Which of the following functions are increasing on \(x \in \left(0,\displaystyle\frac{\pi}{2}\right)\)?
(A) \(f(x) = \sin x\)
(B) \(f(x) = \cos x\)
(C) \(f(x) = \tan x\)
(D) \(f(x) = \cos 3x\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
A function is increasing on the interval when \(f'(x) \gt 0\) throughout it.
(A) \(f(x) = \sin x\) — Correct. \(f'(x) = \cos x \gt 0\) for \(0 \lt x \lt \frac{\pi}{2}\).
(B) \(f(x) = \cos x\) — Incorrect. \(f'(x) = -\sin x \lt 0\) on the interval, so cos x is decreasing.
(C) \(f(x) = \tan x\) — Correct. \(f'(x) = \sec^2 x \gt 0\) on the interval.
(D) \(f(x) = \cos 3x\) — Incorrect. It decreases and then increases:
\(f'(x) = -3\sin 3x\)
For \(0 \lt x \lt \frac{\pi}{3}\), 3x lies in (0, π), so sin 3x > 0 and f decreases. For \(\frac{\pi}{3} \lt x \lt \frac{\pi}{2}\), 3x lies in (π, 3π/2), so sin 3x < 0 and f increases. It is not increasing on the whole interval.