The domain of the function $f(x)=\sqrt{4-x^2}+\sin^{-1}\frac{1+x^2}{2x}$ is
Answer & explanation
Correct answer: option 2
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}