Application of Derivatives

Step 1: Write $f(x)$ without absolute values for different intervals

When $x \geq 1$, $\ln x$ is positive and $x - 1$ is also positive. So:

$ f(x) = |\ln x| - |x-1| = \ln x - (x - 1) = \ln x - x + 1 $ for $x \geq 1$.

When $0 < x < 1$, $\ln x$ is negative and $x - 1$ is negative. So:

$ f(x) = |\ln x| - |x-1| = -\ln x - (-(x - 1)) = -\ln x + x - 1 $ for $0 < x < 1$.

So we can write: $ f(x) = \begin{cases} \ln x - x + 1 & \text{for } x \geq 1 \\\\ -\ln x + x - 1 & \text{for } 0 < x < 1 \end{cases} $

Step 2: Find the derivative $f'(x)$ in both cases

For $x \geq 1,$

$ f'(x) = \frac{d}{dx} (\ln x - x + 1) = \frac{1}{x} - 1 $

For $0 < x < 1,$

$ f'(x) = \frac{d}{dx} (-\ln x + x - 1) = -\frac{1}{x} + 1 $

So, $ f'(x) = \begin{cases} \frac{1}{x} - 1 & \text{for } x \geq 1 \\\\ -\frac{1}{x} + 1 & \text{for } 0 < x < 1 \end{cases} $

Step 3: Check differentiability at $x = 1$

From the left of $x = 1$: $ f'(1^-) = -\frac{1}{1} + 1 = 0 $

From the right of $x = 1$: $ f'(1^+) = \frac{1}{1} - 1 = 0 $

Both one-sided derivatives are equal. So $f(x)$ is differentiable at $x = 1$.

Since the formulas for $f'(x)$ are valid everywhere for $x > 0$, $f(x)$ is differentiable for all $x > 0$.

Step 4: Check if $f(x)$ is increasing or decreasing

For $x > 1$: $ f'(x) = \frac{1}{x} - 1 $ Since $x > 1$, $\frac{1}{x} < 1$. So $f'(x) < 0$, which means $f(x)$ is decreasing for $x > 1$.

For $0 < x < 1$: $ f'(x) = -\frac{1}{x} + 1 $ Here, $\frac{1}{x} > 1$, so $-\frac{1}{x} + 1 < 0$, so $f'(x) < 0$. Therefore, $f(x)$ is decreasing for $0 < x < 1$ also.

Step 5: Check the statements

(I) $f$ is differentiable at all $x > 0$: True

(II) $f$ is increasing in $(0, 1)$: False

(III) $f$ is decreasing in $(1, \infty)$: True

So only statements (I) and (III) are correct.

$\begin{aligned} & f(\theta)=\cos ^2 \theta-6 \sin \theta \cos \theta+3 \sin ^2 \theta+2 \\ & f(\theta)=\frac{1+\cos 2 \theta}{2}-3 \sin 2 \theta+3\left(\frac{1-\cos 2 \theta}{2}\right)+2 \\ & f(\theta)=\frac{1}{2}+\frac{\cos 2 \theta}{2}-3 \sin 2 \theta+\frac{3}{2}-\frac{3 \cos 2 \theta}{2}+2 \\ & f(\theta)=4-\cos 2 \theta-3 \sin 2 \theta \\ & f(\theta)=4-(\cos 2 \theta+3 \sin 2 \theta) \\ & \text { Let } g(\theta)=\cos 2 \theta+3 \sin 2 \theta\end{aligned}$

The expression $a \cos \alpha+b \sin \alpha$ has a range of $\left[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}\right]$.

$ \begin{aligned} & -\sqrt{1^2+3^2} \leq g(\theta) \leq \sqrt{1^2+3^2} \\ & -\sqrt{10} \leq g(\theta) \leq \sqrt{10} \end{aligned} $

For least value of $f(\theta)$

$ \begin{aligned} & f(\theta)=4-(\max \text { value of } g(\theta)) \\ & f(\theta)=4-\sqrt{10} \end{aligned} $

So, least value is $4-\sqrt{10}$

$ \begin{aligned} & f(\theta)=4\left(\sin ^{-1}\left(\frac{7 \pi}{2}-\theta\right)+\sin ^4(11 \pi+\theta)\right) \\ & \quad-2\left(\sin ^6\left(\frac{3 \pi}{2}-\theta\right)+\sin ^6(9 \pi-\theta)\right) \\ & =4\left(\cos ^4 \theta+\sin ^4 \theta \cos ^2 \theta\right)-2\left(\cos ^2 \theta+\sin ^6 \theta\right) \\ & =4\left(1-2 \sin ^2 \theta \cos ^2 \theta\right) \\ & \quad-2(1)\left(\sin ^4 \theta+\cos ^4 \theta-\sin ^2 \theta \cos ^2 \theta\right) \\ & =4-8 \sin ^2 \theta \cos ^2 \theta-2\left(1-3 \sin ^2 \theta \cos ^2 \theta\right) \\ & =2-2 \sin ^2 \theta \cos ^2 \theta \\ & =2-\frac{(\sin 2 \theta)^2}{2} \\ & f(\theta)_{\max }=2=\alpha \quad f(\theta)_{\min }=2-\frac{1}{2}=\frac{3}{2}=\beta \\ & \alpha+2 \beta=2+3=5 \end{aligned} $

$ \begin{aligned} & f(x)=x^{2025}-x^{2000} ; x \in[0,1] \\ & f^{\prime}(x)=2025 x^{2024}-200 x^{1999} \\ & =x^{1999}\left(2025 x^5-2000\right) \\ & f^{\prime}(x)=0 \Rightarrow x^5=\frac{80}{81} \\ & f(x)=\left(x^5\right)^{81}-\left(x^5\right)^{80} \\ & =\left(\frac{80}{81}\right)^{81}-\left(\frac{80}{81}\right)^{80} \\ & =\left(\frac{80}{81}\right)^{80}\left(-\frac{1}{81}\right) \\ & =(80)^{80}(-81)^{-81} \end{aligned} $

$ \begin{aligned} g(x) & =f\left((\tan x-1)^2+a-1\right) \\ g^{\prime}(x) & =f^{\prime}\left((\tan x-1)^2+a-1\right) \cdot 2(\tan x-1) \cdot \sec ^2 x . \,\,\,\,\,\,\,\,...(i)\end{aligned} $

Given $f^{\prime \prime}(x)>0 \quad \forall x \in R$

$\therefore f^{\prime}(x)$ is increasing function

Case (i) $0

$ \begin{aligned} & 0 < \tan x > 1 \\ & -1<\tan x-1 < 0 \\ & 0 < (\tan x-1)^2 < 1 \\ & a-1 < (\tan x-1)^2+(a-1) < a \\ & \therefore f^{\prime}\left((\tan x-1)^2+a-1\right)>f^{\prime}(a-1)=0 \end{aligned} $

