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Relations and Functions: Class 12 Maths Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A delivery company assigns each order a unique tracking code. The tracking system uses a function f : ℝ → ℝ defined by f(x) = (3x + 7)/2, where x ∈ ℝ represents the order number and f(x) represents the tracking code.

A delivery company assigns each order a unique tracking code. The tracking system uses a function f : ℝ → ℝ defined by f(x) = (3x + 7)/2.

(i) A logistics analyst claims that no two different orders can ever receive the same tracking code under this system. Verify whether the analyst's claim is correct by checking if f is one-one. [1]

(ii) The system administrator needs every real number to be a valid tracking code output. Check whether f is onto. [1]

(iii) Based on your results in (i) and (ii), determine whether f is invertible. If yes, find f⁻¹(x) and hence find f⁻¹(13). [2]

Show answer
(i) One-one (Injective):
Let f(a) = f(b) for some a, b ∈ ℝ.
⟹ (3a + 7)/2 = (3b + 7)/2
⟹ 3a + 7 = 3b + 7
⟹ 3a = 3b
⟹ a = b
∴ f is one-one. The analyst's claim is correct.

(ii) Onto (Surjective):
Let y ∈ ℝ (codomain). We need to find x ∈ ℝ such that f(x) = y.
f(x) = y
⟹ (3x + 7)/2 = y
⟹ 3x = 2y − 7
⟹ x = (2y − 7)/3
Since y ∈ ℝ, we have x = (2y − 7)/3 ∈ ℝ, i.e., x ∈ ℝ (domain).
Also, f(x) = f((2y − 7)/3) = (3·(2y − 7)/3 + 7)/2 = (2y − 7 + 7)/2 = 2y/2 = y. ✓
∴ Every y ∈ ℝ has a pre-image in ℝ, so Range = Codomain = ℝ.
∴ f is onto.

(iii) Invertibility and f⁻¹(x):
Since f is both one-one and onto, f is bijective and hence invertible.
Let y = f(x) = (3x + 7)/2.
⟹ 2y = 3x + 7
⟹ x = (2y − 7)/3
∴ f⁻¹(y) = (2y − 7)/3, i.e., f⁻¹(x) = (2x − 7)/3.
Now, f⁻¹(13) = (2(13) − 7)/3 = (26 − 7)/3 = 19/3.
∴ f⁻¹(13) = 19/3.
Q2Case-based4 marks

A school library manages its book collection using a classification system. Two books are said to be 'related' if they belong to the same subject category (e.g., both are Mathematics books, or both are Science books, etc.). Formally, let B be the set of all books in the library, and define a relation R on B as:

R = {(b₁, b₂) : b₁ and b₂ belong to the same subject category, b₁, b₂ ∈ B}

The librarian also defines a function f : {1, 2, 3, 4, 5, 6} → {1, 2, 3, 4, 5, 6} where f(n) represents the shelf number assigned to book-ID n, given by:

f = {(1, 2), (2, 4), (3, 6), (4, 1), (5, 3), (6, 5)}

A school library manages its book collection using a classification system. Two books are said to be 'related' if they belong to the same subject category (e.g., both are Mathematics books, or both are Science books, etc.). Formally, let B be the set of all books in the library, and define a relation R on B as:

R = {(b₁, b₂) : b₁ and b₂ belong to the same subject category, b₁, b₂ ∈ B}

The librarian also defines a function f : {1, 2, 3, 4, 5, 6} → {1, 2, 3, 4, 5, 6} where f(n) represents the shelf number assigned to book-ID n, given by:

f = {(1, 2), (2, 4), (3, 6), (4, 1), (5, 3), (6, 5)}

(i) Show that the relation R defined on the set B is reflexive. [1]
(ii) Show that the relation R defined on the set B is symmetric. [1]
(iii) A student claims that f is a bijective function, and therefore the library can create a perfect 'return-mapping' (i.e., f⁻¹ exists). Verify whether f is one-one and onto, and hence determine whether f⁻¹ exists. [2]

Show answer
(i) Reflexive:

For any book b ∈ B, book b belongs to the same subject category as itself.

∴ (b, b) ∈ R for all b ∈ B.

∴ R is reflexive. [1]

(ii) Symmetric:

Let (b₁, b₂) ∈ R for some b₁, b₂ ∈ B.

⟹ b₁ and b₂ belong to the same subject category.

⟹ b₂ and b₁ belong to the same subject category.

⟹ (b₂, b₁) ∈ R.

∴ R is symmetric. [1]

(iii) One-one (Injective):

The function f is given by:
f(1) = 2, f(2) = 4, f(3) = 6, f(4) = 1, f(5) = 3, f(6) = 5.

All six output values {2, 4, 6, 1, 3, 5} are distinct.

∴ f(a) = f(b) ⟹ a = b for all a, b ∈ {1, 2, 3, 4, 5, 6}.

∴ f is one-one. [½]

Onto (Surjective):

Range of f = {1, 2, 3, 4, 5, 6} = Codomain of f.

∴ Every element of the codomain has a pre-image under f.

∴ f is onto. [½]

Since f is both one-one and onto, f is bijective.

∴ f⁻¹ exists, and the library can create a perfect return-mapping. [1]

The inverse function f⁻¹ : {1, 2, 3, 4, 5, 6} → {1, 2, 3, 4, 5, 6} is:
f⁻¹ = {(2, 1), (4, 2), (6, 3), (1, 4), (3, 5), (5, 6)}.
Q3Case-based4 marks

A school library has a collection of books categorised into 4 genres: Mystery (M), Science Fiction (F), Biography (B), and Romance (R), forming the set G = {M, F, B, R}. The librarian defines a relation T on G as:
T = {(x, y) : x and y belong to the same broad category, where Mystery and Science Fiction are grouped as 'Fiction', and Biography and Romance are grouped as 'Non-Fiction'; x, y ∈ G}.

