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The graph above has
`text(Redrawing the graph in planar form,)`
`text(the graph can be seen to have 6 faces.)`
`text(Alternatively, using Euler’s rule:)`
| `v + f` | `= e + 2` |
| `5 + f` | `= 9 + 2` |
| `:. f` | `= 6` |
`=> C`
The table below shows the time (in minutes) that each of four people, Aiden, Bing, Callum and Dee, would take to complete each of the tasks `U, V, W` and `X.`
If each person is allocated one task only, the minimum total time for this group of people to complete all four tasks is
A. `22\ text(minutes)`
B. `28\ text(minutes)`
C. `29\ text(minutes)`
D. `30\ text(minutes)`
E. `32\ text(minutes)`
`:.\ text(Minimum time)`
`= 9 + 5 + 6 + 8`
`= 28\ text(minutes)`
`=> B`
A project has 12 activities. The network below gives the time (in hours) that it takes to complete each activity.
The critical path for this project is
A. `ADGK`
B. `ADGIL`
C. `BHJL`
D. `CEGIL`
E. `CEHJL`
The network below shows the activities and their completion times (in hours) that are needed to complete a project.
The project is to be crashed by reducing the completion time of one activity only.
This will reduce the completion time of the project by a maximum of
A. 1 hour
B. 2 hours
C. 3 hours
D. 4 hours
E. 5 hours
Five soccer teams played each other once in a tournament. In each game there was a winner and a loser.
A table of one-step and two-step dominances was prepared to summarise the results.
One result in the tournament that must have occurred is that
A. Elephants defeated Bears.
B. Elephants defeated Aardvarks.
C. Aardvarks defeated Donkeys.
D. Donkeys defeated Bears.
E. Bears defeated Chimps.
An undirected connected graph has five vertices.
Three of these vertices are of even degree and two of these vertices are of odd degree.
One extra edge is added. It joins two of the existing vertices.
In the resulting graph, it is not possible to have five vertices that are
A. all of even degree.
B. all of equal degree.
C. one of even degree and four of odd degree.
D. three of even degree and two of odd degree.
E. four of even degree and one of odd degree.
`text(Consider an example of the graph)`
`text{described (below):}`
`A\ text(is possible – join)\ V\ text(and)\ Z`
`B\ text(is possible – join)\ V\ text(and)\ Z`
`C\ text(is possible – join)\ W\ text(and)\ Y`
`D\ text(is possible – join)\ V\ text(and)\ X`
`E\ text(is NOT possible)`
`=> E`
An Euler path through a network commences at vertex `P` and ends at vertex `Q`.
Consider the following five statements about this Euler path and network.
• In the network, there could be three vertices with degree equal to one.
• The path could have passed through an isolated vertex.
• The path could have included vertex `Q` more than once.
• The sum of the degrees of vertices `P` and `Q` could equal seven.
• The sum of the degrees of all vertices in the network could equal seven.
How many of these statements are true?
A. `0`
B. `1`
C. `2`
D. `3`
E. `4`
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
E. 34
John, Ken and Lisa must work together to complete eight activities, `A, B, C, D, E, F, G` and `H`, in minimum time.
The directed network below shows the activities, their completion times in days, and the order in which they must be completed.
Several activities need special skills. Each of these activities may be completed only by a specified person.
With these conditions, the minimum number of days required to complete these eight activities is
A. 14
B. 17
C. 20
D. 21
E. 24
Eight activities, `A, B, C, D, E, F, G\ text(and)\ H`, must be completed for a project.
The graph above shows these activities and their usual duration in hours.
The duration of each activity can be reduced by one hour.
To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is
A. 1
B. 2
C. 3
D. 4
E. 5
Vehicles from a town can drive onto a freeway along a network of one-way and two-way roads, as shown in the network diagram below.
The numbers indicate the maximum number of vehicles per hour that can travel along each road in this network. The arrows represent the permitted direction of travel.
One of the four dotted lines shown on the diagram is the minimum cut for this network.
The maximum number of vehicles per hour that can travel through this network from the town onto the freeway is
A. `310`
B. `330`
C. `350`
D. `370`
E. `390`
The graph above shows five activities, `A, B, C, D\ text(and)\ E`, that must be completed to finish a project.
