Application of Derivatives — Page 17 | JEE Mathematics

Q161

The number of critical points of the function

f(x)=(x-2)^{2 / 3}(2 x+1)

is

A 2

B 1

C 0

D 3

Correct Answer

Option A

Solution

To find the number of critical points of the function

f(x)=(x-2)^{2 / 3}(2 x+1)

, we need to determine where its derivative

f'(x)

is equal to zero or undefined.

Critical points occur where the derivative is zero or does not exist.

First, let's find the derivative of the function:

f(x)=(x-2)^{2 / 3}(2 x+1)

We apply the product rule for differentiation, which states that

(uv)' = u'v + uv'

, where

u = (x-2)^{2/3}

and

v = 2x + 1

. We need the derivatives of

u

and

v

: For

u = (x-2)^{2/3}

, we use the chain rule:

u'(x) = \frac{d}{dx}[(x-2)^{2/3}] = \frac{2}{3}(x-2)^{-1/3} \cdot 1 = \frac{2}{3}(x-2)^{-1/3}

The derivative of

v

is straightforward, as

v = 2x + 1

:

v'(x) = 2

Now we apply the product rule:

f'(x) = u'(x)v(x) + u(x)v'(x)

Substituting

u

,

u'

, and

v

, we get:

f'(x) = \left( \frac{2}{3}(x-2)^{-1/3} \right)(2x+1) + \left( (x-2)^{2/3} \right)(2)

This simplifies to:

f'(x) = \frac{2(2x + 1)}{3(x-2)^{1/3}} + 2(x-2)^{2/3}

For critical points, we need to solve

f'(x) = 0

or where it is undefined. 1. Solve for where the derivative is zero:

\frac{2(2x + 1)}{3(x-2)^{1/3}} + 2(x-2)^{2/3} = 0

Combining like terms in a common denominator, we get:

\frac{2(2x + 1) + 6(x - 2)}{3(x-2)^{1/3}} = 0

Simplifying the numerator:

2(2x + 1) + 6(x-2) = 4x + 2 + 6x - 12 = 10x - 10 = 10(x-1)

So:

\frac{10(x-1)}{3(x-2)^{1/3}} = 0

The numerator is zero when:

10(x-1) = 0

Therefore:

x = 1

2. Solve for where the derivative is undefined: The denominator,

3(x-2)^{1/3}

, is undefined when

(x-2)^{1/3} = 0

, which happens at:

x = 2

From the above analysis, the critical points are at

x = 1

and

x = 2

. Thus, there are 2 critical points. Therefore, the number of critical points of the function

f(x)=(x-2)^{2 / 3}(2 x+1)

is: Option A

Q162

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm , the ice-cream melts at the rate of

81 \mathrm{~cm}^3 / \mathrm{min}

and the thickness of the ice-cream layer decreases at the rate of

\dfrac{1}{4 \pi} \mathrm{~cm} / \mathrm{min}

. The surface area (in

\mathrm{cm}^2

) of the chocolate ball (without the ice-cream layer) is :

A

128 \pi

B

196 \pi

C

225 \pi

D

256 \pi

Correct Answer

Option D

Solution

\begin{aligned} &\begin{aligned} & \mathrm{v}=\frac{4}{3} \pi \mathrm{r}^3 \\ & \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \\ & 81=4 \pi \mathrm{r}^2 \times \frac{1}{4 \pi} \\ & \mathrm{r}^2=81 \\ & \mathrm{r}=9 \end{aligned}\\ &\text{ surface area of chocolate }=4 \pi(r-1)^2=256 \pi \end{aligned}

Q163

For the function

f(x)=(\cos x)-x+1, x \in \mathbb{R}

, between the following two statements (S1)

f(x)=0

for only one value of

x

in

[0, \pi]

. (S2)

f(x)

is decreasing in

\left[0, \dfrac{\pi}{2}\right]

and increasing in

\left[\dfrac{\pi}{2}, \pi\right]

.

A Both (S1) and (S2) are incorrect.

B Only (S1) is correct.

C Only (S2) is correct.

D Both (S1) and (S2) are correct.

Correct Answer

Option B

Solution

Let's analyze the function

f(x) = (\cos x) - x + 1

over the interval

[0, \pi]

and the statements provided. First, let's consider statement (S1): (S1)

f(x)=0

for only one value of

x

in

[0, \pi]

. To examine this statement, we need to explore the zeros of the function

f(x)

within the given interval. Let's define and analyze the function:

f(x) = \cos x - x + 1

We seek to determine if

f(x) = 0

has only one solution in the interval

[0, \pi]

.

