VCE Math Methods 2025 Exam 2
Worked solutions to the 2025 VCE Mathematical Methods 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 Mathematical Methods 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 Mathematical Methods Examination 2, copyright Victorian Curriculum and Assessment Authority (VCAA). Each question is summarised briefly here; open the official examination PDF for the full wording, graphs and answer options.
- Answers are original ExamExplained working. Every multiple-choice answer matches the key in the 2025 Mathematical Methods 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 Mathematical Methods examinations page.
- Where we say "CAS", any approved CAS calculator gives the value shown; the working shows what to enter.
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. That is 1.5 minutes per mark.
- Section A (20 marks): 20 multiple-choice questions. Aim for about 25 minutes.
- Section B (60 marks): 4 extended-response questions (13, 14, 14 and 19 marks). Give exact answers unless told to round, and show the integral, equation or distribution you put into CAS.
Section A: Multiple choice
- Q1
- Which function has range ? Answer: C - the centre is 9 and the amplitude is 3, so runs from 6 to 12. (91% correct.)
- Q2
- Find all vertical asymptotes of . Answer: A - asymptotes occur where , so , (the period is 1).
- Q3
- Given a graph of , choose the graph of . Answer: B - reflect in the -axis, then shift up 2; for example the point moves to and moves to .
- Q4
- For which does the system , have no real solutions? Answer: A - the determinant is zero at or . At the equations are multiples of each other (infinitely many solutions); at they are parallel and inconsistent, so only.
- Q5
- Which set of ordered pairs is a function with an inverse function? Answer: B - is one-to-one; each other set either repeats an -value (not a function) or repeats a -value (not one-to-one).
- Q6
- With two trapeziums on , for which function does the trapezium rule overestimate the area? Answer: B - the rule overestimates when the curve is concave up, and has on ; the other three are concave down there.
- Q7
- A loop starts with , and repeatedly replaces by and prints it while . What is printed? Answer: C - 12, 7, 2. After printing 2 the condition fails and the loop stops.
- Q8
- A 95% confidence interval for a proportion is . Find the sample size. Answer: C - and the margin is , which gives .
- Q9
- Of walkers, 20% take at least 30 minutes; of others, 40% do. Given a student took at least 30 minutes, find the probability they walked. Answer: A - .
- Q10
- With and , when is ? Answer: C - needs (the other branch is impossible), so .
- Q11
- From a 30-day share-price chart, which interval has the greatest average rate of change? Answer: D - day 14 to day 28 rises about $4.40 in 14 days (about $0.31 per day), steeper than the other three chords.
- Q12
- and for a normal . Find and . Answer: D - , so and . Subtracting gives and then .
- Q13
- Given the graphs of a line through the origin with negative gradient and a quartic , which could be ? Answer: C - with is reflected in the -axis and dilated from the -axis. Only C matches.
- Q14
- A pdf is on and on . Find . Answer: B - each piece integrates to , so and .
- Q15
- passes through . Which point must lie on ? Answer: B - set , so and , giving .
- Q16
- and has range . What must be true? Answer: D - . The domain is when and when ; for to be positive, must share the sign of , which has the sign of . So .
- Q17
- Which graph satisfies ? Answer: A - the condition means , so the graph must lie mainly below the -axis between 2 and 3. Only the parabola in A (negative between its roots 1 and 3) does.
- Q18
- Which pair of the four probability mass functions has the same mean? Answer: D - the means are 2.5 (I), 3 (II), 2 (III) and 3 (IV), so II and IV.
- Q19
- is on , is on , and the minimum length of is . Find . Answer: D - the closest point on the curve has gradient 1: gives . Its distance to the line is , so or ; but cuts the curve (distance 0), so .
- Q20
- For , which sequence of transformations does not map to ? Answer: C - dilating by from the -axis gives , then by from the -axis gives , then 1 unit right gives . Options A, B and D all give .
From the report. The hardest multiple-choice items were Q19 (14% correct), Q16 (18% correct), Q20 (36% correct) and Q17 (38% correct). In Q16, 46% chose C ( and ), which is sufficient but not necessary; in Q19 the most popular wrong answer was A.
Section B: Extended response
Question 1 (13 marks)
Let , .
