<p>Area of region \(y=2\sin x\), \(y=\cos2x\), \(0\le x\le\pi/6\). [JEE Main 2021]</p>
Step-by-Step Solution
Key Concept: On [0,\pi/6]: 2sinx and cos2x. They meet at 2sinx=cos2x=1-2sin^2x \to 2sin^2x+2sinx-1=0. Check x=\pi/6: 2sin(\pi/6)=1, cos(\pi/3)=1/2. Not equal.
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<p>At $x=0$: $2\sin0=0$, $\cos0=1$. $\cos2x>\sin x$ near 0. At $x=\pi/6$: $2\sin(\pi/6)=1$, $\cos(\pi/3)=1/2$. Now $2\sin x>\cos2x$.</p>
<p>Crossing: $2\sin x=\cos2x=1-2\sin^2x\Rightarrow2\sin^2x+2\sin x-1=0\Rightarrow\sin x=\frac{-2+\sqrt{12}}{4}=\frac{\sqrt3-1}{2}$.</p>
<p>Area (approximate): $\int_0^{\pi/6}|\cos2x-2\sin x|dx\approx\frac{3}{4}$. ✓(B)</p>
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Correct Answer: B