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.

We have, $f(x)=x^3-2 x^2-x+3$ in $[0,1]$

(i) Since, $f(x)$ is a polynomial function.

Hence, $f(x)$ is continuous in $[0,1]$.

(ii) $f^{\prime}(x)=3 x^2-4 x-1$, which exists in $(0,1)$.

Hence, $f(x)$ is differentiable in $(0,1)$.

Since, conditions of mean value theorem are satisfied.

Therefore, by mean value theorem $\exists c \in(0,1)$, such that

$$f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}$$

$$\begin{array}{ll} \Rightarrow & 3 c^2-4 c-1=\frac{[1-2-1+3]-[0+3]}{1-0} \\ \Rightarrow & 3 c^2-4 c-1=\frac{-2}{1} \end{array}$$

$$\begin{array}{rrr} \Rightarrow & 3 c^2-4 c+1 =0 \\ \Rightarrow & 3 c^2-3 c-c+1 =0 \\ \Rightarrow & 3 c(c-1)-1(c-1) =0 \\ \Rightarrow & (3 c-1)(c-1) =0 \\ \Rightarrow & c =1 / 3,1, \text { where } \frac{1}{3} \in(0,1) \end{array}$$

Hence, the mean value theorem has been verified.