Question

Let $f(x) = \cos x \sin 2x$, then

• A. $min\,f(x)=-\frac{1}{3\sqrt3}$ for $x\,\in \,[-\pi,\pi]$
• B. $min\,f(x)>-\frac{9}{7}$ or $-\frac{7}{9}$ for $x\, \in\,[-\pi,\pi]$
• C. $min\,f(x)>-\frac{1}{9}$ for $x \, \in\,[-\pi,\pi]$
• D. $min\,f(x)>-\frac{2}{9}$ for $x\, \in\,[-\pi,\pi]$

Answer 1

Answer: (B)

$min\,f(x)>-\frac{9}{7}$ or $-\frac{7}{9}$ for $x\, \in\,[-\pi,\pi]$

Answer 2

$f(x) = cos \,x \,sin\, 2x = cos \,x(2sin \,x \,cos \,x)$ $= 2 \,sin \,x (1 - sin^2 \,x) = 2 \,sin \,x - 2 \,sin^3 \,x$ min. $f(x) =$ min.$g(t)$ where $g(t) = 2t - 2t^3$ $x \in [-\pi, \pi]$, $t\,\in [- 1, 1]$ $g'(t) = 2 - 6t^2 = 0$ $\Rightarrow t = \pm \frac{1}{\sqrt{3}}, g''\left(t\right) = -12t$ $\therefore g''\left(\frac{1}{\sqrt{3}}\right) < 0$ and $g'' \left(-\frac{1}{\sqrt{3}} > 0\right)$ Hence min. $g\left(t\right) = g \left(-\frac{1}{\sqrt{3}} \right), t\,\in \left[-1, 1\right]$ $= -\frac{2}{\sqrt{3}}+2\cdot\frac{1}{3\sqrt{3}}$ $=- \frac{4}{3\sqrt{3}} > -\frac{7}{9} > -\frac{9}{7}$