Application of Derivatives

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.

$ \begin{aligned} & \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=3 \\ & \text { since } 3>\frac{7}{3} \Rightarrow 3>f(0) \end{aligned} $

$\Rightarrow \mathrm{x}=0$ is local minima ….option (B) is correct

Now,

$ \begin{aligned} & f(x)=\frac{6 x+\sin x}{2 x+\sin x}=1+\frac{4 x}{2 x+\sin x} \\ & f^{\prime}(x)=\frac{4[(2 x+\sin x) \cdot 1-x(2+\cos x)]}{(2 x+\sin x)^2}=\frac{4(\sin x-x \cos x)}{(2 x+\sin x)^2} \\ & =\frac{4 \cos x(\tan x-x)}{(2 x+\sin x)^2} \end{aligned} $

$ \text { Options (C) & (D) are also correct. } $

Given,

$\alpha = \sum\limits_{k = 1}^\alpha {{{\sin }^{2k}}\left( {{\pi \over 6}} \right)} $

$ = {\sin ^2}\left( {{\pi \over 6}} \right) + {\sin ^4}\left( {{\pi \over 6}} \right) + {\sin ^6}\left( {{\pi \over 6}} \right)\, + \,......$

This is a infinite G.P.

$\therefore$ $\alpha = {{{{\sin }^2}\left( {{\pi \over 6}} \right)} \over {1 - {{\sin }^2}\left( {{\pi \over 6}} \right)}}$

$ = {{{1 \over 4}} \over {1 - {1 \over 4}}}$

$ = {1 \over 4} \times {4 \over 3}$

$ = {1 \over 3}$

$\therefore$ $g(x) = {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}$

Given $x \in [0,1]$

We know,

$AM \ge GM$

For ${2^{{x \over 3}}}$ and ${2^{{1 \over 3}(1 - x)}}$ both are positive in $x \in [0,1]$

${{{x^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}} \over 2} \ge {\left( {{2^{{x \over 3}}}\,.\,{2^{{1 \over 3}(1 - x)}}} \right)^{{1 \over 2}}}$

$ \Rightarrow {{{2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}} \over 2} \ge {\left( {{2^{{1 \over 3}}}} \right)^{{1 \over 2}}}$

$ \Rightarrow {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}} \ge 2\,.\,{2^{{1 \over 6}}}$

$ \Rightarrow g(x) \ge {2^{{7 \over 6}}}$

We know, $AM = GM$ when terms are equal.

$\therefore$ $g(x) = {2^{{7 \over 6}}}$ when

${2^{{x \over 3}}} = {2^{{1 \over 3}(1 - x)}}$

$ \Rightarrow {x \over 3} = {{1 - x} \over 3}$

$ \Rightarrow 2x = 1$

$ \Rightarrow x = {1 \over 2}$

$\therefore$ At $x = {1 \over 2}$, ${\left[ {g(x)} \right]_{\min }} = {2^{{7 \over 6}}}$

$\therefore$ Option (A) is correct.

And option (D) is wrong as at only at a single point $x = {1 \over 2},\,g(x)$ is minimum.

Now, $g(x) = {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}$

$ = {2^{{x \over 3}}} + {{{2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}$

$ = {2^{{x \over 3}}} + {2^{{1 \over 3}}}\,.\,{2^{ - {x \over 3}}}$

$\therefore$ $g'(x) = {2^{{x \over 3}}}\,.\,\log 2\,.\,{1 \over 3} + {2^{{1 \over 3}}}\,.\,{2^{ - {x \over 3}}}\,.\,\log 2\,.\,\left( { - {1 \over 3}} \right)$

$ = {1 \over 3}\log 2\left[ {{2^{{x \over 3}}} - {{{2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right]$

$ = {1 \over 3}\log 2\left[ {{{{2^{{{2x} \over 3}}} + {2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right]$

We already found that at $x = {1 \over 2}$ $g(x)$ is minimum.

