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Exercises and Summary – Relations and Functions (Class 12)

🎓 Class 12 Mathematics CBSE Theory Ch 1 — Relations and Functions ⏱ ~25 min
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This MCQ module is based on: Exercises and Summary – Relations and Functions (Class 12)

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Targeting Class 12 level in Functions, with Advanced difficulty.

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Miscellaneous Examples

Example 18

If R₁ and R₂ are equivalence relations in a set A, show that R₁ ∩ R₂ is also an equivalence relation.

Solution
Since R₁ and R₂ are equivalence relations, \((a, a) \in R_1\) and \((a, a) \in R_2\) for all \(a \in A\). So \((a, a) \in R_1 \cap R_2\). Hence R₁ ∩ R₂ is reflexive.

If \((a, b) \in R_1 \cap R_2\), then \((a, b) \in R_1\) and \((a, b) \in R_2\). Since both are symmetric, \((b, a) \in R_1\) and \((b, a) \in R_2\). Hence \((b, a) \in R_1 \cap R_2\). So R₁ ∩ R₂ is symmetric.

If \((a, b) \in R_1 \cap R_2\) and \((b, c) \in R_1 \cap R_2\), then \((a, b), (b, c) \in R_1\) and \((a, b), (b, c) \in R_2\). By transitivity of each, \((a, c) \in R_1\) and \((a, c) \in R_2\). Hence \((a, c) \in R_1 \cap R_2\). So R₁ ∩ R₂ is transitive.

Therefore, R₁ ∩ R₂ is an equivalence relation.

Example 19

Let R be a relation on the set A of ordered pairs of positive integers defined by \((x, y)\,R\,(u, v)\) if and only if \(xv = yu\). Show that R is an equivalence relation.

Solution
Reflexive: \((x, y)\,R\,(x, y)\) iff \(xy = yx\). True (commutativity).

Symmetric: \((x, y)\,R\,(u, v) \Rightarrow xv = yu \Rightarrow uy = vx \Rightarrow (u, v)\,R\,(x, y)\).

Transitive: If \((x, y)\,R\,(u, v)\) and \((u, v)\,R\,(a, b)\), then \(xv = yu\) and \(ub = va\).
From \(xv = yu\), we get \(\frac{x}{y} = \frac{u}{v}\).
From \(ub = va\), we get \(\frac{u}{v} = \frac{a}{b}\).
So \(\frac{x}{y} = \frac{a}{b}\), giving \(xb = ya\), i.e., \((x, y)\,R\,(a, b)\).

Hence R is an equivalence relation.

Example 20

Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R₁ be a relation in X given by R₁ = {(x, y) : x − y is divisible by 3} and R₂ be another relation on X given by R₂ = {(x, y) : {x, y} ⊂ {1, 4, 7}} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9}. Show that R₁ = R₂.

Solution
The characteristic of the sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is that the difference between any two elements in each set is divisible by 3. Thus \((x, y) \in R_2 \Rightarrow x - y\) is divisible by 3 \(\Rightarrow (x, y) \in R_1\). So \(R_2 \subseteq R_1\).

Similarly, \((x, y) \in R_1 \Rightarrow x - y\) is divisible by 3 \(\Rightarrow x\) and \(y\) leave the same remainder when divided by 3 \(\Rightarrow \{x, y\}\) is contained in one of {1,4,7}, {2,5,8}, or {3,6,9} \(\Rightarrow (x, y) \in R_2\). So \(R_1 \subseteq R_2\).

Hence \(R_1 = R_2\).

Example 21

Let \(f: X \to Y\) be a function. Define a relation R in X given by R = {(a, b) : f(a) = f(b)}. Examine whether R is an equivalence relation or not.

Solution
Reflexive: For every \(a \in X\), \(f(a) = f(a)\). So \((a, a) \in R\).
Symmetric: \((a, b) \in R \Rightarrow f(a) = f(b) \Rightarrow f(b) = f(a) \Rightarrow (b, a) \in R\).
Transitive: \((a, b) \in R\) and \((b, c) \in R \Rightarrow f(a) = f(b)\) and \(f(b) = f(c) \Rightarrow f(a) = f(c) \Rightarrow (a, c) \in R\).

