The largest interval lying in $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ for which the function $$\left[f(x) = 4^{-x^2} + \cos^{-1}\left(\frac{x}{2} - 1\right) + \log(\cos x)\right]$$ is defined, is

The domain of $$f(x) = 4^{-x^{2}} + \cos^{-1}\!\left(\frac{x}{2}-1\right) + \log(\cos x)$$ is the set of all real numbers that satisfy the individual domain conditions of every term appearing in the expression.

Since the question itself restricts us to $$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$, we will enforce this outer restriction at the end. Let us examine each term one by one.

1. The exponential term $$4^{-x^{2}}$$
The base 4 is positive, so $$4^{(\text{anything})}$$ is defined for every real $$x$$. Hence this term imposes no extra restriction.

2. The inverse-cosine term $$\cos^{-1}\!\left(\frac{x}{2}-1\right)$$
For the principal value of $$\cos^{-1}(y)$$ to be defined, its argument must satisfy $$-1 \le y \le 1$$. Therefore
$$-1 \le \frac{x}{2}-1 \le 1.$$

Solving these two inequalities:

Lower bound:
$$\frac{x}{2}-1 \ge -1 \quad\Longrightarrow\quad \frac{x}{2} \ge 0 \quad\Longrightarrow\quad x \ge 0.$$

Upper bound:
$$\frac{x}{2}-1 \le 1 \quad\Longrightarrow\quad \frac{x}{2} \le 2 \quad\Longrightarrow\quad x \le 4.$$

Hence from the inverse-cosine term we obtain the interval
$$0 \le x \le 4.$$

3. The logarithmic term $$\log(\cos x)$$
The natural logarithm is defined only for positive arguments, so we need
$$\cos x \gt 0.$$

Inside the given outer interval $$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$, the cosine function is positive for every point except the endpoints, where $$\cos\!\left(\pm\frac{\pi}{2}\right)=0$$. Therefore
$$\cos x \gt 0 \quad\Longrightarrow\quad x \in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$ with strict inequalities at both ends.

4. Combine all the restrictions
We must honour simultaneously:
• $$0 \le x \le 4$$ (from the inverse-cosine term), and
• $$x \in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$ with open ends (from the logarithmic term).

Intersecting these two intervals gives
$$x \in [0, \frac{\pi}{2})$$ because:
• The lower endpoint 0 is allowed (cos 0 = 1 > 0 and the arccos argument equals -1, which is permitted).
• The upper endpoint $$\frac{\pi}{2}$$ is excluded because $$\cos\!\left(\frac{\pi}{2}\right)=0$$ makes the logarithm undefined.

This is the largest interval contained in $$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$ on which every part of the function is defined.

Hence the required interval is $$\left[0, \frac{\pi}{2}\right).$$

Option D which is: $$\left[0, \frac{\pi}{2}\right)$$