Skill Align ACT BSSS Specialist Methods T Booklet 1 Practice Assessment Pack 0 - 2026 Edition
Original Skill Align Specialist Methods 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
- Writing
- 65 minutes
- Assessment
- 50 marks
Suggested Skill Align Booklet 1 conditions: 10 minutes reading time and 65 minutes working time. This is a restricted-technology booklet: a non-CAS scientific calculator may be used, but CAS, graphics-calculator algebra, stored notes or programs, internet access and external communication are prohibited. A locally supplied formula or reference sheet may be used only when authorised by the administering college. ACT BSSS assessment is school based. The administering college may vary timing, technology, stored-note, program and reference-material conditions. No public PDF download is supplied with Pack 0.
Booklet 1
Booklet 1 assesses exact reasoning, proof, mathematical structure, functions, trigonometry, sequences and calculus under restricted technology. Answer all questions. Show complete working, reasoning and interpretation. State exact values unless an approximation is requested, and include units and context where applicable. Suggested Skill Align Booklet 1 conditions: 10 minutes reading time and 65 minutes working time. This is a restricted-technology booklet: a non-CAS scientific calculator may be used, but CAS, graphics-calculator algebra, stored notes or programs, internet access and external communication are prohibited. A locally supplied formula or reference sheet may be used only when authorised by the administering college. ACT BSSS assessment is school based. The administering college may vary timing, technology, stored-note, program and reference-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) D(4)=8e^(0.6)approx14.58.
Substitute t=4, retain the exponential value and round only the final demand.
(b) t=ln2 / 0.15approx4.62 months.
Set the model equal to twice its initial value and solve the resulting exponential equation.
(c) The instantaneous relative growth rate is 0.15 per month, or 15% per month; finite demand or capacity makes indefinite exponential growth unrealistic.
Connect the coefficient in the exponent to relative growth and challenge the constant-rate assumption.
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, carry unrounded values until the final stated accuracy.
Mark-by-mark evidence:
- Substitutes t=4 to obtain D(4)=8e^(0.15(4)).
- Simplifies the exponent to obtain 8e^(0.6).
- Evaluates and rounds the final value to 14.58.
Acceptable alternatives: For Q1(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches D(4)=8e^(0.6)approx14.58.
Do not credit by itself: Omitting or contradicting this required step: Substitutes t=4 to obtain D(4)=8e^(0.15(4)). Giving the final result without establishing: Evaluates and rounds the final value to 14.58.
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, carry unrounded values until the final stated accuracy.
Mark-by-mark evidence:
- Sets 8e^(0.15t)=16 and divides by 8.
- Takes logarithms to obtain 0.15t=ln2.
- States t=ln2 / 0.15approx4.62 months.
Acceptable alternatives: For Q1(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches t=ln2 / 0.15approx4.62 months.
Do not credit by itself: Omitting or contradicting this required step: Sets 8e^(0.15t)=16 and divides by 8. Giving the final result without establishing: States t=ln2 / 0.15approx4.62 months.
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, include the requested units, context or justification.
Mark-by-mark evidence:
- Identifies 0.15 as the continuous relative growth-rate parameter.
- States the associated rate as 15% per month.
- Identifies a finite-demand, resource or capacity constraint.
- Explains why that constraint invalidates indefinite constant relative growth.
Acceptable alternatives: For Q1(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches The instantaneous relative growth rate is 0.15 per month, or 15% per month; finite demand or capacity makes indefinite exponential growth unrealistic.
Do not credit by itself: Omitting or contradicting this required step: Identifies 0.15 as the continuous relative growth-rate parameter. Giving the final result without establishing: Explains why that constraint invalidates indefinite constant relative growth.
Question 2
(a) f'(x)=xe^(-x)(2-x).
Apply the product and chain rules, then factorise.
(b) The stationary points are (0,0) and (2,4e⁻²).
Solve the factorised derivative equation and evaluate the original function.
(c) f increases on (0,2), decreases on (2,infty), and tends to 0; hence the global maximum is 4e⁻² at x=2.
Use derivative signs, the endpoint and limiting behaviour.
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:
- Applies the product rule to x^2e^(-x).
- Uses d(e^(-x)) / dx=-e^(-x) to obtain 2xe^(-x)-x^2e^(-x).
- Factorises to f'(x)=xe^(-x)(2-x).
Acceptable alternatives: For Q2(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches f'(x)=xe^(-x)(2-x).
