Application of Derivatives — Page 15 | JEE Mathematics

Q141

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

f(x)=\left|x^{2}-5 x+6\right|-3 x+2

in the interval

[-1,3]

is equal to :

A 13

B 24

C 10

D 12

Correct Answer

Option C

Solution

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

f(x) = \left|x^2 - 5x + 6\right| - 3x + 2

in the interval

[-1, 3]

can be found by finding the maximum and minimum values of

f(x)

in this interval and then adding them. First, let's find the critical points of

f(x)

. To do this, we will find the zeros of the expression inside the absolute value:

x^2 - 5x + 6 = 0

Solving this quadratic equation, we find that the zeros are

x = 2

and

x = 3

. These are the critical points of

f(x)

. Next, we evaluate

f(x)

at these critical points and at the endpoints of the interval

[-1, 3]

:

f(-1) = \left|(-1)^2 - 5(-1) + 6\right| - 3(-1) + 2 = 17

f(2) = \left|(2)^2 - 5(2) + 6\right| - 3(2) + 2 = -4

f(3) = \left|(3)^2 - 5(3) + 6\right| - 3(3) + 2 = -7

So, the minimum value of

f(x)

is

-7

and the maximum value is

17

. So, the sum of the absolute maximum and minimum values of the function is

-7 + 17 = 10

.

Note : The absolute maximum and minimum values of a function are the largest and smallest values that the function takes on a given interval, respectively.

These values are also called the "extrema" of the function.

The absolute maximum value is the highest point on the graph of the function, and the absolute minimum value is the lowest point.

Q142

If the functions

f(x)=\dfrac{x^3}{3}+2 b x+\dfrac{a x^2}{2}

and

g(x)=\dfrac{x^3}{3}+a x+b x^2, a \neq 2 b

have a common extreme point, then

a+2 b+7

is equal to :

A 6

B

\dfrac{3}{2}

C 3

D 4

Correct Answer

Option A

Solution

f'(x)=x^2+2b+ax

g'(x)=x^2+a+2bx

\Rightarrow x=1

is common root

a+2b+1=0

Q143

Let

f:(0,1)\to\mathbb{R}

be a function defined

f(x) = {1 \over {1 - {e^{ - x}}}}

, and

g(x) = \left( {f( - x) - f(x)} \right)

. Consider two statements (I) g is an increasing function in (0, 1) (II) g is one-one in (0, 1) Then,

A Both (I) and (II) are true

B Neither (I) nor (II) is true

C Only (II) is true

D Only (I) is true

Correct Answer

Option A

Solution

$g(x)=f(-x)-f(x)$

\begin{aligned} & =\frac{1}{1-e^{x}}-\frac{1}{1-e^{-x}} \\\\ & =\frac{1}{1-e^{x}}-\frac{e^{x}}{e^{x}-1} \\\\ & =\frac{1+e^{x}}{1-e^{x}} \\\\ g^{\prime}(x) & =\frac{\left(1-e^{x}\right) e^{x}-\left(1+e^{x}\right)\left(-e^{x}\right)}{\left(1-e^{x}\right)^{2}} \\\\ & =\frac{e^{x}-2 e^{x}+e^{x}+2 e^{x}}{\left(1-e^{x}\right)^{2}}>0 \end{aligned}

So both statements are correct

Q144

\max \limits_{0 \leq x \leq \pi}\left\{x-2 \sin x \cos x+\frac{1}{3} \sin 3 x\right\}=

A

\dfrac{5 \pi+2+3 \sqrt{3}}{6}

B 0

C

\dfrac{\pi+2-3 \sqrt{3}}{6}

D

\pi

Correct Answer

Option A

Solution

Given the function:

f(x) = x - 2\sin{x}\cos{x} + \frac{1}{3}\sin{3x}

We want to find the maximum value of this expression for

0 \leq x \leq \pi

. Step 1: Rewrite the expression Notice that we can rewrite the expression as:

f(x) = x - \sin{2x} + \frac{1}{3}\sin{3x}

Step 2: Find the first derivative Now, let's find the derivative of this expression with respect to

x

:

f'(x) = 1 - 2\cos{2x} + \cos{3x}

Step 3: Find the critical points To find the critical points, we set

f'(x) = 0

:

1 - 2\cos{2x} + \cos{3x} = 0

Step 4: Solve for x This equation can be rewritten as:

