Mathematics Advanced • Year 11 • Module 1 • Lesson 1
Functions & Relations
HSC-style writing on the function definition, the vertical line test, and justification of restrictions.
1. Short-answer questions
1.1 Explain, with reference to the definition of a function, why the equation y² = x does not define y as a function of x. Use a specific value of x in your justification. 2 marks Band 3
1.2 Given f(x) = 3x² − 2x + 4, evaluate f(2) and f(−1), showing all working. 3 marks Band 3
1.3 The graph of a relation R consists of the union of the line y = x and the lower half of the circle x² + y² = 1. Use the vertical line test to determine whether R defines y as a function of x. Identify all values of x (if any) at which the test fails. 4 marks Band 4
Stuck on 1.3? Sketch both pieces and look for x-values where a vertical line cuts both pieces.2. Extended response
2.1 A national park entry kiosk uses a face-recognition system. For each captured image x, the system outputs F(x) ∈ {0, 1} (refuse / admit).
(a) Show, using the formal definition of a function, that F must be a function from the set of captured images to {0, 1} for the system to behave deterministically. Use a counterexample for the failure case.
(b) The kiosk software is updated so it also accepts a swipe card with ID w. The new decision is written G(x, w). Re-state the function property in terms of an appropriate input set, and explain why the vertical line test is no longer the right diagnostic.
(c) The kiosk introduces a "test mode" where the same image x is fed in twice and the decision must match. Express this requirement as a property of the function F, and explain whether the property is already guaranteed by F being a function, or whether it is an additional condition. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — quotes the definition (each input gives exactly one output) in context of the kiosk.
• 1 mark — gives a concrete counterexample (same image, two different decisions) and links it to non-deterministic / unsafe behaviour.
Part (b) — 3 marks
• 1 mark — identifies that the input set is now ordered pairs (x, w) (Cartesian product / paired set).
• 1 mark — restates the function property: each pair (x, w) must give exactly one output.
• 1 mark — explains the VLT requires a single real variable on the x-axis; with two inputs the test no longer applies (or needs a higher-dimensional analogue).
Part (c) — 2 marks
• 1 mark — expresses the requirement as F(x) = F(x), i.e. the function is well-defined / consistent on repeated inputs.
• 1 mark — concludes this is already guaranteed by the function definition (and explains why: a function maps each input to a unique output, so two queries of the same input give the same answer).
Your response:
Stuck on (c)? Read the definition aloud: "each input gives exactly one output". Is consistency on a repeated input baked in, or extra?How did this worksheet feel?
What I'll revisit before next class:
1.1 — y² = x is not a function of x (2 marks)
Sample response. A function requires each input x to produce exactly one output y. Take x = 4. Then y² = 4 gives y = 2 or y = −2 — two different y-values from one x. Hence y² = x does not define y as a function of x.
Marking notes. 1 mark — quoting the "exactly one output" definition. 1 mark — providing a specific x (e.g. x = 4) with two y-values. Responses that say only "it fails the vertical line test" without producing both y-values: 1/2.
1.2 — Evaluate f(2), f(−1) for f(x) = 3x² − 2x + 4 (3 marks)
Sample response.
f(2) = 3(2)² − 2(2) + 4 = 3(4) − 4 + 4 = 12.
f(−1) = 3(−1)² − 2(−1) + 4 = 3(1) + 2 + 4 = 9.
Marking notes. 1 mark — correct substitution with brackets for each input. 1 mark — correct value of f(2) = 12. 1 mark — correct value of f(−1) = 9. Lose a mark for sign errors on (−1)² (writing −1 instead of +1) or for dropping a bracket.
1.3 — R = (y = x) ∪ (lower half of x² + y² = 1) (4 marks)
Sample response. The line y = x is defined for all real x. The lower semicircle is defined on [−1, 1] with y = −√(1 − x²). On the overlap −1 ≤ x ≤ 1 the relation R contains both points (x, x) and (x, −√(1 − x²)).
For these to give the same y, we need x = −√(1 − x²). Then x ≤ 0 and x² = 1 − x², so 2x² = 1 and x = −1/√2. So the two pieces coincide only at the single point x = −1/√2 (y = −1/√2).
At every other x in [−1, 1] \ {−1/√2}, the vertical line meets both pieces at two distinct y-values. Hence R is not a function of x. The vertical line test fails for every x in [−1, 1] except x = −1/√2.
Marking notes. 1 mark — identifies overlap interval [−1, 1]. 1 mark — writes y on each piece (y = x and y = −√(1 − x²)). 1 mark — solves for the unique x where the two y-values agree (x = −1/√2). 1 mark — concludes "not a function" and states the failure set [−1, 1] \ {−1/√2}.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). A function from a set X to a set Y is a rule that assigns to each x ∈ X exactly one y ∈ Y. Here X is the set of captured images and Y = {0, 1}. [1 mark — definition stated in context.]
Suppose for contradiction the rule is not a function: there exists an image x* with both F(x*) = 1 and F(x*) = 0 (admit on one query, refuse on the next). The kiosk would then admit and refuse the same face on different attempts, defeating its purpose. Hence reliability forces F to satisfy the function definition. [1 mark — concrete counterexample linked to non-deterministic outcome.]
Part (b). The new system has two simultaneous inputs: the image x (from the set of images I) and the card ID w (from the set of valid IDs W). The natural input set is the Cartesian product I × W, whose elements are ordered pairs (x, w). [1 mark — paired input set.]
G is a function from I × W to {0, 1} provided each pair (x, w) gives exactly one decision G(x, w) ∈ {0, 1}. [1 mark — restated function property on pairs.]
The vertical line test diagnoses one-input-one-output for a graph y vs x on the Cartesian plane. With two independent inputs the "graph" would live in three dimensions (x-axis, w-axis, decision-axis), and the analogous test would be a vertical pillar test through each (x, w) point. The plane-based VLT no longer applies. [1 mark — explains why VLT fails for 2-input rules.]
Part (c). "Same image twice gives the same decision" is the requirement F(x) = F(x) — equivalently, F is well-defined (deterministic) on each input. [1 mark — formal statement of the test-mode requirement.]
This is already guaranteed by F being a function: the definition assigns each input exactly one output, so any two evaluations on the same input must return that same output. No additional condition is needed. [1 mark — concludes the property is built into the function definition.]
Total: 7/7.
Band descriptors for marker.
Band 3: Restates definition once; attempts (a) but only with words, no contrast to the failure case. Misses paired input set in (b) or treats it as still single-variable. Doesn't see (c) as already implied. ≈ 2-3 marks.
Band 4: Correct (a). Identifies pair (x, w) in (b) but does not explain why VLT fails. Says (c) is required but does not see it follows from the definition. ≈ 4-5 marks.
Band 5: All of (a) and (b) correct. Argues (c) is implied by the definition but argument is loose. ≈ 5-6 marks.
Band 6: Precise definitions in context, named Cartesian product, mentions higher-dimensional analogue of VLT, and explicitly says "no additional condition needed" with reason. 7/7.