Question

The function f (x) = 4 sin³x – 6 sin²x + 12 sinx + 100 is strictly

Answer 1

Given f (x) = 4 sin³x – 6 sin²x + 12 sin x + 100

Applying the first derivative we get

Applying the derivative,

⇒ f' (x)=12 sin²x (cos x)-12sin x (cos x)+12(cos x)

⇒ f' (x)=12 sin²x cos x-12sin x cos x +12cos⁡x

⇒ f' (x)=12 cos x(sin²x -sin x +1)

Now 1-sin x≥0 and sin²x≥0

Hence sin²x -sin x +1≥0

So the correct option is option B.