One One and Onto Function (Bijection) โ€“ Definition and Examples

Here, you will learn one one and onto function (bijection) with definition and examples.

Letโ€™s begin โ€“

What is Bijection Function (One-One Onto Function) ?

Definition : A function f : A \(\rightarrow\) B is a bijection if it is one-one as well as onto.

In other words, a function f : A \(\rightarrow\) B is a bijection, if it is

(i) one-one i.e. f(x) = f(y) \(\implies\) x = y for all x, y \(\in\) A.

(ii) onto i.e. for all y \(\in\) B, there exist x \(\in\) A such that f(x) = y.

Also Read : Types of Functions in Maths โ€“ Domain and Range

Example : Let f : A \(\rightarrow\) B be a function represented by the following diagram :

bijection function

Solution : Clearly, f is a bijection since it is both one-one (injective) and onto (surjective).

Example : Prove that the function f : Q \(\rightarrow\) Q given by f(x) = 2x โ€“ 3 for all x \(\in\) Q is a bijection.

Solution : We observe the following properties of f.

One-One (Injective) : Let x, y be two arbitrary elements in Q. Then,

f(x) = f(y) \(\implies\) 2x โ€“ 3 = 2y โ€“ 3 \(\implies\) 2x = 2y \(\implies\) x = y

Thus, f(x) = f(y) \(\implies\) x = y for all x, y \(\in\) Q.

So, f is one-one.

Onto (Surjective) :ย Let y be an arbitrary element of Q. Then,

f(x) = y \(\implies\) 2x โ€“ 3 = y \(\implies\) x = \(y + 3\over 2\)

Clearly, for all y \(\in\) Q, x = \(y + 3\over 2\) \(\in\) Q. Thus, for all y \(\in\) Q (co-domain) there exist x \(\in\) Q (domain) given by x = \(y + 3\over 2\) such that f(x) = f(\(y + 3\over 2\)) = 2(\(y + 3\over 2\)) โ€“ 3 = y. That is every element in the co-domain has its pre-image in x.

So, f is onto function.

Hence, f : Q \(\rightarrow\) Q is a bijection.

Example : Let f : R \(\rightarrow\) R be a function defined as f(x) = \(2x^3 + 6x^2\) + 12x +3cosx โ€“ 4sinx; then f is-

Solution : We have f(x) = \(2x^3 โ€“ 6x^2\) + 12x + 3cosx โ€“ 4sinx

\(\implies\) f'(x) = \(6x^2 โ€“ 12x\) + 12 โ€“ 3sinx โ€“ 4cosx

\(\implies\) \(6x^2 โ€“ 12x\) + 12 = 6(\(x-1)^2\) + 6 = g(x) and 3sinx + 4cosx = h(x)

range of g(x) = [6, \(\infty\))

range of h(x) = [-5, 5]

hence f'(x) always lies in the interval [1, \(\infty\))

\(\implies\) f'(x) > 0

Hence f(x) is increasing i.e. one-one

Now x \(\rightarrow\) \(\infty\) \(\implies\) f \(\rightarrow\) \(\infty\) & x \(\rightarrow\) -\(\infty\) \(\implies\) f \(\rightarrow\) -\(\infty\) & f(x) is continous.

hence its range is R \(\implies\) f is onto so f is bijective.

Note :

(i)ย  If a line parallel to x-axis cuts the graph of the functions atmost at one point, then the f is one-one.

(ii)ย  If any line parallel to x-axis cuts the graph of the functions atleast at two points, then f is many-one.

(iii)ย  If continous functions f(x) is always increasing or decreasing in whole domain, then f(x) is one-one.

(iv)ย  All linear functions are one-one.

(v)ย  All trigonometric functions in their domain are many one.

(vi)ย  All even degree polynomials are many one.

(vii)ย  Linear by linear is one-one.

(viii)ย  Quadratic by quadratic with no common factor is many one.

(ix)ย  ย A polynomial function of degree even define from R \(\rightarrow\) R will always be into.

(x)ย  A polynomial function of degree odd defined from R \(\rightarrow\) R will always be onto

(xi)ย  Quadratic by quadratic without any common factor define from R \(\rightarrow\) R is always an into function.

Thus a function can be of these four types :

(i)ย  one-one onto (Injective and Surjective)(Also known as Bijective mapping)

(ii)ย  one-one into (Injective but not surjective)

(iii)ย  many-one onto (surjective but not injective)

(iv)ย  many-one into (neither surjective nor injective)

ย 

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