Functions: one-one/many-one/into/onto

Functions can be classified according to their images and pre-images relationships. Consider the function x → f(x) = y with the domain A and co-domain B.

If for each x ε A there exist only one image y ε B and each y ε B has a unique pre-image x ε A (i.e. no two elements of A have the same image in B), then f is said to be one-one function. Otherwise f is many-to-one function.



e.g. x → x3, x ε R is one-one function

while x → x2, x ε R is many-to-one function. (see figure above)

e.g.  x = + 2, y = x2 = 4


Graphically, if a line parallel to x axis cuts the graph of f(x) at more than one point then f(x) is many-to-one function and if a line parallel to y-axis cuts the graph at more than one place, then it is not a function.

        For a one-to-one function

        If x1 ≠ x2 then f(x1) ≠ f(x2)

        or if (x1) = f(x2) => x1 = x2

One-to-one mapping is called injection (or injective).

Mapping (when a function is represented using Venn-diagrams then it is called mapping), defined between sets X and Y such that Y has at least one element 'y' which is not the f-image of X are called into mappings.

Let a function be defined as: f : X → Y

Where X = {2, 3, 5, 7} and Y = {3, 4, 6, 8, 9, 11}


The mapping is shown in the figure below.


Clearly, element 9 and 11 of Y are not the f-image of any of x ε X

So the mapping is into-mapping

Hence for into mappings:

f[X} c Y and f[X] ≠ Y. => f [X] sym-new Y that is range is not a proper subset of co-domain.

The mapping of 'f' is said to be onto if every element of Y is the f-image of at least one element of X. Onto mapping are also called surjection.

One-one and onto mapping are called bijection.


Check whether y = f(x) = x3; f : R → R is one-one/many-one/into/onto function.

We are given domain and co-domain of 'f' as a set of real numbers.

For one-one function:

        Let x1, x2 ε Df and f(x1) = f(x2)

        =>X13 =X23

        =>  x1 = x2

i.e. f is one-one (injective) function.

For onto-into:

     Ltx→a   y = Ltx→a (x)3 = α

     Ltx→a   y =  Ltx→a  (X)3 =

     Therefore y = x3 is bijective function.




What kind of function does the Venn diagram in figure given below represent?


Solution: This many-one into function

Domain = Df = {a, b, c}

Co-domain = {1, 2, 3}

Range = Rf = {1, 2}

f(a) = 1 ; f(b) = 2; f(c) = 2


Classify the following functions.



(i) Many-one and onto (surjective).

(ii) One-one (injective) and into.

(iii) One-one (injective) and onto (surjective) i.e. Bijective.

(iv) and (v) are not functions.




1. Given the sets A = {1, 2, 3, 4} and B = {a, b, c} construct a

        (i)  Many-one into                    

        (ii) Many-one onto function

2. Given the sets c = {1, 2, 3} and D = {a, b, c}

        (i) How many one-one onto functions can be constructed.

        (ii) How many-one into functions can be constructed.


                        f : A → B                                          f : A → B


2.     (i)     6

        (ii)    33 - 6 = 21


What is the domain and range of the following functions?

(a) y = 3x + 5  (b) y = (x2 +x)/(x2 - x)

Domain of y = f(x) is the set of values of x for which y is real and finite.

Range is the set of values of y for which x is real and finite.


(a) For all real and finite x, y is also real and finite

Therefore Df = R = (-∞, ∞) and Rf = R = (-∞,∞)

(b) y = (x(x+1))/(x(x-1)) = (x+1)/(x-1) , x ≠ 0

when x = 0, y is 0/0  from (i.e. indetermined form)

when x = 1, y = ∞ (infinite)

Therefore Df = R -{0, 1}

also xy - y = x + 1

        => x (y - 1) = y + 1

        x = (y+1)/(y-1)

when y = 1, x = ∞ (infinite)  => y ≠ 1

also, for ≠ 0                     => y ≠ -1

Therefore Rf = R - {-1, 1}




What is the domain of the following functions?

(a) y =√((x-1)(3-x))        (b)   √(((x-1)(x-5))/(x-3))    (c)    y =   √sin x


 (a) y is real and finite if (x - 1)(3 - x) > 0

        or (x - 1)(x - 3) < 0

        i.e. x - 1 < 0 and x - 3 > 0  or      x - 1 > 0 and x - 3 < 0

        => x < 1 and x > 3                      =>  1 < x < 3

        which is not possible                    =>  1 < x < 3

                                                        =>  Df = [1, 3]

(b) Numerator becomes zero for x = 1, x = 5

Denominator becomes zero for x = 3



These three points divide x-axes into four intervals

        (-∞, 1), (1, 3), (3, 5), (5, ∞)

        Therefore Df = [1, 3) υ [5, ∞); at x = 3, we here open interval,

        Because at x = 3, y is infinite.

(c)    y = √sin x

        sin x > 0 ∀   x ε [2n∏, (2n + 1) ∏], n ε I


1. What is domain of the following?

(a)    y =√((x-1)(3-x))       (b)    y = √xsinx           (c)  y = Sin-1((1+x2)/(2x))

2. What is domain and range of the following?

(a)  find-domain-and-range                                                              (b)    y =|x




1.     (a)    Df = [1, 3)

        (b)    Df = [-(2n-1)∏, -2(n-1)∏] υ [2n ∏,   (2n + 1)∏],   n ε N

        (c)    Df = {-1, 1}

2.     (a)    Df = [a, b[ and Rf = [c, d]

        (b)    Df = {0, 1, 2, 3, 4,......}

                Rf = {1, 2, 6, 24, ......}


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