Which of the following functions are strictly decreasing on \( \left(0,\displaystyle\frac{\pi}{2}\right) \)?
(A) \( \cos x \)
(B) \( -\cot x \)
(C) \( \cos 2x \)
(D) \( \sin x \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
A function is strictly decreasing on an interval if its derivative is negative throughout the interval. For \( 0\lt x\lt\frac{\pi}{2} \), both sin x and cos x are positive, and \( 0\lt 2x\lt\pi \), so sin 2x is positive too.
\( \displaystyle \frac{d}{dx}(\cos x)=-\sin x \)
\( \displaystyle \frac{d}{dx}(-\cot x)=\mathrm{cosec}^{2}x \)
\( \displaystyle \frac{d}{dx}(\cos 2x)=-2\sin 2x \)
\( \displaystyle \frac{d}{dx}(\sin x)=\cos x \)
So on \( \left(0,\frac{\pi}{2}\right) \) the derivatives of cos x and cos 2x are negative, and the derivatives of −cot x and sin x are positive.
Checking the statements
(A) \( \cos x \) — Correct. The derivative \( -\sin x \) is negative, so it is strictly decreasing.
(B) \( -\cot x \) — Incorrect. The derivative \( \mathrm{cosec}^{2}x \) is positive, so it is strictly increasing.
(C) \( \cos 2x \) — Correct. The derivative \( -2\sin 2x \) is negative, so it is strictly decreasing.
(D) \( \sin x \) — Incorrect. The derivative \( \cos x \) is positive, so it is strictly increasing.