balcony seats were sold for `180. Find the price of a gallery seat and the price of a balcony seat. 7. If the numerator of a fraction is increased by 2 and the denominator is decreased by 4, then it becomes 2. If the numerator is decreased by 1 and the denominator is increased by 2, 1 then it becomes . Find the fraction. 3 8. Solve:

1 1 1 1 1 1 + = 6, + = 7 and + = 5. x y y z z x

9. Solve:

2 1 5 4 − = 11 and + = 8. x+y x−y x+y x−y

10. Jaydeep starts his job with a certain monthly salary and earns a fixed increment in his monthly salary in the middle of every year, starting from the first year. If his monthly salary was `78000 at the end of 6 years of service and ` 84000 at the end of 12 years of service, find his initial salary and annual increment. 6 7 of a number, but instead, he multiplied it by . As a result, he got an 7 6 answer, which was more than the correct answer by 299. What was the number?

11. Alok was asked to find

IL Foundation Series Class 10

12. Shriya has a certain number of 25 paise and 50 paise coins in her purse. If the total number of coins is 35 and their total value is `15.50, find the number of coins of each denomination. 13. For what value of k will the following pair of linear equations have no solution? 2 x 3y 1 and 3 k 1 x 1 2 k y 2 k 3. 14. Solve:

x y x y + = a 2 + b 2 and 2 + 2 = a + b . a b a b

15. Solve: x − 2y + z = 0, 9 x − 8y + 3z = 0 and 2 x + 3y +5 z = 36. 16. Four friends, P, Q, R and S, have some money. The amount with P equals the total amount with the others. The amount with Q equals one-third of the total amount with the others. The amount with R equals one-fifth of the total amount with the others. The amount with S equals one-eleventh of the total amount with the others. The sum of the smallest and the largest amounts with them is ` 210. Find the sum of the amounts with the other two (in `). 17. The population of a town is 25000. If, in the next year, the number of males were to increase by 5% and that of females by 3%, the population would grow to 26010, find the number of males and females in the town at present. 18. What is the solution set of

12 5 8 6 + = −7 and + = −10? 2 x + 3y 3 x − 2y 2 x + 3y 3 x − 2y

19. Tito purchased two varieties of ice cream cups, vanilla and strawberry, spending a total amount of ` 330. If each vanilla cup costs ` 25 and each strawberry cup costs `40, then in how many different combinations could he have purchased the ice cream cups? 20. While covering a distance of 30 km , Ajeet takes 2 hours more than Amit. If Ajeet doubles his speed, he will take 1 hour less than Amit. Find their speeds of walking. 21. A railway half ticket costs half the full fare, and the reservation charge is the same on a half ticket as on a full ticket. One reserved first-class ticket from Mumbai to Ahmedabad costs `216 and one full and one half reserved first-class ticket costs `327. What is the basic firstclass full fare, and what is the reservation charge? 22. Solve the system of linear equations: i) 5 x − 4y + 4 z = 18

− x + 3y − 2 z = 0

4 x − 2y + 7 z = 3

ii) −5 x + 3y + z = −15

10 x + 2y + 8 z = 18

15 x + 5y + 7 z = 9

23. The sum of three numbers is 26. Twice the first minus the second is 2 less than the third. The third is the second minus three times the first. Find the numbers.

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

24. In triangle ABC , the measure of angle B is twice the measure of angle A. The measure of angle C is 80 more than that of angle A. Find the angle measures. 25. If ab = 15, bc = 12 and ac = 5, then find the values of a, b and c. 26. Find the value of x + y + z , if x + y + xy = 3, y + z + yz = 8 and x + z + xz = 15.

4.1

QUADRATIC EQUATIONS

INTRODUCTION TO QUADRATIC EQUATIONS

If p(x) is a quadratic polynomial, then p(x) = 0 is called a quadratic equation. OR An equation of degree two is called a quadratic equation. The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, c ∈ R and a ≠ 0.

SOLVED EXAMPLES Example 1: Check whether the following are quadratic equations. i) (x - 2)2 + 1 = 2x - 3

3 ii) x + x = x2

iii) x(x + 1) + 8 = (x + 2)(x - 2)

Solution: i) (x 2)2 1 2 x 3 x2 4x 4 1 2x 3 x2 4x 5 2x 3 0 x2 6x 8 0 It is in the form of ax2 + bx + c = 0. Therefore, the given equation is a quadratic equation. 3 ii) x + x = x2 ⇒ x x+ 3 = x2 2

⇒ x2 + 3 = x3 ⇒ x3 -x2 - 3 = 0 The degree of this equation is 3. So, the given equation is not a quadratic equation. iii) x(x + 1)+8 = (x + 2)(x - 2) ⇒ x2+ x + 8 = x 2- 4 ⇒ x2+ x + 8 - x2+ 4 = 0 ⇒ x + 12 = 0 87

QUADRATIC EQUATIONS

It is not in the form of ax2 + bx + c = 0. So, the given equation is not a quadratic equation.

4.2 FORMATION OF QUADRATIC EQUATION SOLVED EXAMPLES Represent the situation in the form of quadratic equations: Example 1: The product of two consecutive positive integers is 306. Formulate the quadratic equation whose roots are these integers. Solution: Let two consecutive positive integers be x and x + 1. The product = x(x + 1) It is given that the product = 306 ∴ x(x + 1) = 306 ⇒ x2 + x = 306 ⇒ x2 + x - 306 = 0 ⇒ x2 + 18x - 17x - 306 = 0 ⇒ x(x + 18) - 17(x + 18) = 0 ⇒ (x +18)(x - 17) = 0 ∴ x = - 18(or )x = 17 If x = - 18 x + 1 = -18 + 1 = -17 Now, the quadratic equation with roots –18,–17. ⇒ x2 - (-18 -17)x + (-18)(-17)=0 ⇒ x2 + 35x + 306 = 0 If x = 17 x + 1 = 17+1 = 18 Now, the quadratic equation with roots 18, 17. ⇒ x2 - (18 + 17)x +(18)(17) = 0 ⇒ x2 - 35x + 306 = 0 Example 2: Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. Formulate the quadratic equation.

IL Foundation Series Class 10

Solution: Let Rohan's present age be 'x' years. Then his mother's age is (x + 26) years. After 3 years Rohan's age = (x + 3) years After 3 years Rohan's mother age = (x + 3 + 26) years = (x + 29) years According to the question, after 3 years, the product of Rohan's and his mother's age = 360 ∴ (x + 3)(x + 29) = 360 ⇒ x2 + 29x + 3x + 87 = 360 ⇒ x2 + 32x - 273 = 0 ∴ This is the required equation.

4.3 SOLVING QUADRATIC EQUATIONS 4.3.1 Roots of a quadratic equation Let p(x) = 0 be a quadratic equation. Then, the zeroes of the polynomial p(x) are called the roots of the quadratic equation. A quadratic equation can have at most two real roots. Finding the roots of a quadratic equation is known as solving the quadratic equation. 4.3.2 Solution of quadratic equation by factorisation method To solve ax2 + bx + c = 0, ax2 + bx + c must be resolvable into factors. i)

Write ax2 + bx + c as a product of its factors.

ii) If this product is zero, then one of the factors should be zero. Recall that the value of x to which the value of a polynomial becomes zero is called a zero of the polynomial. It is also called a solution of the equation. Therefore, finding the zeroes of the factors which give the solutions of the equation ax2 + bx + c = 0 are called roots of the equation.

SOLVED EXAMPLES Example 1: Solve the following quadratic equations by factorisation. i) 2x2 - 5x + 3 = 0

ii) x2 + 6x + 5 = 0

iii) 3x2 - 2 6 x + 2 = 0

Solution: i) 2x2 - 5x + 3 = 0

QUADRATIC EQUATIONS

2x2 - 2x - 3x + 3 = 0 ⇒ 2x ^ x - 1 h - 3 ^ x - 1 h = 0 ⇒ ^ x - 1h^2x - 3h = 0 Thus, either x - 1 = 0 or 2x - 3 = 0 3 x 1 or x 2 3 ∴ x = 1 and 2 are the roots of the equation. ii) x2 + 6x + 5 = 0 ⇒ x2 + 5x + x + 5 = 0 ⇒ x ^ x + 5h + 1 ^ x + 5h = 0 ⇒ ^ x + 1h^ x + 5h = 0 ⇒ x + 1 = 0 or x + 5 = 0 ∴ x =- 1 or x =- 5 ∴ - 1 and - 5 are the roots of the equation. iii) 3x2 - 2 6 x + 2 = 0 ⇒ 3x2 - 6 x - 6 x + 2 = 0 ⇒ 3 x^ 3 x - 2 h - 2 ^ 3 x - 2 h = 0 ⇒ ^ 3 x - 2 h^ 3 x - 2 h = 0 ⇒ 3x- 2 = 0 2 2 = 3 ∴ x= 3 This root is repeated twice, one for each repeated factor, 3 x - 2 . 2 2 Therefore, the roots of 3x2 - 2 6 x + 2 = 0 are 3 , 3 . -3 5 4 Example 2: Solve the quadratic equation by factorisation method. x - 3 = 2x + 3 , x ! 0, 2 Solution: 5 4 We have x - 3 = 2x + 3 4 - 3x 5 x = 2x + 3 ⇒ ^4 - 3xh^2x + 3h = 5x ⇒

⇒ 8x + 12 - 6x2 - 9x = 5x

⇒ 6x2 + 9x - 8x + 5x - 12 = 0 ⇒ 6x2 + 6x - 12 = 0 ⇒ x2 + x - 2 = 0 ⇒ x2 + 2x - x - 2 = 0 ⇒ x ^ x + 2h - 1 ^ x + 2h = 0 90

IL Foundation Series Class 10

⇒ ^ x - 1h^ x + 2h = 0

⇒ x - 1 = 0 or x + 2 = 0 ⇒ x = 1 or x =- 2 ⇒ x = 1 or - 2 Example 3: The length of a rectangular plot is greater than thrice its breadth by 2 m. The area of the plot is 120 m2. Find the length and breadth of the plot. Solution: Let the breadth be 'x' m. Then the length =(3x + 2) m Area of the rectangular plot = 120 m2 ⇒ l × b = 120 ⇒ (3x + 2) # x = 120 ⇒ 3x2 + 2x = 120 ⇒ 3x2 + 2x - 120 = 0 ⇒ 3x2 + 20x - 18x - 120 = 0

⇒ x ^3x + 20h - 6 ^3x + 20h = 0 ⇒ ^ x - 6h^3x + 20h = 0 ⇒ x = 6 or

- 20 3

- 20 Breadth can't be negative, so ignore 3 . Therefore length = (3x + 2) m

= 3 ( 6) 2 m

= 18 + 2 = 20 m and breadth = 6 m

4.3.3 Solution of a quadratic equation by using the quadratic formula (Shreedharacharya's rule) Consider the quadratic equation ax2 + bx + c = 0, a ≠ 0. We have ax2 + bx + c = 0 (Dividing throughout by 'a') ⇒

b c x2 + a x + a = 0

⇒

-c b x2 + a x = a

⇒

-c b x2 + 2 a 2a k x = a

QUADRATIC EQUATIONS

b 2 Adding a 2a k on both sides

b b 2 -c b 2 ⇒ ^ xh2 + 2 a 2a k x + a 2a k = a + a 2a k

b 2 -c b2 ⇒ a x + 2a k = a + 2 4a 2 - 4ac + b2 b ⇒ a x + 2a k = 4a 2 [Taking the square root of both sides and assuming b2 - 4ac ≥ 0] b2 - 4 ac 4a2 b2 - 4ac b ⇒ x + 2a = ! 2a 2 b - 4ac -b ⇒ x = 2a ! 2a 2 - b ! b - 4ac ⇒ x= 2a - b + b2 - 4ac ⇒ x= 2a - b - b2 - 4ac (or) x = 2a b ⇒ x + 2a = !

- b + b2 - 4ac - b - b2 - 4ac ( or ) . 2a 2a This formula for finding the roots of a quadratic equation is known as the quadratic formula. So, the roots of ax 2 bx c 0 are

Note: Here, b2-4ac is known as its discriminant and is generally denoted by ' D '.

SOLVED EXAMPLES Example 1: Find the roots of the following quadratic equations, if they exist, using the quadratic formula. i) 3x2 + 2x - 1 = 0 Solution: i) 3x2 + 2x - 1 = 0 Here, a = 3, b = 2, c =- 1 D = b2 - 4ac = ^2 h2 - 4 ^3 h^- 1h

= 4 + 12 = 16 2 0

So, the given equation has real roots given by

-b ! D - 2 ! 16 - 2 !4 = 2a 6 2 #3 = ⇒ x = - 2 + 4 or - 2 - 4 6 6 x=

ii) x2 + x + 1 = 0

IL Foundation Series Class 10

-6 2 1 ⇒ x = 6 or 6 & x = 3 or - 1 ii)

x2 + x + 1 = 0

Here, a = 1, b = 1, c = 1

D = b2 - 4ac = ^1 h2 - 4 # ^1 h # ^1 h

= 1 - 4 =- 3 1 0 ∴ 0

Hence, the given equation has no real roots.

4.4 NATURE OF ROOTS OF A QUADRATIC EQUATION Nature of Roots: Let the quadratic equation be ax2 + bx + c = 0. Then x =

- b ! b2 - 4ac 2a

So, a quadratic equation ax2 + bx + c = 0 i)

has no real roots if b2 - 4ac 1 0

ii) has two equal roots if b2 - 4ac = 0 iii) has two distinct real roots if b2 - 4ac 2 0 Note: Discriminant, b2 - 4ac is denoted by D.

SOLVED EXAMPLES Example 1: Find the discriminant of the following quadratic equations. i) x2 - 4x + 3 = 0

ii) 3x2 - 5x - 2 = 0

Solution: i) The given quadratic equation is x2 - 4x + 3 = 0 Here, clearly a = 1, b =- 4, c = 3 D = b2 - 4ac = (- 4) 2 - 4 # 1 # 3 = 16 - 12 = 4 ii) The given quadratic equation 3x2 - 5x - 2 = 0 Here, a = 3, b =- 5, c =- 2 D = b2 - 4ac = (- 5) 2 - (4 # 3 # - 2) = 25 + 24 = 49 93

QUADRATIC EQUATIONS

Example 2: Find the nature of the roots of the following quadratic equations. i) 2x2 + x - 1 = 0

ii) x2 - 4x + 4 = 0

iii) x2 + x + 1 = 0

Solution: i) The given quadratic equation is 2x2 + x - 1 = 0 Here, a = 2, b = 1, c =- 1 D = b2 - 4ac = ^1 h2 - 4 ^2 h^- 1h = 1+8 = 9 Since D > 0 ∴ The given equation has real and distinct roots. ii)

x2 4x 4 0

Here, a = 1, b =- 4, c = 4 D = b2 - 4ac = ^- 4h2 - 4 ^1 h^4 h = 16 - 16 = 0 Since D = 0 ∴ The given equation has real and equal roots. iii) x2 + x + 1 = 0 Here, a = 1, b = 1, c = 1 D = b2 - 4ac = ^1 h2 - 4 ^1 h^1 h = 1 - 4 =- 3 Since D 1 0 ∴The given equation has no real roots. Example 3: Find the value of 'k' for which the equation x2 - 2x + k = 0 has equal roots. Solution: The given equation x2 - 2x + k = 0 Here, a = 1, b =- 2, c = k For equal roots, D = b2 - 4ac = 0 ⇒ ^- 2h2 - 4 ^1 h^ k h = 0

⇒ 4 - 4k = 0 & 4k = 4 & k = 1

IL Foundation Series Class 10

Example 4: Determine the values of 'k' for which the quadratic equation kx 2 5 x k 0 has equal roots. Solution: The given equation kx2 - 5x + k = 0 Here, a = k, b =- 5, c = k For equal roots, D = b2 - 4ac = 0 ⇒ ^- 5h2 - 4 ^ k h^ k h = 0

⇒ 25 - 4k2 = 0 & 4k2 = 25 25 5 ⇒ k2 = 4 & k = ! 2 Example 5: If -4 is a root of the quadratic equation x2 + px - 4 = 0 and the quadratic equation x2 + px + k = 0 has equal roots, find the value of k. Solution: Since -4 is a root of the equation x2 + px - 4 = 0 ^- 4h2 + p ^- 4h - 4 = 0

⇒ 16 - 4p - 4 = 0 ⇒ 12 - 4p = 0 ⇒ 4p = 12

⇒ p=3 The given equation x 2 px k 0 has equal roots. Here, x2 + 3x + k = 0 a = 1, b = 3, c = k ` D = b2 - 4ac = 0 (3) 2 - 4 (1) (k) = 0 9 - 4k = 0 4k = 9 9 k= 4

Example 6: Find the value of k for which the equation x2 + 5kx + 16 = 0 has no real roots. Solution: The given equation is x2 + 5kx + 16 = 0 Here, a = 1, b = 5k, c = 16

QUADRATIC EQUATIONS

D = b2 - 4ac = (5k) 2 - 4 # 1 # 16 2 = 25k - 64 The given equation will have no real roots if

D10 25k2 - 64 1 0 64 25 a k2 - 25 k < 0

[a ab < 0 and a > 0 & b < 0]

64 0  x 2 a 2 0 a x a 25 -8 8 5 <k< 5 Example 7: Find the discriminant of the equation 3x2 - 4 3x + 4 = 0 and, hence, find the k2

nature of its roots. Find them if they are real. Solution: The given equation 3x2 - 4 3x + 4 = 0 Here, a = 3, b =- 4 3 , c = 4 D = b2 - 4ac D = (- 4 3 ) 2 - 4 (3) (4) D = 48 - 48 D=0 ∴ The given equation has two equal real roots. - (- 4 3 ) - (- 4 3 ) -b -b The roots are 2a , 2a i.e. 2#3 , 2#3 4 3 4 3 6 , 6 2 3 2 3 3 , 3 OR 2 2 , 3 3

IL Foundation Series Class 10

4.5 SOLVING EQUATIONS REDUCIBLE TO QUADRATIC EQUATIONS Equations reducible to quadratic form: Some equations can be reduced to the quadratic form by proper substitution. Now, let us solve such problems.

SOLVED EXAMPLES Example 1: Solve: ^ x2 - xh2 - 8 ^ x2 - xh + 12 = 0. Solution: Let ^ x2 - xh2 - 8 ^ x2 - xh + 12 = 0

...(1)

Put x2 - x = a (1) & a2 - 8a + 12 = 0 a2 - 6a - 2a + 12 = 0 a (a - 6) - 2 (a - 6) = 0 (a - 2) (a - 6) = 0 a - 2 = 0 (or) a - 6 = 0 ` a = 2 (or) a=6

In case a = 2, we have x2 - x = 2 i.e., x2 - x - 2 = 0 x 2 2 x x 2 0 (x 1)(x 2) 0 ... x =- 1 or 2 In case a = 6, we have x2 - x - 6 = 0 i.e., x 2 x 6 0 x2 3x 2x 6 0 (x 3)(x 2) 0 x 3, 2 ∴ The roots of the given equations are -2,-1, 2, 3. ∴ The solution set = "- 2, - 1, 2, 3 , Example 2: Find the roots of the equation:

x 4

x 7

11 30

QUADRATIC EQUATIONS

Solution: 1 1 11 − = x + 4 x − 7 30 (x − 7) − (x + 4) 11 ⇒ = (x + 4)(x − 7) 30 x − 7 − x − 4 11 ⇒ 2 = x − 3 x − 28 30 11 −11 ⇒ 2 = x − 3 x − 28 30 ⇒ x 2 − 3 x − 28 = −30 0 ⇒ x 2 − 3 x − 28 + 30 = 0 ⇒ x2 − 2x − x + 2 = 0 ⇒ x(x − 2) − 1(x − 2) = 0 ⇒ (x − 1)(x − 2) = ⇒x= 1 or 2

4.6 SOLVING WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS

SOLVED EXAMPLES Example 1: Keerthana can row her boat at a speed of 5 km/h in still water. If it takes her 1 hour more to row the boat 5.25 km upstream than to return downstream, find the speed of the stream. Solution:

/h . Let the speed of the stream be x km/h, speed of the boat upstream = ^5 - xh km km/h. Speed of the boat downstream = ^5 + xh km km/h. /h

Time = Distance Speed

Time taken for going 5.25 km upstream = 5.25 hours 5-x Time taken for going 5.25 km downstream = 5.25 hours 5+x According to the problem,

IL Foundation Series Class 10

5. 25 5. 25 & 5-x - 5+x = 1

1 1 525 5 + x - 5 + x F =1 & 5.25 : 5 - x - 5 + x D = 1 & 100 < (5 - x) (5 + x) 2x 21 21 x =1 & 4 ; 25 - x2 E = 1 & 2 # 25 - x2 21x = 1 & 21x = 50 - 2x2 & 50 - 2x2 2 & 2x + 21x - 50 = 0 & 2x2 + 25x - 4x - 50 = 0 & x (2x + 25) - 2 (2x + 25) = 0 & (x - 2) (2x + 25) = 0 - 25 x = 2 or 2 speed cannot be negative,

Hence, the speed of the stream is 2 km/h. Example 2: A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away on time, it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed. Solution: Let the usual speed of the plane be 'x' km/h, then the new speed of the plane = (x + 250) km/h Distance = 1500 km By the formula Time = Distance Speed 1500 Time taken to cover 1500 km with the usual speed = x hrs Time taken to cover 1500 km with the new speed = According to the condition,

1500 hrs (x + 250)

1500 1500 1 x x 250 2 1500(x 250) 1500 x 1 2 (x)(x 250) 1500 x 375000 1500 x 1 x 2 250 x 2 2 x 250 x 750000 x 2 250 x 750000 0

QUADRATIC EQUATIONS

x 2 1000 x 750 x 750000 0 x(x 1000) 750(x 1000) 0 (x 750)(x 1000) 0 x 1000 or x 750; speed cannot be negative. Hence, the usual speed of the plane = 750 km/h. Example 3: A piece of cloth costs ₹200. If the piece was 5 m longer and each metre of cloth cost ₹ 2 less, the cost of the piece would have remained unchanged. How long is the piece, and what is the original rate per metre? Solution: Let the length of the piece be 'x' metres. 200 Then, rate per metre = ₹ x Now, length = (x + 5) metres 200 ∴ New rate per metre = ₹ x+5 According to the condition, 200 200 2 x x 5 200(x 5) 200 x 2 x(x 5) 1000 2 2 x 5x 2 x 2 10 x 1000 2 x 2 10 x 1000 0 x 2 5 x 500 0 x 2 25 x 20 x 500 0 x(x 25) 20(x 25) 0 (x 25)(x 20) 0 x 20 or 25 Length cannot be negative. 200 Rate per metre = = ₹ 10 20 Length of the piece = 20 metres. Example 4: The sum of the reciprocals of Rehman's ages (in years) 3 years ago and 1 5 years from now is 3 . Find his present age.

100

IL Foundation Series Class 10

Solution: Let Rehman's present age be x years.

Then, 3 years ago, his age was ^ x - 3h years, and 5 years from now, his age will be ^ x + 5h years. According to the question, 1 1 1 x 3 x 5 3 (xx 5) ( x 3) 1 ( x 3)( x 5) 3 2x 2 1 2 x 2 x 15 3 6 x 6 x 2 2 x 15 x 2 2 x 15 6 x 6 0 x 2 4 x 21 0 x 2 7 x 3x 21 0 x ( x 7) 3( x 7) 0 ( x 3)( x 7) 0 x 3 or x 7, age cannot be negative Rehman's present age ( x ) 7 years

3 Example 5: Two water taps together can fill a tank in 9 8 hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. Solution: Let the smaller tap fill the tank in 'x' hrs. Then, the larger tap fills the tank in ^ x - 10h hrs. 1 Work done by the smaller tap in 1 hour = x 1 Work done by the larger tap in 1 hour = x - 10 3 Given that, the time taken by two taps together to fill the tank = 9 8 hrs 1 8 75 Work done by both the taps in 1 hour = 8 = 75

75 = 8 hrs

101

QUADRATIC EQUATIONS

1 1 8 x x 10 75 x 10 x 8 x(x 10) 75 2 x 10 8 2 x 10 x 75 150 x 750 8 x 2 80 x 8 x 2 80 x 150 x 750 0 8 x 2 230 x 750 0 4 x 2 115 x 375 0 4 x 2 100 x 15 x 375 0 4 x(x 25) 15(x 25) 0 (x 25)(4 x 15) 0 15 x 25 or x 4 15 15 x = 4 does not satisfy because if the time taken by smaller tap = x = 4 hrs And time taken by the larger tap = (x - 10) hrs = b 15 - 10 l hrs

= b 15 4 40 l hrs

25 = - 4 < 0, and time taken cannot be negative. Hence, x = 25 ∴ The smaller tap can fill the tank = 25 hrs The larger tap can fill the tank = 15 hrs

4.7 SYMMETRIC FUNCTIONS OF ROOTS If and are the roots of a quadratic equation ax2 + bx + c = 0 , then -b i) The sum of the roots = a + b = a ii) The product of the roots = a b = ac

102

IL Foundation Series Class 10

Example : Find the sum and product of the roots of the equation 5x2 - 2x - 7 = 0. Solution: Here, a = 5, b =- 2, c =- 7 - b - (- 2) 2 =5 ∴ The sum of the roots = a = 5 -7 c The product of the roots = a = 5 4.7.1 Symmetric function 1 1 2 , 2 2 , 2 2, and similar expressions are referred to as 2 functions of the roots and . The expressions , ,

An expression is known as symmetric if it remains unchanged upon interchanging and . In other words, an expression in and , which remains the same when and are interchanged, is called a symmetric function in and . 1 1 2 2 symmetric , , function, 2 , whereas 2 2 , 2 2does not possess this is identified as a 2 1 1 1 1 2 2 2 22 2 2 2 symmetry. The expressions , and termed symmetric functions. , , ,2are 2 specifically 2 , 2 , , , elementary For example,

4.7.2 Symmetry formulae b2 - 2ac a2

a 2 + b 2 = (a + b)2 - 2a b =

1 1 a + b -b a + b = ab = c

| a - b| = (a + b) 2 - 4a b =

1 1 a 2 + b2 b 2 - 2ac + = 2 2 = a 2 b2 ab c2

5. 6. 7. 8.

Δ b2 - 4ac = |a | |a |

3 3 3abc - b a + b = (a + b) - 3a b (a + b) = a3 2 2 2 a b a +b b - 2ac + = = ac b a ab 3

a4 - b = ^a - b h^a + b h a+ b c = a -1 + b -1 a 2

103

QUADRATIC EQUATIONS

SOLVED EXAMPLES Example 1: If and are the roots of the equation x2 + 2x - 15 = 0, then find the values 2

of a 2 + b and a 3 + b . Solution: The sum of the roots = a + b = - 2 =- 2 1 The product of the roots = a × b = - 15 =- 15 1 2 2 2 2 a + b = (a + b) - 2a b = (- 2) - 2 (- 15) = 34 3

a + b =(a + b)3 - 3a b(a + b) = (- 2) 3 - 3 (- 15) (- 2) =- 98 Example 2: If and are the roots of the equation x2 + x - 15 = 0, then find the values of 1 1 1 1 , 2 2 , , and 1 1 Solution: Given equation is x2 + x - 15 = 0 ∴ The sum of the roots = a + b =- 1 The product of the roots = ab =- 15

a2 + b2 = (a + b) 2 - 2ab = (- 1) 2 - 2 (- 15) = 1 + 30 = 31 a+b -1 1 1 1 a + b = ab = - 15 = 15 a2 + b2 1 1 31 31 = 2 + 2 = 2 2 2 = 225 a b ab (- 15) 2 2 a +b - 31 a b +a = = 15 b ab a+b = ab =- 15. a -1 + b -1 Example 3: If and are the roots of the equation ax2 + bx + c = 0 , then show that

1 1 b + = aa + b ab + b ac

104

IL Foundation Series Class 10

Solution: If and are the roots of the quadratic equation ax2 + bx + c = 0, then aα 2 + bα= + c 0 and a β2 + b β= +c 0 c 0 and β(a β + b ) + c 0 α ( aα + b ) + = = −c −c and= (a β + b ) α β 1 1 −α (−β) L.H.S. = + = + aα + b a β + b c c

= aα + b

−(α + β)  b  b =  =  = c  ac  ac

R.H.S.

4.8 TRANSFORMATION OF ROOTS Transformation of equations: If and are the roots of the quadratic equation f ^ xh = ax2 + bx + c = 0, then the quadratic equation whose roots are 1. 2. 3. 4. 5. 6. 7. 8.

1 1 1 the reciprocal (i.e., , ) of f ^ xh = 0 is f a x k = 0 .

the opposite signs (i.e., - a, - b ) of f ^ xh = 0 is f ^- xh = 0 .

increased by k (i.e., k, k) more than that of f ^ xh = 0 is f ^ x - kh = 0. ( 6k ! R )

decreased by k (i.e., k, k) more than that of f ^ xh = 0 is f ^ x - kh = 0. ( 6k ! R ) x multiplied by k (i.e., ka , k b) of f ^ xh = 0 is f ` k j = 0. (for k ! R - {0}h the squares (i.e.,a 2, b h of f ^ xh = 0 is f ( x ) = 0 . 2

the cubes (i.e., a3, b h of f ^ xh = 0 is f (3 x ) = 0 . (i.e., a3 x2 + ^b3 - 3abch x + c3 = 0h 3

b a x in the form of 1 + a , 1 + b of f ^ xh = 0 is f ` 1 - x j = 0 .

x-b in the form of aa + b, a b+ b of f ^ xh = 0 is f a a k = 0 . 1-b 1-x 10. in the form of 1 - a and of f ^ xh = 0 is f a 1 + x k = 0 . 1+a 1+b 9.

SOLVED EXAMPLES Example 1: Find the equation that is formed by the reciprocals of the roots of 3x2 + 5x - 2 = 0 . Solution: 105

QUADRATIC EQUATIONS

Let f ^ xh = 3x2 + 5x - 2 = 0

1 Required equation is f a x k = 0 2

1 1 3 5 2 0 x x 3 5x 2x2 0 x2 2x2 5x 3 0 Example 2: Find the equation that is formed by the negatives of the roots of the equation x2 - 5x + 4 = 0. Solution: Let f ^ xh = x2 - 5x + 4 = 0

Required equation is f ^- xh = ^- xh2 - 5 ^- xh + 4 = 0 x2 5x 4 0

Example 3: Find the equation that is formed by increasing each root of 2x2 - 3x - 1 = 0 by 2. Solution: Let f (x) 2 x 2 3 x 1 0 Required equation is f (x 2) 2(x 2)2 3(x 2) 1 0 f (x 2) 2 x 2 4 x 4 3 x 6 1 0 2 x 2 11x 13 0 Example 4: Find the equation that is formed when each root of x2 - 2x - 15 = 0 is multiplied by 2. Solution: Let f ^ xh = x2 - 2x - 15 = 0

x x 2 x Required equation is f ` 2 j = ` 2 j - 2 ` 2 j - 15 = 0 Simplifying this equation gives: ⇒ x2 − 4x − 60 = 0 Example 5: Find the equation that is formed by squaring each root of the equation x2 - 3x + 2 = 0. Solution: Let f ^ xh = x2 - 3x + 2 = 0 Required equation is f ( x ) = 0

106

IL Foundation Series Class 10

( x )2 3 x 2 0 x 3 x 2 0 x 2 3 x Squaring on both sides (x 2)2 (3 x )2 x2 4x 4 9x x2 5x 4 0 Example 6: Find the equation that is formed by cubing each root of the equation x2 - 3x + 2 = 0 . Solution:

Let f ^ xh = x2 - 3x + 2 = 0 Here, a = 1, b =- 3, c = 2

Required equation is f (3 x ) = 0 a 3 x 2 b 3 - 3abc x c 3 0 13 x 2 (-3)3 - 3 1(-3)2 x 23 0 x 2 - 9 x 8 0 Example 7: If and are the roots of the equation x2 - x - 6 = 0,then find the equation whose roots are

b a . , 1+a 1+b

Solution: Let f ^ xh = x2 - x - 6 = 0

x Required equation is f ` 1 - x j = 0

x j2 ` x j 2 2 `1 x - 1 - x - 6 = 0 & x - x (1 - x) - 6 (1 - x) = 0 & x2 - x + x2 - 6 - 6x2 + 12x = 0 & 4x2 - 11x + 6 = 0

107

QUADRATIC EQUATIONS

4.9

QUADRATIC EXPRESSIONS - GRAPHS, MAXIMUM AND MINIMUM VALUES

4.9.1 Graphs of different quadratic expressions Graph of y = ax2 + bx + c Case i.

Graphical Representation

I f a > 0 and Δ > 0, then the graph of y = ax2 + bx + c intersects the x-axis at two distinct points.

y = f(x) (b, 0) X

(a, 0)

, -Δ {-b 2a 4a{

ii. I f a > 0 and Δ = 0, then the graph of y = ax2 + bx + c touches the x-axis and lies entirely above the x-axis.

y = f(x)

,0 { -b 2a {

iii. I f a > 0 and Δ < 0, then the graph of y = ax2 + bx + c lies entirely above the x-axis.

Y y = f(x) , -Δ {-b 2a 4a{ X

108

IL Foundation Series Class 10

iv. I f a < 0 and Δ > 0, then the graph of y = ax2 + bx + c intersects the x-axis in two distinct points.

, -Δ {-b 2a 4a{ (α, 0)

(β, 0) X y = f(x)

I f α < 0 and Δ = 0, then the graph of y = ax2 + bx + c touches the x-axis and lies entirely below the x-axis.

Y ,0 {-b 2a { X

y = f(x)

vi.

If a < 0 and Δ < 0, then the graph of y = ax2 + bx + c lies entirely below the x-axis.

{

-b , -Δ 2a 4a

y = f(x)

4.9.2 Location of roots Let f ^ xh = ax2 + bx + c = 0 , where a 2 0.

• Conditions for both the roots of f ^ xh = 0 to be greater than a given number k are -b b2 - 4ac $ 0, f (k) > 0, 2a > k. • •

The number k lies between the roots of f ^ xh = 0 if f ^ k h < 0, D > 0.

The condition for exactly one root of f ^ xh = 0, to lie between p and q is f (p), f (q) < 0, D > 0.

• f(x) has roots and confined between the numbers m and n if

b 2 4ac 0, f (m) 0, f (n) 0 (or) m

b n. 2a 109

QUADRATIC EQUATIONS

•

The graph of quadratic function ax2 + bx + c = 0, a ≠ 0. Characteristics of the functions

b2 -4ac < 0

b2 -4ac = 0

b2 -4ac > 0 Y

When a > 0

X Minima

Minima

Minima O

X Max

Max

When a > 0

4.9.3 Maximum and minimum values of quadratic expression (ax2 + bx + c ) b k2 4ac - b2 ax2 + bx + c a = + x a 2a + 4a2

-b 4ac - b2 i) If a > 0, then the minimum value of ax2 + bx + c at x = 2a is 4a . -b 4ac - b2 ii) If a < 0, then the maximum value of ax2 + bx + c at x = 2a is 4a . Example : Find the value of a when the value of 8a − a2 − 15 is maximum. Solution: Let us assume = 8a − a2 − 15

y =- 15 - â2 - 8ah y =- 15 - â2 - 2 # a # 4 + 42 - 42h y =- 15 - â - 4h2 + 16 y = 1 - â - 4h2

110

IL Foundation Series Class 10

Hence, we can clearly see that (a - 4)2 ≥ 0 [Since a is real] Therefore, from y = 1 - (a - 4)2 , we can clearly see that y ≤ 1 and y = 1 when (a - 4)2 = 0 or a = 4. Therefore, when a is 4, then the expression 8a − a2 − 15 reaches the maximum value, and the maximum value is 1.

QUICK REVIEW • A polynomial of degree 2 is called a quadratic polynomial. The general form of a quadratic polynomial is ax 2 bx c 0 where a, b, and c are real numbers such that a ≠ 0 and x is a real variable. • If p ^ xh = ax2 + bx + c , a ≠ 0 is a quadratic polynomial and a is a real number, then p (a ) = a a 2 + ba + c is known as the value of the quadratic polynomial p(x).

• If p ^ xh = ax2 + bx + c is a quadratic polynomial, then p ^ xh = 0 , i.e., ax2 + bx + c = 0, if a ! 0 called a quadratic equation. • The roots of a quadratic equation can also be found by using the method of completing the square. • The roots of the quadratic equation ax2 + bx + c = 0, a ! 0 can be found by using the quadratic - b ! b2 - 4ac formula , provided that b2 - 4ac $ 0. 2a • The nature of the roots of the quadratic equation ax2 + bx + c = 0, a ! 0 depends upon the value of D = b2 - 4ac which is known as the discriminant of the quadratic equation. •

The quadratic equation ax2 + bx + c = 0, a ! 0 has: i) two distinct real roots if D = b2 - 4ac 2 0 . ii) two equal roots, i.e., coincident real roots if D = b2 - 4ac = 0 . iii) no real roots if D = b2 - 4ac 1 0.