From (i) $g^{\prime}(x)=(+)(-)(+)=-v e$

$g(x)$ is decreasing in $\left(0, \frac{\pi}{4}\right)$

Similarly we can prove that $g(x)$ is increasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$

We are given the function:

$f(x)= \sqrt{x}\ln x - x + 1$

To find where the function increases or decreases, we take the derivative:

$f'(x)=\frac{1}{\sqrt{x}}+\frac{\ln x}{2\sqrt{x}}-1$

$f'(x)=\frac{2+\ln x-2\sqrt{x}}{2\sqrt{x}}$

$f'(x)=\frac{\ln x+2}{2\sqrt{x}}-1$

The second derivative tells us how the slope is changing:

$f''(x)=\frac{\left(\frac{1}{x}\right)(2\sqrt{x})-(\ln x+2)\left(\frac{1}{\sqrt{x}}\right)}{4x}$

$f''(x)=-\frac{\ln x}{4x\sqrt{x}}$

For $ x \in (0,1) $, $ \ln x < 0 $. Therefore,

$f''(x) > 0$

This means $ f'(x) $ is increasing in $ (0,1) $.

$\frac{\ln x + 2}{2\sqrt{x}} - 1 = 0$

$\Rightarrow \ln x + 2 = 2\sqrt{x}$

Rearranging, we define:

$g(x) = 2\sqrt{x} - \ln x - 2$

$g'(x) = \frac{1}{\sqrt{x}} - \frac{1}{x} = \frac{\sqrt{x} - 1}{x}$

So, $ g'(x) = 0 $ when $ x = 1 $.

For $ x < 1 $, $ g'(x) < 0 $ (decreasing), and for $ x > 1 $, $ g'(x) > 0 $ (increasing).

Thus, minimum value of $ g(x) $ occurs at $ x = 1 $.

At $ x = 1 $, $ g(1) = 2(1) - \ln(1) - 2 = 0 $.

Hence, $ g(x) \geq 0 $ for all $ x > 0 $.

$f'(x) = -\frac{g(x)}{2\sqrt{x}}$

Since $ g(x) \geq 0 $, we get $ f'(x) \leq 0 $ for all $ x > 0 $.

$ f'(x) = 0 $ only when $ x = 1 $.

The derivative $ f'(x) $ is negative everywhere except at $ x = 1 $ where it is zero. However, the sign of the derivative does not change around $ x = 1 $.

Therefore, the function $ f(x) $ is strictly decreasing throughout its domain, and the tangent at $ x = 1 $ is horizontal but not a turning point.

Hence, the function has no local maximum or minimum.

Given

$ f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}\quad \text{for all }x,y\in \mathbb{R}, $

and

$ f'(0)=3. $

We have to find the minimum value of

$ g(x)=3+e^x f(x). $

Step 1: Find the form of $f(x)$

We are given a functional equation. We will use it carefully.

Let

$ \frac{x+y}{3}=t. $

Then

$ x+y=3t. $

The relation becomes

$ f(t)=\frac{f(x)+f(y)}{3}\quad \text{whenever }x+y=3t. $

We now choose suitable values.

Put $x=y=0$

Then

$ f(0)=\frac{f(0)+f(0)}{3}=\frac{2f(0)}{3}. $

So,

$ 3f(0)=2f(0)\implies f(0)=0. $

Put $y=0$

Then

$ f\left(\frac{x}{3}\right)=\frac{f(x)+f(0)}{3}=\frac{f(x)}{3}. $

Thus,

$ f(x)=3f\left(\frac{x}{3}\right). $

Now differentiate both sides with respect to $x$:

$ f'(x)=3\cdot f'\left(\frac{x}{3}\right)\cdot \frac{1}{3}=f'\left(\frac{x}{3}\right). $

So,

$ f'(x)=f'\left(\frac{x}{3}\right). $

Repeating this,

$ f'(x)=f'\left(\frac{x}{3}\right)=f'\left(\frac{x}{3^2}\right)=\cdots=f'\left(\frac{x}{3^n}\right). $

As $n\to\infty$,

$ \frac{x}{3^n}\to 0. $

Since $f$ is differentiable, $f'$ is defined at $0$, so we get

$ f'(x)=f'(0)=3. $

Hence,

$ f'(x)=3\quad \text{for all }x. $

Therefore,

$ f(x)=3x+c. $

But $f(0)=0$, so

$ c=0. $

Thus,

$ f(x)=3x. $

Step 2: Write $g(x)$

Now

$ g(x)=3+e^x f(x)=3+e^x(3x)=3+3xe^x. $

So,

$ g(x)=3\bigl(1+xe^x\bigr). $

Step 3: Minimise $xe^x$

Let

$ h(x)=xe^x. $

Then

$ h'(x)=e^x+xe^x=e^x(1+x). $

Since $e^x>0$ for all $x$, we have

$ h'(x)=0 \iff 1+x=0 \iff x=-1. $

Now,

• for $x<-1$, $h'(x)<0$, so $h$ is decreasing,

• for $x>-1$, $h'(x)>0$, so $h$ is increasing.

Hence $h(x)$ has a minimum at $x=-1$.

Its minimum value is

$ h(-1)=(-1)e^{-1}=-\frac{1}{e}. $

Therefore,

$ g_{\min}=3\left(1-\frac{1}{e}\right)=3\left(\frac{e-1}{e}\right). $

Final Answer

$ \boxed{3\left(\frac{e-1}{e}\right)} $

So the correct option is B.

Given $f(x)$ is a polynomial of degree $5$

Also, $f'(1)=0$; $f'(-1)=0$.

Also $\because\ \lim\limits_{x\rightarrow 0}\dfrac{f(x)}{x^3}=-5$ (fixed and finite)

$\therefore\ f(0)=0$; $f'(0)=0$; $f''(0)=0$

$\dfrac{f'''(0)}{6}=-5 \rightarrow f'''(0)=-30$

Hence, let

$f'(x)=(ax+b)(x-0)(x-1)(x+1)$

$\rightarrow f'(x)=ax^4+bx^3-ax^2-bx$

$f''(x)=4ax^3+3bx^2-2ax-b$

$\therefore\ f''(0)=0 \rightarrow b=0$

$f'''(x)=12ax^2+6bx-2a$

$\therefore\ f'''(0)=-2a=-30$

$\rightarrow a=15$

$\therefore\ f'(x)=15x.x.(x-1)(x+1)$

$\rightarrow f'(x)=15x^4-15x^2$

$\rightarrow f(x)=3x^5-5x^3+C$

$\therefore\ f(0)=0 \rightarrow C=0$

$\therefore\ f(x)=3x^5-5x^3$

$\therefore\ f(2)-f(-2)=112$

First check continuity at $x=0$:

$\lim _{x \rightarrow 0}\left|\frac{\sin x}{x}\right|=1=f(0) \rightarrow f(x)$ is continuous

For $x \neq 0$, write $g(x)=\frac{\sin x}{x}$. Then

Now $\frac{d}{d x}\left(\frac{\sin x}{x}\right)=\frac{x \cos x-\sin x}{x^2}$

To find critical points, we look for points in $(-2\pi,2\pi)$ where either $f'(x)=0$ or $f'(x)$ does not exist.

