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