CBSE Class 12 Mathematics Relations And Functions Class Test Set 01

XII - Relations And Functions - Test No. 1

Question 1. Define Equivalence relation.
Answer: A relation \(R\) defined on a set \(A\) is called an equivalence relation if and only if it satisfies three properties:
1. Reflexive: \((a, a) \in R\) for all \(a \in A\).
2. Symmetric: If \((a, b) \in R\), then \((b, a) \in R\) for all \(a, b \in A\).
3. Transitive: If \((a, b) \in R\) and \((b, c) \in R\), then \((a, c) \in R\) for all \(a, b, c \in A\).
In simple words: An equivalence relation is a connection between elements that behaves like equality, meaning it is reflexive, symmetric, and transitive.

Exam Tip: Always verify all three conditions independently. Failing any single condition means the relation is not an equivalence relation.

Question 2. Show that the relation \(R\) in the set \(\{1, 2, 3\}\) given by \(R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}\) is reflexive but neither symmetric nor transitive.
Answer: Let the given set be \(A = \{1, 2, 3\}\) and \(R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}\).
1. Reflexive: For every element \(a \in A\), the pair \((a, a)\) must belong to \(R\). Here, \((1, 1) \in R\), \((2, 2) \in R\), and \((3, 3) \in R\). Therefore, \(R\) is reflexive.
2. Symmetric: For symmetry, if \((a, b) \in R\), then \((b, a)\) must also belong to \(R\). Here, \((1, 2) \in R\), but the reverse pair \((2, 1) \notin R\). Therefore, \(R\) is not symmetric.
3. Transitive: For transitivity, if \((a, b) \in R\) and \((b, c) \in R\), then \((a, c)\) must belong to \(R\). Here, \((1, 2) \in R\) and \((2, 3) \in R\), but the pair \((1, 3) \notin R\). Therefore, \(R\) is not transitive.
Hence, \(R\) is reflexive but neither symmetric nor transitive.
In simple words: The relation has identical pairs for every element making it reflexive, but it fails the symmetric test because reversed pairs are missing, and it fails the transitive test because chained connections lack their direct link.

Exam Tip: Quote counter-examples clearly when proving a relation is not symmetric or transitive.

Question 3. Prove that the relation \(R\) in the set \(A = \{1, 2, 3, 4, 5\}\) given by \(R = \{(a, b) : |a - b| \text{ is even}, \forall a, b \in A\}\) is an equivalence relation.
Answer: Given set \(A = \{1, 2, 3, 4, 5\}\) and relation \(R = \{(a, b) : |a - b| \text{ is even}, \forall a, b \in A\}\).
1. Reflexive: For any \(a \in A\), we have \(|a - a| = 0\), which is an even number. Thus, \((a, a) \in R\) for all \(a \in A\). Therefore, \(R\) is reflexive.
2. Symmetric: Let \((a, b) \in R\). Then \(|a - b|\) is even. This implies that \(|b - a| = |-(a - b)| = |a - b|\) is also even. Thus, \((b, a) \in R\). Therefore, \(R\) is symmetric.
3. Transitive: Let \((a, b) \in R\) and \((b, c) \in R\). This means \(|a - b|\) is even and \(|b - c|\) is even. Since the sum of two even integers is even, \((a - b) + (b - c) = a - c\) is even. Thus, \(|a - c|\) is even, which means \((a, c) \in R\). Therefore, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: The absolute difference between a number and itself is zero (even), reversing the order of subtraction does not change parity, and adding two even differences results in another even difference.

Exam Tip: Use properties of absolute values and even integers clearly to demonstrate closure under addition for transitivity.

Question 4. Determine whether each of the following relations are Reflexive, Symmetric, Transitive.
(i) \(R = \{(x, y) : x \text{ is the wife of } y\}\)
Answer:
1. Reflexive: A person cannot be the wife of themselves. Thus, \((x, x) \notin R\). Not reflexive.
2. Symmetric: If \(x\) is the wife of \(y\), then \(y\) is the husband of \(x\), not the wife. Thus, if \((x, y) \in R\), then \((y, x) \notin R\). Not symmetric.
3. Transitive: If \(x\) is the wife of \(y\), then \(y\) is a male. Since a wife must be female, no person can be the wife of \(y\) in a way that forms a chain \((x, y)\) and \((y, z)\). Thus, it is vacuously or logically not transitive.
In simple words: Marriage roles are directional and gender-specific, so the relation fails all three basic properties.

Exam Tip: For family and gender-based relations, carefully check gender constraints to rule out reflexivity and symmetry.

(ii) \(R = \{(x, y) : x \text{ is exactly } 8\text{ cm taller than } y\}\)
Answer:
1. Reflexive: No person can be 8 cm taller than themselves. Thus, \(|x - x| = 0 \neq 8\), so \((x, x) \notin R\). Not reflexive.
2. Symmetric: If \(x\) is 8 cm taller than \(y\), then \(y\) is 8 cm shorter than \(x\), not taller. Thus, if \((x, y) \in R\), then \((y, x) \notin R\). Not symmetric.
3. Transitive: If \(x\) is 8 cm taller than \(y\) (i.e., \(x = y + 8\)), and \(y\) is 8 cm taller than \(z\) (i.e., \(y = z + 8\)), then \(x = (z + 8) + 8 = z + 16\). Thus, \(x\) is 16 cm taller than \(z\), not 8 cm. Therefore, \((x, z) \notin R\). Not transitive.
In simple words: Being taller by a fixed height is a strictly ordered inequality, so it is neither reflexive, symmetric, nor transitive.

Exam Tip: When testing transitivity for height differences, calculate the exact combined difference to disprove the relation.