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Relations and Functions - Introduction-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An ordered pair consists of two objects or elements in a fixed order, written as (a,b)(a, b). Two ordered pairs (a,b)(a, b) and (c,d)(c, d) are equal if and only if their corresponding elements are equal: a=ca = c and b=db = d. This is the foundation for plotting points on a Cartesian plane.

A Cartesian plane showing the ordered pair (3, 2) where the x-coordinate is 3 and y-coordinate is 2.
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The Cartesian Product A×BA \times B of two non-empty sets AA and BB is the set of all ordered pairs (a,b)(a, b) such that a∈Aa \in A and b∈Bb \in B. Visually, if A={1,2,3}A = \{1, 2, 3\} and B={x,y}B = \{x, y\}, the product represents all possible connections from AA to BB.

A mapping diagram showing lines from elements 1, 2, 3 in set A to elements x, y in set B.
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A Relation RR from set AA to set BB is a subset of the Cartesian product A×BA \times B. The set of all first elements of the ordered pairs in RR is the Domain, and the set of all second elements is the Range. The entire set BB is called the Codomain.

Venn diagram showing a relation as a set of mappings between two sets A and B.
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A Function ff from AA to BB is a special type of relation where every element xx in set AA is associated with exactly one element yy in set BB. We write y=f(x)y = f(x), where yy is the image of xx and xx is the pre-image of yy.

📐Formulae

n(A×B)=n(A)×n(B)n(A \times B) = n(A) \times n(B) (where nn represents the number of elements in the set)

Total number of relations from A to B=2n(A)×n(B)\text{Total number of relations from } A \text{ to } B = 2^{n(A) \times n(B)}

Range⊆Codomain\text{Range} \subseteq \text{Codomain}

If (x,y)=(u,v), then x=u and y=v\text{If } (x, y) = (u, v), \text{ then } x = u \text{ and } y = v

💡Examples

Problem 1:

Find the values of xx and yy if (x3+1,y−23)=(53,13)(\frac{x}{3} + 1, y - \frac{2}{3}) = (\frac{5}{3}, \frac{1}{3}).

Solution:

Given (x3+1,y−23)=(53,13)(\frac{x}{3} + 1, y - \frac{2}{3}) = (\frac{5}{3}, \frac{1}{3}). By the equality of ordered pairs: x3+1=53\frac{x}{3} + 1 = \frac{5}{3} x3=53−1=5−33=23\frac{x}{3} = \frac{5}{3} - 1 = \frac{5 - 3}{3} = \frac{2}{3} x=2x = 2

Also, y−23=13y - \frac{2}{3} = \frac{1}{3} y=13+23=33=1y = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1 So, x=2,y=1x = 2, y = 1.

Explanation:

We use the property that two ordered pairs are equal if and only if their corresponding components are equal. This results in two independent linear equations.

Problem 2:

Let A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\}. Define a relation RR from AA to AA by R={(x,y):y=x+1}R = \{(x, y) : y = x + 1\}. Write down the domain and range.

Solution:

Given A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\} and y=x+1y = x + 1 where x,y∈Ax, y \in A. When x=1,y=1+1=2x = 1, y = 1+1 = 2 When x=2,y=3x = 2, y = 3 When x=3,y=4x = 3, y = 4 When x=4,y=5x = 4, y = 5 When x=5,y=6x = 5, y = 6 When x=6,y=7x = 6, y = 7 (but 7∉A7 \notin A, so we exclude this pair). R={(1,2),(2,3),(3,4),(4,5),(5,6)}R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\} Domain ={1,2,3,4,5}= \{1, 2, 3, 4, 5\} Range ={2,3,4,5,6}= \{2, 3, 4, 5, 6\}

Explanation:

To find the domain and range, we first list the elements of the relation by applying the condition y=x+1y = x + 1. The domain is the set of all xx values that have a corresponding yy in AA.

Problem 3:

If n(A)=3n(A) = 3 and B={3,4}B = \{3, 4\}, find the number of elements in A×BA \times B and the total number of possible relations.

Solution:

Given n(A)=3n(A) = 3. Set B={3,4}B = \{3, 4\}, so n(B)=2n(B) = 2. Number of elements in A×B=n(A)×n(B)=3×2=6A \times B = n(A) \times n(B) = 3 \times 2 = 6. Number of relations =2n(A×B)=26=64= 2^{n(A \times B)} = 2^6 = 64.

Explanation:

The number of elements in the Cartesian product is the product of the number of elements in each set. The number of relations is 22 raised to the power of the number of elements in the Cartesian product because every subset of A×BA \times B is a relation.

Problem 4:

Given the set A={1,2,3,4}A = \{1, 2, 3, 4\}, define a relation RR on AA by R={(x,y):y=x2}R = \{(x, y) : y = x^2\}. List the elements of RR and represent it as a mapping diagram.

Mapping diagram from set A to A where 1 maps to 1 and 2 maps to 4, while 3 and 4 have no arrows.

Solution:

  1. Find pairs where x∈Ax \in A and y=x2∈Ay = x^2 \in A.
  2. For x=1,y=12=1x=1, y=1^2=1. (Valid: (1,1)(1, 1))
  3. For x=2,y=22=4x=2, y=2^2=4. (Valid: (2,4)(2, 4))
  4. For x=3,y=32=9x=3, y=3^2=9. (Invalid: 9∉A9 \notin A)
  5. For x=4,y=42=16x=4, y=4^2=16. (Invalid: 16∉A16 \notin A) So, R={(1,1),(2,4)}R = \{(1, 1), (2, 4)\}.

Explanation:

In a relation defined on set AA, both the first and second elements must belong to AA. Since 323^2 and 424^2 are not in the set {1,2,3,4}\{1, 2, 3, 4\}, those inputs do not have related outputs within this set.

Problem 5:

Identify which of the following relations represent a function by analyzing their graphs: R1={(x,y):y=x}R_1 = \{(x, y) : y = x\} and R2={(x,y):x2+y2=4}R_2 = \{(x, y) : x^2 + y^2 = 4\}.

A graph showing a straight line y=x and a circle centered at the origin with radius 2.

Solution:

  1. R1R_1 is a linear equation y=xy = x. For every xx, there is exactly one yy. This is a function.
  2. R2R_2 is a circle x2+y2=22x^2 + y^2 = 2^2. For a single value of xx (like x=0x=0), there are two values of yy (y=2y=2 and y=−2y=-2). This fails the vertical line test and is not a function.

Explanation:

A relation is a function if any vertical line drawn through the graph intersects it at most once. The circle fails this because a vertical line through the center hits the top and bottom of the circle.