Application of Derivatives — Page 8 | JEE Mathematics

Q71

Let ƒ : [0, 2]

\to

R be a twice differentiable function such that ƒ''(x) > 0, for all x

\in

(0, 2). If

\phi

(x) = ƒ(x) + ƒ(2 – x), then

\phi

is :

A decreasing on (0, 2)

B decreasing on (0, 1) and increasing on (1, 2)

C increasing on (0, 2)

D increasing on (0, 1) and decreasing on (1, 2)

Correct Answer

Option B

Solution

$\phi$ (x) = ƒ(x) + ƒ(2 – x) $\Rightarrow$ $\phi$ '(x) = ƒ'(x) - ƒ'(2 – x) Since ƒ''(x) > 0 for all x

\in

(0, 2) $\Rightarrow$ ƒ'(x) is an increasing function for all x

\in

(0, 2).

Case 1 : When $\phi$ (x) is increasing function So $\phi$ '(x) > 0 $\Rightarrow$ ƒ'(x) - ƒ'(2 – x) > 0 $\Rightarrow$ ƒ'(x) > ƒ'(2 – x) $\Rightarrow$ x > 2 – x $\Rightarrow$ x > 1 $\therefore$ $\phi$ (x) is increasing on (1, 2).

Case 2 : When $\phi$ (x) is decreasing function So $\phi$ '(x) < 0 $\Rightarrow$ ƒ'(x) - ƒ'(2 – x) < 0 $\Rightarrow$ ƒ'(x) < ƒ'(2 – x) $\Rightarrow$ x < 2 – x $\Rightarrow$ x < 1 $\therefore$ $\phi$ (x) is decreasing on (0, 1).

Q72

Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is

2y \over x^2

. If the curve passes through the centre of the circle x2 + y2 – 2x – 2y = 0, then its equation is :

A x loge|y| = 2(x – 1)

B x2 loge|y| = –2(x – 1)

C x loge|y| = x – 1

D x loge|y| = –2(x – 1)

Correct Answer

Option A

Solution

Slope,

{{dy} \over {dx}}

=

2y \over x^2

$\Rightarrow$

\int {{{dy} \over y}} = \int {2{{dx} \over {{x^2}}}}

$\Rightarrow$

{\log _e}|y| = - {2 \over x} + C

.......

(1) Center of the circle x2 + y2 – 2x – 2y = 0 is (1, 1) Equation (1) passes through point (1, 1) $\therefore$ 0 = -2 + C $\Rightarrow$ C = 2 $\therefore$

{\log _e}|y| = - {2 \over x} + 2

$\Rightarrow$ x

{\log _e}|y| = - 2 + 2x

$\Rightarrow$ x

{\log _e}|y| = 2(x - 1)

Q73

The function, f(x) = (3x – 7)x2/3, x

\in

R, is increasing for all x lying in :

A

\left( { - \infty ,0} \right) \cup \left( {{3 \over 7},\infty } \right)

B

\left( { - \infty ,0} \right) \cup \left( {{{14} \over {15}},\infty } \right)

C

\left( { - \infty ,{{14} \over {15}}} \right)

D

\left( { - \infty ,{{14} \over {15}}} \right) \cup \left( {0,\infty } \right)

Correct Answer

Option B

Solution

f(x) = (3x – 7)x2/3 f’(x) =

\left( {3x - 7} \right){2 \over {3{x^{1/3}}}} + {x^{{2 \over 3}}}.3

=

{{6x - 14 - 9x} \over {3{x^{1/3}}}}

=

{{15x - 14} \over {3{x^{1/3}}}}

As f(x) increasing so f'(x) > 0 $\therefore$

{{15x - 14} \over {3{x^{1/3}}}}

> 0 $\therefore$ x

\in

\left( { - \infty ,0} \right) \cup \left( {{{14} \over {15}},\infty } \right)

