Question

The domain of $f\left(x\right)={\sin }^{-1}x+\sqrt{16-x^2}$ is

• A. $(-4,4)$
• B. $[4,4]$
• C. $[-1,1]$
• D. $(0,\infty )$

Answer 1

The correct option is C $[-1,1]$

Function $f\left(x\right)={\sin }^{-1}x+\sqrt{16-x^2}$ contains $2$ functions ${\sin }^{-1}x$ and $\sqrt{16-x^2}$

Thus domain of the function is value of $x$ satisfying both the functions

For ${\sin }^{-1}x$ , $-1\le x\le 1$ and

For $\sqrt{16-x^2}$ , $16-x^2\ge 0$

$-4\le x\le 4$

Therefore, $x\in [-1,1]$

Hence, option 'C' is correct.