The function \(f(x) = \cos 2x\) is strictly decreasing in the interval
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\left(0, \displaystyle\frac{\pi}{2}\right)\)
Differentiating:
\( f'(x) = -2\sin 2x \)
f is strictly decreasing where \(f'(x) \lt 0\), that is, where \(\sin 2x \gt 0\).
For \(0 \lt x \lt \frac{\pi}{2}\), we have \(0 \lt 2x \lt \pi\), so \(\sin 2x \gt 0\) and \(f'(x) \lt 0\) throughout.
Why the other options are incorrect
Option (2) \(\left(\frac{\pi}{2}, \pi\right)\) — Incorrect. Here \(\pi \lt 2x \lt 2\pi\), so \(\sin 2x \lt 0\) and f′(x) > 0: f is increasing.
Option (3) \(\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)\) — Incorrect. f decreases up to \(\frac{\pi}{2}\) and then increases, since \(\sin 2x\) changes sign at \(x = \frac{\pi}{2}\).
Option (4) \(\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\) — Incorrect. For \(-\frac{\pi}{4} \lt x \lt 0\), \(\sin 2x \lt 0\), so f is increasing there; it decreases only after x = 0.