Skill Align Mathematics Specialist 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
- 93 marks
Standard items: pens, pencils including coloured pencils, sharpener, correction fluid or tape, eraser, ruler and highlighters. Special items: drawing instruments, templates, notes on two unfolded A4 sheets, and up to three calculators, which may include scientific, graphic and Computer Algebra System (CAS) calculators. Candidates are assumed to have CAS capability. In the official examination a formula sheet is provided by the supervisor. The authorised 2026 Mathematics Specialist formula sheet is not bundled with this Skill Align pack and must be obtained separately from the official SCSA source: https://senior-secondary.scsa.wa.edu.au/__data/assets/pdf_file/0008/1230659/2026-MAS-Formula-Sheet.PDF. An examination changeover period of up to 15 minutes applies, during which candidates are not permitted to work.
Section Two
Answer all questions. Valid working or justification is required for questions or parts worth more than two marks. Working may also be required where directed by words such as show, verify, justify, prove or explain.
Question 1
8 marksQuestion 2
8 marksQuestion 3
7 marksQuestion 4
8 marksQuestion 5
7 marksQuestion 6
8 marksQuestion 7
8 marksQuestion 8
8 marksQuestion 9
7 marksQuestion 10
8 marksQuestion 11
8 marksQuestion 12
8 marksWorked Solutions And Marking Guide
Section Two Question 1
(a) (7,-1,2).
Substitution gives 3+2lambda=9, so lambda=3.
(b) sin⁻¹(((2) / (3sqrt6))).
Use sintheta=((|mathbf d × mathbf n|) / (|mathbf d||mathbf n|)).
(c) No; its direction vector is not parallel to the plane's normal.
(2,-1,1) is not a scalar multiple of (1,2,2).
Mark allocation
- Part (a) (3 marks): method and intermediate evidence (2 marks): Substitution gives 3+2lambda=9, so lambda=3; required conclusion (1 mark): (7,-1,2).
- Part (b) (2 marks): method and intermediate evidence (1 mark): Use sintheta=((|mathbf d × mathbf n|) / (|mathbf d||mathbf n|)); required conclusion (1 mark): sin⁻¹(((2) / (3sqrt6))).
- Part (c) (3 marks): method and intermediate evidence (2 marks): (2,-1,1) is not a scalar multiple of (1,2,2); required conclusion (1 mark): No; its direction vector is not parallel to the plane's normal.
Section Two Question 2
(a) Centre 1-2i, radius 4.
Read the circle equation.
(b) -3-2i and 5-2i.
Move four units left and right of the centre.
(c) 4+sqrt5.
Add the radius to the centre's distance sqrt5 from the origin.
(d) 4-sqrt5.
The origin lies inside the circle, so subtract the centre distance from the radius.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): Read the circle equation; required conclusion (1 mark): Centre 1-2i, radius 4.
- Part (b) (2 marks): method and intermediate evidence (1 mark): Move four units left and right of the centre; required conclusion (1 mark): -3-2i and 5-2i.
- Part (c) (2 marks): method and intermediate evidence (1 mark): Add the radius to the centre's distance sqrt5 from the origin; required conclusion (1 mark): 4+sqrt5.
- Part (d) (2 marks): method and intermediate evidence (1 mark): The origin lies inside the circle, so subtract the centre distance from the radius; required conclusion (1 mark): 4-sqrt5.
Section Two Question 3
(a) 0.7 hours.
Use s / sqrt n=5.6 / √64.
(b) (40.9485,43.2515) hours, or (40.95,43.25) hours.
Calculate 42.1pm1.645(0.7).
(c) The observations are a random independent sample and the sample is large enough for the normal approximation.
Identify the random-sampling, independence and large-sample conditions.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): Use s / sqrt n=5.6 / √64; required conclusion (1 mark): 0.7 hours.
- Part (b) (3 marks): method and intermediate evidence (2 marks): Calculate 42.1pm1.645(0.7); required conclusion (1 mark): (40.9485,43.2515) hours, or (40.95,43.25) hours.
- Part (c) (2 marks): method and intermediate evidence (1 mark): Identify the random-sampling, independence and large-sample conditions; required conclusion (1 mark): The observations are a random independent sample and the sample is large enough for the normal approximation.
Section Two Question 4
(a) 1200.
The logistic factor is zero at P=1200.
(b) 32.
0.12(400)(1-400 / 1200)=32.
