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

The Correct Option is C

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 \(0\).

Answer 2

The smallest value the left side can be is 2 when x is 0. But the biggest value the right side can be is also 2 when x is 2. Since these happen at different x values, there are no solutions where both sides are equal.