VCE Specialist Mathematics 2025 Exam 2
Worked solutions to the 2025 VCE Specialist Mathematics Examination 2 (80 marks, CAS allowed): all 20 multiple-choice answers with reasons and every Section B part, checked against the VCAA external assessment report.
- Marks
- 80
- Time
- 120 min
- Authority
- VCAA
- Updated
Every question from the 2025 VCE Specialist Mathematics Examination 2, the technology-active (CAS) paper. Multiple-choice answers come with a one-line reason; Section B solutions sit behind a Show worked solution toggle. For the no-calculator paper, see the 2025 Examination 1 walkthrough.
How to use this page
- Questions are from the 2025 VCE Specialist Mathematics Examination 2, copyright Victorian Curriculum and Assessment Authority (VCAA). Each question is summarised briefly here; open the official examination PDF for the full wording, diagrams and answer options.
- Answers are original ExamExplained working. Every multiple-choice answer matches the key in the 2025 Specialist Mathematics Examination 2 external assessment report (Word document), and every Section B result was recomputed and compared with the report. Both files are listed on the VCAA Specialist Mathematics examinations page.
Structure and timing
Examination 2 is 80 marks in 120 minutes (plus 15 minutes reading time), with a CAS calculator and one bound reference allowed. Take .
- Section A (20 marks): 20 multiple-choice questions. Aim for about 25 minutes.
- Section B (60 marks): 6 extended-response questions of 10 marks each.
Section A: Multiple choice
- Q1
- Contrapositive of "If I have a tiger, then I have a cat". Answer: C - swap and negate both parts: "If I do not have a cat, then I do not have a tiger".
- Q2
- Which function is a counter-example to "if then has a point of inflection at "? Answer: D - has , which is 0 at but does not change sign, so there is no inflection there. (48% correct; 40% chose C.)
- Q3
- has asymptote and -intercept . Find . Answer: A - dividing gives , so () and (); then gives .
- Q4
- An algorithm adds for left with . What is printed? Answer: B - .
- Q5
- has root . Find . Answer: A - the conjugate is also a root, giving the factor ; the third factor is (since ). Expanding gives and , so .
- Q6
- If and , find . Answer: C - with , and , so the real part is .
- Q7
- Use to rewrite . Answer: D - and , giving .
- Q8
- Which differential equation matches the direction field? Answer: A - the slopes are symmetric about the -axis (so appears squared), and on the positive -axis the slope is negative: .
- Q9
- Surface area when , , , is rotated about the -axis. Answer: C - ; with , , this is .
- Q10
- The region under , , beside is rotated about the -axis, with volume . Find . Answer: D - , which equals the given volume at .
- Q11
- with . Domain of ? Answer: B - separating gives , which needs : .
- Q12
- Constant acceleration, velocity at and at . Velocity at the midpoint of ? Answer: D - and give . (33% correct.)
- Q13
- A ball is thrown up at 20 m/s from 50 m and lands on a tray 1 m above the ground. Time taken? Answer: B - taking up as positive, , so s.
- Q14
- Non-zero with . Angle between them? Answer: B - , so and .
- Q15
- Planes and meet at angle . Equation for ? Answer: A - using normals, , which simplifies to .
- Q16
- ; for which are acceleration and velocity perpendicular at ? Answer: C - at gives , so .
- Q17
- A particle from rest has . Find . Answer: C - integrate and choose constants so : .
- Q18
- Two lines meet at . Find , , . Answer: D - matching coordinates gives , , and so , , .
- Q19
- Area of the triangle cut from the axes by . Answer: B - the intercepts are along each axis, forming an equilateral triangle of side , with area .
- Q20
- , , independent. Find . Answer: A - the mean is and the variance is , so .
Section B: Extended response
Question 1 (10 marks)
- a
- Sketch , labelling asymptotes, the turning point, and the point of inflection (to one decimal place). (3 marks)
- b
- The region bounded by this graph, the coordinate axes and is rotated about the -axis. i. Write a definite integral for the volume. (1 mark)
- ii
- Find the volume to two decimal places. (1 mark)
- c
- Find the vertical asymptotes of . (1 mark)
- d
- For the family : i. if the graph has a stationary point , find its -coordinate in terms of . (1 mark)
- ii
- For one value of there is no stationary point. Find the vertical asymptotes in this case. (1 mark)
- iii
- Find if the graph has a point of inflection at . (2 marks)
Show worked solution
a. [3 marks]. and the quadratic has no real roots, so the only vertical asymptote is ; the horizontal asymptote is .
