Gr-11_M1-JEE_IL-Ranker_Physics_Basic Math in Physics_V2

CHAPTER

1

BASIC MATHEMATICS Chapter Outline

Symbol

1.1

Mensuration Formulas

1.2

Basic Algebra

1.3

Logarithmic Representation and Functions

1.4

Geometry

1.5

Basic Trigonometry

1.6

Exponential and Roots

1.7

Binomial Theorem

1.8

Series Expansion

1.9

Differential Calculus

∫ 1.1

Meaning indefinite integral

MENSURATION FORMULAS

r = radius; d = diameter; V = Volume; A = surface area

Circle

r

D O

1.10 Maxima and Minima 1.11 Integral Calculus 1.12 Some Idefinite Integral Formulae

Symbol

Meaning

=

is equal to

≠

is not equal to

∝

is proportional to

Perimeter or circumference: 2πr = πd 1 Area: r 2 d 2 4

Sphere Surface area = 4πr2 = πd2 4 1 Volume r 3 d 3 3 6

Spherical Shell r

≈

is approximately equal to

∆x

the change in value of x

dx dt

the derivative of x w.r.t. t

Surface area = 4πr2 = πd2 Volume of material used = (4πr2)(dr) dr = thickness

2

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CHAPTER 1: Basic Mathematics

Hollow Sphere

Parallelogram R2

a R1

h θ b

Volume:

A = bh

4 3 4 3 R2 R1 3 3

= absin θ

where R2 and R1 are outer and inner radii.

Cylinder

a = side; h = height; b = base θ = angle between sides a and b

Trapezium

h r

Lateral surface area = 2πrh Volume = πr2h Total area = 2πrh+2πr2 =2πr(h+r)

a h b

h (a + b) 2 a and b are parallel sides.

Area =

Cone h

h = Height

Triangle

r

α

Lateral surface area r r 2 h2 h = height Total area r

r 2 h2 r

b

1 Volume r 2 h 3

c

h

γ

β a

Ellipse

a, b, and c sides are opposite to angles α, β, and γ. b = Base; h = Height b

a a = Semi major axis b = Semi minor axis

Circumference 2 Area = πab

a 2 b2 2

Area =

ah 2

ab sin 2 s(s a)(s b)(s c)

1 s (a b c) 2

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3. Mathematically,

Rectangular Container h b l Lateral area = 2(lb + bh + lh) V = lbh l, b, and h are sides of the container.

1.2

3

BASIC ALGEBRA

Factorisation 1. x(a ± b) = ax ± bx 2. (x ± y)2 = x2 + y2 ± 2xy 3. x2 – y2 = (x + y)(x – y) 4. x3 + y3 = (x + y)(x2 + y2 – xy)

n ! n n(n 1)(n 2)(n 3) 4 3 2 1 n n n 1 4. Examples: 6 6 5 4 3 2 1 720 5 120 0 1 (Must remember) 1 1

Determinant 1. D

ab cd

ad bc

a1 a2 a3 2. D b1 b2 b3 c1 c2 c3

a1 b2 c3 b3c2 a2 b1c3 b3c1 a3 b1c2 b2 c1

5. x3 – y3 = (x – y)(x2 + y2 + xy)

Arithmetic Progression (AP):

Quadratic Equation and Roots

■ 1+2+3+...+n=

If an algebraic expression has 2 as the power of x, then it contains a quadratic nature. If ax 2 + bx + c = 0 (a ≠ 0) is the required expression, then

■ 12 + 22 + 32 + . . . + n2 =

n n 1 2

.

n 2n 1 2n 3 6 2

(a) this expression possesses no real roots if b2 – 4ac < 0

n n 1 ■ 1 +2 +3 +...+n = .

2

(b) this expression possesses identical roots b if b2 – 4ac = 0 and that root is − 2a (c) this expression possesses two distinct roots if b 2 – 4ac > 0 and the roots are

Geometric Progression (GP):

b b2 4ac .

2a

Factorial Representation 1. Factorial is always defined for positive integers, including zero.

2. It is written as n! or n and is read as n factorial.

.

