Application of Derivatives Questions — JEE Mathematics

Q1

A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3 /min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice decreases, is :

A

{5 \over {6\pi }}

B

{1 \over {9\pi }}

C

{1 \over {36\pi }}

D

{1 \over {18\pi }}

Correct Answer

Option D

Solution

V = {4 \over 3}\pi \left( {{{\left( {10 + h} \right)}^3} - {{10}^3}} \right)

\Rightarrow {{dV} \over {dt}} = 4\pi {(10 + h)^2}{{dh} \over {dt}}

\Rightarrow - 50 = 4\pi {\left( {10 + 5} \right)^2}{{dh} \over {dt}}

\Rightarrow {{dh} \over {dt}} = - {1 \over {18\pi }}

Q2

The tangent to the curve, y = xex2 passing through the point (1, e) also passes through the point

A

\left( {{4 \over 3},2e} \right)

B (3, 6e)

C (2, 3e)

D

\left( {{5 \over 3},2e} \right)

Correct Answer

Option A

Solution

y = xex2

{\left. {{{dy} \over {dx}}} \right|_{(1,e)}}{\left. { = \left( {e.e{x^2}.2x + {e^{{x^2}}}} \right)} \right|_{(1,e)}}

= 2 \cdot e + e = 3e

T : y $-$ e = 3e (x $-$ 1) y = 3ex $-$ 3e + e y =

\left( {3e} \right)x - 2e

\left( {{4 \over 3},2e} \right)

lies on it

Q3

If the minimum value of

f(x)=\dfrac{5 x^{2}}{2}+\dfrac{\alpha}{x^{5}}, x>0

, is 14 , then the value of

\alpha

is equal to :

A 32

B 64

C 128

D 256

Correct Answer

Option C

Solution

f(x) = {{5{x^2}} \over 2} + {\alpha \over {{x^5}}}\,\,\{ x > 0\}

f'(x) = 5x - {{5\alpha } \over {{x^6}}} = 0

\Rightarrow x = {(\alpha )^{{1 \over 7}}}

f{(x)_{\min }} = {{5{{(\alpha )}^{{2 \over 7}}}} \over 2} + {\alpha \over {{\alpha ^{{5 \over 7}}}}} = 14

{5 \over 2}{\alpha ^{{2 \over 7}}} + {\alpha ^{{2 \over 7}}} = 14

{7 \over 2}{\alpha ^{{2 \over 7}}} = 14

\alpha = 128

Q4

The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S =

\left\{ {x\, \in R:{x^2} + 30 \le 11x} \right\}

is :

A

-

222

B

-

122

C

122

D 222

Correct Answer

Option C

Solution

S = {x

\in

R, x2 + 30 $-$ 11x $\le$ 0} = {x

\in

R, 5 $\le$ x $\le$ 6} Now f(x) = 3x3 $-$ 18x2 + 27x $-$ 40 $\Rightarrow$ f '(x) = 9(x $-$ 1)(x $-$ 3), which is positive in [5, 6] $\Rightarrow$ f(x) increasing in [5, 6] Hence maximum value = f(6) = 122

Q5

Let

f(x)=4 \cos ^3 x+3 \sqrt{3} \cos ^2 x-10

. The number of points of local maxima of

f

in interval

(0,2 \pi)

is

A 1

B 3

C 4

D 2

Correct Answer

Option D

Solution

\begin{aligned} & f(x)=4 \cos ^3 x+3 \sqrt{3} \cos ^2 x-10 \\ & f^{\prime}(x)=12 \cos ^2 x \cdot(-\sin x)+6 \sqrt{3} \cos x \cdot(-\sin x)=0 \\ & =-6 \sqrt{3} \cos x \cdot \sin x\left(1+\frac{2}{\sqrt{3}} \cos x\right)=0 \\ & \cos x=0, \sin x=0, \cos x=\frac{-\sqrt{3}}{2} \end{aligned}

Sign of

f(x)

