Answer :

It is given that A = R – {3} and B = R – {1}

f: A B defined by



Now, let x, y ϵ A such that f(x) = f(y)



(x-2)(y-3) = (y-2)(x-3)


xy 3x -2y + 6 = xy -3y -2x +6


-3x -2y = -3y -2x


x = y


f is one one.


Let y ϵ B = R – {1}


Then, y ≠ 1.


The function f is onto if there exist x ϵ A such that f(x) = y


Now, f(x) = y



x 2 = xy -3y


x(1-y) = -3y + 2



y ϵ B, there exists ϵ A such that



f is onto.


Therefore, function f is one- one and onto.


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