Question

If the function $$ f(x) = \sin x - \cos^2 x $$ is defined on the interval $ [-\pi, \pi] $, then $ f $ is strictly increasing in the interval:

• A. \( \left(-\frac{5\pi}{6}, -\frac{\pi}{6}\right) \cup \left(-\frac{\pi}{6}, \frac{\pi}{2}\right) \)
• B. \( \left(-\frac{\pi}{2}, -\frac{\pi}{6}\right) \)
• C. \( \left(-\frac{5\pi}{6}, \frac{\pi}{2}\right) \)
• D. \( \left(-\frac{5\pi}{6}, -\frac{\pi}{2}\right) \cup \left(-\frac{\pi}{6}, \frac{\pi}{2}\right) \)

Hint:

Use product rule and sign analysis for increasing intervals.

Answer 1

The Correct Option is D

Find derivative: \[ f'(x) = \cos x + 2 \cos x \sin x = \cos x (1 + 2 \sin x) \] \( f \) is increasing when \( f'(x)>0 \): Case 1: \[ \cos x>0 \implies x \in (-\frac{\pi}{2}, \frac{\pi}{2}) \] Case 2: \[ 1 + 2 \sin x>0 \implies \sin x>-\frac{1}{2} \] On \( [-\pi, \pi] \), \( \sin x>-\frac{1}{2} \) holds for: \[ x \in \left(-\frac{5\pi}{6}, \pi \right) \] Combine intervals to find strictly increasing intervals: \[ (-\frac{5\pi}{6}, -\frac{\pi}{2}) \cup (-\frac{\pi}{6}, \frac{\pi}{2}) \]