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

---

Let us use the AM-GM inequality (Arithmetic Mean ≥ Geometric Mean).

For any two positive numbers $a$ and $b$: $\frac{a + b}{2} \geq \sqrt{ab}$

Applying this to $2^x$ and $2^{-x}$:

$\frac{2^x + 2^{-x}}{2} \geq \sqrt{2^x \cdot 2^{-x}}$

Since $2^x \cdot 2^{-x} = 2^{x + (-x)} = 2^0 = 1$:

$\frac{2^x + 2^{-x}}{2} \geq \sqrt{1} = 1$

Therefore: $2^x + 2^{-x} \geq 2$

---

AM-GM equality happens when $2^x = 2^{-x}$

This means $x = -x$, so $x = 0$.

At $x = 0$: $2^0 + 2^0 = 1 + 1 = 2$

---

Since $(x-2)^2 \geq 0$ for all real numbers $x$:

$2 - (x-2)^2 \leq 2$

Equality occurs when $(x-2)^2 = 0$, which means $x = 2$.

At $x = 2$: $2 - (2-2)^2 = 2 - 0 = 2$

---

Left side: Always $\geq 2$, equals 2 only when $x = 0$

Right side: Always $\leq 2$, equals 2 only when $x = 2$

Since both sides can only equal 2, but this happens at different values of $x$, let us check what happens at these critical points:

At $x = 0$:

• LHS $= 2^0 + 2^0 = 2$
• RHS $= 2 - (0-2)^2 = 2 - 4 = -2$

At $x = 2$:

• LHS $= 2^2 + 2^{-2} = 4 + \frac{1}{4} = 4.25$
• RHS $= 2 - (2-2)^2 = 2$

---

The left side is always greater than or equal to 2, while the right side is always less than or equal to 2. They can both equal 2, but only at different values of $x$. Since there's no single value of $x$ where both sides equal 2, and the left side is always greater than the right side everywhere else, there are no real solutions.

The number of real-valued solutions is 0.