Q86 / 180 · KCET 2009 · Mathematics
The minimum value of $27^{\cos 2 x} 81^{\sin 2 x}$ is
- A$-5$
- B$\frac{1}{5}$
- Correct answerC$\frac{1}{243}$
- D$\frac{1}{27}$
Solution
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, $\quad \sin (\phi+2 \mathrm{x})=-1$
$\therefore \quad \mathrm{f}(\mathrm{x})=3^{5(-1)}=\frac{1}{243}$
Hence, the correct answer is option: 'C'.