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