AP EAMCET · Maths · Application of Derivatives
If the function \(y=\sin x(1+\cos x)\) is defined in the interval \([-\pi, \pi]\), then \(y\) is strictly increasing in the interval
- A \(\left(-\pi,-\frac{\pi}{3}\right) \cup\left(\frac{\pi}{3}, \pi\right)\)
- B \(\left(\frac{\pi}{6}, \frac{\pi}{2}\right)\)
- C \(\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)\)
- D \(\left(-\pi,-\frac{\pi}{6}\right) \cup\left(\frac{\pi}{6}, \pi\right)\)
Answer & Solution
Correct Answer
(C) \(\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)\)
Step-by-step Solution
Detailed explanation
\(y = \sin x + \frac{1}{2}\sin(2x)\) \(y' = \cos x + \cos(2x)\) \(y' = \cos x + 2\cos^2 x - 1\) \(2\cos^2 x + \cos x - 1 > 0\) \((2\cos x - 1)(\cos x + 1) > 0\) \(\cos x > 1/2\) (since \(\cos x \ge -1\), \(\cos x + 1 \ge 0\)) \(x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)\)
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