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Mathematical Methods Units 3&4 Free Online - Pack 0

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Paper 1 Technology-free Question and Response Book Showcase

QCE Mathematical Methods Units 3&4 Free Online Pack 0 — Paper 1 Technology-free Question and Response Book Showcase

Read Paper 1 Technology-free Question and Response Book Showcase online for free, including every question, worked solution, marking note and diagnostic action. No public PDF download or checkout is provided.

QCE Year 12 Final Exam 2026 Edition - Pack 0 v1.0
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This full-length showcase paper is available to read online.

Paper 1 Technology-free Question and Response Book Showcase

19 questions

55 marks

Estimated duration: Perusal time 5 minutes; working time 90 minutes

Reading: 5 minutes perusal · Writing: 90 minutes

Read Paper 1 Technology-free Question and Response Book Showcase online

Skill Align

Skill Align QCE Mathematical Methods Paper 1 - Free Online Pack 0

Full-length Units 3&4 external-assessment-style showcase paper

Paper
Paper 1 Technology-free Question and Response Book Showcase
Reading
5 minutes perusal
Writing
90 minutes
Assessment
55 marks

QCAA formula book provided; calculators and other technology are not permitted for Paper 1. No public PDF or formula-book download is supplied with Pack 0.

Section 1

Questions 1-10 are multiple choice. Select the best answer for each question. The QCAA Mathematical Methods formula book is provided. Calculators and other technology are not permitted.

Question 1

1 mark
The derivative of f(x)=xe^x is
  1. e^x(x+1)
  2. xe^x
  3. e^x(x-1)
  4. x^2e^x

Question 2

1 mark
The domain of g(x)=ln(2x-1) is
  1. xgefrac12
  2. x>frac12
  3. x<frac12
  4. xnefrac12

Question 3

1 mark
The region bounded by y=2x, the x-axis and the line x=3 has area
  1. 6
  2. 12
  3. 9
  4. 18

Question 4

1 mark
The solutions of sintheta=-frac12 for 0lethetale2pi are
  1. ((pi) / (6)),((5pi) / (6))
  2. ((5pi) / (6)),((7pi) / (6))
  3. ((4pi) / (3)),((5pi) / (3))
  4. ((7pi) / (6)),((11pi) / (6))

Question 5

1 mark
The graph of y=-2f(x+3)+1 is obtained from y=f(x) by
  1. shifting 3 left, reflecting in the x-axis, stretching vertically by 2, then shifting 1 up
  2. shifting 3 right, reflecting in the y-axis, stretching horizontally by 2, then shifting 1 up
  3. shifting 3 left, stretching vertically by 2, then shifting 1 down
  4. shifting 1 left, reflecting in the x-axis, then shifting 3 up

Question 6

1 mark
If Xsim B(20,0.3), then operatorname(Var)(X)=
  1. 2.1
  2. 4.2
  3. 6
  4. 14

Question 7

1 mark
The tangent to y=ln x at x=e is
  1. y-e=x-1
  2. y=ex-1
  3. y-1=((x-e) / (e))
  4. y-1=e(x-e)

Question 8

1 mark
Events A and B satisfy P(B)=0.5 and P(Amid B)=0.6. What is P(Acap B)?
  1. 0.55
  2. 0.60
  3. 1.10
  4. 0.30

Question 9

1 mark
For f'(x)=x²-4, the stationary point at x=-2 is
  1. a local maximum
  2. a local minimum
  3. a stationary inflection point
  4. not a stationary point

Question 10

1 mark
A fair coin is tossed three times. The probability of exactly two heads is
  1. frac18
  2. frac38
  3. frac14
  4. frac12

Section 2

Questions 11-19 are short response. Show exact working and relevant mathematical reasoning.

Question 11

5 marks
A circular oil slick has radius r(t)=2+frac t2 metres, where tge0 is measured in minutes. Its area is A=pi r^2.
(a) 1 mark
Express A as a function of t.
(b) 2 marks
Determine the rate of change of area after 4 minutes.
(c) 2 marks
Determine when ((dA) / (dt))=5pi, and find the area at that time.

