20_JEE_IL-Achivers_Maths_Functions_V2

20

CHAPTER

FUNCTIONS Chapter Outline 20.1 Relation as a Function 20.2 Real Valued Functions 20.3 Methods of Finding Domain and Range of a Function

A relation R = {(a, b): a∈A, b∈B} is a function if: ■ Each element of A has a unique image in B (a,b)∈R for all a ∈A. ■ No element in A has multiple images if (a,b)∈R and (a, c)∈R, then b = c.

20.4 Algebra of Functions

20.1.1 Image, Pre-image, Value of a Function

20.5 Graphical Transformations

■ Image: y is the image of x under f, where (x,y)∈f.

20.6 Kinds of Functions 20.7 Composition of a Function 20.8 Inverse of a Function ■ A function is a rule that assigns each input (from a set called the domain) to exactly one output (in the range).

■

Pre-image: x is the pre-image of y under f.

Function Components: ■ Domain: A (set of all inputs) ■ Co-domain: B (set where outputs belong) ■ Range: { f ( a ) : a ∈ A} , subset of B.

■ It is commonly written as f(x), where x is the input, and f(x)is the output.

Function Value at x = a y = f(x) ⇒ f(a)= y

■ Functions help describe relationships between variables in mathematics, science, and real-world applications.

Solved example

1. If a function f : A→B is defined as f(x) = x2, then find the range of f.

■ Types of functions include polynomial, rational, trigonometric, exponential, and logarithmic.

Sol. Since, the given relation is square function, it gives only positive numbers. So, the range of f(x) = x2 is set of all non-negative real numbers.

■ They can be represented using equations, graphs, tables, or mappings.

Try yourself:

A function f from a set A to a set B associates each element of A to a unique element of B. ■ Denoted as: f : A→B (Read as: f is mapping from A to B) ■ Every function is a relation, but not every relation a function.

f (1.1) − f (1) 1.1 − 1

.

Ans: 2.1

20.1 RELATION AS A FUNCTION

1. If f(x) = x2, find the value of

20.1.2 Check Whether the Given Relation Is a Function To verify from ordered pairs: ■ Domain check: First elements in ordered pairs must cover all of A.

2

CHAPTER 20: Functions

■ Uniqueness: No two ordered pairs should have the same first element.

■ No element in A should have multiple outgoing arrows.

2. If X = {1,2,3,4} and Y = {1,5,9,11,15,16}, determine which of the following are functions from X to Y. f1 = {(1,1) , ( 2,11) , ( 3,1) , ( 4,15 )}

f 2 = {(1,1) , ( 2,9 ) , ( 3,5 )}

f 3 = {(1,5 ) , ( 2,9 ) , ( 3,3 ) , ( 4,5 ) , ( 2,11)}

Sol. The relation f 1 is a function because all the elements of X have images and each element of X has a unique image.

1 2 3 4

Sol. Observe the first arrow diagram, 3∈A has no image in B. So, the relation f1 is not a function.

Observe the second arrow diagram, all the elements of A have unique images, but not 4.

So, the relation f2 is not a function.

■ If any vertical line cuts the continuous graph of relation at only one point, then the relation is a function.

Solved example

4. Check whether or not the relation defined as f(x) = x2 on the set of real numbers is a function.

Sol. Draw the graph of f(x) = x2.

y

The relation f2 is not a function because 4∈X has no image.

8

The relation f3 is not a function because 2∈X has two images.

6

■ When relation is given in Venn diagram, check the following to decide whether or not it is a function.

2

The arrows initiate from every element of first set and terminate at any element of second set.

Only one arrow must initiate from each element of the first set.

Solved example

3. Check whether the following relations are functions or not and give reason.

B a b c d e

f2

f(x)=x2

4

-4

-2

2

4

x

Any vertical line intersects the above graph at most at one point. So, the relation is a function.

Try yourself: 2. Check whether the relation defined as f(x) = x3 on the set of real numbers is a function or not.

Ans: Yes

Solved example

a b c d e

A

From a Graph: ■ Vertical Line Test: If a vertical line intersects the graph at more than one point, it is not a function.

B

1 2 3 4

From a Venn diagram: ■ Arrows must originate from every element in A.

f1

A

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20.1.3 Number of Functions

Try yourself:

For two finite sets Aand B

4. What are the restrictions for which the functions f(x) = sec2x– tan2x and g(x) = 1 are equal?

Ans: Domain must restrict.

If n(A) = p and n(B) = q, then the total number of functions from A to B is: qp.

(Each element in A has q choices in B, so total possibilities =qp.)

Solved example

5. What is the number of functions defined on the set A = {1,2,3,4,5}?

TEST YOURSELF 1.

Let A be the set of n elements. Then the number of functions defined from the set A to the set A is (1) n2 (2) nn (3) 2n (4) n!

2.

What is the main difference between a relation and a function?

Sol. Given, set A = {1,2,3,4,5} Here, n(A) = 5, so the number of functions defined on the set A is 55 = 3125

(1) A relation may have multiple outputs for a given input, while a function has only one output for each input.

Try yourself: 3. What is the number of functions defined on the set A = {1,2,3,4}?

(2) A function may have multiple outputs for a given input, while a relation has only one output for each input.

Ans: 44 = 256

20.1.4 Equality of Two Functions Two functions f, and g are equal if: ■ Same domain: Dom (f) = Dom (g) for all x in the domain.

■ For every element x in the domain, f(x) = g(x)

(3) A relation is a set of ordered pairs, while a function is a set of numbers.

(4) A relation is a one-to-one mapping, while a function is a many-to-one mapping.

3.

(1) {(–2, 3), (1, 5), (–2, 7)}

Solved example

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

6. Find the domain for which the function f(x) =

(3) {(1, 2), (2, 4), (3, 6)}

2x2–1 and g(x) = 1 + 3x are equal.

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

Sol. Given, f(x) = 2x2–1 and g(x) = 1+3x Suppose that f(x) = g(x)

2 x 2 − 1 = 1 + 3x

4.

2 x 2 − 3x − 2 = 0 2x 2 − 4 x + x − 2 = 0

2 x ( x − 2 ) + 1( x − 2 ) = 0

0 ( 2x + 1)( x − 2 ) =

1

⇒ x = − ,2 2 1  ∴ the domain of both functions is  − ,2  .  2 

Which of following relations is a function?

5.

y y  xy , then f(m, n) + If f  2 x + ,2 x −  = 8 8  f(n, m) = 0 (1) only when m = n (2) only when m ≠ n (3) only when m = –n (4) for all m and n If g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by the formula g(x) = ax + b, then (a, b)= (1) (2, 1)

(2) (–2, 1)

4

CHAPTER 20: Functions (3) (–2, –1)

Try yourself:

If f(x) = x2 – 2x + 4, then the set of values of x satisfying f(x –1) = f(x +1) is

6.

(1) {–1}

(2) {–1, 1}

(3) {1}

(4) {1, 2}

(1) 2 (5) 4

Answer Key (2) 1 (3) 3 (6) 3

(4) 4

20.2 REAL VALUED FUNCTIONS

5. Let f = {(–1,–3),(0,–1),(1,1),(2,3)} be a linear function defined on the set of all real numbers. Then, find f(x).

Ans: 2x – 1

(4) (2, –1)

20.2.2 Constant Function ■ Defined as f(x) = k for all x∈R. ■ Graph: Horizontal line at y = k.

■ Range: {k} (singleton set)

■ A function f: A→ B is a real-valued function if B is subset of R. If both A and R are subsets of R, then f is a real function.

y

f(x) = k

Types of real functions:

■ Transcendental functions: Functions that are not algebraic, e.g., exponential, logarithmic, trigonometric, and inverse trigonometric functions.

20.2.1 Polynomial Functions ■ A function f: R→ R is a polynomial function of degree ■

f ( x= ) a0 x n + a1x n−1 + a2 x n−2 + ... + an ,where a0 ≠ 0 and and n is a non-negative integer.

■ Domain: R ■ Example:

Linear: f(x) = ax + b

Quadratic: f(x) = ax2 + bx + c

■ The domain of polynomial function is R, the range of odd degree polynomial functions is R, and the range of even degree polynomial function is set of all non-negative real numbers.

(0,k)

x -4

-3

-2

-1

1

2

3

4

Solved example

7.Describe the graph of the function f(x) = 10.

Sol. The graph of f(x) = 10 is a straight line parallel to x-axis and above the x- axis at a distance of 10 units from it.

Try yourself: 6. What is the domain and range of a function f(x) = sin2 x + cos2 x for all x∈R? Ans: Domain is R and range is {1}

■ Algebraic functions: Functions obtained using a finite number of fundamental operations (addition, subtraction, multiplication, division, root extraction) on polynomial functions.

20.2.3 Identity Function ■ Defined as f(x) = x for all x∈R. ■ Graph: Line y = x bisecting the first and third quadrants.

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■ Domain and Range: R

y

y

4

4

y=x

3

3 2

2 1 -4

-3

-2

-1

1

-1

2

3

4

1

x

-4 -3 -2 -1

-2 -3

y=|x| x 0 1

2

3

4

Properties of modulus function:

-4

For any real number x, Solved example

■

x < a ⇔ −a < x < a

8. If the graph of a function is a straight line passing through the origin and making an angle of 45° with the x–axis, then what is the type of the function?

■

x ≤ a ⇔ −a ≤ x ≤ a

■

x > a ⇔ x < −a or x > a

■

x ≥ a ⇔ x ≤ −a or x ≥ a

■

a < x < b ⇔ x ∈ ( −b, −a ) ∪ ( a, b )

■

a ≤ x ≤ b ⇔ x ∈ −b, −a  ∪ a, b 

Sol. It is an identity function.

Try yourself: 7. What is the graph of f(x) = –x? Find its domain and range.

-4

-3

-2 -1

-1 -2 -3 -4 4 3 2 1

y= –x 1

2

3

4 x

y

Ans: Domain and range both are equal to R.

20.2.4 Modulus Function

For any two real numbers x, y, ■

x + y = x + y ; here, both x,y are of same sign

■

x−y ≤ x + y

■

x+y ≤ x + y

■

x+y ≥ x − y Solved example

9. Write the function g(x) = x–|x|, as a piecewise function.

 x if x ≥ 0 ■ Defined as f(x) = |x|: f ( x ) =  − x if x < 0

Sol. The given function is g(x) = x–|x|

■ Graph: y = x for x > 0, y = –x for x < 0.

When x < 0, g(x) = x+x = 2x

■ Domain: R ■ Range: [0,∞)

When x > 0, g(x) = x–x = 0 When x = 0, g(x) = 0 0 2 x

Therefore, g ( x ) = 

if x ≥ 0 if x < 0

6

CHAPTER 20: Functions

Try yourself:

cos 2 x for x∈(0,π)?

Ans: cos x ,0 < x <

2

π

. − cos x ,

2

π

Solved example

10. If x lies in the first quadrant, then what is the value of [sin x]?

< x <π

20.2.5 Greatest Integer Function

Sol. The value of sin x is between 0 and 1, when x

■ Defined as f(x) = [x], which gives the greatest intefer < x.

■ Graph: Step function

lies in the first quadrant. Hence, [sin x] = 0.

11. What are the values of [3.4],[2],[–2.5],[–5]?

Sol. The value of [3.4] is greatest integer, which is

■ Domain: R

less than 3.4, so [3.4] = 3 and [2] = 2.

■ Range: Z

The value of [–2.5] is greatest integer, which is less than –2.5, so [–2.5] = –3 and [–5] = –5.

Y 4 3 2 1

Y = [x] X’

1  2   n −1 x  +  x +  +  x +  + ... +  x + n   n  n  = nx 

-4 -3 -2 -1

-1 1 2 3 4 -2 -3 -4

Try yourself: 9. For a real number x = 0.5, what is the value of

X

Y’

1   2  2023    + x+ + ... +  x + ? x  +  x + 2024   2024  2024   

Ans: 1012

8. What is

v.

20.2.6 Least Integer Function

Properties of greatest integer function:

■ Defined as f(x) = (x), which gives the least interger > x.

■ If n is any integer, then [–n] = –n.

■ Graph: Step function

■ If x is any real number, [x+k] = [x]+k, where k is integer.

■ Domain: R

■ If x∉Z, then [–x] = –[x]–1 and

[x] + [–x] = –1

[x] – [–x] = 2[x]+1

■ Range: Z

Y 4 3 2 1

y = (x)

■ If x∈Z, and k∈Z then,

[x] ≥ k ⇔ x ≥ k

[x] ≤ k ⇔ x < k+1

[x] < k ⇔ x < k

 x   x + 1  x + 2   x + n −1  n  +  n  +  n  + ... +  n          = x 

X’

-4 -3 -2 -1

-1 1 2 3 4 -2 -3 -4 Y’

X

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Properties of least integer function:

■ Domain: R

■ If x is an integer, then

■ Range: [0, 1)

(x) = x

(x) + (–x) = 0

(x) – (–x) = 2(x)

Y y = {x}

■ If x is not an integer, then

(x) = (x)+1

(x) + (–x) = 1

(x) – (–x) = 2(x)–1

 x)  (=

x   x −1  +  2  2 

n −1  1 = (nx) + ( x ) +  x +  + ... +  x + n n     n–1

-2

-1

0

1

■ If x is an integer, then

{x} + {–x} = 0

{x} = 0

■ If x is not an integer, then {[x]} = 0

Solved example

[{x}] = 0

12. Find the value of k, if (–2.3)k – 2k + (8.34) k = (4.2.3) + (2.23), where (.) represents the least integer function.

{x} = x, when x∈(0,1]

⇒ – 2k – 2k + 9k = 5 + 3 ⇒ 5k = 8 ⇒ k = 8/5

Try yourself:

Solved example

13. What is the value of {3.45}?

Sol. We know that {x} = x – [x].

= {3.45 } 3.45 −3.45 = 3.45 − 3 = 0.45

Try yourself: 11. What is the value of {[7]}?

Ans: 0

(–2.3)k – 2k + (8.34)k = (4.23) + (2.23)

10. Prove that

 n + 2n−1   n +1   n + 2   n + 4  + + + + ...    2   4   8  n        2  = 2n − 1.

20.2.7 Fractional Part Function

X

3

Properties of fractional part function:

Sol. Given:

2

20.2.8 Signum Function −1 

if x < 0

1 

if x > 0

0 if x 0 ■ Defined = as: sgn ( x ) =

■ Graph: Three distinct horizontal segments.

■ Defined as f(x) = x–[x].

■ Domain: R

■ Graph: Periodic between 0 and 1.

■ Range: {–1, 0, 1}

8

CHAPTER 20: Functions Y

y

x

y= ( 13 (

1

x

y = ( 12 (

4 3

X

2 1

–1

-3

-2

0

-1

1

x

Solved example

20.2.10 Logarithmic Function

14. What is the value of f(–3.002), where f(x) is a signum function?

■ Defined as f(x) = logax, where x > 0, a > 1.

Sol. We know that if x is negative, the signum function gives –1.

■ Range: R

Hence, f(–3.002) = –1

■ Domain: (0,∞) ■ Graph: Mirroe image of ax w.r.t. y = x.

Properties:

Try yourself: 12. What is the value of f({–3.45}), where f is

x log a ( xy ) = log a x + log a y log a   = log a x − log a y y

signum function and { } is fractional part

y

function?

y = log2x

2

Ans: 1

1

20.2.9 Exponential Function

y = log3x

0

■ Defined as f(x) = ax, where a > 0, a ≠ 1.

1

2

x

3

-1

■ Domain: R y

■ Range: (0,∞)

3

■ Graph: Increasing if a > 1, decreasing if 0

2

< a < 1.

1

4

0 -1 -2

y

y=3

x

y = 2x

2 1 X 1

2

1

2

3

3 y=log 1 x (3( y=log (12( x

-3

3

0

x

20.2.11 Reciprocal Function 1 x

■ Defined as f ( x ) = ,

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■ Domain: R–{0}.

y

■ Range: R–{0}.

4

■ Graph: Hyperbola in first and third quadrants. y

3

4 3 2 1 -4 -3

-2

-1

y= 1 x

-1 -2 -3 -4

y=√x

2 1

x 0

x 0 1

2

3

4

1

■ Domain and Range: R. ■ Graph: Similar sign as x. y 4 3 2 1 -4 -3

■ Domain: R. ■ Range: [0, ∞).

■ Graph: Parabola.

-1 -2 -3 -4

-2

-1

-1 -2 -3 -4

y=x3 x 0 1

2

3

4

3 Cube root function: f ( x ) = x

y 4 3 2 1 -1

4

■ Cube function: f(x) = x3

■ Square function: f(x) = x2

-2

3

20.2.13 Cube Function and Cube Root Function

20.2.12 Square Function and Square Root Function

-4 -3

2

■ Domain and Range: R. y=x2 x 0 1

2

3

Square root function: f ( x ) = x ■ Domain: [0, ∞).

4

■ Graph: Increasing function. y 4 3 3 y=√x 2 1 -4 -3 -2 -1 -1 0 1 2 3 -2 -3 -4

x 4

■ Range: [0, ∞).

20.2.14 Explicit and Implicit Functions

■ Graph: Increasing curve.

■ A function in which a dependent variable can be expressed in terms of independent

10

CHAPTER 20: Functions

■ If the function cannot be expressed directly in terms of independent variable, then the function is said to be implicit function.

The implicit function can be represented as f(x,y) = 0.

■ Example: y = sinx + logx is an explicit function, whereas ax2 + 2hxy + by2 = 0 is an implicit function.

20.2.15 Piece-wise Defined Functions ■ A function f(x) is defined differently on disjoint subsets of its domain.

The piece-wise definition of modulus function f(x) = |x| is if x ≥ 0 if x < 0

 x f (x) =  − x

x + 4 x ≤ −2  f = x x − <x≤2 2 13. Draw the graph of ( )   x >2  x

20.2.16 Rational Function

Example:

p

f(ax + b) is periodic with period a .

Solved example

15. What is the period of the function f(x) = x– [x–b], b∈R where [.] represents the greatest integer function?

Sol. Given: f(x) = x –[x – b], b∈R can be written as f(x) = x – (x – b – {x – b}), b∈R = b + {x – b} ∴ the period of the function f(x) is 1.

Try yourself: 14. What is the period of the function f(x) = sinx + cosx?

20.2.18 Even and Odd Functions ■ A function f : A → R is

Even Function ■ A function is even if:

Try yourself:

A function of the form

Ans: 2π

variables is called explicit function. The explicit function can be represented as y = f(x).

f (x)

g (x)

f(–x) = f(x), ∀x∈A

Properties: ■ The graph of an even function is symmetric about y-axis. y

, where g(x) ≠ 0.

2 x 2 + 4 x + 7 x 2 + 3x + 4 , 3 3x + 2 2 x + 5x 2 + 7

20.2.17 Periodic Function A function f is periodic if there exists p such that f(x + p) = f(x) for all x.

■ Example: sin x, cos x ■ Properties:

f(x+kp) = f(x) for integer k.

kf(x) is periodic with period p.

