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Chapter 11 · Class 12 Mathematics

Relations and Functions — Important Questions

33 questions With answers CBSE format

SUMMARY: This chapter explores the fundamental concepts of relations and functions, including their types and properties.
KEY TOPICS: types of relations, equivalence relation, types of functions, one-one and onto functions, composition of functions, inverse of a function, binary operations, domain and range

Q1 1 Mark

The relation R on the set A = {1, 2, 3} given by R = {(1, 2), (2, 1)} is:

AReflexive
BSymmetric
CTransitive
DEquivalence
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Correct answer: Option 2 — Symmetric
Q2 1 Mark

If f: R → R, f(x) = x² + 1, then f is:

AOne-one and onto
BMany-one and into
COne-one and into
DMany-one and onto
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Correct answer: Option 2 — Many-one and into
Q3 1 Mark

Let f(x) = 2x + 3 and g(x) = x − 1. Then (fog)(2) equals:

A5
B7
C9
D11
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Correct answer: Option 1 — 5
Q4 1 Mark

The number of equivalence relations on the set {1, 2, 3} that contain (1, 2) is:

A1
B2
C3
D4
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Correct answer: Option 2 — 2
Q5 1 Mark

For a one-one function f: A → B, n(A) = 5 and n(B) = 7, the number of one-one functions from A to B is:

A2520
B7⁵
C5⁷
D5040
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Correct answer: Option 1 — 2520
Q6 3 Marks

Define an equivalence relation and give one example.

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A relation R on a set A is an equivalence relation if it is reflexive (xRx ∀ x ∈ A), symmetric (xRy ⇒ yRx), and transitive (xRy and yRz ⇒ xRz). Example: 'is congruent to' on the set of triangles — every triangle is congruent to itself, congruence is symmetric and transitive.
Q7 3 Marks

Show that the function f: N → N defined by f(n) = 2n is one-one but not onto.

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One-one: Suppose f(a) = f(b) ⇒ 2a = 2b ⇒ a = b. So f is injective. Onto: Range = {2, 4, 6, ...} ⊂ N. Odd natural numbers like 1, 3, 5, ... have no preimage. Therefore f is not surjective. Hence one-one but not onto.
Q8 3 Marks

For f(x) = x + 1 and g(x) = x², find (fog)(x) and (gof)(x).

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(fog)(x) = f(g(x)) = f(x²) = x² + 1. (gof)(x) = g(f(x)) = g(x+1) = (x+1)² = x² + 2x + 1. Note that fog ≠ gof in general — composition of functions is not commutative.
Q9 3 Marks

State the necessary and sufficient condition for a function f: A → B to be invertible.

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A function f: A → B is invertible if and only if it is bijective, i.e. one-one (injective) and onto (surjective). Only then does there exist a unique inverse function f⁻¹: B → A such that fof⁻¹ = identity on B and f⁻¹of = identity on A.
Q10 3 Marks

On Z define R: aRb ⇔ a − b is divisible by 3. Show R is reflexive.

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For any a ∈ Z, a − a = 0 which is divisible by 3 (since 0 = 3 × 0). Hence aRa for every a ∈ Z, so R is reflexive on Z.
Q11 6 Marks

Show that the relation R on N × N defined by (a, b) R (c, d) ⇔ a + d = b + c is an equivalence relation.

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Reflexive: (a, b) R (a, b) requires a + b = b + a — true. Symmetric: (a, b) R (c, d) ⇒ a + d = b + c ⇒ c + b = d + a ⇒ (c, d) R (a, b). Transitive: (a, b) R (c, d) and (c, d) R (e, f) ⇒ a + d = b + c and c + f = d + e. Adding: a + d + c + f = b + c + d + e ⇒ a + f = b + e ⇒ (a, b) R (e, f). All three properties hold so R is an equivalence relation.
Q12 6 Marks

Show that f: R − {−4/3} → R − {4/3} given by f(x) = (4x + 3) / (3x + 4) is a bijection.

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One-one: Suppose f(x) = f(y). Then (4x+3)/(3x+4) = (4y+3)/(3y+4). Cross-multiply: (4x+3)(3y+4) = (4y+3)(3x+4) ⇒ 12xy + 16x + 9y + 12 = 12xy + 16y + 9x + 12 ⇒ 7x = 7y ⇒ x = y. Onto: For y ∈ R − {4/3}, set y = (4x+3)/(3x+4). Solve for x: y(3x+4) = 4x+3 ⇒ 3xy + 4y = 4x + 3 ⇒ x(3y − 4) = 3 − 4y ⇒ x = (3 − 4y)/(3y − 4), defined when y ≠ 4/3. Hence every y has a preimage, so f is onto. Therefore f is bijective.
Q13 6 Marks

Show that the relation R on Z given by aRb ⇔ a − b is divisible by 5 is an equivalence relation. Find the equivalence class of 0.

