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

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

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

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

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

$ 5 f(x)+4 f(1 / x)=x^2-2 $ ........(1)

Replace $x$ by $1 / x$

$ 5 f(1 / x)+4 f(x)=\frac{1}{x^2}-2 $ ..........(2)

Multiply equation (1) by 5 and multiply equation (2) by 4 and then subtract equation (2) from (1)

$\begin{aligned} & 25 f(x)-16 f(x)=5 x^2-10-\frac{4}{x^2}+8 \\\\ & 9 f(x)=5 x^2-\frac{4}{x^2}-2 \\\\ & 9 f(x)=\frac{5 x^4-4-2 x^2}{x^2}\end{aligned}$

$\begin{aligned} & y=9 x^2 f(x) \\\\ & y=5 x^4-2 x^2-4 \\\\ & y^{\prime}=20 x^3-4 x \\\\ & \text { Put } y^{\prime}>0 \\\\ & 20 x^3-4 x>0 \\\\ & 5 x^3-x>0 \\\\ & x\left(5 x^2-1\right)>0\end{aligned}$



$x \in\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$

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

$\because \mathrm{f}$ 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 $8 \alpha=8 \times \frac{9}{4}=18$

Given the function:

$ f(x) = x - 2\sin{x}\cos{x} + \frac{1}{3}\sin{3x} $

We want to find the maximum value of this expression for $0 \leq x \leq \pi$.

Step 1: Rewrite the expression

Notice that we can rewrite the expression as:

$ f(x) = x - \sin{2x} + \frac{1}{3}\sin{3x} $

Step 2: Find the first derivative

Now, let's find the derivative of this expression with respect to $x$:

$ f'(x) = 1 - 2\cos{2x} + \cos{3x} $

Step 3: Find the critical points

To find the critical points, we set $f'(x) = 0$:

$ 1 - 2\cos{2x} + \cos{3x} = 0 $

Step 4: Solve for x

This equation can be rewritten as:

$ (2\cos{x} + \sqrt{3})(2\cos{x} - \sqrt{3})(\cos{x} - 1) = 0 $

We get three possible solutions for $x$:

$ \cos{x} = \frac{-\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 1 $

Which gives us: $ x = \frac{5\pi}{6}, \frac{\pi}{6}, 0 $

Step 5: Find the second derivative

Now, let's find the second derivative of the expression:

$ f''(x) = 4\sin{2x} - 3\sin{3x} $

Step 6: Analyze the critical points Evaluate the second derivative at the critical points:

1. $f''\left(\frac{5\pi}{6}\right) = -2\sqrt{3} - \sqrt{3} < 0$, indicating a local maximum.
2. $f''\left(\frac{\pi}{6}\right) = 2\sqrt{3} - \sqrt{3} > 0$, indicating a local minimum.
3. $f''(0) = 0$, inconclusive.

Step 7: Evaluate the function at the local maximum point and the boundary points Since we are looking for the maximum value of $f(x)$, we can now evaluate the function at the local maximum point and the boundary points:

1. $f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{5\pi + 2 + 3\sqrt{3}}{6}$
2. $f(0) = 0$
3. $f(\pi) = \pi$

Step 8: Compare the values The maximum value of $f(x)$ is given by $f\left(\frac{5\pi}{6}\right) = \frac{5\pi + 2 + 3\sqrt{3}}{6}$.

$ \text { Let } y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x} $

$ \ln \mathrm{y}=\sin ^2 \mathrm{x} \cdot \ln \left(\frac{\sqrt{3 \mathrm{e}}}{2 \sin \mathrm{x}}\right) $

$ \frac{1}{y} y^{\prime}=\ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x+\sin ^2 x \frac{2 \sin x}{\sqrt{3 e}} \frac{\sqrt{3 e}}{2}(-\operatorname{cosec} x \cot x) $

For maxima or minima, $ \frac{d y}{d x}=0 $

$ \Rightarrow \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x-\sin x \cos x=0 $

$ \Rightarrow \sin x \cos x\left[2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)-1\right]=0 $

