Skill Align ACT BSSS Specialist Mathematics T Booklet 1 Practice Assessment Pack 0 - 2026 Edition
Original Skill Align Specialist Mathematics T practice-assessment Pack 0, Booklet 1; not an official ACT BSSS examination and not endorsed by ACT BSSS or the ACT Government.
- Paper
- Booklet 1 Showcase
- Reading
- 10 minutes total planning time across both booklets
- Writing
- 115 minutes total working time across both booklets
- Assessment
- 50 marks
Suggested Skill Align conditions for the two-booklet pair: 10 minutes planning time and 115 minutes working time (125 minutes total). Scientific, graphics and CAS calculators are permitted, and a locally supplied formula sheet may be used. Stored notes or programs, internet access, AI tools, messaging and external communication are not permitted. Sufficient mathematical reasoning must be shown even when technology is used. ACT BSSS assessment is school based. Individual ACT colleges may apply different timing, calculator, formula-sheet and stored-material conditions. No public PDF download is supplied with Pack 0.
Booklet 1
Booklet 1 emphasises exact and symbolic reasoning, proof, mathematical structure and non-routine analysis. Answer all questions. Show complete working, proof, reasoning and interpretation. State exact values unless an approximation is requested, and include units and context where applicable. Suggested Skill Align conditions for the two-booklet pair: 10 minutes planning time and 115 minutes working time (125 minutes total). Scientific, graphics and CAS calculators are permitted, and a locally supplied formula sheet may be used. Stored notes or programs, internet access, AI tools, messaging and external communication are not permitted. Sufficient mathematical reasoning must be shown even when technology is used. ACT BSSS assessment is school based. Individual ACT colleges may apply different timing, calculator, formula-sheet and stored-material conditions.
Question 1
10 marksQuestion 2
10 marksQuestion 3
10 marksQuestion 4
10 marksQuestion 5
10 marksWorked Solutions And Marking Guide
Question 1
(a) 16operatorname(cis)π.
The modulus is 16 and the principal argument is π.
(b) 2operatorname(cis)(π / 4), 2operatorname(cis)(3π / 4), 2operatorname(cis)(5π / 4), 2operatorname(cis)(7π / 4).
Use arguments (π+2kpi) / 4, for k=0,1,2,3.
(c) A square centred at the origin with circumradius 2.
All four roots have modulus 2, and consecutive arguments differ by π / 2, giving four equally spaced points on a circle centred at the origin.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Finds modulus 16.
- Uses principal argument π.
- Writes 16operatorname(cis)π.
Acceptable alternatives: For Q1(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches 16operatorname(cis)π.
Do not credit by itself: Omitting or contradicting this required step: Finds modulus 16. Giving the final result without establishing: Writes 16operatorname(cis)π.
Part b (4 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 4. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Takes the fourth root of the modulus to obtain 2.
- Uses arguments (π+2kpi) / 4.
- Uses k=0,1,2,3 without omitting or repeating a root.
- Lists all four roots with arguments π / 4,3π / 4,5π / 4,7π / 4.
Acceptable alternatives: For Q1(b), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches 2operatorname(cis)(π / 4), 2operatorname(cis)(3π / 4), 2operatorname(cis)(5π / 4), 2operatorname(cis)(7π / 4).
Do not credit by itself: Omitting or contradicting this required step: Takes the fourth root of the modulus to obtain 2. Giving the final result without establishing: Lists all four roots with arguments π / 4,3π / 4,5π / 4,7π / 4.
Part c (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, include the requested units, context or justification.
Mark-by-mark evidence:
- Notes equal modulus 2.
- Notes argument spacing π / 2.
- Concludes the roots form a square centred at the origin with circumradius 2.
Acceptable alternatives: For Q1(c), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches A square centred at the origin with circumradius 2.
Do not credit by itself: Omitting or contradicting this required step: Notes equal modulus 2. Giving the final result without establishing: Concludes the roots form a square centred at the origin with circumradius 2.
Question 2
(a) -4.
1(2)+2(-1)+(-2)(2)=-4.
(b) (2,-6,-5).
Evaluate the determinant using i, j and k components.
(c) 2x-6y-5z=0.
The cross product is normal to the plane.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Forms the scalar product 1(2)+2(-1)+(-2)(2).
- Evaluates the three products with signs intact.
- Obtains -4.
Acceptable alternatives: For Q2(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches -4.
Do not credit by itself: Omitting or contradicting this required step: Forms the scalar product 1(2)+2(-1)+(-2)(2). Giving the final result without establishing: Obtains -4.
Part b (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Sets up the determinant for mathbf a × mathbf b.
- Evaluates the three signed components as 2,-6,-5.
