Given f(x) = sinx + 1/2cos2x
⇒ f'(x) = cos x – sin2x
Now, f'(x) = 0 gives cosx – sin2x = 0
⇒ cos x (1 – 2 sinx) = 0
⇒ cos x = 0, (1 – 2 sinx) = 0
⇒ cos x = 0, sinx = 1/2
⇒ x = π/6 , π/2
Now, f(0) = 1/2,
f(π/6) = 1/2 + 1/4 = 3/4,
f(π/2) = 1 – 1/2 = 1/2
Therefore, the absolute max value = 3/4 and absolute min = 1/2.