Question

The minimum value of $27^{\cos 2x}\, 81^{\sin 2x}$ is

• A. $\frac {1}{27}$
• B. $-5$
• C. $\frac {1}{5}$
• D. $\frac {1}{243}$

Answer 1

Answer: (D)

Let $ f(x) =27^{\cos 2 x} 81^{\sin 2 x}=3^{3 \cos 2 x+4 \sin 2 x} $
$=3^{5\left(\frac{3}{5} \cos 2 x+\frac{4}{5} \sin 2 x\right)} $
Let $\frac{3}{5}=\sin \phi $
$\Rightarrow \frac{4}{5}=\cos \phi $
then $ f(x)=3^{5(\sin \phi \cos 2 x+\cos \phi \sin 2 x)} $
$=3^{5(\sin (\phi+2 x))}$
For minimum value of given function, $\sin (\phi+2 x)$ will be minimum,
ie, $\sin (\phi+2 x)=-1$
$\therefore f(x)=3^{5(-1)}=\frac{1}{243} $