Question

The function $f\left(x\right)=4{\sin }^{3}x–6{\sin }^{2}x+12\mathrm{sinx}+100$ is strictly

• A. decreasing in $(\frac{\pi }{2},\frac{3\pi }{2})$
• B. decreasing in $(\frac{\pi }{2},\pi )$
• C. increasing in $(\frac{-\pi }{2},\frac{\pi }{2})$
• D. increasing in $(\frac{\pi }{2},\frac{3\pi }{2})$

Answer 1

Answer: (A), (B), (C)

The correct options are
A decreasing in $(\frac{\pi }{2},\frac{3\pi }{2})$
B decreasing in $(\frac{\pi }{2},\pi )$
C increasing in $(\frac{-\pi }{2},\frac{\pi }{2})$

$\mathrm{We}\mathrm{have},f\left(x\right)=4{\sin }^{3}x–6{\sin }^{2}x+12\sin x+100\therefore f\text{'}\left(x\right)=12{\sin }^{2}x\cos x-12\sin x\cos x+12\cos x\Rightarrow f\text{'}\left(x\right)=12\cos x\left({\sin }^{2}x-\sin x+1\right)\Rightarrow f\text{'}\left(x\right)=12\cos x\left({\sin }^{2}x+\left(1-\sin x\right)\right]\mathrm{Now},{\sin }^{2}x\ge 0\mathrm{and}1-\sin x\ge 0\therefore {\sin }^{2}x+\left(1-\sin x\right)\ge 0\therefore \mathrm{Sign}\mathrm{of}f\text{'}\left(x\right)\mathrm{depends}\mathrm{upon}\cos x.\because \cos x\mathrm{is}\mathrm{positive}\mathrm{when}x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\mathrm{and}\mathrm{negative}\mathrm{when}x\in \left(\frac{\pi }{2},\pi \right)\mathrm{or}x\in \left(\frac{\pi }{2},\frac{3\pi }{2}\right).\therefore f\left(x\right)\mathrm{is}\mathrm{increasing}\mathrm{function}\mathrm{when}x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\mathrm{and}\mathrm{decreasing}\mathrm{function}\mathrm{when}x\in \left(\frac{\pi }{2},\pi \right)\mathrm{or}x\in \left(\frac{\pi }{2},\frac{3\pi }{2}\right).$