A school library has a collection of books categorised into 4 genres: Mystery (M), Science Fiction (F), Biography (B), and Romance (R), forming the set G = {M, F, B, R}. The librarian defines a relation T on G as:
T = {(x, y) : x and y belong to the same broad category, where Mystery and Science Fiction are grouped as 'Fiction', and Biography and Romance are grouped as 'Non-Fiction'; x, y ∈ G}.

Based on the above, answer the following:
(i) List all the elements of the relation T. [1 mark]
(ii) Check whether T is reflexive. Give a reason. [1 mark]
(iii) A student claims that T is an equivalence relation. Verify whether the student's claim is correct by checking symmetry and transitivity as well. [2 marks]

Show answer
(i) The Fiction group is {M, F} and the Non-Fiction group is {B, R}.
All pairs (x, y) such that x and y belong to the same group are:
T = {(M, M), (M, F), (F, M), (F, F), (B, B), (B, R), (R, B), (R, R)}
∴ T has 8 elements as listed above.

(ii) Reflexive: For every x ∈ G, x belongs to the same broad category as itself.
⟹ (M, M), (F, F), (B, B), (R, R) ∈ T.
∴ T is reflexive.

(iii) Symmetric: Let (x, y) ∈ T for some x, y ∈ G.
⟹ x and y belong to the same broad category.
⟹ y and x belong to the same broad category.
⟹ (y, x) ∈ T.
∴ T is symmetric.

Transitive: Let (x, y) ∈ T and (y, z) ∈ T for some x, y, z ∈ G.
⟹ x and y belong to the same broad category, and y and z belong to the same broad category.
⟹ x, y, and z all belong to the same broad category.
⟹ (x, z) ∈ T.
∴ T is transitive.

Since T is reflexive, symmetric and transitive, the student's claim is correct.
∴ T is an equivalence relation.
Q4Case-based4 marks

A school library uses a digital cataloguing system. Every book in the library is assigned a unique ID from the set A = {1, 2, 3, 4, 5}. The librarian defines a function f : A → B that maps each book ID to a 'shelf code' using the rule f(x) = 3x − 1, where B is the range of f.

A school library uses a digital cataloguing system. Every book in the library is assigned a unique ID from the set A = {1, 2, 3, 4, 5}. The librarian defines a function f : A → B that maps each book ID to a 'shelf code' using the rule f(x) = 3x − 1, where B is the range of f.

(i) Write down the set B (range of f). [1 mark]
(ii) Is f a one-one function? Justify your answer. [1 mark]
(iii) The librarian now considers a NEW function g : A → ℕ defined by the same rule g(x) = 3x − 1, where ℕ is the set of all natural numbers. Determine whether g is onto. Justify your answer. [2 marks]

Show answer
(i) Finding set B (range of f):

For each element of A = {1, 2, 3, 4, 5}, applying f(x) = 3x − 1:

f(1) = 3(1) − 1 = 2
f(2) = 3(2) − 1 = 5
f(3) = 3(3) − 1 = 8
f(4) = 3(4) − 1 = 11
f(5) = 3(5) − 1 = 14

∴ B = {2, 5, 8, 11, 14}

(ii) Checking whether f is one-one:

Let f(a) = f(b) for some a, b ∈ A.

⟹ 3a − 1 = 3b − 1
⟹ 3a = 3b
⟹ a = b

∴ f(a) = f(b) ⟹ a = b, so f is one-one.

(iii) Checking whether g : A → ℕ is onto:

For g to be onto, every element of the codomain ℕ must have a pre-image in A.

The range of g = {2, 5, 8, 11, 14} (same calculation as above).

Since the codomain is ℕ = {1, 2, 3, 4, 5, 6, …}, and the range of g = {2, 5, 8, 11, 14} ⊊ ℕ,

there exist elements in ℕ (for example, 1 ∈ ℕ) that have no pre-image in A, since 3x − 1 = 1 ⟹ x = 2/3 ∉ A.

∵ Range of g ≠ Codomain (ℕ),

∴ g is NOT onto.
Q5Case-based4 marks

A school library maintains a record of books issued to students. The librarian defines a relation R on the set S = {1, 2, 3, 4, 5, 6}, where S represents student ID numbers, as follows:

R = {(a, b) : a and b have issued a book from the same genre}

After checking the records:
• Students 1, 2, 3 → 'Science Fiction'
• Students 4, 5 → 'Mystery'
• Student 6 → 'Biography'

A school library maintains a record of books issued to students. The librarian defines a relation R on the set S = {1, 2, 3, 4, 5, 6}, where S represents student ID numbers, as follows:

R = {(a, b) : a and b have issued a book from the same genre}

After checking the records, the librarian finds:
• Students 1, 2, 3 have issued books from the genre 'Science Fiction'
• Students 4, 5 have issued books from the genre 'Mystery'
• Student 6 has issued a book from the genre 'Biography'

Based on this situation, answer the following:

(i) List any four ordered pairs that belong to R.

(ii) Check whether R is reflexive. Give a reason.

(iii) The librarian claims: 'If student a has issued a book from the same genre as student b, and student b has issued a book from the same genre as student c, then student a has also issued a book from the same genre as student c.' Verify whether this claim is mathematically correct by checking if R is transitive.

(iii) OR

A new student (ID = 7) joins and issues a 'Mystery' genre book. The librarian now updates R accordingly. Determine whether the updated relation R (on the new set S' = {1, 2, 3, 4, 5, 6, 7}) is an equivalence relation. Justify your answer by checking all three properties.

Show answer
Ans.

(i) From the given genre groupings:
• Students 1, 2, 3 share 'Science Fiction', so (1,2), (2,1), (1,3), (2,3) ∈ R.
• Students 4, 5 share 'Mystery', so (4,5), (5,4) ∈ R.
• Student 6 is alone in 'Biography', so only (6,6) ∈ R (self-pair).

Four ordered pairs belonging to R: (1, 2), (2, 3), (1, 3), (4, 5).

