Question

Domain of the function,
$f\left(x\right)=\sqrt{\cos \left(\sin x\right)}+{\sin }^{-1}(x^2-1)$ is

• A. $[-1,1]$
• B. $[-2,2]$
• C. $[-\pi ,-\sqrt{2}]\cup [\sqrt{2},\pi ]$
• D. $[-\sqrt{2},\sqrt{2}]$

Answer 1

The correct option is D $[-\sqrt{2},\sqrt{2}]$

$f\left(x\right)=\sqrt{\cos \left(\sin x\right)}+{\sin }^{-1}(x^2-1)$

$-1\le \sin x\le 1$ $\forall x\in R$

$\therefore \cos \left(\sin x\right)>0$ $\forall x\in R$

$\therefore$ domain of $\sqrt{\cos \left(\sin x\right)}$ is $R$ .....(i)

$-1\le x^2-1\le 1$ i.e. $0\le x^2\le 2$

i.e. $x\in [-\sqrt{2},\sqrt{2}]$

$\therefore$ domain of ${\sin }^{-1}(x^2-1)$ is $[-\sqrt{2},\sqrt{2}]$ ....(ii)

From (i) and (ii)

Domain of $f\left(x\right)$ is $[-\sqrt{2},\sqrt{2}]$