3

3

3

3

■ nth term of the series a, a + b, a + 2b, . . . is given by tn = a + (n – 1)b and sum of n the n terms is given by 2a n 1 b 2 ■ If a series is given by a, ar, ar2, ar3, . . . nth term is given by tn = arn – 1.

■ s n = a + ar + ar 2 + . . . upto n terms =

1 r . n

1 r

■ If r < 1 and n is large, then Sn

a .

1 r

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4

1.3

CHAPTER 1: Basic Mathematics

LOGARITHMIC REPRESENTATION AND FUNCTIONS

is d P1P2 r12 r22 2r1r2 cos 1 2 y

P1

1. ln = Logarithm to the base e, where e = 2.718 (also called natural logarithm) q1

3. logex = ln x

x

m

If (x1, y1) are Cartesian coordinates of P1, then x1 = r1 cosq1, y1 = r1 sinq1

x 6. log e log e x log e y y 7. loge(xy) = logex + logey 8. loge(xn) = nlogex 1 9. log e log e x x

x12 y12 r12 r1

and

y 1 tan 1 1 x1

x y 2 1

2 1

3. A linear equation has the general form given by y = mx + c, where m and c are constants called the slope of the line and the intercept of the line on y-axis, respectively (see the figure).

10. If logax = a ⇒ x = aa Similarly, if logex = a,

2

(x

2

,y

)

y

⇒ x = ea

2. Consider y = logax.

This function is only defined for x > 0. and a > 0 and a ≠ 1.

GEOMETRY

1. The distance d between two points P1 and P2 with coordinates (x1, y1) and (x2, y2) is

x2 x1 y2 y1 2

2

.

1

(x

1. Logex or log10x are always defined for x > 0 and so, sometimes, we do write them as loge |x| or log10|x|.

,

1

y

)

Important

d

P2 (r2, q2)

q2

O

5. loge(ex) = x

1.4

d

(q1–q2)

2. log10 = Logarithm to the base 10 4. logex = 2.303 log10x

(r1, q1)

(0,c) θ

∆y = (y2 – y1)

θ

∆x = (x2 – x1) c x

O (0,0)

Slope of the line (m) is also equal to tan q where q is the angle that the line makes with the positive x-axis. slope m

y y2 y1

tan x x2 x1

This formula is popularly known as distance formula.

m and c can have either positive or negative values.

2. The distance between two points P1 and P2 with polar coordinates (r1, q1) and (r2, q2)

(a) If m > 0, the straight line has a positive slope and q is acute.

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5

y

(b) If m < 0, the straight line has a negative slope and q is abtuse. (c) If m = 0, then we have straight line parallel to x-axis.

Three other possible situations are further shown in figure.

(0, b)

y (1) m > 0 c<0

O

x

7. Equation of rectangular hyperbola is xy = constant.

x (2) m < 0 c>0

y

(3) m < 0 c<0

x

4. Equation of a circle of radius a centered at origin is x2 + y2 = a2. y a

1.5 x

O

x2 y2

1. a 2 b2

180 degree

= 57.3

1 = radian 180 = 0.0174 rad.

1. 1 rad =

5. Equation of an ellipse with origin as its centre is

BASIC TRIGONOMETRY

a = semi-major axis

b = semi-minor axis

2. Basic functions of trigonometry are sine, cosine, and tangent functions defined in terms of the ratio of the sides of the

b (0, 0)

x

a 6. Equation of parabola with vertex at (0, b) y = ax2 + b

right angled triangle. These functions are defined as follows.