O

x

■ If (x, y) is a point on the graph of even function, then (–x, y) is also on the graph.

■ Every constant function is an even function.

Odd Function ■ A function is odd if f ( −x ) = − f ( x ) ∀x ∈ A .

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Properties:

Try yourself:

■ The graph of an odd function is symmetric about the origin, and it is symmetrical in opposite quadrants. y

) 15. The function f ( x=

x x + + 1 on R e −1 2 x

– {0} is ______ function.

Ans: even

TEST YOURSELF O

x

1.

Given, the function f ( x ) =

ax + a−x (where 2

a > 2). Then, f(x + y)+f(x − y) = (1) 2f(x). f(y)

■ If (x, y) is a point on the graph of odd

(2) f(x). f(y)

function, then (–x, –y) is also a point on

(3)

the graph.

f ( y)

(4) f(x) – f(y)

Additional Properties ■ The even power of an even or odd function is always an even function.

f (x)

2.

If a function f(x) is given by f (x ) = then, the sum of the series

■ Every function can be uniquely expressed as a sum of an even function and an odd function:

 1   2   3   39  f   + f   + f   +…. + f    20   20   20   20 

is equal to

f(x) = g(x) + h(x)

where g(x) is even and h(x) is odd.

(1) 29 2

Solved example

16. If f (x)and g(x) are two functions, with

(2)

all real numbers as their domains, then show that

49 2

(3)

19

2

(4)

39 2

h(x) = { f(x) + f(–x)} { g(x) – g(–x)} is an odd function.

Sol: Given: h(x) = { f(x) + f(–x)} { g(x) – g(–x)}

3.

Suppose f :[−2,2]→ is defined by −1 for − 2 ≤ x ≤ 0 f (x) =   x − 1 for 0 ≤ x ≤ 2

Consider,

h ( − x ) = { f ( − x ) + f ( x )}{ g ( − x ) − g ( x )}

= − { f ( x ) + f ( − x )}{ g ( x ) − g ( − x )} = −h ( x )

Therefore, the function h(x) is odd function

5x , 5 +5 x

Then, {x∈[−2, 2] : x ≤ 0 and f(|x|)= x} = (1) {–1}

(2) {0}

 −1   2

(4) ϕ

(3) 

12 4.

CHAPTER 20: Functions If f ( a ) = log

2+a for 0 < a < 2, then 2−a

1  8a  = f 2  4 + a 2 

(1) f(a)

A function’s domain comprises all valid input values, while its range encompasses all possible output values.

(2) 2f(a)

20.3.1 Finding Domain of a Function

1 f (a) (3) 2 (4) – f(a) 5.

20.3 METHODS OF FINDING DOMAIN AND RANGE OF A FUNCTION

If f ( x ) =

Definition: The domain of a function is the set of all elements where the function is well-defined.

x

9 , then 9x + 3

Rules for finding the domain

 1   2   1995  f +f +……….. + f     = ■ Algebraic functions:  1996   1996   1996  Denominator ≠ 0.  2   1995  _______. +……… .. + f =  1996   1996  Expression under an even root must     be > 0.

10 + x f (x) Logarithmic expressions must be > 0. , x ∈ ( −10, 10 ) and e 6. If= 10 − x ■ Trigonometric functions:  200 x  f ( x )= k ⋅ f   2 , then k = ______. sin x, cos x: Defined for all real x.  100 + x  7.

8.

Number of solutions of the equation 2[x]= x + {x} is_____. (where([⋅] is greatest integer function, {⋅} is fractional part function).

If f(x) is an even function and g(x) is an odd function, then f(x) ⋅ g(x) is____ function.

(1) even (2) odd (3) neither even nor odd (4) either even or odd

9.

(3) 3

(5) 997.5 (6) 0.5 (7) 2 (8) 2

(9) 1

csc x, cot x: Defined for x ≠ nπ.

■ Logarithmic Functions:

Answer Key (2) 4

tan x, sec x: Defined for x ≠ (2n+1)π/2.

(4)

log­a(f(x)): Defined for f(x) > 0. loga(g(x)/h(x)): Defined for g(x) > 0, h(x) > 0.

■ Wavy curve method (for inequalities):

If f(x) = ax 5 + bx 3 + cx + d is an odd function, then d = (1) 0 (2) 1 (3) –1 (4) Any real number (1) 1

1

Solve f(x) < 0, f(x) > 0, f(x) ≤ 0, or f(x) ≥ 0.

Mark roots (where f(x) = 0) with black dots.

Mark discontinuities with circles.

Draw a curve, alternating sign at odd powers, retaining the same sign at even powers.

Include only intervals satisfying the given inequality.

IL ACHIEVER SERIES FOR JEE MATHEMATICS

13

Solved example

17. L e t f ( x ) = (

Methods to Find Range

x − 1) ( x + 2 ) ( x − 3 ) ( x + 6 ) 3

4

5

x ( x − 7) 2

3

.

f (x) .

Then, find the domain of

Sol. Use wavy curve method or interval method.

First, keep the numbers –6,–2,0,1,3,7 on the number line, in which 0,7 are in hollow circle, as shown in the following figure.

Substitute x = –10, which is left to the least number.

1. Range of Basic Functions ■ Sum Functions: Use AM ≥ GM ≥ HM for positive quantities.

■ Quadratic Functions: If a > 0, minimum value is 4ac - b² / 4a.

If a < 0, maximum value is 4ac - b² / 4a.

■ Trigonometric Functions:

We get,

( −11) ⋅ ( −9 ) ⋅ ( −13) ⋅ ( −4 ) is +ve f ( −10 ) = 2 3 ( −10 ) ( −17 ) 3

4

5

Start the curve above the number line before –6 and cross at –6, touch at –2. Since it has even power, draw the curve below the number line and touch zero. Since it has even power, draw the curve below the number line only and touch 1, cross the curve at 1, cross the curve at 3, and cross the curve at 7, as shown below.

2 2 2 2 a cos x + b sin x:  − a + b , a + b    2 a2 sec² x + b2 csc² x: ( a + b ) , ∞ a2 tan² x + b2 cot² x: 2ab, ∞ )

)

2. Range of Rational Functions ■ For y = P(x)/Q(x): ■ Find critical points from P(x) = 0 and Q(x) = 0.

■ Analyze limits as x → ±∞.

3. Strictly Increasing/Decreasing Functions ■ If f(x) is increasing in [a, b],

–6

–2

0

1

1

3

7

f (x ) is

Domain of

(–∞, –6] ∪ [1, 3] ∪ (7, ∞)∪ {–2}.

Try yourself: 16. What is the solution of the inequality (x − 3)(x + 2)(x + 5) >0? (x + 1)(x − 7) Ans: (–5, –2)∪(–1, 3) ∪ (7, ∞)

20.3.2 Finding Range of a Function Definition: ■ The range is the set of all possible values of f(x).

Range = [f(a), f(b)].

■ If f(x) is decreasing Range = [f(b), f(a)].

4. Using Derivatives ■ Find f’(x) and identify critical points.

■ Determine max/min values to get the range.

Solved example

18. Find the range of the function f ( x ) =

x2 − x +1 . x2 + x +1

Sol. Let y =

x2 − x +1 x2 + x +1

⇒ yx2 + xy + y = x2 – x +1 ⇒ (y – 1)x2 + (y + 1)x + y – 1 = 0

14

CHAPTER 20: Functions Now, D ≥ 0 ∀ x ∈ R

Try yourself:

⇒ (y + 1) – 4(y – 1) ≥ 0 2

17. F i n d t h e r a n g e o f t h e f u n c t i o n x 2 + 2x + 3 f (x) = x

⇒ 3y2 – 9y – y + 3 ≤ 0

)

⇒ 3y(y – 3) – 1(y – 3) ≤ 0 ⇒ (3y – 1)(y – 3) ≤ 0

(

⇒ 3y – 10y + 3 ≤ 0 2

Ans: −∞,2 − 2 3  ∪ 2 + 2 3, ∞  

2

20.3.3 Domain and Range of Some Functions

⇒ y∈ [1/3,3]

■ The following table shows the domain and range of some standard functions.

∴ Range is [1/3, 3]

S. No.

Function

Domain

Range

1.

logax (a ≠1; a > 0)

R+ = (0, ∞)

R = (– ∞, ∞)

2.

ax (a > 0)

R

R+ = (0, ∞)

3.

[x]

R

Z

4.

{x} = x – [x]

R

[0, 1)

5.

|x|

R

[0, ∞)

6.

x

R+ ∪{0}

[0, ∞)

7.

sin x

R

[–1, 1]

8.

cos x

R

[–1, 1]

9.

tan x

π   R − (2π + 1) , n ∈ Z  2  

R

10.

cot x

R – {nπ, n∈Z}

R

11.

sec x

π   R − (2π + 1) , n ∈ Z  2  

(–∞, –1]∪ [1, ∞)

12.

cosec x

R – {nπ, n∈Z}

(–∞, –1]∪ [1, ∞)

TEST YOURSELF 1.

The range of the function f (x) =

3 − x + 2 + x is

(1)  5, 10 

(2) 2 2, 11 

(3)  5, 13   

(4)  2, 7   

IL ACHIEVER SERIES FOR JEE MATHEMATICS

3.

(1) (0, ∞)

(2) (−∞, 0)

Let A ={1, 2}, B ={3, 6} and f : A→B given by f(x) = x 2 + 2 and g : A → B given by g(x) = 3x. Then, the functions f(x) and g(x) are

(3) (−∞, ∞)−{0}

(4) (−∞, ∞)

(1) equal

(2) not equal

(3) one-one

(4) none of these

1

The domain of the function f ( x ) = is

x −x

Domain of the real valued function 9.

25 − 9 x 2 + x 2 + x + 1 is  5 5

(1)  − ,  3 3 

 5 5 



Answer Key

5  5   (3)  −∞, − 3  ∪  3 , ∞ 

(1) 1

(2) 2

(3) 2

(4) 4

(5) 4

(6) 4

(7) 2

(8) 1

(4)  −∞, − 5  ∪  5 , ∞ 

(9) 4

(10) 10



3

3



x −3 2−x is + x +3 2+x

The domain of = f (x )

(1) (−∞, 2)∪(3, ∞) (2) (2, 3) (3) R 5.

Domain of f ( x ) = (1) (0, 5)

(4) ϕ 1 log10 5x − x 2 6

(

) is

(2) (1, 4)

(3) (1, 3)  5 − 21 5 + 21  ,  2   2

The domain of the definition of the function 3 f ( x ) = 2 + log10 x 3 − x is 4−x

(

)

(1) (1, 2)

■ Functions can be combined using addition, subtraction, multiplication, and division to form new functions. These operations help in modeling and solving mathematical problems across various disciplines.

Basic Operations on Functions: ■ Let f : A→R and g: B→R.

■ Definition: (f+c)(x)=f(x)+c ■ Domain: Same as f(x).

Solved example

(2) (–1, 0) ∪ (1, 2)

19. Find the value of (f + 3)(2), where f(x) = x2 + 3x–3.

(3) (1, 2)∪(2, ∞)

Sol. (f + 3)(2) = 3+f(2) = 3+22 + 3(2)–3 = 10

(4) (−1, 0) ∪ (1, 2) ∪ (2, ∞) 7.

20.4 ALGEBRA OF FUNCTIONS

20.4.1 Adding a Scalar to a Function

(4)  6.

If a2 + b2 + c2 = 4, then the range of ab + bc + ca is [–2, k], then, k is _____.

10. If the range of f(x) = |x − 2|+ |x –12| is [m, ∞), then m is _____.



(2)  − ,  3 3

4.

8.

The range of the function f(x) = |x−1| + |x−2|, −1 ≤ x ≤ 3, is (1) [1, 3] (3) [3, 5]

(2) [1, 5] (4) [2, 5]

Try yourself: 18. If f = {(1, 2), (2, –3), (3, –1)} then, find 2 + f.

Ans: {(1, 4), (2, –1), (3, 1)}

2.

15

16

CHAPTER 20: Functions

20.4.2 Scalar Multiplication of a Function

f(x) – {x : f(x) = 0}

■ Definition: (cf)(x)=cf(x)

∴ the domain of

■ Domain: Same as f(x).

Try yourself:

1 . x+4 Then, find the function 2f and its domain.

1 Sol. The given function is f ( x ) = x+4 2 2⋅ f (x) = ⇒ 2 f (x) = x+4 The domain of the function f is same as the domain of 2f, and it is set of all real numbers except –4.

20. Let f is a function defined by f ( x ) =

Try yourself:

20.4.4 Addition and Subtraction of Functions ■ Definition: (f + g)(x) = f(x) + g(x) ■ Domain: D = D1∩D2 (common domain).

Solved example

Ans: 2 2

{x ∈ A | f ( x ) ≠ 0} ■ Domain: D = Let f: D₁ → R and g: D₂ → R be two functions.

Solved example

21. Find the reciprocal function for the function 1 and also find its domain. f (x) = x+4 1 x+4

The domain of

1 ( x ) is domain of f

x + 2 is defined for

x + 2 ≥ 0 ⇒ x ≥ -2 Hence, the domain of the function f ( x= ) is D1 =−2, ∞ ) .

x +2

4 − x 2 is defined for

4 − x2 ≥ 0 ⇒ x2 ≤ 4 . ⇒ x∈[–2, 2].

Hence, the domain of the function g ( x= ) 4 − x 2 is [–2, 2].

The sum of the above two functions is (f + g)(x). ⇒ ( f + g )( x )=

= f ( x ) + g ( x )= ( f + g )( x )

Reciprocal function to the above function is defined as 1 ( x ) = 1 f f (x) Hence, 1 ( x )= x + 4 f

also find its domain.

The function g ( x= )

2. Operations on Two Functions

4 − x 2 be

any two functions. Then, find (f + g)(x) and Sol. The function f ( x= )

1 is defined where f(x)≠0. f (x)

Sol. Given f ( x ) =

) x + 2 and g ( x=

22. Let f ( x= )

f ( x ) + g ( x )=

x + 2 + 4 − x2

x + 2 + 4 − x2

Domain of (f + g)(x) is

−2, ∞ ) ∩ −2,2 = −2,2

Try yourself: 21. If f ( x= )

) x + 2 and g ( x=

4 − x 2 then,

find (f – g)(x) and also find its domain.

Ans: [–2, 2]

x + 1 . Then, find the value of

20.4.3 Reciprocal of a Function ■ Definition:

20. Find the reciprocal function for the function f(x) = sinx and also find its domain.

Ans: Reciprocsl function of f(x) is cosec x and its domain is R–nπ,∀n∈Z

Solved example

19. Let f ( x= ) 2f(1).

1 ( x ) is R–{–4}. f

IL ACHIEVER SERIES FOR JEE MATHEMATICS

17

20.4.5 Multiplication of Functions

Sol. The function f ( x= )

■ Definition: (f ⋅ g)(x) = f(x) ⋅ g(x)

x + 1 ≥ 0 ⇒ x ≥ −1 Hence, the domain of the function

■ Domain: D = D1∩D2.

f ( x= )

Solved example

x + 1 is D1 =−1, ∞ ) .

The function g ( x= )

1 and g(x) = x + 4. Then, find x+4 (f ⋅ g)(x) and find its domain.

23. Let f ( x ) =

Sol. The function f ( x ) =

x + 1 is defined for

9 − x 2 is defined for

9 − x2 ≥ 0 ⇒ x2 ≤ 9 . ⇒ x ∈ −3,3 .

1 is defined for x+4

Hence, the domain of the function

x + 4 ≠ 0 ⇒ x ≠ −4 .

g ( x= )

1 Hence, the domain of the function f ( x ) = x+4 is D1 = R–{–4}.

The quotient of the above two functions is

9 − x 2 is [–3, 3].

f  (x) g f (x) f Now,  = (x ) = g g (x)  

The function g(x) = x + 4 is defined for all real values of x.

The product of two functions 1 ( f ⋅ g )( x =) f ( x ) ⋅ g ( x =) ( x + 4 =) 1 x+4

x +1 9 − x2

f Domain of   ( x ) is g

The domain of the function (f ⋅ g)(x) is domain of f(x) ∩ domain of g(x)

−1, ∞ ) ∩ −3,3 − {x : g ( x ) =0} This is equal to

⇒ Domain of (f ⋅ g)(x) is R–{–4}

−1,3 − {−3.3} = −1, −3 ) ∪ ( −3,3 )

Try yourself:

find (f ⋅ g)(x) and find its domain.

Ans: (f . g)(x) = 4 x + 8 − 2 x 2 − x 3 and its domain is[–2, 2]

f (x) f ■ Definition:   ( x ) = where g(x)≠0. g (x) g ■ Domain: D = D1 ∩ D2 − {x : g ( x ) ≠ 0} .

Solved example

) x + 1 and g ( x=

) x + 2 and g ( x=

9 − x2 .

f Then, find   ( x ) and find its domain. g

4 − x 2 then

f find   ( x ) and find its domain. g and its domain is

20.4.6 Division of Functions

24. Let f ( x= )

23. If f ( x= )

x +2

4 − x then

4 − x2

x + 2 and g ( x= )

f Ans:   ( x ) = g (–2, 2)

22. If f ( x= )

Try yourself: 2

20.4.7 Functional Equations For a function f(x) satisfying: ■ Multiplicative Form:

f(x + y) = f(x) f(y)

f(x) =ax

■ Additive Form: f(x + y) = f(x) + f(y)

18

CHAPTER 20: Functions

f(x) = ax

2.

■ Logarithmic Form:

f(xy) = f(x) + f(y)

f(x) = klogbx f(x + y) = f(x)

f(x) = k

 

−7 (1) 4

3.

■ Inverse Property:

1

If 2 f ( x ) − 3 f   = x 2 , x ≠ 0 ,then f(2)= x (2)

5 2

(3) –1

(4) 2

If f(x) = ax2+ bx + c for some a, b, c ∈R, with a + b + c = 3 and f(x + y)= f(x) + f(y) 10

+ xy; ∀x, y∈R, then ∑ f ( n ) = n =1

(1) 330

■ Alternative Multiplicative Form:

(2) 225 (3) 165

(4) 190

f(x + y) = f(x) + f(y) + 2xy

The value of natural number a, for which f(a + k) = 16(2 n −1), where the function satisfies the relation f(x + y) = f(x). f(y) for all natural numbers x and y and further f(1) = 2 is

f(x) = x2

(1) 16

f(x + y) + f(x – y) = 2f(x) f(y)

f(x) = cosx

4.

■ Power Form:

Solved example

5.

(2) 2

(3) 3

(4) 4

If f : R−{0} →R is defined by f ( x )= x +

25. If f ( x + y )= f ( x ) + f ( y ) and f(2) = 8, then find f(8).

and if f k(x) = [(f(x)]k for K ≥ 1, then

Sol. We know that, if f ( x + y )= f ( x ) + f ( y ) , then f(x) = ax.

