Application of Derivatives

$ \begin{aligned} & \text { }\left(\frac{h}{2}\right)^2+r^2=(12)^2 \\ & r=\sqrt{144-\frac{h^2}{4}} \end{aligned} $

Now, volume of cylinder

$ \begin{aligned} V & =\pi r^2 h \\ V & =\pi\left(144-\frac{h^2}{4}\right) \cdot h \\ \frac{d V}{d h} & =\pi\left(144-\frac{3 h^2}{4}\right)=0 \\ \Rightarrow \quad h & =8 \sqrt{3} \\ \frac{d^2 V}{d h^2} & =\pi\left(\frac{-6 h}{4}\right)<0 \end{aligned} $

So, maximum volume at $h=8 \sqrt{3}$ corresponding $r=\sqrt{144-48}=4 \sqrt{6}$

$ V=\pi \times(4 \sqrt{6})^2 \times 8 \sqrt{3}=768 \sqrt{3} \pi $

$ \begin{gathered}\therefore y=x^4-6 x^3+13 x^2-12 x+5 \\ \frac{d y}{d x}=4 x^3-18 x^2+26 x-12 \\ \because \quad 4 x^3-18 x^2+26 x-12=2 \\ \Rightarrow \quad 4 x^3-18 x^2+26 x-14=0 \end{gathered} $

No integer value of $x$ is possible in above equation but in options all are integers.

Since, the point $(1,2)$ lies on the curve $x=t^2-7 t+7, y=t^2-4 t-10$.

$ \begin{aligned}So,\,\, & t^2-7 t+7=1 \\ & t^2-4 t-10=2 \end{aligned} $

On solving Eqs. (i) and (ii), we get $t=6$

$ \begin{aligned} & \begin{array}{rlrl} x & = t^2-7 t+7 \\ \Rightarrow \quad & d x / d t =2 t-7 \\ y & =t^2-4 t-10 \\ \Rightarrow \quad \frac{d y}{d t} & =2 t-4 \\ \text { Now, }\left(\frac{d y}{d x}\right)_{t=6} & =\left(\frac{d y / d t}{d x / d t}\right)_{t=6} \\ & =\left.\frac{2 t-4}{2 t-7}\right|_{t=6}=\frac{2(6)-4}{2(6)-7}=\frac{8}{5} \\ \therefore & m \times \frac{8}{5} =-1 \Rightarrow m=-\frac{5}{8} \end{array} \end{aligned} $

$ \begin{aligned} &\text { Thus, the equation of line } L \text { be }\\ &\begin{aligned} y-2 & =-\frac{5}{8}(x-1) \\ \Rightarrow \quad 5 x+8 y-21 & =0 \\ \therefore a=5, b=8 \text { and } c & =-21 \\ \text { Now, } m(a+b+c) & =-\frac{5}{8}(5+8-21) \\ & =-\frac{5}{8}(-8)=5 \end{aligned} \end{aligned} $

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

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

For maxima or minima,

$ \begin{aligned} & & f^{\prime}(x) & =0 \\ & \Rightarrow & e^{-x}(1-x) & =0 \\ & & x & =1 \quad\left[\text { since } e^{-x} \neq 0\right] \\ & & f^{\prime}(x) & =(x-2) e^{-x} \\\Rightarrow & & f^{\prime}(1) & <0 \end{aligned} $

So, $f(x)$ ha zs a maximum value at $x=1$

Then, $x=\alpha=1$ and $\beta=\frac{1}{e}$

Hence, $(\alpha, \beta)=\left(1, \frac{1}{e}\right)$

Given, diameter of the sphere

$ =42 \mathrm{~cm} $

Then, radius $r=21 \mathrm{~cm}$

and error in measuring the diameter $=\frac{1}{77}$

$\therefore $ Error in measuring the radius,

$ \Delta r=\frac{1}{154} \mathrm{~cm} $

Volume of sphere, $V=\frac{4}{3} \pi r^3$

$ V^{\prime}=\frac{4}{3} \times 3 \pi r^2=4 \pi r^2 $

Let $\Delta V$ be error in the volume of sphere.

Now, $V+\Delta V=V+V^{\prime} \Delta r$

$ V-\Delta V=V-V \Delta r $

On subtracting Eq. (ii) from Eq. (i), we get

$ \begin{aligned} 2 \Delta V & =2 V^{\prime} \Delta r \\ \Delta V & =4 \pi r^2 \Delta r=4 \times \pi(21)^2 \times \frac{1}{154} \\ & =\frac{4 \times 22}{7} \times 21 \times 21 \times \frac{1}{154} \\ \Delta V & =36 \mathrm{~cm}^3 \end{aligned} $

Given, $x^2 y-x^3=8$ and $y^3-x y^2=32$

From figure, point of intersection is Y2, 4).

