AP EAMCET · Maths · Application of Derivatives
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 \(\left(\frac{-5 \pi}{6}, \frac{-\pi}{2}\right) \cup\left(\frac{-\pi}{6}, \frac{\pi}{2}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{-5 \pi}{6}, \frac{-\pi}{2}\right) \cup\left(\frac{-\pi}{6}, \frac{\pi}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(f'(x) = \cos x - 2 \cos x (-\sin x) = \cos x (1+2\sin x)\) \(f'(x) > 0 \implies \cos x (1+2\sin x) > 0\) Condition 1: \(\cos x > 0\) for \(x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Condition 2: \(1+2\sin x > 0 \implies \sin x > -\frac{1}{2}\) for…
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