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Examination 2 Showcase

VCE Specialist Mathematics Units 3&4 Free Online Pack 0 — Examination 2 Showcase

Read Examination 2 Showcase online for free, including every question, worked solution, marking note and diagnostic action. No public PDF download or checkout is provided.

VCE Year 12 Final Exam 2026 Edition - Pack 0 v1.0
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Examination 2 Showcase

26 questions

80 marks

Reading: 15 minutes · Writing: 2 hours

Read Examination 2 Showcase online

Skill Align

Skill Align VCE Specialist Mathematics Units 3&4 - Free Online Pack 0

Examination 2 showcase | Technology active

Paper
Examination 2 Showcase
Reading
15 minutes
Writing
2 hours
Assessment
80 marks

Approved materials: one approved CAS calculator or CAS software, one scientific calculator, one bound reference, protractors, set squares and aids for curve sketching. Use the standard VCE Specialist Mathematics formula sheet available separately.

Section A - Multiple-Choice Questions

For this free online paper, record one response for each question in your own notes or on a separate answer sheet before checking solutions. Choose exactly one answer for each question. A correct answer scores 1 mark; an incorrect or multiple answer scores 0, with no mark deduction.

Question 1

1 mark
The value of (1+i)⁸ is
  1. 16
  2. -16
  3. 16i
  4. -16i

Question 2

1 mark
The acute angle theta between the planes x+2y+2z=5 and 2x-y+2z=1 satisfies
  1. costheta=frac49
  2. costheta=frac29
  3. costheta=frac13
  4. costheta=frac89

Question 3

1 mark
A curve is defined by x=e^t and y=te^t. The value of ((dy) / (dx)) at t=0 is
  1. 1
  2. 0
  3. e
  4. frac1e

Question 4

1 mark
The solution of ((dy) / (dx))=3-y, with y(0)=5, is
  1. y=3+2e^(-x)
  2. y=3-2e^(-x)
  3. y=5e^(-x)
  4. y=2+3e^(-x)

Question 5

1 mark
A particle has position vector mathbf r(t)=3cos(2t)mathbf i+3sin(2t)mathbf j. Its speed is
  1. 6
  2. 3
  3. 2
  4. 6t

Question 6

1 mark
A population has standard deviation 12. For a random sample of size 36, the 95% confidence-interval margin of error based on z=1.96 is
  1. 3.92
  2. 1.96
  3. 2
  4. 23.52

Question 7

1 mark
On an Argand diagram, the locus |z-2|=|z+2i| has Cartesian equation
  1. y=-x
  2. y=x
  3. x=1
  4. y=-1

Question 8

1 mark
If z=2,mathrm(cis)(2pi / 3), then z³ is
  1. 8
  2. -8
  3. 8i
  4. -8i

Question 9

1 mark
The perpendicular distance from (1,2,3) to the plane 2x-y+2z=4 is
  1. frac23
  2. frac43
  3. 2
  4. 6

Question 10

1 mark
A curve is defined by x=t² and y=t³, where t>0. At t=2, ((dy) / (dx)) is
  1. 3
  2. 6
  3. frac32
  4. 12

Question 11

1 mark
The coefficient of x² in the Maclaurin expansion of e^xcos x is
  1. 0
  2. 1
  3. frac12
  4. -frac12

Question 12

1 mark
A population satisfies ((dy) / (dt))=y(1-y / 10). When y=5, its instantaneous growth rate is
  1. frac52
  2. 5
  3. frac12
  4. 25

Question 13

1 mark
A force F(t)=4t newtons acts for 0leq tleq3 seconds. The impulse delivered is
  1. 18 mathrm(N,s)
  2. 12 mathrm(N,s)
  3. 36 mathrm(N,s)
  4. 6 mathrm(N,s)

Question 14

1 mark
A particle moves at constant speed 10 mathrm(m,s⁻¹) on a circle of radius 5 mathrm m. The magnitude of its acceleration is
  1. 20 mathrm(m,s⁻²)
  2. 2 mathrm(m,s⁻²)
  3. 50 mathrm(m,s⁻²)
  4. 100 mathrm(m,s⁻²)

Question 15

1 mark
Two particles have position vectors mathbf r_A=(t,2t) and mathbf r_B=(4-t,2). Their separation is least when
  1. t=frac32
  2. t=1
  3. t=2
  4. t=frac34

Question 16

1 mark
For Xsimmathrm(Bin)(100,0.3), the mean and standard deviation are
  1. 30 and √21
  2. 30 and 21
  3. 70 and √21
  4. 30 and √30

Question 17

1 mark
A continuous random variable has density f(x)=2x for 0leq xleq1. Then E(X) is
  1. frac23
  2. frac12
  3. frac13
  4. frac34