The turning point is a maximum at . Solving with CAS gives the point of inflection .
- b. i. [1 mark]
- .
- ii. [1 mark]
- cubic units.
- c. [1 mark]
- , so the asymptotes are , and .
- d. i. [1 mark]
- (the terms cancel), so the stationary point is always at and .
- ii. [1 mark]
- There is no stationary point when is not in the domain: , so . Then , giving asymptotes and .
- iii. [2 marks]
- The numerator of is . Setting it to zero at : , so .
From the report. In part a many graphs were inaccurate: match the CAS window to the given axes, give the maximum exactly, do not draw the point of inflection as stationary, and label . In part c each asymptote needed its own equation. Part d.ii was answered correctly by only 46% of students.
Question 2 (10 marks)
- a
- Sketch on the Argand plane. (1 mark)
- b. i
- Show that is the line . (2 marks)
- ii
- Sketch this line on the same diagram. (1 mark)
- c. i
- Find the intersection points of the curves in parts a and b.i, in the form . (2 marks)
- ii
- Label them on the diagram. (1 mark)
- d
- A ray starts at and passes through , with equation . Find and . (1 mark)
- e
- Find the area of the minor segment cut from by the chord , in the form . (2 marks)
Show worked solution
a. [1 mark]. , so : the circle of radius 2 centred at the origin (see the figure below).
b. i. [2 marks]. Let :
(Geometrically: it is the perpendicular bisector of the points and .)
- ii. [1 mark]
- A straight line through the origin at to the real axis.
- c. i. [2 marks]
- Substitute into : , so . The points are and .
- ii. [1 mark]
d. [1 mark]. The ray starts at , so . The direction has argument .
e. [2 marks]. is at argument and at 0 on the circle of radius 3, so the central angle is :
From the report. In part d half the students found but not the principal argument; drawing the ray helps. In part e some used the wrong angle or did not give the required form.
Question 3 (10 marks)
A 3000 L tank initially holds 5 kg of salt. Brine at 0.1 kg/L flows in at 20 L/min and the well-mixed solution drains at 20 L/min. kg is the salt at time min.
- a
- By considering concentration, explain whether the quantity of salt increases. (1 mark)
- b
- Show that . (1 mark)
- c
- Use Euler's method with step 15 minutes to estimate to two decimal places. (2 marks)
- d
- Use calculus to solve for in terms of . (3 marks)
- e
- What does approach as ? (1 mark)
- f
- Find the time for the salt to reach 100 kg. (1 mark)
- g
- At that moment the drain is closed (inflow continues). How many minutes until the concentration reaches kg/L? (1 mark)
Show worked solution
- a. [1 mark]
- The initial concentration, kg/L, is less than the incoming kg/L, so salt enters faster than it leaves and increases.
- b. [1 mark]
- Rate in kg/min. Rate out kg/min (volume stays 3000 L). So .
- c. [2 marks]
- :
- kg
d. [3 marks]. Separate variables:
gives , so .
- e. [1 mark]
- kg.
- f. [1 mark]
- gives , so minutes.
- g. [1 mark]
- From then on the tank gains 20 L and 2 kg of salt per minute. After more minutes:
From the report. Part a was poorly answered (20% correct) because many did not quote the two concentrations. Part b needed the rate in minus rate out to be developed. In part c the Euler steps had to be shown, not just 61.05. In part d a frequent error was not using the initial condition. In part f an exact answer was expected. Part g was answered correctly by only 18%; 25 was the most common wrong answer.
Question 4 (10 marks)
A particle has position , .
- a
- Give the starting point. (1 mark)
- b
- Draw an arrow at showing the direction of motion. (1 mark)
- c
- Find when the particle first returns to its starting point. (1 mark)
- d
- Show that its speed is . (3 marks)
- e
- Find the maximum speed. (1 mark)
- f
- Trace the path for . (1 mark)
- g
- Find the length of that path, to one decimal place. (2 marks)
Show worked solution
- a. [1 mark]
- : the point .