[1 mark]

(ii) Reflexive: For every a ∈ S, (a, a) ∈ R because every student has issued a book from the same genre as themselves.

Specifically:
(1,1), (2,2), (3,3) ∈ R [Science Fiction group]
(4,4), (5,5) ∈ R [Mystery group]
(6,6) ∈ R [Biography group]

∴ (a, a) ∈ R for all a ∈ S.

∴ R is reflexive.

[1 mark]

(iii) Transitive: We need to verify that whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.

Suppose (a, b) ∈ R and (b, c) ∈ R.
⟹ a and b have issued a book from the same genre, and b and c have issued a book from the same genre.
⟹ a, b and c all belong to the same genre group.
⟹ a and c have also issued a book from the same genre.
⟹ (a, c) ∈ R.

Since the genre groups are disjoint and fixed, this holds for all a, b, c ∈ S.

∴ R is transitive.

Hence, the librarian's claim is mathematically correct.

[2 marks]

OR

(iii) When student 7 (Mystery genre) joins, the updated set is S' = {1, 2, 3, 4, 5, 6, 7} and the genre groups become:
• Students 1, 2, 3 → 'Science Fiction'
• Students 4, 5, 7 → 'Mystery'
• Student 6 → 'Biography'

R on S' = {(a, b) : a and b have issued a book from the same genre}.

Reflexive: For every a ∈ S', a belongs to some genre, so (a, a) ∈ R for all a ∈ S'.
∴ R is reflexive. ...(i)

Symmetric: Let (a, b) ∈ R.
⟹ a and b have issued a book from the same genre.
⟹ b and a have issued a book from the same genre.
⟹ (b, a) ∈ R.
∴ R is symmetric. ...(ii)

Transitive: Let (a, b) ∈ R and (b, c) ∈ R.
⟹ a, b belong to the same genre and b, c belong to the same genre.
⟹ a, b, c all belong to the same genre.
⟹ (a, c) ∈ R.
∴ R is transitive. ...(iii)

From (i), (ii) and (iii),
∴ R is reflexive, symmetric and transitive. Hence, R is an equivalence relation.

[2 marks]
Q6Case-based4 marks

A school library system uses a digital cataloguing software. The librarian assigns each book a unique 'Library Code' using a function f : ℕ → ℕ defined by f(n) = 2n + 3, where n ∈ ℕ (ℕ = {1, 2, 3, …}).

A school library system uses a digital cataloguing software. The librarian assigns each book a unique 'Library Code' using a function f : ℕ → ℕ defined by f(n) = 2n + 3, where n ∈ ℕ (ℕ = {1, 2, 3, …}).

Based on the above information, answer the following questions:

(i) Show that f is one-one. [1]
(ii) Determine whether f is onto. Justify your answer. [1]
(iii) The librarian considers a new relation R on ℕ defined as:
R = {(m, n) : m, n ∈ ℕ and |m − n| is even}.
Show that R is an equivalence relation. [2]

Show answer
(i) One-one:
Let f(a) = f(b) for some a, b ∈ ℕ.
⟹ 2a + 3 = 2b + 3
⟹ 2a = 2b
⟹ a = b
∴ f is one-one.

(ii) Onto:
For f to be onto, every element of the codomain ℕ must have a pre-image in ℕ.
Consider y = 1 ∈ ℕ. Then f(n) = 1 ⟹ 2n + 3 = 1 ⟹ n = −1 ∉ ℕ.
∵ n = −1 ∉ ℕ, the element 1 ∈ ℕ has no pre-image in ℕ.
∴ Range of f = {5, 7, 9, …} ⊊ ℕ = Codomain.
∴ f is NOT onto.

(iii) R = {(m, n) : m, n ∈ ℕ and |m − n| is even}.

Reflexive:
For any m ∈ ℕ, |m − m| = 0, which is even.
∴ (m, m) ∈ R for all m ∈ ℕ.
Hence R is reflexive.

Symmetric:
Let (m, n) ∈ R for some m, n ∈ ℕ.
⟹ |m − n| is even.
∵ |n − m| = |m − n|, we have |n − m| is also even.
⟹ (n, m) ∈ R.
Hence R is symmetric.

Transitive:
Let (m, n) ∈ R and (n, p) ∈ R for some m, n, p ∈ ℕ.
⟹ |m − n| is even and |n − p| is even.
⟹ (m − n) is even and (n − p) is even. …(i)
From (i): (m − p) = (m − n) + (n − p) = sum of two even numbers = even.
⟹ |m − p| is even.
⟹ (m, p) ∈ R.
Hence R is transitive.

∴ R is reflexive, symmetric and transitive. Hence, R is an equivalence relation.
Q7Case-based4 marks

A school library system assigns each student a unique Library Card Number. The librarian models this as a function f : S → N, where S = {Aarav, Bela, Chirag, Diya} is the set of students and N = {101, 102, 103, 104} is the set of card numbers, defined by:

f(Aarav) = 101, f(Bela) = 102, f(Chirag) = 103, f(Diya) = 104

A school library system assigns each student a unique Library Card Number. The librarian models this as a function f : S → N, where S = {Aarav, Bela, Chirag, Diya} is the set of students and N = {101, 102, 103, 104} is the set of card numbers, defined by:

f(Aarav) = 101, f(Bela) = 102, f(Chirag) = 103, f(Diya) = 104

(i) Is f one-one? Justify your answer. [1]
(ii) Is f onto? Justify your answer. [1]
(iii) The librarian wants to add a fifth student, Esha, to set S. She considers assigning f(Esha) = 101 (reusing Aarav's card number). Will f remain one-one? Give reason. Also, if instead a brand-new card number 105 is added to N (making N = {101, 102, 103, 104, 105}) but no new student is added to S, will f remain onto? Justify. [2]

Show answer
(i) One-one (Injective):

A function is one-one if distinct elements in the domain have distinct images in the codomain.

