Mathematics - Functions Question with Solution | TestHub

Using $\mathrm{A}.\mathrm{M}\ge \mathrm{G}.\mathrm{M}.$

$\frac{2^{\mathrm{sinx}}+2^{\mathrm{cosx}}}{2}\ge \sqrt{2^{\mathrm{sinx}}·2^{\mathrm{cosx}}}$

$\Rightarrow 2^{\mathrm{sinx}}+2^{\mathrm{cosx}}\ge 2^{1+\frac{\mathrm{sinx}+\mathrm{cosx}}{2}}$

Now    $-\sqrt{2}\le \mathrm{sinx}+\mathrm{cosx}\le \sqrt{2}$

So minimum value of 

$2^{\mathrm{sinx}}+2^{\mathrm{cosx}}=2^{1+\frac{-\sqrt{2}}{2}}=2^{1-\frac{1}{\sqrt{2}}}$