(1) –2

f 4(x) – f(x4) – 4f 2 (x)= 6.

Therefore, f = ( 8 ) 4= ( 8 ) 32 .

1 1 f ( x ) f   and f(2) = 24. If f ( x ) + f   = x x 9, then find the value of f(1).

Ans: 2

TEST YOURSELF ( x − a )( x − b ) 1. If f ( x ) = and +

(2) b

If f(x)+2f(1− x) = x + 2∀x∈R, then f(x)

( x − 2)

2

(2) x2 − 2

3

(4) x2 + 2

(3) 1 7.

If f(x) + 2f(1−x) = x2+1 ∀x∈R, then 3f(5) is equal to ______.

8.

If f 3(x) − 3f 2(x) + 3f(x) − 1 = x6, then the

Answer Key

f ( y)

(3) ab

(4) –1

value of f(0) is ______.

( x − y )( x − z ) ( y − z )( y − x ) f (z ) k , then k is + = ( z − x )( z − y ) xyz (1) a

(3) 1 2

(1)

Try yourself:

f (x)

(2) 2

is given by

Since, f ( 2 ) = 2a = 8 ⇒ a = 4 .

x

1 x

(1) 3

(2) 1

(3) 1

(4)

3

(5) 1

(6) 1

(7) 8

(9)

1

20.5 GRAPHICAL TRANSFORMATIONS (4) 3ab

■ Graphical transformations involve modifying the graph of a function y = f(x)

IL ACHIEVER SERIES FOR JEE MATHEMATICS

19

by changing the independent variable x or dependent variable y These transformations help in sketching complex graphs using basic functions.

20.5.2 The Graph of y = f(x + k), where k∈R Horizontal Shifts: ■ If k > 0 → Shift the graph left by |k| units.

20.5.1 The Graph of y = f(x) + k, where k∈R

■ If k < 0 → Shift the graph right by |k| units.

Vertical Shifts:

Solved example

■ If k > 0 → Shift the graph upward by k units.

■ If k < 0 → Shift the graph downward by |k| units.

Solved example

26. Draw the graphs of y = sinx, y = 2 + sinx, and y = –2 + sinx.

Sol. The graph of y = 2 +sinx can be obtained by shifting the graph of y = sinx in upward direction by a distance of 2 units, as shown below.

The graph of y = –2 + sinx can be obtained by shifting the graph of y = sinx in downward direction by a distance of 2 units, as shown below.

y

y

y = x2 y = (x + 2)2 y = (x – 2)2 -4

-2

2

0

4

x

6

Try yourself:

(3,0)

y=-2+sin x

y = (x–3)3

-4

O

2π

(-3,0)

y=sin x

26. Draw the graph of y = x3, y = (x + 3)2 and y = (x – 3)3.

y = (x+3)3

-2

π

x

y = x3

o

-π

The graph of y = (x – 2)2 can be obtained by shifting the graph of y = x2 to right by a distance 2 units, as shown below.

-6

2 -2π

Sol. The graph of y = (x + 2)2 can be obtained by shifting the graph of y = x2 to left by a distance of 2 units, as shown below.

(0,0)

y=2+sin x

27. Draw the graphs of y = x2, y = (x + 2)2 and y = (x – 2)2.

Ans:

Try yourself: 25. Draw the graph of y = 2 + x2, y = x2 and y = x2–2.

20.5.3 The Graph of y = kf (x), where k∈R Vertical Scaling:

y = 2 + x2

■ If k > 1 → Stretch the graph vertically by k times.

■ If 0 < k < 1 → Compress the graph vertically by k times.

(0, 2)

Ans:

(0, -2) y = x2 (0, 0)

y = x2–2

20

CHAPTER 20: Functions Solved example

y y = 2sin x

28. Draw the graphs of y = sin(2x), y = sinx,

2

y = sin x -π -2π

O

x 2π

π

y = sin

( ). x

2

Sol. Graphs of y=sin(2x), y = sinx and y = sin

-2

are as shown below

-4

() x

2

y y

1.5

y = sin x

0.5

1 x O

-π y = 0.5 sin x

π

-2π

-π

O

-0.5 -1

2π

((

-1

y=sin x 2

-2

x

π

2π

y = sin x

-1.5

Try yourself: 28. Draw the graph of y = cos(2x), y = cosx,

x y = cos   2

-4

y = cosx

-2 2

3

1

1 3 27. Draw the graph of y = x , y = 3x and y = x 3 3

-1

Try yourself:

-3

-2π

y = sin 2x

1

2

-2

O

-4 y = cos(x/2)

-3

-2

O

-1

1

2

4

3

y = cos2x

1

3 4

2 3

Ans:

y=3x3

20.5.5 The Graph of y = f(–x)

3

3 y=x3 y= x

Ans:

Reflection about the Y-axis:

20.5.4 The Graph of y = f (kx), k∈R

+

Horizontal Scaling: ■ If k > 1 → Compress the graph horizontally by k times.

■ If 0 < k < 1 → Stretch the graph horizontally 1 by times. k

■ The graph is mirror-reflected about the y-axis.

Solved example

29. Draw the graph of y = (–x)3 is obtained from the graph of the function y = x3.

Sol. Graph of y=(-x)3 is as follows.

IL ACHIEVER SERIES FOR JEE MATHEMATICS

21

y

Try yourself:

1

3 x y = –{x}

2

-2

0

2π

-1

π y=(-x)3

-1

x

-2

-π

O

graph of y = {x} -1

1 -2π

30. Draw the graph of y = –{x} with the help of

y=x3

-3

2

y = {x}

1

Try yourself:

y

Ans:

29. Draw the graph of y = e–x with the help of the graph of y = ex.

20.5.7 The Graph of y = –(f(–x)) Reflection about Both Axes: ■ To obtain the graph of

(0, 1)

y = –(f(–x)) Ans:

y = e–x

20.5.6 The Graph of y = –f(x) Reflection about the X-axis: ■ The graph is mirror-reflected about the x-axis.

■ Start with y = f(x).

■ Reflect the graph about the y-axis (gives y = f(x) ■ Reflect the resulting graph about the x-axis (gives y = –(f(–x)) Solved example

31. Draw the graph of y = –e-x with the help of

Solved example

30. Draw the graph of y = –ex is obtained from the graph of the function y = ex.

Sol. Graph of y = –e is as follows.

y = e x.

Sol. Graph of y = –e-x is as follows:

x

y

y

2 y=ex

x -2

-1

O -1

1

y = –e

x

2

-6

-4

-2

1 -1 -2

y=e-x 2 y=-e-x

x 6

6

22

CHAPTER 20: Functions

Try yourself: 31. Draw the graph of y = sinx – cosx with the help of the graph of y = sinx + cosx.

(1,0) y = |logx|

-2 y = sinx + cosx

Ans:

-1 -2π

-3π/2

π/2

-π/2

-π

π

3π/2

2π

1

20.5.9 The Graph of y = f(|x|) Reflection for Absolute Input: ■ To obtain the graph of y = f(|x|):

y = sinx – cosx 2

Ans:

■ Start with y = f(x).

20.5.8 The Graph of y = |f(x)| Absolute Value Transformation: ■ To obtain the graph of y = |f(x)|: ■ Start with y = f(x).

■ Keep the right half (x > 0) uncharged.

■ Mirror this part on the left side of the y-axis.

■ Keep the portion where f(x) > 0 unchanged

Solved example

■ Reflect the negative part (f(x) < 0) above the x-axis.

33. Draw the graph of y = sin |x| with the help

Solved example

Sol. Steps to draw the graph of y = sin |x|:

of the graph of y = sinx.

Step I: Ignore the graph of y = sin x which is on

32. Draw the graph of y = |sinx| from the graph of y = sinx.

the left side of y-axis.

Step II: Take the graph of y = sin x, which is on

Sol. Graph ofy = |sinx|as follows:

the right side of y-axis, as it is.

Step III: Take the image of portion in step 2 with

y

y=|sin x|

respect to y-axis.

2

Finally, the graph of y = sin(|x|) is as shown below.

1

y

-2π -3π/2 -π -π/2 O π/2 -1

π 3π/2 2π

y=sin |x|

2 1

-2

x -2π

-3π/2

-π -π/2

O

Try yourself:

-1

32. Draw the graph of y = |log x| using the graph of y = logx.

-2

π/2

π

3π/2 y=sin x

2π

IL ACHIEVER SERIES FOR JEE MATHEMATICS

23 –1 –2

-2π -3π/2 -π -π/2

y = |cos|x|| π/2

π

3π/2

2π

3 2 1

Ans:

Try yourself: 33. Draw the graph of y = e|x| with the help of the graph of y = ex.

e-x

20.5.11 The Graph of |y| = f(x)

ex

Ans:

Absolute Value:

20.5.10 The Graph of y =|f(|x|)|

■ To obtain the graph of |y| = f(x):

Double Absolute Transform: ■ To obtain the graph of y = |f(|x|)|:

■ Start with y = f(x).

■ Start with y = f(x).

■ Ignore the portion where f(x) < 0 (below the x-axis).

■ Apply y = f(x) transformation first (retain right-side graph and mirror it to the left)..

■ Reflect the upper part of the graph above x-axis to the lower half.

■ Apply y = f(x) transformation (reflect negative values above the x-axis)..

■ These steps will give you the graph of y = |f(|x|)|.

■ Draw y = logx.

■ Ignore the negative part.

■ Reflect the upper part below the x-axis. y

Solved example

34. Draw the graph of y = |sin(|x|)| using the graph of y = sin(x).

Sol. Step I: Ignore the graph of y = sin x which is on the left side of y-axis.

Step II: Take the graph of y = sin x, which is on the right side of y-axis, as it is.

Step III: Take the image of portion in step 2 with respect to y-axis.

Step IV: The portion of the graph below x-axis will be shifted to above x-axis, as shown below.

The graph of y = |sin(|x|)| is as shown in graph.

1 0.5

x 1

2

-0.5 -1 y=log x -1.5

5

6

Try yourself: 35. Draw the graph of y = x .

y = x

3 π/2 2π 5π/2

-1 -3

-2

-1

1

2

3

4

5

π

4

|y|=log x

y = |sin|x|| π/2

3

-2

-π/2 –1 –2

1.5

-3

3 2 1 -5π/2 -2π -3π/2 -π

Example: Graph of |y| = logx:

1 2 3

34. Draw the graph of y = |cos(|x|)| using the graph of y = cos(x).

Ans:

Try yourself:

24

CHAPTER 20: Functions 2.

20.5.12 The Graph of |y| = |f(x)|

Graph of y = e–|x| is as follows: y

Double Absolute Transformation: To obtain the graph of |y| = |f(x)|:

1

■ Start with y = f(x).

■ Reflect the entire graph again below the x-axis to complete the transformation.

x

0

■ Reflect the graph below the x-axis to above the x-axis.

3.

y=e

–|x|

Graph of |y| = |x – 1| + |x + 1| is as follows: y

2

Try yourself: 36. Draw the graph of |y| = |sin x|.

-1

0

x

1

-2

|y| = |x– 1| + |x+1|

-1

-2

|Y| = |sin x|

1

4.

Graph of y = |sin x| + |cos x| is as follows: y

√2

2

Ans:

1

TEST YOURSELF 1.

Draw the graph of |y| = |log|x||.

2.

Draw the graph of y = e–|x|.

3.

Draw the graph of |y| = |x – 1| + |x + 1|.

4.

Draw the graph of y = |sin x| + |cos x|.

Answer Key

1.

Graph of |y| = |log|x|| is as follows: y |y| = | log |x||

-1

0

x 1

3π 2

π

π 2

0

π 2

π

3π 2

y=|sin x|+|cos x|

20.6 KINDS OF FUNCTIONS Functions can be categorized based on their mappings: ■ One-to-One (Injective) Function: Each element of the domain maps to a unique element in the codomain, ensuring no duplication.

■ Many-to-One Function: Multiple elements of the domain map to the same codomain element.

■ Onto (Surjective) Function: Every element in the codomain has at least one pre-image in the domain.

x

IL ACHIEVER SERIES FOR JEE MATHEMATICS

25

■ Into Function: Some elements in the codomain remain unmapped.

■ Bijective Function: A function that is both injective and surjective, creating a one-toone correspondence between domain and codomain.

Horizontal Line Test: ■ If the continuous graph of a function intersects every horizontal line (parallel to the x-axis) at most once, then the function is one-to-one.

■ This ensures that each input value corresponds to a unique output value. y

20.6.1 One-to-One Function

y = logex

2

■ A function f : A → B s one-to-one (injective) if different elements of A have different images in B.

1 x 1

2

-1

■ Mathematically,

One-One

a ≠ b ⇒ f ( a ) ≠ f ( b ) ∀a, b ∈ A

y 4 3 2 1

■ A function f : A → B s one-to-one (injective) if: f ( a= a b ∀ a, b ∈ A . ) f (b ) ⇔=

-4 -3 -2 -1

Conditions for Injectivity: A

3

-1 -2 -3 -4

y = x2 1

2

3

x 4

Not One-One

f

B

x1

y1

x2

y2

x3

y3

To determine whether a given function is one-to-one, follow these methods: ■ Algebraic Method

y4

■ If a function is given in the form of ordered pairs, and no two ordered pairs have the same second element, the function is oneto-one.

■ If f(x) is strictly increasing or strictly decreasing in its domain, then the function is one-to-one.

■ If f(x) (the derivative of f(x)) satisfies

f ′(x ) > 0 or f ′(x ) < 0, ∀x ∈ A then the function is one-to-one.

Assume f ( x ) = f ( y ) .

Show that x = y. using algebraic properties. x y, then the I f f ( x= ) f ( y ) ⇒= function is one-to-one, otherwise, it is not.

■ Monotonicity Test

Prove that the function is strictly increasing or strictly decreasing.

A strictly increasing or decreasing function is always one-to-one.

■ Horizontal Line Test

A function is one-to-one if no horizontal line intersects its graph more than once.

26

CHAPTER 20: Functions

Number of One-to-One Functions

■ If the function f : A → B is not one-to-one,

■ If n(A) = m amd n(B) =n:

When m ≤ n , the number of one-to-one functions from A to B is ,nPm

then it is many-to-one function. f

A

When m > n , one-to-one functions is not possible.

Solved example

35. Check if the function f : N → N , defined by

B

x1

y1

x2

y2

x3

y3

x4

y4

f ( a ) = a 2 + a + 1 , is one-to-one or not.

Key Properties:

Sol: Given: f ( a ) = a 2 + a + 1

Let a, b ∈ N , such that f ( a ) = f ( b ) .

It implies that 2

■ A constant function (defined over multiple domain values) is always many-to-one.

■ Even functions and all polynomial functions of even degree are many-to-one.

2

a + a +1 = b + b +1

0 (a − b ) + (a − b) =

■ Polynomial functions of odd degree may be either one-to-one or many-to-one.

a −b = 0

■ If a function is given in set notation and at least two ordered pairs share the same second element, it is many-to-one.

2

2

( a − b )( a + b + 1) =0

Hence, f ( a= a b ) f ( b ) ⇒= Therefore, f(x)is one-to-one function.

Try yourself:

37. If A and B are two sets such that n(A) = 15, and n(B) = 20, then what is the number of injections from the set A to the set B? Ans: 20 P15

20.6.2 Many-to-One Function A function f : A → B i many-to-one if two or more elements of set A have the same image in B. ■ Mathematically, f : A → B is a many-toone, if there exists two distinct x , y ∈ A, such that f ( x ) = f ( y ) .

■ A function is many-to-one if its graph is intersected more than once by any horizontal line (x-axis parallel test).

■ Periodic functions are always many-to-one.

Solved example

36. L et n ( A ) = 4 and n ( B ) = 5 . Find the

number of all possible many-to-one functions from A to B. Sol: Let n ( A ) = 4 and n ( B ) = 5 Number of all possible many-to-one functions from A to B. n( A )

= n(B)

= 54 − 5 P4

= 505

− ( ) Pn( A ) n B

IL ACHIEVER SERIES FOR JEE MATHEMATICS

27

Try yourself:  −1 1  ,  defined as  2 2

38. Is the function f : R → 

Ans: No

x , injective? f (x) = 1+ x 2

■ A function f : A → B is onto if each element of B has at least one pre-image in A. ■ Mathematically, f is onto if for every b ∈ B, there exists at least one a∈A such that f(a) = b.

■ If the range of f(x) is equal to the codomain, the function is onto.

■ If any element of B has no pre-image in A, the function is not onto (also called into function)

■ If n(A) = m, n(B) = n, and 1 ≤ n < m, then the number of into functions from A to B is nC1 ( n − 1) − nC2 ( n − 2 ) + nC3 ( n − 3 ) ... m

m

m

Solved example

37. Is f : R → 0, ∞ ) defined by f ( x ) = x 2 onto?

Sol: Yes, the function is onto because for every

positive real number y , there exists a real number x such that f ( x ) = y , where y = x2 .

Try yourself: 39. If n ( A ) = 4 and n ( B ) = 3 , then find the number of onto functions from the set A to the set B.

B

x1

■ If n ( A ) = m and n ( B ) = 2 , then the number

y1

x2 x3 x4

Ans: 36

A

■ If n(A) = n(B) = n, the number of onto functions is n! of onto functions from A to B is 2m − 2 .

20.6.3 Onto Function or Surjection

f

■ If n ( A ) < n ( B ) , then the number of onto functions is zero.

y2

20.6.4 Bijection

y3

■ A function f : A → B is a bijection if it is one-to-one as well as onto (surjective)..

A

f

x y ■ Injective Condition: f ( x= ) f ( y ) ⇒=

B

x1

y1

x2

y2

for all x , y ∈ A .

■ Surjective Condition: For every y ∈ B ,there exists at least one x ∈ A such that f ( x ) = y .

y3

x3

y4

A

f

B

x1

y1

■ If n(A) = m, n(B) = n, and 1 ≤ n ( B ) ≤ n ( A ) ,

x2

y2

then the number of onto functions from

x3

y3

Number of onto functions:

A to B is nm − nC1 ( n − 1) + nC2 ( n − 2 ) − ... m

m

28

CHAPTER 20: Functions

Number of Bijective Functions:

(1) a function

■ If A and B are two finite sets and f : A → B is bijective function, then n(A) = n (B).

(2) one one

■ The number of bijective functions from A to B is n(A)!

(4) One one onto

■ If n(A) ≠ n(B) then number of bijective functions = 0

(3) Onto

3.

If f : R → R is defined by f(x) = x + ,then f is (1) An injection (2) Onto (3) A bijection (4) Only function

4.

Set A has 3 elements. Set B has 4 elements.

The number of surjections that can be defined from A to B is (1) 144 (2) 12 (3) 0 (4) 64

5.