Here, f(Aarav) = 101, f(Bela) = 102, f(Chirag) = 103, f(Diya) = 104.

Since all four students are mapped to distinct card numbers, no two different students share the same card number.

∴ f is one-one.

(ii) Onto (Surjective):

A function is onto if every element of the codomain has a pre-image in the domain, i.e., Range = Codomain.

Range of f = {101, 102, 103, 104} = N (Codomain).

Since every card number in N is assigned to some student in S, Range = Codomain.

∴ f is onto.

(iii) Part A — Effect on one-one when f(Esha) = 101:

If Esha is added to S and assigned f(Esha) = 101, then both Aarav and Esha map to the same card number 101.

∵ f(Aarav) = f(Esha) = 101, but Aarav ≠ Esha,

∴ f will NOT remain one-one. Two distinct students share the same image, violating the one-one criterion.

Part B — Effect on onto when N = {101, 102, 103, 104, 105} but S is unchanged:

The new codomain is N = {101, 102, 103, 104, 105}.

Range of f = {101, 102, 103, 104} (since S still has only four students).

∵ 105 ∈ N has no pre-image in S, Range ≠ Codomain.

∴ f will NOT remain onto.
Q8Case-based4 marks

A data analytics company assigns each employee a unique 'performance code' using a function f : ℝ − {−1} → ℝ defined by f(x) = (3x − 2)/(x + 1). The HR manager claims that this function is both one-one and onto ℝ, ensuring a perfect one-to-one correspondence between employees and performance codes.

A data analytics company assigns each employee a unique 'performance code' using a function f : ℝ → ℝ defined by f(x) = (3x − 2)/(x + 1), for all x ∈ ℝ − {−1}.

The HR manager claims:
• Assertion (A): The function f is one-one.
• Assertion (B): The function f is onto ℝ.

(i) Verify whether f is one-one. [2 marks]
(ii) Determine the range of f and hence decide whether f is onto ℝ. Justify your answer. [2 marks]

Show answer
(i) One-One:

Let f(a) = f(b) for some a, b ∈ ℝ − {−1}.

⟹ (3a − 2)/(a + 1) = (3b − 2)/(b + 1)

⟹ (3a − 2)(b + 1) = (3b − 2)(a + 1)

⟹ 3ab + 3a − 2b − 2 = 3ab + 3b − 2a − 2

⟹ 3a − 2b = 3b − 2a

⟹ 5a = 5b

⟹ a = b

∴ f(a) = f(b) ⟹ a = b, so f is one-one. ...(1 mark)

The HR manager's claim that f is one-one is CORRECT.

(ii) Range of f and Onto check:

Let y ∈ ℝ be any element in the codomain. We find x such that f(x) = y.

f(x) = y ⟹ (3x − 2)/(x + 1) = y

⟹ 3x − 2 = y(x + 1)

⟹ 3x − 2 = yx + y

⟹ 3x − yx = y + 2

⟹ x(3 − y) = y + 2

⟹ x = (y + 2)/(3 − y), provided y ≠ 3. ...(1/2 mark)

When y = 3: the equation gives 0 = 5, which is impossible.
∴ x is not defined in ℝ − {−1} for y = 3.

For all y ∈ ℝ − {3}, x = (y + 2)/(3 − y) is a real number.
We verify x ≠ −1: if x = −1 then (−1 + 2)/(3 − y) = −1 ⟹ 1 = −(3 − y) ⟹ y = 4 ≠ 3, and checking: f(−1) is undefined, so x = −1 is not in the domain regardless.

∴ Range of f = ℝ − {3} ≠ ℝ (the codomain). ...(1/2 mark)

Since Range of f ⊊ ℝ (y = 3 has no pre-image in the domain), f is NOT onto ℝ. ...(1 mark)

∴ The HR manager's second claim that f is onto ℝ is INCORRECT. The function f is one-one but not onto ℝ, so it is not bijective as a mapping from ℝ − {−1} to ℝ.
Q9MCQ1 mark

A relation R on the set A = {1, 2, 3} is defined as R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}. Which of the following is correct?

Show answer
Option (C) is correct.

Explanation: Reflexive: (1,1), (2,2), (3,3) ∈ R, so R is reflexive. Symmetric: (1,2) ∈ R and (2,1) ∈ R, so R is symmetric. Transitive: (1,2) ∈ R and (2,1) ∈ R ⟹ (1,1) ∈ R ✓; all other pairs trivially satisfy the condition. ∴ R is reflexive, symmetric and transitive, i.e., R is an equivalence relation.
Q10MCQ1 mark

Let f : ℝ − {2} → ℝ be defined by f(x) = (x − 1)/(x − 2). Is f one-one?

Show answer
Option (A) is correct.

Explanation: Let f(a) = f(b) for some a, b ∈ ℝ − {2}.

⟹ (a − 1)/(a − 2) = (b − 1)/(b − 2)

⟹ (a − 1)(b − 2) = (b − 1)(a − 2)

⟹ ab − 2a − b + 2 = ab − 2b − a + 2

⟹ −2a − b = −2b − a

⟹ −a = −b ⟹ a = b

∴ f(a) = f(b) implies a = b, so f is one-one.
Q11Short Answer1 mark

Assertion (A): The function f : ℝ → ℝ defined by f(x) = 2x + 5 is both one-one and onto.

Reason (R): A function f : ℝ → ℝ is onto if and only if for every y ∈ ℝ (Codomain), there exists x ∈ ℝ (Domain) such that f(x) = y.

Show answer
Option (a) is correct.

Explanation: Reason (R) states the standard definition of an onto function, which is TRUE.

For Assertion (A): Let f(a) = f(b) for some a, b ∈ ℝ. Then 2a + 5 = 2b + 5 ⟹ a = b, so f is one-one. For onto: let y ∈ ℝ; then x = (y − 5)/2 ∈ ℝ satisfies f(x) = 2·(y − 5)/2 + 5 = y, so f is onto. ∴ Assertion (A) is TRUE, and Reason (R) directly explains why f is onto.

