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Relations and Functions - Relations

Grade 11CBSE

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

🔑Concepts

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The Cartesian Product of two sets AA and BB is the set of all ordered pairs (a,b)(a, b) where a∈Aa \in A and b∈Bb \in B. Visually, if AA is represented on the x-axis and BB on the y-axis, the product A×BA \times B forms a grid of points.

Cartesian product A x B shown as points on a 2D coordinate plane.
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A Relation RR from set AA to set BB is a subset of the Cartesian product A×BA \times B. It is often depicted using an arrow diagram where arrows connect elements of AA to related elements in BB.

Arrow diagram mapping elements from set A to set B.
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The Domain of a relation is the set of all first elements of the ordered pairs in RR, while the Range is the set of all second elements. The entire set BB is called the Codomain.

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The total number of possible relations from set AA (size pp) to set BB (size qq) is 2p⋅q2^{p \cdot q}, which represents all possible subsets of the Cartesian product.

📐Formulae

AtimesB=(a,b):ainA,binBA \\times B = \\{(a, b) : a \\in A, b \\in B\\}

n(AtimesB)=n(A)timesn(B)n(A \\times B) = n(A) \\times n(B)

Total relations from AA to B=2n(A)cdotn(B)B = 2^{n(A) \\cdot n(B)}

(a,b)=(x,y)iffa=xtextandb=y(a, b) = (x, y) \\iff a = x \\text{ and } b = y

Range(R)=binB:(a,b)inRtextforsomeainARange(R) = \\{b \\in B : (a, b) \\in R \\text{ for some } a \\in A\\}

💡Examples

Problem 1:

If (fracx3+1,y−frac23)=(frac53,frac13)(\\frac{x}{3} + 1, y - \\frac{2}{3}) = (\\frac{5}{3}, \\frac{1}{3}), find the values of xx and yy.

Solution:

Step 1: Since the ordered pairs are equal, equate the corresponding elements. fracx3+1=frac53\\frac{x}{3} + 1 = \\frac{5}{3} y−frac23=frac13y - \\frac{2}{3} = \\frac{1}{3} Step 2: Solve the first equation for xx: fracx3=frac53−1\\frac{x}{3} = \\frac{5}{3} - 1 fracx3=frac23\\frac{x}{3} = \\frac{2}{3} x=2x = 2 Step 3: Solve the second equation for yy: y=frac13+frac23y = \\frac{1}{3} + \\frac{2}{3} y=frac33y = \\frac{3}{3} y=1y = 1 Final Answer: x=2x = 2 and y=1y = 1.

Explanation:

This solution relies on the fundamental property of ordered pairs: two pairs are identical if and only if their first components are equal and their second components are equal.

Problem 2:

Let A=1,2,3,5A = \\{1, 2, 3, 5\\} and B=4,6,9B = \\{4, 6, 9\\}. Define a relation RR from AA to BB by R=(x,y):x−ytextisodd,xinA,yinBR = \\{(x, y) : x - y \\text{ is odd}, x \\in A, y \\in B\\}. Write RR in roster form and find its domain.

Solution:

Step 1: Test the 'difference is odd' condition for all pairs (x,y)(x, y). Recall that a difference is odd if one number is even and the other is odd. Step 2: Check for x=1x=1 (odd): 1−4=−31-4=-3 (odd), 1−6=−51-6=-5 (odd), 1−9=−81-9=-8 (even). Pairs: (1,4),(1,6)(1,4), (1,6). Step 3: Check for x=2x=2 (even): 2−4=−22-4=-2 (even), 2−6=−42-6=-4 (even), 2−9=−72-9=-7 (odd). Pair: (2,9)(2,9). Step 4: Check for x=3x=3 (odd): 3−4=−13-4=-1 (odd), 3−6=−33-6=-3 (odd), 3−9=−63-9=-6 (even). Pairs: (3,4),(3,6)(3,4), (3,6). Step 5: Check for x=5x=5 (odd): 5−4=15-4=1 (odd), 5−6=−15-6=-1 (odd), 5−9=−45-9=-4 (even). Pairs: (5,4),(5,6)(5,4), (5,6). Step 6: Write RR in roster form: R=(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6).R = \\{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\\}. Step 7: Find the domain (set of all first elements): Domain=1,2,3,5.Domain = \\{1, 2, 3, 5\\}.

Explanation:

To solve this, we systematically verify the arithmetic condition (x−ytextisodd)(x-y \\text{ is odd}) for every possible pairing in the Cartesian product AtimesBA \\times B, then extract the unique first elements to define the domain.

Problem 3:

Let A={1,2,3}A = \{1, 2, 3\} and B={1,4,9,10}B = \{1, 4, 9, 10\}. Define a relation RR from AA to BB such that R={(x,y):y=x2,x∈A,y∈B}R = \{(x, y) : y = x^2, x \in A, y \in B\}. List the elements of RR and find the Domain and Range.

Arrow diagram for the relation y = x^2 mapping {1,2,3} to {1,4,9,10}.

Solution:

  1. Calculate y=x2y = x^2 for each x∈Ax \in A:
  • For x=1,y=12=1∈Bx = 1, y = 1^2 = 1 \in B.
  • For x=2,y=22=4∈Bx = 2, y = 2^2 = 4 \in B.
  • For x=3,y=32=9∈Bx = 3, y = 3^2 = 9 \in B.
  1. The relation in roster form is R={(1,1),(2,4),(3,9)}R = \{(1, 1), (2, 4), (3, 9)\}.
  2. Domain = {1,2,3}\{1, 2, 3\}.
  3. Range = {1,4,9}\{1, 4, 9\}.

Explanation:

We check each element of the first set AA to see if its square exists in set BB. The resulting ordered pairs form the relation RR. The first components of these pairs are the domain, and the second components are the range.

Problem 4:

Identify the relation RR shown in the coordinate plot where A={1,2}A = \{1, 2\} and B={1,2,3}B = \{1, 2, 3\}. Express RR in set-builder form.

Coordinate plot showing points (1, 2) and (2, 3) representing a relation.

Solution:

  1. From the diagram, the points plotted are (1,2)(1, 2) and (2,3)(2, 3).
  2. Observe the relationship between xx and yy: 2=1+12 = 1 + 1 and 3=2+13 = 2 + 1.
  3. Thus, y=x+1y = x + 1.
  4. In set-builder form: R={(x,y):y=x+1,x∈A,y∈B}R = \{(x, y) : y = x + 1, x \in A, y \in B\}.

Explanation:

By identifying the coordinates of the points in the Cartesian plane, we can deduce a mathematical rule that links xx (from set AA) to yy (from set BB).