Skill Align SACE Stage 2 Mathematical Methods - Pack 0 - Question Booklet 2
Questions 7-11 | 50 marks | one part of a 100-mark, 130-minute complete examination
- Question booklet
- Question Booklet 2
- Recommended time
- Approximately 65 minutes
- Complete examination
- 130 minutes across both question booklets
- Assessment
- 50 marks
SACE Stage 2 Mathematical Methods — examination conditions
- The supervising centre supplies the current official formula sheet as a separate resource. It is not bundled with this Skill Align practice paper.
- Candidates may use either two approved graphics calculators or one approved graphics calculator and one scientific calculator. Computer algebra system (CAS) calculators are not permitted. Scientific-calculator memory must be cleared; graphics-calculator memory does not need to be cleared.
- Candidates may bring two unfolded A4 sheets, using all four sides, containing their own handwritten notes.
- Show sufficient working and logical steps. Give numerical answers to three significant figures unless a question says otherwise.
- Write in blue or black pen. A sharp dark pencil may be used for diagrams and graphs.
Question Booklet 2
Answer all questions 7-11. Show complete working and steps of logic. Give numerical answers to three significant figures unless otherwise instructed.
Question 7
9 marksQuestion 8
10 marksQuestion 9
10 marksQuestion 10
11 marksQuestion 11
10 marksWorked Solutions And Marking Guide
Question 7
(a) H'(t)=((π) / (3))cos(π t / 6).
Differentiate the sine term using the chain rule.
(b) The maximum is 5 m at t=3.
The sine term is 1 when π t / 6=π / 2, so t=3 and H=5.
(c) π / 3approx1.05 m / month, at t=0,6,12.
The magnitude is greatest when |cos(π t / 6)|=1, giving the stated times.
(d) At t=0 the level is rising; at t=6 it is falling at the same instantaneous speed.
H'(0)=π / 3 and H'(6)=-π / 3.
Detailed marking criteria
Part a (2 marks)
Award the 2 available marks for the following observable evidence: uses the derivative of sine and the inner derivative π / 6; simplifies to H'(t)=(π / 3)cos(π t / 6).
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part b (2 marks)
Award the 2 available marks for the following observable evidence: solves sin(π t / 6)=1 on the stated interval; states H(3)=5 m.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part c (3 marks)
Award the 3 available marks for the following observable evidence: recognises that |H'(t)| is maximised when |cos(π t / 6)|=1; finds t=0,6,12 in the domain; reports the maximum magnitude π / 3approx1.05 m / month.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part d (2 marks)
Award the 2 available marks for the following observable evidence: uses the positive sign at t=0 to identify rising water; uses the negative sign at t=6 to identify falling water.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Question 8
(a) 4.000, 6.121, 7.000, 6.121, 4.000.
Substitute the five times into the rate model.
(b) L_4=3(4+6.121+6.121+4)approx60.7 kWh.
Each rectangle has width 3. Use the smaller endpoint rate on each interval: 4,6.121,6.121,4.
(c) U_4=3(6.121+7+7+6.121)approx78.7 kWh.
Use the larger endpoint rate on each interval: 6.121,7,7,6.121.
(d) 48+((72) / (π))approx70.9 kWh.
An antiderivative is 4t-((36) / (π))cos(π t / 12). Evaluation gives 48+72 / π.
(e) On each interval the rate lies between its smaller and larger endpoint values, so L_4<int_0¹²r(t),dt<U_4.
Monotonicity on each side of the turning point makes the selected rectangle heights genuine lower and upper bounds.
Detailed marking criteria
Part a (2 marks)
Award the 2 available marks for the following observable evidence: evaluates the two endpoints and midpoint correctly; reports the symmetric five-ordinate list to three decimal places.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part b (2 marks)
Award the 2 available marks for the following observable evidence: selects the smaller endpoint height on each monotonic interval; multiplies their sum by width 3 to obtain 60.7 kWh.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part c (2 marks)
Award the 2 available marks for the following observable evidence: selects the larger endpoint height on each monotonic interval; multiplies their sum by width 3 to obtain 78.7 kWh.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part d (3 marks)
Award the 3 available marks for the following observable evidence: uses a correct antiderivative for the sine term; evaluates the bounds to obtain 48+72 / π; reports 70.9 kWh to three significant figures.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part e (1 mark)
Award the 1 available mark for the following observable evidence: uses monotonicity to justify the lower and upper bounds.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Question 9
(a) P(+)=0.1196.
P(+)=0.08(0.92)+0.92(0.05)=0.0736+0.0460=0.1196.
(b) Approximately 0.615.
Bayes' rule gives 0.0736 / 0.1196approx0.615385.
(c) 460 false positives.
10000(0.92)(0.05)=460.
(d) The 92% is sensitivity conditioned on already having the condition; the required probability also depends on prevalence and false positives.
The conditioning direction differs, and the low prevalence makes false positives a substantial share of positive results.
