Mathematics Advanced • Year 11 • Module 1 • Lesson 15
Module Synthesis & Exam Technique
Practise HSC-style mixed questions and a structured Module 1 extended response with full marking criteria.
1. Short-answer questions
1.1 Consider f(x) = x² − 4x + 5 defined for x ≥ 2. (a) Show f is one-to-one on this domain. (b) Find f−1(x). (c) State Dom(f−1) and Ran(f−1). 4 marks Band 4
1.2 A company's daily revenue R(n) = −20(n − 5)² + 800 dollars (n in hundreds of units, n ≥ 0). (a) State the maximum daily revenue and the n that produces it. (b) Find both break-even quantities. (c) Sketch R(n), labelling vertex, R-intercept and n-intercepts. 5 marks Band 4-5
1.3 A student writes: "If f(x) = x² and g(x) = √x, then f(g(x)) = x for all real x." Evaluate this statement, identifying any errors and stating the correct domain on which the equation holds. 3 marks Band 5
Stuck on 1.1(a)? Show that f'(x) ≥ 0 for x ≥ 2, or argue from the vertex of the parabola.2. Extended response
2.1 Let f(x) = √(x − 1).
(a) State the domain and range of f, with justification.
(b) Find f−1(x) using the swap-and-solve method. State its domain and range, justifying via Dom(f−1) = Ran(f) and Ran(f−1) = Dom(f).
(c) Sketch both y = f(x) and y = f−1(x) on the same axes, also drawing the line y = x. Label one corresponding pair of mirror points and the start/end of each curve.
(d) Now consider h(x) = f(2x − 3) + 5. By factoring inside, state the transformations that map f to h, and find the image of the start-point of f (where f begins on its domain) on the graph of h. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• Dom(f) = [1, ∞) (radicand ≥ 0) and Ran(f) = [0, ∞) (output of √ is non-negative).
Part (b) — 2 marks
• 1 mark — swap and solve: x = √(y − 1) ⇒ x² = y − 1 ⇒ f−1(x) = x² + 1, with restriction x ≥ 0.
• 1 mark — Dom(f−1) = [0, ∞), Ran(f−1) = [1, ∞), with the swap-of-domain-and-range principle named.
Part (c) — 2 marks
• 1 mark — both graphs drawn correctly with the line y = x as a dashed line of reflection symmetry.
• 1 mark — one explicit pair of mirror points labelled, e.g. (1, 0) on f mirrors to (0, 1) on f−1.
Part (d) — 3 marks
• 1 mark — factors inside: 2x − 3 = 2(x − 3/2), so h(x) = √(2(x − 3/2)) + 5 (i.e. a = 1, b = 2, h = 3/2, k = 5).
• 1 mark — lists transformations: horizontal dilation by ½ from y-axis, translation 3/2 right, translation 5 up. (Note a = 1 means no vertical dilation/reflection here.)
• 1 mark — image of start point (1, 0): x_new = 1/2 + 3/2 = 2, y_new = 1·0 + 5 = 5, so image (2, 5).
Your response:
Stuck on (d)? Always factor the coefficient of x before reading the horizontal translation.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Inverse of f(x) = x² − 4x + 5, x ≥ 2 (4 marks)
Sample response. (a) f(x) = (x − 2)² + 1, vertex (2, 1). For x ≥ 2 the parabola is increasing (right of the vertex), so f is one-to-one on this domain. (b) Swap: x = (y − 2)² + 1 ⇒ (y − 2)² = x − 1 ⇒ y − 2 = +√(x − 1) (positive root because y ≥ 2) ⇒ f−1(x) = 2 + √(x − 1). (c) Dom(f−1) = Ran(f) = [1, ∞); Ran(f−1) = Dom(f) = [2, ∞).
Marking notes. (a) 1 mark for showing one-to-one with vertex argument or monotonic-increasing argument. (b) 1.5 marks for correct algebra and choice of positive root; 0.5 mark for naming the restriction. (c) 1 mark for both domain and range correct with swap principle invoked.
