Skill Align ACT BSSS Mathematical Methods T Booklet 1 Practice Assessment Pack 0 - 2026 Edition
Original Skill Align Mathematical 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, algebraic fluency and connected calculus and probability arguments 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) 3x²-6x-9.
Differentiate term by term.
(b) x=-1 and x=3.
Solve 3(x-3)(x+1)=0.
(c) Local maximum at (-1,10); local minimum at (3,-22).
Use f''(x)=6x-6 and substitute into f.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Differentiates x³ to 3x^2.
- Differentiates -3x²-9x+5 to -6x-9.
- Combines the terms to obtain f'(x)=3x²-6x-9.
Acceptable alternatives: For Q1(a), award up to 3 marks for an independently valid method that answers the stated task and reaches 3x²-6x-9. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Differentiates x³ to 3x^2. Giving the final result without establishing: Combines the terms to obtain f'(x)=3x²-6x-9.
Part b (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. 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 if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Sets the stationary-point equation 3(x-3)(x+1)=0.
- Solves x+1=0 to obtain x=-1.
- Solves x-3=0 to obtain x=3.
Acceptable alternatives: For Q1(b), award up to 3 marks for an independently valid method that answers the stated task and reaches x=-1 and x=3. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Sets the stationary-point equation 3(x-3)(x+1)=0. Giving the final result without establishing: Solves x-3=0 to obtain x=3.
Part c (4 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 4. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. 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 if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Obtains the stationary-point coordinates (-1,10) and (3,-22).
- Uses a valid first-derivative sign analysis or second-derivative test.
- Justifies the local maximum at (-1,10) from a positive-to-negative derivative change or negative second derivative.
- Justifies the local minimum at (3,-22) from a negative-to-positive derivative change or positive second derivative.
Acceptable alternatives: For Q1(c), award up to 4 marks for an independently valid method that answers the stated task and reaches Local maximum at (-1,10); local minimum at (3,-22). Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit a derivative sign analysis or valid second-derivative test only when it establishes the requested local classification.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Obtains the stationary-point coordinates (-1,10) and (3,-22). Giving the final result without establishing: Justifies the local minimum at (3,-22) from a negative-to-positive derivative change or positive second derivative.
Question 2
(a) 8x(x²+2)^3.
Apply the chain rule.
(b) 216.
8(1)(3³)=216.
(c) y-81=216(x-1).
h(1)=3⁴=81 and the gradient is 216.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Differentiates the outer fourth power to 4(x²+2)^3.
- Multiplies by the inner derivative 2x.
- Obtains h'(x)=8x(x²+2)^3.
Acceptable alternatives: For Q2(a), award up to 3 marks for an independently valid method that answers the stated task and reaches 8x(x²+2)^3. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Differentiates the outer fourth power to 4(x²+2)^3. Giving the final result without establishing: Obtains h'(x)=8x(x²+2)^3.
Part b (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. 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 if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Substitutes x=1 into 8x(x²+2)^3.
- Evaluates x²+2=3 and hence 8(1)(3³).
- Obtains h'(1)=216.
Acceptable alternatives: For Q2(b), award up to 3 marks for an independently valid method that answers the stated task and reaches 216. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Substitutes x=1 into 8x(x²+2)^3. Giving the final result without establishing: Obtains h'(1)=216.
Part c (4 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 4. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. 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 if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Calculates h(1)=3⁴=81.
- Uses the gradient h'(1)=216.
- Writes the tangent as y-81=216(x-1).
- Uses the point (1,81) and slope 216 consistently in an equivalent tangent equation.
Acceptable alternatives: For Q2(c), award up to 4 marks for an independently valid method that answers the stated task and reaches y-81=216(x-1). Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Calculates h(1)=3⁴=81. Giving the final result without establishing: Uses the point (1,81) and slope 216 consistently in an equivalent tangent equation.
Question 3
(a) int_0⁶(6x-x²),dx.
The curve is non-negative on the interval.
(b) 36 square units.
[3x²-x³ / 3]_0⁶=36.
(c) Maximum 9 units; average 6 units.
The vertex is at x=3 and average height is area divided by width.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Identifies the intersection bounds x=0 and x=6.
- Uses the height 6x-x² above the x-axis.
- Writes int_0⁶(6x-x²),dx.
Acceptable alternatives: For Q3(a), award up to 3 marks for an independently valid method that answers the stated task and reaches int_0⁶(6x-x²),dx. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit an algebraically equivalent bounded integral or antiderivative only when the integrand, bounds, sign and required units are preserved.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Identifies the intersection bounds x=0 and x=6. Giving the final result without establishing: Writes int_0⁶(6x-x²),dx.
Part b (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. 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 if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Finds the antiderivative 3x²-x³ / 3.
- Evaluates [3x²-x³ / 3]_0^6.
- Obtains an area of 36 square units.
