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VICSpecialist Mathematics2025Exam 2

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

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 g=9.8 m s−2g = 9.8 \text{ m s}^{-2}.

  • 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 f′′(0)=0f''(0) = 0 then ff has a point of inflection at x=0x = 0"? Answer: D - f(x)=x4−xf(x) = x^4 - x has f′′(x)=12x2f''(x) = 12x^2, which is 0 at x=0x = 0 but does not change sign, so there is no inflection there. (48% correct; 40% chose C.)
Q3
y=x2+abx+cy = \dfrac{x^2 + a}{bx + c} has asymptote y=−12x+14y = -\tfrac12x + \tfrac14 and yy-intercept −2-2. Find a,b,ca, b, c. Answer: A - dividing gives y=xb−cb2+…y = \tfrac{x}{b} - \tfrac{c}{b^2} + \ldots, so 1b=−12\tfrac1b = -\tfrac12 (b=−2b = -2) and −c4=14-\tfrac{c}{4} = \tfrac14 (c=−1c = -1); then ac=−2\tfrac{a}{c} = -2 gives a=2a = 2.
Q4
An algorithm adds π(f(left))2\pi\big(f(\text{left})\big)^2 for left =1,2,3= 1, 2, 3 with f(x)=x+1f(x) = \sqrt{x + 1}. What is printed? Answer: B - π(2+3+4)=9π\pi(2 + 3 + 4) = 9\pi.
Q5
z3+az2+bz−52=0z^3 + az^2 + bz - 52 = 0 has root 2−3i2 - 3i. Find abab. Answer: A - the conjugate is also a root, giving the factor z2−4z+13z^2 - 4z + 13; the third factor is z−4z - 4 (since 13×(−4)=−5213 \times (-4) = -52). Expanding gives a=−8a = -8 and b=29b = 29, so ab=−232ab = -232.
Q6
If ∣z∣=1|z| = 1 and z≠1z \ne 1, find Re(11−z)\text{Re}\left(\dfrac{1}{1 - z}\right). Answer: C - with z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta, 11−z=1−zˉ∣1−z∣2\dfrac{1}{1 - z} = \dfrac{1 - \bar z}{|1 - z|^2} and ∣1−z∣2=2−2cos⁡θ=2(1−cos⁡θ)|1 - z|^2 = 2 - 2\cos\theta = 2(1 - \cos\theta), so the real part is 1−cos⁡θ2(1−cos⁡θ)=12\dfrac{1 - \cos\theta}{2(1 - \cos\theta)} = \dfrac12.
Q7
Use u=cos⁡θu = \cos\theta to rewrite 12∫0π/2sin⁡2θ1+cos⁡θ dθ\tfrac12\int_0^{\pi/2}\frac{\sin 2\theta}{1 + \cos\theta}\,d\theta. Answer: D - sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta and du=−sin⁡θ dθdu = -\sin\theta\,d\theta, giving ∫01u1+u du=∫01(1−11+u)du\int_0^1\frac{u}{1 + u}\,du = \int_0^1\left(1 - \frac{1}{1 + u}\right)du.
Q8
Which differential equation matches the direction field? Answer: A - the slopes are symmetric about the yy-axis (so xx appears squared), and on the positive yy-axis the slope is negative: dydx=x2−y\dfrac{dy}{dx} = x^2 - y.
Q9
Surface area when x=ktx = kt, y=ekty = e^{kt}, a≤t≤ba \le t \le b, is rotated about the xx-axis. Answer: C - 2π∫abektk2+k2e2kt dt2\pi\int_a^b e^{kt}\sqrt{k^2 + k^2e^{2kt}}\,dt; with u=ektu = e^{kt}, du=kekt dtdu = ke^{kt}\,dt, this is 2π∫ekaekb1+u2 du2\pi\int_{e^{ka}}^{e^{kb}}\sqrt{1 + u^2}\,du.
Q10
