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 [AP EAPCET 26 May 2025 S1]

Correct Answer is : (−5π6,−π2)∪(−π6,π2)

(\frac{-5 \pi}{6}, \frac{-\pi}{6}) \cup (\frac{-\pi}{6}, \frac{\pi}{2})

(\frac{-\pi}{2}, \frac{-\pi}{6})

(\frac{-5 \pi}{6}, \frac{\pi}{2})

(\frac{-5 \pi}{6}, \frac{-\pi}{2}) \cup (\frac{-\pi}{6}, \frac{\pi}{2})

AP EAPCET 26 May 2025 Shift 1 Paper

Section: Mathematics

Question : 67 of 160

Marks: +1, -0

If the function f(x)=sin‌x−cos2x is defined on the interval [−π,π], then f is strictly increasing in the interval

[AP EAPCET 26 May 2025 S1]

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