Band 5 – Page 37

Quadratic, EXT1 2018 HSC 12e

The points `P(2ap, ap^2)` and `Q(2aq, aq^2)` lie on the parabola `x^2 = 4ay`. The focus of the parabola is `S(0, a)` and the tangents at `P` and `Q` intersect at `T(a(p + q), apq)`. (Do NOT prove this.)

The tangents at `P` and `Q` meet the `x`-axis at `A` and `B` respectively, as shown.

Show that `/_ PAS = 90^@`. (2 marks)
Explain why `S, B, A, T` are concyclic points. (1 mark)
Show that the diameter of the circle through `S, B, A` and `T` has length

`qquad qquad a sqrt((p^2 + 1)(q^2 + 1))`. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(See Worked Solutions)`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `y=x^2/(4a)\ \ => dy/dx=x/(2a)`

`text(At)\ \ x=2ap,\ \ m_text(tang) = p`

`text(Equation of tangent)\ \ m=p, text(through)\ (2ap, ap^2):`

`y-ap^2`	`=p(x-2ap)`
`y`	`=px-ap^2`

`text(When)\ \ y=0, x=ap`

`:. A(ap,0)`

`m_(AP)`	`= (ap^2 – 0)/(2ap-ap)`
	`= p`

`m_(AS)`	`= (0 – a)/(ap – 0)`
	`= -1/p`

`m_(AP) xx m_(AS) = -1`

`:. /_PAS = 90°\ \ text(… as required)`

(ii) `text{Similarly to part (i):}`

♦ Mean mark 37%.

`=> B(aq,0)`

`=> m_(BQ) xx m_(BS) = -1`

`=> /_QBS = 90°`

`text(S)text(ince)\ \ T\ \ text(is the intersection of both tangents,)`

`/_ PAS = /_ TAS = 90^@`

`/_QBS = /_ TBS = 90^@`

`=> ST\ text(can be considered the extreme points of a circle diameter)`

`text(that is subtending 2 right angles on the circle circumference.)`

`:. SBAT\ text(are concyclic points.)`

(iii) `text(Diameter = distance)\ ST:`

♦ Mean mark 44%.

`S(0,a),\ \ T((a(p + q), apq)`

`d^2`	`= ST^2`
	`=(x_2-x_1)^2 + (y_2-y_1)^2`
	`= (a(p + q) – 0)^2 + (apq – a)^2`
	`= a^2 (p + q)^2 + a^2 (pq – 1)^2`
	`= a^2(p^2 + 2pq + q^2 + p^2q^2 – 2pq + 1)`
	`= a^2(p^2 + q^2 + p^2q^2 + 1)`
	`= a^2(p^2 + 1) (q^2 + 1)`
`:.d`	`= a sqrt((p^2 + 1)(q^2 + 1))\ \ text(… as required)`

Calculus, EXT1 C2 2018 HSC 12c

Let `f(x) = sin^(-1) x + cos^(-1) x`.

Show that `f^{′}(x) = 0` (1 mark)

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Hence, or otherwise, prove

`qquad sin^(-1) x + cos^(-1) x = pi/2`. (1 mark)

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Hence, sketch

`qquad f(x) = sin^(-1) x + cos^(-1) x`. (1 mark)

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Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(See Worked Solutions)`

Show Worked Solution

i. `f^{′}(x) = 1/sqrt(1-x^2) + (-1/sqrt (1-x^2)) = 0`

ii. `text(S)text(ince)\ \ f^{′}(x) = 0\ \ => f(x)\ \ text(is a constant.)`

♦ Mean mark (ii) 37%.

`text(Substituting)\ \ x=1\ \ text{into the equation (any value works)}`

	`sin^(-1) 1 + cos^(-1) 1`	`= pi/2 + 0`
		`= pi/2\ \ text(… as required)`

iii. `text(Domain restrictions require:)\ \ -1<x<1`

♦ Mean mark (iii) 40%.

Networks, STD2 N3 EQ-Bank 19

An engineering project requires activities A to G to be completed, as shown in the table.

The minimum completion time for the project is 25 days and the critical path includes activities B, D, E and F. The float for activity G is two days and the float for activity C is four days.

Find the possible duration for each of the activities A, C, F and G. Include a network diagram in your answer. (5 marks)

Show Answers Only

`text(Duration of)\ F = 1\ text(day)`

`text(Duration of)\ G = 4\ text(days)`

`text(*Understand why there are many possibilities for the duration of)`

`A and C, text(provided they add up to 20 days and)\ A\ text(is not longer)`

`text{than 7 days (or a new critical path is created)}.`

Show Worked Solution

`text(Sketch the network:)`

`text{Critical path information included in the network (where EST = LST)}`

`F\ text(is on critical path ⇒ no float)`

`:.\ text(Duration of)\ F= 25 – 24 = 1\ text(day)`

`text(Duration of)\ G`	`=\ text(LST of next activity − EST of)\ G – text(float)`
	`= 25 – 19 – 2`
	`= 4\ text(days)`

`text(The float of)\ C\ text{is 4 days (given)}`

`:.\ text(Duration of)\ C`	`=\ text(LST of)\ F -\ text(EST of)\ C – 4`
	`= 24 -\ text(EST of)\ C – 4`
	`= 20 -\ text(EST of)\ C`

`text(EST of)\ C =\ text(Duration of)\ A\ \ (A\ text(has no prerequisites))`

`=>\ text(Duration of)\ A +\ text(Duration of)\ C = 20\ text(days)`

`:.\ text(Possible durations of)\ A\ text(and)\ C\ text(are:)`

`A = 5\ text(days), C = 15\ text(days)`

`text(*Understand why there are many possibilities for the duration of)`

`A and C, text(provided they add up to 20 days and)\ A\ text(is not longer)`

`text{than 7 days (or a new critical path is created)}.`

Networks, STD2 N3 EQ-Bank 18

An engineering project requires activities A to G to be completed, as shown in the table.

The minimum completion time for the project is 40 weeks and the critical path includes activities A, C, E and G. The float for activity F is six weeks and the float for activity D is 9 weeks.

Find the possible duration for each of the activities B, D, F and G. Include a network diagram in your answer. (5 marks)

Show Answers Only

`text(Duration of)\ G = 3\ text(weeks)`

`text(Duration of)\ F = 9\ text(weeks)`

`text(There are many possibilities for the duration of)\ B and D`

`text(provided they add up to 28 weeks and)\ B\ text(is not longer)`

`text{than 10 weeks (new critical path)}.`

Show Worked Solution

`text(Sketch network:)`

`text{Critical path added to network (where EST = LST)}`

`G\ text(is on critical path ⇒ no float)`

`:.\ text(Duration of)\ G = 40-37 = 3\ text(weeks)`

`text(Duration of)\ F`	`=\ text(40 − EST of)\ F – text(float)`
	`= 40 – 25 – 6`
	`= 9\ text(weeks)`

`text(Float of)\ D = 9\ text{weeks (given)}`

`:.\ text(Duration of)\ D`	`=\ text(LST of)\ G -\ text(EST of)\ D – 9`
	`= 37 -\ text(EST of)\ D – 9`
	`= 28 -\ text(EST of)\ D`

`text(EST of)\ D =\ text(Duration of)\ B\ \ \ (B\ text(has no prerequisites))`

`text(Duration of)\ B +\ text(Duration of)\ D = 28\ text(weeks)`

`:.\ text(Possible durations of)\ B\ text(and)\ D\ text(are:)`

`B = 1\ text(week), D = 27\ text(weeks)`

`text(*Understand why there are many possibilities for the duration of)`

`B and D, text(provided they add up to 28 weeks and)\ B\ text(is not longer)`

`text{than 10 weeks (or a new critical path is created)}.`

Mechanics, EXT2* M1 2018 HSC 10 MC

A particle is moving in simple harmonic motion. The displacement of the particle is `x` and its velocity, `v`, is given by the equation `v^2 = n^2 (2kx - x^2)`, where `n` and `k` are constants.

The particle is initially at `x = k`.

Which function, in terms of time `t`, could represent the motion of the particle?

A. `x = k cos (nt)`

B. `x = k sin (nt) + k`

C. `x = 2k cos (nt) - k`

D. `x = 2k sin (nt) + k`

Show Answers Only

`B`

Show Worked Solution

`text(Completing the square):`

♦♦ Mean mark 33%.

`v^2 = n^2(2kx-x^2)`

`=n^2(k^2-(x^2-2kx+k^2))`

`=n^2(k^2-(x-k)^2)`

`:.\ text(The centre of motion is)\ \ x = k,\ text(amplitude) = k,`

`⇒ B`

Trig Ratios, EXT1 2018 HSC 9 MC

Which of the following is a general solution of the equation `sin 2x = -1/2`?

A. `x = n pi + (-1)^n pi/12`

B. `x = (n pi)/2 + (-1)^n pi/12`

C. `x = n pi + (-1)^(n + 1) pi/12`

D. `x = (n pi)/2 + (-1)^(n + 1) pi/12`

Show Answers Only

`D`

Show Worked Solution

`text(Using the general solution for sin):`

♦ Mean mark 45%.

`2x`	`= n pi + (-1)^n (-pi/6)`
	`= n pi + (-1)^(n + 1) (pi/6)`
`:. x`	`= (n pi)/2 + (-1)^(n + 1) pi/12`

`⇒ D`

Combinatorics, EXT1 A1 2018 HSC 8 MC

Six men and six women are to be seated at a round table.

In how many different ways can they be seated if men and women alternate?

A. `5!\ 5!`

B. `5!\ 6!`

C. `2!\ 5!\ 5!`

D. `2!\ 5!\ 6!`

Show Answers Only

`B`

Show Worked Solution

`text(Position the 1st man in any seat.)`

♦ Mean mark 42%.

`text(The remaining 5 men can be positioned in 5! ways).`

`text(The 6 women can be positioned in the alternate seats)`

`text(in 6! ways.)`

`:.\ text(Total seating arrangements)\ = 5! xx 6!`

`⇒ B`

Networks, STD2 N3 2007 FUR2 4

A community centre is to be built on the new housing estate.

Nine activities have been identified for this building project.

The directed network below shows the activities and their completion times in weeks.

Determine the minimum time, in weeks, to complete this project. (1 mark)

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Determine the float time, in weeks, for activity `D`. (2 marks)

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The builders of the community centre are able to speed up the project.

Some of the activities can be reduced in time at an additional cost.

The activities that can be reduced in time are `A`, `C`, `E`, `F` and `G`.

Which of these activities, if reduced in time individually, would not result in an earlier completion of the project? (1 mark)

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The owner of the estate is prepared to pay the additional cost to achieve early completion.

The cost of reducing the time of each activity is $5000 per week.

The maximum reduction in time for each one of the five activities, `A`, `C`, `E`, `F`, `G`, is `2` weeks.

Determine the minimum time, in weeks, for the project to be completed now that certain activities can be reduced in time. (1 mark)

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Determine the minimum additional cost of completing the project in this reduced time. (1 mark)

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Show Answers Only

`19\ text(weeks)`
`5\ text(weeks)`
`A, E, G`
`text(15 weeks)`
`$25\ 000`

Show Worked Solution

a. `text(Scanning forwards and backwards:)`

`BCFHI\ \ text(is the critical path.)`

♦ Mean mark of all parts (combined) 40%.

`:.\ text(Minimum time)`	`= 4 + 3 + 4 + 2 + 6`
	`= 19\ text(weeks)`

b.	`text(EST of)\ D`	`= 4`
	`text(LST of)\ D`	`= 9`

`:.\ text(Float time of)\ D`	`= 9 – 4`
	`= 5\ text(weeks)`

c. `A, E,\ text(and)\ G\ text(are not currently on the critical path,)`

`text(therefore reducing their time will not result in an)`

`text(earlier completion time.)`

d. `text(Reduce)\ C\ text(and)\ F\ text(by 2 weeks each.)`

`text(However, a new critical path created:)\ BEHI\ \ text{(16 weeks)}`

`:.\ text(Also reduce)\ E\ text(by 1 week.)`

`:.\ text(Minimum completion time = 15 weeks)`

e.	`text(Additional cost)`	`= 5 xx $5000`
		`= $25\ 000`

Networks, STD2 N3 2006 FUR2 3

The five musicians are to record an album. This will involve nine activities.

The activities and their immediate predecessors are shown in the following table.

The duration of each activity is not yet known.

Use the information in the table above to complete the network below by including activities `G`, `H` and `I`. (2 marks)

There is only one critical path for this project.

How many non-critical activities are there? (1 mark)

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The following table gives the earliest start times (EST) and latest start times (LST) for three of the activities only. All times are in hours.

Write down the critical path for this project. (1 mark)

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The minimum time required for this project to be completed is 19 hours.

What is the duration of activity `I`? (1 mark)

The duration of activity `C` is 3 hours.

