How is the QCE Maths Methods EA structured?

Two papers sat in the November assessment block. Paper 1: 90 minutes plus 5 minutes perusal, technology-free (no calculator), 55 marks (10 MC for 10 marks + 9 SR for 45 marks). Paper 2: 90 minutes plus 5 minutes perusal, technology-active (CAS calculator permitted), 55 marks. Combined total 110 marks; the EA contributes 50 percent of the subject result. Cumulative across Units 3 and 4.

What does Paper 1 test?

By-hand mathematical fluency. Differentiation (chain, product, quotient). Antidifferentiation. Solving equations algebraically. Exact values of trig functions at standard angles. Logarithm and index manipulation. No CAS, no scientific calculator.

What does Paper 2 test?

Calculator-assisted problem solving. Modelling problems. Probability and statistics (binomial, normal, confidence intervals). Calculator-assisted differentiation and integration. Transcendental equation solving.

What CAS commands matter most?

On TI-Nspire or Casio Classpad: solve() for equations, derivative() and integral() for calculus, binomCdf and binomPdf for binomial, normCdf and invNorm for normal distribution, factor and expand for algebra. Practice these in context.

How should time be budgeted?

Paper 1: about 1.6 minutes per mark (55 marks in 90 minutes). Paper 2: about 1.6 minutes per mark (55 marks in 90 minutes), with CAS handling routine algebra so longer modelling questions get more time. Reserve 5 to 10 minutes per paper for review.

QLDMath Methods

QCE Maths Methods EA strategy: 2026 guide

A 2026 guide to QCE Maths Methods External Assessment strategy. The two-paper structure, Paper 1 technology-free, Paper 2 technology-active with CAS, common calculation patterns across Units 3 and 4, step-by-step worked examples, and a six-week preparation routine.

Generated by Claude Opus 4.814 min readQCAA-MM-EAUpdated 2026-05-25

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

How the EA is structured
Topic frequency analysis
Paper 1 calculation patterns
Paper 2 calculation patterns
Common student errors
Six-week preparation routine
QCAA marking criteria
Check your knowledge

How the EA is structured

The QCAA Maths Methods EA is two papers sat in the November assessment block.

Paper 1. 90 minutes plus 5 minutes perusal. 55 marks (Section 1: 10 multiple-choice for 10 marks; Section 2: 9 short response for 45 marks). Technology-free: no CAS, no scientific calculator. Tests by-hand mathematical fluency.

Paper 2. 90 minutes plus 5 minutes perusal. 55 marks. Technology-active: CAS calculator permitted (TI-Nspire or Casio Classpad). Tests problem solving where the CAS handles algebra.

Combined: 50 percent of the subject result. Cumulative across Units 3 and 4 (Year 12 in QCAA terms).

Topic frequency analysis

From Unit 3 (continuous from Year 12 Term 1):

Differentiation by chain, product, quotient rules.
Stationary points and optimisation.
Antidifferentiation (standard plus reverse-chain factor).
Definite integrals and area between curves.
Exponential and logarithmic equations.
Modelling growth and decay.

From Unit 4 (final term of Year 12):

Differentiation of $\sin x$ , $\cos x$ , $\tan x$ , $e^x$ , $\ln x$ extended.
Integration of $\sin(kx)$ , $\cos(kx)$ , $e^{kx}$ .
Discrete random variables, expected value, variance.
Binomial distribution: mean $np$ , variance $np(1-p)$ .
Normal distribution: standardisation, P(X less than x).
Sample proportions: distribution, standard error.
Confidence intervals for proportions.

The QCAA Subject Matter Reference (SMR) lists every dot point. Use it as a checklist.

