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

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

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

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

We have, $\quad x^m \cdot y^n=(x+y)^{m+n}\quad\text{.... (i)}$

(i) Differentiating Eq. (i) w.r.t. $x$, we get

$\frac{d}{d x}\left(x^m \cdot y^n\right)=\frac{d}{d x}(x+y)^{m+n}$

$$\begin{array}{ll} \Rightarrow & x^m \cdot \frac{d}{d y} y^n \cdot \frac{d y}{d x}+y^n \cdot \frac{d}{d x} x^m=(m+n)(x+y)^{m+n-1} \frac{d}{d x}(x+y) \\ \Rightarrow & x^m \cdot n y^{n-1} \frac{d y}{d x}+y^n \cdot m x^{m-1}=(m+n)(x+y)^{m+n-1}\left(1+\frac{d y}{d x}\right) \end{array}$$

$$\begin{array}{ll} \Rightarrow & \frac{d y}{d x}\left[x^m \cdot n y^{n-1}-(m+n) \cdot(x+y)^{m+n-1}\right]=(m+n)(x+y)^{m+n-1}-y^n m x^{m-1} \\ \Rightarrow & \frac{d y}{d x}\left[n x^m y^{n-1}-(m+n)(x+y)^{m+n-1}\right]=(m+n) \cdot(x+y)^{m+n-1}-\frac{y^{n-1} \cdot y \cdot m x^m}{x} \end{array}$$

$$\begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{\frac{(m+n)(x+y)^{m+n}}{(x+y)}-\frac{y^{n-1} \cdot y \cdot m x^m}{x}}{\frac{n x^m y^n}{y}-(m+n)(x+y)^{m+n} \frac{1}{(x+y)}} \\ & =\frac{\frac{x(m+n)(x+y)^{m+n}-(x+y) \cdot y \cdot{ }^{n-1} y \cdot m x^m}{(x+y) \cdot x}}{\frac{(x+y) n x^m y^n-y(m+n)(x+y)^{m+n}}{(x+y) \cdot y}} \end{aligned}$$

$=\frac{\frac{x(m+n) \cdot x^m \cdot y^n-m(x+y) y^n x^m}{(x+y) \cdot x}}{\frac{(x+y) n x^m \cdot y^n-y(m+n) \cdot x^m \cdot y^n}{(x+y) \cdot y}} \quad\left[\because(x+y)^{m+n}=x^m \cdot y^n\right]$

$$\begin{aligned} & =\frac{x^m y^n[m x+n x-m x-m y] \cdot(x+y) y}{x^m y^n[n x+n y-m y-n y] \cdot(x+y) \cdot x} \\ & =\frac{y}{x}\quad\text{.... (ii)} \end{aligned}$$

Hence proved.

(ii) Further, differentiating Eq. (ii) i.e., $\frac{d y}{d x}=\frac{y}{x}$ on both the sides w.r.t. $x$, we get

$$\begin{array}{rlr} \frac{d^2 y}{d x^2} & =\frac{x \cdot \frac{d y}{d x}-y \cdot 1}{x^2} & \\ & =\frac{x \cdot \frac{y}{x}-y}{x^2} & {\left[\because \frac{d y}{d x}=\frac{y}{x}\right]} \\ & =0 & \text { Hence proved. } \end{array}$$

$$\begin{aligned} & \text { We have, } x=\sin t \text { and } y=\sin p t \\ & \therefore \quad \frac{d x}{d t}=\cos t \text { and } \frac{d y}{d t}=\cos p t \cdot p \end{aligned}$$

$\Rightarrow \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{p \cdot \cos p t}{\cos t}\quad\text{.... (i)}$

$$\begin{aligned} &\text { Again, differentiating both sides w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{d^2 y}{d x^2} & =\frac{\cos t \cdot \frac{d}{d t}(p \cdot \cos p t) \frac{d t}{d x}-p \cos p t \cdot \frac{d}{d t} \cos t \cdot \frac{d t}{d x}}{\cos ^2 t} \\ & =\frac{[\cos t \cdot p \cdot(-\sin p t) \cdot p-p \cos p t \cdot(-\sin t)] \frac{d t}{d x}}{\cos ^2 t} \end{aligned} \end{aligned}$$