$\therefore$ $g'(x) = 0$ at $x = {1 \over 2}$

Now, $g'(x) > 0$ when

${1 \over 3}\log \left[ {{{{2^{{{2x} \over 3}}} - {2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right] > 0$

$ \Rightarrow {2^{{{2x} \over 3}}} > {2^{{1 \over 3}}}$

$ \Rightarrow {{2x} \over 3} > {1 \over 3}$

$ \Rightarrow x > {1 \over 2}$

Similarly, $g'(x) < 0$ when $x < {1 \over 2}$

If we put it in number line we get this

We know $g'(x)$ represent slope of curve $g(x)$ and it is negative when $x < {1 \over 2}$ and positive when $x > {1 \over 2}$ and zero when $x = {1 \over 2}$.

$\therefore$ Graph of $g(x)$ is

From graph you can see value of $g(x)$ is maximum either at $x = 0$ or $x = 1$ in the range $x \in [0,1]$.

$\therefore$ $g(0) = {2^0} + {2^{{1 \over 3}}} = 1 + {2^{{1 \over 3}}}$

$g(1) = {2^{{1 \over 3}}} + {2^0} = 1 + {2^{{1 \over 3}}}$

$\therefore$ We get maximum value at $x = 0$ and $x = 1$ both.

$\therefore$ B and C options are correct.

Let f : R $\to$ (0, $\infty$) and g : R $\to$ R.

f(x) > 0 $\forall$x $\in$ R

It is given that f'(2) = 0, g(2) = 0, f''(2) $\ne$ 0 and g'(2) $\ne$ 0.

It is also given that

$\mathop {\lim }\limits_{x \to 2} {{f(x)g(x)} \over {f'(x)g'(x)}} = 1$ $\left( {{0 \over 0}} \right)$

Applying L' Hospital rule, we get

$\mathop {\lim }\limits_{x \to 2} {{f'(x)g(x) + g'(x)f(x)} \over {f''(x)g'(x) + f'(x)g''(x)}} = 1$

For finite limit, we get

${{f'(2)g(2) + g'(2)f(2)} \over {f''(2)g'(2) + f'(2)g''(2)}} = 1$

${{g'(2)f(2)} \over {f''(2)g'(2)}} = 1$

${{f(2)} \over {f''(2)}} = 1 \Rightarrow f''(2) = f(2) > 0$ and f'(2) = 0 which means that f(x) has local minima at x = 2.

Hence, option (A) is correct.

$f(2) - f''(2) = 0$

Therefore, we can say that $f(x) - f''(x) = 0$ has at least one solution in x $\in$ R.

Hence, option (D) is correct.

Let $F(x) = f(x) - 3g(x)$

$\therefore$ $F( - 1) = 3$, $F(0) = 3$ and $F(2) = 3$

So, $F'(x)$ will vanish at least twice in

$( - 1,0) \cup (0,2)$.

$\because$ $F''(x) > 0$ or $ < 0$, $\forall x \in ( - 1,0) \cup (0,2)$

Hence, $f'(x) - 3g'(x) = 0$ has exactly one solution in $( - 1,0)$ and one solution in $(0,2)$.

Given, $f(0)=f(1)=0$

$\begin{aligned} & \text { And } f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^x, x \in[0,1] \\ & \Rightarrow e^{-x} f^{\prime \prime}(x)-e^{-x} f^{\prime}(x)-e^{-x} f^{\prime}(x)+e^{-x} f(x) \geq 1 \\ & \Rightarrow\left(e^{-x} f^{\prime}(x)\right)^{\prime}-\left(e^{-x} f^{\prime}(x)\right)-\left(e^{-x} f(x)\right)^{\prime} \geq 1 \\ & \Rightarrow\left(e^{-x} f^{\prime}(x)-e^{-x} f(x)\right)^{\prime} \geq 1 \\ & \Rightarrow\left(e^{-x} f(x)\right)^{\prime \prime} \geq 1 \end{aligned}$

Let $g(x)=e^{-x} f(x)$

Now, $g(0)=e^{-0} f(0)=0$ and $g(1)=e^{-1} f(1)=0$

Here, $g^{\prime \prime}(x)=\left(e^{-x} f(x)\right)^{\prime \prime} \geq 1>0$

$\therefore$ Concavity of $g(x)$ is up and $g(0)=g(1)=0$

$\begin{aligned} & \Rightarrow g(x)=e^{-x} f(x)<0 \forall x \in(0,1) \\ & \Rightarrow f(x)<0 \end{aligned}$

Hints:

(i) $f^{\prime \prime}(x)>0$ implies the concavity of $f(x)$ is upward and $f^{\prime \prime}(x)<0$ implies the concavity of $f(x)$ is downward.