Paper 1 calculation patterns

Paper 1: differentiate

x^2 \sin(3x)

using the product rule

Step 1. Identify the structure: This is a product of two functions: $f(x) = x^2$ and $g(x) = \sin(3x)$ . Product rule applies.
Step 2. Differentiate each factor: $f'(x) = 2x$ . For $g(x) = \sin(3x)$ the chain rule gives $g'(x) = 3 \cos(3x)$ , since the inner derivative is $3$ .
Step 3. Apply the product rule: Substitute into $f'g + f g'$ :

\frac{d}{dx}\left[x^2 \sin(3x)\right] = 2x \sin(3x) + x^2 \cdot 3 \cos(3x)

Step 4. Tidy. Group the $3$ with $x^2$ :

\frac{d}{dx}\left[x^2 \sin(3x)\right] = 2x \sin(3x) + 3x^2 \cos(3x)

Answer. $2x \sin(3x) + 3x^2 \cos(3x)$ . Both terms required for full marks under the QCAA "mathematical reasoning" criterion.

Paper 1: antidifferentiate

\cos(5x + 1)

by reverse-chain factor

Step 1. Recognise the form: The integrand is $\cos(\text{linear in } x)$ , which has standard antiderivative $\sin$ times a reciprocal-coefficient factor.
Step 2. Write the candidate antiderivative: Try $\sin(5x + 1)$ . Differentiating, $\frac{d}{dx}[\sin(5x + 1)] = 5 \cos(5x + 1)$ , which is $5$ times the integrand.
Step 3. Correct for the chain factor: Divide by $5$ to undo the excess factor:

\int \cos(5x + 1)\, dx = \frac{1}{5} \sin(5x + 1) + C

Step 4. Include $+C$ . The indefinite integral always carries the arbitrary constant.

Answer. $\frac{1}{5} \sin(5x + 1) + C$ .

Exact trig values. Memorise the standard table:

angle	$\sin$	$\cos$	$\tan$
0	0	1	0
$\pi/6$	$1/2$	$\sqrt{3}/2$	$1/\sqrt{3}$
$\pi/4$	$\sqrt{2}/2$	$\sqrt{2}/2$	1
$\pi/3$	$\sqrt{3}/2$	$1/2$	$\sqrt{3}$
$\pi/2$	1	0	undefined

Area between two curves. Find intersection points. Determine which curve is above on each interval. Integrate the difference.

Paper 2 calculation patterns

Paper 2: logistic growth, when does

P(t)

reach 500?

Step 1. Read the scenario: Population $P(t) = \frac{L}{1 + A e^{-kt}}$ with carrying capacity $L = 1000$ , initial population $P(0) = 100$ , and growth constant $k = 0.2$ per year. Find $t$ when $P(t) = 500$ .
Step 2. Use the initial condition to find $A$: Substituting $t = 0$ : $100 = \frac{1000}{1 + A}$ , so $1 + A = 10$ and $A = 9$ .
Step 3. Write the equation to solve: With $A = 9$ :

500 = \frac{1000}{1 + 9 e^{-0.2 t}}

Step 4. Rearrange algebraically (CAS-checked): Cross-multiply: $500 (1 + 9 e^{-0.2 t}) = 1000$ , so $1 + 9 e^{-0.2 t} = 2$ and $9 e^{-0.2 t} = 1$ . Hence $e^{-0.2 t} = 1/9$ .
Step 5. Take logs: $-0.2 t = \ln(1/9) = -\ln 9$ , so $t = \frac{\ln 9}{0.2} = 5 \ln 9$ .
Step 6. Evaluate: $t = 5 \ln 9 \approx 10.99$ years.
Step 7. State with units and reasonableness: The population reaches $500$ (half of capacity, the inflection point) at $t \approx 11.0$ years, which sits between the initial small population and the saturation phase. Reasonable.
Answer: $t \approx 11.0$ years.

Paper 2: binomial,

P(X \le 5)

for

n = 20

p = 0.3

Step 1. Confirm the model: $X \sim \text{Bin}(20, 0.3)$ : $n = 20$ independent trials, success probability $p = 0.3$ , $X$ counts successes.
Step 2. State what is required: Cumulative probability $P(X \le 5)$ .
Step 3. Set up the CAS command: On TI-Nspire: binomCdf(20, 0.3, 0, 5). On Classpad: binomialCDf(0, 5, 20, 0.3). Always report the setup as well as the output.
Step 4. Execute: Output: $0.4164$ (to 4 decimal places).
Step 5. Report supporting statistics: Mean $= np = 20 \times 0.3 = 6$ . Variance $= np(1 - p) = 20 \times 0.3 \times 0.7 = 4.2$ . Standard deviation $\approx 2.05$ .
Step 6. Interpret in context: Probability of at most $5$ successes is about $0.416$ , slightly below half because the mean is $6$ .
Answer: $P(X \le 5) \approx 0.4164$ .

