Application of Derivatives — Page 14 | JEE Mathematics

Q131

The function

f(x)=\dfrac{x}{x^2-6 x-16}, x \in \mathbb{R}-\{-2,8\}

A decreases in

(-\infty,-2) \cup(-2,8) \cup(8, \infty)

B increases in

(-\infty,-2) \cup(-2,8) \cup(8, \infty)

C decreases in

(-2,8)

and increases in

(-\infty,-2) \cup(8, \infty)

D decreases in

(-\infty,-2)

and increases in

(8, \infty)

Correct Answer

Option A

Solution

f(x)=\frac{x}{x^2-6 x-16}

Now,

\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{-\left(\mathrm{x}^2+16\right)}{\left(\mathrm{x}^2-6 \mathrm{x}-16\right)^2} \\ & \mathrm{f}^{\prime}(\mathrm{x}) Thus

f(x)

is decreasing in

(-\infty,-2) \cup(-2,8) \cup(8, \infty)$$

Q132

If the tangent at the point (x1, y1) on the curve

y = {x^3} + 3{x^2} + 5

passes through the origin, then (x1, y1) does NOT lie on the curve :

A

{x^2} + {{{y^2}} \over {81}} = 2

B

{{{y^2}} \over 9} - {x^2} = 8

C

y = 4{x^2} + 5

D

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

Correct Answer

Option D

Solution

Given curve,

y = {x^3} + 3{x^2} + 5

Slope of tangent,

{{dy} \over {dx}} = 3{x^2} + 6x

Slope of line joined by (x1, y1) and (0, 0) is

= {{{y_0} - 0} \over {{x_1} - 0}}

$\therefore$

{{{y_1}} \over {{x_1}}} = 3x_1^2 + 6{x_1}

\Rightarrow {y_1} = 3x_1^3 + 6x_1^2

...... (1) Point (x1, y1) is on the line, $\therefore$

{y_1} = x_1^3 + 3x_1^2 + 5

..... (2) $\therefore$

3x_1^3 + 6x_1^2 = x_1^3 + 3x_1^2 + 6

\Rightarrow 2x_1^3 + 3x_1^2 - 5 = 0

x = 1 satisfy the equation Now,

{y_1} = x_1^3 + 3x_1^2 + 5

= 1 + 3 + 5

= 9

$\therefore$ (x1, y1) = (1, 9) Option D does not satisfy point (1, 9).

Q133

The sum of absolute maximum and absolute minimum values of the function

f(x) = |2{x^2} + 3x - 2| + \sin x\cos x

in the interval [0, 1] is :

A

3 + {{\sin (1){{\cos }^2}\left( {{1 \over 2}} \right)} \over 2}

B

3 + {1 \over 2}(1 + 2\cos (1))\sin (1)

C

5 + {1 \over 2}(\sin (1) + \sin (2))

D

2 + \sin \left( {{1 \over 2}} \right)\cos \left( {{1 \over 2}} \right)

Correct Answer

Option B

Solution

$f(x)=|(2 x-1)(x+2)|+\dfrac{\sin 2 x}{2}$ $0 \leq x0$ So, minima occurs at $x=\dfrac{1}{2}$

\begin{aligned} \left.f(x)\right|_{\min } &=\left|2\left(\frac{1}{2}\right)^{2}+\frac{3}{2}-2\right|+\sin \left(\frac{1}{2}\right) \cdot \cos \left(\frac{1}{2}\right) \\\\ &=\frac{1}{2} \sin 1 \end{aligned}

So, maxima is possible at $x=0$ or $x=1$ Now checking for $x=0$ and $x=1$ , we can see it attains its maximum value at $x=1$

\begin{aligned} \left.f(x)\right|_{\max }=&|2+3-2|+\frac{\sin 2}{2} \\\\ &=3+\frac{1}{2} \sin 2 \end{aligned}

Sum of absolute maximum and minimum value $=3+\dfrac{1}{2}(\sin 1+\sin 2)$ $=3+\dfrac{1}{2}(\sin 1+2\sin 1\cos 1)$ =

3 + {1 \over 2}(1 + 2\cos (1))\sin (1)

