Let the area of the region { ( x , y ) : | 2 x - 1 | y | x 2 - x | , 0 x 1 } be A. Then ( 6 A + 11 ) 2 is equal to __________. [2023]

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Solution

Q.
Let the area of the region ${(x, y) : | 2 x - 1 | \leq y \leq | x^{2} - x |, 0 \leq x \leq 1}$ be A. Then ${(6 A + 11)}^{2}$ is equal to __________. [2023]

Ans.
(125)

$y \geq$ $| 2 x - 1 |$ $,$ $y \leq$ $| x^{2} - x |$

Both curves are symmetric about $x = \frac{1}{2}$

Hence

$A = 2 \int_{\frac{3 - \sqrt{5}}{2}}^{\frac{1}{2}} ((x - x^{2}) - (1 - 2 x)) d x$

$A = 2 \int_{\frac{3 - \sqrt{5}}{2}}^{\frac{1}{2}} (- x^{2} + 3 x - 1) d x = 2 {[\frac{- x^{3}}{3} + \frac{3}{2} x^{2} - x]}_{\frac{3 - \sqrt{5}}{2}}^{\frac{1}{2}}$

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

$On solving, 6 A + 11 = 5 \sqrt{5}, {(6 A + 11)}^{2} = 125$

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