MHT CET · Maths · Mathematical Reasoning
The negation of \(\forall \mathrm{x} \in \mathrm{N}, \mathrm{x}^2+\mathrm{x}\) is even number' is
- A \(\forall \mathrm{x} \in \mathrm{N}, \mathrm{x}^2+\mathrm{x}\) is not an even number
- B \(\forall \mathrm{x} \in \mathrm{N}, \mathrm{x}^2+\mathrm{x}\) is not an odd number
- C \(\exists x \in N\) such that \(x^2+x\) an even number
- D \(\exists \mathrm{x} \in \mathrm{N}\) such that \(\mathrm{x}^2+\mathrm{x}\) is not an even number
Answer & Solution
Correct Answer
(D) \(\exists \mathrm{x} \in \mathrm{N}\) such that \(\mathrm{x}^2+\mathrm{x}\) is not an even number
Step-by-step Solution
Detailed explanation
Let \(\mathrm{p}: \forall \mathrm{x} \in \mathrm{N}\) and \(\mathrm{q}: \mathrm{x}^2+\mathrm{x}\) is even number.
The logical form of given statement is \(\mathrm{p} \wedge \mathrm{q}\).
\(\sim(\mathrm{p} \wedge \mathrm{q}) \equiv \sim \mathrm{p} \vee \sim\) q i.e. \(\exists \mathrm{x} \in \mathrm{N}\) such that \(\mathrm{x}+\mathrm{x}\) is not an even number.
The logical form of given statement is \(\mathrm{p} \wedge \mathrm{q}\).
\(\sim(\mathrm{p} \wedge \mathrm{q}) \equiv \sim \mathrm{p} \vee \sim\) q i.e. \(\exists \mathrm{x} \in \mathrm{N}\) such that \(\mathrm{x}+\mathrm{x}\) is not an even number.
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