Q74

The equation of the normal to the curve y = (1+x)2y + cos 2(sin–1x) at x = 0 is :

A y = 4x + 2

B x + 4y = 8

C y + 4x = 2

D 2y + x = 4

Correct Answer

Option B

Solution

Given equation of curve y = (1+x)2y + cos 2(sin–1x) at x = 0 $\Rightarrow$ y = (1 + 0)2y + cos2(sin–10) $\Rightarrow$ y = 1 + 1 $\Rightarrow$ y = 2 So we have to find the normal at (0, 2) Now, y =

{e^{2y\ln \left( {1 + x} \right)}} + {\cos ^2}\left( {{{\cos }^{ - 1}}\sqrt {1 - {x^2}} } \right)

$\Rightarrow$ y =

{e^{2y\ln \left( {1 + x} \right)}} + {\left( {\sqrt {1 - {x^2}} } \right)^2}

$\Rightarrow$ y =

{e^{2y\ln \left( {1 + x} \right)}} + \left( {1 - {x^2}} \right)

Now differentiate w.r.t. x y' =

{e^{2y\ln \left( {1 + x} \right)}}\left[ {2y.\left( {{1 \over {1 + x}}} \right) + \ln \left( {1 + x} \right).2y'} \right]

- 2x Put x = 0 & y = 2 $\Rightarrow$ y' =

{e^{2y\ln \left( {1 + 0} \right)}}\left[ {2y.\left( {{1 \over {1 + 0}}} \right) + \ln \left( {1 + 0} \right).2y'} \right] - 2 \times 0

$\Rightarrow$ y' = e0 [4 + 0] – 0 $\Rightarrow$ y' = 4 = slope of tangent to the curve so slope of normal to the curve = -

{1 \over 4}

Hence equation of normal at (0, 2) is y - 2 = -

{1 \over 4}

(x - 0) $\Rightarrow$ 4y – 8 = –x $\Rightarrow$ x + 4y = 8

Q75

If the surface area of a cube is increasing at a rate of 3.6 cm2/sec, retaining its shape; then the rate of change of its volume (in cm3/sec), when the length of a side of the cube is 10 cm, is :

A 9

B 10

C 18

D 20

Correct Answer

Option A

Solution

For cube of side 'a' A = 6a2 and V = a3 Given

{{dA} \over {dt}} = 3.6

$\Rightarrow$

12a{{da} \over {dt}}

= 3.6

{{dV} \over {dt}} = 3{a^2}.{{da} \over {dt}} = 3{a^2}\left( {{{3.6} \over {12a}}} \right)

at a = 10

{{dV} \over {dt}} = 9

Q76

Let f be a twice differentiable function on (1, 6). If f(2) = 8, f’(2) = 5, f’(x)

\ge

1 and f''(x)

\ge

4, for all x

\in

(1, 6), then :

A f(5)

\le

10

B f(5) + f'(5)

\ge

28

C f(5) + f'(5)

\le

26

D f'(5) + f''(5)

\le

20

Correct Answer

Option B

Solution

Given,

f'(x) \ge 1

$\therefore$

\int_2^5 {f'(x)} dx\, \ge \,\int_2^5 {dx}

\Rightarrow f(5) - f(2) \ge 3

\Rightarrow f(5) - 8 \ge 3

\Rightarrow f(5) \ge 11

...(1) Also,

f''(x) \ge 4

$\therefore$

\int_2^5 {f''(x)} dx\, \ge \,\int_2^5 {4dx}

\Rightarrow f'(5) - f'(2) \ge 4(3)

\Rightarrow f'(5) - 5 \ge 12

\Rightarrow f'(5) \ge 17

...(2) From (1) and (2),

f'(5) + f'(5) \ge 11 + 17

\Rightarrow f'(5) + f'(5) \ge 28

Q77

The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices C and D lie on the parabola, y = x2–1 below the x-axis, is :

