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} Y and f[X] ≠ Y. => f [X] 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.

Illustration

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 D_f and f(x₁) = f(x₂)

=>X₁³ =_X2³

=> x₁ = x₂

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

For onto-into:

Lt_x→a y = Lt_x→a (x)³ = α

Lt_{x→a y =}Lt_{x→a (X)3 =}-α

Therefore y = x³ is bijective function.

Illustration:

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

Solution: This many-one into function

Domain = D_f = {a, b, c}

Co-domain = {1, 2, 3}

Range = R_f = {1, 2}

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

Examples

Classify the following functions.

Ans.

(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.

Examples:

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.

Ans.1

f : A → B f : A → B

2. (i) 6

(ii) 3³ - 6 = 21

Illustration:

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.

Solution:

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

Therefore D_f = R = (-∞, ∞) and R_f = 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 D_f = 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 R_f = R - {-1, 1}

Illustration:

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

Solution:

(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

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

Because at x = 3, y is infinite.

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

Examples:

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)

(b) y =|x

Ans.

1. (a) D_f = [1, 3)

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

2. (a) D_f = [a, b[ and R_f = [c, d]

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

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

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