Let $f: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}$ be real valued function

defined as $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$.

Then range of $f$ is

Let the plane $\mathrm{P}: 8 x+\alpha_{1} y+\alpha_{2} z+12=0$ be parallel to

the line $\mathrm{L}: \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the intercept of $\mathrm{P}$

on the $y$-axis is 1 , then the distance between $\mathrm{P}$ and $\mathrm{L}$ is :

The equation $\mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{x}+1=0, x \in \mathbb{R}$ has :

Let (a, b) $\subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi)$, holds.

If $\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to :