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CBSE · Class 9 · Mathematics

Relations and Functions (Advanced)

Standard difficulty · 20 questions · 40 marks · Time: 34 minutes

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
If (2x + 7, y − 9) = (15, −8), then:
  1. x = 4, y = 1
  2. x = 1, y = 4
  3. x = −4, y = 1
  4. x = 4, y = 3
2.
If n(A) = 6 and n(B) = 5, how many relations can be defined from A to B?
  1. 211
  2. 302
  3. 30
  4. 230
3.
If (a, b) = (c, d), then:
  1. a + b = c + d
  2. ab = cd
  3. a = d and b = c
  4. a = c and b = d
4.
f : {−3, −1, 0, 3} → Z is f(x) = x2 − 2. The range of f is:
  1. {−2, −1, 7}
  2. {−5, −3, −2, 1}
  3. {−3, −1, 0, 3}
  4. {2, 3, 11}
5.
[5.6] − [−3.1] + |−13| equals (where [x] is the greatest integer not more than x):
  1. −4
  2. 21
  3. 14
  4. 22
6.
A relation from A to B is a function when:
  1. A and B have the same number of elements
  2. every element of B has a pre-image
  3. some element of A has two images
  4. every element of A has exactly one image
7.
R = {(a, b) : a + 3b = 22; a, b ∈ N}. How many ordered pairs are in R?
  1. 7
  2. 9
  3. 22
  4. 8
8.
The graph of y = (x − 6)2 is the graph of y = x2:
  1. shifted 6 units to the right
  2. shifted 6 units down
  3. shifted 6 units to the left
  4. shifted 6 units up
Section B — Short answer8 × 2 = 16 marks
9.
A = {1, 4, 7} and B = {1, 7, 8}. Find (A × B) ∩ (B × A).
10.
A = {4, 5, 6} and B = {5, 7, 8, 9}. R = {(a, b) : a divides b; a ∈ A, b ∈ B}. Write R in roster form, and its domain and range.
11.
A = {p, q, r} and B = {7, 8}. Find (a) n(A × B) (b) the number of relations from A to B (c) the number of those relations that contain the pair (p, 7).
12.
f(x) = 2x2 − 2 on the domain {−2, −1, 0, 1, 2}. Find the range of f and say how many elements it has.
13.
A = {2, 6, 8}, B = {c, e, f, h} and R = {(2, h), (8, h)}. Is R a function from A to B? Give the reason.
14.
Evaluate [6.6] + [−7.4] − |−15| + [−3.1], where [x] is the greatest integer not more than x.
15.
On A = {1, 2, …, 7}, R = {(a, b) : |a − b| = 1}. Write R in roster form and find n(R), its domain and its range.
16.
f is a linear function f(x) = ax + b with f(3) = −23 and f(8) = −53. Find a and b.
Section C — Word problems4 × 4 = 16 marks
17.
A = {3, 4}, B = {3, 4, 6} and C = {4, 5, 7}. Find B ∪ C and A × (B ∪ C), and check that A × (B ∪ C) = (A × B) ∪ (A × C). How many elements does it have?
18.
f(x) = x2 − 9 on the domain {−4, −3, −2, −1, 0, 1, 2, 3, 4}. (a) Write the range of f. (b) Find the pre-images of −8. (c) How many ordered pairs does f contain? (d) Is f a function? Why?
19.
Class XI has 29 students (set A) and Class IX has 25 students (set B). Find (a) n(A × B) (b) the number of relations from A to B, as a power of 2 (c) the number of relations from A to B that contain exactly two ordered pairs.
20.
Find the domain and range of each function: (a) y = 1/(x − 2) (b) y = 7 − |x| (c) y = √(x − 2).
AmbifulWorksheets
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CBSE · Class 9 · Mathematics

Answer key

Relations and Functions (Advanced) · AW-9M-ARF-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(a) x = 4, y = 12x + 7 = 15 and y − 9 = −8
2(d) 230n(A × B) = 6 × 5 = 30; a relation is any subset: 230
3(d) a = c and b = dordered pairs are equal only when matching entries are equal
4(a) {−2, −1, 7}f(−3) = 7, f(−1) = −1, f(0) = −2, f(3) = 7
5(d) 225 − (−4) + 13
6(d) every element of A has exactly one imageno first element appears twice, and none is missing
7(a) 7b = 1, 2, …, 7 keep a = 22 − 3b ≥ 1
8(a) shifted 6 units to the righty = f(x + k) shifts left for k > 0 and right for k < 0
9{(1, 1), (1, 7), (7, 1), (7, 7)}a pair is in both only if each entry is in A ∩ B = {1, 7}; so it is (A ∩ B) × (A ∩ B)
10R = {(4, 8), (5, 5)}; domain {4, 5}; range {5, 8}check every pair of A × B against the rule
11(a) 6; (b) 64; (c) 32(b) 26; (c) that pair is fixed in, the other 5 pairs are free: 25
12{−2, 0, 6}; 3 elementsf(−x) = f(x), so the 5 inputs give only 3 images
13No: 6 has no imagea function needs exactly one image for every element of A
14−216 + (−8) − 15 + (−4)
15R = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6), (6, 5), (6, 7), (7, 6)}; n(R) = 12; domain = range = {1, 2, 3, 4, 5, 6, 7}6 pairs with b = a + 1 and 6 with b = a − 1
16a = −6, b = −5a = (−53 − (−23)) ÷ (8 − 3) = −6; then b = −23 − (−6) × 3 = −5
17B ∪ C = {3, 4, 5, 6, 7}; A × (B ∪ C) = {(3, 3), (3, 4), (3, 5), (3, 6), (3, 7), (4, 3), (4, 4), (4, 5), (4, 6), (4, 7)}; 10 elementsn = n(A) × n(B ∪ C) = 2 × 5; (A × B) ∪ (A × C) gives the same pairs
18(a) {−9, −8, −5, 0, 7}; (b) −1 and 1; (c) 9; (d) yes, every element of the domain has exactly one imagex2 − 9 = −8 gives x2 = 1; one pair per element of the domain
19(a) 725; (b) 2725; (c) 262,450(c) choose 2 of the 725 pairs: 725 × 724 ÷ 2
20(a) domain: all real x except 2; range: all real y except 0; (b) domain: all real x; range: y ≤ 7; (c) domain: x ≥ 2; range: y ≥ 0the reciprocal cannot divide by 0 and is never 0; |x| ≥ 0, so y is at most 7; a square root needs a non-negative number and is never negative