Question

The number of real-valued solutions of the equation $2^x+2^{-x}=2-(x-2)^2$ is

• A. infinite
• B. 1
• C. 0
• D. 2

Answer 1

Answer: (C)

0

Answer 2

The correct option is (C): $0$
$2^x+2^{-x} = 2-(x-2)^2$
The minimum value of $2x+2-x$ is $2$ when $x=0$
But $x = 0;2-(x-2)^2=-2$
The maximum value of $2-(x-2)^2$ is $2$ when $x=2$
But $x=2 2^x+2^{-x} = \frac{17}{4}$
Hence there is no value of $x,2^x+2^{-x}=2-(x-2)^2$
The number of solutions is $0$.