∴ Both (A) and (R) are true, and (R) is the correct explanation of (A).
Q12MCQ1 mark

Let f : {1, 2, 3, 4} → {1, 2, 3, 4} be defined by f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 1. Which of the following is true?

Show answer
Option (C) is correct.

Explanation: A function f is one-one if f(a) = f(b) ⟹ a = b for all a, b in the domain, and onto if Range = Codomain.

The given mapping is f(1)=2, f(2)=3, f(3)=4, f(4)=1.

All four images 2, 3, 4, 1 are distinct ⟹ f is one-one.
Range = {1, 2, 3, 4} = Codomain ⟹ f is onto.

∴ f is both one-one and onto (bijective).
Q13MCQ1 mark

The function f : ℝ → ℝ defined by f(x) = 3x + 7 is:

Show answer
Option (C) is correct.

Explanation: One-one — Let f(a) = f(b) for some a, b ∈ ℝ. Then 3a + 7 = 3b + 7 ⟹ 3a = 3b ⟹ a = b. ∴ f is one-one. Onto — Let y ∈ ℝ. Then x = (y − 7)/3 ∈ ℝ and f(x) = 3·(y − 7)/3 + 7 = y. ∴ Range = ℝ = Codomain, so f is onto. ∴ f is both one-one and onto.
Q14Short Answer1 mark

Assertion (A): The relation R = {(x, y) : x + y = 10 and x, y ∈ ℕ} is not a reflexive relation.

Reason (R): For reflexivity, (x, x) must belong to R for every x ∈ ℕ, but x + x = 2x ≠ 10 for all x ∈ ℕ.

Show answer
Option (a) is correct.

Explanation: For R to be reflexive, (x, x) ∈ R for every x ∈ ℕ, i.e., x + x = 2x = 10, which gives x = 5. Since this holds only for x = 5 and not for every x ∈ ℕ, R is not reflexive. ∴ A is true.

R correctly states that (x, x) ∈ R requires x + x = 10, i.e., 2x = 10, which is not satisfied by all x ∈ ℕ. ∴ R is true and is the correct explanation of A.
Q15Short Answer1 mark

Assertion (A): The relation R = {(a, b) : a + b is even, a, b ∈ ℕ} is a reflexive relation on ℕ.

Reason (R): For any n ∈ ℕ, n + n = 2n, which is always even.

Show answer
Option (a) is correct.

Explanation: For R to be reflexive, (n, n) ∈ R for every n ∈ ℕ, i.e., n + n must be even. Since n + n = 2n for all n ∈ ℕ, and 2n is always even, we have (n, n) ∈ R for every n ∈ ℕ. ∴ A is true. R states exactly this fact — n + n = 2n is always even — which directly establishes reflexivity. ∴ R is true and is the correct explanation of A.
Q16MCQ1 mark

Let R be a relation on ℝ defined by R = {(a, b) : a ≤ b²}. Check whether R is reflexive.

Show answer
Option (B) is correct.

Explanation: For R to be reflexive, we need (a, a) ∈ R for every a ∈ ℝ, i.e., a ≤ a² must hold for all a ∈ ℝ.

Take a = 1/2. Then a² = 1/4, and 1/2 ≤ 1/4 is false.

∴ R is not reflexive.
Q17Short Answer1 mark

Assertion (A): The function f : ℕ → ℕ defined by f(x) = 2x + 1, ∀ x ∈ ℕ, is one-one but not onto.

Reason (R): For y = 2 ∈ ℕ (Codomain), there does not exist any x ∈ ℕ (Domain) such that f(x) = 2x + 1 = 2.

Show answer
Option (a) is correct.

Explanation: A function f : ℕ → ℕ is defined by f(x) = 2x + 1.

One-one: Let f(a) = f(b) for some a, b ∈ ℕ. Then 2a + 1 = 2b + 1 ⟹ a = b. ∴ f is one-one.

Not onto: For y = 2 ∈ ℕ (Codomain), f(x) = 2 ⟹ 2x + 1 = 2 ⟹ x = 1/2 ∉ ℕ. ∴ f is not onto.

∴ A is true. R correctly states that y = 2 has no pre-image in ℕ, which is precisely the reason f is not onto. ∴ R is true and R is the correct explanation of A.
Q18MCQ1 mark

Let f : ℝ → ℝ be defined by f(x) = 2x + 7. Which of the following is correct?

Show answer
Option (D) is correct.

Explanation: One-one: Let f(a) = f(b) for some a, b ∈ ℝ. Then 2a + 7 = 2b + 7 ⟹ a = b. ∴ f is one-one. Onto: For any y ∈ ℝ, x = (y − 7)/2 ∈ ℝ gives f(x) = y. ∴ Range = ℝ = Codomain, so f is onto. Hence f is both one-one and onto.
Q19Short Answer2 marks

Check whether the function f : ℝ → ℝ defined by f(x) = 3x + 5 is one-one (injective) or not.

Show answer
Let f(a) = f(b) for some a, b ∈ ℝ.

⟹ 3a + 5 = 3b + 5

⟹ 3a = 3b

⟹ a = b

∴ f(a) = f(b) ⟹ a = b, so f is one-one (injective).
Q20Short Answer2 marks

Check whether the function f : ℝ → ℝ defined by f(x) = 3x + 5 is one-one (injective) or not.

Show answer
Let f(a) = f(b) for some a, b ∈ ℝ.

⟹ 3a + 5 = 3b + 5

⟹ 3a = 3b

⟹ a = b

∴ f(a) = f(b) ⟹ a = b for all a, b ∈ ℝ, so f is one-one (injective).
Q21Short Answer2 marks

A function f : ℝ → ℝ is defined as f(x) = 5x − 3. Check whether f is one-one.

Show answer
Let f(a) = f(b) for some a, b ∈ ℝ.