The region under y=3cos⁡−1(x)y = 3\cos^{-1}(x), 0≤y≤a0 \le y \le a, beside x=0x = 0 is rotated about the yy-axis, with volume π(4π+33)8\dfrac{\pi(4\pi + 3\sqrt3)}{8}. Find aa. Answer: D - V=π∫0acos⁡2 ⁣(y3)dy=π2(a+32sin⁡2a3)V = \pi\int_0^a\cos^2\!\left(\tfrac y3\right)dy = \tfrac{\pi}{2}\left(a + \tfrac32\sin\tfrac{2a}{3}\right), which equals the given volume at a=πa = \pi.
Q11
dydx=x2y3\dfrac{dy}{dx} = x^2y^3 with y(1)=3y(1) = 3. Domain of yy? Answer: B - separating gives y2=97−6x3y^2 = \dfrac{9}{7 - 6x^3}, which needs 7−6x3>07 - 6x^3 > 0: x<(76)1/3x < \left(\tfrac76\right)^{1/3}.
Q12
Constant acceleration, velocity uu at AA and vv at BB. Velocity at the midpoint of ABAB? Answer: D - vm2=u2+2as2v_m^2 = u^2 + 2a\tfrac{s}{2} and v2=u2+2asv^2 = u^2 + 2as give vm=u2+v22v_m = \sqrt{\dfrac{u^2 + v^2}{2}}. (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, −49=20t−4.9t2-49 = 20t - 4.9t^2, so t≈5.80t \approx 5.80 s.
Q14
Non-zero a,b\mathbf{a}, \mathbf{b} with a⋅b=∣a×b∣\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}\times\mathbf{b}|. Angle between them? Answer: B - ∣a∣∣b∣cos⁡θ=∣a∣∣b∣sin⁡θ|\mathbf a||\mathbf b|\cos\theta = |\mathbf a||\mathbf b|\sin\theta, so tan⁡θ=1\tan\theta = 1 and θ=π4\theta = \tfrac{\pi}{4}.
Q15
Planes 2x+2y+z=22x + 2y + z = 2 and ax+4z=1ax + 4z = 1 meet at angle cos⁡−123\cos^{-1}\tfrac23. Equation for aa? Answer: A - using normals, 2a+43a2+16=23\dfrac{2a + 4}{3\sqrt{a^2 + 16}} = \dfrac23, which simplifies to a+2=a2+16a + 2 = \sqrt{a^2 + 16}.
Q16
r(t)=ne−2ti−t2j\mathbf{r}(t) = ne^{-2t}\mathbf{i} - t^2\mathbf{j}; for which nn are acceleration and velocity perpendicular at t=12t = \tfrac12? Answer: C - v⋅a=−8n2e−4t+4t=0\mathbf{v}\cdot\mathbf{a} = -8n^2e^{-4t} + 4t = 0 at t=12t = \tfrac12 gives n2=e24n^2 = \tfrac{e^2}{4}, so n=e2n = \tfrac{e}{2}.
Q17
A particle from rest has a(t)=4cos⁡(2t)i+10sin⁡(2t)j−6e−2tk\mathbf{a}(t) = 4\cos(2t)\mathbf{i} + 10\sin(2t)\mathbf{j} - 6e^{-2t}\mathbf{k}. Find v(t)\mathbf{v}(t). Answer: C - integrate and choose constants so v(0)=0\mathbf{v}(0) = \mathbf 0: 2sin⁡(2t)i−5(cos⁡(2t)−1)j+3(e−2t−1)k2\sin(2t)\mathbf{i} - 5\big(\cos(2t) - 1\big)\mathbf{j} + 3\big(e^{-2t} - 1\big)\mathbf{k}.
Q18
Two lines meet at (4,3,t)(4, 3, t). Find rr, ss, tt. Answer: D - matching coordinates gives λ=2\lambda = 2, μ=3\mu = 3, and so r=5r = 5, s=8s = 8, t=5t = 5.
Q19
Area of the triangle cut from the axes by x+y+z=ax + y + z = a. Answer: B - the intercepts are aa along each axis, forming an equilateral triangle of side a2a\sqrt2, with area 34(2a2)=3a22\tfrac{\sqrt3}{4}(2a^2) = \tfrac{\sqrt3a^2}{2}.
Q20
P∼N(−2,22)P \sim N(-2, 2^2), Q∼N(3,32)Q \sim N(3, 3^2), R∼N(5,62)R \sim N(5, 6^2) independent. Find Pr⁡(3P+2Q−R>25)\Pr(3P + 2Q - R > 25). Answer: A - the mean is −6+6−5=−5-6 + 6 - 5 = -5 and the variance is 36+36+36=10836 + 36 + 36 = 108, so Pr⁡ ⁣(Z>3063)=Pr⁡ ⁣(Z>533)\Pr\!\left(Z > \tfrac{30}{6\sqrt3}\right) = \Pr\!\left(Z > \tfrac{5\sqrt3}{3}\right).

