Skill Align Mathematics Methods 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
- 47 marks
Formula sheet provided by the supervisor. No special candidate-supplied items are permitted.
Section One
Answer all questions. Calculators, CAS technology and personal notes are not permitted.
Question 1
6 marksQuestion 2
7 marksQuestion 3
6 marksQuestion 4
7 marksQuestion 5
7 marksQuestion 6
7 marksQuestion 7
7 marksWorked Solutions And Marking Guide
Section One Question 1
(a) f'(x)=e^(2x)(1+2x).
Apply the product rule and the chain rule.
(b) 1.
f'(0)=e⁰(1)=1.
(c) y=x.
The tangent passes through the origin with gradient 1.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response f'(x)=e^(2x)(1+2x). Working is not required; accept mathematically equivalent answers.
- Part (b) (2 marks): award full marks for the correct response 1. Working is not required; accept mathematically equivalent answers.
- Part (c) (2 marks): award full marks for the correct response y=x. Working is not required; accept mathematically equivalent answers.
Section One Question 2
(a) G(x)=x⁴-3x²+2x.
Integrate each term.
(b) 8.
[x⁴-3x²+2x]_0²=16-12+4=8.
(c) A(2)=13.
The total change is A(2)-A(0)=int_0^2g(x),dx=8, so A(2)=5+8=13.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response G(x)=x⁴-3x²+2x. Working is not required; accept mathematically equivalent answers.
- Part (b) (3 marks): uses a correct antiderivative for the stated integrand - 1 mark; substitutes the stated bounds and preserves the required area sign - 1 mark; simplifies to 8 - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (c) (2 marks): award full marks for the correct response A(2)=13. Working is not required; accept mathematically equivalent answers.
Section One Question 3
(a) x>1.
Both logarithm arguments must be positive.
(b) (x-1)(x+2)=18, so x=4 or x=-5.
The equation becomes x²+x-20=0.
(c) x=4.
The domain restriction excludes x=-5.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response x>1. Working is not required; accept mathematically equivalent answers.
- Part (b) (2 marks): award full marks for the correct response (x-1)(x+2)=18, so x=4 or x=-5. Working is not required; accept mathematically equivalent answers.
- Part (c) (2 marks): award full marks for the correct response x=4. Working is not required; accept mathematically equivalent answers.
Section One Question 4
(a) 0.2, 0.5, 0.3, respectively.
Read the three bar heights.
(b) 1.4.
0(0.2)+1(0.5)+3(0.3)=1.4.
(c) 1.24.
E(X²)=3.2, so operatorname(Var)(X)=3.2-1.4²=1.24.
Mark allocation
- Part (a) (2 marks): award full marks for the correct response 0.2, 0.5, 0.3, respectively. Working is not required; accept mathematically equivalent answers.
- Part (b) (2 marks): award full marks for the correct response 1.4. Working is not required; accept mathematically equivalent answers.
- Part (c) (3 marks): obtains the required second moment or transformed variance - 1 mark; uses operatorname(Var)(X)=E(X²)-[E(X)]², or the applicable variance rule - 1 mark; obtains 1.24 - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
Section One Question 5
(a) z=1.5.
z=((81-72) / (6))=1.5.
(b) 1-Phi(1.5).
Use the upper tail of the standard normal distribution.
(c) Approximately (60.24,83.76).
Calculate 72pm1.96(6).
Mark allocation
- Part (a) (2 marks): award full marks for the correct response z=1.5. Working is not required; accept mathematically equivalent answers.
- Part (b) (2 marks): award full marks for the correct response 1-Phi(1.5). Working is not required; accept mathematically equivalent answers.
- Part (c) (3 marks): uses the stated mean, standard deviation and central standard-score cut-offs - 1 mark; calculates the lower and upper deviations from the mean - 1 mark; states the two endpoints as Approximately (60.24,83.76) - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
Section One Question 6
(a) Xsimoperatorname(Bin)(4,frac13) and P(X=0)=((16) / (81)).
There are four independent trials with constant success probability frac13, so P(X=0)=(frac23)⁴=((16) / (81)).
(b) P(Xle1)=((16) / (27)).
P(X=1)=4(frac13)(frac23)³=((32) / (81)), so P(Xle1)=((16) / (81))+((32) / (81))=((16) / (27)).
(c) P(Xge2)=((11) / (27)).
Use the complement: P(Xge2)=1-P(Xle1)=1-((16) / (27))=((11) / (27)).
Mark allocation
- Part (a) (2 marks): award full marks for the correct response Xsimoperatorname(Bin)(4,frac13) and P(X=0)=((16) / (81)). Working is not required; accept mathematically equivalent answers.
- Part (b) (3 marks): forms the requested event using the stated distribution, CDF, complement or standardisation - 1 mark; substitutes and evaluates the relevant boundary or component probabilities - 1 mark; obtains P(Xle1)=((16) / (27)) - 1 mark. Allow consequential marking only after a correct setup is shown. Accept mathematically equivalent answers.
- Part (c) (2 marks): award full marks for the correct response P(Xge2)=((11) / (27)). Working is not required; accept mathematically equivalent answers.
Section One Question 7
(a) -3le x<-1 and 3<xle5.
The function increases where f'(x)>0.
(b) x=-1 and x=3.
Stationary points occur where f'(x)=0.
(c) A local maximum.
The derivative changes from positive to negative at x=-1.
Mark allocation
- Part (a) (2 marks): The function increases where f'(x)>0 - 1 mark; states -3le x<-1 and 3<xle5 - 1 mark.
- Part (b) (2 marks): award full marks for the correct response x=-1 and x=3. Working is not required; accept mathematically equivalent answers.
- Part (c) (3 marks): identifies the derivative behaviour immediately to the left of the stationary point - 1 mark; identifies the derivative behaviour immediately to the right of the stationary point - 1 mark; uses the change to classify it as A local maximum - 1 mark. Accept mathematically equivalent answers.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Calculus | Q1-Q2, Q7 | 20 | ___ | Review differentiation, exact integration and derivative-sign interpretation. |
| Functions | Q3 | 6 | ___ | Review logarithm laws and domain restrictions. |
| Probability and inference | Q4-Q6 | 21 | ___ | Review discrete random variables, normal models and exact binomial probabilities. |