$ \Rightarrow \ln \left(\frac{3 \mathrm{e}}{4 \sin ^2 \mathrm{x}}\right)=1 $

$ \Rightarrow \frac{3 e}{4 \sin ^2 x}=e $

$ \Rightarrow \sin ^2 x=\frac{3}{4} $

$ \Rightarrow \sin \mathrm{x}=\frac{\sqrt{3}}{2} \quad\left(\text { as } \mathrm{x} \in\left(0, \frac{\pi}{2}\right)\right) $

$ \Rightarrow \text { Local max value }=\left(\frac{\sqrt{3 \mathrm{e}}}{\sqrt{3}}\right)^{3 / 4}=\mathrm{e}^{3 / 8}=\frac{\mathrm{k}}{\mathrm{e}} $

$ \Rightarrow \mathrm{k}^8=\mathrm{e}^{11} $

$ \therefore $ $ \left(\frac{k}{e}\right)^8+\frac{k^8}{e^5}+k^8=e^3+e^6+e^{11} $

Given, $\left(x \log _{e} x\right) f^{\prime}(x)+\left(\log _{e} x\right) f(x)+f(x) \geq 1, x \in[2,4]$

$ \left(x\log _e x\right) f^{\prime}(x)+f(x)\left[\log _e x+1\right] \geq 1$

$ \Rightarrow $ $ \frac{d}{d x}\left[x \log _e x f(x)\right] \geq 1$

$ \begin{aligned} &\Rightarrow \frac{d}{d x}\left[x \log _e x f(x)-x\right] \geq 0 \quad\left[\because \frac{d}{d x}(x)=1\right] \\\\ & \forall x \in[2,4] \end{aligned} $

Let $g(x)=x \log _e x f(x)-x$

As $g(x) \geq 0, \forall x \in[2,4], g(x)$ is an increasing function in $[2,4]$

$ \begin{aligned} g(2) & =2 \log _e 2 f(2)-2 \\\\ & =\log _e 2-2 \quad\left[\because f(x)=\frac{1}{2}\right] \end{aligned} $

$ \begin{aligned} & g(4)=4 \log _e 4 f(4)-4=\log _e 4-4 \\\\ &=2\left(\log _e 2-2\right)\\\\ & {\left[\therefore f(4)=\frac{1}{4}\right] } \end{aligned} $

As, $g(x)$ is an increasing function,

$ \begin{aligned} & g(2) \leq g(x) \leq g(4) \\\\ & \log _e 2-2 \leq g(x) \leq 2\left(\log _e 2-2\right) \\\\ & \log _e 2-2 \leq x \log _e x f(x)-x \leq 2\left(\log _e 2-2\right) \end{aligned} $

$ \frac{\log _e 2-2+x}{x \log _e x} \leq f(x) \leq \frac{2\left(\log _e 2-2\right)+x}{x \log _e x} $

$ \begin{aligned} & \text {Now for } x \in[2,4] \\\\ & \begin{aligned} \frac{\log _e 2-2+x}{x \log _e x} & \leq \frac{2\left(\log _e 2-2\right)+e^2}{2 \log _e 2} =1-\frac{1}{\log _e 2}<1 \end{aligned} \end{aligned} $

$ \Rightarrow \mathrm{f}(\mathrm{x}) \leq 1 \text { for } \mathrm{x} \in[2,4] $

$ \text { Also for } \mathrm{x} \in[2,4] \text { : } $

Now,

$ \begin{aligned} \frac{2\left(\log _e 2-2\right)+2}{2 \log _e 2} & \geq \frac{\log _e 2-2+4}{4 \log _e 4} =\frac{1}{8}+\frac{1}{2 \log _e 2}>\frac{1}{8} \end{aligned} $

$ \therefore f(x) \geq \frac{1}{8}, \forall x \in[2,4] $

Hence, both statements $A$ and $B$ are true.

Note : LMVT on $(\mathrm{yx}(\ln \mathrm{x}))$ not satisfied.

Hence no such function exists.

Therefore it should be bonus.

We have, $g(x)=f(x)+f(1-x)$

Differentiating both side, we get

$g^{\prime}(x)=f^{\prime}(x)-f^{\prime}(1-x)$

As $f^{\prime \prime}(x)>0, f^{\prime}(x)$ is an increasing function.

