Application of Derivatives

As we know that, volume of sphere

$ =v=\frac{4}{3} \pi r^3 $

and Surface area of sphere $=s=4 \pi r^2$

Now, $\frac{d \nu}{d t}=4 \pi r^2 \frac{d r}{d t}$

$ \Rightarrow \quad \frac{d r}{d t}=\frac{1}{4 \pi r^2} \times \frac{d v}{d t} $

Putting $r=6$ and $\frac{d v}{d t}=12$

( $\because$ Diameter of sphere $=12 \mathrm{~cm} \quad$ (given)

$\therefore$ Radius of sphere $=6 \mathrm{~cm}$ )

$ \frac{d r}{d t}=\frac{12}{4 \pi(6)^2}=\frac{1}{12 \pi} $

Now, $\frac{d s}{d t}=8 \pi r \cdot \frac{d r}{d t}$

$ \therefore \quad \frac{d s}{d t}=8 \pi \times 6 \times \frac{1}{12 \pi}=4 \mathrm{~cm}^2 $

(putting $r=6$ and $\frac{d r}{d t}=\frac{1}{12 \pi}$ )

Given equation of curve is

$ \begin{aligned} y & =x^2+x-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ \Rightarrow \quad \frac{d y}{d x} & =2 x+1 \\ \left.\therefore \frac{d y}{d x}\right|_{x=1} & =2 \times 1+1=3=m \end{aligned} $

i.e. slope of tangent $=m=3$. and point is $(1,1)$ (by putting $x=1$ in Eq. (i))

Now, length of tangent

$ \begin{aligned} & =a=\left|\frac{y \sqrt{1+m^2}}{m}\right| \\ & =\left|\frac{1 \sqrt{1+3^2}}{3}\right|=\frac{\sqrt{10}}{3} \end{aligned} $

length of subtangent

$ =b=\left|\frac{y}{m}\right|=\left|\frac{1}{3}\right|=\frac{1}{3} $

length of normal

$ =c=\left|y \sqrt{1+m^2}\right|=|1 \times \sqrt{10}|=\sqrt{10} $

length of subnormal

$ =d=|y \cdot m|=|3|=3 $

Now, from above result, we conclude that $b

$ \begin{aligned} & \text { } y=x^4-2 x^3+x^2+5 x \\ & \Rightarrow \frac{d y}{d x}=4 x^3-6 x^2+2 x+5 \end{aligned} $

The slope of tangent line at $\left(x_1, y_1\right)$ is

$ m=4 x_1^3-6 x_1^2+2 x_1+5 $

The equation of tangent line is

$ y-y_1=\left(4 x_1^3-6 x_1^2+2 x_1+5\right)\left(x-x_1\right) $

Since, the tangent line passes through $(0,0)$

$ \begin{aligned} \Rightarrow 0-y_1 & =\left(4 x_1^3-6 x_1^2+2 x_1+5\right)\left(0-x_1\right) \\ y_1 & =\left(4 x_1^3-6 x_1^2+2 x_1+5\right) x_1 \end{aligned} $

Since, ( $x_1, y_1$ ) is on the curve

$ y_1=x_1^4-2 x_1^3+x_1^2+5 x_1 $

Substituting this into the previous equation, we get

$ \begin{aligned} & x_1^4-2 x_1^3+x_1^2+5 x_1 \\ & =\left(4 x_1^3-6 x_1^2+2 x_1+5\right) x_1 \\ \Rightarrow & x_1^4-2 x_1^3+x_1^2+5 x_1 \\ & =4 x_1^4-6 x_1^3+2 x_1^2+5 x_1 \\ \Rightarrow & 3 x_1^4-4 x_1^3+x_1^2=0 \\ \Rightarrow & x_1^2\left(3 x_1^2-4 x_1+1\right)=0 \\ \Rightarrow & x_1^2\left(3 x_1-1\right)\left(x_1-1\right)=0 \\ \Rightarrow & x_1=0, x_1=\frac{1}{3}, x_1=1 \\ & x_1=1, y_1=5 \Rightarrow x_1+y_1=1+5=6 \end{aligned} $

From the given, we have

$ f(x)=\log (1+x)-x+\frac{x^2}{2} $

Differentiating both sides, we get $\Rightarrow f^{\prime}(x)=\frac{1}{(1+x)}-1+x$

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

Hence, $f(x)$ is monotonically increasing in its domain.