(ii) $f^{\prime \prime}(x)>0 \quad \forall x \in(a, b)$ and $f(a)=f(b)=0$ implies that the concavity of the function is upward and $f(x)$ lies below the $x$-axis in the interval $x \in(a, b)$

Given, $e^{-x} f(x)$ has point of minima at $x=\frac{1}{4}$ in the interval $x \in[0,1]$

Also given $f(0)=f(1)=0$

$\begin{aligned} & \text { And } f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^x, x \in[0,1] \\ & \Rightarrow e^{-x} f^{\prime \prime}(x)-e^{-x} f^{\prime}(x)-e^{-x} f^{\prime}(x)+e^{-x} f(x) \geq 1 \\ & \Rightarrow\left(e^{-x} f^{\prime}(x)\right)^{\prime}-\left(e^{-x} f(x)\right)^{\prime} \geq 1 \\ & \Rightarrow\left(e^{-x} f^{\prime}(x)-e^{-x} f(x)\right)^{\prime} \geq 1 \\ & \Rightarrow\left(e^{-x} f(x)\right)^{\prime \prime} \geq 1>0 \end{aligned}$

Hence, the concavity of $e^{-x} f(x)$ is upward and given $x=\frac{1}{4}$ is the point of local minima of $e^{-x} f(x)$ in $x \in[0,1]$

$\therefore e^{-x} f(x)$ is decreasing in $x \in\left(0, \frac{1}{4}\right)$ and increasing in $x \in\left(\frac{1}{4}, 1\right)$

$\Rightarrow\left(e^{-x} f(x)\right)^{\prime}<0$ in $x \in\left(0, \frac{1}{4}\right)$ and $\left(e^{-x} f(x)\right)^{\prime}>0$ in $x \in\left(\frac{1}{4}, 1\right)$

$\Rightarrow e^{-x} f^{\prime}(x)-e^{-x} f(x)<0$ in $x \in$

$\therefore e^{-x} f(x)$ is decreasing in $x \in\left(0, \frac{1}{4}\right)$ and increasing in $x \in\left(\frac{1}{4}, 1\right)$

$\Rightarrow\left(e^{-x} f(x)\right)^{\prime}<0, x \in\left(0, \frac{1}{4}\right)$ and

$\left(e^{-x} f(x)\right)^{\prime}>0, x \in\left(\frac{1}{4}, 1\right)$

$\Rightarrow e^{-x} f^{\prime}(x)-e^{-x} f(x)<0, x \in\left(0, \frac{1}{4}\right)$ and

$e^{-x} f^{\prime}(x)-e^{-x} f(x)>0, \quad x \in\left(\frac{1}{4}, 1\right)$

$\Rightarrow f^{\prime}(x) < f(x), x \in\left(0, \frac{1}{4}\right)$ and

$f^{\prime}(x) > f(x), x \in\left(\frac{1}{4}, 1\right)$

Hints :

$\Rightarrow e^{-x} f(x)$ is minimum at $x=\frac{1}{4}$ in $x \in\left[0, \frac{1}{4}\right]$ implies $e^{-x} f(x)$ is decreasing in $x \in\left[0, \frac{1}{4}\right]$ and $e^{-x} f(x)$ is increasing in $x \in\left(\frac{1}{4}, 1\right)$.

