The minimum value of $
Question:

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

  1. (1) $2^{-1+\frac{1}{\sqrt{2}}}$

  2. (2) $2^{-1+\sqrt{2}}$

  3. (3) $2^{1-\sqrt{2}}$

  4. (4) $2^{1-\frac{1}{\sqrt{2}}}$


Correct Option: , 4

Solution:

$\frac{2^{\sin x}+2^{\cos x}}{2} \geq\left(2^{\sin x+\cos x}\right)^{\frac{1}{2}} \quad(\because \mathrm{AM} \geq \mathrm{GM})$

$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2 \cdot 2^{\frac{\sin x+\cos x}{2}}$

Since, $-2 \leq \sin x+\cos x \leq \sqrt{2}$

$\therefore$ Minimum value of $2^{\frac{\sin x+\cos x}{2}}=2^{-\frac{1}{\sqrt{2}}}$

$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2^{1-\frac{1}{\sqrt{2}}}$

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