Skill Align Mathematics Specialist Year 12 Section One Calculator-free for WACE - 2026 Edition
Original Skill Align practice examination content
- Paper
- Section One Calculator-free Question and Response Book Showcase
- Reading
- 5 minutes reading time
- Writing
- 50 minutes
- Assessment
- 45 marks
Standard items: pens, pencils including coloured pencils, sharpener, correction fluid or tape, eraser, ruler and highlighters. Special items: nil. 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 One
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
6 marksQuestion 2
6 marksQuestion 3
6 marksQuestion 4
7 marksQuestion 5
6 marksQuestion 6
7 marksQuestion 7
7 marksWorked Solutions And Marking Guide
Section One Question 1
(a) z=2operatorname(cis)((pi) / (6)).
The modulus is 2 and the argument is pi / 6.
(b) -64.
By De Moivre's theorem, z⁶=64operatorname(cis)pi=-64.
(c) sqrt3-i.
Change the sign of the imaginary part.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): The modulus is 2 and the argument is pi / 6; required conclusion (1 mark): z=2operatorname(cis)((pi) / (6)).
- Part (b) (2 marks): method and intermediate evidence (1 mark): By De Moivre's theorem, z⁶=64operatorname(cis)pi=-64; required conclusion (1 mark): -64.
- Part (c) (2 marks): method and intermediate evidence (1 mark): Change the sign of the imaginary part; required conclusion (1 mark): sqrt3-i.
Section One Question 2
(a) ((dy) / (dx))=((3t²-3) / (2t)).
Use ((dy / dt) / (dx / dt)).
(b) (2,2) and (2,-2).
Horizontal tangents occur at t=-1 and t=1.
(c) (1,0).
At t=0, dx / dt=0 while dy / dtne0.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): Use ((dy / dt) / (dx / dt)); required conclusion (1 mark): ((dy) / (dx))=((3t²-3) / (2t)).
- Part (b) (2 marks): method and intermediate evidence (1 mark): Horizontal tangents occur at t=-1 and t=1; required conclusion (1 mark): (2,2) and (2,-2).
- Part (c) (2 marks): method and intermediate evidence (1 mark): At t=0, dx / dt=0 while dy / dtne0; required conclusion (1 mark): (1,0).
Section One Question 3
(a) overrightarrow(AB)=(2,-3,3), overrightarrow(AC)=(-2,-2,2).
Subtract the coordinates of A from B and C.
(b) A normal vector is (0,1,1).
overrightarrow(AB) × overrightarrow(AC)=(0,-10,-10), which is parallel to (0,1,1).
(c) y+z=1.
Use normal (0,1,1) and point A.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): Subtract the coordinates of A from B and C; required conclusion (1 mark): overrightarrow(AB)=(2,-3,3), overrightarrow(AC)=(-2,-2,2).
- Part (b) (2 marks): method and intermediate evidence (1 mark): overrightarrow(AB) × overrightarrow(AC)=(0,-10,-10), which is parallel to (0,1,1); required conclusion (1 mark): A normal vector is (0,1,1).
- Part (c) (2 marks): method and intermediate evidence (1 mark): Use normal (0,1,1) and point A; required conclusion (1 mark): y+z=1.
Section One Question 4
(a) f⁻¹(x)=((1+2x) / (1-x)), where xne1.
Solve y=(x-1) / (x+2) for x and use the range exclusion yne1.
(b) Substitution simplifies to x for xne-2.
Substitute (x-1) / (x+2) into (1+2x) / (1-x) and simplify.
(c) The value -2 makes f undefined, while 1 is not attained by f and makes f⁻¹ undefined.
Use the denominator restrictions and the horizontal asymptote.
Mark allocation
- Part (a) (3 marks): method and intermediate evidence (2 marks): Solve y=(x-1) / (x+2) for x and use the range exclusion yne1; required conclusion (1 mark): f⁻¹(x)=((1+2x) / (1-x)), where xne1.
- Part (b) (2 marks): method and intermediate evidence (1 mark): Substitute (x-1) / (x+2) into (1+2x) / (1-x) and simplify; required conclusion (1 mark): Substitution simplifies to x for xne-2.