Section B: Extended response

Question 1 (10 marks)

a
Sketch y=3xx3+x+2y = \dfrac{3x}{x^3 + x + 2}, 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 x=2x = 2 is rotated about the xx-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 y=3xx3−5x+2y = \dfrac{3x}{x^3 - 5x + 2}. (1 mark)
d
For the family y=3xx3+ax+2y = \dfrac{3x}{x^3 + ax + 2}: i. if the graph has a stationary point PP, find its yy-coordinate in terms of aa. (1 mark)
ii
For one value of aa there is no stationary point. Find the vertical asymptotes in this case. (1 mark)
iii
Find aa if the graph has a point of inflection at x=2x = 2. (2 marks)
Show worked solution

a. [3 marks]. x3+x+2=(x+1)(x2−x+2)x^3 + x + 2 = (x + 1)(x^2 - x + 2) and the quadratic has no real roots, so the only vertical asymptote is x=−1x = -1; the horizontal asymptote is y=0y = 0.

dydx=3(x3+x+2)−3x(3x2+1)(x3+x+2)2=6(1−x3)(x3+x+2)2=0  ⟹  x=1.\frac{dy}{dx} = \frac{3(x^3 + x + 2) - 3x(3x^2 + 1)}{(x^3 + x + 2)^2} = \frac{6(1 - x^3)}{(x^3 + x + 2)^2} = 0 \implies x = 1.

The turning point is a maximum at (1,34)\left(1, \tfrac34\right). Solving d2ydx2=0\tfrac{d^2y}{dx^2} = 0 with CAS gives the point of inflection (1.7,0.6)(1.7, 0.6).

Graph of y = 3x / (x cubed + x + 2) A vertical asymptote at x = -1 and the x-axis (y = 0) as horizontal asymptote. Left of x = -1 the curve rises from just above the x-axis towards positive infinity; right of x = -1 it rises from negative infinity through the origin to a maximum at (1, 3/4), passes a point of inflection near (1.7, 0.6), then decays towards the x-axis. x y -4 -3 -2 2 3 4 -2 -1 1 2 (1, 3/4) (1.7, 0.6) x = -1 y = 0

b. i. [1 mark]
V=π∫02(3xx3+x+2)2dxV = \pi\displaystyle\int_0^2\left(\frac{3x}{x^3 + x + 2}\right)^2dx.
ii. [1 mark]
V≈2.29V \approx 2.29 cubic units.
c. [1 mark]
x3−5x+2=(x−2)(x2+2x−1)x^3 - 5x + 2 = (x - 2)(x^2 + 2x - 1), so the asymptotes are x=2x = 2, x=−1+2x = -1 + \sqrt2 and x=−1−2x = -1 - \sqrt2.
d. i. [1 mark]
dydx=6(1−x3)(x3+ax+2)2\dfrac{dy}{dx} = \dfrac{6(1 - x^3)}{(x^3 + ax + 2)^2} (the aa terms cancel), so the stationary point is always at x=1x = 1 and y=3a+3y = \dfrac{3}{a + 3}.
ii. [1 mark]
There is no stationary point when x=1x = 1 is not in the domain: 1+a+2=01 + a + 2 = 0, so a=−3a = -3. Then x3−3x+2=(x−1)2(x+2)x^3 - 3x + 2 = (x - 1)^2(x + 2), giving asymptotes x=1x = 1 and x=−2x = -2.
iii. [2 marks]
The numerator of d2ydx2\tfrac{d^2y}{dx^2} is 6(3x5−ax3−12x2−2a)6\left(3x^5 - ax^3 - 12x^2 - 2a\right). Setting it to zero at x=2x = 2: 96−8a−48−2a=096 - 8a - 48 - 2a = 0, so a=245a = \dfrac{24}{5}.

