MHT CET · Maths · Mathematical Reasoning
The negation of a statement ' \(\mathrm{x} \in \mathrm{A} \cap \mathrm{B}\) \(\rightarrow(\mathrm{x} \in \mathrm{A} \text { and } \mathrm{x} \in \mathrm{B})^{\prime}\) is
- A \(\mathrm{x} \in \mathrm{A} \cap \mathrm{B} \rightarrow(\mathrm{x} \in \mathrm{A}\) or \(\mathrm{x} \in \mathrm{B})\)
- B \(\mathrm{x} \in \mathrm{A} \cap \mathrm{B}\) and \((\mathrm{x} \notin \mathrm{A}\) or \(\mathrm{x} \notin \mathrm{B})\)
- C \(\mathrm{x} \in \mathrm{A} \cap \mathrm{B}\) or \((\mathrm{x} \in \mathrm{A}\) and \(\mathrm{x} \in \mathrm{B})\)
- D \(\mathrm{x} \notin \mathrm{A} \cap \mathrm{B}\) and \((\mathrm{x} \in \mathrm{A}\) and \(\mathrm{x} \in \mathrm{B})\)
Answer & Solution
Correct Answer
(B) \(\mathrm{x} \in \mathrm{A} \cap \mathrm{B}\) and \((\mathrm{x} \notin \mathrm{A}\) or \(\mathrm{x} \notin \mathrm{B})\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{p}: \mathrm{x} \in \mathrm{A} \cap \mathrm{B}\) and \(\mathrm{q}: \mathrm{x} \in \mathrm{A}\) and \(\mathrm{x} \in \mathrm{B}\). \(\therefore\) Logical form of given statement is \(\mathrm{p} \rightarrow \mathrm{q}\). Now \(\mathrm{p} \rightarrow \mathrm{q} \equiv \sim \mathrm{p} \wedge \sim \mathrm{q}\), which is stated as \(\mathrm{x} \in \mathrm{A} \cap \mathrm{B}\) and \((\mathrm{x} \notin \mathrm{A}\) or \(\mathrm{x} \notin \mathrm{B})\).
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