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

We are given the equation: 

\(2^x + 2^{-x} = 2 - (x - 2)^2\)

Let’s analyze each side of the equation.

Left-Hand Side (LHS): \(2^x + 2^{-x}\)

• Both \(2^x\) and \(2^{-x}\) are always positive for any real \(x\).
• Using the AM ≥ GM inequality: 
  \[ 2^x + 2^{-x} \geq 2\sqrt{2^x \cdot 2^{-x}} = 2 \] with equality when \(2^x = 2^{-x} \Rightarrow x = 0\).
• So, LHS ≥ 2, and it is equal to 2 only when \(x = 0\).

Right-Hand Side (RHS): \(2 - (x - 2)^2\)

• \((x - 2)^2\) is always non-negative for real \(x\).
• Maximum value of RHS is when \((x - 2)^2 = 0 \Rightarrow x = 2\).
• Then, RHS = 2 at \(x = 2\), and for all other values, RHS < 2.

Comparing LHS and RHS

• LHS ≥ 2 and equality occurs at \(x = 0\).
• RHS ≤ 2 and equality occurs at \(x = 2\).
• There is no value of \(x\) for which both sides are equal.

∴ There is no real solution to the equation.

Final Answer: \(0\) real-valued solutions.