Here, $f(x) = 2\left| x \right| + \left| {x + 2} \right| - \left| {\left| {x + 2} \right| - 2\left| x \right|} \right|$

$ = \left\{ {\matrix{ { - 2x - (x + 2) + (x - 2),} & {when\,x \le - 2} \cr { - 2x + x + 2 + 3x + 2,} & {when\, - 2 < x \le - {2 \over 3}} \cr { - 4x,} & {when\, - {2 \over 3} < x \le 0} \cr {4x,} & {when\,0 < x \le 2} \cr {2x + 4,} & {when\,x > 2} \cr } } \right.$

$ = \left\{ {\matrix{ { - 2x - 4,} & {x \le - 2} \cr {2x + 4,} & { - 2 < x \le - 2/3} \cr { - 4x,} & { - {2 \over 3} < x \le 0} \cr {4x,} & {0 < x \le 2} \cr {2x + 4,} & {x > 2} \cr } } \right.$

Graph for y = f(x) is shown as

Let $a$ rectangular sheet with sides $15 a$ and $8 a$ and four squares of side length $b$ removing from each corners of the rectangular sheet

Given, the perimeter of rectangular sheet is constant.

$\therefore 15 a+8 a=$ Constant

$\Rightarrow a$ is constant

Let $V$ be the volume of open rectangular box

$\begin{aligned} & \Rightarrow \quad \mathrm{V}=(15 a-2 b) b(8 a-2 b) \\ & \Rightarrow \quad \mathrm{V}=4 b^3-46 a b^2+120 a^2 b \\ & \Rightarrow \quad \frac{d \mathrm{~V}}{d b}=12 b^2-92 a b+120 a^2=0 \\ & \Rightarrow \quad \frac{d \mathrm{~V}}{d b}=4(3 b-5 a)(b-6 a)=0 \\ & \Rightarrow \quad b=\frac{5 a}{3}, 6 a \\ \end{aligned}$

Now $\frac{d^2 v}{d b^2}=24 b-92 a<0$ at $b=\frac{5 a}{3}$

So, $b=\frac{5 a}{3}$ is the point of local maxima

Given, the area of removed squares is 100

$\begin{aligned} & \therefore & 4 b^2 & =100 \\ & \Rightarrow & b^2 & =25 \\ & \Rightarrow & \left(\frac{5 a}{3}\right)^2 & =25 \\ & \Rightarrow & a & =3 \end{aligned}$

Hence, the side length of the given rectangular sheet are $15 a=45$ and $8 a=24$.

Hints:

$\Rightarrow$ If $x=x_0$ is the point of local maxima of a function $y=f(x)$, then $f^{\prime}\left(x_0\right)=0$ and $f^{\prime \prime}\left(x_0\right)<0$

Given,

$\begin{aligned} g(x) & =\int_1^x\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t \\ \text { and } f(x) & =(1-x)^2 \sin ^2 x+x^2 \\ \Rightarrow g(x) & =\int_1^x\left(\frac{2(t-1)}{t+1}-\ln t\right)\left((1-t)^2 \sin ^2 t+t^2\right) d t \end{aligned}$

On differentiating the above equation w.r.t. $x$

$\begin{aligned} & \Rightarrow g^{\prime}(x)=\left(\frac{2(x-1)}{x+1}-\ln x\right)\left((1-x)^2 \sin ^2 x+x^2\right) \\ & \Rightarrow g^{\prime}(x)=\left(2-\frac{4}{x+1}-\ln x\right)\left((1-x)^2 \sin ^2 x+x^2\right) \end{aligned}$

Here $(1-x)^2 \sin ^2 x+x^2>0$ for all $x \in \mathrm{R}$

Now, draw the graph of $y=2-\frac{4}{x+1}$ and $y=-\ln x$

For $x>0$

$\text { Add the graph of } y=-\ln x \text { and } y=2-\frac{4}{x+1}$

$\Rightarrow g^{\prime}(x)<0 \text { for } x \in(1, \infty)$

Hence, $g(x)$ is decreasing function for $x \in(1, \infty)$

For statement $P$

$\begin{aligned} & f(x)+2 x=2\left(1+x^2\right) \\ \Rightarrow & (1-x)^2 \sin ^2 x+x^2+2 x=2+2 x^2 \\ \Rightarrow & (1-x)^2 \sin ^2 x=x^2-2 x+1+1 \\ \Rightarrow & (1-x)^2 \sin ^2 x=(1-x)^2+1 \\ \Rightarrow & -(1-x)^2\left(1-\sin ^2 x\right)=1 \\ \Rightarrow & -\cos ^2 x=\frac{1}{(1-x)^2}>0 \\ \Rightarrow & \cos ^2 x<0 \end{aligned}$

Here, no value of $x$ that satisfy the above equation.