Step 1: Find the Derivative and Slope of Normal

The curve is $x^2 + 3y^2 = 9$.

Differentiate both sides with respect to $x$:

$2x + 3 \cdot 2y \cdot y' = 0$

Solve for $y'$:

$2x + 6y y' = 0 \implies 6y y' = -2x \implies y' = -\frac{2x}{6y} = -\frac{x}{3y}$

The slope of the tangent is $y' = -\frac{x}{3y}$, so the slope of the normal is $-\frac{1}{y'} = \frac{3y}{x}$.

Step 2: Find the Slope of Normals at Given Points

First point: $(3\cos\theta,\, \sqrt{3}\sin\theta)$

Slope at this point:

$m_1 = \frac{3 \cdot \sqrt{3} \sin\theta}{3\cos\theta} = \sqrt{3}\tan\theta$

Second point: $(-3\sin\theta,\, \sqrt{3}\cos\theta)$

Slope at this point:

$m_2 = \frac{3 \cdot \sqrt{3} \cos\theta}{-3\sin\theta} = -\sqrt{3}\cot\theta$

Step 3: Find the Angle Between Normals ($\beta$)

Use the formula for the tangent of the angle between two lines:

$\tan\beta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$

$\tan\beta = \left|\frac{\sqrt{3}\tan\theta - (-\sqrt{3}\cot\theta)}{1 + (\sqrt{3}\tan\theta)(-\sqrt{3}\cot\theta)}\right|$

$\tan\beta = \left|\frac{\sqrt{3}(\tan\theta + \cot\theta)}{1 - 3}\right|$

$\tan\beta = \left|\frac{\sqrt{3}(\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta})}{-2}\right|$

$\tan\beta = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sin\theta\cos\theta}$

Recall $2\sin\theta\cos\theta = \sin2\theta$, so:

$\tan\beta = \frac{\sqrt{3}}{\sin2\theta}$

This means $\frac{1}{\cot\beta} = \frac{\sqrt{3}}{\sin2\theta}$ or $\sqrt{3}\cot\beta = \sin2\theta$.

$ \begin{aligned} & x^2+y^2=5^2 \\ & y^2=5^2-x^2 \Rightarrow y=\sqrt{5^2-x^2} \\ & 0 \leq x \leq 5 \end{aligned} $

$ \begin{aligned} & \text { Area }=\frac{1}{2} \text { base } \times \text { height } \\ & \begin{aligned} & A=\frac{1}{2} x y=\frac{1}{2} x \sqrt{25-x^2} \\ & \frac{d A}{d x}=\frac{1}{2}\left(25-x^2\right)^{-\frac{1}{2}}\left(25-2 x^2\right) \\ &=\frac{25-2 x^2}{2 \sqrt{25-x^2}} \end{aligned} \\ & \Rightarrow \quad \frac{d A}{d x}=0 \Rightarrow \frac{25-2 x^2}{2 \sqrt{25-x^2}}=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 25-2 x^2=0 \\ & \Rightarrow \quad 25=2 x^2 \\ & \Rightarrow \quad x=\frac{5}{\sqrt{2}}, y=\frac{5}{\sqrt{2}} \\ & \Rightarrow \quad A B=\frac{5}{\sqrt{2}}, B C=\frac{5}{\sqrt{2}}, A C=5 \\ & \text { Perimeter }=A B+B C+A C \\ & =\frac{5}{\sqrt{2}}+\frac{5}{\sqrt{2}}+5=5 \sqrt{2}+5 \\ & =5(\sqrt{2}+1) \end{aligned} $

$ y=|\cos x-\sin x|+|\tan x-\cot x| $

For $^{\prime \prime} x=\frac{\pi}{3}$

$ \begin{aligned} y & =-(\cos x-\sin x)+(\tan x-\cot x) \\ & =\cos x+\sin x+\tan x-\cot x \end{aligned} $

