Domain and Range

Naeem Gurukul

2 years ago

Today our topic of discussion is Domain and Range.

Domain and Range

For ordered pairs of any relation the set of first elements is said to be domain and the set of second elements is said to be range.

Let R be a relation from set A to the set B, that is, R ⊆ AB . For the ordered pairs of R the set of first elements will be the domain of R and the set of second elements will be called range of R. The domain of R is denoted by Dom R and range by Range R.

Example 20. Find domain S = {(2, 1), (2, 2), (3, 2), (4, 5)}

Solution: Given, S = {(2, 1), (2, 2), (3, 2), (4,5))

The first elements of the ordered pairs of S are 2, 2, 3, 4 and the second elements are 1,2,2,5

∴ Dom S = {2, 3, 4} and Range S = {1, 2, 5}

Example 21. If A = {0, 1, 2, 3} and R={ (x, y) 😡 ∈ A,y ∈ A and y = x + 1 ] then express R in tabular method and determine Dom R and Range R.

Solution: Given, R ={ (x, y) 😡 ∈ A,y ∈ A y=x+ 1}

From the conditions of R we get y = x + 1

Now for each x in A determine the value of y = x + 1 .

Since 4 A, (3, 4) R

… R = {(0, 1), (1, 2), (2, 3)}

.. Dom R = {0, 1, 2} and Range R = {1, 2, 3}

Work:

1) If S = {(- 3, 8), (- 2, 3), (- 1, 0), (0, – 1), (1, 0), (2, 3)} then determine domain and range of S.

2) Let S={ (x, y) / x ,y in A and y – x = 1 } , where A = {- 3, – 2, – 1, 0} Determine Dom S and Range S.

Graph of a Function

Visual display of a function is said to be graph. In order to clarify the concept of a function significance of graph is important. French philosopher and mathematician Rene Descartes (1596-1650) was the first to play pioneering role in introducing relations between algebra and geometry.

He introduced a modern concept in geometry by defining the position of a point in two dimensional space by drawing two straight lines perpendicularly crossing each other. He termed the perpendicular lines as axes, and their point of intersection as origin.

Let two perpendicularly crossing straight lines XOX’ and YOY’ be drawn on a plane. Any point lying on this plane can be uniquely determined through these two straight lines. Each of these perpendicular straight lines is called axis. The line XOX’ parallel to horizon is called z-axis, and vertical line YOY’ is called y-axis and point of intersection of these lines is called Origin.

The signed perpendicular distance of a point, located in the plane determined by two axes, from the axes are called coordinates of the point. Let P be any point on the plane determined by two axes. Let us draw perpendiculars PN and PM from P to XOX’ and YOY’.

Then PM = ON which is perpendicular distance to YOY’ from P and PN = OM which is perpendicular distance from P to XOX’. If PM = x and PN = y then coordinates of point P are (x,y).

Here is called abscissa and y is called ordinate. These coordinates are called Cartesian coordinates. It is very easy to depict a function using Cartesian coordinate system. Usually values of independent variable is set along z axis, and those of dependent variable along y axis.

In order to draw a graph for the function y = f(x) some values of independent variable are chosen from domain, and with corresponding values of dependent variables ordered pairs are formed. After that these ordered pairs are placed on the plane. The resultant points are then connected with line which is known as graph of the function y = f(x)

Example 22. Draw the graph for the function y = 2x where – 3 <= x <= 3

Solution: Calculate values of y for some values of r from the domain – 3 <= x <= 3 and draw a list.

In the graph paper each considering side of each smallest square as of length points of the list are marked and added. This results in the graph of the function.

Example 24. Let Universal set be U ={ x:x ∈ N and x <= 6 } , A={ x : x ′ and x <= 5 \ , B=\ x / x even number and x <= 6 \ and C =A \ B

1) Determine Aº.

2) Prove that, A ∪B=(A \B) ∪(B \ A) ∪(A ∩B)

3) Prove that, (A ∩C)* B= (AB) ∩(C* B)

Solution:

1) Given U=\ x:x in N and x <= 6 \ =\ 1,2,3,4,5,6\ A = {x: x′ and <5} = {2,3,5) .. A^ c =U \ A = \{1, 2, 3, 4, 5, 6\} – \{2, 3, 5\} = \{1, 4, 6\}

2) Given,

B = {z:x ′and x<=6\ = \{2, 4, 6\}

.. A ∪B=\ 2,3,5\ ∪\ 2,4,6\ = \{2, 3, 4, 5, 6\} ****(1)

A B = \{2, 3, 5\} – \{2, 4, 6\} = \{3, 5\} B \ A = \{2, 4, 6\} – \{2, 3, 5\} = \{4, 6\}

A ∩ B=\ 2,3,5\ ∩\{2, 4, 6\} = \{2\}

. (A \ B) ∪(B \A) ∪(A ∩ B)=\ 3,5\ ∪\ 4,6\ ∪\ 2\ = \{2, 3, 4, 5, 6\} ***(2)

Therefore comparing (1) and (2) we get,

B =(A \B) ∪(B \A) ∪(A ∩ B)

3) From (2) we get,

C=A \B = \{3, 5\}

A ∩C=\ 2,3,5\ ∩ \{3, 5\} = \{3, 5\}

( A ∩C)* B = \{3, 5\}\{2, 4, 6\} = \{(3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6)\} *****(3)

AX B = \{2, 3, 5\}\{2, 4, 6\}

= \{(2, 2), (2, 4), (2, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6)\}

B = \{3, 5\}\{2, 4, 6\}

= \{(3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6)\}

. (Ax B) (Cx B)

=\ (2, 2), (2, 4), (2, 6), (3, 2), (3, 4), (3, 6), (5, 2) , (5, 4), (5,6)}

∩\{(3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6)\}

= \{(3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6)\} *****(4)

Therefore, comparing (3) and (4) we get,

(A ∩ C)* B= (AB) ∩ (C* B)

Example 25. Let A = \{4, 5, 6, 7\} B = \{0, 1, 2, 3\} and R=\ (x, y) 😡 ∈ A,y in

A and y = x + 1 ] 1) Prove that, sets A and B are mutually disjoint.

2) Determine P(B) to show that number of elements of P(B) is 2 ^ n where n is the number of elements of B.

3) Express the relation R in tabular method and determine its domain.

Solution:

1) Given, A = {4, 5, 6, 7} and B = {0, 1, 2, 3} .. A ∩ B = {4, 5, 6, 7} n {0, 1, 2, 3} = ∅ Since A ∩B= ∅

Therefore, A and B are mutually disjoint.

2) Given,

B = {0, 1, 2, 3}

. P(B)={{0}, {1}, {2}, {3}, {0, 1}, {0, 2}, {0.3}, {1, 2}, {1, 3}, {2, 3}

{(0,1,2), (0, 1, 3), (0,2,3), (1,2,3), (0, 1,2,3), 0}

Here number of elements of B is 4 and number of elements of the power set is 2 ^ 4 = 16

∴ if n is the number of elements of B_{i} then number of elements of power set is 2 ^ n

∴ number of elements of P(B) satisfies 2 ^ n formula.

3) Given, R={(x, y) 😡 ∈A,y in A and y = x + 1 } and A = {4, 5, 6, 7}

From the conditions of R we get y = x + 1

Now, for each x in A draw a list by finding values of y = x + 1

Since 8 notin A therefore (7, 8) notin R R = {(4, 5), (5, 6), (6, 7)}

Dom R = {4, 5, 6}

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