- States mathbf a × mathbf b=(2,-6,-5).
Acceptable alternatives: For Q2(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches (2,-6,-5).
Do not credit by itself: Omitting or contradicting this required step: Sets up the determinant for mathbf a × mathbf b. Giving the final result without establishing: States mathbf a × mathbf b=(2,-6,-5).
Part c (4 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 4. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Uses mathbf a × mathbf b as a plane normal.
- Uses that the plane passes through the origin.
- Forms 2x-6y-5z=0.
- Checks both mathbf a and mathbf b satisfy the plane equation.
Acceptable alternatives: For Q2(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches 2x-6y-5z=0.
Do not credit by itself: Omitting or contradicting this required step: Uses mathbf a × mathbf b as a plane normal. Giving the final result without establishing: Checks both mathbf a and mathbf b satisfy the plane equation.
Question 3
(a) f(x)=x+1+2 / (x-1).
Use polynomial division.
(b) x=1 and y=x+1.
The term 2 / (x-1) is unbounded as xto1, giving the vertical asymptote x=1. As |x|toinfty, f(x)-(x+1)=2 / (x-1)to0, giving the oblique asymptote y=x+1.
(c) x=1pmsqrt2.
Solve f'(x)=1-2 / (x-1)²=0.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Performs polynomial division of x²+1 by x-1.
- Obtains quotient x+1 and remainder 2.
- Writes f(x)=x+1+2 / (x-1).
Acceptable alternatives: For Q3(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches f(x)=x+1+2 / (x-1).
Do not credit by itself: Omitting or contradicting this required step: Performs polynomial division of x²+1 by x-1. Giving the final result without establishing: Writes f(x)=x+1+2 / (x-1).
Part b (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, include the requested units, context or justification.
Mark-by-mark evidence:
- Uses the unbounded term 2 / (x-1) as xto1 to justify the vertical asymptote x=1.
- Uses the form from part (a) and 2 / (x-1)to0 as |x|toinfty to justify the oblique asymptote.
- States the oblique asymptote y=x+1.
Acceptable alternatives: For Q3(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches x=1 and y=x+1.
Do not credit by itself: Omitting or contradicting this required step: Uses the unbounded term 2 / (x-1) as xto1 to justify the vertical asymptote x=1. Giving the final result without establishing: States the oblique asymptote y=x+1.
Part c (4 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 4. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Differentiates to f'(x)=1-2 / (x-1)^2.
- Sets f'(x)=0.
- Solves (x-1)²=2.
- States x=1pmsqrt2, both in the domain.
Acceptable alternatives: For Q3(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches x=1pmsqrt2.
Do not credit by itself: Omitting or contradicting this required step: Differentiates to f'(x)=1-2 / (x-1)^2. Giving the final result without establishing: States x=1pmsqrt2, both in the domain.
Question 4
(a) S_1=2=1(2)(3) / 3.
Substitute n equals 1 into both sides.
(b) S_(k+1)=S_k+(k+1)(k+2)=frac(k(k+1)(k+2))3+(k+1)(k+2)=(k+1)(k+2)(k / 3+1)=frac((k+1)(k+2)(k+3))3, the required formula for n=k+1.
Use the induction hypothesis, add the next term and simplify to the formula with index k+1.
(c) S_n=frac(n(n+1)(n+2))3 for every integer ngeq1.
A complete quantified conclusion earns the mark; no repetition of the earlier proof is required.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Substitutes n=1 into the sum.
- Calculates S_1=1(2)=2.
- Checks 1(2)(3) / 3=2.
Acceptable alternatives: For Q4(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches S_1=2=1(2)(3) / 3.
Do not credit by itself: Omitting or contradicting this required step: Substitutes n=1 into the sum. Giving the final result without establishing: Checks 1(2)(3) / 3=2.
Part b (6 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 6. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, include the requested units, context or justification.
Mark-by-mark evidence:
- States the induction hypothesis S_k=k(k+1)(k+2) / 3 for arbitrary integer kgeq1.
- Writes S_(k+1)=S_k+(k+1)(k+2).
- Substitutes the induction hypothesis for S_k.
- Factors to (k+1)(k+2)(k / 3+1).
- Simplifies to (k+1)(k+2)(k+3) / 3.
- Identifies this as the required formula for n=k+1, completing the implication.
Acceptable alternatives: For Q4(b), credit an equivalent mathematically correct method only when it establishes all 6 required checkpoints and reaches S_(k+1)=S_k+(k+1)(k+2)=frac(k(k+1)(k+2))3+(k+1)(k+2)=(k+1)(k+2)(k / 3+1)=frac((k+1)(k+2)(k+3))3, the required formula for n=k+1.