Also, $g(x)=f(x)+f(2 a-x)$ is always symmetric about $x=a$

So, $g(x)=f(x)+f(1-x)$ is also symmetric about $x=1 / 2$

$\therefore g$ is decreasing in the interval $(0,1 / 2)$ and increasing in the interval $(1 / 2,1)$.

Now,

$ \begin{aligned} & \tan ^{-1} 2 \alpha+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) \\\\ & =\tan ^{-1} 1+\tan ^{-1}(2)+\tan ^{-1} 3=\pi \end{aligned} $

Let the slope of tangent at any point

$(x, y)$ on a curve $y=y(x)$ is $\frac{d y}{d x}$

According to the question, $\frac{d y}{d x}=\frac{x^2+y^2}{2 x y}(x>0)$ [Given]

$ \begin{aligned} &\text { Let } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\\\ &\Rightarrow v+x \frac{d v}{d x}=\frac{x^2+v^2 x^2}{2 v x^2}=\frac{1+v^2}{2 v} \\\\ &\Rightarrow x \frac{d v}{d x}=\frac{1+v^2}{2 v}-v=\frac{1-v^2}{2 v} \end{aligned} $

$ \begin{aligned} & \Rightarrow \int \frac{2 v d v}{1-v^2}=\int \frac{d x}{x} \text { [Taking integration on both sides] } \\\\ & \Rightarrow-\ln \left|1-v^2\right|=\ln x-\ln c \\\\ & \Rightarrow \ln \left(1-v^2\right) x=\ln c \quad(c=\text { constant }) \\\\ & \Rightarrow \left(1-y^2 / x^2\right) \cdot x=c \\\\ & \Rightarrow \left(x^2-y^2\right)=c x \end{aligned} $

Put $y(2)=0 \Rightarrow c=2$

$ \begin{array}{rlrl} & \therefore x^2-y^2 =2 x \\\\ & \therefore y^2 =x^2-2 x \\\\ & \Rightarrow y(x) = \pm \sqrt{x^2-2 x} \end{array} $

Put $x=8$, we get

$ \Rightarrow y(8)= \pm \sqrt{64-16}= \pm \sqrt{48}= \pm 4 \sqrt{3} $

Given, length of square sheet of side is $30 \mathrm{~cm}$.



Let, side of small square be $x \mathrm{~cm}$.

Since, a square piece of tin is to be made into a box without top by cutting square from each corner and folding up the flaps to from a box. Then, this shape formed a cuboidal shape.

Thus, volume of the box $(V)$

$ \begin{aligned} & =(30-2 x)(30-2 x) x \\\\ &\Rightarrow V =x(30-2 x)^2 \end{aligned} $

On differentiating both side with respect to $x$, we get

$ \begin{aligned} \frac{d V}{d x} & =(30-2 x)^2+x\{2(30-2 x) \cdot(-2)\} \\\\ & =(30-2 x)^2-4 x(30-2 x) \\\\ & =(30-2 x)\{30-6 x\} \end{aligned} $

Therefore, maximum of $V, \frac{d V}{d x}=0$

$ \begin{array}{ll} &\Rightarrow (30-2 x)(30-6 x)=0 \\\\ &\Rightarrow x=15 \text { or } x=5 \quad[x=15 \text { (not taken) }] \end{array} $

Thus, $x=5$

So, the surface area

$ \begin{array}{ll} =4 \times(30-2 x) \times x+(30-2 x)^2 & \\\\ =4 \times(30-10) \times 5+(30-10)^2 & (\text { Put } x=5) \\\\ =400+400=800 & \end{array} $

The sum of the absolute maximum and minimum values of the function $f(x) = \left|x^2 - 5x + 6\right| - 3x + 2$ in the interval $[-1, 3]$ can be found by finding the maximum and minimum values of $f(x)$ in this interval and then adding them.

First, let's find the critical points of $f(x)$. To do this, we will find the zeros of the expression inside the absolute value:

$x^2 - 5x + 6 = 0$

Solving this quadratic equation, we find that the zeros are $x = 2$ and $x = 3$. These are the critical points of $f(x)$.