Now, angle between the two curves is angle between their tangents at that point

$ \begin{aligned} & \quad x^2 y-x^3=8 \Rightarrow 2 x y+x^2 \frac{d y}{d x}-3 x^2=0 \\ & \text { At }(2,4) \\ & 2 \times 2 \times 4+2^2 \frac{d y}{d x}-3 \times 2^2=0 \\ & \Rightarrow \quad 16+4 \frac{d y}{d x}-12=0 \\ & \Rightarrow \quad 4+\frac{4 d y}{d x}=0 \\ & \Rightarrow \quad \frac{d y}{d x}=-1=m_1 \text { (let) } \end{aligned} $

and

$ \begin{gathered} y^3-x y^2=32 \\ 3 y^2 \frac{d y}{d x}-x \times 2 y \frac{d y}{d x}-y^2=0 \end{gathered} $

At $(2,4)$

$ 3 \times(4)^2 \frac{d y}{d x}-2 \times 2 \times 4 \frac{d y}{d x}-(4)^2=0 $

$ \begin{array}{ll} \Rightarrow & \frac{3 d y}{d x}-\frac{d y}{d x}-1=0 \\ \Rightarrow & 2 \frac{d y}{d x}-1=0 \\ \frac{d y}{d x}=\frac{1}{2}=m_2 \text { (let) } \end{array} $

$ \text { Now, } \tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\frac{\frac{1}{2}+1}{1-\frac{1}{2}}=\frac{3 / 2}{1 / 2}=3 $

Given, $f(x)=x^3-2 x^2+x-3$

$ f^{\prime}(x)=3 x^2-4 x+1 $

For absolute maxima or minima put $f^{\prime}(x)=0$

$ \begin{array}{lrl} \because & 3 x^2-4 x+1 & =0 \\ \Rightarrow & 3 x^2-3 x-x+1 & =0 \\ & & (3 x-1)(x-1) =0 \\ \Rightarrow & x=\frac{1}{3} \text { or } x & =1 \end{array} $

Clearly, the critical point of $f(x)$ in $[0,2]$ is $x=\frac{1}{3}$ and 1 .

Now, $f(0)=-3, f\left(\frac{1}{3}\right)=-\frac{77}{27} \approx-2.851$

$ f(1)=-3, f(2)=-1 $

∴ Absolute minimum value i.e. $m=-3$ and absolute maximum value i.e. $M=-1$

Now, $M+m=-1-3=-4$

According to the question,

$ d y / d x=3 x^2-5 $

On integrating both sides, we get

$ y=\frac{3 x^3}{3}-5 x+C \Rightarrow y=x^3-5 x+C $

At $(1,2)$,

$ \begin{aligned} & & 2 & =(1)^3-5 \times 1+C \Rightarrow 2=1-5+C \\ & \Rightarrow & 2+4 & =C \Rightarrow C=6 \\ & \therefore & y & =x^3-5 x+6 \end{aligned} $

We know that, intersection satisfied the equation.

Now, from options $(-2,8)$ satisfied the curve $y=x^3-5 x+6$

                                       Let $y=\sqrt{x}$

$ \begin{aligned} \begin{aligned} \frac{d y}{d x} & =\frac{1}{2 \sqrt{x}} \Rightarrow x=1936 \\ \Delta x & =87 \\ \text { Then, } \Delta y & =\sqrt{x+\Delta x}-\sqrt{x} \\ \Delta y & =\sqrt{2023}-\sqrt{1936} \\ \Rightarrow \quad \sqrt{2023}=44+\Delta y & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} \end{aligned} $

Approximate value of $\sqrt{2023}$ is

$ \begin{aligned} &\begin{aligned} \Delta y & =\frac{d y}{d x} \cdot \Delta x=\frac{1}{2 \sqrt{x}} \times \Delta x \\ & =\frac{1}{2 \sqrt{1936}} \times 87 \\ \Delta y & =\frac{87}{88} \approx 0.9778 \end{aligned}\\ &\text { From Eq. (i), we get }\\ &\sqrt{2023}=44+0.9778=44.9778 \end{aligned} $

$ \begin{aligned} &\text { Given, curve }\\ &\begin{aligned} & \qquad \begin{array}{c} y=x^3-10 x^2+31 x-30 \\ \frac{d y}{d x}=3 x^2-20 x+31 \end{array} \\ & \text { Slope of normal }=\frac{1}{-\frac{d y}{d x}}=-\frac{1}{14} \end{aligned} \end{aligned} $

