Question

Let $g: R \rightarrow(0, \pi / 3]$ is defined by $g(x)=\cos ^{-1}\left(\frac{x^2-k}{1+x^2}\right)$. Then find the possible values of ' $k$ ' for which $g$ is surjective.

Answer 1

For $g(x)$ to be surjective $\forall x \in R \Rightarrow 0<\cos ^{-1}\left(\frac{x^2-k}{1+x^2}\right) \leq \pi / 3 \Rightarrow \frac{1}{2} \leq \frac{x^2-k}{x^2+1}<1$

$\because \quad \mathrm{x}^2+1>0 \quad \forall \mathrm{x} \in \mathrm{R} \\$

$\frac{1}{2}\left(\mathrm{x}^2+1\right) \leq \mathrm{x}^2-\mathrm{k}<\mathrm{x}^2+1$

From Eq. (1), taking RHS

$x^2-k<x^2+1 \quad \Rightarrow \quad k>-1$

From Eq. (1), taking LHS

$\mathrm{x}^2+1 \leq 2 \mathrm{x}^2-2 \mathrm{k}  \Rightarrow \quad \mathrm{x}^2 \geq 2 \mathrm{k}+1  \forall \mathrm{x} \in \mathrm{R} \\$

$2 \mathrm{k}+1 \leq 0  \Rightarrow \quad \mathrm{k} \leq \frac{-1}{2}  \Rightarrow$