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VICMath Methods2025Exam 2

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 [6,12][6, 12]? Answer: C - the centre is 9 and the amplitude is 3, so f(x)=9−3cos⁡(6x)f(x) = 9 - 3\cos(6x) runs from 6 to 12. (91% correct.)
Q2
Find all vertical asymptotes of y=2tan⁡ ⁣(π(x+12))y = 2\tan\!\left(\pi\left(x + \tfrac12\right)\right). Answer: A - asymptotes occur where π(x+12)=π2+kπ\pi\left(x + \tfrac12\right) = \tfrac{\pi}{2} + k\pi, so x=kx = k, k∈Zk \in Z (the period is 1).
Q3
Given a graph of y=f(x)y = f(x), choose the graph of y=f(−x)+2y = f(-x) + 2. Answer: B - reflect in the yy-axis, then shift up 2; for example the point (0,1)(0, 1) moves to (0,3)(0, 3) and (4,2)(4, 2) moves to (−4,4)(-4, 4).
Q4
For which kk does the system kx+3y=k2kx + 3y = k^2, 2x+(2k+1)y=6−2k2x + (2k+1)y = 6 - 2k have no real solutions? Answer: A - the determinant k(2k+1)−6=(2k−3)(k+2)k(2k+1) - 6 = (2k-3)(k+2) is zero at k=32k = \tfrac32 or k=−2k = -2. At k=32k = \tfrac32 the equations are multiples of each other (infinitely many solutions); at k=−2k = -2 they are parallel and inconsistent, so k=−2k = -2 only.
Q5
Which set of ordered pairs is a function with an inverse function? Answer: B - {(−1,3),(2,2),(3,1)}\{(-1, 3), (2, 2), (3, 1)\} is one-to-one; each other set either repeats an xx-value (not a function) or repeats a yy-value (not one-to-one).
Q6
With two trapeziums on [0,1][0, 1], for which function does the trapezium rule overestimate the area? Answer: B - the rule overestimates when the curve is concave up, and x3+1x^3 + 1 has f′′(x)=6x≥0f''(x) = 6x \ge 0 on [0,1][0, 1]; the other three are concave down there.
Q7
A loop starts with n=17n = 17, k=5k = 5 and repeatedly replaces nn by n−kn - k and prints it while n>kn > k. What is printed? Answer: C - 12, 7, 2. After printing 2 the condition 2>52 > 5 fails and the loop stops.
Q8
A 95% confidence interval for a proportion is (0.248,0.552)(0.248, 0.552). Find the sample size. Answer: C - p^=0.4\hat p = 0.4 and the margin is 0.152=1.960.4×0.6n0.152 = 1.96\sqrt{\tfrac{0.4 \times 0.6}{n}}, which gives n≈40n \approx 40.
Q9
Of mm walkers, 20% take at least 30 minutes; of nn others, 40% do. Given a student took at least 30 minutes, find the probability they walked. Answer: A - 0.2m0.2m+0.4n=mm+2n\dfrac{0.2m}{0.2m + 0.4n} = \dfrac{m}{m + 2n}.
Q10
With f(x)=2x2+x−1f(x) = 2x^2 + x - 1 and g(x)=sin⁡(x)g(x) = \sin(x), when is (f∘g)(x)>0(f \circ g)(x) > 0? Answer: C - 2sin⁡2x+sin⁡x−1=(2sin⁡x−1)(sin⁡x+1)>02\sin^2 x + \sin x - 1 = (2\sin x - 1)(\sin x + 1) > 0 needs sin⁡x>12\sin x > \tfrac12 (the other branch sin⁡x<−1\sin x < -1 is impossible), so 12<sin⁡(x)≤1\tfrac12 < \sin(x) \le 1.
