# Let A = R – {3} a

We have f : A B where A = R – {3} and B = R – {1} First, we will prove that f is one-one.

Let x1, x2ϵ A (domain) such that f(x1) = f(x2) (x1 – 2)(x2 – 3) = (x1 – 3)(x2 – 2)

x1x2 – 3x1 – 2x2 + 6 = x1x2 – 2x1 – 3x2 + 6

–3x1 – 2x2 = –2x1 – 3x2

–3x1 + 2x1 = 2x2 – 3x2

–x1 = –x2

x1 = x2

So, we have f(x1) = f(x2) x1 = x2.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ B (co-domain) such that f(x) = y       Clearly, for every y ϵ B, there exists x ϵ A (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y x = f-1(y)

But, we found f(x) = y Hence, Thus, f(x) is invertible and Rate this question :

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