Class 12th PCMB

Q101. Choose the correct answer from the given four options: The function f(x) = 2x – 3x – 12x + 4, has:

1. Two points of local maximum.

2. Two points of local minimum.

3. One maxima and one minima.

4. No maxima or minima.

( )

( )

[ ]

[ ]

( )

( )

( )

( )

( )

( )

3 2

Ans: 3. One maxima and one minima.

Solution:

We have, f(x) = 2x – 3x – 12x + 4

⇒ f'(x) = 6x – 6x – 12

⇒ f'(x) = 6(x – x – 2)

⇒ f'(x) = 6(x + 1)(x – 2)

Find the critical points by equating f'(x) to 0.

∴ f'(x) = 0

⇒ 6(x + 1)(x – 2) = 0

⇒ x = -1 and x = +2

From the above number line, we can conclude that, x = -1 is point of local maxima and x = 2 is point of local minima.

Thus, f(x) has one maxima and one minima.

Q102. The function f(x) = x e is monotonic increasing when:

1. x ∈ R − [0, 2]

2. 0 < x < 2

3. 2 < x < ∞

4. x < 0

Ans: 2. 0 < x < 2

Solution:

f(x) = x e

⇒ f'(x) = -x e + 2xe

⇒ f'(x) = -e x(x – 2)

Given that function is monotonically increasing.

-e x(x – 2) > 0

x(x – 2) < 0

0 < x < 2

Q103. Choose the correct answer from the given four options:

At x =

5π

6 , f(x) = 2sin3x + 3cosx is 6:

1. Maximum.

2. Minimum.

3. Zero.

4. Neither maximum nor minimum.

Ans: 4. Neither maximum nor minimum.

Solution:

We have, f(x) = 2sin3x + 3cos3x

∴ f ′ (x) = 2 ⋅ cos3x3 + 3( − sin3x) ⋅ 3

⇒ f ′ (x) = 6cos3x − 9sin3x …(i)

Now, f ″ (x) = − 18sin3x − 27cos3x

= − 9(2sin3x + 3cos3x)

∴ f ′ 5π

6 = 6cos 3 ⋅

5π

6 − 9sin 3 ⋅

5π

6

= 6cos

5π

2 − 9sin

5π

2

= 6cos 2π +

π

2 − 9sin 2π +

π

2

= − 9 ≠ 0

So, x =

5π

6 cannot be point of maxima or minima.

Hence, f(x) at x =

5π

6 is neither maximum nor minimum.

Q104. The point on the curve 9y = x , where the normal to the curve makes equal intercepts with the axes is:

1. 4,

8

3

3 2

2

2

2 -x

2 -x

2 -x -x

-x

-x

( ) ( ) ( )

( ) ( )

2 3

( )

2. − 4,

8

3

3. 4, −

8

3

4. None of these.

Ans: 1. 4,

8

3

Solution:

9y = x

⇒ 18y

dy

dx = 3×2

⇒

dy

dx =

x2

6y = slope of tangent

Slope of normal =

− 6y

x2

Given that normal makes equal intercept on axes.

⇒ Slope of normal = ± 1

− 6y

x2 = 1 or

− 6y

x2 =

= − 1

⇒ y =

− x2

6 or y =

x2

6

y =

− x2

6

9y2 = x3

9

− x2

6

2 = x3

⇒ x = 0 or 4

⇒ y = 0, ±

8

3

Point are 4,

8

3 and 4,

− 8

3

Q105. The diameter of a circle is increasing at the rate of 1cm/sec. When its radius is π the rate of increase of its area is:

1. πcm2 / sec.

2. 2πcm2 / sec.

3. π2cm2 / sec.

4. 2π2cm2 / sec2.

Ans: 3. π2cm2 / sec.

Solution:

Let D be the diameter and A be the area of the cricle at any time t. Then,

A = πr2 (where r is the radius of the circle)

⇒ A = π

D2

4 ∴ r =

D

2

⇒

dA

dt = 2π

D

4

dD

dt ∴

dD

dt = 1cm/ sec

⇒

dA

dt = π2cm2 / sec.

Q106. The distance moved by the particle in time t is given by x = t – 12t + 6t + 8. At the instant when its acceleration is

zero, the velocity is:

1. 42

2. -42

3. 48

4. -48

Ans: 2. -42

Solution:

x = t3 − 12t2 + 6t + 8

⇒

dx

dt = 3t2 − 24t + 8

⇒

d2x

dt2 = 6t − 24

⇒ 6t − 24 = 0 [ ∴ acceleration is zero]

( )

( )

( )

2 3

( )

( ) ( )

[ ]

[ ]

3 2

⇒ t = 4

So, Velocity at t = 4

⇒

dx

dt = 3(4)2 − 24 × 4 + 6

⇒

dx

dt = 48 − 96 + 6

⇒

dx

dt = − 42

Q107. The equation of the normal to the curve 3x – y = 8 which is parallel to x + 3y = 8 is:

1. x + 3y = 8

2. x + 3y + 8 = 0

3. x + 3y ± 8 = 0

4. x + 3y = 0

Ans: 3. x + 3y ± 8 = 0

Solution:

Since the normal is parallel to the given line, the equation of normal will be of the given form.

x+3y = k

3×2 − y2 = 8

Let (x , y ) be the point of intersection of the two curves.