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
Pr⁡(X>200)=0.325\Pr(X > 200) = 0.325 and Pr⁡(180<X<200)=0.589\Pr(180 < X < 200) = 0.589 for a normal XX. Find μ\mu and σ\sigma. Answer: D - Pr⁡(X<180)=0.086\Pr(X < 180) = 0.086, so 200−μσ=0.4538\tfrac{200 - \mu}{\sigma} = 0.4538 and 180−μσ=−1.3658\tfrac{180 - \mu}{\sigma} = -1.3658. Subtracting gives σ≈11\sigma \approx 11 and then μ≈195\mu \approx 195.
Q13
Given the graphs of a line y=f(x)y = f(x) through the origin with negative gradient and a quartic y=g(x)y = g(x), which could be y=(g∘f)(x)y = (g \circ f)(x)? Answer: C - g(f(x))=g(−ax)g(f(x)) = g(-ax) with a>0a > 0 is gg reflected in the yy-axis and dilated from the yy-axis. Only C matches.
Q14
A pdf is ksin⁡(x)k\sin(x) on [0,π4)\left[0, \tfrac{\pi}{4}\right) and kcos⁡(x)k\cos(x) on [π4,π2]\left[\tfrac{\pi}{4}, \tfrac{\pi}{2}\right]. Find kk. Answer: B - each piece integrates to k(1−22)k\left(1 - \tfrac{\sqrt2}{2}\right), so k(2−2)=1k(2 - \sqrt2) = 1 and k=12−2k = \dfrac{1}{2 - \sqrt2}.
Q15
y=g(x)y = g(x) passes through (1,3)(1, 3). Which point must lie on y=1−g(2x−3)y = 1 - g(2x - 3)? Answer: B - set 2x−3=12x - 3 = 1, so x=2x = 2 and y=1−3=−2y = 1 - 3 = -2, giving (2,−2)(2, -2).
Q16
h(x)=alog⁡e(bx)h(x) = a\log_e(bx) and h′(x)h'(x) has range (0,∞)(0, \infty). What must be true? Answer: D - h′(x)=axh'(x) = \tfrac{a}{x}. The domain is x>0x > 0 when b>0b > 0 and x<0x < 0 when b<0b < 0; for ax\tfrac{a}{x} to be positive, aa must share the sign of xx, which has the sign of bb. So ab>0ab > 0.
Q17
Which graph satisfies ∫12f(x) dx>∫13f(x) dx\int_1^2 f(x)\,dx > \int_1^3 f(x)\,dx? Answer: A - the condition means ∫23f(x) dx<0\int_2^3 f(x)\,dx < 0, so the graph must lie mainly below the xx-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
AA is on y=x+cy = x + c, BB is on y=log⁡e(x−1)y = \log_e(x - 1), and the minimum length of ABAB is 2\sqrt2. Find cc. Answer: D - the closest point on the curve has gradient 1: 1x−1=1\tfrac{1}{x-1} = 1 gives B(2,0)B(2, 0). Its distance to the line is ∣2+c∣2=2\tfrac{|2 + c|}{\sqrt2} = \sqrt2, so c=0c = 0 or c=−4c = -4; but y=x−4y = x - 4 cuts the curve (distance 0), so c=0c = 0.
Q20
For a>1a > 1, which sequence of transformations does not map f(x)=axf(x) = a^x to g(x)=a2x+2g(x) = a^{2x+2}? Answer: C - dilating by aa from the xx-axis gives ax+1a^{x+1}, then by 12\tfrac12 from the yy-axis gives a2x+1a^{2x+1}, then 1 unit right gives a2(x−1)+1=a2x−1≠g(x)a^{2(x-1)+1} = a^{2x-1} \ne g(x). Options A, B and D all give a2x+2a^{2x+2}.