1.2 — Bakery revenue (5 marks)
Sample response. (a) Vertex (5, 800): max $800 at n = 5 (i.e. 500 units). (b) 0 = −20(n − 5)² + 800 ⇒ (n − 5)² = 40 ⇒ n = 5 ± 2√10, both valid for n ≥ 0: approximately n ≈ 0.68 and n ≈ 9.32 (68 units and 932 units). (c) Downward parabola, vertex (5, 800), R-intercept (0, 300), n-intercepts at 5 ± 2√10. Label all four features.
Marking notes. (a) 1 mark (vertex read correctly). (b) 2 marks (correct quadratic solve; both roots stated; correctly rejected if any). (c) 2 marks (sketch with vertex, R-intercept and both n-intercepts labelled).
1.3 — Critique f(g(x)) = x claim (3 marks)
Sample response. The student's statement is incorrect. f(g(x)) = (√x)² = x, but this is only valid where g(x) is defined, i.e. for x ≥ 0. For x < 0, g(x) = √x is undefined, so the composite is undefined. The correct statement is "f(g(x)) = x for x ≥ 0".
Marking notes. 1 mark — identifies the statement as incorrect with the domain reason named. 1 mark — shows f(g(x)) = x algebraically. 1 mark — states the correct domain x ≥ 0. Common slip: just writing "incorrect" without naming the domain error scores 1/3.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). f(x) = √(x − 1) requires the radicand to be non-negative: x − 1 ≥ 0 ⇒ Dom(f) = [1, ∞). The output of √ is non-negative, so Ran(f) = [0, ∞). [1 mark]
Part (b). Let y = √(x − 1). Swap x and y: x = √(y − 1). Square both sides (valid because x ≥ 0): x² = y − 1, so y = x² + 1. The restriction x ≥ 0 is needed because the original f outputs only non-negative values. So f−1(x) = x² + 1, x ≥ 0. [1 mark] By the principle that domain and range swap, Dom(f−1) = Ran(f) = [0, ∞) and Ran(f−1) = Dom(f) = [1, ∞). [1 mark]
Part (c). Sketch shows f starting at (1, 0) and curving up-right (slope decreasing); f−1 starting at (0, 1) and curving up-right (slope increasing, faster than linear). The dashed line y = x bisects them: the pair (1, 0) on f mirrors to (0, 1) on f−1, and the pair (2, 1) on f mirrors to (1, 2) on f−1. [1 mark for both graphs + line y = x; 1 mark for labelled mirror pair]
Part (d). Factor inside h(x) = √(2x − 3) + 5: 2x − 3 = 2(x − 3/2). So h(x) = √(2(x − 3/2)) + 5, which has a = 1, b = 2, h = 3/2, k = 5. [1 mark for factorisation] Transformations: (1) horizontal dilation by factor ½ from y-axis (b = 2), (2) translation 3/2 right (h = 3/2), (3) translation 5 up (k = 5). No reflections or vertical dilation. [1 mark]
Start point of f is (1, 0). Image: x_new = 1/b + h = 1/2 + 3/2 = 2, y_new = a·0 + k = 0 + 5 = 5. Image (2, 5). [1 mark]
Total: 8/8. ▮
Band descriptors for marker.
Band 3: Dom and Ran of f correct; inverse formula correct but no domain restriction; sketch present but no y = x line or no labelled mirror pair; (d) attempted with arithmetic slip. ≈ 4 marks.
Band 4: All of (a), (b), (c) correct including the mirror line; (d) factors inside correctly but misses one transformation or makes a slip in image arithmetic. ≈ 5−6 marks.
Band 5: All parts substantially correct; (b) explicitly invokes the swap principle; (d) lists all transformations and computes image correctly. ≈ 7 marks.
Band 6: Full response with precise domain/range justification, swap principle named in (b), mirror line dashed and labelled pair in (c), factorisation step shown in (d) before transformations are read, and image computed with a, b, h, k explicitly identified. 8/8.