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 +12cosx
⇒ 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