The function f ( x ) = 4 sin 3 x - 6 sin 2 x + 12 sin x + 100 is strictly :

Question

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

Smart Achievers · Accepted Answer

(2)

Ans. Decreasing in $(\frac{π}{2}, π)$

Explanation:

We have,

$f (x) = 4 \sin^{3} x - 6 \sin^{2} x + 12 \sin x + 100$

$∴$ $f' (x) = 12 \sin^{2} x \cdot \cos x - 12 \sin x \cdot \cos x + 12 \cos x$

$= 12 [\sin^{2} x \cdot \cos x - \sin x \cdot \cos x + \cos x]$

$= 12 \cos x [\sin^{2} x - \sin x + 1]$

$\Rightarrow$ $f' (x) = 12 \cos x [\sin^{2} x + (1 - \sin x)]$ ...(i)

$∵$ $1 - \sin x \geq 0$ and $\sin^{2} x \geq 0$

$∴$ $\sin^{2} x + 1 - \sin x \geq 0$

Hence, $f' (x) > 0,$ when $\cos x > 0,$ i.e., $x \in (- \frac{π}{2}, \frac{π}{2})$

So, $f (x)$ is increasing when $x \in (- \frac{π}{2}, \frac{π}{2})$

and $f' (x) < 0,$ when $\cos x < 0,$ i.e., $x \in (\frac{π}{2}, \frac{3 π}{2})$

Hence, $f (x)$ is decreasing when $x \in (\frac{π}{2}, \frac{3 π}{2})$

Since, $(\frac{π}{2}, π) \in (\frac{π}{2}, \frac{3 π}{2})$

Hence, $f (x)$ is decreasing in $(\frac{π}{2}, π)$

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