Skill Align Mathematics Methods Year 12 Section Two Calculator-assumed for WACE - 2026 Edition
Original Skill Align practice examination content
- Paper
- Section Two Calculator-assumed Question and Response Book Showcase
- Reading
- 10 minutes reading time
- Writing
- 100 minutes
- Assessment
- 102 marks
Formula sheet retained from Section One. Drawing instruments, templates, notes on two unfolded A4 sheets, and up to three approved scientific, graphic or CAS calculators may be used.
Section Two
Answer all questions. For any question or part question worth more than two marks, valid working or justification is required to receive full marks.
Question 1
9 marksQuestion 2
10 marksQuestion 3
9 marksQuestion 4
10 marksQuestion 5
10 marksQuestion 6
9 marksQuestion 7
10 marksQuestion 8
10 marksQuestion 9
13 marksQuestion 10
12 marksWorked Solutions And Marking Guide
Section Two Question 1
(a) 60-2x.
Two widths use 2x metres of fencing.
(b) A(x)=x(60-2x).
Area is width multiplied by length.
(c) 450 m^2.
The quadratic 60x-2x² has its maximum at x=15.
(d) 15 m by 30 m.
The length is 60-2(15)=30.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response 60-2x. Working is not required; accept mathematically equivalent answers.
- Part (b) (2 marks): award full marks for the correct response A(x)=x(60-2x). Working is not required; accept mathematically equivalent answers.
- Part (c) (3 marks): differentiates A(x)=60x-2x² and solves A'(x)=60-4x=0 to obtain x=15 - 1 mark; confirms a maximum using A''(x)=-4<0 and the physical domain - 1 mark; evaluates A(15)=450 m² - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (d) (2 marks): award full marks for the correct response 15 m by 30 m. Working is not required; accept mathematically equivalent answers.
Section Two Question 2
(a) 100%.
P(0)=10+90=100.
(b) Graph estimate: approximately 3 hours. Algebraic confirmation: t=((ln2) / (0.25))approx2.77 hours.
Reading the crossing from the graph gives about 3 hours. Solving 55=10+90e^(-0.25t) gives e^(-0.25t)=0.5, then t=((ln2) / (0.25))approx2.77.
(c) P=10; the model approaches a residual 10 percent.
The exponential term tends to zero.
(d) P'(t)=-22.5e^(-0.25t)<0.
The negative derivative means the model decreases over time.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response 100%. Working is not required; accept mathematically equivalent answers.
- Part (b) (3 marks): reads an estimate of approximately 3 hours from the graph - 1 mark; forms and rearranges 55=10+90e^(-0.25t) to obtain e^(-0.25t)=0.5 - 1 mark; obtains t=((ln2) / (0.25))approx2.77 hours - 1 mark. Accept mathematically equivalent answers.
- Part (c) (2 marks): The exponential term tends to zero - 1 mark; states P=10; the model approaches a residual 10 percent - 1 mark.
- Part (d) (3 marks): gives the requested decision, interpretation or effect: P'(t)=-22.5e^(-0.25t)<0 - 1 mark; cites the decisive mathematical evidence: The negative derivative means the model decreases over time - 1 mark; links that evidence to the parameter, event or context named in the question - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
Section Two Question 3
(a) k=frac18.
Since int_0^4kx,dx=8k=1, k=1 / 8.
(b) frac12.
int_1³((x) / (8)),dx=((9-1) / (16))=frac12.
(c) frac83.
E(X)=int_0^4x((x) / (8)),dx=frac18[((x³) / (3))]_0⁴=frac83.
Mark allocation
- Part (a) (3 marks): uses normalisation to write int_0^4kx,dx=1 - 1 mark; evaluates the integral to obtain 8k=1 - 1 mark; solves k=frac18 - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (b) (3 marks): forms the requested event using the stated distribution, CDF, complement or standardisation - 1 mark; substitutes and evaluates the relevant boundary or component probabilities - 1 mark; obtains frac12 - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (c) (3 marks): writes E(X)=int_0^4x((x) / (8)),dx - 1 mark; evaluates frac18[((x³) / (3))]_0⁴ - 1 mark; obtains E(X)=frac83 - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
Section Two Question 4
(a) z=((512-500) / (8))=1.5.
Subtract the mean and divide by the standard deviation.
(b) P(-1<Z<1.5)approx0.7745.
