JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 16)
Let the area of the region
$\left\{(x, y):|2 x-1| \leq y \leq\left|x^{2}-x\right|, 0 \leq x \leq 1\right\}$ be $\mathrm{A}$.
Then $(6 \mathrm{~A}+11)^{2}$ is equal to
$\left\{(x, y):|2 x-1| \leq y \leq\left|x^{2}-x\right|, 0 \leq x \leq 1\right\}$ be $\mathrm{A}$.
Then $(6 \mathrm{~A}+11)^{2}$ is equal to
Answer
125
Explanation
For $B$,
$$ \begin{aligned} & x-x^{2}=2 x-1 \\\\ & x^{2}+x-1=0 \\\\ & x=\frac{-1+\sqrt{5}}{2} \end{aligned} $$_31st_January_Evening_Shift_en_16_1.png)
Area $=2($ area of $B C E)$
$A=2 \int_{\frac{1}{2}}^{\frac{\sqrt{5}-1}{2}}\left(x-x^{2}\right)-(2 x-1) d x$
$=2 \int_{\frac{1}{2}}^{\frac{\sqrt{5}-1}{2}} (1-x-x^{2}) d x=\left.2\left(x-\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)\right|_{\frac{1}{2}} ^{]_{5}^{\frac{\sqrt{5}-1}{2}}}$
$$ \begin{aligned} =2\left\{\left(\frac{\sqrt{5}-1}{2}-\frac{1}{2}\right)-\left\{\left(\frac{\sqrt{5}-1}{2}\right)^{2}\right.\right. & \left.-\frac{1}{4}\right\} \frac{1}{2} \\\\ & \left.-\left[\left(\frac{\sqrt{5}-1}{2}\right)^{3}-\frac{1}{8}\right] \frac{1}{3}\right\} \end{aligned} $$
$$ A=\frac{-11+5 \sqrt{5}}{6} $$
$\Rightarrow(6 A+11)^{2}=125$
$$ \begin{aligned} & x-x^{2}=2 x-1 \\\\ & x^{2}+x-1=0 \\\\ & x=\frac{-1+\sqrt{5}}{2} \end{aligned} $$
Area $=2($ area of $B C E)$
$A=2 \int_{\frac{1}{2}}^{\frac{\sqrt{5}-1}{2}}\left(x-x^{2}\right)-(2 x-1) d x$
$=2 \int_{\frac{1}{2}}^{\frac{\sqrt{5}-1}{2}} (1-x-x^{2}) d x=\left.2\left(x-\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)\right|_{\frac{1}{2}} ^{]_{5}^{\frac{\sqrt{5}-1}{2}}}$
$$ \begin{aligned} =2\left\{\left(\frac{\sqrt{5}-1}{2}-\frac{1}{2}\right)-\left\{\left(\frac{\sqrt{5}-1}{2}\right)^{2}\right.\right. & \left.-\frac{1}{4}\right\} \frac{1}{2} \\\\ & \left.-\left[\left(\frac{\sqrt{5}-1}{2}\right)^{3}-\frac{1}{8}\right] \frac{1}{3}\right\} \end{aligned} $$
$$ A=\frac{-11+5 \sqrt{5}}{6} $$
$\Rightarrow(6 A+11)^{2}=125$
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