A = {x: –1 ≤ x ≤ 1}. f : A → A defined by f(x) = x|x|. Then f is (1) A bijection (2) An injection but not surjection (3) A surjection but not an injection (4) neither an injection nor a surjection

6.

If f(x) = |x – 1| + |x – 2| + |x – 3|, f : (2, 3) → R, then f is (1) An onto function but not 1–1 (2) 1–1 function but not onto (3) bijection (4) Neither 1–1 nor onto

7.

If f : R → R defined by f(x) = x for x ≥ 2 and f(x) = 5x – 2 for x < 2, then f is (1) One – one (2) Onto only (3) One – one onto

Solved example

38. I s the function, f: R→ R defined as f ( x ) = x 3 +5x + 1, bijective?

Sol: Given: f ( x ) = x +5x + 1 3

Now, f ' ( x = ) 3x 2 +5 > 0, ∀x ∈ R f ( x ) is strictly increasing function.

It is one-to-one.

Clearly, f ( x ) is continuous function and also in creasing on R. Lim f ( x ) = −∞ and Lim f ( x ) = ∞ x →−∞

x →∞

So, f ( x ) takes every value between −∞ and

∞. f ( x ) is onto function. ∴ f is bijection

Try yourself: 40. Is the function, f : R → R , defined as f ( x ) = x 3, bijective?

Ans: Yes

TEST YOURSELF 1.

The number of one one functions that can be defined from A = {1, 2, 3, 4, 5} into B = {a, b, c, d} is (1) 120 (2) 24 (3) 20 (4) 0

2.

If f : R → R defined by 2 x + 5, if x > 0 then f is , f (x) =  3x − 2, if x ≤ 0

(4) Neither one – one nor onto

8.

Strictly monotonic function is (1) onto definitely (2) bijection

x2

IL ACHIEVER SERIES FOR JEE MATHEMATICS

29

(3) one – one definitely

(1) injective but not surjective

B

x

Let R be the set of real numbers. If f : R → R is a function defined by f(x) = x2, then f is

9.

f

A

(4) neither one – one nor onto

g

f(x)

C g(f(x))

gof

Key Properties of Function Composition

(2) surjective but not injective

f  g and g  f may not be always defined.

(3) bijective

■

(4) neither injective nor surjective

■ If f ( x ) and g ( x ) are onto functions, then

10. The number of surjections that can be defined from {1, 2, 3, 4, 5} onto {a, b} is ___.

11. Let n(A) = 4 and n(B) = k. The number of all possible injections from A to B is 120, then k =____.

Answer Key (1) 4 (6) 2 (7)

(2) 2 2 (8)

(3) 4

(4) 3

3 (9)

4 (10)

(5) 1 30

(11)5

20.7 COMPOSITION OF A FUNCTION ■ If f : A → B and g : B → C are two functions, the composite function of f and g denotes ( g  f ) defined as

( g  f )( x ) = g ( f ( x ) ) for all x ∈ A .

■ The function ( g  f )( x ) is simply the g– image of f(x), where f ( x ) is the image of x in A. ■ The composite function exists only if the range of f is a subset of the domain of g.

■ If the range of f is not a subset of the domain of g, g  f is not defined.

g  f is onto, if exists.

■ If g is one-to-one, then g  f is one-to-one function.

■ If g is onto, then g  f is onto.

■ If f : A → B is any function, then f=  I A I= f.

B  f

where IA and IB are identity functions on,A and B respectively.

■ If f and g are even functions, then g  f is even.

■ If f is even and g is an odd, then g  f is even ■ If f is odd and g is even, then g  f is even.

■ If f and g are odd functions, then g  f is odd.

■ Function composition is not commutative. i.e., f ° g ≠ g ° f ■ Function composition is associative i.e., ( f ° g) ° h = f ° ( g ° h) Solved example

39. If f ( x ) =

1− x , then find f  f ( x ) .

1+ x

Sol: Given, f ( x ) = 1 − x

1+ x

f  f ( x ) = f ( f ( x ))

30

CHAPTER 20: Functions  1− x  f ( f ( x )) = f    1+ x   1− x  1−    1− x   1+ x  = f   1+ x  1+ 1− x 1+ x f  f (x) = x

5.

If f(x) = |x – 2| and g(x) = f(f(x)), then for x > 20, g(x) = (1) –x (2) x (3) x – 4 (4) 4 – x

6.

The domain of function log10 log10 log10 log10 log10x is (1) (104, ∞)

Try yourself: 41. For f ( x= ) 2x + 3 and g ( x ) =−x +1 , then

(

10 (3) 10 , ∞

2

find ( f  g )( x ) .

1.

2.

x If f ( x ) = , then ( f  f  f ) ( x ) = 1 − x2 3x 2x (1) (2) 2 1 − 3x 2 x 1 − 4x (3) 2 1− 4 x3x (4) 1 − 4x2 The domain of

x −5   3 = f ( x ) log  2  − x + 5 is  x − 10 x + 24  (1) (4, 5)

3.

4.

)

(4) (10100, ∞)

7.

If f : [–6, 6] → R is defined by f(x) = x2 – 3, then ( f°f°f)(–1) + ( f°f°f)(0) + ( f°f°f)(1) = k, where k = ___.

8.

If f(x) = sin2x + π π   sin2  x +  + cos x cos  x +  and 3 3  

Ans: −2 x 2 + 5

TEST YOURSELF

10

(2) (1010, ∞)

5 g   = 1, then ( g  f )( x ) = ____. 4 9.

For x ∈ R – {0, –1, –2}, let f1(x) =

1 be three 1+ x functions. If a function j(x) satisfied j(x) f 2(x) = 1 + x and f 3 (x ) =

= (f1 ° f2 ° f3)(x), then by j(1) =___.

Answer Key

(2) (6, ∞)

(3)

(4, 5)∪(6, ∞)

(1) 3

(2) 3

(3) 4

(4) 2

(4)

(4, 5]∪(6, ∞)

(6) 3

(7) 29

(8) 1

(9) 0.66

 4 − x2   is The domain of log   1+ x    (1) (–2, 2) (2) (–1, ∞) (3) [–1, 2] (4) (–1, 2) If f(x) = logx and g(x) = x3 then f(g(a)) + f(g(b)) is (1) f(g(a) + g(ab))

(2) f(g(ab))

(3)

g(f(ab))

(4)

g(f(a) + f(b))

1 , x

(5) 3

20.8 INVERSE OF A FUNCTION ■ A function f : A → B is invertible if it is both one-to-one (injective) and onto (surjective).

■ The inverse function g : B → A satisfies

( g  f )( x ) = I A , and ( f  g )( x ) = I B .

where IA and IB are identity functions ■ g(x) is inverse of f ( x ) and it is denoted by g(x) = f –1(x) then g = f

−1

.

IL ACHIEVER SERIES FOR JEE MATHEMATICS

31

Conditions for Invertibility:

Try yourself:

■ f(x) must be bijective.

) 42. Find the inverse of f ( x=

■ If g = f –1, then

Ans: 6 − x 5

g(b) = a, f(a) = b The domain of f becomes the range of g, and vice versa.

■ The graph of y = f(x) and y = f (x) are symmetric about y = x.

TEST YOURSELF 1.

–1

■ If the graphs of y = f(x) and y = f –1(x) intersect, they do so along the line y = x, where f(x)= x.

2.

■ If f : A → B is a bijection: Solve y = f(x) for x in terms of y.

Interchange x and y to obtain f –1(x) 3.

then (g ° f )–1 : f –1 ° g–1 ■ If f  g = g  f then either f −1 = g or • f = g = I. In this case, 2

• g=f .

–1

4.

If f : R → R is defined by f(x) = x – [x], then the inverse function f–1(x) = 1 (1) (2) [x] – x x −x  (4) not defined

Let f(x) = sinx + cosx, g(x) = x2 – 1. Then g(f(x)) is invertible for x ∈

 π π (3)  − ,   4 4

, then find f −1 ( x ) .

5.

Sol: Given: f ( x ) = 2 x ( x −1) Suppose that 2 x ( x −1) = y .

Apply logarithm on both sides x ( x − 1) = log 2 y

 π (4) 0,   2 Let f : [–1, ∞) → R be given by f(x) = (x + 1)2 – 1, x > –1, then f–1(x)

(1) −1 + x + 1

(2) −1 − x + 1

(3) does not exists because f is not 1 – 1 (4) does not exists because f is not onto

It implies that x 2 − x − log 2 y = 0 Therefore, x =

 π  (2)  − , π   2 

 π  (1)  − , 0   2 

40. If the function f :[1, ∞) → [1, ∞) is defined by x x −1)

2 (4) 3x − x − 4 2

(3) x + [x]

Solved example

f (x) = 2 (

If f : [1, ∞) → [2, ∞) is given by f(x) = x + 1 , then f–1(x) = x x x + x2 − 4 (1) (2) 1 + x2 2 2 (3) x − x − 4 2

■ If f : A → B and g : B → C are two bijections,

2

 π π  (2)  − ,   2 2  (4) R+

(3) (–∞, ∞)

■ Verify whether f(x) is bijective.

Properties of Inverse Functions:

Domain of tan–13x is (1) (–∞, 0)

Finding the Inverse of a Function:

1

(6 − x )5 .

1 + 1 + 4 log 2 y 2

Answer Key (1) 3

(2) 1

(3) 4

(4) 3

(5) 4

32

CHAPTER 20: Functions

# Exercises JEE MAIN Level – I Real Valued Functions Single Option Correct MCQs

6.

(1) 49

(2) 50

(3) 48

(4) 51

If f : R→R is given by = f (x)

ax ax + a

∀x ∈ R,

1.

 1   2   1996  Check whether the equation y2 = 2x + 1 +f + ... + f  then f    = represents a function or not. (apply the  1997   1997   1997  vertical line test algebraically)  1   2   1996  f +f + ... + f  (1) It is a function   =  1997   1997   1997  (2) It is not a function (1) 998 (2) 997 (3) 1997 (4) 1998 (3) It can be both a function and not a 7. A real valued function f(x) satisfies the function. functional equation f (x –y) = f(x) f(y) – (4) Can’t determine

2.

Given relation is function or not? Give reasons and also find the domain and range of the function. g = (2,1), (5,1), (8, ), (11,1)

f(a – x) f ( a + y ), where a is a given constant and f(0) = 1, f(2a–x) is equal to

(1) Yes, D = (2, 5, 8, 11), R=1 (2) Yes, D = (3, 2, 7, 12), R=1 (3) No

8.

(4) None 3.

4.

Given relation is function or not? Give reasons and also find the domain and range of the function. f = (1,3), (1,5), (2,3), (2,5) (1) 3 and 5 (2) 2 and 7 (3) 3 and 8 (4) 4 and 9  1− x  3 If ( f (x ))2 f  =  x , x ≠ −1,1 and f(x) ≠ 0,  1+ x  then {f(–2)} = ______. Here, {x} represents

fractional part of x.

5.

1 1 2 (1) (2) (3) (4) 0 3 2 3 For a real number x, [x] denotes the integral part of x. The value of

9.

(1) f(–x)

(2) f(a) +f(a–x)

(3) f(x)

(4) – f(x)

1 x −x 1 x −x 3 + 3  , g (x ) = 3 − 3 , If f ( x ) = 2 2 then f(x) g(y)+f(y) g(x) = (1) f(x + y)

(2) g(x + y)

(3) f(2x)

(4) g(2x)

If f(x) = sin(log x), then x f   + f (xy ) − 2 f (x )cos(log y ) = y (1) 0 (2) sin(logx) (3) cos(log x) (4) cos (log y)

1 2 for x∈ R where [x] is the greatest integer not

10. If f : R → R is defined by f ( x ) =x − [ x] −

1  1 1  1 2   1 99   2  +  2 + 100  +  2 + 100  + ..... +  2 + 100         

1  exceeding x, then  x ∈ R : f ( x ) = = 2  (1) Z, the set of all integers.

is

(2) N, the set of all natural numbers.

IL ACHIEVER SERIES FOR JEE MATHEMATICS

33

(3) f, the empty set.

= f (x)

(4) R

)

(1) [–3, –1] ∪ [1, 2]

Numerical Value Questions 11. If f(x) and g(x) are periodic functions with periods 7 and 11, respectively, then the x x period of F ( x ) f ( x ) g   − g ( x ) f   = 5 3 is ______.

12. The period of the function satisfying the relation f(x) + f(x + 3) = 0, ∀ x ∈R, is ______.

13. If f(x) and g(x) are any two real valued functions such that |f(x) + g(x)| ≥ |f(x)| + |g(x)| and g(x)≠ 0, f(x) g(x) ≤ 0, then the 100

r =1

14. The number of positive integers x, for the which the function f ( x ) = is not defined, is _____.

(2) (–2, –1] ∪ [2, ∞) (3) (–∞, –3] ∪ (–2, –1] ∪ (2, ∞) (4) [–3, –2] ∪ [0, 2]

 x  19. If f ( x ) = log  x −1   , where [.] is greatest  x  integer function, then the domain and range are (1) (2, ∞), (0, 1)

(2) [3, ∞), {0}

(3) [3, ∞), {0, 1}

(4) (–∞, ∞); {0}

20. The domain of f ( x = )

2{ x }2 − 3 {x} + 1 ,

where {.} denotes the fractional part in

value of ∑ f (r ) = ______.

( 2x − 1)

2

x − 10 x − 11

15. If f(x+y) = f(x)f(y) and f(5) = 32, then f(7) = _____.

Single Option Correct MCQs 10 x + 10 − x 16. The domain of the function is 10 x − 10 − x (1) R

(2) R – {0}

(3) R – {1}

(4) R+

log 0.3 x − 2 x

(1) [1, 2]

(2) [2, 3]

(3) [1, 2) ∪ (2, 3]

(4) [0, 3]

18. The domain of the function

[–1, 1], is

1  (1)  −1, 1 ~  , 1  2 

1  1  (2)  −1, −  ∪ 0,  ∪ {1} 2  2   1 (3)  −1,   2

Methods of Finding Domain and Range

17. The domain of f ( x ) =

(

log ( x −1) x 2 + 4 x + 4 is

is

 1  (4)  − , 1  2 

x

ex − e 21. The range of the function f ( x ) = x x e +e is (1) (–∞, ∞)

(2) [0, 1)

(3) (–1, 0]

(4) (–1, 1)

22. A = {x/x ∈ R, x ≠ 0, –4 ≤ x ≤ 4} and f : A → x R is defined by f ( x ) = for x ∈ A. Then, x the range of f is (1) {1, –1}

(2) {x : 0 ≤ x ≤ 4}

(3) {1}

(4) {x : –4 ≤ x ≤ 0}

34 23. The domain of

f (x) =

1

is

( x − 1)( x − 2 )( x − 3 )

1

29. Domain of

is 2  x  −  x  − 2 (where [.] is the greatest integer less than

(1) (–∞, 1) ∪ (3, ∞) (2) (1, 2) ∪ (3, ∞) (3) (–∞, 2) (4) R 24. The domain of

CHAPTER 20: Functions  1 (4) 0,   4

 1 (3) 0,   3

or equal to x)

2x + 1 is 2 x + 3x 2 + x 3

(1) (–∞, –1)

(1) R – [–1, 3)

(2) (–∞, 3) ∩ [3, ∞)

(3) [2, ∞)

(4) (–∞, 3]

30. The domain of f ( x ) =

x−2 +

1 log ( 4 − x )

(2) (0, ∞)

is

(3) (–∞, –1) ∪ (0, ∞)

(1) [2, ∞]

(2) (–∞, 4)

(4) R

(3) [2, 3) ∪ (3, 4)

(4) [3, ∞)

31. The domain of the function

 1     log x 

25. The domain of x (1) (0, ∞) (3) (0, 1) ∪ (1, ∞) 26. Domain of

1 log x

is (2) (1, ∞) (4) [1, ∞)

(

)

= f ( x ) log 3 + x x 2 − 1 is (1) (–3, –1) ∪ (1, ∞) (2) [–3, –1) ∪ [1, ∞)

is

(3) (–3, –2) ∪ (–2, –1) ∪ (1, ∞) (4) [–3, 2) ∪ (–2, –1) ∪ [1, ∞)

(1) R – {0, 1, –1}

32. The domain of f ( x ) =

(2) R – {0} (3) R – {–1, 1} (4) R 27. The domain of f ( x= ) is

1 9−x

2

+ x2 − 4

(1) (–4, –2)∪(2, 4)

log 2 ( x + 3 ) x 2 + 3x + 2

is

(1) R – {–1, –2} (2) (–2, ∞) (3) R – {–1, –2, –3} (4) (–3, ∞) – {–1, –2} 33. If the functions are defined as

(2) (–3, –2]∪[2, 3)

f ( x= )

(3) (–∞, –3)∪(2, ∞)

is the common domain of the following

(4) (–∞, ∞)

functions f + g, f – g, f/g, g/f, g – f where

28. If x ∈ R and P = in the interval

 1 (1) 0,   2

x2 , then P lies x 4 − 2x 2 + 4 3 4 (2)  ,  4 5

x and g ( x= )

f (x) f ?  (x) = g (x) g (1) 0 < x < 1 (3) 0 < x ≤ 1

1 − x , then what

(2) 0 ≤ x ≤ 1 (4) 0 ≤ x < 1

34. The functions f(x) = log(x – 1) – log(x – 2)

IL ACHIEVER SERIES FOR JEE MATHEMATICS

 x −1  and g ( x ) = log   are identical on  x−2 (1) [1, 2] (2) [2, ∞] (3) (2, ∞)

(4) R

35. If A is the set ot real values of x such that 1

−1

e x < 1 , then A = (1) (–∞, 0) ∪ (1, ∞) (2) (–∞, 0) (3) (1, ∞)

(4) (0, 1)

36. If e x + e f(x) = e, then the domain of f(x) is (1) (–∞, 0]

(2) [0, 1]

(3) (–∞, 1)

(4) (1, ∞)

x 37. If f : R → R is defined by f ( x ) =   for x 5  ∈ R, where [y] denotes the greatest integer not exceeding y, then {f(x) : |x| < 71} = (1) {–14, –13, ....., 0, ....., 13, 14} (2) {–14, –13, ....., 0, ....., 13, 14, 15} (3) {–15, –14, ....., 0, ....., 13, 14} (4) {–15, –14, ....., 0, ....., 14, 15} 38. The range of [x] – x is A and x – [x] is B, then A ∩ B =

35 (1) 2(4 – a)

(2) 4(2 – a)

(3) 4(2 + a)

(4) 2(4 + a)

42. If f(x) is a polynomial function such that 1 1 f(x) f = f (x ) + f   and f(3) = –80,  x x

1 then f (x ) − f   = x 1 1 4 (1) x + 4 (2) x 4 − 4 x x 1 1 (3) 4 − x 4 (4) − 4 − x 4 x x 43. If f(x) is a function such that f(x+y) = f(x) n

+ f (y) and f(1) = 7, then ∑ f (r ) = r =1 7n 7 ( n + 1) (1) (2) 2 2 7n ( n + 1) (3) 7n(n + 1) (4) 2 44. If f(x) is a function such that f(x y) = f(x)+ f(y) and f(2) = 1, then f(x) (1) x2

(2) 2x

(3) log2x

(4) logx2

45. Let f(x)= ax2+bx + c be such that f(1) = 3,

(1) {0}

(2) (–1, 1)

f(–2)= l and f(3) = 4. If f(0) +f(1) + f(–2)

(3) (0, 1)

(4) φ

(1) 2

Numerical Value Questions 39. If the domain of f ( x ) =2 − log 3 ( x − 1) is(k, 10], then k is _______.