Acceptable alternatives: For Q3(b), award up to 3 marks for an independently valid method that answers the stated task and reaches 36 square units. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit an algebraically equivalent bounded integral or antiderivative only when the integrand, bounds, sign and required units are preserved.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Finds the antiderivative 3x²-x³ / 3. Giving the final result without establishing: Obtains an area of 36 square units.
Part c (4 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 4. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. 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 if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Locates the vertex at x=3.
- Calculates the maximum height 6(3)-3²=9 units.
- Uses average height =area / width=36 / 6.
- States the average height as 6 units and distinguishes it from the maximum.
Acceptable alternatives: For Q3(c), award up to 4 marks for an independently valid method that answers the stated task and reaches Maximum 9 units; average 6 units. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Locates the vertex at x=3. Giving the final result without establishing: States the average height as 6 units and distinguishes it from the maximum.
Question 4
(a) 0.60.
Add the last two probabilities.
(b) 1.7.
0(0.1)+1(0.3)+2(0.4)+3(0.2)=1.7.
(c) 0.81.
E(X²)=3.70, so variance is 3.70-1.7^2.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Identifies Xgeq2 as the outcomes 2 and 3.
- Adds 0.40+0.20.
- Obtains P(Xgeq2)=0.60.
Acceptable alternatives: For Q4(a), award up to 3 marks for an independently valid method that answers the stated task and reaches 0.60. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit direct summation or the complement of the excluded outcomes only when it represents the event correctly.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Identifies Xgeq2 as the outcomes 2 and 3. Giving the final result without establishing: Obtains P(Xgeq2)=0.60.
Part b (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. 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 if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Writes E(X)=sum xP(X=x).
- Substitutes 0(0.10)+1(0.30)+2(0.40)+3(0.20).
- Obtains E(X)=1.7.
Acceptable alternatives: For Q4(b), award up to 3 marks for an independently valid method that answers the stated task and reaches 1.7. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit an equivalent weighted-sum, distribution-moment, symmetry or linear-transformation method only when it establishes every requested moment and interpretation.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Writes E(X)=sum xP(X=x). Giving the final result without establishing: Obtains E(X)=1.7.
Part c (4 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 4. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. 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 if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Calculates E(X²)=0²(0.10)+1²(0.30)+2²(0.40)+3²(0.20)=3.70.
- Uses operatorname(Var)(X)=E(X²)-[E(X)]^2.
- Substitutes 3.70-1.7^2.
- Obtains operatorname(Var)(X)=0.81.
Acceptable alternatives: For Q4(c), award up to 4 marks for an independently valid method that answers the stated task and reaches 0.81. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit an equivalent weighted-sum, distribution-moment, symmetry or linear-transformation method only when it establishes every requested moment and interpretation.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Calculates E(X²)=0²(0.10)+1²(0.30)+2²(0.40)+3²(0.20)=3.70. Giving the final result without establishing: Obtains operatorname(Var)(X)=0.81.
Question 5
(a) Domain x>1 / 3; asymptote x=1 / 3.
The logarithm input must be positive.
(b) 3 / (3x-1).
Differentiate the logarithm using the chain rule.
(c) x=(e+1) / 3.
ln(3x-1)=1, so 3x-1=e.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Requires 3x-1>0 for the logarithm.
- Solves the inequality to obtain the domain x>1 / 3.
- States the vertical asymptote x=1 / 3.
Acceptable alternatives: For Q5(a), award up to 3 marks for an independently valid method that answers the stated task and reaches Domain x>1 / 3; asymptote x=1 / 3. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Requires 3x-1>0 for the logarithm. Giving the final result without establishing: States the vertical asymptote x=1 / 3.
Part b (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. 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 if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Uses d(ln u) / dx=u' / u.
- Identifies u=3x-1 and u'=3.
- Obtains g'(x)=3 / (3x-1).
Acceptable alternatives: For Q5(b), award up to 3 marks for an independently valid method that answers the stated task and reaches 3 / (3x-1). Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Uses d(ln u) / dx=u' / u. Giving the final result without establishing: Obtains g'(x)=3 / (3x-1).
Part c (4 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 4. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. 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 rather than replacing it with a decimal.
Mark-by-mark evidence:
- Rearranges 2+ln(3x-1)=3 to ln(3x-1)=1.
- Exponentiates to obtain 3x-1=e.
- Solves 3x=e+1.
- Obtains the exact solution x=(e+1) / 3, which lies in the domain.