(2\cos{x} + \sqrt{3})(2\cos{x} - \sqrt{3})(\cos{x} - 1) = 0

We get three possible solutions for

x

:

\cos{x} = \frac{-\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 1

Which gives us:

x = \frac{5\pi}{6}, \frac{\pi}{6}, 0

Step 5: Find the second derivative Now, let's find the second derivative of the expression:

f''(x) = 4\sin{2x} - 3\sin{3x}

Step 6: Analyze the critical points Evaluate the second derivative at the critical points:

f''\left(\frac{5\pi}{6}\right) = -2\sqrt{3} - \sqrt{3} < 0

, indicating a local maximum.

f''\left(\frac{\pi}{6}\right) = 2\sqrt{3} - \sqrt{3} > 0

, indicating a local minimum.

f''(0) = 0

, inconclusive.

Step 7: Evaluate the function at the local maximum point and the boundary points Since we are looking for the maximum value of

f(x)

, we can now evaluate the function at the local maximum point and the boundary points:

f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{5\pi + 2 + 3\sqrt{3}}{6}

f(0) = 0

f(\pi) = \pi

Step 8: Compare the values The maximum value of

f(x)

is given by

f\left(\frac{5\pi}{6}\right) = \frac{5\pi + 2 + 3\sqrt{3}}{6}

.

Q145

If the local maximum value of the function

f(x)=\left(\dfrac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \dfrac{\pi}{2}\right)

, is

\dfrac{k}{e}

, then

\left(\dfrac{k}{e}\right)^{8}+\dfrac{k^{8}}{e^{5}}+k^{8}

is equal to

A

e^{3}+e^{6}+e^{10}

B

e^{3}+e^{5}+e^{11}

C

e^{3}+e^{6}+e^{11}

D

e^{5}+e^{6}+e^{11}

Correct Answer

Option C

Solution

\text{ Let } y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}

\ln \mathrm{y}=\sin ^2 \mathrm{x} \cdot \ln \left(\frac{\sqrt{3 \mathrm{e}}}{2 \sin \mathrm{x}}\right)

\frac{1}{y} y^{\prime}=\ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x+\sin ^2 x \frac{2 \sin x}{\sqrt{3 e}} \frac{\sqrt{3 e}}{2}(-\operatorname{cosec} x \cot x)

For maxima or minima,

\frac{d y}{d x}=0

\Rightarrow \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x-\sin x \cos x=0

\Rightarrow \sin x \cos x\left[2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)-1\right]=0

\Rightarrow \ln \left(\frac{3 \mathrm{e}}{4 \sin ^2 \mathrm{x}}\right)=1

\Rightarrow \frac{3 e}{4 \sin ^2 x}=e

\Rightarrow \sin ^2 x=\frac{3}{4}

\Rightarrow \sin \mathrm{x}=\frac{\sqrt{3}}{2} \quad\left(\text{ as } \mathrm{x} \in\left(0, \frac{\pi}{2}\right)\right)

\Rightarrow \text{ Local max value }=\left(\frac{\sqrt{3 \mathrm{e}}}{\sqrt{3}}\right)^{3 / 4}=\mathrm{e}^{3 / 8}=\frac{\mathrm{k}}{\mathrm{e}}

\Rightarrow \mathrm{k}^8=\mathrm{e}^{11}

$\therefore$

\left(\frac{k}{e}\right)^8+\frac{k^8}{e^5}+k^8=e^3+e^6+e^{11}

Q146

If the function

f:(-\infty,-1] \rightarrow(a, b]

defined by

f(x)=e^{x^3-3 x+1}

is one - one and onto, then the distance of the point

P(2 b+4, a+2)

from the line

x+e^{-3} y=4

is :

A

2 \sqrt{1+e^6}

B

\sqrt{1+e^6}

C

3 \sqrt{1+e^6}

D

4 \sqrt{1+e^6}

Correct Answer

Option A

Solution

\begin{aligned} & f(x)=e^{x^3-3 x+1} \\ & f^{\prime}(x)=e^{x^3-3 x+1} \cdot\left(3 x^2-3\right) \\ & =e^{x^3-3 x+1} \cdot 3(x-1)(x+1) \end{aligned}

