Question

Let \(S=\){\(n∈NIn^3+3n^2+5n+3\) is not divisible by \(3\)}.Then, which of the following statements is true about \(S\)

• A. \(S=ϕ\)
• B. \(|S|≥2\) and \(|S|\)\(\) is a multiple of \(5\)
• C. \(|S|\) is infinite
• D. \(S\) is non empty and \(|S|\) is a multiple of \(3\)
• E. \(S\) is non empty and \(|S|\) is a multiple of \(3\).

Answer 1

Answer: (E)

Given that:

 \(S=\){\(n∈NIn^3+3n^2+5n+3\) is not divisible by \(3\)}.

Here as \(n∈N\)

let us test:

take, \(n=1\)

\(⇒1^3+3.1^2+5.1+3=12\)

take \(n=2\)

\(⇒2^3+3.2^2+5.2+3=33\)

take, \(n=3\)

\(⇒3^3+3.3^2+5.3+3=72\)

All these above cases proves that S is divisible by \(3\)

Hence,\(S={n∈N : n^3+3n^2+5n+3}\) is divisible by \(3\)

So, S is non empty and \(|S|\)is a multiple of \(3\) (_Ans.)