⟹ 5a − 3 = 5b − 3

⟹ 5a = 5b

⟹ a = b

∴ f is one-one.
Q22Short Answer2 marks

Check whether the function f : ℝ − {2} → ℝ − {1} defined as f(x) = (x − 1)/(x − 2) is one-one and onto.

Show answer
One-one: Let f(a) = f(b) for some a, b ∈ ℝ − {2}.

(a − 1)/(a − 2) = (b − 1)/(b − 2)

⟹ (a − 1)(b − 2) = (b − 1)(a − 2)

⟹ ab − 2a − b + 2 = ab − 2b − a + 2

⟹ −2a − b = −2b − a

⟹ −a = −b ⟹ a = b

∴ f is one-one.

Onto: Let y ∈ ℝ − {1}. We need x ∈ ℝ − {2} such that f(x) = y.

(x − 1)/(x − 2) = y ⟹ x − 1 = y(x − 2) ⟹ x − yx = 1 − 2y ⟹ x = (1 − 2y)/(1 − y)

Since y ≠ 1, x is well-defined. Also, x = 2 would require 1 − 2y = 2(1 − y) = 2 − 2y ⟹ 1 = 2, a contradiction, so x ≠ 2.

∴ Every y ∈ ℝ − {1} has a pre-image in ℝ − {2}, so f is onto.

∴ f is one-one and onto.
Q23Short Answer3 marks

Show that f : [0, ∞) → [−3, ∞) defined by f(x) = 9x² + 6x − 3 is both one-one and onto. Also, find q ∈ [0, ∞) such that f(q) = 6.

Show answer
One-one:
Let f(a) = f(b) for some a, b ∈ [0, ∞).
⟹ 9a² + 6a − 3 = 9b² + 6b − 3
⟹ 9(a² − b²) + 6(a − b) = 0
⟹ (a − b)(9(a + b) + 6) = 0

Since a, b ∈ [0, ∞), we have a + b ≥ 0,
∴ 9(a + b) + 6 ≥ 6 > 0.

∴ a − b = 0, i.e., a = b.
∴ f is one-one. ...(1)

Onto:
Let y ∈ [−3, ∞). We need to find x ∈ [0, ∞) such that f(x) = y.

f(x) = y
⟹ 9x² + 6x − 3 = y
⟹ 9x² + 6x − (3 + y) = 0
⟹ x = [−6 + √(36 + 36(3 + y))] / 18 (taking positive root as x ≥ 0)
⟹ x = [−6 + √(36(4 + y))] / 18
⟹ x = [−6 + 6√(4 + y)] / 18
⟹ x = [−1 + √(4 + y)] / 3

Since y ≥ −3, we have 4 + y ≥ 1 > 0, so √(4 + y) ≥ 1,
∴ x = (−1 + √(4 + y))/3 ≥ 0, i.e., x ∈ [0, ∞). ...(2)

Also, Range of f = [−3, ∞) = Codomain.
∴ f is onto. ...(3)

From (1) and (3), f is one-one and onto.
∴ f is bijective. Hence proved.

Finding q such that f(q) = 6:
f(q) = 6
⟹ 9q² + 6q − 3 = 6
⟹ 9q² + 6q − 9 = 0
⟹ 3q² + 2q − 3 = 0
⟹ q = [−2 + √(4 + 36)] / 6 (taking positive root as q ∈ [0, ∞))
⟹ q = [−2 + √40] / 6
⟹ q = [−2 + 2√10] / 6
⟹ q = (−1 + √10) / 3

∴ q = (√10 − 1)/3 ∈ [0, ∞).
Q24Short Answer3 marks

Let R be a relation on ℕ defined by R = {(a, b) : a, b ∈ ℕ and 3a + b = 30}. Find the domain and range of R. Also check whether R is reflexive, symmetric and transitive. Hence, state whether R is an equivalence relation or not.

Show answer
Given: R = {(a, b) : a, b ∈ ℕ and 3a + b = 30}.

For a, b ∈ ℕ, we have b = 30 − 3a > 0, so a < 10, and since a ∈ ℕ, a ∈ {1, 2, 3, ..., 9}.

The corresponding values of b = 30 − 3a are:
a = 1 ⟹ b = 27, a = 2 ⟹ b = 24, a = 3 ⟹ b = 21,
a = 4 ⟹ b = 18, a = 5 ⟹ b = 15, a = 6 ⟹ b = 12,
a = 7 ⟹ b = 9, a = 8 ⟹ b = 6, a = 9 ⟹ b = 3.

∴ R = {(1,27),(2,24),(3,21),(4,18),(5,15),(6,12),(7,9),(8,6),(9,3)}.

∴ Domain of R = {1, 2, 3, 4, 5, 6, 7, 8, 9}.

∴ Range of R = {3, 6, 9, 12, 15, 18, 21, 24, 27}.

Reflexive: For R to be reflexive, (a, a) ∈ R for every a ∈ ℕ.
For (a, a) ∈ R, we need 3a + a = 30 ⟹ 4a = 30 ⟹ a = 7.5 ∉ ℕ.
∴ (1, 1) ∉ R (since 3(1) + 1 = 4 ≠ 30).
∴ R is NOT reflexive.

Symmetric: For R to be symmetric, (a, b) ∈ R ⟹ (b, a) ∈ R.
Consider (1, 27) ∈ R. For (27, 1) ∈ R, we need 3(27) + 1 = 82 ≠ 30.
∴ (27, 1) ∉ R.
∴ R is NOT symmetric.

Transitive: For R to be transitive, (a, b) ∈ R and (b, c) ∈ R ⟹ (a, c) ∈ R.
Now (a, b) ∈ R ⟹ 3a + b = 30 ...(i)
For (b, c) ∈ R, we need b ∈ Domain of R, i.e., b ∈ {1, 2, ..., 9}.
But from (i), b = 30 − 3a. For a ∈ {1,...,9}, b ∈ {3, 6, ..., 27}, and b ∈ Domain of R only if b ≤ 9.
This requires 30 − 3a ≤ 9 ⟹ a ≥ 7, so a ∈ {7, 8, 9}, giving b ∈ {9, 6, 3}.
For (7, 9) ∈ R and (9, 3) ∈ R: check if (7, 3) ∈ R: 3(7) + 3 = 24 ≠ 30. ∴ (7, 3) ∉ R.
∴ R is NOT transitive.

