Question

The minimum value of $2^{\sin x}+2^{\cos x}$ is

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

Answer 1

The correct option is B $2^{1-\frac{1}{\sqrt{2}}}$

Let $2^{\sin x},2^{\cos x}$ be the two positive numbers.
Then, using A.M. $\ge$ G.M., we have

$\frac{2^{\sin x}+2^{\cos x}}{2}\ge \sqrt{2^{\sin x}\cdot 2^{\cos x}}$
$2^{\sin x}+2^{\cos x}\ge 2\cdot \sqrt{2^{\sin x+\cos x}}\cdots \left(1\right)$
Now, $2^{\sin x+\cos x}=2^{\sqrt{2}\cos (\frac{\pi }{4}-x)}$
we know that $\cos x\in [-1,1]$
$\Rightarrow 2^{\sqrt{2}\cos (\frac{\pi }{4}-x)}\ge 2^{-\sqrt{2}}$
From $\left(1\right),$ we have
$2^{\sin x}+2^{\cos x}\ge 2\cdot 2^{\frac{-\sqrt{2}}{2}}$
$\therefore 2^{\sin x}+2^{\cos x}\ge 2^{1-\frac{1}{\sqrt{2}}}$
Hence, the minimum value of $2^{\sin x}+2^{\cos x}$ is $2^{1-\frac{1}{\sqrt{2}}}.$