We have, $$y=-x^3+3 x^2+9 x-27$$

$\frac{d y}{d x}=-3 x^2+6 x+9=$ Slope of tangent to the curve

Now, $$\frac{d^2 y}{d x^2}=-6 x+6$$

For $\frac{d}{d x}\left(\frac{d y}{d x}\right)=0$

$$\begin{aligned} -6 x+6 & =0 \\ \Rightarrow\quad x & =\frac{-6}{-6}=1 \end{aligned}$$

$\therefore \quad \frac{d}{d x}\left(\frac{d^2 y}{d x^2}\right)=-6<0$

So, the slope of tangent to the curve is maximum, when $x=1$.

For $x=1$,

$$\left(\frac{d y}{d x}\right)_{(x=1)}=-3 \cdot 1^2+6 \cdot 1+9=12$$

which is maximum slope.

Also, for

$$\begin{aligned} x=1, y & =-1^3+3 \cdot 1^2+9 \cdot 1-27 \\ & =-1+3+9-27 \\ & =-16 \end{aligned}$$

So, the required point is $(1,-16)$.

$$\begin{aligned} & \begin{aligned} \text { We have, } \quad f(x) & =\sin x+\sqrt{3} \cos x \\ \therefore \quad f^{\prime}(x) & =\cos x+\sqrt{3}(-\sin x) \\ & =\cos x-\sqrt{3} \sin x \end{aligned} \end{aligned}$$

$$\begin{aligned} \text { For } f^{\prime}(x)=0, & & \cos x & =\sqrt{3} \sin x \\ \Rightarrow & & \tan x & =\frac{1}{\sqrt{3}}=\tan \frac{\pi}{6} \\ \Rightarrow & & x & =\frac{\pi}{6} \end{aligned}$$

Again, differentiating $f^{\prime}(x)$, we get

$$f^{\prime \prime}(x)=-\sin x-\sqrt{3} \cos x$$

At $x=\frac{\pi}{6}, \quad f^{\prime \prime}(x)=-\sin \frac{\pi}{6}-\sqrt{3} \cos \frac{\pi}{6}$

$$\begin{aligned} & =-\frac{1}{2}-\sqrt{3} \cdot \frac{\sqrt{3}}{2} \\ & =-\frac{1}{2}-\frac{3}{2}=-2<0 \end{aligned}$$

Hence, at $x=\frac{\pi}{6}, f(x)$ has maximum value at $\frac{\pi}{6}$ is the point of local maxima.

Let $A B C$ be a triangle with $A C=h, A B=x$ and $B C=y$.

$$\begin{aligned} \text{Also,}\quad & \angle C A B=\theta \\ \text{Let}\quad & h+x=k\quad\text{.... (i)} \end{aligned}$$

$$\begin{array}{lr} \therefore & \cos \theta=\frac{x}{h} \\ \Rightarrow & x=h \cos \theta \\ \Rightarrow & h+h \cos \theta=k\quad \text{[using Eq. (i)]} \end{array}$$

$$\begin{array}{lr} \Rightarrow & h(1+\cos \theta)=k \\ \Rightarrow & h=\frac{k}{(1+\cos \theta)}\quad\text{.... (ii)} \end{array}$$

Also, $\quad$ area of $\triangle A B C=\frac{1}{2}(A B \cdot B C)$

$$\begin{aligned} A & =\frac{1}{2} \cdot x \cdot y \\ & =\frac{1}{2} h \cos \theta \cdot h \sin \theta \quad\left[\because \quad \sin \theta=\frac{y}{h}\right]\\ & =\frac{1}{2} h^2 \sin \theta \cdot \cos \theta \\ & =\frac{2 h^2}{4} \sin \theta \cdot \cos \theta \\ & =\frac{1}{4} h^2 \sin 2 \theta\quad\text{.... (iii)} \end{aligned}$$

$$\begin{array}{ll} \text { Since, } & h=\frac{k}{1+\cos \theta} \\ \therefore & A=\frac{1}{4}\left(\frac{k}{1+\cos \theta}\right)^2 \cdot \sin 2 \theta \\ \Rightarrow & A=\frac{k^2}{4} \cdot \frac{\sin 2 \theta}{(1+\cos \theta)^2}\quad\text{.... (iv)} \end{array}$$