The number next to each letter shows the completion time, in hours, for the activity.
Each of the five activities can have its completion time reduced by a maximum of one hour at a cost of $100 per hour.
The least cost to achieve the greatest reduction in the time taken to finish the project is
A. $100
B. $200
C. $300
D. $400
E. $500
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.
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`.
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.
As an attraction for young children, a miniature railway runs throughout the 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.
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.
a. `text(The capacity of Cut 1)`
`=14 + 8 + 13 + 8`
`= 43`
| b. |  |
| `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.)`
A network of tracks connects two car parks in a festival venue to the exit, as shown in the directed graph below.
The arrows show the direction that cars can travel along each of the tracks and the numbers show each track’s capacity in cars per minute.
Five cuts are drawn on the diagram.
The maximum number of cars per minute that will reach the exit is given by the capacity of
A. Cut A
B. Cut B
C. Cut C
D. Cut D
E. Cut E
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.
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.
Write down the path that group 1 will take. (1 mark)
Complete the six missing entries shaded in the table below. (2 marks)
| a. | `text(Maximum flow)` | `=\ text(minimum cut through)\ CD and ED` |
| `= 24 + 13` | ||
| `= 37` |
`:.\ 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:)`
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.
Determine the longest float time, in weeks, on the supervisor’s list. (1 mark)
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.
Determine the total overall time, in weeks, for the completion of this building project. (1 mark)
a. `7\ text(weeks)`
b. `BDFGIK`
c. `H\ text(or)\ J\ text(can be delayed for a maximum)`
`text(of 3 weeks.)`
| d. | |
e. `text(The new critical path is)\ BLEGIK.`
`L\ text(now takes 7 weeks.)`
`:.\ text(Time for completion)`
`= 4 + 7 + 1 + 5 + 2 + 6`
`= 25\ text(weeks)`
Stormwater enters a network of pipes at either Dunlop North (Source 1) or Dunlop South (Source 2) and flows into the ocean at either Outlet 1 or Outlet 2.
On the network diagram below, the pipes are represented by straight lines with arrows that indicate the direction of the flow of water. Water cannot flow through a pipe in the opposite direction.
The numbers next to the arrows represent the maximum rate, in kilolitres per minute, at which stormwater can flow through each pipe.
A length of pipe, show in bold on the network diagram below, has been damaged and will be replaced with a larger pipe.
What minimum rate of flow through the pipe, in kilolitres per minute, will achieve this? (1 mark)
a. `text(Storm water from Source 2 cannot reach Outlet 1)`
| b. |  |
`text(The minimum cut includes the 200 kL/min pipe from Source 1.)`
`:.\ text(Maximum rates are)`
`text(Outlet 1: 700 kL/min)`
`text(Outlet 2: 700 kL/min)`
c. `text(The next smallest cut in the lower pipe system is 800.)`
`:.\ text(The minimum flow through the new pipe that will achieve)`
`text(this is 300 kL/min.)`
To restore a vintage train, 13 activities need to be completed.
The network below shows these 13 activities and their completion times in hours.
The minimum time in which all 13 activities can be completed is 21 hours.
Just before they started restoring the train, the members of the club needed to add another activity, `X`, to the project.
Activity `X` will take seven hours to complete.
Activity `X` has no predecessors, but must be completed before activity `G` starts.
Activity `A` can be crashed by up to four hours at an additional cost of $90 per our.
This may reduce the minimum completion time for the project, including activity `X`.
Market researchers claim that the ideal number of bookshops (`x`), sports shoe shops (`y`) and music stores (`z`) for a shopping centre can be determined by solving the equations
`2x + y + z = 12`
`x-y+z=1`
`2y-z=6`
`qquad[(qquadqquadqquad),(),()][(quad),(quad),(quad)] = [(quad),(quad),(quad)]`
To study the life-and-death cycle of an insect population, a number of insect eggs (`E`), juvenile insects (`J`) and adult insects (`A`) are placed in a closed environment.
The initial state of this population can be described by the column matrix
`S_0 = [(400),(200),(100),(0)]{:(E),(J),(A),(D):}`
A row has been included in the state matrix to allow for insects and eggs that die (`D`).