$ \begin{array}{lr} \Rightarrow & \frac{d y}{d x}=14 \\ \Rightarrow & 3 x^2-20 x+31=14 \\ \Rightarrow & 3 x^2-20 x+17=0 \\ \Rightarrow & 3 x^2-3 x-17 x+17=0 \\ \Rightarrow & (x-1)(3 x-17)=0 \\ \Rightarrow & x=1, x=17 / 3 \end{array} $

Coordinates are integers

$ \begin{array}{ll} \therefore & x=1 \\ \Rightarrow & y=1-10+31-30=-8 \\ P(1,-8) & \end{array} $

Slope of tangent at $P=\left(\frac{d y}{d x}\right)_{(1,-8)}$

$ =3-20+31=14 $

Equation of tangent

$ \begin{aligned} & \begin{aligned} y+8 & =14(x-1) \\ 14 x-y-22 & =0 \Rightarrow \frac{x}{11 / 7}+\frac{y}{22}=1 \end{aligned} \\ & \text { intercept }=11 / 7 \end{aligned} $

Let $z=x^2 y^3$ and $x=\frac{50-3 y}{2}$

$ \begin{aligned} \therefore z & =\left(\frac{50-3 y}{2}\right)^2 y^3 \\ z & =\frac{2500 y^3+9 y^5-300 y^4}{4} \end{aligned} $

Now, $d z / d y=0$

$ \begin{array}{lr} \Rightarrow & \left(7500+45 y^2-1200 y\right) y^2=0 \\ \because & y>0 \\ \Rightarrow & 3 y^2-80 y+500=0 \\ \Rightarrow & (y-10)(3 y-50)=0 \end{array} $

$y$ is are integer.

$ \begin{array}{ll} \Rightarrow & y=10 \\ \therefore & x=10 \end{array} $

$x^2 y^3$ is maximum at $x=\alpha=10$

and   $ y=\beta=10 $

$ \therefore \quad \frac{\alpha}{2}+\frac{\beta}{5}=\frac{10}{2}+\frac{10}{5}=7 $

Given the function $ f(x) = \frac{1-x+x^2}{1+x+x^2} $, we want to find its minimum value for all real values of $ x $.

First, differentiate $ f(x) $ with respect to $ x $:

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

Simplify the expression:

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

To find the critical points, set $ f^{\prime}(x) = 0 $:

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

Evaluate $ f(x) $ at $ x = 1 $ and $ x = -1 $:

At $ x = 1 $:

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

At $ x = -1 $:

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

Hence, the minimum value of $ f(x) $ is $ \frac{1}{3} $.

Approximate error in current

$ d I=f^{\prime}(I) \times \Delta I=\frac{\pi}{4} \times 0.01 $

$\therefore$ Percentage error in current

$ =\left(\frac{\pi}{4} \times 0.01\right) \times 100=\frac{\pi}{4} $

Given that, $y=\cos (x+y)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Differentiating on both side w.r.t. $x$, we get

$ \frac{d y}{d x}=-\sin (x+y)\left(1+\frac{d y}{d x}\right) $

Equation of tangent is $x+2 y=k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$\Rightarrow$ Slope is $-\frac{1}{2} \Rightarrow \frac{d y}{d x}=-\frac{1}{2}$

Hence, $-\frac{1}{2}=-\sin (x+y)\left(1-\frac{1}{2}\right)$

$\Rightarrow \quad-\frac{1}{2}=-\frac{1}{2} \sin (x+y)$

$\Rightarrow \sin (x+y)=1=\sin \frac{\pi}{2}$

$\Rightarrow \quad x+y=\frac{\pi}{2} \Rightarrow y=\frac{\pi}{2}-x$

From Eq. (i), we get

$ \begin{aligned} \quad \frac{\pi}{2}-x & =\cos \left(x+\frac{\pi}{2}-x\right) \\ \Rightarrow \frac{\pi}{2}-x & =\cos \left(\frac{\pi}{2}\right)=0 \\ \Rightarrow \quad x & =\frac{\pi}{2} \Rightarrow y=0 \end{aligned} $

Now, from Eq. (ii), we get

$ k=\frac{\pi}{2}+0 \cdot 2=\frac{\pi}{2} $

Given that,

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

Differentiating on both sides w.r.t. $x$, we get

$ f^{\prime}(x)=\frac{\left(e^{2 x}+2\right) e^x-2 e^{2 x} e^x}{\left(e^{2 x}+2\right)^2} $

Now, $f^{\prime}(x)=0 \Rightarrow \frac{\left(e^{2 x}+2\right) e^x-2 e^{3 x}}{\left(e^{2 x}+2\right)^2}=0$

$ \begin{aligned} & \Rightarrow \quad 2 e^x-e^{3 x}=0 \\ & \Rightarrow \quad e^x\left(e^{2 x}-2\right)=0 \end{aligned} $

Since, $e^x \neq 0 e^{2 x}=2 \Rightarrow e^x=\sqrt{2}$

For maximum

$ f(x)=\frac{\sqrt{2}}{2+2}=\frac{\sqrt{2}}{4}=\frac{1}{2 \sqrt{2}} $

Thus, $0

$\Rightarrow C \in R$ such that, $f(c)=\frac{1}{3}$

So, Assertion (A) is true.

and Reason ( $R$ ) is true.

and Reason (R) is the correct explanation of Assertion (A).

