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

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

Applying the first derivative we get

Applying the sum rule of differentiation, so we get

Applying the power rule 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

Therefore, f’(x)>0, when cos x>0, i.e.,

So f(x) is increasing when

and f’(x)<0, when cos x<0, i.e.,

Hence f(x) is decreasing when

Now

Therefore, f(x) is decreasing in

So the correct option is option B