In this population
In this population, the adult insects have been sterilised so that no new eggs are produced. In these circumstances, the life-and-death cycle of the insects can be modelled by the transition matrix
`{:(qquadqquadqquadqquadquadtext(this week)),((qquadqquadqquadE,quad\ J,quadA,\ D)),(T = [(0.4,0,0,0),(0.5,0.4,0,0),(0,0.5,0.8,0),(0.1,0.1,0.2,1)]{:(E),(J),(A),(D):}):}`
Assuming 30% of adults lay eggs each week, the population matrix after one week, `S_1`, is now given by
`qquad S_1 = TS_0 + BS_0`
where `B = [(0,0,0.3,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]` and `S_0 = [(400),(200),(100),(0)]{:(E),(J),(A),(D):}`
This pattern continues. The population matrix after `n` weeks, `S_n`, is given by
`qquad qquad qquad S_n = TS_(n - 1) + BS_(n - 1)`
A breeding program is started in the wetlands. It is aimed at establishing a colony of native ducks.
The matrix `W_0` displays the number of juvenile female ducks (`J`) and the number of adult female ducks (`A`) that are introduced to the wetlands at the start of the breeding program.
`W_0 = [(32),(64)]{:(J),(A):}`
The number of juvenile female ducks (`J`) and the number of adult female ducks (`A`) in the colony at the end of Year 1 of the breeding program is determined using the matrix equation
`W_1 = BW_0`
In this equation, `B` is the breeding matrix
`{:((qquadqquadqquad\ J,qquadA)),(B = [(0,2),(0.25,0.5)]{:(J),(A):}):}`
The number of juvenile female ducks (`J`) and the number of adult female ducks (`A`) in the colony at the end of Year `n` of the breeding program is determined using the matrix equation
`W_n = BW_(n - 1)`
The graph below is incomplete because the points for the end of Year 3 of the breeding program are missing.
Plot the corresponding points on the graph. (2 marks)
Write your answer correct to the nearest whole number. (1 mark)
The breeding matrix `B` assumes that, on average, each adult female duck lays and hatches two female eggs for each year of the breeding program.
If each adult female duck lays and hatches only one female egg each year, it is expected that the duck colony in the wetland will not be self-sustaining and will, in the long run, die out.
The matrix equation
`W_n = PW_(n - 1)`
with a different breeding matrix
`{:((qquadqquadqquad\ J,qquadA)),(P = [(0,1),(0.25,0.5)]{:(J),(A):}):}`
and the initial state matrix
`W_0 = [(32),(64)]{:(J),(A):}`
models this situation.
Changing the number of juvenile and adult female ducks at the start of the breeding program will also change the expected size of the colony.
a. `text(Total female ducks introduced)`
`= 32 + 64`
`= 96`
| b. | `W_1` | `= BW_0` |
| `= [(0,2),(0.25,0.5)][(32),(64)]` | ||
| `= [(128),(40)]` |
| c.i. | |
`W_n = BW_(n – 1)`
`W_2 = [(0,2),(0.25,0.5)][(128),(40)] = [(80),(52)]`
`W_3 = [(0,2),(0.25,0.5)][(80),(52)] = [(104),(46)]`
`:.\ text{Plot (3, 104) and (3, 46)}`
c.ii. `text(Consider)\ n = text(50 and 51,)`
`W_50 = [(0,2),(0.25,0.5)]^49[(32),(64)] = [(96),(48)]`
`W_51 = [(0,2),(0.25,0.5)]^50[(32),(64)] = [(96),(48)]`
`:.\ text(Total female ducks in the long term)`
`= 96 + 48`
`= 144`
d. `text(Initial female ducks = 96)`
`W_n = PW_(n – 1)`
`W_4 = [(0,1),(0.25,0.5)]^3[(32),(64)] = [(28),(33)]`
`:. 51\ text(ducks at end of year 4.)`
`W_5 = [(0,1),(0.25,0.5)]^4[(32),(64)] = [(23),(18.5)]`
`:. 41.5\ text(ducks at end of year 5.)`
`:.\ text{Numbers halve (drop below 48) in year 5.}`
| e. | `text(Let )` | `a = text(initial juvenile ducks)` |
| `b = text(initial adult ducks)` |
`text(Find)\ a\ text(and)\ b\ text(such that,)`
`[(0,1),(0.25,0.5)]^2[(a),(b)]= [(100),(50)]`
`[(0.25,0.5),(0.125,0.5)][(a),(b)] = [(100),(50)]`
| `:. [(a),(b)]` | `= [(0.25,0.5),(0.125,0.5)]^(-1)[(100),(50)]` |
| `= [(400),(0)]` |
`:. 400\ text(juvenile ducks and 0 adult ducks)`
`text(should be introduced.)`
10 000 trout eggs, 1000 baby trout and 800 adult trout are placed in a pond to establish a trout population.