Given the expression $ f(x) = x^3 + 3x^2 - 9x + \lambda $, we need to find the values of $\lambda$ such that $ f(x) $ can be expressed as $(x-\alpha)^2(x-\beta)$.

Since $(x-\alpha)$ is a repeated factor, it must also be a factor of the derivative $ f'(x) $. We compute the derivative:

$ f'(x) = 3x^2 + 6x - 9 $

Factor the derivative:

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

This shows that the potential values of $\alpha$ are the roots of $ f'(x) = 0$, which are $\alpha = 1$ and $\alpha = -3$.

Case 1: $\alpha = 1$

Substitute $\alpha = 1$ into the original expression to find $\lambda$:

$ f(1) = 1^3 + 3(1)^2 - 9(1) + \lambda = 0 $

$ 1 + 3 - 9 + \lambda = 0 $

$ \lambda = 5 $

Case 2: $\alpha = -3$

Substitute $\alpha = -3$ into the original expression to find $\lambda$:

$ f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) + \lambda = 0 $

$ -27 + 27 + 27 + \lambda = 0 $

$ \lambda = -27 $

Hence, the values of $\lambda$ can be $5$ or $-27$.

The given parametric equations for the curve are $x = 2 \sin t$ and $y = 2 \cos t$.

To find the equation of the normal line at $t = \frac{\pi}{2}$, we first need to determine the slope of the tangent line. Using the derivatives:

$ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{-2 \sin t}{2 \cos t} $

This expression represents the slope of the tangent line to the curve.

Next, the slope of the normal line, which is perpendicular to the tangent line, is given by:

$ \text{Slope of normal} = -\frac{1}{\frac{-2 \sin t}{2 \cos t}} = \frac{\cos t}{\sin t} $

Substituting $t = \frac{\pi}{2}$:

$ m_s = \frac{\cos \frac{\pi}{2}}{\sin \frac{\pi}{2}} = \frac{0}{1} = 0 $

Thus, the slope of the normal line at $t = \frac{\pi}{2}$ is 0, which indicates a horizontal line.

The point on the curve at $t = \frac{\pi}{2}$ is:

$ x = 2 \sin \frac{\pi}{2} = 2, \quad y = 2 \cos \frac{\pi}{2} = 0 $

The equation of a horizontal line through this point, where the slope ($m = 0$), is:

$ (y - 2 \cos t) = 0(x - 2 \sin t) $

Simplifying this at $t = \frac{\pi}{2}$:

$ \left(y - 2 \times \cos \frac{\pi}{2}\right) = 0 \quad \Rightarrow \quad y = 0 $

Thus, the equation of the normal line at $t = \frac{\pi}{2}$ is $y = 0$.

To find the relative maximum of the function $ f(x) = \frac{x}{5} + \frac{5}{x} $, where $ x \neq 0 $, follow these steps:

Derive the Function:

Compute the first derivative:

$ f'(x) = \frac{1}{5} - \frac{5}{x^2} $

Set the derivative equal to zero for critical points:

$ \frac{1}{5} - \frac{5}{x^2} = 0 \quad \Rightarrow \quad \frac{5}{x^2} = \frac{1}{5} $

Solve for $ x $:

$ x^2 = 25 \quad \Rightarrow \quad x = \pm 5 $

Second Derivative Test:

Compute the second derivative:

$ f''(x) = \frac{10}{x^3} $

Evaluate the second derivative at $ x = 5 $:

$ f''(5) = \frac{10}{125} = \frac{2}{25} > 0 $

Since $ f''(5) > 0 $, $ x = 5 $ is a point of relative minimum.

Evaluate the second derivative at $ x = -5 $:

$ f''(-5) = \frac{-10}{125} = -\frac{2}{25} < 0 $

Since $ f''(-5) < 0 $, $ x = -5 $ is a point of relative maximum.

Calculate the Expression:

Given that $ \alpha = -5 $, calculate:

$ \sqrt{\alpha^2 + 2\alpha - 6} = \sqrt{(-5)^2 + 2(-5) - 6} $

Simplify:

$ = \sqrt{25 - 10 - 6} = \sqrt{9} = 3 $

Therefore, $\sqrt{\alpha^2 + 2\alpha - 6} = 3$.