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 (a>0a > 0 and b>0b > 0), which is sufficient but not necessary; in Q19 the most popular wrong answer was A.

Section B: Extended response

Question 1 (13 marks)

Let g(x)=4x3−3x4g(x) = 4x^3 - 3x^4, x∈Rx \in R.

a
Find the coordinates of both stationary points of gg. (2 marks)
b
Sketch y=g(x)y = g(x), labelling stationary points and axial intercepts with coordinates. (2 marks)
c
Complete a gradient table of xx and g′(x)g'(x) values that shows gg has a stationary point of inflection. (2 marks)
d
Find the average value of gg between x=0x = 0 and x=2x = 2. (2 marks)
e
Give a sequence of three transformations mapping gg to a function hh with a stationary point of inflection at (1,0)(1, 0) and a local maximum at (−1,1)(-1, 1). (3 marks)
f
For X∼Bi(4,p)X \sim \text{Bi}(4, p), show that Pr⁡(X≥3)=g(p)\Pr(X \ge 3) = g(p) for all p∈[0,1]p \in [0, 1]. (2 marks)
Show worked solution

a. [2 marks]. g′(x)=12x2−12x3=12x2(1−x)=0g'(x) = 12x^2 - 12x^3 = 12x^2(1 - x) = 0 gives x=0x = 0 or x=1x = 1. The stationary points are (0,0)(0, 0) and (1,1)(1, 1).

b. [2 marks]. g(x)=x3(4−3x)g(x) = x^3(4 - 3x), so the xx-intercepts are (0,0)(0, 0) and (43,0)\left(\tfrac43, 0\right). The graph has a stationary point of inflection at the origin, a local maximum at (1,1)(1, 1), and falls steeply either side (a negative quartic).

Graph of y = 4x cubed minus 3x to the fourth The quartic rises from the lower left, flattens at a stationary point of inflection at the origin, climbs to a local maximum at (1, 1), then falls through the x-intercept (4/3, 0) and down to the lower right. x y -2 -1 1 2 -2 -1 1 2 (0, 0) (1, 1) (4/3, 0)

c. [2 marks]. Choose xx-values either side of 0 but less than 1 (the other stationary point):

xx −1-1 00 12\tfrac12
g′(x)g'(x) 2424 00 1.51.5

The gradient is positive, zero, positive: the sign does not change, so (0,0)(0, 0) is a stationary point of inflection.

d. [2 marks].

average value=12−0∫02(4x3−3x4)dx=12[x4−35x5]02=12(16−965)=−85.\text{average value} = \frac{1}{2 - 0}\int_0^2\left(4x^3 - 3x^4\right)dx = \frac12\left[x^4 - \tfrac35x^5\right]_0^2 = \frac12\left(16 - \frac{96}{5}\right) = -\frac85.

e. [3 marks]. The inflection point must go (0,0)→(1,0)(0, 0) \to (1, 0) and the maximum (1,1)→(−1,1)(1, 1) \to (-1, 1). The yy-values are unchanged and x↦1−2xx \mapsto 1 - 2x. One valid order:

  1. Dilate by a factor of 2 from the yy-axis.
  2. Reflect in the yy-axis.
  3. Translate 1 unit in the positive direction of the xx-axis.

Check: (1,1)→(2,1)→(−2,1)→(−1,1)(1, 1) \to (2, 1) \to (-2, 1) \to (-1, 1) and (0,0)→(1,0)(0, 0) \to (1, 0).

f. [2 marks].

Pr⁡(X≥3)=(43)p3(1−p)+(44)p4=4p3−4p4+p4=4p3−3p4=g(p).\Pr(X \ge 3) = \binom43p^3(1-p) + \binom44p^4 = 4p^3 - 4p^4 + p^4 = 4p^3 - 3p^4 = g(p).

From the report. In part a some gave only the xx-values. In part b the intercept (43,0)\left(\tfrac43, 0\right) was often drawn too close to 2, and the inflection at the origin was not always shown. In part c many put (0,0)(0, 0) in the wrong column, chose xx-values of 1 or more, or entered g(x)g(x) instead of g′(x)g'(x). 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 Pr⁡(X≤3)\Pr(X \le 3) instead of Pr⁡(X≥3)\Pr(X \ge 3) or left out the Pr⁡(X=4)\Pr(X = 4) term.

Question 2 (14 marks)

Let f(x)=x2+7f(x) = \tfrac{x}{2} + 7 and g(x)=Aekxg(x) = Ae^{kx}, whose graphs meet at (−12,1)(-12, 1) and (2,8)(2, 8).

