Question

If $f:R\to [\frac{\pi }{6},\frac{\pi }{2}),f\left(x\right)={\sin }^{-1}\left(\frac{x^2-a}{x^2+1}\right)$ is a onto function, then set of values of $a$ is

• A. $\{-\frac{1}{2}\}$
• B. $[-\frac{1}{2},-1)$
• C. $(-1,\infty )$
• D. none of these

Answer 1

Answer: (C)

$(-1,\infty )$

Given, $f\left(x\right)$ is onto.
$\therefore \frac{\pi }{6}\le {\sin }^{-1}\left(\frac{-x^2-a}{x^2+1}\right)<\frac{\pi }{2}$
$\Rightarrow \frac{1}{2}\le \frac{x^2-a}{x^2+1}<1$
$\Rightarrow \frac{1}{2}\le 1-\frac{(a+1)}{x^2+1}<1,\forall xϵR$
$\Rightarrow a+1>0$
$\Rightarrow a\in (-1,\infty )$
Hence, (c) is the correct answer.