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 y=0y = 0. 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 {z:zzˉ=4}\{z : z\bar z = 4\} on the Argand plane. (1 mark)
b. i
Show that {z:∣z−2i∣=∣z−3−i∣}\{z : |z - 2i| = |z - \sqrt3 - i|\} is the line y=3xy = \sqrt3x. (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 a+bia + bi. (2 marks)
ii
Label them on the diagram. (1 mark)
d
A ray starts at P(32,332)P\left(\tfrac32, \tfrac{3\sqrt3}{2}\right) and passes through Q(3,0)Q(3, 0), with equation Arg(z−z0)=θ\text{Arg}(z - z_0) = \theta. Find z0z_0 and θ\theta. (1 mark)
e
Find the area of the minor segment cut from ∣z∣=3|z| = 3 by the chord PQPQ, in the form cπ+dc\pi + d. (2 marks)
Show worked solution

a. [1 mark]. zzˉ=∣z∣2=4z\bar z = |z|^2 = 4, so ∣z∣=2|z| = 2: the circle of radius 2 centred at the origin (see the figure below).

b. i. [2 marks]. Let z=x+yiz = x + yi:

x2+(y−2)2=(x−3)2+(y−1)2  ⟹  −4y+4=−23x+3−2y+1  ⟹  y=3x.x^2 + (y - 2)^2 = (x - \sqrt3)^2 + (y - 1)^2 \implies -4y + 4 = -2\sqrt3x + 3 - 2y + 1 \implies y = \sqrt3x.

(Geometrically: it is the perpendicular bisector of the points 2i2i and 3+i\sqrt3 + i.)

ii. [1 mark]
A straight line through the origin at 60∘60^\circ to the real axis.
c. i. [2 marks]
Substitute y=3xy = \sqrt3x into x2+y2=4x^2 + y^2 = 4: 4x2=44x^2 = 4, so x=±1x = \pm1. The points are 1+3i1 + \sqrt3i and −1−3i-1 - \sqrt3i.
ii. [1 mark]

Argand plane: the circle |z| = 2 and the line y = root 3 x A circle of radius 2 centred at the origin, and the straight line through the origin at 60 degrees to the real axis. They meet at 1 + root 3 i in the first quadrant and -1 - root 3 i in the third quadrant. Re(z) Im(z) -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 1 + √3 i -1 - √3 i

d. [1 mark]. The ray starts at PP, so z0=32+332iz_0 = \tfrac32 + \tfrac{3\sqrt3}{2}i. The direction Q−P=32−332iQ - P = \tfrac32 - \tfrac{3\sqrt3}{2}i has argument θ=−π3\theta = -\dfrac{\pi}{3}.

e. [2 marks]. PP is at argument π3\tfrac{\pi}{3} and QQ at 0 on the circle of radius 3, so the central angle is π3\tfrac{\pi}{3}:

Area=12r2(θ−sin⁡θ)=92(π3−32)=3π2−934.\text{Area} = \tfrac12r^2(\theta - \sin\theta) = \tfrac92\left(\frac{\pi}{3} - \frac{\sqrt3}{2}\right) = \frac{3\pi}{2} - \frac{9\sqrt3}{4}.