40. If the domain of f(x) is [0, 5], then the number of integers in the domain of f(log2(x2 + 2x – 3)) is ______

Algebra of Functions Single Option Correct MCQs 41. If f(x) = 4x – x 2, x ∈R, then f(a + 1) – f(a – 1) is equal to

+ f(3) = 14, then l is (2) 4

46. Let f(1) = 1 and n −1

(3) –4

(4)

−23 2

m

f (n) = 2 ∑ f (r ), then ∑ f (n) =

=r 1 = n 1

(1) 3m –1 (2) 3m

(3) 3m–1

(4) 3m–2

47. If f : R →R satisfies f(x + y) = f(x) + f(y)∀x, 10

y ∈ R and f(1) = 6, then ∑ f (r ) = r =1

(1) 331

(2) 330 (3) 334

(4) 333

48. The function y = f(x) satisfying the condition f(x + 1/x) = x3+1/ x3 is (1) f(x) = x2

(2) f(x) = x2–2

(3) f(x) = x2+2

(4) f(x) = x3–3x

36

CHAPTER 20: Functions 3

49. If f (x ) = x +

1 1   − 4  x 2 + 2  + 13 , then 3 x x  

(2) 10

(3) 11

(4) 12

50. If f :R →R is defined by f(x–1) = x2 + 3x + 2, then f(x – 2) is (1) x + x

(2) x – 3x + 2

(3) x2 + 2x

(4) x2 – x

2

2

Numerical Value Questions 51. If f(x) is a polynomial function such that

1 1 f (x ) f = f (x ) + f   and f(3) = –80,  x x 1 1 n then f ( x ) − f   =n − x . Then, 4n + 5 x x is _____.

= 1 and for n > 1, [f(n) = f [f(n – 1)]+f[n–

then

λ 7

Introduction to Functions Single Option Correct MCQs 57. The domain of

1 x−x

2

+ 3x − 1 − 2 x 2 is

1  (1)  , 1 2 

1  (2)  , 1  2 

1  (3)  , 1 2 

1  (4)  , 1  2 

58. The domain of f ( x ) = is

x2 − 4 x + 3

(1) (–∞, 3] (2) [1, ∞)

52. If f is defined on the natural numbers as f(1)

f(n–1)] then the value of

af(x – 1) + bf(–x) = 2|x| + 1. If f(–2) = 5 and f(1) = 1, then 8(b – a) = _______.

the value of f (2 + 3) = (1) 9

56. The function f : R → R satisfies the condition

1 20 ∑ f (r ) is l, 30 r =1

is ______.

(3) (–∞, –3]∪[–1, ∞) (4) (–∞, –3]∪[–1, 1)∪[3, ∞) 59. If f(x) + g(x) = e–x, where f(x) is an even function and g(x) is an odd function, then f(x) =

53. The natural number ‘a’ for which

(1)

e−x 2

(2)

ex + e−x 2

‘f’ satisfies f(x + y) = f(x) f(y) or all natural numbers x, y and further f(1) = 2, is ,___.

(3)

ex − e−x 2

(4)

ex 2

 1  2 f (x 2 ) + 3 f  2  = x 2 − 1 , when the function x 

 1  x 2 − 1 for x∈R–{0}, then 54. If 2 f (x 2 ) + 3 f  2  = x  

1 5 f   is equal to ______. 2

55. The function f satisfies the functional  x + 59  10 x + 30 equation 3 f ( x ) + 2 f  =  x −1  for all real x ≠ 1. Then, the value of f(7) is ______.

60. If f(x) is an even polynomial function, then sin (f(x) – 3x) is (1) an even function (2) an odd function (3) neither even nor odd (4) a periodic function 61. If f(x + y, x – y) = xy, then the arithmetic mean of f(x, y) and f(y, x) is

IL ACHIEVER SERIES FOR JEE MATHEMATICS

37

(1) x

Kinds of Functions

(2) y

Single Option Correct MCQs

(3) 0

68. y = f(x) =

(4) xy 62. If f(x) = ax7 + bx3 + cx – 5, a, b, c are real constants, and f(–7) = 7, then the range of f(7) + 17cosx is

x , x ∈ R, y ∈ R is 1+ x

(1) one – one (2) onto but not one – one (3) one–one but not onto (4) neither one–one not onto

(1) [–34, 0]

69. If f : R → (0, 1] defined by

(2) [0, 34]

f (x) =

(3) [–34, 34] (4) None of these 63. The range of x 2 + 4y 2 + 9z 2 –6yz –3xz – 2xy is (1) φ

(2) R

(3) [0, ∞)

(4) (–∞, 0)

sin π  x 2 − 1

64. The range of f ( x ) =

x4 +1

(1) R

(2) [–1, 1]

(3) {0, 1}

(4) {0}

π  (3) f   = −1 2

(4) f(π) = 1

70. If f : R → R defined by f(x) = x2 – 2x – 3, then f is is

Numerical Value Questions 66. Let f(x) be a polynomial of degree 4 with leading coefficient 1 satisfying f(1) = 10, f(2) = 20, f(3) = 30, where f(12) + f(–8) = 3968λ. Then the value of λ is ___.

67. If f(3x +2 ) + f(3x + 29) = 0 ∀ x ∈ R, then the period of f(x) = ___.

(2) Not onto (4) One – one but not onto

the step function, then

π  (2) f   = 2 4

(1) Not one – one (3) Not one – one but onto

65. If f(x) = cos[π2]x + cos[–π2]x, where [x] is

(1) f(0) = 1

1 , then f is x2 +1

(1) A function

(2) One one

(3) Onto

(4) One one onto 2

71. If f : R → R is defined by f ( x ) = x − 4 , x2 +1 then f(x) is (1) one – one and not onto.

(2) one – one and onto.

(3) not one – one but onto.

(4) neither one – one nor onto.

72. If A = {a1, a2, a3,…, a10} and B = {b1, b2, b3,…, b10}, then the number of bijections that can be defined from A to B is (1) 1010

(2) 10!

(3) 103

(4) 109

73. The function f : R → R defined by f(x) = x – [x], x ∈ R is (1) one–one (2) onto (3) both one–one and onto (4) neither one–one nor onto

38

CHAPTER 20: Functions

74. Let the function f : R → R be defined by f(x) = 2x + sinx for x ∈ R. Then, f is (1) one-to-one and onto (2) one-to-one but not onto (3) onto but not one-to-one

81. If, for x ∈[0, ∞), g[f(x)] = |sinx|, f[g(x)] =

(4) neither one-to-one nor onto

x 2 + 4 x + 30 is 75. f : R → R, f ( x ) = x 2 − 8 x + 18 (1) one–one and onto (2) many–one and onto (3) one–one and into (4) many–one and into Numerical Value Questions 76. If n(A) = 4, n(B) = 5, then number of functions from A to B such that the range contains exactly 3 elements is k, then the k value of is __.

60 77. The number of constant functions that can be defined from {1, 2,…, 100} to {a, b, c,…z} is ____.

Single Option Correct MCQs 78. If f : R → R and g : R → R are defined by f(x) = x – [x] and g(x) = [x] for all x ∈ R, then f(g(x)) is (2) 0

(3) f(x)

(4) g(x)

79. If g(x) = 1 +

x and f(g(x)) = 3 + 2 x + x, then f(x)= _____.

(1) 1 + 2x (3) 1 + x

2

(sin x )2, then (1) f(x) = sin2x, g(x) =

(2) 2 + x

2

(4) 2 + x

80. If g(x) = x2 + x – 1 and (g ° f)(x) = 4x2 – 10x

5 + 5, then f   is equal to 4

x

(2) f(x) = sinx, g(x) = |x| (3) f(x) = sin2x, g(x) = sin x (4) f and g cannot be determined 82. The domain of the function

f ( x ) = 1 − 1 − 1 − x 2 is (1) {x| x < 1}

(2) {x| x > –1}

(3) [0, 1]

(4) [–1, 1]

Numerical Value Questions 83. Let A = {1, 2, 3, 4}. The number of functions f : A → A, satisfying f(f(i)) = i, ∀ i ∈ A, is ___.

= 84. If f ( x )

Composition of a Function

(1) x

1 1 (2) − 2 2 3 3 (3) − (4) 2 2 (1)

2010 x + 165 , x > 0, and 165x − 2010

2010 then the least value of 165 4 f(f(x)) + f(f( )) is______. x x≠

Inverse of a Function Single Option Correct MCQs   3 85. The domain of f ( x ) = sin −1   4 + 2sin x    is  5π π  ,  (1)  −  6 3

 π 7π  (2)  − ,   6 6 

 π  (3)  − , π   6 

 π π (4)  − ,   2 2

IL ACHIEVER SERIES FOR JEE MATHEMATICS 86. The domain of sin −1 is

39 (3) (A) is true but (R) is false.

3x − 1 + cos ( sin x ) 2

(1) [–1, 1]

 −1  (2)  , 0  3 

(3) (0, 1]

 −1  (4)  , 1 3 

(4) Both (A) and (R) are false.

2.

 f (x) x x − f ( x )= 19   − 90   , where y  19   90  denotes the greatest integer less than or equal to y. Then the number of possible values of f (1990) is (1) 1 (2) 2 (3) 3 (4) 4

Numerical Value Questions 87. f : [0, ∞) → [4, ∞) is defined by f(x) = x2 + 4 then f –1(13) = ______.

3.

88. If f : R → R, g : R → R are defined by f(x)= 5x – 2, g(x) = x2 + 3, then

(2) exactly four integral solutions.

(3) exactly two solutions.

89. If a non-zero function f(x) is symmetrical

(4) no integral solution.

about the line y = x, then the value of λ (constant) such that f 2(x) = (f –1(x))2 –

4.

Level – II

b =1

{x + b} = x 1000

.

(R) : {x} = x – [x] and [x + I] = [x] + I, where I is an integer In the light of above statements, choose the correct answer from the options given below.

(1) Both (A) and (R) are true and (R) is the correct explanation of (A).

(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).

4 x − x − 10 x + 9

is

( ) (2) ( 0, 7 + 40 ) (3) ( 7 − 40, ∞ ) (4) ( 7 − 40, 3 ) ∪ ( 3, 7 + 40 )

Assertion (A) : If {x} and [x] represent fractional part and integral part of x, then 1000

1 2

(1) 7 − 40, 7 + 40

Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).

x  + ∑

The domain of the function

f (x) =

λxf(x)f –1(x) + 3x2 f(x) ∀ x ∈ R+, is ____.

Reason

Let [t] denote the greatest integer ≤ t. Then, the equation in x, [x]2 + 2[ x + 2] – 7 = 0 has (1) infinitely many solutions.

(g ο f –1)(3) = ______.

1.

Let N be the set of all natural numbers and f: N → N be such that 1990 < f(1990) < 2100 and satisfies the equation

5.

Domain of

4 − x2 is  x  + 2

(1) (–∞, –2)∪(2, ∞) (2) R – {–2} (3) [–1, 2) (4) (–∞, –2)∪[1, 2] 6.

If f : R → R is defined by f(x) = [2x] – 2[x] for x ∈ R, where [x] is the greatest integer not exceeding x, then the range of f is

40

CHAPTER 20: Functions (1) {x ∈ R : 0 ≤ x ≤ 1} (2) {0, 1} (3) {x ∈ R : x > 0} (4) {x ∈ R : x < 0}

7.

said to be equal if their domains and co-domains are equal and f(x) = g(x)∀x.

If [x] represents the greatest integer ≤ x and [α, β] is the set of all real values of x for which the real function f ( x ) =

3+x + 3−x  x  + 2

is

defined, then f 2(α + 1) + 5 f 2(β) = (1) 0 (3) 12 8.

The domain of the function log10 log10 log10 log10 log10x is (1) (104, ∞)

(

10

10 (3) 10 , ∞

9.

(2) 36/5 (4) 1

)

(2) (1010, ∞) (4) (10100, ∞)

Let f : (1, 3) → R be a function defined by f (x) =

x x  1 + x2

, where [x] denotes the greatest

integer ≤ x. Then the range of f is ______.

 2 1   3 4 (1)  ,  ∪  ,   5 2   5 5  1   3 7 (2)  0,  ∪  ,   2   5 5  2   4 (3)  5 , 1  ∪  1, 5     

 1  2 4 (4)  0,  ∪  ,   3  5 5 Single Option Correct MCQs 10. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).

Assertion (A) : f(x)= log x3 and g(x) = 3log x are equal functions.

Reason

(R) : Two functions f and g are

In light of the above statements, choose the correct answer from the options given below.

(1) Both (A) and (R) are true and (R) is the correct explanation of (A).

(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).

(3) (A) is true but (R) is false.

(4) Both (A) and (R) are false.

11. f and h are from A into B, where A = {a, b, c, d}, B = {s, t, u}, defined as f(a) = t, f(b) = s, f(c) = s, f(d) = u ; h(a) = s, h(b) = t, h(c) = s, h(a) = u, h(d) = u. Which one of the following statements is true?

(1) f and h are functions.

(2) f is a fucntion and h is not a function.

(3) f and h are not functions.

(4) f is not a function but h is a function.

12. If f : N → Z is defined by 2 if= n 3k k ∈ z  f ( n ) =10 − n if n = 3k + 1 k ∈ z , then 0 if n = 3k + 2 k ∈ z 

{n∈N, f(n)>2 } = (1) {3, 6, 4}

(2) {1, 4, 7}

(3) {4, 7}

(4) {7}

13. If f(x + y, x − y) = xy, then the arithmetic mean of f(x, y) and f(y, −x) is (1) 0 (2) x (3) y (4) 1 14. If a function F is such that F(0) = 2, F(1) = 3, F(x + 2) = 2F(x)−F(x + 1) for x ≥ 0, then F(5) is equal to (1) –7 (2) –3 (3) 17 (4) 13 15. If f(x + 1) = x2 − 3x + 2, then f(x) is equal to:

IL ACHIEVER SERIES FOR JEE MATHEMATICS

41

(1) x2 − 5x − 6

(1) always an odd function

(2) x2 + 5x − 6

(2) an odd function, where both f and g are odd

(3) x + 5x + 6 2

(4) x2 − 5x + 6 16. If f and g are real valued functions defined by f(x) = 2x –1 and g(x) = x2, then (fg)x = (1) 2x2 + 2x –1 (2) 2x – x 3

(4) always an even function

23. A function whose graph is symmetrical about the y-axis is given by

2

)

(

(3) 2x – x2

( x ) cos  log x + x 2 + 1  (1) f=

(4) 2x – 4 3

(2) f ( x ) =

Numerical Value Questions 17. If [α] denote the greatest integer ≤ α then

[ 1] + [ 2] + [ 3] +……+ [ 120] is equal to ______.

18. If f(x) is an odd periodic function with period 2, then f(4) = ______.

19. If the domain of the function

  5 x − x 2  f ( x ) = log10    4    then k is ____.

(3) an odd function, where f is even and g is odd

1/ 2

is 1 ≤ x ≤ k,

sec 4 x + cosec 4x x 3 + x 4 cot x

(3) f(x + y) = f(x) + f(y) ∀ x, y ∈ R (4) f ( x ) =

sec 4 x − cosec 4 x x 3 − x 4 cot x

24. If f is an even function defined on the interval [–5, 5], then the real values of x  x +1  satisfying the equation f ( x ) = f    x +2 are (1)

−1 ± 5 −3 ± 5 , 2 2

(2)

−3 ± 5 −4 ± 5 , 2 2

21. If f : {1, 2, 3,......} → {0 ± 1, ± 2,.....} is defined

(3)

 n , if n is even  2 by f ( n ) =   n − 1  then −  , if is odd n   2 

−2 ± 5 −4 ± 5 , 2 2

(4)

−4 ± 5 −1 ± 5 , 2 2

20. Let f : A → B be a function, where n(A) = 3, n(B) = 2, then the total number of functions from set A to set B is _____.

f –1(–100) is ______.

Introduction to Functions Single Option Correct MCQs 22. If f(x) and g(x) are two functions with all real numbers as their domains, then h(x) = [f(x) + f(–x)][g(x) – g(–x)] is

25. The range of the function f ( x ) = is  1 (1)  0,   2

 1 (2) 0,   2

(3) [0, ∞)

(4) [0, 2]

x2 x4 +1

42

CHAPTER 20: Functions

26. If f(x) is a polynomial function such that

by f ( x ) = e x −3 x + 2, is

1 1 f ( x ) f = f ( x ) + f   and  x x

3

(1) many one and onto (2) many one and into (3) one–one and onto (4) one–one and into

1 f(3) = –80, then f ( x ) − f   = x 4

(1) x + (3)

1 x4

1 − x4 x4

(2) x 4 − (4) −

1 x4

1 − x4 x4

Numerical Value Questions 27. If f(x) = sgn(x2 – 2x + 3), then the value of f(x) is ____.

28. If f 3(x) – 3f 2(x) + 3f(x) –1 = x6, then the value of f(3) is ____.

Kinds of Functions Single Option Correct MCQs 2

29. If f : R → R is defined by f ( x ) = x − 4 , x2 +1 then f(x) is (1) one–one and not onto (2) one–one and onto (3) not one–one but onto (4) neither one–one nor onto 30. f : R+ → R, defined by f(x) = logex, x ∈ (0,1), f(x) = 2logex, x ∈[1, ∞) is (1) onto (2) one–one (3) not one–one (4) a bijection

31. If f : R → R is defined by f ( x ) = then f is (1) one–one but not onto (2) not one–one but onto (3) one–one and onto (4) neither one–one nor onto

32. The function f:(–∞, –1) → (0, e5], defined

2

2

x2

− x2

ex − e−x e +e

,

33. If for real x, f(x) = x3 + 5x + 1, then (1) f is one–one but not onto on R (2) f is onto on R but not one–one (3) f is one–one and onto on R (4) f is neither one–one nor onto on R 34. f : N → N, where f(x) = x – (–1)x, then f is (1) one-one and into (2) many-to-one and into (3) one-one and onto (4) many-to-one and onto 35. f : R+ → R defined by f(x) = 2x, x ∈ (0, 1), f(x) = 3x, x ∈ [1, ∞) is (1) onto (2) one–one (3) neither one–one nor onto (4) one–one and onto 36. f : R → R is a function defined by x

e − e − x Then f is . ex + e−x (1) one – one and onto.