Hence R is an equivalence relation.

Example 22

Find the number of all one-one functions from set A = {1, 2, 3} to itself.

Solution
A one-one function from {1, 2, 3} to itself is simply a permutation of the three symbols 1, 2, 3. The total number of such permutations is \(3! = 6\).

Example 23

Let A = {1, 2, 3}. Then show that the number of relations containing (1, 2) and (2, 3) which are reflexive and transitive but not symmetric is three.

Solution
The smallest relation R₁ containing (1, 2) and (2, 3) which is reflexive and transitive must include (1, 1), (2, 2), (3, 3) (reflexive), and (1, 3) (by transitivity from (1, 2) and (2, 3)). So R₁ = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)}.

We can enlarge R₁ by adding pairs (2, 1), (3, 2), or (3, 1) — but we cannot add both members of a symmetric pair without making it symmetric. Adding (3, 2) to R₁ gives R₂, which requires also adding (3, 1) for transitivity (already present). Adding (3, 1) alone to R₁ gives another relation. But if we add (2, 1) to R₁, we must also add (3, 1) for transitivity — still not symmetric since (2, 3) has no (3, 2).

The total number of such relations is three.

Example 24

Show that the number of equivalence relations in the set {1, 2, 3} containing (1, 2) and (2, 1) is two.

Solution
The smallest equivalence relation R₁ containing (1, 2) and (2, 1) is: {(1,1), (2,2), (3,3), (1,2), (2,1)}.

The only larger equivalence relation containing these pairs would need to be the universal relation A × A = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}. Any intermediate relation would violate one of the three properties.

Hence the total number of equivalence relations containing (1, 2) and (2, 1) is two.

Example 25

Consider the identity function \(I_\mathbf{N}: \mathbf{N} \to \mathbf{N}\) defined as \(I_\mathbf{N}(x) = x\) for all \(x \in \mathbf{N}\). Show that although \(I_\mathbf{N}\) is onto, \(I_\mathbf{N} + I_\mathbf{N}: \mathbf{N} \to \mathbf{N}\) defined as \((I_\mathbf{N} + I_\mathbf{N})(x) = I_\mathbf{N}(x) + I_\mathbf{N}(x) = x + x = 2x\) is not onto.

Solution
\(I_\mathbf{N}\) is onto because for any \(y \in \mathbf{N}\), \(I_\mathbf{N}(y) = y\).

But \(I_\mathbf{N} + I_\mathbf{N}\) maps \(x\) to \(2x\). Consider \(y = 3 \in \mathbf{N}\). We need \(x\) such that \(2x = 3\), giving \(x = \frac{3}{2} \notin \mathbf{N}\). So 3 has no pre-image. Hence \(I_\mathbf{N} + I_\mathbf{N}\) is not onto.

Example 26

Consider a function \(f: \left[0, \frac{\pi}{2}\right] \to \mathbf{R}\) given by \(f(x) = \sin x\) and \(g: \left[0, \frac{\pi}{2}\right] \to \mathbf{R}\) given by \(g(x) = \cos x\). Show that \(f\) and \(g\) are one-one, but \(f + g\) is not one-one.

Solution
On \(\left[0, \frac{\pi}{2}\right]\), \(\sin x\) is strictly increasing, so \(f\) is one-one. Similarly, \(\cos x\) is strictly decreasing, so \(g\) is one-one.

But \((f + g)(0) = \sin 0 + \cos 0 = 0 + 1 = 1\), and \((f + g)\left(\frac{\pi}{2}\right) = \sin\frac{\pi}{2} + \cos\frac{\pi}{2} = 1 + 0 = 1\).

Since \(0 \neq \frac{\pi}{2}\) but \((f+g)(0) = (f+g)\left(\frac{\pi}{2}\right) = 1\), \(f + g\) is not one-one.