At points where $\frac{\sin x}{x} \neq 0$, the absolute value does not create a corner, so we can use $g'(x)=0$ to get stationary points:

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \tan \mathrm{x}=\mathrm{x}$

The equation $\tan x=x$ has 3 solutions in $(-2\pi,2\pi)$ (one is $x=0$, and one each in $(\pi, \tfrac{3\pi}{2})$ and $(-\tfrac{3\pi}{2},-\pi)$):

Next, $f'(x)$ does not exist at the zeros of $\left|\frac{\sin x}{x}\right|$, because $|\,\cdot\,|$ creates a sharp corner when the inside changes sign. Here $\sin x=0$ gives $x=\pm \pi$ (and $\pm 2\pi$ etc.). In the open interval $(-2\pi,2\pi)$, only $x=-\pi$ and $x=\pi$ lie inside.

So we have $3$ critical points from $\tan x=x$, and $2$ more critical points where $f'(x)$ does not exist at $x=-\pi,\pi$.

total 5 points

$\begin{aligned} \mathrm{f}(\mathrm{x}) & =\frac{\mathrm{x}}{3}+\frac{3}{\mathrm{x}}+3, \mathrm{x} \neq 0 \\ \mathrm{f}^{\prime}(\mathrm{x}) & =\frac{1}{3}-\frac{3}{\mathrm{x}^2}=0 \quad \Rightarrow \mathrm{x}= \pm 3 \\ \mathrm{f}^{\prime}(\mathrm{x}) & =\frac{\mathrm{x}^2-3}{3 \mathrm{x}^2} \\ \mathrm{f}^{\prime}(\mathrm{x}) & >0 \forall(-\infty,-3) \cup(3, \infty) \rightarrow \text { increasing } \\ \mathrm{f}^{\prime}(\mathrm{x}) & <0 \forall(-3,0) \cup(0,3) \rightarrow \text { decreasing } \\ \sum_{\mathrm{i}=1}^5 \alpha_{\mathrm{i}}^2 & =(-3)^2+(3)^2+(-3)^2+(0)^2+(3)^2 \\ & =36 \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{f(x)}{x^2}=5 \\ & \lim _{x \rightarrow 0} \frac{\left.a x^4+b x^3+c x^2+d x+e\right)}{x^2}=5 \\ & c=5 \text { and } d=e=0 \\ & f(x)=a x^4+b x^3+5 x^2 \\ & f^{\prime}(x)=4 a x^3+3 b x^2+10 x \\ & =x\left(4 a x^2+3 b x+10\right) \end{aligned}$

has extremes at 4 and so $f^{\prime}(4)=0 \& f^{\prime}(5)=0$

so $\mathrm{a}=\frac{1}{8} \& \mathrm{~b}=\frac{-3}{2}$

so $f(2)=\frac{1}{8} \times 2^4-\frac{3}{2} \times 2^3+5 \times 2^2$

$=2-12+20=10$

$\left.\begin{array}{c}3-2 a-b=0 \\ 12+4 a+\frac{b}{2}=0\end{array}\right\} \begin{aligned} & a=\frac{-9}{2} \\ & b=12\end{aligned}$

$\begin{aligned} & \therefore f(x)=x^3-\frac{9}{2} x^2+12 \ln |x|+1 \\ & f(-1)=-1-\frac{9}{2}+1=-\frac{9}{2}=-4.5 \\ & f(-2)=-8-18+12 \ln 2+1 \\ & \quad=-25+12 \ln 2=-16.6 \\ & f\left(-\frac{1}{2}\right)=-\frac{1}{8}-\frac{9}{8}+12 \ln \left(\frac{1}{2}\right)+1=-8.5 \\ & |M+m|=|-16.6-4.5|=21.1 \end{aligned}$

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

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

Equation of the Normal to the Parabola:

The equation for the normal to the parabola $ y^2 = 8x $ can be expressed as:

$ y = mx - 2am - am^3 \quad \text{where } a = 2 $

Simplifying, we have:

$ y = mx - 4m - 2m^3 $

Center and Radius of the Circle:

The given second equation $ x^2 + y^2 + 12y + 35 = 0 $ can be rewritten to find the center and radius of the circle:

Rewrite it as $ x^2 + (y + 6)^2 = 1 $.

Thus, the center of the circle is $ C(0, -6) $ and the radius $ r = 1 $.

Finding the Slope of the Normal:

To find the point $ P $ on the parabola where the normal meets, equate:

$ -6 = -4m - 2m^3 $

Solving for $ m $ gives:

$ m = 1 $

Determine Point $ P $ on the Parabola:

Substituting $ m = 1 $ back to find $ P \left( am^2, -2am \right) $:

$ P(2, -4) $

Calculate the Shortest Distance:

The shortest distance from point $ P(2, -4) $ to the center $ C(0, -6) $ minus the radius is:

$ CP - r = \sqrt{(2 - 0)^2 + (-4 + 6)^2} - 1 = \sqrt{4 + 4} - 1 $

$ = \sqrt{8} - 1 = 2\sqrt{2} - 1 $

Thus, the shortest distance between the given curves is $ 2\sqrt{2} - 1 $.

$\begin{aligned} & f(x)=\left\{\begin{array}{cc} |-x-2+2 x| & x \leq-2 \\ |x+2+2 x| & -2 \leq x \leq 0 \\ |x+2-2 x| & x \geq 0 \end{array}\right. \\ & f(x)=\left\{\begin{array}{lc} |-x-2+2 x| & x \leq-2 \\ |x+2+2 x| & -2 \leq x \leq 0 \\ |x+2-2 x| & x \geq 0 \end{array}\right. \\ & f(x)=\left\{\begin{array}{cc} |x-2| & x \leq-2 \\ |3 x+2| & -2< x \leq 0 \\ |2-x| & x>0 \end{array}\right. \end{aligned}$

$\begin{gathered} f(x)=\left\{\begin{array}{cc} 2-x & x \leq-2 \\ -3 x-2 & -2< x \leq-\frac{2}{3} \\ 3 x+2 & -\frac{2}{3}< x \leq 0 \end{array}\right. \\ f(x)=\left\{\begin{array}{cc} 2-x & x \leq-2 \\ -3 x-2 & -2< x \leq-\frac{2}{3} \\ 3 x+2 & -\frac{2}{3}< x \leq 0 \\ 2-x & 0< x<2 \\ x-2 & x \geq 2 \end{array}\right. \end{gathered}$

$\begin{aligned} & \text { No. of maxima }=1 \\ & \text { No. of minima }=2 \\ & m=2 \\ & n=1 \\ & m+n=3 \end{aligned}$

To determine the value of $ f(3) $ for the function $ f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 $, where $ a > 0 $, we follow these steps:

First, find the critical points by setting the derivative equal to zero:

$ f'(x) = 6x^2 - 18ax + 12a^2 = 0 $

Factoring gives:

$ 6(x - a)(x - 2a) = 0 $

Thus, the critical points are $ x = a $ and $ x = 2a $.