$\therefore \quad \frac{d A}{d \theta}=\frac{k^2}{4}\left[\frac{(1+\cos \theta)^2 \cdot \cos 2 \theta \cdot 2-\sin 2 \theta \cdot 2(1+\cos \theta) \cdot(0-\sin \theta)}{(1+\cos \theta)^4}\right]$

$$\begin{aligned} & =\frac{k^2}{4}\left\{\frac{2(1+\cos \theta)[(1+\cos \theta) \cdot \cos 2 \theta+\sin 2 \theta(\sin \theta)}{(1+\cos \theta)^4}\right\} \\ & =\frac{k^2}{4} \cdot \frac{2}{(1+\cos \theta)^3}\left[(1+\cos \theta) \cdot \cos 2 \theta+2 \sin ^2 \theta \cdot \cos \theta\right] \\ & =\frac{k^2}{2(1+\cos \theta)^3}\left[(1+\cos \theta)\left(1-2 \sin ^2 \theta\right)+2 \sin ^2 \theta \cdot \cos \theta\right] \\ & =\frac{k^2}{2(1+\cos \theta)^3}\left[1+\cos \theta-2 \sin ^2 \theta-2 \sin ^2 \theta \cdot \cos \theta+2 \sin ^2 \theta \cdot \cos \theta\right] \\ & =\frac{k^2}{2(1+\cos \theta)^3}\left[(1+\cos \theta)-2 \sin ^2 \theta\right] \\ & =\frac{k^2}{2(1+\cos \theta)^3}\left[1+\cos \theta-2+2 \cos ^2 \theta\right] \\ & =\frac{k^2}{2(1+\cos \theta)^3}\left(2 \cos ^2 \theta+\cos \theta-1\right)\quad\text{.... (v)} \end{aligned}$$

For $\frac{d A}{d \theta}=0$,

$$\begin{array}{rr} & \frac{k^2}{2(1+\cos \theta)^3}\left(2 \cos ^2 \theta+\cos \theta-1\right)=0 \\ \Rightarrow & 2 \cos ^2 \theta+\cos \theta-1=0 \\ \Rightarrow & 2 \cos ^2 \theta+2 \cos \theta-\cos \theta-1=0 \\ \Rightarrow & 2 \cos \theta(\cos \theta+1)-1(\cos \theta+1)=0 \\ \Rightarrow & (2 \cos \theta-1)(\cos \theta+1)=0 \end{array}$$

$$\begin{array}{ll} \Rightarrow & \cos \theta=\frac{1}{2} \text { or } \cos \theta=-1 \\ \Rightarrow & \theta=\frac{\pi}{3} \quad\text{[possible]}\\ \text { or } & \theta=2 n \pi \pm \pi \quad\text{[not possible]}\\ \therefore & \theta=\frac{\pi}{3} \end{array}$$

