Assertion (A): $x + \dfrac{1}{x} = 4 \text{ and } \dfrac{1}{x} = 2 + \sqrt{3}, \text{ then x } = 2 - \sqrt{3}$

Reason (R):

$\Rightarrow \dfrac{1}{x} = 2 + \sqrt{3}\\[1em] \Rightarrow x = \dfrac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Given, $\dfrac{1}{x} = 2 + \sqrt{3}\text{ and } x + \dfrac{1}{x} = 4$

And,

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

So, assertion (A) is true.

If, $\dfrac{1}{x} = 2 + \sqrt{3}$

$\Rightarrow x = \dfrac{1}{2 + \sqrt{3}}\\[1em] = \dfrac{1}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} \text{ (On rationalizing)} \\[1em] = \dfrac{2 - \sqrt{3}}{(2)^2 - (\sqrt{3})^2}\\[1em] = \dfrac{2 - \sqrt{3}}{4 - 3}\\[1em] = \dfrac{2 - \sqrt{3}}{1}\\[1em] = 2 - \sqrt{3}.$

So, reason (R) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is correct option.