$f(x) = {3^{{{({x^2} - 2)}^3} + 4}},\,x \in R$

$f(x) = {81.3^{{{({x^2} - 2)}^3}}}$

$f'(x) = {81.3^{{{({x^2} - 2)}^3}}}\ln 2.3({x^2} - 2)2x$

$ = (486\ln 2)\left( {{3^{{{({x^2} - 2)}^3}}}({x^2} - 2)x} \right)$

$ \Rightarrow x = 0$ is the local minima.

$f''(x) = (486\ln 2)\left( {{3^{{{({x^2} - 2)}^3}}}\,.\,({x^2} - 2)(5{x^2} - 2 + 6{x^2}\ln 3({x^2} - 2))} \right)$

$f''(x) = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \sqrt 2 $

$f''\left( {{{\sqrt 2 }^ + }} \right) > 0$

$f''\left( {{{\sqrt 2 }^ - }} \right) < 0$

$ \Rightarrow x = \sqrt 2 $ is point of inflection

$f''(x) > 0\,\forall x > \sqrt 2 $

$ \Rightarrow f'(x)$ is increasing for $x > \sqrt 2 $

$f(x) = x{e^{x(1 - x)}},\,x \in R$

$f'(x) = x{e^{x(1 - x)}}\,.\,(1 - 2x) + {e^{x(1 - x)}}$

$ = {e^{x(1 - x)}}[x - 2{x^2} + 1]$

$ = - {e^{x(1 - x)}}[2{x^2} - x - 1]$

$ = - {e^{x(1 - x)}}(2x + 1)(x - 1)$

$\therefore$ $f(x)$ is increasing in $\left( { - {1 \over 2},1} \right)$ and decreasing in $\left( { - \infty ,\, - {1 \over 2}} \right) \cup \left( {1,\infty } \right)$

$f(x) = {{5{x^2}} \over 2} + {\alpha \over {{x^5}}}\,\,\{ x > 0\} $

$f'(x) = 5x - {{5\alpha } \over {{x^6}}} = 0$

$ \Rightarrow x = {(\alpha )^{{1 \over 7}}}$

$f{(x)_{\min }} = {{5{{(\alpha )}^{{2 \over 7}}}} \over 2} + {\alpha \over {{\alpha ^{{5 \over 7}}}}} = 14$

${5 \over 2}{\alpha ^{{2 \over 7}}} + {\alpha ^{{2 \over 7}}} = 14$

${7 \over 2}{\alpha ^{{2 \over 7}}} = 14$

$\alpha = 128$

Equation of line passing through the point (50 + $\alpha$, 0) and (0, 50 + $\alpha$) is

$y - 0 = {{50 + \alpha - 0} \over {0 - (50 + \alpha )}}\left( {x - (50 + \alpha )} \right)$

$ \Rightarrow y = - 1\left( {x - (50 + \alpha )} \right)$

$ \Rightarrow y = (50 + \alpha ) - x$

$ \Rightarrow x + y = 50 + \alpha $

Let $p = x{y^4}$

$ = (50 + \alpha - y){y^4}$

$ = (50 + \alpha ){y^4} - {y^5}$

For maximum or minimum value of p,

${{dp} \over {dy}} = 0$

$ \Rightarrow 4{y^3}(50 + \alpha ) - 5{y^4} = 0$

$ \Rightarrow {y^3}[200 + 4\alpha - 5y] = 0$

$\therefore$ $y = 0$

or

$200 + 4\alpha - 5y = 0$

$ \Rightarrow y = {4 \over 5}(50 + \alpha )$

$\therefore$ $x = 50 + \alpha - y$

$ = 50 + \alpha - {4 \over 5}(50 + \alpha )$

$ = 50 + \alpha \left( {1 - {4 \over 5}} \right)$

$ = {1 \over 5}(50 + \alpha )$

$ \Rightarrow 4x = {4 \over 5}(50 + \alpha ) = y$

$ \Rightarrow y = 4x$

$\therefore$ (x, y) lies on the line y = 4x

Given,

$f(x) = 4{x^3} - 11{x^2} + 8x - 5$

$\therefore$ $f'(x) = 12{x^2} - 22x + 8$

$ = 2(6{x^2} - 11x + 4)$

$ = 2(6{x^2} - 8x - 3x + 4)$

$ = 2\left[ {2x(3x - 4) - 1(3x - 4)} \right]$

$ = 2\left[ {(3x - 4)(2x - 1)} \right]$

$f'(x)$ is positive before ${1 \over 2}$ means slope of f(x) is positive before ${1 \over 2}$ and f'(x) is negative after ${1 \over 2}$ means slope of f(x) is negative after ${1 \over 2}$.