Q134

If xy4 attains maximum value at the point (x, y) on the line passing through the points (50 +

\alpha

, 0) and (0, 50 +

\alpha

),

\alpha

> 0, then (x, y) also lies on the line :

A y = 4x

B x = 4y

C y = 4x +

\alpha

D x = 4y

-

\alpha

Correct Answer

Option A

Solution

Equation of line passing through the point (50 + $\alpha$ , 0) and (0, 50 + $\alpha$ ) is

y - 0 = {{50 + \alpha - 0} \over {0 - (50 + \alpha )}}\left( {x - (50 + \alpha )} \right)

\Rightarrow y = - 1\left( {x - (50 + \alpha )} \right)

\Rightarrow y = (50 + \alpha ) - x

\Rightarrow x + y = 50 + \alpha

Let

p = x{y^4}

= (50 + \alpha - y){y^4}

= (50 + \alpha ){y^4} - {y^5}

For maximum or minimum value of p,

{{dp} \over {dy}} = 0

\Rightarrow 4{y^3}(50 + \alpha ) - 5{y^4} = 0

\Rightarrow {y^3}[200 + 4\alpha - 5y] = 0

$\therefore$

y = 0

or

200 + 4\alpha - 5y = 0

\Rightarrow y = {4 \over 5}(50 + \alpha )

$\therefore$

x = 50 + \alpha - y

= 50 + \alpha - {4 \over 5}(50 + \alpha )

= 50 + \alpha \left( {1 - {4 \over 5}} \right)

= {1 \over 5}(50 + \alpha )

\Rightarrow 4x = {4 \over 5}(50 + \alpha ) = y

\Rightarrow y = 4x

$\therefore$ (x, y) lies on the line y = 4x

Q135

Let

f(x) = 4{x^3} - 11{x^2} + 8x - 5,\,x \in R

. Then f :

A has a local minina at

x = {1 \over 2}

B has a local minima at

x = {3 \over 4}

C is increasing in

\left( {{1 \over 2},{3 \over 4}} \right)

D is decreasing in

\left( {{1 \over 2},{4 \over 3}} \right)

Correct Answer

Option D

Solution

Given,

f(x) = 4{x^3} - 11{x^2} + 8x - 5

$\therefore$

f'(x) = 12{x^2} - 22x + 8

= 2(6{x^2} - 11x + 4)

= 2(6{x^2} - 8x - 3x + 4)

= 2\left[ {2x(3x - 4) - 1(3x - 4)} \right]

= 2\left[ {(3x - 4)(2x - 1)} \right]

f'(x)

is positive before

{1 \over 2}

means slope of f(x) is positive before

{1 \over 2}

and f'(x) is negative after

{1 \over 2}

means slope of f(x) is negative after

{1 \over 2}

. $\therefore$ At

{1 \over 2}

, f(x) is local maximum Also,

f'(x)

is negative before

{4 \over 3}

means slope of f(x) is negative before

{4 \over 3}

and f'(x) is positive after

{4 \over 3}

means slope of f(x) is positive after

{4 \over 3}

. $\therefore$ At

{4 \over 3},\,f(x)

is local minimum From wavy curve method,

f'(x) > 0,\,\forall x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)

$\therefore$ f(x) increasing in

x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)

f'(x) < 0\,\forall x \in \left( {{1 \over 2},{4 \over 3}} \right)

$\therefore$ f(x) decreasing in

x \in \left( {{1 \over 2},{4 \over 3}} \right)

Q136

If the absolute maximum value of the function

f(x)=\left(x^{2}-2 x+7\right) \mathrm{e}^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}

in the interval

[-3,0]

is

f(\alpha)

, then :

A

\alpha=0

B

\alpha=-3

C

\alpha \in(-1,0)

D

\alpha \in(-3,-1]

