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

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The function f(x) = 4 sin3x – 6 sin2x + 12 sinx + 100 is strictly decreasing in (pi/2, pi). Explanation: Here, f(x) = 4 sin3x – 6 sin2x + 12 sin x + 100 f'(x) = 12 sin2x · cos x – 12 sin x cos x + 12 cos = 12 cos x [sin2x – sin x + 1] = 12 cos x [sin2x + (1 – sin x)] ∵ 1 – sin x $\geq$ 0 and sin2x $\geq$ 0 $\therefore$ sin2x + 1 – sin x $\geq$ 0 .....(when cos x > 0) Hence, f'(x) > 0, when cos x > 0 i.e., x ∈ ((-pi)/2, pi/2) So, f(x) is increasing where x ∈ ((-pi)/2, pi/2) and f'(x) < 0 When cos x < 0 i.e. x ∈ (pi/2, (3pi)/2) Hence, (x) is decreasing when x ∈ (pi/2, (3pi)/2) As (pi/2, pi) ∈ (pi/2, (3pi)/2) So f(x) is decreasing in (pi/2, pi).

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