Then,

x1 + 3y1 = k . . . (1)

3×21

− y21

= 8 . . . (2)

Now, 3×2 − y2 = 8

On diffierentiating both sides w.r.t.x, we get

6x − 2y

dy

dx = 0

⇒

dy

dx =

6x

2y =

3x

y

Slope of the normal, =

dy

dx ( x1 , y1 ) =

3×1

y1

Slope of the normal, m =

− 1

3x

y

=

− y1

3×1

Given:

Slope of the normal = Slope of the given line

⇒

− y1

3×1

=

− 1

3

⇒ y1 = x1 . . . (3)

From (2), we get

3×1

2 − x1

2 = 8

⇒ 2×1

2 = 8

⇒ x1

2 = 4

⇒ x1 = ± 2

Case 1

when x1 = 2

From (3), we get

y1 = x1 = 2

From (3), we get

2 + 3 (2) = k

⇒ 2 + 6 = k

⇒ k = 8

∴ Equation of the normal from (1)

⇒ x + 3y = 8

⇒ x + 3y – 8 = 0

⇒ -2 – 6 = k

⇒ k = -8

∴ Equation of the normal from (1)

⇒ x + 3y = -8

2 2

1 1

( )

( )

⇒ x + 3y – 8 = 0

From both the case, we get the equation of the normal as:

x+3y ± 8 = 0

Q108. Choose the correct answer:

The rate of change of the area of a circle with respect to its radius r at r = 6 cm is:

1. 10π

2. 12π

3. 8π

4. 11π

Ans: Area of circle (A) = πr2 ⇒

dA

dr = 2πr = 2π × 6 = 12π

Therefore, option (B) is correct.

Q109. The volume of a sphere is increasing at 3cm /sec. The rate at which the radius increases when radius is 2cm, is:

1.

3

32π cm/ sec.

2.

3

16π cm/ sec.

3.

3

48π cm/ sec.

4.

1

24π cm/ sec.

Ans: 2.

3

16π cm/ sec.

Solution:

Let r be the radius and V be the volume of the sphere at any time t. Then, v =

4

3 πr3

⇒

dv

dt = 4πr2 dr

dt

⇒

dr

dt =

1

4πr2

dv

dt

⇒

dr

dt =

3

4π ( 2 )2

⇒

dr

dt =

3

16π cm/ sec.

Q110. Choose the correct answer from the given four options:

If y = x(x – 3) decreases for the values of x given by:

1. 1 < x < 3

2. x < 0

3. x > 0

4. 0 < x <

3

2

Ans: 1. 1 < x < 3

Solution:

We have, y = x(x – 3)

∴

dy

dx =2(x – 3).1 + (x – 3) .1

= 2x – 6x + x + 9 = 6x = 3x – 12x + 9

= 3(x – 3x – x + 3) = 3(x – 3)(x – 1)

So, y = x(x – 3) decreases for (1, 3).

[since, y’ < 0 for all x ∈ (1, 3), hence y is decreasing on (1, 3)]

Q111. Let f(x) = 2x – 3x – 12x + 5 on [-2, 4]. The relative maximum occurs at x =

1. -2

2. -1

3. 2

4. 4

3

2

2

2

2 2 2

2

2

3 2

Ans: 3. 2

Solution :

Given, f(x) = 2x – 3x – 12x + 5

⇒ f'(x) = 6x – 6x – 12

For a local maxima or a local minima, we must have f'(x) = 0

⇒ 6x – 6x – 12 = 0

⇒ x – x – 2 = 0

⇒ (x – 2)(x + 1) = 0

⇒ x = 2, -1

Now, f”(x) = 12x – 6

⇒ f”(-1) = -12 – 6 = 18 > 0

So, x = 1 is a local maxima.

Also, f”(2) = 24 – 6 = 18 > 0

So, x = 2 is a local minima.

Q112. Let ϕ(x) = f(x) + f(2a − x) and f'(x) > 0 for all x ∈ [0, a]. Then, ϕ(x):

1. Increases on [0, a]

2. Decreases on [0, a]

3. Increases on [-a, 0]

4. Decreases on [a, 2a]

Ans: 2. Decreases on [0, a]

Solution:

ϕ(x) = f(x) + f(2a − x)

ϕ ′ (x) = f ′ (x) − f ′ (2a − x)

f ″ (x) > 0 as f ′ (x) > 0

Considering x ∈ [0, a]

x ≤ 2a − x

f ′ (x) ≤ f(2a − x)

Also, ϕ(x) = f ′ (x) − f ′ (2a − x)

ϕ(x) is decreasing on [0, a]

Q113. The distance moved by a particle travelling in straight line in t seconds is given by s = 45t + 11t – t The time

taken by the particle to come to rest is:

1. 9 sec.

2.

5

3 sec.

3.

3

5 sec.

4. 2 sec.

Ans: 1. 9 sec.

Solution:

s = 45t + 11t2 − t3

ds

dt = 45 + 22t − 3t2

Given that particle moves in a straight line.

⇒

ds

dt = 0

⇒ 3t2 − 22t − 45 = 0

⇒ t = 9 or t ≠

− 5

3

t = 9 sec.

Q114. The minimum value of xlogex is equal to :

1. e

2.

1

e

3.

− 1

e

4. 2e

3 2

2

2

2

2 3 .

Ans: 3.

− 1

e

Solution :

Here, f(x) = xlogex

lmplies that f ′ (x) = logex + 1

For a local maxima or a local minima, we must have f'(x) = 0

lmplies that logex + 1 = 0

lmplies that logex = − 1

lmplies that x = e − 1

Now, f ″ (x) =

1

x

lmplies that f ″ (e − 1) = e > 0

Threrfore, f(e − 1) is a local minima.