90 – θ

h

p θ b

6

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CHAPTER 1: Basic Mathematics

(a) sin

Perpendicular p

Hypotenuse h

(b) cos

Base b

Hypotenuse h

(c) tan

Perpendicular p

Base b

6. Trigonometric ratios:

3. The cosecant (cosec), secant (sec) and cotangent (cot) functions are defined as 1 cosec sin 1 sec cos 1 cot = tan

Quadrant I (90° – q) sin (90° – q) = cos q

Quadrant I (360° + q) sin (360° + q) = sin q

cos (90° – q) = sin q

cos (360° + q) = cos q

tan (90° – q) = cot q

tan (360° + q) = tan q

Quadrant II (90° + q) sin (90° + q) = cos q

Quadrant II (180° – q) sin (180° – q) = sin q

cos (90° + q) = –sin q

cos (180° – q) = –cos q

tan (90° + q) = –cot q

tan (180° – q) = –tan q

Quadrant III Quadrant III (180° + q) (270° – q) sin (180° + q) = –sin q sin (270° – q) = –cos q cos (180° + q) = –cos q cos (270° – q) = –sin q

4. Quadrant I has all of sin, cos, tan, etc. as

tan (180° + q) = tan q

tan (270° – q) = cot q

Quadrant II has sine and cosec as positive

Quadrant IV Quadrant IV (270° + q) (360° – q) sin (270° + q) = –cos q sin (360° – q) = –sin q

and all others as negative.

cos (270° + q) = sin q

Quadrant III has tan and cot as positive

tan (270° + q) = –cot q tan (360° – q) = –tan q

positive.

and all others as negative.

Quadrant IV has cosine and secant as positive and all others as negative.

90° Quadrant II Sine and cosec positive Sugar

Quadrant I All positive Add

180°

0° (360°) Quadrant III Tan and cot positive To

Quadrant IV Cos and sec positive Coffee 270°

5. sin (–q) = –sin q

(ODD Function)

cos (–q) = cos q

(EVEN Function)

tan (–q) = –tan q (ODD Function)

cos (360° – q) = cos q

CAUTION (i) While writing the above formulae, we must see the quadrant in which the angle lies, e.g., (90° – q) lies in quadrant I, (90° + q) and (180° – q) lie in quadrant II, (180° + q) and (270° – q) lie in quadrant III, (270° + q) and (360° – q) lie in quadrant IV. Finally (360° + q) lies back in quadrant I. According to location of the angle in a Quadrant, the values of sine, cosine, and tangent have to be assigned a sign. (ii) Further, if the principle value of an angle is odd multiple of 90°, like (90° – q), (90° + q), (270° + q), then sine has to be changed to cosine, cosine to

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sine, tangent to cotangent, and so on.

Similarly, no changes are done when principal value is an even multiple of 90°, like (180° – q), (180° + q), (360° – q), (360° + q).

7. Law of cosines for a triangle: γ a

b

β

α c

a2 = b2 + c2 – 2bc cos a b2 = a2 + c2 – 2ac cos b c2 = a2 + b2 – 2ab cos g 8. Law of sines (Lami’s theorem): sin sin sin

a b c 9. Some useful trigonometric identities: 1. sin2 q + cos2 q = 1 2. sec2 q = 1 + tan2 q 3. cosec2 q = 1 + cot2 q 4. sin (2q) = 2sin q cos q

5. sin 2 sin cos 2 2 2 2 6. cos 2 cos sin 2 2 7. cos cos sin 2 2 1 cos 2 2 8. sin 2 2 9. 1 cos 2 sin 2

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2 10. 1 cos 2 cos 2 11. sin A B sin A cosB cosA sinB

12. cos A B cos A cosB  sinA sinB A B A B cos 13. sin A + sinB = 2sin

2 2 A B A B 14. sin A sinB = 2cos sin

2 2 A B A B 15. cos A cosB = 2cos cos

2 2 A B A B sin

2 2

16. cos A cos B 2 sin 17. tan 2

2 tan 1 tan2

2 tan 2 18. tan

1 tan2 2 2

19. cos

1 cos 2 2

1 cos

20. tan 1 cos 2 1.6

EXPONENTIAL AND ROOTS

1 x n n x 1/n n x x 1 x n/m m n x x 0 1;

n

x xn n n n x y y y

x x n

m

n m 1/n

1/n

x y

1/n

x (xy ) ; y 1/n

x1/n 1/n y

7

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8

1.7

CHAPTER 1: Basic Mathematics

BINOMIAL THEOREM

d n x nx n 1 dx

3.

dy d Ky K where K is a constant dx dx

4.

d Kx n K nx n 1 dx

5.

d du dv dw u v w ... ... dx dx dx dx

6.

d udv vdu uv dx dx dx

7.

du du v u d u dx 2 dx dx v v

8.

dy dy . du

Chain rule dx du dx

9.

d sin ax acos ax dx

10.

d cos ax asin ax dx

11.

d ax e ae ax dx

12.

d 1 1 log e (ax ) a dx ax x

13.

d 1 log a x dx x log e a

14.

d dy d 2 y

dx dx dx 2

2

n(n 1)x

2! n(n 1) 2 x (1 x ) n 1  nx 2!

