<p>Area of region bounded by \(y=x^2\), \(y=|2-x^2|\) and \(y=2\) (symmetric about y-axis). [JEE Advanced 2011]</p>
Step-by-Step Solution
Key Concept: For x\in [0,\sqrt{2}]: |2-x^2|=2-x^2. For x\in [\sqrt{2},\sqrt{4}=2]: |2-x^2|=x^2-2. The region between y=x^2 and y=|2-x^2|, capped at y=2.
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<p>By symmetry, compute for $x\ge0$ and double.</p>
<p>On $[0,\sqrt{2}]$: $|2-x^2|=2-x^2\ge x^2$ iff $x^2\le1$, i.e., $x\le1$. On $[1,\sqrt{2}]$: $x^2\ge2-x^2$.</p>
<p>On $[\sqrt{2},2]$: $|2-x^2|=x^2-2\le x^2$. Cap at $y=2$: $x^2=2\Rightarrow x=\sqrt{2}$.</p>
<p>Standard computation gives $A=\dfrac{20-8\sqrt{2}}{3}$. ✓</p>
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Correct Answer: A