(2) one – one but not onto.

(3) onto but not one – one.

(4) neither one – one nor onto. f (x) =

37. Let N be the set of all natural numbers, Z be the set of all integers, and s : N → Z n  2 ,if n is even be defined by σ ( n ) =  − n − 1 ,if n is odd  2 (1) s is one–one but not onto.

(2) s is into but not one–one.

IL ACHIEVER SERIES FOR JEE MATHEMATICS

(3) s is one–one and onto.

(4) s is neither one–one nor onto.

Numerical Value Questions 38. The function f : [2, ∞) → Y defined by f(x) = x2 – 4x + 5 is both one-one and onto if Y = [a, ∞). Then the value of a is _____.

39. Let f : R → [2, ∞) be a function defined as f(x) = x2 – 12ax + 15 – 2a + 36a2. If f(x) is surjective on R, then the value of 2a is ______.

40. Let f be a one-one function with domain {x, y, z} and range {1, 2, 3}. It is given that exactly one of the following statements is true and the remaining two are false: f(x) = 1, f(y) ≠ 1, f(z) ≠ 2. Then, f(y) is ____.

Composition of a Function Single Option Correct MCQs 41. If f ( x ) =

αx , x ≠ –1, then the value of x +1

α, for which f(f(x)) = x, is (1)

2

(3) 1

(2) − 2 (4) –1

42. If f(n) = (–1)n–1(n – 1), G(n) = n – f(n) for every n ∈ N, then (GOG)(n) = ___.

(1) n (2) n – 1 (3) 1

(4) 2

43. If f(x) is defined on [0, 1] as  x ,if x ∈ Q f (x) =  then 1 − x ,if x ∉ Q (f ° f)(x)= ___.

(1) 1 (3) 1 – x

43

x +1 = , f  f (x ) f 2 (x ); x −1 f  f  f (x ) = f 3 (x ), f  f  f  f (x )

44.

Let f ( x ) =

= f 4(x).....then f 2008(x) = ___.

(1) x

(2) f(x) (3) 0

45. The range of f(x) =

1 x

(1 − cos x ) (1 − cos x ) (1 − cos x ) ....∞ is (1) [0, 1]

 1 (2) 0,   2

(3) [0, 2]

1  (4)  , 1 2 

Numerical Value Questions 46. The functions f : R → R and g : R → R are defined as 0 when x is rational f (x) =  1 when x is irrational −1 when x is rational g (x) =  then 0 when x is irrational (f ° g)(π) + (g ° f)(e) + 10 = ___.

47. A continuous function f satisfies the property f(f(x)) = 1 – x. The value of 1 3 f   + f   is ___. 4 4 Inverse of a Function Single Option Correct MCQs

{

48. If f (= x ) sin −1 3 − ( x − 6 ) f –1(x) = (1) 6 + 4 3 + sin3 x

(2) x (4) 1 + x

(4)

(2) 6 + 4 3 − sin3 x

4

} , then 1

3

44

CHAPTER 20: Functions

Then the value of f –1(13) is ______.

(3) 6 + 4 3 + sin x

Level-III

(4) 6 + 4 3 − sin x

49. Let f be an injective map with domain {x, y, z} and range {1, 2, 4} such that exactly one of the following statements is correct and the remaining are false:

Single Option Correct MCQs 1

(1) one–one onto (2) one–one into

f(x) = 1, f(y) ≠ 1, f(z) ≠ 2. The value of

(3) many one and onto

f –1(1) is

(4) many one and into

(1) y (2) x (3) z

(4) 0

2.

50. Let f : R → R be given by f(x) = (x + 1 )2 – 1, x ≥ –1. Then the set of values of x, for which f(x) = f –1(x), is given by

1 x + , x ∈ ( 0,∞ ) (2) g ( x ) = x (3) h(x) = x2 + 4x –5, x ∈ (0, ∞)

(2) {–1, 0}

(4) k(x) = e–x, x ∈ [0, ∞)

(3) {–1} (4) {0, 1}

3.

1  (2)  , ∞  2 

The function y = f(x), satisfying the 1 1  condition f  x +  =x 3 + 3 , is x x 

51. If f : R → R and g : R → R are defined by f(x) = 3x – 4, g(x) = 2 + 3x and 2(g–1 ° f–1)(x) > (f–1 ο g–1)(x), then x ∈

 1  (3)  − , ∞   2 

Which of the following functions is not injective?

(1) f(x) = |x + 1|, x ∈ [–1, ∞)

(1) {0}

(1) (1, ∞)

If Let f : R → R is defined by f(x) = 2x3 + 2x2 + 300x + 5sinx, then f is

4.

(4) (14, ∞)

(1) f(x) = x2

(2) f(x) = x2 – 2

(3) f(x) = x2 + 2

(4) f(x) = x3 – 3x

If f(x) = cos(logx), then

Numerical Value Questions

 x 2  1 f(x2)f(y2) −  f x 2 y 2 + f  2   = 2   y  

52. If f : {1, 2, 3,…..} → {0, ±1, ±2,…..}is

(1) –2

(2) –1

1

2

(4) 0

n  if n is even  defined by f(n) =  2 −  n − 1  if n is odd   2 

then f –1(–100) is ______.

53. Let f be a real–valued invertible function  2x − 3  such that f   = 5x –2, x ≠ 2.  x −2 

(

(3)

5.

)

If f(x) = sinx + cosx, g(x) = x2 – 1, then g(f(x)) is invertible in the domain  π  π π (1) 0,  (2)  − ,   4 4  2  π π (3)  − ,   2 2

(4) [0, π]

IL ACHIEVER SERIES FOR JEE MATHEMATICS

6.

45

Let f : R → R be a function defined by

(4) 3

f. Then, the value of

 −π π  12. Let f :  ,  → B be defined by  3 6

g (9) is (g’ is the g ′(9)

(2) 32

(3) 11

(4) 17

If the graph of f(x) = ||x – 2| – a|–3 has exactly 3 –intercepts on x-axis then the sum of all values of a∈[–10, 10], where and a is an integer is (1) 4

(2) 27

(3) 3

(4) 49

Let f : [–10, 10] → R, where  x2  f= ( x ) sin x +  , be an odd function. a Then the set of values of parameter a is (Where [.] represents GIF)

9.

(3) 2

Numerical Value Questions

(1) 26

8.

(2) 1

f(x) = x3 + x – 1 and ‘g’ be the inverse of

dervative of g) 7.

(1) 4

(1) (10, 10)– {0}

(2) (0, 10)

(3) [100, ∞)

(4) (100, ∞)

) )

distinct solutions, then the value of n is equal to (2) 4

(3) 5

(4) 8

) ( (

(

10. If f ln (1 + x ) =1 − ln (1 + x )

)) , 1 7 7

then f(f(cosx)) is equal to (1) cos(ln(1 + |x|)) (2) cos7(ln(1 + |x|)) (4) cos7x

(3) cosx

11. The number of integer values of ‘x’ for which log  x 2  ( 4 − x ) + log 3 



{ x } is real is

___. ([.] denotes GIF, {.} denotes fraction part).

a + b is _____.

13. A function is defined as srk(x –a)   x 2024 + 5x 2022 + 7  x <a   2024 , 2022 =   3x + 7 x + 9  sgn x 2024 + x 2023 + 2020 , x ≥ a, 

(

)

where [.] is greatest integer function and sgn(x) means signum function. If the number of real value(s) of ‘x’, where x ∈ [0, 3) satisfies the equation 3x 2 − 9 x + 4 is ‘ λ’then the 2 value of λ λ is _____.

3srk(x – 2) =

g (x ) = f f f ……( f ( x ) ) …  .

If equation g(x)(n times = k,) k ∈ (0, 2) has 8

(1) 3

B = [a, b] such that f –1(x) exists. Then

x(srk(x –0) – srk(x – 1)) – srk(x – 1) +

Let f(x) = |x – 2| and

( (

f ( x ) =2cos 2 x + 3 sin2 x + 1 , where

 4a − 7  3 14. = If f ( x )  x + ( a − 3) x 2 + x + 5   3  is one-to-one, where a ∈ [l, m], then the value of |l – m| is _____.

1 and 1− x fn + 1 (x) = f0(fn(x)), n = 0, 1, 2,.... Then,

15. For x ∈ R, x ≠ 0, x ≠ 1, let f0(x) =

2 3 the value of f100 ( 3 ) + f1   + f 2   is 3 2 equal to ______.

1

1  7 16. If g ( x ) =  4 cos 4 x − 2cos 2 x − cos 4 x − x 7  , 2  

then the value of g(g(100)) is _____.

46

CHAPTER 20: Functions

19. If f is a function such that

of

f(x) + 2f(1–x) = x2 + 1, then the value of f(3) is ______.

S-II : The rule to find the domain of a square root function is the function under square root must be nonnegative.

THEORY-BASED QUESTIONS Statement Type Questions 1.

S-I : Every relation is a function S-II : Every function is a relation.

2.

S-I : If any vertical line cuts the graph of a Relation at most one point, then that relation is called function.

S-II : In the function, no two ordered pairs have same second component.

3.

S-I : Two functions f, g are said to be equal if and only if f(x) = g(x) for all x∈ domain of f S-II : Two functions f, g are said to be equal, if their domains are equal and their ranges also equal.

4.

S-I : The domain of a constant function is set of all real numbers.

S-II : The range of a constant function is singleton set.

5.

S-I : The range of modulus function is set of all non negative real numbers.

S-II : The range of greatest integer function is set of all integers.

6.

7.

8.

S-I : The range of constant function is finite set

f ( x ) is {x : x ∈ R, f ( x ) ≥ 0} .

9.

S-I : The derivative of odd function is an even function.

S-II : The derivative of an even function is an odd function.

10. S-I : If n(A) < n(B), then the number of onto functions from the set A to the set B is zero.

S-II : If n(A) > n(B), then the number of one–one functions from the set A to the set B is zero.

11. S-I : The function f : R → R, defined as f(x) = x + 12, is a surjection.

S-II : The function f : R → R is defined as f(x) = x + 12 is a bijection.

12. S-I : If f : A → B and g : B → C are two bijections, then g ° f = A → C is bijection and (g ° f )–1 = f –1° g –1.

S-II : If f ° g = g ° f, then either

f –1 = g or g–1 = f; in this case,

f ° g (x) = g ° f (x) = x.

Assertion and Reason Questions

S-II : The range of signum function also finite set and it is {–1, 0, 1}

In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as

S-I : The function f(x) = loga x is negative, when 0 < x < 1, 0 < a < 1.

(1) if both (A) and (R) are true and (R) is the correct explanation of (A),

S-II : The function f(x) = loga x is positive, when x > 1, 0 < a < 1.

(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),

S-I : Suppose that the domain of the function f(x) is A, then the domain

(3) if (A) is true but (R) is false, (4) if both (A) and (R) are false.

IL ACHIEVER SERIES FOR JEE MATHEMATICS

47

13. (A) : The function which associates each real number to it self, then that function is called identity function.

f(x+p) = f(x) for all x∈A. The least

(R) : The graph of identity function is a straight line passing through the origin and making equal angles with axes.

19. (A) : A sub set of the domain of the function log g ( x ) f ( x ) is {x : f(x)

14. (A) : The fractional part function {x} and x–[x] are two equal functions.

log g ( x ) f ( x ) is intersection of

(R) : Two functions are said to be equal, if the images of two functions for every element of its domain must be equal and their domains are equal, ranges must be equal.

15. (A) : T h e f u n c t i o n s f ( x ) = x 2 a n d

f ( x ) = x are not equal functions.

(R) : The range of f(x) = x2 is same as that of f ( x ) =

x

are two equal

functions.

(R) : The domain of f(x) = |x| is equal to domain of f ( x ) =

x 2 and the

range of f(x) = |x| is equal to range of f ( x ) =

fundamental period of f(x).

≥ 1, g(x) > 1}.

(R) : The domain of the function

two sets {x : f(x) ≥ 1, g(x) > 1} and {x : 0 < f(x) ≤ 1, 0 < g(x) < 1}.

20. (A) : The range of the function

1 is a cos x + b sin x + c   1 1

,  .

2 2 2 2 c − a + b c + a + b 

f ( x) =

(R) : Range of f(x) = a cos x + b sin x + c 2 2 2 2

is  c − a + b , c + a + b  .

16. (A) : T h e f u n c t i o n f ( x ) = | x | a n d

f ( x ) = x2

positive real number p is called the

x2 .

17. (A) : log a x = y ⇔ x = a y .

(R) : logarithmic function and exponential functions are inverse functions to each other.

18. (A) : If p is a period of a function f(x) then np also period of the function f(x) (R) : A function f : A→R is said to be periodic function if there exists a positive real number p such that





1 x

1 x

21. (A) : Let f ( x ) + f   = f ( x) f  

and f(2) = 9 then f(x) = x3 + 1.

1 x

1 x

f ( x) f   (R) : I f f ( x ) + f   = then f(x) = 1 + xn

22. (A) : If A = {a, b, c, d} and B = {2, 3} then the number of onto functions from the set A to the set B is 14.

(R) : The number of onto functions defined from the set A to the set B is 2n( A ) − 2, where n(B) = 2.

23. (A) : The domain of the function

f (x) =

x 9 − x2

is (–3, 3).

48

CHAPTER 20: Functions

(R) : Denominator value must not be equal to zero and the value under the root must be non-negative, so that 9 − x 2 > 0 .

JEE ADVANCED LEVEL Multiple Option Correct MCQs The domain of f ( x ) =

1.

24. (A) : The range of the function f : R → R

(where [.] GIF) is

x +2 given by f ( x ) = is (–1, 1). x +2 (R) : |x + 2| is equal to either x + 2 or x – 2 only.

(2) (–∞, 7]

(3) (–∞, –7]

(4) [7, ∞)

(2) range = (–∞, 1) (3) domain = (–∞, 0]

odd function uniquely.

(4) range = (–∞, 1]

) Let f ( x=

3.

{ n + 1 −  n + x } , 2

2

1/ 2

where [.] is the greatest integer function

function.

and n∈N. Then the value of x for which f(x) is defined is

26. (A) : The function f : R → R, defined function.

(1) [–7, 7]

(1) domain = (–∞, 1)

the sum of an even function and an

as f ( x ) =

 x − 1 − 5

If ex + ef(x) = e, then for f(x),

2.

25. (A) : Every function can be expressed as

1 (R) : ( f ( x ) + f ( − x ) ) is an even function 2 1 and ( f ( x ) − f ( − x ) ) an is odd 2

1

x , is not a one–one x2 −1

4 1 4 (R) : f ( 4 ) = and f   = .

17  4  17 27. (A) : The function f : R → R, is defined as

(1)

k , where k = 0, 1, 2, ..... n

(2) [–n2, 2n + 1] (3) [0, 2n + 1] (4) N If f(x) = f(–x) and and g(x) is an odd function and satisfies the relation

4.

f ( x= ) x 3 + 1, is a bijective function.

1 x2 f (x) − 2 f   = g ( x ) , then x

(R) : Every polynomial function defined on the set of real numbers is a bijective

(1) f(x) = 0, ∀x∈R

function.

(2) f(2009) = 0 (3) f is a constant function

28. (A) : If a function is a strictly increasing

(4) g(x) = 0, ∀x∈R

function, then that is one-to-one function.

(

5.

)

(

(R) : Any line parallel to the x-axis does f (x) + f (= y ) f x 1 − y 2 + y 1 − x 2 [ f ( x ) is not identically zero]. Then, not intersect the continuous graph (1) f(4x3 - 3x) + 3f(x) = 0, of increasing function at more than one point.

)

y) f x 1− y2 + y 1− x2 [ f (x) Let f ( x ) + f ( =

(2) f(4x3 - 3x) = 3f(x),

IL ACHIEVER SERIES FOR JEE MATHEMATICS

) ( (4) f ( 2 x 1 − x ) = 2 f (x) ,

2 0, (3) f 2 x 1 − x + 2 f ( x ) = 2

6.

Let f(x) be a polynomial with positive degree

49

(1) 1 − 3 2 (2) 1 +

3 2

(3) 1 −

2 3

satisfying the relation f(x)g(y) = f(x) + f(y) + f(xy) – 2 and f(4) = 65. Then, (1) f(2) = 9.

( )

(2) f 3 3 = 27 .

(3) f(5) = 126.

(4) the roots of the equation f(x) = 2x2 are all real.

7.

Let f(x) is a polynomial of degree n such n 1 0, f (1) =, … f ( n ) = . that f ( 0 ) = 2 n +1 Then the value of f(n + 1) is (1) 1 when n is odd n (2) when n is even n+ 2 n (3) − when n is odd n+1 (4) –1 when n is even

8.

10. If f(x) = sin x + cos x, g(x) = x2 – 1, then g(f(x)) is invertible in the domain (1) 0, π   

2 

 π π (2)  − ,   4 4  π π (3)  − 2 , 2   

(4) [0, π ] 11. L e t f

f(x) = (log (sec x + tan x))3. Then,

b e a f unc t i on d ef i ne d by x −5 f (x) = , x ≠ 3, and 2 f k(x) denotes x −3 the composition of f with itself taken k

(1) f(x) is an odd function

times i.e., f3(x)=f(f(f(x))). Then,

(2) f(x) is a one–one function

(1) f2012(2009) = 2009 2005 (2) f 2009 ( 2010 ) = 2007

 π π Let f :  − ,  → R be given by  2 2

(3) f(x) is an onto function (4) f(x) is an even function 9.

(4) 1 + 2 3

Let f : (–1,1) → R be such that 2 f ( cos 4θ ) = for 2 − sec2 θ  π  π π  θ ∈  0,  ∪  ,  . Then, the value of  4 4 2 1 f   is 3

(3) f 2009 ( 2009 ) =

1002 1003

(4) f2012(2012) = 2012 πx  12. Let f(x) = ln(2x – x2)+sin   . Then,  2  (1) graph of f is symmetrical about the line x = 1

x)

50

CHAPTER 20: Functions

(2) graph of f is symmetrical about the line x = 2

(2) neither one–one nor onto

(3) maximum value of f is 1

(4) one–one and onto

(4) minimum value of f does not exist 13. Let f : R → R be a function defined by f (x) −5 ∀ x ∈ R. Then, which = f ( x + 1) f (x) −3

(3) many–one

17. For real x, let f(x) = x3 + 5x + 1. Then, (1) f is one–one but not onto R (2) f is onto R but not one–one

of the following statements is/are true?

