Given that A = R - {3} , B = R - {1}
Consider the function
f : A → B defined by f (x) = (

x−2

x−3

)
Let x, y ∊ A such that f (x) = f (y)
⇒

x−2

x−3

=

y−2

y−3

⇒ (x - 2) (y - 3) = (y - 2) (x - 3)
⇒ xy - 3x - 2y + 6 = xy - 3y - 2x + 6
⇒ - 3x - 2y = - 3y - 2x
⇒ 3x - 2x = 3y - 2y
⇒ x = y
∴ f is one - one
Let y ₹ B = R - {1}
Then, y ≠ 1. The function f is onto if
there exists x ∊ A such that f (x) = y.
Now, f (x) = y
⇒

x−2

x−3

= y
⇒ x - 2 = y (x - 3)
⇒ x - 2 = xy - 3y
⇒ x - xy = 2 - 3y
⇒ x (1 - y) = 2 - 3y
⇒ x =

2−3y

1−y

∊ A [y ≠ 1] ... (1)
Thus, for any y ∊ B, there exists

2−3y

1−y

∊ A
such that
r (

2−3y

1−y

) =

2−3y 1−y −2

2−3y 1−y −3

=

2−3y−2+2y

2−3y−3+3y

=

−y

−1

= y
∴ f is onto.
Hence, the function is one-one and onto.
Therefore, f−1 exists.
Consider equation (1).
x =

2−3y

1−y

∊ A [y ≠ 1]
Replace y by x and x by f−1 (x) in the above equation,
we have,
f−1 (x) =

2−3x

1−x

, x ≠ 1