If a = x + $\dfrac{1}{\text{x}}$ and b = x - $\dfrac{1}{\text{x}}$, then a² - b² is:

1. x² + $\dfrac{1}{\text{x}^2}$

2. x² - $\dfrac{1}{\text{x}^2}$

3. 4

4. $2\Big(x^2 + \dfrac{1}{\text{x}^2}\Big)$

Statement 1: x > 0 and $x^2 + \dfrac{1}{x^2}$ = 2, then $x^2 - \dfrac{1}{x^2}$ = 0

Statement 2: $(x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + 2$ = 2 + 2 = 4

$(x - \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} - 2$ = 2 - 2 = 0

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

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Assertion (A): $\Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4$

Reason (R): (a + b)² - (a - b)² = 4ab

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.

Assertion (A): If x > y, x + y = 6 and x - y = 2 then x² + y² = 40

Reason (R): (x + y)² + (x - y)² = 2(x² + y²)

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.