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 $(\pi,-2)$.

We have, $y=x(x-4), x \in[0,4]$

(i) $y$ is a continuous function since $x(x-4)$ is a polynomial function.

Hence, $y=x(x-4)$ is continuous in $[0,4]$.

(ii) $y^{\prime}=(x-4) \cdot 1+x \cdot 1=2 x-4$ which exists in (0,4).

Hence, $y$ is differentiable in $(0,4)$.

(iii) $y(0)=0(0-4)=0$

$$\begin{array}{ll} \text { and } & y(4)=4(4-4)=0 \\ \Rightarrow & y(0)=y(4) \end{array}$$

Since, conditions of Rolle's theorem are satisfied.

Hence, there exists a point $c$ such that

$f^{\prime}(c)=0$ in $(0,4) \quad\left[\because f(x)=y^{\prime}\right]$

$$\begin{array}{rrl} \Rightarrow & 2 c-4 =0 \\ \Rightarrow & c =2 \\ \Rightarrow & x=2 ; y =2(2-4)=-4 \end{array}$$

Thus, $(2,-4)$ is the point on the curve at which the tangent drawn is parallel to $X$-axis.

We have, $f(x)=\frac{1}{4 x-1}$ in $[1,4]$

(i) $f(x)$ is continuous in $[1,4]$.

Also, at $x=\frac{1}{4}, f(x)$ is discontinuous.

Hence, $f(x)$ is continuous in $[1,4]$.

(ii) $f^{\prime}(x)=-\frac{4}{(4 x-1)^2}$, which exists in (1, 4).

Since, conditions of mean value theorem are satisfied.

Hence, there exists a real number $c \in] 1,4$ [ such that

$f^{\prime}(c)=\frac{f(4)-f(1)}{4-1}$

$\Rightarrow \quad \frac{-4}{(4 c-1)^2}=\frac{\frac{1}{16-1}-\frac{1}{4-1}}{4-1}=\frac{\frac{1}{15}-\frac{1}{3}}{3}$

$$\begin{array}{ll} \Rightarrow & \frac{-4}{(4 c-1)^2}=\frac{1-5}{45}=\frac{-4}{45} \\ \Rightarrow & (4 c-1)^2=45 \\ \Rightarrow & 4 c-1= \pm 3 \sqrt{5} \\ \Rightarrow & c=\frac{3 \sqrt{5}+1}{4} \in(1,4)\quad \text{[neglecting ( -ve ) value]} \end{array}$$

Hence, mean value theorem has been verified.