a
Write two simultaneous equations in AA and kk and solve them algebraically to show A=2187A = 2^{\frac{18}{7}} and k=314log⁡e(2)k = \tfrac{3}{14}\log_e(2). (3 marks)
b
Find bb such that g(x)=A×2bxg(x) = A \times 2^{bx}. (1 mark)
c
Use a definite integral to find the area between the graphs for x∈[−12,2]x \in [-12, 2], to two decimal places. (2 marks)
d
Let h(x)=f(x)−g(x)h(x) = f(x) - g(x). i. Write down h′(x)h'(x). (1 mark)
ii
Find the maximum value of hh on [−12,2][-12, 2], to two decimal places. (1 mark)
e
Find the points where y=g−1(x)y = g^{-1}(x) meets y=2(x−7)y = 2(x - 7). (2 marks)
f
FF is an antiderivative of ff through (0,c)(0, c). i. Show that y=F(x)y = F(x) cannot pass through both (−12,1)(-12, 1) and (2,8)(2, 8). (2 marks)
ii
If y=F(x)y = F(x) is dilated by a factor of mm from the xx-axis so that the image passes through both points, find mm and cc. (2 marks)
Show worked solution

a. [3 marks]. Substitute each point into gg:

Ae−12k=1(1)Ae2k=8(2).Ae^{-12k} = 1 \quad (1) \qquad Ae^{2k} = 8 \quad (2).

Divide (2) by (1): e14k=8=23e^{14k} = 8 = 2^3, so 14k=3log⁡e(2)14k = 3\log_e(2) and k=314log⁡e(2)k = \tfrac{3}{14}\log_e(2).

Then from (1): A=e12k=e3614log⁡e2=2187A = e^{12k} = e^{\frac{36}{14}\log_e 2} = 2^{\frac{18}{7}}.

b. [1 mark]. ekx=e314xlog⁡e2=2314xe^{kx} = e^{\frac{3}{14}x\log_e 2} = 2^{\frac{3}{14}x}, so b=314b = \dfrac{3}{14}.

c. [2 marks]. The line is above the curve between the intersections:

Area=∫−122(x2+7−2187e3log⁡e214x)dx≈15.87 square units.\text{Area} = \int_{-12}^{2}\left(\frac{x}{2} + 7 - 2^{\frac{18}{7}}e^{\frac{3\log_e 2}{14}x}\right)dx \approx 15.87 \text{ square units}.

(The exact value is 63−983log⁡e263 - \dfrac{98}{3\log_e 2}.)

d. i. [1 mark].

h′(x)=12−Ak ekx=12−3log⁡e(2)14 2187 e3log⁡e(2)14x.h'(x) = \frac12 - Ak\,e^{kx} = \frac12 - \frac{3\log_e(2)}{14}\, 2^{\frac{18}{7}}\, e^{\frac{3\log_e(2)}{14}x}.

ii. [1 mark]
Solve h′(x)=0h'(x) = 0 with CAS: x≈−3.828x \approx -3.828. The maximum value is h(−3.828…)≈1.72h(-3.828\ldots) \approx 1.72.
e. [2 marks]
y=2(x−7)y = 2(x - 7) is the inverse of ff (swap xx and yy in y=x2+7y = \tfrac{x}{2} + 7). The graphs of ff and gg meet at (−12,1)(-12, 1) and (2,8)(2, 8), so their inverses meet at the reflected points (1,−12)(1, -12) and (8,2)(8, 2).
f. i. [2 marks]
F(x)=x24+7x+cF(x) = \dfrac{x^2}{4} + 7x + c. Then F(−12)=36−84+c=c−48F(-12) = 36 - 84 + c = c - 48 and F(2)=1+14+c=c+15F(2) = 1 + 14 + c = c + 15.

Passing through (−12,1)(-12, 1) needs c=49c = 49, but passing through (2,8)(2, 8) needs c=−7c = -7. One constant cannot take two values, so it is not possible.

ii. [2 marks]. The image is y=mF(x)y = mF(x):

m(c−48)=1,m(c+15)=8.m(c - 48) = 1, \qquad m(c + 15) = 8.

Dividing, c+15=8(c−48)c + 15 = 8(c - 48), so 7c=3997c = 399 and c=57c = 57. Then m=157−48=19m = \dfrac{1}{57 - 48} = \dfrac19.

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, b=34b = \tfrac34 and b=34log⁡e(2)b = \tfrac34\log_e(2) were common wrong answers. In d.ii many gave only the xx-value −3.83-3.83 or the turning point's coordinates, not the maximum value, and 0.35 was a common wrong answer. In part e some gave only xx-values or just the rule for the inverse. In f.ii students multiplied the wrong expression by mm or made algebra errors solving by hand.

