Let A be the area of the region $left{(mathrm{x}, mathrm{y}): mathrm{y} geq mathrm{x}^2, mathrm{y} geq(1-mathrm{x})^2, mathrm{y} leq 2 mathrm{x}(1-mathrm{x})right}$. Then 540 is equal to

Let be the area of the region <img alt="\mathrm{\left\{(x, y): y \geq x^{2}, y \geq(1-x)^{2}, y \leq

Let $\mathrm{A}$ be the area of the region $\mathrm{\left\{(x, y): y \geq x^{2}, y \geq(1-x)^{2}, y \leq 2 x(1-x)\right\}}$ . Then 540 is equal to_____.
Option: 1
25

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

$\mathrm{x^{2}=(1-x)^{2} }$        $\mathrm{x^{2}=2 x-2 x^{2} }$                  $\mathrm{(1-x)^{2}+2 x-2 x^{2}}$
$\mathrm{x^{3}=1+x^{2}-2 x}$    $\mathrm{ 3 x^{1}=2 x}$ $\mathrm{ 1+x^{2}-2 x=2 x-2 x^{2} }$
$\mathrm{ x=\frac{1}{2} }$                               $\mathrm{ x(3 x-2)=0 }$                   $\mathrm{ \Rightarrow 3 x^{2}-4 x+1=0 }$
$\mathrm{ x=0, \frac{2}{3} }$                             $\mathrm{ \Rightarrow 3 x^{2}-3 x-x+1=0 }$
                                             $\mathrm{ \Rightarrow 3 x(x-1)-1(x-1)=0 }$
                                              $\mathrm{x=1, \frac{1}{3} }$

Required Area
$\mathrm{A=\int_{\frac{1}{3}}^{\frac{1}{2}}\left\{\left(2 x-2 x^{2}\right)-(1-x)^{2}\right\} d x+\int_{\frac{1}{2}}^{\frac{2}{3}}\left(\left(2 x-2 x^{2}\right)-x^{2}\right) d x}$
$\mathrm{\Rightarrow\left[x^{2}-\frac{2 x^{3}}{3}+\frac{(1-x)^{3}}{3}\right]_{\frac{1}{3}}^{\frac{1}{2}}+\left(x^{2}-x^{3}\right)_{1 / 2}^{2 / 3}}$
$\mathrm{\Rightarrow\left(\frac{1}{4}-\frac{2}{3} \cdot \frac{1}{8}+\frac{1}{8.3}\right)-\left(\frac{1}{9}-\frac{2}{3} \cdot \frac{1}{27}+\frac{8}{27.3}\right)+\left(\frac{4}{9}-\frac{8}{27}\right)-\left(\frac{1}{4}-\frac{1}{8}\right)}$
$\mathrm{\Rightarrow-\frac{1}{24}-\frac{1}{9}-\frac{6}{3 \times 27}+\frac{4}{9}-\frac{8}{27}+\frac{1}{8}}$
$\mathrm{\Rightarrow-\frac{1}{24}+\frac{3}{9}-\frac{10}{27}+\frac{3}{24}=\frac{-27+216-240+81}{24 \times 27}=\frac{297-267}{24 \times 27}=\mathrm{A}}$
$\mathrm{540 \mathrm{~A}=540 \times \frac{30}{24 \times 27}=25}$