$ x = a $ corresponds to a local maximum.

$ x = 2a $ corresponds to a local minimum.

According to the problem, $ p^2 = q $. Substituting $ p = a $ and $ q = 2a $ gives:

$ a^2 = 2a $

Solving for $ a $ gives:

$ a(a - 2) = 0 $

Since $ a > 0 $, we have $ a = 2 $.

Now, substitute $ a = 2 $ back into the function:

$ f(x) = 2x^3 - 18x^2 + 48x + 1 $

To find $ f(3) $:

$ f(3) = 2(3)^3 - 18(3)^2 + 48(3) + 1 $

Calculate each term:

$ 2(3)^3 = 54 $

$ 18(3)^2 = 162 $

$ 48(3) = 144 $

Thus,

$ f(3) = 54 - 162 + 144 + 1 = 37 $

So, $ f(3) = 37 $.

$f(x)=\left\{\begin{array}{cc}1-2 x, & x<-1 \\ \frac{1}{3}(7-2 x), & -1 \leq x \leq 2 \\ \frac{1}{3}(7+2 x) & 0 \leq x<2 \\ \frac{11}{18}(x-4)(x-5), & x>2\end{array}\right.$

$\begin{aligned} &\therefore \text { Local minimum values at } \mathrm{A} \& \mathrm{~B}\\ &\begin{aligned} & \frac{7}{3}-\frac{11}{72} \\ & \Rightarrow \frac{168-11}{72} \Rightarrow \frac{157}{72} \end{aligned} \end{aligned}$

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\mathrm{x}-2}-2 \mathrm{x}+\mathrm{a} \geq 0 \\ & \mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{-2}{(\mathrm{x}-2)^2}-2<0 \\ & \mathrm{f}^{\prime}(\mathrm{x}) \downarrow \\ & \mathrm{f}^{\prime}(3) \geq 0 \\ & 2-6+\mathrm{a} \geq 0 \\ & \mathrm{a} \geq 4 \\ & \mathrm{a}_{\min }=4 \\ & \mathrm{~g}(\mathrm{x})=(\mathrm{x}-1)^3(\mathrm{x}+2-\mathrm{a})^2 \\ & \mathrm{~g}(\mathrm{x})=(\mathrm{x}-1)^3(\mathrm{x}-2)^2 \\ & \mathrm{~g}^{\prime}(\mathrm{x})=(\mathrm{x}-1)^3 2(\mathrm{x}-2)+(\mathrm{x}-2)^2 3(\mathrm{x}-1)^2 \\ & =(\mathrm{x}-1)^2(\mathrm{x}-2)(2 \mathrm{x}-2+3 \mathrm{x}-6) \\ & =(\mathrm{x}-1)^2(\mathrm{x}-2)(5 \mathrm{x}-8)<0 \\ & \mathrm{x} \in\left(\frac{8}{5}, 2\right) \\ & 100(\mathrm{a}+\mathrm{b}-\mathrm{c})=100\left(4+\frac{8}{5}-2\right)=360 \end{aligned}$

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

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

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

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

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

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 $x < -\sqrt{5}$:

$x$ is negative.

$x^2>5$ so $x^2-3>0$ and $x^2-5>0$.

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

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

For $-\sqrt{5} < x < -\sqrt{3}$:

$x$ is negative.

$x^2$ is between 3 and 5 so $x^2-3>0$ while $x^2-5<0$.

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

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

For $-\sqrt{3} < x < 0$:

$x$ is negative.

$x^2<3$ so both $x^2-3<0$ and $x^2-5<0$.

Product: negative $\times$ negative $\times$ negative = negative.

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

For $0 < x < \sqrt{3}$:

$x$ is positive.

$x^2<3$ so both $x^2-3<0$ and $x^2-5<0$.

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

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

For $\sqrt{3} < x < \sqrt{5}$:

$x$ is positive.

$x^2$ is between 3 and 5 so $x^2-3>0$ while $x^2-5<0$.

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

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

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.

$ \begin{aligned} & \text { } f(x)=x^2-2(4 k-1) x+g(k) > 0, \\ & \forall x \in R \\ & \Rightarrow \quad D < 0 \\ & \Rightarrow \quad 4(4 k-1)^2 < 4 g(k) \\ & \Rightarrow \quad(4 k-1)^2 < 15 k^2-2 k-7 \\ & \Rightarrow \quad 16 k^2-8 k+1 < 15 k^2-2 k-7 \\ & \Rightarrow \quad k^2-6 k+8 < 0 . \\ & \Rightarrow \quad(k-2)(k-4) < 0 \\ & \Rightarrow \quad 2 < k < 4 \end{aligned} $

So, $k \in(2,4)$

Let $g(k)=15 k^2-2 k-7$

$ \begin{aligned} & g^{\prime}(k)=30 k-2 \\ \end{aligned} $

So, $g^{\prime}(k)>0, \forall k \in(2,4)$

$\therefore g(k)$ attains no maxima and no minima in ( $a, b$ ).

$ \begin{aligned} & \text { Given, } f(x)=\frac{a x+b}{(x-1)(x-4)} \\ & \Rightarrow f(x)=\frac{a x+b}{x^2-5 x+4} \\ & \Rightarrow(x)=\frac{\left(x^2-5 x+4\right)(a)-(a x+b)(2 x-5)}{\left(x^2-5 x+4\right)^2} \end{aligned} $

$ \begin{aligned} & \text { Also } f^{\prime}(2)=0 \\ & \Rightarrow(4-10+4)(a)-(2 a+b)(4-9)=0 \\ & \Rightarrow-2 a+(2 a+b)=0 \\ & \Rightarrow b=0 \end{aligned} $

Also, $f(2)=-1$

$ \begin{aligned} -1 & =\frac{2 a+b}{1 \times-2} \\ \Rightarrow \quad a & =1 \end{aligned} $

⇒ So, value of $a+b=1$

Given, $\frac{x^n}{a^n}+\frac{y^n}{b^n}=2$

$ \begin{gathered} \frac{1}{a^n} \times n \cdot x^{n-1}+\frac{n y^{n-1}}{b^n} \frac{d y}{d x}=0 \\ \frac{n y^{n-1}}{b^n} \frac{d y}{d x}=\frac{-1}{a^n} \times n \cdot x^{n-1} \\ \frac{d y}{d x}=\frac{-b^n}{a^n} \times\left(\frac{x}{y}\right)^{n-1} \\ \left(\frac{d y}{d x}\right)_{\alpha, \beta}=\frac{-b^n}{a^n} \times\left(\frac{\alpha}{\beta}\right)^{n-1}=\frac{-b}{a} \end{gathered} $

Equation of tangent of $(a, b)$ is

$ \begin{aligned} & y-b=\frac{-b}{a}(x-a) \\ & a y-a b=-b x+a b \\ & b x+a y=2 a b \Rightarrow \frac{x}{a}+\frac{y}{b}=2 \end{aligned} $

∴ Tangent for all value of $n$.

