Question

For how many real values of n the equation \(a^{2n^2+2}=1\) has a solution?
1. 1
2. zero
3. 2
4. 4

Answer 1

Correct Answer - Option 2 : zero

Concept:

From the law of exponent,

a⁰ = 1

Calculation:

We have \(a^{2n^2+2}=1\)

\(⇒ a^{2n^2+2}=a^0\)

⇒ 2n² + 2 = 0

⇒ 2n² = -2

⇒ n² = -1

⇒ n = i

Thus, there is no real value of n for which the given equation has a solution.