$\therefore$ Statement P is false

For Statement Q

$\begin{aligned} & 2 f(x)+1=2 x(1+x) \\ \Rightarrow \quad & 2(1-x)^2 \sin ^2 x+2 x^2+1=2 x+2 x^2 \\ \Rightarrow \quad & 2(x-1)^2 \sin ^2 x=2 x-1 \\ \Rightarrow \quad & \sin ^2 x=\frac{2 x-1}{2(x-1)^2} \\ \Rightarrow \quad & \sin ^2 x=\frac{1}{x-1}+\frac{1}{2(x-1)^2} \quad \text{... (i)} \end{aligned}$

$\begin{aligned} \text { Let } B(x) & =\frac{1}{x-1}+\frac{1}{2(x-1)^2} \\ \Rightarrow \quad B^{\prime}(x) & =\frac{-1}{(x-1)^2}-\frac{1}{(x-1)^3}=\frac{-x}{(x-1)^3} \text { and } \\ B^{\prime \prime}(x) & =\frac{2 x+1}{(x-1)^4} \end{aligned}$

Hence, $\mathrm{B}(x)$ is increasing function on $x \in(0,1)$ and decreasing on $x \in(-\infty, 0) \cup(1, \infty), \mathrm{B}^{\prime \prime}(x)=0$ at $x=-\frac{1}{2}$ and $\mathrm{B}^{\prime \prime}(x) \geq 0$ for $x \in \mathrm{R}-\left\{\frac{-1}{2}, 1\right\} \text {. }$

Draw the graph of the $y=\sin ^2 x$ and $y=\frac{1}{x+1}+\frac{1}{2(x-1)^2}$

The graph of $y=\sin ^2 x$ and $y=\frac{1}{x+1}+\frac{1}{2(x-1)^2}$ intersect at some $x$.

Hence, statement $Q$ is true.

$\begin{aligned} & \text { Given, } f(x)=\int_0^x e^{t^2}(t-2)(t-3) d t, x \in(0, \infty) \\ & \Rightarrow \quad f^{\prime}(x)=e^{x^2}(x-2)(x-3) \end{aligned}$

Here, $f^{\prime}(x)$ changes its sign $+v e$ to $-v e$ about $x=2$ and $f^{\prime}(x)$ changes its sign $-v \mathrm{e}$ to $+v \mathrm{e}$ about $x=3$

Hence, $x=2$ is the point of local maxima and $x=3$ is the point of local minima

$\because f^{\prime}(x)<0$ for $x \in(2,3)$

$\therefore f(x)$ is decreasing on $x \in(2,3)$

$\because \quad f^{\prime}(x)$ is continuous and differentiable for all $x(0, \infty)$ and $f^{\prime}(2)=f^{\prime}(3)=0$

$\therefore$ According to Rolle's theorem, $f^{\prime \prime}(c)=0$ must have at least one root $\in(2,3)$

We have for

$f(x) = x\cos \left( {{1 \over x}} \right),x \ge 1$

$f'(x) = \cos \left( {{1 \over x}} \right) + {1 \over x}\sin \left( {{1 \over x}} \right) \to 1$ for $x \to \infty $

Also,

$f'(x) = \left( {{1 \over x}} \right) + {1 \over x}\sin \left( {{1 \over x}} \right) - {1 \over {{x^2}}}\sin \left( {{1 \over x}} \right) - {1 \over {{x^3}}}\cos \left( {{1 \over x}} \right)$

$ = - {1 \over {{x^3}}}\cos \left( {{1 \over x}} \right) < 0$ for $x \ge 1$

$ \Rightarrow f'(x)$ is decreasing for $[1,\infty )$

$ \Rightarrow f'(x + 2) < f'(x)$. Also,

$\mathop {\lim }\limits_{x \to \infty } f(x + 2) - f(x) = \mathop {\lim }\limits_{x \to \infty } \left[ {(x + 2)\cos {1 \over {x + 2}} - x\cos {1 \over x}} \right]=2$

Hence, $f(x + 2) - f(x) > 2\forall x \ge 1$.

