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Let $\mathrm{A}=\{\mathrm{x} \in \mathrm{R} \mid-9 \leq \mathrm{x} < 4\} ; \mathrm{B}=\{\mathrm{x} \in \mathrm{R} \mid-13 < \mathrm{x} \leq 5\}$ and
$\mathrm{C}=\{\mathrm{x} \in \mathrm{R} \mid-7 \leq \mathrm{x} \leq 8\}$
Then, which one of the following is correct?
Options:
$\mathrm{C}=\{\mathrm{x} \in \mathrm{R} \mid-7 \leq \mathrm{x} \leq 8\}$
Then, which one of the following is correct?
Solution:
1607 Upvotes
Verified Answer
The correct answer is:
$-7 \in(A \cap B \cap C)$
Given sets in set builder form are:
$A=\{x \in R:-9 \leq x < 4\}$
$B=\{x \in R:-13 < x 5\}$
and $C=\{x \in R:-7 \leq x \leq 8\}$
$\Rightarrow A \cap B \cap C=\{x \in R:-7 \leq x < 4\}$
Hence, $-7 \in(A \cap B \cap C)$
$A=\{x \in R:-9 \leq x < 4\}$
$B=\{x \in R:-13 < x 5\}$
and $C=\{x \in R:-7 \leq x \leq 8\}$
$\Rightarrow A \cap B \cap C=\{x \in R:-7 \leq x < 4\}$
Hence, $-7 \in(A \cap B \cap C)$
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