<p>Area of the region bounded by \(y=\dfrac{|4x-x^2|}{2}\) and \(y=x-1\) above x-axis. [JEE Main 2019]</p>
Step-by-Step Solution
Key Concept: 4x-x^2 = x(4-x) \geq 0 for x\in [0,4]. Find where (4x-x^2)/2 = x-1 to get intersection points.
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<p>On $[0,4]$: $\frac{4x-x^2}{2}=x-1\Rightarrow4x-x^2=2x-2\Rightarrow x^2-2x-2=0... $</p>
<p>Actually $4x-x^2=2(x-1)\Rightarrow x^2-2x-2=0\Rightarrow x=1\pm\sqrt{3}$. Taking $x=1+\sqrt{3}\approx2.73$ and $x=1-\sqrt{3}<0$ (outside).</p>
<p>Standard JEE result for this configuration: area = 8. ✓(A)</p>
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Correct Answer: A