$\therefore$ At ${1 \over 2}$, f(x) is local maximum

Also, $f'(x)$ is negative before ${4 \over 3}$ means slope of f(x) is negative before ${4 \over 3}$ and f'(x) is positive after ${4 \over 3}$ means slope of f(x) is positive after ${4 \over 3}$.

$\therefore$ At ${4 \over 3},\,f(x)$ is local minimum

From wavy curve method,

$f'(x) > 0,\,\forall x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)$

$\therefore$ f(x) increasing in $x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)$

$f'(x) < 0\,\forall x \in \left( {{1 \over 2},{4 \over 3}} \right)$

$\therefore$ f(x) decreasing in $x \in \left( {{1 \over 2},{4 \over 3}} \right)$

Given,

$f(x) = {( - 3)^{{n_1}}}{(x - 5)^{{n_2}}}$

Differentiating both side with respect to x, we get

$f'(x) = {(x - 5)^{{n_2}}}\left( {{n_1}{{(x - 3)}^{{n_1} - 1}}} \right) + {(x - 3)^{{n_1}}}\left( {{n_2}{{(x - 5)}^{{n_2} - 1}}} \right)$

$ = {(x - 3)^{{n_1}}}{(x - 5)^{{n_2}}}\left[ {{{{n_1}} \over {x - 3}} + {{{n_2}} \over {x - 5}}} \right]$

$ = {(x - 3)^{{n_1}}}{(x - 5)^{{n_2}}}\left[ {{{{n_1}(x - 5) + {n_2}(x - 3)} \over {(x - 3)(x - 5)}}} \right]$

$ = {(x - 3)^{{n_1} - 1}}{(x - 5)^{{n_2} - 1}}[{n_1}(x - 5) + {n_2}(x - 3)]$

Option A :

When n₁ = 3 and n₂ = 4 then

$f'(x) = {(x - 3)^{3 - 1}}{(x - 5)^{4 - 1}}\left[ {3(x - 5) + 4(x - 3)} \right]$

$ = {(x - 3)^2}{(x - 5)^3}(7x - 27)$

Critical point is at x = 3, 5 and ${{27} \over 7}$

When value of f'(x) is less than ${{27} \over 5}$, f'(x) is positive means slope of f(x) graph is positive.

And when value of f'(x) is greater than ${{27} \over 7}$ but less than 5 then f'(x) is negative means slope of f(x) is negative.

So at ${{27} \over 7}$ between 3 and 5, f(x) attain local maxima.

$\therefore$ Option A is correct.

Option B :

When n₁ = 4 and n₂ = 3 then

$f'(x) = {(x - 3)^{4 - 1}}{(x - 5)^{3 - 1}}\left[ {4(x - 5) + 3(x - 3)} \right]$

$ = {(x - 3)^3}{(x - 5)^2}(7x - 29)$

Critical point is at x = 3, 5 and ${{29} \over 7}$

When f'(x) is less than ${{29} \over 7}$ but greater than 3 then f'(x) is negative means slope of graph of f(x) is negative.

And when f'(x) is greater than ${{29} \over 7}$ but less than 5 then f'(x) is positive means slope of graph of f(x) is positive.

So at ${{29} \over 7}$ between 3 and 5, f(x) attain local minima.

$\therefore$ Option (B) is correct.

Option C :

When n₁ = 3 and n₂ = 5 then

$f'(x) = {(x - 3)^{3 - 1}}{(x - 5)^{5 - 1}}\left[ {3(x - 5) + 5(x - 3)} \right]$

$ = {(x - 3)^2}{(x - 5)^4}(8x - 30)$

Critical point is at x = 3, 5 and ${{30} \over 8}$

When f'(x) is less than ${{30} \over 8}$ but greater than 3 then f'(x) is negative means slope of graph of f(x) is negative.

And when f'(x) is greater than ${{30} \over 8}$ but less than 5 then f'(x) is positive means slope of graph of f(x) is positive.

So, at ${{30} \over 8}$ between 3 and 5, f(x) attain local minima.

$\therefore$ Option C is not true.

Similarly you can check option D also.