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Determine the maximum combined duration of activities `F` and `H`. (1 mark)
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Show Answers Only

`5`
`B E G I`
`text(7 hours)`
`text(8 hours)`

Show Worked Solution

b. `text(S)text(ince no time information, possible critical paths are:)`

`ADGI, BEGI\ text(or)\ CFHI\ \ text{(all have 4 activities)}`

`:.\ text(Non-critical activities)`

`= 9 – 4 = 5`

c. `text(Critical activities have zero float time.)`

♦ Mean mark of parts (c)-(e) (combined) was 36%.

`=> A\ text(and)\ C\ text(are non-critical.)`

`:. B E G I\ text(is the critical path.)`

d.	`text(Duration of)\ \ I`	`= 19 – 12`
		`= 7\ text(hours)`

e. `text(Maximum time for)\ F\ text(and)\ H`

♦♦ MARKER’S COMMENT: Many students incorrectly answered 9 hours in part (e).

`=\ text(LST of)\ I – text(duration)\ C – text(slack time of)\ C`

`= 12 – 3 – 1`

`= 8\ text(hours)`

Networks, STD2 N3 2013 FUR2 2

A project will be undertaken in the wildlife park. This project involves the 13 activities shown in the table below. The duration, in hours, and predecessor(s) of each activity are also included in the table.

Activity `G` is missing from the network diagram for this project, which is shown below.

Complete the network diagram above by inserting activity `G`. (1 mark)
Determine the earliest starting time of activity `H`. (1 mark)

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Given that activity `G` is not on the critical path
1. write down the activities that are on the critical path in the order that they are completed (1 mark)
  
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2. find the latest starting time for activity `D`. (1 mark)
  
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Consider the following statement.

‘If the time to complete just one of the activities in this project is reduced by one hour, then the minimum time to complete the entire project will be reduced by one hour.’

Explain the circumstances under which this statement will be true for this project. (1 mark)

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Assume activity `F` is reduced by two hours.
What will be the minimum completion time for the project? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

b. `7\ text(hours)`

c.i. `AFIM`

c.ii. `14\ text(hours)`

d. `text(The statement will only be true if the crashed activity)`
`text(is on the critical path)\ \ A F I M.`

e. `text(36 hours)`

Show Worked Solution

b. `text(Scanning forwards and backwards:)`

`text(EST for Activity)\ H`

`= 4 + 3`

`= 7\ text(hours)`

c.i. `A F I M`

♦♦ Mean mark of parts (c)-(e) (combined) was 40%.

c.ii. `text(LST of)\ G = 20 – 4 = 16\ text(hours)`

`text(LST of)\ D = 16 – 2 = 14\ text(hours)`

d. `text(The statement will only be true if the time reduced activity)`

MARKER’S COMMENT: Most students struggled with part (d).

`text(is on the critical path)\ \ A F I M.`

e. `A F I M\ text(is 37 hours.)`

`text(If)\ F\ text(is reduced by 2 hours, the new critical)`

`text(path is)\ \ C E H G I M\ text{(36 hours)}`

`:.\ text(Minimum completion time = 36 hours)`

Networks, STD2 N3 2012 FUR2 2

Thirteen activities must be completed before the produce grown on a farm can be harvested.

The directed network below shows these activities and their completion times in days.

Determine the earliest starting time, in days, for activity `E`. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
An activity with zero duration starts at the end of activity `B`.

Explain why this activity is used on the network diagram. (1 mark)

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Determine the earliest starting time, in days, for activity `H`. (1 mark)

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In order, list the activities on the critical path. (1 mark)

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Determine the latest starting time, in days, for activity `J`. (1 mark)

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Show Answers Only

`12\ text(days)`
`F\ text(has)\ B\ text(as a predecessor while)\ G\ text(and)\ H`
`text(have)\ B\ text(and)\ C\ text(as predecessors.)`
`text(S)text(ince there cannot be 2 activities called)\ B, text(a zero)`
`text{duration activity is drawn as an extension of}\ B\ text(to)`
`B\ text(show that it is also a predecessor of)\ G\ text(and)\ H.`
`15\ text(days)`
`ABHILM`
`25\ text(days)`

Show Worked Solution

a.	`text(EST of)\ E`	`= 10 + 2`
		`= 12\ text(days)`

♦ Mean mark of all parts (combined) 47%.

b. `F\ text(has)\ B\ text(as a predecessor while)\ G\ text(and)\ H`

`text(have)\ B\ text(and)\ C\ text(as predecessors.)`

`text(S)text(ince there cannot be 2 activities called)\ B, text(a zero)`

`text{duration activity is drawn as an extension of}\ B\ text(to)`

`text(show that it is also a predecessor of)\ G\ text(and)\ H.`

♦♦ “Few students” were able to correctly deal with the zero duration activity in part (c).

c. `text(Scanning forwards:)`

`text(EST for)\ H = 15\ text(days)`

d. `text(The critical path is)\ \ ABHILM`

e. `text(The shortest time to complete all the activities)`

MARKER’S COMMENT: A correct calculation based on an incorrect critical path in part (d) gained a consequential mark here. Show your working!.

`= 10 + 5 + 4 + 3 + 4 + 2`

`= 28\ text(days)`

`:.\ text(LST of)\ J`	`= 28 − 3`
	`= 25\ text(days)`

Networks, STD2 N3 2009 FUR2 4

A walkway is to be built across the lake.

Eleven activities must be completed for this building project.

The directed network below shows the activities and their completion times in weeks.

What is the earliest start time for activity E? (1 mark)

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Write down the critical path for this project. (1 mark)

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The project supervisor correctly writes down the float time for each activity that can be delayed and makes a list of these times.

Determine the longest float time, in weeks, on the supervisor’s list. (1 mark)

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A twelfth activity, L, with duration three weeks, is to be added without altering the critical path.

Activity L has an earliest start time of four weeks and a latest start time of five weeks.

Draw in activity L on the network diagram above. (1 mark)
Activity L starts, but then takes four weeks longer than originally planned.

Determine the total overall time, in weeks, for the completion of this building project. (1 mark)

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Show Answers Only

`7`
`BDFGIK`
`H\ text(or)\ J\ text(can be delayed for)`
`text(a maximum of 3 weeks.)`
`text(25 weeks)`

Show Worked Solution

a. `7\ text(weeks)`

♦ Mean mark of all parts (combined): 44%.

b. `text(Scanning forwards and backwards)`

`text(Critical Path is)\ BDFGIK`

c. `H\ text(or)\ J\ text(can be delayed for a maximum)`

`text(of 3 weeks.)`

e. `text(The new critical path is)\ BLEGIK.`

`=>\ text(Activity)\ L\ text(now takes 7 weeks.)`

`:.\ text(Time for completion)`

`= 4 + 7 + 1 + 5 + 2 + 6`

`= 25\ text(weeks)`

Networks, STD2 N3 2011 FUR1 7 MC

Andy, Brian and Caleb must complete three activities in total (K, L and M)

The table shows the person selected to complete each activity, the time it will take to complete the activity in minutes and the immediate predecessor for each activity.

All three activities must be completed in a total of 40 minutes.

The instant that Andy starts his activity, Caleb gets a telephone call.

The maximum time, in minutes, that Caleb can speak on the telephone before he must start his allocated activity is

A. `5`

B. `13`

C. `18`

D. `24`

Show Answers Only

`D`

Show Worked Solution

`text(Maximum speaking time on phone)`

♦♦ Mean mark 33%.
MARKER’S COMMENT: Many students incorrectly answered the earliest starting time, C.

`= 40 – text(duration of)\ M`

`= 40 – 16`

`= 24\ text(minutes)`

`=> D`

Networks, STD2 N3 2014 FUR1 8 MC

Which one of the following statements about critical paths is true?

There can be only one critical path in a project.
A critical path will always include the activity that takes the longest time to complete.
Reducing the time of any activity on a critical path for a project will always reduce the minimum completion time for the project.
If there are no other changes, increasing the time of any activity on a critical path will always increase the completion time of a project.

Show Answers Only

`D`

Show Worked Solution

`text(If any activity on a critical path takes longer, the project)`

♦ Mean mark 40%.

`text{completion time increases by the equivalent amount}`

`text{of time (although the reverse is not true).}`

`=> D`

Networks, STD2 N3 SM-Bank 49

The directed graph below shows the sequence of activities required to complete a project.

The time to complete each activity, in hours, is also shown.

Find the earliest starting time, in hours, for activity `N`. (2 marks)

To complete the project in minimum time, some activities cannot be delayed.

Calculate the number of activities that cannot be delayed. (1 mark)

Show Answers Only

`12\ text(hours)`
`4`

Show Worked Solution

i. `text{Scanning forward:}`

`:.\ text(EST for)\ N`	`= CGJ`
	`= 4 +3+5`
	`= 12`

ii. `text(Critical path is:)\ CFHM`

`:. 4\ text(activities can’t be delayed.)`

Networks, STD2 N3 SM-Bank 47

The directed graph below shows the sequence of activities required to complete a project.

All times are in hours.

Find the number of activities that have exactly two immediate predecessors. (1 mark)

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Identify the critical path for this project. (3 marks)
If Activity E is reduced by one hour, identify the two new critical paths. (1 mark)

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Show Answers Only

`2`
`BEIL`
`ADIL and CHKL`

Show Worked Solution

i. `I\ text(and)\ J`

`:.\ text(2 activities have exactly two immediate predecessors.)`

ii. `text(Scanning forwards:)`

`:.\ text(Initial critical path)\ BEIL`

iii. `text(If)\ E\ text(reduced by 1 hour, critical path)\ BEIL`

`text(reduces to 19 hours.)`

`:.\ text(Other critical paths of 19 hours are:)`

`ADIL = 5 + 2 + 4 + 8 = 19`

`CHKL = 2 + 6 + 3 + 8 = 19`

Networks, STD2 N3 SM-Bank 48

The network shows the activities that are needed to complete a particular project.

Find the total number of activities that need to be completed before activity L can begin. (1 mark)

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The duration of every activity is initially 5 hours.

If the completion times of both activity F and activity K are reduced to 3 hours each, calculate the effect on the completion time for the project. (2 marks)

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Show Answers Only

`7`
`text(Completion time is unchanged.)`

Show Worked Solution

i. `A, B, C, D, E, H\ text(and)\ I\ text(must be completed before)\ L.`

`:.\ text(7 activities need to be completed.)`

ii. `text{Scanning forward (all activities take 5 hours):}`

`text(Activity)\ F\ text(and)\ K\ text(are not on any critical path.)`

`:.\ text(Reducing either will not change the completion time)`

`text(for the project.)`

Networks, STD2 N3 2008 FUR1 8-9 MC

The network below shows the activities that are needed to finish a particular project and their completion times (in days).

Part 1

The earliest start time for Activity K, in days, is

A. `7`

B. `15`

C. `16`

D. `19`

Part 2

This project currently has one critical path.

A second critical path, in addition to the first, would be created by

A. increasing the completion time of D by 7 days.

B. increasing the completion time of G by 1 day.

C. increasing the completion time of I by 2 days.

D. decreasing the completion time of C by 1 day.

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ A`

Show Worked Solution

`text(Part 1)`

`text(Scanning forwards:)`

`text(EST for Activity)\ K`

`=\ text(Duration)\ ACFI`

`= 2 + 5 + 6 + 3`

`= 16`

`=> C`

`text(Part 2)`

♦♦ Mean mark of Part 2 was 35%.

`text(Original critical path:)\ ACFHJL\ text{(22 days)}`

`text(Consider option)\ A,`

`text(New critical path:)\ ABDJL\ text{(22 days)}`

`=> A`

Networks, FUR2 2007 VCE 3

As an attraction for young children, a miniature railway runs throughout a new housing estate.

The trains travel through stations that are represented by nodes on the directed network diagram below.

The number of seats available for children, between each pair of stations, is indicated beside the corresponding edge.

Cut 1, through the network, is shown in the diagram above.

Determine the capacity of Cut 1. (1 mark)
Determine the maximum number of seats available for children for a journey that begins at the West Terminal and ends at the East Terminal. (1 mark)

On one particular train, 10 children set out from the West Terminal.

No new passengers board the train on the journey to the East Terminal.

Determine the maximum number of children who can arrive at the East Terminal on this train. (1 mark)

Show Answers Only

`43`
`22`
`7`

Show Worked Solution

a. `text(The capacity of Cut 1)`

♦♦ Mean mark for all parts (combined) was 33%.
MARKER’S COMMENT: A common error was counting the edge with “10” in the reverse direction (in part a).

`=14 + 8 + 13 + 8`

`= 43`

`text(Maximum seats)`	`=\ text(minimum cut)`
	`= 6 + 7 + 9`
	`= 22`

c. `text{The path (edge weights) of the train setting out with}`

`text(10 children starts with: 11 → 13.)`

`text(At the next station, a maximum of 7 seats are available)`

`text(which remain until the East Terminal.)`

`:.\ text(Maximum number of children arriving is 7.)`

Networks, FUR2 2013 VCE 3

The rangers at the wildlife park restrict access to the walking tracks through areas where the animals breed.

The edges on the directed network diagram below represent one-way tracks through the breeding areas. The direction of travel on each track is shown by an arrow. The numbers on the edges indicate the maximum number of people who are permitted to walk along each track each day.