Hence, the minimum value of f(x) = f(e − 1)

lmplies that e − 1loge(e − 1) = − e − 1 =

− 1

e

Q115. The minimum value of f(x) = x – x – 2x + 6 is.

1. 6

2. 4

3. 8

4. None of these.

Ans: 2. 4

Solution:

Given, f(x) = x – x – 2x + 6

⇒ f'(x) = 4x – 2x – 2

⇒ f'(x) = (x – 1)(4x + 4x + 2)

For a local maxima or a local minima, we must have f'(x) = 0

⇒ (x – 1)(4x + 4x + 2) = 0

⇒ (x – 1) = 0

⇒ x = 1

Now, f”(x) = 12x – 2

⇒ f”(x) = 12 – 2 = 10 > 0

So, x = 1 is a local minima.

The local minimum value is given by

f(1) = 1 – 1 -2 + 6 = 4

Q116. In the interval (1, 2), function f(x) = 2|x – 1| + 3|x – 2| is:

1. Monotonically increasing.

2. Monotonically decreasing.

3. Not monotonic.

4. Constant.

Ans: 2. Monotonically decreasing.

Solution:

f(x) = 2|x – 1| + 3|x – 2|

x ∈ (1, 2)

x > 1 and x < 2

⇒ x – 1 > 0 and x – 2 < 0

⇒ f(x) = 2|x – 1| + 3|x – 2|

⇒ f(x) = 2(x – 1) – 3(x – 2)

⇒ f(x) = 2x – 2 – 3x + 6

⇒ f(x) = -x + 4

⇒ f'(x) = -1

Hence, function is monotonically decreasing.

Q117. For the function f(x) = x +

1

x

1. x = 1 is a point of maximum.

4 2

4 2

3

2

2

2

2. x = -1 is a point of minimum.

3. maximum value > minimum value.

4. maximum value < minimum value.

Ans: 4. maximum value < minimum value.

Solution:

Given, f(x) = x +

1

x

lmplies that f ″ (x) = x −

1

x

For a local maxima or a local minima, we must have f'(x) = 0

lmplies that 1 −

1

x2 = 0

lmplies that x2 − 1 = 0

lmplies that x2 = 0

lmplies that x = ± 1

Now, f ′ (x) =

2

x3

lmplie that f ′ (1) =

2

1 = 2 < 0

Threrefore, x = 1 is a local minima.

Also, f”(1) – 2 < 0

Threrefore, x = -1 is a local maxima.

The local minimum value is given by

f(1) = 2

The local maximum value is given by

f(-1) = -2

∴ maximum value < minimum value.

Q118. Choose the correct answer from the given four options:

The function f(x) = tanx − x

1. Always increases.

2. Always decreases.

3. Never increases.

4. Sometimes increases and sometimes decreases.

Ans: 1. Always increases.

Solution:

We have, f(x) = tanx − x

∴ f ′ (x) = sec2x − 1

Since, f ′ (x) > 0, ∀ x ∈ R

Hence, f(x) always increases.

Q119. Choose the correct answer

The maximum value of [x(x − 1) + 1]

1

3 , 0 ≤ x ≤ 1]is:

1.

1

3

1

3

2.

1

2

3. 1

4. 0

Ans: 1. 1

Let f(x) = [x(x − 1) + 1]

1

3 = (x2 − x + 1)

1

3 , 0 ≤ x ≤ 1 …(i)

∴ f ′ (x) =

1

3 (x2 − x + 1)

− 2

3

d

dx (x2 − x + 1) =

( 2x − 1 )

3 ( x2 − x + 1 )

2

3

Now f ′ (x) = 0 ⇒

( 2x − 1 )

3 ( x2 − x + 1 )

2

3

= 0 ⇒ 2x − 1 = 0

⇒ x =

1

2 [Turning point] and it belongs to the given enclosed interval 0 ≤ x ≤ 1 i.e.,[0, 1].

At x =

1

2 , from eq.(i), f

1

2 =

1

4 −

1

2 + 1

1

3 =

1 − 2 + 4

4

1

3 =

3

4

1

3 < 1

( )

( ) ( ) ( ) ( )

At x = 0, form eq.(i), f(0) = (1)

1

3 = 1

At x = 1, from eq.(i), f(1) = (1 − 1 + 1)

1

3 = (1)

1

3 = 1

∴ Maximum value of f(x) is 1.

Q120. The function f(x) = 2x – 15x + 36x + 4 is maximum at x =

1. 3

2. 0

3. 4

4. 2

Ans: 4. 2

Soluctio :

Given, f(x) = 2x – 15x + 36x + 4

lmplies that f'(x) = 6x – 30x + 36

For a local maxima or a local minima, we must have f'(x) = 0

lmplies that 6x – 30x + 36 = 0

lmplies that x – 5x + 6 = 0

(x – 2)(x – 3) = 0

lmplies that x = 2, 3

Now, f”(x) = 12x – 30

lmplies that f”(2) = 24 – 30 = 6 < 0

Therefore, x = 1 is a local maxima.

Also, f”(3) = 36 – 30 = 6 > 0

Therefore, x = 2 is a local maxima.

Q121. Choose the correct answer from the given four options:

The curve y = x

1

5 has at (0, 0)

1. A vertical tangent (parallel to y-axis).

2. A horizontal tangent (parallel to x-axis).

3. An oblique tangent.

4. No tangen.

Ans: 1. A vertical tangent (parallel to y-axis).