Question 18

1 mark
Independent observations have population standard deviation 6. The standard deviation of the sample mean for samples of size 9 is
  1. 2
  2. 6
  3. frac23
  4. 18

Question 19

1 mark
If z(t)=e^(it), then ((dz) / (dt)) at t=pi / 2 is
  1. -1
  2. 1
  3. i
  4. -i

Question 20

1 mark
Euler's method with step size 0.5 is applied to ((dy) / (dx))=x+y, y(0)=1. The approximation to y(1) is
  1. 2.5
  2. 2.0
  3. 2.75
  4. 3.0

Section B - Extended-Response Questions

Answer all questions. Unless otherwise specified, an exact answer is required for each question. In questions where more than one mark is available, appropriate working must be shown. The structured graphs contain information required to solve the questions.

Question 1

7 marks
The velocity-time graph shows the vertical velocity of a particle projected from a platform. Upwards is positive. Use the graph for every part.
Graph Preview
048200-20(0, 20)(4, 0)(8, -20)time t (s)vertical velocity v (m/s)
(a) 1 mark
Find the constant acceleration.
(b) 2 marks
Find the greatest height reached above the platform.
(c) 2 marks
Explain why the particle returns to platform height at t=8.
(d) 2 marks
Find the total distance travelled during the first 8 seconds.

Question 2

7 marks
The graph shows the probability density function of a continuous random variable X. The density is zero outside the interval shown.
Graph Preview(0, 0)(2, 0.5)(4, 0)xf(x)
(a) 1 mark
Verify that the total area under the graph is 1.
(b) 2 marks
Find Pr(Xleq1).
(c) 1 mark
State E(X), giving a reason.
(d) 3 marks
Show that operatorname(Var)(X)=frac23.

Question 3

11 marks
The solutions of z³=-8 are represented by points on an Argand diagram.
(a) 4 marks
Find all three solutions in exact Cartesian form.
(b) 3 marks
Show that the three points form an equilateral triangle and find its exact area.
(c) 4 marks
For each root z, define w=z+1. Find the monic cubic polynomial with real coefficients whose roots are the three possible values of w.

Question 4

12 marks
A function y satisfies ((dy) / (dx))=x(4-y), y(0)=1, for xgeq0.
(a) 4 marks
Solve the differential equation for y as a function of x.
(b) 2 marks
Find lim_(xtoinfty)y and interpret it for the solution curve.
(c) 4 marks
Find the value of x at which y is increasing most rapidly, and find this maximum rate.
(d) 2 marks
Find the equation of the tangent to the solution curve at x=0.

Question 5

11 marks
A particle moves on a straight line. Its acceleration is a(t)=6t-12 metres per second squared. Initially v(0)=9 mathrm(m,s⁻¹) and x(0)=0. Consider 0leq tleq4.
(a) 2 marks
Find the velocity v(t).
(b) 2 marks
Find the times at which the particle is at rest.
(c) 2 marks
Find the position x(t).
(d) 3 marks
Find the total distance travelled from t=0 to t=4.
(e) 2 marks
Find the maximum speed on the interval, and state when it occurs.

Question 6

12 marks
A continuous random variable X has probability density f(x)=kx(2-x) for 0leq xleq2, and f(x)=0 otherwise.
(a) 2 marks
Find k.
(b) 3 marks
Find the cumulative distribution function F(x) for 0leq xleq2.
(c) 2 marks
Find Pr(X>frac32).
(d) 3 marks
Find E(X) and operatorname(Var)(X).
(e) 2 marks
For a random sample of size 25, use a normal approximation to estimate Pr(overline X>1.1).

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Worked Solutions And Marking Guide

Section A Question 1

Answer: 16

1+i=sqrt2,mathrm(cis)(pi / 4), so the eighth power is 16,mathrm(cis)(2pi)=16.

Section A Question 2

Answer: costheta=frac49

The normals are (1,2,2) and (2,-1,2). Their dot product is 4 and each has magnitude 3.

Section A Question 3

Answer: 1

((dy) / (dx))=((e^t(1+t)) / (e^t))=1+t, which equals 1 at t=0.

Section A Question 4

Answer: y=3+2e^(-x)

The equilibrium is 3, so y=3+Ce^(-x). The initial condition gives C=2.

Section A Question 5

Answer: 6

The velocity is -6sin(2t)mathbf i+6cos(2t)mathbf j, whose magnitude is 6.

Section A Question 6

Answer: 3.92

The standard error is 12 / √36=2, so the margin of error is 1.96(2)=3.92.

Section A Question 7

Answer: y=-x

Let z=x+iy. Equating squared distances gives (x-2)²+y²=x²+(y+2)², which simplifies to y=-x.

Section A Question 8

Answer: 8

By De Moivre's theorem, z³=8,mathrm(cis)(2pi)=8.