For $x=\frac{\pi}{6}$

$ \begin{aligned} y & =(\cos x-\sin x)-(\tan x-\cot x) \\ & =\cos x-\sin x-\tan x+\cot x \\ \frac{d y}{d x} & ]_x=\frac{\pi}{3} \\ = & \sin \frac{\pi}{3}+\cos \frac{\pi}{3}+\sec ^2 \frac{\pi}{3}+\operatorname{cosec}^2 \frac{\pi}{3} \\ = & \frac{35+3 \sqrt{3}}{6} \end{aligned} $

$ \begin{aligned} & \left.\frac{d y}{d x}\right]_{x=\frac{\pi}{6}} \\ & =-\sin \frac{\pi}{6}-\cos \frac{\pi}{6}-\sec ^2 \frac{\pi}{6}-\operatorname{cosec}^2 \frac{\pi}{6} \\ & =\frac{-35-3 \sqrt{3}}{6} \\ & \left.\left.\frac{d y}{d x}\right]_{x=\frac{\pi}{3}}+\frac{d y}{d x}\right]_{x=\frac{\pi}{6}}=0 \end{aligned} $

Given curve, $x^{\frac{2}{3}}+y^{\frac{2}{3}}=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and line $y=-\sqrt{3} x+1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i), $\frac{2}{3} x^{-\frac{1}{3}}+\frac{2}{3} y^{-\frac{1}{3}} y^{\prime}=0$

$ \begin{aligned} y^{\prime} l_{(\alpha, \beta)} & =-\left(\frac{\beta}{\alpha}\right)^{\frac{1}{3}} \\ -\left(\frac{\beta}{\alpha}\right)^{\frac{1}{3}} & =-\sqrt{3} \Rightarrow \frac{\beta}{\alpha}=3 \sqrt{3} \\ \beta & =3 \sqrt{3} \alpha \end{aligned} $

From Eq. (i) $\alpha^{\frac{2}{3}}+3 \alpha^{\frac{2}{3}}=4 \Rightarrow \alpha= \pm 1$

$ \begin{aligned} \beta & = \pm 3 \sqrt{3} \\ \therefore \quad \alpha^2+\beta^2 & =28 \end{aligned} $

$S=t^3-16 t^2+64 t-6$

$ \Rightarrow \frac{d S}{d t}=3 t^2-32 t+64 $

For maxima/minima $\frac{d S}{d t}=0$

$ \begin{aligned} \Rightarrow \quad & \left(3 t^2-8\right)(t-8)=0 \Rightarrow t=8, \frac{8}{3} \\ & \left.\frac{d^2 s}{d t^2}=6 t-32 \Rightarrow \frac{d^2 s}{d t^2}\right]_{t=\frac{8}{3}}<0 \end{aligned} $

So, required time, $t=\frac{8}{3}$

$ \begin{aligned} & \text { } f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x} \\ & \Rightarrow f^{\prime}(x)=\frac{-4 \cos x}{\sin ^2 x}+\frac{\cos x}{(1-\sin x)^2} \\ & f^{\prime}(x)=0 \\ & \Rightarrow \cos x\left[\frac{1}{(1-\sin x)^2}-\frac{4}{\sin ^2 x}\right]=0 \\ & \Rightarrow \cos x\left[\frac{\sin ^2 x-4(1-\sin x)^2}{\left(\sin ^2 x\right)(1-\sin x)^2}\right]=0 \\ & \Rightarrow \cos x[(\sin x+2-2 \sin x) (\sin x-2+2 \sin x)] \end{aligned} $

$ \begin{aligned} & =0 \\ \Rightarrow \quad x & =\frac{\pi}{2}, \sin x=2 \sin x=\frac{2}{3} \\ f(0) & =\text { Not possible } \\ f\left(\frac{\pi}{2}\right) & =\text { Not possible } \\ \sin x & \neq 2 \quad(\because \sin x \in[-1,1]) \end{aligned} $