From the report. In part d half the students found z0z_0 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. QQ kg is the salt at time tt min.

a
By considering concentration, explain whether the quantity of salt increases. (1 mark)
b
Show that dQdt=300−Q150\dfrac{dQ}{dt} = \dfrac{300 - Q}{150}. (1 mark)
c
Use Euler's method with step 15 minutes to estimate Q(30)Q(30) to two decimal places. (2 marks)
d
Use calculus to solve for QQ in terms of tt. (3 marks)
e
What does QQ approach as t→∞t \to \infty? (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 120\tfrac{1}{20} kg/L? (1 mark)
Show worked solution
a. [1 mark]
The initial concentration, 53000≈0.0017\tfrac{5}{3000} \approx 0.0017 kg/L, is less than the incoming 0.10.1 kg/L, so salt enters faster than it leaves and QQ increases.
b. [1 mark]
Rate in =20×0.1=2= 20 \times 0.1 = 2 kg/min. Rate out =20×Q3000=Q150= 20 \times \tfrac{Q}{3000} = \tfrac{Q}{150} kg/min (volume stays 3000 L). So dQdt=2−Q150=300−Q150\dfrac{dQ}{dt} = 2 - \dfrac{Q}{150} = \dfrac{300 - Q}{150}.
c. [2 marks]
Qn+1=Qn+15×300−Qn150Q_{n+1} = Q_n + 15 \times \dfrac{300 - Q_n}{150}:
  • Q(15)≈5+15×295150=34.5Q(15) \approx 5 + 15 \times \tfrac{295}{150} = 34.5
  • Q(30)≈34.5+15×265.5150=61.05Q(30) \approx 34.5 + 15 \times \tfrac{265.5}{150} = 61.05 kg

d. [3 marks]. Separate variables:

∫dQ300−Q=∫dt150  ⟹  −log⁡e(300−Q)=t150+c  ⟹  300−Q=Ae−t/150.\int\frac{dQ}{300 - Q} = \int\frac{dt}{150} \implies -\log_e(300 - Q) = \frac{t}{150} + c \implies 300 - Q = Ae^{-t/150}.

Q(0)=5Q(0) = 5 gives A=295A = 295, so Q=300−295e−t/150Q = 300 - 295e^{-t/150}.

e. [1 mark]
300300 kg.
f. [1 mark]
300−295e−t/150=100300 - 295e^{-t/150} = 100 gives e−t/150=4059e^{-t/150} = \tfrac{40}{59}, so t=150log⁡e ⁣(5940)≈58.3t = 150\log_e\!\left(\tfrac{59}{40}\right) \approx 58.3 minutes.
g. [1 mark]
From then on the tank gains 20 L and 2 kg of salt per minute. After τ\tau more minutes:

100+2τ3000+20τ=120  ⟹  2000+40τ=3000+20τ  ⟹  τ=50 minutes.\frac{100 + 2\tau}{3000 + 20\tau} = \frac{1}{20} \implies 2000 + 40\tau = 3000 + 20\tau \implies \tau = 50 \text{ 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 r(t)=(5cos⁡t−4cos⁡5t2)i+(5sin⁡t−4sin⁡5t2)j\mathbf{r}(t) = \left(5\cos t - 4\cos\tfrac{5t}{2}\right)\mathbf{i} + \left(5\sin t - 4\sin\tfrac{5t}{2}\right)\mathbf{j}, t≥0t \ge 0.

a
Give the starting point. (1 mark)
b
Draw an arrow at (9,0)(9, 0) 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 125−100cos⁡(3t2)\sqrt{125 - 100\cos\left(\tfrac{3t}{2}\right)}. (3 marks)
e
Find the maximum speed. (1 mark)
f
Trace the path for t∈[0,π]t \in [0, \pi]. (1 mark)
g
Find the length of that path, to one decimal place. (2 marks)
Show worked solution
a. [1 mark]
r(0)=(5−4)i+0j\mathbf{r}(0) = (5 - 4)\mathbf{i} + 0\mathbf{j}: the point (1,0)(1, 0).
b. [1 mark]
The particle is at (9,0)(9, 0) when t=2πt = 2\pi. There x˙=−5sin⁡t+10sin⁡5t2=0\dot x = -5\sin t + 10\sin\tfrac{5t}{2} = 0 and y˙=5cos⁡t−10cos⁡5t2=5+10=15>0\dot y = 5\cos t - 10\cos\tfrac{5t}{2} = 5 + 10 = 15 > 0, so the arrow points straight up: the motion is anticlockwise.
c. [1 mark]
The two terms have periods 2π2\pi and 4π5\tfrac{4\pi}{5}; their lowest common multiple is t=4πt = 4\pi.
d. [3 marks]