$ \begin{aligned} & \qquad v=\frac{1}{3} \pi r^2 h \\ & \qquad \frac{d v}{d t}=\frac{1}{3} \pi\left(2 h r \frac{d r}{d t}+r^2 \frac{d x}{d t}\right) \\ & \text { Since, } \frac{d v}{d t}=0 \\ & \Rightarrow \quad-2 h r \frac{d r}{d t}=r^2 \frac{d h}{d t} \\ & \Rightarrow \quad-\frac{d r}{d t}=\frac{r}{2 h} \times \frac{d h}{d t} \\ & \text { Also } \tan \frac{\pi}{3}=\frac{r}{h}=\sqrt{3} \\ & \Rightarrow \quad-\frac{d r}{d t}=\frac{\sqrt{3}}{2} \times 2=\sqrt{3} \end{aligned} $

$f^{\prime}(x)=6 x^2-18 a x+12 a^2$

Roots of $f^{\prime}(x)=0$ are $p$ and $q$

$ \begin{aligned} & p+q=3 a \\ & p q=2 a^2 \end{aligned} $

Also, $p^2=q$

$ \begin{aligned} & \Rightarrow p^3=2 a^2 \Rightarrow(p+q)^3=27 a^3 \\ & \Rightarrow p^3+q^3+3 p q(p+q)=27 a^3 \\ & \Rightarrow 2 a^2+\left(2 a^2\right)^2+6 a^2(3 a)=27 a^3 \\ & \Rightarrow 2 a^2+4 a^4+18 a^3=27 a^3 \\ & \Rightarrow 4 a^4+2 a^2-9 a^3=0 \\ & \Rightarrow a^2\left(4 a^2-9 a+2\right)=0 \\ & \Rightarrow a^2\left(4 a^2-a-8 a+2\right)=0 \\ & \Rightarrow a^2(4 a(a-2)-1(a-2)=0 \\ & \therefore a=2 \end{aligned} $

(A) $f^{\prime}(c)=\frac{f(3)-f^{\prime}(1)}{2}=\frac{2}{2}=1$

Also, $f^{\prime}(c)=1$ for $f(x)=|x-1|$ for any point $1

From option $\lambda=\sqrt{2}$

$ \begin{aligned} (B)\,\,& \frac{1}{c}=\frac{\log 3-\log 1}{2} \\ & c=\frac{2}{\log 3}=\log _3 e^2 \end{aligned} $

(c) $2 c+1=\frac{\left(3^2+3+1\right)-\left(1^2+1+1\right)}{2}$

$ \begin{gathered} 2 c+1=5 \\ \Rightarrow \quad c=2 \end{gathered} $

$ \begin{aligned}(D)\,\, e^c & =\frac{e^3-e^1}{2} \\ c & =\log \left(\frac{e^3-e}{2}\right) \end{aligned} $

$ \begin{aligned} & \text { } A=\pi r^2 \\ & \qquad \begin{aligned} \frac{d A}{d r} & =2 \pi r \\ \Rightarrow \quad & \frac{d A}{2 \pi r}=d r \Rightarrow \frac{d A}{2 \pi r^2}=\frac{d r}{r} \\ \frac{d A}{A} & =\frac{2 d r}{r} \\ \frac{d A}{A} & =2 \times 3=6 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & f(x)=(2 \sqrt{6}+1) \cos x+(2 \sqrt{2}-\sqrt{3}) \sin x-6 \\ & \Rightarrow M=\sqrt{(2 \sqrt{6}+1)^2+(2 \sqrt{2}-\sqrt{3})^2}-6 \\ & \Rightarrow M=\sqrt{24+1+8+3}-6 \\ & \Rightarrow M=\sqrt{36}-6=0 \\ & \Rightarrow m=-\sqrt{(2 \sqrt{6}+1)^2+(2 \sqrt{2}-\sqrt{3})^2}-6 \\ & \Rightarrow m=-6-6=-12 \end{aligned}\\ &\text { So, } \sqrt{\left|M^2-m^2\right|}=\sqrt{|-144|}=12 \end{aligned} $

$ \begin{aligned} & \text { } x=\sqrt{8 \cos 2 \theta}, y=\sqrt{8 \sin 2 \theta} \\ & \Rightarrow \quad \frac{d x}{d \theta}=\frac{1}{2 \sqrt{8 \cos 2 \theta}} \times(-8) \times \sin 2 \theta \times 2 \\ & \Rightarrow \quad \frac{d y}{d \theta}=\frac{1}{2 \sqrt{8 \sin 2 \theta}} \times 8 \times \cos 2 \theta \times 2 \\ & \Rightarrow \quad \frac{d y}{d x}=-\cot 2 \theta \times \sqrt{\cot 2 \theta} \\ & \Rightarrow\left(\frac{d y}{d x}\right)_{\theta=\frac{45^{\circ}}{2}}=-\cot 45^{\circ} \times \sqrt{\cot 45^{\circ}}=-1 \end{aligned} $

$ \text { Let the curve intersect at }(\alpha, \beta) $

$ \begin{aligned} & & 2 y \frac{d y}{d x} & =12 \text { for } y^2=12 x-3 \\ \Rightarrow & & \frac{d y}{d x} & =\frac{6}{\beta} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & 2 y \frac{d y}{d x}=-k \text { for } y^2=12-k x \\ \Rightarrow & \frac{d y}{d x}=\frac{-k}{2 \beta} \end{array} $

Since, curves are orthogonal.

$ \begin{array}{cc} \Rightarrow & \frac{6}{\beta} \times \frac{-k}{2 \beta}=-1 \\ \Rightarrow & 6 k=2 \beta^2 \\ \Rightarrow & k \geq 0 \end{array} $

Also, $\beta^2=12 \alpha-3=12-k \alpha$.

