Mathematics Advanced • Year 11 • Module 1 • Lesson 15

Module Synthesis & Exam Technique

Build mixed-topic fluency: rapidly switch between domain, inverse, composite, transformation and sketch tasks.

Build · Skill Drill

1. Quick recall — the Module 1 toolkit

Answer each part in the space provided. 1 mark each

Q1.1 Complete each definition:

Inverse: swap ____ and ____, then solve for ____.

Composite (f ○ g)(x) = ________________.

Q1.2 Even function: f(−x) = ____________. Odd function: f(−x) = ____________.

Q1.3 "Spend ____ to ____ minutes per mark" is the standard exam pacing rule from the lesson.

Stuck? Revisit lesson § Formula Reference and § Time Allocation.

2. Worked example — inverse + domain (4-mark template)

Find f⁻¹(x) for f(x) = √(x − 1), then state the domains and ranges of both. This is the "Think First" question from the lesson.

Step 1 — Domain & range of f.

Radicand ≥ 0 ⇒ x ≥ 1. So Dom(f) = [1, ∞). Output of √ is ≥ 0, so Ran(f) = [0, ∞).

Step 2 — Swap and solve.

y = √(x − 1) swap → x = √(y − 1)

Square: x² = y − 1 ⇒ y = x² + 1

Step 3 — Restrict the inverse to match Ran(f).

Since Ran(f) = [0, ∞), we restrict the inverse to x ≥ 0.

Step 4 — Domain and range swap.

Dom(f⁻¹) = Ran(f) = [0, ∞). Ran(f⁻¹) = Dom(f) = [1, ∞).

Conclusion. f⁻¹(x) = x² + 1 with x ≥ 0. Sketch both curves with the line y = x as the axis of reflection.

3. Faded example — composite domain

Let f(x) = √(x + 1) and g(x) = 2x − 3. Find the domain of f(g(x)). Fill in each blank. 4 marks

Step 1 — Domain of f. Radicand ≥ 0 ⇒ x + 1 ≥ 0 ⇒ x ≥ ____. So Dom(f) = ____________.

Step 2 — Domain of g. Linear, no restrictions. Dom(g) = ____________.

Step 3 — Substitute g into f.

f(g(x)) = √((2x − 3) + 1) = √( ____________ )

Step 4 — Apply the radical condition.

Radicand ≥ 0 ⇒ ____________ ≥ 0 ⇒ x ≥ ____.

Conclusion. Dom(f ○ g) = ____________.

Stuck? Revisit lesson § IQ2 summary and § Common Mistakes.

4. Graduated mixed-topic practice

Foundation — single-topic recall (4 questions)

Q	Task	Answer
4.1 1	State the domain of h(x) = 1/(x − 4).
4.2 1	If f(x) = 3x + 2, find f⁻¹(x).
4.3 1	If f(x) = x² and g(x) = x − 1, write f(g(x)) as a single polynomial.
4.4 1	Is f(x) = x³ − x even, odd, or neither? (state only)

Standard — two-topic synthesis (6 questions)

Show one line of working at minimum.

4.5 Find f⁻¹(x) for f(x) = 2/(x − 1) + 3. State Dom(f⁻¹) and Ran(f⁻¹). 3 marks

4.6 Let f(x) = √(x + 1) and g(x) = 2x − 3. Find Dom(f ○ g) and Dom(g ○ f). 3 marks

4.7 Sketch y = −2 f(x − 1) + 4 given that y = f(x) passes through (0, 2), (2, 0) and has max at (1, 3). Label the image of each point. 3 marks

4.8 Determine whether f(x) = (x − 2)² + 3 is even, odd, or neither. Provide the algebraic check. 2 marks

4.9 A function f has zeros at x = 1 and x = 4 and a minimum at (2.5, −2). Sketch y = 1/f(x), labelling the vertical asymptotes, the image of the minimum, and the sign on each interval. 3 marks

4.10 If f(x) = x² (defined for all real x) and g(x) = √x (defined for x ≥ 0), find f(g(x)) and state the domain for which "f(g(x)) = x" is actually true. 2 marks

Extension — multi-topic synthesis (2 questions)

4.11 Consider f(x) = x² − 4x + 5 defined for x ≥ 2. Show f is one-to-one on this domain, find f⁻¹(x), and state the domain and range of f⁻¹. 4 marks

4.12 A company's daily revenue R = −20(n − 5)² + 800 (R in dollars, n in hundreds of units, n ≥ 0). (a) State maximum daily revenue and the n at which it occurs. (b) Find both break-even quantities. (c) Sketch R(n) for n ≥ 0, labelling the vertex, R-intercept and n-intercepts. 4 marks

Stuck on 4.11? Show derivative or "as x increases past the vertex, f increases" to argue one-to-one.

