Question

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

• A. $minf(x)=−\frac{1}{3\sqrt{3}}$ for $x∈[−π,Ï€]$
• B. $minf(x)>−\frac{9}{7}$ or $−\frac{7}{9}$ for $x∈[−π,Ï€]$
• C. $minf(x)>−\frac{1}{9}$ for $x∈[−π,Ï€]$
• D. $minf(x)>−\frac{2}{9}$ for $x∈[−π,Ï€]$

Answer 1

Answer: (B)

$f(x)=cosxsin2x=cosx(2sinxcosx)$
$=2sinx(1−si{n}^{2}x)=2sinx−2si{n}^{3}x$
min. $f(x)=$ min.$g(t)$ where $g(t)=2t−2t^3$
$x∈[−π,π]$, $t∈[−1,1]$
$g^{′}(t)=2−6t^2=0$
$⇒t=Â\pm \frac{1}{\sqrt{3}},g^{′′}\left(t\right)=−12t$
$∴g^{′′}\left(\frac{1}{\sqrt{3}}\right)<0$ and $g^{′′}\left(−\frac{1}{\sqrt{3}}>0\right)$
Hence min. $g\left(t\right)=g\left(−\frac{1}{\sqrt{3}}\right),t∈\left[−1,1\right]$
$=−\frac{2}{\sqrt{3}}+2â‹\dots \frac{1}{3\sqrt{3}}$
$=−\frac{4}{3\sqrt{3}}>−\frac{7}{9}>−\frac{9}{7}$