QJEEM23Ma2324: JEE Main 2023 Mathematics - Class 12 JEEM23Ma2324 Solution: Let the area of the region be . Then is equal to

Step-by-Step Verified Solution

For $B$ ,

$$\begin{array}{rl}& x-x^2=2x-1\\ \\ & x^2+x-1=0\\ \\ & x=\frac{-1+\sqrt{5}}{2}\end{array}$$

Area $=2($ area of $B C E)$

$A=2 \int_{\frac{1}{2}}^{\frac{\sqrt{5}-1}{2}}\left(x-x^{2}\right)-(2 x-1) d x$

$=2 \int_{\frac{1}{2}}^{\frac{\sqrt{5}-1}{2}} (1-x-x^{2}) d x=\left.2\left(x-\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)\right|_{\frac{1}{2}} ^{]_{5}^{\frac{\sqrt{5}-1}{2}}}$

$$\begin{array}{rl}=2\{(\frac{\sqrt{5}-1}{2}-\frac{1}{2})-\{{\left(\frac{\sqrt{5}-1}{2}\right)}^2& -\frac{1}{4}\}\frac{1}{2}\\ \\ & -[{\left(\frac{\sqrt{5}-1}{2}\right)}^3-\frac{1}{8}]\frac{1}{3}\}\end{array}$$

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

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