- Part (c) (2 marks): method and intermediate evidence (1 mark): Use the denominator restrictions and the horizontal asymptote; required conclusion (1 mark): The value -2 makes f undefined, while 1 is not attained by f and makes f⁻¹ undefined.
Section One Question 5
(a) y=3e^(x²).
Separate variables to obtain ln y=x²+C, then use y(0)=3.
(b) 3e.
Substitute x=1 into the solution.
(c) It determines the constant of integration.
Without it, the differential equation has a family of solutions.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): Separate variables to obtain ln y=x²+C, then use y(0)=3; required conclusion (1 mark): y=3e^(x²).
- Part (b) (2 marks): method and intermediate evidence (1 mark): Substitute x=1 into the solution; required conclusion (1 mark): 3e.
- Part (c) (2 marks): method and intermediate evidence (1 mark): Without it, the differential equation has a family of solutions; required conclusion (1 mark): It determines the constant of integration.
Section One Question 6
(a) int_1^2u^(1 / 2),du.
Use du=2x,dx; the bounds x=0,1 become u=1,2.
(b) frac23(2sqrt2-1).
Apply frac23u^(3 / 2) at u=2 and u=1.
(c) The value is positive and less than 2sqrt2.
On [0,1], the integrand is non-negative and no greater than 2sqrt2.
Mark allocation
- Part (a) (3 marks): method and intermediate evidence (2 marks): Use du=2x,dx; the bounds x=0,1 become u=1,2; required conclusion (1 mark): int_1^2u^(1 / 2),du.
- Part (b) (2 marks): method and intermediate evidence (1 mark): Apply frac23u^(3 / 2) at u=2 and u=1; required conclusion (1 mark): frac23(2sqrt2-1).
- Part (c) (2 marks): method and intermediate evidence (1 mark): On [0,1], the integrand is non-negative and no greater than 2sqrt2; required conclusion (1 mark): The value is positive and less than 2sqrt2.
Section One Question 7
(a) 0.4.
Calculate s / sqrt n=3.2 / √64.
(b) (17.616,19.184).
Substitute into 18.4pm1.96(0.4) and calculate both endpoints.
(c) The observations are independent, and the population distribution and sample size justify the normal approximation.
Give one valid condition for each mark; do not repeat the random-sampling fact supplied in the stem.
Mark allocation
- Part (a) (2 marks): method and intermediate evidence (1 mark): Calculate s / sqrt n=3.2 / √64; required conclusion (1 mark): 0.4.
- Part (b) (3 marks): method and intermediate evidence (2 marks): Substitute into 18.4pm1.96(0.4) and calculate both endpoints; required conclusion (1 mark): (17.616,19.184).
- Part (c) (2 marks): method and intermediate evidence (1 mark): Give one valid condition for each mark; do not repeat the random-sampling fact supplied in the stem; required conclusion (1 mark): The observations are independent, and the population distribution and sample size justify the normal approximation.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Complex Numbers | Q1 | 6 | ___ | Rework the exact complex numbers setup, intermediate evidence and conclusion assessed in Q1, using the worked solutions and calculator-free exact working. |
| Functions and Graphs | Q2, Q4 | 13 | ___ | Rework the exact functions and graphs setup, intermediate evidence and conclusion assessed in Q2, Q4, using the worked solutions and calculator-free exact working. |
| Vectors | Q3 | 6 | ___ | Rework the exact vectors setup, intermediate evidence and conclusion assessed in Q3, using the worked solutions and calculator-free exact working. |
| Differential Equations | Q5 | 6 | ___ | Rework the exact differential equations setup, intermediate evidence and conclusion assessed in Q5, using the worked solutions and calculator-free exact working. |
| Integration | Q6 | 7 | ___ | Rework the exact integration setup, intermediate evidence and conclusion assessed in Q6, using the worked solutions and calculator-free exact working. |
| Statistical Inference | Q7 | 7 | ___ | Rework the exact statistical inference setup, intermediate evidence and conclusion assessed in Q7, using the worked solutions and calculator-free exact working. |