Let f(x) = 3sin^4x + 10sin^3x + 6sin^2x – 3, x ∈ [-π/6, π/2]. then f is: (1) increasing in (-π/2, π/2) ​ (2) decreasing in (0,π/2)

Question

Let f(x) = 3sin⁴x + 10sin³x + 6sin²x – 3,

x ∈ \(\left[ - \frac{\pi}{6}, \frac{\pi}{2}\right]\). Then, f is:

(1) increasing in \(\left(- \frac{\pi}{2}, \frac{\pi}{2}\right)\)

(2) decreasing in \(\left(0, \frac{\pi}{2} \right)\)

(3) increasing in \(\left(- \frac{\pi}{6}, 0\right)\)

(4) decreasing in \(\left(- \frac{\pi}{6}, 0\right)\)

Answer 1

Correct option (4) decreasing in \(\left(- \frac{\pi}{6}, 0\right)\) 

ƒ(x) = 3sin⁴x + 10sin³x + 6sin²x – 3,  x ∈ \(\left[-\frac{\pi}{6}, \frac{\pi}{2}\right]\)

ƒ'(x) = 12sin³ xcosx + 30sin² xcosx + 12sinxcosx

= 6 sinx cosx (2sin² x + 5 sinx + 2)

= 6 sinx cosx (2sinx + 1)(sin + 2)

Decreasing in \(\left(- \frac{\pi}{6}, 0\right)\)