Application of Derivatives

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } f(x)=\frac{x}{\sin x} \\ & \therefore f^{\prime}(x)=\frac{\sin x-x \cos x}{\sin ^2 x}=\frac{g(x)}{\sin ^2 x} \text {, where } g(x) \\ & =\sin x-x \cos x \\ & \text { Now, } g(x)=\sin x-x \cos x \\ & g^{\prime}(x)=\cos x-\cos x+x \sin x=x \sin x \\ & \Rightarrow g^{\prime}(x)>0 \text { for all } x \in\left(0, \frac{\pi}{2}\right) \\ & \qquad\left[\because \sin x>0 \text { for } x \in\left(0, \frac{\pi}{2}\right)\right] \end{aligned} \end{aligned} $

$\Rightarrow g(x)$ is an increasing function on $\left(0, \frac{\pi}{2}\right)$

$ \begin{aligned} & \Rightarrow g(x)>g(0), \forall x \in\left(0, \frac{\pi}{2}\right) \\ & \Rightarrow g(x)>0, \forall x \in\left(0, \frac{\pi}{2}\right) \\ & \Rightarrow \frac{g(x)}{\sin ^2 x}>0, \forall x \in\left(0, \frac{\pi}{2}\right) \\ & \quad \quad\left[\because \sin ^2 x>0 \text { for } x \in\left(0, \frac{\pi}{2}\right)\right] \\ & \Rightarrow f^{\prime}(x)>0, \forall x \in\left(0, \frac{\pi}{2}\right) \end{aligned} $

Hence, $f(x)$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$.

We have,

$ \begin{aligned} f(x) & =x^3+3(a-7) x^2+3\left(a^2-9\right) x-1 \\ f^{\prime}(x) & =3 x^2+6(a-7) x+3\left(a^2-9\right) \\ & =3\left[x^2+2(a-7) x+\left(a^2-9\right)\right] \end{aligned} $

For local maximum, we must have $f^{\prime}(x)=0$

$ \begin{aligned} & \Rightarrow x^2+2(a-7) x+a^2-9=0 \\ & \Rightarrow x=-(a-7) \pm \sqrt{58-14 a} \end{aligned} $

Let $x_1=-(a-7)+\sqrt{58-14 a}$ and

$ x_2=-(a-7)-\sqrt{58-14 a} $

For $x_1, x_2$ to be real, we must have

$ 58-14 a>0 \Rightarrow a<\frac{58}{14}=\frac{29}{7} $

Now, $f^{\prime \prime}(x)=3[2 x+2(a-7)]=6[x+a-7]$

Clearly, $f^{\prime \prime}\left(x_2\right)=6\left(x_2+a-7\right)=-6 \sqrt{58-14 a}<0$

So, $x_2=-(a-7)-\sqrt{58-14 a}$ is a point of local maximum.

Now, $x_2>0$

$ \begin{aligned} & \Rightarrow-(a-7)-\sqrt{58-14 a}>0 \\ & \Rightarrow 7-a>\sqrt{58-14 a} \Rightarrow(7-a)^2>58-14 a \\ & \Rightarrow a^2-9>0 \Rightarrow a<-3 \text { or } a>3 \\ & \text { Hence, } a \in(-\infty,-3) \cup\left(3, \frac{29}{7}\right) \end{aligned} $

$ \text { The equation of the curve is } 2 y^2=a x^2+b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Differentiate w.r.t. $x$, we get

$ \begin{aligned} & 4 y \frac{d y}{d x}=2 a x \Rightarrow \frac{d y}{d x}=\frac{a x}{2 y} \\ \therefore \quad & \left(\frac{d y}{d x}\right)_{(1,-1)}=\frac{-a}{2} \end{aligned} $

Also, since the slope of the tangent at $(1,-1)$ is -1 .

$ \therefore \quad \frac{-a}{2}=-1 \Rightarrow a=2 $

Since, the point $(1,-1)$ lies on Eq. (i)

$ \begin{array}{ll} \therefore & 2(-1)^2=a(1)^2+b=a+b=2 \\ \Rightarrow & 2+b=2 \Rightarrow b=0 \end{array} $

According to question,

Diameter $ =\boldsymbol{R} $

Diagonals, $ d_{1}=d_{2}=R $

$ \begin{aligned} \text { Max Area of rectangle (any quadrilateral) } & =\frac{1}{2} d_{1} \times d_{2} \\ & =\frac{1}{2} \times R \times R=\frac{R^{2}}{2} \end{aligned} $

Given, $ f(x)=\left\{\begin{array}{cl}\frac{x-1}{x^{2}} ; & x \geq 1 \\ \frac{-x+1}{x^{2}} ; & x < 1\end{array}\right. $

$ f(x)=\left\{\begin{array}{cl} \frac{1}{x}-\frac{1}{x^{2}} ; & x \geq 1 \\ -\frac{1}{x}+\frac{1}{x^{2}} ; & x < 1 \end{array}\right. $

$ f^{\prime}(x)=\left\{\begin{array}{ll} \frac{-1}{x^{2}}+\frac{2}{x^{3}} ; & x \geq 1 \\ \frac{1}{x^{2}}-\frac{2}{x^{3}} ; & x < 1 \end{array}\right. $

$ f^{\prime}(x)=\left\{\begin{array}{ll} \frac{2-x}{x^{3}} ; & x \geq 1 \\ \frac{x-2}{x^{3}} ; & x < 1 \end{array}\right. $

By observation, $ f^{\prime}(x) $ will be

Decreasing in $ (0,1) \cup(2, \infty) $

and increasing in $ (-\infty, 0) \cup(1,2) $

Let $r$ be the radius of the cylinder.