Starting at `A`, how many people, in total, are permitted to walk to `D` each day? (1 mark)

One day, all the available walking tracks will be used by students on a school excursion.

The students will start at `A` and walk in four separate groups to `D`.

Students must remain in the same groups throughout the walk.

1. Group 1 will have 17 students. This is the maximum group size that can walk together from `A` to `D`.
  Write down the path that group 1 will take. (1 mark)
2. Groups 2, 3 and 4 will each take different paths from `A` to `D`.
  Complete the six missing entries shaded in the table below. (2 marks)

Show Answers Only

`37`
1. `A B E C D`
2. `text{One possible solution is:}`

Show Worked Solution

a.	`text(Maximum flow)`	`=\ text(Minimum cut through)\ CD and ED`
		`= 24 + 13`
		`= 37`

♦ Mean mark of all parts (combined) was 41%.

`:.\ text(A maximum of 37 people can walk)`

`text(to)\ D\ text(from)\ A.`

b.i. `A B E C D`

b.ii. `text(One solution using the second possible largest)`

`text(group of 11 students and two groups from the)`

`text(remaining 9 students is:)`

Networks, FUR1 2017 VCE 8 MC

The flow of oil through a series of pipelines, in litres per minute, is shown in the network below.

The weightings of three of the edges are labelled `x`.

Four cuts labelled A–D are shown on the network.

The maximum flow of oil from the source to the sink, in litres per minute, is given by the capacity of

`text(Cut A if)\ \ x = 1`
`text(Cut B if)\ \ x = 2`
`text(Cut C if)\ \ x = 2`
`text(Cut D if)\ \ x = 3`

Show Answers Only

`B`

Show Worked Solution

`text(Consider the value of each cut for different)\ xtext(-values:)`

`text(When)\ \ x=1,`

♦ Mean mark 44%.

`text{Cut A = 26, Cut B = 24, Cut C = 25, Cut D = 32}`

`text(When)\ \ x=2,`

`text(Cut A = 27, Cut B = 25, Cut C = 26, Cut D = 32)`

`text(When)\ \ x=3,`

`text(Cut A = 28, Cut B = 26, Cut C = 27, Cut D = 32)`

`:.\ text(If)\ \ x=2,\ text(minimum cut / maximum flow is given by)`

`text(the capacity of Cut B).`

`=> B`

Networks, STD2 N3 2009 FUR1 3 MC

The maximum flow from source to sink through the network shown above is

A. `6`

B. `7`

C. `11`

D. `16`

Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 44%.

`text(Maximum flow)`	`=\ text(minimum cut)`
	`= 1 + 4 + 2`
	`= 7`

`=> B`

Networks, STD2 N3 2008 FUR1 6 MC

For the graph above, the capacity of the cut shown is

A. `36`

B. `30`

C. `42`

D. `46`

Show Answers Only

`=> C`

Show Worked Solution

`text(Ignoring any flows that cross the cut from the)`

♦♦ Mean mark 35%.
MARKER’S COMMENT: For an individual flow to contribute to the “cut”, it must flow from the source to the sink.

`text(sink side to the source side.)`

`text(Capacity of the cut)`

`= 4 + 2 + 7 + 9 + 8 + 6 + 6`

`= 42`

`=> C`

Networks, STD2 N3 SM-Bank 36

In the network below, the values on the edges give the maximum flow possible between each pair of vertices. The arrows show the direction of flow. A cut that separates the source from the sink in the network is also shown.

Calculate the capacity of the cut shown in the diagram. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Calculate the maximum flow between source and sink. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`23`
`10`

Show Worked Solution

i. `text(Capacity of the cut)`

♦ Mean mark part (a) 50%.
COMMENT: A quarter of students incorrectly included the “8” which is flowing in the opposite direction.

`= 11 + 5 + 7`

`= 23`

ii.

`text(The maximum flow)`

♦♦ Mean mark part (b) 24%.

`=\ text{minimum cut (see above)}`

`= 4 + 2 + 3 + 1`

`= 10`

Networks, FUR2 2016 VCE 1

A map of the roads connecting five suburbs of a city, Alooma (`A`), Beachton (`B`), Campville (`C`), Dovenest (`D`) and Easyside (`E`), is shown below.

Starting at Beachton, which two suburbs can be driven to using only one road? (1 mark)

A graph that represents the map of the roads is shown below.

One of the edges that connects to vertex `E` is missing from the graph.

1. Add the missing edge to the graph above. (1 mark)
  
  (Answer on the graph above.)
2. Explain what the loop at `D` represents in terms of a driver who is departing from Dovenest. (1 mark)

Show Answers Only

`text(Alooma and Easyside.)`
2. `text(The loop represents that a driver can take a route out of)`
  `text(Dovenest and return home without going through another)`
  `text(suburb or turning back.)`

Show Worked Solution

a. `text(Alooma and Easyside.)`

b.i.

`text(Draw a third edge between Easyside and Dovenest.)`

b.ii. `text(The loop represents that a driver can take a)`

♦♦ Mean mark 30%.

`text(route out of Dovenest and return home without)`

`text(going through another suburb or turning back.)`

Networks, FUR1 2017 VCE 3 MC

Consider the following graph.

The matrix for this graph that records the number of direct routes between points, is shown below.

This adjacency table contains 16 values when complete.

Of the 12 missing elements

eight are ‘1’ and four are ‘2’.
four are ‘1’ and eight are ‘2’.
six are ‘1’ and six are ‘2’.
two are ‘0’, six are ‘1’ and four are ‘2’.

Show Answers Only

`A`

Show Worked Solution

`text(Completing the adjacency matrix:)`

`=> A`

Networks, STD2 N2 2016 FUR1 4 MC

The minimum spanning tree for the network below includes the edge with weight labelled `k`.

The total weight of all edges for the minimum spanning tree is 33.

The value of `k` is

Show Answers Only

`D`

Show Worked Solution

`text(Include)\ k\ text(as a given edge before using Kruskal’s Algorithm,)`

`text(Edge 1:)\ k`

`text(Edge 2: 1)`

`text(Edges 3-4: 2)`

`text(Edge 4: 3 etc…)`

`:.\ text(Minimum spanning tree)`

`text(Total weight)`	`= k + 1+2+2+3+4+5+5+6`
`33`	`= k + 28`
`:. k`	`= 5`

`=> D`

Networks, FUR1 2013 VCE 6 MC

The map above shows the road connections between three towns, `P, Q\ text(and)\ R`.

The graph that could be used to model these road connections is

Show Answers Only

`C`

Show Worked Solution

♦ Mean mark 40%.

`text(Each town has 2 different routes of getting to the)`

`text(the other 2 towns.)`

`=> C`

Geometry, NAP-K3-CA03

Patchouli draws a shape on some grid paper.

She puts a pin through the dot.

She then rotates the paper a half turn clockwise.

Which of the following shows Patchouli's shape after the rotation?

Show Answers Only

Show Worked Solution

Number, NAP-K3-CA05

The entry fees to visit the Toorak Aquarium are $15 for an adult and $7 for a child.

A group of 8 people paid $88 in total to visit the aquarium.

How many adults are in the group?

`2`	`3`	`4`	`5`

Show Answers Only

`2`

Show Worked Solution

`text(An entry fee of $88 would consist of:)`

`text(4 adults and 4 children)`

`4 xx 15 + 4 xx 7 = $88`

Geometry, NAP-K4-CA01

Which of these shows the top view of this stack of boxes?

Show Answers Only

Show Worked Solution

Calculus, MET1 SM-Bank 17

The diagram shows a point `T` on the unit circle `x^2+y^2=1` at an angle `theta` from the positive `x`-axis, where `0<theta<pi/2`.

The tangent to the circle at `T` is perpendicular to `OT`, and intersects the `x`-axis at `P`, and the line `y=1` intersects the `y`-axis at `B`.

Show that the equation of the line `PT` is `xcostheta+ysin theta=1`. (2 marks)
Find the length of `BQ` in terms of `theta`. (1 mark)
Show that the area, `A`, of the trapezium `OPQB` is given by
`A=(2-sintheta)/(2costheta)` (2 marks)
Find the angle `theta` that gives the minimum area of the trapezium. (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`(1-sin theta)/cos theta`
`text{Proof (See Worked Solutions)}`
`theta=pi/6\ \ text(radians)`

Show Worked Solution

`text(Find)\ T:`

♦♦ Mean mark 20%

`text(S)text(ince)\ \ cos theta=x/1\ \ \ text(and)\ \ \ sin theta=y/1`

`:. T\ (cos theta, sin theta)`

`text(Gradient of)\ OT=sin theta/cos theta`

`:.\ text(Gradient)\ PT=-cos theta/sin theta\ \ text{(} _|_ text{lines)}`

`text(Equation of)\ PT\ text(where)`

`m=-cos theta/sin theta,\ \ text(and through)\ \ (cos theta, sin theta)`

`text(Using)\ \ y-y_1`	`=m(x-x_1)`
`y-sin theta`	`=-cos theta/sin theta(x-cos theta)`
`y sin theta-sin^2 theta`	`=-x cos theta+cos^2 theta`
`x cos theta+y sin theta`	`=sin^2 theta+cos^2 theta`
`x cos theta+y sin theta`	`=1\ \ \ \ \ text(… as required)`

ii. `text(Find)\ Q:`

`Q\ => text(intersection of)\ xcos theta+y sin theta=1\ \ text(and)\ \ y=1`

`x cos theta+sin theta`	`=1`
`x cos theta`	`=1-sin theta`
`x`	`=(1-sin theta)/cos theta`

`:.\ text(Length of)\ BQ\ text(is)\ \ (1-sin theta)/cos theta\ text(units)`

iii. `text(Show Area)\ \ OPQB=(2-sin theta)/(2cos theta)`

`A=1/2h(a+b)\ \ text(where)\ \ h=OB=1\ \ a=OP\ \ text(and)`

`b=BQ=(1-sin theta)/cos theta`

`text(Find length)\ OP:`

`P => xcos theta+ysin theta=1 \ text(cuts)\ \ x text(-axis)`

♦♦ Mean mark 24%

`xcos theta`	`=1`
`x`	`=1/cos theta`
`=>\ text(Length)\ OP`	`=1/cos theta`

`text(Area)\ OPQB`	`=1/2xx1(1/cos theta+(1-sin theta)/cos theta)`
	`=1/2((2-sin theta)/cos theta)`
	`=(2-sin theta)/(2cos theta)\ \ text(u²)\ \ \ text(… as required)`

iv. `text(Find)\ theta\ text(such that Area)\ OPQB\ text(is a MIN)`

`A`	`=(2-sin theta)/(2cos theta)`
`(dA)/(d theta)`	`=(2cos theta(-cos theta)-(2- sin theta)(-2 sin theta))/(4cos^2 theta)`
	`=(4 sin theta-2sin^2 theta-2 cos^2 theta)/(4 cos^2 theta)`
	`=(4sin theta-2(sin^2 theta+cos^2 theta))/(4 cos^2 theta)`
	`=(4sin theta-2)/(4cos^2 theta)`
	`=(2sin theta-1)/(2 cos^2 theta)`

Mean mark 19%
IMPORTANT: Look for any opportunity to use the identity `sin^2 theta“+cos^2 theta=1` as it is an examiner’s favourite and can often be the key to simplifying difficult trig equations.

`text(MAX or MIN when)\ (dA)/(d theta)=0`

`=>2sin theta-1`	`=0`
`sin theta`	`=1/2`
`theta`	`=pi/6\ \ \ \ \0<theta<pi/2`

`text(Test for MAX/MIN)`

IMPORTANT: Is the 1st or 2nd derivative test easier here? Students must evaluate and choose, knowing that examiners have often made one significantly easier than the other.

`text(If)\ theta=pi/12\ \ (dA)/(d theta)<0`

`text(If)\ theta=pi/3\ \ (dA)/(d theta)>0\ \ =>text(MIN)`

`:.\text(Area)\ OPQB\ text(is a MIN when)\ theta=pi/6`.

Calculus, MET1 SM-Bank 27

A cone is inscribed in a sphere of radius `a`, centred at `O`. The height of the cone is `x` and the radius of the base is `r`, as shown in the diagram.