If x << 1, then (1 ± x)n (1 x )n 1 nx

= 1 ± nx (neglecting higher terms) (1 ± x)–n = 1 ± (– n)x = 1  nx

Key Insights: ■ When n is a positive integer, then expansion will have (n + 1) terms.

■ When n is a negative integer, expansion will have infinite terms.

■ When n is a fraction, expansion will have infinite terms.

1.8

SERIES EXPANSION

n(n 1) 2 x 2! n(n 1)(n 2) 3

x 3! x2 x3 2. e x 1 x 2 ! 3!

1. (1 x )n 1 nx

3. log e (1 x ) x

x2 x3 x4 x5 ..

2 3 4 5

x3 x5 4. sin x x , x is in radian 3! 5! x2 x4 5. cos x 1 , x is in radian 2! 4!

For x << 1 n

(1 + x) ≅ 1 + nx

sin x ≅ x

ex ≅ 1 + x

cos x ≅ 1

ln (1 ± x) ≅ + x

tan x ≅ x

1.9 1.

DIFFERENTIAL CALCULUS d constant 0 dx

2.

[Read as ‘dee square y by dee x square’] [also written as y']

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Geometrical Significance of

dy dx

dy is actually the rate measurer, which dx measures the change in value of y, a dependent variable, according to the change in the value of x, an independent variable.

Geometrically, it gives us the instantaneous slope of a curve.

y

9

So, for the maximum or minimum value dy =0. of y , dx Just before the minimum point, the slope is negative. At the minimum point, it is zero, and just after the minimum point, it is positive. dy Thus, increases at the minimum point, dx dy i.e., the rate of change of is positive at dx the minimum point. So, d dy 0 dx dx d2 y 0 or, dx 2

P(x,y) θ

O 1.10

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Hence, condition for the minimum value of y is dy d2 y 0 and 0 dx dx 2

x

Similarly, the condition for the maximum value of y is

MAXIMA AND MINIMA

In Physics, we usually deal with those functions that have a continuous and smooth graph.

Let us suppose that a quantity y depends upon another quantity x in a manner shown in the figure below. It becomes minimum at x1 and maximum at x2. At these points, the tangent to the curve is parallel to the x-axis and its slope is tan θ = 0.

Remark: While doing physics problems, when we are asked to find the condition of either maxima or of minima, we just have to equate the derivative (of the function to be minimised or maximised) to zero; rest can be visualised physically.

1.11

y

dy d2 y 0 and 0 dx dx 2

INTEGRAL CALCULUS

Consider a function y = f(x) If

d [ f (x )] = g (x ) , dx

then g (x )dx f (x ) O

x1

x

x2

The slope of the curve =

dy . dx

Integration is just the reverse process of differentiation.

10

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CHAPTER 1: Basic Mathematics b

Geometrical significance of ∫a g(x)dx

1.12

dy gives us the dx instantaneous slope of the curve. y g(x) We have already studied that

a

b

x

Similarly, ba g (x )dx f (b) f (a) = area of shaded portion

This gives the area under the curve g(x) between the lower limit ‘a’ to the upper limit ‘b’.