Again, differentiating w.r.t. $\theta$ in Eq. (v), we get

$$\begin{aligned} \frac{d}{d \theta}\left(\frac{d A}{d \theta}\right) & =\frac{d}{d \theta}\left[\frac{k^2}{2(1+\cos \theta)^3}\left(2 \cos ^2 \theta+\cos \theta-1\right)\right] \\ \therefore \quad \frac{d^2 A}{d \theta^2} & =\frac{d}{d \theta}\left[\frac{k^2(2 \cos \theta-1)(1+\cos \theta)}{2(1+\cos \theta)^3}\right]=\frac{d}{d \theta}\left[\frac{k^2}{2} \cdot \frac{(2 \cos \theta-1)}{(1+\cos \theta)^2}\right] \\ & =\frac{k^2}{2}\left[\frac{(1+\cos \theta)^2 \cdot(-2 \sin \theta)-2(1+\cos \theta) \cdot(-\sin \theta)(2 \cos \theta-1)}{(1+\cos \theta)^4}\right] \\ & =\frac{k^2}{2}\left[\frac{(1+\cos \theta) \cdot[1+\cos \theta](-2 \sin \theta)+2 \sin \theta(2 \cos \theta-1)}{(1+\cos \theta)^4}\right] \\ & =\frac{k^2}{2}\left[\frac{-2 \sin \theta-2 \sin \theta \cdot \cos \theta+4 \sin \theta \cdot \cos \theta-2 \sin \theta}{(1+\cos \theta)^3}\right] \\ & =\frac{k^2}{2}\left[\frac{-4 \sin \theta-\sin 2 \theta+2 \sin 2 \theta}{(1+\cos \theta)^3}\right]=\frac{k^2}{2}\left[\frac{\sin 2 \theta-4 \sin \theta}{(1+\cos \theta)^3}\right] \end{aligned}$$

$$\begin{aligned} \therefore \quad\left(\frac{d^2 A}{d \theta^2}\right)_{\text {at } \theta=\frac{\pi}{3}} & =\frac{k^2}{2}\left[\frac{\sin \frac{2 \pi}{3}-4 \sin \frac{\pi}{3}}{\left(1+\cos \frac{\pi}{3}\right)^3}\right]=\frac{k^2}{2}\left[\frac{\frac{\sqrt{3}}{2}-\frac{4 \sqrt{3}}{2}}{\left(1+\frac{1}{2}\right)^3}\right] \\ & =\frac{k^2}{2}\left[\frac{-3 \sqrt{3} \cdot 8}{2 \cdot 27}\right]=-k^2\left(\frac{2 \sqrt{3}}{9}\right) \end{aligned}$$

which is less than zero.

Hence, area of the right angled triangle is maximum, when the angle between them is $\frac{\pi}{3}$.

Given that, $$f(x)=x^5-5 x^4+5 x^3-1$$

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

$f^{\prime}(x)=5 x^4-20 x^3+15 x^2$

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

$$\begin{aligned} \Rightarrow \quad& 5 x^4-20 x^3+15 x^2 =0 \\ \Rightarrow \quad& 5 x^2\left(x^2-4 x+3\right) =0 \\ \Rightarrow \quad& 5 x^2\left(x^2-3 x-x+3\right) =0 \\ \Rightarrow \quad& 5 x^2[x(x-3)-1(x-3)] =0 \\ \Rightarrow \quad& 5 x^2[(x-1)(x-3)] =0 \\ \therefore \quad& x =0,1,3 \end{aligned}$$

Sign scheme for $\frac{d y}{d x}=5 x^2(x-1)(x-3)$

So, $y$ has maximum value at $x=1$ and minimum value at $x=3$.

At $x=0, y$ has neither maximum nor minimum value.

$$\begin{aligned} & \therefore \quad \text { Maximum value of } y=1-5+5-1=0 \\ & \text { and } \\ & \text { minimum value }=(3)^5-5(3)^4+5(3)^3-1 \\ & =243-81 \times 5-27 \times 5-1=-298 \end{aligned}$$

Consider that company increases the annual subscription by ₹ $x$.

So, $x$ subscribes will discontinue the service.

$\therefore$ Total revenue of company after the increment is given by

$$\begin{aligned} R(x) & =(500-x)(300+x) \\ & =15 \times 10^4+500 x-300 x-x^2 \\ & =-x^2+200 x+150000 \end{aligned}$$

$$\begin{aligned} &\text { On differentiating both sides w.r.t. } x \text {, we get }\\ &\begin{array}{rlrl} & R^{\prime}(x) =-2 x+200 \\ \text { Now, } & R^{\prime}(x) =0 \\ \Rightarrow & 2 x =200 \Rightarrow x=100 \\ \therefore & R^{\prime \prime}(x) =-2<0 \\ \text { So, } R(x) \text { is maximum when } \quad x & =100 \end{array} \end{aligned}$$

Hence, the company should increase the subscription fee by ₹ $100$, so that it has maximum profit.