Correct Answer

Option B

Solution

Given, $f(x)=\underbrace{\left(x^{2}-2 x+7\right)}_{f_{1}(x)} \underbrace{e^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}}_{f_{2}(x)}$ $f_{1}(x)=x^{2}-2 x+7$ $f_{1}^{\prime}(x)=2 x-2$ , so $f(x)$ is decreasing in $[-3,0]$ and positive also $f_{2}(x)=e^{4 x^{3}-12 x^{2}-180 x+31}$ $f_{2}^{\prime}(x)=e^{4 x^{3}-12 x^{2}-180 x+31} \cdot 12 x^{2}-24 x-180$ $=12(x-5)(x+3) e^{4 x^{3}-12 x^{2}-180 x+31}$ So, $f_{2}(x)$ is also decreasing and positive in $\{-3,0\}$ $\therefore$ absolute maximum value of $f(x)$ occurs at $x=-3$ $\therefore \quad \alpha=-3$

Q137

A wire of length

20 \mathrm{~m}

is to be cut into two pieces. A piece of length

l_{1}

is bent to make a square of area

A_{1}

and the other piece of length

l_{2}

is made into a circle of area

A_{2}

. If

2 A_{1}+3 A_{2}

is minimum then

\left(\pi l_{1}\right): l_{2}

is equal to :

A 6 : 1

B 1 : 6

C 4 : 1

D 3 : 1

Correct Answer

Option A

Solution

$\ell_{1}+\ell_{2}=20 \Rightarrow \dfrac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=-1$ $\mathrm{A}_{1}=\left(\dfrac{\ell_{1}}{4}\right)^{2}$ and $\mathrm{A}_{2}=\pi\left(\dfrac{\ell_{2}}{2 \pi}\right)^{2}$ Let $\mathrm{S}=2 \mathrm{~A}_{1}+3 \mathrm{~A}_{2}=\dfrac{\ell_{1}^{2}}{8}+\dfrac{3 \ell_{2}^{2}}{4 \pi}$ $\dfrac{\mathrm{ds}}{\mathrm{d} \ell_1}=0 \Rightarrow \dfrac{2 \ell_{1}}{8}+\dfrac{6 \ell_{2}}{4 \pi} \cdot \dfrac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=0$ $\Rightarrow \dfrac{\ell_{1}}{4}=\dfrac{6 \ell_{2}}{4 \pi}$ $\Rightarrow \dfrac{\pi \ell_{1}}{\ell_{2}}=6$

Q138

The curve

y(x)=a x^{3}+b x^{2}+c x+5

touches the

x

-axis at the point

\mathrm{P}(-2,0)

and cuts the

y

-axis at the point

Q

, where

y^{\prime}

is equal to 3 . Then the local maximum value of

y(x)

is:

A

\dfrac{27}{4}

B

\dfrac{29}{4}

C

\dfrac{37}{4}

D

\dfrac{9}{2}

Correct Answer

Option A

Solution

$f(x)=y=a x^{3}+b x^{2}+c x+5 \quad \ldots$ (i)

\frac{d y}{d x}=3 a x^{2}+2 b x+c \quad \ldots (ii)

Touches $x$ -axis at $P(-2,0)$ $\left.\Rightarrow y\right|_{x=-2}=0 \Rightarrow-8 a+4 b-2 c+5=0 \quad \ldots ...(iii)$ Touches $x$ -axis at $P(-2,0)$ also implies

\left.\frac{d y}{d x}\right|_{x=-2}=0 \Rightarrow 12 a-4 b+c=0 \quad \ldots...(iv)

$y=f(x)$ cuts $y$ -axis at $(0,5)$ Given, $\left.\dfrac{d y}{d x}\right|_{x=0}=c=3$ From (iii), (iv) and (v)

\begin{aligned} &a=-\frac{1}{2}, b=-\frac{3}{4}, c=3 \\\\ &\Rightarrow f(x)=\frac{-x^{2}}{2}-\frac{3}{4} x^{2}+3 x+5 \\\\ &f^{\prime}(x)=\frac{-3}{2} x^{2}-\frac{3}{2} x+3 \\\\ &=\frac{-3}{2}(x+2)(x-1) \\\\ &f^{\prime}(x)=0 \text{ at } x=-2 \text{ and } x=1 \end{aligned}

By first derivative test $x=1$ in point of local maximum Hence local maximum value of $f(x)$ is $f(1)$ i.e., $\dfrac{27}{4}$