(c) P=600.
The logistic growth rate is greatest at half the carrying capacity.
(d) The limiting factor reduces growth as P approaches 1200.
Logistic growth includes a carrying-capacity constraint.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): The logistic factor is zero at P=1200; required conclusion (1 mark): 1200.
- Part (b) (2 marks): method and intermediate evidence (1 mark): 0.12(400)(1-400 / 1200)=32; required conclusion (1 mark): 32.
- Part (c) (2 marks): method and intermediate evidence (1 mark): The logistic growth rate is greatest at half the carrying capacity; required conclusion (1 mark): P=600.
- Part (d) (2 marks): method and intermediate evidence (1 mark): Logistic growth includes a carrying-capacity constraint; required conclusion (1 mark): The limiting factor reduces growth as P approaches 1200.
Section Two Question 5
(a) V=piint_0³ 4x,dx.
The disk radius is y=2sqrt(x), so y²=4x.
(b) 18pi cubic units.
V=pi[2x²]_0³=18pi.
(c) It is expressed in terms of pi without decimal approximation.
Exact form avoids calculator rounding.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): The disk radius is y=2sqrt(x), so y²=4x; required conclusion (1 mark): V=piint_0³ 4x,dx.
- Part (b) (3 marks): method and intermediate evidence (2 marks): V=pi[2x²]_0³=18pi; required conclusion (1 mark): 18pi cubic units.
- Part (c) (2 marks): method and intermediate evidence (1 mark): Exact form avoids calculator rounding; required conclusion (1 mark): It is expressed in terms of pi without decimal approximation.
Section Two Question 6
(a) (x-1)²+4=x²-2x+5.
Multiply the two conjugate linear factors.
(b) x³+x²-x+15.
Multiply (x+3)(x²-2x+5).
(c) A polynomial with real coefficients has non-real roots in conjugate pairs.
This is the conjugate-root theorem.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): Multiply the two conjugate linear factors; required conclusion (1 mark): (x-1)²+4=x²-2x+5.
- Part (b) (3 marks): method and intermediate evidence (2 marks): Multiply (x+3)(x²-2x+5); required conclusion (1 mark): x³+x²-x+15.
- Part (c) (3 marks): method and intermediate evidence (2 marks): This is the conjugate-root theorem; required conclusion (1 mark): A polynomial with real coefficients has non-real roots in conjugate pairs.
Section Two Question 7
(a) mathbf v(t)=(2t,6t²-3).
Differentiate each component.
(b) √13.
At t=1 the velocity is (2,3).
(c) mathbf a(t)=(2,12t).
Differentiate the velocity.
(d) t=0 or t=pmfrac23.
Set mathbf v × mathbf a=8t(9t²-4)=0.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): Differentiate each component; required conclusion (1 mark): mathbf v(t)=(2t,6t²-3).
- Part (b) (2 marks): method and intermediate evidence (1 mark): At t=1 the velocity is (2,3); required conclusion (1 mark): √13.
- Part (c) (2 marks): method and intermediate evidence (1 mark): Differentiate the velocity; required conclusion (1 mark): mathbf a(t)=(2,12t).
- Part (d) (2 marks): method and intermediate evidence (1 mark): Set mathbf v × mathbf a=8t(9t²-4)=0; required conclusion (1 mark): t=0 or t=pmfrac23.
Section Two Question 8
(a) 6.
3(2)-2(1)+1(2)=6.
(b) 9.
2²+1²+2²=9.
(c) 2.
The scalar projection is 6 / 3=2.
(d) (frac43,frac23,frac43).
The vector projection is frac69mathbf b.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): 3(2)-2(1)+1(2)=6; required conclusion (1 mark): 6.
- Part (b) (2 marks): method and intermediate evidence (1 mark): 2²+1²+2²=9; required conclusion (1 mark): 9.
- Part (c) (2 marks): method and intermediate evidence (1 mark): The scalar projection is 6 / 3=2; required conclusion (1 mark): 2.
- Part (d) (2 marks): method and intermediate evidence (1 mark): The vector projection is frac69mathbf b; required conclusion (1 mark): (frac43,frac23,frac43).
Section Two Question 9
(a) int_1^2frac1u,du.
Since du=sec^2x,dx, the bounds become u=1 and u=2.
(b) ln2.
An antiderivative is ln u, so the value is ln2-ln1.
(c) Since 1+tan xge1 on [0,pi / 4], the denominator is never zero.