Show that the volume, `V`, of the cone is given by
`V = 1/3 pi(2ax^2 - x^3)`. (2 marks)
Find the value of `x` for which the volume of the cone is a maximum. You must give reasons why your value of `x` gives the maximum volume. (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`x = 4/3 a`

Show Worked Solution

i. `text(Show)\ V = 1/3 pi (2ax^2 – x^3)`

`V = 1/3 pi r^2 h`

`text(Using Pythagoras)`

`(x – a)^2 + r^2`	`= a^2`
`r^2`	`= a^2 – (x – a)^2`
	`= a^2 – x^2 + 2ax – a^2`
	`= 2ax – x^2`

`:. V`	`= 1/3 xx pi xx (2ax – x^2) xx x`
	`= 1/3 pi (2ax^2 – x^3)\ …\ text(as required)`

ii. `(dV)/(dx) = 1/3 pi (4ax – 3x^2)`

`(d^2V)/(dx^2) = 1/3 pi (4a – 6x)`

`text(Max or min when)\ (dV)/(dx) = 0`

`1/3 pi (4ax – 3x^2)`	`= 0`
`4ax – 3x^2`	`= 0`
`x(4a – 3x)`	`= 0`

`3x`	`= 4a,`	` \ \ \ \ x ≠ 0`
`x`	` =4/3 a`

`text(When)\ \ x = 4/3 a`

`(d^2V)/(dx^2)`	`= 1/3 pi (4a – 6 xx 4/3 a)`
	`= 1/3 pi (-4a) < 0`
`=>\ text(MAX)`

`:.\ text(Cone volume is a maximum when)\ \ x = 4/3 a.`

Calculus, MET1 SM-Bank 35

The diagram shows two parallel brick walls `KJ` and `MN` joined by a fence from `J` to `M`. The wall `KJ` is `s` metres long and `/_KJM=alpha`. The fence `JM` is `l` metres long.

A new fence is to be built from `K` to a point `P` somewhere on `MN`. The new fence `KP` will cross the original fence `JM` at `O`.

Let `OJ=x` metres, where `0<x<l`.

Show that the total area, `A` square metres, enclosed by `DeltaOKJ` and `DeltaOMP` is given by
`A=s(x-l+l^2/(2x))sin alpha`. (3 marks)
Find the value of `x` that makes `A` as small as possible. Justify the fact that this value of `x` gives the minimum value for `A`. (3 marks)
Hence, find the length of `MP` when `A` is as small as possible. (1 mark)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`l/sqrt2`
`(sqrt2-1)s\ \ text(metres)`

Show Worked Solution

`A=text(Area)\ Delta OJK+text(Area)\ Delta OMP`

♦♦♦ Although mean mark data for question parts is not available before 2009, students found this question extremely challenging.

`text(Using sine rule)`

`text(Area)\ Delta OJK=1/2\ x s sin alpha`

`text(Area)\ DeltaOMP =>text(Need to find)\ \ MP`

`/_OKJ`	`=/_MPO\ \ text{(alternate angles,}\ MP\ text(\|\|)\ KJtext{)}`
`/_PMO`	`=/_OJK=alpha\ \ text{(alternate angles,}\ MP\ text(\|\|)\ KJtext{)}`

`:.\ DeltaOJK\ text(|||)\ Delta OMP\ \ text{(equiangular)}`

`=>x/s`	`=(l-x)/(MP)\ `	` text{(corresponding sides of similar triangles)}`
`MP`	`=(l-x)/x *s`

`text(Area)\ Delta\ OMP`	`=1/2 (l-x)* MP * sin alpha`
	`=1/2 (l-x)((l-x))/x s sin alpha`

`:. A`	`=1/2 xs sin alpha+1/2 (l-x)((l-x))/x* s sin alpha`
	`=s sin alpha(1/2 x+1/2 (l-x)*((l-x))/x)`
	`=s sin alpha(1/2 x+(l-x)^2/(2x))`
	`=s sin alpha(1/2 x+l^2/(2x)-l+1/2 x)`
	`=s(x-l+l^2/(2x))sin alpha\ \ \ \ text(… as required)`

ii. `text(Find)\ x\ text(such that)\ A\ text(is a minimum)`

MARKER’S COMMENT: Students who could not complete part (i) are reminded that they can still proceed to part (ii) and attempt to differentiate the result given.
Note that `l` and `alpha` are constants when differentiating.

`A`	`=s(x-l+l^2/(2x))sin alpha`
`(dA)/(dx)`	`=s(1-l^2/(2x^2))sin alpha`

`text(MAX/MIN when)\ (dA)/(dx)=0`

`s(1-l^2/(2x^2))sin alpha`	`=0`
`l^2/(2x^2)`	`=1`
`2x^2`	`=l^2`
`x^2`	`=l^2/2`
`x`	`=l/sqrt2,\ \ \ x>0`

`(d^2A)/(dx^2)=s((l^2)/(2x^3))sin alpha`

`text(S)text(ince)\ \ 0<alpha<90°\ \ =>\ sin alpha>0,\ \ l>0\ \ text(and)\ \ x>0`

`(d^2A)/(dx^2)>0\ \ \ =>text(MIN at)\ \ x=l/sqrt2`

iii. `text(S)text(ince)\ \ MP=((l-x))/x s\ \ text(and MIN when)\ \ x=l/sqrt2`

`MP`	`=((l-l/sqrt2)/(l/sqrt2))s xx sqrt2/sqrt2`
	`=((sqrt2 l-l))/l s`
	`=(sqrt2-1)s\ \ text(metres)`

`:.\ MP=(sqrt2-1)s\ \ text(metres when)\ A\ text(is a MIN.)`

Calculus, MET1 SM-Bank 34

A cylinder of radius `x` and height `2h` is to be inscribed in a sphere of radius `R` centred at `O` as shown.

Show that the volume of the cylinder is given by
`V = 2pih(R^2 − h^2).` (1 mark)
Hence, or otherwise, show that the cylinder has a maximum volume when
`h = R/sqrt3.` (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions).}`
`text{Proof (See Worked Solutions).}`

Show Worked Solution

i.	`V`	`= pir^2h\ \ \ \ text{(general form)}`
		`= pix^2 xx 2h`

`text(Using Pythagoras:)`

`R^2`	`= h^2 + x^2`
`x^2`	`= R^2 − h^2`
`:.V`	`= 2pih(R^2 − h^2)\ \ …text(as required.)`

ii.	`V`	`= 2pih(R^2 − h^2)`
		`= 2piR^2h − 2pih^3`
	`(dV)/(dh)`	`= 2piR^2 − 3 xx 2pih^2`
		`= 2piR^2 − 6pih^2`
	`(d^2V)/(dh^2)`	`= −12pih`

`text(Max or min when)\ (dV)/(dh) = 0`

`2piR^2 − 6pih^2`	`= 0`
`6pih^2`	`= 2piR^2`
`h^2`	`= R^2/3`
`h`	`= R/sqrt3,\ \ h > 0`

`text(When)\ h = R/sqrt3`

`(d^2V)/(dh^2) = −12pi xx R/sqrt3 < 0`

`:.\ text(Volume is a maximum when)\ h = R/sqrt3\ text(units.)`

Calculus, MET1 SM-Bank 24

The rule for function `f` is `f(x) = x^3 - 3x^2 + kx + 8`, where `k` is a constant.

Find the values of `k` for which `f(x)` is an increasing function. (2 marks)

Show Answers Only

`k>3`

Show Worked Solution

`f(x)`	`= x^3 – 3x^2 + kx + 8`
`f prime (x)`	`= 3x^2 – 6x + k`

`f(x)\ text(is increasing when)\ f prime (x) > 0`

`=> 3x^2 – 6x + k > 0`

♦♦ Mean mark (HSC) 28%.
MARKER’S COMMENT: The arithmetic required to solve `36-12k“<0` proved the undoing of too many students in this question. TAKE CARE!

`text(Note)\ \ \ f prime (x)\ text(is always positive if)`

`f prime (x)\ text(is a positive definite.)`

`text(i.e. when)\ \ a > 0\ text(and)\ Delta < 0`

`a=3>0`

`Delta = b^2\ – 4ac`

`:. (–6)^2 – (4 xx 3 xx k)`	`<0`
`36-12k`	`<0`
`12k`	`>36`
`k`	`>3`

`:.\ f(x)\ text(is increasing when)\ k > 3.`

Calculus, MET1 SM-Bank 30

A function is given by `f(x) = 3x^4 + 4x^3 - 12x^2`.

Find the coordinates of the stationary points of `f(x)` and determine their nature. (3 marks)
Hence, sketch the graph `y = f(x)` showing the stationary points. (2 marks)
For what values of `x` is the function increasing? (1 mark)
For what values of `k` will `f(x) = 3x^4 + 4x^3 - 12x^2 + k = 0` have no solution? (1 mark)

Show Answers Only

`text(MAX at)\ (0,0)`
`text(MIN at)\ text{(1,–5)}`
`text(MIN at)\ text{(–2,–32)}`
`f(x)\ text(is increasing for)\ -2 < x < 0\ text(and)\ x > 1`
`text(No solution when)\ k > 32`

Show Worked Solution

i.	`f(x)`	`= 3x^4 + 4x^3 -12x^2`
	`f prime (x)`	`= 12x^3 + 12x^2 – 24x`
	`f″(x)`	`= 36x^2 + 24x – 24`

`text(Stationary points when)\ f prime (x) = 0`

`=> 12x^3 + 12x^2 – 24x`	`=0`
`12x(x^2 + x – 2)`	`=0`
`12x (x+2) (x -1)`	`=0`

`:.\ text(Stationary points at)\ x=0,\ 1\ text(or)\ –2`

`text(When)\ x=0,\ \ \ \ f(0)=0`
`f″(0)`	`= -24 < 0`
`:.\ text{MAX at (0,0)}`

`text(When)\ x=1`

`f(1)`	`= 3+4\ – 12 = -5`
`f″(1)`	`= 36 + 24\ – 24 = 36 > 0`
`:.\ text{MIN at (1,–5)}`

`text(When)\ x=–2`

`f(–2)`	`=3(–2)^4 + 4(–2)^3 – 12(–2)^2`
	`= 48 -32\ – 48`
	`= -32`
`f″(–2)`	`= 36(–2)^2 + 24(–2) -24`
	`=144 – 48 – 24 = 72 > 0`
`:.\ text{MIN at (–2,–32)}`

ii.

♦ Mean mark (HSC) 42%
MARKER’S COMMENT: Be careful to use the correct inequality signs, and not carelessly include `>=` or `<=` by mistake.

iii.	`f(x)\ text(is increasing for)`
	`-2 < x < 0\ text(and)\ x > 1`

iv. `text(Find)\ k\ text(such that)`

♦♦♦ Mean mark (HSC) 12%.

`3x^4 + 4x^3 – 12x^2 + k = 0\ \ text(has no solution)`

`k\ text(is the vertical shift of)\ \ y = 3x^4 + 4x^3 – 12x^2`

`=>\ text(No solution if it does not cross the)\ x text(-axis.)`

`:.\ text(No solution when)\ \ k > 32`

Calculus, MET1 2016 VCAA 6b

Let `f : [–π, π] → R`, where `f (x) = 2 sin (2x) - 1`.

Calculate the average value of `f` over the interval `−pi/3 <= x <= pi/6`. (3 marks)

Show Answers Only

`−2/pi – 1`

Show Worked Solution

`text(Average Value)`

`= 1/(pi/6 – (−pi/3)) int_(−pi/3)^(pi/6) (2sin(2x) – 1)\ dx`

`= 1/(pi/2) [−cos(2x) – x]_(−pi/3)^(pi/6)`

`= 2/pi [(−cos(pi/3) – pi/6) – (−cos((−2pi)/3) – ((−pi)/3))]`

`= 2/pi [(−1/2 – pi/6) – (1/2 + pi/3)]`

♦ Mean mark 44%.

`= 2/pi [−1 – pi/2]`

`= −2/pi – 1`

Calculus, 2ADV C3 SM-Bank 13

The figure shown represents a wire frame where `ABCE` is a convex quadrilateral. The point `D` is on line segment `EC` with `AB = ED = 2\ text(cm)` and `BC = a\ text(cm)`, where `a` is a positive constant.

`/_ BAE = /_ CEA = pi/2`

Let `/_ CBD = theta` where `0 < theta < pi/2.`

Find `BD` and `CD` in terms of `a` and `theta`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Find the length, `L` cm, of the wire in the frame, including length `BD`, in terms of `a` and `theta`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Find `(dL)/(d theta)`, and hence show that `(dL)/(d theta) = 0` when `BD = 2CD`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Find the maximum value of `L` if `a = 3 sqrt 5`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`BD = a cos theta,\ \ \ CD = a sin theta`
`L = 4 + a + 2 a cos theta + a sin theta`
`text(Proof)\ text{(See Worked Solutions)}`
`L_max = 19 + 3 sqrt 5`

Show Worked Solution

i. `text(In)\ \ Delta BCD,`

`cos theta`	`= (BD)/a`
`:. BD`	`= a cos theta`
`sin theta`	`= (CD)/a`
`:. CD`	`= a sin theta`

ii.	`L`	`= 4 + 2 BD + CD + a`
		`= 4 + 2a cos theta + a sin theta + a`
		`= 4 + a + 2a cos theta + a sin theta`

iii. `text(Noting that)\ a\ text(is a constant:)`

♦ Mean mark (Vic) 35%.

`(dL)/(d theta)= – 2 a sin theta + a cos theta`

`text(When)\ \ (dL)/(d theta) = 0`,

`- 2 a sin theta+ a cos theta`	`= 0`
`a cos theta`	`= 2 a sin theta`
`:. BD`	`= 2CD\ \ text{(using part (a))}`

iv. `text(SP’s when)\ \ (dL)/(d theta)=0,`

♦♦♦ Mean mark (Vic) 5%.