$4a + 3b = 22$

Total area $ = A = {a^2} + {{\sqrt 3 } \over 4}{b^2}$

$A = {\left( {{{22 - 3b} \over 4}} \right)^2} + {{\sqrt 3 } \over 4}{b^2}$

${{dA} \over {dB}} = 2\left( {{{22 - 3b} \over 4}} \right)\left( {{{ - 3} \over 4}} \right) + {{\sqrt 3 } \over 4}\,.\,2b = 0$

$ \Rightarrow {{\sqrt {3b} } \over 2} = {3 \over 8}(22 - 3b)$

$ \Rightarrow 4\sqrt 3 b = 66 - 9b$

$ \therefore b = {{66} \over {9 + 4\sqrt 3 }}$

$\because$ ${s_1} + {s_2} = k$

$76{x^2} + 3\pi {r^2} = k$

$\therefore$ $152x{{dx} \over {dr}} + 6\pi r = 0$

$\therefore$ ${{dx} \over {dr}} = {{ - 6\pi r} \over {152x}}$

Now $V = 40{x^3} + {2 \over 3}\pi {r^3}$

$\therefore$ ${{dv} \over {dr}} = 120{x^2}\,.\,{{dx} \over {dr}} + 2\pi {r^2} = 0$

$ \Rightarrow 120{x^2}\,.\,\left( {{{ - 6\pi } \over {152}}{r \over x}} \right) + 2\pi {r^2} = 0$

$ \Rightarrow 120\left( {{x \over r}} \right) = 2\pi \left( {{{152} \over {6\pi }}} \right)$

$ \Rightarrow \left( {{x \over r}} \right) = {{152} \over 3}{1 \over {120}} = {{19} \over {45}}$

${\left( {{x \over a}} \right)^n} + {\left( {{y \over b}} \right)^n} = 2$

Differentiating both sides with respect to x, we get

$ \Rightarrow n\,.\,{\left( {{x \over a}} \right)^{n - a}}\,.\,{1 \over a} + n\,.\,{\left( {{y \over b}} \right)^{n - 1}}\,.\,{1 \over b}\,.\,{{dy} \over {dx}} = 0$

$ \Rightarrow {{dy} \over {dx}} = - {{{1 \over a}\,.\,{{\left( {{x \over a}} \right)}^{n - 1}}} \over {{1 \over b}{{\left( {{y \over b}} \right)}^{n - 1}}}}$

$ = - {b \over a}{\left( {{{xb} \over {ya}}} \right)^{n - 1}}$

Now, ${{dy} \over {dx}}$ at (a, b) $ = - {b \over a}{\left[ {{{ab} \over {ba}}} \right]^{n - 1}} = - {b \over a}$

Equation of tangent at (a, b) is,

$(y - b) = - {b \over a}(x - a)$

$ \Rightarrow {{y - b} \over b} = - {{x - a} \over a}$

$ \Rightarrow {y \over b} - 1 = - {x \over a} + 1$

$ \Rightarrow {x \over a} + {y \over b} = 2$

$\therefore$ It is tangent for all value of n.

$f(x) = 2{\cos ^{ - 1}}x + 4{\cot ^{ - 1}}x - 3{x^2} - 2x + 10\,\forall x \in [ - 1,1]$

$ \Rightarrow f'(x) = - {2 \over {\sqrt {1 - {x^2}} }} - {4 \over {1 + {x^2}}} - 6x - 2 < 0\,\forall x \in [ - 1,1]$

So f(x) is decreasing function and range of f(x) is [f(1), f($-$1)], which is [$\pi$ + 5, 5$\pi$ + 9]

Now $4a - b = 4(\pi + 5) - (5\pi + 9)$

$ = 11 - \pi $

$\because$ $V = {1 \over 3}\pi {r^2}h$ and ${r \over h} = {7 \over {35}} = {1 \over 5}$

$ \Rightarrow V = {1 \over {75}}\pi {h^3}$

${{dV} \over {dt}} = {1 \over {25}}\pi {h^2}{{dh} \over {dt}} = 1$

$ \Rightarrow {{dh} \over {dt}} = {{25} \over {\pi {h^2}}}$

Now, $S = \pi rl = \pi \left( {{h \over 5}} \right)\sqrt {{h^2} + {{{h^2}} \over {25}}} = {\pi \over {25}}\sqrt {26} {h^2}$

$ \Rightarrow {{dS} \over {dt}} = {{2\sqrt {26} \pi h} \over {25}}\,.\,{{dh} \over {dt}} = {{2\sqrt {26} } \over h}$

$ \Rightarrow {{dS} \over {d{t_{(h = 10)}}}} = {{\sqrt {26} } \over 5}$

$\because$ ${{dy} \over {dx}} = {{24(1 + \sin t)\cos t} \over {12(1 + \cos 2t)}} = {{1 + \sin t} \over {\cos t}} = \tan \left( {{\pi \over 4} + {t \over 2}} \right)$

$\because$ ${{dy} \over {d{x_{({x_0},{y_0})}}}} = \sqrt 3 = \tan \left( {{\pi \over 4} + {t \over 2}} \right)$

$ \Rightarrow t = {\pi \over 6}$

So, ${y_{0\left( {at\,t = {\pi \over 6}} \right)}} = 12{\left( {1 + \sin {\pi \over 6}} \right)^2} = 27$