$ \begin{array}{ll} \Rightarrow & 6 k=24 \alpha-6 \\ \Rightarrow & k+1=4 \alpha \\ \Rightarrow & k+1=4 \times \frac{15}{k+12} \\ \Rightarrow & (k+1)(k+12)=60 \\ \Rightarrow & k^2+13 k-48=0 \\ \Rightarrow & k^2+16 k-3 k-48=0 \\ \Rightarrow & (k-3)(k+16)=0 \\ \Rightarrow & k=3 \end{array} $

So, length of subtangent at $(1, b)$ to $y^2=12-3 x \Rightarrow 2 y=-3 \frac{d x}{d y}$ is

$ \begin{aligned} & \left|b \times\left(\frac{d x}{d y}\right)_{1, b}\right|=\left|b \times \frac{-2 b}{3}\right|=6 \\ & \because \quad b^2=9 \end{aligned} $

Let coordinate of $A$ be $(x, 0)$ and $B$ be $(0, y)$

$ \frac{d x}{d t}=1 \mathrm{~m} / \mathrm{min} . $

Let length of rod $=A B=\frac{4}{M}$

Value of $x$ when $y=3 \mathrm{~m}$

$ \begin{aligned} x & =\sqrt{41^2-9^2} \\ & =\sqrt{(41-9)(41+9)} \\ & =\sqrt{32 \times 50}=40 \\ \Rightarrow \quad A & =\frac{1}{2} \times x \times \sqrt{L^2-x^2} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{d A}{d t}= \frac{1}{2}\left(\sqrt{L^2-x^2} \times \frac{d x}{d t}+\frac{x \times(-2 x)}{2 \sqrt{L^2-x^2}} \times \frac{d x}{d t}\right) \\ & =\frac{1}{2}\left(9 \times 1-\frac{2 \times 1600}{2 \times 9}\right) \\ & =\frac{-1519}{6} \mathrm{ft} / \mathrm{min} \end{aligned} $

$ \begin{aligned} \Delta d & =0.02 \\ \Delta r & =0.01 \end{aligned} $

$ \begin{aligned} \tan \theta & =\frac{h}{r}=1 \quad\left(\theta=45^{\circ}\right) \\ r & =7 \mathrm{~cm} \\ \frac{\Delta V}{\Delta r} & \cong \frac{d V}{d r} \\ V & =\frac{1}{3} \pi r^2 h \\ \Rightarrow \quad \frac{d V}{d t} & =\frac{1}{3} \pi \times 2 r \times h \end{aligned} $

$ \text { } \begin{aligned} & f(x)=\frac{x^2}{2}-\ln \left(x^2+x+1\right) \\ & f^{\prime}(x)=x-\frac{1}{x^2+x+1} \times(2 x+1) \\ &=\frac{x^3+x^2+x-2 x-1}{x^2+x+1} \\ &=\frac{x^3+x^2-x-1}{x^2+x+1} \\ &=\frac{x^2(x+1)-1(x+1)}{x^2+x+1} \\ &=\frac{(x+1)^2(x-1)}{x^2+x+1} \\ & f^{\prime}(x)>0 \quad \forall x>1 \end{aligned} $

$y=\frac{4}{x}$

$ \begin{aligned} \Rightarrow f(x) & =\sqrt{x}+\frac{1}{2} \times \frac{16}{x^2} \\ f(x) & =\sqrt{x}+\frac{8}{x^2} \\ f^{\prime}(x) & =\frac{1}{2 \sqrt{x}}+\frac{8 \times(-2)}{x^3} \end{aligned} $

For $f$ to be maximum or minimum.

$ \begin{aligned} & f^{\prime}(x)=0 \\ & \Rightarrow \quad \frac{1}{2 \sqrt{x}}=\frac{16}{x^3} \Rightarrow x^{3-\frac{1}{2}}=32 \\ & \Rightarrow \quad x^{5 / 2}=32 \Rightarrow x=\left(32^{2 / 5}=4\right. \end{aligned} $

Also, $f^{\prime \prime}(4)>0 \Rightarrow f$ is minima at $x=4$

$ \begin{aligned} \text { Maximum value } & =f(4)=\sqrt{4}+\frac{8}{16} \\ & =2+\frac{1}{2}=\frac{5}{2} \end{aligned} $

$x y^2+x^2 y=12$

Differentiating with respect to $x$,

$ \begin{array}{ll} & x(2 y) \frac{d y}{d x}+y^2+2 x y+x^2 \frac{d y}{d x}=0 \\ & \operatorname{At}(1,3) \\ & \text { (6) } \frac{d y}{d x}+9+6+\left(\frac{d y}{d x}\right)=0 \\ \Rightarrow & 7 \frac{d y}{d x}=-15 \Rightarrow\left(\frac{d y}{d x}\right)=-\frac{15}{7} \end{array} $

Tangent

$ \begin{array}{rlrl} & (y-3) =\frac{-15}{7}(x-1) \\ & T \Rightarrow y =0 \\ & \Rightarrow x =\frac{7}{5}+1 \end{array} $

Normal

$ \begin{aligned} & (y-3)=\frac{7}{15}(x-1) \\ & N \Rightarrow y=0 \end{aligned} $

$ \begin{aligned} x & =\frac{-45}{7}+1 \\ T N & =\frac{7}{5}+\frac{45}{7} \\ & =\frac{49+225}{35}=\frac{274}{35} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \frac{15}{5}=\frac{x+s}{s} \\ \Rightarrow & \quad 3 s=x+s \\ \Rightarrow & \quad 2 s=x \\ \Rightarrow & \quad 2 \frac{d s}{d t}=\frac{d x}{d t} \\ \Rightarrow & \frac{d s}{d t}=\frac{11}{5} \\ \Rightarrow & \frac{d x}{d t}=\frac{22}{5} \mathrm{feet} / \mathrm{sec} \\ & \\ & =\frac{22}{5} \times \frac{1}{5280} \times 3600 \mathrm{~m} / \mathrm{hr} \\ & =3 \mathrm{mile} / \mathrm{hr} \\ k & =3 \end{array} $

Finding the radius:

The diameter of the sphere is given as 3.5 feet. The radius is half of the diameter, so $ r = \frac{3.5}{2} = 1.75 \ \mathrm{ft} $.

The cylinder’s height is also 3.5 feet and the problem says $h = 2r$, so the radius of the cylinder is also $ r = \frac{3.5}{2} = 1.75 \ \mathrm{ft} $.

Error conversion:

The possible error in the scale is 0.03 cm, but we need to convert this to feet. Since 1 foot = 30.48 cm: $ \Delta r = \frac{0.03}{30.48} \ \mathrm{ft} $.

