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:

If $f(x) = \frac{x}{{x - 1}} = \frac{1}{y}$, then $f(y) = $

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 $........$

If $f(x)$ satisfies the relation $f\left( {\frac{{5x - 3y}}{2}} \right)\, = \,\frac{{5f(x) - 3f(y)}}{2}\,\forall x,y\in R$ $f(0) = 1, f '(0) = 2$ then period of $sin \ (f(x))$ is

The domain of the function

$f(x)=\frac{\cos ^{-1}\left(\frac{x^{2}-5 x+6}{x^{2}-9}\right)}{\log _{e}\left(x^{2}-3 x+2\right)} \text { is }$