Q. 73.7( 3 Votes )

Show that the function f : R – {3} R {1} given by is a bijection.

Answer :

TIP: One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.


So, is One – One function


a≠b


f(a)≠f(b) for all


f(a) = f(b)


a = b for all


Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.


So, is Surjection iff for each , there exists such that f(a) = b


Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.


Now, f: R R given by


To Prove: – is a bijection


Check for Injectivity:


Let x,y be elements belongs to R i.e. such that


f(x) = f(y)



(x – 2)(y – 3) = (x – 3)(y – 2)


xy – 3x – 2y + 6 = xy – 2x – 3y + 6


– 3x – 2y + 2x + 3y = 0


– x + y = 0


x = y


Hence, f is One – One function


Check for Surjectivity:


Let y be element belongs to R i.e be arbitrary, then


f(x) = y



x – 2 = xy – 3y


x – xy = 2 – 3y



is a real number for all y ≠ 1.


Also, for any y


Therefore for each element in R (co – domain), there exists an element in domain R.


Hence, f is onto function


Thus, Bijective function


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