Let g(t) = ∫cos[π/4 t+f(x)]dx x ∈ [-x/2,x/2], where ƒ(x) = loge (x+√x^2+1), x ∈ R. Then which one of the following is correct? (1) g(1) = g(0)

Question

Let g(t) = \(\int \limits_{-x/2}^{x/2}\)cos[π/4 t+f(x)]dx, where ƒ(x) = log_e (x+\(\sqrt{x^2+1}\)), x ∈ R. Then which one of the following is correct?

(1) g(1) = g(0)

(2) √2g(1) = g(0)

(3) g(1) = √2g(0)

(4) g(1)+g(0) = 0

Answer 1

Answer is: (2) √2g(1) = g(0)