The function \(f(x) = \sin 3x,\ x \in \left[0, \displaystyle\frac{\pi}{2}\right]\)
(A) is increasing on \(\left[0, \displaystyle\frac{\pi}{6}\right]\)
(B) is decreasing on \(\left[\displaystyle\frac{\pi}{6}, \frac{\pi}{2}\right]\)
(C) is increasing on \(\left[0, \displaystyle\frac{\pi}{2}\right]\)
(D) is decreasing on \(\left[0, \displaystyle\frac{\pi}{2}\right]\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (B) only
The sign of the derivative decides where f increases or decreases:
\( \displaystyle f'(x) = 3\cos 3x \)
\( \displaystyle 0 \le x \le \frac{\pi}{6} \Rightarrow 0 \le 3x \le \frac{\pi}{2} \Rightarrow \cos 3x \ge 0 \)
\( \displaystyle \frac{\pi}{6} \le x \le \frac{\pi}{2} \Rightarrow \frac{\pi}{2} \le 3x \le \frac{3\pi}{2} \Rightarrow \cos 3x \le 0 \)
In each case \(f'\) is zero only at the end points.
So f is increasing on \(\left[0, \frac{\pi}{6}\right]\) and decreasing on \(\left[\frac{\pi}{6}, \frac{\pi}{2}\right]\), with \(f(0) = 0\), \(f\left(\frac{\pi}{6}\right) = 1\) and \(f\left(\frac{\pi}{2}\right) = -1\).
Checking the statements
(A) is increasing on \(\left[0, \frac{\pi}{6}\right]\) — Correct. \(f' \ge 0\) there.
(B) is decreasing on \(\left[\frac{\pi}{6}, \frac{\pi}{2}\right]\) — Correct. \(f' \le 0\) there.
(C) is increasing on \(\left[0, \frac{\pi}{2}\right]\) — Incorrect. f falls from 1 to −1 after \(\frac{\pi}{6}\).
(D) is decreasing on \(\left[0, \frac{\pi}{2}\right]\) — Incorrect. f first rises from 0 to 1.
f is not monotonic on the whole interval.
NCERT (2026-27 edition):
Increasing and Decreasing Functions