Q139

If the maximum value of

a

, for which the function

f_{a}(x)=\tan ^{-1} 2 x-3 a x+7

is non-decreasing in

\left(-\dfrac{\pi}{6}, \dfrac{\pi}{6}\right)

, is

\bar{a}

, then

f_{\bar{a}}\left(\dfrac{\pi}{8}\right)

is equal to :

A

8-\dfrac{9 \pi}{4\left(9+\pi^{2}\right)}

B

8-\dfrac{4 \pi}{9\left(4+\pi^{2}\right)}

C

8\left(\dfrac{1+\pi^{2}}{9+\pi^{2}}\right)

D

8-\dfrac{\pi}{4}

Correct Answer

Option A

Solution

$\text{Given, }$

\begin{aligned} f_a(x) & =\tan ^{-1} 2 x-3 a x+7 \\\\ f_a^{\prime}(x) & =\frac{2}{1+4 x^2}-3 a \end{aligned}

As the function $f_a^{\prime}(x)$ is non-decreasing

\begin{aligned} & \text{ in }\left(-\frac{\pi}{6}, \frac{\pi}{6}\right), \\\\ & f_a^{\prime}(x) \geq 0 \end{aligned}

\begin{aligned} & \Rightarrow \frac{2}{1+4 x^2}-3 a \geq 0 \Rightarrow \frac{2}{1+4 x^2} \geq 3 a \\\\ & \Rightarrow a \leq \frac{2}{3\left(1+4 x^2\right)}, \text{ when } x \in\left(-\frac{\pi}{6}, \frac{\pi}{6}\right) \\\\ & \because a \text{ is maximum when } x^2=\frac{\pi^2}{36}, \end{aligned}

\begin{aligned} a_{\max } & =\frac{2}{3\left(1+\frac{4 \pi^2}{36}\right)}=\frac{2 \times 12}{36+4 \pi^2} \\\\ & =\frac{6}{9+\pi^2} \\\\ \therefore \bar{a} & =\frac{6}{9+\pi^2} \end{aligned}

\begin{aligned} & \therefore f_a(x)=\tan ^{-1} 2 x-\frac{18}{9+\pi^2} x+7 \\\\ & f_\pi\left(\frac{\pi}{8}\right)=\tan ^{-1} 2\left(\frac{\pi}{8}\right)-\frac{18}{9+\pi^2} \times \frac{\pi}{8}+7 \\\\ & =\tan ^{-1} \frac{\pi}{4}-\frac{9 \pi}{36+4 \pi^2}+7 \\\\ & =8-\frac{9 \pi}{36+4 \pi^2}=8-\frac{9 \pi}{4\left(9+\pi^2\right)} \end{aligned}

Q140

Let

f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in \mathrm{R}

. Then which of the following statements are true?

\mathrm{P}: x=0

is a point of local minima of

f

\mathrm{Q}: x=\sqrt{2}

is a point of inflection of

f

R: f^{\prime}

is increasing for

x>\sqrt{2}

A Only P and Q

B Only P and R

C Only Q and R

D All P, Q and R

Correct Answer

Option D

Solution

f(x) = {3^{{{({x^2} - 2)}^3} + 4}},\,x \in R

f(x) = {81.3^{{{({x^2} - 2)}^3}}}

f'(x) = {81.3^{{{({x^2} - 2)}^3}}}\ln 2.3({x^2} - 2)2x

= (486\ln 2)\left( {{3^{{{({x^2} - 2)}^3}}}({x^2} - 2)x} \right)

\Rightarrow x = 0

is the local minima.

f''(x) = (486\ln 2)\left( {{3^{{{({x^2} - 2)}^3}}}\,.\,({x^2} - 2)(5{x^2} - 2 + 6{x^2}\ln 3({x^2} - 2))} \right)

f''(x) = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \sqrt 2

f''\left( {{{\sqrt 2 }^ + }} \right) > 0

f''\left( {{{\sqrt 2 }^ - }} \right) < 0

\Rightarrow x = \sqrt 2

is point of inflection

f''(x) > 0\,\forall x > \sqrt 2

\Rightarrow f'(x)

is increasing for

x > \sqrt 2