Question 12

4 marks
A random sample of 64 households contains 32 that support a proposal. Use z=2 to construct an approximate confidence interval for the population proportion.
(a) 1 mark
Determine the sample proportion hat p.
(b) 1 mark
Determine the standard error √(((hat p(1-hat p)) / (n))).
(c) 1 mark
State the approximate confidence interval in simplified fractional form.
(d) 1 mark
Interpret the interval in context.

Question 13

4 marks
Let f(x)=ln(3-x).
(a) 1 mark
State the domain of f.
(b) 1 mark
Determine the x-intercept of the graph of f.
(c) 2 marks
Determine f⁻¹(x) and state its domain.

Question 14

4 marks
Let X be the number of heads obtained when a fair coin is tossed five times.
(a) 2 marks
Determine P(X=2) exactly.
(b) 1 mark
Determine P(Xge3) exactly.
(c) 1 mark
Determine the expected number of heads.

Question 15

6 marks
The function f(x)=begin(cases)ax²+b,&xle1 ln x+3,&x>1end(cases) is differentiable at x=1.
(a) 3 marks
Determine a and b.
(b) 1 mark
Determine the tangent equation at x=1.
(c) 2 marks
Evaluate int_0^e f(x),dx exactly.

Question 16

5 marks
Two irrigation plans have water-delivery rates A(t)=6t-t² and B(t)=frac72t litres per minute for 0le tle4. A suitable plan must deliver at least 26 litres in total and must never exceed 10 litres per minute. Determine which plan, if either, is suitable. Show the mathematical evidence supporting your decision.
(a) 2 marks
Determine the total volume delivered by each plan.
(b) 2 marks
Determine the maximum delivery rate of each plan on the interval.
(c) 1 mark
State and justify the suitable plan.

Question 17

6 marks
Let g(x)=2sin(2x)-1, for 0le xle2pi.
(a) 2 marks
State the period and range of g.
(b) 3 marks
Solve g(x)=0 exactly on the stated interval.
(c) 1 mark
Determine the first x-value at which g reaches its maximum.

Question 18

6 marks
A delivery time X, in hours, has probability density f(x)=kx for 0le xle4, and f(x)=0 otherwise. A compensation plan must have an expected cost no greater than 4 per delivery and must offer compensation on at least half of deliveries.
(a) 1 mark
Determine k.
(b) 2 marks
Determine the median delivery time exactly.
(c) 3 marks
Plan A pays 12 when X>3. Plan B pays 4 when X>2. Determine which plan, if either, meets both company requirements.

Question 19

5 marks
The line y=kx, where k>0, is tangent to the curve y=ln x.
(a) 3 marks
Determine the point of tangency and the value of k.
(b) 2 marks
Determine the exact area enclosed by the line, the curve and the line x=1.

Queensland Certificate of Education (QCE) subjects and external assessments are administered by the Queensland Curriculum and Assessment Authority (QCAA). Skill Align is an independent publisher and is not affiliated with, authorised by, sponsored by, approved by, or endorsed by QCAA or the Queensland Government.

Copyright (c) 2026 Skill Align. Free for personal, non-commercial online viewing at https://skillalign.au. You may share the Skill Align page link. Except as permitted by law or with Skill Align's prior written permission, the pack itself must not be resold, copied, redistributed, republished, automatically extracted, or uploaded to a question bank.

Worked Solutions And Marking Guide

Section 1 Question 1

Answer: e^x(x+1)

Apply the product rule: f'(x)=e^x+xe^x.

Section 1 Question 2

Answer: x>frac12

The logarithm requires 2x-1>0.

Section 1 Question 3

Answer: 9

The region is a triangle with base 3 and height 6, so its area is frac12(3)(6)=9.

Section 1 Question 4

Answer: ((7pi) / (6)),((11pi) / (6))

Sine is negative in quadrants III and IV with reference angle pi / 6.

Section 1 Question 5

Answer: shifting 3 left, reflecting in the x-axis, stretching vertically by 2, then shifting 1 up

Read horizontal changes inside the function and vertical changes outside it.

