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MHT CET · Maths · Mathematical Reasoning

The negation of \(\forall \mathrm{x} \in \mathrm{N}, \mathrm{x}^2+\mathrm{x}\) is even number' is

  1. A \(\forall \mathrm{x} \in \mathrm{N}, \mathrm{x}^2+\mathrm{x}\) is not an even number
  2. B \(\forall \mathrm{x} \in \mathrm{N}, \mathrm{x}^2+\mathrm{x}\) is not an odd number
  3. C \(\exists x \in N\) such that \(x^2+x\) an even number
  4. D \(\exists \mathrm{x} \in \mathrm{N}\) such that \(\mathrm{x}^2+\mathrm{x}\) is not an even number
Verified Solution

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.