To do this, we can use the Intermediate Value Theorem and the behavior of the function's derivative.

First, compute the derivative of

f(x)

:

f'(x) = \frac{d}{dx}(\cos x - x + 1) = -\sin x - 1

The critical points occur when

f'(x) = 0

:

- \sin x - 1 = 0 \Rightarrow \sin x = -1

. The equation

\sin x = -1

does not hold for any

x

in

[0, \pi]

. Note that: For

x \in [0, \frac{\pi}{2}]

,

\sin x

ranges from 0 to 1. For

x \in [\frac{\pi}{2}, \pi]

,

\sin x

ranges from 1 to 0. Since

f'(x)

is always negative (i.e.,

f'(x)

f(0) = \cos 0 - 0 + 1 = 1 + 1 = 2

f(\pi) = \cos \pi - \pi + 1 = -1 - \pi + 1 = -\pi

Given the continuous and strictly decreasing nature of

f(x)

in

[0, \pi]

, by the Intermediate Value Theorem, there is exactly one value of

x

in the interval

[0, \pi]

where

f(x) = 0

, confirming (S1). Now, let's consider statement (S2): (S2)

f(x)

is decreasing in

\left[0, \frac{\pi}{2}\right]

and increasing in

\left[\frac{\pi}{2}, \pi\right]

. We already analyzed that

f'(x)

shows that

f'(x) = -\sin x - 1

is always less than zero in

[0, \pi]

. No interval exists where the derivative is positive. This means that

f(x)

is strictly decreasing throughout the entire interval of

[0, \pi]$$, invalidating (S2). Therefore, the correct option is: Option B: Only (S1) is correct.

Q164

Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a+b)

^2

is equal to :

A 64

B 80

C 60

D 72

Correct Answer

Option D

Solution

\begin{aligned} & P D=4 \cos \theta \Rightarrow P S=4 \cos \theta+2 \sin \theta \\ & D S=2 \sin \theta \\ & A P=4 \sin \theta \\ & Q A=2 \cos \theta \Rightarrow P Q=2 \cos \theta+4 \sin \theta \\ & \Rightarrow \text{ Area of } P Q R S=4(2 \cos \theta+\sin \theta)(\cos \theta+2 \sin \theta) \\ &=4\left[2 \cos ^2 \theta+2 \sin ^2 \theta+5 \sin \theta \cos \theta\right] \\ &=8+10 \sin 2 \theta \end{aligned}

Area is maximum when

\sin 2 \theta=1 \Rightarrow \theta=45^{\circ}

$\Rightarrow$ Maximum area

=8+10=18

\therefore P S=4 \times \frac{1}{\sqrt{2}}+2 \times \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}}

\begin{aligned} & P Q=2 \times \frac{1}{\sqrt{2}}+4 \times \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}} \\ \therefore \quad & (a+b)^2=\left(\frac{6}{\sqrt{2}}+\frac{6}{\sqrt{2}}\right)^2=\left(\frac{12}{\sqrt{2}}\right)=(6 \sqrt{2})^2=72 \end{aligned}

Q165

Let

f(x)=x^5+2 x^3+3 x+1, x \in \mathbf{R}

, and

g(x)

be a function such that

g(f(x))=x

for all

x \in \mathbf{R}

. Then

\dfrac{g(7)}{g^{\prime}(7)}

is equal to :

A 42

B 7

C 1

D 14

Correct Answer

Option D

Solution

\begin{aligned} & f(x)=x^5+2 x^3+3 x+1 \\ & g(f(x))=x . \quad \Rightarrow g^{\prime}(f(x)) f^{\prime}(x)=1 \\ \end{aligned}

\begin{aligned} &\begin{aligned} & \text{ Now } \frac{g(7)}{g^{\prime}(7)} \\ & g(7) \Rightarrow f(x)=7 \\ & x^5+2 x^3+3 x+1=7 \\ & \Rightarrow x\left(x^4+2 x^2+3\right)=0 \\ & \Rightarrow x=1 \\ & \therefore g(7) \Rightarrow g(f(1))=1 \\ & \& ~g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \\ & g^{\prime}(7) \\ & \Rightarrow f(x)=7 \Rightarrow x=1 \\ & \therefore g^{\prime}(7)=\frac{1}{f^{\prime}(1)} \\ & =\frac{1}{5 x^4+6 x^2+3} \\ & =\frac{1}{14} \\ & \therefore \frac{g(7)}{g^{\prime}(7)}=\frac{1}{\frac{1}{14}} \quad 14 \end{aligned}\\ \end{aligned}

Q166

For the function

f(x)=\sin x+3 x-\dfrac{2}{\pi}\left(x^2+x\right), \text{ where } x \in\left[0, \dfrac{\pi}{2}\right],

consider the following two statements : (I)

f

is increasing in

\left(0, \dfrac{\pi}{2}\right)

. (II)

f^{\prime}

is decreasing in

\left(0, \dfrac{\pi}{2}\right)

. Between the above two statements,

A only (I) is true.

