We have, $$f(x)=\log \left(x^2+2\right)-\log 3$$

(i) Logarithmic functions are continuous in their domain. Hence, $f(x)=\log \left(x^2+2\right)-\log 3$ is continuous in $[-1,1]$.

$$\begin{aligned} \text{(ii)}\quad f^{\prime}(x) & =\frac{1}{x^2+2} \cdot 2 x-0 \\ & =\frac{2 x}{x^2+2}, \text { which exists in }(-1,1) \end{aligned}$$

Hence, $f(x)$ is differentiable in $(-1,1)$.

$$\begin{aligned} & \text { (iii) } f(-1)=\log \left[(-1)^2+2\right]-\log 3=\log 3-\log 3=0 \text { and } \\ & f(1)=\log \left(1^2+2\right)-\log 3=\log 3-\log 3=0 \\ & \Rightarrow \quad f(-1)=f(1) \end{aligned}$$

Conditions of Rolle's theorem are satisfied.

Hence, there exists a real number c such that

$$\begin{aligned} f^{\prime}(c) =0 \\ \Rightarrow \quad \frac{2 c}{c^2+2} & =0 \\ \Rightarrow \quad c =0 \in(-1,1) \end{aligned}$$

Hence, Rolle's theorem has been verified.

We have, $$f(x)=x(x+3) \mathrm{e}^{-x / 2}$$

(i) $f(x)$ is a continuous function. [since, it is a combination of polynomial functions $x(x+3)$ and an exponential function $\mathrm{e}^{-x / 2}$ which are continuous functions]

So, $f(x)=x(x+3) \mathrm{e}^{-x / 2}$ is continuous in $[-3,0]$.

(ii) $\therefore \quad \quad f(x)=\left(x^2+3 x\right) \cdot \frac{d}{d x} \mathrm{e}^{-x / 2}+\mathrm{e}^{-x / 2} \cdot \frac{d}{d x}\left(x^2+3 x\right)$

$$\begin{aligned} & =\left(x^2+3 x\right) \cdot \mathrm{e}^{-x / 2} \cdot\left(-\frac{1}{2}\right)+\mathrm{e}^{-x / 2} \cdot(2 x+3) \\ & =\mathrm{e}^{-x / 2}\left[2 x+3-\frac{1}{2} \cdot\left(x^2+3 x\right)\right] \\ & =\mathrm{e}^{-x / 2}\left[\frac{4 x+6-x^2-3 x}{2}\right] \\ & =\mathrm{e}^{-x / 2} \cdot \frac{1}{2}\left[-x^2+x+6\right] \\ & =\frac{-1}{2} \mathrm{e}^{-x / 2}\left[x^2-x-6\right] \\ & =\frac{-1}{2} \mathrm{e}^{-x / 2}\left[x^2-3 x+2 x-6\right] \\ & =\frac{-1}{2} \mathrm{e}^{-x / 2}[(x+2)(x-3)] \text { which exists in }(-3,0) . \end{aligned}$$

Hence, $f(x)$ is differentiable in $(-3,0)$. (iii) $\therefore$ $$f(-3)=-3(-3+3) e^{-3 / 2}=0$$

and $$f(0)=0(0+3) e^{-0 / 2}=0$$

$$\Rightarrow \quad f(-3)=f(0)$$

Since, conditions of Rolle's theorem are satisfied.

Hence, there exists a real number $c$ such that $f(c)=0$

$$\begin{array}{ll} \Rightarrow & -\frac{1}{2} e^{-c / 2}(c+2)(c-3)=0 \\ \Rightarrow & c=-2,3 \text { where }-2 \in(-3,0) \end{array}$$

Therefore, Rolle's theorem has been verified.

We have, $f(x)=\sqrt{4-x^2}=\left(4-x^2\right)^{1 / 2}$

(i) $f(x)=\sqrt{4-x^2}$ is a continuous function.