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Reflexive: a − a = 0 = 5 · 0, divisible by 5. Symmetric: a − b = 5k ⇒ b − a = 5(−k), divisible by 5. Transitive: a − b = 5k and b − c = 5m ⇒ a − c = 5(k + m), divisible by 5. Hence equivalence relation. Equivalence class of 0 = {a ∈ Z : a − 0 is divisible by 5} = {..., −10, −5, 0, 5, 10, ...} = 5Z.
Q14 6 Marks

Let f: R → R, f(x) = x³ + 4. Show that f is bijective and find f⁻¹.

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One-one: f(x) = f(y) ⇒ x³ + 4 = y³ + 4 ⇒ x³ = y³ ⇒ x = y. Onto: For any y ∈ R, set y = x³ + 4 ⇒ x = (y − 4)^(1/3) ∈ R. Hence every y has a real cube-root preimage. So f is bijective. f⁻¹(y) = (y − 4)^(1/3).
Q15 6 Marks

Let A = {1, 2, 3, ..., 9} and R be the relation in A × A defined by (a, b) R (c, d) iff a + d = b + c. Show that R is an equivalence relation. Also find equivalence class of (2, 5).

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Equivalence (analogous proof to the standard a + d = b + c relation on N × N). For (2, 5): (a, b) R (2, 5) ⇔ a + 5 = b + 2 ⇔ b − a = 3. Listing pairs (a, b) ∈ A × A with b = a + 3: (1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9). So [(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.
Q16 6 Marks

Differentiate between reflexive symmetric and transitive relations in tabular form with one example each.

Q17 1 Mark

Assertion (A): The identity relation is an equivalence relation.

Reason (R): It is reflexive, symmetric and transitive — every element is related only to itself.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q18 1 Mark

Assertion (A): The function f: R → R defined by f(x) = 2x + 3 is bijective.

Reason (R): A linear function with non-zero slope is one-one and onto.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q19 1 Mark

Assertion (A): Composition of functions is commutative for all f and g.

Reason (R): Composition is associative: (f o g) o h = f o (g o h).

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Correct answer: Option 3 — A is true, but R is false.
Q20 1 Mark

Assertion (A): A function has an inverse if and only if it is bijective.

Reason (R): Bijection guarantees a unique correspondence between domain and range.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q21 1 Mark

Assertion (A): The relation (a, b) R (c, d) ⇔ a + d = b + c on N × N is an equivalence relation.

Reason (R): It is reflexive, symmetric and transitive on the Cartesian product.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q22 1 Mark

Statement 1: Every reflexive relation is symmetric.

Statement 2: Every symmetric relation is reflexive.

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Correct answer: Option 2 — Only Statement 1 is true.
Q23 1 Mark

Statement 1: The function sin x is many-one on R.

Statement 2: The function sin x restricted to [−π/2, π/2] is one-one.

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Correct answer: Option 1 — Both statements are true.
Q24 1 Mark

Statement 1: Composition of functions is associative.

Statement 2: Composition of functions is commutative.

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Correct answer: Option 3 — Only Statement 2 is true.
Q25 1 Mark

Statement 1: The inverse of the identity function is itself.

Statement 2: The inverse of a bijective function is also bijective.

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Correct answer: Option 1 — Both statements are true.
Q26 1 Mark

Statement 1: A set with n elements has 2^(n²) total relations.

Statement 2: The number of reflexive relations on such a set is 2^(n² − n).