Solution:

We are given that y = x

1

5

⇒

dy

dx ⇒

dy

dx =

1

5 x

1

3 − 1 ∵

d

dx (xn) = nxn − 1

⇒

dy

dx =

1

5 x

− 4

5

⇒

dy

dx =

1

5x

4

5

⇒

dy

dx ( 0 , 0 ) =

1

5 ( 0 )

4

5

= ∞

So, the curve y = x

1

5 has a vertical tangent at (0, 0), which is parallel to Y-axis.

Q122. If the rate of chage of volume of sphere is equal to the rate of change of its radius, then its radius is equal to:

1. 1 unit

2. √2π units

3. √2π units

4.

1

2√π

unit

Ans: 4.

1

2√π

unit

Solution:

Let r be the radius and V be the volume of the sphere at any time t. Then,v =

4

3 πr3

⇒

dv

dt =

4

3 (3πr2)

dr

dt

3 2

3 2

2

2

2

[ ]

( )

⇒

dv

dt = 4πr2 dr

dt

⇒ 4πr2 = 1 [ ∴

dv

dt =

dr

dt ]

⇒ r2 =

1

4π

⇒ r =

1

4π

⇒ r =

1

2√π

unit

Q123. The minimum value of (x2 +

250

x ) is:

1. 75

2. 50

3. 25

4. 55

Ans: 1. 75

Sloution:

f(x) = x2 +

250

x

f ′ (x) = 2x −

250

x2

For the local minima a or maxima. We must have f'(x) = 0

= 2x −

250

x2 = 0

⇒ x = 5

= 2x −

250

x2 = 0

f ″ (x) = 2 +

500

x3

f ″ (x) = 2 +

500

125 > 0

function has minima at x = 5

f(5) = 75.

Q124. The radius of a circular plate is increasing at the rate of 0.01cm/sec. The rate of increase of its area when the

radius is 12cm, is:

1. 144πcm2 / sec.

2. 2.4πcm2 / sec.

3. 0.24πcm2 / sec.

4. 0.024πcm2 / sec.

Ans: 3. 0.24πcm2 / sec.

Solution:

A = πr2

dA

dt = 2πr

dr

dt

⇒

dA

dt = 2π × 12 × 0.01

= 0.24πcm2 / sec.

Q125. The function f(x) = x + 3x + 64 is increasing on:

1. R

2. ( − ∞, 0)

3. (0, ∞)

4. R0

Ans: 1. R

Solution:

f(x) = x9 + 3×7 + 64

f ′ (x) = 9×8 + 21×6 > 0, ∀ x ∈ R

√

9 7

So, f(x) is increasing on R.

Q126. f(x) = sin + √3cosx is maximum when x =

1.

π

3

2.

π

4

3.

π

6

4. 0

Ans: 3.

π

6

Solution:

f(x) = sin + √3cosx

⇒ f ′ (x) = cosx − √3sinx

For maxima or maxima,

f'(x) = 0

cosx − √3sinx = 0

⇒ tanx =

1

√3

⇒ x =

π

6

f ″ π

6 = − sin

π

6 − √3cos

π

6 =

− 1 − √3

2 < 0

function has local maima at x =

π

6

Q127. In the interval (1, 2), function f(x) = 2|x – 1| + 3|x – 2| is:

1. Increasing.

2. Decreasing.

3. Constant.

4. None of these.

Ans: 2. Decreasing.

Solution:

f(x) = 2|x – 1| + 3|x – 2|

In the interval (1, 2)

⇒ |x -1| = x – 1 and |x – 2| = -(x – 2)

⇒ f(x) = 2(x – 1) – 3(x – 2)

⇒ f(x) = -x + 4

⇒ f'(x) = -1

⇒ function is decreasing on (1, 2).

Q128. The function f(x) = x decreases on the interval:

1. (0, e)

2. (0, e)

3. 0,

1

e

4. None of these

Ans: 3. 0,

1

e

Solution:

Given, f(x) = xx

Applying log with base e on both sides, we get

log(f(x)) = xlogex

f ′ ( x )

f ( x ) = 1 + logex

f ′ (x) = f(x)(1 + logex)

= xx(1 + logex)

For f(x) to be decreasing, we must have

f ′ (x) < 0

( )

x

( )

( )

⇒ xx(1 + logex) < 0

Here, logarithmic function is defined for positive values of x.

⇒ xx > 0

⇒ 1 + logex < 0

[Since xx(1 + logex) < 0 ⇒ 1 + logex < 0]

⇒ logex < − 1

⇒ x < e − 1 [ ∵ logax < N ⇒ aN for a > 1]

Here,

e > 1

⇒ logex < − 1

⇒ x < e − 1

⇒ x ∈ (0, e − 1 )

So, f(x) is decreasing on 0,

1

e .

Q129. The slope of the tangent to the curve x = t + 3 t – 8, y = 2t – 2t – 5 at point (2, -1) is:

1.

22

7

2.

6

7

3. −6

4. None of these

Ans: 2.

6

7

Solution:

x = t2 + 3t − and y = 2t2 − 2t2 − 2t − 5

dx

dt = 2t + 3 and

dy

dt = 4t − 2

∴

dy

dx =

dy

dt

dx

dt

=

4t − 2

2t + 3

The given point is (2, -1).

∴ x = 2 and y=−1

Now,

t2 + 3t − 8 = 2 and 2t2 − 2t − 5 = − 1

Let u solve one of these to get the value of t.

t2 + 3t − 10 = 0 and 2t2 − 2t − 4 = 0

⇒ (t + 5)(t − 2) = 0 and (2t − 2)(t − 2) = 0

⇒ t = − 5 or t = 2

These two have t = 2 as a comman solution.

