Range of the function

$f(x) = \sqrt {\left| {{{\sin }^{ - 1}}\left| {\sin x} \right|} \right| - {{\cos }^{ - 1}}\left| {\cos x} \right|} $ is

The range of the function $f(x){ = ^{7 - x}}{\kern 1pt} {P_{x - 3}}$ is

The range of the function $f(x) = \frac{{\sqrt {1 - {x^2}} }}{{1 + \left| x \right|}}$ is 

Let $\quad E_1=\left\{x \in R : x \neq 1\right.$ and $\left.\frac{x}{x-1}>0\right\}$ and $\quad E_2=\left\{x \in E_1: \sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)\right.$ is a real number $\}$.

(Here, the inverse trigonometric function $\sin ^{-1} x$ assumes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ )

Let $f : E _1 \rightarrow R$ be the function defined by $f(x)=\log _c\left(\frac{x}{x-1}\right)$ and $g: E_2 \rightarrow R$ be the function defined by $g(x)=\sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)$

The correct option is:

Let $f ( x )$ be a quadratic polynomial with leading coefficient $1$ such that $f(0)=p, p \neq 0$ and $f(1)=\frac{1}{3}$. If the equation $f(x)=0$ and $fofofof (x)=0$ have a common real root, then $f(-3)$ is equal to $........$

Let $f(x) = sin\,x,\,\,g(x) = x.$

Statement $1:$ $f(x)\, \le \,g\,(x)$ for $x$ in $(0,\infty )$

Statement $2:$ $f(x)\, \le \,1$ for $(x)$ in $(0,\infty )$ but $g(x)\,\to \infty$ as $x\,\to \infty$