Question

Let $g:R\to [0,\frac{\pi }{3}]$ be a function defined by $g\left(x\right)={\cos }^{-1}\left(\frac{x^2-k}{1+x^2}\right)$. If the absolute value of $k$ for which $g$ is surjective function is $\frac{1}{a}$, then the value of $a$ is

Answer 1

$\frac{1}{2}\le \frac{x^2-k}{x^2+1}\le 1$
Let, $y=\frac{x^2-k}{x^2+1}$
For $g\left(x\right)$ to be surjective $y\in [\frac{1}{2},1]$
$\Rightarrow x^2(1-y)-k-y=0$
$\because x\in R$
$\Rightarrow 4(1-y)(k+y)\ge 0$
$\Rightarrow (y-1)(y+k)\le 0$
For $y\in [\frac{1}{2},1]$
$k=-\frac{1}{2}$
$\Rightarrow |k|=\frac{1}{2}=\frac{1}{a}$
$\therefore a=2$