Summary

Key Points — Chapter 1
  • Empty relation: R = φ ⊂ A × A (no element relates to any element).
  • Universal relation: R = A × A (every element relates to every element).
  • Reflexive: \((a, a) \in R\) for every \(a \in X\).
  • Symmetric: \((a, b) \in R\) implies \((b, a) \in R\).
  • Transitive: \((a, b) \in R\) and \((b, c) \in R\) implies \((a, c) \in R\).
  • Equivalence relation: reflexive + symmetric + transitive.
  • Equivalence class [a]: subset of X containing all elements related to a.
  • One-one (injective): \(f(x_1) = f(x_2) \Rightarrow x_1 = x_2\).
  • Onto (surjective): for every \(y \in Y\), there exists \(x \in X\) with \(f(x) = y\).
  • Bijective: both one-one and onto.
  • For finite sets: \(f: X \to X\) is one-one iff it is onto (and vice versa). This is NOT true for infinite sets.
  • Composition: \(gof(x) = g(f(x))\). Generally \(gof \neq fog\).
  • Invertible function: \(f\) is invertible iff \(f\) is bijective.

Miscellaneous Exercise on Chapter 1

Q1. Show that the function \(f: \mathbf{R} \to \{x \in \mathbf{R} : -1 < x < 1\}\) defined by \(f(x) = \frac{x}{1 + |x|}\), \(x \in \mathbf{R}\) is one-one and onto function.
One-one: Suppose \(f(x_1) = f(x_2)\), i.e., \(\frac{x_1}{1+|x_1|} = \frac{x_2}{1+|x_2|}\).
Case 1: Both \(x_1, x_2 \geq 0\). Then \(\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}\) gives \(x_1(1+x_2) = x_2(1+x_1)\), so \(x_1 + x_1 x_2 = x_2 + x_1 x_2\), hence \(x_1 = x_2\).
Case 2: Both negative — similar reasoning gives \(x_1 = x_2\).
Case 3: \(x_1 \geq 0, x_2 < 0\). Then \(f(x_1) \geq 0\) but \(f(x_2) < 0\). Contradiction.
Hence \(f\) is one-one.

Onto: Let \(y \in (-1, 1)\). If \(y \geq 0\), take \(x = \frac{y}{1-y} \geq 0\). Then \(f(x) = \frac{y/(1-y)}{1 + y/(1-y)} = \frac{y}{1} = y\).
If \(y < 0\), take \(x = \frac{y}{1+y}\). Then \(f(x) = y\). Hence \(f\) is onto.
Q2. Show that the function \(f: \mathbf{R} \to \mathbf{R}\) given by \(f(x) = x^3\) is injective.
Suppose \(f(x_1) = f(x_2)\), i.e., \(x_1^3 = x_2^3\). Then \(x_1^3 - x_2^3 = 0\), i.e., \((x_1 - x_2)(x_1^2 + x_1 x_2 + x_2^2) = 0\).

The expression \(x_1^2 + x_1 x_2 + x_2^2 = \frac{1}{2}[(x_1 + x_2)^2 + x_1^2 + x_2^2]\) is always positive unless \(x_1 = x_2 = 0\).

So either \(x_1 = x_2\) or \(x_1 = x_2 = 0\). In both cases, \(x_1 = x_2\). Hence \(f\) is injective.
Q3. Given a non-empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.
Reflexive: For every \(A \in P(X)\), \(A \subseteq A\). So \((A, A) \in R\). Yes.
Symmetric: Let A = {1} and B = {1, 2}. Then \(A \subseteq B\), so \((A, B) \in R\). But \(B \not\subseteq A\), so \((B, A) \notin R\). Not symmetric.
Transitive: If \(A \subseteq B\) and \(B \subseteq C\), then \(A \subseteq C\). Yes.

Since R is not symmetric, R is NOT an equivalence relation.
Q4. Find the number of all onto functions from the set {1, 2, 3, ..., n} to itself.
Any onto function from a finite set to itself must also be one-one (by the pigeonhole principle). So we are counting bijections, which are just permutations.

