Mathematics · 2022
JEE Main · 30 June 2022, Shift 1 · Q1
Let S_1={x ∈ R -{1,2}: ((x+2)(x^2+3 x+5))/(-2+3 x-x^2) ≥ 0} and S_2={x ∈ R: 3^2 x-3^x+1-3^x+2+27 ≤ 0}. Then, S_1 ∪ S_2 is equal to:
Let $\displaystyle \mathrm{S}_1=\left\{x \in \mathbf{R}-\{1,2\}: \frac{(x+2)\left(x^2+3 x+5\right)}{-2+3 x-x^2} \geqslant 0\right\}$ and $\displaystyle \mathrm{S}_2=\left\{x \in \mathbf{R}: 3^{2 x}-3^{x+1}-3^{x+2}+27 \leq 0\right\}$. Then, $\displaystyle \mathrm{S}_1 \cup \mathrm{~S}_2$ is equal to :
Official answer
(2)
$\displaystyle (-\infty,-2] \cup[1,2]$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.