Detailed marking criteria
Part a (3 marks)
Award the 3 available marks for the following observable evidence: calculates the true-positive pathway 0.08(0.92)=0.0736; calculates the false-positive pathway 0.92(0.05)=0.0460; adds the disjoint pathways to obtain 0.1196.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part b (3 marks)
Award the 3 available marks for the following observable evidence: uses the joint true-positive probability 0.0736; divides by the total positive probability 0.1196; reports the conditional probability as 0.615.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part c (2 marks)
Award the 2 available marks for the following observable evidence: uses the false-positive probability 0.0460; multiplies by 10000 to obtain 460.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part d (2 marks)
Award the 2 available marks for the following observable evidence: distinguishes P(+midcondition) from P(conditionmid+); explains the effect of prevalence or false positives.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Question 10
(a) V(x)=x(24-2x)(16-2x), for 0<x<8.
The height is x, and the base dimensions are 24-2x and 16-2x. Both must be positive.
(b) x=((20pm4sqrt7) / (3)); the physical value is ((20-4sqrt7) / (3))approx3.14.
V'=12x²-160x+384=4(3x²-40x+96). Solving gives the two roots; only the smaller lies in 0<x<8.
(c) It is a local maximum.
V''(x)=24x-160, which is negative at x=(20-4sqrt7) / 3.
(d) Approximately 541 cm^3.
Substitution of x=(20-4sqrt7) / 3 gives Vapprox540.829 cm^3.
(e) It ignores material thickness or loss at the folds.
The geometric model assumes perfect cuts and folds with negligible thickness.
Detailed marking criteria
Part a (3 marks)
Award the 3 available marks for the following observable evidence: identifies the tray dimensions x, 24-2x and 16-2x; forms their product for the volume; states the physical domain 0<x<8.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part b (3 marks)
Award the 3 available marks for the following observable evidence: differentiates to obtain V'=12x²-160x+384; solves the quadratic to obtain (20pm4sqrt7) / 3; rejects the larger root using the physical domain.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part c (2 marks)
Award the 2 available marks for the following observable evidence: finds V''(x)=24x-160; shows that V''<0 at the physical stationary point.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation.
Part d (2 marks)
Award the 2 available marks for the following observable evidence: substitutes the unrounded physical stationary value into V(x); reports 541 cm³ to three significant figures.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part e (1 mark)
Award the 1 available mark for the following observable evidence: states a relevant physical limitation of the tray model.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation.
Question 11
(a) SE(bar X)=0.600 minutes.
SE(bar X)=s / sqrt n=4.8 / √64=0.6.
(b) Approximately (17.2,19.6) minutes.
18.4pm1.96(0.6)=(17.224,19.576), which rounds to (17.2,19.6).
(c) The data do not support the claim because the whole interval lies below 20 minutes.
The upper endpoint is approximately 19.6<20.
(d) n=139.
Require 1.96(4.8) / sqrt nleq0.8. Thus ngeq(1.96(4.8) / 0.8)²=138.298ldots, so round up to 139.
Detailed marking criteria
Part a (2 marks)
Award the 2 available marks for the following observable evidence: uses s / sqrt n; evaluates the standard error as 0.600 minutes.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part b (3 marks)
Award the 3 available marks for the following observable evidence: uses the 95% coefficient 1.96; forms 18.4pm1.96(0.6); reports (17.2,19.6) minutes to three significant figures.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part c (2 marks)
Award the 2 available marks for the following observable evidence: compares 20 with the confidence interval; states a population-level conclusion consistent with the comparison.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Part d (3 marks)
Award the 3 available marks for the following observable evidence: forms the margin-of-error inequality 1.96(4.8) / sqrt nleq0.8; obtains ngeq138.298ldots; rounds upward to the minimum integer 139.
Acceptable alternatives: Accept algebraically equivalent exact forms and any logically equivalent sequence of valid steps. Accept a calculator-supported value that rounds to the stated answer using unrounded intermediate values.
Do not credit by itself: Do not award evidence that is only asserted when the prompt requires working, reasoning or interpretation. Prematurely rounded intermediate values that change the final three-significant-figure result are not acceptable.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Further Differentiation and Applications | Q7 | 9 | ___ | Rework Q7: practise trigonometric differentiation extrema steepest rate and contextual interpretation. Check every labelled part against the evidence-specific criterion and retain unrounded values until the final answer. |
| Integral Calculus | Q8 | 10 | ___ | Rework Q8: practise equal width lower and upper rectangle sums exact integration and bounding. Check every labelled part against the evidence-specific criterion and retain unrounded values until the final answer. |
| Discrete Random Variables | Q9 | 10 | ___ | Rework Q9: practise conditional probability total probability false positives and predictive value. Check every labelled part against the evidence-specific criterion and retain unrounded values until the final answer. |
| Further Differentiation and Applications | Q10 | 11 | ___ | Rework Q10: practise optimisation domain cubic volume stationary point second derivative and limitation. Check every labelled part against the evidence-specific criterion and retain unrounded values until the final answer. |
| Sampling and Confidence Intervals | Q11 | 10 | ___ | Rework Q11: practise standard error population mean confidence interval interpretation and sample size. Check every labelled part against the evidence-specific criterion and retain unrounded values until the final answer. |