KCET · Maths · Sets and Relations
If \(A=\left\{x: x\right.\) is an integer and \(\left.x^2-9=0\right\}\)
\(B=\{x: x\) is a natural number and \(2 \leq x \lt 5\}\)
\(\mathrm{C}=\{\mathrm{x}: \mathrm{x}\) is a prime number \(\leq 4\}\)
Then \((B-C) \cup A\) is,
- A \(\{-3,3,4\}\)
- B \(\{2,3,4\}\)
- C \(\{3,4,5\}\)
- D \(\{2,3,5\}\)
Answer & Solution
Correct Answer
(A) \(\{-3,3,4\}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& A=\left\{x: x \text { is an integer and } x^2-9=0\right\} \\
& x^2=9 \Rightarrow x= \pm 3=\{-3,3\}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{B}=\{\mathrm{x}: \mathrm{x} \text { is a natural number and } 2 \leq \mathrm{x} \lt 5\} \\
& =\{2,3,4\} \\
& \mathrm{C}=\{\mathrm{x}: \mathrm{x} \text { is a prime number } \leq 4\} \\
& =\{2,3\} \\
& (\mathrm{B}-\mathrm{C}) \cup \mathrm{A}=\{4\} \cup\{-3,3\}=\{-3,3,4\}
\end{aligned}\)
& A=\left\{x: x \text { is an integer and } x^2-9=0\right\} \\
& x^2=9 \Rightarrow x= \pm 3=\{-3,3\}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{B}=\{\mathrm{x}: \mathrm{x} \text { is a natural number and } 2 \leq \mathrm{x} \lt 5\} \\
& =\{2,3,4\} \\
& \mathrm{C}=\{\mathrm{x}: \mathrm{x} \text { is a prime number } \leq 4\} \\
& =\{2,3\} \\
& (\mathrm{B}-\mathrm{C}) \cup \mathrm{A}=\{4\} \cup\{-3,3\}=\{-3,3,4\}
\end{aligned}\)
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