Acceptable alternatives: For Q5(c), award up to 4 marks for an independently valid method that answers the stated task and reaches x=(e+1) / 3. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Rearranges 2+ln(3x-1)=3 to ln(3x-1)=1. Giving the final result without establishing: Obtains the exact solution x=(e+1) / 3, which lies in the domain.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Unit 3 — Further differentiation and applications — Find f'(x) | Q1(a) | 3 | ___ | Q1(a): practise this worked approach, or an independently valid equivalent method: Differentiates x³ to 3x²; then Differentiates -3x²-9x+5 to -6x-9; then Combines the terms to obtain f'(x)=3x²-6x-9. |
| Unit 3 — Further differentiation and applications — Find the x-coordinates of the stationary points | Q1(b) | 3 | ___ | Q1(b): practise this worked approach, or an independently valid equivalent method: Sets the stationary-point equation 3(x-3)(x+1)=0; then Solves x+1=0 to obtain x=-1; then Solves x-3=0 to obtain x=3. |
| Unit 3 — Further differentiation and applications — Classify both stationary points and give their coordinates | Q1(c) | 4 | ___ | Q1(c): practise this worked approach, or an independently valid equivalent method: Obtains the stationary-point coordinates (-1,10) and (3,-22); then Uses a valid first-derivative sign analysis or second-derivative test; then Justifies the local maximum at (-1,10) from a positive-to-negative derivative change or negative second derivative; then Justifies the local minimum at (3,-22) from a negative-to-positive derivative change or positive second derivative. |
| Unit 3 — Further differentiation and applications — Find h'(x) | Q2(a) | 3 | ___ | Q2(a): practise this worked approach, or an independently valid equivalent method: Differentiates the outer fourth power to 4(x²+2)³; then Multiplies by the inner derivative 2x; then Obtains h'(x)=8x(x²+2)^3. |
| Unit 3 — Further differentiation and applications — Find h'(1) | Q2(b) | 3 | ___ | Q2(b): practise this worked approach, or an independently valid equivalent method: Substitutes x=1 into 8x(x²+2)³; then Evaluates x²+2=3 and hence 8(1)(3³); then Obtains h'(1)=216. |
| Unit 3 — Further differentiation and applications — Find the equation of the tangent at x=1 | Q2(c) | 4 | ___ | Q2(c): practise this worked approach, or an independently valid equivalent method: Calculates h(1)=3⁴=81; then Uses the gradient h'(1)=216; then Writes the tangent as y-81=216(x-1); then Uses the point (1,81) and slope 216 consistently in an equivalent tangent equation. |
| Unit 3 — Integrals — Write a definite integral for the area | Q3(a) | 3 | ___ | Q3(a): practise this worked approach, or an independently valid equivalent method: Identifies the intersection bounds x=0 and x=6; then Uses the height 6x-x² above the x-axis; then Writes int_0⁶(6x-x²),dx. |
| Unit 3 — Integrals — Evaluate the area | Q3(b) | 3 | ___ | Q3(b): practise this worked approach, or an independently valid equivalent method: Finds the antiderivative 3x²-x³ / 3; then Evaluates [3x²-x³ / 3]_0⁶; then Obtains an area of 36 square units. |
| Unit 3 — Integrals — Find the maximum and average heights | Q3(c) | 4 | ___ | Q3(c): practise this worked approach, or an independently valid equivalent method: Locates the vertex at x=3; then Calculates the maximum height 6(3)-3²=9 units; then Uses average height =area / width=36 / 6; then States the average height as 6 units and distinguishes it from the maximum. |
| Unit 3 — Discrete random variables — Find P(X at least 2) | Q4(a) | 3 | ___ | Q4(a): practise this worked approach, or an independently valid equivalent method: Identifies Xgeq2 as the outcomes 2 and 3; then Adds 0.40+0.20; then Obtains P(Xgeq2)=0.60. |
| Unit 3 — Discrete random variables — Find E(X) | Q4(b) | 3 | ___ | Q4(b): practise this worked approach, or an independently valid equivalent method: Writes E(X)=sum xP(X=x); then Substitutes 0(0.10)+1(0.30)+2(0.40)+3(0.20); then Obtains E(X)=1.7. |
| Unit 3 — Discrete random variables — Find Var(X) | Q4(c) | 4 | ___ | Q4(c): practise this worked approach, or an independently valid equivalent method: Calculates E(X²)=0²(0.10)+1²(0.30)+2²(0.40)+3²(0.20)=3.70; then Uses operatorname(Var)(X)=E(X²)-[E(X)]²; then Substitutes 3.70-1.7²; then Obtains operatorname(Var)(X)=0.81. |
| Unit 4 — The logarithmic function — State the domain and vertical asymptote | Q5(a) | 3 | ___ | Q5(a): practise this worked approach, or an independently valid equivalent method: Requires 3x-1>0 for the logarithm; then Solves the inequality to obtain the domain x>1 / 3; then States the vertical asymptote x=1 / 3. |
| Unit 4 — The logarithmic function — Find g'(x) | Q5(b) | 3 | ___ | Q5(b): practise this worked approach, or an independently valid equivalent method: Uses d(ln u) / dx=u' / u; then Identifies u=3x-1 and u'=3; then Obtains g'(x)=3 / (3x-1). |
| Unit 4 — The logarithmic function — Solve g(x)=3 exactly | Q5(c) | 4 | ___ | Q5(c): practise this worked approach, or an independently valid equivalent method: Rearranges 2+ln(3x-1)=3 to ln(3x-1)=1; then Exponentiates to obtain 3x-1=e; then Solves 3x=e+1; then Obtains the exact solution x=(e+1) / 3, which lies in the domain. |