Question

The minimum value of $8^{sin\left(x/8\right)} + 8^{cos\left(x/8\right)}$ is

• A. $2^{\frac{\frac{1}{3-\sqrt{2}}}{\sqrt{2}}}$
• B. $2^{\frac{3+\sqrt{2}}{\sqrt{2}}}$
• C. $2^{ \frac{\frac{1}{3+\sqrt{2}}}{\sqrt{2}}}$
• D. $2^{\frac{3-\sqrt{2}}{\sqrt{2}}}$

Answer 1

Answer: (B)

We know that $A.M. \ge G.M$.
$\Rightarrow \frac{ 2^{3sin \frac{x}{8}} + 2^{3 cos\, \frac{x}{8}} }{2} \ge \sqrt{2^{3\sqrt{2}\,cos \left(\frac{x}{8} - \frac{\pi}{4}\right) }} $
Now maximum of $\sqrt{ 2^{3\sqrt{2} \,cos \left(\frac{x}{8} -\frac{\pi}{4}\right)}}$
$ = \sqrt{ 2^{3\sqrt{2}\cdot1}}$
$ = 2^{ \frac{3}{\sqrt{2}} }$
So $A. M. \ge 2^{ {3}/{\sqrt{2}} } $
$ \Rightarrow 8^{ sin\, {x}/{8}} + 8^{ cos\,{x}/{8} } \ge 2^{\left(\frac{3}{\sqrt{2}}\right)+1}$