∴ Slope of the tangent =

dy

dx t = 2 =

8 − 2

4 + 3 =

6

7

Q130. if ax+

b

x ≥ c for all positive x where a, b, > 0, then.

1. ab <

c2

4

2. ab >

c2

4

3. ab >

c

4

4. None of these

Ans: 2. ab >

c2

4

Solution:

= f(x) = ax +

b

x

f(x) = 0

⇒ a −

b

x2 = 0

( )

2 2

( )

⇒ x = ±

b

a

f ″ (x) =

2b

x3

f ″ b

a =

2b

b

c

3

> 0

⇒ x =

b

a has a minima.

f ″ b

a = 2√ab ≥ c

c

2 ≤ √ab

⇒

c

4 ≤ ab

Q131. Any tangent to the curve y = 2x + 3x + 5:

1. Is parallel to x-axis.

2. Is parallel to y-axis.

3. Makes an acute angle with x-axis.

4. Makes an obtuse angle with x-axis.

Ans: 3. Makes an acute angle with x-axis.

Solution:

We have, y = 2x + 3x + 5

dy

dx = 14×6 + 3

⇒

dy

dx > 3 ( ∵ x is always positive for any real value of x)

⇒

dy

dx > 0

So, tanθ > 0

Hence, θ lies in first quadrant.

Thus, the tangent to the curve makes an acute angle with x-axis.

Q132. The normal at the point (1, 1) on the curve 2y + x = 3 is:

1. x + y = 0

2. x – y = 0

3. x + y + 1 = 0

4. x – y = 1

Ans: 2. x − y = 0

Solution:

2y + x = 3

2

dy

dx + 2x = 0

dy

dx = − x

dy

dx ( 1 , 1 ) = − 1

Slope of the normal = 1

Equation of the normal

y – 1 = x – 1

y = x

x – y = 0

Q133. The function f(x) = loge x3 + √x6 + 1 is of the following type:

1. Even and increasing.

2. Odd and increasing.

3. Even and decreasing.

4. Odd and decreasing.

√

(√ ) (√ )

√

(√ )

7

7

6

2

2

( )

( )

Ans: 2. Odd and increasing.

Solution:

f(x) = loge x3 + √x6 + 1

⇒ f( − x) = loge − x3 + √x6 + 1

= loge

( − x3 + √x6 + 1 ) ( x3 + √x6 + 1 )

x3 + √x6 + 1

= loge

x6 + 1 − x6

x3 + √x6 + 1

= loge

1

x3 + √x6 + 1

= − loge x3 + √x6 + 1

= − f(x)

Hence, f(-x) = -f(x)

Therefore, it is an odd function.

f(x) = loge x3 + √x6 + 1

d

dx {f(x)} =

1

x3 + √x6 + 1

× 3×2 +

1

2√x6 + 1

× 6×5

=

1

x3 + √x6 + 1

×

6×2√x6 + 1 + 6×5

2√x6 + 1

=

1

x3 + √x6 + 1

×

6×2 ( √x6 + 1 + x3 )

2√x6 + 1

=

6×2

2√x6 + 1

> 0

Therefore the given function is an increasing function.

Q134. The circumference of a circle is measured as 28cm with an error of 0.01cm. The percentage error in the area is:

1.

1

14

2. 0.01

3.

1

7

4. None of these

Ans: 1.

1

14

Solution:

Let x be the radius of the circle and y be its circumference.

x = 28cm

△x = 0.01cm

x = 2πr

y = πr2 = π ×

x2

4π2 =

x2

4π

⇒

dy

dx =

x

2π

⇒

△y

y =

x

2πy dx =

2

x × 0.01

⇒

△y

y × 100 =

2

x =

1

14

Hence, the percentage error in the area is

1

14

Q135. The minimum value of

x

loge x is .

1. e

2.

1

e

3. 1

4. None of these

( )

( )

{ }

( )

( )

( )

( )

( ) ( )

( ) ( )

( ) { }

( )

Ans: 1. e

Solution:

Given, f(x) =

x

loge x

⇒ f ′ (x) =

loge x − 1

( loge x )2

For a local maximum or a local minima, we must have f ′ (x) = 0

⇒

loge x − 1

( loge x )2 = 0

⇒ logex − 1 = 0

⇒ logex = 1

⇒ x = e

Now, ⇒ f ″ (x) =

− 1

x ( loge x )2 +

2

x ( loge x )3

⇒ f ″ (e) =

− 1

e +

2

e =

1

e > 0

So, x = e is a local minima.

∴ minimum value of f(x) =

e

loge e = e

Q136. Choose the correct answer

The line y = x + 1 is a tangent to the curve y = 4x at the point:

1. (1, 2)

2. (2, 1)

3. (1, -2)

4. (-1, 2)

Ans: 1. (1, 2)

The equation of the given curve is y = 4x.

Differentiating with respect to x, we have:

2y

dy

dx = 4 ⇒

dy

dx =

2

y

Therefore, the slope of the tangent to the given curve at any point (x, y) is given by,

dy

dx =

2

y

The given line is y = x + 1 (which is of the form y = mx + c)

∴ Slope of the line = 1

The line y = x + 1 is a tangent to the given curve if the slope of the line is equal to the slope of the tangent. Also, the line must

intersect the curve,

Thus, we must have:

2

y = 1

⇒ y = 2

Now, y = x + 1 ⇒ x = y -1 ⇒ x = 2 -1 = 1

Hence, the line y = x + 1 is a tangent to the given curve at the point (1, 2).