Do not credit by itself: Omitting or contradicting this required step: States the induction hypothesis S_k=k(k+1)(k+2) / 3 for arbitrary integer kgeq1. Giving the final result without establishing: Identifies this as the required formula for n=k+1, completing the implication.
Part c (1 mark)
Award one mark for each numbered, task-specific statement below, to a maximum of 1. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, include the requested units, context or justification. Award the conclusion mark for the complete quantified statement alone; do not require or award repetition of the earlier proof.
Mark-by-mark evidence:
- States S_n=n(n+1)(n+2) / 3 for every integer ngeq1.
Acceptable alternatives: For Q4(c), credit an equivalent mathematically correct method only when it establishes all 1 required checkpoints and reaches S_n=frac(n(n+1)(n+2))3 for every integer ngeq1.
Do not credit by itself: Omitting or contradicting this required step: States S_n=n(n+1)(n+2) / 3 for every integer ngeq1. Giving the final result without establishing: States S_n=n(n+1)(n+2) / 3 for every integer ngeq1.
Question 5
(a) The upper curve is y=2x; the intersections are (0,0) and (2,4).
Solving 2x=x² gives x=0,2. For 0<x<2, 2x-x²=x(2-x)>0, so y=2x is above y=x^2.
(b) V=piint_0²[(2x)²-(x²)²]dx.
The outer and inner radii are 2x and x squared, respectively.
(c) 64π / 15 cubic units.
piint_0²(4x²-x⁴)dx=π[4x³ / 3-x⁵ / 5]_0²=64π / 15.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. For the final-result mark, include the requested units, context or justification.
Mark-by-mark evidence:
- Solves 2x=x² to get x=0,2.
- Identifies intersections (0,0) and (2,4).
- Establishes 2x>x² for 0<x<2.
Acceptable alternatives: For Q5(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches The upper curve is y=2x; the intersections are (0,0) and (2,4).
Do not credit by itself: Omitting or contradicting this required step: Solves 2x=x² to get x=0,2. Giving the final result without establishing: Establishes 2x>x² for 0<x<2.
Part b (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Uses washers about the x-axis.
- Identifies outer radius 2x and inner radius x^2.
- Writes V=piint_0²[(2x)²-(x²)²]dx.
Acceptable alternatives: For Q5(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches V=piint_0²[(2x)²-(x²)²]dx.
Do not credit by itself: Omitting or contradicting this required step: Uses washers about the x-axis. Giving the final result without establishing: Writes V=piint_0²[(2x)²-(x²)²]dx.
Part c (4 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 4. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, retain the requested exact form.
Mark-by-mark evidence:
- Expands the integrand to 4x²-x^4.
- Integrates to 4x³ / 3-x⁵ / 5.
- Applies limits 0 and 2.
- Obtains 64π / 15 cubic units exactly.
Acceptable alternatives: For Q5(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches 64π / 15 cubic units.
Do not credit by itself: Omitting or contradicting this required step: Expands the integrand to 4x²-x^4. Giving the final result without establishing: Obtains 64π / 15 cubic units exactly.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Unit 3: Specialist Mathematics — Complex numbers — complex-numbers: Write \(-16\) in polar form using principal argument \(\pi\) | Q1(a) | 3 | ___ | Q1(a): reproduce this exact process without the solution: Finds modulus 16; then Uses principal argument π; then Writes 16operatorname(cis)π. |
| Unit 3: Specialist Mathematics — Complex numbers — complex-numbers: Find all four roots in polar form | Q1(b) | 4 | ___ | Q1(b): reproduce this exact process without the solution: Takes the fourth root of the modulus to obtain 2; then Uses arguments (π+2kpi) / 4; then Uses k=0,1,2,3 without omitting or repeating a root; then Lists all four roots with arguments π / 4,3π / 4,5π / 4,7π / 4. |
| Unit 3: Specialist Mathematics — Complex numbers — complex-numbers: Describe and justify the quadrilateral formed by the roots, including its centre and circumradius | Q1(c) | 3 | ___ | Q1(c): reproduce this exact process without the solution: Notes equal modulus 2; then Notes argument spacing π / 2; then Concludes the roots form a square centred at the origin with circumradius 2. |