`- 2 a sin theta+ a cos theta`	`= 0`
`sin theta`	`=1/2 cos theta`
`tan theta`	`=1/2`

`text(If)\ \ tan theta=1/2,\ \ cos theta = 2/sqrt5,\ \ sin theta = 1/sqrt5`

`L_(max)`	`= 4 + a + 2a cos theta + a sin theta`
	`= 4 + (3 sqrt 5) + 2 (3 sqrt 5) (2/sqrt 5) + (3 sqrt 5) (1/sqrt 5)`
	`= 4 + 3 sqrt 5 + 12 + 3`
	`= 19 + 3 sqrt 5\ text(cm)`

Calculus, 2ADV C4 SM-Bank 12

Let `f(x) = 2e^(-x/5)\ \ \ text(for)\ \ x>=0`

A right-angled triangle `OQP` has vertex `O` at the origin, vertex `Q` on the `x`-axis and vertex `P` on the graph of `f`, as shown. The coordinates of `P` are `(x, f(x)).`

Find the area, `A`, of the triangle `OPQ` in terms of `x`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Find the maximum area of triangle `OQP` and the value of `x` for which the maximum occurs. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---
Let `S` be the point on the graph of `f` on the `y`-axis and let `T` be the point on the graph of `f` with the `y`-coordinate `1/2`.Find the area of the region bounded by the graph of `f` and the line segment `ST`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`x e^(-x/5)`
`5/e\ text(u²)`
`25/4 log_e (4) – 15/2\ text(u²)`

Show Worked Solution

i.	`text(Area)`	`= 1/2 xx b xx h`
		`= 1/2x(2e^(-x/5))`
	`:. A`	`= xe^(-x/5)`

ii. `text(Stationary point when)\ \ (dA)/(dx) = 0,`

♦ Mean mark (Vic) 35%.

`x(-1/5 e^(-x/5)) + e^(-x/5)`	`= 0`
`e^(-x/5)(1 – x/5)`	`= 0`
`:. x`	`= 5\ \ \ \ (e^(-x/5) >0,\ \ text(for all)\ x)`

`text(When)\ \ x = 5,\ \ A`	`= xe^(-x/5)= 5e^-1`

`:. A_max = 5/e\ text(u², when)\ \ x = 5`

iii. `text(Find)\ \ S:\ F(0) = 2`

`=>S(0, 2)`

♦♦ Mean mark (Vic) 32%.

`text(Find)\ \ T:\ \ \ `	`2e^(-x/5)`	`= 1/2`
	`e^(-x/5)`	`= 1/4`
	`-x/5`	`= log_e (1/4)`
	`x`	`= 5 log_e (4)`

`=> T(5log_e(4), 1/2)`

`:.\ text(Area)`	`= text(Area)\ SOAT – int_0^(5 log_e(4)) (2e^(-x/5)) dx`
	`=1/2h(a+b) + 10 [e^(-x/5)]_0^(5 log_e (4))`
	`= 5/2 log_e (4) (2 + 1/2) + 10 [e^(-log_e (4)) – e^0]`
	`= 25/4 log_e (4) +10 (1/4 – 1)`
	`= 25/4 log_e (4) – 15/2\ text(u²)`

GRAPHS, FUR2 2017 VCAA 3

Lifeguards are required to ensure the safety of swimmers at the beach.

Let `x` be the number of junior lifeguards required.

Let `y` be the number of senior lifeguards required.

The inequality below represents the constraint on the relationship between the number of senior lifeguards required and the number of junior lifeguards required.

Constraint 1 `y >= x/4`

If eight junior lifeguards are required, what is the minimum number of senior lifeguards required? (1 mark)

There are three other constraints.

Constraint 2 `x ≥ 6`

Constraint 3 `y ≥ 4`

Constraint 4 `x + y ≥ 12`

Interpret Constraint 4 in terms of the number of junior lifeguards and senior lifeguards required. (1 mark)

The shaded region of the graph below contains the points that satisfy Constraints 1 to 4.

All lifeguards receive a meal allowance per day.

Junior lifeguards receive $15 per day and senior lifeguards receive $25 per day.

The total meal allowance cost per day, `$C`, for the lifeguards is given by

`C = 15x + 25y`

Determine the minimum total meal allowance cost per day for the lifeguards. (2 marks)
On rainy days there will be no set minimum number of junior lifeguards or senior lifeguards required, therefore:

• Constraint 2 `(x ≥ 6)` and Constraint 3 `(y ≥ 4)` are removed

• Constraint 1 and Constraint 4 are to remain.

Constraint 1 `y >= x/4`

Constraint 4 `x + y >= 12`

The total meal allowance cost per day, `$C`, for the lifeguards remains as

`C = 15x + 25y`

How many junior lifeguards and senior lifeguards work on a rainy day if the total meal allowance cost is to be a minimum?

Write your answers in the boxes provided below. (1 mark)

Show Answers Only

a. `2`

b. `text(The total combined number of junior and)`

`text(senior lifeguards must be at least 12.)`

c. `$220`

Show Worked Solution

a. `text(Minimum senior lifeguards) = 8/4 = 2`

b. `text(The total combined number of junior and)`

`text(senior lifeguards must be at least 12.)`

c. `text(Minimum cost occurs at (8, 4))`

`:. C_text(min)`	`= 15 xx 8 + 25 xx 4`
	`= $220`

d. `text(Consider the graph without the restrictions)`

♦♦ Mean mark 24%.
MARKER’S COMMENT: A common incorrect answer was 10 and 3.

`x >= 6quadtext(and)quady >= 4:`

`text(By inspection, intersection around (9.5, 2.4))`

`text(⇒ Minimum allowance when)`

GRAPHS, FUR2 2017 VCAA 2

A swimming race is being held in the ocean.

From the shore, competitors swim 500 m out to a buoy in the ocean and then return to the shore.

One competitor, Edgar, reaches the buoy after 12.5 minutes and completes the race in 25 minutes.

The graph below shows his distance from shore, in metres, `t` minutes after the race begins.

What distance, in metres, has Edgar swum after 15 minutes? (1 mark)
Let `E` be Edgar’s distance from the shore, in metres, `t` minutes after the race begins.

The linear relation that represents his swim out to the buoy is of the form

`E = kt,` where `0 < t ≤ 12.5`

The slope of the line `k` is the speed at which Edgar is swimming, in metres per minute.

Show that `k = 40`. (1 mark)

A second competitor, Zlatko, began the race at the same time as Edgar.

Below is the relation that describes Zlatko’s swim, where `Z` is his distance from the shore, in metres, `t` minutes after the race begins and `F` is the time it took Zlatko to finish the race.

`Z = {(qquadqquadqquad50t,0 < t <= 10),(−62.5t + 1125, 10 < t <= F):}`

The graph below again shows the relation representing Edgar’s swim.

Sketch the relation representing Zlatko’s swim on the graph below. (2 marks)
How many minutes after the start of the race were Zlatko and Edgar the same distance from the shore?

Round your answer to two decimal places. (1 mark)

Show Answers Only

a. `600\ text(metres)`

b. `text(See Worked Solutions)`

d. `10.98\ text(minutes)`

Show Worked Solution

a.	`text(Distance)`	`= 500 + 100`
		`= 600\ text(metres)`

b. `text(When)\ t = 12.5,quadE = 500`

♦ Mean mark 49%.
MARKER’S COMMENT: Note that it was not sufficient to type in the equation and write “Solve”.

`500`	`= k xx 12.5`
`:. k`	`= 500/12.5`
	`= 40\ \ text(… as required)`

c. `text(Zlatko reaches 500 m buoy when)`

♦ Mean mark 47%

`t = 500/50 = 10\ text(min)`

`text(Zlatko returns to shore when)`

`0`	`=-62.5t + 1125`
`:. t`	`= 1125/62.5= 18\ text(min)`

d. `text(Same distance from shore when:)`

`40t`	`= −62.5t + 1125`
`102.5t`	`= 1125`
`:. t`	`= 1125/102.5`
	`= 10.975…`
	`= 10.98\ text(minutes)`

GEOMETRY, FUR2 2017 VCAA 3

Some hostel buildings are arranged around a grassed area.

The grassed area is shown shaded in the diagram below.

The grassed area is made up of a square overlapping a circle.

The square has side lengths of 65 m.

The circle has a radius of 50 m.

An angle, `theta`, is also shown on the diagram.

Use the cosine rule to show that the angle `theta`, correct to the nearest degree, is equal to 81°. (1 mark)
What is the perimeter, in metres, of the entire grassed area?

Round your answer to the nearest metre. (1 mark)
The hostel’s management is planning to build a pathway from point A to point B, as shown on the diagram below.

Calculate the length, in metres, of the planned pathway.

Round your answer to the nearest metre. (2 marks)

Show Answers Only

a. `81^@`

b. `438\ text(m (nearest metre))`

c. `153\ text(m)`

Show Worked Solution

a.	`costheta`	`= (50^2 + 50^2 – 65^2)/(2 xx 50 xx 50)`
		`= 0.155`
		`= 81.08…`
		`= 81^@\ \ text{(nearest degree) … as required)`

♦♦ Mean mark part (b) 34%.

b.	`text(Perimeter)`	`= (3 xx 65) + ((360-81))/360 xx 2 xx pi xx 50`
		`= 195 + 279/360 xx 2 xx pi xx 50`
		`= 438.47…`
		`= 438\ text(m (nearest metre))`

♦♦ Mean mark part (c) 34%.

`text(Using Pythagoras,)`

`x = sqrt(50^2 – 32.5^2) = 37.99…`

`:. text(Length)\ AB`	`= 50 + 37.99… + 65`
	`= 152.99…`
	`= 153\ text(m)`

GEOMETRY, FUR2 2017 VCAA 2

Miki will travel from Melbourne (38° S, 145° E) to Tokyo (36° N, 140° E) on Wednesday, 20 December.

The flight will leave Melbourne at 11.20 am, and will take 10 hours and 40 minutes to reach Tokyo.

The time difference between Melbourne and Tokyo is two hours at that time of year.

On what day and at what time will Miki arrive in Tokyo? (1 mark)

Miki will travel by train from Tokyo to Nemuro and she will stay in a hostel when she arrives.

The hostel is located 186 m north and 50 m west of the Nemuro railway station.

1. What distance will Miki have to walk if she were to walk in a straight line from the Nemuro railway station to the hostel?
  
  Round your answer to the nearest metre. (1 mark)
2. What is the three-figure bearing of the hostel from the Nemuro railway station?
  
  Round your answer to the nearest degree. (1 mark)

The city of Nemuro is located 43° N, 145° E.

Assume that the radius of Earth is 6400 km.

The small circle of Earth at latitude 43° N is shown in the diagram below.

What is the radius of the small circle of Earth at latitude 43° N?

Round your answer to the nearest kilometre. (1 mark)
Find the shortest great circle distance between Melbourne (38° S, 145° E) and Nemuro (43° N, 145° E).

Round your answer to the nearest kilometre. (1 mark)

Show Answers Only

`8\ text(pm (Wed))`
i. `193\ text(m (nearest metre))`
ii. `345^@`
`4681\ text(km (nearest km))`
`9048\ text(km (nearest km))`

Show Worked Solution

a. `text{Flight arrival (in Melb time) = 11:20 + 10:40 = 22:00 (Wed)}`

♦ Mean mark 40%.
COMMENT: A surprisingly poor result for this standard question.

`text(Tokyo time)`	`=\ text(Melb time less 2 hrs)`
	`= 20:00\ (text(Wed))`
	`= 8\ text(pm (Wed))`

b.i. `text(Using Pythagoras:)`

`d`	`= sqrt(186^2 + 50^2)`
	`= 192.603…`
	`= 193\ text(m (nearest metre))`

♦♦ Mean mark part (b)(ii) 27%.
MARKER’S COMMENT: N15°W is not a 3-figure bearing and received no marks.

b.ii.	`tan theta`	`= 50/186`
	`theta`	`= 15.04…^@`

`:. text(Bearing of)\ H\ text(from)\ N`

`= 360 – 15`

`= 345^@`

c. `text(Let)\ \ x = text(radius of small circle)`

♦ Mean mark part (c) 43%

`sin47^@`	`= x/6400`
`:.x`	`= 6400 xx sin47^@`
	`= 4680.66…`
	`= 4681\ text(km (nearest km))`

`text(Shortest distance)`

♦ Mean mark part (d) 38%

`= text(Arc length)\ NM`

`= 81/360 xx 2 xx pi xx 6400`

`= 9047.78…`

`= 9048\ text(km (nearest km))`

GEOMETRY, FUR2 2017 VCAA 1

Miki is planning a gap year in Japan.

She will store some of her belongings in a small storage box while she is away.

This small storage box is in the shape of a rectangular prism.

The diagram below shows that the dimensions of the small storage box are 40 cm × 19 cm × 32 cm.

The lid of the small storage box is labelled on the diagram above.

