Q. 135.0( 2 Votes )

Let A = R – {3} a

Answer :

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


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