Q137. The equation of motion of a particle is s = 2t2 + sin2t, where s is in metres and t is in seconds. The velocity of the

particle when its acceleration is 2m/sec , is:

1. π + √3m/ sec.

2.

π

3 + √3m/ sec.

3.

2π

3 + √3m/ sec.

4.

π

3 +

1

√3

m/ sec.

Ans: 2.

π

3 + √3m/ sec.

Solution:

s = 2t2 + sin2t

v =

ds

dt = 4t + 2cos2t

2

2

2

a =

d2s

dt2 = 4 − 4sin2t

Given that a = 2m/ sec2

⇒ 4 − 4sin2t = 2

⇒ 2 − 2sin2t = 1

⇒ 2sin2t = 1

⇒ sin2t =

1

2

⇒ 2t =

π

6

⇒ t =

π

12

v = 4t + 2cos2t at t =

π

12 ,

v = 4t ×

π

12 + 2cos

π

6 =

π

3 + √3m/ sec.

Q138. Choose the correct answer from the given four options:

The sides of an equilateral triangle are increasing at the rate of 2cm/ sec. The rate at which the area increases,

when side is 10cm is:

1. 10cm2/s

2. √3cm2/s

3. 10√3cm2/s

4.

10

3 cm2/s

Ans: 3. 10√3cm2/s

Solution:

Let the side of an equilateral triangle be x cm,

∴ Area of equilateral triangle, A =

√3

4 x2 …(i)

Also,

dx

dt = 2cm/s

On differentiating Eq. (i) w.r.t. t, we get

dA

dt =

√3

2 ⋅ 2x ⋅

dx

dt

=

√3

4 ⋅ 2 ⋅ 10 ⋅ 2 ∵ x = 10 and

dx

dt = 2

= 10√3cm2/s

Q139. If the ratio of base radius and height of a cone is 1 : 2 and percentage error in radius is λ%, then the error in its

volume is:

1. λ%

2. 2λ%

3. 3λ%

4. None of these

Ans: 3. 3λ%

Solution:

Let x be the radius of the cone and V be the volume.

Given that

△r

r × 100 = λ

V =

1

3 πr2h

⇒ V =

2

3 πr3 ∵

r

h =

1

2 ⇒ h = 2r

⇒

dV

dr = 2πr2

⇒

dV

V =

2πr2

2

3 πr3

×

λr

100

⇒

dv

V = 3λ%

Q140. The interval in which y = x e is increasing is:

[ ]

( )

2 -x

1. ( − ∞, ∞)

2. ( − 2, 0)

3. (2, ∞)

4. (0, 2)

Ans: 4. (0, 2)

Given: f(x) = [y = x2e − x]

⇒

dy

dx = x2 d

dx e − x + e − x d

dx x2 = x2e − x( − 1) + e − x(2x)

⇒

dy

dx = − x2e − x + 2xe − x = xe − x( − x + 2)

⇒

dy

dx =

x ( 2 − x )

ex

In option (D),

dy

dx > 0 for all x in the interval (0, 2).

Q141. While measuring the side of an equilateral triangle an error of k % is made, the percentage error in its area is:

1. k%

2. 2k%

3.

K

2 %

4. 3k%

Ans: 2. 2K%

Solution:

Area of equilateral triangle is,

A =

√3

4 a2

Given that

da

a × 100 = k

and

dA

da =

√3

2 a

⇒

△A

a =

√3

2 da

√3

4 a2

⇒

△A

A =

2

a ×

Ka

100 = 2k

The error in the area of the triangle is 2K%

Q142. The sum of two non-zero number is 8, the minimum value of the sum of the reciprohcle is :

1.

1

4

2.

1

2

3.

1

8

4. None of these.

Ans: 2.

1

2

Solution:

Let the two non-zero number be x and y. Then,

x + y = 8

⇒ y = 8 – x …(i)

Now, f(x) =

1

x +

1

y

⇒ f(x) =

1

x +

1

8 − x [from eq.(i)]

⇒ f ′ (x) =

1

x +

1

8 − x2

For a local minima or a local maxima, we must have f'(x) = 0

⇒

− 1

x2 +

1

8 − x2 = 0

⇒

− ( 8 − x2 ) + x2

( x )2 ( 8 − x )2 = 0

⇒ − 64 − x2 + 16x + x2 = 0

⇒ 16x − 64 = 0

⇒ x = 4

f ″ (x) =

2

x3 −

2

( 8 − x )3

⇒ f ″ (4) =

2

x3 −

2

( 8 − 4 )3

⇒ f ″ (4) =

2

43 −

2

( 8 − 4 )3

⇒ f ″ (4) =

2

64 −

2

64 = 0

∴ minimum value =

1

4 +

1

4 =

1

2

Q143. Side of an equilateral triangle expands at the rate of 2cm/ sec. The rate of increase of its area when each side

is 10cm is:

1. 10√2cm2 / sec.

2. 10√3cm2 / sec.

3. 10cm2 / sec.

4. 5cm2 / sec.

Ans: 2. 10√3cm2 / sec.

Solution:

A =

√3

4 x2

dA

dt =

√3

2 x

dx

dt

⇒

dA

dt =

√3

2 × 2 × 10 = 10√3cm2 / sec.

Q144. If the function f(x) = cos | x | − 2ax + b increases along entire number scale, then:

1. a = b

2. a =

1

2 b

3. a ≤ −

1

2

4. a > −

3

2

Ans: 3. a ≤ −

1

2

Solution:

Given:

f(x) = cos | x | − 2ax + b

Now, | x | =

x, x ≥ 0

−x, x < 0

And cos | x | =

cos(x), x ≥ 0

cos( − x) = cos(x), x < 0

∴ cos | x | = cosx, ∀ x ∈ R

∴ f(x) = cosx − 2ax + b

⇒ f ′ (x) = − sinx − 2a

It is given that f(x) is increasing.