Section A Question 9

Answer: frac23

The distance is ((|2(1)-2+2(3)-4|) / (√(2²+(-1)²+2²)))=frac23.

Section A Question 10

Answer: 3

((dy) / (dx))=((3t²) / (2t))=((3t) / (2)), which equals 3 at t=2.

Section A Question 11

Answer: 0

Using e^x=1+x+x² / 2+ × s and cos x=1-x² / 2+ × s, the two x² / 2 contributions cancel.

Section A Question 12

Answer: frac52

Substitute y=5: 5(1-5 / 10)=5 / 2.

Section A Question 13

Answer: 18 mathrm(N,s)

Impulse is int_0³ 4t,dt=2t^2big|_0³=18 mathrm(N,s).

Section A Question 14

Answer: 20 mathrm(m,s⁻²)

Centripetal acceleration has magnitude v² / r=100 / 5=20 mathrm(m,s⁻²).

Section A Question 15

Answer: t=frac32

The squared separation is (2t-4)²+(2t-2)²=8t²-24t+20, whose vertex occurs at t=24 / 16=3 / 2.

Section A Question 16

Answer: 30 and √21

For a binomial variable, E(X)=np=30 and operatorname(sd)(X)=√(np(1-p))=√21.

Section A Question 17

Answer: frac23

E(X)=int_0¹ x(2x),dx=2 / 3.

Section A Question 18

Answer: 2

The standard deviation of the sample mean is sigma / sqrt n=6 / 3=2.

Section A Question 19

Answer: -1

((dz) / (dt))=ie^(it). At t=pi / 2, this is i(i)=-1.

Section A Question 20

Answer: 2.5

The two Euler steps give y(0.5)=1+0.5(1)=1.5, then y(1)=1.5+0.5(0.5+1.5)=2.5.

Section B Question 1

(a) -5 mathrm(m,s⁻²).

The gradient is ((-20-20) / (8-0))=-5.

(b) 40 mathrm m.

The upward displacement is the triangular area from t=0 to t=4: frac12(4)(20)=40.

(c) The signed areas from 0 to 4 and from 4 to 8 are +40 and -40 square units, so the net displacement at t=8 is zero.

Displacement is signed area under a velocity-time graph. Equal areas above and below the time axis cancel.

(d) 80 mathrm m.

Distance uses absolute area, so 40+40=80.

Mark allocation

  • Award 1 mark for the signed gradient with units.
  • Award 1 mark for identifying area as displacement and 1 mark for the 40 m height.
  • Award 1 mark for comparing the signed areas and 1 mark for the zero-displacement conclusion.
  • Award 1 mark for using absolute areas and 1 mark for 80 m.

Section B Question 2

(a) frac12(4)(0.5)=1.

The density graph is a triangle with base 4 and height 0.5.

(b) frac18.

On 0leq xleq2, f(x)=x / 4. Hence Pr(Xleq1)=int_0¹ x / 4,dx=1 / 8.

(c) E(X)=2, by symmetry about x=2.

The triangular density is symmetric about x=2.

(d) E(X²)=((14) / (3)), so operatorname(Var)(X)=E(X²)-[E(X)]²=((14) / (3))-4=frac23.

Use f(x)=x / 4 on [0,2] and f(x)=(4-x) / 4 on [2,4]. Then E(X²)=int_0² x³ / 4,dx+int_2⁴ x²(4-x) / 4,dx=14 / 3.

Mark allocation

  • Award 1 mark for the correctly evaluated triangular area.
  • Award 1 mark for the left-branch density or equivalent triangle reasoning and 1 mark for the probability.
  • Award 1 mark for the mean with a valid symmetry reason.
  • Award 1 mark for a correct piecewise second-moment setup, 1 mark for E(X²) = 14 / 3, and 1 mark for the variance.

Section B Question 3

(a) -2, 1+isqrt3, 1-isqrt3.

Write -8=8,mathrm(cis)(pi+2kpi). The cube roots have modulus 2 and arguments pi / 3, pi, 5pi / 3, giving the stated Cartesian forms.

(b) Each side has length 2sqrt3, so the area is 3sqrt3 square units.

The distance between 1+isqrt3 and 1-isqrt3 is 2sqrt3, and each distance to -2 is also √(3²+(sqrt3)²)=2sqrt3. The equilateral-triangle area is ((sqrt3) / (4))(2sqrt3)²=3sqrt3.

(c) w³-3w²+3w+7.

Since z=w-1 and z³+8=0, substitute to obtain (w-1)³+8=w³-3w²+3w+7=0.

Mark allocation

  • Award 1 mark for the cube-root modulus and arguments, then up to 3 marks for the complete exact root set.
  • Award 2 marks for verifying all equal side lengths and 1 mark for the exact area.
  • Award 1 mark for the substitution, 2 marks for correct expansion and 1 mark for the monic polynomial.