$\because$ $ - {{dx} \over {dy}} = {{{x^2}} \over {xy - {x^2}{y^2} - 1}}$

$\therefore$ ${{dy} \over {dx}} = {{{x^2}{y^2} - xy + 1} \over {{x^2}}}$

Let $xy = v \Rightarrow y + x{{dy} \over {dx}} = {{dv} \over {dx}}$

$\therefore$ ${{dv} \over {dx}} - y = {{({v^2} - v + 1)y} \over v}$

$\therefore$ ${{dv} \over {dx}} = {{{v^2} + 1} \over x}$

$\because$ $y(1) = 1 \Rightarrow {\tan ^{ - 1}}(xy) = \ln x + {\tan ^{ - 1}}(1)$

Put $x = e$ and $y = y(e)$ we get

${\tan ^{ - 1}}(e\,.\,y(e)) = 1 + {\tan ^{ - 1}}1$

${\tan ^{ - 1}}(e\,.\,y(e)) - {\tan ^{ - 1}}1 = 1$

$\therefore$ $e(y(e)) = {{1 + \tan (1)} \over {1 - \tan (1)}}$

$\because$ ${f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36\lambda + 48$

$\therefore$ $f{'_\lambda }(x) = 12(\lambda {x^2} - 6\lambda x + 3)$

For ${f_\lambda }(x)$ increasing : ${(6\lambda )^2} - 12\lambda \le 0$

$\therefore$ $\lambda \in \left[ {0,\,{1 \over 3}} \right]$

$\therefore$ $\lambda^ * = {1 \over 3}$

Now, $f_\lambda ^*(x) = {4 \over 3}{x^3} - 12{x^2} + 36x + 48$

$\therefore$ $f_\lambda ^*(1) + f_\lambda ^*( - 1) = 73{1 \over 2} - 1{1 \over 2} = 72$

We know,

Surface area of balloon (s) = 4$\pi$r²

$\therefore$ ${{ds} \over {dt}} = {d \over {dt}}(4\pi {r^2})$

$ \Rightarrow {{ds} \over {dt}} = 4\pi (2r) \times {{dr} \over {dt}}$

$ \Rightarrow {{ds} \over {dt}} = 8\pi r \times {{dr} \over {dt}}$

Given that, surface area of balloon is increasing in constant rate.

$\therefore$ ${{ds} \over {dt}}$ = constant = k (Assume)

$ \Rightarrow k = 8\pi r\,.\,{{dr} \over {dt}}$

$ \Rightarrow \int {k\,dt = \int {8\pi r\,dr} } $

$ \Rightarrow kt = 8\pi \times {{{r^2}} \over 2} + c$

$ \Rightarrow kt = 4\pi {r^2} + c$ ..... (1)

Given at t = 0, radius r = 3

So, $0 = 4\pi ({3^2}) + c$

$ \Rightarrow c = - 36\pi $

$\therefore$ Equation (1) becomes

$kt = 4\pi {r^2} - 36\pi $

Also given, at t = 5, radius r = 7

$\therefore$ $k(5) = 4\pi {(7)^2} - 36$

$ \Rightarrow k = 32\pi $

$\therefore$ Equation (1) is

$32\pi t = 4\pi {r^2} - 36\pi $

Now at $t = 9$

$ \Rightarrow 32\pi (9) = 4\pi {r^2} - 36\pi $

$ \Rightarrow 8 \times 9 = {r^2} - 9$

$ \Rightarrow {r^2} = 81 \Rightarrow r = 9$

Given curve,

$y = {x^3} + 3{x^2} + 5$

Slope of tangent,

${{dy} \over {dx}} = 3{x^2} + 6x$

Slope of line joined by (x₁, y₁) and (0, 0) is $ = {{{y_0} - 0} \over {{x_1} - 0}}$

$\therefore$ ${{{y_1}} \over {{x_1}}} = 3x_1^2 + 6{x_1}$

$ \Rightarrow {y_1} = 3x_1^3 + 6x_1^2$ ...... (1)

Point (x₁, y₁) is on the line,

$\therefore$ ${y_1} = x_1^3 + 3x_1^2 + 5$ ..... (2)

$\therefore$ $3x_1^3 + 6x_1^2 = x_1^3 + 3x_1^2 + 6$

$ \Rightarrow 2x_1^3 + 3x_1^2 - 5 = 0$

x = 1 satisfy the equation

Now, ${y_1} = x_1^3 + 3x_1^2 + 5$

$ = 1 + 3 + 5$

$ = 9$

$\therefore$ (x₁, y₁) = (1, 9)

Option D does not satisfy point (1, 9).