Assertion (A): (x^-1 - y^-1) × (x - y)^-1 = x^-1y^-1

Reason (R): For any non-zero number $x,\ x^{-1} = \dfrac{1}{x}$

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,

(x^-1 - y^-1) × (x - y)^-1 = x^-1y^-1

Solving L.H.S,

$\Rightarrow (x^{-1} - y^{-1}) \times (x - y)^{-1} \\[1em] \Rightarrow \Big(\dfrac{1}{x} - \dfrac{1}{y}\Big) \times \Big(\dfrac{1}{x - y}\Big) \\[1em] \Rightarrow \Big(\dfrac{y - x}{xy}\Big) \times \dfrac{1}{x - y} \\[1em] \Rightarrow \dfrac{-(x - y)}{xy} \times \dfrac{1}{x - y} \\[1em] \Rightarrow \dfrac{-1}{xy} \\[1em] \Rightarrow -x^{-1}y^{-1}.$

L.H.S ≠ R.H.S

Thus, Assertion (A) is false.

For any non-zero number x, the notation x⁻¹ is defined as its multiplicative inverse, which is equal to $\dfrac{1}{x}$

Thus, Reason (R) is true.

A is false, R is true

Hence, option 2 is the correct option.