Do not credit by itself: Omitting or contradicting this required step: Applies the product rule to x^2e^(-x). Giving the final result without establishing: Factorises to f'(x)=xe^(-x)(2-x).
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 xe^(-x)(2-x)=0.
- Uses e^(-x)>0 to obtain x=0 or x=2.
- Evaluates f to state (0,0) and (2,4e⁻²).
Acceptable alternatives: For Q2(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches The stationary points are (0,0) and (2,4e⁻²).
Do not credit by itself: Omitting or contradicting this required step: Sets xe^(-x)(2-x)=0. Giving the final result without establishing: Evaluates f to state (0,0) and (2,4e⁻²).
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, include the requested units, context or justification.
Mark-by-mark evidence:
- Shows f'(x)>0 for 0<x<2.
- Shows f'(x)<0 for x>2.
- Checks f(0)=0 and lim_(xtoinfty)x^2e^(-x)=0.
- Concludes that the global maximum is f(2)=4e⁻².
Acceptable alternatives: For Q2(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches f increases on (0,2), decreases on (2,infty), and tends to 0; hence the global maximum is 4e⁻² at x=2.
Do not credit by itself: Omitting or contradicting this required step: Shows f'(x)>0 for 0<x<2. Giving the final result without establishing: Concludes that the global maximum is f(2)=4e⁻².
Question 3
(a) S'(r)=4π r-1000π / r^2.
Write the reciprocal term as a negative power before differentiating.
(b) r=√3(250) cm.
Solve the positive stationary-point equation.
(c) h=2sqrt(3)(250)approx12.60 cm; Stoinfty as rto0^+ and as rtoinfty, so the stationary point is the global minimum.
Use the volume constraint and endpoint behaviour on the positive domain.
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:
- Rewrites 1000π / r as 1000π r⁻¹.
- Differentiates 2π r² to 4π r.
- Obtains S'(r)=4π r-1000π r⁻².
Acceptable alternatives: For Q3(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches S'(r)=4π r-1000π / r^2.
Do not credit by itself: Omitting or contradicting this required step: Rewrites 1000π / r as 1000π r⁻¹. Giving the final result without establishing: Obtains S'(r)=4π r-1000π r⁻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. For the final-result mark, retain the requested exact form; include the requested units, context or justification.
Mark-by-mark evidence:
- Sets 4π r-1000π / r²=0.
- Multiplies by r² / (4π) to obtain r³=250.
- Uses r>0 to state r=√3(250) cm.
Acceptable alternatives: For Q3(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches r=√3(250) cm.
Do not credit by itself: Omitting or contradicting this required step: Sets 4π r-1000π / r²=0. Giving the final result without establishing: Uses r>0 to state r=√3(250) cm.
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, include the requested units, context or justification.
Mark-by-mark evidence:
- Uses π r^2h=500π to obtain h=500 / r^2.
- Uses r³=250 to simplify h=2r=2sqrt(3)(250)approx12.60 cm.
- Shows S(r)toinfty as rto0^+.
- Shows S(r)toinfty as rtoinfty and concludes the sole stationary point is the global minimum.
Acceptable alternatives: For Q3(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches h=2sqrt(3)(250)approx12.60 cm; Stoinfty as rto0^+ and as rtoinfty, so the stationary point is the global minimum.
Do not credit by itself: Omitting or contradicting this required step: Uses π r^2h=500π to obtain h=500 / r^2. Giving the final result without establishing: Shows S(r)toinfty as rtoinfty and concludes the sole stationary point is the global minimum.
Question 4
(a) F'(x)=3-x², so the stationary value is x=sqrt3.
Apply the fundamental theorem and solve on the stated interval.
(b) F(x)=3x-x³ / 3-8 / 3 and F(sqrt3)=2sqrt3-8 / 3.
Evaluate the prescribed polynomial antiderivative at variable and fixed limits.
(c) F increases on [1,sqrt3], decreases on [sqrt3,3], and the endpoint values are F(1)=0 and F(3)=-8 / 3; hence F(sqrt3) is the global maximum.
Use the derivative sign and both endpoints rather than an unprescribed integration technique.
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:
- Uses the fundamental theorem to state F'(x)=3-x^2.
- Sets 3-x²=0.
- Uses 1leq xleq3 to obtain x=sqrt3.
Acceptable alternatives: For Q4(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches F'(x)=3-x², so the stationary value is x=sqrt3.
Do not credit by itself: Omitting or contradicting this required step: Uses the fundamental theorem to state F'(x)=3-x^2. Giving the final result without establishing: Uses 1leq xleq3 to obtain x=sqrt3.