Question 3 (14 marks)

A driver's travel time TT (minutes) has pdf f(t)=11 215 000(t−29)(59−t)3f(t) = \dfrac{1}{1\,215\,000}(t - 29)(59 - t)^3 for 29≤t≤5929 \le t \le 59, 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 kk minutes; days are independent. i. For k=47k = 47, write a definite integral showing that the probability of being late is 0.08704. (1 mark)
ii
For k=47k = 47, find the probability of being late at least once in a five-day week, to four decimal places. (2 marks)
iii
For k=47k = 47, with P^\hat P the proportion of late days in a five-day week, find Pr⁡(0.4≤P^≤0.6)\Pr(0.4 \le \hat P \le 0.6) to four decimal places. (2 marks)
iv
Find the integer kk 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 σ\sigma. i. If σ=0.6\sigma = 0.6, find the probability the wait is under 3.5 minutes, to two decimal places. (1 mark)
ii
Find σ\sigma so that there is a 2% chance of a wait longer than 3.5 minutes, to two decimal places. (1 mark)
d
Lights AA, BB, CC are red with probabilities 0.2, 0.3 and 0.1, independently. Complete the probability distribution of YY, the number of red lights. (2 marks)
Show worked solution

a. i. [1 mark]. E(T)=∫2959t f(t) dt=39\text{E}(T) = \displaystyle\int_{29}^{59} t\,f(t)\,dt = 39 minutes.

ii. [2 marks].

var(T)=∫2959(t−39)2f(t) dt=2007,sd(T)=2007=10147≈5.35 minutes.\text{var}(T) = \int_{29}^{59}(t - 39)^2 f(t)\,dt = \frac{200}{7}, \qquad \text{sd}(T) = \sqrt{\frac{200}{7}} = \frac{10\sqrt{14}}{7} \approx 5.35 \text{ minutes}.

Give the exact value: the report notes that a decimal answer alone did not get full marks.

b. i. [1 mark]. Pr⁡(T>47)=∫475911 215 000(t−29)(59−t)3 dt=0.08704\Pr(T > 47) = \displaystyle\int_{47}^{59}\frac{1}{1\,215\,000}(t - 29)(59 - t)^3\,dt = 0.08704.

ii. [2 marks]. Let L∼Bi(5,0.08704)L \sim \text{Bi}(5, 0.08704) be the number of late days.

Pr⁡(L≥1)=1−(1−0.08704)5=1−0.912965≈0.3658.\Pr(L \ge 1) = 1 - (1 - 0.08704)^5 = 1 - 0.91296^5 \approx 0.3658.

iii. [2 marks]. P^=L5\hat P = \tfrac{L}{5}, so 0.4≤P^≤0.60.4 \le \hat P \le 0.6 means L=2L = 2 or L=3L = 3:

Pr⁡(L=2)+Pr⁡(L=3)=(52)(0.08704)2(0.91296)3+(53)(0.08704)3(0.91296)2≈0.0631.\Pr(L = 2) + \Pr(L = 3) = \binom52(0.08704)^2(0.91296)^3 + \binom53(0.08704)^3(0.91296)^2 \approx 0.0631.

iv. [2 marks]. Let q=Pr⁡(T>k)q = \Pr(T > k). We need 1−(1−q)51 - (1 - q)^5 to round to 0.2. Test integer values with CAS:

kk q=Pr⁡(T>k)q = \Pr(T > k) 1−(1−q)51 - (1-q)^5
48 0.0639 0.281
49 0.0453 0.207
50 0.0308 0.145

Only k=49k = 49 gives a probability that rounds to 0.2. (Solving 1−(1−q)5=0.21 - (1 - q)^5 = 0.2 exactly gives k≈49.1k \approx 49.1, which is not an integer.)