• For and , roots of a quadratic expression, an expression is known as symmetric if it remains unchanged upon interchanging and . •

p ^ xh =( ax2 + bx + c ): Maximum and Minimum values of quadratic expression 2 2 2 ax + bx + c a b 4ac - b = x + 2a k + a 4a 2 - b 4ac - b2 i) If a > 0, then the minimum value of ax2 + bx + c at x = 2a is 4a . - b 4ac - b2 ii) If a < 0, then the maximum value of ax2 + bx + c at x = 2a is 4a .

111

QUADRATIC EQUATIONS

WORKSHEET - 1 I.

INTRODUCTION TO QUADRATIC EQUATIONS

Check whether the following are quadratic equations. i) ^ x + 1h2 = 2 ^ x - 3h

ii) ^ x - 2h^ x + 1h = ^ x - 1h^ x + 3h iii) ^ x - 3h^2x + 1h = x ^ x + 5h

iv) ^2x - 1h^ x - 3h = ^ x + 5h^ x - 1h v) x2 + 3x + 1 = ^ x - 2h2

vi) (x + 2) 3 = 2x ^ x2 - 1h II. FORMATION OF A QUADRATIC EQUATION 1.

Represent the following situations in the form of quadratic equations. i) T he area of a rectangular plot is 528 m2 . If the length of the plot (in metres) is one more

than twice its breadth, then find the length and breadth of the plot.

ii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train. iii) The sum of the areas of two squares is 468 m m22. If the difference of their perimeters is 24 m, m, formulate the quadratic equation to find the sides of the two squares. iv) The height of a right-angled triangle is 7 cm cm less than its base. If the hypotenuse is 13 cm, then form the quadratic equation. 2. A child purchases a number of notebooks for `80. If he had purchased four more notebooks for the same amount from the other bookseller, then a notebook would have cost `1 less. Represent the above situation in the form of a quadratic equation. 3. An electric cable costs `200. If the cable was 3 metres longer and each metre of cable cost `2 less, the cost of cable would have remained unchanged. Represent the above situation in the form of a quadratic equation. III. SOLVING QUADRATIC EQUATIONS 1.

Find the roots of the following quadratic equations by factorisation. i) x2 - 3x - 10 = 0 ii) 2x2 + x - 6 = 0

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IL Foundation Series Class 10

iii)

2 x2 + 7x + 5 2 = 0

1 iv) 2x2 - x + 8 = 0 v) 100x2 - 20x + 1 = 0 vi) 4 3 x2 + 5x - 2 3 = 0 x x + 1 34 vii) x + 1 + x = 15 , x ! 0, x ! - 1 viii) 4x2 - 2 ^a2 + b2h x + a2 b2 = 0

1 1 1 1 ix) a + b + x = a + b + x , a + b ! 0 x) a2 b2 x2 + b2 x - a2 x - 1 = 0 2.

Find the roots of the following quadratic equations. i) 2x2 + x - 4 = 0 ii) 4x2 + 4 3 x2 + 3 = 0 iii) 2x2 + x + 4 = 0 iv) a2 x2 - 3abx + 2b2 = 0 v) x2 - ( 3 + 1) x + 3 = 0

Find the roots of the quadratic equations, if they exist, by the method of quadratic formula. i) 3x2 - 2x + 2 = 0 ii) x2 - 2x + 1 = 0 iii) 2x2 + 5 3 x + 6 = 0 iv) x2 + 2x + 4 = 0 v) 3a2 x2 + 8abx + 4b2 = 0, a ! 0 vi) abx2 + ^b2 - ach x - bc = 0

Find the roots of the following equations. 1 i) x - x = 3, x ! 0 x-1 x-3 1 ii) x - 2 + x - 4 = 3 3 , x ! 2, 4 1 2 4 iii) x + 1 + x + 2 = x + 4 , x ! 1, - 2, - 4 x x + 1 34 iv) x + 1 + x = 15 , x ! 1, 0 v) x 4 - 5x2 + 6 = 0

113

QUADRATIC EQUATIONS 2

vi) x 3 x 3 2 0 vii) 3 x + 3 -x - 2 = 0 viii) 71 + x + 71 - x = 50 x x-3 5 x =2 x-3 + 1 1 x) a x2 + 2 k - 5 a x + x k + 6 = 0 x xi) x 4 - 2x3 - x2 - 2x + 1 = 0 ix)

xii) ^ x + 1h^ x + 2h^ x + 3h^ x + 4h = 120 xiii) x + 1 + 2x - 5 = 3 5.

Solve the equation z2 = z , where z = x + iy.

IV. NATURE OF ROOTS OF A QUADRATIC EQUATION 1.

Find the discriminant of the following quadratic equations i) 2x2 - 3x + 4 = 0 ii) x2 - x + 1 = 0 iii) x2 + 2x + 4 = 0 iv) 3 x2 + 2 2 x - 2 3 = 0

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them, too. i) 2x2 - 3x + 5 = 0 ii) 2x2 - 6x + 3 = 0 iii) 6x2 + x - 2 = 0 iv) 6x2 + 23x + 20 = 0

Find the values of 'k' for which the roots are real and equal in each of the following equations. i) kx 4 x 1 0 2

ii) 4x + kx + 9 = 0 2

iii) 4x - 3kx + 1 = 0 2

iv) ^ k + 1h x2 - 2 ^ k - 1h x + 1 = 0 v) x2 - 2 ^ k + 1h x + k2 = 0

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4. If -5 is a root of the quadratic equation 2x2 + px - 15 = 0 and the quadratic equation p ^ x2 + xh + k = has equal roots, find the values of 'p' and 'k'.

5. If the equation ^1 + m2h x2 + 2mcx + ^c2 - a2h = 0 has equal roots, prove that c2 = a2 ^1 + m2h . 6. If the roots of the equation ^b - ch x2 + ^c - ah x + ^a - bh = 0 are equal, then prove that 2b = a + c .

7. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area 800 m2,. If so, find its length and breadth. 8. Is the following situation possible? If so, determine their present ages. The sum of the present ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. 9. Is it possible to design a rectangular park with a perimeter of 80m m and an area of 400 m22 ? If so, find its length or breadth. 10. Is the following situation possible? If so, find their present ages: One year ago, a man was 8 times as old as his son. Now, his age is equal to the square of his son's age. 11. If b = 0, c 1 0 in the quadratic equation x2 + bx + c = 0, then what will be the nature of the roots? 12. Find the value of p for which the quadratic equation x ^ x - 4h + p = 0 has real roots. 13. Find the values of p for which the following quadratic equation has two equal roots : ^ p - 12h x2 + 2 ^ p - 12h x + 2 = 0 14. Find whether the quadratic equation x2 - x - 2 = 0 has real roots or not. If yes, find the roots. V. SOLVING WORD PROBLEMS 1.

Find two numbers whose sum is 27 and whose product is 182.

Find two consecutive positive integers when the sum of their squares is 365.

3. The altitude of a right-angled triangle is 7 cm less than its base. If the hypotenuse is 13 cm , find the other two sides. 4. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article. 5. Find two positive numbers whose squares have a difference of 48, and the sum of the numbers is 12. 6. Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46. Find the integers. 7.

Two numbers differ by 3, and their product is 504. Find the numbers.

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QUADRATIC EQUATIONS

If one root of the equation 4x2 - 8kx - 9 = 0 is negative of the other, then find the value of k.

Find k if 3 is a solution of 3x2 + ^ k - 3h x + 9 = 0.

10. Solve for x : x 2

pq qx

11. The sum of the ages of a boy and his brother is 25 years and the product of their ages (in years) is 126. Find their present ages. 12. Seven years ago, Varun's age was five times the square of Swati's age. Three years hence Swati's age will be two-fifth of Varun's age. Find their present ages. 13. One year ago, a man was 8 times as old as his son. Now, his age is equal to the square of his son's age. Find their present ages. 14. In a class test, the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects. 15. The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side, find the sides of the field. 16. The difference between the squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers. 17. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken an hour less for the same journey. Find the speed of the train. 18. In a flight of 600 km, an aircraft faced a slowdown due to bad weather. Its average speed for the trip was reduced by 200 km/h and the time was increased by 30 minutes. Find the duration of the flight. 19. 300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of students. 20. Some students planned a picnic. The budget for food was ₹ 500. But, 5 of them failed to go and, thus, the cost of food for each number increased by ₹ 5. How many students attended the picnic? 21. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains. 22. The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares. 23. A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km/h less than that of the fast train, find the speeds of the two trains. 24. One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to the mountains, and the remaining 15 camels were seen on the bank of a river. Find the total number of camels. 116

IL Foundation Series Class 10

25. If the list price of a toy is reduced by ₹ 2, a person can buy 2 toys more for ₹360. Find the original price of the toy. 26. Five years ago, a woman's age (in years) was the square of her son's age. Ten years later from now, her age will be twice that of her son's age. Find: i) The age of the son five years ago. ii) The present age of the woman. 1

2 3 2 3 27. If and are the roots of the equation 8x - 3x + 27 = 0, then find the value of . 2

VI. SYMMETRIC FUNCTIONS & TRANSFORMATION OF ROOTS AND QUADRATIC EXPRESSIONS - GRAPHS, MAXIMUM AND MINIMUM VALUES 1.

If and are the roots of ax2 + bx + c = 0, then find the values of i) a2 + b2 1 1 ii) a + b 3

iii) a + b

2 iv) a-2 + b-2 a +b 2 v) c a - b m b a vi) a4 b7 + a7 b4

b a 2 + b a2 2 viii) c 12 - 1 2 m a b ix) (aa + b) -2 + (a b+ b) -2 vii)

Find the equation that is formed by squaring each root of the equation x2 + 2x - 3 = 0.

3. If a, b are the roots of 3x2 + 2x + 1 = 0, then find the equation whose roots are

1 - a and 1 - b . 1+a 1+b 4.

If and are the roots of 3x2 - 4x + 1 = 0, then find the equation whose roots are

5 2 7 Find the minimum value of a x - 3 k + 2 .

2 a2 and b . b a

6. Find the value of a such that the sum of the squares of the roots of the equation x2 - ^a - 2h x - ^a + 1h = 0 is the least.

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QUADRATIC EQUATIONS

WORKSHEET -2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER

Which of the following equations has the product of its roots as 4? a) x2 + 4x + 4 = 0

2 x2 -

b) 2x2 - 7x + 6 = 0

3 x + 1 = 0 d) 3x2 - 3x + 3 = 0 2

c) 10x2 - 3x - 1 = 0

d) 35x2 + 12x + 1 = 0

b) 2

c) 3

d) -3

c) 6x2 - x - 2 = 0

1 d) 10x - x = 3

Which of the following equations has equal roots? a) x2 + 6x + 5 = 0

d) x2 + 4x - 24 = 0

The value of k for which 3 is a root of the equation kx2 - 7x + 3 = 0 is a) -2

b) - x2 + 3x - 3 = 0

1 Which of the following equations has 5 as a root? a) 35x2 - 2x - 1 = 0

c) - x2 + 4x + 4 = 0

Which of the following has the sum of the roots as 3? a) 2x2 - 3x + 6 = 0

b) x2 + 4x - 4 = 0

b) x2 - 8x + 16 = 0

If ax2 + bx + c = 0 has equal roots, then c = -b a) 2a

b b) 2a

- b2 c) 4a If p and q are the roots of the equation x2 - px + q = 0 , then

b2 d) 4a

a) p = 1, q =- 2

b) p = 0, q = 1

d) p =- 2, q = 1

c) p = 1, q = 0

8. If the sum and product of the roots of the equation kx 2 + 6x + 4k = 0 are equal, then k = 3 a) - 2 9.

3 b) 2

2 c) 3

2 d) - 3

If a and b are roots of the equation x2 + ax + b = 0 , then a + b = a) 1

b) 2

c) -2

d) -1

10. If one root of the equation 2x2 + kx + 4 = 0 is 2, then the other root is a) 6

b) -6

c) -1

d) 1

11. The quadratic equation 2x2 - 5 x + 1 = 0 has a) Two distinct real roots

b) Two equal roots

c) No real roots

d) More than two real roots

12. The roots of the equation a) x = 6

118

2x2 + 9 = 9 are b) x = ! 6

c) x =- 6

d) x = 0

IL Foundation Series Class 10

13. If 3 is a solution of 3x2 + ^ k - 1h x + 9 = 0, then the value of k is a) 11

b) -11

c) 13

d) -13

14. If the equation x2 + 4x + k = 0 has real and distinct roots, then a) k < 4

b) k > 4

c) k $ 4

d) k # 4

15. If the equation x2 - ax + 1 = 0 has distinct roots, then a) a = 2

b) a 1 2

c) a 2 2

d) None of these

16. If the sum of the roots of the equation x2 - x = m (2x - 1) is zero, then m = a) -2

1 c) - 2

b) 2

1 d) 2

17. The positive value of k for which the equation x2 + kx + 64 = 0 and x2 - 8x + k = 0 will both have real roots, is a) 4

b) 8

c) 12

d) 16

18. If the roots of the equation ^a2 + b2h x2 - 2b (a + c) x + ^b2 + c2h = 0 are equal, then a) 2b = a + c

2ac c) b = a + c

b) b2 = ac

d) b = ac

19. The solutions of x2 - x - 6 = x + 2 where x is real are a) -4, 2,-2, 2

b) 4, 2, -2, -2

c) 2, 3, -3, -2

d) None

20. If sin θ, and cosθ are the roots of the equation ax2 + bx + c = 0, then a) a2 - b2 + 2ac = 0

b) a2 + b2 + 2ac = 0

c) a - b + 2ac = 0

d) a + b + 2ac = 0

21. If x2 - ax - 3 = 0 and x2 + ax - 15 = 0 have a common root, then a = a) 1

b) 2

c) 3

d) None

22. The number of real solutions of x 2 - 3 x + 2 = 0 is a) 0

b) 2

c) 3 2 m

d) 4

7 m

23. If m is a positive integer, solutions of x 5 x 4 0 are a) 2 m, 1

b) 1, 4 -m

c) 1, 4 m

d) None

c) -3 < a < 3

d) a < -2

24. If the roots of x2 + x + a = 0 exceed a, then a) 2 < a < 3

b) a > 3

25. The value of a for which each one of the roots of x2 - 4ax + 2a2 - 3a + 5 = 0 is greater than 2, are a) a ! ^1, 3h

b) a = 1

c) a ! ^- 3, 1h

9 d) a ! a 2 , 3 k

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QUADRATIC EQUATIONS n

26. Let and be the roots of equation x2 - 6x - 2 = 0. If an = a - b ,for n $ 1, then the value of a10 - 2a8 is equal to 2a9 a) 3 b) -3 c) 6 d) -6 27. The sum of all real values of x satisfying the equation ^ x2 - 5x + 5hx + 4x - 60 = 1 is 2

a) 5

b) 3

c) -4

28. The graph of y = x2 + x + 1 a) Intersects the x-axis at two distinct points b) Lies entirely above the x-axis. c) Touches the x-axis and lies entirely below the x-axis. d) Lies entirely below the x-axis. 29. The graph of y = x2 - 4x + 1 a) Intersects the x-axis at two distinct points. b) Touches the x-axis. c) Lies entirely below the x-axis. d) Lies entirely above the x-axis. 30. The graph of y = x2 - 4x + 4 a) Touches the x-axis and lies entirely below the x-axis. b) Lies entirely below the x-axis. c) Touches the x-axis and lies entirely above the x-axis. d) Lies entirely above the x-axis. 31. The graph of y =- 2x2 + 4x - 3 a) Lies entirely above the x-axis. b) Touches the x-axis and lies entirely below the x-axis. c) Touches the x-axis and lies entirely above the x-axis. d) Lies entirely below the x-axis.

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d) 6

IL Foundation Series Class 10

32. Statement (A): In the figure, we must have a 2 0, c 2 0, b 2 0 Y y = ax2 +bx+c

Statement (B): In the figure, we must have a 2 0, c 2 0, b 2 0 Y y = ax2 +bx+c

a) Both A and B are true b) Both A and B are false c) A is true, and B is false d) A is false, and B is true 33. The maximum value of a2 - abx - b2 x2 is 5a 4a a) 4 b) 5 34. 3x - 5x2 + 12 has maximum at x = a,then a is

4a2 c) 5

5a2 d) 4

-2 b) 5 35. The minimum value of x2 -8x + 17 is

3 c) 10

-3 d) 10

c) 1

d) 2

2 a) 5 a) 17

b) -1

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QUADRATIC EQUATIONS

36. Assertion (A): 3y2 + 17y - 30 = 0 has distinct roots. Reason (R): The quadratic equation ax2 + bx + c = 0 has distinct roots (real roots) if D 2 0.

a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

37. Assertion (A): 5x2 + 14x + 10 = 0 has no real roots. Reason (R): ax 2 bx c 0 has no real roots if b2 1 4ac.

a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

b) Both Assertion (A) and Reason (R) are true, but Reason(R) is not the correct explanation of Assertion (A). c) Assertion (A) is true, but Reason (R) is false. d) Assertion (A) is false, but Reason (R) is true. 38. Assertion (A): The graph of y = 3x2 + 6x + 3 touches the x-axis and lies entirely below the x-axis. Reason (R): If a > 0, D = 0, then the graph of y = ax2 + bx + c touches the x-axis and lies entirely above the x-axis. a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

39. Assertion (A): The graph of y = 2x2 - 3x - 4 intersects the x-axis at two distinct points.

Reason (R): If D > 0, then the graph of y = ax2 + bx + c intersects the x-axis at two distinct points.

a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

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IL Foundation Series Class 10

1 1 a+b =

a) - b c ii) a + b = b a 2 a) b + 2ac ac iii) 12 + 12 = a b 2 a) b - 2ac c

b) c b

b) b - 2ac ac 2

c) - c b

d) b - 22ac a

c) b - 22ac c

d) b - 22ac a

2 2 b) b - 22ac c) b - 2ac d) - b c 2 c c b roots are the 41. Passage: If a, b are the roots of the Q.E f(x) = ax2 + bx + c = 0, then then the a +Q.E b =whose a and squares of the roots of f ^ xh = 0 is f ( x) = 0

1 1 is , a3 b3 d) x2 - 2x + 1 = 0

i) If a, b are the roots of x2 + x + 1 = 0, then the equation whose roots are a) 2x2 + x + 1 = 0

b) 2x2 - x + 1 = 0

c) x2 - x + 1 = 0

ii) The equation that is formed by squaring each root of the equation x2 + 5x - 9 = 0 is a) x2 + 43x + 81 = 0

b) x2 - 43x - 81 = 0

c) x2 - 43x + 81 = 0 d) none

II. FILL IN THE BLANKS 1. If y = 1 is the common root of ly2 + ly + 3 = 0 and y2 + y + m = 0 , then the value of l − m is ___________. 2. What will be the constant term of the quadratic equation whose roots are -5 and -4? ____________. 3. If 6x + x2 = 1, then the value of c x 4.

1 m is _________________. x If ^ x + 1h2 + 6 = 6, then the value of x is _____________.

The product of the roots of the equation 13 - x2 = x + 5 is ________.

One root of x2 - ^ p - 1h x + 10 = 0 is 5 , then the value of p is _______.

x 9 If 3 + x = 4, then the values of x is ________.

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QUADRATIC EQUATIONS

8. If the ratio of the roots of the equation a1 x2 + b1 x + c1 = 0 is equal to the ratio of roots of a2 x2 + b2 x + c2 = 0 , then ___________. 9.

The roots of the quadratic equation ax2 + bx = 0 are _________.

10. ^ x2 + 1h2 - x2 = 0 has __________ real roots. 11. If the discriminant of the equation 6x2 - bx + 2 = 0 is 1, then the value of 'b' is ______________. 12. The value of k for which the roots of the quadratic equation k2 + 2x + 3 = 0 are equal is __________. 13. The roots of x2 - 2x - ^r2 - 1h = 0 are _________.

14. If 3 is a solution of 3x2 + ^ k - 1h x + 9 = 0, then k = ________. 15. If ax2 + bx + c = 0 has equal roots, then c is _____________. 16. The value of c for which the equation ax2 + 2bx + c = 0 has equal roots is ________. 17. If the quadratic equation kx ^ x - 2h + 6 = 0 has two equal roots, then the value of k is ________. 18. The common root of the equation x2 - 7x + 10 = 0 and x2 - 10x + 16 = 0 is _______. 19. The quadratic equation, one of whose roots is 3 + 2 5 , is ___________________. 20. The value of

7 7 7 7f.. is ____________.

III. SUBJECTIVE QUESTIONS 1.

If x = 3 is one root of the quadratic equation x2 - 2kx - 6 = 0, then find the value of k.

2. If the quadratic equation px2 - 2 5 px + 15 = 0 has two equal roots, then find the value of p. 3.

Check whether ^2x - 1h^ x - 3h = ^ x + 5h^ x - 1h is a quadratic equation or not.

4. The product of two consecutive positive integers is 306. Represent this situation in the form of a quadratic equation. 5.

Check whether -3 is a solution of the equation 3x2 + 5x + 2 = 0.

6. Write the condition to be satisfied by a, b, and c so that the quadratic equation ax2 + bx + c = 0 has two distinct real roots. 7.

2 If one root of the equation 3x2 + px + 4 = 0 is 3 , then find the value of p.

Write the discriminant for the quadratic equation 2x2 - 3x + 5 = 0.

Determine the nature of the roots of the quadratic equation 2x2 - 6x + 3 = 0.

1 5 10. If 2 is a root of the equation x2 + kx - 4 = 0, then find the value of k.

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INEQUATIONS

5.1 INTRODUCTION If x and y are two quantities, then both quantities will satisfy any one of the following four conditions (relations): i.e. either (i) x > y (ii) x ≥ y (iii) x < y or (iv) x ≤ y Each of the four conditions given above is an inequation. In the same way, each of the following also represents an inequation: x < 8, x ≥ 5, − x + 4 ≤ 3, x + 8 > 4, etc.

5.2 LINEAR INEQUATIONS If a,b and c are real numbers, then each of the following is called a linear inequation in one variable: (i) ax + b > c. Read as: ax + b is greater than c. (ii) ax + b < c. Read as: ax + b is less than c. (iii) ax + b ≥ c. Read as: ax + b is greater than or equal to c. (iv) ax + b ≤ c . Read as: ax + b is less than or equal to c. Note: In an inequation, the signs '>', '<', ' ≥ ' and ' ≤ ' are called signs of inequality. 5.2.1 Solving a linear inequation algebraically To solve a given linear inequation means to find the value or values of the variable used in it. Thus; (i) to solve the inequation 3x + 5 > 8 means to find the value or values of x that satisfies the given inequation. (ii) to solve the inequation 8 - 5y ≤ 3 means to find the value or values of y that satisfies the given inequation and so on. The following working rules must be adopted for solving a given linear inequation: Rule 1: On transferring a positive term from one side of an inequation to its other side, the sign of the term becomes negative. Example: 2x + 3 > 6 ⇒ 2x > 6 – 3, 7x + 4 ≤ 13 ⇒ 7x ≤ 13 – 4, and so on.

125

INEQUATIONS

Rule 2: On transferring a negative term from one side of an inequation to its other side, the sign of the term becomes positive. Example: 4x - 5 > 8 ⇒ 4x > 8 + 5, 25 ≥ 4x - 15 ⇒ 25 + 15 ≥ 4x, and so on. Rule 3: If each term of an inequation is multiplied or divided by the same positive number, the sign of inequality remains the same. That is, if p is positive x y (i) x < y ⇒ px < py and   <   ,  p  p x y (ii) x > y ⇒ px > py and   >   ,  p  p x y (iii) x ≤ y ⇒ px ≤ py and   ≤   , and  p  p x y (iv) x ≥ y ⇒ px ≥ py and   ≥   .  p  p Rule 4: If each term of an inequation is multiplied or divided by the same negative number, the sign of inequality reverses. That is, if p is negative x y (i) x < y ⇒ px > py and   >    p  p x y (ii) x ≥ y ⇒ px ≤ py and   ≤    p  p Rule 5: If the sign of each term on both sides of an inequation is changed, the sign of inequality gets reversed. i.e. (i) − x > 5 ⇒ x < −5 (ii) 3y ≤ 15 ⇒ −3y ≥ −15 (iii) −2y < −7 ⇒ 2y > 7 and so on. Rule 6: If both the sides of an inequation are positive or both are negative, then on taking their reciprocals, the sign of inequality reverses. i.e. if x and y both are either positive or both are negative, then

1 1 ≥ x y 1 1 (ii) x ≤ y ⇒ ≥ x y (i) x ≤ y ⇒

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IL Foundation Series Class 10

5.2.2 Replacement set and solution set The set from which the value of the variable x is to be chosen is called the replacement set, and its subset, whose elements satisfy the given inequation, is called solution set. Example: Let the given inequation be x < 3, if: i) The replacement set = N, the set of natural numbers; The solution set ={1, 2} ii) The replacement set = W, the set of whole numbers; The solution set ={0, 1, 2} iii) The replacement set = Z or I, the set of integers; The solution set ={...-2, -1, 0, 1, 2} But, if the replacement set is the set of real numbers, the solution set can only be described in set-builder form, i.e. x : x ∈ R and x < 3.

SOLVED EXAMPLES Example 1: If the replacement set is the set of natural numbers (N), find the solution set of 3x + 4 < 16. Solution: 3x + 4 < 16 ⇒ 3x < 16 − 4

[Using rule (1)]

⇒ 3x < 12 12 [Using rule (3)] 3 ⇒x<4 ⇒x<

Since the replacement set = N (set of natural numbers) ∴ Solution set = {1, 2, 3} Example 2: If the replacement set is the set of whole numbers (W), find the solution set of 5x + 4 ≤ 24. Solution: 5x + 4 ≤ 24 ⇒ 5x ≤ 24 − 4 ⇒ 5x ≤ 20 and x ≤

20 i.e. x ≤ 4 5

Since the replacement set is the set of whole numbers ∴ Solution set = { 0, 1, 2, 3, 4 }

127

INEQUATIONS

Example 3: If the replacement set is the set of integers (I or Z) between -6 and 8, find the solution set of 6x − 1 ≥ 9 + x . Solution: 6x − 1 ≥ 9 + x ⇒ 6x − x ≥ 9 + 1 ⇒ 5x ≥ 10 ⇒ x≥2 Since the replacement set is the set of integers between -6 and 8 ∴ Solution set = {2, 3, 4, 5, 6, 7 } Example 4: If the replacement set is the set of real numbers (R), find the solution set of 5 - 3x < 11. Solution: 5 - 3x < 11 ⇒ −3x < 11 − 5 ⇒ −3x < 6 −3x 6 > −3 −3 ⇒ x > −2

⇒

[Using rule 4]

Since the replacement set is the set of real numbers (R) ∴ Solution set = {x : x > -2 and x ∈ R} Example 5: Solve:

x x − 5 ≤ − 4,where x is a positive odd integer. 2 3

x x −5≤ −4 2 3 x x ⇒ − ≤ −4 + 5 2 3

Solution:

⇒

3x − 2x ≤ 1⇒ x ≤ 6 6

Since x is a positive odd integer ∴ Solution set = {1, 3, 5}

5.3

QUADRATIC INEQUATIONS

Firstly, let us recall what a quadratic equation is. A quadratic equation is an equation of the form: ax2 + bx + c = 0

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IL Foundation Series Class 10

Here, a, b, and c are constants, and x is the variable. The solutions to this equation are given by the quadratic formula: −b ± b2 − 4ac x= 2a Now, let us talk about quadratic inequalities. Quadratic inequalities involve quadratic expressions, but instead of asking for the specific values of x, they ask for the range of values of x that satisfy the inequality. The symbol ’<’, ’>’, ’≤’, or ’≥’ is used to represent the relationship between the quadratic expression and zero. Some examples of quadratic inequalities in one variable are: • x2 + x - 1 > 0 • 2x2 - 5x - 2 ≥ 0 • x2 + 2x - 1 < 0 To solve this kind of inequality, you can use a method similar to solving quadratic equations. Notations used in quadratic inequalities: Let us look at some of the important notations used in quadratic inequalities. () → Open Brackets [ ] → Closed Brackets o → Open Value(x cannot take this value) • → Closed Value (x can take this value) (-1, 1) → x cannot take value –1 and 1. [-1, 1) → x can take value –1 but not 1. (-1, 1] → x cannot take the value –1, but it can take the value 1. [-1, 1] → x can take both –1 and 1 values. 5.3.1 How to solve quadratic inequalities? Solving a quadratic inequality means to find the values of x which satisfy the given condition of the question. Now consider a quadratic expression ax2 + bx + c. We can write the quadratic expression in the form of ( x − α)( x − β) and α < β. Using the number line method, the solution for the quadratic inequality can be expressed as follows. -∞

+∞

It means that if ax2 + bx + c > 0, then x can take values between – ∞ to α and β to + ∞.

129

INEQUATIONS

If x2 + bx + c > 0, then x ∈ (−∞, α) ∪ (β, + ∞) If x2 + bx + c < 0, then x can take values between α and β. If x2 + bx + c < 0, then x ∈ (α, β) Let's take a quadratic inequality x2 – 1 > 0. Here the expression x2 – 1 > 0 can be factorised as (x – 1)(x + 1) > 0. This gives the values of α = -1 and β =1. Hence, we obtain the range of x as x ∈ (−∞, −1) ∪ (1, +∞). -∞

-1

+∞

If the quadratic inequality is x2 – 1 < 0, the expression x2 – 1 < 0, can be factorised as (x – 1)(x + 1) < 0. This gives α = -1 and β = 1. Therefore, the range of x is x ∈ (-1, 1). -∞

-1

+∞

If the quadratic inequality is x2 − 1 ≥ 0 (where it shows the quadratic inequality is greater than or equal to zero), the expression x2 − 1 ≥ 0 can be factorised as ( x − 1)( x + 1) ≥ 0. Here we obtain α = -1 and β =1 and the range of x is x ∈ ( −∞, −1) ∪ (1, +∞ ) . -∞

-1

+∞

If the quadratic inequality is x2 − 1 ≤ 0 (where it shows the quadratic inequality is less than or equal to zero), the expression x2 − 1 ≤ 0 is factorised as (x – 1)(x + 1)≤ 0. Here, the roots of the expression are α = −1 and β =1, and the range of x is x ∈ [-1, +1]. -∞

-1

+∞

Example: Find the range of values of x which satisfy the quadratic inequality x2 − 7x + 10 ≤ 0. Solution: First, let's factorise the quadratic expression x2 – 7x + 10. x2 − 7x + 10 ≤ 0 x2 − 5x − 2x + 10 ≤ 0 x ( x − 5) − 2( x − 5) ≤ 0 ( x − 2)( x − 5) ≤ 0 Hence, the values of x that satisfy the quadratic inequality are x ∈[2, 5]. -∞

Therefore, we have x ∈[2, 5].

130

+∞

IL Foundation Series Class 10

5.4 POLYNOMIAL INEQUATIONS - WAVY CURVE METHOD The wavy curve method or method of intervals is a strategy used to solve inequalities of the form f ( x) > 0, (>, <, ≥, ≤). g( x ) Where f(x) and g(x) are functions of x. Examples: 2 (i) x − 12x + 35 ≥ 0

(ii) ( x − 2) ⋅ ( x + x − 42) < 0 2

( x − 3)3 ( x + 2)9 ( x + 5)5 <0 (iii) ( x + 1)( x − 7)7 Here is a step-by-step explanation of how the wavy curve method works: 1) Write the inequality in standard form: Make sure the inequality is in the standard form f ( x ) > 0 or f ( x ) < 0 , where f (x) is a polynomial function. If it's not in this form, rearrange it accordingly. 2) F ind the roots (critical points) of the polynomial: The roots are the values of x that make the polynomial function equal to zero. To find the roots, solve the equation f (x) = 0. These roots will help us in dividing the number line into intervals. 3) Plot the critical points on the number line: Mark the critical points of the polynomial on the number line. This divides the number line into intervals. 4) Determine the sign of the polynomial in each interval: Pick a test point from each interval and evaluate the polynomial at that point. If the value is positive, the polynomial is positive in that interval; if it's negative, the polynomial is negative in that interval. 5) Draw a "wavy curve": Based on the signs of the polynomial in each interval, draw a curve that alternates between above and below the x-axis. For example, if the polynomial is positive in one interval and negative in the next, the curve should "wave" from above the x-axis to below and vice versa. 6) D etermine the solution: Look for the intervals where the curve lies above (for f(x) > 0) or below (for f(x) < 0) the x-axis. These intervals represent the solutions to the inequality. 7) E xpress the solution in interval notation: Write down the solution using interval notation. For example, if the solution is f ( x ) < 0, it means the values of x that satisfy the inequality lie between a and b but do not include a and b. By following these steps, you can effectively solve polynomial inequalities using the wavy curve method. Let us understand this with an example:

131

INEQUATIONS

Example: Solve the inequality x2 + 35 > 12x. Solution: Step 1: Bring everything to the L.H.S. and make R.H.S. = 0 x2 − 12x + 35 > 0 Step 2: Factorise the expression into as many linear factors as possible. x2 − 12x + 35 > 0 ⇒ ( x − 5)( x − 7) > 0 Step 3: Identify the critical points. (The value of x at which each individual factor is equal to zero.) (x – 5)(x – 7) > 0

( x − 5) = 0 ⇒ x = 5 ( x − 7) = 0 ⇒ x = 7 So, the critical points are 5, 7. Step 4: Plot the critical points on the number line and determine the regions.

7 I

III

Step 5: Determine the sign of the expression in each region. • If the critical point satisfies the given expression, then mention a solid circle on a number line; else, mention a hollow circle on the number line. • Take any real value greater than the extreme right critical point and substitute it in the expression to get the sign for region I. • Repeat the same to get the sign of the expression in II and III.

+ 5

III

7 I

Step 6: Generate the wave diagram starting from the rightmost region and moving towards the leftmost region and hence write the solution as per the given question.

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IL Foundation Series Class 10

+ III

+ 7

∴ x ∈ (−∞, 5) ∪ (7, ∞)

5.5 RATIONAL INEQUATIONS A rational inequality is an inequality that contains a rational expression. 1 2 3 3 2x 2x − 3 > 1, < 4, ≥ x, and − 2 ≤ are rational inequalities as they 4 x x 2x x −3 x −6 each contain a rational expression. Inequalities such as

Solving a rational inequality is somehow similar to finding solutions to linear inequalities. 5.5.1 How to solve rational inequalities? Below are the summarised steps in order to find rational inequalities and solve them. Step 1: Write the expression of inequality as one quotient on the left and zero (0) on the right. Step 2: Identify the critical points - the points where the rational expression will either be undefined or zero. Step 3: Use the critical points for dividing the number line into intervals. Step 4: Test the value of each interval. The number line displays the sign of each factor of the numerator as well as the denominator in each interval. The number line also shows the sign of the quotient. Step 5: Identify the intervals where inequality is appropriate. Step 6: Write the solution in the form of interval notation. Example: Solve and write the solution in interval notation:

x −3 ≤ 0. x +1

x −3 ≤0 x +1 Step 2: This rational expression will be equal to zero when the numerator is equal to zero. Solution: Step 1: x–3=0 ⇒x=3 3 is a critical point. This rational expression will be undefined when the denominator is equal to zero. 133

INEQUATIONS

x+1=0 ⇒ x = -1 -1 is a critical point. The critical points are -1 and 3. Step 3: Use the critical points to divide the number line into intervals.

-6

-5

-4

-3

-2

-1

The number line is divided into three intervals: (−∞, − 1), (−1, 3), (3, ∞). −2 − 3 −5 −2 : = = 5 (Positive) Step 4: Test at x = −2 + 1 −1 So, this interval is not a solution because the inequality needs to be less than or equal to zero. 1 − 3 −2 = = −1 (Negative) 1+ 1 1 This interval is a solution because the inequality needs to be less than or equal to zero. Test at x = 1 :

4−3 1 Test = at x 4= : (Positive) 4+1 5 This interval is not a solution because the inequality needs to be less than or equal to zero. Step 5: The answer to this rational inequality is [-1, 3]. The middle interval of this number line is the only interval that is a solution because the test value from this interval is negative and this inequality is supposed to be negative according to the original inequality. Parentheses are used in front of -1 because the fraction is undefined when x = -1. + ve -6

-5

-4

-3

- ve -2

-1

+ ve

Therefore, -1 cannot be included in the solution. A square bracket is beside the 3 because 3 can be included in the solution.

