JEE Main 2023 — Area Under Curves Question with Solution

Given,

The area of the region $\left\{\left(x,y\right):\left|2x-1\right|\le y\le \left|x^2-x\right|,0\le x\le 1\right\}$ be $A$,

Now solving,

$x-x^2=2x-1$ 

$\Rightarrow x^2+x-1=0$

$\Rightarrow x=\frac{-1\pm \sqrt{5}}{2}$

And equating $x-x^2=1-2x$, we get $x=\frac{3\pm \sqrt{5}}{2}$

Now plotting the diagram of  $y\ge \left|2x-1\right|$ and $y\le \left|x^2-x\right|$,

Now from diagram, both curve are symmetric about $x=\frac{1}{2}$

Hence, area of the bounded region is given by,

$A=\left|2{\int }_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}\left(\left(x-x^2\right)-\left(1-2x\right)\right)dx\right|$

$\Rightarrow A=\left|2{\int }_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}\left(-x^2+3x-1\right)dx\right|$

$\Rightarrow A=\left|2{\left(\frac{-x^3}{3}+\frac{3}{2}{x}^2-x\right)}_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}\right|$

$\Rightarrow A=\frac{5\sqrt{5}-11}{6}$

$\Rightarrow 6A+11=5\sqrt{5}$

$\Rightarrow {\left(6A+11\right)}^2=125$.