Section B Question 4

(a) y=4-3e^(-x² / 2).

Separate variables to obtain int((1) / (4-y)),dy=int x,dx. Hence -ln(4-y)=x² / 2+C. Applying y(0)=1 gives 4-y=3e^(-x² / 2).

(b) The limit is 4, so y=4 is a horizontal asymptote approached from below.

The exponential factor tends to zero as x increases.

(c) The maximum occurs at x=1, and the maximum rate is 3e^(-1 / 2).

From the solution, y'=3xe^(-x² / 2). Differentiating gives y''=3e^(-x² / 2)(1-x²), so the positive rate is maximised at x=1.

(d) y=1.

The point is (0,1) and y'(0)=0, so the tangent is horizontal.

Mark allocation

  • Award 2 marks for separation and integration and 2 marks for applying the initial condition correctly.
  • Award 1 mark for the limit and 1 mark for the asymptote interpretation.
  • Award 2 marks for the rate and its derivative, 1 mark for x=1, and 1 mark for the exact maximum rate.
  • Award 1 mark for the tangent gradient and 1 mark for its equation.

Section B Question 5

(a) v(t)=3t²-12t+9=3(t-1)(t-3).

Integrate acceleration and use v(0)=9 to determine the constant.

(b) t=1 and t=3.

Solve 3(t-1)(t-3)=0.

(c) x(t)=t³-6t²+9t.

Integrate velocity and use x(0)=0.

(d) 12 mathrm m.

The direction changes at t=1 and t=3. Since x(0)=0, x(1)=4, x(3)=0 and x(4)=4, the distance is 4+4+4=12.

(e) The maximum speed is 9 mathrm(m,s⁻¹), occurring at t=0 and t=4.

Check the endpoints and the velocity turning point t=2: the velocities are 9,-3,9, so the largest magnitude is 9.

Mark allocation

  • Award 1 mark for integration and 1 mark for the velocity constant and factorisation.
  • Award 1 mark for solving the rest condition and 1 mark for both times.
  • Award 1 mark for integration and 1 mark for the position constant.
  • Award 1 mark for splitting at both direction changes, 1 mark for the positions, and 1 mark for the total distance.
  • Award 1 mark for checking the velocity turning point and endpoints and 1 mark for the maximum speed with times.

Section B Question 6

(a) k=frac34.

Set kint_0^2x(2-x),dx=1. The integral is 4 / 3, so k=3 / 4.

(b) F(x)=((3x²) / (4))-((x³) / (4)).

Integrate the density from 0 to x: frac34int_0^x(2u-u²),du=frac34(x²-x³ / 3).

(c) frac5(32).

F(3 / 2)=27 / 32, so the upper-tail probability is 1-27 / 32=5 / 32.

(d) E(X)=1 and operatorname(Var)(X)=frac15.

The density is symmetric about 1, so E(X)=1. Also E(X²)=frac34int_0^2x³(2-x),dx=6 / 5, hence the variance is 6 / 5-1=1 / 5.

(e) Approximately 0.132.

The sample mean has mean 1 and standard deviation √((1 / 5) / 25)=1 / (5sqrt5). Thus z=(1.1-1) / (1 / (5sqrt5))approx1.118, giving an upper-tail probability of about 0.132.

Mark allocation

  • Award 1 mark for the normalising integral and 1 mark for k.
  • Award 2 marks for a correct integral and 1 mark for the simplified CDF.
  • Award 1 mark for evaluating the CDF and 1 mark for the upper tail.
  • Award 1 mark for the mean, 1 mark for the second moment and 1 mark for the variance.
  • Award 1 mark for the sample-mean standard deviation and z-score and 1 mark for the approximate probability.

Diagnostic Checklist

TopicQuestionsMarksMarks LostAction
Multiple-choice foundations Section A Q1-Q20 20 ___ Review any missed algebra, complex-number, vector, calculus, mechanics and probability foundations.
Mechanics Section B Q1 and Q5 18 ___ Review signed area, acceleration, velocity sign changes and total distance.
Probability and statistics Section B Q2 and Q6 19 ___ Review continuous densities, moments, symmetry and sampling distributions.
Complex numbers and geometry Section B Q3 11 ___ Review roots in polar form, Argand geometry and transformations of roots.
Calculus and differential equations Section B Q4 12 ___ Review separation of variables, limiting behaviour, stationary rates and tangents.

What is included

Examination 1 Showcase questions (40 marks)

Examination 2 Showcase questions (80 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 Specialist Mathematics 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.

Can I download Pack 0 as a PDF?

No. Pack 0 is intentionally online-only and no downloadable PDF is provided.

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