The number of permutations of \(n\) elements is \(n!\).
Answer: \(n!\)
Q5. Let A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2} and \(f, g: A \to B\) be functions defined by \(f(x) = x^2 - x\), \(x \in A\) and \(g(x) = 2\left|x - \frac{1}{2}\right| - 1\), \(x \in A\). Are \(f\) and \(g\) equal? Justify your answer.
Compute \(f\) and \(g\) for all elements of A:

\(f(-1) = (-1)^2 - (-1) = 1 + 1 = 2\), \(g(-1) = 2|{-1} - \frac{1}{2}| - 1 = 2 \cdot \frac{3}{2} - 1 = 2\). Equal.
\(f(0) = 0 - 0 = 0\), \(g(0) = 2|0 - \frac{1}{2}| - 1 = 2 \cdot \frac{1}{2} - 1 = 0\). Equal.
\(f(1) = 1 - 1 = 0\), \(g(1) = 2|1 - \frac{1}{2}| - 1 = 2 \cdot \frac{1}{2} - 1 = 0\). Equal.
\(f(2) = 4 - 2 = 2\), \(g(2) = 2|2 - \frac{1}{2}| - 1 = 2 \cdot \frac{3}{2} - 1 = 2\). Equal.

Since \(f(a) = g(a)\) for all \(a \in A\), \(f\) and \(g\) are equal.
Q6. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(A) 1   (B) 2   (C) 3   (D) 4
Required pairs: (1,1), (2,2), (3,3) (reflexive), (1,2), (2,1), (1,3), (3,1) (symmetric closure).
For transitivity to fail, we need some chain that does not close. Check: (2,1) and (1,3) are in R, so if (2,3) were in R, transitivity would require (2,3) — but we can choose NOT to include (2,3) and (3,2).
The answer is (A) 1.
Q7. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
(A) 1   (B) 2   (C) 3   (D) 4
The smallest equivalence relation containing (1, 2) is: {(1,1), (2,2), (3,3), (1,2), (2,1)}.
The only larger one is the universal relation A × A.
Answer: (B) 2
Historical Note
The concept of function has evolved over centuries. The word "function" was first used by Leibniz in 1673. Leonhard Euler (1707–1783) introduced the notation \(f(x)\) in 1734. The modern set-theoretic definition of a function was given by Lejeune Dirichlet (1805–1859), and later formalized by Georg Cantor (1845–1918).
Activity: Counting Equivalence Relations
Predict: How many equivalence relations are possible on the set {1, 2, 3}? (Hint: equivalence relations correspond to partitions of the set.)
  1. List all possible partitions of {1, 2, 3}:
    • {{1}, {2}, {3}} — each element in its own class
    • {{1, 2}, {3}} — 1 and 2 grouped together
    • {{1, 3}, {2}} — 1 and 3 grouped together
    • {{2, 3}, {1}} — 2 and 3 grouped together
    • {{1, 2, 3}} — all elements in one class
  2. Write the equivalence relation for each partition. For example, {{1, 2}, {3}} gives R = {(1,1), (2,2), (3,3), (1,2), (2,1)}.
  3. Count: there are 5 partitions.

Observation: There are exactly 5 equivalence relations on a 3-element set. This number is called a Bell number: B(3) = 5.

Explanation: Every equivalence relation on a set corresponds to a unique partition of that set, and vice versa. The number of partitions of an n-element set is given by the nth Bell number: B(1) = 1, B(2) = 2, B(3) = 5, B(4) = 15, B(5) = 52, ...