∴ Since R is neither reflexive, nor symmetric, nor transitive, R is NOT an equivalence relation.
Q25Short Answer3 marks

A relation R is defined on ℕ × ℕ (where ℕ is the set of natural numbers) as:
(a, b) R (c, d) ⟺ ad = bc
Show that R is an equivalence relation.

Show answer
We need to show that R is reflexive, symmetric and transitive.

Reflexive:
For any (a, b) ∈ ℕ × ℕ,
ab = ba (multiplication is commutative in ℕ)
∴ (a, b) R (a, b).
Hence R is reflexive.

Symmetric:
Let (a, b) R (c, d) for some (a, b), (c, d) ∈ ℕ × ℕ.
⟹ ad = bc
⟹ cb = da
∴ (c, d) R (a, b).
Hence R is symmetric.

Transitive:
Let (a, b) R (c, d) and (c, d) R (e, f) for some (a, b), (c, d), (e, f) ∈ ℕ × ℕ.
⟹ ad = bc ...(i)
⟹ cf = de ...(ii)
Multiplying (i) and (ii),
(ad)(cf) = (bc)(de)
⟹ af · cd = be · cd
∵ c, d ∈ ℕ, so cd ≠ 0; dividing both sides by cd,
⟹ af = be
∴ (a, b) R (e, f).
Hence R is transitive.

∴ R is reflexive, symmetric and transitive. Hence, R is an equivalence relation.
Q26Short Answer3 marks

Let R be a relation defined on ℤ (the set of all integers) by R = {(a, b) : (a − b) is divisible by 5, a, b ∈ ℤ}. Show that R is an equivalence relation.

Show answer
We need to show R is reflexive, symmetric and transitive.

Reflexive:
For any a ∈ ℤ,
a − a = 0 = 0 × 5, which is divisible by 5.
⟹ (a, a) ∈ R for all a ∈ ℤ.
∴ R is reflexive. …(½)

Symmetric:
Let (a, b) ∈ R for some a, b ∈ ℤ.
⟹ (a − b) is divisible by 5
⟹ (a − b) = 5k for some k ∈ ℤ
⟹ (b − a) = 5(−k), where −k ∈ ℤ
⟹ (b − a) is divisible by 5
⟹ (b, a) ∈ R.
∴ R is symmetric. …(1)

Transitive:
Let (a, b) ∈ R and (b, c) ∈ R for some a, b, c ∈ ℤ.
⟹ (a − b) = 5k₁ for some k₁ ∈ ℤ …(i)
and (b − c) = 5k₂ for some k₂ ∈ ℤ …(ii)
Adding (i) and (ii),
(a − b) + (b − c) = 5k₁ + 5k₂
⟹ (a − c) = 5(k₁ + k₂), where (k₁ + k₂) ∈ ℤ
⟹ (a − c) is divisible by 5
⟹ (a, c) ∈ R.
∴ R is transitive. …(1)

∴ R is reflexive, symmetric and transitive. Hence, R is an equivalence relation. …(½)
Q27Short Answer3 marks

Let f : ℝ − {2} → ℝ − {1} be defined by f(x) = (3x − 1)/(3x − 6). Show that f is one-one and onto.

Show answer
One-one:

Let f(a) = f(b) for some a, b ∈ ℝ − {2}.

⟹ (3a − 1)/(3a − 6) = (3b − 1)/(3b − 6)

⟹ (3a − 1)(3b − 6) = (3b − 1)(3a − 6)

⟹ 9ab − 18a − 3b + 6 = 9ab − 18b − 3a + 6

⟹ −18a − 3b = −18b − 3a

⟹ −15a = −15b

⟹ a = b

∴ f is one-one. …(1 mark)

Onto:

Let y ∈ ℝ − {1}. We need to find x ∈ ℝ − {2} such that f(x) = y.

f(x) = y

⟹ (3x − 1)/(3x − 6) = y

⟹ 3x − 1 = y(3x − 6)

⟹ 3x − 1 = 3xy − 6y

⟹ 3x − 3xy = 1 − 6y

⟹ 3x(1 − y) = 1 − 6y

⟹ x = (1 − 6y) / (3(1 − y)) …(i)

Since y ≠ 1, we have (1 − y) ≠ 0, so x is well-defined.

Also, x = 2 would require 1 − 6y = 6(1 − y) = 6 − 6y, i.e. 1 = 6, which is a contradiction.

∴ x ≠ 2, so x ∈ ℝ − {2}.

Now, f(x) = f((1 − 6y)/(3(1 − y)))

= (3·(1 − 6y)/(3(1 − y)) − 1) / (3·(1 − 6y)/(3(1 − y)) − 6)

= ((1 − 6y)/(1 − y) − 1) / ((1 − 6y)/(1 − y) − 6)

= ((1 − 6y − (1 − y))/(1 − y)) / ((1 − 6y − 6(1 − y))/(1 − y))

= (1 − 6y − 1 + y) / (1 − 6y − 6 + 6y)

= (−5y) / (−5)

= y …(ii)

∴ For every y ∈ ℝ − {1}, there exists x = (1 − 6y)/(3(1 − y)) ∈ ℝ − {2} such that f(x) = y.

∴ Range of f = ℝ − {1} = Codomain, so f is onto. …(1 mark)

∴ f is one-one and onto, hence f is bijective. …(½ + ½ mark for conclusions)
Q28Short Answer3 marks

Let A = ℝ − {2} and B = ℝ − {1}. A function f : A → B is defined by f(x) = (x − 1)/(x − 2). Show that f is bijective (one-one and onto).

