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