CBSE Class 12 Mathematics Relations And Functions Class Test Set 02

XII - Relations And Functions Test No. 1

Question 1. Define Equivalence relation
Answer: A relation \(R\) defined on a set \(A\) is said to be an equivalence relation if it satisfies three conditions simultaneously:
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 connects elements in a way that is self-referential, reversible, and chain-linked.

Exam Tip: Always check all three properties (reflexive, symmetric, transitive) in sequence when proving a relation is 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 set be \(A = \{1, 2, 3\}\) and the relation be \(R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}\).
1. Reflexive: Since \((1, 1) \in R\), \((2, 2) \in R\), and \((3, 3) \in R\), for every element \(a \in A\), \((a, a) \in R\). Therefore, \(R\) is reflexive.
2. Symmetric: We have \((1, 2) \in R\), but the reverse pair \((2, 1) \notin R\). Also, \((2, 3) \in R\) but \((3, 2) \notin R\). Therefore, \(R\) is not symmetric.
3. Transitive: We have \((1, 2) \in R\) and \((2, 3) \in R\), but the combined pair \((1, 3) \notin R\). Therefore, \(R\) is not transitive.
Hence, the relation \(R\) is reflexive but neither symmetric nor transitive.
In simple words: Every element relates to itself, but reverse pairs and skip-links are missing.

Exam Tip: To disprove symmetry, find a single pair \((a, b)\) where \((a, b) \in R\) but \((b, a) \notin R\). To disprove transitivity, find \((a, b)\) and \((b, c)\) in \(R\) such that \((a, c)\) is missing.

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: Let \(A = \{1, 2, 3, 4, 5\}\) and \(R = \{(a, b) : |a - b| \text{ is even }\}\).
1. Reflexive: For any \(a \in A\), we have \(|a - a| = 0\), which is an even number. Therefore, \((a, a) \in R\) for all \(a \in A\). Thus, \(R\) is reflexive.
2. Symmetric: Let \((a, b) \in R\). Then \(|a - b|\) is even.
This means \(|a - b| = 2k\) for some integer \(k\).
Then \(|b - a| = |-(a - b)| = |a - b| = 2k\), which is also even.
Therefore, \((b, a) \in R\). Thus, \(R\) is symmetric.
3. Transitive: Let \((a, b) \in R\) and \((b, c) \in R\).
Then \(|a - b|\) is even and \(|b - c|\) is even.
This means \(a - b\) and \(b - c\) are both even integers.
The sum of two even integers is even, so \((a - b) + (b - c) = a - c\) is even.
Therefore, \(|a - c|\) is even, which means \((a, c) \in R\). Thus, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: The absolute difference between two numbers being even satisfies all three conditions of equivalence.

Exam Tip: Use the property that the difference of two numbers with the same parity (both even or both odd) is always even.

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. Therefore, \((x, x) \notin R\). Hence, \(R\) is not reflexive.
2. Symmetric: If \(x\) is the wife of \(y\), then \(y\) must be the husband of \(x\), so \(y\) cannot be the wife of \(x\). Therefore, if \((x, y) \in R\), then \((y, x) \notin R\). Hence, \(R\) is not symmetric.
3. Transitive: If \(x\) is the wife of \(y\), then \(x\) is a female and \(y\) is a male. There can be no element \(z\) such that \(y\) is the wife of \(z\) (since \(y\) is male). Thus, the condition for transitivity is vacuously true, but conventionally in school mathematics, it is stated as not transitive because no chain \((x, y)\) and \((y, z)\) can be formed. Formally, it is not transitive.
In simple words: Marriage relations are strictly one-way and cannot connect back to self or form transitive chains.

Exam Tip: For family and gender-based relations, test simple real-world scenarios to instantly check 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, \(|x - x| = 0 \neq 8\). Therefore, \((x, x) \notin R\). Hence, \(R\) is not reflexive.
2. Symmetric: If \(x\) is 8 cm taller than \(y\), then \(y\) is 8 cm shorter than \(x\), not 8 cm taller. Therefore, if \((x, y) \in R\), then \((y, x) \notin R\). Hence, \(R\) is not symmetric.
3. Transitive: If \(x\) is 8 cm taller than \(y\) (so \(x = y + 8\)) and \(y\) is 8 cm taller than \(z\) (so \(y = z + 8\)), then \(x = (z + 8) + 8 = z + 16\). Thus, \(x\) is 16 cm taller than \(z\), not 8 cm taller. Therefore, \((x, z) \notin R\). Hence, \(R\) is not transitive.
In simple words: A height difference of 8 cm is directional and does not apply to oneself, reverse pairs, or summed pairs.

Exam Tip: For height or length comparisons with a fixed non-zero difference, the relation is never reflexive, never symmetric, and never transitive.