⇒ f ′ (x) ≥ 0

⇒ − sinx − 2a ≥ 0

⇒ sinx + 2a ≤ 0

⇒ 2a ≤ − sinx

The least value of −sinx is -1.

⇒ 2a ≤ − 1

⇒ a ≤

− 1

2

Q145. At what point the slope of the tangent to the curve x + y – 2x – 3 = 0 is zero:

{

{

2 2

1. (3, 0), (-1, 0)

2. (3, 0), (1, 2)

3. (-1, 0), (1, 2)

4. (1, 2), (1, -2)

Ans: 4. (1, 2), (1, 2)

Solution:

Let (x , y ) be the required point.

Since, the point lie on the curve.

Hence, x21

+ y21

− 2×1 − 3 = 0 . . . (1)

Now, x2 + y2 − 2x − 3 = 0

⇒ 2x + 2y

dy

dx − 2 = 0

∴

dy

dx =

2 − 2x

2y =

1 − x

y

Now,

Slope of the tangent =

dy

dx ( x1 , y1 ) =

1 − x1

y1

Slope of tangent = 0 (Given)

∴

1 − x1

y1

= 0

⇒ 1 − x1 = 0

⇒ x1 = 1

From (1), we get

x21

+ y21

− 2×1 − 3 = 0

⇒ 1 + y21

− 4 = 0

⇒ y1 = ± 2

So, the points are (1, 2) and (1, -2).

Q146. The radius of the base of a cone is increasing at the rate of 3cm/minute and the altitude is decreasing at the rate

of 4cm/minute. The rate of change of lateral surface when the radius = 7cm and altitude 24cm is:

1. 54πcm2 / min

2. 7πcm2 / min

3. 27cm2 / min

4. none of these

Ans: 1. 54πcm2 / min

Solution:

Given that

dr

dt = 3cm/ min,

dh

dt = − 4cm/ min

h = 24cm, r = 7cm

l2 = h2 + r2

⇒ l2 = 242 + 72

⇒ l = 25

s = πrl

⇒ s2 = πr2l2

⇒ s2 = πr2(h2 + r2)

⇒ s = πr2h2 + πr4

⇒ 2s

ds

dt = 2π2r2h

dh

dt + 2π2h2r

dr

dt + 4π2r3 dr

dt

⇒ 2πrl

ds

dt = 2π2rh r +

dh

dt + h

dr

dt +

2r2

h

dr

dt

1 1

( )

( )

⇒ πrl

ds

dt = π2rh r +

dh

dt + h

dr

dt +

2r2

h

dr

dt

⇒ 25

ds

dt = πrh r +

dh

dt + h

dr

dt +

2r2

h

dr

dt

⇒ 25

ds

dt = πh r +

dh

dt + h

dr

dt +

2r2

h

dr

dt

⇒ 25

ds

dt = 24π 7 × ( − 4) + 24 × 3 +

2 × ( 7 )2

24 × 3

⇒

ds

dt = 54πcm2 / sec

Q147. Choose the correct answer from the given four options:

A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides

downwards at the rate of 10cm/ sec, then the rate at which the angle between the floor and the ladder is

decreasing when lower end of ladder is 2 metres from the wall is:

1.

1

10 radian/sec

2.

1

20 radian/sec

3. 20 radian/sec

4. 10 radian/sec

Ans: 2.

1

20 radian/sec

Solution:

Let the angle between floor and the ladder be θ.

Let at any time ‘t’ AB = x cm and BC = y cm

∴ sinθ =

x

500 and cosθ =

y

500

⇒ x = 500sinθ and y = 500cosθ

Also it is given that

dx

dt = 10cm/s

⇒ 500 ⋅ cosθ ⋅

dθ

dt = 10cm/ s

⇒

dθ

dt =

10

500cos θ =

1

50cos θ

For y = 2m = 20cm,

dθ

dt =

1

50 ⋅

y

500

=

10

y =

10

200 =

1

20 radian/sec

Q148. The point on the curve y = x – 3x + 2 where tangent is perpendicular to y = x is:

1. (0, 2)

2. (1, 0)

3. (-1, 6)

4. (2, -2)

Ans: 2. (1, 0)

Solution:

y = x

⇒

dy

dx = 1

Let (x y ) be the required point.

Since, the point lies on the curve,

Hence, y1 = x21

− 3×1 + 2

Now, y = x2 − 3x + 2

( )

( )

( )

( )

2

2, 2

∴

dy

dx = 2x − 3

Slope of the perpendicular to this line.

∴ Slope of the tangent =

− 1

slope of the line =

− 1

1 = − 1

Now,

2×1 − 3 = − 1

⇒ 2×1 = 2

⇒ x1 = 2

and

y1 = x21

− 3×1 + 2 = 1 − 3 + 2 = 0

∴ (x1, y1) = (1, 0)

Q149. A man of height 6ft walks at a uniform speed of 9ft/sec. from a lamp fixed at 15ft height. The length of his

shadow is increasing at the rate of:

1. 15ft/ sec.

2. 9ft/ sec.

3. 6ft/ sec.

4. None of these.

Ans: 3. 6ft/ sec.

