Q50 / 75 · JEE Main 2020 (04 Sep Shift 2) · Mathematics

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

Solution

Using $A . M \geq G . M .$

$\frac{2^{sinx} + 2^{cosx}}{2} \geq \sqrt{2^{sinx} \cdot 2^{cosx}}$

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

Now $- \sqrt{2} \leq sinx + cosx \leq \sqrt{2}$

So minimum value of

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

Hence, the correct answer is option: 'D'.