$ \begin{aligned} \therefore \quad \sin x & =\frac{2}{3} \\ f(x) l_{\max } & =6+3=9=m \\ \Rightarrow \quad \cos k & =\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\sqrt{\frac{5}{m}} \end{aligned} $

Let $P(h, k)$

$ \begin{aligned} & y=x \log x \\ \Rightarrow \quad & \frac{d y}{d x}=1+\log x \\ \therefore & \text { Slope of normal }=-\frac{1}{1+\log h} \end{aligned} $

And slope of the given line $2 x-2 y=3$ is 1

$ \begin{aligned} & \Rightarrow \quad-\frac{1}{1+\log h}=1 \\ & \Rightarrow \quad 1+\ln h=-1 \\ & \Rightarrow \ln h=-2 \\ & h=e^{-2} \\ & k=h \ln h \\ & k=e^{-2} \ln e^{-2}=e^{-2}(-2)=-\frac{2}{e^2} \end{aligned} $

$\therefore$ The point $P$ is $\left(\frac{1}{e^2},-\frac{2}{e^2}\right)$

We have the curve

$ y^2=16 x \text { and } 9 x^2+\alpha y^2=25 $

On differentiating, we get slopes

$ 2 y y^{\prime}=16 \text { and } 18 x+2 \alpha y y^{\prime}=0 $

$ y^{\prime}=\frac{8}{y}, y^{\prime}=-\frac{9 x}{\alpha y} $

Clearly, $m_1=\frac{8}{y}$ and $m_2=-\frac{9 x}{\alpha y}$

$ \begin{aligned} & m_1 m_2=-1 \\ & \quad \frac{8}{y} \times\left(\frac{-9 x}{\alpha y}\right)=-1 \Rightarrow-\frac{72 x}{\alpha y^2}=-1 \\ & \quad 72 x=\alpha y^2=\alpha \cdot 16 x \\ & 16 \alpha=72 \\ & \quad \alpha=\frac{72}{16}=\frac{9}{2} \\ & \quad \therefore \quad \alpha=\frac{9}{2} \end{aligned} $

We have, $y=\sin x(1+\cos x)$

$ \begin{aligned} \frac{d y}{d x} & =\sin x(-\sin x)+(1+\cos x) \cos x \\ & =-\sin ^2 x+\cos x+\cos ^2 x \\ & =-1+\cos ^2 x+\cos x+\cos ^2 x \\ & =2 \cos ^2 x+\cos x-1 \\ \Rightarrow \frac{d y}{d x} & =(\cos x+1)(2 \cos x-1) \end{aligned} $

Given $y$ is strictly increasing

$ \begin{aligned} & \therefore \quad \frac{d y}{d x}=(\cos x+1)(2 \cos x-1)>0 \\ & \Rightarrow \cos x<-1 \text { or } \cos x>\frac{1}{2} \\ & \Rightarrow x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right) \text { in the interval }(-\pi, \pi) \\ & \therefore x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right) \end{aligned} $

Given:

Velocity $ v $ is proportional to the cube root of the displacement $ x $.

$ v \propto x^{1/3} $

This means we can write:

$ v = k x^{1/3} $

where $ k $ is a constant of proportionality.

We are asked to find how acceleration $ a $ depends on other quantities.

We know that:

$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} $

Step 1: Find $ \frac{dv}{dx} $

From $ v = k x^{1/3} $,

$ \frac{dv}{dx} = k \cdot \frac{1}{3} x^{-2/3} = \frac{k}{3} x^{-2/3} $

Step 2: Substitute in acceleration formula:

$ a = v \frac{dv}{dx} = (k x^{1/3}) \cdot \left(\frac{k}{3} x^{-2/3}\right) $

Simplify:

$ a = \frac{k^2}{3} x^{1/3 - 2/3} = \frac{k^2}{3} x^{-1/3} $

Step 3: Express in terms of velocity if desired

Since $ v = k x^{1/3} \Rightarrow x^{1/3} = \frac{v}{k} $,

$ x^{-1/3} = \frac{k}{v} $

So:

$ a = \frac{k^2}{3} \cdot \frac{k}{v} = \frac{k^3}{3v} $

Therefore:

$ a \propto \frac{1}{v} $

✅ Final Answer:

Option B: inversely proportional to its velocity

We have, $f(x)=4 x^4+4 x^3-23 x^2$$ -12 x+36=0 $

Using concept a polynomial $f(x)$ has a multiple root $r$, then $f(r)=0$ and

$ \begin{aligned} & f^{\prime}(r)=0 \\ & f^{\prime}(x)=16 x^3+12 x^2-46 x-12 \end{aligned} $

Now, using hit and trial

By taking $x=-2$

$ \begin{gathered} f(-2)=4(-2)^4+4(-2)^3-23(-2)^2 \\ -12(-2)+36 \\ =64-32-92+24+36=0 \\ \Rightarrow f^{\prime}(-2)=16(-2)^3+12(-2)^2 \\ -46(-2)-12 \\ =-128+48+92-12=0 \end{gathered} $

$\because \quad x=-2$ is a multiple root.

By dividing the equation by $(x+2)^2$, we get another multiple root $x=\frac{3}{2}$

$ \begin{aligned} & \text { Given, } \alpha>\beta \\ & \therefore \quad \alpha=\frac{3}{2} \text { and } \beta=-2 \\ & \therefore 2 \alpha-\beta=3+2=5 \end{aligned} $

$ \begin{aligned} &\text { We have, the curve }\\ &x^3+y^3=2 x y \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 3 x^2+3 y^2 y^{\prime}=2 x y^{\prime}+y \cdot 2 \\ & \Rightarrow \quad y^{\prime}\left(3 y^2-2 x\right)=2 y-3 x^2 \\ & \Rightarrow \quad y^{\prime}=\frac{2 y-3 x^2}{3 y^2-2 x} \end{aligned} $

Slope of tangent $=\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{2-3}{3-2}$

$ =\frac{-1}{1}=-1 $

Slope of normal $=1$

Equation of tangent $\Rightarrow y-1=-1(x-1)$

$ \begin{aligned} & y-1=-x+1 \\ & x+y=2 \end{aligned} $

$x$-intercept of tangent $=(2,0)$

Equation of normal,

$ \begin{gathered} y-1=1(x-1) \\ y=x \end{gathered} $

$\because$ Area of $\triangle P T N=\frac{1}{2} \times N T \times P S$

$ =\frac{1}{2} \times 2 \times 1=1 \text { sq. unit } $

$f(x)=2 \sin x+\sin 2 x$

$f(x)$ is continuous and differentiable

$ f(0)=f(\pi) $

$\therefore$ Rolle's Theorem is applicable

$ \begin{aligned} & f^{\prime}(x)=2 \cos x+\cos 2 x \cdot 2 \\ & \begin{aligned} \Rightarrow f^{\prime}(c) & =2 \cos c+2\left(2 \cos ^2 c-1\right)=0 \\ & =2\left[2 \cos ^2 c+\cos c-1\right]=0 \\ & =2 \cos ^2 c+2 \cos c-\cos c-1=0 \end{aligned} \\ & \Rightarrow 2 \cos c(\cos c+1)-1(\cos c+1)=0 \\ & \Rightarrow(\cos c+1)(2 \cos c-1)=0 \\ & \Rightarrow \cos c=-1 \text { and } \cos c=\frac{1}{2} \\ & \because \quad c=\frac{\pi}{3} \end{aligned} $

$y=3 x^4-5 x^3-12 x^2+18 x+3$

Slope of tangent

$ \begin{aligned} g(x) & =\frac{d y}{d x}=12 x^3-15 x^2-24 x+18 \\ g^{\prime}(x) & =36 x^2-30 x-24 \\ & =6\left(6 x^2-5 x-4\right) \end{aligned} $

$\therefore g(x)$ is increasing         $ \left[\because g^{\prime}(x)>0\right] $

$ \begin{aligned} & x<-\frac{1}{2} \text { or } x>\frac{4}{3} \\ \therefore \quad & R-\left[-\frac{1}{2}, \frac{4}{3}\right] \end{aligned} $

$\begin{aligned} & f(x)=6 x^2-18 a x+12 a^2 \\ & =6\left(x^2-3 a+2 a^2\right) \\ & =6(x-a)(x-2 a)=0 \\ & x=a, 2 a \end{aligned}$

$a=\alpha, \quad 2 a=\alpha^2 \quad \Rightarrow \alpha=0,2$

$\begin{array}{lll} a>0 & \therefore & \alpha=2 \\ & & \alpha^2=4 \end{array}$

$\therefore x^2-6 x+8=0$ is the required quadratic equation.