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, retain the requested exact form.
Mark-by-mark evidence:
- Uses the antiderivative 3t-t³ / 3.
- Evaluates the lower limit to obtain F(x)=3x-x³ / 3-8 / 3.
- Substitutes x=sqrt3 to obtain 2sqrt3-8 / 3.
Acceptable alternatives: For Q4(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches F(x)=3x-x³ / 3-8 / 3 and F(sqrt3)=2sqrt3-8 / 3.
Do not credit by itself: Omitting or contradicting this required step: Uses the antiderivative 3t-t³ / 3. Giving the final result without establishing: Substitutes x=sqrt3 to obtain 2sqrt3-8 / 3.
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, include the requested units, context or justification.
Mark-by-mark evidence:
- Shows F'(x)>0 for 1leq x<sqrt3.
- Shows F'(x)<0 for sqrt3<xleq3.
- Evaluates F(1)=0 and F(3)=-8 / 3.
- Compares the stationary and endpoint values to prove the global maximum occurs at x=sqrt3.
Acceptable alternatives: For Q4(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches F increases on [1,sqrt3], decreases on [sqrt3,3], and the endpoint values are F(1)=0 and F(3)=-8 / 3; hence F(sqrt3) is the global maximum.
Do not credit by itself: Omitting or contradicting this required step: Shows F'(x)>0 for 1leq x<sqrt3. Giving the final result without establishing: Compares the stationary and endpoint values to prove the global maximum occurs at x=sqrt3.
Question 5
(a) Since f'(x)=1+1 / x>0 for x>0, f is strictly increasing and one-to-one.
Use derivative sign to establish strict monotonicity.
(b) The unique solution is x=1.
Verify the solution and use strict monotonicity for uniqueness.
(c) g(1)=1 and g'(1)=1 / 2.
Use the inverse relation and differentiate f(g(y))=y.
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:
- Differentiates to obtain f'(x)=1+1 / x.
- Shows 1+1 / x>0 for every x>0.
- Concludes that f is strictly increasing and therefore one-to-one.
Acceptable alternatives: For Q5(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches Since f'(x)=1+1 / x>0 for x>0, f is strictly increasing and one-to-one.
Do not credit by itself: Omitting or contradicting this required step: Differentiates to obtain f'(x)=1+1 / x. Giving the final result without establishing: Concludes that f is strictly increasing and therefore one-to-one.
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:
- Substitutes x=1 to verify 1+ln1=1.
- Uses the strict increase proved in part (a).
- Concludes that x=1 is the unique solution.
Acceptable alternatives: For Q5(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches The unique solution is x=1.
Do not credit by itself: Omitting or contradicting this required step: Substitutes x=1 to verify 1+ln1=1. Giving the final result without establishing: Concludes that x=1 is the unique solution.
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 f(1)=1 to state g(1)=1.
- Differentiates f(g(y))=y to obtain f'(g(y))g'(y)=1.
- Substitutes y=1 and g(1)=1.
- Uses f'(1)=2 to obtain g'(1)=1 / 2.
Acceptable alternatives: For Q5(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches g(1)=1 and g'(1)=1 / 2.