Given that, $f(x) = \left\{ {\matrix{ {{{(2 + x)}^3},} & { - 3 < x \le - 1} \cr {{x^{2/3}},} & { - 1 < x < 2} \cr } } \right.$

$f'(x) = \left\{ {\matrix{ {3{{(2 + x)}^2},} & { - 3 < x < - 1} \cr {{2 \over 3}{x^{{{ - 1} \over 3}}},} & { - 1 < x < 2} \cr } } \right.$

Clearly, $f'(x)$ changes signs from positive to negative as x passes through x $=-1$

$f'(x) < 0$ for all $x\in(-3,-1)$

Also, $f'(x)>0$ for all $x\in$

$f'(x) < 0$ for all $x\in(-1,0)$

But $f'(0)$ does not exist

So, $f(x)$ attains a local minimum at x = 0

Hence, the total number of local maxima and local minima is 2.

We have,

$\mathop {\lim }\limits_{t \to x} {{{t^2}f(x) - {x^2}f(t)} \over {t - x}} = 1$

$ \Rightarrow \mathop {\lim }\limits_{t \to x} {{2tf(x) - {x^2}f'(t)} \over {1 - 0}} = 1$

$ \Rightarrow 2xf(x) - {x^2}f'(x) = 1$

$ \Rightarrow {{{x^2}f'(x) - 2xf(x)} \over {{x^4}}} = {{ - 1} \over {{x^4}}}$

$ \Rightarrow {d \over {dx}}\left( {{{f(x)} \over {{x^2}}}} \right) = {{ - 1} \over {{x^4}}}$ ...... (i)

On integrating both sides of this equation, we get

${{f(x)} \over {{x^2}}} = {1 \over {3{x^3}}} + c$

Now, since

$f(1) = 1 \Rightarrow c = {2 \over 3}$

We get

$f(x) = {1 \over {3x}} + {2 \over 3}{x^2}$

We know that equation of tangent

$ (\mathrm{Y}-y)=\frac{d y}{d x}(\mathrm{X}-x) $

Intersection of above line with $x$-axis

i.e, $\quad \mathrm{Y}=0$

$ \begin{aligned} \Rightarrow & & (0-y) & =\frac{d y}{d x}(\mathrm{X}-x) \\ \Rightarrow & & \mathrm{X} \frac{d y}{d x} & =x \frac{d y}{d x}-y \\ \Rightarrow & & \mathrm{X} & =x-\frac{y}{\frac{d y}{d x}} \end{aligned} $

Similarly for intersection with $Y-$ axis $X=0$

$ \begin{array}{ll} \Rightarrow & \mathrm{Y}=y+\frac{d y}{d x}(0-x) \\ \Rightarrow & \mathrm{Y}=y-x \frac{d y}{d x} \\ \text { Here, } & \frac{\mathrm{BP}}{\mathrm{AP}}=\frac{3}{1} \\ \Rightarrow & \frac{-d y}{3 y}=\frac{d x}{x} \end{array} $

Integrating on both sides

$ \begin{aligned} \Rightarrow & & -\int \frac{d y}{3 y} & =\int \frac{d x}{x} \\ \Rightarrow & & \frac{-1}{3} \ln y & =\ln x+\ln c \\ \Rightarrow & & \ln y & =-3 \ln x-3 \ln c \\ \Rightarrow & & \ln y & =\ln \left(x^{-3}\right)+\ln c \\ \Rightarrow & & \ln y & =\ln \left(x^{-3} \cdot c\right) \end{aligned} $

(using properties of logarithm)

