The function \(f(x) = x + \cos 2x\), \(0 \le x \le \displaystyle\frac{\pi}{2}\), is strictly decreasing in
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\left(\displaystyle\frac{\pi}{12}, \frac{5\pi}{12}\right)\)
Differentiating,
\(f'(x) = 1 - 2\sin 2x\)
f is strictly decreasing where \(f'(x) \lt 0\):
\( \displaystyle \sin 2x \gt \frac{1}{2} \)
For \(0 \le x \le \frac{\pi}{2}\), the angle 2x runs over \([0, \pi]\), and \(\sin 2x \gt \frac{1}{2}\) exactly when
\( \displaystyle \frac{\pi}{6} \lt 2x \lt \frac{5\pi}{6} \)
\( \displaystyle \frac{\pi}{12} \lt x \lt \frac{5\pi}{12} \)
Outside this interval \(\sin 2x \lt \frac{1}{2}\), so f is increasing in \(\left(0, \frac{\pi}{12}\right)\) and in \(\left(\frac{5\pi}{12}, \frac{\pi}{2}\right)\). Each of the other options contains one of these pieces.