JEE Main 2023 — Mathematics Calculus

Given,

$A$A be the area of the region

$${(x,y):y\geq {x}^{2},y\geq (1-x{)}^{2},y\leq 2x(1-x)}$$(x,y):y≥x2,y≥(1−x)2,y≤2x(1−x)

Now solving y={x}^{2}&y=2x(1-x) we get, $x=0,\frac{2}{3}$x=0,32​

And solving y={(1-x)}^{2}&y=2x(1-x) we get,

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

$$\Rightarrow 3{x}^{2}-4x+1=0$$⇒3x2−4x+1=0

$$\Rightarrow x=1,\frac{1}{3}$$⇒x=1,31​

Now on plotting the diagram of the above region we get,

Now from the above diagram area of the shaded region will be,

$$A={\int }_{\frac{1}{3}}^{\frac{2}{3}}(2x-2{x}^{2})dx-{{\int }_{\frac{1}{3}}^{\frac{1}{2}}{(1-x)}^{2}dx+{\int }_{\frac{1}{2}}^{\frac{2}{3}}{x}^{2}dx}$$A=∫31​32​​(2x−2x2)dx−∫31​21​​(1−x)2dx+∫21​32​​x2dx

$$\Rightarrow A={[{x}^{2}-\frac{2{x}^{3}}{3}]}_{\frac{1}{3}}^{\frac{2}{3}}-{{[\frac{{(x-1)}^{3}}{3}]}_{\frac{1}{3}}^{\frac{1}{2}}+{[\frac{{x}^{3}}{3}]}_{\frac{1}{2}}^{\frac{2}{3}}}$$⇒A=[x2−32x3​]31​32​​−[3(x−1)3​]31​21​​+[3x3​]21​32​​

$$\Rightarrow A=\frac{5}{108}$$⇒A=1085​

$$\therefore A=\frac{5}{108}\Rightarrow 540A=\frac{5}{108}\times 540=25$$∴A=1085​⇒540A=1085​×540=25