Question

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

Answer 1

Answer: (B)

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