- b. [1 mark]
- The particle is at when . There and , so the arrow points straight up: the motion is anticlockwise.
- c. [1 mark]
- The two terms have periods and ; their lowest common multiple is .
- d. [3 marks]
Using with :
e. [1 mark]. Maximum when : m/s.
f. [1 mark]. From to , shown in orange below.
g. [2 marks]. m.
From the report. In part b many did not draw the arrow at as instructed. In part c an exact answer was expected. Part d is a "show that": responses that skipped steps lost marks. In part f the trace had to finish at . In part g the definite integral had to be written, not just 36.6.
Question 5 (10 marks)
Planes , , .
- a
- Find the point of intersection of the three planes. (1 mark)
- b. i
- Find a direction vector for the line of intersection of and . (2 marks)
- ii
- Give parametric equations for this line. (1 mark)
- c
- Find the shortest distance from to . (2 marks)
- d
- is the family of planes , . i. Show that is parallel to every member of . (1 mark)
- ii
- Find all for which the distance between and is . (3 marks)
Show worked solution
a. [1 mark]. Solving the three equations simultaneously gives .
b. i. [2 marks]. The line lies in both planes, so it is perpendicular to both normals:
A direction vector is (any non-zero multiple).
ii. [1 mark]. The point lies on both planes, so , , , .
c. [2 marks]. Distance from a point to a plane:
d. i. [1 mark]. has normal and each member of has normal . Parallel normals mean parallel planes.
ii. [3 marks]. Write as . The distance between and is
From the report. In b.i some solved with CAS and left the line in parametric form without naming the direction vector. In b.ii a Cartesian equation was not accepted. In d.i students needed to say they were comparing normal vectors, and some stated the scalar multiple incorrectly (it is 3). In d.ii many missed the modulus and found only one value of .
Question 6 (10 marks)
Water drunk per day, mL, is normal with mean 1000 and standard deviation 80.
a. i. Give the mean and standard deviation of the sample mean for samples of 25 students. (1 mark)
ii. Find the probability that such a sample mean exceeds 970 mL, to four decimal places. (1 mark)
Wasser bottle volumes have mL; a sample of 30 bottles has mean 750 mL.
- b
- Find a 95% confidence interval for the mean, to one decimal place. (1 mark)
- c
- Of 300 independent such 95% intervals, how many would be expected to contain the true mean? (1 mark)
- d
- Find the minimum sample size so that the sample mean is within 1 mL of the true mean at the 95% level. (1 mark)
Apa bottles are normal with mean 750 mL and mL. After a service, a sample of 50 has mean 748 mL; the company claims the mean is now less than 750 mL. A one-tailed test at the 1% level is proposed.
- e
- State and . (1 mark)
- f. i
- Find the value to four decimal places. (1 mark)
- ii
- Is the claim correct? Explain using the value. (1 mark)
- g
- Find the critical sample mean for at the 1% level, to three decimal places. (1 mark)
- h
- If the true mean is actually 747.5 mL, find the probability of concluding the mean has not been reduced, to three decimal places. (1 mark)
Show worked solution
- a. i. [1 mark]
- Mean 1000 mL; standard deviation mL.
- ii. [1 mark]
- .
- b. [1 mark]
- gives mL.
- c. [1 mark]
- .
- d. [1 mark]
- gives , so the minimum is .
- e. [1 mark]
- and .
- f. i. [1 mark]
- with standard deviation : .
- ii. [1 mark]
- Yes: , so is rejected at the 1% level and the claim that the mean is now less than 750 mL is supported.
- g. [1 mark]
- Solve : mL.
- h. [1 mark]
- This is a type II error: .
From the report. In a.i some gave the variance instead of the standard deviation. In part d some rounded down to 96, which would allow an error of more than 1 mL. In f.ii the answer had to comment on the claim and quote the significance level. In part g some used the wrong tail.
Use this paper well
- Sit the paper under exam conditions (120 minutes, 80 marks).
- Mark yourself against the official VCAA marking notes.
- Compare against the Specialist Mathematics hub to find the syllabus dot points this paper tested.
