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

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

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

We have, $y=(x-3)^2$, which is continuous in $x_1=3$ and $x_2=4$ i.e., $[3,4]$.

Also, $y^{\prime}=2(x-3) \cdot 1=2(x-3)$ which exists in $(3,4)$.

Hence, by mean value theorem there exists a point on the curve at which tangent drawn is parallel to the chord joining the points $(3,0)$ and $(4,1)$.

Thus, $$f(c)=\frac{f(4)-f(3)}{4-3}$$

$$\begin{array}{lr} \Rightarrow & 2(c-3)=\frac{(4-3)^2-(3-3)^2}{4-3} \\ \Rightarrow & 2 c-6=\frac{1-0}{1} \Rightarrow c=\frac{7}{2} \\ \text { For } x=\frac{7}{2}, & \quad y=\left(\frac{7}{2}-3\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4} \end{array}$$

So, $\left(\frac{7}{2}, \frac{1}{4}\right)$ is the point on the curve at which tangent drawn is parallel to the chord joining the points $(3,0)$ and $(4,1)$.

We have, $y=2 x^2-5 x+3$, which is continuous in $[1,2]$ as it is a polynomial function.

Also, $y^{\prime}=4 x-5$, which exists in $(1,2)$.

By mean value theorem, $\exists c \in(1,2)$ at which drawn tangent is parallel to the chord $A B$, where $A$ and $B$ are $(1,0)$ and $(2,1)$, respectively.

$$\therefore\quad f(c)=\frac{f(2)-f(1)}{2-1}$$

$$\begin{array}{lrl} \Rightarrow & 4 c-5 & =\frac{(8-10+3)-(2-5+3)}{1} \\ \Rightarrow & 4 c-5 & =1 \\ \therefore & c & =\frac{6}{4}=\frac{3}{2} \in(1,2) \\ \text { For } x=\frac{3}{2}, & y & =2\left(\frac{3}{2}\right)^2-5\left(\frac{3}{2}\right)+3 \\ & & =2 \times \frac{9}{4}-\frac{15}{2}+3=\frac{9-15+6}{2}=0 \end{array}$$

Hence, $\left(\frac{3}{2}, 0\right)$ is the point on the curve $y=2 x^2-5 x+3$ between the points $A(1,0)$ and $B(2,1)$, where tangent is parallel to the chord $A B$.