5.6 IRRATIONAL INEQUATIONS Irrational inequalities involve expressions containing irrational numbers, which are numbers that cannot be expressed as a simple fraction or decimal and have non-repeating, non-terminating decimal expansions. These inequalities are similar to rational inequalities, but they involve irrational expressions like square roots, cube roots, or other roots. 134

IL Foundation Series Class 10

Here is a simple example of an irrational inequality:

x −3 > 2

5.6.1 How to solve irrational inequalities? To solve an irrational inequality, we follow these steps: Step 1: Isolate the radical: Get the radical term (the term with the square root) on one side of the inequality. In this case, it's already isolated. x −3 > 2 Step 2: Square both sides: This step is often necessary when dealing with square roots. Squaring both sides eliminates the square root. ( x − 3)2 > 22 x–3>4 Step 3: Solve for x: Now, we solve the resulting rational inequality. x>7 So, the solution to the given irrational inequality is x > 7. This means any value of x greater than 7 will satisfy the inequality. Important point: It is important to note that when squaring both sides of an inequality, extraneous solutions might be introduced. Therefore, it's crucial to check the solutions in the original inequality to ensure they are valid. However, in this example, no extraneous solutions are introduced.

5.7 MODULUS INEQUATIONS Inequalities of the form |ax + b|< k or |ax + b| > k are called the modulus or absolute inequalities. To solve these inequalities, keep the following rules in mind: • If |x| < a, then –a < x < a • If |x| > a, then either x > a or x < –a • If |x – 1| < a, then | –a < x < 1 + a • If |x – 1| > a, then either x > 1 + a or x < 1 – a. Let us understand modulus inequality with some examples: Example 1: Solve |x – 3| < 5. Solution: We have |x – 3| < 5. ⇒ −5 < x − 3 < 5 (If | x |< a, then − a < x < a )

135

INEQUATIONS

⇒ 3 − 5 < x < 3 + 5 (The same number can be added on both sides of the inequality.) ⇒ −2 < x < 8 , which is the required solution. Example 2: Solve |8x + 5| < 9. Solution: We have |8x + 5| < 9. ⇒ −9 < 8x + 5 < 9 ( If | x |< a, then − a < x < a ) ⇒ −5 − 9 < 8x < 9 − 5 (Same number can be added on both sides of the inequality) ⇒ −14 < 8x < 4 −14 4 ⇒ <x< 8 8 −7 1 ⇒ < x < which is the required solution. 4 2

SOLVED EXAMPLES Example 1: Solve 4x + 3 < 6x + 7. Solution: We have, 4x + 3 < 6x + 7 ⇒ 4x − 6x < 6x + 4 − 6x ⇒ −2x < 4 or x > −2 i.e., all the real numbers which are greater than –2, are the solutions to the given inequality. Hence, the solution set is (−2, ∞). Example 2: Solve the inequality

2x + 1 < 1. x+3

2x + 1 <1 x+3 Here, the right-hand side is not equal to zero. So, transpose 1 to the left-hand side and solve. Solution: We have

We get: 2x + 1 <1 x+3 ⇒

2x + 1 −1< 0 x+3

⇒

2x + 1 − x − 3 <0 x+3

⇒

x −2 <0 x+3

The critical points are x = 2, –3. Plot these points on the number line, we get: 136

IL Foundation Series Class 10

+ve

-ve

+ve

-3

Since the given inequality is negative, the solution is −3 < x < 2. Example 3: Solve the inequality

x−4 > 2. 2x − 1

x−4 > 2. 2x − 1 Here, the right–hand side is not equal to zero. So, transpose 1 to the left–hand side and solve. Solution: We have We get: x−4 −2> 0 2x − 1 ⇒

x − 4 − 2(2x − 1) >0 2x − 1

⇒

x − 4 − 4x + 2 >0 2x − 1

⇒

−3x − 2 >0 2x − 1

⇒

3x + 2 <0 2x − 1

2 1 The critical points are x = − , . 3 2

+ve

-ve

2 3

+ve

1 2

2 1 So, the solution is − < x < . 3 2 Example 4: Solve | -3x + 7 | +8 < 15. Solution: We have | -3x + 7 | + 8 < 15. ⇒| −3x + 7 |< 15 − 8 ⇒| −3x + 7 |< 7

137

INEQUATIONS

⇒ −7 < −3x + 7 < 7 ⇒ −7 − 7 < −3x < 0 ⇒ −14 < −3x < 0 ⇒ 0 < 3x < 14 ⇒ 0< x <

14 , which is the required solution. 3

QUICK REVIEW • Two real numbers or two algebraic expressions related by the symbols <, >, ≤ or ≥ form an inequality. • Equal numbers may be added to (or subtracted from) both sides of an inequality. • Both sides of an inequality can be multiplied (or divided) by the same positive number. But, when both sides are multiplied (or divided) by a negative number, then the inequality is reversed. • The values of x, which make an inequality a true statement, are called solutions of the inequality. • A quadratic inequality is an equation of second degree that uses an inequality sign (<, >, ≤, ≥) instead of an equal sign. • A rational inequality is an inequality that contains a rational expression. • Inequalities of the form | ax + b |< k or | ax + b | > k are called the modulus or absolute inequalities.

WORKSHEET - 1 I.

LINEAR INEQUATIONS 1. Solve 24x < 100, when x is a natural number. 2. Solve 24x < 100, when x is an integer. 3. Solve –12x > 30, when x is a natural number. 4. Solve –12x > 30, when x is an integer. 5. Solve 3x + 8 > 2, when x is an integer. 6. Solve 3x + 8 > 2, x is a real number. 7. Solve for x: 3x – 7 > 5x – 1 8. Solve for x : 3 ( x − 1) ≤ 2 ( x − 3)

x x + < 11 2 3 3( x − 2) 5(2 − x ) . ≤ 10. Solve for x : 5 3 9. Solve for x : x +

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IL Foundation Series Class 10

1  3x  1  + 4  ≥ ( x − 6) 2 5  3 12. Solve 2(2x + 3) –10 < 6(x – 2). 11. Solve for x :

13. Solve 37 − (3x + 5) ≥ 9x − 8( x − 3). x (5x − 2) (7x − 3) < − . 4 3 5 (2x − 1) (3x − 2) (2 − x ) ≥ − 15. Solve . 3 4 5 14. Solve

II. QUADRATIC INEQUATIONS 1. Solve for x: x2 – 5x + 6 ≥ 0 2. Solve for x: 4x2 − 4x + 1 ≤ 0 3. Solve for x: x2 + 6x + 5 ≤ 0 4. Solve for x: x2 + 7x + 10 < 0. 5. Solve for x: x2 + 3x + 10 ≤ 0. 6. List the integers that satisfy x2 < 16. ( x − 3)( x − 2) >0 ( x − 1) (3x − 1)( x + 4) ≤0 8. Solve: ( x − 2) 7. Solve:

9. Solve for x and select the correct option given below: ( x − 1)( x + 5) ≥ 0 a) x ∈ (−∞, −5) ∪ [-2, 0] ∪ [1, ∞)

x ( x + 2)

b) x ∈ (−∞, −5) ∪ (-2,0) ∪ [1,∞) c) x ∈ (−∞, −5) ∪ [1,∞) d) x ∈ (−∞, −5) ∪ (-2,0) ∪ (1,∞) 10. Solve for x and select the correct option given below: ( x + 2)( x − 1)2 ( x − 5) ≥ 0 a) x ∈ (−∞, −2] ∪ (5,∞) b) x ∈ (−∞, −2] ∪ (5,∞) ∪ {2} c) x ∈ [−2, −5] d) None of these 11. Solve for x: 12. If

(2 − x )3 ( x − 21)6 ≥0 (2x − 3)2

( x − 2)( x ) ≥ 0, find the values of x. ( x − 1)2 ( x2 − 5x + 6) 139

INEQUATIONS

13. Solve for x and select the correct option given below: a) x ∈ [−3, −1] ∪ [2, )

( x + 1)4 ( x + 3) ≥0 ( x − 2)3 ( x2 + 1)

b) x ∈ (−∞, −3] ∪ [2, ∞) c) x ∈ (−∞, −3] ∪ (2, ) − 1 d) x ∈ [−3, −1] ∪ [−1, 2] 14. If

(2x + 3)(4 − 3x )3 ( x − 4) ≤ 0, find the values of x. ( x − 2)2 x5

( x − 1)2001 ( x − 2)100 >0 15. Solver for x: ( x − 3)23 ( x − 4) 16. Solve for x:

( x − 1)( x − 2)3 ( x − 3)4 ≤0 ( x − 4)6 ( x − 5)5

17. Solve for x:

( x − 1)( x − 2)3 ( x − 3)4 >0 ( x − 4)6 ( x − 5)5

x2 ( x + 1)10 ( x − 2)4 ≥0 18. Solve for x and select the correct option given below: ( x + 4)7 ( x + 7)5 ( x − 5)3 a) x ∈ (−7, −4) ∪ (5, ∞) ∪ −1, 0, 2 b) x ∈ (−∞, −4) ∪ (5, −∞) ∪ −1, 0, 2 c) x ∈ (−∞, −4) ∪ (5, ∞) d) None of these 19. Find the set of all x for which

2x 1 > . 2x + 5x + 2 x + 1 2

III. MODULUS INEQUATIONS 1. Solve for x: |x – 2| = 6 2. Solve x2 − 16 = 0 , and choose the correct option: a) {0, 4}

b) [-4, 4]

c) (-4, 4)

d) {-4, 4}

3. The solution set of the inequality | 2x + 1 |< 0 is æ1 ö æ 1 ö æ 1ö a) çç-¥, - ÷÷ b) èççç , ¥ø÷÷÷ c) çç- , ¥÷÷ 2 çè 2 ÷ø èç 2ø÷ 4. The solution set of the inequality | 2x − 3 |≥ 0 is a) φ

b) (3, ∞)

c) R

5. The solution set of the inequality |x - 5| < 9 is a) (0, 14) b) (-4, 14) c) (-4, 0)

140

d) φ

d) None of these d) (9, 14)

IL Foundation Series Class 10

IV. IRRATIONAL INEQUATIONS 1. Solve:

2x − 5 < 3

a) (−∞, 3) 2. Solve:

b) (−∞, 14)

c) (−∞, 7)

5  d)  , 7  2 

c) (4, +∞)

d) (5, + ∞)

c) x ∈ (−∞, −24)

d) x ∈ (−∞, +∞)

3x − 7 > 3

16   16 a)  , + ∞  b)  −∞,  3   3  3. Solve: 7 − 3x < −9 a) x ∈ (−24, +∞) b) x ∈∅

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. The set of all real numbers lying between 3 and 4 is represented as

a) [3,4)

b) (3,4]

c) (3,4)

d) [3,4]

2. The set of all real numbers lying between 5 and 6, including 5 is represented as a) (5,6)

b) [5,6)

c) [5,6]

d) (5,6]

c) ± x

d) –x

3. When x < 0, then |x|= a) 0

b) x

4. If a < b and c < 0, then which of the following is true? a b < c) ac > bc c c 5. All real numbers less than or equal to 5 are included in

a) ac < bc

a) (−∞, 5)

b) (−∞, 5)

c) (−∞, − 3) ∪ (5, ∞)

d) None of these d) (5, ∞)

II. SUBJECTIVE QUESTIONS 1. Solve 5x - 3 < 3x + 1 when x is an integer. 2. Solve 5x - 3 < 3x + 1 when x is a real number. 3. Solve 4x + 3 < 6x + 7. 5 − 2x x ≤ − 5. 3 6 3x − 4 x + 1 ≥ − 1. 5. Solve 2 4 6. Solve the inequation 2y − 3 < y + 1 ≤ 4y + 7; if y ∈ {Integers}. 4. Solve

7. Solve the inequation 2y − 3 < y + 1 ≤ 4y + 7; if y ∈ R (real numbers).

141

INEQUATIONS

x x − 5 ≤ − 4 , where x is a positive odd integer. 2 3 9. Harshit obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks. 8. Solve:

10. To receive a Grade 'A' in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita's marks in the first four examinations are 87, 92, 94 and 95, find the minimum marks that Sunita must obtain in the fifth examination to get Grade 'A' in the course. 2

11. Solve x − 6x + 8 ≥ 0. 12. List the integers that satisfy 2x2 + 9x + 4 ≤ 0. 13. Solve x2 + 13x + 30 < 0. 14. Solve for x: |x + 1| = 0 15. Solve for x: |3x + 2| = 10

142

6.1

ARITHMETIC AND GEOMETRIC PROGRESSIONS

INTRODUCTION TO ARITHMETIC PROGRESSIONS

In earlier classes, you might have come across various patterns of numbers, like 1, 3, 5, 7, 9, … 0, -2, -4, -6, -8, … 1, 4, 9, 16, 25, … These patterns are generally known as sequences. In this chapter, we aim to study a particular type of sequence known as arithmetic progressions. Sequence: A sequence consists of numbers arranged in a specific order based on a defined rule, and its individual numbers are referred to as terms. We denote the terms of a sequence by a1 , a2 , a3 , …... etc. (or) x1 , x2 …….. etc. Here, the subscripts denote the positions of the terms. The number in the first place is called its first term, and it is denoted by a1. The number in the second place is called the second term, and it is denoted by a2. The number at the n th place is called the n th term of the sequence, and it is denoted by an . The n th term is also called the general term of the sequence. For example, 2, 4, 6, 8, 10 is a sequence whose a1 2, a2 4, a3 6, Progression: The sequence of numbers under a certain rule is known as progression. Example: 1. 2, 4, 6, 8, 10, 12, …. 2. −1, −3, −5, −7, −9, …. Arithmetic Progression (AP): A sequence a1 , a2 , a3 , …….., an is called an arithmetic progression, if there exists a constant number d such that: a2 = a1 + d a3 = a2 + d a4 = a3 + d  an = an−1 + d and so on.

143

ARITHMETIC AND GEOMETRIC PROGRESSIONS

The constant d is called the common difference of the AP. Thus, if the first term is a and the common difference is d, then, a, a + d, a + 2d, a + 3d, a + 4d ….. is an arithmetic progression. In other words, a sequence a1 , a2 …. an is called an arithmetic progression if the difference of a term and the preceding term is always constant. This constant is called the common difference of the AP. Thus, if a1 , a2 , a3 , ……….. an , is an AP, with common difference d, then, a2 − a1 = d a3 − a2 = d a4 − a3 = d  an − an−1 = d and so on. 6.1.1 Properties of arithmetic progression The properties of an arithmetic sequence are as follows: •

hen a constant value is either added to or subtracted from every term in an AP, the resulting W sequence remains an AP with the same common difference.

•

I f each term in an AP is divided or multiplied by the same non-zero number, then the resulting sequence is also an AP.

•

Three consecutive numbers x, y, and z are in arithmetic progression if and only if 2y = x + z.

•

sequence is an AP if and only if its n th term is a linear expression in n, i.e., an = An + B , where A A and B are two constant quantities. Here, the coefficient of n in An is the common difference (C.D.) of the AP.

•

sequence is an AP if and only if the sum of its first n terms is of the form An 2 + Bn, where A A and B are two constant quantities that are independent of n. Here, the common difference is 2A, that is 2 times the coefficient of n 2.

SOLVED EXAMPLES Example 1: Write an AP whose first term is 10 and the common difference is 3.

144

IL Foundation Series Class 10

Solution: We know that if a is the first term and d is the common difference, then the arithmetic progression is: a, a + d, a + 2d, a + 3d, … Here, a = 10, d = 3 So, the arithmetic progression is a 10 a d 10 3 13 a 2d 10 2 3 16 a 3d 10 3 3 19 Therefore, the required progression is 10, 13, 16, 19, … Example 2: For the following arithmetic progressions, write the first term and common difference. 1 5 9 13 a) , , , , … 3 3 3 3

b) 0.6, 1.7, 2.8, 3.9, …

Solution: 1 5 9 13 a) , , , , … 3 3 3 3 1 Here, a = = First term 3 Common difference = d = t2 − t1 = a2 − a1 =

5 1 4 − = 3 3 3

b) 0.6, 1.7, 2.8, 3.9, … Here, a = 0.6 = First term Common difference = d = t2 − t1 = a2 − a1 = 1.7 − 0.6 = 1.1

6.2

nth TERM (OR) GENERAL TERM OF AN AP

Terms of AP are: •

First term ( a1 ) = a

•

Second term ( a2 ) = a + d

•

Third term ( a3 ) = a + 2 d 145

ARITHMETIC AND GEOMETRIC PROGRESSIONS

•

Fourth term ( a4 ) = a + 3 d

•

n th term (or) general term of an AP = a + ( n − 1 ) d tn = a + ( n − 1 ) d = an

If there are m terms in the AP, then am represents the last term, which is sometimes also denoted by l. The nth term of an AP from the end = l − ( n − 1 ) d

SOLVED EXAMPLES Example 1: Find the 6th term from the end of the AP 17, 14, 11, ...., -40. Solution: We have, l = last term = −40. d = common difference = −3 6th term from the end = l − ( n − 1 ) d = −40 − ( 6 − 1 ) ( −3 ) = −40 + 15 = −25 Example 2: If the 8th term of an AP is 31 and the 15th term is 16 more than the 11th term, then find the AP. Solution: Let a be the first term and d be the common difference of the AP. We have, a8 = 31 and a15 = 16 + a11 a 7 d 31 and a 14d 16 a 10 d a 7 d 31 and a 14d 16 a 10d a 7 d 31 and 4d 16 a 7 d 31 and d 4 a 7 4 31 ⇒ a + 28 = 31 ⇒a=3 Hence, the AP is a, a + d, a + 2d, a + 3d, … i.e., 3, 7, 11, 15, 19, …

146

IL Foundation Series Class 10

Example 3: Which term of the arithmetic progression 5, 15, 25, … will be 130 more than its 31st term? Solution: We have a = 5 and d = 10. ∴ a31 = a + 30 d = 5 + 30 × 10 = 305 Let the nth term of the given AP be 130 more than its 31st term then, an 130 a31 a n 1 d 130 305

5 n 1 10 435 n 1 10 430 n 1 43 n 1 43 44 ∴ The 44th term of the given AP will be 130 more than its 31st term.

6.3

THE SUM OF FIRST 'n' TERMS OF AP

Let the first n terms of the AP be a, a + d, a + 2d, a + 3d, … a + (n - 1)d. S denotes the sum of the first n terms of the AP. ∴ S = a + ( a + d ) + ( a + 2d ) + …+ a + ( n − 1 ) d …(1) Rewriting the terms in reverse order, we get S = a + ( n − 1 ) d + a + ( n − 2 ) d + …. ( a + d ) + a …(2) On adding (1) and (2) term-wise, we get, ⎡ 2a + ( n − 1 ) d ] + [ 2a + ( n − 1 ) d ] +…+ [ 2a + ( n − 1 ) d ⎤⎦ + ⎣⎡ 2a + (n − 1)d ⎤⎦ 2S = ⎣ n times (or) 2 S = n ⎡⎣ 2a + ( n − 1 ) d ⎤⎦ (since there are n terms) n ⎡ 2a + ( n − 1 ) d ⎤⎦ 2⎣ So, the sum of the first n terms of AP is given by: n Sn = ⎡⎣ 2a + ( n − 1 ) d ⎤⎦ 2 (or) S =

We can write this as:

n a a n 1 d 2 147

ARITHMETIC AND GEOMETRIC PROGRESSIONS

Sn =

n n a + an ] = [ a + l ] [ 2 2

Where 'l’ is the last term of the AP. 6.3.1 Series of natural numbers 1. Sum of n natural numbers: Sn =

n ( n + 1) 2

2. Sum of squares of n natural numbers: Sn =

n ( n + 1 ) ( 2n + 1 ) 6

3. Sum of cubes of n natural numbers: ⎡ n ( n + 1) ⎤ Sn = ⎢ ⎥ 2 ⎣ ⎦ Note: In a sequence, if Sn is the sum of the first n terms and Sn−1 is the sum of the first (n − 1) terms, then the nth term is given by Tn = Sn − Sn −1 . 2

Example: Find the sum of the first 50 positive integers (terms) of AP. Solution: 50 ( 50 + 1 ) 2 50 ( 51 ) ⇒ S50 = 2 S50 =

⇒ S50 = 25 × 51 ⇒ S50 = 1275 6.3.2 Selection of terms in an AP Sometimes, it is required to select a finite number of terms in AP. It is always convenient if we select the terms in the following manner:

148

Number of terms

Terms

Common difference

a − d, a, a + d

a − 3 d, a − d, a + d, a + 3 d

a − 2 d, a − d, a, a + d, a + 2 d

a 5 d, a 3 d, a d, a d, a 3 d, a 5 d

IL Foundation Series Class 10

It is important to observe that when dealing with an odd number of terms, the middle term is a, and the common difference is d. Conversely, when dealing with an even number of terms, the middle terms are a - d and a + d, with a common difference of 2d.

6.4

ARITHMETIC MEAN (AM)

If a, x, and b are in AP, then x is called A.M. of a and b. ∴x =

a+b 2

6.4.1 Single A.M. of 'n’ positive numbers Let a1 , a2 , …. an be n positive numbers, then A.M. of these numbers, A=

a1 + a2 + …. + an . n

Note: First n Natural No.

First n Even Natural No.

First n Odd Natural No.

n ( n + 1) 2

n ( n + 1)

( n + 1)

Sum Mean

( n + 1) 2

SOLVED EXAMPLES Example 1: The sum of the 5th and 9th terms of an AP is 72, and the sum of the 7th and 12th terms is 97. Find the AP. Solution: Let a be the first term and d be the common difference of the AP. Given, a5 + a9 = 72 and a7 + a12 = 97 . a 4d a 8d 72 and a 6d a 11d 97 Thus, we have, ⇒ 2a + 12d = 72 …(1)

149

ARITHMETIC AND GEOMETRIC PROGRESSIONS

⇒ 2a + 17 d = 97 …(2) Subtracting (1) from (2), we get, 5d = 25 ⇒d=5 Now, from equation (1), 2a + 12(5) = 72 ⇒ 2a = 72 - 60 ⇒ 2a = 12 ⇒ a=6 Therefore, a = 6 and d = 5 Hence, the AP is 6, 11, 16, 21, 26, … Example 2: If the p th term of an AP is q and the q th term is p, prove that its n th term is ( p + q − n ). Solution:

Let a be the first term and d be the common difference of the given AP. Given, p th term q a p 1 d q ...(1) q th term p a q 1 d p ...(2) Subtracting equation (2) from equation (1), we get:

( p − q )d = ( q − p ) ⇒ d = −1 Putting d = −1 in equation (1), we get, a + ( p − 1 ) ( −1 ) = q ⇒ a = ( p + q − 1) n th term a n 1 d = ( p + q − 1 ) + ( n − 1 ) ( −1 ) = ( p+q −n) Hence proved. Example 3: T he sum of three numbers in AP is -3, and their product is 8. Find the numbers.

150

IL Foundation Series Class 10

Solution: Let the numbers be (a - d), a, (a + d). Sum of numbers = −3 ⇒ ( a − d ) + a + ( a + d ) = −3 ⇒ 3a = −3 ⇒ a = −1 Now, Product of numbers = 8 ⇒ ( a − d )( a ) ( a + d ) = 8

(

)

⇒ a a2 − d2 = 8

(

)

⇒ ( −1 ) 1 − d 2 = 8 [ a = −1 ] ⇒ d2 = 9 ⇒ d = ±3 If d = 3, the numbers are -4, -1, 2 If d = −3, the numbers are 2, -1, -4 Thus, the numbers are: −4, − 1, 2 (or) 2, − 1, − 4. Example 4: Find the sum of all the three-digit natural numbers, which are divisible by 7. Solution: he smallest and the largest numbers of three digits, which are divisible by 7, are 105 and 994, T respectively. So, the sequence of three-digit numbers which are divisible by 7 are: 105, 112, 119, …994 This forms an AP with the first term a = 105 and common difference d = 7. Let there be n terms in this sequence, then, an = a + ( n − 1 ) 7 = 994 Now, ⇒ 105 + ( n − 1 ) 7 = 994 ⇒ ( n − 1 ) 7 = 889 ⇒ ( n − 1 ) = 127 ⇒ n = 128

151

ARITHMETIC AND GEOMETRIC PROGRESSIONS

n Therefore, the required sum = ⎡⎣ 2a + ( n − 1 ) d ⎤⎦ 2 128 = ⎡⎣ 2 × 105 + ( 128 − 1 ) 7 ⎤⎦ 2 = 64 [ 210 + 889 ] = 64 [ 1099 ] = 70, 336 Example 5: The test scores of five students in a mathematics class are 78, 85, 92, 88, and 95. Find the arithmetic mean of these scores. Solution: To find the arithmetic mean, sum up all the scores and divide by the number of students. 78 + 85 + 92 + 88 + 95 Arithmetic Mean = 5 438 = 5 = 87.6 Therefore, the arithmetic mean of the test scores is 87.6. Example 6: Find the AP whose sum to n terms is 2n 2 + n. Solution: Given Sn = 2n 2 + n Put n = 1, 2, 3, 4, …, in succession, we get S1 = 2(1)2 + 1 = 2 + 1 = 3 S2 = 2(2)2 + 2 = 8 + 2 = 10 S3 = 2(3)2 + 3 = 18 + 3 = 21 S4 = 2(4)2 + 4 = 32 + 4 = 36 and so on. Therefore, a1 = S1 = 3 a2 = S2 − S1 = 10 − 3 = 7 a3 = S3 − S2 = 21 − 10 = 11 a4 = S4 − S3 = 36 − 21 = 15 and so on. Hence, the required AP is 3, 7, 11, 15, …

152

IL Foundation Series Class 10

Example 7: A sum of ` 280 is to be used to award four prizes. If each prize after the first is ` 20 less than its preceding prize, find the value of each of the prizes. Solution: he value of four prizes forms an AP with a common difference d = -20, and the sum of the terms is T 280. Let the value of the first prize be ` a. Then the sum (S ) = ` 280 We know that: n Sn = ⎡⎣ 2a + ( n − 1 ) d ⎤⎦ 2 4 ⇒ 280 = ⎡⎣ 2a + ( 4 − 1 ) × −20 ⎤⎦ 2 ⇒ 280 = 2 [ 2a − 60 ] ⇒ 140 = 2a − 60 ⇒ 2a = 140 + 60 = 200 ⇒ a = 100 Hence, the values of 4 prizes are `100, `80, `60, and `40. Example 8: I f the contractor exceeds the specified completion date on a construction project, the penalty is structured as follows: ` 200 for the initial day, ` 250 for the second day, ` 300 for the third day, and so on, with each subsequent day incurring a ` 50 higher penalty than the preceding day. What is the penalty amount the contractor must pay for a 30-day delay in completing the work? Solution:

Since the penalty for each succeeding day is ` 50 more than for the preceding day, the amount of penalty for different days forms an AP with first term a = 200 and common difference d = 50 Sum up to n terms Sn =

n ⎡ 2a + ( n − 1 ) d ⎤⎦ 2⎣

The amount of penalty that must be paid if he has delayed the work by 30 days = S30 Therefore, the required sum for the delay of 30 days 30 = ⎡ 2 × 200 + ( 30 − 1 ) × 50 ⎤⎦ 2 ⎣ = 15 [ 400 + 29 × 50 ] = 15 [ 400 + 1450 ] = 15 [ 1850 ] = 27, 750 Thus, a delay of 30 days will cost the contractor ` 27,750.

153

ARITHMETIC AND GEOMETRIC PROGRESSIONS

Example 9: In a school, students thought of planting trees in and around the school to reduce noise pollution and air pollution. It was decided that the number of trees in each section of each class would be the same as the class they are studying. E.g., a section of class I will plant 1 tree; a section of class II will plant 2 trees, and so on till class XII. There are three sections of each class. How many trees will be planted by the students? Solution:

Since each section of each class plants the same number of trees as the class number and there are three sections for each class, the total number of trees planted by students is given by = 3 [ 1 + 2 + 3 + …12 ] ⎡ 12 ⎤ = 3 ⎢ { 2 × 1 + ( 12 − 1 ) × 1} ⎥ ⎣ 2 ⎦ = 3 ⎡⎣ 6 ( 2 + 11 ) ⎤⎦

n  Sn 2 2a n 1 d

= 18 × 13 = 234 Example 10: A ladder has rungs 2.5 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2.5 metres apart, what is the length of the wood required for the rungs? 25 cm

2.5 m

2.5 cm 45 cm

Solution:

It is given that the gap between two consecutive rungs is 2.5 cm, and the top and bottom rungs are 2.5 m, i.e., 250 cm apart. 250 1 11 25 It is given that the rungs are decreasing uniformly in length from 45 cm at the bottom to 25 cm at the top. Therefore, lengths of the rungs form an AP with first term a = 45 cm and 11th term = 25 cm

Number of rungs

∴Length of the wood required for rungs = Sum of the 11 terms of an AP with first term 45 and last term 25 cm

154

IL Foundation Series Class 10

11 n 45 25 cm  Sn a l 2 2

= 385 = cm 3.85 metres Therefore, the length of the wood required for the rungs is 3.85 metres.

6.5

INTRODUCTION TO GEOMETRIC PROGRESSION (GP)

Let us consider the following sequences: (i) 2, 6, 18, 54,… 1 1 1 1 , , , 9 18 36 72 How do their terms progress in each of these sequences? We note that each term, except the first, progresses in a definite order. a a a In (i), we have = a1 2= , 2 3= , 3 3, 4 = 3 and so on. a1 a2 a3 1 a 1 a 1 a 1 In (ii), we observe, a1 , 2 , 3 , 4 and so on. 9 a1 2 a2 2 a3 2 (ii)

It is observed that in each case, every term except the first term bears a constant ratio to the term 1 immediately preceding it. In (i), this constant ratio is 3, and in (ii), the constant ratio is − . 2 Such sequences are called geometric sequences or geometric progression, abbreviated as GP. A sequence a1 , a2 , a3 , …, an , … is called geometric progression if each term is non-zero and (constant), for k ≥ 1.

ak +1 =r ak

By letting a1 = a, we obtain a geometric progression, a, ar, ar 2 , ar 3 , …, where a is called the first term and r is called the common ratio of the GP.

6.6

nth TERM (OR) GENERAL TERM OF A GP

Let us consider a GP with the first non-zero term a and common ratio r. Write a few terms of it. The second term is obtained by multiplying a by r, thus a2 = ar. Similarly, the third term is obtained by multiplying a2 by r. Thus, a3 = a2 r = ar 2, and so on. We write these below and a few more terms.

155

ARITHMETIC AND GEOMETRIC PROGRESSIONS

1st term a1 a ar1 1 , 2nd term a2 ar ar 2 1 , 3rd term a3 ar 2 ar 3 1 4th term a4 ar 3 ar 4 1 , 5th term a5 ar 4 ar 5 1 Do you see the pattern? What will be the 16th term? a16 = ar16−1 = ar15 Therefore, the pattern suggests that the n th term of a GP is given by: an = ar n −1

6.7

SUM TO 'n' TERMS OF A GEOMETRIC PROGRESSION

Let the first term of a GP be a and the common ratio be r. Let us denote by Sn the sum to the first n terms of GP then Sn = a + ar + ar 2 + …+ ar n −1 …(1) Case 1: If r = 1, we have Sn = a + a + a + …+ a ( n terms) = na Case 2: If r ≠ 1, multiplying (1) by r, we have rSn = ar + ar 2 + ar 3 + …+ ar n …(2) Subtracting (2) from (1), we get

( 1 − r ) Sn = a − ar n = a ( 1 − r n )

This gives, Sn =

(

a rn − 1 r −1

)

6.7.1 Sum of an infinite GP If a + ar + ar 2 + … is an infinite series with r < 1, then the sum is S∞ =

SOLVED EXAMPLES Example 1: Find the 11th and n th terms of the GP 5, 25, 125, … Solution: Here a = 5 and r = 5 Thus, a11 = 5(5)11−1 = 5(5)10 = 511 and an = ar n −1 = 5(5)n −1 = 5n

156

a . 1− r

IL Foundation Series Class 10

Example 2: In a GP, the 3 rd term is 24 and the 6th term is 192. Find the 10th term. Solution: Here, a3 = ar 2 = 24 …(1) And a6 = ar 5 = 192 …(2) Dividing (2) by (1), we get r = 2. Substituting r = 2 in (1), we get a = 6. Hence, a10 = 6(2)9 = 3072. Example 3: Find the sum of the first n terms and the sum of the first 5 terms of the geometric series 2 4 1+ + +… 3 9 Solution: 2 Here a = 1 and r = . 3 Therefore ⎡ ⎛ 2⎞n ⎤ ⎢1 − ⎜ ⎟ ⎥ a 1 − rn ⎡ ⎛ 2⎞n ⎤ ⎢⎣ ⎝ 3 ⎠ ⎥⎦ = Sn = = 3 ⎢1 − ⎜ ⎟ ⎥ 2 1− r ⎢⎣ ⎝ 3 ⎠ ⎥⎦ 1− 3

(

)

⎡ ⎛ 2 ⎞5 ⎤ 211 211 In particular, S5 = 3 ⎢ 1 − ⎜ ⎟ ⎥ = 3 × = . 243 81 ⎢⎣ ⎝ 3 ⎠ ⎥⎦ Example 4: Find the sum to infinity of the series Solution:

1 1 1 + + + …. 3 9 27

1 1 1 3 1 Here, a = , r = 9 = × = 1 9 1 3 3 3 Since r < 1, so S∞ exists. 1 a 1 3 1 S∞ = = 3 = × = 1 3 2 2 1− r 1− 3 6.7.2 Selection of terms in a GP Sometimes, it is required to select a finite number of terms in GP. It is always convenient if we select the terms in the following manner:

157

ARITHMETIC AND GEOMETRIC PROGRESSIONS

6.8

No. of Terms

Terms

Common Ratio

a , a, ar r

a a , , ar, ar 3 3 r r

a a , , a, ar, ar 2 2 r r

GEOMETRIC MEAN

The geometric mean of two positive numbers, a and b, is the number ab . Therefore, the geometric mean of 2 and 8 is 4. We observe that the three numbers 2, 4, and 8 are consecutive terms of a GP. This leads to a generalisation of the concept of geometric means of two numbers.

6.9

RELATIONSHIP BETWEEN A.M. AND G.M.

Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively. Then a b A and G ab 2 Thus, we have A G

a b ab 2

A G

a b 2 ab 2

A G

( a b )2 0 …(1) 2

From (1), we obtain the relationship A ≥ G .

158

IL Foundation Series Class 10

SOLVED EXAMPLES Example 1: What is the 21st term of the sequence defined by an = ( n − 1 ) ( 2 − n ) ( 3 + n ) ? Solution:

Putting n = 21, we obtain a20 = ( 21 − 1 ) ( 2 − 21 ) ( 3 + 21 ) ⇒ a20 = 20 × ( −19 ) × ( 24 ) = −9,120 Example 2: Which term of the GP, 2, 8, 32,… up to n terms is 131072? Solution:

Let 131072 be the n th term of the given GP. Here, a = 2 and r = 4. Therefore 131072 = an = 2(4)n −1 or 65536 = 4n −1 This gives 48 = 4n −1 So that, n − 1 = 8, i.e., n = 9. Hence, 131072 is the 9th term of the GP. 2 4 Example 3: F ind the sum of the first 5 terms of the geometric series 1 + + + … 3 9 Solution: 2 Here a = 1 and r = . 3 a 1 − rn Therefore, Sn = 1− r

(

)

⎡ ⎛ 2 ⎞5 ⎤ 211 211 ⇒ S5 = 3 ⎢ 1 − ⎜ ⎟ ⎥ = 3 × = 243 81 ⎢⎣ ⎝ 3 ⎠ ⎥⎦ Example 4: Find the sum of the sequence 8, 88, 888, 8888, … to n terms. Solution: This is not a GP. However, we can relate it to a GP by writing the terms as Sn 8 88 888 8888 to n terms

8 9 99 999 9999 to n term 9

159

ARITHMETIC AND GEOMETRIC PROGRESSIONS

8 10 1 102 1 103 1 104 1 n terms 9

8 10 102 103 n terms 1 1 1 n terms 9

(

)

(

)

n ⎤ 8 ⎡⎢ 10 10 − 1 −n⎥ = 9 ⎢ 10 − 1 ⎥ ⎣ ⎦ n ⎤ 8 ⎡⎢ 10 10 − 1 = −n⎥ 9⎢ 9 ⎥ ⎣ ⎦

Example 5: I f A.M. and G.M. of two positive numbers a and b are 10 and 8, respectively, find the numbers. Solution: a b 10 …(1) 2 = ab 8 …(2)

Given that A.M. and G.M. =

From (1) and (2), we get, a + b = 20 …(3) ab = 64 …(4) utting the value of a and b from (3), (4) in the identity (a − b)2 = (a + b)2 − 4ab , we get P (a − b)2 = 400 − 256 = 144 or a − b = ±12 …(5) By solving (3) and (5), we obtain = a 4= , b 16 or = a 16 = ,b 4 Thus, the numbers a and b are 4, 16 or 16, 4 respectively.

