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The minimum value of $2^{\sin x}+2^{\cos x}$ is
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Verified Answer
The correct answer is:
$2^{1-1 / \sqrt{2}}$
We know that
$\mathrm{AM} \geq \mathrm{GM}$
$\therefore \quad \frac{2^{\sin x}+2^{\cos x}}{2} \geq \sqrt{2^{\sin x} 2^{\cos x}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2 \sqrt{2^{\sin x+\cos x}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2 \times 2^{\frac{\sin x+\cos x}{2}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2^{1+\frac{\sin x+\cos x}{2}}$
But $\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \geq-\sqrt{2}$
$\therefore \quad 2^{\sin x}+2^{\cos x} \geq 2^{1-\frac{\sqrt{2}}{2}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2^{1-\frac{1}{\sqrt{2}}}, \forall x \in R$
Hence, minimum value is $2^{1-\frac{1}{\sqrt{2}}}$.
$\mathrm{AM} \geq \mathrm{GM}$
$\therefore \quad \frac{2^{\sin x}+2^{\cos x}}{2} \geq \sqrt{2^{\sin x} 2^{\cos x}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2 \sqrt{2^{\sin x+\cos x}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2 \times 2^{\frac{\sin x+\cos x}{2}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2^{1+\frac{\sin x+\cos x}{2}}$
But $\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \geq-\sqrt{2}$
$\therefore \quad 2^{\sin x}+2^{\cos x} \geq 2^{1-\frac{\sqrt{2}}{2}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2^{1-\frac{1}{\sqrt{2}}}, \forall x \in R$
Hence, minimum value is $2^{1-\frac{1}{\sqrt{2}}}$.
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