QJEEAdv16Mae1ce: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Mae1ce Solution: Area of the region is equal to

Here, $\{ (x,y) \in {R^2}:y \ge \sqrt {\left| {x + 3} \right|} ,\,5y \le (x + 9) \le 15\}$

$\therefore$ $y \ge \sqrt {x + 3}$

$$\Rightarrow y\ge \{\begin{array}{c}\sqrt{x+3},whenx\ge -3\hfill \\ \sqrt{-x-3},whenx\le -3\hfill \end{array}$$

or, $$y^2\ge \{\begin{array}{c}x+3,whenx\ge -3\hfill \\ -3-x,whenx\le -3\hfill \end{array}$$

Shown as

Also, $5y \le (x + 9) \le 15$

$\Rightarrow (x + 9) \ge 5y$ and $x \le 6$

Shown as

$\therefore$ $\{ (x,y) \in {R^2}:y \ge \sqrt {\left| {x + 3} \right|} ,\,5y \le (x + 9) \le 15\}$

$\therefore$ Required area = Area of trapezium ABCD $-$ Area of ABE under parabola $-$ Area of CDE under parabola

$= {1 \over 2}(1 + 2)(5) - \int_{ - 4}^{ - 3} {\sqrt { - (x + 3)} dx - \int_{ - 3}^1 {\sqrt {(x + 3)} dx} }$

$= {{15} \over 2} - \left[ {{{{{( - 3 - x)}^{3/2}}} \over { - {3 \over 2}}}} \right]_{ - 4}^{ - 3} - \left[ {{{{{(x + 3)}^{3/2}}} \over {{3 \over 2}}}} \right]_{ - 3}^1$

$= {{15} \over 2} + {2 \over 3}[0 - 1] - {2 \over 3}[8 - 0]$

$= {{15} \over 2} - {2 \over 3} - {{16} \over 3} = {{15} \over 2} - {{18} \over 3} = {3 \over 2}$