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

Rate this question :

Fill in the blanks in each of the

Let f :R → R be defined by. Then (f o f o f) (x) = _______

Mathematics - ExemplarLet f : [2, ∞) → R be the function defined by f (x) = x^{2}–4x+5, then the range of f is

Let f : N → R be the function defined byand g : Q → R be another function defined by g (x) = x + 2. Then (g o f)3/2 is

Mathematics - ExemplarFill in the blanks in each of the

Let f = {(1, 2), (3, 5), (4, 1) and g = {(2, 3), (5, 1), (1, 3)}. Then g o f = ______and f o g = ______.

Mathematics - ExemplarLet f :R → R be defined by

Then f (– 1) + f (2) + f (4) is

Mathematics - ExemplarLet f : [0, 1] → [0, 1] be defined by

Then (f o f) x is

Mathematics - ExemplarWhich of the following functions from Z into Z are bijections?

Mathematics - Exemplar