Question

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

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

Answer 1

Answer: (B)

$f(x)-4si{n}^{3}x-6si{n}^{2}x+12sinx+100$
$f^{\prime }\left(x\right)=12si{n}^{2}xcosx-12sinxcosx+12cosx$
$f^{\prime }\left(x\right)=12cosx\left[si{n}^{2}x-sinx+1\right]$
$\therefore -1\le sinx\le 1$ and $0\le si{n}^{2}x\le 1$
$\Rightarrow -1\le si{n}^{2}x-sinx\le 0$
$\Rightarrow 0\le si{n}^{2}x-sinx+1\le 1$
$\Rightarrow si{n}^{2}x-sinx+1$ is always positive.
$cosx$ is negative in $\left(\frac{\pi }{2},\pi \right)$
so $f^{\prime }\left(x\right)\le 0$ for all $x\in \left(\frac{\pi }{2},\pi \right)$
$\Rightarrow f\left(x\right)$ is decreasing in $\left(\frac{\pi }{2},\pi \right)$