Paper 2: 95 percent confidence interval for a proportion

Step 1. Identify the parameter: $p$ = unknown population proportion. Sample size $n = 400$ , observed successes $144$ .
Step 2. Compute the sample proportion: $\hat{p} = 144 / 400 = 0.36$ .
Step 3. Choose the critical value: For a 95 percent CI, $z^* = 1.96$ .
Step 4. Compute the standard error: $SE = \sqrt{\hat{p}(1 - \hat{p}) / n} = \sqrt{0.36 \times 0.64 / 400}$ .

SE = \sqrt{0.2304 / 400} = \sqrt{0.000576} = 0.024

Step 5. Compute the margin of error: $E = z^* \cdot SE = 1.96 \times 0.024 \approx 0.047$ .
Step 6. Form the interval: $\hat{p} \pm E = 0.36 \pm 0.047$ , giving $(0.313, 0.407)$ .
Step 7. Communicate in context: We are 95 percent confident that the true population proportion lies between $0.313$ and $0.407$ . Note the QCAA EA expects this interpretation as part of the "communicate" descriptor.
Answer: 95 percent CI: $(0.313, 0.407)$ .

Common student errors

Factor in trig antiderivatives. $\int \sin(3x)\, dx = -\frac{1}{3} \cos(3x) + C$ .

Missing +C. Always include in indefinite integrals.

Domain check in log equations. Verify the argument is positive at the solution.

Wrong exact trig values. Memorise.

CAS without setup. Report the setup (the equation or function) plus the CAS output. Method marks reward setup.

Significant figures. 3 sig fig unless told otherwise. Exact form when asked.

Confusing binomial and normal. Discrete versus continuous. Don't mix.

Reading question carelessly. "At least" includes the equality; "more than" excludes it.

Six-week preparation routine

Weeks 1 to 2. Key knowledge review using the QCAA Maths Methods Subject Matter Reference as a checklist. Map each subject matter point to your notes.

Weeks 3 to 4. Paper 1 by-hand drills. 30 minutes per day on differentiation, antidifferentiation, exact trig values, exponential and log equations.

Week 5. Paper 2 CAS drills. Modelling problems. Probability distributions. Confidence intervals. Build a personal cheat-sheet of CAS commands.

Week 6. Full timed exam pairs. Two pairs (Paper 1 plus Paper 2) per week. Mark against QCAA reports. Identify topics with persistent errors and revisit.

QCAA marking criteria

Marks are awarded for:

Correct mathematics (right concept, right formula, right answer).
Show working (method marks even if arithmetic slips).
Significant figures and exact form (as appropriate).
Notation and presentation (correct use of variables, equations).
Clear communication including context units in modelling problems.

Check your knowledge

(Paper 1, by hand) Differentiate $y = \ln(x^2 + 1)$ with respect to $x$ .
(Paper 1, by hand) Find $\int (3x - 2)^4\, dx$ .
(Paper 1, by hand) Solve $2 \ln(x) = \ln(x + 6)$ exactly, including any domain check.
(Paper 1, by hand) Evaluate $\int_0^{\pi/2} \sin(2x)\, dx$ exactly.
(Paper 2, CAS) Let $X \sim N(170, 10^2)$ . Find $P(180 < X < 195)$ to 4 decimal places.
(Paper 2, CAS) The number of defective items in a batch of $50$ follows $\text{Bin}(50, 0.04)$ . Find $P(X \ge 2)$ .
(Paper 2, CAS) A sample of $250$ voters has $130$ supporting a policy. Construct a 90 percent confidence interval for the population proportion. Use $z^* = 1.645$ .
(Paper 2, modelling) A radioactive isotope decays according to $A(t) = A_0 e^{-kt}$ with half-life $5.27$ years. Find $k$ and the time for the substance to decay to $10\%$ of $A_0$ .