Do not credit by itself: Omitting or contradicting this required step: Uses f(1)=1 to state g(1)=1. Giving the final result without establishing: Uses f'(1)=2 to obtain g'(1)=1 / 2.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Unit 2 — Exponential functions — Evaluate an exponential model at a stated time | Q1(a) | 3 | ___ | Q1(a): reproduce this exact process without the solution: Substitutes t=4 to obtain D(4)=8e^(0.15(4)); then Simplifies the exponent to obtain 8e^(0.6); then Evaluates and rounds the final value to 14.58. |
| Unit 3 — The logarithmic function — Solve an exponential doubling-time equation | Q1(b) | 3 | ___ | Q1(b): reproduce this exact process without the solution: Sets 8e^(0.15t)=16 and divides by 8; then Takes logarithms to obtain 0.15t=ln2; then States t=ln2 / 0.15approx4.62 months. |
| Unit 2 — Exponential functions — Interpret a continuous growth parameter and critique extrapolation | Q1(c) | 4 | ___ | Q1(c): reproduce this exact process without the solution: Identifies 0.15 as the continuous relative growth-rate parameter; then States the associated rate as 15% per month; then Identifies a finite-demand, resource or capacity constraint; then Explains why that constraint invalidates indefinite constant relative growth. |
| Unit 3 — Further differentiation and applications — Differentiate and factorise an exponential product | Q2(a) | 3 | ___ | Q2(a): reproduce this exact process without the solution: Applies the product rule to x^2e^(-x); then Uses d(e^(-x)) / dx=-e^(-x) to obtain 2xe^(-x)-x^2e^(-x); then Factorises to f'(x)=xe^(-x)(2-x). |
| Unit 3 — Further differentiation and applications — Solve a factorised stationary-point equation | Q2(b) | 3 | ___ | Q2(b): reproduce this exact process without the solution: Sets xe^(-x)(2-x)=0; then Uses e^(-x)>0 to obtain x=0 or x=2; then Evaluates f to state (0,0) and (2,4e⁻²). |
| Unit 3 — Further differentiation and applications — Justify a global maximum on an unbounded domain | Q2(c) | 4 | ___ | Q2(c): reproduce this exact process without the solution: Shows f'(x)>0 for 0<x<2; then Shows f'(x)<0 for x>2; then Checks f(0)=0 and lim_(xtoinfty)x^2e^(-x)=0; then Concludes that the global maximum is f(2)=4e⁻². |
| Unit 3 — Further differentiation and applications — Differentiate a reciprocal surface-area model | Q3(a) | 3 | ___ | Q3(a): reproduce this exact process without the solution: Rewrites 1000π / r as 1000π r⁻¹; then Differentiates 2π r² to 4π r; then Obtains S'(r)=4π r-1000π r⁻². |
| Unit 3 — Further differentiation and applications — Solve a constrained stationary-point equation exactly | Q3(b) | 3 | ___ | Q3(b): reproduce this exact process without the solution: Sets 4π r-1000π / r²=0; then Multiplies by r² / (4π) to obtain r³=250; then Uses r>0 to state r=√3(250) cm. |
| Unit 3 — Further differentiation and applications — Recover a constrained dimension and prove global minimality | Q3(c) | 4 | ___ | Q3(c): reproduce this exact process without the solution: Uses π r^2h=500π to obtain h=500 / r²; then Uses r³=250 to simplify h=2r=2sqrt(3)(250)approx12.60 cm; then Shows S(r)toinfty as rto0^+; then Shows S(r)toinfty as rtoinfty and concludes the sole stationary point is the global minimum. |
| Unit 3 — Integrals — Apply the fundamental theorem to an accumulation function | Q4(a) | 3 | ___ | Q4(a): reproduce this exact process without the solution: Uses the fundamental theorem to state F'(x)=3-x²; then Sets 3-x²=0; then Uses 1leq xleq3 to obtain x=sqrt3. |
| Unit 3 — Integrals — Evaluate a variable-limit polynomial integral exactly | Q4(b) | 3 | ___ | Q4(b): reproduce this exact process without the solution: Uses the antiderivative 3t-t³ / 3; then Evaluates the lower limit to obtain F(x)=3x-x³ / 3-8 / 3; then Substitutes x=sqrt3 to obtain 2sqrt3-8 / 3. |
| Unit 3 — Integrals — Establish a global extremum of an accumulation function | Q4(c) | 4 | ___ | Q4(c): reproduce this exact process without the solution: Shows F'(x)>0 for 1leq x<sqrt3; then Shows F'(x)<0 for sqrt3<xleq3; then Evaluates F(1)=0 and F(3)=-8 / 3; then Compares the stationary and endpoint values to prove the global maximum occurs at x=sqrt3. |
| Unit 1 — Functions and graphs — Prove that a function is one-to-one | Q5(a) | 3 | ___ | Q5(a): reproduce this exact process without the solution: Differentiates to obtain f'(x)=1+1 / x; then Shows 1+1 / x>0 for every x>0; then Concludes that f is strictly increasing and therefore one-to-one. |
| Unit 3 — The logarithmic function — Solve and justify a logarithmic equation structurally | Q5(b) | 3 | ___ | Q5(b): reproduce this exact process without the solution: Substitutes x=1 to verify 1+ln1=1; then Uses the strict increase proved in part (a); then Concludes that x=1 is the unique solution. |
| Unit 1 — Functions and graphs — Differentiate an inverse relation at a known point | Q5(c) | 4 | ___ | Q5(c): reproduce this exact process without the solution: Uses f(1)=1 to state g(1)=1; then Differentiates f(g(y))=y to obtain f'(g(y))g'(y)=1; then Substitutes y=1 and g(1)=1; then Uses f'(1)=2 to obtain g'(1)=1 / 2. |