QUICK REVIEW

160

•

nth term of an AP: Tn = a + ( n − 1 ) d

•

Sum of n terms in an AP: S =

•

Sum of n terms in an AP when first and last terms are known: S =

•

Arithmetic mean (A.M.) of any two numbers a and b is

n ⎡ 2a + ( n − 1 ) d ⎤⎦ 2⎣

( a + b ). 2

n (a + l ) 2

IL Foundation Series Class 10

nth term of a GP: Tn = ar n −1

• •

Sum of n terms in a GP when r < 1: Sn =

(

a 1 − rn

(

1− r

a rn − 1

) )

•

Sum of n terms in a GP when r > 1: Sn =

•

Sum of n terms in a GP when r = 1: Sn = na

•

Sum of an infinite GP: S∞ =

•

Geometric mean of (G.M.) of any two positive numbers a and b is ab .

r −1

a 1− r

WORKSHEET - 1 I.

nth TERM (OR) GENERAL TERM OF AN AP 1. Write an AP whose first term is 12 and the common difference is 4. 2. Find the 6th term from the end of the AP 17, 14, 11, ..., -40. 3. How many multiples of 4 lie between 10 and 250? 4. The 17 th term of an AP exceeds its 10th term by 7. Find the common difference. 5. Check whether -150 is a term of the AP 11, 8,5, 2,… 6. The sum of the 4th and the 8th terms of an AP is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms of AP. 7. In an AP, if the common difference ( d ) = −4 and the seventh term ( a7 ) is 4, then find the first term. 8. What is the common difference of an AP in which a21 − a7 = 84? 9. For what value of k, will k 9, 2 k 1 and 2 k + 7 are the consecutive terms of an AP?

10. Which term of an AP: 21, 42, 63, 84,… is 210? 11. Write the first four terms of an AP whose first term is 3x + y , and the common difference is x − y. 12. Find the next term of the AP 7 , 28 , 63 , … 1 13. Find the 11th term of AP −3, − , 2, 2 14. The first term of an AP is p and its common difference is q. Find the 10th term. 15. Assertion (A): 184 is the 50th term of the sequence 3, 7, 11, 15 , …

161

ARITHMETIC AND GEOMETRIC PROGRESSIONS

Reason (R): The nth term of AP is given by an = a + ( n − 1 ) d a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). b) Both Assertion (A) and Reason (R) are true, but Reason(R) is not the correct explanation of Assertion (A). c) Assertion (A) is true, but Reason (R) is false. d) Assertion (A) is false, but Reason (R) is true. II. SUM OF FIRST 'n' TERMS OF AN AP 1. Find the sum of the first five positive integers, which are divisible by 6. 2. If Sn is the sum of the first n terms of an AP and Sn = 5n 2 + 3n, then find the n th term of AP. 3. How many terms of the series 54, 51, 48,… be taken so that their sum is 513? Explain the double answer. 4. If the sum of first the n terms of an AP is given by Sn = 4n 2 + n, then find its n th term. 3n 2 5n 5. In an AP, the sum of the first n terms is + . Find its 25th term. 2 2 6. If in an AP, the sum of m terms is equal to n and the sum of n terms is equal to m, then prove that the sum of ( m + n ) terms is − ( m + n ). 7. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the number of the houses following it. Find this value of x. 8. 200 logs are stacked in the following manner. 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are 200 logs placed, and how many logs are in the top row?

III. nth TERM (OR) GENERAL TERM OF A GP 1. If the 4th, 7th, and 10th terms of a GP are p, q, and r respectively, then a) p2 = q 2 + r 2

162

b) q 2 = pr

c) p2 = qr

d) pqr + pq + 1 = 0

IL Foundation Series Class 10

2. If x, 2 x + 2, 3 x + 3 are in GP then the fourth term is a) 27

b) −

27 2

⎛ 1⎞ 3. The common ratio of the series an = 3 ⎜ ⎟ ⎝ 5⎠ a) 0.1

27 2

d) -27

c) 0.2

d) 0.5

c) n −1

b) 3

4. If x, 2 x + 2, 3 x + 3 are in GP, then 5 x, 10 x + 10, 15 x + 15 forms a) GP

b) AP

c) a constant sequence

d) neither AP nor GP

13 5. The sum of the first three terms of a GP is and their product is -1. Find the common ratio 12 and the terms. 6. Statement (A): If tn = 6n + 5, then the common difference is 6 . Statement (B): The number of terms of the AP: 3, 6, 9, ......, 111 is 37. a) Both A and B are true

b) Both A and B are false

c) A is true and B is false

d) A is false and B is true

7. Statement (A): 12 , 22 , 32 , 42 , ….. are in AP. Statement (B): If the n th term of the list of numbers is 2n 2 − 5, then it forms an AP. a) Both A and B are true

b) Both A and B are false

c) A is true and B is false

d) A is false and B is true

IV. SUM OF FIRST 'n' TERMS OF A GP 1. Statement (A): If a, b, c, d are in GP, then (b − c)2 + (c − a)2 + (d − b)2 = (a − d)2. Statement (B): The sum of all natural numbers from 1 to 100 is 5050. a) Both A and B are true

b) Both A and B are false

c) A is true and B is false

d) A is false and B is true

2. Assertion (A): The 13th term of 2, 2 2 , 4, ….. is 128. Reason (R): The n th term of GP is ar n−1. a) Both A and R are true, and R is the correct explanation of A b) Both A and R are true, but R is not the correct explanation of A c) A is correct and R is incorrect d) A is incorrect and R is correct 3. Assertion (A): 2 , 8 , 18 , 32 , …….. are in GP.

163

ARITHMETIC AND GEOMETRIC PROGRESSIONS

Reason (R): a, b, c are in GP, then b 2 = ac . a) Both A and R are true, and R is the correct explanation of A b) Both A and R are true, but R is not the correct explanation of A c) A is correct and R is incorrect d) A is incorrect and R is correct 3069 3 3 ? 4. How many terms of the GP 3, , , … are needed to give the sum 512 2 4 5. A person has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own. 6. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 … 7. How many terms of GP 3, 32 , 33 , … are needed to give the sum 120? 8. The sum of the first three terms of a GP is 16, and the sum of the next three terms is 128. Determine the first term, the common ratio, and the sum to n terms of the GP 9. If AM and GM of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation. 10. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio 3+2 2 : 3−2 2 .

(

)(

)

WORKSHEET - 2 I.

FILL IN THE BLANKS 1. If in AP = a 15 = , d 3, and an = 60, then n is __________. 2. The first negative term of 84, 80, 76,….. is __________. 3. The middle term of three consecutive terms of an AP when their sum is 24 is __________. 4. The first three terms of an AP, respectively, are 3y 1, 3y 5 and 5y + 1. Then, y equals __________. 5. If

1 1 1 , , are in AP, then x is __________. x+2 x+3 x+5

6. If the n th term of an AP is 2n + 1, then the sum of the first n terms of the AP is __________. 7. The common difference of an AP in which a24 − a17 = −28 is __________. 8. If the sum of n terms of an AP is 2n 2 + 5n , then its 2nd term is __________. 9. If the sum of the first k terms of an AP is 3 k2 − k , then the common difference of AP is __________. 10. In an AP, if d 4, n 7 and an = 4, then a is equal to __________.

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IL Foundation Series Class 10

11. In an AP, = if a 1= , an 20 and Sn = 399, then n is equal to __________. 12. If 8th term of an AP is zero, then the 38th term is __________ of its 18th term. 13. In an AP, 5 times of the 5th term is equal to 20 times of the 20th term, then the 25th term of AP is __________. 14. If the equation 9 x 2 + 6 px + 4 = 0 has equal roots, then p = __________. 15. The common difference of the AP

1 1 2q 1 4q , , , is __________. 2q 2q 2q

II. MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. Which term of the AP 5, 2, 1, . is -49? a) 19th

b) 15th

c) 16th

d) 20th

2. The 21st term of the AP, where the first two terms are -3 and 4 respectively, is a) 77

b) 137

c) 143

d) -143

c) 11

d) 15

3. If 18, a, b, −3 are in AP, then a + b = a) 19

b) 7

4. For what value of k, k 2, 4 k 6, 3 k 2 are three consecutive terms of an AP? a) 1

b) -1

c) 3

d) -3

5. If the common difference of an AP is 5, then the value of a18 − a13 is a) 5

b) 20

c) 25

d) 30

c) 9th

d) 11th

6. Which term of the AP 27, 24, 21, …. is zero? a) 8th

b) 10th

7. The sum of first n terms of the series a, 3a, 5a, …. is a) na

b) ( 2n − 1 ) a

c) n 2a

d) n 2a 2

8. If the first term of an AP is 2 and the common difference is 4, then the sum of its 40 terms is a) 3200

b) 1600

c) 200

d) 2800

9. The number of terms of the AP 3, 7, 11, 15, … to be taken so that the sum is 406 is a) 5

b) 10

c) 12

d) 14

10. The n th term of an AP, the sum of whose n terms is Sn , is a) Sn + Sn −1

b) Sn − Sn −1

c) Sn + Sn +1

d) Sn − Sn +1

11. The sum of the first 16 terms of the AP: 10, 6, 2, ….. is a) -320

b) 320

c) -352

d) -400

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ARITHMETIC AND GEOMETRIC PROGRESSIONS

12. If

3 5 7 . n terms 7, then the value of n is: 5 8 11 .. 10 terms

a) 35 13. If

b) 36

c) 37

d) 40

c) 7

d) 5

1 3 5 n terms 12 , then n = 1 2 3 ...n terms 7

a) 6

b) 4

14. If the first term of an AP is 2 and the sum of the first five terms is equal to one-fourth of the sum of the next five terms, then the sum of the first 30 terms is a) 2550

b) 3000

c) -2550

d) -3000

1 1 and the q th term of an AP is , then the sum of the first pq terms is q p 1 1 p+q pq a) b) c) ( pq − 1 ) d) ( pq + 1 ) 2 2 pq p+q 1 1 1 + …. is + + 16. The sum to n terms of the series 1+ 3 3+ 5 5+ 7 15. If the p th term of an AP is

{

}

1 ⎛ 1⎞ b) ⎜ ⎟ 2n + 1 c) 2n + 1 − 1 d) 2n + 1 − 1 ⎝ 2⎠ 2 17. If the sum of 5 terms of an AP is the same as the sum of its 11 terms, then the sum of 16 terms is a) 0 b) 16 c) -16 d) 32 4 18. The sum of an infinite GP is 2. If the sum of their squares is then, the third term is 3 1 1 1 a) b) 1 c) d) 2 4 8 1 1 1 19. If a1 , a2 , a3 , …… an be an AP of non-zero terms, then + + …….. + = a1a2 a2a3 an −1an a) 2n + 1

n −1 a1 + an

n −1 a1 ⋅ an

1− n a1 ⋅ an

20. Among the following, the term is not in AP; 3, 11, 19, 27, .... is a) 83

b) 163

c) 243

d) 137

c) 42

d) 39

21. If 1 + 4 + 7 + 10 + ... + x = 287, then x = a) 40

b) 41

22. Two APs have the same common difference. The difference between their 100th terms is 100, then the difference between their 1000th terms is a) 100 166

b) 1000

c) 1

d) 0

IL Foundation Series Class 10

a−b = b−c a b a a) b) c) b c c 24. In an AP, = if a 1= , an 20 and Sn = 399, then n =

d) 1

a) 19

d) 42

23. If a, b, c are in AP, then

b) 21

c) 38

25. If the n th term of 24, 20, 16, 12, …. is same as the n th term of 11, 8, 5, then n = a) 4

b) 6

c) 8

d) 7

c) 0

d) m − n

c) 59

d) 62

26. In an AP, if mam = nan , then am + n = a) m + n

b) m2 + n 2

3 1 27. In an AP, a2 7 , a29 1, an 6 , then n = 4 2 a) 48 b) 58

28. The number of terms in the AP, 6, 3, 0, 3, , 36 is a) 13 b) 14 2 5k 29. If , k, are in AP, then k = 3 8 1 2 a) b) 3 3

c) 15

d) 16

8 11

16 33

5 5 , 0, , . is 70. 2 2 Statement (B): In an AP, if a16 = P , then the sum of the first 31 terms of it is 16P. 30. Statement (A): The 25th term of an AP 5,

a) Both A and B are true

b) Both A and B are false

c) A is true and B is false

d) A is false and B is true

167

TRIANGLES

7.1 INTRODUCTION TO TRIANGLES In our previous class, we studied about congruence of figures, particularly triangles. In this chapter, we shall be introduced to the concept of the similarity of figures and triangles and its application in proving the Pythagoras theorem. Similar figures: Two figures are said to be similar if and only if they have the same shape. Similarity is symbolised as ~. Examples: Two equilateral triangles, two circles, etc. Note: Congruent figures are always similar, but similar figures are not necessarily congruent. Similar polygons: Two polygons with the same number of sides are similar if: i) Their corresponding angles are equal. ii) Their corresponding sides are proportional (in the same ratio). Examples: T 2

E 1.5

105°

D 100°

120°

135° 140°

1.5

C 1

120°

U 2

120° 135°

S 100°

105°

140° 3

R 2

In hexagons ABCDEF and PQRSTU (as shown in the figure), i) ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R, ∠D = ∠S, ∠E = ∠T, ∠F = ∠U ii)

AB BC CD DE EF FA = = = = = PQ QR RS ST TU UP

Hence, the two hexagons are similar. Note: While concluding about similarity, it is important that the corresponding vertices occupy the same positions. In the above example, the correct conclusion is ⬡ABCDEF ~ ⬡PQRSTU.

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IL Foundation Series Class 10

7.2 BASIC PROPORTIONALITY THEOREM (THALES THEOREM) Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio. A M N E

In the above DABC , line l ||BC, and it intersects AB and AC at D and E, respectively. So, by the Basic Proportionality Theorem (BPT), AD AE = DB EC Note: The basic proportionality theorem is also applicable when a line is drawn parallel to one side of a triangle intersecting the other two sides externally at distinct points. Case - (i)

Case - (ii)

A E

In both the above cases, DE||BC in ΔABC. AD AE = DB EC AD AE ⇒ = AB AC DB EC . ⇒ = AB AC

∴ by BPT,

7.2.1 Converse of the basic proportionality theorem (converse of the Thales theorem) Statement: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. 169

TRIANGLES

Let us consider DABC and the line l intersecting sides AB and AC at D and E AD AE = . DB EC So, by the converse of BPT,

such that

l || BC or DE || BC. D

E B

SOLVED EXAMPLES Example 1: In the given figures, ABCD and PQRS are two rectangles. Verify whether they are similar. D

1 A

C 1

Solution: ere, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R, ∠D = ∠S. H AB 3 CD AD 1 BC = = = = and PQ 4 RS PS 2 QR Since the corresponding sides are not in proportion, ABCD and PQRS are not similar. Example 2: In the figure below, DE=  BC, AD 1= .5 cm, DB 3 cm, AE = 1 cm . Find EC. A 1.5 cm

1 cm

3 cm

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IL Foundation Series Class 10

Solution: Given: DE || BC By BPT, AD AE DB EC 1.5 cm 1 cm 3 cm EC EC 2 cm Example 3: In ΔABC, D and E are points on AB and AC, respectively, such that DE || BC. If AD 4 x 3 cm, AE 8 x 7 cm, BD 3 x 1 cm , and CE = ( 5 x − 3 ) cm , then find the value of x. Solution: A

Given that DE || BC, ∴ By BPT AD AE DB EC 4x 3 8x 7 3x 1 5x 3 4x 3 5x 3 3x 1 8x 7 2

20 x 12 x 15 x 9 24 x 21x 8 x 7 4 x 2 2 x 2 0 2x2 x 1 0 2x2 2x x 1 0 2 x x 1 1 x 1 0 2x 1 x 1 0 2 x 1 0 or x 1 0 1 orr x 1 x 2 But x > 0, because for x =

−1 ; AD, DB, AE, and EC give negative values. 2

∴x =1

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TRIANGLES

Example 4: In the given figure, DE || AC and DF || AE. Prove that

BF BE = . FE EC

Solution: In DABE, DF || AE . ∴ By BPT,

BF BD = …(1) FE DA

And, in DABC, DE || AC . BD BE = …(2) DA EC Therefore, from (1) and (2),

∴ By BPT,

BF BE = FE EC Example 5: If D and E are points on sides AB and AC, respectively, of ΔABC such that DE || BC and BD = CE . Prove that ΔABC is isosceles. Solution: In ΔABC, it is given that DE || BC and BD = CE . A

Therefore, by BPT: AD AE = BD CE AD AE ⇒ = [Since BD = CE] CE CE ⇒ AD = AE ∴AB = AD + BD ⇒ AB = AE + CE ⇒ AB = AC ⇒ ΔABC is isosceles. Hence, proved.

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IL Foundation Series Class 10

7.3 INTERNAL AND EXTERNAL ANGLE BISECTOR THEOREM 7.3.1 Interior angle bisector theorem In triangle ABC, the angle bisector intersects the side BC at point D, as shown in the figure below. B

According to the angle bisector theorem, the ratio of the line segment BD to DC is equal to the ratio of the lengths of the sides AB to AC. BD AB = DC AC Example: Seema drew a triangle PQR on the board where PX is the line drawn on side QR, where PQ = 4 cm, PR = 6 cm, QX = 1.6 cm, and XR = 2.4 cm. She wants to know whether PX is the angle bisector of ∠P. Can you help her? Solution: To show whether PX is the angle bisector or not, let us use the angle bisector theorem. So, we need QX PQ to prove that = XR PR Let us find the ratio

PQ . PR

R 2.4 cm

PQ 4 2 = = PR 6 3 Let us find the ratio

6 cm

X 1.6 cm

QX . XR

QX 1.6 2 = = XR 2.4 3 Both ratios are equal.

4 cm

Therefore, in the triangle drawn by Seema, PX bisects ∠P.

173

TRIANGLES

7.3.2 Exterior angle bisector theorem The external angle bisector of a triangle divides the opposite side externally in the ratio of the sides containing the angle. This condition usually occurs in non-equilateral triangles. In ΔABC, AD is the external bisector of ∠BAC and intersects BC produced at D. P

A E

Therefore, by the external angle bisector theorem: BD AB = DC AC Example: In DABC, AE is the external bisector of ∠A , meeting BC produced at E. If AB = 10 cm, AC = 6 cm, and BC = 12 cm, then find CE. D

A 6c

12 cm

Solution: In DABC, AE is the external bisector of ∠A meeting BC produced at E. Let CE = x cm. Now, by the angle bisector theorem: BE AB = CE AC 12 + x 10 ⇒ = x 6 ⇒ 3 ( 12 + x ) = 5 x ⇒ 36 + 3 x = 5 x ⇒ x = 18 Hence, CE = 18 cm.

174

IL Foundation Series Class 10

7.4 CRITERIA FOR SIMILARITY OF TRIANGLES To test the similarity of two triangles, we need to check whether the corresponding angles are equal and the corresponding sides are proportional. As per the definition, this requires ( 6 + 6 = 12 ) measurements to be known. There are certain criteria for similarity of triangles which do not require these 12 measurements, but they require a little information. The similarity criteria are: i) AAA similarity ii) SAS similarity iii) SSS similarity 7.4.1 Angle-Angle-Angle (AAA) similarity In two triangles, if the corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion), and hence, the two triangles are similar. This criterion is referred to as the AAA (Angle-Angle-Angle) criterion for the similarity of two triangles. Consider two triangles, ABC and DEF, such that ∠A = ∠D, ∠B = ∠E, and ∠C = ∠F, as shown in the figure. A D

E B

Therefore, AB BC AC = = DE EF DF Hence, ΔABC ~ ΔDEF. Remark: If two angles of a triangle are respectively equal to any two angles of another triangle, then by the angle sum property of a triangle, their third angles will also be equal. Therefore, the AAA similarity criterion can also be stated as follows: If two angles of one triangle are respectively equal to any two angles of another triangle, then the two triangles are similar. This may be referred to as the AA similarity criterion for two triangles. 175

TRIANGLES

7.4.2 Side-Angle-Side (SAS) Similarity If one angle of a triangle is equal to one angle of the other triangle, and the sides, including these angles, are proportional, then the two triangles are similar. This criterion is referred to as the SAS (Side - Angle - Side) similarity criterion for two triangles. Consider two triangles, ABC and PQR. A

In DABC and DPQR, If ∠B = ∠Q and

AB BC = , then the triangle ABC is similar to the triangle PQR. PQ QR

Hence, ΔABC ~ ΔPQR . 7.4.3 Side-Side-Side (SSS) Similarity In two triangles, if the sides of one triangle are proportional to the sides of the other triangle (i.e., are in the same ratio), then their corresponding angles are also equal, and hence, the two triangles are similar. This criterion is referred to as the SSS (Side-Side-Side) similarity criterion for two triangles. Consider two triangles, ABC and DEF. A

In ΔABC and ΔDEF, If

AB AC BC = = , then ∠A = ∠D, ∠B = ∠E, and ∠C = ∠F DE DF EF

Hence, ΔABC ~ ΔDEF.

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IL Foundation Series Class 10

SOLVED EXAMPLES Example 1: Prove that the following pair of triangles are similar. A

60°

80°

40°

Solution: In ΔABC and ΔPQR, ∠A = ∠P = 60° ∠B = ∠Q = 80° ∠C = ∠R = 40° Therefore, by AAA similarity, ΔABC ∼ ΔPQR. Example 2 : Prove that the following pair of triangles are similar. P A 2 B

2.5

Solution: I n ΔABC and ΔPQR, AB 2 1 = = QR 4 2 BC 2.5 1 = = 5 2 RP AC 3 1 = = PQ 6 2 AB BC CA = = QR RP PQ Therefore, by SSS similarity, ΔABC ∼ ΔQRP. Since

177

TRIANGLES

Example 3 : In the figure, ∆ ODC ∼ ∆ OBA, ∠BOC = 125° and ∠CDO = 70°. Find ÐDOC, ÐDCO and ÐOAB . D

C 70° O

125°

Solution: Given: ΔODC ∼ ΔOBA, ⇒ ∠ODC = ∠OBA = 70 ∠COD and ∠BOC form a linear pair. ∴ ∠COD + ∠BOC = 180 ⇒ ∠COD = 180 − ∠BOC = 180 − 125 = 55 ∴ ∠DOC = 55

(

)

nd ∠DCO = 180 − 70 + 55 = 180 − 125 = 55 {By the angle sum property of a triangle} A Now, AOB COD 55 {Vertically opposite angles}

(

)

∴ ∠OAB = 180 − 55 + 70 = 180 − 125 = 55 {By the angle sum property of a triangle} Example 4: In the figure,

QR QT = and ∠1 = ∠2. Show that PQS TQR. QS PR T P

1 Q

Solution: In ΔPQR, ∠ 1 = ∠2 [Given] ∴ PR = PQ QR QT Now, = [Given] QS PR 178

2 S

IL Foundation Series Class 10

QR QT = [Since PR = PQ] QS PQ Now, in ΔPQS and ΔTQR, QR QT ∠Q = ∠Q and = QS PQ ∴ ΔPQS ∼ ΔTQR [By the SAS similarity criterion]

⇒

Example 5: S and T are points on sides PR and QR of ∆ PQR, such that ∠P = ∠RTS. Show that ΔRPQ ∼ ΔRTS. R

T S

Solution:

In ΔRPQ and ΔRTS, ∠ R = ∠R [Common] ∠ P = ∠RTS [Given] ∴ΔRPQ ∼ ΔRTS [By AA similarity criterion] Example 6: In the given figure, ΔABE ≅ ΔACD , then show that ΔADE ∼ ΔABC. Solution: A Given: ΔABE ≅ ΔACD ∴ AE = AD and AB = AC [By CPCT] ⇒ AB - AD = AC - AE ⇒ BD = CE AD AE ∴ = DB EC DE  BC [By converse of BPT]

∴∠ ADE = ∠ABC [Corresponding angles] So, in ΔADE and ΔABC, ∠A = ∠A ∠ADE = ∠ABC ∴ ΔADE ∼ ΔABC [By AA similarity criterion]

179

TRIANGLES

Example 7: The perimeters of two similar triangles are 30 cm and 20 cm, respectively. If one side of the first triangle is 12 cm, then find the corresponding side of the second triangle. Solution: Let the two triangles be ABC PQR.

AB BC CA PQ QR RP

AB BC CA AB BC CA Perimeter of ABC PQ QR RP PQ QR RP Perimeter of PQR

12 cm 30 cm (where x is required side) x 20 cm

20 12 cm 30

x 8 cm

∴ The required side length = 8 cm

7.5 AREA OF SIMILAR TRIANGLES Statement: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. P A

Given two triangles ABC and PQR, such that ΔABC ∼ ΔPQR ...(1) AB BC CA 2 PQ QR RP B Q (3) In ΔAMB and ΔPNQ, B Q [ from (3)] Draw altitudes AM and PN in the triangles ABC and PQR, respectively. And, AMB PNQ 90

180

IL Foundation Series Class 10

AMB PNQ [By AA similarity] Therefore, AB AM = PQ PN By applying the formula for the area of a triangle: 1 BC AM Area of triangle ABC 2 1 Area of triangle PQR QR PN 2 Area of triangle ABC BC AM Area of triangle PQR QR PN

( Since

AB AM AB BC ) and PQ PN PQ QR

Area of triangle ABC BC BC Area of triangle PQR QR QR Area of triangle ABC BC Area of triangle PQR QR

Area of triangle ABC AM Also, Area of triangle PQR PN

Similarly, we can prove for medians, angle bisectors, and perimeter. Hence, for any two similar triangles, 2  ( Altitude )1  2  ( Median )1  2 Area of ∆1  ( Side )1    = = =  ( Altitude )2   ( Median )2  Area of ∆2  ( Side )2     

 ( Angle bisector )1  2  ( Peerimeter)1  2  =  =  ( Angle bisector )2   ( Perimeter )2     

7.6 APOLLONIUS THEOREM Statement: The sum of the squares of any two sides of a triangle is equal to twice the square of half of the third side, along with twice the square of the median bisecting the third side.

181

TRIANGLES Y

L(b, c)

M(-a, 0)

N(a, 0)

If O is the midpoint of MN, one of the sides of the triangle LMN, then, by Apollonius theorem, LN2 + LM2 = 2(MO2 + LO2)

SOLVED EXAMPLES Example 1: Let ΔABC ∼ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, then find BC. Solution: Given: ΔABC ∼ ΔDEF ar ( ΔABC ) ⎛ BC ⎞ ∴ =⎜ ⎟ ar ( ΔDEF ) ⎝ EF ⎠ ⇒

64 cm2 ⎛ BC ⎞ =⎜ ⎟ 121 cm2 ⎝ 15.4 cm ⎠

64 × (15.4)2 ⎡ 8 × 15.4 ⎤ ⇒ BC = =⎢ 121 ⎣ 11 ⎥⎦

∴ BC =

8 × 15.4 cm = 8 × 1.4 cm = 11.2 cm 11

Example 2: Diagonals of a trapezium ABCD with AB||CD intersect each other at the point O. If AB = 2CD, then find the ratio of the areas of ΔAOB and ΔCOD. D

C O

A 182

IL Foundation Series Class 10

Solution: In ΔAOB and ΔOCD, ∠AOB = ∠COD(vertically opposite angles) ∠OAB = ∠OCD (alternate angles as AB  CD ) By AA similarity, OAB OCD ar OAB AB 2 2CD 2 4 1 ar OCD CD CD ar AOB : ar COD 4 : 1 Example 3: If the areas of two similar triangles are equal, prove that they are congruent. A

Solution: Let ΔABC ∼ ΔPQR and ar ABC ar PQR …(1) ar ABC AB BC CA 2 ar PQR PQ QR RP 2

ar PQR AB BC CA 2 [Using eq(1)] ar PQR PQ QR RP 2

AB BC CA 1 PQ QR RP 1

AB BC CA PQ QR RP

AB PQ, BC QR, and CA RP ABC PQR [By SSS congruency] Hence, proved.

183

TRIANGLES

Example 4: D, E, and F are, respectively, the midpoints of sides AB, BC, and CA of ΔABC. Find the ratio of the areas of DDEF and DABC . A

Solution: D and E are the midpoints of BA and BC, respectively. ∴ DE  AC and DE

1 AC 2

⇒ DE  AF and FC and DE = AF = FC ∴ ADEF and CFDE are parallelograms. A DEF The opposite angles in a parallelogram. and C EDF Therefore, by AA similarity: ΔABC ∼ ΔEFD 2

ar ABC AC 2 AC 2 2 4 ar EFD DE 1 AC 1 1 2 ar DEF : ar ABC 1 : 4 Example 5: In the given figure, PB and QA are perpendicular to the line segment AB. If PO = 5 cm, QO = 7 cm, and ar POB 150 cm2, then find ar QOA . Q

P 184

IL Foundation Series Class 10

Solution: In ΔPOB and ΔQOA, ∠ POB = ∠QOA [Vertically opposite angle] ∠OBP = ∠OAQ = 90 ∴ ΔPOB ∼ ΔQOA [By AA similarity] ar POB PO 5 2 25 ar QOA QO 7 49 2

150 cm2 25 49 ar QOA 150 cm2 294 cm2 ar QOA 49 25 Example 6: In the figure below, ΔACB ∼ ΔAPQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm, and AP = 2.8 cm. Find CA and AQ. P

Solution: Given: ΔACB ∼ ΔAPQ

AC BC AB AP QP AQ

CA 10 cm 6.5 cm 2.8 cm 5 cm AQ

CA 2 6.5 cm 2.8 cm 1 AQ

CA 2 2 6.5 cm and 2.8 cm 1 1 AQ

CA 2 2.8 cm 5.6 cm and AQ

6.5 3.25 cm 2

185

TRIANGLES

Example 7: A triangle has sides measuring 8 cm, 7 cm, and 6 cm. Find the length of the median to the side of the length 8 cm. Solution: Given: a = 8 cm, b = 7 cm, c = 6 cm As the 8 cm side is bisected, we have, m=

a = 4 cm. Let the length of the median be d. 2

By Apollonius theorem formula,

(

c2 + b 2 = 2 m2 + d 2

)

Substituting the required values, we have:

(

62 + 7 2 = 2 42 + d 2 Which gives,

36 49 2 16 d 2

)

85 32 2d 2 2d 2 53 53 2 d 5.14 cm d

Thus, the length of the median is 5.14 cm.

QUICK REVIEW •

Two figures having the same shape but not necessarily the same size are called similar figures.

•

All the congruent figures are similar, but the converse is not true.

•

wo polygons of the same number of sides are similar if (i) their corresponding angles are T equal, and (ii) their corresponding sides are in the same ratio (i.e., proportion).

•

I f a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio.

•

I f a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

•

Test for similar triangles: AAA - If three of their corresponding angles are equal. SSS - If three of their corresponding sides are in the same ratio.

SAS - The ratio of two pairs of corresponding sides are equal, and their included angles are equal. 186

IL Foundation Series Class 10

WORKSHEET - 1 I.

INTRODUCTION TO TRIANGLES 1. Explain why two squares are always similar. 2. In the figure, ABCD is a square, and PQRS is a rhombus. Check whether they are similar. S D

2 cm

2 cm A

2 cm

3 cm

2 cm

3 cm

80°

100°

3 cm 100°

80°

3 cm

3. In the given figure, verify whether quadrilaterals ABCD and PQRS are similar or not. R

3.4 cm

70°

2.5 cm

2.6 cm

1.7 cm

85°

70°

85°

1.3 cm

5 cm

105° 100° A 1 cm B

100°

105° P

2 cm

4. State whether the following quadrilaterals are similar or not: D S

1.5 cm

1.5 cm P

R 1.5 cm

1.5 cm

3 cm

Q A

3 cm

187

TRIANGLES

II. BASIC PROPORTIONALITY THEOREM 1. If a line intersects sides AB and AC of ΔABC at D and E, respectively and is parallel to BC, AD AE then prove that = . AB AC 2. ABCD is a trapezium with AB||CD. E and F are points on non-parallel sides AD and BC, AE BF = respectively, such that EF || AB. Show that . ED FC A

3. In the given figure,

PS PT = and ∠PST = ∠PRQ. Prove that ΔPQR is an isosceles triangle. SQ TR P

4. In the given figure, DE || BC. If AD x, DB x 2, AE x 2 and EC x 1 , then find the value of x. C E

188

IL Foundation Series Class 10

5. ABCD is a parallelogram, P is a point on side BC, and DP, when produced, meets AB produced at L. Prove that: i)

DP DC DL AL = ii) = PL BL DP DC

6. In the given figure, DE || BC, and CD || EF. Prove that AD2 = AB.AF. A F

7. D and E are, respectively, the points on the sides AB and AC of ΔABC, such that AB = 5.6 cm, = AD 1= .4 cm, AC 7.2 cm and AE = 1.8 cm, then show that DE || BC. 8. Any point X inside ΔDEF is joined to its vertices. From a point P in DX, PQ is drawn parallel to DE meeting XE at Q and QR is drawn parallel to EF meeting XF in R. Prove that PR || DF. 9. In DABC, D and E are points on sides AB and AC, respectively, such that BD = CE . If ∠B = ∠C, then show that DE || BC. 10. In DABC, D and E are points on sides AB and AC, respectively, such that DE || BC. If AD = 6 cm, DB = 9 cm, and AE = 8 cm, then find AC. 11. In the figure, AB  CD, OA 4 cm, OB x 1, OC 4 x 2 , and OD 2 x 4. Find the value of x. A

4 x+1

2x + 4

O 4x - 2

189

TRIANGLES

III. INTERNAL AND EXTERNAL ANGLE BISECTOR THEOREM 1. Find the value of x for the given triangle using the angle bisector theorem. C

12 cm

9 cm

6 cm

2. In ∆ ABC, AB = 5 cm, BC = 6 cm, and AC = 7 cm. The angle bisector AD intersects BC at D. Find BD. A

6-x

3. ABCD is a quadrilateral in which the bisectors of angle B and angle D intersect on AC AB AD = . at point E. Show that BC DC C

4. In a triangle, AE is the bisector of the exterior angle ∠CAD, meeting BC at point E. If the lengths of AB, AC, and BC are 10 cm, 6 cm, and 12 cm, respectively, then determine the length of CE.

190

IL Foundation Series Class 10

IV. CRITERIA FOR SIMILARITY OF TRIANGLES 1. In the given figure, PQ || RS. Prove that ΔPOQ ∼ ΔSOR. R

O Q S

2. From the figure given below, find ∠P. R A 3.8

80°

6√3

3√3

7.6

60°

3. In the given figure, OA × OB = OC × OD. Show that ∠A = ∠C and ∠B = ∠D. C A

O D B

4. A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds. 5. In the given figure, CM and RN are, respectively, the medians of DABC and DPQR . CM AB = If ΔABC ∼ ΔPQR, then prove that . RN PQ

191

TRIANGLES

M C

6. In the figure, QA and PB are perpendicular to AB. If AO = 10 cm, BO = 6 cm, and PB = 9 cm, then find AQ. P O

7. In the given figure, if ∠ADE = ∠B, then show that ΔADE ∼ ΔABC. If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, and BC = 4.2 cm, then find DE. A

E D

8. In the figure, E is a point on CB produced from an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC. Prove that ΔABD ∼ ΔECF . A

192

IL Foundation Series Class 10

9. D is a point on the side BC of ΔABC, such that ∠ADC = ∠BAC. Show that CA2 = CB ⋅ CD. 10. In ΔABC, P and Q are points on sides AB and AC, respectively, such that PQ || BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm, then find AB and PQ. 11. The legs (sides other than the hypotenuse) of a right triangle are of lengths 16 cm and 8 cm. Find the length of the side of the largest square that can be inscribed in the triangle. 12. Sides AB and AC and median AD of a triangle ABC are, respectively, proportional to the sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ∼ ΔPQR. 13. In a triangle ABC, it is known that AB = AC. Suppose D is the midpoint of AC and BD = BC = 2. Find the area of the triangle ABC. 14. In the given figure, ABC is a triangle in which AD bisects ∠A, AC = BC, ∠B = 72, and CD = 1 cm. Find the length of BD. C

15. In the given figure, AB  PQ  CD, AB = x, CD = y, and PQ = z. Prove that

1 1 1 + = . x y z

C P

y z B

193

TRIANGLES

16. The perimeter of this triangle is 25 units. An angle bisector divides one side of the triangle into lengths of 2 and 3. Find the lengths of the remaining two sides.