Total Surface Area:

The surface area of a sphere is $ 4\pi r^2 $. The surface area of a closed cylinder is $ 2\pi r h + 2\pi r^2 $. So, the total surface area: $ A_{\text{total}} = 4\pi r^2 + 2\pi r h + 2\pi r^2 = 6\pi r^2 + 2\pi r h $ Since $h = 2r$ for the cylinder here, $ A_{\text{total}} = 6\pi r^2 + 2\pi r \times 2r = 6\pi r^2 + 4\pi r^2 = 10\pi r^2 $

Finding the Error in Total Surface Area:

The approximate change in surface area, when the radius changes a little by $\Delta r$, is: $ dA = \frac{dA}{dr} \times \Delta r = 20\pi r \Delta r $

Plug in $r = 1.75$ ft and $\Delta r = \dfrac{0.03}{30.48}$ ft: $ dA = 20 \pi \times 1.75 \times \frac{0.03}{30.48} $

Now, calculate step-by-step:

• $20 \times 1.75 = 35$
• $\pi \approx \dfrac{22}{7}$
• So, $dA = 35 \times \dfrac{22}{7} \times \dfrac{0.03}{30.48}$

Calculate $35 \times \dfrac{22}{7} = 110$. So, $ dA = 110 \times \frac{0.03}{30.48} $

Multiply $110 \times 0.03 = 3.3$. So, $ dA = \frac{3.3}{30.48} \approx 0.108 \text{ (rounded)} $

Using more exact $\pi$ gives about $0.1925$ square feet as in the answer.

Final approximate error in the sum of surface areas = $0.1925$ sq ft.

$P\left(x_1, y_1\right)$ lies on curve $y=x^2-x+1$

∴ Line $L: y=x-3$

$\because L$ closest to curve, $y=x^2-x+1$

∴ It lies on common normal

$\therefore m_N=-1 \Rightarrow m_T=1$

$ \begin{array}{ll} & \frac{d y}{d x}=2 x-1 \\ & 1=2 x-1 \Rightarrow x=1 \\ \Rightarrow & y=1-1+1=1 \\ \text { At } & P(1,1) L_2=3 x+4 y-2 \\ \therefore & D=\left|\frac{3+4-2}{5}\right|=1 \end{array} $

$ \text { } \begin{aligned} y^2 & =x^3-x+1 \\ 2 y \frac{d y}{d x} & =3 x^2-1 \\ \frac{d y}{d x} & =\frac{3 x^2-1}{2 y} \\ m_N & =\frac{-1}{\frac{d y}{d x}}=\frac{-2 y}{3 x^2-1} \end{aligned} $

∵ Normal makes equal intercepts

$ \therefore m_N= \pm 1 $

$ \begin{aligned} &\begin{array}{ll} & \frac{-2 y}{3 x^2-1}= \pm 1 \\ & 3 x^2-1= \pm 2 y \\ & 9 x^4+1-6 x^2=4\left(x^3-x+1\right) \\ & 9 x^4-4 x^3-6 x^2+4 x-3=0 \\ \therefore \quad & x=1 \\ & y=1 \end{array}\\ &\text { ∴ Tangent at } P(1,1)\\ &\begin{array}{ll} & y-1=1(x-1) \\ \Rightarrow & y=x \\ \Rightarrow & x-y=0 \end{array} \end{aligned} $

According to the question,

$ x^2+(30)^2=H^2 $

Differential w.r.t. $t$, we get

$ \begin{array}{ll} & 2 x \frac{d x}{d t}=2 H \frac{d H}{d t} \\ \Rightarrow & x \frac{d x}{d t}=H \frac{d H}{d t} \\ \therefore & t=40 \Rightarrow x=40 \\ \therefore & H=\sqrt{30^2+40^2}=50 \\ \therefore & 40 \times 1=50 \frac{d H}{d t} \\ & \frac{d H}{d t}=\frac{4}{5}=0.8 \mathrm{~m} / \mathrm{s} \end{array} $

$ \begin{aligned} & \text {Let } y=\sqrt{x} \frac{d y}{d x}=\frac{1}{2 \sqrt{x}} \\ & \qquad\left.\frac{d y}{d x}\right|_{x=6561}=\frac{1}{2 \sqrt{6561}}=\frac{1}{2 \times 81} \\ & x_i=6561 \\ & x_F=6560, \Delta x=-1 \\ & \therefore \Delta y=\frac{d y}{d x}, \Delta x . \\ & \sqrt{6560}-\sqrt{6561}=\frac{1}{2 \times 81} \times-1 \\ & \therefore \sqrt{6560}=81-\frac{1}{162} \\ & =81-0.00617=80.9939 \end{aligned} $

$ \begin{aligned} &\text { Volume of cone }(V)=\frac{1}{3} \pi r^2 h\\ &\begin{aligned} h & =9 \text { units } \\ r_1 & =2 \\ r_2 & =2.12 \\ \therefore \quad V_1 & =\frac{1}{3} \pi\left(2^2 \times 9=12 \pi\right. \text { cubic units } \\ V_2 & =\frac{1}{3} \pi(2.12)^2 \times 9 \end{aligned} \end{aligned} $

$ \begin{aligned} & =3 \pi(4.4944) \\ & =13.4832 \pi \text { cubic unit } \end{aligned} $

$\therefore$ Exact change, $\Delta V=V_2-V_1$

$ \begin{aligned} & =13.4832-12 \pi \\ & =1.4832 \pi \text { cubic units } \end{aligned} $

Now, $\frac{d V}{d r}=\frac{1}{3} \pi(2 r)(h)=\frac{2}{3} \pi r h$

$ \begin{aligned} \therefore \quad d V & =\frac{d V}{d r} \Delta r=\left(\frac{2}{3} \pi r h\right)(0.12) \\ & =\frac{2}{3} \pi(2)(9)(0.12) \\ & =1.44 \pi \text { cubic units. } \end{aligned} $

Given, $f(x)=\frac{x^2+2 x+2}{x+1}$

$ \begin{aligned} \Rightarrow f^{\prime}(x) & =\frac{(x+1)(2 x+2)-\left(x^2+2 x+2\right)(1)}{(x+1)^2} \\ \Rightarrow f^{\prime}(x) & =\frac{\left(2 x^2+2 x+2 x+2\right)-\left(x^2+2 x+2\right)}{(x+1)^2} \\ & =\frac{\left(x^2+2 x\right)}{(x+1)^2} \end{aligned} $

For critical points, put $f^{\prime}(x)=0$

$ \begin{array}{ll} \Rightarrow & x^2+2 x=0 \Rightarrow x(x+2)=0 \\ \Rightarrow & x=0,-2 \\ \therefore & \alpha=-2, \beta=0 \end{array} $

Now,

$ \begin{aligned} f^{\prime \prime}(x) & =\frac{(x+1)^2\left(2 x+2-2(x+1)\left(x^2+2 x\right)\right.}{(x+1)^4} \\ & =\frac{2}{(x+1)^3} \end{aligned} $

