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Given that, $2^{x} + 2^{-x} = 2 – (x – 2)^{2}$

$\Rightarrow 2^{x} + \frac{1}{2^{x}} = 2 – (x – 2)^{2} \quad \longrightarrow (1)$

We know that, $\text{AM} \geq \text{GM}$

$\Rightarrow \frac{2^{x} + \frac{1}{2^{x}}}{2} = \sqrt{ 2^{x} \cdot \frac{1}{2^{x}}}$

$\Rightarrow \boxed{ 2^{x} + \frac{1}{2^{x}} \geq 2}$

The minimum value of $2^{x} + \frac{1}{2^{x}}$ is $2,$ when $x=0.$

So, it follows $\text{LHS} \geq 2.$

And, $2 – (x-2)^{2} \leq 2$

$\Rightarrow – (x-2)^{2} \leq 0$

$\Rightarrow (x-2)^{2} \geq 0$

$\Rightarrow \boxed{x \geq 2}$

The maximum value of $2 – (x-2)^{2}$ is $2,$ when $x=2.$

Hence, there is no value of $x.$

Correct Answer$: \text{C}$

$\Rightarrow 2^{x} + \frac{1}{2^{x}} = 2 – (x – 2)^{2} \quad \longrightarrow (1)$

We know that, $\text{AM} \geq \text{GM}$

$\Rightarrow \frac{2^{x} + \frac{1}{2^{x}}}{2} = \sqrt{ 2^{x} \cdot \frac{1}{2^{x}}}$

$\Rightarrow \boxed{ 2^{x} + \frac{1}{2^{x}} \geq 2}$

The minimum value of $2^{x} + \frac{1}{2^{x}}$ is $2,$ when $x=0.$

So, it follows $\text{LHS} \geq 2.$

And, $2 – (x-2)^{2} \leq 2$

$\Rightarrow – (x-2)^{2} \leq 0$

$\Rightarrow (x-2)^{2} \geq 0$

$\Rightarrow \boxed{x \geq 2}$

The maximum value of $2 – (x-2)^{2}$ is $2,$ when $x=2.$

Hence, there is no value of $x.$

Correct Answer$: \text{C}$