MHT CET · Maths · Mathematical Reasoning
If \(p: \forall \in N, n^2+n\) is an even number
\(q: \forall n \in N, n^2-n\) is an odd number,
then the truth values of \(p \wedge q, p \wedge q\) and \(p \rightarrow q\) are respectively
- A \(\mathrm{F}, \mathrm{T}, \mathrm{T}\)
- B \(\mathrm{F}, \mathrm{F}, \mathrm{T}\)
- C F, T, F
- D \(\mathrm{T}, \mathrm{T}, \mathrm{F}\)
Answer & Solution
Correct Answer
(C) F, T, F
Step-by-step Solution
Detailed explanation
\(\because\) product of two natural numbers is an even natural number
Hence, \(n^2+n=n(n+1)\) and \(n^2-n=n(n-1)\) are even numbers
So, \(p\) is true and \(q\) is false
\(\Rightarrow p \wedge q\) is false
and \(p \vee q\) is true
and \(p \rightarrow q\) is false
Hence, \(n^2+n=n(n+1)\) and \(n^2-n=n(n-1)\) are even numbers
So, \(p\) is true and \(q\) is false
\(\Rightarrow p \wedge q\) is false
and \(p \vee q\) is true
and \(p \rightarrow q\) is false
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