If x = 2 - $\sqrt{2}$, then $x - \dfrac{1}{x}$ =

1. 4

2. -4

3. $\dfrac{2 - 3\sqrt{2}}{2}$

4. $\dfrac{2 + 3\sqrt{2}}{2}$

Given,

⇒ x = 2 - $\sqrt{2}$

$\Rightarrow \dfrac{1}{x} = \dfrac{1}{(2 - \sqrt{2})}$

Rationalizing the denominator, we get :

$\Rightarrow \dfrac{1}{(2 - \sqrt{2})} \times \dfrac{(2 + \sqrt{2})}{(2 + \sqrt{2})} \\[1em] \Rightarrow \dfrac{2 + \sqrt{2}}{2^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{2 + \sqrt{2}}{4 - 2} \\[1em] \Rightarrow \dfrac{2 + \sqrt{2}}{2}$

Substituting values in $x - \dfrac{1}{x}$, we get :

$\Rightarrow 2 - \sqrt{2} - \dfrac{2 + \sqrt{2}}{2} \\[1em] \Rightarrow \dfrac{2(2 - \sqrt{2})-(2 + \sqrt{2})}{2} \\[1em] \Rightarrow \dfrac{4 - 2\sqrt{2} - 2 - \sqrt{2}}{2} \\[1em] \Rightarrow \dfrac{2 - 3\sqrt{2}}{2} \\[1em]$

$x - \dfrac{1}{x} = \dfrac{2 - 3\sqrt{2}}{2}$

Hence, Option 3 is the correct option.