If the value of the integral I = ∫ 0 2 π s g n e x d x is equal to k π , then the smallest prime number greater than 2 k is (where, s g n x represents the signum function of x )

Step-by-step solution

As e x > 0   ∀ x ∈ R ∴ s g n e x = 1 ⇒ I = ∫ 0 2 π 1 d x = 2 π ⇒ k = 2 ∴ 2 k = 4 Hence, the required prime number = 5