QJEEM22Ma837f: JEE Main 2022 Mathematics - Class 11 JEEM22Ma837f Solution: Let and . Then, is equal to :

Given,

${{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0$

${x^2} + 3x + 5$ is a quadratic equation

a = 1 > 0 and D = {( - 3)^2} - 4\,.\,1\,.\,5 = - 11 < 0

$\therefore$ {x^2} + 3x + 5 > 0 (always)

So, we can ignore this quadratic term

${{(x + 2)} \over { - 2 + 3x - {x^2}}} \ge 0$

$\Rightarrow {{x + 2} \over { - ({x^2} - 3x + 2)}} \ge 0$

$\Rightarrow {{x + 2} \over {{x^2} - 3x + 2}} \le 0$

$\Rightarrow {{x + 2} \over {{x^2} - 2x - x + 2}} \le 0$

$\Rightarrow {{x + 2} \over {(x - 1)(x - 2)}} \le 0$

$\therefore$ $x \in ( - \alpha , - 2] \cup (1,2)$

$\therefore$ ${S_1} = ( - \alpha , - 2] \cup (1,2)$

Now,

${3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0$

$\Rightarrow {({3^x})^2} - 3\,.\,{3^x} - {3^2}\,.\,{3^x} + 27 \le 0$

Let ${3^x} = t$

$\Rightarrow {t^2} - 3\,.\,t - {3^2}\,.\,t + 27 \le 0$

$\Rightarrow t(t - 3) - 9(t - 3) \le 0$

$\Rightarrow (t - 3)(t - 9) \le 0$

$\therefore$ $3 \le t \le 9$

$\Rightarrow {3^1} \le {3^x} \le {3^2}$

$\Rightarrow 1 \le x \le 2$

$\therefore$ $x \in [1,2]$

$\therefore$ ${S_2} = [1,2]$

$\therefore$ ${S_1} \cup {S_2} = ( - \alpha ,2] \cup (1,2) \cup [1,2]$

$= ( - \alpha ,2] \cup [1,2]$