1. What is the surface area of the lid, in square centimetres? (1 mark)
2. What is the total outside surface area of this storage box, including the lid and base, in square centimetres? (1 mark)
Miki has a large storage box that is also a rectangular prism.



The large storage box and the small storage box are similar in shape.



The volume of the large storage box is eight times the volume of the small storage box.



The length of the small storage box is 40 cm.

What is the length of the large storage box, in centimetres? (1 mark)

Show Answers Only

a.i. `760\ text(cm²)`

a.ii. `5296\ text(cm²)`

b. `80\ text(cm)`

Show Worked Solution

a.i.	`text{Area (lid)}`	`= 40 xx 19`
		`= 760\ text(cm²)`

a.ii.	`text(Total S.A.)`	`= 2 xx (32 xx 19) + 2 xx (40 xx 32) + 2 xx 760`
		`= 5296\ text(cm²)`

b. `text(If volume scale factor = 8,)`

♦♦ Mean mark 32%.
MARKER’S COMMENT: A lack of understanding between linear and volume scale factors was again notable.

`text(⇒ Linear scale factor) = root(3)8 = 2`

`:.\ text(Length of large storage box)`

`= 2 xx 40`

`= 80\ text(cm)`

NETWORKS, FUR2 2017 VCAA 4

The rides at the theme park are set up at the beginning of each holiday season.

This project involves activities A to O.

The directed network below shows these activities and their completion times in days.

Write down the two immediate predecessors of activity I. (1 mark)
The minimum completion time for the project is 19 days.
1. There are two critical paths. One of the critical paths is A–E–J–L–N.
  
  Write down the other critical path. (1 mark)
2. Determine the float time, in days, for activity F. (1 mark)
The project could finish earlier if some activities were crashed.

Six activities, B, D, G, I, J and L, can all be reduced by one day.

The cost of this crashing is $1000 per activity.
1. What is the minimum number of days in which the project could now be completed? (1 mark)
2. What is the minimum cost of completing the project in this time? (1 mark)

Show Answers Only

a. `D\ text(and)\ E`

b.i. `A E I L N`

b.ii. `6\ text(days)`

c.i. `17`

c.ii. `$4000`

Show Worked Solution

a. `D\ text(and)\ E\ (text(note the dummy is not an activity.))`

b.i. `A E I L N`

♦♦ Mean mark part (b)(i) 44% and part (b)(ii) 28%.

b.ii.	`text(Float time)`	`= 19 – (2 + 3 + 3 + 3 + 2)`
		`= 6\ text(days)`

c.i. `text(Reduce activities:)\ I, J, L\ \ (text(on critical path))`

♦♦ Mean mark part (c)(i) 35%.

`text(New critical path)\ \ A C G N\ \ text(takes 18 days.)`

`:. text(Reduce activity)\ G\ text(also.)`

`text(⇒ this critical path reduces to 17 days.)`

`text(⇒ Minimum Days = 17)`

c.ii. `text(Minimum time requires crashing)\ \ I, J, L\ text(and)\ G`

♦♦♦ Mean mark part (c)(ii) 15%.

`:.\ text(Minimum Cost)`	`= 4 xx 1000`
	`= $4000`

NETWORKS, FUR2 2017 VCAA 2

Bai joins his friends Agatha, Colin and Diane when he arrives for a holiday in Seatown.

Each person will plan one tour that the group will take.

Table 1 shows the time, in minutes, it would take each person to plan each of the four tours.

The aim is to minimise the total time it takes to plan the four tours.

Agatha applies the Hungarian algorithm to Table 1 to produce Table 2.

Table 2 shows the final result of all her steps of the Hungarian algorithm.

In Table 2 there is a zero in the column for Colin.
When all values in the table are considered, what conclusion about minimum total planning time can be made from this zero? (1 mark)
Determine the minimum total planning time, in minutes, for all four tours. (1 mark)

Show Answers Only

a. `text(Colin must plan tour 2.)`

b. `43\ text(minutes)`

Show Worked Solution

a. `text(Colin must plan tour 2.)`

♦ Mean mark part (a) 37%.
MARKER’S COMMENT: The answer “Colin will plan Tour 2 because he is the fastest” received no marks.

`(text(No additional information is required.))`

b. `text{Allocations: Colin (T2), Diane (T3), Bai (T1), Agatha (T4)}`

`:.\ text(Minimum time)`	`= 8 + 18 + 7 + 10`
	`= 43\ text(minutes)`

MATRICES, FUR2 2017 VCAA 3

Senior students at a school choose one elective activity in each of the four terms in 2018.

Their choices are communication (`C`), investigation (`I`), problem-solving (`P`) and service (`S`).

The transition matrix `T` shows the way in which senior students are expected to change their choice of elective activity from term to term.

`{:(qquadqquadqquadqquadquadtext(this term)),(qquadqquadqquad\ CqquadquadIqquadquadPqquad\ S),(T = [(0.4,0.2,0.3,0.1),(0.2,0.4,0.1,0.3),(0.2,0.3,0.3,0.4),(0.2,0.1,0.3,0.2)]{:(C),(I),(P),(S):}qquadtext(next term)):}`

Let `S_n` be the state matrix for the number of senior students expected to choose each elective activity in Term `n`.

For the given matrix `S_1`, a matrix rule that can be used to predict the number of senior students in each elective activity in Terms 2, 3 and 4 is

`S_1 = [(300),(200),(200),(300)],qquadS_(n + 1) = TS_n`

How many senior students will not change their elective activity from Term 1 to Term 2? (1 mark)
Complete `S_2`, the state matrix for Term 2, below. (1 mark)
Of the senior students expected to choose investigation (`I`) in Term 3, what percentage chose service (`S`) in Term 2? (2 marks)
What is the maximum number of senior students expected in investigation (`I`) at any time during 2018? (1 mark)

Show Answers Only

`320`
`S_2 = [(250),(250),(300),(200)]`
`text(25%)`
`250`

Show Worked Solution

a. `text(Students who do not change)`

♦ Mean mark 47%.

`= 0.4 xx 300 + 0.4 xx 200 + 0.3 xx 200 + 0.2 xx 300`

`= 120 + 80 + 60 + 60`

`= 320`

b.	`S_2 = TS_1`	`= [(0.4,0.2,0.3,0.1),(0.2,0.4,0.1,0.3),(0.2,0.3,0.3,0.4),(0.2,0.1,0.3,0.2)][(300),(200),(200),(300)]`
		`= [(250),(250),(300),(200)]`

c.	`S_3`	`= TS_2`
		`= [(260),(240),(295),(205)]`

♦♦♦ Mean mark 13%.
MARKER’S COMMENT: A poorly understood and answered question worthy of careful attention.

`text(Number)\ (I)\ text(in Term 3 = 240)`

`text(Number)\ (S)\ text(in Term 2 = 200)`

`text(S)text(ince 30% move from)\ S\ text(to)\ I\ text(each term:)`

`text(Percentage)`	`= (0.3 xx 200)/240`
	`= 60/240`
	`= 25text(%)`

d.	`S_4`	`= TS_3`
		`= [(261),(239),(294.5),(205.5)]`

♦ Mean mark 43%.

`:. text(Max number of)\ (I)\ text(students is 250.)`

`(text(During term 2))`

MATRICES, FUR2 2017 VCAA 2

Junior students at a school must choose one elective activity in each of the four terms in 2018.

Students can choose from the areas of performance (`P`), sport (`S`) and technology (`T`).

The transition diagram below shows the way in which junior students are expected to change their choice of elective activity from term to term.

Of the junior students who choose performance (`P`) in one term, what percentage are expected to choose sport (`S`) the next term? (1 mark)

Matrix `J_1` lists the number of junior students who will be in each elective activity in Term 1.

`J_1 = [(300),(240),(210)]{:(P),(S),(T):}`

306 junior students are expected to choose sport (`S`) in Term 2.

Complete the calculation below to show this. (1 mark)
In Term 4, how many junior students in total are expected to participate in performance (`P`) or sport (`S`) or technology (`T`)? (1 mark)

Show Answers Only

`text(40%)`
`300 xx 0.4 + 240 xx 0.6 + 210 xx 0.2 = 306`
`750`

Show Worked Solution

a. `text(40%)`

b. `300 xx 0.4 + 240 xx 0.6 + 210 xx 0.2 = 306`

♦♦ Mean mark part (c) 30%.

MARKER’S COMMENT: No matrix calculations were required here.

c. `text(Each term, every student will do)\ P\ text(or)\ S\ text(or)\ T.`

`:. text(Total students (Term 4))`	`=\ text(Total students (Term 1))`
	`= 300 + 240 + 210`
	`= 750`

CORE, FUR2 2017 VCAA 7

Alex sold his mechanics’ business for $360 000 and invested this amount in a perpetuity.

The perpetuity earns interest at the rate of 5.2% per annum.

Interest is calculated and paid monthly.

What monthly payment will Alex receive from this investment? (1 mark)
Later, Alex converts the perpetuity to an annuity investment.

This annuity investment earns interest at the rate of 3.8% per annum, compounding monthly.

For the first four years Alex makes a further payment each month of $500 to his investment.

This monthly payment is made immediately after the interest is added.

After four years of these regular monthly payments, Alex increases the monthly payment.

This new monthly payment gives Alex a balance of $500 000 in his annuity after a further two years.

What is the value of Alex’s new monthly payment?

Round your answer to the nearest cent. (2 marks)

Show Answers Only

`$1560`
`$805.65 (text(nearest cent))`

Show Worked Solution

a.	`text(Monthly payment)`	`= 360\ 000 xx 0.052/12`
		`= $1560`

♦ Mean mark part (a) 50%.

b. `text(By TVM Solver,)`

`text(Find balance after 4 years:)`

`N`	`= 4 xx 12 = 48`
`I(%)`	`= 3.8`
`PV`	`= −360\ 000`
`PMT`	`= −500`
`FV`	`= ?`
`text(PY)`	`=\ text(CY) = 12`

`=> FV`	`= 444\ 872.9445`

`:.\ text(Balance is $444 872.9445)`

♦♦ Mean mark 29%.
MARKER’S COMMENT: A common error was entering the $500 payment as a positive value. Know why this is incorrect!

`text(Find)\ \ PMT\ \ text(when)\ \ FV = 500\ 000\ \ text(and)\ \ N = 24:`

`N`	`= 24`
`I(%)`	`= 3.8`
`PV`	`= −444\ 872.9445`
`PMT`	`= ?`
`FV`	`= 500\ 000`
`text(PY)`	`= text(CY) = 12`

`=> PMT`	`= −805.6505…`

`:. text(Alex’s new monthly payment = $805.65 (nearest cent))`

CORE, FUR2 2017 VCAA 6

Alex sends a bill to his customers after repairs are completed.

If a customer does not pay the bill by the due date, interest is charged.

Alex charges interest after the due date at the rate of 1.5% per month on the amount of an unpaid bill.

The interest on this amount will compound monthly.

Alex sent Marcus a bill of $200 for repairs to his car.

Marcus paid the full amount one month after the due date.

How much did Marcus pay? (1 mark)

Alex sent Lily a bill of $428 for repairs to her car.

Lily did not pay the bill by the due date.

Let `A_n` be the amount of this bill `n` months after the due date.

Write down a recurrence relation, in terms of `A_0`, `A_(n + 1)` and `A_n`, that models the amount of the bill. (2 marks)
Lily paid the full amount of her bill four months after the due date.

How much interest was Lily charged?

Round your answer to the nearest cent. (1 mark)

Show Answers Only

`$203`
`A_o = 428,qquadA_(n + 1) = 1.015A_n`
`$26.26\ \ (text(nearest cent))`

Show Worked Solution

a.	`text(Amount paid)`	`= 200 + 200 xx 1.5text(%)`
		`= 1.015 xx 200`
		`= $203`

♦ Mean mark part (b) 47%.
MARKER’S COMMENT: A recurrence relation has the initial value written first. Know why `A_n=428 xx 1.015^n` is incorrect.

b. `A_o = 428,qquadA_(n + 1) = 1.015A_n`

c.	`text(Total paid)\ (A_4)`	`= 1.015^4 xx 428`
		`= $454.26`

♦♦ Mean mark part (c) 29%.

`:.\ text(Total Interest)`	`= 454.26 – 428`
	`= $26.26\ \ (text(nearest cent))`

CORE, FUR2 2017 VCAA 4

The eggs laid by the female moths hatch and become caterpillars.

The following time series plot shows the total area, in hectares, of forest eaten by the caterpillars in a rural area during the period 1900 to 1980.

The data used to generate this plot is also given.

The association between area of forest eaten by the caterpillars and year is non-linear.

A log₁₀ transformation can be applied to the variable area to linearise the data.

When the equation of the least squares line that can be used to predict log₁₀ (area) from year is determined, the slope of this line is approximately 0.0085385

Round this value to three significant figures. (1 mark)
Perform the log₁₀ transformation to the variable area and determine the equation of the least squares line that can be used to predict log₁₀ (area) from year.

Write the values of the intercept and slope of this least squares line in the appropriate boxes provided below.

