**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 → x_{3}, x ε R is one-one function

while 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** (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) = 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} = α

Lt_{x→a y = }Lt_{x→a (X)3 = }-α

Therefore 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 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) 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 x

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+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, ......}

AskIITians is unique platform which offers you one year and two years online classroom programmes for IIT JEE, AIEEE and other engineering examinations. You can be a part of these programmes even from home and for that you need not travel down to any other place.

**To read more, Buy study materials of ****Set Relations and Functions**** comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics ****here**.