The function $f (x) = 4 {sin}^{3} x - 6 {sin}^{2} x + 12 sinx + 100$ is strictly

decreasing in

(\frac{π}{2}, \frac{3 π}{2})

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decreasing in

(\frac{π}{2}, π)

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increasing in

(\frac{- π}{2}, \frac{π}{2})

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increasing in

(\frac{π}{2}, \frac{3 π}{2})

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Solution

The correct option is C increasing in $(\frac{- π}{2}, \frac{π}{2})$
$We have, f (x) = 4 {sin}^{3} x - 6 {sin}^{2} x + 12 sin x + 100 ∴ f' (x) = 12 {sin}^{2} x cos x - 12 sin x cos x + 12 cos x \Rightarrow f' (x) = 12 cos x ({sin}^{2} x - sin x + 1) \Rightarrow f' (x) = 12 cos x ({sin}^{2} x + (1 - sin x)] Now, {sin}^{2} x \geq 0 and 1 - sin x \geq 0 ∴ {sin}^{2} x + (1 - sin x) \geq 0 ∴ Sign of f' (x) depends upon cos x . ∵ cos x is positive when x \in (- \frac{π}{2}, \frac{π}{2}) and negative when x \in (\frac{π}{2}, π) or x \in (\frac{π}{2}, \frac{3 π}{2}) . ∴ f (x) is increasing function when x \in (- \frac{π}{2}, \frac{π}{2}) and decreasing function when x \in (\frac{π}{2}, π) or x \in (\frac{π}{2}, \frac{3 π}{2}) .$

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