V. AREA OF SIMILAR TRIANGLES 1. If ΔABC ∼ ΔDEF , such that AB = 1.2 cm and DE = 1.4 cm, then find the ratio of the areas of ΔABC and DDEF. 2. In trapezium ABCD, AB || CD and AB = 2CD. If the area of ΔAOB = 84 cm2, then find the area of ΔCOD. 3. Two isosceles triangles have equal vertical angles, and their areas are in the ratio 16 : 25. Find the ratio of their corresponding heights. 4. The areas of two similar triangles are 81 cm2 and 49 cm2, respectively. Find the ratio of their corresponding heights. 5. The areas of two similar triangles are 169 cm2 and 121 cm2, respectively. If the longest side of the larger triangle is 26 cm, then find the longest side of the smaller triangle. 6. The corresponding altitudes of two similar triangles are 6 cm and 9 cm, respectively. Find the ratio of their areas. 7. In ΔABC, D and E are the midpoints of AB and AC, respectively. Find the ratio of the areas of ΔADE and ΔABC. 8. If the areas of two similar triangles, ABC and PQR, are in the ratio 9 : 16. If BC = 4.5 cm, then find the length of QR. 9. If D is a point on the side AB of ΔABC, such that AD : DB = 3 : 2 and E is a point on BC, such that DE || AC, then find the ratio of areas of ΔABC and ΔBDE. 10. In the given figure, XY || AC and divides ΔABC into two parts of equal areas. Find the ratio of AX and AB. A X

194

IL Foundation Series Class 10

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. A vertical stick 30 m long casts a shadow 15 m long on the ground. At the same time, a tower casts a shadow 75 m long on the ground. The height of the tower is:

a) 150 m

b) 100 m

c) 25 m

d) 200 m

2. In the figure given below, if DE  BC , then x equals: A

3 cm D

4 cm

2 cm x

a) 3 cm

b) 2 cm

c) 4 cm

d) 6.7 cm

3. In the given ΔXYZ, DE || YZ, such that the lengths of sides XD, XE, and EZ (in cm) are 2.4, 3.2, and 4.8, respectively. The length of XY (in cm) is: X 2.4 D

3.2 E 4.8

a) 3.6

b) 6

c) 6.4

d) 1.6

4. If ΔABC ∼ ΔPQR, perimeter of ΔABC = 20 cm, perimeter of ΔPQR = 40 cm, and PR = 8 cm, then the length of AC is: a) 8 cm

b) 6 cm

c) 4 cm

d) 5 cm

AB BC = , if then they will be similar when: DE FD b) ∠A = ∠D c) ∠B = ∠D d) ∠A = ∠F

5. In triangles ABC and EDF, a) ∠B = ∠E

195

TRIANGLES

6. In the given figure, if ΔABC ∼ ΔPQR, then the value of x is: A 6 cm B

a) 2.5 cm

5 cm

4 cm

3.75 cm

Q 4.5 cm

b) 3.5 cm

7. In the figure, PQ  BC , if

c) 2.75 cm

d) 3 cm

PQ 2 AP = , then is: BC 5 PB A P

2 5

2 3

8. In the given figure, XY  QR,

3 2

3 5

PQ 7 = and PR = 6.3 cm, then YR equals: XQ 3 P

a) 2.7 cm

b) 18.9 cm

c) 2.1 cm

d) 0.9 cm

9. In triangles ABC and DEF, A E 40 , AB : ED AC : EF and ∠F = 65, then ∠B =? a) 35

b) 65

c) 75

d) 85

10. If in ΔDEF and ΔPQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true? a)

196

EF DF = PR PQ

DE EF = PQ RP

DE DF = QR PQ

EF DE = RP QR

IL Foundation Series Class 10

11. ΔABC ∼ ΔPQR. If a) 14 cm

ar ABC 9 and AB = 18 cm, then the length of PQ is: ar PQR 4 b) 8 cm

c) 10 cm

d) 12 cm

12. ΔDEF ∼ ΔABC ; if DE : AB = 2 : 3 and ar ( ΔDEF ) is equal to 44 square units, then ar ( ΔABC ) in square units is: 176 9 13. The length of the altitude of an equilateral triangle of side a is:

a) 99

2a 3

b) 120

3 2a

3a 2

d) 66

a 2 3

14. If sides of two similar triangles are in the ratio 4 : 9. Then areas of these triangles are in the ratio: a) 2 : 3

b) 4 : 9

15. If ΔABC and ΔDEF are two triangles such that

( ΔDEF ) = ? a) 2 : 5

b) 4 : 25

c) 81 : 16

d) 16 : 81

AB BC CA 2 = = = , then area ( ΔABC ) : Area DE EF FD 5 c) 4 : 15

d) 8 : 125

= cm, EF 4 cm and ar ( ΔABC ) = 54 cm2, then ar ( ΔDEF ) = ? 16. ΔABC ∼ ΔDEF . If BC 3= a) 108 cm2 17. If ABC QRP, a) 10 cm

b) 96 cm2

c) 48 cm2

d) 100 cm2

ar ABC 9 , AB 18 cm , and BC = 15 cm, then PQ is equal to: ar PQR 4 b) 12 cm

20 cm 3

d) 8 cm

18. If ABC DEF, BC 4 cm, EF 5 cm , and the area of ΔABC = 80 cm2 , then the area of ΔDEF is: a) 100 cm2

b) 125 cm2

c) 150 cm2

d) 200 cm2

II. ASSERTION AND REASON 1. Assertion (A): If ΔABC and ΔPQR are congruent triangles, then they are also similar triangles. Reason (R): All congruent triangles are similar, but the similar triangles need not be congruent. a) Both Assertion A and Reason R are true, and Reason R is the correct explanation of Assertion A. b) Both Assertion A and Reason R are true, but Reason R is not the correct explanation of Assertion A. 197

TRIANGLES

c) Assertion A is true, but Reason R is false. d) Assertion A is false, but Reason R is true. 2. Assertion (A): In the given figures, ΔABC ~ ΔGHI. Reason (R): If the corresponding sides of two triangles are proportional, then they are similar. B 20 A

22 24

G 30 C

H 33

a) Both Assertion A and Reason R are true, and Reason R is the correct explanation of Assertion A. b) Both Assertion A and Reason R are true, but Reason R is not the correct explanation of Assertion A. c) Assertion A is true, but Reason R is false. d) Assertion A is false, but Reason R is true. III. FILL IN THE BLANKS 1. In a triangle, if a line divides any two sides in the same ratio, then it is __________ to the third side. 2. Two circles with different radii are always __________. 3. In a ΔABC, DE  BC, D is the midpoint of the AB, then AE = __________. 4. In an equilateral triangle of side 3 3 cm, the length of attitude is __________. 5. In a rectangle ABCD, E is a point on AB, such = that AE value of DE = __________.

2 = AB, AB 6 m, AD = 3 m , the 3

6. If areas of two similar triangles are equal, then the triangles are __________. 7. D, E, and F are the midpoint of AB, BC, and CA of ΔABC, respectively, then the ratio of ΔABC and ΔDEF is equal to __________. 8. The ratio of the area of two similar triangles is 9 : 144. The ratio of their perimeter is __________. 9. If ΔACB ~ ΔAPQ, BC = 10 cm, PQ = 5 cm, BA = 6.5 cm, AP = 2.8 cm, then the ratio of the ar ( ΔACB ) : ( ΔAPQ ) = __________. 10. In ABC, AB BC CA 3a, AD BC . AD = __________.

198

IL Foundation Series Class 10

IV. SUBJECTIVE QUESTIONS 1. In ΔABC if DE  AB, AD 8 x 9, CD x 3, BE 3 x 4 , and CE = x, then find the value of x. AM AN 2. In the given figure, LM  CB and LN  CD. Prove that = . AB AD B M A

N D

3. In the given figure, DE  BC , then find x. A 1.5 cm

1 cm

E x

3 cm

= cm, BC 8 cm, and PQ = 4 cm, then find AQ. 4. In the given figure, ΔACB ∼ ΔAPQ . If AB 6= P

5. If triangle ABC is similar to triangle DEF such that 2AB = DE and BC = 8 cm, then find EF. 6. In the given figure, find the value of x (in cm). P 3.2 cm

2.4 cm A

2 cm

3.6 cm

B 4.8 cm

x cm Q

199

TRIANGLES

7. If ΔPQR ∼ ΔXYZ, ∠Q = 50 and ∠R = 70, then find ∠X + ∠Y. 8. Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, then find the distance between their tops. 9. P Q is drawn parallel to the base BC of a cutting AB at P and AC at Q of ΔABC. If AB = 4BP and CQ = 2 cm, then find AQ. 10. ΔABC is such that AB = 3 cm, BC = 2 cm , and CA = 2.5 cm. If ΔDEF ∼ ΔABC and EF = 4 cm, then find the perimeter of ΔDEF.

200

COORDINATE GEOMETRY

8.1 CARTESIAN COORDINATE SYSTEM In class IX, we have learnt that to locate the position of a point on a plane, we require a pair of mutually perpendicular lines known as coordinate axes. The horizontal line is known as the x-axis, and the vertical line is known as the y-axis. The intersection point of the coordinate axes is known as the origin. The perpendicular distance from the y-axis measured along the x-axis is called its x-coordinate or abscissa, and the perpendicular distance of a point from the x-axis measured along the y-axis is called y-coordinate or ordinate. Y Q

2 (-, +)

X’

-5 -4 -3 -2 -1-1 Q

3 (-, -)

5 4 3 2 1

1 (+, +)

1 2 3 4 5

-2 -3 -4 -5

4 (+, -)

Y’

The horizontal line X'OX is called the x-axis, and the vertical line YOY' is called the y-axis. The point of intersection of these two lines is called the origin. Let XOX' and YOY' be the coordinate axes. These two axes divide the XY-plane into four parts or regions, namely XOY, X'OY, X'OY', and Y'OX. Each part is called a quadrant and is referred to as Q1, Q2, Q3, and Q4, respectively. For every point P(x, y) if it lies: In the first quadrant, the x-coordinate is +ve, and the y-coordinate is +ve. In the second quadrant, the x-coordinate is -ve, and the y-coordinate is +ve. In the third quadrant, the x-coordinate is -ve, and the y-coordinate is -ve. In the fourth quadrant, the x-coordinate is +ve, and the y-coordinate is -ve. 201

COORDINATE GEOMETRY

The coordinates of the origin are (0, 0). Example: Determine the quadrant for the point (-4, -8). Solution: The x-coordinate of the given point is negative (-4). The y-coordinate of the given point is negative (-8). Therefore, the point (-4, -8) lies in the third quadrant.

8.2 DISTANCE FORMULA 8.2.1 Distance formula Let A x1 , y1 , and B x2 , y2 be any two points on a line which are not parallel to the axes. From the figure below, we have the right-angled triangle ABC. Y

} }

B (x₂, y₂)

y₂-y₁

A (x₁, y₁)

} x₁

x₂-x₁

y₂

y₁

} O

x₂

According to Pythagoras theorem, AB 2 =AC2 +BC2 But, AC x2 x1 , BC y2 y1 AB 2 x2 x1 y2 y1 2

x2 x1 2 y2 y1 2

Note: The distance to the point P x1 , y1 from the origin is 202

x12 + y12 .

IL Foundation Series Class 10

Points to remember: i) Square: All four sides are equal. Each angle is a right angle. Diagonals are equal and bisect each other at right angles. ii) Rhombus: All the four sides are equal. Its diagonals bisect each other at right angles but are not equal to each other. iii) Rectangle: Its opposite sides are equal. Diagonals are equal and bisect each other. Each angle is a right angle. iv) Parallelogram: Its opposite sides are equal. Diagonals are not equal but bisect each other. Example : Find the distance between the points 2, 1 and 3, 2 . Solution: Let A x1 , yl 2, 1 and B x2 , y2 3, 2 . ∴ The distance between A and B, i.e., AB

x2 x1 2 y2 y1 2

(3 2)2 (2 1)2 12 12 2 units.

Example : Find the distance to the point 1, 2 from the origin. Solution: Let A x1 , y1 1, 2 . The distance from the origin to the point (1, 2), denoted as OA, is given by: OA x12 y12 12 22 5 units ∴ The distance to the point (1, 2) from the origin is

5 units.

8.2.2 Collinear Points The points that lie on the same line are called collinear points. If the three points A, B, and C are collinear, then AB + BC = AC or, AC + CB = AB

or, BA + AC = BC Example: Show that the points 1, 7 , 3, 5 , and 4, 8 are collinear. Solution: Let A 1, 7 , B 3, 5 , and C 4, 8 be the given points.

203

COORDINATE GEOMETRY

The distance between two points x1 , y1 , and x2 , y2 is

x2 x1 2 y2 y1 2 .

Now, AB (3 1)2 ( 5 7)2 42 ( 12)2 16 144 160 4 10 units BC (4 3)2 ( 8 5)2 12 ( 3)2 10 units CA ( 1 4)2 (7 8)2 25 225 250 5 10 units

Since AB+BC 4 10 10 5 10 AC ∴A, B, and C points are collinear.

SOLVED EXAMPLES Example 1: Find the distance between the points (-5, -4) and (-8, 4). Solution:

x2 x1 2 y2 y1 2 . Here, x1 , y1 5, 4 and x2 , y2 8, 4 The distance between two points is given as

Distance, d

8 5 2 4 4 2

d 9 64 d = 73 units Example 2: I f (1, x) is at 10 units from (0, 0), then find the value of x. Solution: Let A = (1, x) Given that OA = 10 Now, OA x 2 y 2 10 12 x 2 10 1 x 2 x2 = 9

x 3 Therefore, the value of x is ± 3. Example 3: Show the points (1, 2), (2, 1), and (-2, 5) are collinear. Solution: Let A(1, 2), B(2, 1), and C(-2, 5). The distance between two points is given as 204

x2 x1 2 y2 y1 2 .

IL Foundation Series Class 10

AB =

2 1 2 1 2 2 2 units

BC =

2 2 2 5 1 2 32 4 2 units

AC =

2 1 2 5 2 2 18 3 2 units

Here,

2 3 2 4 2

So, AB + AC = BC Therefore, points (1, 2), (2, 1), and (-2, 5) are collinear. Example 4: Show that A 6, 4 , B 5, 2 , and C 7, 2 are the vertices of an isosceles triangle. Solution: We are given, A(6, 4), B(5, -2), and C (7, -2). Therefore, AB (5 6)2 ( 2 4)2 ( 1)2 ( 6)2 1 36 37 units AC (7 6)2 ( 2 4)2 ( 1)2 ( 6)2 1 36 37 units

AB AC

Thus, ∆ABC is an isosceles triangle. Example 5: If P 2, 1 , Q 3, 4 , R 2, 3 , and S 3, 2 are four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus. Solution: The given points are: P 2, 1 , Q 3, 4 , R 2, 3 , S 3, 2 PQ (3 2)2 (4 1)2 1 25 26 units QR

( 2 3)2 (3 4)2

25 1

26 units

( 3 2)2 ( 2 3)2

1 25

26 un nits

( 3 2)2 ( 2 1)2

25 1

26 units

( 2 2)2 (3 1)2

16 16

4 2 units

205

COORDINATE GEOMETRY

and QS ( 3 3)2 ( 2 4)2 36 36 72 6 2 units PQ = QR = RS = SP 26 units and PR QS This means that PQRS is a quadrilateral whose sides are equal, but the diagonals are unequal. Thus, PQRS is a rhombus but not a square. 1 2

Now, the area of the rhombus PQRS (product of lengths of the diagonals) 1 PR×QS 2 1 4 2 6 2 2 24 sq. units

8.3 Section Formula Let A x1 , y1 and B x2 , y2 be two points in the Cartesian plane. Let M(x, y) be the point that divides the line segment AB internally in the ratio m: n. PA, MN, and QR are perpendicular to the x-axis, and PS and MB are parallel to the x-axis. Y

Q(x₂, y₂)

n M(x, y)

m P(x₁, y₁)

In ∆PMS and ∆QMB, MPS QMB (Corresponding angles) MSP QBM 90 By AA similarity criterion, ∆PMS ~ ∆MQB PM PS MS m …(1) MQ MB QB n 206

IL Foundation Series Class 10

PS = AN = ON - OA = x − x1 MB = NR = OR- ON = x2 − x MS = MN - SN = y − y1 QB = RQ - RB = y2 − y From (1), m x x1 y y1 n x2 x y2 y x

mx2 nx1 m n

Similarly, y

my2 ny1 m n

mx2 nx1 my2 ny1 So, the coordinates of the point M (x, y) are , . m n m n 8.3.1 Division of a line segment in a given ratio (Section formulae)

(i)

n P

(ii)

(P divides AB in the ratio m : n internally)

(P divides AB in the ratio m : n externally)

The point P that divides the line segment joining the points A x1 , y1 , and B x2 , y2 in the ratio m:n. mx2 nx1 my2 ny1 i) Internally is , ; m n 0 m n m n mx2 nx1 my2 ny1 ii) Externally is , ; m n m n m n Example: Find the points which divide the line segment joining the points 1, 3 , 3, 9 in the ratio 1 : 3 internally and externally. Solution: Let A x1 , y1 1, 3 and B x2 , y2 3, 9 . Given ratio m : n = 1 : 3 The point which divides AB in the ratio 1 : 3 internally mx2 nx1 my2 ny1 , m n m n 1 ( 3) 3 1 1 9 3 ( 3) 3 3 9 9 , 4 , 4 0, 0 1 3 1 3 207

COORDINATE GEOMETRY

The point which divides AB in the ratio 1 : 3 externally mx2 nx1 my2 ny1 , m n m n 1 ( 3) 3 1 1 9 3 ( 3) 3 3 9 9 6 18 , 2 , 2 2 , 2 3, 9 . 1 3 1 3 8.3.2 Mid-point of a line segment: Let A x1 , y1 , B x2 , y2

1 P Mid-point

If P is the mid-point of the line segment AB, then P divides AB in the ratio 1 : 1 internally. Let m:n=1:1 mx2 nx1 my2 ny1 1 x2 1 x1 1 y2 1 y1 x1 x2 y1 y2 Now, P , , , 2 1 1 1 1 m n m n 2 x x y y The mid-point of the line segment joining A x1 , y1 and B x2 , y2 is 1 2 , 1 2 . 2 2 Example: Find the mid-point of the line segment joining 1, 2 and 3, 4 . Solution: Let A x1 , y1 1, 2 and B x2 , y2 3, 4 . x x y y 1 3 2 4 Now, the mid-point of AB 1 2 , 1 2 , 2, 3 2 2 2 2

SOLVED EXAMPLES Example 1: Find the coordinates of the point which divides the line segment joining the points 6, 3 and 4, 5 in the ratio 3 : 2 internally.

3 A (6, 3) Solution: Let P x, y be the required point.

208

2 P

B (-4, 5)

IL Foundation Series Class 10

mx nx1 my2 ny1 3 ( 4) 2 6 3 5 2 3 , , P 2 3 2 3 2 m n m n 12 12 15 6 , 5 5 0 21 21 , 0, 5 5 5 x 0 and y

21 5

Example 2: Find the coordinates of the points which trisect the line segment joining 1, 2 and 3, 4 . Solution: Given points: A 1, 2 and B 3, 4 . Let the points of trisection be P and Q. Then, AP PQ QB . (say)

λ A(1, -2)

λ Q

B (-3, 4)

PB PQ QB 2 AQ AP PQ 2 AP : PB : 2 1 : 2 AQ : QB 2 : 2 : 1 So, P divides AB internally in the ratio 1 : 2 while Q divides internally in the ratio 2 : 1. Thus, the coordinates of P and Q are: mx nx1 my2 ny1 P 2 , m n m n 2 1 3 2 1 1 4 2 , 1 2 1 2 3 2 4 4 1 , , 0 3 3 3 mx nx1 my2 ny1 , Q 2 m n m n 2 3 1 1 2 4 1 2 , 2 1 2 1 6 1 8 2 5 , , 2 3 3 3

209

1 2

3 2 4 4 3 3

1 3

1 2

, COORDINATE GEOMETRY , 0 mx nx1 my2 ny1 , Q 2 m n m n 2 3 1 1 2 4 1 2 , 2 1 2 1 6 1 8 2 5 , , 2 3 3 3

1 5 ∴ The two points of trisection are , 0 and , 2 . 3 3 3 11 Example 3: In what ratio does the point C , divide the line segment joining the points 5 5 A 3, 5 and B 3, 2 ? Solution: Let the point C divide AB in the ratio λ : 1 .

λ A(3, 5)

1 C

So, the coordinates of C are given by mx nx1 my2 ny1 3 1 3 2 1 5 2 , , 1 1 m n m n 3 3 2 5 , 1 1 3 11 But, the coordinates of C are given as , . 5 5 Hence, 3 3 2 5 3 11 , , 1 5 5 1 3 3 3 2 5 11 and 1 1 5 5 15 15 3 3 and 10 25 11 11 18 12 and 21 14

12 2 14 2 and 18 3 21 3

Hence, the point C divides AB in the ratio 2 : 3.

210

B(-3, -2)

IL Foundation Series Class 10

Example 4: Determine the ratio in which the line 3 x y 9 0 divides the line segment joining A 1, 3 and B 2, 7 . Solution: Suppose the line 3 x y 9 0 divides the line segment joining A 1, 3 and B 2, 7 in the ratio k : 1 at point C . 2k 1 7 k 3 , Then, the coordinates of C are . k 1 k 1 But, C lies on 3 x y 9 0 . 2k 1 7 k 3 3 9 0 k 1 k 1 3 2k 1 7 k 3 9 k 1 0 6k 3 7 k 3 9k 9 0 4k 3 0 4k 3 k

3 4

Example 5: Point P divides the line segment joining the points A 2, 1 and B 5, 8 such AP 1 = . If P lies on the line 2 x y k 0 , find the value of k . AB 3 Solution: that

We have, AP 1 AB 3 AP 1 AP+PB 3 3AP AP+PB 2AP BP AP 1 BP 2 So, P divides AB in the ratio 1 : 2.

1 A(2, 1)

2 P

B(5, -8)

∴ Coordinates of P are

211

COORDINATE GEOMETRY

1 5 2 2 1 8 2 1 , 1 2 1 2 5 4 8 2 9 6 , , 3 3 3 3 3, 2 Since P 3, 2 lies on the line 2 x y k 0. 2 3 2 k 0 6 2 k 0 8 k 0 k 8 Example 6: If the coordinates of the mid-points of the sides of a triangle are 1, 2 , 0, 1 , and 2, 1 . Find the coordinates of its vertices. Solution: et A x1 , y1 , B x2 , y2 , and C x3 , y3 be the vertices of ∆ABC . Let D 1, 2 , E 0, 1 , and L F 2, 1 be the mid-points of sides BC, CA , and AB , respectively. A (x1, y1)

F (2, -1)

B (x2, y2)

E (0, -1)

D (1, 2)

C (x3, y3)

Since D is the mid-point of BC: x + x3 y +y ∴ 2 = 1 and 2 3 = 2 2 2 ⇒ x2 + x3 = 2 and y2 + y3 = 4 ...(1) Similarly, E and F are the mid-points of CA and AB, respectively. x1 x3 y y 0 and 2 3 1 2 2 x1 x3 0 and y1 y3 2 ...(2)

y y x x 1 2 2 and 1 2 1 2 2 212

IL Foundation Series Class 10

x1 x2 4 and y1 y2 2 ...(3) On adding (1), (2), and (3), we get,

x2 x3 x1 x3 x1 x2 2 0 4, y2 y3 y1 y3 y1 y2 4 2 2 2 x1 x2 x1 6 and 2 y1 y2 y3 0 x1 x2 x3 3 and y1 y2 y3 0 ...(4) From (1) and (4) we get, x1 2 3 and y1 4 0

x1 1 and y1 4

So, the coordinates of A are 1, 4 . From (2) and (4) we get, x2 0 3 and y2 2 0 x2 3 and y2 2 So, the coordinates of B are 3, 2 . From (3) and (4) we get, x3 4 3 and y3 2 0 x3 1 and y3 2 So, the coordinates of C are 1, 2 . Therefore, the vertices of the triangle ABC are A 1, 4 , B 3, 2 , and C 1, 2 .

8.4

AREA OF TRIANGLE AND QUADRILATERAL

8.4.1 Area of triangle The area of a triangle, with vertices at x1 , y1 , x2 , y2 , and x3 , y3 is given by the formula: 1 Area x1 y2 y3 x2 y3 y1 x3 y1 y2 sq. units. 2 Note: Three given points will be collinear if the area of the triangle formed by these points is zero. Example: Find the area of a triangle whose vertices are A 3, 2 , B 11, 8 , and C 8, 12 . Solution: Let the coordinates of the vertices of triangle ABC be

213

COORDINATE GEOMETRY

A x1 , y1 3, 2 B x2 , y2 11, 8 C x3 , y3 8, 12 Area of ABC

1 x1 y2 y3 x2 y3 y1 x3 y1 y2 2

1 3 8 12 11 12 2 8 2 8 2

1 12 110 48 2

1 50 25 sq. units 2 8.4.2 Area of quadrilateral The area of a quadrilateral can be found by dividing the quadrilateral into two triangles by drawing a diagonal. Now, calculate the area of each of these triangles separately and the sum of their areas will be equal to the area of the quadrilateral. Hence, the area of the quadrilateral ABCD area ABC area ADC

C D

Note: The area of the quadrilateral ABCD formed by the vertices A x1 , y1 , B x2 , y2 , C x3 , y3 , and D x4 , y4 is given by the formula Area of quadrilateral =

1 x1y2 x2 y1 x2 y3 x3 y2 x3 y4 x4 y3 x4 y1 x1y4 sq. units. 2

Example: Find the area of the quadrilateral ABCD whose vertices are A (1, 1), B(7, -3), C(12, 2), and D (7, 21). Solution: Area of the quadrilateral ABCD Area of ABC Area of ACD

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IL Foundation Series Class 10

D (7,21)

A (1,1)

C (12,2)

B (7,-3)

We know that, Area of triangle Area of ABC

1 x1 y2 y3 x2 y3 y1 x3 y1 y2 2

1 1 3 2 7 2 1 12 1 3 2 1 5 7 48 25 sq. units 2

1 1 1 2 21 12 21 1 7 1 2 19 240 7 2 2 1 214 1007 sq. units 2

Area of ACD

∴ Area of the quadrilateral ABCD 25 107 132 sq. units.

SOLVED EXAMPLES Example 1: Prove that the points a, b c , b, c a , and c, a b are collinear. Solution: Let the points be A x1 , y1 a, b c B x2 , y2 b, c a C x3 , y3 c, a b Area of the triangle ABC

215

COORDINATE GEOMETRY

1 x1 y2 y3 x2 y3 y1 x3 y1 y2 2

1 a c a a b b a b b c c b c c a 2

1 a c b b a c c b a 2

1 ac ab ab bc bc ac 2

1 0 0 sq. units. 2

Since the area of triangle ABC = 0, the given points are collinear. Example 2: Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are 0, 1 , 2, 1 , and 0, 3 . Find the ratio of this area to the area of the given triangle. Solution: Let A 0, 1 , B 2, 1 , and C 0, 3 be the vertices of ∆ABC . Let D, E, and F be the mid-points of sides BC, CA, and AB, respectively. Then, the coordinates of D, E, and F are 1, 2 , 0, 1 , and 1, 0 , respectively.

A(0, -1) F(1, 0)

B (2, 1) Now, the area of ABC ⇒ Area of ABC

E(0, 1)

D (1, 2)

C (0, 3)

1 x1 y2 y3 x2 y3 y1 x3 y1 y2 2

1 0 1 3 2 3 1 0 1 1 2 1 0 8 0 2

4 sq. units Area of DEF 216

1 x1 y2 y3 x2 y3 y1 x3 y1 y2 2

IL Foundation Series Class 10 ⇒ Area of DEF

1 1 1 0 0 0 2 1 2 1 2 1 1 1 1 sq. unit 2

∴ Area of ∆DEF : Area of ABC 1 : 4

Example 3: If A 4, 6 , B 3, 2 , and C 5, 2 are the vertices of ∆ABC , verify that a median of a triangle ABC divides it into two triangles of equal areas. Solution: Area of ABC

1 x1 y2 y3 x2 y3 y1 x3 y1 y2 2

A (4, -6)

B (3, -2) ∴ Area of ∆ABC

D (4, 0)

C (5, 2)

1 4 2 2 3 2 6 5 6 2 2 1 16 24 20 2

6 sq. units

Also, area of ABD

1 4 2 0 3 0 6 4 6 2 2 1 8 18 16 3 sq. units 2

Area of ABC 6 2 Area of ABD 3 1

⇒ Area of ABC 2 (Area of ∆ABD)

Example 4: For what values of k are the points k, 2 2 k , k 1, 2 k , and 4 k, 6 2 k collinear? Solution: Given points are collinear if the area of the triangle formed by them is zero. We have, A k, 2 2 k , B k 1, 2 k , and C 4 k, 6 2 k

217

COORDINATE GEOMETRY

1 | x1 y2 y3 x2 y3 y1 x3 y1 y2 | 2 1 Area of ABC | k(2 k 6 2 k k 1 6 2 k 2 2 k 4 k 2 2k 2 k | 2

Area of ABC

1 k 4 k 6 k 1 4 4 k 2 4 k 2

1 2 1 4 k 6 k 4 k 4 8 16 k 2 k 4 k2 8 k2 4 k 4 2 2 Since A, B, and C are collinear, i.e., area of ABC 0

1 2 8k 4k 4 0 2 8k2 4k 4 0

2k2 k 1 0 2k k 1 k 1 0 k 1 2k 1 0 k 1 0 or 2 k 1 0 1 k 1 or k 2

8.5

GEOMETRIC POINT OF A TRIANGLE

8.5.1 Centroid of a triangle Definition: The point of intersection of the medians of a triangle is called the centroid of the triangle, and it divides the median internally in the ratio 2 : 1 from the vertex. The coordinates of the centroid of the triangle whose vertices are x1 , y1 , x2 , y2 , and x3 , y3 are

x1 x2 x3 y1 y2 y3 , . This also deduces that the medians of a triangle are concurrent. 3 3

Example: Find the centroid of the triangle whose vertices are 4, 0 , 0, 6 , and 2, 6 . Solution: Let A x1 , y1 4, 0 , B x2 , y2 0, 6 , C x3 , y3 2, 6 x x x y y y 4 0 2 0 6 6 6 0 The centroid G 1 2 3 , 1 2 3 , , 2, 0 . 3 3 3 3 3 3

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8.5.2 Incentre of a triangle Incentre: The point of concurrence of internal angle bisectors of a triangle is called the incentre of the triangle. A (x1, y1)

B (x2, y2)

b E

I a D

C (x3, y3)

The coordinates of the incentre of a triangle with vertices A x1 , y1 , B x2 , y2 , and C x3 , y3 are: ax1 bx2 cx3 ay1 by2 cy3 , a b c a b c Where a, b, and c are the lengths of sides BC, CA, and AB, respectively. 3 3 Example: If , 0 , , 6 , and 1, 6 are mid-points of the sides of a triangle, then find 2 2 incentre of the triangle.

Solution: 3  3  Given,  , 0  ,  , 6  , and ( −1, 6 ) are mid-points of the sides of a triangle. 2  2  In the given triangle, all the sides are divided in the ratio of 1 : 1. By using mid-point formula,

( x, y ) =  x1 + x2 , y1 + y2  

 B(x2, y2)

(1.5, 0)

A (x1, y1)

(1.5, 6)

(-1, 6)

C (x3, y3)

Taking mid-points

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COORDINATE GEOMETRY

x + x y +y  (1.5, 0) =  2 1 , 2 1  2   2 1.5 =

x2 + x1 y +y ,0= 2 1 2 2

x2 + x1 = 1.5 × 2 = 3 ------------- (1) y2 + y1 = 2 × 0 = 0 ------------- (2) Similarly, for mid-points (1.5, 6) x2 + x3 = 3 ------------- (3) y2 + y3 = 12 ------------- (4) For mid-points (-1, 6) x1 + x3 = -2 ------------- (5) y1 + y3 = 12 ------------- (6) On solving, x2 + x1 = 3, x2 + x3 = 3, x1 + x3 = −2, we get x1 −= 1, x2 4, x3 = = −1 Similarly, On solving, y2 + y1= 0, y2 + y3= 12, y1 + y3= 12 , we get = y1 0,= y2 0,= y3 12 (4, 0) (x2, y2) a

(-1, 0) (x1, y1)

a, b, and c are the distances. Distance formula = d

220

x2 x1 2 y2 y1 2

(−4 − 1)2 + (0 − 0)2 = 5

(4 + 1)2 + (0 − 12)2 = 13

(−1 + 1)2 + (12 − 0)2 = 12

(-1, 12) (x3, y3)

IL Foundation Series Class 10

 ax1 + bx2 + cx3 ay1 + by2 + cy3  , ∴ The incentre of ∆ABC is   a+b+c   a+b+c

 5 × ( −1) + 13 × ( −1) + 12 × 4 5 × 12 + 13 × 0 + 12 × 0  , or   , i.e., (1, 2). 5 + 13 + 12 5 + 13 + 12  

8.5.3 Excentre of triangle This is the point of intersection of one internal bisector and two external bisectors of the angles in a triangle. The circle opposite to the vertex A is called the escribed circle opposite to the point A or the circle escribed to the side BC. If I1 is the point of intersection of the internal bisector of ∠BAC and the external bisector of ∠ABC and ∠ACB , then, ax1 bx2 cx3 ay1 by2 cy3 , I1 a b c a b c Example: If the coordinates of the mid-points of the sides of a triangle are 1, 1 , 2, 3 , and 3, 4 , then find the excentre opposite to the vertex A. Solution: Let D 1, 1 , E 2, 3 , and F 3, 4 be the mid-points of the sides BC, CA, and AB, respectively. B (2, 8)

F (3, 4) D (1, 1)

A (4, 0)

E (2, -3) C (0, -6)

Let A= , Then B = 6 , 8 and C 4 , 6 Also, D is the mid-point of B and C, then, 6 4 4 2 8 6 0 1 2 1

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COORDINATE GEOMETRY

A 4, 0 , B 2, 8 and C 0, 6 . Then, a BC (2 0)2 (8 6)2 200 10 2 ; b CA (0 4)2 ( 6 0)2 52 2 13 And, c AB (4 2)2 (0 8)2 68 2 17 ∴ The coordinates of the excentre opposite to A are ax1 bx2 cx3 ay1 by2 cy3 , a b c a b c 10 2 4 2 13 2 2 17 0 10 2 0 2 13 8 2 17 6 , 10 2 2 13 2 17 10 2 2 13 2 17 20 2 2 13 8 13 6 17 , 5 2 13 17 5 2 13 17 8.5.4 Circumcentre of triangle The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle (i.e., the lines through the mid-point of a side and perpendicular to it). A(x¹, y¹)

b E

a B(x², y²)

C(x³, y³)

The coordinates are: x sin 2 A x2 sin 2 B x3 sin 2C y1 sin 2 A y2 sin 2 B y2 sin 2C S 1 , or sin 2 A sin 2 B sin 2C sin 2 A sin 2 B sin 2C ax cos A bx2 cos B cx3 cos C ay1 cos A by2 cos B cy3 cos C S 1 , a cos A b cos B c cos C a cos A b cos B c cos C = BC a= , CA b , and AB = c . Where

Example: If a triangle ABC has vertices A 1, 2 , B 2, 3 , and C 3, 1 , 1 4 A cos 1 , and B C cos 1 , then find the circumcentre of the triangle ABC. 5 10

222

IL Foundation Series Class 10

Solution: 4 Since A cos 1 5 cos A

4 3 then sin A = 5 5

3 4 24 sin 2A 2 sin A cos A 2 5 5 25 1 Now, B C cos 1 10 Then, cos = B cos = C

1 1 3 and sin B sin C 1 10 10 10

sin 2B 2 sin B cos B 2

3 1 3 sin 2C 10 10 5

Let the circumcentre be x, y . Then, x sin 2A x2 sin 2B x3 sin 2C x 1 sin 2A sin 2B sin 2C

3 3 24 2 3 5 25 5 11 and 24 3 3 6 25 5 5

y sin 2A y2 sin 2B y3 sin 2C and y 1 sin 2A sin 2B sin 2C

3 3 24 3 1 5 25 5 2 24 3 3 25 5 5

11 Hence, the coordinates of the circumcentre are , 2 . 6 8.5.5 Orthocentre of triangle The altitudes of a triangle are concurrent, and their point of concurrence is called the orthocentre, which is generally denoted by H or O. A(x¹, y¹) F

E O

B(x², y²)

C(x³, y³)

ax sec A bx2 sec B cx3 sec C ay1 sec A by2 sec B cy3 sec C O 1 , a sec A b sec B c sec C a sec A b sec B c sec C

223

COORDINATE GEOMETRY

8.6 EQUATION OF A STRAIGHT LINE IN VARIOUS FORMS Inclination of a line: If a line l makes an angle θ with the positive direction of the x-axis in the anticlockwise direction, then θ is called the inclination of the line l. The range of θ is 0 . If 0 , then the line l is parallel to the x-axis. If 90 , then the line l is perpendicular to the x-axis. Two lines have the same inclination if and only if they are parallel. Slope of a line: If θ is the inclination of a line l, then tanθ is called the slope (or gradient) of the line l. The slope of a line l is usually denoted by m.

m tan 90

Note: 1. If the inclination of a line is 45, then its slope is 1. 2. If the slope of a line is 3, then its inclination is 60.