SOME INDEFINITE INTEGRAL FORMULAE x n 1 (n 1) n 1

1.

x ndx

2.

dx log e x ln x x

3.

dx 1 log e | a bx | a bx b

4.

ax 2

(ax b)dx bx 2

5.

dx 1 x a log e 2 a 2 x a x a 2

6. ∫sin x dx = – cos x 7. ∫cos x dx = sin x

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11

Exercises JEE MAIN 1. If a + b = 15 and ab = 30, then the value of a2 + b2 is (1) 195 (2) 225 (3) 165 (4) 123

1 1    2. If  x +  = 5, then  x 2 + 2  = x x     (1) 25 (3) 23

(2) 27 (4) 1/25

3. The positive real root of the equation 81x2 –1 = 0 is (1) 1/3 (2) 1/9 (3) 1/18 (4) 1/27 4. If the roots of the quadratic equation x2 + px + 16 = 0 are equal, then the value of p is (1) ±9 (3) ±7

(2) ±6 (4) ±8 −2

 h 5. Find  1 +  if h << R.  R (1) z = xy

2h R

(2) 1 +

2h R

h R 6. What will be the binomial approximation 1 of when x << = 1?

(1 − x ) (3) 1 −

(4) 1 −

x (1) 1 − 2

x (2) 1 + 2

(3) 1 – x

(4) 1 + x

7. Find the 13th term of the following series: 1, 3, 5, 7, 9, 11.......

(1) 21 (3) 25

(2) 28 (4) 29

1 1 1 8. The value of 1 + + + + .... up to 4 16 64 n . Find the value of n. ∞ is 3 (1) 4 (2) 2 (3) 5 (4) 7 9. Find the 10th term of the given series: 2, 4, 8, 16, 32, 64, ……

(1) 512

(2) 1024

(3) 2048

(4) 128

10. In GP 4, 8, 16, 32, …………, find the sum up to 5th term.

(1) 16 (2) 64 (3) 124 (4) 128 11. The value of the sum of first hundred natural numbers 1 + 2 + 3 + .... + 100 is (1) 4950 (2) 5050 (3) 5150 (4) 52050 12. If log2 a = 5, then the value of a is (where a > 0) (1) 64 (2) 32 (3) 25 (4) e5 13. The distance of the point (9, –12) from the origin will be (1) 13 units (2) 14 units (3) 15 units (4) 16 units 14. The distance between (p, –q) and (q, –p) is (p > q) (1) p – q (3)

2 ( p − q)

(2) p + q (4)

2 ( p + q)

15. The equation of the line having inclination 45° and y–intercept –2 is (1) x + y + 2 = 0 (2) x – y + 2 = 0 (3) x – y – 2 = 0 (4) x + y – 2 = 0

12

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CHAPTER 1: Basic Mathematics

16. The equation of the line having slope a/b and passing through the origin is −b (1) y = (2) bx – ay = 0 x a (3) ax + by = 0 (4) ax – by = 0 17. Find the sum of the following series: 12 + 22 + 32 + .... + 102.

(1) 285 (2) 385 (3) 429 (4) 324

23. If the curved surface area of a cylinder of height 21 cm is 660 cm2, find its radius.

(1) 4 cm (2) 6 cm (3) 5 cm (4) 7 cm 24. Total surface area of a hemisphere of radius r is (1) 2πr2 (2) 4πr2 (3) 3πr2 (4) πr2

1 , then cos2q =

18. Find the value of x in the equation 10x2 – 27 + 5 = 0.

(1) 1/5 (2) 1/6 (3) 1/8 (4) 1/9

25. If sin θ = (1) 1/2

(2) 1/3

19. The slope of the graph, as shown in the figure, at points 1, 2 and 3 is m1, m2 and m3, respectively. Then,

(3) 1/5

(4) 1/8

y

2

(4) 2cosx cosy 27. tan (180° + q) =

x

(1) tan q

(2) cot q

(2) m1 < m2 < m3 (4) m1 = m2 > m3

(3) –tan q

(4) –cot q

20. Find the sum of the series 1 + 2 + 3 + 4 + ... + n3.

3

(2) (3) (4)

(1) 2sinx cosy (3) 2sinx siny

3

(1) m1 > m2 > m3 (3) m1 = m2 = m3

n2 ( n + 1)

3

3

3

2

4 2

26. sin(x + y) + sin(x – y) = (2) 2cosx siny

1

(1)

3

2

n (n + 1) 6

n ( n + 1)( 2n + 1) 6

n ( n + 1)( 2n + 3 )

6 21. If the surface area of a sphere is (144π) m2, find its volume.

