Question

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

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

Answer 1

The correct option is A

$2^{\left(1-\frac{1}{\sqrt{2}}\right)}$

Explanation for the correct option:

Compute the minimum value of the given expression.

An expression $2^{\sin x}+2^{\cos x}$ is given.

We know that the Arithmetic mean is greater than the geometric mean.

Therefore,

$$\begin{gathered} \frac{2^{\sin x}+2^{\cos x}}{2}\ge \sqrt{2^{\sin x}·2^{\cos x}} \\ \Rightarrow 2^{\sin x}+2^{\cos x}\ge 2·{\left(2^{\sin x+\cos x}\right)}^{\frac{1}{2}} \end{gathered}$$

We know that the minimum value of $a\cos x+b\sin x$ type expressions is given by $-\sqrt{a^2+b^2}$.

Therefore,

$$\begin{gathered} 2^{\sin x}+2^{\cos x}\ge 2·{\left(2^{-\sqrt{2}}\right)}^{\frac{1}{2}} \\ \Rightarrow 2^{\sin x}+2^{\cos x}\ge 2·{\left(2\right)}^{-\frac{1}{\sqrt{2}}} \\ \Rightarrow 2^{\sin x}+2^{\cos x}\ge {\left(2\right)}^{1-\frac{1}{\sqrt{2}}} \end{gathered}$$

So, the minimum value of the given expression is $2^{\left(1-\frac{1}{\sqrt{2}}\right)}$.

Hence, option $A$ is the correct answer.