The domain of the function $f(x)=\sqrt{4-x^2}+\sin^{-1}\frac{1+x^2}{2x}$ is

{−1, 1}

For f(x) to be defined

(i) $4 − x^2 ≥ 0 ⇒ −2 ≤ x ≤ 2$ …(i)

(ii) $-1 ≤\frac{1+x^2}{2x}≤1⇒\left|\frac{1+x^2}{2x}\right|≤1$

$⇒\frac{1+x^2}{2|x|}≤1⇒1+x^2≤2|x|⇒1+|x|^2-2|x|≤0$

$⇒(1-|x|)^2≤0⇒(1-|x|)^2=0$

$⇒ | x | = 1 ⇒ x = ± 1$ …(ii)

From (i) and (ii), domain of f = {−1, 1}