Round your answers to three significant figures. (2 marks)
1. The least squares line predicts that the log₁₀ (area) of forest eaten by the caterpillars by the year 2020 will be approximately 2.85
  
  Using this value of 2.85, calculate the expected area of forest that will be eaten by the caterpillars by the year 2020.
  
  Round your answer to the nearest hectare. (1 mark)
2. Give a reason why this prediction may have limited reliability. (1 mark)

Show Answers Only

a. `0.00854\ (text(3 sig fig))`

b. `log_10(text(area)) = −14.4 + 0.000854 xx text(year)`

c.i. `708\ text(hectares)`

c.ii. `text(This prediction extrapolates significantly from the given)`
`text(data range and as a result, its reliability decreases.)`

Show Worked Solution

a. `0.0085385 = 0.00854\ (text(3 sig fig))`

♦ Mean marks of part (a) and (b) 44%.

`log_10(text(area))`

`= −14.4 + 0.000854 xx text(year)`

♦♦ Mean mark part (c)(i) 29%.
COMMENT: When the question specifies using the value 2.85, use it!

c.i.	`log_10(text(Area))`	`= 2.85`
	`:.\ text(Area)`	`= 10^2.85`
		`= 707.94…`
		`= 708\ text(hectares)`

c.ii. `text(This prediction extrapolates significantly from the given)`

`text(data range and as a result, its reliability decreases.)`

Calculus, 2ADV C3 SM-Bank 7

The graph of `f(x) = sqrt x (1 - x)` for `0<=x<=1` is shown below.

Calculate the area between the graph of `f(x)` and the `x`-axis. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
For `x` in the interval `(0, 1)`, show that the gradient of the tangent to the graph of `f(x)` is `(1 - 3x)/(2 sqrt x)`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

The edges of the right-angled triangle `ABC` are the line segments `AC` and `BC`, which are tangent to the graph of `f(x)`, and the line segment `AB`, which is part of the horizontal axis, as shown below.

Let `theta` be the angle that `AC` makes with the positive direction of the horizontal axis.

Find the equation of the line through `B` and `C` in the form `y = mx + c`, for `theta = 45^@`. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`4/15\ text(units)^2`
`text(Proof)\ \ text{(See Workes Solutions)}`
`y = -x + 1`

Show Worked Solution

i.	`text(Area)`	`= int_0^1 (sqrt x – x sqrt x)\ dx`
		`= int_0^1 (x^(1/2) – x^(3/2))\ dx`
		`= [2/3 x^(3/2) – 2/5 x^(5/2)]_0^1`
		`= (2/3 – 2/5) – (0 – 0)`
		`= 10/15 – 6/15`
		`= 4/15\ text(units)^2`

ii.	`f (x)`	`= x^(1/2) – x^(3/2)`
	`f prime (x)`	`= 1/2 x^(-1/2) – 3/2 x^(1/2)`
		`= 1/(2 sqrt x) – (3 sqrt x)/2`
		`= (1 – 3x)/(2 sqrt x)\ \ text(.. as required.)`

iii. `m_(AC) = tan 45^@=1`

♦♦♦ Mean mark (Vic) part (iii) 20%.
MARKER’S COMMENT: Most successful answers introduced a pronumeral such as `a=sqrtx` to solve.

`=> m_(BC) = -1\ \ (m_text(BC) _|_ m_(AC))`

`text(At point of tangency of)\ BC,\ f prime(x) = -1`

`(1 – 3x)/(2 sqrt x)`	`=-1`
`1-3x`	`=-2sqrtx`
`3x-2sqrt x-1`	`=0`

`text(Let)\ \ a=sqrtx,`

`3a^2-2a-1`	`=0`
`(3a+1)(a-1)`	`=0`
`a=1 or -1/3`

`:. sqrt x`	`=1`	`or`	`sqrt x=- 1/3\ \ text{(no solution)}`
`x`	`=1`

`f(1)=sqrt1(1-1)=0\ \ =>B(1,0)`

`text(Equation of)\ \ BC, \ m=-1, text{through (1,0):}`

`y-0`	`=-1(x-1)`
`y`	`=-x+1`

Trigonometry, 2ADV T2 SM-Bank 8

Let `(tantheta - 1) (sin theta - sqrt 3 cos theta) (sin theta + sqrt 3 costheta) = 0`.

State all possible values of `tan theta`. (1 mark)
Hence, find all possible solutions for `(tan theta - 1) (sin^2 theta - 3 cos^2 theta) = 0`,
where `0 <= theta <= pi`. (2 marks)

Show Answers Only

`tan theta = 1 or tan theta = +- sqrt 3`
`theta = pi/4, pi/3 or (2 pi)/3`

Show Worked Solution

i. `(tantheta – 1) (sin theta – sqrt 3 cos theta) (sin theta + sqrt 3 costheta) = 0`

`=> tan theta = 1`

♦ Mean mark (Vic) 42%.

`=>sin theta – sqrt 3 cos theta`	`=0`
`sin theta`	`=sqrt3 cos theta`
`tan theta`	`=sqrt3`

`=>sin theta + sqrt 3 cos theta`	`=0`
`sin theta`	`=-sqrt3 cos theta`
`tan theta`	`=-sqrt3`

`:. tan theta = 1 or tan theta = +- sqrt 3`

ii. `(tan theta – 1) (sin^2 theta – 3 cos^2 theta) = 0`

`text{Using part (i):}`

♦ Mean mark (Vic) 42%.

`(tan theta – 1) (sin theta – sqrt 3 cos theta) (sin theta + sqrt 3 cos theta) = 0`

`=> tan theta`	`= 1`	`qquad or qquad`	`tan theta`	`= +- sqrt 3`
`theta`	`= pi/4`		`theta`	`= pi/3, (2 pi)/3\ \ text(for)\ \ 0<=theta <= pi`

`:. theta = pi/4, pi/3 or (2 pi)/3`

CORE, FUR2 2017 VCAA 3

The number of male moths caught in a trap set in a forest and the egg density (eggs per square metre) in the forest are shown in the table below.

Determine the equation of the least squares line that can be used to predict the egg density in the forest from the number of male moths caught in the trap.

Write the values of the intercept and slope of this least squares line in the appropriate boxes provided below.

Round your answers to one decimal place. (2 marks)
The number of female moths caught in a trap set in a forest and the egg density (eggs per square metre) in the forest can also be examined.

A scatterplot of the data is shown below.

The equation of the least squares line is

egg density = 191 + 31.3 × number of female moths
1. Draw the graph of this least squares line on the scatterplot (provided above). (1 mark)
2. Interpret the slope of the regression line in terms of the variables egg density and number of female moths caught in the trap. (1 mark)
3. The egg density is 1500 when the number of female moths caught is 55.
  
  Determine the residual value if the least squares line is used to predict the egg density for this number of female moths. (1 mark)
4. The correlation coefficient is `r = 0.862`
  
  Determine the percentage of the variation in egg density in the forest explained by the variation in the number of female moths caught in the trap.
  
  Round your answer to one decimal place. (1 mark)

Show Answers Only

`text(egg density)\ = −46.8 + 18.9 xx text(number of male moths)`
`text(Egg density per square metre increases)`
`text(by 31.3 eggs for every extra female moth)`
`text(caught in the trap.)`
`−412.5`
`text(74.3%)`

Show Worked Solution

a. `text(By calculator)`

`text(egg density)\ = −46.8 + 18.9 xx text(number of male moths)`

b.i. `text(Calculating extreme points on graph.)`

♦♦ Mean mark 26%.
MARKER’S COMMENT: Not well answered! Many students did not realise the graph started at 10 on the `x`-axis and many did not use a ruler!

`x = 10, y = 191 + 31.3 xx 10 = 504`

`x = 60, y = 191 + 31.3 xx 60 = 2069`

♦ Mean mark 39%.
MARKER’S COMMENT: Students must clearly refer to the increase in egg density for every one-unit increase in female moths.

b.ii. `text(Egg density per square metre increases)`

`text(by 31.3 eggs for every extra female moth)`

`text(caught in the trap.)`

b.iii. `text(Predicted egg density)`

♦ Mean mark 48%.

`= 191 + 31.3 xx 55`

`= 1912.5`

`:.\ text(Residual value)`	`= 1500 – 1912.5`
	`= −412.5`

b.iv. `r = 0.862`

♦ Mean mark 47%.

`r^2 = 0.862^2 = 0.7430… = 74.3text{% (1 d.p.)}`

`:.\ text(74.3% is explained.)`

Algebra, MET2 2017 VCAA 4

Let `f : R → R :\ f (x) = 2^(x + 1)-2`. Part of the graph of `f` is shown below.

The transformation `T: R^2 -> R^2, \ T([(x),(y)]) = [(x),(y)] + [(c),(d)]` maps the graph of `y = 2^x` onto the graph of `f`.

State the values of `c` and `d`. (2 marks)
Find the rule and domain for `f^(−1)`, the inverse function of `f`. (2 marks)
Find the area bounded by the graphs of `f` and `f^(−1)`. (3 marks)
Part of the graphs of `f` and `f^(−1)` are shown below.

Find the gradient of `f` and the gradient of `f^(−1)` at `x = 0`. (2 marks)

The functions of `g_k`, where `k ∈ R^+`, are defined with domain `R` such that `g_k(x) = 2e^(kx)-2`.

Find the value of `k` such that `g_k(x) = f(x)`. (1 mark)
Find the rule for the inverse functions `g_k^(−1)` of `g_k`, where `k ∈ R^+`. (1 mark)
1. Describe the transformation that maps the graph of `g_1` onto the graph of `g_k`. (1 mark)
2. Describe the transformation that maps the graph of `g_1^(–1)` onto the graph of `g_k^(–1)`. (1 mark)
The lines `L_1` and `L_2` are the tangents at the origin to the graphs of `g_k` and `g_k^(–1)` respectively.
Find the value(s) of `k` for which the angle between `L_1` and `L_2` is 30°. (2 marks)
Let `p` be the value of `k` for which `g_k(x) = g_k^(−1)(x)` has only one solution.
i. Find `p`. (2 marks)
ii. Let `A(k)` be the area bounded by the graphs of `g_k` and `g_k^(–1)` for all `k > p`.
State the smallest value of `b` such that `A(k) < b`. (1 mark)

Show Answers Only

a. `c = −1, \ d = −2`

b. `f^(−1)(x) = log_2 (x + 2)-1, \ x ∈ (−2,∞)`

c. `3-2/(log_e(2))\ text(units)²`

d. `f′(0)= 2log_e(2) and f^(-1′)(0)= 1/(2log_e(2))`

e. `k = log_e(2)`

f. `g_k^(−1)(x)== 1/k log_e((x + 2)/2)`

g.i. `text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

g.ii. `text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

h. `k=sqrt3/6\ or\ sqrt3/2`

i.i. `p = 1/2`

i.ii. `b=4`

Show Worked Solution

a. `text(Using the matrix transformation:)`

`x′`	`=x+c\ \ => x=x′-c`
`y′`	`=y+d\ \ =>y= y′-d`

`y′-d`	`=2^(x′-c)`
`y′`	`= 2^(x′-c)+d`

`:. c = −1, \ d = −2`

b. `text(Let)\ \ y = f(x)`

`text(Inverse : swap)\ x\ ↔ \ y`

`x`	`= 2^(y + 1)-2`
`x + 2`	`= 2^(y + 1)`
`y + 1`	`= log_2(x + 2)`
`y`	`= log_2(x + 2)-1`

`text(dom)(f^(−1)) = text(ran)(f)`

`:. f^(−1)(x) = log_2 (x + 2)-1, \ x ∈ (−2,∞)`

c. `text(Intersection points occur when)\ \ f(x)=f^(−1)(x)`

MARKER’S COMMENT: Specifically recommends using `f^(-1)(x)-f(x)` in this type of integral to avoid errors in stating the integral.

`x`

`= −1, 0`

`text(Area)`	`= int_(−1)^0 (f^(−1)(x)-f(x))\ dx`
	`= 3-2/(log_e(2))\ text(units)²`

d.	`f′(0)`	`= 2log_e(2)`
	`f^(-1′)(0)`	`= 1/(2log_e(2))`

e. `g_k(x) = 2e^(kx)-2`

`text(Solve:)\ \ g_k(x) = f(x)quad text(for)\ k ∈ R^+`

`:. k = log_e(2)`

f. `text(Let)\ \ y = g_k(x)`

`text(Inverse : swap)\ x\ text(and)\ y`

`x`	`= 2e^(ky)-2`
`e^(ky)`	`=(x+2)/2`
`ky`	`=log_e((x+2)/2)`
`:. g_k^(−1)(x)`	`= 1/k log_e((x + 2)/2)`

♦♦ Mean mark part (g)(i) 31%.

g.i.	`g_1(x)`	`= 2e^x-2`
	`g_k(x)`	`= 2e^(kx)-2`

`:. text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

♦♦ Mean mark part (g)(ii) 30%.

g.ii.	`g_1^(−1)(x)`	`= log_e((x + 2)/2)`
	`g_k^(−1)(x)`	`= 1/klog_e((x + 2)/2)`

`:. text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

h. `text(When)\ \ x=0,`

♦♦♦ Mean mark part (h) 13%.