- a
- Find the coordinates of both stationary points of . (2 marks)
- b
- Sketch , labelling stationary points and axial intercepts with coordinates. (2 marks)
- c
- Complete a gradient table of and values that shows has a stationary point of inflection. (2 marks)
- d
- Find the average value of between and . (2 marks)
- e
- Give a sequence of three transformations mapping to a function with a stationary point of inflection at and a local maximum at . (3 marks)
- f
- For , show that for all . (2 marks)
Show worked solution
a. [2 marks]. gives or . The stationary points are and .
b. [2 marks]. , so the -intercepts are and . The graph has a stationary point of inflection at the origin, a local maximum at , and falls steeply either side (a negative quartic).
c. [2 marks]. Choose -values either side of 0 but less than 1 (the other stationary point):
The gradient is positive, zero, positive: the sign does not change, so is a stationary point of inflection.
d. [2 marks].
e. [3 marks]. The inflection point must go and the maximum . The -values are unchanged and . One valid order:
- Dilate by a factor of 2 from the -axis.
- Reflect in the -axis.
- Translate 1 unit in the positive direction of the -axis.
Check: and .
f. [2 marks].
From the report. In part a some gave only the -values. In part b the intercept was often drawn too close to 2, and the inflection at the origin was not always shown. In part c many put in the wrong column, chose -values of 1 or more, or entered instead of . Part d was sometimes confused with average rate of change. In part e the order of transformations and the exact wording both mattered. In part f (a "show that") some worked out instead of or left out the term.
Question 2 (14 marks)
Let and , whose graphs meet at and .
- a
- Write two simultaneous equations in and and solve them algebraically to show and . (3 marks)
- b
- Find such that . (1 mark)
- c
- Use a definite integral to find the area between the graphs for , to two decimal places. (2 marks)
- d
- Let . i. Write down . (1 mark)
- ii
- Find the maximum value of on , to two decimal places. (1 mark)
- e
- Find the points where meets . (2 marks)
- f
- is an antiderivative of through . i. Show that cannot pass through both and . (2 marks)
- ii
- If is dilated by a factor of from the -axis so that the image passes through both points, find and . (2 marks)
Show worked solution
a. [3 marks]. Substitute each point into :
Divide (2) by (1): , so and .
Then from (1): .
b. [1 mark]. , so .
c. [2 marks]. The line is above the curve between the intersections:
(The exact value is .)
d. i. [1 mark].
- ii. [1 mark]
- Solve with CAS: . The maximum value is .
- e. [2 marks]
- is the inverse of (swap and in ). The graphs of and meet at and , so their inverses meet at the reflected points and .
- f. i. [2 marks]
- . Then and .
Passing through needs , but passing through needs . One constant cannot take two values, so it is not possible.
ii. [2 marks]. The image is :
Dividing, , so and . Then .
From the report. Part a needed visible algebra: using CAS for the solving step, or re-solving for the intersection points already given on the diagram, cost marks. In part b, and were common wrong answers. In d.ii many gave only the -value or the turning point's coordinates, not the maximum value, and 0.35 was a common wrong answer. In part e some gave only -values or just the rule for the inverse. In f.ii students multiplied the wrong expression by or made algebra errors solving by hand.
Question 3 (14 marks)
A driver's travel time (minutes) has pdf for , and 0 elsewhere.
- a. i
- Find the mean travel time. (1 mark)
- ii
- Find the standard deviation of the travel time. (2 marks)
- b
- The driver is late if the trip takes longer than minutes; days are independent. i. For , write a definite integral showing that the probability of being late is 0.08704. (1 mark)
- ii
- For , find the probability of being late at least once in a five-day week, to four decimal places. (2 marks)
- iii
- For , with the proportion of late days in a five-day week, find to four decimal places. (2 marks)
- iv
- Find the integer for which the probability of being late at least once in a five-day week is 0.2, correct to one decimal place. (2 marks)
- c
- A traffic-light wait is normal with mean 2.5 minutes and standard deviation . i. If , find the probability the wait is under 3.5 minutes, to two decimal places. (1 mark)
- ii
- Find so that there is a 2% chance of a wait longer than 3.5 minutes, to two decimal places. (1 mark)
- d
- Lights , , are red with probabilities 0.2, 0.3 and 0.1, independently. Complete the probability distribution of , the number of red lights. (2 marks)
Show worked solution
a. i. [1 mark]. minutes.
ii. [2 marks].
