Given equation of the curve is

$$3 x^2-y^2=8\quad\text{.... (i)}$$

On differentiating both sides w.r.t. $x$, we get

$$\begin{aligned} 6 x-2 y \frac{d y}{d x} & =0 \\ \Rightarrow\quad \frac{d y}{d x}=\frac{6 x}{2 y} & =\frac{3 x}{y} \\ \Rightarrow\quad m_1 & =\frac{3 x}{y}\quad\text{[say]} \end{aligned}$$

and slope of normal $\left(m_2\right)=\frac{-1}{m_1}=\frac{-y}{3 x}\quad\text{.... (ii)}$

Since, slope of normal to the curve should be equal to the slope of line $x+3 y=4$, which is parallel to curve.

For line, $$y=\frac{4-x}{3}=\frac{-x}{3}+\frac{4}{3}$$

$\Rightarrow \quad$ Slope of the line $\left(m_3\right)=\frac{-1}{3}$

$$\begin{array}{lr} \therefore & m_2=m_3 \\ \Rightarrow & \frac{-y}{3 x}=-\frac{1}{3} \\ \Rightarrow & -3 y=-3 x \\ \Rightarrow & y=x\quad\text{.... (iii)} \end{array}$$

$$\begin{aligned} &\text { On substituting the value of } y \text { in Eq. (i), we get }\\ &\begin{array}{rlrl} & 3 x^2-x^2 =8 \\ \Rightarrow & x^2 =4 \\ \Rightarrow & x = \pm 2 \end{array} \end{aligned}$$

For $x=2$, $y=2\quad$ [using Eq. (iii)]

and for $x=-2$, $y=-2\quad$ [using Eq. (iii)]

Thus,the points at which normal to the curve are parallel to the line $x+3 y=4$ are $(2,2)$ and $(-2,-2)$.

Required equations of normal are

$$y-2=m_2(x-2) \text { and } y+2=m_2(x+2)$$

$$\begin{array}{llll} \Rightarrow & y-2=\frac{-2}{6}(x-2) \quad \text { and } y+2=\frac{-2}{6}(x+2) \\ \Rightarrow & 3 y-6=-x+2 \quad \text { and } 3 y+6=-x-2 \\ \Rightarrow & 3 y+x=+8 \quad \text { and } 3 y+x=-8 \end{array}$$

So, the required equations are $3 y+x= \pm 8$.

$$\begin{aligned} &\text { Given, equation of curve which is }\\ &\begin{aligned} & x^2+y^2-2 x-4 y+1=0 \quad\text{.... (i)}\\ \Rightarrow \quad & 2 x+2 y \frac{d y}{d x}-2-4 \frac{d y}{d x}=0 \end{aligned} \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad & \frac{d y}{d x}(2 y-4) =2-2 x \\ \Rightarrow \quad & \frac{d y}{d x} =\frac{2(1-x)}{2(y-2)} \end{aligned}$$

$$\begin{aligned} &\text { Since, the tangents are parallel to the } Y \text {-axis i.e., } \tan \theta=\tan 90^{\circ}=\frac{d y}{d x} \text {. }\\ &\begin{array}{lr} \therefore & \frac{1-x}{y-2}=\frac{1}{0} \\ \Rightarrow & y-2=0 \\ \Rightarrow & y=2 \end{array} \end{aligned}$$

For $y=2$ from Eq. (i), we get

$$\begin{array}{rr} & x^2+2^2-2 x-4 \times 2+1=0 \\ \Rightarrow & x^2-2 x-3=0 \\ \Rightarrow & x^2-3 x+x-3=0 \\ \Rightarrow & x(x-3)+1(x-3)=0 \\ \Rightarrow & (x+1)(x-3)=0 \\ \therefore & x=-1, x=3 \end{array}$$

So, the required points are $(-1,2)$ and $(3,2)$.