For

\mathrm{f}^{\prime}(\mathrm{x}) \geq 0

\therefore \mathrm{f}(\mathrm{x})

is increasing function

\begin{aligned} & \therefore \mathrm{a}=\mathrm{e}^{-\infty}=0=\mathrm{f}(-\infty) \\ & \mathrm{b}=\mathrm{e}^{-1+3+1}=\mathrm{e}^3=\mathrm{f}(-1) \\ & \mathrm{P}(2 \mathrm{~b}+4, \mathrm{a}+2) \\ & \therefore \mathrm{P}\left(2 \mathrm{e}^3+4,2\right) \end{aligned}

\mathrm{d}=\frac{\left(2 \mathrm{e}^3+4\right)+2 \mathrm{e}^{-3}-4}{\sqrt{1+\mathrm{e}^{-6}}}=2 \sqrt{1+\mathrm{e}^6}

Q147

Let

f:[2,4] \rightarrow \mathbb{R}

be a differentiable function such that

\left(x \log _{e} x\right) f^{\prime}(x)+\left(\log _{e} x\right) f(x)+f(x) \geq 1, x \in[2,4]

with

f(2)=\dfrac{1}{2}

and

f(4)=\dfrac{1}{4}

. Consider the following two statements : (A) :

f(x) \leq 1

, for all

x \in[2,4]

(B) :

f(x) \geq \dfrac{1}{8}

, for all

x \in[2,4]

Then,

A Neither statement (A) nor statement (B) is true

B Only statement (A) is true

C Only statement (B) is true

D Both the statements

(\mathrm{A})

and (B) are true

Correct Answer

Option D

Solution

Given,

\left(x \log _{e} x\right) f^{\prime}(x)+\left(\log _{e} x\right) f(x)+f(x) \geq 1, x \in[2,4]

\left(x\log _e x\right) f^{\prime}(x)+f(x)\left[\log _e x+1\right] \geq 1

$\Rightarrow$

\frac{d}{d x}\left[x \log _e x f(x)\right] \geq 1

\begin{aligned} &\Rightarrow \frac{d}{d x}\left[x \log _e x f(x)-x\right] \geq 0 \quad\left[\because \frac{d}{d x}(x)=1\right] \\\\ & \forall x \in[2,4] \end{aligned}

Let $g(x)=x \log _e x f(x)-x$ As $g(x) \geq 0, \forall x \in[2,4], g(x)$ is an increasing function in $[2,4]$

\begin{aligned} g(2) & =2 \log _e 2 f(2)-2 \\\\ & =\log _e 2-2 \quad\left[\because f(x)=\frac{1}{2}\right] \end{aligned}

\begin{aligned} & g(4)=4 \log _e 4 f(4)-4=\log _e 4-4 \\\\ &=2\left(\log _e 2-2\right)\\\\ & {\left[\therefore f(4)=\frac{1}{4}\right] } \end{aligned}

As, $g(x)$ is an increasing function,

\begin{aligned} & g(2) \leq g(x) \leq g(4) \\\\ & \log _e 2-2 \leq g(x) \leq 2\left(\log _e 2-2\right) \\\\ & \log _e 2-2 \leq x \log _e x f(x)-x \leq 2\left(\log _e 2-2\right) \end{aligned}

\frac{\log _e 2-2+x}{x \log _e x} \leq f(x) \leq \frac{2\left(\log _e 2-2\right)+x}{x \log _e x}

\begin{aligned} & \text{Now for } x \in[2,4] \\\\ & \begin{aligned} \frac{\log _e 2-2+x}{x \log _e x} & \leq \frac{2\left(\log _e 2-2\right)+e^2}{2 \log _e 2} =1-\frac{1}{\log _e 2}\frac{1}{8} \end{aligned}

\therefore f(x) \geq \frac{1}{8}, \forall x \in[2,4]

Hence, both statements $A$ and $B$ are true.

Note : LMVT on $(\mathrm{yx}(\ln \mathrm{x}))$ not satisfied.

Hence no such function exists.

Therefore it should be bonus.

Q148

Let

\mathrm{g}(x)=f(x)+f(1-x)

and

f^{\prime \prime}(x) > 0, x \in(0,1)

. If

\mathrm{g}

is decreasing in the interval

(0, a)

and increasing in the interval

(\alpha, 1)

, then

\tan ^{-1}(2 \alpha)+\tan ^{-1}\left(\dfrac{1}{\alpha}\right)+\tan ^{-1}\left(\dfrac{\alpha+1}{\alpha}\right)

is equal to :

A

\dfrac{3 \pi}{4}

B

\pi

C

\dfrac{5 \pi}{4}

D

\dfrac{3 \pi}{2}

Correct Answer

Option B

Solution

We have, $g(x)=f(x)+f(1-x)$ Differentiating both side, we get $g^{\prime}(x)=f^{\prime}(x)-f^{\prime}(1-x)$ As $f^{\prime \prime}(x)>0, f^{\prime}(x)$ is an increasing function.

