QJEEM22Maaafd: JEE Main 2022 Mathematics - Class 12 JEEM22Maaafd Solution: If for some > 0, the area of the region is equal to , then the…

Point A is the intersection of y = x + $\alpha$ and y = 2 $-$ x lines.

$\therefore$ $y = 2 - y + \alpha$

$\Rightarrow 2y = 2 + \alpha$

$\Rightarrow y = {{2 + \alpha } \over 2}$

and $x = 2 - {{2 + \alpha } \over 2} = {{2 - \alpha } \over 2}$

$\therefore$ Point $A = \left( {{{2 - \alpha } \over 2},{{2 + \alpha } \over 2}} \right)$

$\therefore$ AN = y-coordinate of point $A = {{2 + \alpha } \over 2}$

Point B is the intersection of $y = 2 + x$ and $y = - x - \alpha$ lines.

$\therefore$ $y = 2 - y - \alpha$

$\Rightarrow 2y = 2 - \alpha$

$\Rightarrow y = {{2 - \alpha } \over 2}$

$\therefore$ $x = {{2 - \alpha } \over 2} - 2 = {{2 - \alpha - 4} \over 2} = {{ - \alpha - 2} \over 2}$

$\therefore$ Point $B = \left( {{{ - \alpha - 2} \over 2},{{2 - \alpha } \over 2}} \right)$

$\therefore$ $BM = y -$ coordinate of point $B = {{2 - \alpha } \over 2}$

Area of the common region $BRAE$

$= \Delta CDE - \left( {\Delta BCR + \Delta ARD} \right)$

$= {1 \over 2} \times 4 \times 2 - \left( {{1 \over 2}( - \alpha + 2) \times BM + {1 \over 2} \times (2 + \alpha ) \times AN} \right)$

$= 4 - \left( {{1 \over 2} \times (2 - \alpha ) \times {{(2 - \alpha )} \over 2} + {1 \over 2} \times (2 + \alpha ) \times {{2 + \alpha } \over 2}} \right)$

$= 4 - \left[ {{{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4}} \right]$

Given, $4 - \left[ {{{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4}} \right] = {3 \over 2}$

$\Rightarrow {{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4} = {5 \over 2}$

$\Rightarrow {(2 - \alpha )^2} + {(2 + \alpha )^2} = 10$

$\Rightarrow 4 + {\alpha ^2} - 4\alpha + 4 + {\alpha ^2} + 4\alpha = 10$

$\Rightarrow 2{\alpha ^2} + 8 = 10$

$\Rightarrow 2{\alpha ^2} = 2$

$\Rightarrow {\alpha ^2} = 1$

$\Rightarrow \alpha = \, \pm \,1$

Given that \alpha > 0 so accepted value of $\alpha = + \,1$ .

Now, $0 \le y \le x + 2\alpha$ and $|x| \le 1$

$\Rightarrow 0 \le y \le x + 2$ and $- 1 \le x \le 1$

Area of $ABCD = {1 \over 2}(1 + 3) \times (1 - ( - 1))$

$= {1 \over 2} \times 4 \times 2$

$= 4$ sq. unit