$=\frac{\left[-p^2 \sin p t \cdot \cos t+p \sin t \cdot \cos p t\right] \cdot \frac{1}{\cos t}}{\cos ^2 t}$

$\Rightarrow \quad \frac{d^2 y}{d x^2}=\frac{-p^2 \sin p t \cdot \cos t+p \cos p t \cdot \sin t}{\cos ^3 t}\quad\text{.... (ii)}$

Since, we have to prove

$\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+p^2 y=0$

$\therefore \quad \mathrm{LHS}=\left(1-\sin ^2 t\right) \frac{\left[-p^2 \sin p t \cdot \cos t+p \cos p t \cdot \sin t\right]}{\cos ^3 t} -\sin t \cdot \frac{p \cos p t}{\cos t}+p^2 \sin p t$

$$\begin{aligned} & =\frac{1}{\cos ^3 t}\left[\begin{array}{l} \left(1-\sin ^2 t\right)\left(-p^2 \sin p t \cdot \cos t+p \cos p t \cdot \sin t\right) \\ -p \cos p t \cdot \sin t \cdot \cos ^2 t+p^2 \sin p t \cdot \cos ^3 t \end{array}\right] \\ & =\frac{1}{\cos ^3 t}\left[\begin{array}{l} -p^2 \sin p t \cdot \cos ^3 t+p \cos p t \cdot \sin t \cdot \cos ^2 t \\ -p \cos p t \cdot \sin t \cdot \cos ^2 t+p^2 \sin p t \cdot \cos ^3 t \end{array}\right]\left[\because 1-\sin ^2 t=\cos ^2 t\right] \\ & =\frac{1}{\cos ^3 t} \cdot 0 \\ & =0 \quad \quad \text { Hence proved. } \end{aligned}$$

We have, $y=x^{\tan x}+\sqrt{\frac{x^2+1}{2}}\quad\text{.... (i)}$

Taking $u=x^{\tan x}$ and $v=\sqrt{\frac{x^2+1}{2}}$

$$\begin{aligned} \log u & =\tan x \log x \quad\text{.... (ii)}\\ \text{and}\quad v^2 & =\frac{x^2+1}{2}\quad\text{.... (iii)} \end{aligned}$$

$$\begin{aligned} &\text { On, differentiating Eq. (ii) w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{1}{u} \cdot \frac{d u}{d x} & =\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^2 x \\ \Rightarrow\quad \frac{d u}{d x} & =u\left[\frac{\tan x}{x}+\log x \cdot \sec ^2 x\right] \\ & =x^{\tan x}\left[\frac{\tan x}{x}+\log x \cdot \sec ^2 x\right]\quad\text{.... (iv)} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { Also, differentiating Eq. (iii) w.r.t. } x \text {, we get }\\ &2 v \cdot \frac{d v}{d x}=\frac{1}{2}(2 x) \Rightarrow \frac{d v}{d x}=\frac{1}{4 v} \cdot(2 x) \end{aligned}$$

$\Rightarrow \quad \frac{d v}{d x}=\frac{1}{4 \cdot \sqrt{\frac{x^2+1}{2}}} \cdot 2 x=\frac{x \cdot \sqrt{2}}{2 \sqrt{x^2+1}}$

$\Rightarrow \quad \frac{d v}{d x}=\frac{x}{\sqrt{2\left(x^2+1\right)}}\quad\text{.... (v)}$

$$\begin{aligned} \text { Now, } \quad y & =u+v \\ \therefore \quad \frac{d y}{d x} & =\frac{d u}{d x}+\frac{d v}{d x} \\ & =x^{\tan x}\left[\frac{\tan x}{x}+\log x \cdot \sec ^2 x\right]+\frac{x}{\sqrt{2\left(x^2+1\right)}} \end{aligned}$$