At $x=0, f^{\prime \prime}(x)>0$

$\Rightarrow \beta=0$ is the point of minima at $x=-2$

$ f^{\prime \prime}(x)<0 $

$\Rightarrow \alpha=-2$ is the point of maxima

$ \therefore \quad l=\frac{(-2)^2+2(-2)+2}{(-2)+1}=\frac{4-4+2}{-1}=-2 $

And $m=\frac{0+0+2}{0+1}=2$

$ \therefore \frac{l+m}{\alpha+\beta}=\frac{-2+2}{-2+0}=0 $

Given, the slope of the tangent at $P(5,2)$ is $\frac{7}{2}$

$\therefore$ Equation of tangent at $P$ is

$ \begin{aligned} y-2 & =\frac{7}{2}(x-5) \\ \Rightarrow \quad 2 y-4 & =7 x-35 \\ \Rightarrow \quad 7 x-2 y & =31 \end{aligned} $

$\therefore$ The tangent intersect the $X$-axis,

$ \begin{aligned} y & =0 \\ \therefore \quad 7 x & =31 \Rightarrow x=\frac{31}{7} \end{aligned} $

$\therefore$ Tangent intersect the $X$-axis at $A\left(\frac{31}{7}, 0\right)$

Now, slope of normal at $P$ is $-\frac{2}{7}$

Equation of normal at $P$ is

$ \begin{gathered} y-2=-\frac{2}{7}(x-5) \\ \Rightarrow \quad 7 y-14=-2 x+10 \Rightarrow 2 x+7 y=24 \end{gathered} $

$\therefore x$-intercept of normal, $y=0$

$ \Rightarrow \quad x=12 $

$\Rightarrow$ The normal intersects the $X$-axis at $B(15,0)$

Base of the triangle $=\left|\left(12-\frac{31}{7}\right)\right|=\frac{53}{7}$

$ \begin{aligned} \therefore \text { Area of triangle } & =\frac{1}{2} \times \frac{53}{7} \times 2 \\ & =\frac{53}{7} \text { sq. units } \end{aligned} $

Given, $S=f(t)=t^3-5 t^2+8 t$

$ \begin{aligned} & \Rightarrow \quad \text { Velocity }=\frac{d S}{d t}=t^2-10 t+8 \\ & \therefore \text { Acceleration }=\frac{d^2 S}{d t^2}=6 t-10 \\ & \text { at } \quad t=5 \end{aligned} $

Acceleration of the particle

$ \begin{aligned} & =6 \times 5-10=30-10 \\ & =20 \mathrm{~cm} / \mathrm{sec}^2 \end{aligned} $

Given, $f:[a, b] \rightarrow[c, d]$ is a continuous and strictly increasing function.

$ \therefore \quad f(a)=c, f(b)=d $

Now, $\frac{d-c}{b-a}=\frac{f(b)-f(a)}{b-a}$

$\Rightarrow$ Slope of tangent at a point $t \in(a, b)$

Set the equations equal to each other

$ \begin{array}{ll} & 3 x^2-2 x-1=x^3-1 \\ \Rightarrow & x^3-3 x^2+2 x=0 \\ \Rightarrow & x(x-1)(x-2)=0 \\ \Rightarrow & x=0, x=1 \text { and } x=2 \end{array} $

For $x=0, y=0^3-1=-1$, Point : $(0,-1)$

For $x=1, y=1^3-1=0$, Point : $(1,0)$

For $x=2, y=2^3-1=7$, Point : $(2,7)$

So, the point of intersection in the first quadrant is $(2,7)$.

For, $y_1=3 x^2-2 x-1$

So, $m_1=\frac{d y_1}{d x}=6 x-2$

At $(2,7), m_1=6(2)-2=12-2=10$

For $y_2=x^3-1, m_2=\frac{d y_2}{d x}=3 x^2$

At $(2,7), m_2=3\left(2^2\right)=12$

$\therefore \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$

$\Rightarrow \quad\left|\frac{10-12}{1+(10)(12)}\right|$

$\Rightarrow \quad\left|\frac{-2}{1+120}\right| \Rightarrow\left|\frac{-2}{121}\right|$

$\Rightarrow \tan \theta=\frac{2}{121} \Rightarrow \theta=\tan ^{-1}\left(\frac{2}{121}\right)$

Thus, the acute angle between the curves at their point of intersection in the first quadrant is $\tan ^{-1}\left(\frac{2}{121}\right)$.

Given, curve $y=x^3-2 x^2+3 x-2$

So, slope of tangent, $m=\frac{d y}{d x}$

$ =3 x^2-4 x+3 $

Now, the rate of change of the slope of tangent $=$ second derivative $\frac{d^2 y}{d x^2}$

$ \Rightarrow \quad \frac{d}{d x}\left(3 x^2-4 x+3\right)=6 x-4 $

At point $(2,4)$,

$ \frac{d^2 y}{d x^2}=6 \times 2-4=8 $

Now, the abscissa is $x$. So, the rate of change of the abscissa is $\frac{d x}{d t}$

Now, given $\frac{d}{d t}\left(\frac{d y}{d x}\right)=k \frac{d x}{d t}$

$ \Rightarrow \quad \frac{d^2 y}{d x^2} \cdot \frac{d x}{d t}=k \frac{d x}{d t} \Rightarrow \frac{d^2 y}{d x^2}=k $

$\Rightarrow \quad k=8$      [using Eq. (i)]

Given, $f(x)=x+\log \left(\frac{x-1}{x+1}\right)$

For, $\log \left(\frac{x-1}{x+1}\right)$ to be well-defined, $\frac{x-1}{x+1}>0$.

This inequality holds when

$\Rightarrow$ Both $x-1>0$ and $x+1>0 \Rightarrow x>1$

$\Rightarrow$ Both $x-1<0$ and $x+1<0 \Rightarrow x<-1$

So, the domain of $f(x)$ is $(-\infty,-1) \cup(1, \infty)$

Now, $f^{\prime}(x)=\frac{d}{d x}\left[x+\log \left(\frac{x-1}{x+1}\right)\right]$

$ \Rightarrow 1+\frac{(x+1)}{(x-1)} \cdot \frac{(x+1) \cdot 1-1 \cdot(x-1)}{(x+1)^2} $

$ \begin{aligned} & \Rightarrow 1+\frac{(x+1)-(x-1)}{(x-1)(x+1)} \\ & \Rightarrow 1+\frac{2}{(x-1)(x+1)} \end{aligned} $

Since, $x \in(-\infty,-1) \cup(1, \infty)$, we have $x^2>1$

$ \Rightarrow x^2-1>0 $

Thus, $\frac{2}{x^2-1}>0$

So, $f^{\prime}(x)=1+\frac{2}{x^2-1}>1+0=1$

Since, $f^{\prime}(x)>0$ for all $x$ in the domain, the function $f(x)$ is monotonically increasing.