We have the equation of line given by $\frac{x}{a}+\frac{y}{b}=1$, which touches the curve $y=b \cdot e^{-x / a}$ at the point, where the curve intersects the axis of $Y$ i.e., $x=0$.

$$\therefore \quad y=b \cdot e^{-0 / a}=b \quad\left[\because e^0=1\right]$$

So, the point of intersection of the curve with $Y$-axis is $(0, b)$.

Now, slope of the given line at $(0, b)$ is given by

$$\begin{array}{rlrl} \Rightarrow & \frac{1}{a} \cdot 1+\frac{1}{b} \cdot \frac{d y}{d x} =0 \\ \Rightarrow & \frac{d y}{d x} =\frac{-1}{a} \cdot b \\ \Rightarrow & \frac{d y}{d x} =-\frac{1}{a} \cdot b=\frac{-b}{a}=m_1\quad\text{[say]} \end{array}$$

Also, the slope of the curve at $(0, b)$ is

$$\begin{aligned} \frac{d y}{d x} & =b \cdot e^{-x / a} \cdot \frac{-1}{a} \\ \frac{d y}{d x} & =\frac{-b}{a} e^{-x / a} \\ \left(\frac{d y}{d x}\right)_{(0, b)} & =\frac{-b}{a} e^{-0}=\frac{-b}{a}=m_2\quad\text{[say]} \end{aligned}$$

Since, $$m_1=m_2=\frac{-b}{a}$$

$$\begin{aligned} &\begin{aligned} \text { We have, }\quad f(x) & =2 x+\cot ^{-1} x+\log \left(\sqrt{1+x^2}-x\right) \\ \therefore\quad f^{\prime}(x) & =2+\left(\frac{-1}{1+x^2}\right)+\frac{1}{\left(\sqrt{1+x^2}-x\right)}\left(\frac{1}{2 \sqrt{1+x^2}} \cdot 2 x-1\right) \\ & =2-\frac{1}{1+x^2}+\frac{1}{\left(\sqrt{1+x^2}-x\right)} \cdot \frac{\left(x-\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \\ & =2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}} \\ & =\frac{2+2 x^2-1-\sqrt{1+x^2}}{1+x^2}=\frac{1+2 x^2-\sqrt{1+x^2}}{1+x^2} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { For increasing function, }f^{\prime}(x) \geq 0\\ &\begin{array}{rr} \Rightarrow & \frac{1+2 x^2-\sqrt{1+x^2}}{1+x^2} \geq 0 \\ \Rightarrow & 1+2 x^2 \geq \sqrt{1+x^2} \\ \Rightarrow & \left(1+2 x^2\right)^2 \geq 1+x^2 \\ \Rightarrow & 1+4 x^4+4 x^2 \geq 1+x^2 \\ \Rightarrow & 4 x^4+3 x^2 \geq 0 \\ \Rightarrow & x^2\left(4 x^2+3\right) \geq 0 \end{array} \end{aligned}$$

which is true for any real value of $x$.

Hence, $f(x)$ is increasing in R.

We have, $a \geq 1$, $$f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$$

$\therefore\quad f^{\prime}(x)=\sqrt{3} \cos x-(-\sin x)-2 a$

$$ \begin{aligned} & =\sqrt{3} \cos x+\sin x-2 a \\ & =2\left[\frac{\sqrt{3}}{2} \cdot \cos x+\frac{1}{2} \cdot \sin x\right]-2 a \\ & =2\left[\cos \frac{\pi}{6} \cdot \cos x+\sin \frac{\pi}{6} \cdot \sin x\right]-2 a \\ & =2\left(\cos \frac{\pi}{6}-x\right)-2 a \quad [\because \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B]\\ & =2\left[\left(\cos \frac{\pi}{6}-x\right)-a\right] \end{aligned}$$

We know that, $$\cos x \in[-1,1]$$

and $$a \geq 1$$

$$\begin{aligned} \text{So, }\quad 2\left[\cos \left(\frac{\pi}{6}-x\right)-a\right] & \leq 0 \\ \therefore\quad f^{\prime}(x) & \leq 0 \end{aligned}$$

Hence, $f(x)$ is a decreasing function in $R$.