$$\begin{aligned} &\begin{aligned} \text { We have, }\quad f(x) & =2 x+\cot ^{-1} x+\log \left(\sqrt{1+x^2}-x\right) \\ \therefore\quad f^{\prime}(x) & =2+\left(\frac{-1}{1+x^2}\right)+\frac{1}{\left(\sqrt{1+x^2}-x\right)}\left(\frac{1}{2 \sqrt{1+x^2}} \cdot 2 x-1\right) \\ & =2-\frac{1}{1+x^2}+\frac{1}{\left(\sqrt{1+x^2}-x\right)} \cdot \frac{\left(x-\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \\ & =2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}} \\ & =\frac{2+2 x^2-1-\sqrt{1+x^2}}{1+x^2}=\frac{1+2 x^2-\sqrt{1+x^2}}{1+x^2} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { For increasing function, }f^{\prime}(x) \geq 0\\ &\begin{array}{rr} \Rightarrow & \frac{1+2 x^2-\sqrt{1+x^2}}{1+x^2} \geq 0 \\ \Rightarrow & 1+2 x^2 \geq \sqrt{1+x^2} \\ \Rightarrow & \left(1+2 x^2\right)^2 \geq 1+x^2 \\ \Rightarrow & 1+4 x^4+4 x^2 \geq 1+x^2 \\ \Rightarrow & 4 x^4+3 x^2 \geq 0 \\ \Rightarrow & x^2\left(4 x^2+3\right) \geq 0 \end{array} \end{aligned}$$

which is true for any real value of $x$.

Hence, $f(x)$ is increasing in R.

We have, $a \geq 1$, $$f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$$

$\therefore\quad f^{\prime}(x)=\sqrt{3} \cos x-(-\sin x)-2 a$

$$ \begin{aligned} & =\sqrt{3} \cos x+\sin x-2 a \\ & =2\left[\frac{\sqrt{3}}{2} \cdot \cos x+\frac{1}{2} \cdot \sin x\right]-2 a \\ & =2\left[\cos \frac{\pi}{6} \cdot \cos x+\sin \frac{\pi}{6} \cdot \sin x\right]-2 a \\ & =2\left(\cos \frac{\pi}{6}-x\right)-2 a \quad [\because \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B]\\ & =2\left[\left(\cos \frac{\pi}{6}-x\right)-a\right] \end{aligned}$$

We know that, $$\cos x \in[-1,1]$$

and $$a \geq 1$$

$$\begin{aligned} \text{So, }\quad 2\left[\cos \left(\frac{\pi}{6}-x\right)-a\right] & \leq 0 \\ \therefore\quad f^{\prime}(x) & \leq 0 \end{aligned}$$

Hence, $f(x)$ is a decreasing function in $R$.

$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} f(x) & =\tan ^{-1}(\sin x+\cos x) \\ \therefore\quad f^{\prime}(x) & =\frac{1}{1+(\sin x+\cos x)^2} \cdot(\cos x-\sin x) \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\frac{1}{1+\sin ^2 x+\cos ^2 x+2 \sin x \cdot \cos x}(\cos x-\sin x) \\ & =\frac{1}{(2+\sin 2 x)}(\cos x-\sin x) \\ & \quad\left[\because \sin 2 x=2 \sin x \cos x \text { and } \sin ^2 x+\cos ^2 x=1\right] \end{aligned}$$

$$\begin{aligned} &\text { For } f^{\prime}(x) \geq 0 \text {, }\\ &\frac{1}{(2+\sin 2 x)} \cdot(\cos x-\sin x) \geq 0 \end{aligned}$$

$\Rightarrow \quad \cos x-\sin x \geq 0 \quad\left[\because(2+\sin 2 x) \geq 0\right.$ in $\left.\left(0, \frac{\pi}{4}\right)\right]$

$$\Rightarrow \quad \cos x \geq \sin x$$

which is true, if $x \in\left(0, \frac{\pi}{4}\right)$.

Hence, $f(x)$ is an increasing function in $\left(0, \frac{\pi}{4}\right)$.

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.