`g_k′(0)=2k\ \ => m_(L_1)=2k`

`g_k^(-1′)(0)=1/(2k)\ \ => m_(L_2)=1/(2k)`

`text(Using)\ \ tan 30^@=|(m_1-m_2)/(1+m_1m_2)|,`

`text(Solve:)\ \ 1/sqrt3`

`=+- ((2k-1/(2k)))/2\ \ text(for)\ k`

`:. k=sqrt3/6\ or\ sqrt3/2`

i.i `text(By inspection, graphs will touch once if at)\ \ x=0,`

♦♦♦ Mean mark part (i)(i) 4%.

`m_(L_1)`	`=m_(L_2)`
`2k`	`=1/(2k)`
`k`	`=1/2,\ \ (k>0)`

`:. p = 1/2`

i.ii `text(As)\ k→oo, text(the graph of)\ g_k\ text(approaches)\ \ x=0`

♦♦♦ Mean mark part (i)(ii) 2%.

`text{(vertically) and}\ \ y=–2\ \ text{(horizontally).}`

`text(Similarly,)\ \ g_k^(-1)\ \ text(approaches)\ \ x=–2`

`text{(vertically) and}\ \ y=0\ \ text{(horizontally).}`

`:. lim_(k→oo) A(k) = 4`

`:.b=4`

Probability, MET2 2017 VCAA 3

The time Jennifer spends on her homework each day varies, but she does some homework every day.

The continuous random variable `T`, which models the time, `t,` in minutes, that Jennifer spends each day on her homework, has a probability density function `f`, where

`f(t) = {{:(1/625 (t - 20)),(1/625 (70 - t)),(0):}qquad{:(20 <= t < 45),(45 <= t <= 70),(text(elsewhere)):}:}`

Sketch the graph of `f` on the axes provided below. (3 marks)
Find `text(Pr)(25 ≤ T ≤ 55)`. (2 marks)
Find `text(Pr)(T ≤ 25 | T ≤ 55)`. (2 marks)
Find `a` such that `text(Pr)(T ≥ a) = 0.7`, correct to four decimal places. (2 marks)
The probability that Jennifer spends more than 50 minutes on her homework on any given day is `8/25`. Assume that the amount of time spent on her homework on any day is independent of the time spent on her homework on any other day.
1. Find the probability that Jennifer spends more than 50 minutes on her homework on more than three of seven randomly chosen days, correct to four decimal places. (2 marks)
2. Find the probability that Jennifer spends more than 50 minutes on her homework on at least two of seven randomly chosen days, given that she spends more than 50 minutes on her homework on at least one of those days, correct to four decimal places. (2 marks)

Let `p` be the probability that on any given day Jennifer spends more than `d` minutes on her homework.

Let `q` be the probability that on two or three days out of seven randomly chosen days she spends more than `d` minutes on her homework.

Express `q` as a polynomial in terms of `p`. (2 marks)
1. Find the maximum value of `q`, correct to four decimal places, and the value of `p` for which this maximum occurs, correct to four decimal places. (2 marks)
2. Find the value of `d` for which the maximum found in part g.i. occurs, correct to the nearest minute. (2 marks)

Show Answers Only

`4/5`
`1/41`
`39.3649`
1. `0.1534`
2. `0.7626`
`q =7p^2(1-p)^4(2p+3)`
1. `p = 0.3539quadtext(and)quadq = 0.5665`
2. `49\ text(min)`

Show Worked Solution

MARKER’S COMMENT: Many did not draw graph along `t`-axis between 0 and 20 and for `t>70`.

b. `text(Pr)(25 <= T <= 55)`

`= int_25^45 1/625 (t – 20)\ dt + int_45^55 1/625 (70 – t)\ dt`

`= 4/5`

c. `text(Pr)(T ≤ 25 | T ≤ 55)`

`=(text(Pr)(T <= 25))/(text(Pr)( T <= 55))`

`= (int_20^25 1/625(t – 20)\ dt)/(1 – int_55^70 1/625(70 – t)\ dt)`

`= (1/50)/(1 – 9/50)`

`= 1/41`

d. `text(Pr)(T ≥ a) = 0.7`

♦ Mean mark part (d) 36%.

`=>\ text(Pr)(T <= a) = 0.3`

`text(Solve:)`

`int_20^a 1/625(t – 20)\ dt`

`= 0.3quadtext(for)quada ∈ (20, 45)`

`:. a == 39.3649`

e.i. `text(Let)\ X =\ text(Number of days Jenn studies more than 50 min)`

`X ~\ text(Bi) (7, 8/25)`

`text(Pr)(X >= 4) = 0.1534`

e.ii.	`text(Pr)(X >= 2 \| X >= 1)`	`= (text(Pr)(X >= 2))/(text(Pr)(X >= 1))`
		`= (0.7113…)/(0.9327…)`
		`= 0.7626\ \ text{(to 4 d.p.)}`

f. `text(Let)\ Y =\ text(Number of days Jenn spends more than)\ d\ text(min)`

`Y ~\ text(Bi)(7,p)`

♦ Mean mark part (f) 36%.

`q`	`= text(Pr)(Y = 2) + text(Pr)(Y = 3)`
	`= ((7),(2))p^2(1 – p)^5 + ((7),(3))p^3(1 – p)^4`
	`= 21p^2(1 – p)^5 + 35p^3(1 – p)^4`
	`=7p^2(1-p)^4[3(1-p)+5p]`
	`=7p^2(1-p)^4(2p+3)`

g.i. `text(Solve)\ \ q′(p) = 0,`

♦♦ Mean mark part (g)(i) 30%.

`p`	`=0.35388…`
	`=0.3539\ \ text{(to 4 d.p.)}`

`:. q_text(max)= 0.5665\ \ text{(to 4 d.p.)}`

g.ii. `text(Pr)(T > d) = p= 0.35388…`

♦♦♦ Mean mark part (g)(ii) 8%.

`text(Solve:)`

`int_d^70 (1/625(70 – t))dt`

`= 0.35388… quadtext(for)quadd ∈ (45,70)`

`:. d`	`=48.967…`
	`=49\ text(mins)`

Algebra, MET2 2017 VCAA 2

Sammy visits a giant Ferris wheel. Sammy enters a capsule on the Ferris wheel from a platform above the ground. The Ferris wheel is rotating anticlockwise. The capsule is attached to the Ferris wheel at point `P`. The height of `P` above the ground, `h`, is modelled by `h(t) = 65 - 55cos((pit)/15)`, where `t` is the time in minutes after Sammy enters the capsule and `h` is measured in metres.

Sammy exits the capsule after one complete rotation of the Ferris wheel.

State the minimum and maximum heights of `P` above the ground. (1 mark)
For how much time is Sammy in the capsule? (1 mark)
Find the rate of change of `h` with respect to `t` and, hence, state the value of `t` at which the rate of change of `h` is at its maximum. (2 marks)

As the Ferris wheel rotates, a stationary boat at `B`, on a nearby river, first becomes visible at point `P_1`. `B` is 500 m horizontally from the vertical axis through the centre `C` of the Ferris wheel and angle `CBO = theta`, as shown below.

Find `theta` in degrees, correct to two decimal places. (1 mark)

Part of the path of `P` is given by `y = sqrt(3025 - x^2) + 65, x ∈ [−55,55]`, where `x` and `y` are in metres.

Find `(dy)/(dx)`. (1 mark)

As the Ferris wheel continues to rotate, the boat at `B` is no longer visible from the point `P_2(u, v)` onwards. The line through `B` and `P_2` is tangent to the path of `P`, where angle `OBP_2 = alpha`.

Find the gradient of the line segment `P_2B` in terms of `u` and, hence, find the coordinates of `P_2`, correct to two decimal places. (3 marks)
Find `alpha` in degrees, correct to two decimal places. (1 mark)
Hence or otherwise, find the length of time, to the nearest minute, during which the boat at `B` is visible. (2 marks)

Show Answers Only

`h_text(min) = 10\ text(m), h_text(max) = 120\ text(m)`
`30\ text(min)`
`t = 7.5`
`7.41^@`
`(−x)/(sqrt(3025 – x^2))`
`P_2(13.00, 118.44)`
`13.67^@`
`7\ text(min)`

Show Worked Solution

a.	`h_text(min)`	`= 65 – 55`	`h_text(max)`	`= 65 + 55`
		`= 10\ text(m)`		`= 120\ text(m)`

b. `text(Period) = (2pi)/(pi/15) = 30\ text(min)`

c. `h′(t) = (11pi)/3\ sin(pi/15 t)`

♦ Mean mark 50%.
MARKER’S COMMENT: A number of commons errors here – 2 answers given, calc not in radian mode, etc …

`text(Solve)\ h″(t) = 0, t ∈ (0,30)`

`t = 15/2\ \ text{(max)}`

`text(or)`

`t = 45/2\ \ text{(min – descending)}`

`:. t = 7.5`

♦ Mean mark 36%.
MARKER’S COMMENT: Choosing degrees vs radians in the correct context was critical here.

`tan(theta)`	`= 65/500`
`:. theta`	`=7.406…`
	`= 7.41^@`

`(dy)/(dx)`

`= (−x)/(sqrt(3025 – x^2))`

`P_2(u,sqrt(3025-u^2) + 65),\ \ B(500,0)`

`:. m_(P_2B)`

`= (sqrt(3025 – u^2) + 65)/(u – 500)`

`text{Using part (e), when}\ \ x=u,`

♦♦♦ Mean mark part (f) 18%.
MARKER’S COMMENT: Many students were unable to use the rise over run information to calculate the second gradient.

`dy/dx=(−u)/(sqrt(3025 – u^2))`

`text{Solve (by CAS):}`

`(sqrt(3025 – u^2) + 65)/(u – 500)`

`= (−u)/(sqrt(3025 – u^2))\ \ text(for)\ u`

`u=12.9975…=13.00\ \ text{(2 d.p.)}`

`:. v`	`= sqrt(3025 – (12.9975…)^2) + 65`
	`= 118.4421…`
	`= 118.44\ \ text{(2 d.p.)}`

`:.P_2(13.00, 118.44)`

♦♦♦ Mean mark part (g) 7%.

g.	`tan alpha`	`=v/(500-u)`
		`= (118.442…)/(500 – 12.9975…)`
	`:. alpha`	`= 13.67^@\ \ text{(2 d.p.)}`

♦♦♦ Mean mark 5%.

`text(Find the rotation between)\ P_1 and P_2:`

`text(Rotation to)\ P_1 = 90-7.41=82.59^@`

`text(Rotation to)\ P_2 = 180-13.67=166.33^@`

`text(Rotation)\ \ P_1 → P_2 = 166.33-82.59 = 83.74^@`

`:.\ text(Time visible)`	`= 83.74/360 xx 30\ text(min)`
	`=6.978…`
	`= 7\ text{min (nearest degree)}`

Calculus, MET2 2017 VCAA 20 MC

The graphs of `f:[0,pi/2] -> R, f(x) = cos(x)` and `g:[0,pi/2] -> R, g(x) = sqrt3 sin(x)` are shown below.

The graph intersect at `B`.

The ratio of the area of the shaded region to the area of triangle `OAB` is

`9:8`
`sqrt3 - 1 : (sqrt3 pi)/8`
`8sqrt3 - 3 : 3pi`
`sqrt3 - 1 : (sqrt3 pi)/4`
`1 : (sqrt3 pi)/8`

Show Answers Only

`B`

Show Worked Solution

`text(Find intersection point)\ B:`

`cos(x)`	`=sqrt3 sin(x)`
`tan x`	`=1/sqrt3`
`x`	`=pi/6`

`:. B(pi/6,sqrt3/2)`

`A_text(shaded) : A_Δ`

`int_0^(pi/6) sqrt3 sin(x)\ dx + int_(pi/6)^(pi/2) cos(x)\ dx\ :\ 1/2 xx pi/2 xx sqrt3/2`

`sqrt3-1\ :\ (sqrt3 pi)/8`

`=> B`

Probability, MET2 2017 VCAA 18 MC

Let `X` be a discrete random variable with binomial distribution `X ~\ text(Bi)(n, p)`. The mean and the standard deviation of this distribution are equal.

Given that `0 < p < 1`, the smallest number of trials, `n`, such that `p ≤ 0.01` is

`37`
`49`
`98`
`99`
`101`

Show Answers Only

`D`

Show Worked Solution

`E(X)`	`= σ_x`
`np`	`= sqrt(np(1 – p))`
`n^2p^2`	`= np(1 – p)`
`np(np-1+p)`	`=0`
`np-1+p`	`=0,\ \ \ (np!=0)`
`p(n+1)`	`=1`
`p`	`=1/(n+1)`

`:.\ text(If)\ \ 1/(n + 1)`	`<= 0.01`
`n`	`>= 99`

`=> D`