Then, volume $(V)$ of the cylinder is given by,

$\begin{aligned} & V=\pi(\text { radius })^2 \times \text { height } \\ & \Rightarrow V=\pi(10)^2 h \\ & \Rightarrow V=100 \pi h \end{aligned}$

On differentiating w.r.t. $t$, we get

$\begin{aligned} & \frac{d V}{d t}=100 \pi \frac{d h}{d t} \\ & \Rightarrow \quad 200 \pi=100 \pi \frac{d h}{d t} \\ & {\left[\because \text { Given, } \frac{d V}{d t}=200 \pi \mathrm{m}^3 / \mathrm{h}\right]} \\ & \Rightarrow \quad \frac{d h}{d t}=\frac{200 \pi}{100 \pi}=2 \mathrm{~m} / \mathrm{h} \\ \end{aligned}$

Let $r$ be the radius and $h$ be the height of cone.

$\therefore \quad \frac{r}{h}=\frac{7}{35}=\frac{1}{5} \Rightarrow r=\frac{h}{5} \mathrm{~cm}$

Now, volume of cone $=V=\frac{1}{3} \pi r^2 h=\frac{1}{75} \pi h^3$

$\Rightarrow \quad \frac{d \dot{V}}{d t}=\frac{1}{75} \pi\left(3 h^2\right) \frac{d h}{d t}$

$\Rightarrow \quad \frac{d h}{d t}=\frac{1 \times 75}{\pi \times 3 h^2}$

$ \left[\because \text { Given, } \frac{d V}{d t}=1 \mathrm{~cm}^3 / \mathrm{sec}\right] $

$ \begin{aligned} & \text { Now, } A=\pi r l=\pi r \sqrt{r^2+h^2} \\ & \Rightarrow \quad A=\pi r^2 \sqrt{1+\frac{h^2}{r^2}}=\frac{\pi h^2}{25} \sqrt{1+25} \\ & \Rightarrow \quad A=\frac{\sqrt{26}}{25} \pi h^2 \\ & \therefore \frac{d A}{d t}=\frac{\sqrt{26}}{25} \pi \cdot 2 h \frac{d h}{d t} \\ & \left.\Rightarrow \quad \frac{d A}{d t}\right|_{h=10}=\frac{\sqrt{26}}{25} \pi \times 2 \times 10 \times \frac{1 \times 75}{\pi \times 3 \times 10^2} \\ & =\frac{\sqrt{26}}{5} \mathrm{~cm}^2 / \mathrm{s} \\ & \end{aligned}$

Perimeter = 440 ft

$\Rightarrow$ 2x + $\pi$r + $\pi$r = 440

2x + 2$\pi$r = 440

A = Area of the rectangular portion = x2r

A = $x\frac{(440-2x)}{\pi}=\frac{1}{\pi}(440x-2x^2)$

Let $\frac{dA}{dx}=\frac{1}{\pi}(440-4x)=0$

$\Rightarrow$ x = 110 for which $\frac{d^2A}{dx^2} < 0$

$\Rightarrow$ A is maximum when x = 110

$\Rightarrow$ $2r=\frac{440-2x}{\pi}=\frac{440-220}{22/7}=70$

$\therefore$ r = 35 ft and x = 110 ft

V = $\frac{4}{3}\pi r^3\Rightarrow4500\pi=\frac{4\pi r^3}{3}$

$\Rightarrow$ r = 15 m

After 49 min = (4500 $-$ 49.72) $\pi$

= 972 $\pi$m³

$\Rightarrow$ 972$\pi=\frac{4}{3}\pi r^3$

$\Rightarrow$ r³ = 3 $\times$ 243 = 3 $\times$ 3⁵

r = 9

$\frac{dv}{dt}=4 \pi r^2\left(\frac{dr}{dt}\right)$

72$\pi$ = 4$\pi$ $\times$ 9 $\times$ $\left(\frac{dr}{dt}\right)$

$\frac{dr}{dt}=\left(\frac{2}{9}\right)$

We have, ${{dx} \over {dt}} = 2t + 3$ and ${{dy} \over {dt}} = 4t - 2$

$\therefore$ ${{dy} \over {dx}} = {{dy/dt} \over {dx/dt}} = {{4t - 2} \over {2t + 3}}$

Thus, slope of the tangent to the curve at the point t = 2 is

${\left[ {{{dy} \over {dx}}} \right]_{t = 2}} = {{4(2) - 2} \over {2(2) + 3}} = {6 \over 7}$

We have,

$f(x) = {a_0} + {a_1}{x^2} + {a_2}{x^4} + {a_3}{x^6} + ... + {a_n}{x^{2n}}$

$f'(x) = 2{a_1}x + 4{a_2}{x^3} + 6{a_3}{x^5} + .... + 2n{a_n}{x^{2n - 1}}$

$f'(x) = 2x({a_1} + 2{a_2}{x^2} + 3{a_3}{x^4} + .... + n{a_n}{x^{2n - 2}})$

For maxima or minima

Put f'(x) = 0 $\Rightarrow$ x = 0

$f''(x) = 2({a_1} + 6{a_2}{x^2} + 15{a_3}{x^4} + ... + n(2n - 1){a_n}{x^{2n - 2}})$

${(f''(x))_{x = 0}} = 2{a_1} > 0$

$\therefore$ f(x) has only one minima at x = 0.