QJEEAdv22Mabfcc: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Mabfcc Solution: Consider the functions defined by If is the area of the region…

$f(x) = {x^2} + {5 \over {12}}$

This represent upward parabola.

$g(x) = 2\left( {1 - {{4|x|} \over 3}} \right)$

At $x = 0$

$g(0) = 2$

At $x = {3 \over 4}$

$g\left( {{3 \over 4}} \right) = 0$

$\therefore$ Graph is

Intersection point of f(x) and g(x) at first quadrant,

$f(x) = g(x)$

$\Rightarrow {x^2} + {5 \over {12}} = 2\left( {1 - {{4x} \over 3}} \right)$ [ $|x| = x$ as in first quadrant x > 0 ]

$\Rightarrow 12{x^2} + 5 = 24 - 32x$

$\Rightarrow 12{x^2} + 32x - 19 = 0$

$\Rightarrow 12{x^2} + 38x - 6x - 19 = 0$

$\Rightarrow 2x(6x + 19) - 1(6x + 19) = 0$

$\Rightarrow (2x - 1)(6x + 19) = 0$

$\therefore$ $x = {1 \over 2},\, - {{19} \over 6}$

In first quadrant $x = {1 \over 2}$

When $x = {1 \over 2}$ ,

$y = {\left( {{1 \over 2}} \right)^2} + {5 \over {12}} = {2 \over 3}$

$\therefore$ Area $A = 2$ ( $\int\limits_0^{{1 \over 2}} {\left( {{x^2} + {5 \over {12}}} \right)dx + }$ Area of $\Delta ABC$ )

$= 2\left[ {\int\limits_0^{{1 \over 2}} {\left( {{x^2} + {5 \over {12}}} \right)dx + {1 \over 2} \times \left( {{3 \over 4} - {1 \over 2}} \right) \times {2 \over 3}} } \right]$

$= 2\left[ {\left[ {{{{x^3}} \over 3} + {{5x} \over {12}}} \right]_0^{{1 \over 2}} + {1 \over 2} \times {1 \over 4} \times {2 \over 3}} \right]$

$= 2\left[ {\left( {{1 \over {24}} + {5 \over {12}} \times {1 \over 2}} \right) + {1 \over {12}}} \right]$

$= 2\left[ {{1 \over {24}} + {5 \over {24}} + {2 \over {24}}} \right]$

$= 2\left[ {{{1 + 5 + 2} \over {24}}} \right]$

$= 2 \times {8 \over {24}}$

$= {2 \over 3} = \alpha$

$\therefore$ $= 9\alpha = {2 \over 3} \times 9 = 6$