Use technology to evaluate Phi(1.5)-Phi(-1).
(c) xapprox489.7 mL.
Using z_(0.10)approx-1.2816, x=500+8(-1.2816)approx489.7.
(d) About 10% of bottles have a fill volume below 489.7 mL.
A tenth percentile has 10 percent of the modelled population below it.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response z=((512-500) / (8))=1.5. Working is not required; accept mathematically equivalent answers.
- Part (b) (3 marks): forms the requested event using the stated distribution, CDF, complement or standardisation - 1 mark; substitutes and evaluates the relevant boundary or component probabilities - 1 mark; obtains P(-1<Z<1.5)approx0.7745 - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (c) (3 marks): forms the requested event using the stated distribution, CDF, complement or standardisation - 1 mark; substitutes and evaluates the relevant boundary or component probabilities - 1 mark; obtains xapprox489.7 mL - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (d) (2 marks): A tenth percentile has 10 percent of the modelled population below it - 1 mark; states About 10% of bottles have a fill volume below 489.7 mL - 1 mark.
Section Two Question 5
(a) 0.58.
The midpoint is (0.54+0.62) / 2=0.58.
(b) 0.04.
It is half the interval width.
(c) We are 95% confident that between 54% and 62% of all Western Australian Year 12 students use public transport to travel to school.
The interpretation states the confidence level, population and attribute being estimated.
(d) Yes; every plausible value in the 95% confidence interval is above 0.5.
The lower endpoint is 0.54.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response 0.58. Working is not required; accept mathematically equivalent answers.
- Part (b) (2 marks): award full marks for the correct response 0.04. Working is not required; accept mathematically equivalent answers.
- Part (c) (3 marks): states the confidence level - 1 mark; identifies the population being estimated - 1 mark; identifies the attribute and gives the interval in context - 1 mark. Accept mathematically equivalent answers.
- Part (d) (3 marks): gives the requested decision, interpretation or effect: Yes; every plausible value in the 95% confidence interval is above 0.5 - 1 mark; cites the decisive mathematical evidence: The lower endpoint is 0.54 - 1 mark; links that evidence to the parameter, event or context named in the question - 1 mark. Accept mathematically equivalent answers.
Section Two Question 6
(a) y=4x-x^2.
Their difference is 3x-x²>0 for 0<x<3.
(b) int_0³(3x-x²),dx.
Subtract the lower curve from the upper curve.
(c) frac92 square units.
[frac32x²-frac13x³]_0³=((27) / (2))-9=frac92.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response y=4x-x^2. Working is not required; accept mathematically equivalent answers.
- Part (b) (3 marks): identifies the correct limits of integration - 1 mark; forms the required integrand, including the correct sign or upper-minus-lower order - 1 mark; writes int_0³(3x-x²),dx - 1 mark. Accept mathematically equivalent answers.
- Part (c) (4 marks): uses the antiderivative frac32x²-frac13x³ - 1 mark; substitutes the bounds to obtain ((27) / (2))-9 - 1 mark; simplifies the exact value to frac92 - 1 mark; states the area as frac92 square units - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
Section Two Question 7
(a) xinmathbb R.
An exponential function is defined for all real inputs.
(b) y=3.
The exponential term tends to zero as x increases.
(c) (0,2e+3).
Substitute x=0, then simplify y=2e^(-(0-1))+3=2e+3; hence the intercept is (0,2e+3).
(d) Reflect in the y-axis, shift right 1, dilate vertically by factor 2, then shift up 3.
Read the transformations from 2e^(-(x-1))+3.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response xinmathbb R. Working is not required; accept mathematically equivalent answers.
- Part (b) (2 marks): award full marks for the correct response y=3. Working is not required; accept mathematically equivalent answers.
- Part (c) (3 marks): substitutes x=0 into the function - 1 mark; simplifies exactly to y=2e+3 - 1 mark; states the intercept as (0,2e+3) - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (d) (3 marks): identifies the reflection in the y-axis and the shift right by 1 from the exponent - 1 mark; identifies the vertical dilation by factor 2 - 1 mark; identifies the upward shift by 3 - 1 mark. Accept mathematically equivalent answers.
Section Two Question 8
(a) V=frac23π r^3.
Substitute h=2r.
(b) 2π r^2.
Differentiate with respect to r.
