Find the maximum value of the function f (x) = 1 + 2 sin x + 3 cos^2 x , 0 ≤ x ≤ 2π/3.

f(x)=1+2sin x+3cos ²x
f(x)=1+2sin x+ 3 (1 - sin2x)
f(x) = -3sin2x + 2 sin x + 4
df(x)/dx = -6sinxcosx + 2cosx = 0
hence, sin x = 1/3 or cosx = 0
d/dx(df(x)/dx) = -3cos2x - 2sinx
It is negative for sin x = 1/3 and positive for cos x = 0
Hence, we get a maxima at x = sin-1 (1/3)





Therefore, max value of f(x) is 13/3.
Thanks & Regards
Bharat Bajaj
askIITians Faculty
IIT Delhi