\therefore \text{ Maxima at } \frac{5 \pi}{6}, \frac{7 \pi}{6}

Q6

Let

\mathrm{a}>0

. If the function

f(x)=6 x^3-45 \mathrm{a} x^2+108 \mathrm{a}^2 x+1

attains its local maximum and minimum values at the points

x_1

and

x_2

respectively such that

x_1 x_2=54

, then

\mathrm{a}+x_1+x_2

is equal to :

A 15

B 13

C 24

D 18

Correct Answer

Option D

Solution

\begin{aligned} &f(x)=6 x^3-45 a x^2+108 a^2 x+1\\ &\text{ For maxima or minima } f^{\prime}(x)=0 \end{aligned}

\begin{aligned} & x_1 x_2=\frac{108 a^2}{18}=54 \\ & \Rightarrow a^2=9 \Rightarrow a=3 \\ & \text{ Now, } a+x_1+x_2=3+\frac{90}{6}=3+15=18 \end{aligned}

Q7

If

2a+3b+6c=0,

\left( {a,b,c \in R} \right)

then the quadratic equation

a{x^2} + bx + c = 0

has

A at least one root in

\left[ {0,1} \right]

B at least one root in

\left[ {2,3} \right]

C at least one root in

\left[ {4,5} \right]

D none of these

Correct Answer

Option A

Solution

Let

f\left( x \right) = {{a{x^3}} \over 3} + {{b{x^2}} \over 2} + cx \Rightarrow f\left( 0 \right) = 0

and

f(1)

= {a \over 3} + {b \over 2} + c = {{2a + 3b + 6c} \over 6} = 0

Also

f(x)

is continuous and differentiable in

\left[ {0,1} \right]

and

\left[ {0,1\left[ {.\,\,} \right.} \right.

So by Rolle's theorem

f'\left( x \right) = 0.

i.e

\,\,a{x^2} + bx + c = 0

has at least one root in

\left[ {0,1} \right].

Q8

The real number

x

when added to its inverse gives the minimum sum at

x

equal :

A -2

B 2

C 1

D -1

Correct Answer

Option C

Solution

f(x) = y = x + {1 \over x}

or

{{dy} \over {dx}} = 1 - {1 \over {{x^2}}}

For max. or min,

1 - {1 \over {{x^2}}} = 0 \Rightarrow x = \pm 1

{{{d^2}y} \over {d{x^2}}} = {2 \over {{x^3}}}

{\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = 1}} = 2

(

+ve

) $\therefore$ f(x) will be minima at

x=1

.

Q9

A spherical iron ball

10

cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of

50

cm

^3

/min. When the thickness of ice is

5

cm, then the rate at which the thickness of ice decreases is

A

{1 \over {36\pi }}

cm/min

B

{1 \over {18\pi }}

cm/min

C

{1 \over {54\pi }}

cm/min

D

{5 \over {6\pi }}

cm/min

Correct Answer

Option B

Solution

Given that

{{dv} \over {dt}} = 50\,c{m^3}/\min

\Rightarrow {d \over {dt}}\left( {{4 \over 3}\pi {r^3}} \right) = 50

\Rightarrow 4\pi {r^2}{{dr} \over {dt}} = 50

\Rightarrow {{dr} \over {dt}} = {{50} \over {4\pi {{\left( {15} \right)}^2}}} = {1 \over {18\pi }}\,\,cm/\min \,\,

(here

r=10+5)

Q10

Let f be differentiable for all x. If f(1) = -2 and f'(x)

\ge

2 for x

\in \left[ {1,6} \right]

, then

A f(6)

\ge

8

B f(6) < 8

C f(6) < 5

D f(6) = 5

Correct Answer

Option A

Solution

As

\,\,f\left( 1 \right) = - 2\,\,\& \;\,f'\left( x \right) \ge 2\,\forall x \in \left[ {1,6} \right]

Applying Lagrange's mean value theorem

{{f\left( 6 \right) - f\left( 1 \right)} \over 5} = f'\left( c \right) \ge 2

\Rightarrow f\left( 6 \right) \ge 10 + f\left( 1 \right)

\Rightarrow f\left( 6 \right) \ge 10 - 2

\Rightarrow f\left( 6 \right) \ge 8.