3. The slope of the x-axis is 0  0 .

4. The slope of the y-axis is not defined ∵ 90 .

Theorem: The slope of the line passing through the points A x1 , y1 and B x2 , y2 is y2 − y1 , x1 ≠ x2. x2 − x1 Example: The slope of the line passing through 1, 2 and 3, 7 is −7 − 2 = − 9 . 3 −1 2 8.6.1 Slope-intercept form of equation of a line Intercepts of a line

If a line cuts the x-axis at A a, 0 and the y-axis at B 0, b then a is called the x-intercept, and b is called the y-intercept of the line. Y B(0, b)

A(a, 0) O

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IL Foundation Series Class 10

Example: i) If a line cuts the x-axis at 3, 0 , then its x-intercept is 3. ii) If a line cuts the y-axis at 0, 5 , then its y-intercept is -5. Slope intercept form

Theorem: The equation of the line having slope m and y-intercept c is y = mx + c . Proof: Since the y-intercept of the line is c, the line cuts the y-axis at 0, c . ∴ The equation of the line is y c m x 0 y mx c Note: i) The equation of line passing through the origin and having slope m is y = mx. ii) The equation of the line having slope m and x-intercept a is y = m ( x − a ). Example: The equation of the line having slope

2 2 and y-intercept 2 is y = x + 2. 3 3

8.6.2 Point-slope form of equation of a line Theorem: The equation of the line passing through A x1 , y1 and having slope m is y - y1 = m (x - x1 ). 3 3 Example: The equation of the line passing through 2, 3 and having slope is y − 3 = ( x + 2 ) or 4 4 4y 12 3 x 6 3 x 4y 18 0 8.6.3 Two-point form of equation of a line Theorem: The equation of the line passing through A x1 , y1 , B x2 , y2 is

y y1 x2 x1 (x x1 )(y2 y1 ) . Example: The equation of the line passing through 2, 3 , 4, 7 is y + 3 = ⇒ y + 3 = −2 ( x − 2 )

−7 + 3 ( x − 2 ) or 4−2

⇒ 2x + y − 1 = 0 8.6.4 Intercept form of equation of a line Theorem: The equation of the line having x-intercept a 0 and y-intercept b ( ≠ 0 ) is Example: The equation of the line having x-intercept 3, y-intercept 5 is

x y + = 1. a b

x y + = 1. 3 5

225

COORDINATE GEOMETRY

SOLVED EXAMPLES Example 1: Find the slope of the line 2y = 3 x. Solution: Let O, A, B, C,… are the points on the line with O 0, 0 , A 2, 3 , B 4, 6 , C 8, 12 . Y l C B A O

Now, for any two points of O, A, B, and C The difference of y coordinates 3 0 6 3 12 6 3 , a constant. The difference of x coordinates 2 0 4 2 8 4 2 This constant is called the slope of the given line. 3 m . 2

2 3 Example 2: Find the equation of the line having slope and y-intercept . 3 4 Solution: The slope-intercept form of the line is y = mx + c Here slope of the line, m = − y-intercept of the line, c =

3 4

2 3

3 2 The equation of the line: y = − x + 4 3 −9 x + 8 ⇒y= 12 ⇒ 12y = −9 x + 8 ⇒ 9 x + 12y − 8 = 0

226

IL Foundation Series Class 10

Example 3: Show that the lines x 2y 0, 2 x 4y 9 are parallel. Solution: First line is x + 2y = 0 Slope m1

coefficient of x 1 2 coefficient of y

Second line is 2 x + 4y = 9 Slope m2

coefficient of x 2 1 4 2 coefficient of y

m1 m2

1 , which means the given lines are parallel. 2

Example 4: Show that the lines 2 x + y + 1 = 0 and x − 2y + 1 = 0 are perpendicular. Solution: First line is 2 x + y + 1 = 0 Slope m1

coefficient of x 2 1 coefficient of y

Second line is x 2y 1 0 Slope m2

coefficient of x 1 1 coefficient of y 2 2

1 Now, m1 m2 2 1 . 2 So, the given lines are perpendicular. Example 5: Find the intercepts made by the line 4 x − 5y + 20 = 0 on the coordinate axes. Solution: Reduce the given equation into the form 4 x − 5y + 20 = 0

x y + = 1. a b

⇒ 4 x − 5y = −20 4x 5y ⇒ − =1 −20 ( −20 ) x y ⇒ + =1 ( −5 ) 4 x intercept = 5 and y − intercept = 4.

227

COORDINATE GEOMETRY

Example 6: Find the equation of the line passing through the points 3, 4 and 5, 1 . Solution: Given points x1 , y1 3, 4 and x2 , y2 5, 1 . The equation of a line is x − x1 y − y1 = x1 − x2 y1 − y2 x −3 y −4 ⇒ = 3 + 5 4 −1 x −3 y −4 ⇒ = 8 3 ⇒ 3 x − 9 = 8y − 32 ⇒ 3 x − 8y + 23 = 0 Example 7: Find the value of k if the points k, 2 2 k , k 1, 2 k , and 4 k, 6 2 k are collinear. Solution: If the points are collinear, then the slopes are equal. y2 − y1 y3 − y2 = x2 − x1 x3 − x2 2k − 2 + 2k 6 − 2k − 2k 4k − 2 6 − 4k ⇒ = ⇒ = − k + 1 − k −4 − k + k − 1 −2 k + 1 −5 20 k 10 12 k 6 8 k2 4 k 4k 8k2 4 8k2 4k 4 0 2k2 k 1 0 k

1 1 8 1 9 4 4

1 3 4 1 ⇒ k = (or) − 1 2 k

QUICK REVIEW

228

•

The coordinates of any point on the x-axis are of the form x, 0 .

•

The coordinates of any point on the y-axis are of the form 0, y .

•

The distance between points P x1 , y1 and Q x2 , y2 is PQ

•

Distance of a point P x, y from the origin O 0, 0 is given by OP x 2 y 2 .

x2 x1 2 y2 y1 2

IL Foundation Series Class 10

•

The coordinates of the point which divides the line segment joining the points P x1 , y1 and Q x2 , y2 internally in the ratio m : n are mx2 nx1 , my2 ny1 . m n m n The coordinates of the mid-point of the line segment joining the points x1 , y1 and x2 , y2 x x y y are 1 2 , 1 2 . 2 2 The coordinates of the centroid of a triangle formed by the points A x1 , y1 , B x2 , y2 , and x x x y y y C x3 , y3 are 1 2 3 , 1 2 3 . 3 3

•

The coordinates of the incentre of a triangle with vertices A ( x1 , y1 ) , B ( x2 , y2 ) , and C ( x3 , y3 )  ax + bx2 + cx3 ay1 + by2 + cy3  , are  1 , where a, b, and c are the lengths of the sides BC , CA, and a + b + c   a+b+c AB, respectively.

•

If A ( x1 , y1 ) , B ( x2 , y2 ) , and C ( x3 , y3 ) are the vertices of a triangle ABC , then  −ax1 + bx2 + cx3 −ay1 + by2 + cy3  , the coordinates of the excentre are given by I1 =   −a + b + c   −a + b + c  ax − bx2 + cx3 ay1 − by2 + cy3   ax1 + bx2 − cx3 ay1 + by2 − cy3  I2 =  1 , ,  , and I3 =  . a−b+c  a+b−c   a−b+c  a+b−c

•

The coordinates of the circumcentre of a triangle are

 x sin2A + x2 sin2B + x3 sin2C y1 sin2A + y2 sin2B + y2 sin2C  S 1 ,  or sin2A + sin2B + sin2C sin2A + sin2B + sin2C    ax cosA + bx2 cosB + cx3 cosC ay1cosA + by2 cosB + cy2 cosC  S 1 , = BC a= , CA b,  , where acosA + bcosB + ccosC   acosA + bcosB + ccosC

and AB = c . •

The coordinates of the orthocentre of a triangle are  ax secA + bx2 secB + cx3 secC ay1secA + by2 secB + cy3 secC  , where O= 1 ,  asecA + bsecB + csecC   asecA + bsecB + csecC = BC a= , CA b, and AB = c .

•

The area of the triangle formed by the points A x1 , y1 , B x2 , y2 , and C (x3 , y3 ) is

1 x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) . 2 •

If points A x1 , y1 , B x2 , y2 , and C x3 , y3 are collinear, then

x1 y2 y3 x2 y3 y1 x3 y1 y2 0 229

COORDINATE GEOMETRY

−a . b

•

The slope of a line ax by c 0 is

•

The slope of the line passing through A x1 , y1 and B (x2 , y2 ) is

•

If a line makes an angle 0 180 with the x-axis in the anticlockwise (positive)

y2 − y1 , where x1 ≠ x2. x2 − x1

direction, then the slope m tan . The slope of the x- axis is 0 . The slope of the y-axis is undefined. •

The two non-vertical lines are parallel if their slopes are equal.

•

The two non-vertical lines are perpendicular if the product of their slopes is −1.

•

S lope form or gradient form: The equation of a line passing through the origin and having slope m is y = mx.

•

lope point form of a line: The equation of a line passing through A x1 , y1 and having slope m S is y y1 m x x1 .

•

Two-point form of a line: The equation of a line passing through A x1 , y1 , B x2 , y2 is

•

y y1

y2 y1 x x1 . x2 x1

Intercept form of a line: The equation of the line with x-intercept a and y-intercept b is

x y + = 1. a b

WORKSHEET - 1 I.

DISTANCE FORMULA 1. Find the distance of a point P(x, y) from the origin. 2. Find the distance between the points (2, 3) and (4, 1). 3. If AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0), and B(5, 0), then find the length of its diagonal. 4. If R (5, 6) is the mid-point of the line segment AB, joining the points A(6, 5) and B(4, y), then find y. 5. Name the triangle formed by the points (-4, 0), (4, 0), and (0, 3). 6. Write the condition of collinearity for the points (x1, y1), (x2, y2), and (x3, y3).

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IL Foundation Series Class 10

7. If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then find the coordinates of the other end of the diameter. II. SECTION FORMULA 1. If the coordinates of the midpoints of the sides of a triangle are (3, 4), (4, 6), and (5, 7), find its vertices. 2. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that 3 AP = AB and P lies on the line segment AB. 7 3. Find the coordinates of the points which divide the line segment joining A(-2, 4) and B(2, 8) into four equal parts. 4. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4), and ( -2, -1) taken in order. 5. Find the ratio in which the line segment joining the points A (1, -5) and B( -4, 5) is divided by the x-axis. Also, find the coordinates of the points of division. III. AREA OF TRIANGLES AND QUADRILATERALS 1. Find the area of the quadrilateral whose vertices taken in order are ( -4, -2), ( -3, -5), (3,2), and (2, 3). 2. P(5, 2), Q(4, 7), and R(7, -4) are the vertices of a triangle PQR. Q(4, 7), R(7, -4), and S(6, 5)are the vertices of another triangle on the same base. Find the ratio of the areas of the two triangles. 3. D(7, 9), E(1, 1), and F( -3, -7 ) are the vertices of a DEF. L 4, 5 , M 1, 3 , and N(2, 1) are the midpoints of DE, EF, and FD, respectively. Prove that

area of DEF 4 area of LMN 1

IV. EQUATION OF A STRAIGHT LINE IN VARIOUS FORMS 1. Find the equation of a line having slope 5 and passing through the origin. 2. Find the equation of a line passing through (4, 3) with a slope 3. 3. Find the equation of a line having slope 2 and y-intercept 4. 4. Find the equation of a line having slope 2 and passing through (2, 2). 5. Find the equation of a line making intercepts 4 and -7 on the x and the y axes, respectively. 6. Find the equation of a line making intercepts a and b on the x and the y axes such that a + b = 3 and ab = 2. 7. Find the equation of a line parallel to the x-axis and passing through the point (-3, -4). 8. Find the equation of a line perpendicular to the x-axis and passing through the point (3, 4). 9. Find the equation of a line having inclination 60°and passing through the point (0, 5). 231

COORDINATE GEOMETRY

10. Find the equation of a line having inclination 45° and y-intercept -2.

3 , 0, 0 and 2, 0 . , 12. Find the equations of a line having intercepts a and b on the axes such that a + b = 5, ab = 6. 11. Find the incentre of the triangle formed with vertices 1,

13. Find the equation of a line having slope 5 and x-intercept 3. 14. Find the equation of a line parallel to the x-axis at a distance of 5 units from it. 15. If the points (k, 2), (2, 3), and (3, 4) are collinear, then find the value of k. 16. Find the equation of a line passing through (-2, 3) and having an intercept equal in magnitude but opposite in sign. 17. Find the equation of the line passing through the points (a, 0) and (0, b). 18. Find the inclination of the line (in radian) passing through the points (-2, 3) and (-1, 4). 19. Find the equation of the line parallel to the y-axis and passing through (-3, 2). 20. Find the equation of a line parallel to the y-axis at a distance of 10 units from it. 21. Find the equation of a line with x-intercept of 2 and parallel to 2 x + y + 1 = 0 . 22. Find the equation of a line passing through the points A 1, 2 , and B 3, 4 . 23. Find the equation of a line passing through the point 2, 3 and parallel to x + y + 1 = 0. 24. Find the equation of a line passing through the point 1, 2 and perpendicular to x − y + 2 = 0. 25. Find the equation of a line passing through the points A 1, 5 , and B 3, 5 .

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. The distance between the points cos , sin and sin , cos is:

a) 3

b) 2

c) 2

d) 1

2. If the points k, 2 k , 3 k, 3 k ,and 3, 1 are collinear, then k = a)

1 3

b) −

1 3

2 3

d) −

2 3

3. If 1, 2 , 2, 1 , 3, 1 , and a, b , taken in order, are vertices of a parallelogram, then a)= a 2= ,b 0

b) a 2, b 0

a 0= ,b 4 c)=

4. If points a, 0 , 0, b , and 1, 1 are collinear, then a) 1

232

b) 2

c) 0

a 6= ,b 2 d)=

1 1 + = a b d) -1

IL Foundation Series Class 10

5. The line segment joining points 3, 4 and 1, 2 is divided by the y-axis in the ratio: a) 1 : 3

b) 2 : 3

c) 3 : 1

d) 2 : 3

6. If the centroid of the triangle formed by the points a, b , b, c , and c, a is at the origin, then a 3 + b 3 + c3 = a) abc

b) 0

c) a + b + c

d) 3abc

7. In the figure, the area of ∆ABC (in square units) is: Y

A(1, 3) B

X’

-4-3-2-1

1 2 3 4 5 6

Y’

a) 15

b) 10

c) 7.5

d) 2.5

8. The distance of the point P 6, 8 from the origin is: a) 8

b) 2 7

c) 10

d) 6

9. The perimeter of a triangle with vertices 0, 4 , 0, 0 , and 3, 0 is: a) 5

b) 12

c) 11

d) 7 + 5

10. The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2, 3), B(6, 7), and C(8, 3) is: a) (0, 1)

b) (0, -1)

c) (-1, 0)

d) (1, 0)

11. Assertion(A): The value of y is 6 for which the distance between the points P (2, 3) and Q(10, y) is 10. Reason (R): Distance between two given points A(x1, y1) and B(x2, y2) is given, AB

x2 x1 2 y2 y1 2

233

COORDINATE GEOMETRY

a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). b) Both Assertion (A) and Reason (R) are true, but Reason(R) is not the correct explanation of Assertion (A). c) Assertion (A) is true, but Reason (R) is false. d) Assertion (A) is false, but Reason (R) is true. 12. Assertion(A): The points which divide the line segment joining the points 0, 3 and 14 7

7, 1 in the ratio 2 : 3 is 5 , 5 . eason (R): The mid-point of a line segment joining of x1 , y1 and x2 , y2 is R x1 x2 y1 y2 , 2 . 2 a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). b) Both Assertion (A) and Reason (R) are true, but Reason(R) is not the correct explanation of Assertion (A). c) Assertion (A) is true, but Reason (R) is false. d) Assertion (A) is false, but Reason (R) is true. 13. Inclination of a vertical line is: a) Not defined

b) 0

π 2

d) π

⎛x y ⎞ ⎛x y ⎞ 14. If the slope of the line ⎜ + − 1 ⎟ + k ⎜ + − 1 ⎟ = 0 is -1 , then the value of 'k' is: ⎝a b ⎠ ⎝b a ⎠ a) -1

b) 1

c) 0

d) 2

15. The equation of the line passing through 2, 3 and perpendicular to the line joining 5, 6 and 6, 5 is: a) x + y + 5 = 0

b) x − y + 5 = 0

c) x − y − 5 = 0

d) x + y − 5 = 0

16. In a rhombus ABCD , the diagonals AC and BD intersect at the point 3, 4 . If the point A is 1, 2 , then the diagonal BD has the equation: a) x − y − 1 = 0

b) x + y − 1 = 0

c) x − y + 1 = 0

d) x + y − 7 = 0

17. If the straight line ( 3 x + 4y + 5 ) + k ( x + 2y − 3 ) = 0 is parallel to the x-axis, then the value of k = a) 1

b) -3

c) 4

d) 2

18. If the line ( x − y + 1 ) + K ( y − 2 x + 4 ) = 0 makes equal intercepts on the axes, then the value of K is: 1 3 1 2 a) b) c) d) 3 4 2 3 234

IL Foundation Series Class 10

19. The equation of a line passing through the point (1, 2) with the slope a) 4 x − 3y + 17 = 0

b) 5 x + 3y − 11 = 0

2 is 5

c) 7 x + 2y − 27 = 0

d) 2 x − 5y + 8 = 0

3 and y-intercept 5 is 4 b) 4 x + 3y + 20 = 0 c) 3 x 4y 20 0

d) 3 x 4y 20 0

20. The equation of a line having slope a) 3 x + 4y + 20 = 0

21. The equation of a line parallel to the line ax + by + c = 0 and passing through the origin is a) ax by 0

b) y =

b x a

c) ax − by = 0

d) y =

a x b

22. The equation of a line parallel to the x-axis and passing through 3, 4 is a) x 3 0

b) x + 3 = 0

c) y 4 0

d) y + 4 = 0

23. The equation of a line perpendicular to the x-axis and passing through the point 3, 4 is a) x + 3 = 0

b) x 3 0

c) y + 4 = 0

d) y − 4 = 0

24. The intercepts of the line 2 x − 3y − 5 = 0 are a)

5 5 , 2 3

−5 5 , 2 3

−5 −5 , 2 3

5 −5 , 2 3

25. The distance between the points cos , sin and sin , cos is a) 1

b) 2

c) 2

d) 0

26. The x-axis divides the line segment joining 2, 3 and 5, 7 in the ratio a) −3 : 7

b) 3 : 7

c) −2 : 5

d) 2 : 5

27. The y-axis divides the line segment joining 3, 5 , and 4, 7 in the ratio a) 5 : 7

b) −5 : 7

c) −3 : 4

d) 3 : 4

28. If 1, 2 , 4, 3 , and 2, 4 are the midpoints of the sides of a triangle, then its centroid is a) (1, 1)

b) (1, 0)

c) (1, 2)

d) (2, 1)

29. The incentre of the triangle formed by the points 0, 8 , 6, 0 , and 0, 0 is a) 1, 1

b) 2, 2

c) 3, 3

d) 4, 4

30. The excentre of the triangle formed by the points 0, 3 , 4, 0 , and 0, 0 , which is opposite to 0, 0 , is

a) 3, 1

b) 6, 6

c) 3, 3

d) 4, 4

31. If the orthocentre and circumcentre of a triangle are 2, 3 , and 5, 6 , then the centroid is a) 2, 7

4 b) 3, 3

c) 4, 3

d) 1, 3

235

COORDINATE GEOMETRY

32. If 0, 1 is the orthocentre and 2, 3 is the centroid of a triangle, then its circumcentre is b) 1, 0

a) 3, 2

c) 4, 3

d) 3, 4

33. If G is the centroid of ∆ ABC and if the area of DAGB is 5 sq. units, then, the area of ∆ ABC is a) 20 sq. units

b) 15 sq. units

34. The area of the triangle formed by 0, 0 ,

c) 10 sq. units

d) 25 sq. unit

a ,0 , and 0,a is 2a1 sq. units then x = x2

a) 1 or 5 b) -1 or 5 c) 1 or -5 2π , is 35. The slope of the line, whose inclination 3 1 a) 1 b) 3 b) 3

d) -1 or -5 d) − 3

36. The vertices of a ∆ABC are A 1, 1 , B 3, 4 , and C 2, 5 . The equation of the altitude through the vertex A is a) 5 x + 9y + 4 = 0

b) 5 x + 9y − 4 = 0

c) 5 x − 9y + 4 = 0

d) 5 x − 9y − 4 = 0

37. The equation of the perpendicular bisector of the line segment joining 1, 2 , and 3, 4 is a) 2 x − y + 5 = 0

b) x + y − 5 = 0

c) 3 x − 2y + 5 = 0

d) x + y − 4 = 0

38. If a line cuts the y-axis at 0, 3 , then the y-intercept is a) 3

b) -3

c) 0

d) 6

4 2 and x-intercept − is 3 5 b) 2 x − 3y − 14 = 0 c) 8 x + 12y − 9 = 0

39. The equation of the line having slope − a) 3 x − 4y + 20 = 0

d) 20 x + 15y + 8 = 0

40. Reduce 9 x − 3y + 5 = 0 into slope intercept form. a) y = 3 x +

5 3

b) y =

3x +2 4

c) y =

5x +5 2

d) y =

−4 x 5 + 3 2

II. FILL IN THE BLANKS 1. The area of the triangle formed by the line passing through the points 3, 4 , and 5, 6 with the coordinate axes is ___________. 2. The ratio in which 4, 6 divides the line segment joining the points A 6, 10 and B 3, 8 is _________. 3. The ratio in which the straight line x y 2 divides the line segment joining 3, 1 and 8, 9 is _________. 4. The x-coordinates of a point P is twice its y-coordinate. If P is equidistanct form Q 2, 5 and R 3, 6 , then the coordinates of P is _________. 5. If A, B, and C are collinear points, then ar ABC _________.

236

IL Foundation Series Class 10

6. The equation of a line having slope ( tan ) and passing through the point a2sin , a2cos is _________.

7. The equation of a line passing through the points (at 12 , 2at1 , and at22 , 2at2 is _________. 8. The equation of a line parallel to the line ax + by + c = 0 and passing through the point a, b is _________. 9. If A 1, 3 , B 2, p , and C 5, 1 are collinear, then the value of p is _______.

10. The points a, a , a, a , and 3a,

3a forms ______________ triangle.

237

TRIGONOMETRY

9.1 TRIGONOMETRIC RATIOS 9.1.1 Trigonometric ratios definition Let ∆ABC be right-angled at B. Then, ∠A and ∠C are acute angles. For each angle, there are six trigonometric ratios defined as: C

side opp. to A BC i) sine of A sin A AC hypotenuse

ii) cosine of A cos A

side adj. to A AB AC hypotenuse

iii) tangent of A tan A

side opp. to A BC side adj. to A AB

iv) cosecant of A cosec A v) secant of A sec A

AC hypotenuse side opp. to A BC

hypotenuse AC side adj. to A AB

vi) cotangent of A cot A side adj. to A AB side opp. to A BC Similarly, the 6 trigonometric ratios for ∠ C can be defined. Example: From the given right-angled triangle ABC, find all the trigonometric ratios with respect to angle A.

Solution: sin= A

BC 4 AC 5 = ; cosec= A = AC 5 BC 4

A cos=

AB 3 AC 5 = ; sec= A = AC 5 AB 3

BC 4 AB 3 A= ; cot A = = tan= AB 3 BC 4

238

IL Foundation Series Class 10

9.1.2 Relation between trigonometric ratios i) cosec A

hypotenuse 1 side opp. to A sin A

∴ sin A ⋅ cosec A = 1 ii) sec A

hypotenuse 1 side adj. to A cos A

cos A sec A 1 iii) cot A

side adj. to A 1 side opp. to A tan A

tan A cot A 1 iv) tan A =

side opp. to ∠A side adj. to ∠A

side opp. to A sin A hypotenuse side adj. to A cos A hypotenuse sin A cos A 1 1 cos A v) cot A tan A sin A sin A cos A

tan A

cot A Note:

cos A sin A

i) sin A is a symbol, and sin A does not mean sin × A. It is also true for all other trigonometric ratios. ii) The trigonometric ratios are real numbers and have no units. iii) sin A is a meaningful symbol, but sin is not. The same is true for all other trigonometric ratios. iv) (sin θ)n is symbolised as sin n θ . Example: In PQR right-angled at Q, where PR QR 25 cm and PQ = 5 cm. Determine the values of sin P, cos P, tan P . Solution: By Pythagoras theorem, PR2 QR2 PQ2

239

TRIGONOMETRY

PR QR PR QR ( PQ)2

25 PR QR (5)2

PR QR 1

PR QR 25 1 PR QR 1 2

By adding (1) and (2), 2 PR 26

∴ PR=13 cm Therefore, we get: 13 cm QR 25 cm (By substituting the value PR 13 cm)

QR 12 cm sin P

opp QR 12 cm 12 hyp PR 13 cm 13

cos P

adj PQ 5 cm 5 hyp PR 13 cm 13

tan P

sin P 12 / 13 12 cos P 5 / 13 5

9.1.3 Trigonometric ratios of some specific angles

240

ÐA

0

30

45

60

90

sin A

1 2

3 2

cos A

3 2

1 2

tan A

1 3

Not defined

cosec A

Not defined

2 3

sec A

2 3

Not defined

IL Foundation Series Class 10

ÐA

0

30

45

60

90

cot A

Not defined

1 3

Example: Evaluate 2tan2 45 cos2 30 sin2 60. Solution: 2tan2 45 + cos2 30 − sin2 60 2

⎛ 3⎞ ⎛ 3⎞ = 2(1) + ⎜ ⎟ −⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠

= 2+

3 3 − =2 4 4

SOLVED EXAMPLES Example 1: If cot Solution:

1 sin 1 sin 7 , evaluate 1 cos 1 cos 8

Let ∆ABC be right-angled at B, with C . Given cot

7 8

side adj. to 7 side opp. to 8

BC 7 AB 8

let BC 7 K and AB 8 K where K 0 AC2 AB 2 BC2 (By Pythagoras Theorem) (8 K)2 (7 K)2 64 K2 49 K2 113K2 AC 113K

241

TRIGONOMETRY

8K 8 side opp. to AB AC hyp 113.K 113 side adj. to BC 7K 7 and cos hypotenuse AC 113 K 113 2 1 sin 1 sin 1 sin sin

64 49 8 1 1 113 113 113 and

1 cos 1 cos 1 cos2 2

49 64 7 1 1 113 113 113

49 1 sin 1 sin 113 1 cos 1 cos 64 113

Example 2: In ∆ABC, right-angled at B, the ratio of AB to AC is 1 : 2 . Find the values of: i)

2 tan A 1 + tan2 A

ii)

2 tan A 1 − tan2 A

Solution: Given AB : AC = 1 : 2 AC AB 2 BC 2 AC 2 AB 2 2

( AB 2 ) AB AB

∴ BC 2 = AB 2 ⇒ BC = AB opp BC AB 1 adj AB AB Hence, by substituting the value of tan A, we get:

tan A tan A 1 i)

242

2 1 2 tan A 1 2 1 tan A 1 (1)2

IL Foundation Series Class 10

ii)

2 1 2 tan A 2 (not defined) 2 2 0 1 tan A 1 (1)

Example 3: If tan A B 3 and tan A B Solution: t an A B 3 and tan A B

1 ; 0 A B 90 ; A B 0 , find A and B. 3

1 1 but, tan60 = 3 and tan 30 = 3 3

A B 60 ...(1) and A B 30 ...(2) 1 2 2 A 90 A 45 We have A B 60, 45 B 60

B 60 45 15

Example 4: Find the value of θ in each of the following. i) 2 sin 2 3 ;  45 ii) 2 cos 3 1; 0 30 Solution: i) Given: 2 sin 2 3 sin 2 

But, sin 60 =

3 2

sin2 sin60

2 60

Hence,

30

ii) Given:

2cos3 1

cos3

1 2

But, cos60

1 2

3 60

20 243

TRIGONOMETRY

9.1.4 Trigonometric ratios of complementary angles In a right-angled triangle, the two non-right angles are complementary. The trigonometric ratios of complementary angles are related to each other in a specific way. sin 90 cos

cos 90 sin

tan 90 cot

cot 90 tan

sec 90 cosec

cosec(90 ) = sec

Example: Prove that

A 900 -θ

cos 70 cos 59 8 sin2 30 0 . sin 20 sin 31

Solution:

cos 59 cos 90 31 sin 31

cos 70 cos 90 20 sin 20 



L.H.S. cos 70 cos 59 sin 20 sin 31 1 2  8 sin 30 8     sin 20 sin 31 sin 20 sin 31 2 1 1 2 0 R.H.S.

Hence, proved. 9.1.5 Variations of trigonometric ratios

Y In the I quadrant, the

In the II quadrant, the sine cosine tangent contangent secant cosecant

X’

In the III quadrant, the sine

cosine tangent contangent secant cosecant

sine cosine tangent contangent secant cosecant

from 1 to 0 from 0 to -1 from -∞ to 0 from 0 to -∞ from-∞ to -1 from 1 to ∞

from 0 to -1 from -1 to 0 from 0 to -∞ from ∞ to 0 from -1 to -∞ from -∞ to -1

Y’ 244

from 0 to 1 from 1 to 0 from 0 to -∞ from ∞ to 0 from 1 to ∞ from ∞ to 1

X In the IV quadrant, the sine cosine

from -1 to 0

tangent contangent secant cosecant

from -∞ to 0 from 0 to -∞

from 0 to 1

from ∞ to 1 from -1 to -∞

IL Foundation Series Class 10

1 Example: If sec tan , then find the value of sinθ and determine the quadrant in which θ 5 lies.

Solution: 1 ...(1) 5 Since, sec tan and sec tan are reciprocal to each other, sec tan 5 ...(2) sec tan

1 2 2sec

1 26 13 5 5 sec and cos 0 5 5 5 13

24 12 1 (1) 2 2tan 5 tan 0 5 5 5

12 5 12 Also sin tan cos 0 5 13 13 Since sin 0 and cos 0 , the angle θ lies in the 4th quadrant. 9.1.6 Trigonometric ratios of the angle in terms of q Let the given angle be A and A n , where n Z,0 . 2 2 i) sin n sin , if n is even 2 cos , if n is odd ii) cos n cos , if n is even 2 sin , if n is odd iii) tan n tan , if n is even 2 cot , if n is odd iv) cot n cot , if n is even 2 tan , if n is odd v) sec n sec , if n is even 2 cosec , if n is odd vi) cosec n cosec , if n is even 2 sec , if n is odd

Example: Prove that: cot A tan 180 A tan 90 A tan 360 A 0.

245

TRIGONOMETRY

Solution:

L.H.S cot A tan 180 A tan 90 A tan 360 A

cot A tan A cot A tan A 0 R.H.S. 9.1.7 Trigonometric Identities Identity I: sin2 cos2 1 R 

Consider a right-angled triangle ABC, with B 90 and CAB sin

BC AB , cos AC AC

Now, by Pythagoras theorem, AB 2 BC 2 AC 2 Dividing both sides with AC 2 , we get:

AB 2 BC2 AC2 AC2 AC2 AC2 2

AB BC 1 AC AC (cos )2 (sin )2 1 cos2 sin2 1 or

sin2 cos2 1

Identity II: sec2 tan2 1 (where θ is not an odd multiple of

π ) 2

From the first identity: sin2 cos2 1

Dividing both sides with cos2θ , we get: sin2 cos2 1 2 2 cos cos cos2 2

sin 1 1 cos cos

(tan )2 1 (sec )2 tan2 1 sec2 or sec2 tan2 1

Identity III: cosec2 cot 2 1 (where θ is not an integral multiple of π ) From the first identity: sin2 cos2 1.

246

IL Foundation Series Class 10

Dividing both sides with sin2θ , we get: sin2 cos2 1 2 2 sin sin sin2 1 cot 2 cosec2 or cosec2 cot 2 1

Useful formulae from identities: 1. sin2 cos2 1 , where R i) sin2 1 cos2 ii) sin 1 cos2 iii) cos2 1 sin2 iv) cos 1 sin2

2. sec2 tan2 1 , where 2n 1 , n Z or cos 0 2 2 2 i) sec 1 tan ii) sec 1 tan2 iii) tan2 sec2 1 iv) tan sec2 1 v) sec tan sec tan 1 vi) sec tan

1 sec tan

vii) sec tan

1 sec tan

3. cosec2 cot 2 1 , where n , n Z and sin 0 i) cosec2 1 cot 2 2

ii) cosec 1 cot iii) cot 2 cosec2 1 2

iv) cot cosec 1 v) cosec cot cosec cot 1 vi) cosec cot

1 cosec cot

247

TRIGONOMETRY

Example: Express the trigonometric ratios sin A, sec A , and tan A in terms of cot A Solution: i)

(

sin2 A = 1 − cos2 A  sin2 A + cos2 A = 1

)

= 1−

1 sec2 A

= 1−

1  sec2 A − tan2 A = 1 2 1 + tan A

= 1−

(

)

1 1+

1 cot 2 A

cot 2 A cot 2 A + 1

cot 2 A + 1 − cot 2 A = 1 + cot 2 A =

1 1 + cot 2 A 1 1 = 1 + cot 2 A 1 + cot 2 A

∴ sin A =

ii) sec2 A 1 tan2 A,  sec2 A tan2 A 1 1

248

1 cot 2 A

cot 2 A 1 cot 2 A

1 cot 2 A cot 2 A

sec A

1 cot 2 A 1 cot 2 A 1 cot 2 A cot A cot 2 A cot 2 A

iii) tan A =

1 cot A

IL Foundation Series Class 10

SOLVED EXAMPLES cos2 40 cos2 50 Example 1: Evaluate: cos 40 sin 50 sin2 40 sin2 50

Solution:



cos 40 sin 90 40 sin 90 40 sin 50

sin 50 cos 90 50 cos 40

cos 50 sin 90 50 sin 40 



cos2 40 cos2 50 sin2 40 sin2 50

cos2 40 sin2 40 1 sin2 40 cos2 40

cos 40 sin 50



sin 50 sin 50

Example 2: Prove that: sin 50 cos 40 tan 1 tan 10 tan 20 tan 70 tan 80 tan 89 1 . Solution:

Consider sin 50 cos 90 50

cos 40

cos 90 50



Consider tan 1 tan 10 tan 20 tan 70 tan 80 tan 89

tan 1 tan 10 tan 20 tan 90 20 tan 90 10 tan 90 1

tan 1 tan 10 tan 20 cot 20 cot 10 cot 1 tan 1 tan 10 tan 20

1 1 1   tan 20 tan 10 tan 1

sin 5 sin 5 1

L.H.S. sin 50 cos 40 tan 1 tan 10 tan 70 tan 80 tan 89 



1 R.H.S. L.H.S. R.H.S.

249

TRIGONOMETRY

Example 3: Evaluate: sin2 63 + sin2 27  i) cos2 17  + cos2 73

ii) sin 25 cos 65 + cos 25 sin 65

Solution:

cos 73 cos 90 17 sin 17

i) sin 63 sin 90 27  cos 27  



∴ sin2 63 + sin2 27 = cos2 27 + sin2 27 = sin2 27 + cos2 27

(

= 1 ∵ sin2 A + cos2 A = 1

)

cos2 17 + cos2 73 = cos2 17 + sin2 17 = sin2 17 + cos2 17

(

= 1 ∵sin2 A + cos2 A = 1

)

sin2 63 sin2 27  1 1 cos2 17 cos2 73 1

sin 65 cos 90 65 cos 25

ii) cos 65 cos 90 25 sin 25 



sin 25 cos 65 cos 25 sin 65 sin 25 . sin 25 cos 25 . cos 25 sin2 25 cos2 25

1 ∵ sin2 A cos2 A 1

Example 4: Prove the following identities when the angles involved are acute angles for which the expressions are defined: cos A 1 sin A 1 cos 2 sec A i) (cosec cos )2 i i) 1 sin A cos A 1 cos Solution

cos 1 i) (cosec cos ) sin sin 2

250

IL Foundation Series Class 10

1 cos sin

(1 cos )2 1 cos2

1 cos 1 cos 1 cos 1 cos 1 cos 1 cos

sin A cos A 1 2

ii)

cos A 1 sin A cos A 1 sin A 1 sin A cos A 1 sin A 1 sin A 1 sin A cos A

cos A 1 sin A 1 sin A cos A 1 sin2 A

cos A 1 sin A 1 sin A cos A cos3 A

1 sin A 1 sin A cos A cos A

sin A cos A 1 2

1 sin A 1 sin A cos A 2 = cos A

1 cos A = 2sec A =2

9.2

TRIGONOMETRIC RATIOS FOR COMPOUND ANGLES

9.2.1 Compound angles Definition: The algebraic sum of two or more angles is called a compound angle. If A, B, and C are angles, then A B , A B , A B C , A B C etc., are compound angles. Theorem: cos A B cos A cos B sin A sin B for called all A, B ∈ R.

