Functions: one-one/many-one/into/ontoFunctions 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-onefunction.

e.g. x → x

_{3}, x ε R is one-one functionwhile x → x

_{2}, x ε R is many-to-one function. (see figure above)e.g. x = + 2, y = x

_{2}= 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 ≠ x

_{2}then f(x_{1}) ≠ f(x_{2})or if (x

_{1}) = f(x_{2}) => x_{1}= x_{2}One-to-one mapping is called

injection(orinjective).Mapping (when a function is represented using

Venn-diagramsthen 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-mappingHence 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.

Ontomapping are also calledsurjection.

One-one and ontomapping are calledbijection.

IllustrationCheck whether y = f(x) = x

^{3}; 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 x

_{1}, x_{2}ε D_{f}and f(x_{1}) = f(x_{2})=>X

_{1}^{3}=_{X2}^{3}=> x

_{1}= x_{2}i.e. f is one-one (injective) function.

For onto-into:Lt

_{x→a}y= Lt_{x→a}(x)^{3}= αLt

_{x→a y = }Lt_{x→a (X)3 = }-αTherefore y = x

^{3}is bijective function.

## Illustration:

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

Solution:This many-one into functionDomain = D

_{f}= {a, b, c}Co-domain = {1, 2, 3}

Range = R

_{f}= {1, 2}f(a) = 1 ; f(b) = 2; f(c) = 2

ExamplesClassify 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) 3

^{3}- 6 = 21

Illustration:What is the domain and range of the following functions?

(a) y = 3x + 5 (b) y = (x

^{2}+x)/(x^{2}- 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.

(c) y = √sin

xsin 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+x^{2})/(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(c) D

_{f}= {-1, 1}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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