Competency-Based Questions

A school database stores student records. Each student has a unique Aadhaar number (a 12-digit number). The school assigns roll numbers from 1 to N. Let \(f\) be the function mapping each student to their Aadhaar number, and \(g\) be the function mapping each student to their roll number.
Q1. Is the function \(f\) (student to Aadhaar) one-one? Is it onto \(\mathbf{N}\)?
L2 Understand
One-one: Yes, since each student has a unique Aadhaar number. Different students map to different numbers.
Onto N: No, because there are infinitely many 12-digit numbers that are not assigned to any student in this school.
Q2. If the school has exactly 50 students, and roll numbers are from 1 to 50, is the roll-number function \(g\) bijective?
L3 Apply
Yes, \(g\) is bijective. One-one: no two students share a roll number. Onto: every number from 1 to 50 is assigned to exactly one student. The domain and codomain are finite sets of the same size, and \(g\) is one-one, so by the pigeonhole principle it must also be onto.
Q3. Can we define an inverse function \(g^{-1}\) that maps each roll number back to the student? What condition makes this possible?
L4 Analyse
Yes, since \(g\) is bijective. The inverse \(g^{-1}: \{1, ..., 50\} \to \text{Students}\) maps each roll number to the unique student holding that number. The necessary and sufficient condition is that \(g\) is bijective (one-one and onto). If \(g\) were not one-one (two students sharing a roll number), the inverse would be undefined. If not onto (some roll numbers unassigned), the inverse would have gaps.
Q4. Suppose the school introduces a new coding: house number \(h = g(s) \mod 4 + 1\) (giving houses 1–4). Is this composite function \(h \circ g^{-1}\) injective from roll numbers to house numbers? Justify.
L5 Evaluate
No, it is not injective. The function maps 50 roll numbers to only 4 house numbers. By the pigeonhole principle, at least \(\lceil 50/4 \rceil = 13\) students map to the same house. Multiple roll numbers produce the same house number, making the composite function many-one.

Assertion–Reason Questions

Assertion (A): The number of one-one functions from {a, b, c} to {1, 2, 3, 4} is 24.
Reason (R): The number of one-one functions from a set of m elements to a set of n elements (n ≥ m) is \(\frac{n!}{(n-m)!}\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Using R: \(\frac{4!}{(4-3)!} = \frac{24}{1} = 24\). Both A and R are true, and R explains A.
Assertion (A): The intersection of two equivalence relations on a set is always an equivalence relation.
Reason (R): Reflexivity, symmetry, and transitivity are each preserved under intersection.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Both are true. If R₁ and R₂ are equivalence relations, R₁ ∩ R₂ inherits reflexivity (both contain (a,a)), symmetry (both swap pairs), and transitivity (both close chains). R correctly explains A.

Frequently Asked Questions

How to solve equivalence relation problems?

Check reflexive: (a,a) in R for all a. Check symmetric: (a,b) implies (b,a). Check transitive: (a,b) and (b,c) imply (a,c). Document each check for full marks.

What exercises are in NCERT Class 12 Chapter 1?

Chapter 1 contains Exercise 1.1 (types of relations), Exercise 1.2 (one-one, onto), Exercise 1.3 (composition and inverse), Exercise 1.4 (binary operations), and Miscellaneous Exercise.

How to prove a function is bijective?

Prove one-one: assume f(x1) = f(x2) and show x1 = x2. Prove onto: for any y in codomain, find x such that f(x) = y. Both together establish bijectivity.

How to find the inverse of a function?

Write y = f(x), solve for x in terms of y, replace x with f-inverse(y). Verify f(f-inverse(y)) = y and f-inverse(f(x)) = x.

What are common mistakes in Class 12 Relations and Functions?

Confusing reflexive with symmetric, not checking all three equivalence properties, assuming composition is commutative, and errors in binary operation closure checks.

Frequently Asked Questions — Relations and Functions

What is Exercises and Summary - Relations and Functions (Class 12) in NCERT Class 12 Mathematics?

Exercises and Summary - Relations and Functions (Class 12) is a key concept covered in NCERT Class 12 Mathematics, Chapter 1: Relations and Functions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Exercises and Summary - Relations and Functions (Class 12) step by step?

To solve problems on Exercises and Summary - Relations and Functions (Class 12), follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 1: Relations and Functions?

The essential formulas of Chapter 1 (Relations and Functions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Exercises and Summary - Relations and Functions (Class 12) important for the Class 12 board exam?

Exercises and Summary - Relations and Functions (Class 12) is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Exercises and Summary - Relations and Functions (Class 12)?

Common mistakes in Exercises and Summary - Relations and Functions (Class 12) include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Exercises and Summary - Relations and Functions (Class 12)?

End-of-chapter NCERT exercises for Exercises and Summary - Relations and Functions (Class 12) cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.

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