Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Find the answer

5. Given the sets
$\begin{array}{l} M_{1}=\{x \mid(x+2)(x-1)>0\}, \\ M_{2}=\{x \mid(x-2)(x+1)>0\} . \end{array}$

Then the set equal to $M_{1} \cup M_{2}$ is:

Pick one

Solution

5.B.

From $M_{1}$ we have $x1$ .
From $M_{2}$ we have $x2$ .
Since the union of $x>1$ and $x2$ are respectively
$\begin{array}{l} |x|>1, \\ |x|>2, \end{array}$

thus, the union of equations (1) and (2) is $|x|>1$ , which is $x^{2}-1>0$ .
Therefore, $\left(x^{2}+4\right)\left(x^{2}-1\right)>0$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.