(1) 288π m3 (2) 188π m3 3 (3) 300π m (4) 316π m3 22. Find the volume of right circular cone having base diameter 16 cm and height 15 cm.

(1) 60π cm3 (2) 68π cm3 (3) 149π cm3 (4) 320π cm3

28. If A and B are acute angles such that sinA = cosB, then A + B = (1) π/2

(2) π/4

(3) π

(4) π/3

3 (1 − cos θ ) = 29. If tan θ = then 4 (1 + cos θ ) (1) 9

(2) 1/9

(3) 4

(4) 1/4

30. sin15°. cos30°+cos15° . sin30°= (1) 1

(2)

3 (3) 2

(4) 0

1 2

31. For the equation y = –2x2 + 3, find the nature of the graph.

(1) Parabola passing through origin (2) Parabola not passing through origin (3) Parabola passing through (2, 1) (4) Hyperbola passing through origin

IL RANKER SERIES FOR JEE 32. Graph of the function y = 4 + x2 is shown in figure. What is the value of a? y

(0,a) (0, 0)

(1) 2

(2) 4

x

(3) 0

(4) 2 2

(4) 4.474 cm

(1) 5 cm

(2) 8 cm

(3) 12 cm

(4) 7 cm

1 1 (1) − (2) − 2 5

B

(3)

(3, 3)

(2) −

3 2

(3) + 2 3

(4) −

2 3

1

2

(4) 1

38. If y = x3 + 2x2 + 7x + 8, then

x

(1, 0)

3 2

(1) +

(3) 3.744 cm

37. Find the value of cos120°.

y

A

(2) 2.743 cm

13

36. The area of a trapezium is 24 cm2 and the distance between its parallel sides is 4 cm.

If one of the parallel sides is 7 cm, then the other parallel side is

33. The slope of straight line AB is

(0, 0)

(1) 2.744 cm

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(1) 3x2 + 2x + 15 (2) 3x2+ 4x + 7 (3) x3+ 2x2 + 15 (4) x3+ 4x + 7 39. If y = sinx, then

34. The equation of shown straight line is

(1) cosx (3) –sinx

dy will be dx

d2 y will be dx 2 (2) sinx (4) sinx + c

40. Given below are two statements.

y (0, 5)

S-I : Slope of horizontal line is zero and slope of vertical line is undefined.

(0, 0)

S-II : Two lines whose slopes are m1 and m2 are perpendicular if and only if m1m2 = –1 In light of the given statements, choose the correct answer from the options given below.

(1) Both S-I and S-II are correct.

(2) Both S-I and S-II are incorrect.

(3) S-I is correct but S-II is incorrect.

(4) S-I is incorrect but S-II is correct.

(1)

x y 1 + = 4 5

(3) 4x + 5y = 1

(4, 0) (2)

x

x y 1 + = 5 4

(4) 5x + 4y = 1

35. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Then, the height of the cylinder is

dy will be dt (2) cos x (4) –cos x

41. If y = sin x and x = 3t, then (1) 3 cos (x) (3) cos (3t)

14

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CHAPTER 1: Basic Mathematics

42. The area A of a blot of ink is growing such that, after t seconds, its area is given by A = 5t2 + 9 cm2. Calculate the rate of increase of area at t = 5 sec.

(1) 30 cm2/s

(2) 134 cm2/s

49. The point for the curve y = xex (1) x = –1 is minimum (2) x = 0 is minimum (3) x = 0 is maximum (4) x = –1 is maximum

(3) 50 cm2/s

(4) 40 cm2/s

50. Find the derivative of log10x2

43. A particle is moving on the x-axis. Its position x at time t is given by x = 4t – t2.

Find its position at the time when its velocity dx

is zero. Velocity = dt

(1) 1 m

(2) 2 m (3) 3 m

(4) 4 m

44. Differentiate the following functions with respect to x, (2x + 3)6 (1) 12(2x + 3)6

(2) 6(2x + 3)5

(3) 6(2x + 3)6

(4) 12(2x + 3)5

45. A particle is moving along the x-axis.

At any instant, its x-coordinate is dx x = 4 + 3t + 5t2. What is the value of dt at t = 1?