Show answer
Given: f : A → B defined by f(x) = (x − 1)/(x − 2), where A = ℝ − {2} and B = ℝ − {1}.

One-one:
Let f(a) = f(b) for some a, b ∈ A = ℝ − {2}.
⟹ (a − 1)/(a − 2) = (b − 1)/(b − 2)
⟹ (a − 1)(b − 2) = (b − 1)(a − 2)
⟹ ab − 2a − b + 2 = ab − 2b − a + 2
⟹ −2a − b = −2b − a
⟹ −2a + a = −2b + b
⟹ −a = −b
⟹ a = b
∴ f(a) = f(b) ⟹ a = b, so f is one-one.

Onto:
Let y ∈ B = ℝ − {1}. We need to find x ∈ A such that f(x) = y.
f(x) = y
⟹ (x − 1)/(x − 2) = y
⟹ x − 1 = y(x − 2)
⟹ x − 1 = xy − 2y
⟹ x − xy = 1 − 2y
⟹ x(1 − y) = 1 − 2y
⟹ x = (1 − 2y)/(1 − y) ...(i)

Since y ≠ 1, (1 − y) ≠ 0, so x is well defined.
We check that x ≠ 2:
If x = 2, then from (i): 2(1 − y) = 1 − 2y ⟹ 2 − 2y = 1 − 2y ⟹ 2 = 1, which is a contradiction.
∴ x ≠ 2, so x ∈ A = ℝ − {2}.

Now verify: f(x) = (x − 1)/(x − 2) = ((1 − 2y)/(1 − y) − 1) / ((1 − 2y)/(1 − y) − 2)
= ((1 − 2y − 1 + y)/(1 − y)) / ((1 − 2y − 2 + 2y)/(1 − y))
= (−y)/(−1)
= y ✓

Thus, for every y ∈ B, there exists x = (1 − 2y)/(1 − y) ∈ A such that f(x) = y.
∴ f is onto.

∴ f is one-one and onto. Hence, f is bijective.
Q29Case-based4 marks

A school library uses a system to assign unique ID codes to books. The librarian defines a function f : ℕ → ℕ by f(n) = 2n, where ℕ is the set of natural numbers and f(n) represents the ID code assigned to book number n.

A school library uses a system to assign unique ID codes to books. The librarian defines a function f : ℕ → ℕ by f(n) = 2n, where ℕ is the set of natural numbers and f(n) represents the ID code assigned to book number n.

Based on the above information, answer the following questions:
(i) Show that f is one-one. [1]
(ii) Is f onto? Justify your answer. [1]
(iii) The librarian considers a new coding function g : ℕ → ℕ defined by g(n) = n + 1. Check whether g is one-one and onto. Hence determine whether g is bijective. [2]

Show answer
(i) One-one check for f(n) = 2n:

Let f(a) = f(b) for some a, b ∈ ℕ.

⟹ 2a = 2b

⟹ a = b

∴ f is one-one.

(ii) Onto check for f(n) = 2n:

Let y ∈ ℕ (codomain). For f to be onto, we need n ∈ ℕ such that f(n) = y, i.e., 2n = y ⟹ n = y/2.

However, if y = 3 (an odd natural number), then n = 3/2 ∉ ℕ.

∵ not every element of the codomain has a pre-image in ℕ, Range = {2, 4, 6, …} ≠ ℕ.

∴ f is NOT onto.

(iii) One-one check for g(n) = n + 1:

Let g(a) = g(b) for some a, b ∈ ℕ.

⟹ a + 1 = b + 1

⟹ a = b

∴ g is one-one.

Onto check for g(n) = n + 1:

Let y ∈ ℕ (codomain). We need n ∈ ℕ such that g(n) = y, i.e., n + 1 = y ⟹ n = y − 1.

For y = 1 ∈ ℕ, n = 0 ∉ ℕ.

∵ the element 1 ∈ ℕ has no pre-image in ℕ, Range = {2, 3, 4, …} ≠ ℕ.

∴ g is NOT onto.

∵ g is one-one but not onto, g is NOT bijective.
Q30Case-based4 marks

A school library system assigns each book a unique digital code. The librarian defines a function f : ℕ → ℕ by f(n) = 2n, where ℕ is the set of natural numbers, representing that every book with catalogue number n is assigned the digital code 2n. A second librarian proposes a new coding rule g : ℕ → ℕ defined by g(n) = n + 1.

A school library system assigns each book a unique digital code. The librarian defines a function f : ℕ → ℕ by f(n) = 2n, where ℕ is the set of natural numbers, representing that every book with catalogue number n is assigned the digital code 2n.

(i) Show that f is a one-one function. [1]
(ii) Determine whether f is an onto function. Justify your answer. [1]
(iii) A second librarian proposes a new coding rule g : ℕ → ℕ defined by g(n) = n + 1. Find the composite function (g ∘ f)(n) and determine whether (g ∘ f) is one-one. [2]

Show answer
(i) One-one (Injective):

Let f(a) = f(b) for some a, b ∈ ℕ.

⟹ 2a = 2b

⟹ a = b

∴ f is one-one. [1]

(ii) Onto check:

For f to be onto, every element of the codomain ℕ must have a pre-image in ℕ.

Consider the element 3 ∈ ℕ (codomain). Then f(n) = 3 ⟹ 2n = 3 ⟹ n = 3/2, which does not belong to ℕ.

∴ There exists an element in the codomain that has no pre-image in the domain.

∴ f is NOT onto. [1]

(iii) Composite function (g ∘ f)(n):

(g ∘ f)(n) = g(f(n)) = g(2n) = 2n + 1

One-one check for (g ∘ f):

Let (g ∘ f)(a) = (g ∘ f)(b) for some a, b ∈ ℕ.

⟹ 2a + 1 = 2b + 1

⟹ 2a = 2b

⟹ a = b

∴ (g ∘ f) is one-one. [1 + 1]

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