Next, we evaluate $f(x)$ at these critical points and at the endpoints of the interval $[-1, 3]$:

$f(-1) = \left|(-1)^2 - 5(-1) + 6\right| - 3(-1) + 2 = 17$

$f(2) = \left|(2)^2 - 5(2) + 6\right| - 3(2) + 2 = -4$

$f(3) = \left|(3)^2 - 5(3) + 6\right| - 3(3) + 2 = -7$

So, the minimum value of $f(x)$ is $-7$ and the maximum value is $17$.

So, the sum of the absolute maximum and minimum values of the function is $-7 + 17 = 10$.

Note : The absolute maximum and minimum values of a function are the largest and smallest values that the function takes on a given interval, respectively. These values are also called the "extrema" of the function. The absolute maximum value is the highest point on the graph of the function, and the absolute minimum value is the lowest point.

$ \ell_{1}+\ell_{2}=20 \Rightarrow \frac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=-1$

$\mathrm{A}_{1}=\left(\frac{\ell_{1}}{4}\right)^{2}$ and $\mathrm{A}_{2}=\pi\left(\frac{\ell_{2}}{2 \pi}\right)^{2}$

Let $\mathrm{S}=2 \mathrm{~A}_{1}+3 \mathrm{~A}_{2}=\frac{\ell_{1}^{2}}{8}+\frac{3 \ell_{2}^{2}}{4 \pi}$

$\frac{\mathrm{ds}}{\mathrm{d} \ell_1}=0 \Rightarrow \frac{2 \ell_{1}}{8}+\frac{6 \ell_{2}}{4 \pi} \cdot \frac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=0$

$\Rightarrow \frac{\ell_{1}}{4}=\frac{6 \ell_{2}}{4 \pi} $

$\Rightarrow \frac{\pi \ell_{1}}{\ell_{2}}=6$

$f'(x)=x^2+2b+ax$

$g'(x)=x^2+a+2bx$

$\Rightarrow x=1$ is common root

$a+2b+1=0$

$y'=270x^4-540x^3-210x^2+360x+210$

Slope of normal $=-\frac{1}{90}$

$\therefore$ Slope of tangent = 90

$\therefore$ Number of normal will be number of solutions of

$270x^4-540x^3-210x^2+360x+210=90$

$\Rightarrow 9x^4-18x^3-7x^2+12x+4=0$

$\therefore x=1,2,-\frac{1}{3},-\frac{2}{3}$ are roots

$f^{\prime}(x)=6 x^{2}+2 x(2 p-7)+3(2 p-9)$

$x_{1}<0, x_{2}>0$

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

$\Rightarrow p<\frac{9}{2}$

$p \in \left( { - \infty ,{9 \over 2}} \right)$

$ \begin{aligned} & f(x)=8 x^3-36 x+8 \\\\ & =4\left(2 x^3-9 x+2\right) \\\\ & =4(x-2)\left(2 x^2+4 x-1\right) \\\\ & =4(x-2)\left(x-\frac{-2+\sqrt{6}}{2}\right)\left(x-\frac{-2 \sqrt{6}}{2}\right) \end{aligned} $

Local maxima occurs at $x=\frac{-2+\sqrt{6}}{2}=x_0$

$ f\left(x_0\right)=12 \sqrt{6}-\frac{33}{2} $

$g(x)=f(-x)-f(x)$

$ \begin{aligned} & =\frac{1}{1-e^{x}}-\frac{1}{1-e^{-x}} \\\\ & =\frac{1}{1-e^{x}}-\frac{e^{x}}{e^{x}-1} \\\\ & =\frac{1+e^{x}}{1-e^{x}} \\\\ g^{\prime}(x) & =\frac{\left(1-e^{x}\right) e^{x}-\left(1+e^{x}\right)\left(-e^{x}\right)}{\left(1-e^{x}\right)^{2}} \\\\ & =\frac{e^{x}-2 e^{x}+e^{x}+2 e^{x}}{\left(1-e^{x}\right)^{2}}>0 \end{aligned} $

So both statements are correct