B both (I) and (II) are true.

C only (II) is true.

D neither (I) nor (II) is true.

Correct Answer

Option B

Solution

\begin{aligned} & f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right), \text{ where } x \in\left[0, \frac{\pi}{2}\right] \\ & f^{\prime}(x)=\cos x+3-\frac{2}{\pi}(2 x+1) \\ & =\cos x-\frac{4 x}{\pi}-\frac{2}{\pi}+3 \\ & \text{ as } x \in\left[0, \frac{\pi}{2}\right] \\ & \frac{4 x}{\pi} \in[0,2] \end{aligned}

\Rightarrow 3-\frac{2}{\pi}-\frac{4 x}{\pi}>0

and also

\cos x>0

when

x \in\left[0, \frac{\pi}{2}\right]

\Rightarrow f^{\prime}(x)>0

\Rightarrow f(x)

is increasing Now,

f^{\prime \prime}(x)=-\sin x-\frac{4}{\pi} Hence,

f^{\prime}(x)

is decreasing

\therefore \quad$$ Both statements (I) and (II) are true

Q167

The interval in which the function

f(x)=x^x, x>0

, is strictly increasing is

A

(0, \infty)

B

\left(0, \dfrac{1}{e}\right]

C

\left[\dfrac{1}{e^2}, 1\right)

D

\left[\dfrac{1}{e}, \infty\right)

Correct Answer

Option D

Solution

First note that

f(x)=x^x=e^{\,x\ln x}\,,\quad x>0.

Differentiating gives

f'(x)=x^x\bigl(\ln x+1\bigr)\,.

Since $x^x>0$ , the sign of $f'(x)$ is the sign of $\ln x+1$ .

Hence $f'(x)>0\quad\Longleftrightarrow\quad \ln x>-1\quad\Longleftrightarrow\quad x>\frac1e.$ So $f$ is strictly increasing for all $x>1/e$ .

Among the given choices that corresponds to Option D: $\displaystyle\bigl[\tfrac1e,\infty\bigr)$ .

Q168

Let

f(x)=\int_0^{x^2} \dfrac{\mathrm{t}^2-8 \mathrm{t}+15}{\mathrm{e}^{\mathrm{t}}} \mathrm{dt}, x \in \mathbf{R}

. Then the numbers of local maximum and local minimum points of

f

, respectively, are :

A 3 and 2

B 2 and 2

C 2 and 3

D 1 and 3

Correct Answer

Option C

Solution

We are given

f(x)=\int_0^{x^2} \frac{t^2-8t+15}{e^t}\,dt, \quad x\in\mathbb{R}.

To find the local extrema, we first compute the derivative using the Fundamental Theorem of Calculus and the chain rule.

Step 1.

Rewrite the derivative: Let

F(u)=\int_0^u \frac{t^2-8t+15}{e^t}\,dt,

so that

f(x)=F(x^2).

Then by the chain rule,

f'(x)=F'(x^2)\cdot 2x.

Since

F'(u)=\frac{u^2-8u+15}{e^u},

we substitute $u=x^2$ to get

f'(x)=\frac{(x^2)^2-8x^2+15}{e^{x^2}}\cdot 2x = \frac{2x\,(x^4-8x^2+15)}{e^{x^2}}.

Step 2.

Factor and Identify Critical Points: Factor the polynomial $x^4-8x^2+15$ by writing it in terms of $x^2$ .

Let $y=x^2$ , then

y^2-8y+15=(y-3)(y-5)

so that

x^4-8x^2+15=(x^2-3)(x^2-5).

Thus,

f'(x)=\frac{2x\,(x^2-3)(x^2-5)}{e^{x^2}}.

Since $e^{x^2}>0$ for all $x$ , the zeros of $f'(x)$ are determined by

2x\,(x^2-3)(x^2-5)=0.

That gives the critical points: $2x=0 \Rightarrow x=0$ , $x^2-3=0 \Rightarrow x=\pm\sqrt{3}$ , $x^2-5=0 \Rightarrow x=\pm\sqrt{5}$ .

