Indian Institute of Foreign Trade 2010 Solved Paper

Using AM ≥.GM, We can say that
$\frac{3^{\mathrm{sinx}}+3^{\mathrm{cosx}}}{2}\ge \sqrt{3^{\sin \mathit{x}}\times 3^{\cos \mathit{x}}}$
$\Rightarrow 3^{\sin \mathit{x}}+3^{\cos \mathit{x}}\ge 2\times 3^{\frac{\sin \mathit{x}+\cos \mathit{x}}{2}}$.....(1)
We know that - $\sqrt{{\mathit{A}}^2+{\mathit{B}}^2}\le \mathit{A}\mathrm{sinx}+\mathit{B}\mathrm{cosx}\le \sqrt{{\mathit{A}}^2+{\mathit{B}}^2}$
Therefore,$-\sqrt{1^2+1^2}\le \sin \mathit{x}+\cos \mathit{x}\le \sqrt{1^2+1^2}$
$\Rightarrow -\sqrt{2}\le \sin \mathit{x}+\cos \mathit{x}$
Hence, we can say that the minimum value of $\sin \mathit{x}+\cos \mathit{x}=-\sqrt{2}$...(2)
From equation (1) and (2) we can say that,
$\Rightarrow 3^{\sin \mathit{x}}+3^{\cos \mathit{x}}\ge .2\times 3^{\frac{-\sqrt{2}}{2}}$
$\Rightarrow 3^{\sin \mathit{x}}+3^{\cos \mathit{x}}\ge .2\times 3^{\frac{-\sqrt{1}}{2}}$
Therefore, option B is the correct answer.