r˙=(−5sin⁡t+10sin⁡5t2)i+(5cos⁡t−10cos⁡5t2)j.\dot{\mathbf{r}} = \left(-5\sin t + 10\sin\tfrac{5t}{2}\right)\mathbf{i} + \left(5\cos t - 10\cos\tfrac{5t}{2}\right)\mathbf{j}.

∣r˙∣2=25(sin⁡2t+cos⁡2t)+100(sin⁡25t2+cos⁡25t2)−100(sin⁡tsin⁡5t2+cos⁡tcos⁡5t2).|\dot{\mathbf{r}}|^2 = 25\left(\sin^2 t + \cos^2 t\right) + 100\left(\sin^2\tfrac{5t}{2} + \cos^2\tfrac{5t}{2}\right) - 100\left(\sin t\sin\tfrac{5t}{2} + \cos t\cos\tfrac{5t}{2}\right).

Using cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A\cos B + \sin A\sin B with A−B=3t2A - B = \tfrac{3t}{2}:

∣r˙∣2=125−100cos⁡(3t2),so speed=125−100cos⁡(3t2).|\dot{\mathbf{r}}|^2 = 125 - 100\cos\left(\tfrac{3t}{2}\right), \quad\text{so speed} = \sqrt{125 - 100\cos\left(\tfrac{3t}{2}\right)}.

e. [1 mark]. Maximum when cos⁡3t2=−1\cos\tfrac{3t}{2} = -1: 225=15\sqrt{225} = 15 m/s.

f. [1 mark]. From (1,0)(1, 0) to r(π)=(−5,−4)\mathbf{r}(\pi) = (-5, -4), shown in orange below.

Path of the particle with the section for t from 0 to pi highlighted The full closed path (grey) loops around the origin, reaching out to radius 9. The highlighted section for t from 0 to pi starts at (1, 0), swings out to the right and round anticlockwise, and ends at (-5, -4). At (9, 0) the particle is moving straight up, so the motion is anticlockwise. x y -10 -5 5 10 -10 -5 5 10 (1, 0) (-5, -4) (9, 0)

g. [2 marks]. L=∫0π125−100cos⁡(3t2) dt≈36.6L = \displaystyle\int_0^{\pi}\sqrt{125 - 100\cos\left(\tfrac{3t}{2}\right)}\,dt \approx 36.6 m.

From the report. In part b many did not draw the arrow at (9,0)(9, 0) 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 (−5,−4)(-5, -4). In part g the definite integral had to be written, not just 36.6.

Question 5 (10 marks)

Planes Π1:2x+9z=8\Pi_1: 2x + 9z = 8, Π2:3x+6y+5z=7\Pi_2: 3x + 6y + 5z = 7, Π3:x+9y−3z=7\Pi_3: x + 9y - 3z = 7.

a
Find the point of intersection of the three planes. (1 mark)
b. i
Find a direction vector for the line of intersection of Π2\Pi_2 and Π3\Pi_3. (2 marks)
ii
Give parametric equations for this line. (1 mark)
c
Find the shortest distance from (1,1,2)(1, 1, 2) to Π3\Pi_3. (2 marks)
d
Ψ\Psi is the family of planes 6x+27z=m6x + 27z = m, m∈Nm \in N. i. Show that Π1\Pi_1 is parallel to every member of Ψ\Psi. (1 mark)
ii
Find all mm for which the distance between Π1\Pi_1 and 6x+27z=m6x + 27z = m is 23385\dfrac{23}{3\sqrt{85}}. (3 marks)
Show worked solution

a. [1 mark]. Solving the three equations simultaneously gives (−5,2,2)(-5, 2, 2).

b. i. [2 marks]. The line lies in both planes, so it is perpendicular to both normals:

(3,6,5)×(1,9,−3)=(−63,14,21)=7(−9,2,3).(3, 6, 5) \times (1, 9, -3) = (-63, 14, 21) = 7(-9, 2, 3).