251

TRIGONOMETRY

Proof: R Q

B A B

P(1,0)

Consider a rectangular cartesian coordinate system OXY. Consider the unit circle in the coordinate plane with the centre at the origin. Let the coordinates of P be 1, 0 . Choose points Q, R and S on the unit circle such that ∠ POQ, ∠ POR and ∠ POS measured in the anti-clockwise sense are A, A + B and 2 B , respectively. Therefore, the coordinates of: BA , sin BA , sin cosA A ,, Rsin Acos , R A cos Q cosQA, sin B , Asin B , Asin B A B B2 B B , sin , cos B 2 cos sinBB, sin B S cosS 2 cos B ,2 sin APOR A QOS B (numerically QOS (numerically ) POR B )

2 2 2 QS 2 PR PR QS

(In ∆POR and QOS , OR 1 OS , OP 1 OQ , and OPR A B QOS numerically) PR2 QS 2 (c os A B 1)2 (sin A B 0)2 cos B cos A ( sin B sin A)2 2

cos2 A B 1 2 cos A B sin2 A B cos2 B cos2 A 2 cos B cos A sin2 B sin2 A 2 sin A sin B 2 2 cos A B 2 2 cos A cos B sin A sin B cos A B cos A cos B sin A sin B

Theorem: For all A, B R,cos A B cos A cos B sin A sin B . Proof: We know that cos A B cos A cos B sin A sin B Replacing B with −B in the above equation, we get:

252

IL Foundation Series Class 10

cos A B cos A cos B sin A sin B

cos B cos B,sin B sin B

cos A B cos A cos B sin A sin B

Theorem: sin A B sin A cos B cos A sin B for all A, B ∈ R . Proof: We know that cos A B cos A cos B sin A sin B ⇒ On substituting A for A in the above equation, we get: 2

cos A B cos A cos B sin A sin B 2 2 2 cos A B sin A cos B cos A sin B sin A B sin A cos B cos A sin B 2

sin A B sin A cos B cos A sin B

sin A B sin A cos B cos A sin B for all A, B R. Theorem: sin A B sin A cos B cos A sin B for all A, B ∈ R . Proof: We know that sin A B sin A cos B cos A sin B Replacing B with −B in the above equation, we get: sin A B sin A cos B cos A sin B

sin A B sin A cos B cos A sin B

sin A B sin A cos B coos A sin B for all A, B R

Theorem: i) If none of A, B, A + B is an odd multiple of

π tanA tanB , then tan A B . 2 1 tanAtanB

ii) If none of A, B, A + B is an integral multiple of π , then cot A B

cot B cot A 1 . cot B cot A

253

TRIGONOMETRY

Proof: Suppose that none of A, B, A + B is an odd multiple of

π , then cos A,cos B and cos A B are not 2

equal to zero. tan A B

sin A B sin A cos B cos A sin B cos A B cos A cos B sin A sin B

On dividing the numerator and the denominator by cosAcosB, we get: sin A cos B cos A sin B cos A cos B cos A cos B tan A tan B tan A B cos A cos B sin A sin B 1 tan A tan B cos A cos B cos A cos B Similarly, we can also prove the result (ii). Note: i) The formula for tan A B in terms of tan A, tan B is valid if tan A, tan B are defined and tan A tan B ≠ 1. ii) The formula for cot A B in terms of cot A, cot B is valid if cot A, cot B are defined and cot A cot B 0. Theorem: i) If none of A, B, A - B is an odd multiple of

π tan A tan B , then tan A B . 1 tan A tan B 2

ii) If none of A, B, A − B is an integral multiple of π , then cot A B Proof: From (i) of the above theorem, we get tan A B

cot B cot A 1 . cot B cot A

tan A tan B . 1 tan A tan B

On replacing B with −B in the above result, we get: tan A B

tan A B

tan A tan B 1 tan A tan B tan A tan B 1 tan A tan B

Similarly, we can also prove the result (ii) Theorem : i) sin A B sin A B sin2 A sin2 B cos2 B cos2 A. ii) cos A B cos A B cos2 A sin2 B cos2 B sin2 A. for all A, B ∈ R.

254

IL Foundation Series Class 10

Proof: i) sin ( A + B ) sin ( A − B ) = ( sin A cos B + cos A sin B ) ( sin A cos B − cos A sin B )

(

) (

)

= sin2 A cos2 B − cos2 A sin2 B = sin2 A 1 − sin2 B − 1 − sin2 A sin2 B

(

) (

)

= sin2 A − sin2 B = 1 − cos2 A − 1 − cos2 B = cos2 B − cos2 A Similarly, we can also prove the result (ii) Note: i) tan A B tan A B

tan2 A tan2 B 1 tan2 A tan2 B

ii) cot A B cot A B

cot 2 B cot 2 A 1 cot 2 B cot 2 A

9.3

MULTIPLE AND SUBMULTIPLE ANGLES

A A 3A 4A 5A If A is an angle, then 2 A, 3 A, 4 A, …… are called multiple angles of A, and , , , , ,… 2 3 2 5 6 are called submultiple angles of A. 9.3.1 Trigonometric ratios of an angle 2A in term of angle A To find the trigonometric ratios of an angle 2A in terms of those of the angle A: i) sin 2 A 2 sin A cos A A R ii) A R, cos 2 A cos2 A sin2 A 2 cos2 A 1 1 2 sin2 A iii) tan 2 A

2 tan A π (if A, 2 A are not odd multiples of ) 2 1 tan A 2

cot 2 A 1 (if A, 2 A are not equal to n , n Z ) 2 cot A v) 2 cot 2 A cot A tan A (2 A is not an integral multiple of π , and A is neither an integral multiple of π nor an odd multiple of π ). 2 Proof: iv) cot 2 A

i) sin A B sin A cos B cos A sin B B A sin A A sin A cos A cos A sin A sin 2 A 2 sin A cos A

255

TRIGONOMETRY

cos A B cosAcosB sinAsinB

ii)

B A cos A A cosAcosA sinAsinA

cos2 A cos2 A sin2 A...(1)

cos2 A 1 cos2 A 2cos2 A 1...(2)

cos2 A cos2 sin2 A 1 sin2 A sin2 A 1 2sin2 A...(3)

From (2), we have 1 cos2 A 2cos2 A From (3), we have 1 cos2 A 2sin2 A tan A tan B iii) tan A B 1 tan A tan B B A tan A A iv) cot A B

cot B cot A 1 cot B cot A

B A cot A A v) cot 2 A

tan A tan A 2 tan A tan 2 A 1 tan A tan A 1 tan2 A

cot A cot A 1 cot 2 A 1 cot 2 A cot A cot A 2 cot A

cot 2 A 1 1 cot A tan A 2 cot 2 A cot A tan A 2 cot A 2

Example: Show that: a)

cos sin 1 sin2 tan cot cos sin 1 sin2 4 4

Solution:

1 sin2 (cos sin )2 cos sin 1 tan tan 2 1 sin2 cos sin 1 tan (cos sin ) 4 tan cot 4 2 4 Similarly, we can show the result for (b). A Note: On substituting in place of A in the ratios given in 9.3.1, we get the following relations: 2 A A i) sin A = 2 sin cos 2 2 256

IL Foundation Series Class 10

ii) cos A cos 2 cos2

2 A

sin2

A 2

A A 1 1 2 sin2 2 2

2 cos2

A 1 cos A 2

A 1 cos A 2  A  A   , A are not odd  2 tan    2 2  tan A =    p 2 A  1− tan  multiples of   2 2      2 sin2

iii)

 A   , A are not integral  cot − 1  2   2 cot A =   iv) A   p 2 cot   multiples of  2 2   A A v) 2 cot A cot tan 2 2 A (A is not an integral multiple of π , and is neither an integral multiple of π nor an odd 2 π multiple of ). 2 9.3.2 sin 2A, cos 2A, tan 2A in terms of tan A 2 A

π To express: sin 2 A, cos 2 A, tan 2 A in terms of tan A. Suppose that A is not an odd multiple of . 2 Then, 2 tan A i) sin 2 A 1 tan2 A 1 tan2 A ii) cos 2 A 1 tan2 A 2 tan A π (Here, 2A is also not an odd multiple of ) 2 1 tan A 2 2 sin A 2 tan A 2 tan A cos2 A Proof: i) sin 2 A 2 sin A cos A 2 cos A sec A 1 tan2 A Similarly, we can also prove the results (ii), (iii). A Note: On substituting in 11.3.2, we have the following relations: 2 A 2 tan A is not an odd 2 i) sin A A 1 tan2 multiple of . 2 iii) tan 2 A

257

TRIGONOMETRY

A 2 ii) cos A A 1 tan2 2 1 tan2

iii) tan A

2 tan

A is not an odd multiple of .

      A is not an odd multiple of p       and A is not an     p    odd multiple of .    2   

A 2

1 tan2

A 2

Example: Show that tan 9 tan 27  tan 63 tan 81 4. Solution:

(

) (

)

⎛ tan2 9 + 1 ⎞ ⎛ tan2 27  + 1 ⎞ =⎜ ⎟ ⎟ −⎜ ⎝ tan 9 ⎠ ⎝ tan 27  ⎠

L.H.S. = tan 9 − tan 27  − tan 63 + tan 81 = tan 9 + tan 81 − tan 27 + tan 63

(

) (

= tan 9 + cot 9 − tan 27  + cot 27 



⎛ 1 + tan2 9 ⎞ ⎛ tan2 27  + 1 ⎞ − = 2⎜ = 2 ⎡⎣ cosec18 − cosec 54 ⎤⎦ 2 ⎜  ⎟  ⎟ ⎝ 2 tan 9 ⎠ ⎝ 2 tan 27 ⎠ 1 ⎤ 4 ⎤ ⎡ 4 ⎡ 1 = 2⎢ − = 8⎢ − ⎥ 5 + 1 ⎥⎦ 5 +1⎦ ⎣ 5 −1 ⎣ 5 −1 ⎡ = 8⎢ ⎢ ⎣

( 5 + 1 ) − ( 5 − 1 ) ⎥⎤ = 2[1 + 1 ] = 2 ( 2 ) = 4 = R.H.SS. 4

⎥ ⎦

9.3.3 Trigonometric ratios of 3A in term of A To find the trigonometric ratios of 3A in terms of A: i) sin 3 A 3 sin A 4 sin3 A A R ii) cos 3 A 4 cos3 A 3 cos A A R iii) tan 3 A

π 3 tan A tan3 A (3 A, A are not odd multiples of 2 1 3 tan A 2

iv) cot 3 A

3 cot A cot 3 A (3 A, A are not odd multiples of π ) 1 3 cot 2 A

Proof: i) sin 3 A sin A 2 A sin A cos 2 A cos A sin 2 A

258

)

IL Foundation Series Class 10

sinA 1 2sin2 A cosA 2sinAcosA sinA 2sin3 A 2sinA 1 sin2 A

sinA 2sin3 A 2sinA 2sin3 A 3sinA 4sin3 A

Similarly, we can also prove result (ii).

π iii) Suppose A, 2A, 3A are not odd multiples of , 2 2 tan A tan A tan 2 A 1 tan2 A tan 3 A tan A 2 A 1 tan A tan 2 A 1 tan A 2 tan A 1 tan2 A tan A

tan A 1 tan2 A 2 tan A

1 tan A 2 tan A 2

3 tan A tan3 A 1 3 tan2 A

Similarly, we can also prove result (iv). Example: Simplify cos5θ .

cos 3 2 cos 3 cos 2 sin 3 sin 2

2 cos 1 3 sin 4 sin 2 sin cos 8 cos 10 cos 3 cos 2 cos sin  3 4 sin 8 cos 10 cos 3 cos 2 cos 1 cos 4 cos 1 8 cos 10 cos 3 cos 2 cos 5 cos 4 cos 1 4 cos3 3 cos 5

16 cos5 20 cos3 5 cos

SOLVED EXAMPLES π Example 1: If 3θ is not an odd multiple of , then prove that tan tan 60 tan 60 tan 3 2    and deduce that tan 6 tan 42 tan 66 tan 78 = 1.

Solution: 3 tan 3 tan tan tan 60 tan 60 tan 1 3 tan 1 3 tan

259

TRIGONOMETRY

3 tan2 3 tan tan3 tan 2 1 3 tan2 tan 3 R..H.S. 1 3 tan

tan tan 60 tan 60 tan 3 ...(1)

Put 6 and 18 in (1), we get tan 6 tan 66 tan 54 = tan 18...(2) tan 18 tan 78 tan 42 = tan 54 ...(3)

From (2) and (3), we get tan 6 tan 66 tan 54

tan18 tan 78 tan 42 tan18 tan 54 



tan 6 tan 66 tan 78 tan 42 1 tan 6 tan 42 tan 66 tan 78 1 Example 2: Show that

1 3 4.  sin 10 cos 10

Solution:

1 3   10 10 cos 2 sin 2 2 cos 10 3 sin 10 L.H.S.   1  sin 10 cos 10 sin 20 2

4 sin 30 cos 10 cos 30 sin 10

sin 20

4 sin 30 10 sin 20

4 R.H.S.

Example 3: If x, y are acute angles and cos 2 x

3 cos 2y 1 , then prove that tan x : tan y = 2 : 1. 3 cos 2y

Solution: 1 3 cos 2y on applying componendo and dividendo. cos 2 x 3 cos 2y 1

260

1 cos 2 x 3 cos 2y 3 cos 2y 1 1 cos 2 x 3 cos 2y 3 cos 2y 1 2 1 cos 2y 1 2 cot y 4 1 cos 2y 2

cot 2 x

tan2 x : tan2 y 2 : 1 tan x : tan y 2 : 1

IL Foundation Series Class 10

9.4

HEIGHT AND DISTANCE

9.4.1 Line of Sight Line of Sight (Line of Vision): The line of sight is an imaginary line drawn from the eye of the observer to the object when a person is looking at the object.

ne o

ig ht

Object

Observer

Horizontal level

9.4.2 Angle of Elevation

of sig h

Angle of Elevation: The angle formed by the line of sight with the horizontal level when the point being viewed is above the horizontal level.

vation

ele Angle of

Observer

Horizontal level

9.4.3 Angle of Depression Angle of Depression: The angle formed by the line of sight with the horizontal level when the point being viewed is below the horizontal level.

261

TRIGONOMETRY

Horizontal level Angle of depression

Observer

Lin

f si

Object Theodolite: The angle of elevation or angle of depression of the objects is measured by an instrument called a theodolite. Theodolite is based on the principles of trigonometry, which is used to measure angles with a rotating telescope. Example: From an aeroplane vertically flying over a straight road, the angles of depression of two consecutive kilometre-stones on the same side of the road are 45 and 60. Find the height of the aeroplane. Solution: Let B and C be the two consecutive kilometre-stones. BC = 1 kilometre. Let A be the position of the aeroplane at a certain time. XAB 45 ABD and AD h and CD x. XAC 60 ACD Now, from ABD, tan45 or 1

h or 1 x h or x h 1 ...(1) 1 x

And, from ACD, tan60

262

h h AD BD BC CD 1 x

AD h CD x

3 1 3 [from (1)]

or 3

h or 3 h 3 h or 3 h h 3 or h h 1

or h

3 3 1 3 3 3 1 1 3 3 2.366 km 2 2 3 1 3 1 3 1

IL Foundation Series Class 10

SOLVED EXAMPLES Example 1: A tree breaks due to a storm, and the broken part bends so that the top of the tree touches the ground making an angle 30 with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. Solution: et us assume that AB = h m is the height of the tree. AC is the part of the broken tree, CDB 30 L and DB = 8 m. A

In the right-angled ∆CBD, we have: CB = tan30 DB CB 1 = 8 3 CB =

8 3

Also, in the right-angled ∆CBD, we have: DB = cos30 CD

300 D

8 3 = CD 2 CD =

16 3

Height of the tree AB AC CB AC CD CD CB

8 24 3 24 3 16 8 3m 3 3 3 3 3

∴ The height of the tree = 8 3 m. Example 2: From the top of a hill 200 m high, the angles of depression of the top and the bottom of a pillar are 30 and 60, respectively. Find the height of the pillar and its distance from the hill. Solution: AB is the height of the hill. et the height PQ of the tower = h and AP = x . By the question, AB = 200 m, and from the top B, the L angles of depression of Q and P are, respectively, 30 and 60. 263

TRIGONOMETRY

Draw QC perpendicular to AB. Now, CQB QBX 30



X 60 0

300

and APB PBX 60. From APB,

200 tan 60 3 x

200 m

200 200 3 200 1.732 115.5 m. x 3 3 3 BC BC From BCQ,tan30 CQ x BC x tan 30

300

200 3 1 200 3 3 3

600 x

1 1 1 h AB BC 200 200 400 133 m. 3 3 3 Example 3: A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30, which is approaching the foot of the tower at a uniform speed. Six seconds later, the angle of depression of the car is found to be 60. Find the time taken by the car to reach the foot of the tower from this point. Solution: Let us assume that the height of the tower AB is h, and C is the position of the car. XAC ACB 30 , ADB 60 XAD

In the right-angled ∆ADB, AB We have = tan60 DB h 3 DB

300 600 h

h 3 DB ...(1) Again, consider the right ∆ACB , AB tan 30 CB 1 h 3 h CB CB 3 h

264

BC ...(2) 3

300 C

600 D

IL Foundation Series Class 10

From (1) & (2), we have: CB DB 3 3 3 DB CB 3 DB CD DB 2 DB CD The speed of the car when it covers a distance CD= in 6 sec

Distance CD = Time 6 Distance The time taken by the car to cover the distance DB Speed

DB DB 6 CD CD 6

6 DB 2 DB

3 sec. So, the time taken by the car to reach the foot of the tower from this point is 3 seconds. Example 4: The angles of elevation of the top of a tower from two points on the ground, at distances a metres and b metres from the base of the tower and in the same straight line, are complementary. Prove that the height of the tower is ab metres. Solution: et us assume that the height of tower AB is h and C and D are the two positions of the two L observers, such that AC = a and AD = b . ACB , ADB 90 .

In the right ∆ABC, we have h tan a

B AB tan AC h

h a tan ...(1) Again, in the right ∆BAD, we have: AB tan 90 DA

900 - θ

θ C

h cot b h b cot ...(2) 265

TRIGONOMETRY

Multiply (1) & (2), we get: h h a tan b cot

h 2 ab h ab

∴ The height of the tower = ab metres. Example 5: If the angle of elevation of a cloud from a point h metres above a lake is α and the angle depression of its reflection in the lake is β . Prove that the height of the cloud is

h(tan tan ) . tan tan

Solution: Let AB be the surface of the lake and P be a point of observation such that AP = h metres. Let C be the position of the cloud and E be its reflection in the lake. Then, CB = BE DPC and DPE

Let CD = x . The height of the cloud CB CD BD CD PA x h In the right PCD,

x tan PD

x ...(1) tan

In the right PDE ,

266

CD tan PD

DE tan PD

DE tan

DB BE CB BE tan

h x h ... 2 tan

P A

α ꞵ

D B

IL Foundation Series Class 10

From (1) & (2), x x 2h tan tan

x tan x 2 h tan

x tan x tan 2 h tan

x tan x tan 2 h tan x tan tan 2 h tan

2 h tan tan tan

The height of the cloud x h =

2h tan α + h ( tan β − tan α ) ( tan β − tan α )

2h tan α + h tan β − h tan α tan β − tan α

h tan α + h tan β tan β − tan α

h ( tan α + tan β ) tan β − tan α

2h tan h tan tan

QUICK REVIEW •

Trigonometric Ratios:

If ∆ABC is right-angled at B and BAC , then with reference to angle θ, we have: sin

Perpendicular Hypotenuse

cos

Base Hypotenuse

tan

Perpendicular Base

cosec

Hypotenuse Perpendicular

267

TRIGONOMETRY

•

sec

Hypotenuse Base

cot

Base Perpendicular

Reciprocal of trigonometric ratios: 1 1 1 and cot , sec tan sin cos cos sec Also, tan and cot sin cos We have, cosec

•

Trigonometric ratios of complementary angles The trigonometric ratios for angles 0 , 30 , 45 , 60 and 90:

The values of sinθ and cosθ never exceed 1, whereas the values of secθ and cosecθ are always greater than or equal to 1. If θ is an acute angle, then,

tan 90 cot , cot 90 tan sec 90 cosec , cosec 90 sec sin 90 cos , cos 90 sin 



•



Trigonometric identities

An equation is called an identity if it is true for all values of the variable(s) involved. Following are some trigonometric identities: i) sin2 cos2 1 or, 1 cos2 sin2 or, 1 sin2 cos2 ii) 1 tan2 sec2 or, sec2 tan2 1 iii) 1 cot 2 cosec2 or, cosec2 cot 2 1 •

Combined angles sin A B sin A cos B cos A sin B cos A B cos A cos B sin A sin B sin A B sin A cos B cos A sin B cos A B cos A cos B sin A sin B tan A B

268

tan A tan B where A n ; B n 1 tan A tan B 2 2

IL Foundation Series Class 10

tan A B

tan A tan B ; where A n , n Z , B n , n Z 1 tan A tan B 2 2

cot A B

cot B cot A 1 ; where A n , B n cot B cot A

cot A B

cot B cot A 1 ; where A n , B n cot B cot A

sin A B sin A B 2 sin A cos B sin A B sin A B 2 cos A sin B cos A B cos A B 2 cos A cos B cos A B cos A B 2 sin A sin B (or) cos A B cos A B 2 sin A sin B sin A B sin A B sin2 A sin2 B (oor) cos2 B cos2 A cos A B cos A B cos2 A sin2 B (or) cos2 B sin2 A tan2 A tan2 B 1 tan2 A tan2 B cot 2 A cot 2 B 1 cot A B cot A B cot 2 B cot 2 A

tan A B tan A B

•

Multiple angles 2q

2 tan 2 tan 2 , sin 2 2 sin cos ,sin 2 sin cos 2 2 2 1 tan2 1 tan 2 1 tan2 cos 2 cos2 sin2 2 cos2 1 1 2 sin2 1 tan2 1 tan2 2, cos cos2 sin2 2 cos2 1 1 2 sin2 2 2 2 2 1 tan2 2 2 tan , where and 2 are not odd multiples of . tan 2 2 2 1 tan

2 , where θ and θ are not odd multiples of π . tan 2 2 1 tan2 2 2tan

cot2

cot 2 1 , where θ and 2θ are not integral multiples of π . 2cot

269

TRIGONOMETRY

1 2 , where θ and θ are not integral multiples of π . cot 2 2cot 2 cot 2

•

Multiple Angles 3q sin3 3sin 4sin3 cos3 4cos3 3cos tan3

3tan tan3 1 3tan2

WORKSHEET - 1 I.

TRIGONOMETRIC RATIOS 1. In figure, = AB 4= cm, BC 3 cm, then cot θ equals, A

θ 4 cm

3 4

3 cm

5 4

4 3

3 5

13 12

= cm, AB 5 cm , then sin A =? 2. In ∆ABC, right-angled = at B, AC 13

5 13

5 12

4 3

3. If tan , then the value of a) 5

12 13

3 sin 2 cos is: 3 sin 2 cos

b) 8

c) 3

d) 6

b) -1

c) θ

4. cos2 sec2 ? a) 1

1 2

5. If (sin cosec )2 (cos sec )2 tan2 cot 2 kk, then k equals: a) 9 270

b) 7

c) 5

d) 8

IL Foundation Series Class 10

6. cos 1 cos 2 cos 3 cos 90 is equal to: a) 0

b) 1

7. The expression

c) -1

1 2

tan A cot A can be written as: 1 cot A 1 tan A

a) sec A cosec A + 1

b) tan A + cot A

c) sec A + cosec A

d) sin A cos A + 1

8. The value of sin2 30 − cos2 30 is: a) −

1 2

3 2

2 3

9. If sin sin2 1 , then cos12 3 cos10 3 cos8 cos6 is equal to: a) 0 10. If

b) 2

c) 1

d) 4

c) 3 − 4

d) 3 − 1

cos x 3 = 2 and cos x y , then tan y = cos y 2

a) 3 + 4

b) 3 + 1

11. If cos A =

7 , then find the value of tan A + cot A. 25 2

12. Prove the identity

1 tan2 A 1 tan A 2 tan A. 2 1 cot A 1 cot A

II. TRIGONOMETRIC RATIOS FOR COMPOUND ANGLES 1. If sin A cos B a and sin B cos A b, then sin A B is equal to: a)

a2 + b2 2

a2 b2 2 2

a2 − b2 − 2 2

2. If tan A B p,tan A B q , then find cot2A .  3. = If A 35 = , B 15 and C = 40, then tan A tan B tan B tan C tan C tan A ?

a) 0

b) 1

c) 2

d) 3

4. If ∆ABC is right-angled at C, then the value of cos A B is _______. 5. In ABC , sin B C A sin C A B sin A B C k sin A sin B sin C , where k equals ________. B C A cot . 2 2

6. If A, B, C are the interior angles of ∆ABC, then prove that tan

271

TRIGONOMETRY

7. Prove the identity: 8. If secθ = x +

sin A cos A sin A cos A 2 2 . 2 2 2 sin A cos A sin A cos A sin A cos A 2 sin A 1

1 1 . , then prove that sec tan 2x or 4x 2x

9. If cos A B

3 and tan A tan B 2, then find cos A cos B . 5

III. MULTIPLE AND SUBMULTIPLE ANGLES 1. If

2 sin 1 cos sin is also equal to y. y, then prove that 1 cos sin 1 sin

2. If tan x =

b , then find the value of a

a b a b . a b a b

3. Prove that sin 4 A 4 sin A cos3 A 4 cos A sin3 A. 1 sin cos ? 1 sin cos

a) sin

θ 2

b) cos

θ 2

c) tan d) cot

θ 2

sin sin2 ? 1 cos cos2

a) sinθ

b) cosθ

c) tanθ

d) cotθ

a) cot2θ

b) cot 4θ

c) cot3θ

d) 2cotθ

cos 3 ? 2 cos 2 1 a) sinθ

b) cos θ

c) tan θ

d) cot θ

3cos cos3 ? 3sin sin3

8. If sin cosec 2 , then the value of sin100 cosec100 is _________. 9. If cosec10 3 sec10 k, then k equals _______. 2 sin x tan x ________. 10. sin 3 x tan 3 x

272

11. Prove that

sec 8 A 1 tan 8 A . sec 4 A 1 tan 2 A

12. Prove that:

sin3 3 A cos3 3 A 8 cos 2 A. sin2 A cos2 A

IL Foundation Series Class 10

13. Evaluate:

cot x tan x . cot x cot 3 x tan x tan 3 x

IV. HEIGHTS AND DISTANCES 1. The height of a vertical tower is 200 m. It casts a shadow on the horizontal ground. When observed from the end point of the shadow, the angle of elevation of the top of the tower is 30. The length of its shadow is ______. a) 100 3m

b) 200 3m

c) 300 3m

d) 200 m

2. The angle of elevation of the top of a tower at a point on the ground is 30. If the height of the tower is tripled, then the angle of elevation is ________. 3. It is found that on walking x meters towards a chimney in a horizontal line through its base, the elevation of its top changes from 30 to 60. The height of the chimney is: a) 3 2x

b) 2 3x

3 2 x d) x 2 3

4. The angles of elevation of the top of a tower from two points a and b units away from the base in the same straight line are 60 and 30. The height of the tower is ______. a) ab

b) ab

a b

d) 2ab

5. A tower AB is 20 m high. Its shadow on the ground is 20 3 m long. Find the Sun’s altitude. 1 6. The ratio of the length of a tree and its shadow is 1 : . Find the angle of the Sun’s elevation. 3 7. The shadow of a tower standing on a level plane is found to be 50 m longer when the Sun’s elevation is 30 than when it is 60. Find the height of the tower. 8. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the h tan flagstaff are α and β , respectively. Prove that the height of the tower is . tan tan 9. The angle of elevation of the top of a vertical tower from a point on the ground is 60. From another point, 10 m vertically above the first, its angle of elevation is 45. Find the height of the tower. 10. A window of a house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be α and β , respectively. Prove that the height of the other house is h 1 tan cot metres.

273

TRIGONOMETRY

11. The lower window of a house is at a height of 2 m above the ground, and its upper window is 4 m vertically above the lower window. At certain instants, the angles of elevation of a balloon from these windows are observed to be 60 and 30, respectively. Find the height of the balloon above the ground.

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER

1. If sin 2 A cos A 18 , then the value of A is: a) 18

b) 36

c) 24

d) 27 

4 2. In the figure below, ∆ABC is right-angled at B and tan A = . If AC = 15 cm, the length of AB 3 is: C

15 cm

a) 4 cm

b) 3 cm

c) 12 cm

d) 9 cm

c) sec90

d) sin90

3. Which of the following is not defined? a) cos 0



b) tan 45

4. If sin cos , then the value of cosecθ is: a) 2 5. The value of 6 tan2 a) 1

2 3

b) 1

6 is: cos2 b) 36

c) 6

d) 2

d) -6

6. If 3 cot 2 , then the value of tanθ is: a)

2 3

2 13

3 13

3 2

7. The value of 2 sin2 A 4 sec2 A 5 cot 2 A 2 cos2 A 4 tan2 A 5 cosec2 A is: a) 0

274

b) 1

c) 2

d) 3

IL Foundation Series Class 10

8. If cot A = 7, then the value of a)

3 2

cosec2 A sec2 A is: cosec2 A sec2 A

3 4

4 3

c) 2

d) 1

2 3

9. If tan cot 2 , then tan2 cot 2 is: a) 4

b) 6

10. sec A tan A 1 sin A upon simplification gives: a) tan2 A 11. If tan a)

b) sec2 A

c) cos A

d) sin A

1 1 and tan , then the value of is: 3 2

π 6

b) π

π 4

c) 0

b) cos15

c) sin 15 cos 15

d) sin 15 cos 75

b) cot 3θ tan θ

c) cot 3θ cot θ

d) tan 3θ tan θ

12. Which of the following is rational? a) sin15



tan2 2 tan2 ? 1 tan2 2 tan2

13.

a) tan 3θ cot θ

II. FILL IN THE BLANKS 1. Given 15 cot 8 , then sin _____. 2. The value of

2 tan 30 is _____. 1 + tan2 30

3. The value of sec2 1 .cot 2 is _____. 4. The simplified value of

1 1 ______. 2 sec cosec2

5. Let 0 ,

1 1 . If tan and sin , then the value of tan 2 is _______. 7 2 10

6. The value of

sin 50 is ______. sin 130

7. In a triangle ABC with C 90 , the equation whose roots are tan A and tan B is ______.

275

TRIGONOMETRY

III. MATCH THE FOLLOWING 1.

Column - I

Column - II 1 sin A 2

A i) tan 45 2

A A ii) sin2 sin2 8 2 8 2

A B q) 4 cos2 2

iii) (cos A cos B )2 (sin A sin B )2

r) sec A − tan A

iv) (cos A cos B )2 (sin A sin B )2

A B s) 4 sin2 2

a) i-p, ii-q, iii-r, iv-s

b) i-q, ii-p, iii-s, iv-r

c) i-r, ii-s, iii-q, iv-p

d) i-r, ii-p, iii-q, iv-s

IV. SUBJECTIVE QUESTIONS 1. The triangle ABC is right-angled at B , AB = 5 cm and ACB 30 . Determine the lengths of the sides BC and AC.

2. If sec 4 A cosec A 20 , where 4 A is an acute angle, find the value of A. 3. What is the value of sin2

1 ? 1 tan2

4. If sin A sin2 A 1, then find the value of cos2 A + cos 4 A . 5. Prove that tan2 A tan2 B

cos2 B cos2 A sin2 A sin2 B . cos2 B cos2 A cos2 A cos2 B

6. If cos sin 2 cos , then show that cos sin 2 sin . 7. If sec tan P , then show that 8. Evaluate:

P2 1 sin . P2 1

sin2 20 sin2 70 sin 90 sin cos 90 cos . tan cot cos2 20 cos2 70

1 9. If sin 1 and sin , where , 0, , then what is the value of 2 2 tan 2 ?

10. If tan 40 2 tan 10 cot x , then find x. 11. If triangle ABC, if sin A cos B =

276

1 and 3tan A = tan B , then what is the value of cot 2 A ? 4

IL Foundation Series Class 10

12. Prove that:

1 sin cos tan . 1 sin cos 2

13. Prove that

cos 3 cos . 2 cos 2 1

14. A tower stands near an airport. The angle of elevation θ of the tower from a point on the 5 ground is such that its tangent is . Find the height of the tower if the distance of the 12 observer from the tower is 120 metres. 15. From the top of a rock 50 3 m high, the angle of depression of a car on the ground is observed to be 30. Find the distance of the car from the rock. 16. An observer 1.5 m tall is 28.5 m away from a tower 30 m high. Find the angle of elevation from his eye to the top of the tower. 1 cos 17. If cosec cot , then prove that I or II quadrants. 1 cos

277

ANSWER KEY 1. REAL NUMBERS

Worksheet 1 I. The fundamental theorem of arithmetic 1. i) 156 = 22 × 3 × 13 ii) 7429 = 17 × 19 × 23 2. H.C.F =2, L.C.M =60 3. H.C.F = 2, L.C.M =23640 4. H.C.F =3, L.C.M = 420 5. 22338 6. No natural number 7. 2 8. 435 9. 16 10. 999720 II. Revisiting irrational numbers 1. NA 2. NA 3. NA 4. NA 5. NA 6. i) Terminating ii) Terminating iii) Non-terminating repeating iv) Terminating 7. NA 8. i) 3 ii) 3 iii) 4 1423 1048 9. i) ii) 33 41 iii) 333

10. i) 18 − 𝑖𝑖𝑖𝑖

iii) 6 − 4𝑖𝑖𝑖𝑖

8 + 0𝑖𝑖𝑖𝑖 13 1 𝑖𝑖𝑖𝑖 vii) + tan 𝜃𝜃𝜃𝜃 2 2

v) −

III. Interval notation 1. [−1,3] 2. [−5, ∞) 3. [2,8) 4. (−∞, −3) 5. [−2,3) 6. [−4,0] IV. Absolute value 1. 7 2. 9, −9 278

333

ii) −11 − 2𝑖𝑖𝑖𝑖

𝑎𝑎𝑎𝑎 2 −𝑏𝑏𝑏𝑏2 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 � − 𝑖𝑖𝑖𝑖 𝑖 2 +𝑏𝑏𝑏𝑏2 � 𝑎𝑎𝑎𝑎 +𝑏𝑏𝑏𝑏2 𝑎𝑎𝑎𝑎 18 25 vi) = + 𝑖𝑖𝑖𝑖 𝑖 � 13 13

iv) � 2

3. 8 4. 9, −3 V. Complex numbers 1. 2 − 2𝑖𝑖𝑖𝑖 2. 2 + 5𝑖𝑖𝑖𝑖 3. 8 - real, 6 - imaginary 4. −2 − 2𝑖𝑖𝑖𝑖 5. 6 – real , 9 - imaginary 6. 5 − 12𝑖𝑖𝑖𝑖 7. −8 + 8𝑖𝑖𝑖𝑖 8. NA 9. 14 + 5i 10. 4 + 19i 11. 1 − 13𝑖𝑖𝑖𝑖 12. 𝑟𝑟𝑟𝑟𝑟 = −1 + 4𝑖𝑖𝑖𝑖, 𝑟𝑟𝑟𝑟 + 𝑟𝑟𝑟𝑟𝑟 = −2 9 12 13. − 𝑖𝑖𝑖𝑖 5 5 1 5 14. − 𝑖𝑖𝑖𝑖 2 2

15. −1 − 𝑖𝑖𝑖𝑖 16. 0 17. 0 18. 0 19. 4 20. −1 Worksheet 2 I. Multiple choice questions with single correct answer 1. b 2. b 3. c 4. b 5. a 6. c 7. c 8. c 9. c 10. b 11. d 12. b 13. d 14. b 15. a 16. d 17. a 18. c 19. b 20. a II. Subjective questions 1. True 3277081 ; prime factors of 𝑞𝑞𝑞𝑞 are 2 and 5. 2. 10000 3. 𝑞𝑞𝑞𝑞 is not expressed in the form of 2𝑛𝑛𝑛𝑛 × 5𝑚𝑚𝑚𝑚 where n and m are non-negative integers. 4. 1 5. 2𝑚𝑚𝑚𝑚 × 5𝑛𝑛𝑛𝑛, where m and n are nonnegative integers. 6. 4 7. 2 8. 𝑥𝑥𝑥𝑥 3 𝑦𝑦𝑦𝑦 2 9. 150 10. 4 decimal places 11. [−3, 5) 12. (−∞, −2] 13. (−1, 4]

ANSWER KEY 14. (7, ∞) 15. 11, −11 16. 8, −12 17. 4 + 2i 18. 8 − 4𝑖𝑖𝑖𝑖 19. 4 − 6𝑖𝑖𝑖𝑖 20. – 9 +6i 21. 3 − 4𝑖𝑖𝑖𝑖 22. −2 + 7𝑖𝑖𝑖𝑖 23. −2 24. −1 − 𝑖𝑖𝑖𝑖 25. 2𝑤𝑤𝑤𝑤 26. NA 27. 128

13.