$ \Rightarrow \quad y=\frac{c}{x^3} $

$ \begin{aligned} &\begin{array}{lrl} \text { Since, } & f(1) & =1 \\ \Rightarrow & c & =1 \end{array}\\ &\text { Therefore, } y=\frac{1}{x^3} \end{aligned} $

Since, $f(x)$ has local maxima at $x=-1$

and $f^{\prime}(x)$ has local maxima at $x=0$

and we know if $f(x)$ has local maxima then $f^{\prime \prime}(x)<0$

also $f^{\prime}(x)$ has local maxima then $f^{\prime \prime}(x)=0$

So, Let us assume $f^{\prime \prime}(x)=\lambda x$

$ \begin{array}{r} (\because \lambda x=0 \text { at } x=0 \text { and } \\ \lambda x<0 \text { at } x=-1) \end{array} $

$ \begin{aligned} & \Rightarrow \quad f^{\prime}(x)=\lambda \frac{x^2}{2}+c \\ & \text { As } \quad f^{\prime}(-1)=0 \\ & \Rightarrow \quad \frac{\lambda}{2}=-c \Rightarrow \quad \lambda=-2 c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Again integrating $f^{\prime}(x)$, we have

$ f(x)=\frac{\lambda x^3}{6}+c x+d $

Since $\quad f(2)=18$ and $f(1)=-1$

$ \lambda\left(\frac{8}{6}\right)+2 c+d=18 \text { and } \frac{\lambda}{6}+c+d=-1 $

Use $\quad \lambda=-2 c$ in above equations.

$ \Rightarrow-2 c \times \frac{4}{3}+2 c+d=18 \text { and } \frac{-2 c}{6}+c+d=-1 $

$ \begin{aligned} &\begin{aligned} \Rightarrow & -2 c+3 d=54 \text { and } 4 c+6 d=-6 \text { or } \\ & 2 c+3 d=-3 \end{aligned}\\ &\text { Adding above equation. }\\ &\begin{aligned} 6 d & =51 \Rightarrow d=\frac{17}{2} \\ \text { and } c & =\frac{54-3 d}{-2}=\frac{-57}{4} \\ \lambda & =-2 c \\ & =\frac{57}{2} \\ \therefore \quad f(x) & =\frac{57}{2} \times \frac{x^3}{6}-\frac{57}{4} x+\frac{17}{2} \\ f(x) & =\frac{1}{4}\left(19 x^3-57 x+34\right) \\ f^{\prime}(x) & =\frac{1}{4}\left(57 x^2-57\right) \end{aligned} \end{aligned} $

for $\max$ or $\min \frac{57 x^2-57}{4}=0$

$ x^2-1=0 \quad \Rightarrow \quad x= \pm 1 $

$\therefore \quad$ Critical points are $x=-1$ and $x=1$

Now, $f^{\prime \prime}(x)=\frac{114 x}{4}$

$f^{\prime \prime}(-1)<0 \Rightarrow$ At $x=-1 f(x)$ is maximum

$f^{\prime \prime}(1)>0 \Rightarrow$ At $x=1 \quad f(x)$ is minimum

And $f(x)$ is increasing for $x>1$

Therefore $f(x)$ is increasing for $x \in[1,2 \sqrt{5}]$

$ \begin{aligned} g(x) & =\int_0^x f(t) d t \quad x \in[1,3] \\ \therefore & g^{\prime}(x)=f(x)=\left\{\begin{array}{cc} e^x & 0 \leq x \leq 1 \\ 2-e^{x-1} & 1 < x \leq 2 \\ x-e & 2 < x \leq 3 \end{array}\right. \end{aligned} $

$ \begin{aligned} &\text { Now, calculating critical points }\\ &\begin{aligned} g^{\prime}(x) & =0 \\ \text { i.e, } \quad e^x & =0,2-e^{x-1}=0, \text { and } x-e=0 \end{aligned} \end{aligned} $

$ \Rightarrow \quad x=-\infty, 2=e^{x-1} \text { and } x=e $

$ \begin{aligned} \ln 2 & =(x-1) \ln e \\ x-1 & =\ln 2 \\ x & =1+\ln 2 \end{aligned} $

Therefore, critical points

$ \begin{aligned} x & =1+\ln 2 \text { and } x=e \\ g^{\prime \prime}(1+\ln 2) & =-e^{\ln 2}<0 \\ \Rightarrow \quad g(x) \text { is maximum at } x & =1+\ln 2 \\ g^{\prime \prime}(e) & =1>0 \\ \Rightarrow \quad g(x) \text { is minimum at } x & =e \end{aligned} $

Since, from graph $f(x)$ is discontinuous at $x=1$

Then we get local max at $x=1$ and

Local minima at $x=2$.

