JEE Main 2022 · 30 Jun · Shift 1 · Mathematics Q1

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Let $S_1=\left\{x\in\mathbf R-\{1,2\}:\dfrac{(x+2)\left(x^2+3x+5\right)}{-2+3x-x^2}\ge0\right\}$ and $S_2=\left\{x\in\mathbf R:3^{2x}-3^{x+1}-3^{x+2}+27\le0\right\}$ . Then, $S_1\cup S_2$ is equal to :

(-\infty,-2]\cup(1,2)

(-\infty,-2]\cup[1,2]

(-2,1]\cup[2,\infty)

(-\infty,2]

Answer and worked solution

Official answer: B

$S_1$ . $x^2+3x+5>0$ always and $-2+3x-x^2=-(x-1)(x-2)$ , so $\dfrac{x+2}{(x-1)(x-2)}\le0$ : $x\le-2$ or $1<x<2$ .

$S_2$ . With $t=3^x$ : $t^2-12t+27\le0$ , $3\le t\le9$ , so $1\le x\le2$ .

Union $(-\infty,-2]\cup[1,2]$ .

(B)

Topics:Inequalities and modulus Logarithms

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JEE Main 2022 · 30 Jun · Shift 1, Mathematics question 1