c. i. [1 mark]
Pr⁡(W<3.5)\Pr(W < 3.5) with W∼N(2.5,0.62)W \sim N(2.5, 0.6^2) is 0.950.95.
ii. [1 mark]
Pr⁡(W>3.5)=0.02\Pr(W > 3.5) = 0.02 means 3.5−2.5σ=z\tfrac{3.5 - 2.5}{\sigma} = z where Pr⁡(Z<z)=0.98\Pr(Z < z) = 0.98, so z≈2.0537z \approx 2.0537 and σ=12.0537≈0.49\sigma = \tfrac{1}{2.0537} \approx 0.49.
d. [2 marks]
Multiply the red (R) and not-red probabilities for each combination:
  • Pr⁡(Y=0)=0.8×0.7×0.9=0.504\Pr(Y = 0) = 0.8 \times 0.7 \times 0.9 = 0.504
  • Pr⁡(Y=1)=0.2(0.7)(0.9)+0.8(0.3)(0.9)+0.8(0.7)(0.1)=0.126+0.216+0.056=0.398\Pr(Y = 1) = 0.2(0.7)(0.9) + 0.8(0.3)(0.9) + 0.8(0.7)(0.1) = 0.126 + 0.216 + 0.056 = 0.398
  • Pr⁡(Y=2)=0.2(0.3)(0.9)+0.2(0.7)(0.1)+0.8(0.3)(0.1)=0.054+0.014+0.024=0.092\Pr(Y = 2) = 0.2(0.3)(0.9) + 0.2(0.7)(0.1) + 0.8(0.3)(0.1) = 0.054 + 0.014 + 0.024 = 0.092
  • Pr⁡(Y=3)=0.2×0.3×0.1=0.006\Pr(Y = 3) = 0.2 \times 0.3 \times 0.1 = 0.006
yy 0 1 2 3
Pr⁡(Y=y)\Pr(Y = y) 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 ∫k59f(t) dt=0.2\int_k^{59} f(t)\,dt = 0.2 instead. In c.ii (48% correct), 0.48 and 1.19 were common wrong answers. In part d the probabilities for Y=1Y = 1 and Y=2Y = 2 were often swapped.

Question 4 (19 marks)

Let f:[0,5π2]→Rf:\left[0, \tfrac{5\pi}{2}\right] \to R, f(x)=sin⁡(x)+1f(x) = \sin(x) + 1.

a
Evaluate f ⁣(2π3)f\!\left(\tfrac{2\pi}{3}\right). (1 mark)
b
Find the exact values of xx for which f(x)=32f(x) = \tfrac32. (1 mark)
c
Real numbers aa and kk in (0,5π2)\left(0, \tfrac{5\pi}{2}\right) satisfy f(x+k)=f(x)f(x + k) = f(x) for all x∈[0,a]x \in [0, a]. Find kk and the largest possible aa. (2 marks)
d
Find the equation of the tangent to y=f(x)y = f(x) at x=2π3x = \tfrac{2\pi}{3} (point AA). (1 mark)
e
Apply two iterations of Newton's method to ff with x0=2π3x_0 = \tfrac{2\pi}{3}. i. Give x2x_2 to one decimal place. (1 mark)
ii
Draw the tangent to y=f(x)y = f(x) at x=x1x = x_1. (1 mark)
f
y=t(x)y = t(x) is the tangent at (p,f(p))(p, f(p)), p∈(0,5π2)p \in \left(0, \tfrac{5\pi}{2}\right). i. Show that t(x)=cos⁡(p)(x−p)+sin⁡(p)+1t(x) = \cos(p)(x - p) + \sin(p) + 1. (2 marks)
ii
Find the minimum and maximum possible yy-intercepts of y=t(x)y = t(x). (2 marks)
iii
Find pp (to two decimal places) for which y=t(x)y = t(x) has a unique xx-intercept equal to the xx-intercept of y=f(x)y = f(x). (2 marks)
g
g(x)=ax3+bx2+cx+dg(x) = ax^3 + bx^2 + cx + d on [0,5π2]\left[0, \tfrac{5\pi}{2}\right] with g(0)=f(0)g(0) = f(0) and g′(0)=f′(0)g'(0) = f'(0). i. Show c=1c = 1 and d=1d = 1. (2 marks)
ii
If also g(2π)=f(2π)g(2\pi) = f(2\pi) and g′(2π)=f′(2π)g'(2\pi) = f'(2\pi), find the area between the graphs for x∈[0,2π]x \in [0, 2\pi] to two decimal places. (2 marks)
iii
With a=0a = 0, c=1c = 1, d=1d = 1, find bb and r∈(0,5π2)r \in \left(0, \tfrac{5\pi}{2}\right) such that g(r)=f(r)g(r) = f(r) and g′(r)=f′(r)g'(r) = f'(r). (2 marks)
Show worked solution
a. [1 mark]
f ⁣(2π3)=sin⁡ ⁣(2π3)+1=32+1=2+32f\!\left(\tfrac{2\pi}{3}\right) = \sin\!\left(\tfrac{2\pi}{3}\right) + 1 = \dfrac{\sqrt3}{2} + 1 = \dfrac{2 + \sqrt3}{2}.
b. [1 mark]
sin⁡(x)=12\sin(x) = \tfrac12 on [0,5π2]\left[0, \tfrac{5\pi}{2}\right]: x=π6, 5π6, 13π6x = \dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{13\pi}{6}. (The next solution, 17π6\tfrac{17\pi}{6}, is outside the domain.)
c. [2 marks]
The period of sin⁡(x)+1\sin(x) + 1 is 2π2\pi, so k=2πk = 2\pi. The shifted input x+2πx + 2\pi must stay in the domain, so x+2π≤5π2x + 2\pi \le \tfrac{5\pi}{2}, giving x≤π2x \le \tfrac{\pi}{2}. The largest aa is π2\dfrac{\pi}{2}.
d. [1 mark]
f′(x)=cos⁡(x)f'(x) = \cos(x), so the gradient is cos⁡ ⁣(2π3)=−12\cos\!\left(\tfrac{2\pi}{3}\right) = -\tfrac12:

y=−12(x−2π3)+32+1that isy=−x2+π3+32+1.y = -\frac12\left(x - \frac{2\pi}{3}\right) + \frac{\sqrt3}{2} + 1 \quad\text{that is}\quad y = -\frac{x}{2} + \frac{\pi}{3} + \frac{\sqrt3}{2} + 1.

e. i. [1 mark]. Newton's method: xn+1=xn−sin⁡(xn)+1cos⁡(xn)x_{n+1} = x_n - \dfrac{\sin(x_n) + 1}{\cos(x_n)}.

  • x1=2π3+2+3≈5.826x_1 = \tfrac{2\pi}{3} + 2 + \sqrt3 \approx 5.826 (this is where the tangent from part d meets the xx-axis).
  • x2=x1−sin⁡(x1)+1cos⁡(x1)≈5.826−0.5590.897≈5.2x_2 = x_1 - \dfrac{\sin(x_1) + 1}{\cos(x_1)} \approx 5.826 - \dfrac{0.559}{0.897} \approx 5.2.

ii. [1 mark]. At x1≈5.83x_1 \approx 5.83 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 xx-axis at x2≈5.2x_2 \approx 5.2.

Newton method tangents for y = sin(x) + 1 The curve y = sin(x) + 1 on [0, 5pi/2]. The dashed tangent at A, where x = 2pi/3, meets the x-axis at x1, about 5.83. The solid tangent drawn at the point on the curve above x1 has positive gradient and meets the x-axis at x2, about 5.20. x y π 2π 1 2 A x₁ x₂

f. i. [2 marks]. The gradient at x=px = p is f′(p)=cos⁡(p)f'(p) = \cos(p) and the point is (p,sin⁡(p)+1)(p, \sin(p) + 1). Using y−y1=m(x−x1)y - y_1 = m(x - x_1):

t(x)−(sin⁡(p)+1)=cos⁡(p)(x−p)  ⟹  t(x)=cos⁡(p)(x−p)+sin⁡(p)+1.t(x) - (\sin(p) + 1) = \cos(p)(x - p) \implies t(x) = \cos(p)(x - p) + \sin(p) + 1.

ii. [2 marks]. The yy-intercept is Y(p)=t(0)=sin⁡(p)−pcos⁡(p)+1Y(p) = t(0) = \sin(p) - p\cos(p) + 1. Then