| Unit 3: Specialist Mathematics — Vectors in three dimensions — vectors-three-dimensions: Find \(\mathbf a\cdot\mathbf b\) | Q2(a) | 3 | ___ | Q2(a): reproduce this exact process without the solution: Forms the scalar product 1(2)+2(-1)+(-2)(2); then Evaluates the three products with signs intact; then Obtains -4. |
| Unit 3: Specialist Mathematics — Vectors in three dimensions — vectors-three-dimensions: Find \(\mathbf a\times\mathbf b\) | Q2(b) | 3 | ___ | Q2(b): reproduce this exact process without the solution: Sets up the determinant for mathbf a × mathbf b; then Evaluates the three signed components as 2,-6,-5; then States mathbf a × mathbf b=(2,-6,-5). |
| Unit 3: Specialist Mathematics — Vectors in three dimensions — vectors-three-dimensions: Hence find an equation of the plane through the origin containing \(\mathbf a\) and \(\mathbf b\) | Q2(c) | 4 | ___ | Q2(c): reproduce this exact process without the solution: Uses mathbf a × mathbf b as a plane normal; then Uses that the plane passes through the origin; then Forms 2x-6y-5z=0; then Checks both mathbf a and mathbf b satisfy the plane equation. |
| Unit 3: Specialist Mathematics — Functions and sketching graphs — functions-sketching-graphs: Express \(f(x)\) in the form \(x+1+\frac{2}{x-1}\) | Q3(a) | 3 | ___ | Q3(a): reproduce this exact process without the solution: Performs polynomial division of x²+1 by x-1; then Obtains quotient x+1 and remainder 2; then Writes f(x)=x+1+2 / (x-1). |
| Unit 3: Specialist Mathematics — Functions and sketching graphs — functions-sketching-graphs: Using part (a), identify and justify the vertical and oblique asymptotes | Q3(b) | 3 | ___ | Q3(b): reproduce this exact process without the solution: Uses the unbounded term 2 / (x-1) as xto1 to justify the vertical asymptote x=1; then Uses the form from part (a) and 2 / (x-1)to0 as |x|toinfty to justify the oblique asymptote; then States the oblique asymptote y=x+1. |
| Unit 3: Specialist Mathematics — Functions and sketching graphs — functions-sketching-graphs: Find the stationary x-values | Q3(c) | 4 | ___ | Q3(c): reproduce this exact process without the solution: Differentiates to f'(x)=1-2 / (x-1)²; then Sets f'(x)=0; then Solves (x-1)²=2; then States x=1pmsqrt2, both in the domain. |
| Unit 2: Specialist Mathematics — Real and complex numbers — proof: Verify \(S_n=\frac{n(n+1)(n+2)}3\) for \(n=1\) | Q4(a) | 3 | ___ | Q4(a): reproduce this exact process without the solution: Substitutes n=1 into the sum; then Calculates S_1=1(2)=2; then Checks 1(2)(3) / 3=2. |
| Unit 2: Specialist Mathematics — Real and complex numbers — proof: Assume the result for an arbitrary integer \(k\geq1\), and prove it for \(n=k+1\), showing the algebra | Q4(b) | 6 | ___ | Q4(b): reproduce this exact process without the solution: States the induction hypothesis S_k=k(k+1)(k+2) / 3 for arbitrary integer kgeq1; then Writes S_(k+1)=S_k+(k+1)(k+2); then Substitutes the induction hypothesis for S_k; then Factors to (k+1)(k+2)(k / 3+1); then Simplifies to (k+1)(k+2)(k+3) / 3; then Identifies this as the required formula for n=k+1, completing the implication. |
| Unit 2: Specialist Mathematics — Real and complex numbers — proof: State the identity and the positive-integer domain for which it has now been proved | Q4(c) | 1 | ___ | Q4(c): reproduce this exact process without the solution: States S_n=n(n+1)(n+2) / 3 for every integer ngeq1. |
| Unit 4: Specialist Mathematics — Integration and applications of integration — integration-applications: Find the two intersection points and identify the upper curve between them. Justify your choice of upper curve | Q5(a) | 3 | ___ | Q5(a): reproduce this exact process without the solution: Solves 2x=x² to get x=0,2; then Identifies intersections (0,0) and (2,4); then Establishes 2x>x² for 0<x<2. |
| Unit 4: Specialist Mathematics — Integration and applications of integration — integration-applications: Write a washer integral for the volume of the solid | Q5(b) | 3 | ___ | Q5(b): reproduce this exact process without the solution: Uses washers about the x-axis; then Identifies outer radius 2x and inner radius x²; then Writes V=piint_0²[(2x)²-(x²)²]dx. |
| Unit 4: Specialist Mathematics — Integration and applications of integration — integration-applications: Evaluate the exact volume | Q5(c) | 4 | ___ | Q5(c): reproduce this exact process without the solution: Expands the integrand to 4x²-x⁴; then Integrates to 4x³ / 3-x⁵ / 5; then Applies limits 0 and 2; then Obtains 64π / 15 cubic units exactly. |