Question

If f(x) = 1 + 2sinx + 3cos²x, 0 ≤ x ≤ (2/3) then

(a) Minimum value of x = (π/2) 

(b) Maximum value of x = sin^–1(1/√3) 

(c) Minimum value of x = (π/6) 

(d) Maximum value of x = sin^–1(1/6)

Answer 1

The correct option (a) Minimum value of x = (π/2)

Explanation:

f(x) = 1 + 2 sinx + 3 cos²x = 0

∴ f'(x) = 0 + 2 cosx + 6 cosx (– sinx) = 0

∴ 2 cosx(1 – 3 sinx) = 0

∴ cosx = 0 or 1 – 3 sinx = 0

∴ cos x = 0 or sinx = (1/3)

∴ x = (π/2) or x = sin^–1(1/3)

∴ Minimum value of x = (π/2)