[since every polynomial function is a continuous function]

Hence, $f(x)$ is continuous in $[-2,2]$.

(ii) $f^{\prime}(x)=\frac{1}{2}\left(4-x^2\right)^{-1 / 2} \cdot(-2 x)$

$=-x \cdot \frac{1}{\sqrt{4-x^2}}$, which exists everywhere except at $x= \pm 2$.

Hence, $f(x)$ is differentiable in $(-2,2)$.

$$\begin{aligned} \text{(iii)}\quad f(-2)=\sqrt{(4-4)}=0 \text { and } f(2)=\sqrt{(4-4)} & =0 \\ \Rightarrow \quad f(-2) & =f(2) \end{aligned}$$

conditions of Rolle's theorem are satisfied.

Hence, there exists a real number $c$ such that $f^{\prime}(c)=0$.

$$\begin{aligned} &\begin{aligned} & \Rightarrow \quad-c \frac{1}{\sqrt{4-c^2}}=0 \\ & \Rightarrow \quad c=0 \in(-2,2) \end{aligned}\\ &\text { Hence, Rolle's theorem has been verified. } \end{aligned}$$

We have, $$f(x)=\left\{\begin{array}{l} x^2+1, \text { if } 0 \leq x \leq 1 \\ 3-x, \text { if } 1 \leq x \leq 2 \end{array}\right.$$

We know that, polynomial function is everywhere continuous and differentiability.

So, $f(x)$ is continuous and differentiable at all points except possibly at $x=1$.

Now, check the differentiability at $x=1$,

At $x=1$,

$$\begin{aligned} & \mathrm{LDH}=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ &=\lim _{x \rightarrow 1} \frac{\left(x^2+1\right)-(1+1)}{x-1} \quad \quad\left[\because f(x)=x^2+1, \forall 0 \leq x \leq 1\right] \\ &=\lim _{x \rightarrow 1} \frac{x^2-1}{x-1}=\lim _{x \rightarrow 1} \frac{(x+1)(x-1)}{x-1} \\ & \quad=2 \\ \text{and}\quad & \mathrm{RDH}=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1} \frac{(3-x) f(1+1)}{(x-1)} \\ &=\lim _{x \rightarrow 1} \frac{3-x-2}{x-1}=\lim _{x \rightarrow 1} \frac{-(x-1)}{x-1}=-1 \\ \therefore\quad & \mathrm{LHD} \neq \mathrm{RHD} \end{aligned}$$

So, $f(x)$ is not differentiable at $x=1$.

Hence, Polle's theorem is not applicable on the interval $[0,2]$.

The equation of the curve is $y=\cos x-1$.

Now, we have to find a point on the curve in $[0,2 \pi]$.

where the tangent is parallel to $X$-axis $i . e$., the tangent to the curve at $x=c$ has a slope o , wherec $\in] 0,2 \pi[$.

Let us apply Rolle's theorem to get the point.

(i) $y=\cos x-1$ is a continuous function in $[0,2 \pi]$.

[since it is a combination of cosine function and a constant function]

(ii) $y^{\prime}=-\sin x$, which exists in $(0,2 \pi)$.

Hence, $y$ is differentiable in $(0,2 \pi)$.

(iii) $y(0)=\cos 0-1=0$ and $y(2 \pi)=\cos 2 \pi-1=0$,

$$\therefore \quad y(0)=y(2 \pi)$$

Since, conditions of Rolle's theorem are satisfied.

Hence, there exists a real number c such that

$$\begin{aligned} f^{\prime}(c) & =0 \\ \Rightarrow\quad -\sin c & =0 \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & c=\pi \text { or } 0, \text { where } \pi \in(0,2 \pi) \\ \Rightarrow & x=\pi \\ \therefore & y=\cos \pi-1=-2 \end{array}$$

Hence, the required point on the curve, where the tangent drawn is parallel to the X-axis is $(\pi,-2)$.