37 4 2

14. 𝑥𝑥𝑥𝑥 − 3√2𝑥𝑥𝑥𝑥 + 4 15. 15𝑥𝑥𝑥𝑥 2 + 7𝑥𝑥𝑥𝑥 𝑥 4 √5 √5 ,− 2 2 2

16. +

Worksheet 1 I. Geometrical meaning of the zeroes of a polynomial 1. 2 2. True 3. 1 II. Relationship between zeroes and coefficients of a polynomial 1. 24 1 1 2 2

2. i) Zeroes = ,

ii) Zeroes = −1, 3

iii) Zeroes = √2, −3 √2 3. i) 4𝑥𝑥𝑥𝑥 2 – 𝑥𝑥𝑥𝑥 − 4 ii) 3𝑥𝑥𝑥𝑥 2 − 3√2𝑥𝑥𝑥𝑥 + 1 iii) 𝑥𝑥𝑥𝑥 2 + √5 4. i) 𝑥𝑥𝑥𝑥 2 − 7𝑥𝑥𝑥𝑥 + 12 ii) 3𝑥𝑥𝑥𝑥 2 + 𝑥𝑥𝑥𝑥 − 2 iii) 𝑥𝑥𝑥𝑥

√

12. 6

2. POLYNOMIALS

−7

11. 𝑥𝑥𝑥𝑥 = −√3, 3

1 + � � 𝑥𝑥𝑥𝑥 – 4√3 2

iv) 𝑥𝑥𝑥𝑥² − 6𝑥𝑥𝑥𝑥 + 2 = 0 5. 3𝑥𝑥𝑥𝑥 3 − 5𝑥𝑥𝑥𝑥 2 − 11𝑥𝑥𝑥𝑥 − 3 6. 𝑥𝑥𝑥𝑥 3 − 2𝑥𝑥𝑥𝑥 2 − 7𝑥𝑥𝑥𝑥 + 14 7. i) −1 ii) 15 iii) −50 8. 𝑥𝑥𝑥𝑥 2 − 4𝑥𝑥𝑥𝑥 + (4√5 − 5) 9. 8 10. Other zero = 1, p = -6

17. 𝑥𝑥𝑥𝑥 – 4𝑥𝑥𝑥𝑥 + 1 18. 8𝑥𝑥𝑥𝑥 2 – 21𝑥𝑥𝑥𝑥 + 20 19. 𝑥𝑥𝑥𝑥 2 – 𝑥𝑥𝑥𝑥 – 56 20. 𝑎𝑎𝑎𝑎 = 3 21. 𝑝𝑝𝑝𝑝 = −5, 𝑞𝑞𝑞𝑞 = −6 𝑐𝑐𝑐𝑐 𝑏𝑏𝑏𝑏 145 23. − 12

22. 𝑥𝑥𝑥𝑥 = or 𝑥𝑥𝑥𝑥 = −

𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎

24. 𝑘𝑘𝑘𝑘 = 9, other zero = 5 5√3 � 𝑥𝑥𝑥𝑥 + 3 2

25. 𝑥𝑥𝑥𝑥 2 + �

26. −

3 5

27. 2 28. 3 III. Graph, identifying sign of a, b, c if the graph is given, and the maximum and minimum value of quadratic polynomial 1. 𝑎𝑎𝑎𝑎 > 0, 𝑏𝑏𝑏𝑏 < 0 and 𝑐𝑐𝑐𝑐 < 0. 2. 𝑎𝑎𝑎𝑎 > 0, 𝑏𝑏𝑏𝑏 < 0, and 𝑐𝑐𝑐𝑐 > 0. 3. 𝑎𝑎𝑎𝑎 < 0, 𝑏𝑏𝑏𝑏 > 0 and 𝑐𝑐𝑐𝑐 < 0. 4. For all 𝑥𝑥𝑥𝑥 ∈ 𝑅𝑅𝑅𝑅 1 57 � 7 7 1 25 �− , � 3 3

5. �− , − 6.

Worksheet 2 I. Multiple choice questions with single correct answer 1. a 2. b 3. c 4. d 5. d 6. a 7. b 8. a 9. a 10. b 11. b 12. d 13. b 14. b 15. a 16. c 17. b 18. d 19. b 20. b 21. a 22. b 23. a 24. c 25. a 26. c 27. a 28. a 29. c 30. d 31. a II. Fill in the blanks 1. −1 2. 24 279

ANSWER KEY 5. i) (1, 2)

3. 2 − √3 4. 3 4 5. 3 −3

6. 2 7. Parabola 8. 3 9. 5 10. 8 11. 9 12. 44 −𝑏𝑏𝑏𝑏 13. 𝑎𝑎𝑎𝑎 14. −2 15. 3 16. 𝑥𝑥𝑥𝑥 2 − 2√3𝑥𝑥𝑥𝑥 + 2 17. −2 and 3 18. 1 19. −3 20. 7 III. Subjective questions 1. i) 1 ii) 2 iii) 1 iv) 1 v) 3 vi) 3 2. 𝑥𝑥𝑥𝑥 2 − 3𝑥𝑥𝑥𝑥 𝑥 2 3. 𝑎𝑎𝑎𝑎 = 2 4. 𝑥𝑥𝑥𝑥 2 − 2𝑥𝑥𝑥𝑥 + 5 has no real zeroes. 5. 𝑐𝑐𝑐𝑐 = 1 6. no zeroes of 𝑝𝑝𝑝𝑝(𝑥𝑥𝑥𝑥) 7. 𝑥𝑥𝑥𝑥 2 + 𝑥𝑥𝑥𝑥 𝑥 20 8. 𝑎𝑎𝑎𝑎 = −1 9. 8 10. 4𝑥𝑥𝑥𝑥 2 + 𝑥𝑥𝑥𝑥 + 1 11. 3

3. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Worksheet 1 I. Introduction, graphical method of solution of a pair of linear equation in two variables 1. i) Yes ii) No iii) No iv) Yes v) No vi) No 2. NA 1 2

1 2

3. i) 1, , 0, − , −1

4. i) (3, −1) iii) (−1, −5)

280

ii) 2, 3, 4, 5, 6 ii) (2, 1)

ii) (−5, −9) 1 5 2 2

iv) �− , �

iii) (1, 2)

v) (−2, −9)

vi) (5, −4)

1 13 vii) � , � 2 2 3 ix) �5, � 2 3 5 xi) � , � 2 2

viii) (3, 2)

x) (5, 8)

16 20 ,− � 9 9

xii) �

6. i) NA ii) The lines are parallel 7. i) inconsistent ii) consistent and independent iii) consistent and independent iv) consistent and dependent v) inconsistent vi) consistent and dependent 8. (−3, 1) 9. (1, 3), (−2, −3), (2, −1) 10. 𝐴𝐴𝐴𝐴 = 4, 𝐵𝐵𝐵𝐵 = −4 11. NA 12. 𝑚𝑚𝑚𝑚 =

17 11 , 𝑛𝑛𝑛𝑛 = 4 5

13. No solution for 𝑘𝑘𝑘𝑘 = −1 14. i) 𝜆𝜆𝜆𝜆 = −1 ii) 𝜆𝜆𝜆𝜆 = 1 iii) All real values of 𝜆𝜆𝜆𝜆 except ±1 15. 𝛼𝛼𝛼𝛼 = −6 16. (0, 0), (4, 4), (6, 2) 17. 6 sq. units II. Algebraic methods of solving a pair of linear equations in two variables 1. i) (2, −1) ii) (1, 1) iii) (−4, 3) iv) (−3, −15) v) (2, −2) vi) (2, −7) 2. i) (1, 2) ii) (3, 0) iii) (7, −2) iv) (−2, −6) 1 2 10 11 ii) � , � 21 14

v) (−1, 2)

vi) � , −5�

iii) (6, 2) v) No solution

iv) (1, 3) vi) No solution

49 11

3. i) � , −

12 � 11

4. 𝑚𝑚𝑚𝑚 = − 2 , 𝑏𝑏𝑏𝑏 = 2

5. 𝑝𝑝𝑝𝑝 = 2, 𝑞𝑞𝑞𝑞 = −

6. 𝑎𝑎𝑎𝑎 = 5, 𝑏𝑏𝑏𝑏 = 2

1 3

ANSWER KEY 7. 𝑥𝑥𝑥𝑥 =

𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 5𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 , 𝑦𝑦𝑦𝑦 = − 7 7

8. Weight of carbon atom = 12.011u Weight of hydrogen atom = 1.008u 9. inline skating = 25 min, swimming = 15 min 10. Number of men = 285 11. Number of children = 680 12. Number of children = 364 13. Number of DVDs = 12 14. Aaron’s age = 36 years Sejuti’s age = 12 years 15. Sweety’s age = 16 years 16. ₹10,000 in scheme A, ₹12,000 in scheme B 17. Bat = ₹500, Ball = ₹50 18. 64 19. 39, 14 20.

4 7

III. Equations reducible to a pair of linear equations, simultaneous linear equations in three variables 1 1 5 10 1 1 ii) 𝑥𝑥𝑥𝑥 = − , 𝑦𝑦𝑦𝑦 = − 4 2

1. i) 𝑥𝑥𝑥𝑥 = − , 𝑦𝑦𝑦𝑦 =

2. 3. 4. 5. 6. 7. 8.

6 kmph, 4kmph 10 kmph, 4kmph 100 kmph, 80 kmph 36 days, 18 days 10, 15 100, 80 i) (−3, 2, 5) ii) (2, 1, −1) iii) (4, 3, −3) 9. She should invest ₹2000 in saving, ₹12000 in RDs and ₹6000 in bonds 10. 7 first-place finishers, 10 second-place finishers, and 3 third-place finishers 11. 8, 15, 62 12. 32°, 96°, 52° Worksheet 2 I. Multiple choice questions with single correct answer

1. a 2. a 3. b 4. c 6. c 7. a 8. b 9. d 11. a 12. b 13. c 14. a 16. b 17. d 18. d 19. c 21. a 22. b 23. d 24. c 26. b 27. a 28. a 29. b 31. a 32. b 33. a 34. c 36. c 37. b 38. c 39. c II. Fill in the blanks 1. 5 2. Zero 3. 150 cm 4. 6 5. Does not exist 6. Positive 7. Negative 8. 10 9. Cannot be determined 10. 3 11. 4 and 1 12. 5 13. 4 14. 6 III. Subjective questions 1. ₹15, ₹3 2. ₹10, ₹15 3. 36 4. 45 5. 𝑎𝑎𝑎𝑎 = 3 and 𝑏𝑏𝑏𝑏 = −1 6. ₹15, ₹24 7.

4 7

1 2

8. 𝑥𝑥𝑥𝑥 = , 𝑦𝑦𝑦𝑦 = 1

1 4

5. b 10. a 15. c 20. a 25. d 30. c 35. b 40. b

9. 𝑥𝑥𝑥𝑥 = − 24 , 𝑦𝑦𝑦𝑦 = 24

10. ₹72000, ₹1000 11. 966 12. Number of 25-paise coins is 8 and the number of 50-paise coins is 27. 13.

5 13

14. 𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑎𝑎3 , 𝑦𝑦𝑦𝑦 = 𝑏𝑏𝑏𝑏 3 15. 𝑥𝑥𝑥𝑥 = 1, 𝑦𝑦𝑦𝑦 = 3, 𝑧𝑧𝑧𝑧 = 5 16. 𝑥𝑥𝑥𝑥 = 4, 𝑦𝑦𝑦𝑦 = 5 17. 13000, 12000

281

ANSWER KEY 1 2

√3

18. � , 1�

19. 2 20. Ajit’s speed = 5 kmph, Amit’s speed = 7.5 kmph 21. Fare = ₹210, Reservation charge = ₹6 22. i) (6, 0, −3) ii) (1, −4, 2) 23. 8, 21, −3 24. 𝐴𝐴𝐴𝐴 = 25°, 𝐵𝐵𝐵𝐵 = 50°, 𝐶𝐶𝐶𝐶 = 105° 25. 𝑎𝑎𝑎𝑎 = 2.5, 𝑏𝑏𝑏𝑏 = 6 and 𝑐𝑐𝑐𝑐 = 2 43

26. 6

4. QUADRATIC EQUATIONS

Worksheet 1 I. Introduction to quadratic equations 1. i) Yes ii) No iii) Yes iv) Yes v) No vi) No II. Formation of a quadratic equation 1. i) 2𝑏𝑏𝑏𝑏 2 + 𝑏𝑏𝑏𝑏 – 528 = 0, 𝑙𝑙𝑙𝑙 = 33 & 𝑏𝑏𝑏𝑏 = 16 ii) 𝑥𝑥𝑥𝑥 2 − 8𝑥𝑥𝑥𝑥 − 1280 = 0, speed = 40 km/h iii) 𝑥𝑥𝑥𝑥 2 + 6𝑥𝑥𝑥𝑥 − 216 = 0, sides = 12 & 18 iv) 𝑥𝑥𝑥𝑥 2 − 7𝑥𝑥𝑥𝑥 − 60 = 0 2. 𝑥𝑥𝑥𝑥 2 + 4𝑥𝑥𝑥𝑥 – 320 = 0 3. 𝑥𝑥𝑥𝑥 2 + 3𝑥𝑥𝑥𝑥 – 300 = 0 III. Solving quadratic equations 1. i) 5, -2 ii) 1.5, -2 5√2

1 1 4 4 1 1 2 √3 v) , vi) − 3 , 4 10 10 √ 5 3 𝑎𝑎𝑎𝑎 2 𝑏𝑏𝑏𝑏2 viii) 2 , 2 vii) − , 2 2 1 1 ix) – 𝑎𝑎𝑎𝑎, −𝑏𝑏𝑏𝑏 x) − 2 , 2 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 −1+√33 −1−√33 i) 𝑥𝑥𝑥𝑥 = 4 , 4 √3 √3 ii) − 4 , − 4 −1±√−31 iii) 4 2𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 iv) , 𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎

iii) -√2, - 2

iv) ,

v) 1, √3 3. i) No real roots exist ii) 1, 1

282

iii) -2√3, − 2

iv) No real roots exist −2𝑏𝑏𝑏𝑏 −2𝑏𝑏𝑏𝑏 , 𝑎𝑎𝑎𝑎 3𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐 𝑏𝑏𝑏𝑏 vi) , − 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎 3±√13 i) 2 5 ii) 5, 2

iii) 2 ± 2√3 5 3 3 2

iv) − ,

v) √3, √2 vi) −8, 1 vii) 0 viii) ±1 ix) 4, −1

−1 ± √3𝑖𝑖𝑖𝑖 2 3 ± √5 −1 ± √3i xi) 2 , 2 −5±√39𝑖𝑖𝑖𝑖 xii) 1, -6, 2

x) −2 ± 2√3 , xiii) 3

5. 0, 1,

−1±√3𝑖𝑖𝑖𝑖 2

IV. Nature of roots of a quadratic equation 1. i) −23 ii) − 3 iii) −12 iv) 32 2. i) No real roots 3±√3 2 1 2 iii) Two distinct real roots, & − 2 3 4 5 iv) Two distinct real roots, − & − 3 2

ii) Two distinct real roots,

3. i) 4

4 iii) ± 3 1 v) − 2

4. 𝑝𝑝𝑝𝑝 = 7, 𝑘𝑘𝑘𝑘 =

ii) ±12

iv) 0, 3

7 4

5. NA 6. NA 7. Yes, 𝑙𝑙𝑙𝑙 = 40 m & 𝑏𝑏𝑏𝑏 = 20 m 8. Not possible 9. Yes, 𝑙𝑙𝑙𝑙 = 20 m & 𝑏𝑏𝑏𝑏 = 20 m 10. Yes, son’s age = 7 yr & man’s age = 49yr 11. The roots are equal and opposite in sign

ANSWER KEY 12. 𝑝𝑝𝑝𝑝 = 4 13. 𝑝𝑝𝑝𝑝 = 14 14. Two distinct real roots, 𝑥𝑥𝑥𝑥 = −1 & 2 V. Solving word problems 1. 13 and 14 2. 13 and 14 3. Base = 12 cm and Altitude = 5cm 4. No. of articles = 6 and cost of each article = ₹ 15 5. 8 and 4 6. 4, 5, and 6 7. (21 and 24) or (−21 and −24). 8. 𝑘𝑘𝑘𝑘 = 0 9. 𝑘𝑘𝑘𝑘 = −9 10. 𝑝𝑝𝑝𝑝, 𝑞𝑞𝑞𝑞 11. Present ages of boy and his brother are 18 years and 7 years respectively 12. Swati’s present age = 9 years and Varun’s present age = 27 years. 13. present age of son = 7 years and present age of man = 49 years 14. If math marks = 12, then English marks = 18. Or, if math marks = 13, then English marks = 17. 15. Shorter side = 60 m and Longer side = 90 m 16. 12 and 18 17. 40 km/hr 18. 1 hr 19. 50 20. 20 21. 44 km/hr 22. 18m and 12 m 23. 40 km/hr and 50km/hr 24. 36 25. ₹ 20 26. i) Son’s age 5 years ago = 5yrs ii) Woman’s age = 30 yrs 27.

1 4

VI. Symmetric functions & transformation of roots and quadratic expressions - graphs, maximum and minimum values 𝑏𝑏𝑏𝑏2 −2𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐 𝑎𝑎𝑎𝑎 2 −𝑏𝑏𝑏𝑏3 +3𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏𝑐𝑐𝑐𝑐 iii) 𝑎𝑎𝑎𝑎 3 𝑏𝑏𝑏𝑏4 −4ac𝑏𝑏𝑏𝑏2 v) 𝑎𝑎𝑎𝑎 2 𝑐𝑐𝑐𝑐 2 −𝑏𝑏𝑏𝑏3 +3𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 vii) 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 2 𝑏𝑏𝑏𝑏2 −2𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐 ix) (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)2 2

1. i)

ii) –

𝑏𝑏𝑏𝑏 𝑐𝑐𝑐𝑐

𝑐𝑐𝑐𝑐 2

v) 𝑎𝑎𝑎𝑎2

−𝑏𝑏𝑏𝑏3 𝑐𝑐𝑐𝑐 4 +3ab𝑐𝑐𝑐𝑐 5 𝑎𝑎𝑎𝑎 7 −𝑏𝑏𝑏𝑏3 +4𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 viii) 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 2

vi)

2. 𝑥𝑥𝑥𝑥 − 10𝑥𝑥𝑥𝑥 + 9 = 0 3. 𝑥𝑥𝑥𝑥 2 − 2𝑥𝑥𝑥𝑥 + 3 = 0 4. 9𝑥𝑥𝑥𝑥 2 − 28𝑥𝑥𝑥𝑥 + 3 = 0 5.

7 2

6. 1 Worksheet 2 I. Multiple choice questions with single correct answer 1. a 2. b 3. a 4. b 5. b 6. d 7. c 8. a 9. d 10. d 11. c 12. b 13. b 14. a 15. c 16. c 17. d 18. b 19. b 20. a 21. b 22. d 23. c 24. d 25. a 26. a 27. b 28. b 29. a 30. c 31. d 32. b 33. d 34. c 35. c 36. a 37. a 38. d 39. a 40. i) a ii) b iii) b 41. i) d ii) c II. Fill in the blanks 1. 3 2. 20 3. 2 and -2 4. -1 5. 6 6. 8 7. 3 and 9 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 𝑐𝑐𝑐𝑐 8. 𝑎𝑎𝑎𝑎1 = 𝑏𝑏𝑏𝑏1 = 𝑐𝑐𝑐𝑐1 2

−𝑏𝑏𝑏𝑏

9. 0, 𝑎𝑎𝑎𝑎 10. no 11. ±7 1 12. 3 13. 1 − 𝑟𝑟𝑟𝑟, 𝑟𝑟𝑟𝑟 + 1 14. -11

283

ANSWER KEY 𝑏𝑏𝑏𝑏2

15. 4𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏2

16. 𝑎𝑎𝑎𝑎 17. 6 18. 2 19. 𝑥𝑥𝑥𝑥 2 − 6𝑥𝑥𝑥𝑥 𝑥 11 = 0 20. 7 III. Subjective questions 1

1. 2 2. 𝑝𝑝𝑝𝑝 = 3, But 𝑝𝑝𝑝𝑝 𝑝 0 because if p = 0, then the given equation is not a quadratic equation. 3. 𝑥𝑥𝑥𝑥 2 − 11𝑥𝑥𝑥𝑥 + 8 = 0 4. 𝑥𝑥𝑥𝑥 2 + 𝑥𝑥𝑥𝑥 𝑥 360 = 0 5. no 6. 𝑏𝑏𝑏𝑏 2 > 4𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐 7. -8 8. -31 9. Real and distinct 10. 2

5. INEQUATIONS

Worksheet 1 I. Linear inequations 1. {1, 2, 3, 4} 2. {…, -1, 0, 1, 2, 3, 4} 3. No solution 4. {…, -5, -4, -3} 5. -1, 0, 1, 2, 3, … 6. 𝑥𝑥𝑥𝑥 𝑥 (−2, ∞) 7. (−∞, −3) 8. (−∞, −3] 9. (−∞, 6) 10. (−∞, 2] 11. 120 ≥ 𝑥𝑥𝑥𝑥 12. (4, ∞) 13. (−∞, 2] 14. (4, ∞) 15. (−∞, 2] II. Quadratic inequations 1. 𝑥𝑥𝑥𝑥 𝑥 2 or 𝑥𝑥𝑥𝑥 𝑥 3 2. 𝑥𝑥𝑥𝑥 =

1 2

3. −5 ≤ 𝑥𝑥𝑥𝑥 𝑥 𝑥1 4. −5 < 𝑥𝑥𝑥𝑥 < −2 5. −5 ≤ 𝑥𝑥𝑥𝑥 𝑥 2

284

6. -3, -2, -1, 0, 1, 2, 3 7. 𝑥𝑥𝑥𝑥 𝑥 (1,2) ∪ (3, ∞)

1 3

8. 𝑥𝑥𝑥𝑥 𝑥 (−∞, −4] ∪ � , 2�

9. b 10. d

11. 𝑥𝑥𝑥𝑥 𝑥 𝑥𝑥𝑥, � ∪ � � ∪ {21} 2 2,2 12. 𝑥𝑥𝑥𝑥 𝑥 (−∞, 0] ∪ (3, ∞) 13. c 14. 𝑥𝑥𝑥𝑥 𝑥 𝑥𝑥𝑥,

−3 4 � ∪ �0, 3� ∪ [4, ∞) 2

15. 𝑥𝑥𝑥𝑥 𝑥 (1,2) ∪ (2,3) ∪ (4, ∞) 16. 𝑥𝑥𝑥𝑥 𝑥 (−∞, 1] ∪ [2,4) ∪ (4,5) 17. 𝑥𝑥𝑥𝑥 𝑥 (1,2) ∪ (5, +∞) 18. a −2 −1 , � 3 2

19. 𝑥𝑥𝑥𝑥 𝑥 (−2, −1) ∪ �

III. Modulus inequations 1. {-4, 8} 2. d 3. d 4. c 5. b IV. Irrational inequations 1. d 2. a 3. b Worksheet 2 I. Multiple choice questions with single correct answer 1. c 2. b 3. d 4. c 5. a II. Subjective questions 1. {…, -4, -3, -2, -1, 0, 1} 2. 𝑥𝑥𝑥𝑥 𝑥 (−∞, 2) 3. (−2, ∞) 4. 𝑥𝑥𝑥𝑥 𝑥 [8, ∞) 5. 𝑥𝑥𝑥𝑥 𝑥 1 6. Solution set = {-2, -1, 0, 1, 2, 3} 7. Solution set = {𝑦𝑦𝑦𝑦: −2 ≤ 𝑦𝑦𝑦𝑦 < 4 and 𝑦𝑦𝑦𝑦 𝑦 𝑦𝑦𝑦𝑦} 8. Solution set = {1, 3, 5} 9. 35 marks

ANSWER KEY 10. 82 marks 11. 𝑥𝑥𝑥𝑥 𝑥 4 and 𝑥𝑥𝑥𝑥 𝑥 2 12. -4, -3, -2, -1 13. −10 < 𝑥𝑥𝑥𝑥 < −3 14. {-1} 8 15. �−4, � 3

6. ARITHMETIC AND GEOMETRIC PROGRESSIONS

Worksheet 1 I. nth term (or) general term of an AP 1. 12, 16, 20, 24, … 2. -25 3. 60 4. 1 5. No 6. -13, -8, -3 7. 28 8. 6 9. 𝐾𝐾𝐾𝐾 = 18 10. 10th term 11. 3𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑦𝑦, 4𝑥𝑥𝑥𝑥, 5𝑥𝑥𝑥𝑥 𝑥𝑥𝑥𝑥𝑥, 6𝑥𝑥𝑥𝑥 12. √112 13. 22 14. 𝑝𝑝𝑝𝑝 + 9𝑞𝑞𝑞𝑞 15. d II. Sum of first ′𝑛𝑛𝑛𝑛𝑛 terms of an AP 1. 90 2. 10𝑛𝑛𝑛𝑛 – 2 3. 18 terms and 19 terms 4. 8𝑛𝑛𝑛𝑛 – 3 5. 76 6. NA 7. 𝑥𝑥𝑥𝑥 = 35 8. 16 rows, 5 logs in the top row III. nth term (or) general term of a GP 1. 𝑞𝑞𝑞𝑞 2 = 𝑝𝑝𝑝𝑝𝑟𝑟𝑟𝑟 27 2. − 2

3. 0.2 4. G.P

3 4

4 3 3 4 3 = − : , −1, 4 3 4 4 3 4 = − : , −1, 3 4 3

5. 𝑟𝑟𝑟𝑟 = − or − For 𝑟𝑟𝑟𝑟

For 𝑟𝑟𝑟𝑟

6. 𝑎𝑎𝑎𝑎 7. 𝑏𝑏𝑏𝑏 IV. Sum of first ‘𝑛𝑛𝑛𝑛’ terms of a GP 1. 𝑎𝑎𝑎𝑎 2. 𝑎𝑎𝑎𝑎 3. 𝑎𝑎𝑎𝑎 4. 𝑛𝑛𝑛𝑛 = 10 5. 2046 6.

1 [1 − (0.1)20 ] 6

7. 𝑛𝑛𝑛𝑛 = 4 8.

16 (2𝑛𝑛𝑛𝑛 − 1)] 7 2

9. 𝑥𝑥𝑥𝑥 − 16𝑥𝑥𝑥𝑥 + 25 = 0 10. NA Worksheet 2 I. Fill in the blanks 1. 16 2. 23 3. 8 4. 5 5. 1 6. 3 7. -4 8. 11 9. 6 10. 28 11. 38 12. Triple 13. 0 14. ±2 15. -1 II. Multiple choice questions with single correct answer 1. a 2. b 3. d 4. c 5. c 6. b 7. c 8. a 9. d 10. b 11. a 12. a 13. a 14. c 15. d 16. d 17. a 18. c 19. c 20. d 21. a 22. a 23. d 24. c 25. b 26. c 27. c 28. c 29. d 30. b 285

ANSWER KEY 7. TRIANGLES

Worksheet 1 I. Introduction to triangles 1. Because all the angles in a square are right angles and all the sides are equal. 2. Not similar 3. Similar 4. Not similar II. Basic proportionality theorem 1. NA 2. NA 3. NA 4. 𝑥𝑥𝑥𝑥 = 4 5. NA 6. NA 7. NA 8. NA 9. NA 10. AC = 20 cm 11. 𝑥𝑥𝑥𝑥 = 3 III. Internal and external angle bisector theorem 1. 𝑥𝑥𝑥𝑥 = 16 2. 𝑥𝑥𝑥𝑥 = 2.5 3. NA 4. CE = 18 cm IV. Criteria for similarity of triangles 1. NA 2. 40° 3. NA 4. 160 cm 5. NA 6. 15 cm 7. 2.8 cm 8. NA 9. NA 10. AB = 6 cm, PQ = 2.4 cm 11.

16 3

14.

√5−1 cm 2

12. NA 13. √7 sq. units 15. NA

286

16. 8 and 12 units V. Area of similar triangles 1. 36:47 2. 21 cm2 3. 4:5 4. 9:7 5. 22 cm 6. 4:9 7. 1:4 8. 6 cm 9. 25:4 10.

2−√2 2

10.

3√3 𝑎𝑎𝑎𝑎 units 2

Worksheet 2 I. Multiple choice questions with single correct answer 1. a 2. d 3. b 4. c 5. c 6. d 7. b 8. a 9. c 10. b 11. d 12. a 13. c 14. a 15. b 16. b 17. b 18. b II. Assertion and reason 1. a 2. a III. Fill in the blanks 1. parallel 2. similar 3. EC 4. 4.5 cm 5. 5 m 6. congruent 7. 4:1 8. 1:4 9. 4:1 IV. Subjective questions 1. 2 units 2. By BPT 3. 𝑥𝑥𝑥𝑥 = 2 cm 4. AQ = 3 cm 5. 16 cm 6. 𝑥𝑥𝑥𝑥 = 3 cm 7. 110° 8. 13 m 9. 6 cm

ANSWER KEY 10. 15 cm

8. COORDINATE GEOMETRY Worksheet 1 I. Distance formula 1. �𝑥𝑥𝑥𝑥 2 + 𝑦𝑦𝑦𝑦 2

2. 2√2 units

3. √34 units 4. 𝑦𝑦𝑦𝑦 = 7 5. Isosceles triangle 6. |𝐴𝐴𝐴𝐴𝐵𝐵𝐵𝐵 ± 𝐵𝐵𝐵𝐵𝐶𝐶𝐶𝐶| = 𝐴𝐴𝐴𝐴𝐶𝐶𝐶𝐶 (or) ar(∆𝐴𝐴𝐴𝐴𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵) = 0 7. (−6, 7) II. Section formula 1. (1,2), (2,4), (3,5) 2. (10,15), (−2, −3), (8,11) 3. (-1, 3.5), (0, 5), (1, 6.5) 4. 13 sq. units 5. 1:1, (-1.5, 0) III. Area of triangles and quadrilaterals 1. 28 sq. units 2. NA 3. NA IV. Equation of a straight line in various forms 1. 𝑦𝑦𝑦𝑦 = 5𝑥𝑥𝑥𝑥 2. 𝑦𝑦𝑦𝑦 = 3𝑥𝑥𝑥𝑥 − 9 3. 𝑦𝑦𝑦𝑦 = 2𝑥𝑥𝑥𝑥 + 4 4. 𝑦𝑦𝑦𝑦 = 2𝑥𝑥𝑥𝑥 − 2 5. 7𝑥𝑥𝑥𝑥 – 4𝑦𝑦𝑦𝑦 = 28 6. 𝑥𝑥𝑥𝑥 + 2𝑦𝑦𝑦𝑦 – 2 = 0 7. 𝑦𝑦𝑦𝑦 = − 4 8. 𝑥𝑥𝑥𝑥 = 3 9. 𝑦𝑦𝑦𝑦 = √3𝑥𝑥𝑥𝑥 + 5 10. 𝑦𝑦𝑦𝑦 = 𝑥𝑥𝑥𝑥 – 2 1

11. �1, 3� √

12. 2𝑥𝑥𝑥𝑥 + 3𝑦𝑦𝑦𝑦 – 6 = 0 or 3𝑥𝑥𝑥𝑥 + 2𝑦𝑦𝑦𝑦 – 6 = 0 13. 𝑦𝑦𝑦𝑦 = 5(𝑥𝑥𝑥𝑥 − 3) 14. 𝑦𝑦𝑦𝑦 = 5 15. 𝑘𝑘𝑘𝑘 = 1 16. 𝑥𝑥𝑥𝑥 – 𝑦𝑦𝑦𝑦 + 5 = 0 17. 𝑏𝑏𝑏𝑏𝑥𝑥𝑥𝑥 + 𝑎𝑎𝑎𝑎𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 𝜋𝜋𝜋𝜋 18. 4

19. 𝑥𝑥𝑥𝑥 + 3 = 0

20. 𝑥𝑥𝑥𝑥 = 10 21. 2𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑦𝑦 – 4 = 0 22. 𝑥𝑥𝑥𝑥 – 𝑦𝑦𝑦𝑦 + 1 = 0 23. 𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑦𝑦 – 1 = 0 24. 𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑦𝑦 – 1 = 0 25. 𝑦𝑦𝑦𝑦 = 5 Worksheet 2 I. Multiple choice questions with single correct answer 1. b 2. b 3. c 4. a 5. c 6. d 7. c 8. c 9. b 10. b 11. d 12. b 13. c 14. b 15. d 16. d 17. b 18. d 19. d 20. c 21. a 22. d 23. b 24. d 25. c 26. b 27. d 28. a 29. b 30. b 31. c 32. d 33. b 34. d 35. d 36. c 37. b 38. a 39. d 40. a II. Fill in the blanks 1.

1 sq. unit 2

2. 2:7 3. 2:3 4. (16,8) 5. 0 6. 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑖𝑖𝑖𝑖𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 + 𝑦𝑦𝑦𝑦𝑐𝑐𝑐𝑐𝑦𝑦𝑦𝑦𝑥𝑥𝑥𝑥𝜃𝜃𝜃𝜃 = 𝑎𝑎𝑎𝑎2 7. 𝑦𝑦𝑦𝑦(𝑡𝑡𝑡𝑡1 + 𝑡𝑡𝑡𝑡2 ) = 2𝑥𝑥𝑥𝑥 + 2𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎1 𝑡𝑡𝑡𝑡2 8. 𝑎𝑎𝑎𝑎𝑥𝑥𝑥𝑥 + 𝑏𝑏𝑏𝑏𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑎𝑎2 + 𝑏𝑏𝑏𝑏 2 9. 𝑝𝑝𝑝𝑝 = 1 10. Equilateral

9. TRIGONOMETRY

Worksheet 1 I. Trigonometric ratios 1. a 2. c 3. c 4. a 5. b 6. a 7. a 8. a 9. c 10. c 287

ANSWER KEY 11.

625 168

12. NA II. Trigonometric ratios for compound angles 1. c 1−𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 2. cot 2𝐴𝐴𝐴𝐴 = 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

3. b 4. 0 5. 5 6. NA 7. NA 8. NA 1 9. 5 III. Multiple and submultiple angles 1. NA 2cos𝑥𝑥𝑥𝑥 2. √cos 2𝑥𝑥𝑥𝑥

3. NA 4. c 5. c 6. c 7. b 8. 2 9. 4 10. 1 11. NA 12. NA 13. 1 IV. Heights and distances 1. b 2. 60° 3. c 4. b 5. 30° 6. 60° 7. 25√3 8. NA 9. 5√3 �√3 + 1� 𝑚𝑚𝑚𝑚 10. NA 11. 8 m Worksheet 2 I. Multiple choice questions with single correct answer 1. b 2. c 3. c 4. d 5. d 6. d 7. b 8. b 9. c 10. c 11. d 12. c 13. d

288

II. Fill in the blanks 1. sin 𝐴𝐴𝐴𝐴 =

15 17

2. √3 3. 1 4. 1 𝜋𝜋𝜋𝜋 5. 3 6. 1 2 𝑥𝑥𝑥𝑥 + 1 7. 𝑥𝑥𝑥𝑥 2 − sin 2𝐴𝐴𝐴𝐴 III. Match the following 1. d IV. Subjective questions 1. 𝐵𝐵𝐵𝐵𝐶𝐶𝐶𝐶 = 5√3 𝑐𝑐𝑐𝑐𝑚𝑚𝑚𝑚, 𝐴𝐴𝐴𝐴𝐶𝐶𝐶𝐶 = 10𝑐𝑐𝑐𝑐𝑚𝑚𝑚𝑚 2. ∠A = 22° 3. 1 4. 1 5. NA 6. NA 7. NA 8. 2 −1 9. 3 √

10. 40° 11. 3 12. NA 13. NA 14. 50 m 15. 150 m 16. 45° 17. NA