Solution:

15

6 =

u+v

u

⇒

15

6 =

v

u + 1

⇒

v

u =

3

2

⇒ u =

2v

3

⇒

du

du =

2

3

dv

dt

⇒

du

dt =

2

3 × 9

= 6ft/ sec.

Q150. The point on the curve y = x where tangent makes 45° angle with x-axis is:

1.

1

2 ,

1

4

2.

1

4 ,

1

2

3. (4, 2)

4. (1, 1)

Ans: 2.

1

4 ,

1

2

Solution:

Let the required point be(x , y ).

The tangent makes an angle of 45° with the x-axis,

∴ Slope of the tangen = tan 45° = 1

Since, the point lies on the curve.

Hence, y2 = x1

Now, y2 = x

⇒ 2y

dy

dx = 1

dy

dx =

1

2y

2

( )

( )

( )

1 1

Slope of the tangent =

dy

dx ( x1 , y1 ) =

1

2y1

Given:

1

2y1

= 1

⇒ 2y1 = 1

⇒ y1 =

1

2

Now,

x1 = y21

=

1

2

2 =

1

4

∴ (x1, y21

) =

1

4 ,

1

2

Q151. Choose the correct answer from the given four options:

The points at which the tangents to the curve y = x – 12x + 18 are parallel to x-axis are:

1. (2, -2), (-2, -34)

2. (2, 34), (-2, 0)

3. (0, 34), (-2, 0)

4. (2, 2), (-2, 34)

Ans: 4. (2, 2), (-2, 34)

Solution:

The given equation of curve is

y = x – 12x + 18

∴

dy

dx = 3×2 − 12 [on differenttiating w.r.t.x]

So, the slope of line parallel to the X-axis

∴

dy

dx = 0

⇒ 3×2 − 12 = 0

⇒ x2 =

12

3 = 4

∴ x = ± 2

For x = 2, y = 2 – 12 × 2 + 18 = 2

and for x = -2, y = (-2) -12(-2) + 18 = 34

So, the points are (2, 2) and (-2, 34).

Q152. If f(x) =

1

4×2 + 2x + 1

, then its maximum value is :

1.

4

3

2.

2

3

3. 1

4.

3

4

Ans: 1.

4

3

Solution:

Maximum value of

1

4×2 + 2x + 1

= Minimum value of 4×2 + 2x + 1

Now, f(x) = 4×2 + 2x + 1

lmplies that f ′ (x) = 8x + 2

For a local maxima or a local minima, We must have f'(x) = 0

lmplies that 8x + 2 = 0

lmplies that 8x = − 2

lmplies that x = − 14

Now, f ″ (x) = 8

lmplies that f ″ (1) = 8 > 0

Therefore, x =

− 1

4 is a local minima.

Thus,

1

4×2 + 2x + 1

is maximum at x =

− 1

4

( )

( )

( )

3

3

( )

3

3

lmplies that maximum value of

1

4×2 + 2x + 1

=

1

4

− 1

4

2 + 2

− 1

4 + 1

=

16

12 =

4

3

Q153. If the function f(x) =

− x

2 + sinx defined on

− π

3 ,

π

3 is:

1. Increasing.

2. Decreasing.

3. Constant.

4. None of these.

Ans: 1. Increasing.

Solution:

f(x) =

− x

2 + sinx defined on

− π

3 ,

π

3

∴ f ′ (x) =

− 1

2 + cosx

⇒ f ′ (x) ≥ 0, ∀ x ∈

− π

3 ,

π

3

∵ for x ∈

− π

3 ,

− π

3 , cos ≥

1

2

Hence, the given function is increasing.

Q154. If s = t – 4t + 5 describes the motion of a particle, then its velocity when the acceleration vanishes, is:

1.

16

2 unit/ sec.

2.

-32

3 unit/ sec.

3.

4

3 unit/ sec.

4. −

16

3 unit/ sec.

Ans: 4. −

− 16

3 unit/ sec.

Solution:

According to the question,

s = t3 − 4t2 + 5

⇒

ds

dt = 3t2 − 8t

⇒

d2s

dt2 = 6t − 8

⇒ 6t − 8 = 0 As velocity diminish, then

d2s

dt2 = 0

⇒ t =

4

3

Now,

ds

dt t =

4

3

= 3

4

3

2 − 8

4

3

⇒

ds

dt =

16

3 −

32

3

⇒

ds

dt = −

16

3 unit/ sec.

Q155. Choose the correct answer from the given four options:

f(x) = x has a stationary point at:

1. x = e

2. x =

1

e

3. x = 1

4. x = √e

Ans: 2. x =

1

e

Solution:

We have, f(x) = x

Let us suppose y = x

( ) ( )

[ ]

[ ]

[ ]

[ [ ] ]

3 2

[ ]

( ) ( ) ( )

x

x

x

Taking logarithm on both sides, we get

logy = xlogx

∴

1

y ⋅

dy

dx = x ⋅

1

x + logx ⋅ 1 [ ∵ (fg) ′ = fg ′ + gf ′ ]

⇒

dy

dx = (1 + logx) ⋅ xx

Find the critical points by equating

dy

dx to 0.

∴

dy

dx = 0

⇒ (1 + logx)xx = 0

⇒ logx = − 1 as xx ≠ 0

⇒ logx = loge − 1

⇒ x = e − 1

⇒ x =

1

e

Hence, f(x) has a stationary point at x =

1

e .

Q156. Every invertible function is:

1. Monotonic function.

2. Constant function.

3. Identity function.

4. Not necessarily monotonic function.

Ans: 1. Monotonic function.

Solution:

We know that “every invertible function is a monotonic function”.