Use the range of tangent on the interval.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): Since du=sec^2x,dx, the bounds become u=1 and u=2; required conclusion (1 mark): int_1^2frac1u,du.
- Part (b) (3 marks): method and intermediate evidence (2 marks): An antiderivative is ln u, so the value is ln2-ln1; required conclusion (1 mark): ln2.
- Part (c) (2 marks): method and intermediate evidence (1 mark): Use the range of tangent on the interval; required conclusion (1 mark): Since 1+tan xge1 on [0,pi / 4], the denominator is never zero.
Section Two Question 10
(a) 1.
x+2y=1.
(b) -2.
x+2y=-2.
(c) y=-((x) / (2)).
Set x+2y=0.
(d) It is a set of points where solution curves have the same gradient.
Here each line x+2y=c has gradient c.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): x+2y=1; required conclusion (1 mark): 1.
- Part (b) (2 marks): method and intermediate evidence (1 mark): x+2y=-2; required conclusion (1 mark): -2.
- Part (c) (2 marks): method and intermediate evidence (1 mark): Set x+2y=0; required conclusion (1 mark): y=-((x) / (2)).
- Part (d) (2 marks): method and intermediate evidence (1 mark): Here each line x+2y=c has gradient c; required conclusion (1 mark): It is a set of points where solution curves have the same gradient.
Section Two Question 11
(a) Since zbar z=|z|²=1, division by z gives bar z=1 / z.
Use the modulus identity zbar z=|z|², the unit-modulus condition and zne0.
(b) z+frac1z=z+bar z=2operatorname(Re)(z)inmathbb R.
Replace 1 / z by bar z and use the sum of a complex number and its conjugate.
Mark allocation
- Part (a) (4 marks): method and intermediate evidence (3 marks): Use the modulus identity zbar z=|z|², the unit-modulus condition and zne0; required conclusion (1 mark): Since zbar z=|z|²=1, division by z gives bar z=1 / z.
- Part (b) (4 marks): method and intermediate evidence (3 marks): Replace 1 / z by bar z and use the sum of a complex number and its conjugate; required conclusion (1 mark): z+frac1z=z+bar z=2operatorname(Re)(z)inmathbb R.
Section Two Question 12
(a) 80.
The sample mean is an unbiased estimator.
(b) 2.
12 / √36=2.
(c) 1-Phi(1.5).
Standardise with z=(83-80) / 2=1.5.
(d) The sampling distribution becomes narrower.
The standard error decreases as the sample size increases.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): The sample mean is an unbiased estimator; required conclusion (1 mark): 80.
- Part (b) (2 marks): method and intermediate evidence (1 mark): 12 / √36=2; required conclusion (1 mark): 2.
- Part (c) (2 marks): method and intermediate evidence (1 mark): Standardise with z=(83-80) / 2=1.5; required conclusion (1 mark): 1-Phi(1.5).
- Part (d) (2 marks): method and intermediate evidence (1 mark): The standard error decreases as the sample size increases; required conclusion (1 mark): The sampling distribution becomes narrower.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Vectors | Q1, Q8 | 16 | ___ | Rework the exact vectors setup, intermediate evidence and conclusion assessed in Q1, Q8, using the worked solutions and calculator-assumed method selection and interpretation. |
| Complex Numbers | Q2, Q6, Q11 | 24 | ___ | Rework the exact complex numbers setup, intermediate evidence and conclusion assessed in Q2, Q6, Q11, using the worked solutions and calculator-assumed method selection and interpretation. |
| Statistical Inference | Q3, Q12 | 15 | ___ | Rework the exact statistical inference setup, intermediate evidence and conclusion assessed in Q3, Q12, using the worked solutions and calculator-assumed method selection and interpretation. |
| Differential Equations | Q4, Q10 | 16 | ___ | Rework the exact differential equations setup, intermediate evidence and conclusion assessed in Q4, Q10, using the worked solutions and calculator-assumed method selection and interpretation. |
| Integration | Q5, Q9 | 14 | ___ | Rework the exact integration setup, intermediate evidence and conclusion assessed in Q5, Q9, using the worked solutions and calculator-assumed method selection and interpretation. |
| Rates of Change | Q7 | 8 | ___ | Rework the exact rates of change setup, intermediate evidence and conclusion assessed in Q7, using the worked solutions and calculator-assumed method selection and interpretation. |