JEE PYQ: Sequence And Series Question 27
Question 27 - 2020 (04 Sep Shift 2)
The minimum value of $2^{\sin x} + 2^{\cos x}$ is:
(1) $2^{1-\frac{1}{\sqrt{2}}}$
(2) $2^{-1+\sqrt{2}}$
(3) $2^{1-\sqrt{2}}$
(4) $2^{\frac{1}{\sqrt{2}}}$
Show Answer
Answer: (4)
Solution
$\frac{2^{\sin x} + 2^{\cos x}}{2} \geq (2^{\sin x + \cos x})^{1/2}$ ($\because$ AM $\geq$ GM)
$\Rightarrow 2^{\sin x} + 2^{\cos x} \geq 2 \cdot 2^{\frac{\sin x + \cos x}{2}}$
Since, $-\sqrt{2} \leq \sin x + \cos x \leq \sqrt{2}$
$\therefore$ Minimum value of $2^{\frac{\sin x + \cos x}{2}} = 2^{-\frac{1}{\sqrt{2}}}$
$\Rightarrow 2^{\sin x} + 2^{\cos x} \geq 2 \cdot 2^{-\frac{1}{\sqrt{2}}} = 2^{1-\frac{1}{\sqrt{2}}}$
BETA
