WAEC 2023 · Paper 1 · Q14

Sets & logic

Given that $P = \{x : 2 \le x \le 8\}$ and $Q = \{x : 4 < x \le 12\}$ are subsets of the universal set $\mu = \{x : x \in \mathbb{R}\}$ , find $P \cap Q'$ .

Worked solution (try it first)

$Q'$ is every real number not in $Q$ : $x \le 4$ or $x > 12$ .
$P \cap Q'$ is the part of $P$ (from 2 to 8) that is also in $Q'$ , which is from 2 up to 4.
2 is in $P$ , and 4 is in $Q'$ because $Q$ starts just after 4, so both ends are included: $\{x : 2 \le x \le 4\}$ , option B.

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