Mathematics Advanced • Year 11 • Module 3 • Lesson 15
Module 3 Review
HSC-style writing across the Module 3 toolkit: differentiation rules, stationary points, optimisation, definite integrals, areas, and volumes of revolution — finishing with an extended response on a closed-interval optimisation.
1. Short-answer questions
1.1 Differentiate y = (3x² + 1)⁴ · cos(x) with respect to x. Give your answer in fully factored form. 3 marks Band 3-4
1.2 Find the area of the region enclosed by the curve y = x² − 4x and the x-axis. 3 marks Band 3-4
Stuck on 1.2? The curve dips below the x-axis between its roots — signed integral will be negative; take absolute value for area.1.3 The region bounded by y = x² + 1, the x-axis, x = 0 and x = 2 is rotated about the x-axis. Find the exact volume of the solid formed. 4 marks Band 4-5
Stuck on 1.3? Expand [f(x)]² before integrating: (x² + 1)² = x⁴ + 2x² + 1.2. Extended response
2.1 A rectangular open-topped box is to be made by cutting equal squares of side x cm from each corner of a 24 cm × 18 cm rectangular sheet of cardboard, and folding up the sides.
(a) Show that the volume V of the box (in cm³) is V(x) = x(24 − 2x)(18 − 2x), and state the physically meaningful domain for x.
(b) Expand V(x) and find V′(x).
(c) Find the value of x in the interior of the domain that maximises V. Confirm your answer is a maximum using an appropriate test.
(d) State the maximum volume of the box in cm³, rounded to the nearest whole number, and the corresponding dimensions of the box. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — identifies base (24 − 2x) by (18 − 2x), height x, writes V(x) = x(24 − 2x)(18 − 2x), and states 0 < x < 9 (since 18 − 2x > 0).
Part (b) — 2 marks
• 1 mark — correctly expands V(x) = 4x³ − 84x² + 432x.
• 1 mark — V′(x) = 12x² − 168x + 432 (or equivalent factored form 12(x² − 14x + 36)).
Part (c) — 3 marks
• 1 mark — sets V′(x) = 0 and solves x² − 14x + 36 = 0 to get x = 7 ± √13.
• 1 mark — selects x = 7 − √13 ≈ 3.39 as the value in (0, 9); rejects x = 7 + √13 ≈ 10.61 as outside the domain.
• 1 mark — confirms maximum: e.g. V″(x) = 24x − 168 ⇒ V″(7 − √13) = 24(7 − √13) − 168 = −24√13 < 0, so x = 7 − √13 gives a local maximum (or via first-derivative sign change).
Part (d) — 2 marks
• 1 mark — evaluates V(7 − √13) ≈ 655 cm³ (rounded to the nearest whole number).
• 1 mark — states box dimensions: base ≈ 17.21 cm × 11.21 cm, height ≈ 3.39 cm (or exact: (10 + 2√13) × (4 + 2√13) × (7 − √13)).
Your response:
Stuck on (c)? The quadratic 12x² − 168x + 432 = 0 simplifies (divide by 12) to x² − 14x + 36 = 0. The discriminant is 196 − 144 = 52 = 4·13, so x = (14 ± 2√13)/2 = 7 ± √13.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Differentiate y = (3x² + 1)⁴ · cos(x) (3 marks)
Sample response. Product rule with u = (3x² + 1)⁴ and v = cos(x). For u′, chain rule: u′ = 4(3x² + 1)³ · (6x) = 24x(3x² + 1)³. v′ = −sin(x).
y′ = u′v + uv′ = 24x(3x² + 1)³ cos(x) + (3x² + 1)⁴ · (−sin(x))
Factor out (3x² + 1)³: y′ = (3x² + 1)³ · [24x cos(x) − (3x² + 1) sin(x)].
Marking notes. 1 mark — correctly identifies need for product + chain rule and computes u′ = 24x(3x² + 1)³. 1 mark — correct unfactored derivative. 1 mark — fully factored form with (3x² + 1)³ extracted. Common error: forgetting the inner derivative 6x in the chain rule, or losing the negative sign on sin(x).
1.2 — Area enclosed by y = x² − 4x and the x-axis (3 marks)
Sample response. Roots: x² − 4x = x(x − 4) = 0 ⇒ x = 0 and x = 4. Test x = 2: y = 4 − 8 = −4 < 0, so the curve lies below the x-axis on (0, 4).
Signed integral = ∫₀⁴ (x² − 4x) dx = [x³/3 − 2x²]₀⁴ = (64/3 − 32) − 0 = 64/3 − 96/3 = −32/3.
Area = |−32/3| = 32/3 sq units ≈ 10.67 sq units.
Marking notes. 1 mark — finds roots 0 and 4 and identifies curve below x-axis. 1 mark — correct signed integral = −32/3. 1 mark — takes absolute value for total area = 32/3. Common error: leaving the answer as −32/3 (negative area) — "signed area" ≠ "total area".
