Let A = R – {3}, B = R – {1}. Let f : A → B be defined

Given,

A = R – {3}, B = R – {1}

And,

f : A → B be defined by f (x) = x – 2/ x – 3 ∀ x ∈ A

Hence, f (x) = (x – 3 + 1)/ (x – 3) = 1 + 1/ (x – 3)

Let f(x1) = f (x2)

$1+\frac{1}{x_{1}-3}=1+\frac{1}{x_{2}-3}$

$\frac{1}{x_{1}-3}=\frac{1}{x_{2}-3}$

$x_{1}=x_{2}$

So, f (x) is an injective function.

Now let y = (x – 2)/ (x -3)

x – 2 = xy – 3y

x(1 – y) = 2 – 3y

x = (3y – 2)/ (y – 1)

y ∈ R – {1} = B

Thus, f (x) is onto or subjective.

Therefore, f(x) is a bijective function.