Step 3.

Analyzing the Sign of $f'(x)$ : We need to determine the nature (maximum or minimum) by looking at the sign changes of $f'(x)$ on the intervals determined by the critical points $x=-\sqrt{5}$ , $x=-\sqrt{3}$ , $x=0$ , $x=\sqrt{3}$ , and $x=\sqrt{5}$ .

Notice that the factor $2x\,(x^2-3)(x^2-5)$ will dictate the sign.

Let’s define:

h(x)=x\,(x^2-3)(x^2-5).

Examine the sign of $h(x)$ in each interval: For $x$ x $is negative.$ x^2>5 $so$ x^2-3>0 $and$ x^2-5>0 $. Product: negative$ \times $positive$ \times $positive = negative. Thus,$ f'(x) For $-\sqrt{5}$ x $is negative.$ x^2 $is between 3 and 5 so$ x^2-3>0 $while$ x^2-5 Product: negative $\times$ positive $\times$ negative = positive.

Thus, $f'(x)>0$ .

For $-\sqrt{3}$ x $is negative.$ x^2 Product: negative $\times$ negative $\times$ negative = negative.

Thus, $f'(x) For$ 0 $x$ is positive. $x^2 Product: positive$ \times $negative$ \times $negative = positive. Thus,$ f'(x)>0 $. For$ \sqrt{3} $x$ is positive. $x^2$ is between 3 and 5 so $x^2-3>0$ while $x^2-5 Product: positive$ \times $positive$ \times $negative = negative. Thus,$ f'(x) For $x > \sqrt{5}$ : $x$ is positive. $x^2>5$ so both $x^2-3>0$ and $x^2-5>0$ .

Product: positive $\times$ positive $\times$ positive = positive.

Thus, $f'(x)>0$ .

Step 4.

Classify the Critical Points: At $x=-\sqrt{5}$ : $f'(x)$ changes from negative to positive → local minimum.

At $x=-\sqrt{3}$ : $f'(x)$ changes from positive to negative → local maximum.

At $x=0$ : $f'(x)$ changes from negative to positive → local minimum.

At $x=\sqrt{3}$ : $f'(x)$ changes from positive to negative → local maximum.

At $x=\sqrt{5}$ : $f'(x)$ changes from negative to positive → local minimum.

Step 5.

Conclusion: There are local maximum points at $x=-\sqrt{3}$ and $x=\sqrt{3}$ (2 points in total), and local minimum points at $x=-\sqrt{5}$ , $x=0$ , and $x=\sqrt{5}$ (3 points in total).

Thus, the numbers of local maximum and local minimum points of $f$ are 2 and 3, respectively.

The correct option is Option C.

Q169

Consider the region

R=\left\{(x, y): x \leq y \leq 9-\dfrac{11}{3} x^2, x \geq 0\right\}

. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R , is:

A

\dfrac{821}{123}

B

\dfrac{567}{121}

C

\dfrac{730}{119}

D

\dfrac{625}{111}

Correct Answer

Option B

Solution

$\mathrm{t} .\left(9-\dfrac{11 \mathrm{t}^2}{3}-\mathrm{t}\right)$

\begin{aligned} & A=9 t-t^2-\frac{11}{3} t^3 \\ & \frac{d A}{d t}=9-2 t-11 t^2 \\ & \Rightarrow 11 t^2+2 t-9=0 \\ & 11 t^2+11 t-9 t-9=0 \\ & t=-1 \& t=\frac{9}{11} \end{aligned}

$\therefore \dfrac{\mathrm{dA}}{\mathrm{dt}}=$

\begin{aligned} & \therefore \text{ largest area }=\frac{9}{11}\left(9-\frac{11}{3}, \frac{81}{121}-\frac{9}{11}\right) \\ & =\frac{9}{11} \cdot \frac{63}{11}=\frac{567}{121} \end{aligned}

Q170

Let

(2,3)

be the largest open interval in which the function

f(x)=2 \log _{\mathrm{e}}(x-2)-x^2+a x+1

is strictly increasing and (b, c) be the largest open interval, in which the function

\mathrm{g}(x)=(x-1)^3(x+2-\mathrm{a})^2

is strictly decreasing. Then

100(\mathrm{a}+\mathrm{b}-\mathrm{c})

is equal to :

A 360

B 420

C 160

D 280

Correct Answer

Option A

Solution

$$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\mathrm{x}-2}-2 \mathrm{x}+\mathrm{a} \geq 0 \\ & \mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{-2}{(\mathrm{x}-2)^2}-2