A

{1 \over {3\sqrt 3 }}

B

{2 \over {3\sqrt 3 }}

C

{4 \over {3\sqrt 3 }}

D

{4 \over 3}

Correct Answer

Option C

Solution

Area (A) = 2t. (1 $-$ t2) (0 < t < 1) A = 2t $-$ 2t3

{{dA} \over {dt}} = 2 - 6{t^2}

= 0 $\Rightarrow$

t =

$\pm$

{1 \over {\sqrt 3 }}

$\therefore$

{A_{\max }} = |{2 \over {\sqrt 3 }}\left( {1 - {1 \over 3}} \right)| = {4 \over {3\sqrt 3 }}

Q78

If the point P on the curve, 4x2 + 5y2 = 20 is farthest from the point Q(0, -4), then PQ2 is equal to:

A 36

B 48

C 21

D 29

Correct Answer

Option A

Solution

Given ellipse is

{{{x^2}} \over 5} + {{{y^2}} \over 4} = 1

Let point P is

(\sqrt 5 \cos \theta ,\,2\sin \theta )

{(PQ)^2}=5{\cos ^2}\theta + {(2\sin \theta + 4)^2}

$\Rightarrow$ (PQ)2 =

5{\cos ^2}\theta + 4{\sin ^2}\theta + 16\sin \theta + 16

$\Rightarrow$ (PQ)2 =

{\cos ^2}\theta + 4{\cos ^2}\theta + 4{\sin ^2}\theta + 16\sin \theta + 16

$\Rightarrow$ (PQ)2 =

{\cos ^2}\theta + 4 + 16\sin \theta + 16

$\Rightarrow$

{(PQ)^2} = {\cos ^2}\theta + 16\sin \theta + 20

$\Rightarrow$

{(PQ)^2} = - {\sin ^2}\theta + 16\sin \theta + 21

=

- \left( {{{\sin }^2}\theta - 2.8\sin \theta + 64} \right) + 64 + 21

=

85 - {(\sin \theta - 8)^2}

will be maximum when sin $\theta$ = 1

\Rightarrow {(PQ)^2}_{\max } = 85 - 49 = 36

Q79

Which of the following points lies on the tangent to the curve x4ey + 2

\sqrt {y + 1}

= 3 at the point (1, 0)?

A (2, 2)

B (–2, 4)

C (2, 6)

D (–2, 6)

Correct Answer

Option D

Solution

x4ey + 2

\sqrt {y + 1}

= 3 Differentiating w.r.t. x, we get x4eyy' + ey4x3 +

{{2y'} \over {2\sqrt {y + 1} }}

= 0 At P(1, 0)

{y{'_P}}

+ 4 +

{y{'_P}}

= 0 $\Rightarrow$

{y{'_P}}

= -2 Tangent at P(1, 0) is y – 0 = – 2 (x – 1) 2x + y = 2 Only (–2, 6) lies on it.

Q80

If x = 1 is a critical point of the function f(x) = (3x2 + ax – 2 – a)ex , then :

A x = 1 is a local maxima and x =

- {2 \over 3}

is a local minima of f.

B x = 1 and x =

- {2 \over 3}

are local maxima of f.

C x = 1 and x =

- {2 \over 3}

are local minima of f.

D x = 1 is a local minima and x =

- {2 \over 3}

is a local maxima of f.

Correct Answer

Option D

Solution

f(x) = (3x2 + ax – 2 – a)ex $\therefore$ f'(x) = ex(6x + a) + (3x2 + ax – 2 – a)ex = ex(3x2 + x(6 + a) – 2) f '(x) = 0 at x = 1 $\Rightarrow$ 3 + (6 + a) – 2 = 0 $\Rightarrow$ a = -7 $\therefore$ f'(x) = ex(3x2 – x – 2) = ex(x – 1) (3x + 2) x = 1 is a local minima and x =

- {2 \over 3}

is a local maxima of f.