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.

We have, $f(x)=\sin x-\sin 2 x$ in $[0, \pi]$

(i) Since, we know that sine functions are continuous functions hence $f(x)=\sin x-\sin 2 x$ is a continuous function in $[0, \pi]$.

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

So, $f(x)$ is differentiable in $(0, \pi)$.

Conditions of mean value theorem are satisfied. Hence, $\exists c \in(0, \pi)$ such that, $f(c)=\frac{f(\pi)-f(0)}{\pi-0}$

$$\begin{array}{ll} \Rightarrow & \cos c-2 \cos 2 c=\frac{\sin \pi-\sin 2 \pi-\sin 0+\sin 2 \cdot 0}{\pi-0} \\ \Rightarrow & 2 \cos 2 c-\cos c=\frac{0}{\pi} \\ \Rightarrow & 2 \cdot\left(2 \cos ^2 c-1\right)-\cos c=0 \\ \Rightarrow & 4 \cos ^2 c-2-\cos c=0 \\ \Rightarrow & 4 \cos ^2 c-\cos c-2=0 \end{array}$$

$$\begin{array}{ll} \Rightarrow & \cos c=\frac{1 \pm \sqrt{1+32}}{8}=\frac{1 \pm \sqrt{33}}{8} \\ \therefore & c=\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \end{array}$$

Also, $$\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \in(0, \pi)$$

Hence, mean value theorem has been verified.

We have, $f(x)=\sqrt{25-x^2}$ in $[1,5]$

(i) Since, $f(x)=\left(25-x^2\right)^{1 / 2}$, where $25-x^2 \geq 0$

$$\Rightarrow \quad x^2 \leq \pm 5 \Rightarrow-5 \leq x \leq 5$$

Hence, $f(x)$ is continuous in $[1,5]$.

(ii) $f^{\prime}(x)=\frac{1}{2}\left(25-x^2\right)^{-1 / 2}-2 x=\frac{-x}{\sqrt{25-x^2}}$, which exists in $(1,5)$.

Hence, $f^{\prime}(x)$ is differentiable in $(1,5)$.

Since, conditions of mean value theorem are satisfied. By mean value theorem $\exists c \in(1,5)$ such that

$f(c)=\frac{f(5)-f(1)}{5-1} \Rightarrow \frac{-c}{\sqrt{25-c^2}}=\frac{0-\sqrt{24}}{4}$

$$\begin{aligned} &\begin{array}{lr} \Rightarrow & \frac{c^2}{25-c^2}=\frac{24}{16} \\ \Rightarrow & 16 c^2=600-24 c^2 \\ \Rightarrow & c^2=\frac{600}{40}=15 \\ \therefore & c= \pm \sqrt{15} \\ \text { Also, } & c=\sqrt{15} \in(1,5) \end{array}\\ &\text { Hence, the mean value theorem has been verified. } \end{aligned}$$