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QLDMath Methods2023Paper 1

QCE Math Methods 2023 Paper 1

Walkthrough of the 2023 QCE math methods exam: what it assessed, strategy tips, and the common errors flagged in the official marker report.

Marks
55
Time
90 min
Authority
QCAA
Updated
Open the official PDF Official marking notes
Key fact

This paper was set on the 2019 QCAA syllabus. From the 2026 external assessments the 2025 syllabus applies, so some content and the exam format have changed. It is still useful practice, but check the current QCAA Mathematical Methods syllabus and QCAA's 2026 key changes before relying on it. The mark and timing figures shown on this page are approximate study guides, not QCAA's official paper structure.

What this paper assessed

QCE Mathematical Methods is assessed by two external papers; Paper 1 is the technology-free paper (this file), testing fluency with technique by hand. It draws on the Unit 3 and Unit 4 content of the QCAA syllabus:

  • Calculus: differentiation (chain, product, quotient rules), integration and the fundamental theorem, applications to rates of change and area.
  • Functions: exponential, logarithmic and trigonometric functions, their graphs and transformations.
  • Probability and statistics: discrete and continuous random variables, the binomial and normal distributions, and sampling.

Several items required students to interpret a context, set up a model, then execute a technique cleanly without a calculator. The paper rewarded showing the rule or relationship before substituting values.

Structure and timing

Paper 1 is 55 marks in 90 minutes (technology-free), plus 5 minutes perusal.

  • Section 1 - multiple choice (~10 marks): target ~12 minutes (about 1.2 min/question).
  • Section 2 - short response (~45 marks): target ~72 minutes at roughly 1.6 minutes per mark.

A 4-mark item deserves ~6 minutes; show every line of working since method marks reward the correct setup. Reserve ~8 minutes to check signs, domains and arithmetic. Bank the multiple choice first.

Worked practice questions (exam-style)

Worked example

Question 1 (4 marks): Differentiate y=(2x2−5)4y = (2x^2 - 5)^4 and evaluate the gradient at x=1x = 1.

This is a composite function, so we apply the chain rule. Identify the inner and outer functions to keep the substitution tidy.

Step 1: Set up the chain rule substitution

Let u=2x2−5u = 2x^2 - 5 so that y=u4y = u^4. The chain rule says dydx=dydu⋅dudx\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}. We need both derivatives:

dydu=4u3,dudx=4x.\frac{dy}{du} = 4u^3, \qquad \frac{du}{dx} = 4x.

Step 2: Multiply to get dydx\dfrac{dy}{dx}

Substitute back to write everything in terms of xx:

dydx=4(2x2−5)3⋅4x=16x(2x2−5)3.\frac{dy}{dx} = 4(2x^2 - 5)^3 \cdot 4x = 16x(2x^2 - 5)^3.

Step 3: Evaluate the gradient at x=1x = 1

First compute the inner value: 2(1)2−5=−32(1)^2 - 5 = -3. Then substitute into the derivative:

dydx∣x=1=16(1)(−3)3=16×(−27)=−432.\left.\frac{dy}{dx}\right|_{x=1} = 16(1)(-3)^3 = 16 \times (-27) = -432.

Final answer: the gradient at x=1x = 1 is −432-432.

Worked example

Question 2 (5 marks): Evaluate ∫02(3x2−4x+1) dx\displaystyle\int_{0}^{2} (3x^2 - 4x + 1)\, dx and state what the result represents.

We apply the fundamental theorem of calculus: antidifferentiate the integrand, then substitute the upper and lower limits and subtract. This is a technology-free paper, so each step of the antidifferentiation must be shown clearly.

Step 1: Find the antiderivative term by term

For each term, raise the power by one and divide by the new power. There are three terms to handle:

∫(3x2−4x+1) dx=x3−2x2+x+C.\int (3x^2 - 4x + 1)\, dx = x^3 - 2x^2 + x + C.

We drop the constant CC when computing a definite integral because it cancels out when we subtract the lower limit from the upper limit.

Step 2: Substitute the limits and subtract

Write the antiderivative in square brackets with the limits, evaluate at the upper limit first, then subtract the value at the lower limit:

[x3−2x2+x]02=(8−8+2)−(0−0+0)=2.\big[x^3 - 2x^2 + x\big]_0^2 = \big(8 - 8 + 2\big) - \big(0 - 0 + 0\big) = 2.

The lower limit evaluates to zero here, which simplifies the arithmetic.

Step 3: Interpret the result in context

The definite integral equals 22. Geometrically, this is the signed area enclosed between the curve and the horizontal axis over the interval from zero to two. Regions where the curve sits below the axis contribute a negative amount, so this result is the net signed area rather than the total enclosed area. To find the total (unsigned) area you would need to split the integral at each root of the integrand.

Final answer: ∫02(3x2−4x+1) dx=2\displaystyle\int_{0}^{2} (3x^2 - 4x + 1)\, dx = 2.

Worked example

Question 3 (5 marks): A fair coin is tossed 6 times. Let XX be the number of heads. Find P(X=4)P(X = 4) and explain why XX follows a binomial distribution.

Before calculating, we must confirm the distribution applies, then use the binomial probability formula.

Step 1: Confirm the binomial conditions

XX follows a binomial distribution because all four conditions are met: there is a fixed number of trials (n=6n = 6), each toss is independent of the others, each toss has exactly two outcomes (head or tail), and the probability of success (head) is constant at p=0.5p = 0.5 for every toss.

Step 2: Apply the binomial formula

The binomial probability formula is P(X=k)=(nk)pk(1−p)n−kP(X = k) = \dbinom{n}{k} p^k (1-p)^{n-k}. Substituting n=6n = 6, k=4k = 4, and p=0.5p = 0.5:

P(X=4)=(64)(0.5)4(0.5)2=15×(0.5)6.P(X = 4) = \binom{6}{4}(0.5)^4(0.5)^2 = 15 \times (0.5)^6.

Step 3: Evaluate the result

Compute the binomial coefficient (64)=6!4! 2!=15\binom{6}{4} = \dfrac{6!}{4!\,2!} = 15, and note that (0.5)6=164(0.5)^6 = \dfrac{1}{64}:

P(X=4)=15×164=1564≈0.234.P(X = 4) = 15 \times \frac{1}{64} = \frac{15}{64} \approx 0.234.

Final answer: P(X=4)=1564≈0.234P(X = 4) = \dfrac{15}{64} \approx 0.234.

Common errors students made

Recurring errors in QCE Methods Paper 1 responses include:

  • Chain-rule slips - forgetting to multiply by the derivative of the inner function.
  • Sign errors when integrating or evaluating definite integrals at the limits.
  • Skipping domain or restriction statements (e.g. when finding inverse functions or solving over a given interval).
  • Setting up probability questions with the wrong distribution or wrong conditional.

Add these subject-specific traps:

  • Omitting the constant of integration in indefinite integrals.
  • Errors with the binomial coefficient (nk)\binom{n}{k} when computing by hand.
  • Confusing radians and degrees in trigonometric calculus.
  • Writing a final numerical answer without the working that earns method marks.

How to use this paper

Sit the full paper in 90 minutes, technology-free, under timed conditions with only the QCAA formula sheet. Mark against the QCAA marking guide and published assessment report (linked at the top of this page), which show where method marks are allocated. Build a personal error log noting whether each slip was algebraic, conceptual or careless. After a 48-hour gap, re-attempt every multi-step item from a clean page and re-mark it, focusing on the by-hand fluency that Paper 1 specifically tests.

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