Section 1 Question 6

Answer: 4.2

The variance is np(1-p)=20(0.3)(0.7)=4.2.

Section 1 Question 7

Answer: y-1=((x-e) / (e))

The point is (e,1) and the gradient is 1 / e.

Section 1 Question 8

Answer: 0.30

Use P(Acap B)=P(Amid B)P(B)=0.6(0.5)=0.30.

Section 1 Question 9

Answer: a local maximum

Since f''(x)=2x, f''(-2)=-4<0.

Section 1 Question 10

Answer: frac38

There are three arrangements with two heads among eight equally likely outcomes.

Section 2 Question 11

(a) A(t)=pi(2+frac t2)^2.

Substitute r(t)=2+t / 2 into A=pi r^2.

(b) ((dA) / (dt))=4pitext( m)^2text( per minute).

By the chain rule, dA / dt=2pi r,dr / dt=2pi r(1 / 2)=pi r. At t=4, r=4, so dA / dt=4pi.

(c) At t=6 minutes, and the area is 25pitext( m)^2.

Since dA / dt=pi r, the required rate occurs when r=5. Solving 2+t / 2=5 gives t=6, and then A=pi(5)²=25pi.

Detailed marking criteria

Part Part (a) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (2 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (2 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 12

(a) hat p=frac12.

The sample proportion is 32 / 64=1 / 2.

(b) frac1(16).

Substitution gives √((((1 / 2)(1 / 2)) / (64)))=√(1 / 256)=1 / 16.

(c) [frac38,frac58].

The margin is 2(1 / 16)=1 / 8, so 1 / 2pm1 / 8 gives [3 / 8,5 / 8].

(d) The population proportion of households supporting the proposal is estimated to lie between 3 / 8 and 5 / 8.

The interval estimates the unknown population proportion, not the observed sample proportion.

Detailed marking criteria

Part Part (a) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (d) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 13

(a) x<3.

The logarithm requires 3-x>0, hence x<3.

(b) (2,0).

Set ln(3-x)=0. Then 3-x=e⁰=1, so x=2.

(c) f⁻¹(x)=3-e^x, with domain xinmathbb R.

From y=ln(3-x), exponentiation gives e^y=3-x, so x=3-e^y. Interchanging x and y gives f⁻¹(x)=3-e^x. The range of f is all real numbers, so the inverse has domain mathbb R.

Detailed marking criteria

Part Part (a) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (2 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 14

(a) P(X=2)=binom52(frac12)⁵=frac5(16).

There are binom52=10 arrangements with two heads, each having probability (1 / 2)^5. Thus P(X=2)=10 / 32=5 / 16.

(b) P(Xge3)=frac12.

For five fair tosses, symmetry gives P(Xge3)=P(Xle2)=1 / 2. Equivalently, sum the probabilities for 3, 4 and 5 heads.

(c) E(X)=frac52.

For Xsim B(5,1 / 2), E(X)=np=5 / 2.

Detailed marking criteria

Part Part (a) (2 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 15

(a) a=frac12 and b=frac52.

Continuity at 1 requires a+b=3. Equal one-sided derivatives require 2a=1, so a=1 / 2. Substitution into the continuity equation gives b=5 / 2.

(b) y-3=x-1, or y=x+2.

The point is (1,3) and differentiability gives gradient 1, so y-3=x-1.

(c) 3e+frac23.

Split the integral at 1. Using the values from part (a), int_0¹(frac12x²+frac52),dx=8 / 3. Also int_1^e(ln x+3),dx=3e-2. Their sum is 3e+2 / 3.

Detailed marking criteria

Part Part (a) (3 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (2 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 16

(a) Plan A delivers ((80) / (3)) L and Plan B delivers 28 L.

Total volume is the integral of delivery rate. Thus int_0⁴(6t-t²),dt=[3t²-t³ / 3]_0⁴=80 / 3, while int_0⁴(7t / 2),dt=[7t² / 4]_0⁴=28.