$\begin{aligned} & f(x)=4 \cos ^3 x+3 \sqrt{3} \cos ^2 x-10 \\ & f^{\prime}(x)=12 \cos ^2 x \cdot(-\sin x)+6 \sqrt{3} \cos x \cdot(-\sin x)=0 \\ & =-6 \sqrt{3} \cos x \cdot \sin x\left(1+\frac{2}{\sqrt{3}} \cos x\right)=0 \\ & \cos x=0, \sin x=0, \cos x=\frac{-\sqrt{3}}{2} \end{aligned}$

Sign of $f(x)$

$\therefore \text { Maxima at } \frac{5 \pi}{6}, \frac{7 \pi}{6}$

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

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., $f'(x) < 0$), $f(x)$ is a strictly decreasing function in $[0, \pi]$. Moreover, we evaluate:

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

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

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.

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: $\displaystyle\bigl[\tfrac1e,\infty\bigr)$.

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

Area is maximum when $\sin 2 \theta=1 \Rightarrow \theta=45^{\circ}$

$\Rightarrow$ Maximum area $=8+10=18$

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

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

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

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

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

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

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, $f^{\prime \prime}(x)=-\sin x-\frac{4}{\pi}<0 \forall x \in\left[0, \frac{\pi}{2}\right]$

Hence, $f^{\prime}(x)$ is decreasing

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

$\begin{aligned} & f(x)=3 \sqrt{x-2}+\sqrt{4-x} \\ & \text { Let } x=2 \sin ^2 \theta+4 \cos ^2 \theta \\ & =3 \sqrt{2 \sin ^2 \theta+4 \cos ^2 \theta-2}+\sqrt{4-2 \sin ^2 \theta-4 \cos ^2 \theta} \\ & =3 \sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta} \\ & =3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \end{aligned}$

$\begin{aligned} & \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{18+2} \\ & \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{20} \end{aligned}$

Minimum value exists when $\theta=\frac{\pi}{2}$

$\begin{aligned} & \text { So, minimum value }=\sqrt{2} \\ & \Rightarrow \alpha=\sqrt{2} \text { and } \beta=\sqrt{20} \\ & \Rightarrow \alpha^2+2 \beta^2=2+40 \\ & =42 \end{aligned}$

$\begin{aligned} & f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}=y, 2 x^2+3 x+8>0 \quad \forall x \in \mathbb{R} \\ & \Rightarrow \quad x^2(2 y-2)+x(3 y+3)+8 y-8=0 \end{aligned}$

Since $x \in \mathbb{R}$, the equation has real roots

$\begin{aligned} & \Rightarrow \quad D \geq 0 \\ & \Rightarrow(3 y+3)^2-4(2 y-2)(8 y-8) \geq 0 \\ & \Rightarrow 9(y+1)^2-64 y(y-1)^2 \geq 0 \\ & \Rightarrow(3 y+3)^2-(8 y-8)^2 \geq 0 \\ & \Rightarrow(11 y-5)(-5 y+11) \geq 0 \\ & \Rightarrow\left(y-\frac{5}{11}\right)\left(y-\frac{11}{5}\right) \leq 0 \\ & \Rightarrow y \in\left[\frac{5}{11}, \frac{11}{5}\right] \end{aligned}$

Sum of maximum and minimum value

$\begin{aligned} & y_{\max }+y_{\min }=\frac{5}{11}+\frac{11}{5}=\frac{25+121}{55} \\ & =\frac{146}{55}=\frac{m}{n} \Rightarrow m+n=201 \end{aligned}$