The curves $y=4 x^2+2 x-8$ and $y=x^3-x+13$ touch each other at the point $(3,34)$. Given, equation of curves are $y=4 x^2+2 x-8$ and $y=x^3-x+13$

$$\begin{array}{ll} \therefore & \frac{d y}{d x}=8 x+2 \\ \text { and } & \frac{d y}{d x}=3 x^2-1 \end{array}$$

So, the slope of both curves should be same

$\therefore \quad 8 x+2=3 x^2-1$

$$\begin{array}{lr} \Rightarrow & 3 x^2-8 x-3=0 \\ \Rightarrow & 3 x^2-9 x+x-3=0 \\ \Rightarrow & 3 x(x-3)+1(x-3)=0 \\ \Rightarrow & (3 x+1)(x-3)=0 \\ \therefore & x=-\frac{1}{3} \text { and } x=3, \end{array}$$

For $x=-\frac{1}{3}$, $$y=4 \cdot\left(-\frac{1}{3}\right)^2+2 \cdot\left(\frac{-1}{3}\right)-8$$

$$\begin{aligned} & =\frac{4}{9}-\frac{2}{3}-8=\frac{4-6-72}{9} \\ & =-\frac{74}{9} \end{aligned}$$

and for $x=3, y=4 \cdot(3)^2+2 \cdot(3)-8$

$$=36+6-8=34$$

So, the required points are $(3,34)$ and $\left(-\frac{1}{3}, \frac{-74}{9}\right)$.

The equation of normal to the curve $y=\tan x$ at $(0,0)$ is $x+y=0$.

$\because\quad y=\tan x \Rightarrow \frac{d y}{d x}=\sec ^2 x$

$\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(0,0)}=\sec ^2 0=1$ and $-\frac{1}{\left(\frac{d y}{d x}\right)}=-\frac{1}{1}$

$$\begin{aligned} &\therefore \text { Equation of normal to the curve } y=\tan x \text { at }(0,0) \text { is }\\ &\begin{array}{ll} & y-0=-1(x-0) \\ \Rightarrow & y+x=0 \end{array} \end{aligned}$$

The values of $a$ for which the function $f(x)=\sin x-a x+b$ increases on $R$ are $(-\infty,-1)$.

$$\begin{aligned} \because\quad f^{\prime}(x) & =\cos x-a \\ \text{and}\quad f^{\prime}(x) & >0 \Rightarrow \cos x>a \end{aligned}$$

$$\begin{aligned} &\text { Since, } \quad\cos x \in[-1,1]\\ &\Rightarrow \quad a<-1 \Rightarrow a \in(-\infty,-1) \end{aligned}$$

The function $f(x)=\frac{2 x^2-1}{x^4}$, where $x>0$, decreases in the interval $(1, \infty)$.

$\because \quad f^{\prime}(x)=\frac{x^4 \cdot 4 x-\left(2 x^2-1\right) \cdot 4 x^3}{x^8}=\frac{4 x^5-8 x^5+4 x^3}{x^8}$

$=\frac{-4 x^5+4 x^3}{x^8}=\frac{4 x^3\left(-x^2+1\right)}{x^8}$

Also, $f^{\prime}(x)<0$

$$\begin{array}{lc} \Rightarrow & \frac{4 x^3\left(1-x^2\right)}{x^8}<0 \Rightarrow x^2>1 \\ \Rightarrow & x> \pm 1 \\ \therefore & x \in(1, \infty) \end{array}$$

The least value of function $f(x)=a x+\frac{b}{x}$ (where, $a>0, b>0, x>0)$ is $2 \sqrt{a b}$.

$\because\quad f^{\prime}(x)=a-\frac{b}{x^2}$ and $f^{\prime}(x)=0$

$$\begin{array}{ll} \Rightarrow & a=\frac{b}{x^2} \\ \Rightarrow & x^2=\frac{b}{a} \Rightarrow x= \pm \sqrt{\frac{b}{a}} \end{array}$$

$$\begin{aligned} &\begin{aligned} \text { Now, }\quad & f^{\prime \prime}(x)=-b \cdot \frac{(-2)}{x^3}=+\frac{2 b}{x^3} \\ \text { At } x=\sqrt{\frac{b}{a}},\quad & f^{\prime \prime}(x)=+\frac{2 b}{\left(\frac{b}{a}\right)^{3 / 2}}=\frac{+2 b \cdot a^{3 / 2}}{b^{3 / 2}} \end{aligned} \end{aligned}$$

$=+2 b^{-1 / 2} \cdot a^{3 / 2}=+2 \sqrt{\frac{a^3}{b}}>0 \quad[\because a, b>0]$

$\therefore$ Least value of $f(x), \quad f\left(\sqrt{\frac{b}{a}}\right)=a \cdot \sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}$

$$\begin{aligned} & =a \cdot a^{-1 / 2} \cdot b^{1 / 2}+b \cdot b^{-1 / 2} \cdot a^{1 / 2} \\ & =\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b} \end{aligned}$$