Question

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

• A. increasing in $(\pi ,\frac{3\pi }{2})$
• B. decreasing in $(\frac{\pi }{2},\pi )$
• C. decreasing in $(\frac{-\pi }{2},\frac{\pi }{2})$
• D. decreasing in $(0,\frac{\pi }{2})$

Answer 1

The correct option is B decreasing in $(\frac{\pi }{2},\pi )$

we have $f\left(x\right)=4si{n}^{3}x-6si{n}^{2}x+12sinx+100$
$\therefore f^{\prime }\left(x\right)=12si{n}^{2}xcosx-12sinxcosx+12cosx$
$=12[si{n}^{2}xcosx-sinxcosx+cosx]$
$=12cosx[si{n}^{2}x-sinx+1]$
$\Rightarrow f^{\prime }\left(x\right)=12cosx[si{n}^{2}x+(1-sinx\left)\right]...\left(i\right)$
$\because 1-sinx\ge 0$ and $si{n}^{2}x\ge 0$
$\therefore si{n}^{2}x+-sinx\ge 0$
Hence $f^{\prime }\left(x\right)>0$ when $cosx>0$ i.e. $xϵ(-\frac{\pi }{2},\frac{\pi }{2})$

So $f\left(x\right)$ is increasing when $xϵ(-\frac{\pi }{2},\frac{\pi }{2})$

and $f^{\prime }\left(x\right)<0$ when $cosx<0$ i.e. $xϵ(\frac{\pi }{2},\frac{3\pi }{2})$

So $f\left(x\right)$ is decreasing when $xϵ(\frac{\pi }{2},\frac{3\pi }{2})$

hence $f\left(x\right)$ is decreasing when $xϵ(\frac{\pi }{2},\frac{3\pi }{2})$
Since $(\frac{\pi }{2},\pi )ϵ(\frac{\pi }{2},\frac{3\pi }{2})$
hence $f\left(x\right)$ is decreasing in $(\frac{\pi }{2},\pi )$