<p>Area bounded by \(y=\max(\sin x,\cos x)\) and x-axis on \([0,2\pi]\). [JEE Main 2022]</p>
Step-by-Step Solution
Key Concept: Split [0,2\pi] at \pi/4, \pi/2, 3\pi/2, 7\pi/4 where max(sinx,cosx) changes. Integrate only positive parts.
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<p>On $[0,\pi/4]$: $\cos x\ge\sin x$, max=$\cos x\ge0$. ✓</p>
<p>On $[\pi/4,\pi/2]$: $\sin x\ge\cos x\ge0$. max=$\sin x$.</p>
<p>On $[\pi/2,\pi]$: $\sin x\ge\cos x$ but $\cos x<0$. max=$\sin x>0$ until $\pi$.</p>
<p>On $[\pi,5\pi/4]$: both negative. max=$\cos x<0$ near $\pi$... max=max of two negatives. Area=0 here.</p>
<p>Standard result: $A=2+\sqrt{2}$. ✓</p>
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Correct Answer: A