dy 1 1 + 3 . Find dx x x

dy . dx

(3) sec2x

1 x

(2) cosec2x (4) cot2x

52. The derivative of f(x)=esinx w.r.t g (x) = sinx is (1) e2sinx

(2) esinx

(3) ecosx

(4) 1

dy , where y = sec2 x dx (1) 2 secx tanx (2) 2 sec2x tanx

53. Find

(4) sec2x tanx

(1) No maxima and minima (2) One maximum (3) One minima (4) Two minima

(1) 1 + 2x −

1 3 − 4 2 x x

(2) 1 + 2x −

1 2 + 4 2 x x

(3) 1 − 2x −

1 3 + 4 2 x x

(4) 1 + 2x −

1 3 − x2 x3

48. Find derivative of sin22x.

(1) 2 sin2x cos2x (2) 2 sin4x (3) 2 cos2x (4) –2 sin2x cos2x

(1) sin2x

(4) log e 10

54. The function f(x) = 2x3−3x2−12x+4 has

46. Find the derivative of 3x2+2x+5 at x = 2 (1) 12 (2) 16 (3) 8 (4) 14 2

51. If y = tanx, find

(2) 2x logex

(3) secx tanx

(1) 12 (2) 13 (3) 7 (4) 8

47. y = x + x +

1 x 1 (3) log10 e 2 x (1)

55. The function x5–5x4+5x3–10 has a maxima, when x = (1) 3

(2) 2

(3) 1

(4) 0

dy sin x = , then dx x cos x − sin x x sin x − cos x (1) (2) 2 x x2

56. If y =

(3)

x cos x − sin x x2

(4)

x cos x − sin x x3

IL RANKER SERIES FOR JEE 57. x = 3cosθ, y = 3sinθ, then (1) −cotθ

dy = dx

R

61. The value of ∫ ∞ (1)

(2) −sinθ (3) −tanθ

(3) −

(4) −secθ 58. Find

GMm R 2GM R

(4)

3

dy , where = y dx

( 3x + 4 x + 5 ) 2

(2)

5 2 ( ax + b ) 2 + C 5a

3x + 2 2 3x + 4 x + 5

(3)

5 5 ( ax + b ) 2 + C 2a

3x − 2 (3) 2 3x + 4 x + 5

(4)

3 3a ( ax + b ) 2 + C 5

(4)

3x − 2 2

3x + 4 x + 5

63. ∫ e ax +b dx

3x + 2 3x 2 + 4 x + 5

59. Differentiate the following with respect to x,

x−

2GM R

62. Find the value of ∫ ( ax + b ) 2 dx −5 2 ( ax + b ) 2 + C 3a

(2)

15

GMm dr is equal to: r2 GMm (2) − R

(1)

(1)

|

1 x

(1)

1  1 1 +  2 x x

(2)

1  1 1 −  2 x x

(3)

1  1 1 +  x x

(4)

1  1  1 + 2x  2 x 

(1) e ax +b + k

(2)

(3) ae ax +b + k

(4) 5

1 ax +b e +k a a e

ax + b

+k

64. The integral ∫ 1x dx is equal to (1)

2

125 3

124 3 1 (3) 3 (2)

(4) 45 65. Integrate the following functions with respect

60. Find the integration of 5x + 7.

to x ∫(6 x + 2)3 dx

(1)

7 2 x + 7x 2

(2)

5 2 x + 7x 2

(1)

(6 x + 2)4 + c2 24

(2)

(6 x + 2)4 + c2 25

(3)

5 x 2

(4)

3 2 x + 7x 2

(3)

(6 x + 2)2 + c2 24

(4)