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Correct answer: Option 1 — Both statements are true.
Q27 3 Marks
On the set of integers Z a relation R is defined by aRb if and only if (a − b) is divisible by 5. Five students are asked to verify whether this relation is reflexive, symmetric and transitive on Z.
  1. Which of the following properties is satisfied by R?
    AReflexive
    BSymmetric
    CTransitive
    DAll of these
  2. The equivalence class of 0, denoted [0], is best described as which of the listed sets?
    A{0, 5, 10}
    B{1, 6, 11}
    C{2, 7, 12}
    DAll sets above
  3. Explain why R is an equivalence relation and describe the partition of Z it creates.
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1. Option 4 — All of these
2. Option 4 — All sets above
3. R partitions Z into 5 disjoint equivalence classes [0], [1], [2], [3], [4] depending on the remainder when divided by 5. Two integers are related iff they leave the same remainder mod 5.
Q28 3 Marks
A school assigns each student a unique 6-digit roll number. Let A be the set of all students of the school and B be the set of all 6-digit numbers used. Define f: A → B as f(student) = his/her roll number.
  1. Since each student has a unique roll number and every roll number is used the function f is:
    AOne-one but not onto
    BOnto but not one-one
    CBoth one-one and onto
    DNeither
  2. Because f is bijective:
    AInverse exists
    BInverse does not exist
    CInverse is many-valued
    DCannot decide
  3. Explain in words why a bijection guarantees the existence of an inverse function.
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1. Option 3 — Both one-one and onto
2. Option 1 — Inverse exists
3. A bijective function (one-one and onto) sets up a perfect pairing between two sets so the inverse function is well defined. Here knowing the roll number uniquely determines the student and vice-versa.
Q29 3 Marks
In a factory a raw material is first processed by machine f which doubles its weight and then by machine g which adds 5 kg to whatever weight it receives. Let f(x) = 2x and g(x) = x + 5 where x is the original weight in kg.
  1. The composite (g ∘ f)(x) is:
    A2x + 5
    B2x + 10
    C2(x + 5)
    Dx + 10
  2. The composite (f ∘ g)(x) is:
    A2x + 5
    B2x + 10
    Cx + 5
    D4x + 5
  3. Compute (g ∘ f)(3) and (f ∘ g)(3) and comment on whether composition is commutative.
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1. Option 1 — 2x + 5
2. Option 2 — 2x + 10
3. In general (g ∘ f) ≠ (f ∘ g). The order of operations matters: (g ∘ f) means apply f first then g; (f ∘ g) means apply g first then f. Composition of functions is not commutative.
Q30 3 Marks

Study the table of relations on set A = {1, 2, 3} and answer:

RelationReflexiveSymmetricTransitive
R1 = {(1,1),(2,2),(3,3)}YesYesYes
R2 = {(1,2),(2,1)}NoYesNo
R3 = {(1,2),(2,3),(1,3)}NoNoYes
R4 = {(1,1),(2,2),(1,2),(2,1)}NoYesYes
  1. Which relation is an equivalence relation on A?
    AR1
    BR2
    CR3
    DR4
  2. Which relation is symmetric and transitive but not reflexive?
    AR2
    BR3
    CR4
    D
  3. Explain why R4 fails to be an equivalence relation despite being symmetric and transitive.
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1. Option 1 — R1
2. Option 3 — R4
3. For an equivalence relation all three properties — reflexivity, symmetry and transitivity — must hold simultaneously. Missing reflexivity (as in R4) is enough to disqualify a relation from being an equivalence relation.
Q31 3 Marks

Study the function values and answer:

xf(x) = x²g(x) = x + 1(f ∘ g)(x)
0011
1124
2439
−1100
  1. The composite (f ∘ g)(x) equals:
    A(x + 1)²
    Bx² + 1
    Cx²(x + 1)
    Dx + 1
  2. Considering f: R → R defined by f(x) = x² the function is:
    AOne-one onto
    BOne-one not onto
    CMany-one onto
    DMany-one not onto
  3. Explain whether f(x) = x², considered as a function from R to R, is one-one or onto.
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1. Option 1 — (x + 1)²
2. Option 4 — Many-one not onto
3. f(x) = x² maps both x and −x to the same value so it is many-one. Its range is [0,∞) which is a strict subset of the codomain R so it is not onto either.
Q32 6 Marks

For each relation R defined on A = {1, 2, 3} below, identify whether it is reflexive, symmetric or transitive.

Relation RPairs
R1{(1,1),(2,2),(3,3),(1,2)}
R2{(1,2),(2,1)}
R3{(1,1),(2,2),(3,3),(1,2),(2,1)}
R4{(1,2),(2,3)}
Q33 3 Marks

Study the graphs of f(x) = x² and g(x) = x³ on R and answer:

Relations and Functions figure
  1. As a function from R to R, f(x) = x² is:
    AOne-one and onto
    BOne-one but not onto
    CMany-one but onto
    DMany-one and not onto
  2. As a function from R to R, g(x) = x³ is:
    AOne-one and onto
    BOne-one but not onto
    CMany-one but onto
    DMany-one and not onto
  3. Use the horizontal-line test on each graph to justify your answers.
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1. Option 4 — Many-one and not onto
2. Option 1 — One-one and onto
3. A horizontal line y = c crosses x² at two points (for c > 0) showing many-one. Its range is [0, ∞), not all of R, so it is not onto. By contrast x³ is strictly increasing — every horizontal line meets it exactly once — so it is one-one and its range is all of R, making it onto.

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