The mean proportion between $3 + 2\sqrt{2} \text{ and } 3 - 2\sqrt{2}$ is :

1. 1

2. -1

3. $2\sqrt{2}$

4. 3

Let mean proportion be x.

$\Rightarrow \dfrac{3 + 2\sqrt{2}}{x} = \dfrac{x}{3 - 2\sqrt{2}} \\[1em] \Rightarrow x^2 = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) \\[1em] \Rightarrow x^2 = 3^2 - (2\sqrt{2})^2 \\[1em] \Rightarrow x^2 = 9 - (4 \times 2) \\[1em] \Rightarrow x^2 = 9 - 8 \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \sqrt{1} = \pm 1.$

Since, geometrical mean cannot be negative,

∴ x = 1.

Hence, Option 1 is the correct option.