Standard difficulty · 20 questions · 40 marks
Show your working for Sections B and C. Answers are on the last page — try every question first!
Relations and Functions · AW-11M-RNF-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.
| Q | Answer | Working |
|---|---|---|
| 1 | (d) 26 | subsets of A × B, which has 6 elements |
| 2 | (c) (2, 9) | equal ordered pairs: first = first, second = second |
| 3 | (b) {2, 4, 5, 8} | — |
| 4 | (c) 0 | — |
| 5 | (d) ordered pairs (a, b) with a ∈ A and b ∈ B | — |
| 6 | (a) {(3, 7), (4, 9), (2, 8), (1, 2)} | no first element may appear twice |
| 7 | (a) −22 | f(−3) = −5, g(−3) = −17 |
| 8 | (c) A is true, but R is false. | A is true: 4 × 3 = 12. R is false. |
| 9 | x = 3/4, y = 1/2 | equate first and second elements |
| 10 | {(4, 3), (4, 5), (5, 3), (5, 5)}; 4 elements | — |
| 11 | R = {(6, 10), (7, 11), (8, 12)}; domain {6, 7, 8}; range {10, 11, 12} | — |
| 12 | f(4) = −9; x = −5/2 | −2x − 1 = 4 |
| 13 | (f + g)(x) = x2 + x − 3; (f − g)(x) = x2 − x − 5; (fg)(x) = x3 + x2 − 4x − 4 | add, subtract or multiply the expressions |
| 14 | R − {1, 5} | the denominator is (x − 1)(x − 5), zero at x = 1, 5 |
| 15 | (−∞, −1] | x2 ≥ 0, so f(x) ≤ −1 |
| 16 | 3; 23 = 8 | every subset of A × B is a relation |
| 17 | R = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 4), (2, 6), (3, 6)}; domain {1, 2, 3}; codomain {1, 2, 3, 4, 5, 6}; range {2, 3, 4, 5, 6} | — |
| 18 | m = −3, c = 5; f(x) = −3x + 5; f(−6) = 23 | m = (17 − 20) ÷ (−4 − (−5)) |
| 19 | B ∩ C = {5}, so A × (B ∩ C) = {(1, 5), (4, 5)}. (A × B) ∩ (A × C) keeps the pairs whose second element is in both B and C, which is the same set. | — |
| 20 | f(−3) = 9, f(2) = 6, f(4) = 9; range [0, ∞) | use the rule for the interval each x lies in; x2 covers all positive values |