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

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