(3x + 2)4 + c2 24

16

|

CHAPTER 1: Basic Mathematics

66. Integrate the following functions with respect to x ∫ dx 4x + 5 (1)

70. Determine the average value of y = 2x+3 in the interval 0 ≤ x ≤ 1.

(1) 1

1 ln ( 4 x + 5 ) + c2 5

71. ∫

1 (2) ln ( 4 x + 5 ) + c2 4 (3) ln(4x+5)+c2 1 (4) ln ( 4 x + 5 ) + c 4

of f(x).

dx

( 2ax − x ) 2

1/2

(3) 3

(2) 1

(3) 2

(4) 1/2

e x sin x + cos x + x sin x dx sin x

(1) e x − cos ec 2 x +

x2 +c 2 x2 +c 2

(1)

5x 3 3x 2 + −x+c 3 2

(2) e x + cos ec 2 x +

(2)

5x 3 3x 2 + 3 2

(3)

(3)

5x 3 3x 2 + − 2x + c 4 2

(4) e x − log sin x +

(4)

5x 3 3x 2 + − 2x + c 3 2 1

1 dx, then the value of l is 2x + 3 0

68. If l = ∫

5 (1) ln 3 (3)

(2) ln 3

1 ln5 2

(4)

1 5 ln 2 3

constants.

0

A (1) (1 + sin ωt ) ω (3)

A (1 + cosωt ) ω

A (1 − sin ωt ) (2) ω (4)

A (1 − cosωt ) ω

1 − cos ec 2 x + 1 + c x x2 +c 2

73. The instantaneous velocity of a particle moving in a straight line is given as v = αt + βt2, where α and β are constants.

The distance travelled by the particle between t2

1 s and 2 s is (where distance = ∫ vdt ) t1

(1) 3α + 7β

(2)

3 7 α+ β 2 3

α β + 2 3

(4)

3 7 α+ β 2 2

t

69. Evaluate ∫A sin ωt dt where A and ω are

(4) 4

x  = an sin −1  − 1  , the value a 

of n is (1) 0

72. ∫

67. If f(x) = 5x2 + 3x – 2, find the antiderivative

(2) 5

(3)

74. An object, moving with a speed of 6.25 m/s, dv = −2.5 v , is decelerated at a rate given by dt where v is the instantaneous speed. The time taken by the object to come to rest would be

IL RANKER SERIES FOR JEE (1) 2 s (3) 8 s (4) 1 s 75. The velocity of a particle is v = v 0+ gt +

77. Calculate the area enclosed under the curve f(x)=x2 between the limits x=2 and x=3

ft2. If its position is x = 0 at t = 0, then its

y

displacement after unit time (t = 1) is (where

(1) v0 + 2g + 3f

17

Numerical Value Questions 5 4 3 76. If y= 4 x − 2 x + + 9 then find slope of x curve at x = 1

(2) 4 s

displacement =

|

t2

∫ vdt )

t1

(2) v0 + g/2 + f/3

0

(3) v0 + g + f

1

2

3

x

2π

(4) v0 + g/2 + f

78. Find value of ∫ cos xdx = ____ 0

ANSWER KEY (1) 3

(2) 3

(3) 2

(4) 4

(5) 3

(6) 2

(7) 3

(8) 1

(9) 2

(10) 3

(11) 2

(12) 2

(13) 3

(14) 3

(15) 3

(16) 4

(17) 2

(18) 1

(19) 2

(20) 1

(21) 1

(22) 4

(23) 3

(24) 3

(25) 2

(26) 1

(27) 1

(28) 1

(29) 2

(30) 2

(31) 2

(32) 2

(33) 1

(34) 1

(35) 1

(36) 1

(37) 1

(38) 2

(39) 3

(40) 1

(41) 1

(42) 3

(43) 4

(44) 4

(45) 2

(46) 4

(47) 1

(48) 2

(49) 1

(50) 3

(51) 3

(52) 2

(53) 2

(54) 2,3 (55) 3

(56) 3

(57) 1

(58) 4

(59) 1

(60) 2

(61) 2

(62) 2

(63) 2

(64) 2

(65) 1

(66) 2

(67) 4

(68) 4

(69) 4

(70) 4

(71) 1

(72) 4

(73) 2

(74) 1

(75) 2

(76) 5

(77) 6.33 (78) 0