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
- 54 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. For any question or part question worth more than two marks, valid working or justification is required to receive full marks.
Question 1
8 marksQuestion 2
9 marksQuestion 3
6 marksQuestion 4
7 marksQuestion 5
7 marksQuestion 6
7 marksQuestion 7
10 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.
(d) f''(0)=4; concave up.
f''(x)=4(1+x)e^(2x).
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.
- Part (d) (2 marks): mark 1: differentiates again and obtains f''(0)=4; mark 2: connects its positive sign to concave-up curvature.
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.
(d) -18.
Reverse the limits after using linearity: -[2(8)+2]=-18.
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.
- Part (d) (2 marks): mark 1: uses linearity to obtain 18 with limits 0 to 2; mark 2: reverses the limits to obtain -18.
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.
(d) -8; f(3) is 8 less than f(-1).
The two triangles below the axis each have area 4, so the signed integral is -8.
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.
- Part (d) (3 marks): mark 1: uses the signed area under f' for the function change; mark 2: obtains the two triangular areas and total -8; mark 3: interprets the negative change.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Calculus | Q1-Q2, Q7 | 27 | ___ | 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. |