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)

$2^{1-1/\sqrt{2}}$

Answer 2

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