Whichis greater (1/2)e or 1/e2 if there is a given function sinx(sinx), 0 < x < π?

Whichis greater (1/2)e or 1/e2 if there is a given function sinx(sinx), 0 < x < π? Correct Answer (1/2)e

Let, f(x) = sinx(sinx) So, f(x) = esinx log sinx So, on differentiating it we get, f’(x) = sinx(sinx) So, f’(x) = 0 when, x = sin-1(1/e) or x = π/2 Also, f”(x) = sinx(sinx) So that, f”(sin-1(1/e)) = (e – 1/e)e-1/e And, f”(π/2) = -1 Hence, f(x) has local as well as global minimum at x = sin-1(1/e) also have local and global maximum at x = π/2 Global minimum value of f(x) is (1/e)1/e It therefore follows that , sin π/6 (sinπ/6) > (1/e)1/e =>(1/2)1/2 > (1/e)1/e =>(1/2)e > 1/e2