5. Self-check the easy 3

Tick once you've verified your method on the foundation row.

For 4.1 I wrote interval notation: x ≠ 4, equivalently (−∞, 4) ∪ (4, ∞).

For 4.2 I obtained f⁻¹(x) = (x − 2)/3 by swap-and-solve, not by guessing.

For 4.4 I showed at least one line of algebra (not "by inspection").

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Definitions

Inverse: swap x and y, then solve for y. (f ○ g)(x) = f(g(x)).

Q1.2 — Even & odd

Even: f(−x) = f(x). Odd: f(−x) = −f(x).

Q1.3 — Exam pacing

1 to 1.5 minutes per mark.

Q3 — Faded example, Dom(f ○ g)

Step 1: x ≥ −1, Dom(f) = [−1, ∞).
Step 2: Dom(g) = all real x.
Step 3: f(g(x)) = √(2x − 2).
Step 4: 2x − 2 ≥ 0 ⇒ x ≥ 1.
Conclusion: Dom(f ○ g) = [1, ∞).

Q4.1−4.4 — Foundation

4.1 Dom = (−∞, 4) ∪ (4, ∞) (i.e. x ≠ 4). 4.2 f⁻¹(x) = (x − 2)/3. 4.3 f(g(x)) = (x − 1)² = x² − 2x + 1. 4.4 f(−x) = (−x)³ − (−x) = −x³ + x = −(x³ − x) = −f(x), so odd.

Q4.5 — Inverse of f(x) = 2/(x − 1) + 3

Swap: x = 2/(y − 1) + 3 ⇒ x − 3 = 2/(y − 1) ⇒ y − 1 = 2/(x − 3) ⇒ f⁻¹(x) = 2/(x − 3) + 1. Dom(f⁻¹) = Ran(f) = {y : y ≠ 3} = (−∞, 3) ∪ (3, ∞). Ran(f⁻¹) = Dom(f) = (−∞, 1) ∪ (1, ∞).

Q4.6 — Composite domains

f(g(x)) = √(2x − 2): need 2x − 2 ≥ 0 ⇒ Dom(f ○ g) = [1, ∞). g(f(x)) = 2√(x + 1) − 3: need x + 1 ≥ 0 ⇒ Dom(g ○ f) = [−1, ∞).

Q4.7 — Transformed sketch

a = −2, b = 1, h = 1, k = 4. (0, 2) → (1, 0); (2, 0) → (3, 4); (1, 3) → (2, −2). The graph is reflected in the x-axis, vertically dilated by 2, shifted 1 right and 4 up.

Q4.8 — Even/odd test

f(−x) = (−x − 2)² + 3 = (x + 2)² + 3. This is not equal to f(x) = (x − 2)² + 3 (different vertex offset) and not equal to −f(x) = −(x − 2)² − 3 (different sign). So f is neither even nor odd.

Q4.9 — Sketch y = 1/f(x)

VAs x = 1 and x = 4. Min of f at (2.5, −2) → max of 1/f at (2.5, −0.5). Sign: f > 0 outside [1, 4] (or so, depending on f) ⇒ 1/f > 0 there; f < 0 inside [1, 4] (between zeros, where min is) ⇒ 1/f < 0 inside with max −0.5 at (2.5, −0.5). HA y = 0 at infinity.

Q4.10 — f(g(x)) and where it equals x

f(g(x)) = (√x)² = x, but this expression is defined only when x is in the domain of g, i.e. x ≥ 0. So f(g(x)) = x for x ≥ 0; for x < 0 the composite is undefined. The "for all real x" claim is incorrect.

Q4.11 — Inverse of f(x) = x² − 4x + 5, x ≥ 2

f(x) = (x − 2)² + 1, vertex (2, 1). For x ≥ 2 the parabola is increasing, so f is one-to-one. Swap: x = (y − 2)² + 1 ⇒ (y − 2)² = x − 1 ⇒ y − 2 = +√(x − 1) (positive root since y ≥ 2) ⇒ f⁻¹(x) = 2 + √(x − 1). Dom(f⁻¹) = Ran(f) = [1, ∞). Ran(f⁻¹) = Dom(f) = [2, ∞).

Q4.12 — Revenue model

(a) Vertex (5, 800): max revenue $800 at n = 5 (500 units). (b) 0 = −20(n − 5)² + 800 ⇒ (n − 5)² = 40 ⇒ n = 5 ± 2√10 ≈ 0.68 or 9.32 hundred units (about 68 or 932 units). (c) Downward parabola, vertex (5, 800), R-intercept R(0) = −20(25) + 800 = 300, n-intercepts at 5 ± 2√10. Label all.