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 $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$

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.

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

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

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

$\therefore x^2-6 x+8=0$ 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

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

$\begin{aligned} & f: R \rightarrow(0, \infty) \\ & \lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1 \end{aligned}$

$\because \mathrm{f}$ is increasing

$\begin{aligned} & \therefore \mathrm{f}(\mathrm{x})<\mathrm{f}(5 \mathrm{x})<\mathrm{f}(7 \mathrm{x}) \\ & \because \frac{\mathrm{f}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(7 \mathrm{x})}{\mathrm{f}(\mathrm{x})} \\ & 1<\lim _{\mathrm{x} \rightarrow \infty} \frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<1 \\ & \therefore\left[\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}-1\right] \\ & \Rightarrow 1-1=0 \end{aligned}$

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

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

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

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

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

$\begin{aligned} & f(0)=\left|\begin{array}{ccc} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{array}\right|=12 \\ & f^{\prime}(x)=\left|\begin{array}{ccc} 3 x^2 & 4 x & 3 \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{array}\right|+ \end{aligned}$

$\begin{aligned} & \left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 6 x & 2 & 3 x^2 \\ x^3-x & 4 & x^2-2 \end{array}\right|+ \\ & \left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ 3 x^2-1 & 0 & 2 x \end{array}\right| \\ & \therefore f^{\prime}(0)=\left|\begin{array}{ccc} 0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{array}\right|+\left|\begin{array}{ccc} 0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & -2 \end{array}\right|+\left|\begin{array}{ccc} 0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0 \end{array}\right| \\ & =24-6=18 \\ & \therefore 2 f(0)+f^{\prime}(0)=42 \end{aligned}$

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}+3)^2 \cdot 3(\mathrm{x}-2)^2+(\mathrm{x}-2)^3 2(\mathrm{x}+3) \\ & =5(\mathrm{x}+3)(\mathrm{x}-2)^2(\mathrm{x}+1) \\ & \mathrm{f}^{\prime}(\mathrm{x})=0, \mathrm{x}=-3,-1,2 \end{aligned}$

$\begin{aligned} & f(-4)=-216 \\ & f(-3)=0, f(4)=49 \times 8=392 \\ & M=392, m=-216 \\ & M-m=392+216=608 \\ & \text { Ans }=\text { '3' } \end{aligned}$

Area of $\Delta$

$\begin{aligned} & =\frac{1}{2}\left|\begin{array}{ccc} 0 & 0 & 1 \\ x & y & 1 \\ -x & y & 1 \end{array}\right| \\ & \Rightarrow\left|\frac{1}{2}(x y+x y)\right|=|x y| \\ & \operatorname{Area}(\Delta)=|x y|=\left|x\left(-2 x^2+54\right)\right| \\ & \frac{d(\Delta)}{d x}=\left|\left(-6 x^2+54\right)\right| \Rightarrow \frac{d \Delta}{d x}=0 \text { at } x=3 \\ & \text { Area }=3(-2 \times 9+54)=108 \end{aligned}$

$f(x)=\frac{x}{x^2-6 x-16}$

Now,

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{-\left(\mathrm{x}^2+16\right)}{\left(\mathrm{x}^2-6 \mathrm{x}-16\right)^2} \\ & \mathrm{f}^{\prime}(\mathrm{x})<0 \end{aligned}$

Thus $f(x)$ is decreasing in

$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$

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

So, maxima (M) at x = $-$1 & minima (m) at x = 0

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=12 \sqrt{2} \mathrm{x}^2-3 \sqrt{2} \geq 0 \text { for }\left[\frac{1}{2}, 1\right] \\ & \mathrm{f}\left(\frac{1}{2}\right)<0 \end{aligned}$

$\mathrm{f}(1)>0 \Rightarrow(\mathrm{A})$ is correct.

$f(x)=\sqrt{2}\left(4 x^3-3 x\right)-1=0$

Let $\cos \alpha=\mathrm{x}$,

$\cos 3 \alpha=\cos \frac{\pi}{4} \Rightarrow \alpha=\frac{\pi}{12}$

$\mathrm{x}=\cos \frac{\pi}{12}$

(4) is correct.

$g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \text { and } f^{\prime \prime}(x)>0 \forall x \in(0,3)$

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

$\begin{aligned} & g^{\prime}(x)=3 \times \frac{1}{3} \cdot f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \\ & =f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \end{aligned}$

If $\mathrm{g}$ is decreasing in $(0, \alpha)$

$\begin{aligned} & \mathrm{g}^{\prime}(\mathrm{x})<0 \\ & \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})<0 \\ & \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)<\mathrm{f}^{\prime}(3-\mathrm{x}) \\ & \Rightarrow \frac{\mathrm{x}}{3}<3-\mathrm{x} \\ & \Rightarrow \mathrm{x}<\frac{9}{4} \end{aligned}$

Therefore $\alpha=\frac{9}{4}$

Then $8 \alpha=8 \times \frac{9}{4}=18$