In establishing this population
From year to year, this situation can be represented by the transition matrix `T`, where
`{:(qquadqquadqquadqquadqquadtext(this year)),((qquadqquadqquadE,quad\ B,quad\ A,\ D)),(T = [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)]):}{:(),(),(E),(B),(A),(D):}{:(),(),(qquadtext(next year)):}`
The initial state matrix for this trout population, `S_0`, can be written as
`S_0 = [(10\ 000),(1000),(800),(0)]{:(E),(B),(A),(D):}`
Let `S_n` represent the state matrix describing the trout population after `n` years.
The rule `S_n = T S_(n – 1)` that was used to describe the development of the trout in this pond does not take into account new eggs added to the population when the adult trout begin to breed.
The matrix describing the population after one year, `S_1`, is now given by the new rule
`S_1 = T S_0 + 500\ M\ S_0`
where `T=[(0,0,0,0),(0.40,0,0,0),(0,0.25,0.50,0),(0.60,0.75,0.50,1.0)], M=[(0,0,0.50,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]\ text(and)\ S_0=[(10\ 000),(1000),(800),(0)]`
This pattern continues so that the matrix describing the population after `n` years, `S_n`, is given by the rule
`S_n = T\ S_(n – 1) + 500\ M\ S_(n – 1)`
a.i. `text(60% of eggs die in 1st year,)`
`:.\ text(Eggs that die in year 1)`
`= 0.60 xx 10\ 000`
`= 6000`
| a.ii. |
|
| b.i. | `S_1` | `= TS_0` |
| `= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)][(10\ 000),(1000),(800),(0)]` | ||
| `= [(0),(4000),(650),(7150)]` |
| b.ii. | `S_4` | `= T^4S_0` |
| `= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)]^4[(10\ 000),(1000),(800),(0)]` | ||
| `= [(0),(0),(331.25),(11\ 468.75)]` |
`:. 331\ text(trout is the predicted population)`
`text(after 4 years.)`
| c.i. | `S_12 = T^12S_0 = [(0),(0),(1.29),(11\ 791)]` |
`S_13 = T^13S_0 = [(0),(0),(0.65),(11\ 799)]`
`:.\ text{It will take 13 years (when the trout}`
`text{population drops below 1).}`
| c.ii. | `S_1 = TS_0 = [(0),(4000),(650),(7150)]` |
`text(After 1 year, 650 adult trout.)`
`text(Similarly,)`
`S_2 = T^2S_0 = [(0),(0),(1325),(10\ 475)]`
`S_3 = T^3S_0 = [(0),(0),(662.5),(11\ 137.5)]`
`S_4 = T^4S_0 = [(0),(0),(331),(11\ 469)]`
`:.\ text(Largest number of adult trout = 1325.)`
| d. | `S_0 – S_1 = [(10\ 000),(1000),(800),(0)] – [(0),(4000),(650),(7150)] = [(10\ 000),(−3000),(150),(−7150)]` |
`:.\ text(Add 10 000 eggs, remove 3000 baby trout)`
`text(and add 150 adult trout to keep the)`
`text(population constant.)`
| e.i. | `S_1` | `= TS_0 + 500MS_0` |
| `= [(0),(4000),(650),(7150)] + 500 xx [(0,0,0.5,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)][(10\ 000),(1000),(800),(0)]` | ||
| `= [(0),(4000),(650),(7150)] + 500[(400),(0),(0),(0)]` | ||
| `= [(200\ 000),(4000),(650),(7150)]` |
| e.ii. | `S_2` | `= TS_1 + 500MS_1` |
|
`= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)][(200\ 000),(4000),(650),(7150)]` `+ 500 xx [(0,0,0.5,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)][(200\ 000),(4000),(650),(7150)]` |
||
| `= [(162\ 500),(80\ 000),(1325),(130\ 475)]` |
A large population of mutton birds migrates each year to a remote island to nest and breed. There are four nesting sites on the island, A, B, C and D.