(c) 4.8π m³ min⁻¹.
Use ((dV) / (dt))=2π r²((dr) / (dt))=2π(16)(0.15).
(d) The rate ((dV) / (dt)) is proportional to r^2.
The same radial increase adds more volume at a larger radius.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response V=frac23π r^3. Working is not required; accept mathematically equivalent answers.
- Part (b) (2 marks): award full marks for the correct response 2π r^2. Working is not required; accept mathematically equivalent answers.
- Part (c) (3 marks): substitutes r=4 and dr / dt=0.15 into dV / dt=2π r²,dr / dt - 1 mark; evaluates 2π(4²)(0.15) - 1 mark; obtains dV / dt=4.8π m³ min⁻¹ - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (d) (3 marks): gives the requested decision, interpretation or effect: The rate ((dV) / (dt)) is proportional to r² - 1 mark; cites the decisive mathematical evidence: The same radial increase adds more volume at a larger radius - 1 mark; links that evidence to the parameter, event or context named in the question - 1 mark. Accept mathematically equivalent answers.
Section Two Question 9
(a) -3,1,6 with probabilities 0.5,0.3,0.2.
Read the categories and bar heights.
(b) 0.
(-3)(0.5)+1(0.3)+6(0.2)=0.
(c) 12.
9(0.5)+1(0.3)+36(0.2)=12.
(d) operatorname(Var)(X)=12; the game is fair in the long run because E(X)=0.
Variance is E(X²)-[E(X)]²=12.
(e) E(Y)=-1, operatorname(Var)(Y)=48, P(Y>0)=0.5.
Scale the first two moments and note that X=1 or 6 gives a positive revised return.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response -3,1,6 with probabilities 0.5,0.3,0.2. Working is not required; accept mathematically equivalent answers.
- Part (b) (3 marks): writes E(X)=(-3)(0.5)+1(0.3)+6(0.2) - 1 mark; evaluates the contributions as -1.5+0.3+1.2 - 1 mark; obtains E(X)=0 - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (c) (3 marks): writes E(X²)=9(0.5)+1(0.3)+36(0.2) - 1 mark; evaluates the contributions as 4.5+0.3+7.2 - 1 mark; obtains E(X²)=12 - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (d) (2 marks): Variance is E(X²)-[E(X)]²=12 - 1 mark; states operatorname(Var)(X)=12; the game is fair in the long run because E(X)=0 - 1 mark.
- Part (e) (3 marks): mark 1: obtains the revised expectation -1; mark 2: obtains the revised variance 48; mark 3: adds the two positive-return probabilities to obtain 0.5.
Section Two Question 10
(a) h'(x)=e^(-x)(2x-x²).
Use the product rule.
(b) x=2.
Solving x(2-x)=0 gives x=0 and x=2, but only x=2 is interior to the domain.
(c) A local maximum.
The derivative changes from positive to negative.
(d) 2-10e⁻².
An antiderivative is -e^(-x)(x²+2x+2); apply the bounds.
(e) The limit is 0. The absolute maximum is 4e⁻² at x=2.
The exponential decay dominates the polynomial; the derivative changes from positive to negative and h(0)=0.
Mark allocation
- Part (a) (3 marks): applies the product rule to the two stated factors - 1 mark; differentiates both factors, including any inner exponential or logarithmic function - 1 mark; combines and simplifies the derivative to h'(x)=e^(-x)(2x-x²) - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (b) (2 marks): award full marks for the correct response x=2. Working is not required; accept mathematically equivalent answers.
- Part (c) (2 marks): The derivative changes from positive to negative - 1 mark; states A local maximum - 1 mark.
- Part (d) (3 marks): uses a correct antiderivative for the stated integrand - 1 mark; substitutes the stated bounds and preserves the required area sign - 1 mark; simplifies to 2-10e⁻² - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (e) (2 marks): mark 1: gives the limiting value 0; mark 2: uses the full-domain behaviour and endpoint to establish the absolute maximum.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Models and calculus | Q1-Q2, Q6, Q8, Q10 | 50 | ___ | Review optimisation, rates, exponential models and integral applications. |
| Probability and inference | Q3-Q5, Q9 | 42 | ___ | Review densities, normal distributions, intervals and random-variable calculations. |
| Functions | Q7 | 10 | ___ | Review exponential transformations, intercepts and asymptotes. |