The minimum value of 2^{sin x} + 2^{cos x} is | vedclass

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(A) Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for two positive numbers $2^{\sin x}$ and $2^{\cos x}$:$\frac{2^{\sin x} + 2^{\

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for two positive numbers $2^{\sin x}$ and $2^{\cos x}$:$\frac{2^{\sin x} + 2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}}$$\Rightarrow 2^{\sin x} + 2^{\cos x} \geq 2 \cdot 2^{\frac{\sin x + \cos x}{2}}$$\Rightarrow 2^{\sin x} + 2^{\cos x} \geq 2^{1 + \frac{\sin x + \cos x}{2}}$We know that the minimum value of $\sin x + \cos x$ is $-\sqrt{2}$.Substituting this value:$\min(2^{\sin x} + 2^{\cos x}) = 2^{1 + \frac{-\sqrt{2}}{2}} = 2^{1 - \frac{\sqrt{2}}{2}} = 2^{1 - \frac{1}{\sqrt{2}}}$

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

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