Question

The function $f(x)=4{\sin }^{3}x-6{\sin }^{2}x+12\sin x+100$ 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 $\left[0,\frac{\pi }{2}\right]$

Answer 1

Answer: (B)

$\because f(x)=4{\sin }^{3}x-6{\sin }^{2}x+12\sin x+100$
$\Rightarrow f^{\prime }(x)=12{\sin }^{2}x\cdot \cos x-12\sin x\cdot \cos x+12\cos x+0$
$=12\cos x\left({\sin }^{2}x-\sin x+1\right)$
Since, in second quadrant $\sin x$ is $+ve$ and the cos $x$ is $-ve$.
So, $f^{\prime }(x)<0$ for all $x\in \left(\frac{\pi }{2},\pi \right)$