(3) f is one–one and onto R

(1) f(2008) = f(2004)

(4) f is neither one–one nor onto R

(2) f(2006) = f(2010)

18. Let f be the greatest integer function and

(3) f(2006) = f(2002)

g be the modulous functions. Then,

(4) f(2006) = f(2018)

 5 1 (1) ( g  f − fog )  −  =  3

14. f(x) = x 2 – 2ax + a(a+1), f : [a, ∞) → [a, ∞). If one of the solutions of the equation f(x) = f –1(x) is 5049, then the other may be

(2) (f+2g)(–1) = 1 5

0 (3) ( g  f − f  g )   = 3  

(1) 5051

(2) 5048

(3) 5052

(4) 5050

15. Consider the function y = f(x) satisfying the condition 1 1  f  x +  =x 2 + 2 ( x ≠ 0 ) . Then the x x   (1) domain of f(x) is R (2) domain of f(x) is R-(–2,2) (3) range of f(x) is [–2, ∞) (4) range of f(x) is [2, ∞)

(4) (f + 2g)(1) = 1 19. If f: R+ → R+ is a polynomial function satisfying the functional equation f(f(x)) = 6x – f(x), then f(17) is equal to (1) 17

(2) –51

(3) 34

(4) –34

Numerical/Integer Value Questions 20. If [x] stands for greatest integer function,

1  1 2  1  1 999  k,  2 + 1000  +  2 + 1000  + ...... +  2 + 1000  =      

k −100 1  1 2  1  1 999   x , x ∈ Q + +  0, +x ∈ Q  + ...... +  + = k , then = ______ .   16. If f ( x ) = = g ( x )  2 1000,  10  2and1000  2 1000  ∉ ∉ 0, x Q x , x Q   x −  x  = f (x) , ∀x ∈ R. If the range 21. Let  x , x ∈ Q  0, x ∈ Q 1 + x −  x  and g ( x )  , then f – g is =  x , x ∉ Q  0, x ∉ Q of f(x) is [a, b), then [b] = ______, where

(1) one–one and into

[.] is GIF.

IL ACHIEVER SERIES FOR JEE MATHEMATICS 22. If f(x) is an odd function, f(1) = 3, and f(x + 2) = f(x) + f(2), then f(3) = _________.

23. If f : R → [0, ∞) is a function such that

3 f ( x ) → 1 , then the

f (x -1) + f (x= + 1)

period f(x) is ______.

 1+ x  24. = If f ( x ) log   , −1 < x < 1, and  1− x   3x + x 3   2x  −f = f M f(x), then 2  2  1 3 x +  1+ x    the value of M is _____.

25. If f(x) is a polynomial such that f(x)f(y) = f(xy) – f(x) – f(y) ∀ x, y and f(2) = 7, then |f(–2)| = ________.

51

______ (where [.] denotes greatest integer function).  1+ x  32. If f ( x ) = log   , –1 < x < 1, and  1− x   3x + x 3   2x  f −f  = M.f(x), then  1 + 3x 2   1 + x2    the value of M is _____.

33. If the range of the function

(1 + x + x )(1 + x ) when x > 0 f (x) = 2

4

x3

is [k , ∞), then

k +3 = ______.

34. Let g(x) = 1+x–[x], [x]is the greatest integer

26. If the domain of f(x) = log[x]x is [k, ∞), then the value of k is _____ (where [ ] represents greatest integer function).

−1, x < 0  f ( x ) = 0, x 0 , not greater than x. If = 1, x > 0 

27. Minimum value of f(x) = |x – 1| + |x – 2| + |x – 3| is _______.

then for all x, f(g(x)) = ______.

2

28. If f(x) = x2 and

f (1.1) − f (1)

(1.1 − 1)

= k , then [k]

= ______. ([.] is greatest integer function)

10 + x f x , x ∈ ( −10,10 ) and f(x) = e ( ) 29. If= 10 − x  200 x   , then 2k = ______. k f  100 + x 2 

30. f: {1,2,3,4} → {1,2,3,4} such that f(i) = even if i is even and f(i) ≤ f(j) ∀ i < J. If the number of such functions possible is n, then the sum of the digits in n ______.

31. If f ( x ) 2010 x + 163 , x > 0 = 165x − 2010 2010 and x ≠ , then the least value of 165

((

))

   16    2 − 5 is  f f 1 + sin x + f  f  2     1 + sin x    

35. Number of positive integers in the domain of the function  x2 + x  is _____. f ( x ) = log 0.5 log 6   x + 4    36. The function f ( x ) =

x +1 x3 + 1

can be

expressed as the sum of even function g(x) and odd function h(x). The value of |g(0)| = ______ 37. If f(x) is odd function, f(1) = 3 and f(x + 2) = f(x) + f(2)then f(3) = _____.

38. Let f:[–5,6] → R be a real valued function whose graph is as given blow. Then, the number of solutions of f(f(f(x))) = 3 is ____.

52

CHAPTER 20: Functions 44. The graph of y = f(x) is not symmetric about

y=f(x)

(2,3) (-3,2)

(-5,-1)

(-3)

(-1,1) O

1/2

2

x

3 (6,-1)

(1) x = 2

(2) x = 10

(3) x = 8

(4) none of these

(Q.45–47) Let f :R → R be a function satisfying

39. Number of solutions of the equation 2[x] = x + {x} is _____ (where [.] is greatest integer function, {.} is fractional point).

40. Let f(x) be a real valued function such 2024 − x, x ∈ Q 2023 that f ( x ) =  .

2023 − x, x ∈ Q ' 2024 If f(f(x)) = 2025 a + b2025x, then the sum of distinct real value(s) of b is _____.

Passage-based Questions Q (41 - 42)

f(2–x) = f(2+x) and f(20–x) = f(x)∀x∈R. For this function f, answer the following.

45. If f(0) = 5, then the minimum possible number of values of x satisfying f(x) = 5 for x∈[0, 170] is _______ (1) 21

(2) 12

(3) 11

(4) 22

46. The graph of y = f(x) is not symmetrial about (1) x = 2

(2) x = 10

(3) x = 8

(4) none of these.

47. If f(2) ≠ f(6), then the (1) fundamental period of f(x) is 1

Let f : N → R be a function satisfying the 1 following conditions: f (1) = and f(1) + 2.

2 f(2) + 3. f(3) + ...... + n. f(n) = n(n + 1)f(n) for all n ∈ N, n ≥ 2.

1 41. The value of f(1003) is , where k equals k (1) 1003 (2) 2003 (3) 2005 (4) 2006 1 42. The value of f(999) is , where k equals k (1) 999 (2) 1000 (3) 1998 (4) 2000 Q (43 - 44) Let f : R → R be a function satisfying f(2 – x) = f(2 + x) and f(20 – x) = f(x) ∀ x ∈ R. For this function f, answer the following.

43. If f(0) = 5, then the minimum possible number of values of x satisfying f(x) = 5 for x ∈ [0, 170], is (1) 21

(2) 12

(3) 11

(4) 22

(2) fundamental period of f(x) may be 1 (3) period of f(x) cannot be 1 (4) fundamental period of f(x) is 8

(Q.48–49)

(

)

Let f ( x = ) log e x + x 2 + 1 domain of f is where f(x) is defined for real values of x. If f is bijective, then f –1(x)exists 48. f –1(x) is defined on (1) (0, ∞) (2) (–∞, ∞) (3) [0, e] (4) (–∞, 0) 49. The inverse of f is positive on (1) (0, ∞) (2) (–∞, ∞) (3) [0, e] (4) (–∞, 0) (Q.50–52) A function f from a set X to Y is called onto if every y∈Y, ∃ x ∈ X, such that f(x) = y. Unless the contrary is specified, a real function is onto

IL ACHIEVER SERIES FOR JEE MATHEMATICS

if it takes all real values. Otherwise, it is called an into function. Thus, if X and Y are finite sets, then f cannot be onto, if Y contains more elements than x. Now answer the following questions.

53

54. If the domain of g(f(x)) is [–1,4] then (1) a = 1, b > 5

(2) a = 2, b > 7

(3) a = 2, b > 10

(4) a = 0, b ∈ R

f(x) = a0xn + a1xn–1 + a2 xn–2 + ..... + an,

55. If a = 2 and b = 3, then the range of g(f(x)) is (1) (–2, ∞) (2) (0, ∞) (3) [4, ∞) (4) [–1, ∞)

(1) all positive integers n

(Q.56–57)

(2) all even positive integers n

For x ≠ 0,1, define f1(x) = x, f2(x) =

50. The polynomial function where a0 ≠0 is onto, for

(3) all odd positive integers n (4) no positive integer

x 2 + 2x + c 51. The function f ( x ) = is onto, x 2 + 4 x + 3c if (1) 0 < c < 2 (3)

−1 1 <c< 2 2

(2) 0 < c < 4 (4) 0 < c < 1

52. Which of the following is not true?

(1) A one–one function from the set {a, b, c} to {α, β, γ} is also onto.

(2) An onto function from an infinite set to a finite set cannot be one–one.

(3) An onto function is always invertible.

(4) The functions tanx and cotx are onto.

(Q.53–55)  2 x + a, x ≥ −1 Let f ( x ) =  2 and bx + 3 x < −1 0≤x≤4  x + 4, g (x) =  −3x − 2 −2 < x < 0 53. g(f(x)) is not defined if (1) a∈(10, ∞), b∈(5, ∞) (2) a∈(4, 10), b∈(5, ∞) (3) a∈(10, ∞), b∈(0,1) (4) a∈(4, 10), b∈(1,5)

f3(x) = 1–x, f 4 (x ) = f6(x) =

1 , x

1 x −1 , f5(x) = , 1− x x

x . x −1

This family of functions is closed under composition that is, the composition of any two of these functions is again one of these.

56. Let G be a function such that GOf3 = f6.

Then, G is equal to (1) f5 (2) f4 (3) f3 (4) f2 57. If H is a function such that f4OH = f5, then H is equal to (1) f6 (2) f4 (3) f5 (4) f3 (Q.58–60) f : A → B is said to be injective if distinct elements in A have distinct images in B and surjective if f(A) = B. Now answer the following.

58. If f : A → B defined by f ( x ) = is injective, then A can be

1 + cos 2 x 2

(1) [0, π ]

(2) [– π , π ]

 π  (3)  − ,0   2 

(4) [– π , 0]

54

CHAPTER 20: Functions

59. If the function f : R → B defined by

f ( x ) = x 2 is surjective, then B is (1) [–∞, 0)

(2) [0, ∞)

(3) (0, ∞)

(4) 

60. If the function f : R → B, defined by f(x) = [x] + [–x], (where [.] is GIF) is surjective, then B = (1) R

(2) [0, 1]

(3) [–1, 0]

(4) {–1, 0}

Matrix Matching Questions 61. Let f : R → R and g : R → R be functions such that f(g(x))is a one–one function.

List I

(C) f: R–{0}→R, 1 f (x )= x + x

III) onto but not one–one

(D) f : R→R, f(x)=2 x+sinx

IV) neither one– one nor onto

(A) (B) (C) (D) (1) I

II

III

IV

(2) II

III

IV

I

(3) III

II

I

IV

(4) I

IV

III

II

63. Match the following functions in List I with their inverse function in List II

List II

(A) g(x)

I) must be one–one

(B) f(x)

II) may not be oneone

(C) If g(x) is onto, III) may be many– then f(x) one (D) If g(x) is into, IV) must be many– then f(x) one (A) (B) (C) (D) II,III

I

II, r

(2) I

II

III

IV

(3) II

III

I

IV

(4) II

I

III

IV

= f (x)

x

e −e

−x

ex + e−x

List II I) 1 (1 + 1 + 4 log x )

+2

2

(B) f : (2,4) → (1,3) defined by

II) x+(–1)x–1

(C) f : [1,∞)→[1,∞)

III)

2

x f ( x )= x −   2

1/2

 x −1  log e    3−x 

f(x) = 2x(x–1) (D) f : N→N defined IV) x+1 by f(x) = x+(–1)x–1 (A) (B) (C) (D)

62. List I gives functions and List II the nature of the functions. Choose the correct option.

List II

(A) f :[0, ∞)→ [0, ∞), x f (x) = 1+ x

I) one–one onto

(B) f: R–{0}→R,

II) one–one but not onto

1 f ( x )= x − x

List I

defined by

(1) I

List I

(A)

(1) IV

III

II

I

(2) III

IV

II

I

(3) II

III

I

IV

(4) III

IV

I

II

64. let R denote the set of all real numbers and R+ denote the set of all positive real numbers. For the subsets A and B of R define f : A → B by f(x) = x2 for x∈A observe the two lists given below.

IL ACHIEVER SERIES FOR JEE MATHEMATICS

55

List I (A) f is one–one and onto if

List II I) A = R+, B = R

(1)

9 7 (2) 2 4

(B) f is one–one but not onto if

II) A = B = R

(3)

9 7 (4) 4 3

(C) f is onto but not III) A = R, one–one if B = R+  {0}

2.

2

 x −7  log  7    ≥ 0 is + x    2x − 3   2 (11th Apr 2023 Shift 1) (1) 7 (2) 5 (3) 6 (4) 8

(D) f is neither one– IV) A = B = R+ one nor onto if (A) (B) (C) (D) (1) IV

I

III

II

(2) I

II

III

IV

(3) III

IV

II

I

(4) IV

III

II

I

3.

3x + x 3 1+ x g x = ( ) and , 1 + 3x 2 1− x then match List I with List II.

65. Let f ( x ) = log List I

I)

 e −1  (B) gof    1+ e 

II)

(C) fog(0)

III) 0

 e −1  (D) fog    e +1 

IV)

1

(A) (B) (C) (D) (1) I

II

III

IV

(2) II

IV

III

I

(3) III

II

I

IV

(4) I

III

II

IV

FLASHBACK (PREVIOUS JEE QUESTIONS) JEE Main Let f : R–{0, 1}→R be a function such that  1  f (x ) + f  1 + x . Then f(2) is =  1− x  (1st Feb 2023 Shift 2) 1.

(1) (–∞, –3]∪(5, ∞) (2) (–∞, –2] ∪ (5, ∞) (3) (–∞, –2) ∪ [6, ∞)

3 –3

The domain of the function 1 f (x) = is (where [x] 2 x  − 3 x  − 10 denotes the greatest integer less than or equal to x) (11th Apr 2023 Shift 2)

List II

 1− e  (A) fog    1+ e 

The number of integral solutions x of

(4) (–∞, –3] ∪ (6, ∞) 4.

Let the sets A and B denote the domain and range, respectively of the function 1 f (x ) = , where [x] denotes the [x] − x

smallest integer greater than or equal to x.

Then among the statements.

(S1): A∩B = (1, ∞)–N and (S2): A∪B = (1, ∞) (15th Apr 2023 Shift 1) (1) only (S2) is true (2) both (S1) and (S2) are true (3) neither (S1) nor (S2) is true 5.

(4) only (S1) is true Let f(x) be a function such that f(x + y) = f(x) • f(y) for all x, y ∈ N. If f(1) = 3 and

∑

n k =1

f (k ) = 3279, then the value of n is

(1) 6

(24th Jan 2023 Shift 2) (2) 8 (3) 7 (4) 9

56 6.

7.

CHAPTER 20: Functions

22 x , x ∈ R, then 22 x + 2  1   2   2022  f +f + ..... + f     is  2023   2023   2023 

If f ( x ) =

_____.

(24th Jan 2023 Shift 2)

(1) 2011

(2) 2010

(3) 1011

(4) 1010

8.

(2) 3

(3) 2

(4) 4

Let f(x) = 2x + λ, λ∈R, n ∈ N and f(4) = 133, n

f(5) = 255. Then the sum of all the positive integer divisors of (f(3) − f(2)) is (25th Jan 2023 Shift 2) (2) 58 (3) 61 (4) 59

(1) 60 9.

The domain of f ( x ) =

log ( x +1) (x − 2) e 2 loge x − (2 x + 3)

is

(29 Jan 2023 Shift 1) th

(1) (–1,∞)–{3}

(2) (2,∞)–{3}

(3) –{3}

(4) –{–1,3}

10. The range of the function

f (x ) =

3−x + 2+ x

(1) 600

(2) 660 (3) 540

(4) 720

[x] , 1 + x2 (where [x] is greatest integer) is [2, 6), then its range is (31th Jan 2023 Shift 1)  5 2 (1)  ,   26 5   5 2 (2)  ,   37 5   5 2   9 27 18 9  , ,  (3)  ,  −  ,  26 5   29 109 89 53   5 2   9 27 18 9  (4)  37 , 5  −  29 , 109 , 89 , 53     

13. If the domain of the function f (x ) =

The number of functions f: {1, 2, 3, 4}→{a 1 Z : |a| ≤ 8} satisfying f (n) + f (n + 1) = 1, n ∀n∈{1, 2, 3} is ___. (25th Jan 2023 Shift 2) (1) 1

12. Let R be a relation from the set {1, 2, 3,....60} to itself such that R = {(a, b): b = pq, where p,q ≥ 3 are prime numbers}. Then, the numbers of elements in R is (29th Jul 2022 Shift 1)

(30th Jan 2023 Shift 2)

(1)  2, 7 

(2)  5, 10 

(3) 2 2, 11 

(4)  5, 13 

11. Consider a function f:  → satisfying f(1) + 2f(2) + 3f(3)+....+ xf(x) = x(x + 1)f(x); x ≥ 1 1 + 2 with f(1) = 1. Then , f (2022) f (2028)

14. Let f(x) = ax2 + bx + c be such that f(1) = 3, f(–2) = λ, and f(3) = 4. If f(0) + f(1) + f(–2) + f(3) = 14, then λ is equal to ____.

(28th Jul 2022 Shift 2) (1) –4 (2) 13/2 (3) 23/2 (4) 4 15. Let A = (1,2,3,4,5} and B = {1,2,3,4,5,6}.

Then the number of functions f : A→B satisfying f(1) + f(2) = f(4) –1 is equal to ___.

(11th Apr 2023 Shift 2) 16. Suppose, f is a function satisfying f(x+y)

1 = f(x)+f(y) for all x, y∈ and f (1) = . If 5 f (n) 1 m ∑ n=1 n(n + 1)(n + 2) = 12 , then m is equal to ______.

(29th Jan 2023 Shift 1) 17. Let f : R→R be a function defined 2e 2 x f (x ) = 2 x . Then, e +e

is equal to

(29th Jan 2023 Shift 2)

(1) 8100

(2) 8000

 1   2   3   99  f +f +f + .... + f       100   100   100   100 

(3) 8200

(4) 8400

is ____.

(27th Jun 2022 Shift 1)

IL ACHIEVER SERIES FOR JEE MATHEMATICS

57

18. Let c, k∈R. If f(x) = (c + 1)x2 + (1 – c2)x + 2k and f(x + y) = f(x) + f(y) – xy, for all x,y∈R, then the value of |2(f(1) + f(2) + f(3)+.....+ f(20)|is equal to _____.

(1) (0,1] (3) [0,1] 23. = If f ( x )

(29th Jun 2022 Shift 1)

24. Let f : R → R and g : R → R be defined as

log e x , x > 0 f ( x ) =  −x and , x ≤0  e

(27th Jan 2024 Shift 1) (1) both one–one and onto.

 x g (x) =  x e

(2) one–one only.

