Question

Find the domain of $f\left(x\right)=\sqrt{\sin x}+\sqrt{16-x^2}$.

Answer 1

$f\left(x\right)=\sqrt{\sin x}+\sqrt{16-x^2}$

Let $f\left(x\right)=f_1\left(x\right)+f_2\left(x\right)$

Here,

$f_1\left(x\right)=\sqrt{\sin x}$

$f_2\left(x\right)=\sqrt{16-x^2}$

Now,

Domain of $f_2\left(x\right)$, i.e., $\sqrt{16-x^2}$-

$16-x^2\ge 0$

$\Rightarrow x^2\le 16$

$\Rightarrow -4\le x\le 4$

$\Rightarrow$ Domain of $f_2\left(x\right)$ is $[-4,4]$.

Domain of $f_1\left(x\right)$, i.e., $\sqrt{\sin x}$-

$\sin x\ge 0$

$\Rightarrow 0\le \sin x\le 1$

$\Rightarrow {\sin }^{-1}0\le x\le {\sin }^{-1}1$

$\Rightarrow 2n\pi \le x\le (2n+1)\pi$

$\Rightarrow$ Domain of $f_1\left(x\right)$ is $[2n\pi ,(2n+1)\pi ]$

For $n=0\Rightarrow 0\le x\le \pi$

For $n=1\Rightarrow 2\pi \le x\le 3\pi$

Now domain of the function $f\left(x\right)$ will be-

Domain of $f\left(x\right)\equiv$ Domein of $f_1\left(x\right)\cap$ Domain of $f_2\left(x\right)$

$\therefore$ Domain of $f\left(x\right)\equiv -4\le 0\le x\le \pi \le 4$

$\Rightarrow$ Domain of $f\left(x\right)\equiv 0\le x\le \pi$