Function f(x)=sin 3x,x∈≤ft[0,(π)/2] is decreasing in the interval

The correct answer is Option (3) → \( \left[\displaystyle\frac{\pi}{6},\displaystyle\frac{\pi}{2}\right] \)

f is decreasing where \( f'(x)\le 0 \):

    \( \displaystyle f'(x)=3\cos 3x \)

For \( x\in\left[0,\frac{\pi}{2}\right] \), 3x runs over \( \left[0,\frac{3\pi}{2}\right] \), and \( \cos 3x\le 0 \) when

    \( \displaystyle \frac{\pi}{2}\le 3x\le\frac{3\pi}{2} \)

    \( \displaystyle \frac{\pi}{6}\le x\le\frac{\pi}{2} \)

So f is decreasing in \( \left[\frac{\pi}{6},\frac{\pi}{2}\right] \).

Why the other options are incorrect

• Option (1) \( \left[0,\displaystyle\frac{\pi}{2}\right] \) — Incorrect. On \( \left[0,\frac{\pi}{6}\right] \) f rises from 0 to 1, so f is not decreasing on the whole of \( \left[0,\frac{\pi}{2}\right] \).

• Option (2) \( \left[0,\displaystyle\frac{\pi}{6}\right] \) — Incorrect. This is the interval where f is increasing, since \( \cos 3x\ge 0 \) there.

• Option (4) \( \left[0,\displaystyle\frac{\pi}{4}\right] \) — Incorrect. f increases on \( \left[0,\frac{\pi}{6}\right] \), which is part of this interval.

NCERT (2026-27 edition):
Increasing and Decreasing Functions