Also, $g(x)=f(x)+f(2 a-x)$ is always symmetric about $x=a$ So, $g(x)=f(x)+f(1-x)$ is also symmetric about $x=1 / 2$ $\therefore g$ is decreasing in the interval $(0,1 / 2)$ and increasing in the interval $(1 / 2,1)$ .

Now,

\begin{aligned} & \tan ^{-1} 2 \alpha+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) \\\\ & =\tan ^{-1} 1+\tan ^{-1}(2)+\tan ^{-1} 3=\pi \end{aligned}

Q149

The slope of tangent at any point (x, y) on a curve

y=y(x)

is

{{{x^2} + {y^2}} \over {2xy}},x > 0

. If

y(2) = 0

, then a value of

y(8)

is :

A

- 4\sqrt 2

B

2\sqrt 3

C

4\sqrt 3

D

- 2\sqrt 3

Correct Answer

Option C

Solution

Let the slope of tangent at any point $(x, y)$ on a curve $y=y(x)$ is $\dfrac{d y}{d x}$ According to the question, $\dfrac{d y}{d x}=\dfrac{x^2+y^2}{2 x y}(x>0)$ [Given]

\begin{aligned} &\text{ Let } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\\\ &\Rightarrow v+x \frac{d v}{d x}=\frac{x^2+v^2 x^2}{2 v x^2}=\frac{1+v^2}{2 v} \\\\ &\Rightarrow x \frac{d v}{d x}=\frac{1+v^2}{2 v}-v=\frac{1-v^2}{2 v} \end{aligned}

\begin{aligned} & \Rightarrow \int \frac{2 v d v}{1-v^2}=\int \frac{d x}{x} \text{ [Taking integration on both sides] } \\\\ & \Rightarrow-\ln \left|1-v^2\right|=\ln x-\ln c \\\\ & \Rightarrow \ln \left(1-v^2\right) x=\ln c \quad(c=\text{ constant }) \\\\ & \Rightarrow \left(1-y^2 / x^2\right) \cdot x=c \\\\ & \Rightarrow \left(x^2-y^2\right)=c x \end{aligned}

Put $y(2)=0 \Rightarrow c=2$

\begin{array}{rlrl} & \therefore x^2-y^2 =2 x \\\\ & \therefore y^2 =x^2-2 x \\\\ & \Rightarrow y(x) = \pm \sqrt{x^2-2 x} \end{array}

Put $x=8$ , we get

\Rightarrow y(8)= \pm \sqrt{64-16}= \pm \sqrt{48}= \pm 4 \sqrt{3}

Q150

A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in cm

^2

) is equal to :

A 1025

B 900

C 800

D 675

Correct Answer

Option C

Solution

Given, length of square sheet of side is $30 \mathrm{~cm}$ .

Let, side of small square be $x \mathrm{~cm}$ .

Since, a square piece of tin is to be made into a box without top by cutting square from each corner and folding up the flaps to from a box.

Then, this shape formed a cuboidal shape.

Thus, volume of the box $(V)$

\begin{aligned} & =(30-2 x)(30-2 x) x \\\\ &\Rightarrow V =x(30-2 x)^2 \end{aligned}

On differentiating both side with respect to $x$ , we get

\begin{aligned} \frac{d V}{d x} & =(30-2 x)^2+x\{2(30-2 x) \cdot(-2)\} \\\\ & =(30-2 x)^2-4 x(30-2 x) \\\\ & =(30-2 x)\{30-6 x\} \end{aligned}

Therefore, maximum of $V, \dfrac{d V}{d x}=0$

\begin{array}{ll} &\Rightarrow (30-2 x)(30-6 x)=0 \\\\ &\Rightarrow x=15 \text{ or } x=5 \quad[x=15 \text{ (not taken) }] \end{array}

Thus, $x=5$ So, the surface area

\begin{array}{ll} =4 \times(30-2 x) \times x+(30-2 x)^2 & \\\\ =4 \times(30-10) \times 5+(30-10)^2 & (\text{ Put } x=5) \\\\ =400+400=800 & \end{array}