Give the exact value: the report notes that a decimal answer alone did not get full marks.
b. i. [1 mark]. .
ii. [2 marks]. Let be the number of late days.
iii. [2 marks]. , so means or :
iv. [2 marks]. Let . We need to round to 0.2. Test integer values with CAS:
| 48 | 0.0639 | 0.281 |
| 49 | 0.0453 | 0.207 |
| 50 | 0.0308 | 0.145 |
Only gives a probability that rounds to 0.2. (Solving exactly gives , which is not an integer.)
- c. i. [1 mark]
- with is .
- ii. [1 mark]
- means where , so and .
- d. [2 marks]
- Multiply the red (R) and not-red probabilities for each combination:
| 0 | 1 | 2 | 3 | |
|---|---|---|---|---|
| 0.504 | 0.398 | 0.092 | 0.006 |
The probabilities add to 1, a quick check.
From the report. In a.ii some stopped at the variance, gave no working, or gave only 5.34 instead of an exact value. In b.ii some rounded incorrectly or gave the answer without working. Part b.iv was poorly done (average 0.5 out of 2): 49.1 was a common wrong answer because an integer was required, and many tried to solve instead. In c.ii (48% correct), 0.48 and 1.19 were common wrong answers. In part d the probabilities for and were often swapped.
Question 4 (19 marks)
Let , .
- a
- Evaluate . (1 mark)
- b
- Find the exact values of for which . (1 mark)
- c
- Real numbers and in satisfy for all . Find and the largest possible . (2 marks)
- d
- Find the equation of the tangent to at (point ). (1 mark)
- e
- Apply two iterations of Newton's method to with . i. Give to one decimal place. (1 mark)
- ii
- Draw the tangent to at . (1 mark)
- f
- is the tangent at , . i. Show that . (2 marks)
- ii
- Find the minimum and maximum possible -intercepts of . (2 marks)
- iii
- Find (to two decimal places) for which has a unique -intercept equal to the -intercept of . (2 marks)
- g
- on with and . i. Show and . (2 marks)
- ii
- If also and , find the area between the graphs for to two decimal places. (2 marks)
- iii
- With , , , find and such that and . (2 marks)
Show worked solution
- a. [1 mark]
- .
- b. [1 mark]
- on : . (The next solution, , is outside the domain.)
- c. [2 marks]
- The period of is , so . The shifted input must stay in the domain, so , giving . The largest is .
- d. [1 mark]
- , so the gradient is :
e. i. [1 mark]. Newton's method: .
- (this is where the tangent from part d meets the -axis).
- .
ii. [1 mark]. At the curve is at height about 0.56 and has gradient about 0.90. Draw a line with a ruler that touches the curve there (it does not cross it nearby) and meets the -axis at .
f. i. [2 marks]. The gradient at is and the point is . Using :
ii. [2 marks]. The -intercept is . Then
which is zero at and in the interval. and , while near the open endpoints approaches 1 and 2. So the minimum is (at ) and the maximum is (at ).
iii. [2 marks]. only at . Solve , that is , with CAS: , or .
Reject : there the tangent is the -axis itself, which does not have a unique -intercept. So or .
g. i. [2 marks]. and , so . , so , and , so .
ii. [2 marks]. Now . With and , solve
to get and . The graphs also cross at , so split the area there:
iii. [2 marks]. With , the conditions are
Solving with CAS on gives the exact values and . Check: and .
From the report. In part b some gave extra or general solutions ignoring the domain. Part c was weak (15% full marks): many found but gave , or . Part d needed an equation, and CAS avoided algebra slips. In e.ii (28% correct) many lines were not tangent to the curve at ; use a ruler. In f.ii the minimum and maximum were often swapped or given as coordinates. In f.iii the invalid was often included. In g.ii a common error was integrating over without splitting at the crossing. In g.iii exact answers were needed.
General advice from the 2025 report
- Show the set-up before using CAS. Write the integral, equation or binomial expression you entered; bare answers on multi-mark parts lost marks.
- Exact versus rounded. Read each part for "exact" or "correct to decimal places" and give exactly that; both an unrounded decimal where exact was needed and an exact value where a decimal was requested lost marks.
- Answer the actual question. A maximum value is not a coordinate pair, and an integer answer must be an integer.
- Use a ruler for tangents and lines on graphs.
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 Math Methods hub to find the syllabus dot points this paper tested.