(b) Plan A has maximum rate 9 L / min; Plan B has maximum rate 14 L / min.

For Plan A, A'(t)=6-2t=0 at t=3, and comparison with the endpoints gives a maximum of A(3)=9. Plan B is increasing, so its maximum is B(4)=14.

(c) Plan A is suitable: it delivers 80 / 3 L, which exceeds 26 L, and its maximum rate is 9 L / min, which is below 10 L / min. Plan B exceeds the rate limit.

Use both requirements. Plan A meets the total-volume and maximum-rate constraints; Plan B meets the volume requirement but fails the maximum-rate requirement.

Detailed marking criteria

Part Part (a) (2 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (2 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 17

(a) The period is pi and the range is [-3,1].

The coefficient 2 inside the sine gives period 2pi / 2=pi. Since -1lesin(2x)le1, -3le g(x)le1.

(b) x=fracpi(12),((5pi) / (12)),((13pi) / (12)),((17pi) / (12)).

The equation becomes sin(2x)=1 / 2. For 0le2xle4pi, 2x=pi / 6,5pi / 6,13pi / 6,17pi / 6. Dividing by 2 gives the four solutions.

(c) x=fracpi4.

The maximum occurs when sin(2x)=1. The first solution is 2x=pi / 2, hence x=pi / 4.

Detailed marking criteria

Part Part (a) (2 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (3 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 18

(a) k=frac18.

A density integrates to 1, so int_0^4kx,dx=8k=1, giving k=1 / 8.

(b) The median is 2sqrt2 hours.

For median m, P(Xle m)=1 / 2. Thus int_0^m x / 8,dx=m² / 16=1 / 2, so m²=8 and m=2sqrt2 within the support.

(c) Plan B meets both requirements. Plan A pays on 7 / 16 of deliveries and has expected cost 21 / 4, while Plan B pays on 3 / 4 of deliveries and has expected cost 3.

Using F(x)=x² / 16, Plan A pays with probability 1-F(3)=7 / 16, so its expected cost is 12(7 / 16)=21 / 4 dollars; it fails both requirements. Plan B pays with probability 1-F(2)=3 / 4, so its expected cost is 4(3 / 4)=3 dollars; it meets both requirements.

Detailed marking criteria

Part Part (a) (1 mark)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (2 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (3 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 19

(a) The point of tangency is (e,1) and k=frac1e.

Let the tangency occur at x=a. Equal gradients give k=1 / a, while equal y-values give ln a=ka. Hence ln a=1, so a=e, the point is (e,1), and k=1 / e.

(b) frac e2-1-frac1(2e) square units.

On [1,e], the tangent line lies above ln x. The area is int_1^e(x / e-ln x),dx. Using antiderivatives x² / (2e) and xln x-x gives e / 2-1-1 / (2e).

Detailed marking criteria

Part Part (a) (3 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (2 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Diagnostic Checklist

TopicQuestionsMarksMarks LostAction
Functions, Graphs and Algebra Q2, Q5, Q13, Q19 11 ___ Review domains, inverse functions, transformations and connections between logarithmic curves and tangents.
Calculus and Applications Q1, Q3, Q7, Q9, Q11, Q15-Q16 20 ___ Review differentiation, accumulation from rates, related rates, piecewise smoothness and exact integration.
Trigonometry, Probability and Inference Q4, Q6, Q8, Q10, Q12, Q14, Q17-Q18 24 ___ Review exact trigonometry, binomial and continuous distributions, independence and confidence intervals.

What is included

Paper 1 Technology-free Question and Response Book Showcase questions (55 marks)

Paper 2 Technology-active Question and Response Book Showcase questions (55 marks)

Worked solutions and marking guidance shown online

Diagnostic checklist shown online

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Questions about this exam pack

What is included in Mathematical Methods Units 3&4 Free Online - Pack 0?

Pack 0 includes 2 full-length showcase papers, worked solutions, marking guidance and diagnostic checklists, all shown online.

Is Pack 0 really free?

Yes. Pack 0 can be read online without checkout or a monthly subscription.

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