Question:
The function \(\sin x(1+\cos x)\), \(0 \le x \le \pi/2\), has maximum value when \(x\) is
The function \(\sin x(1+\cos x)\), \(0 \le x \le \pi/2\), has maximum value when \(x\) is
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When trig functions are involved, convert to \(\cos x\) or \(\sin x\) to solve easily.
Updated On: Apr 18, 2026
- 0
- \(\pi/2\)
- \(\pi/6\)
- None of these
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The Correct Option is D
Solution and Explanation
Concept:
Use differentiation to find maxima.
Step 1: Define the function. \[ f(x) = \sin x(1+\cos x) = \sin x + \sin x \cos x \]
Step 2: Differentiate. \[ f'(x) = \cos x + (\cos^2 x - \sin^2 x) = \cos x + \cos 2x \]
Step 3: Set \(f'(x)=0\). \[ \cos x + \cos 2x = 0 \] Using \(\cos 2x = 2\cos^2 x - 1\): \[ \cos x + 2\cos^2 x - 1 = 0 \] Let \(t = \cos x\): \[ 2t^2 + t - 1 = 0 \Rightarrow (2t-1)(t+1) = 0 \] \[ t = \frac{1}{2} \Rightarrow \cos x = \frac{1}{2} \Rightarrow x = \frac{\pi}{3} \] \(t = -1 \Rightarrow \cos x = -1 \Rightarrow x = \pi\) (not in given interval)
Step 4: Check interval. \(x = \frac{\pi}{3}\) lies in \([0, \pi/2]\) and is valid.
Step 5: Compare with options. \[ \frac{\pi}{3} \notin \{0, \frac{\pi}{2}, \frac{\pi}{6}\} \]
Step 1: Define the function. \[ f(x) = \sin x(1+\cos x) = \sin x + \sin x \cos x \]
Step 2: Differentiate. \[ f'(x) = \cos x + (\cos^2 x - \sin^2 x) = \cos x + \cos 2x \]
Step 3: Set \(f'(x)=0\). \[ \cos x + \cos 2x = 0 \] Using \(\cos 2x = 2\cos^2 x - 1\): \[ \cos x + 2\cos^2 x - 1 = 0 \] Let \(t = \cos x\): \[ 2t^2 + t - 1 = 0 \Rightarrow (2t-1)(t+1) = 0 \] \[ t = \frac{1}{2} \Rightarrow \cos x = \frac{1}{2} \Rightarrow x = \frac{\pi}{3} \] \(t = -1 \Rightarrow \cos x = -1 \Rightarrow x = \pi\) (not in given interval)
Step 4: Check interval. \(x = \frac{\pi}{3}\) lies in \([0, \pi/2]\) and is valid.
Step 5: Compare with options. \[ \frac{\pi}{3} \notin \{0, \frac{\pi}{2}, \frac{\pi}{6}\} \]
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