If \( f(x)=\cos 2x,0\lt x\lt\pi \), then \( f(x) \) is
(A) Increasing on \( (0,\pi/2) \)
(B) Decreasing on \( (\pi/2,\pi) \)
(C) Increasing on \( (\pi/2,\pi) \)
(D) Decreasing on \( (0,\pi/2) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (C) and (D) only
The sign of the derivative decides where f increases or decreases.
\( \displaystyle f'(x)=-2\sin 2x \)
For \( 0\lt x\lt\frac{\pi}{2} \), \( 0\lt 2x\lt\pi \), so \( \sin 2x\gt 0 \) and f′(x) < 0.
For \( \frac{\pi}{2}\lt x\lt\pi \), \( \pi\lt 2x\lt 2\pi \), so \( \sin 2x\lt 0 \) and f′(x) > 0.
So f is decreasing on \( (0,\pi/2) \) and increasing on \( (\pi/2,\pi) \).
Checking the statements
(A) Increasing on \( (0,\pi/2) \) — Incorrect. f′(x) < 0 there, so f decreases.
(B) Decreasing on \( (\pi/2,\pi) \) — Incorrect. f′(x) > 0 there, so f increases.
(C) Increasing on \( (\pi/2,\pi) \) — Correct. f′(x) > 0 there.
(D) Decreasing on \( (0,\pi/2) \) — Correct. f′(x) < 0 there.