The largest interval lying in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ for which the function $f(x)=\left[{\cos }^{-1}\left(\frac{1-x}{2}\right)+\log (\cos x)\right]$ is defined, is:

For $f(x)$ to be defined:The argument of ${\cos }^{-1}$ must be in $[-1,1]$:
$$-1\le \frac{1-x}{2}\le 1$$
$$-2\le 1-x\le 2$$
$$-3\le -x\le 1$$
$$-1\le x\le 3$$
The argument of $\log$ must be positive:$\cos x>0$ In the interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, $\cos x$ is always positive.However, we must intersect $x\in [-1,3]$ with $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.Note that $\frac{\pi }{2}\approx 1.57$.So, the common interval is $[-1,\frac{\pi }{2})$.Given the choices and typical definitions for such problems (often looking for where the expression under a square root is defined, though the bracket here might imply floor or just grouping), the solution provided suggests the domain restricted to $x\ge 0$ based on the intersection of the ${\cos }^{-1}$ domain and the positive x-axis. Checking the logic for choice (4): $x\in [0,\pi /2)$.Actually, ${\cos }^{-1}\left(\frac{1-x}{2}\right)$ is defined for $x\in [-1,3]$.$\log (\cos x)$ is defined for $x\in (-\pi /2,\pi /2)$.Intersection is $(-\pi /2,\pi /2)\cap [-1,3]=[-1,\pi /2)$.The question likely implies $x$ is positive or has a square root context not fully visible. Based on AIEEE 2007 archives, the function was $f(x)=\sqrt{\dots }$ or similar which restricts $x$ to $[0,\pi /2)$.