Question

The minimum value of \(2^{\sin x} + 2^{\cos x}\) is :

• A. \(2^{-1+\sqrt2}\)
• B. \(2^{1-\frac{1}{\sqrt2}}\)
• C. \(2^{1-\sqrt2}\)
• D. \(2^{-1+\frac{1}{\sqrt2}}\)

Hint:

Apply AM-GM: \(2^{\sin x}+2^{\cos x} \ge 2\sqrt{2^{\sin x+\cos x}}\), then use the minimum of \(\sin x + \cos x\).

Answer 1

The Correct Option is B

By AM-GM: \(2^{\sin x} + 2^{\cos x} \ge 2\sqrt{2^{\sin x}\cdot 2^{\cos x}} = 2\cdot 2^{\frac{\sin x+\cos x}{2}}\).
The minimum of \(\sin x + \cos x\) is \(-\sqrt2\), so the right side is at least \(2\cdot 2^{-\frac{\sqrt2}{2}} = 2^{1-\frac{1}{\sqrt2}}\).
Equality needs \(\sin x = \cos x\) and \(\sin x + \cos x = -\sqrt2\), both true at \(x = \frac{5\pi}{4}\).
Minimum value \(= 2^{1-\frac{1}{\sqrt2}}\).
Answer: (2) \(2^{1-\frac{1}{\sqrt2}}\)