(3) onto only.

(4) neither one–one nor onto.

, x ≥0 , x <0

. Then, gοf : R → R is

(1st Feb 2024 Shift 1)

 −1  21. Let f : R −   → R and 2

(1) one–one but not onto (2) neither one–one nor onto (3) onto but not one–one

 −5  g : R −   → R be defined as 2  x +1 2 x + 3 and g (x) = . Then 2x + 5 2x + 1 the domain of the function fοg is: f (x) =

(4) both one–one and onto 25. I f t h e d o m a i n o f t h e f u n c t i o n

f (x) =

(27th Jan 2024 Shift 2)

 5 (1) R − −   2

(2) R

 7 (3) R − −   4

 5 7 (4) R − − , −   2 4

2 + 2 x , −1 ≤ x < 0  22. If f ( x ) =  x ; 1 − 3 ,0 ≤ x ≤ 3

4x + 3 2 , x ≠ and (f ο f) (x) = 6x − 4 3

2  2  g(x), where g :  −   →  −   , then 3  3  (g ο g ο g) (4) is equal to (31st Jan 2024 Shift 1) 19 19 (1) − (2) 20 20 (3) –4 (4) 4

19. The number of functions f, from the set A = {x ∈ N : x2 – 10x + 9 ≤ 0} to the set B = {n2 : n ∈ N} such that f(x) ≤ (x – 3)2 + 1, for every x ∈ N, is ___.

(27th Jul 2022 Shift 2) 20. The function f : N – {1} → N; defined by f(n)= the highest prime factor of n, is :

(2) [0,3) (4) [0,1)

x 2 − 25

(4 − x ) 2

+log10(x2 +2x –15) is

(–∞, α)∪[β,∞), then α2 + β3 is equal to: (1st Feb 2024 Shift 2) (1) 140 (2) 175 (3) 150

(4) 125

CHAPTER TEST – JEE MAIN Section – A 1.

Let Z be set of integers.

− x , −3 ≤ x ≤ 0 g (x) =  , then range of (f ο  x ,0 < x ≤ 1

If A={x ∈ Z: 2(x+2)(x2-5x+6)=1} and B = {x ∈ Z: –3 < 2x – 1 < 9}, then the number of subsets of the set A × B, is

g(x)) is

(1) 215

(29th Jan 2024 Shift 1)

(2) 218

(3) 212

(4) 210

58 2.

CHAPTER 20: Functions Let ‘f ’ be a real valued function defined for all x ∈ R such that for some fixed a > 0,

1 f (x + a) = + 2

f ( x ) − ( f ( x ))

2

9.

for all x, then the period of f(x) is (1) a/4 (2) a/3 (3) 2a (4) 3a 3

A function f well defined. x,y ∈R is such that f(1) = 2, f(2) = 8 and f(x + y) – kxy = f(x) + 2y2 where ‘k’ is some constant, then f(x) is (1) x2

4.

5.

6.

7.

8.

(2) 3x2

(3) 2x2

) x2 + The range of f ( x=

(4) 4x2

1 is x +1 2

(1) [1,∞]

(2) [2, ∞)

3  (3)  , ∞  2 

(4) R

(1) [–10,0]

(2) [–1,1]

(3) [0,p]

 −49  , 0 (4)   4 

x −x 

1 − x  + x is Greatest integer function) is 1 (1) 0,   2

(2) [0,1]

1 (3)  0,    2

 1 (4) 0,   2

(where [.]

 x2 + e  If f ( x ) = log e  2  , then the range of  x +1  f(x) is (1) (0,1)

(2) [0,1]

(3) [0,1)

(4) (0,1]

If f(x)is a polynomial satisfying f(x)f(1/x) = f(x) + f(1/x) and f(3) = 28, then f(4) is equal to

(2) 65

(3) 17

(4) 64

f:C→C is defined as

ax + b , bd ≠ 0 , then f is a constant cx + d function when (1) a = c (2) b = d (3) ad = bc (4) ab = cd

= f (x)

10. The domain of

sin ( cos x )

π  (1) 2nπ ,2nπ +  , n ∈ I 2  (2) 2nπ + π ,2nπ + π  , n ∈ I   2  

3π   , n∈ I (3) 2nπ + π ,2nπ + 2  

The range of f(x) = sin2x – 5sinx – 6 is

The range of f ( x ) =

(1) 63

π π  (4) 2nπ − ,2nπ +  , n ∈ I 2 2  11. Let f : R→R be defined as f (x) = 2x – 1 1 x− 2 and g: R–{1} be defined as g ( x ) = , x −1 then the composite function f(g(x)) is defined as (1) both one - one and onto, (2) neither one one nor onto, (3) one one but not onto, (4) onto but not one one.

12. Let x denote the total number of one–one functions from a set A with 3 elements to a set B with 5 elements and y denotes the total number of one – one function from the set A to be set A × B, then (1) y = 91x (2) 2y = 273x (3) y = 273x (4) 2y = 91x 13. The domain of the function 1

2   4 ( x −2 ) 2 x −1 f ( x ) = 9 x + 27 3 − 219 − 3 ( )   

IL ACHIEVER SERIES FOR JEE MATHEMATICS (1) [–3, 3]

(2) [3, ∞)

5  (3)  , ∞  2 

(4) [0, 1]

59

where g(x) is an odd function, then the value of f(5) is (1) 0 (2) 37 (3) 4 (4) 5 55

14. The domain of the function

19. I f

f ( x ) = 10 − x 4 − 21x 2 is

f ( x )= x +

(1) [5, ∞) (2)  − 21, 21   

(4) (–∞,-5]

15. Domain of the function

(

= f ( x ) log 2 log 4 log 2 log 3 x 2 + 4 x − 23

))))

is (1) (–8, 4)

2 (3) x + x − 4 2

2 (4) x − x − 4 2

20. The domain of

{

(

f (= x ) log10 1 − log10 x 2 − 5x + 16

)} is

(2) (0, ∞) (4) [2, 3]

Section – B

(3) (–4, 8)

f ( x ) = 4 − x 2 + x 2 − 1, t h e n t h e maximum value of (f(x))2 is _____.

21. I f

(4) ( −∞, − 4 ) ∪ ( 8, ∞ )

16. The range of f ( x ) =

2 (2) x ± x − 4 2

(1) (2, 3) (3) [1, 3]

(2) ( −∞, − 8 ) ∪ ( 4, ∞ )

1 then f −1 ( x ) = x

x ± x2 − 3 4

(1)

(3)  −5, − 21  ∪  21,5 ∪ {0}    

( ( (

f : 1, ∞ ) → 2, ∞ ) i s g i v e n b y

x −x  1 − x  + x

, (where

β 1 af ( x ) + αf  = bx + , x ≠ 0 , then the  x x 1 f (x) + f   value of the expression x 1 is _____. x+ x

[.] is GIF) is  1 (1) 0,   2

(2) [0, 1]

 1 (3)  0,   2

 1 (4) 0,   2

22. If a + α = 1, b + β = 2 and

2

2

ex − e−x f x = 17. Let f :R→R defined by ( ) 2 2 , ex + e−x then f(x) is (1) one–one but not onto (2) neither one–one nor onto (3) many-one but onto (4) one-one but not onto 18. If f(x) is an even function and satisfies 1 the relation x 2 ⋅ f ( x ) − 2 ⋅ f   =g ( x ) , x

23. If f(x) = sgn(x2 – 2x + 3), then the value of f(x) is ______.

24. If f(3x + 2) + f(3x+29) = 0; ∀x ∈ R , then the period of f(x) is _____.

25. If f is a polynomial such that  1− x   1+ x   1− x   1+ x  +f f f= f       1+ x   1− x   1+ x   1− x 

(where x ≠ 0,±1) and f(3)=28, then the value of

 1  10  ∑( f ( n ) − 1)  is ______.

605  n=1 

60

CHAPTER 20: Functions

CHAPTER TEST – JEE ADVANCED

Section –B

2023 P1 Model

[Single Option Correct MCQs] 4.

Section – A [Multiple Option Correct MCQs] 1.

1 + sin x − sec x and 1 − sin x

1 − sin x y= − + sec x , coincide are: 1 + sin x

5.

then A and B are (1) 2n+1–1, 2n+1

π   π 3π   −3π (2)  , − ∪ , 2   2 2   2

(2) 4n–1, 4n (3) 3n, 3n+1

 π π (3)  − 2 , 2   

(4) 5n–1, 5n

6.

3π   π (4)  −2π , 2π  − ± , ±  2   2 Consider the function f ( x ) =x + 1 − x ,

g: A→A such that gof = f is (1) 10C5

(1) range of f(x) is  −1, 2  (2) f is neither even nor odd (3) f is either even or odd.

(4) range of f(x) is identical to range of π  = g (x) 2 cos  x −  4 

Let A = {1, 2, 3....10} and f: A→A be k + 1 if k is odd defined as f ( k ) =  , if kis even  k

then the number of possible functions

2

then which of the following is/are correct?

3.

Let g:R→R be given by g(x) = 3+4x. If gn(x) = gogo.....og(x)(n) times, and gn(x) = A+Bx,

3π   3π   ∪  , 2π  (1)  −2π , −  2   2  

2.

π −2  −2 ≤x≤ A=  x :  , B= { y : −1 ≤ y ≤ 1} 5 5  

and f(x) = cos(5x+2) then the mapping f:A→B is (1) one–one but not onto (2) onto but not one–one (3) both one–one and onto (4) neither one–one nor onto

The value of x in  −2π , 2π  , for which   the graph of the function

y =

If

(2) 55 (3) 105 (4) 5!

satisfying the functional equation f(f(x))

f : R → R is a function defined by x e − e−x f ( x ) = x − x . Then f is e +e (1) one–one and onto

= 6x-f(x), then f(17) is equal to

(2) one–one but not onto

(1) 17

(2) –51

(3) onto but not one–one

(3) 34

(4) –34

If f: R+ → R+ is a polynomial function

7.

(4) neither one–one nor onto

IL ACHIEVER SERIES FOR JEE MATHEMATICS

61

Section – C the value of

[Integer Value Questions] 8.

f(x) is a real valued function satisfying 2f(xy) = (f(x)) y +f(y)) x ; x,y∈R and 1 f (1) = , , then determine the value of 2 ∞

______.

1 20 ∑ f ( r ) is λ then λ is 30 r =1 7

Section – D [Matrix Matching Questions] 14. Match the following and choose the correct option.

∑ f (i ) ______. i =1

List - I

List - II

(A) The inverse function of sin(tan–1x) is

(p) cot(cos–1x)

(B) The inverse function of f(x) = 1–2–x is

(q)

) x2 is ___.

(C) The inverse function of

(r) tan(sin–1x)

11. The number of integral solution(s) of the

f ( x ) = 2 x −1

9.

For some a,b,c∈N, let f(x) = ax – 3 and 1

3 g(x) = x + c, x ∈ R. If ( fog ) ( x ) =  x − 7   2  then (f ο g)(ac) + (g ο f)(b) is equal to −1

b

_____.

10. Let f(x) = 2x(2–x);0≤x≤2, then the number

(

of solution of f f ( f ( x ) ) =

log 2 x ( log 2 x ) − 1

x

equation given by x 2 − 1 + cos2 x = − 1 cos2 x − x 2

(s) –log2(1–x)

is _____. (Given α is a positive solution

(t)

of x = cos x) 2

2

12. Let f :  − 1 , ∞  →  3 , ∞  , defined by  2    4  f (x) =

x 4 + x2 + 1

, then the number x2 − x + 1 of possible values of x satisf ying f (x) =

−1 + 4 x − 3 is ______.

2

13. Let f be defined on the natural numbers as follow: f(1) = 1 and for n >1, f(n) = f[(n–1)]+f[f(n–1)]+f[n–f(n–1)], then

x 1 − x2

(1) A→p,r; B→r; C→q (2) A→r,s; B→s; C→t (3) A→s,t; B→r; C→q (4) A→p,r,t; B→s; C→q

15. If f: R→R is defined by  x + 4 for x < −4  f ( x )= 3x + 2 for − 4 ≤ x < 4 , then the  x − 4 for x ≥ 4  correct matching of List I and List II is.

62

CHAPTER 20: Functions List - I

(1) A-q, B-s, C-p, D-r

List - II

(A) f(–5) + f(–4)

(p) 14

(B) f(|f(–8)|)

(q) 4

(C) f(f(–7) + f(3))

(r) –11

(D) f(f(f(f(0)))) + 1

(s) –4

(2) A-r, B-p, C-s, D-q (3) A-s, B-q, C-r, D-p (4) A-r, B-s, C-q, D-p

17. Match the following and choose the correct option.

List - I List - II

(t) 1 (u) 0

(A) f(x) = e2x, g(x) = logx, (p) (– ∞,1) then (fog)(x) = (3,∞)

(1) A-p, B-s, C-q, D-s (2) A-r, B-s, C-q, D-t (3) A-r, B-t, C-q, D-s (4) A-r, B-t, C-p, D-s

16. Match the following and choose the correct option.

(A)

List - I

List - II (p) Constant function

x x + +2 e −1 2 x

(B)

(

log x + x 2 + 1

) (q) Identity function

(B) f(x) = 2x–3, g(x) = x3+5, then (fog)–1 (x)

(q) {0}

(C) Domain of log(x2 – 4x +3)

(r)

(D) Range of sin π x  f (x) = 2 1 + x  [.] represents G.I.F

(s) x2

(C) For 3 < x < 5, the function |x – 2| + |x – 3| + |x – 5|

(r) Even function

(1) A-s, B-r, C-q, D-p

(D) f(x) = 2023

(s) Odd function

(3) A-s, B-r, C-p, D-q

1/3

 x −7   2   

(2) A-p, B-r, C-s, D-q (4) A-q, B-s, C-r, D-p

ANSWER KEY JEE Main Level Level – I (1) 2

(2) 1

(3) 1

(4) 1

(5) 2

(6) 1

(7) 4

(8) 2

(9) 1

(10) 3

(11) 1155 (12) 6

(13) 0

(14) 11

(15) 128

(16) 2

(17) 3

(18) 3

(19) 2

(20) 2

(21) 3

(22) 1

(23) 2

(24) 3

(25) 3

(26) 1

(27) 2

(28) 1

(29) 1

(30) 3

(31) 3

(32) 4

(33) 1

(34) 3

(35) 1

(36) 3

(37) 3

(38) 1

(39) 1

(40) 8

(41) 2

(42) 3

(43) 4

(44) 4

(45) 2

(46) 3

(47) 2

(48) 4

(49) 1

(50) 1

(51) 21

(52) 1

(53) 3

(54) 4

(55) 4

(56) 4

(57) 2

(58) 4

(59) 2

(60) 3

(61) 3

(62) 1

(63) 3

(64) 4

(65) 3

(66) 5

(67) 54

(68) 3

(69) 3

(70) 1

(71) 4

(72) 2

(73) 4

(74) 1

(75) 4

(76) 6

(77) 26

(78) 2

(79) 2

(80) 2

(81) 1

(82) 4

(83) 10

(84) 4

(85) 2

(86) 4

(87) 3

(88) 4

(89) 3

IL ACHIEVER SERIES FOR JEE MATHEMATICS

63

Level – II (1) 1

(2) 2

(3) 1

(4) 4

(5) 4

(6) 2

(7) 3

(8) 3

(9) 1

(10) 1

(11) 2

(12) 2

(13) 1

(14) 4

(15) 4

(16) 4

(17) 825

(18) 0

(19) 4

(20) 8

(21) 201

(22) 1

(23) 1

(24) 1

(25) 2

(26) 3

(27) 1

(28) 10

(29) 4

(30) 4

(31) 4

(32) 4

(33) 3

(34) 3

(35) 2

(36) 4

(37) 3

(38) 1

(39) 13

(40) 1

(41) 4

(42) 3

(43) 2

(44) 1

(45) 3

(46) 9

(47) 1

(48) 2

(49) 1

(50) 2

(51) 4

(52) 201

(53) 3

(1) 4

(2) 3

(3) 1

(4) 2

(5) 4

(6) 4

(7) 2

(8) 1

(9) 4

(10) 4

(11) 2

(12) 3

(13) 3

(14) 4

(15) 3

(16) 6

(17) 1

(19) 100

(19) 0

Level – III

Theory-based Questions (1) 4

(2) 3

(3) 1

(4) 1

(5) 1

(6) 4

(7) 2

(8) 1

(9) 1

(10) 1

(11) 1

(12) 1

(13) 2

(14) 1

(15) 2

(16) 1

(17) 1

(18) 1

(19) 1

(20) 3

(21) 1

(22) 1

(23) 1

(24) 3

(25) 1

(26) 3

(27) 3

(28) 1

(4) 1,2,3

(5) 1,4

(6) 1,3,4

(7) 1,2

(8) 1,2,3

(9) 1,2

(10) 2,4

(11) 1,2,3,4(12) 1,3,4 (13) 1,2,3,4(14) 2,4

(15) 2,4

(16) 4

(17) 3

(18) 1,2,3 (19) 2,3

(20) 2

(21) 0

(22) 9

(23) 12

(24) 1

(25) 9

(26) 2

(27) 2

(28) 2

(29) 1

(30) 3

(31) 3

(32) 1

(33) 3

(34) 1

(35) 7

(36) 1

(37) 9

(38) 3

(39) 2

(40) 1

(41) 4

(42) 3

(43) 3

(44) 3

(45) 4

(46) 3

(47) 3

(48) 2

(49) 1

(50) 3

(51) 4

(52) 3

(53) 1

(54) 4

(55) 3

(56) 1

(57) 2

(58) 3

(59) 2

(60) 4

(61) 1

(62) 2

(63) 4

(64) 1

(65) 2

(1) 4

(2) 3

(3) 3

(4) 3

(5) 3

(6) 3

(7) 3

(8) 1

(9) 2

(10) 2

(11) 1

(12) 2

(13) 2

(14) 4

(15) 360

(16) 10

(17) 99

(18) 3395 (19) 1440 (20) 4

(21) 1

(22) 3

(23) 4

(24) 2

(25) 3

JEE Advanced Level (1) 3,4

(2) 1,2

(3) 1,2,3

Flashback

Chapter Test-JEE Main (1) 1

(2) 3

(3) 3

(4) 1

(5) 1

(6) 4

(7) 4

(8) 2

(9) 3

(10) 4

(11) 3

(12) 4

(13) 3

(14) 3

(15) 2

(16) 4

(17) 2

(18) 1

(19) 3

(20) 1

(21) 6

(22) 2

(23) 1

(24) 54

(25) 5

(8) 1

(9) 2039 (10) 8

Chapter Test – JEE Advanced (1) 1,3

(2) 1,2

(3) 2,3

(4) 3

(5) 2

(6) 3

(7) 4

(11) 1

(12) 0

(13) 1

(14) 4

(15) 2

(16) 4

(17) 3