Y′(p)=cos⁡(p)−cos⁡(p)+psin⁡(p)=psin⁡(p),Y'(p) = \cos(p) - \cos(p) + p\sin(p) = p\sin(p),

which is zero at p=πp = \pi and p=2πp = 2\pi in the interval. Y(π)=1+πY(\pi) = 1 + \pi and Y(2π)=1−2πY(2\pi) = 1 - 2\pi, while near the open endpoints YY approaches 1 and 2. So the minimum is 1−2π1 - 2\pi (at p=2πp = 2\pi) and the maximum is 1+π1 + \pi (at p=πp = \pi).

iii. [2 marks]. f(x)=0f(x) = 0 only at x=3π2x = \tfrac{3\pi}{2}. Solve t ⁣(3π2)=0t\!\left(\tfrac{3\pi}{2}\right) = 0, that is cos⁡(p)(3π2−p)+sin⁡(p)+1=0\cos(p)\left(\tfrac{3\pi}{2} - p\right) + \sin(p) + 1 = 0, with CAS: p=3π2p = \tfrac{3\pi}{2}, p≈2.38p \approx 2.38 or p≈7.04p \approx 7.04.

Reject p=3π2≈4.71p = \tfrac{3\pi}{2} \approx 4.71: there the tangent is the xx-axis itself, which does not have a unique xx-intercept. So p≈2.38p \approx 2.38 or p≈7.04p \approx 7.04.

g. i. [2 marks]. g(0)=dg(0) = d and f(0)=sin⁡(0)+1=1f(0) = \sin(0) + 1 = 1, so d=1d = 1. g′(x)=3ax2+2bx+cg'(x) = 3ax^2 + 2bx + c, so g′(0)=cg'(0) = c, and f′(0)=cos⁡(0)=1f'(0) = \cos(0) = 1, so c=1c = 1.

ii. [2 marks]. Now g(x)=ax3+bx2+x+1g(x) = ax^3 + bx^2 + x + 1. With f(2π)=1f(2\pi) = 1 and f′(2π)=1f'(2\pi) = 1, solve

8π3a+4π2b+2π+1=1,12π2a+4πb+1=18\pi^3a + 4\pi^2b + 2\pi + 1 = 1, \qquad 12\pi^2a + 4\pi b + 1 = 1

to get a=12π2a = \dfrac{1}{2\pi^2} and b=−32πb = -\dfrac{3}{2\pi}. The graphs also cross at x=πx = \pi, so split the area there:

Area=∫0π(f(x)−g(x))dx+∫π2π(g(x)−f(x))dx=∫02π∣f(x)−g(x)∣dx≈1.53.\text{Area} = \int_0^{\pi}\big(f(x) - g(x)\big)dx + \int_{\pi}^{2\pi}\big(g(x) - f(x)\big)dx = \int_0^{2\pi}\left|f(x) - g(x)\right|dx \approx 1.53.

iii. [2 marks]. With g(x)=bx2+x+1g(x) = bx^2 + x + 1, the conditions are

br2+r+1=sin⁡(r)+1and2br+1=cos⁡(r).br^2 + r + 1 = \sin(r) + 1 \quad\text{and}\quad 2br + 1 = \cos(r).

Solving with CAS on (0,5π2)\left(0, \tfrac{5\pi}{2}\right) gives the exact values r=πr = \pi and b=−1πb = -\dfrac{1}{\pi}. Check: g(π)=−π+π+1=1=f(π)g(\pi) = -\pi + \pi + 1 = 1 = f(\pi) and g′(π)=−2+1=−1=cos⁡(π)g'(\pi) = -2 + 1 = -1 = \cos(\pi).

From the report. In part b some gave extra or general solutions ignoring the domain. Part c was weak (15% full marks): many found kk but gave a=5π2a = \tfrac{5\pi}{2}, 9π2\tfrac{9\pi}{2} or 2π2\pi. Part d needed an equation, and CAS avoided algebra slips. In e.ii (28% correct) many lines were not tangent to the curve at x1x_1; use a ruler. In f.ii the minimum and maximum were often swapped or given as coordinates. In f.iii the invalid p=4.71p = 4.71 was often included. In g.ii a common error was integrating f−gf - g over [0,2π][0, 2\pi] 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 nn 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.

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