Question

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

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

Answer 1

Answer: (A)

We know that
$AM\ge GM$
$\therefore \frac{2^{\sin x}+2^{\cos x}}{2}\ge \sqrt{2^{\sin x}{2}^{\cos x}}$
$\Rightarrow 2^{\sin x}+2^{\cos x}\ge 2\sqrt{2^{\sin x+\cos x}}$
$\Rightarrow 2^{\sin x}+2^{\cos x}\ge 2\times 2^{\frac{\sin x+\cos x}{2}}$
$\Rightarrow 2^{\sin x}+2^{\cos x}\ge 2^{1+\frac{\sin x+\cos x}{2}}$
But $\sin x+\cos x=\sqrt{2}\sin \left(x+\frac{\pi }{4}\right)\ge -\sqrt{2}$
$\therefore 2^{\sin x}+2^{\cos x}\ge 2^{1-\frac{\sqrt{2}}{2}}$
$\Rightarrow 2^{\sin x}+2^{\cos x}\ge 2^{1-\frac{1}{\sqrt{2}}},\forall x\in R$
Hence, minimum value is$2^{1-\frac{1}{2}}$