Q. The minimum value of $4^{x} + 4^{1 - x}, x \in R$ is

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A

2

0%

B

0

7%

C

$\frac{1}{4}$

7%

D

4

87%

Solution:

A. $M \geq G \cdot M$
$\Rightarrow \frac{4 ^{x} + 4 ^{1 - x}}{2} \geq 4^{x} \cdot 4^{1 - x}$
$\frac{4 ^{x} + 4 ^{1 - x}}{2} \geq 4^{x} \cdot \frac{4}{4 ^{x}}$
$\Rightarrow 4^{x} + 4^{1 - x} \geq 4$