Mathematics Advanced • Year 11 • Module 1 • Lesson 15
Module Synthesis & Exam Technique
Apply combined Module 1 ideas to realistic exam-style multi-part problems with time-pressure pacing.
Problem 1 — Tutoring fee inverse (financial)
A tutoring service charges f(h) = 80 + 60h dollars for h hours of tutoring (h ≥ 0). A parent asks: "If I have a budget of $D, how many hours can I afford?"
Set up: What are we solving for?
(i) Find f−1(D) and explain in one sentence what the inverse represents in this context. 2 marks
(ii) State the domain and range of f−1, justified by the physical situation (no negative tutoring hours; minimum fee is $80). 2 marks
(iii) A parent has $500. How many hours can they afford? Verify by computing f(your-answer) = 500. 2 marks
Stuck on (ii)? Range of f is [80, ∞), which is the domain of the inverse.Problem 2 — Currency conversion + tax (composite)
A traveller converts AUD to USD using u(a) = 0.65 a, then pays a US sales tax of 8.25%, modelled by t(x) = 1.0825 x.
Set up: What are we solving for?
(i) Write the composite (t ○ u)(a) representing "amount paid including tax, in USD" as a function of AUD input a. 2 marks
(ii) Find the AUD amount the traveller must start with to end up paying exactly USD 100.00 including tax. (Use the composite.) 3 marks
(iii) Does it matter whether the tax is applied before or after conversion? Compute (u ○ t)(a) and compare to (t ○ u)(a). Explain in one line. 2 marks
Problem 3 — Revenue with break-even (modelling + sketch)
A small bakery's daily revenue R(n) (dollars) when it sells n hundred loaves is
R(n) = −20(n − 5)² + 800, n ≥ 0
Set up: What are we solving for?
(i) State maximum revenue and the number of loaves that produce it. 2 marks
(ii) Find both break-even quantities (where R = 0) and interpret physically. 3 marks
(iii) Sketch R(n) for n ≥ 0, clearly showing the vertex, R-intercept and both n-intercepts. 3 marks
Stuck on (ii)? Set R = 0 and solve the quadratic; expect √40 = 2√10.Problem 4 — Reciprocal fuel economy (further transformations)
Fuel consumption (L/100 km) of a car at speed v (km/h) is modelled by C(v) = 0.005(v − 80)² + 6 for 30 ≤ v ≤ 120. Fuel economy E (km/L) is the reciprocal of consumption per kilometre, related by E(v) = 100 / C(v).
Set up: What are we solving for?
(i) Find the speed that minimises consumption C, and state that minimum value. 2 marks
(ii) Find the corresponding maximum economy E (using E = 100/C). Verify using the reciprocal rule "max of reciprocal occurs where parent has min". 3 marks
(iii) Explain why C has no vertical asymptote on the domain 30 ≤ v ≤ 120 (i.e. why economy E is bounded above). 2 marks
Problem 5 — Critique an exam answer (exam technique)
A student writes: "If f(x) = x² and g(x) = √x, then f(g(x)) = x for all real x."
Set up: What are we solving for?
(i) Compute f(g(x)) algebraically. 1 mark
(ii) State the domain restriction the student missed, and give the correct domain on which "f(g(x)) = x" holds. 2 marks
(iii) Compute g(f(x)) and explain why "g(f(x)) = x" is also not universally true — identify the value(s) of x for which the equation fails, and explain in terms of |x|. 3 marks
Stuck on (iii)? Recall that √(x²) = |x|, not x.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Tutoring fee inverse
Set up. We are inverting a linear pricing formula and giving its physical meaning.
(i) y = 80 + 60h. Swap: h = 80 + 60y ⇒ y = (h − 80)/60. So f−1(D) = (D − 80)/60. Interpretation: given a budget D dollars, the inverse returns the number of hours of tutoring affordable.
(ii) Range of f: since h ≥ 0, f(0) = 80 is the minimum, so Ran(f) = [80, ∞). Thus Dom(f−1) = [80, ∞) (you cannot pay less than $80). Ran(f−1) = [0, ∞) (cannot buy negative hours).
(iii) f−1(500) = (500 − 80)/60 = 420/60 = 7 hours. Verify: f(7) = 80 + 60(7) = 80 + 420 = 500. ✓
Problem 2 — Currency + tax
Set up. We are composing two linear functions, inverting one composition to recover the input, and checking commutativity.
(i) (t ○ u)(a) = t(u(a)) = t(0.65 a) = 1.0825 · 0.65 a = 0.703625 a.
(ii) Set 0.703625 a = 100 ⇒ a = 100/0.703625 ≈ $142.12 AUD.
(iii) (u ○ t)(a) = u(t(a)) = u(1.0825 a) = 0.65 · 1.0825 a = 0.703625 a. Same as (t ○ u)(a). Both are scalar multiplications, and scalar multiplication is commutative, so order doesn't matter here. (This is the exception, not the rule — in general (f ○ g) ≠ (g ○ f).)
Problem 3 — Bakery revenue
Set up. We are reading a vertex-form quadratic to find optimum, then solving for break-even.
(i) Vertex (5, 800), so max revenue $800 at n = 5 (i.e. 500 loaves).
(ii) −20(n − 5)² + 800 = 0 ⇒ (n − 5)² = 40 ⇒ n = 5 ± 2√10 ≈ 0.68 or 9.32 hundred loaves (about 68 or 932 loaves). Below 68 loaves the bakery loses money (or makes none, in this simplified model); above 932 loaves price drops outweigh volume.
(iii) Sketch: downward parabola, vertex (5, 800), R-intercept R(0) = −20(25) + 800 = $300, n-intercepts at 5 − 2√10 and 5 + 2√10. Label all four features.
Problem 4 — Reciprocal fuel economy
Set up. We are using the lesson's reciprocal rule "max of 1/f occurs where f is min" in a real context.
(i) C(v) = 0.005(v − 80)² + 6 has vertex (80, 6). Min consumption 6 L/100 km at v = 80 km/h.
(ii) Max economy E_max = 100/C_min = 100/6 ≈ 16.67 km/L at v = 80 km/h. Verification: the reciprocal rule says where C is min (at v = 80), E = 100/C is max. ✓
(iii) C is bounded below by 6 (minimum value), so C > 0 throughout the domain [30, 120] — never reaches 0, never produces a vertical asymptote in E. Economy E is bounded above by 100/6 (and bounded below by 100/C(120) and 100/C(30)).
Problem 5 — Critique student claim
Set up. We are evaluating an exam-style claim, identifying a domain error, and exploring the reverse composition with absolute value.
(i) f(g(x)) = f(√x) = (√x)² = x.
(ii) The student forgot that g(x) = √x requires x ≥ 0. So f(g(x)) = x only for x ≥ 0; for x < 0, g(x) is undefined and so is the composite.
(iii) g(f(x)) = g(x²) = √(x²) = |x|. This equals x only when x ≥ 0; for x < 0, |x| = −x ≠ x. So the equation "g(f(x)) = x" fails for all x < 0. This is a direct application of the rule √(x²) = |x|, a common Module 1 trap.