1.3 — Volume from rotating y = x² + 1 on [0, 2] about x-axis (4 marks)
Sample response. Disc method: V = π ∫₀² [f(x)]² dx = π ∫₀² (x² + 1)² dx.
Expand: (x² + 1)² = x⁴ + 2x² + 1.
V = π ∫₀² (x⁴ + 2x² + 1) dx = π [x⁵/5 + 2x³/3 + x]₀²
= π · (32/5 + 16/3 + 2) − 0
= π · (96/15 + 80/15 + 30/15) = π · 206/15
= 206π/15 cubic units ≈ 43.14 cubic units.
Marking notes. 1 mark — correctly sets up V = π ∫₀² (x² + 1)² dx (disc method). 1 mark — correctly expands integrand to x⁴ + 2x² + 1. 1 mark — correct antiderivative x⁵/5 + 2x³/3 + x. 1 mark — correct evaluation 206π/15. Common error: writing V = π ∫(x² + 1) dx (forgetting to square the radius f(x)), or expanding (x² + 1)² as x⁴ + 1.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Cutting squares of side x from each corner leaves a base of (24 − 2x) by (18 − 2x), and folding up the sides gives a box of height x. So
V(x) = x · (24 − 2x) · (18 − 2x).
For the box to exist, all three dimensions must be positive: x > 0, 24 − 2x > 0 (i.e. x < 12), and 18 − 2x > 0 (i.e. x < 9). The binding constraint is the shorter side, so the domain is 0 < x < 9. [1 mark — V derived and domain stated.]
Part (b). Expand step by step. First, (24 − 2x)(18 − 2x) = 432 − 48x − 36x + 4x² = 4x² − 84x + 432. Then
V(x) = x(4x² − 84x + 432) = 4x³ − 84x² + 432x.
[1 mark — correct expansion.]
Differentiate:
V′(x) = 12x² − 168x + 432 = 12(x² − 14x + 36).
[1 mark — correct V′.]
Part (c). Set V′(x) = 0: x² − 14x + 36 = 0. Discriminant Δ = 196 − 144 = 52 = 4 · 13, so
x = (14 ± √52)/2 = (14 ± 2√13)/2 = 7 ± √13.
[1 mark — both stationary points identified.]
Since √13 ≈ 3.606, the two roots are x ≈ 10.61 and x ≈ 3.39. Only x = 7 − √13 ≈ 3.39 lies in the domain 0 < x < 9; reject x = 7 + √13 ≈ 10.61. [1 mark — correct root selected with domain reasoning.]
Second-derivative test: V″(x) = 24x − 168. At x = 7 − √13,
V″(7 − √13) = 24(7 − √13) − 168 = 168 − 24√13 − 168 = −24√13 < 0.
So V has a local maximum at x = 7 − √13. [1 mark — second-derivative test confirms maximum.]
Part (d). Evaluate V at x = 7 − √13. Using the factored form (less arithmetic):
24 − 2x = 24 − 2(7 − √13) = 10 + 2√13
18 − 2x = 18 − 2(7 − √13) = 4 + 2√13.
Then V = (7 − √13)(10 + 2√13)(4 + 2√13). Compute (10 + 2√13)(4 + 2√13) = 40 + 20√13 + 8√13 + 4·13 = 40 + 28√13 + 52 = 92 + 28√13. So
V = (7 − √13)(92 + 28√13) = 644 + 196√13 − 92√13 − 28·13 = 644 + 104√13 − 364 = 280 + 104√13.
Numerically: 104√13 ≈ 104 · 3.6056 ≈ 374.98, so V ≈ 280 + 374.98 ≈ 655 cm³. [1 mark — correct maximum volume.]
Box dimensions: height ≈ 3.39 cm, base ≈ (10 + 2√13) × (4 + 2√13) ≈ 17.21 cm × 11.21 cm. [1 mark — correct dimensions stated.]
Total: 8/8.
Band descriptors for marker.
Band 3: Sets up V(x) correctly and may state the domain. Expands V(x) but with errors in V′(x), or attempts the second derivative without solving the quadratic correctly. ≈ 3-4 marks.
Band 4: Sets up V, expands cleanly, finds V′ and solves the quadratic to get x = 7 ± √13, but selects the wrong root (e.g. x = 7 + √13), or fails to test for maximum, or stops before computing V. ≈ 5-6 marks.
Band 5: Completes (a)–(c) correctly with the right root selected and maximum confirmed. In (d), either evaluates V numerically without exact form, or finds V but does not state final box dimensions. ≈ 6-7 marks.
Band 6: All four parts complete: V derived with domain justified, expansion correct, V′ and V″ used, both stationary points found with the in-domain one selected, maximum confirmed via V″ < 0, maximum volume ≈ 655 cm³ stated, and the corresponding box dimensions given. 8/8.