Required equation of tangent is $y-2=0$.

As $\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \leq\left(x_{1}-x_{2}\right)^{2} \forall x_{1}, x_{2} \in \mathrm{R}$

$\Rightarrow\left|\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}\right| \leq\left|x_{1}-x_{2}\right|\left(x^{2}=|x|^{2}\right)$

$\therefore \quad \lim_\limits{x_{1} \rightarrow x_{2}}\left|\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}\right| \leq \lim_\limits{x_{1}-x_{2}}\left|x_{1}-x_{2}\right|$

$\Rightarrow\left|f^{\prime}\left(x_{1}\right)\right| \leq 0, \forall x_{1} \in \mathrm{R}$

$\therefore\left|f^{\prime}(x)\right| \leq 0$ But $\left|f^{\prime}(x)\right| \geq 0$

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

$\Rightarrow f(x)$ is a constant function.

$\Rightarrow$ Equation of tangent at $(1,2)$ is

$\frac{y-2}{x-1}=f^{\prime}(x) \text { or } y-2=0 \quad\left(\because f^{\prime}(x)=0\right) $

i.e., $y=2$ or $y-2=0$ is required tangent.

Let the polynomial be $p(x) = a{x^3} + b{x^2} + cx + d$.

We know $p(-1) = 10$. So, $-a + b - c + d = 10$.

We also know $p(1) = -6$. So, $a + b + c + d = -6$.

It is given that $p(x)$ has a maximum at $x = -1$. For a maximum or minimum, the first derivative is zero: $p'(-1) = 0$.

First, find the derivative: $p'(x) = 3a{x^2} + 2bx + c$.

Plug in $x = -1$ in the derivative: $3a(-1)^2 + 2b(-1) + c = 0$, which means $3a - 2b + c = 0$.

It is given that $p'(x)$ has a minimum at $x = 1$. For a minimum or maximum of the derivative, the second derivative is zero: $p''(1) = 0$.

Find the second derivative: $p''(x) = 6a x + 2b$.

Plug in $x = 1$: $6a(1) + 2b = 0$, which gives $6a + 2b = 0$.

Now, we have four equations:
1) $-a + b - c + d = 10$
2) $a + b + c + d = -6$
3) $3a - 2b + c = 0$
4) $6a + 2b = 0$

Solving these equations, we find:
$a = 1$, $b = -3$, $c = -9$, $d = 5$

So the polynomial is $p(x) = x^3 - 3x^2 - 9x + 5$.

Find where $p'(x) = 0$ (for local maximum and minimum points).
$p'(x) = 3x^2 - 6x - 9$

Set this equal to zero: $3x^2 - 6x - 9 = 0$

Divide both sides by 3: $x^2 - 2x - 3 = 0$

Factorize: $(x + 1)(x - 3) = 0$

So $x = -1$ and $x = 3$.

At $x = -1$, we have a maximum and at $x = 3$, we have a minimum.

Calculate the points:
At $x = -1$, $p(-1) = 10$, so the point is $(-1, 10)$.
At $x = 3$, $p(3) = (3)^3 - 3(3)^2 - 9 \times 3 + 5 = 27 - 27 - 27 + 5 = -22$, so the point is $(3, -22)$.

The distance between these two points is:
$\sqrt{(3 - (-1))^2 + ((-22) - 10)^2}$

$= \sqrt{(4)^2 + (-32)^2}$

$= \sqrt{16 + 1024} = \sqrt{1040} = 4\sqrt{65}$