CUET · MATHS · PYQ PAPER 2023
The function \(f(x)=\sin x(1+\cos x) ; x \in\left[0, \frac{3 \pi}{2}\right]\) will have:
- A Maximum value \(\frac{3 \sqrt{3}}{4}\) at \(x=\frac{\pi}{3}\)
- B Minimum value = 0, at \(x=\pi\)
- C Minimum value = 0 at \(x=\frac{3 \pi}{2}\)
- D Maximum value = \(\frac{3 \sqrt{3}}{4}\) at \(x=\frac{3 \pi}{2}\)
Answer & Solution
Correct Answer
(A) Maximum value \(\frac{3 \sqrt{3}}{4}\) at \(x=\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
\(f'(x) = \frac{d}{dx}(\sin x + \sin x \cos x) = \frac{d}{dx}(\sin x + \frac{1}{2}\sin 2x) = \cos x + \cos 2x\) \(f'(x) = 0 \implies \cos x + (2\cos^2 x - 1) = 0 \implies 2\cos^2 x + \cos x - 1 = 0\) \((2\cos x - 1)(\cos x + 1) = 0 \implies \cos x = \frac{1}{2}\) or…
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