JEE Main 2023 — Mathematics Calculus

Given,

The area of the region ${(x,y):|2x-1|\leq y\leq |{x}^{2}-x|,0\leq x\leq 1}$(x,y):∣2x−1∣≤y≤∣x2−x∣,0≤x≤1 be $A$A,

Now solving,

$$x-{x}^{2}=2x-1$$x−x2=2x−1

$$\Rightarrow {x}^{2}+x-1=0$$⇒x2+x−1=0

$$\Rightarrow x=\frac{-1\pm \sqrt{5}}{2}$$⇒x=2−1±5​​

And equating $x-{x}^{2}=1-2x$x−x2=1−2x, we get $x=\frac{3\pm \sqrt{5}}{2}$x=23±5​​

Now plotting the diagram of $y\geq |2x-1|$y≥∣2x−1∣ and $y\leq |{x}^{2}-x|$y≤∣x2−x∣,

Now from diagram, both curve are symmetric about $x=\frac{1}{2}$x=21​

Hence, area of the bounded region is given by,

$$A=|2{\int }_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}((x-{x}^{2})-(1-2x))dx|$$A=∣2∫23−5​​21​​((x−x2)−(1−2x))dx∣

$$\Rightarrow A=|2{\int }_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}(-{x}^{2}+3x-1)dx|$$⇒A=∣2∫23−5​​21​​(−x2+3x−1)dx∣

$$\Rightarrow A=|2{(\frac{-{x}^{3}}{3}+\frac{3}{2}{x}^{2}-x)}_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}|$$⇒A=∣2(3−x3​+23​x2−x)23−5​​21​​∣

$$\Rightarrow A=\frac{5\sqrt{5}-11}{6}$$⇒A=655​−11​

$$\Rightarrow 6A+11=5\sqrt{5}$$⇒6A+11=55​

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