Solutions to check-your-knowledge questions

These eight questions cover the chain rule for logs, reverse-chain antidifferentiation, solving log equations, definite integrals, CAS normal and binomial calculations, confidence intervals, and exponential decay.

Step 1: Q1. Differentiate $y = \ln(x^2 + 1)$ using the chain rule

The derivative of $\ln(u)$ is $\frac{1}{u} \cdot u'$ . Here $u = x^2 + 1$ , so $u' = 2x$ .

\frac{dy}{dx} = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}

Step 2: Q2. Antidifferentiate $(3x - 2)^4$ by reverse chain

Try raising the power by 1 and dividing by the new power: $\frac{(3x - 2)^5}{5}$ . Differentiating this gives $(3x - 2)^4 \cdot 3$ , which is $3$ times too large, so divide by $3$ to compensate.

\int (3x - 2)^4\, dx = \frac{(3x - 2)^5}{15} + C

Step 3: Q3. Solve $2\ln x = \ln(x + 6)$ exactly

Use the log power rule to write $2\ln x = \ln x^2$ . When two logs with the same base are equal, their arguments must be equal. Then solve the resulting quadratic and check the domain.

\ln x^2 = \ln(x + 6) \implies x^2 = x + 6 \implies x^2 - x - 6 = (x - 3)(x + 2) = 0

Candidates: $x = 3$ or $x = -2$ . Since $\ln x$ requires $x > 0$ , reject $x = -2$ .

The solution is $x = 3$ .

Step 4: Q4. Evaluate $\displaystyle\int_0^{\pi/2} \sin(2x)\, dx$

The antiderivative of $\sin(2x)$ is $-\frac{1}{2}\cos(2x)$ (the factor of $\frac{1}{2}$ compensates for the inner derivative of $2x$ ). Apply the fundamental theorem.

\int_0^{\pi/2} \sin(2x)\, dx = \left[-\frac{1}{2} \cos(2x)\right]_0^{\pi/2} = -\frac{1}{2} \cos(\pi) + \frac{1}{2} \cos 0 = \frac{1}{2} + \frac{1}{2} = 1

Step 5: Q5. CAS: probability for $X \sim N(170, 10^2)$ between 180 and 195

On the CAS use the normal cumulative distribution function with lower bound 180, upper bound 195, mean 170, and standard deviation 10.

\text{normCdf}(180, 195, 170, 10) \approx 0.1525

About $15.25\%$ of values fall in this interval.

Step 6: Q6. CAS: $P(X \ge 2)$ for $X \sim \text{Binomial}(50, 0.04)$

Use the complement rule: $P(X \ge 2) = 1 - P(X \le 1)$ . Evaluate the cumulative binomial on the CAS.

P(X \le 1) = \text{binomCdf}(50, 0.04, 0, 1) \approx 0.4005

P(X \ge 2) = 1 - 0.4005 \approx 0.5995

Step 7: Q7. 90% confidence interval for a proportion

Compute the sample proportion, the standard error, and the margin of error using $z^* = 1.645$ for 90%.

\hat{p} = \frac{130}{250} = 0.52, \qquad SE = \sqrt{\frac{0.52 \times 0.48}{250}} \approx 0.0316

\text{Margin} = 1.645 \times 0.0316 \approx 0.0520

90% confidence interval: $(0.52 - 0.052,\ 0.52 + 0.052) = (0.468, 0.572)$ .

Step 8: Q8. Exponential decay: find $k$ and then the time to reach $10\%$

First find the decay constant $k$ from the half-life condition $A_0/2 = A_0 e^{-5.27k}$ . Then use $k$ to find the time for the amount to fall to $10\%$ of the original.

e^{-5.27k} = \frac{1}{2} \implies k = \frac{\ln 2}{5.27} \approx 0.1315 \text{ per year}

For $10\%$ remaining: $0.1 = e^{-kt}$ , so

t = \frac{\ln 10}{k} = \frac{\ln 10}{0.1315} \approx 17.5 \text{ years}

Final answer: the substance falls to $10\%$ of its original amount after approximately $17.5$ years.