Which of the following functions is strictly decreasing on the interval \(\left(0, \displaystyle\frac{\pi}{2}\right)\)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\cos x\)
A function is strictly decreasing on an interval when its derivative is negative there.
\(\displaystyle \frac{d}{dx}(\cos x) = -\sin x\)
For \(0 \lt x \lt \displaystyle\frac{\pi}{2}\), \(\sin x \gt 0\), so \(-\sin x \lt 0\). Hence \(\cos x\) is strictly decreasing on this interval.
Why the other options are incorrect
Option (1) \(\sin x\) — Incorrect. Its derivative \(\cos x\) is positive on the interval, so it is increasing.
Option (2) \(\tan x\) — Incorrect. Its derivative \(\sec^2 x\) is positive, so it is increasing.
Option (3) \(\sin 2x\) — Incorrect. Its derivative \(2\cos 2x\) is positive on \(\left(0, \frac{\pi}{4}\right)\), so it rises before it falls.