Researchers suggest that the following transition matrix can be used to predict the number of mutton birds nesting at each of the four sites in subsequent years. An equivalent transition diagram is also given.
| `{:(qquad qquad qquad qquad {:text(this year):}), (qquad qquad quad quad \ {:(A,\ \ B,\ \ C,\ D):}), (T = [(0.4, 0, 0.2, 0),(0.35, 1, 0.15, 0), (0.15, 0, 0.55, 0), (0.1, 0, 0.1, 1)] {:(A), (B), (C), (D):} quad {:text(next year):}):}` |  |
Part 1
Two thousand eight hundred mutton birds nest at site C in 2008.
Of these 2800 mutton birds, the number that nest at site A in 2009 is predicted to be
A. `560`
B. `980`
C. `1680`
D. `2800`
E. `3360`
Part 2
This transition matrix predicts that, in the long term, the mutton birds will
A. nest only at site A.
B. nest only at site B.
C. nest only at site A and C.
D. nest only at site B and D.
E. continue to nest at all four sites.
Part 3
Six thousand mutton birds nest at site B in 2008.
Assume that an equal number of mutton birds nested at each of the four sites in 2007. The same transition matrix applies.
The total number of mutton birds that nested on the island in 2007 was
A. `6000`
B. `8000`
C. `12\ 000`
D. `16\ 000`
E. `24\ 000`
Robbie completed a test of four multiple-choice questions.
Each question had four alternatives, A, B, C or D.
Robbie randomly guessed the answer to the first question.
He then determined his answers to the remaining three questions by following the transition matrix
`{:(qquad qquad qquad {:text(this question):}), (qquad qquad quad \ {:(A, B, C, D):}), (T = [(1,\ 0,\ 0,\ 0), (0,\ 0,\ 1,\ 0), (0,\ 0,\ 0,\ 1), (0,\ 1,\ 0,\ 0)] {:(A), (B), (C), (D):} quad {:text(next question):}):}`
Which of the following statements is true?
A. It is impossible for Robbie to give the same answer to all four questions.
B. Robbie would always give the same answer to the first and fourth questions
C. Robbie would always give the same answer to the second and third questions.
D. If Robbie answered A for question one, he would have answered B for question two
E. It is possible that Robbie gave the same answer to exactly three of the four questions.
Vince, Nev and Rani all service office equipment.
The matrix `T` shows the time that it takes (in minutes) for each of Vince (V), Nev (N) and Rani (R) to service a photocopier (P) a fax machine (F) and a scanner (S).
`{:(qquad qquad qquad {:(V,\ N,\ R):}), (T = [(12, 15, 14),(8, 7, 8), (20, 19, 17)] {:(P), (F), (S):}):}`
The matrix `U` below displays the number of photocopiers, fax machines and scanners to be serviced in three schools, Alton (A), Borton (B) and Carlon (C).
`{:(qquad qquad qquad {:(P, F, S):}), (U = [(5,\ 3,\ 2),(4,\ 4,\ 3), (6,\ 1,\ 2)] {:(A), (B), (C):}):}`
A matrix that displays the time that it would take each of Vince, Nev and Rani, working alone, to service the photocopiers, fax machines and scanners in each of the three schools is
| A. | `[(17, 18, 16), (12, 11, 11), (26, 20, 19)]` | B. | `[(204, 110, 97), (116, 60, 53), (278, 153, 131)]` |
| C. | `[(124, 134, 128), (140, 145, 139), (120, 135, 126)]` | D. | `[(7, 12, 12), (4, 3, 5), (14, 18, 15)]` |
| E. | `[(60, 15, 28), (32, 35, 24), (120, 19, 34)]` |
The total cost of one ice cream and three soft drinks at Catherine’s shop is $9.
The total cost of two ice creams and five soft drinks is $16.
