Skill Align SACE Stage 2 Mathematical Methods - Pack 0 - Question Booklet 1
Questions 1-6 | 50 marks | one part of a 100-mark, 130-minute complete examination
- Question booklet
- Question Booklet 1
- 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 1
Answer all questions 1-6. Show complete working and steps of logic. Give numerical answers to three significant figures unless otherwise instructed.
Question 1
8 marksQuestion 2
8 marksQuestion 3
8 marksQuestion 4
8 marksQuestion 5
9 marksQuestion 6
9 marksWorked Solutions And Marking Guide
Question 1
(a) R'(t)=18e^(-0.5t)(1-0.5t).
The product rule gives 18e^(-0.5t)-9te^(-0.5t)=18e^(-0.5t)(1-0.5t).
(b) t=2 hours.
Since the exponential factor is positive, R'(t)=0 gives 1-0.5t=0, so t=2. The derivative changes from positive to negative.
(c) R(2)=36e⁻¹approx13.2.
Substitution gives R(2)=36 / eapprox13.243, hence 13.2 to three significant figures.
(d) The exponential factor approaches zero faster than the linear factor grows.
As ttoinfty, te^(-0.5t)to0; the model therefore approaches zero after its maximum.
Detailed marking criteria
Part a (2 marks)
Award the 2 available marks for the following observable evidence: differentiates 18t and e^(-0.5t) using the product and chain rules; factors the result as 18e^(-0.5t)(1-0.5t).
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 1-0.5t=0 to obtain t=2; uses a sign change or equivalent reasoning to identify a maximum.
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: substitutes t=2 into the original response model; reports 13.2 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 d (2 marks)
Award the 2 available marks for the following observable evidence: compares the linear growth of t with the exponential decay of e^(-0.5t); concludes that R(t)to0 as ttoinfty.
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 2
(a) 0.45.
P(Xgeq120)=0.30+0.15=0.45.
(b) 81.
E(X)=0(0.55)+120(0.30)+300(0.15)=81.
(c) operatorname(Var)(X)=11259text( dollars)^2.
E(X²)=120²(0.30)+300²(0.15)=17820. Hence operatorname(Var)(X)=17820-81²=11259.
Detailed marking criteria
Part a (2 marks)
Award the 2 available marks for the following observable evidence: identifies the outcomes 120 and 300; adds their probabilities to obtain 0.45.
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: forms the weighted sum 0(0.55)+120(0.30)+300(0.15); evaluates the non-zero contributions as 36 and 45; states the expected repair cost as 81.
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: calculates E(X²)=17820; uses operatorname(Var)(X)=E(X²)-[E(X)]²; evaluates the variance as 11259text( dollars)^2.
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 3
(a) The maximum depth is 9text( m) at x=3.
y'=6-2x=0 gives x=3, and y(3)=9.
(b) int_0⁶(6x-x²),dx=36text( m)^2.
[3x²-x³ / 3]_0⁶=108-72=36.
(c) 90.0text( m)^3.
Volume =36(2.5)=90.0text( m)^3.
(d) It assumes the cross-section is unchanged along the full 2.5 m length.
The prismatic-volume calculation is valid only if the cross-section is constant.
Detailed marking criteria
Part a (2 marks)
Award the 2 available marks for the following observable evidence: solves 6-2x=0 to obtain x=3; substitutes into the model to obtain a maximum depth of 9text( 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 b (3 marks)
Award the 3 available marks for the following observable evidence: forms the definite integral int_0⁶(6x-x²),dx; uses the antiderivative 3x²-x³ / 3; evaluates the bounds to obtain 36text( m)^2.
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: multiplies the cross-sectional area by 2.5text( m); states the volume as 90.0text( m)^3.
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 (1 mark)
Award the 1 available mark for the following observable evidence: identifies the constant-cross-section assumption.
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 4
(a) tgeq0.
The physical model specifies non-negative time; this also satisfies t+1>0.
(b) L'(t)=frac4(t+1).
Differentiate 4ln(t+1) by the chain rule.
(c) t=e²-1approx6.39 minutes.
