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The function f (x) = 4 sin^3x - 6 sin^2x + 12 sinx + 100 is strictly
A. increasing in\left ( \pi,\frac{3\pi}{2} \right )
B. decreasing in \left ( \frac{\pi}{2},\pi \right )
C. decreasing in \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )
D. decreasing in \left ( 0,\frac{\pi}{2} \right )

Answers (1)

Given f (x) = 4 sin^3x - 6 sin^2x + 12 sinx + 100

Apply the first derivative and get

\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(4 \sin ^{3} \mathrm{x}-6 \sin ^{2} \mathrm{x}+12 \sin \mathrm{x}+100\right)}{\mathrm{dx}}
Apply sum rule and get
\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(4 \sin ^{3} \mathrm{x}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(6 \sin ^{2} \mathrm{x}\right)}{\mathrm{dx}}+\frac{\mathrm{d}(12 \sin \mathrm{x})}{\mathrm{dx}}+0
Then apply power rule and get
\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=12 \sin ^{2} \mathrm{x} \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}-12 \sin \mathrm{x} \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}+12 \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}

Now apply the derivative,
\\ { \Rightarrow f' (x)=12 sin\textsuperscript{2}x (cos x)-12sin x (cos x)+12(cos x)}\\ { \Rightarrow f' (x)=12 sin\textsuperscript{2}x cos x-12sin x cos x +12 \cos x}\\ { \Rightarrow f' (x)=12 cos x(sin\textsuperscript{2}x -sin x +1)}\\

Now,1-sin x \geq 0 and sin\textsuperscript{2}x \geq 0

Hence sin\textsuperscript{2}x -sin x +1 \geq 0

Therefore, f'(x) > 0, \text{ when } \cos x > 0, \text{ i.e., } x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
Hence f(x) is increasing when \mathrm{x} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)
and f^{\prime}(x)<0, when \cos x<0,$ i.e. $\mathrm{x} \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)
Hence f(x) is decreasing when \mathrm{X} \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)
Now \left(\frac{\pi}{2}, \pi\right) \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)
Hence, f(x) is decreasing in \left(\frac{\pi}{2}, \pi\right)
So the correct answer is option B

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