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 $A B$.

$$\begin{aligned} &\text { We have, } f(x)=\left\{\begin{array}{ll} x^2+3 x+p, & \text { if } x \leq 1 \\ q x+2, & \text { if } x>1 \end{array} \text { is differentiable at } x=1\right. \text {. }\\ &\therefore \quad L f^{\prime}(1)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \end{aligned}$$

$$\begin{aligned} & =\lim _{x \rightarrow 1^{-}} \frac{\left(x^2+3 x+p\right)-(1+3+p)}{x-1} \\ & =\lim _{h \rightarrow 0} \frac{\left[(1-h)^2+3(1-h)+p\right]-[1+3+p]}{(1-h)-1} \\ & =\lim _{h \rightarrow 0} \frac{\left[1+h^2-2 h+3-3 h+p\right]-[4+p]}{-h} \\ & =\lim _{h \rightarrow 0} \frac{\left[h^2-5 h+p+4-4-p\right]}{-h}=\lim _{h \rightarrow 0} \frac{h[h-5]}{-h} \\ & =\lim _{h \rightarrow 0}-[h-5]=5 \end{aligned}$$

$$\begin{aligned} R f^{\prime}(1) & =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{(q x+2)-(1+3+p)}{x-1} \\ & =\lim _{h \rightarrow 0} \frac{[q(1+h)+2]-(4+p)}{1+h-1} \\ & =\lim _{h \rightarrow 0} \frac{[q+q h+2-4-p]}{h}=\lim _{h \rightarrow 0} \frac{q h+(q-2-p)}{h} \end{aligned}$$

$$ \begin{aligned} &\begin{array}{ll} \Rightarrow & q-2-p=0 \Rightarrow p-q=-2 \quad\text{.... (i)}\\ \Rightarrow & \lim _\limits{h \rightarrow 0} \frac{q h+0}{h}=q\quad\text{[for existing the limit]} \end{array}\\ &\text { If } L f^{\prime}(1)=R f^{\prime}(1) \text {, then } 5=q\\ &\begin{aligned} \Rightarrow & \quad p-5 =-2 \Rightarrow p=3 \\ \therefore & \quad p =3 \text { and } q=5 \end{aligned} \end{aligned}$$