12+4ln(t+1)=20 gives ln(t+1)=2, so t=e²-1approx6.389.
(d) The response continues to increase, but at a decreasing rate that approaches zero.
For tgeq0, 4 / (t+1)>0 and decreases towards zero.
Detailed marking criteria
Part a (1 mark)
Award the 1 available mark for the following observable evidence: states tgeq0.
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: uses ((d) / (dt))ln(t+1)=1 / (t+1); states L'(t)=4 / (t+1).
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: isolates ln(t+1)=2; uses the inverse exponential to obtain t=e²-1; reports 6.39 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 d (2 marks)
Award the 2 available marks for the following observable evidence: notes that L'(t) remains positive; explains that the response rate decreases towards zero.
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 5
(a) P(X<45)approx0.122.
z=(45-52) / 6=-1.1667, so Phi(z)approx0.1217.
(b) Approximately 59.7 hours.
The 90th percentile has zapprox1.28155, giving 52+1.28155(6)approx59.689.
(c) P(bar X>55)approx0.0668.
bar Xsim N(52,(6 / sqrt9)²)=N(52,2²). Thus z=(55-52) / 2=1.5, and P(Z>1.5)approx0.0668.
Detailed marking criteria
Part a (2 marks)
Award the 2 available marks for the following observable evidence: standardises 45 to z=-1.1667ldots; uses the lower normal tail to obtain 0.122.
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: identifies the required cumulative probability as 0.90; uses z_(0.90)approx1.28155; transforms back to obtain 59.7 hours.
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 (4 marks)
Award the 4 available marks for the following observable evidence: states that the sample mean has mean 52; calculates the standard error 6 / sqrt9=2; standardises 55 to z=1.5; uses the upper tail to obtain 0.0668.
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 6
(a) hat p=128 / 220approx0.582.
The sample proportion is 128 / 220=0.5818ldots.
(b) SE(hat p)approx0.0332.
√(hat p(1-hat p) / 220)approx0.03324.
(c) Approximately (0.517,0.647).
0.5818ldotspm1.96(0.03324ldots) gives (0.5167ldots,0.6470ldots).
(d) Yes; the entire 95% confidence interval lies above 0.50.
The lower endpoint is approximately 0.517>0.50, so the data support a population majority at the 95% confidence level.
Detailed marking criteria
Part a (1 mark)
Award the 1 available mark for the following observable evidence: calculates hat p=128 / 220approx0.582.
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: uses SE(hat p)=√(hat p(1-hat p) / n) with the unrounded sample proportion; evaluates the standard error as 0.0332.
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: uses the 95% coefficient 1.96; forms hat ppm1.96SE(hat p) without premature rounding; reports the interval as (0.517,0.647).
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: compares the lower endpoint with 0.50; observes that every plausible value in the interval exceeds 0.50; states a conclusion about the population of students, not just the sample.
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 | Q1 | 8 | ___ | Rework Q1: practise product rule stationary point maximum and long term exponential behaviour. Check every labelled part against the evidence-specific criterion and retain unrounded values until the final answer. |
| Discrete Random Variables | Q2 | 8 | ___ | Rework Q2: practise discrete probability expectation second moment and variance. Check every labelled part against the evidence-specific criterion and retain unrounded values until the final answer. |
| Integral Calculus | Q3 | 8 | ___ | Rework Q3: practise stationary point exact definite integral volume and model limitation. Check every labelled part against the evidence-specific criterion and retain unrounded values until the final answer. |
| Logarithmic Functions | Q4 | 8 | ___ | Rework Q4: practise logarithmic domain derivative threshold and rate interpretation. Check every labelled part against the evidence-specific criterion and retain unrounded values until the final answer. |
| Continuous Random Variables and the Normal Distribution | Q5 | 9 | ___ | Rework Q5: practise normal standardisation inverse normal and sample mean distribution. Check every labelled part against the evidence-specific criterion and retain unrounded values until the final answer. |
| Sampling and Confidence Intervals | Q6 | 9 | ___ | Rework Q6: practise sample proportion standard error confidence interval and contextual conclusion. Check every labelled part against the evidence-specific criterion and retain unrounded values until the final answer. |