Assertion (A): If $x^2 - 2x - 1 = 0$, then $x^2 + \dfrac{1}{x^2} = 6$.
Reason (R): x² - 2x - 1 can be written as (x - 1)².

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Given,

$\Rightarrow x^2 - 2x - 1 = 0 \\[1em] \Rightarrow x^2 - 1 = 2x \\[1em] \Rightarrow \dfrac{x^2 - 1}{x} = 2 \\[1em] \Rightarrow \dfrac{x^2}{x} - \dfrac{1}{x} = 2 \\[1em] \Rightarrow x - \dfrac{1}{x} = 2 \\[1em]$

Using identity,

$\Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2$

Substituting,

$\Rightarrow (2)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow 4 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 4 + 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 6.$

Assertion (A) is true.

⇒ (x - 1)² = x² + 1² - 2(x)(1)

⇒ (x - 1)² = x² - 2x + 1

Reason (R) is false.

A is true, R is false

Hence, Option 1 is the correct option.