The domain of function $f(x)=\sqrt{\sec^{-1}\left(\frac{1-|x|}{2}\right)}$ is

(−∞, −3]∪[3, ∞)

For f(x) to be defined

(i) (a) $\frac{1-|x|}{2}≤-1⇒1-|x|≤-2⇒|x|≥3⇒x≤-3$ or $x≥3$

(b) or  $\frac{1-|x|}{2}≥1⇒1-|x|≥2⇒|x|≤-1$ (not possible)

$∴\frac{1-|x|}{2}≥1$ for no real value of x. Hence x ≤ −3 or x ≥ 3

(ii) $\sec^{-1}\left(\frac{1-|x|}{2}\right)≥0$ this is always true as $0≤\sec^{-1}x≤π,\sec^{-1}x≠\frac{π}{2}$

Hence domain of f = (−∞, −3]∪[3, ∞)