A direction vector is −9i+2j+3k-9\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} (any non-zero multiple).

ii. [1 mark]. The point (−5,2,2)(-5, 2, 2) lies on both planes, so x=−5−9tx = -5 - 9t, y=2+2ty = 2 + 2t, z=2+3tz = 2 + 3t, t∈Rt \in R.

c. [2 marks]. Distance from a point to a plane:

d=∣1+9(1)−3(2)−7∣12+92+32=391=39191.d = \frac{|1 + 9(1) - 3(2) - 7|}{\sqrt{1^2 + 9^2 + 3^2}} = \frac{3}{\sqrt{91}} = \frac{3\sqrt{91}}{91}.

d. i. [1 mark]. Π1\Pi_1 has normal (2,0,9)(2, 0, 9) and each member of Ψ\Psi has normal (6,0,27)=3(2,0,9)(6, 0, 27) = 3(2, 0, 9). Parallel normals mean parallel planes.

ii. [3 marks]. Write Π1\Pi_1 as 6x+27z=246x + 27z = 24. The distance between 6x+27z=246x + 27z = 24 and 6x+27z=m6x + 27z = m is

∣m−24∣36+729=∣m−24∣385=23385  ⟹  ∣m−24∣=23  ⟹  m=1 or m=47.\frac{|m - 24|}{\sqrt{36 + 729}} = \frac{|m - 24|}{3\sqrt{85}} = \frac{23}{3\sqrt{85}} \implies |m - 24| = 23 \implies m = 1 \text{ or } m = 47.

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 mm.

Question 6 (10 marks)

Water drunk per day, VV 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 σ=5\sigma = 5 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 σ=5\sigma = 5 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 H0H_0 and H1H_1. (1 mark)
f. i
Find the pp value to four decimal places. (1 mark)
ii
Is the claim correct? Explain using the pp value. (1 mark)
g
Find the critical sample mean for n=50n = 50 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 8025=16\tfrac{80}{\sqrt{25}} = 16 mL.
ii. [1 mark]
Pr⁡(Vˉ>970)≈0.9696\Pr(\bar V > 970) \approx 0.9696.
b. [1 mark]
750±1.96×530750 \pm 1.96 \times \tfrac{5}{\sqrt{30}} gives (748.2,751.8)(748.2, 751.8) mL.
c. [1 mark]
0.95×300=2850.95 \times 300 = 285.
d. [1 mark]
1.96×5n≤11.96 \times \tfrac{5}{\sqrt n} \le 1 gives n≥96.04n \ge 96.04, so the minimum is n=97n = 97.
e. [1 mark]
H0:μ=750H_0: \mu = 750 and H1:μ<750H_1: \mu < 750.
f. i. [1 mark]
p=Pr⁡ ⁣(Xˉ≤748∣μ=750)p = \Pr\!\left(\bar X \le 748 \mid \mu = 750\right) with standard deviation 550\tfrac{5}{\sqrt{50}}: p≈0.0023p \approx 0.0023.
ii. [1 mark]
Yes: p=0.0023<0.01p = 0.0023 < 0.01, so H0H_0 is rejected at the 1% level and the claim that the mean is now less than 750 mL is supported.
g. [1 mark]
Solve Pr⁡ ⁣(Xˉ<c∣μ=750)=0.01\Pr\!\left(\bar X < c \mid \mu = 750\right) = 0.01: c≈748.355c \approx 748.355 mL.
h. [1 mark]
This is a type II error: Pr⁡ ⁣(Xˉ>748.355∣μ=747.5)≈0.113\Pr\!\left(\bar X > 748.355 \mid \mu = 747.5\right) \approx 0.113.

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.

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