The function f(x) = log(sin 2x) , x ∈ ≤ft(0, (π)/2) , is strictly inc…

The correct answer is Option (4) → \(\left(0, \displaystyle\frac{\pi}{4}\right)\)

By the chain rule:

    \( \displaystyle f'(x) = \frac{1}{\sin 2x} \times 2\cos 2x = 2\cot 2x \)

For x in \(\left(0, \frac{\pi}{2}\right)\), the angle 2x lies in \((0, \pi)\), where sin 2x > 0. So the sign of f′(x) is the sign of cos 2x:

    \( \cos 2x \gt 0 \) for \( \displaystyle 0 \lt 2x \lt \frac{\pi}{2} \), that is \( \displaystyle 0 \lt x \lt \frac{\pi}{4} \)
    \( \cos 2x \lt 0 \) for \( \displaystyle \frac{\pi}{4} \lt x \lt \frac{\pi}{2} \)

So f is strictly increasing in \(\left(0, \frac{\pi}{4}\right)\) and strictly decreasing in \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\).