Let `x` be the cost of an ice cream and `y` be the cost of a soft drink
The matrix `[(x), (y)]` is equal to
A. `[(1, 3), (2, 5)] [(x), (y)]`
B. `[(1, 3), (2, 5)] [(9), (16)]`
C. `[(1, 2), (3, 5)] [(9), (16)]`
D. `[(– 5, 2), (3, – 1)] [(9), (16)]`
E. `[(– 5, 3), (2, – 1)] [(9), (16)]`
`P, Q, R` and `S` are matrices such that the matrix product `P = QRS` is defined.
Matrix `Q` and matrix `S` are square, non-zero matrices for which `Q + S` is not defined.
Which one of the following matrix expressions is defined?
A. `R - S`
B. `Q + R`
C. `P^2`
D. `R^-1`
E. `P × S`
If `A = [(1,3), (6,4), (0,0)]` and the matrix product `XA = [(4,1),(1,4), (3,5)],` then the order matrix `X` is
A. `(2 xx 2)`
B. `(2 xx 3)`
C. `(3 xx 1)`
D. `(3 xx 2)`
E. `(3 xx 3)`
Matrix `A` is a 3 x 3 matrix. Seven of the elements in matrix `A` are zero.
Matrix `B` contains six elements, none of which are zero.
Assuming the matrix product `AB` is defined, the minimum number of zero elements in the product matrix `AB` is
A. `0`
B. `1`
C. `2`
D. `4`
E. `6`
A new model for the number of students in the school after each assessment takes into account the number of students who are expected to leave the school after each assessment.
After each assessment, students are classified as beginner (`B`), intermediate (`I`), advanced (`A`) or left the school (`L`).
Let matrix `T_2` be the transition matrix for this new model.
Matrix `T_2`, shown below, contains the percentages of students who are expected to change their ability level or leave the school after each assessment.
`{:(qquadqquadqquadquadtext(before assessment)),(qquadqquadqquadquad{:(B,\ qquadI,qquadA,quadL):}),(T_2 = [(0.30,0,0,0),(0.40,0.70,0,0),(0.05,0.20,0.75,0),(0.25,0.10,0.25,1)]{:(B),(I),(A),(L):}qquadtext(after assessment)):}`
The number of students at each level, immediately before the first assessment of the year, is shown in matrix `R_0` below.
`R_0 = [(20),(60),(40),(0)]{:(B),(I),(A),(L):}`
Matrix `T_2`, repeated below, contains the percentages of students who are expected to change their ability level or leave the school after each assessment.
`{:(qquadqquadqquadquadtext(before assessment)),(qquadqquadqquadquad{:(B,\ qquadI,qquadA,quadL):}),(T_2 = [(0.30,0,0,0),(0.40,0.70,0,0),(0.05,0.20,0.75,0),(0.25,0.10,0.25,1)]{:(B),(I),(A),(L):}qquadtext(after assessment)):}`
Write your answer correct to the nearest whole number. (1 mark)
Another model for the number of students in the school after each assessment takes into account the number of students who are expected to join the school after each assessment.
Let `R_n` be the state matrix that contains the number of students in the school immediately after `n` assessments.
Let `V` be the matrix that contains the number of students who join the school after each assessment.
Matrix `V` is shown below.
`V = [(4),(2),(3),(0)]{:(B),(I),(A),(L):}`
The expected number of students in the school after `n` assessments can be determined using the matrix equation
`R_(n + 1) = T_2 xx R_n + V`
where
`R_0 = [(20),(60),(40),(0)]{:(B),(I),(A),(L):}`
How many are expected to become advanced-level students after the next assessment?
Write your answer correct to the nearest whole number. (2 marks)
| a. | |
| b. | `T_2 R_0` | `= [(0.3, 0, 0, 0), (0.4, 0.7, 0, 0), (0.05, 0.2, 0.75, 0), (0.25, 0.1, 0.25, 1)] [(20), (60), (40), (0)]` |
| `= [(6), (50), (43), (21)]` |
`:.\ text(Percentage that leave school)`
`= 21/120`
`= 17.5 text(%)`
c. `text(After 2 assessments,)`
| `(T_2)^2 R_0` | `= [(0.3, 0, 0, 0), (0.4, 0.7, 0, 0), (0.05, 0.2, 0.75, 0), (0.25, 0.1, 0.25, 1)]^2 [(20), (60), (40), (0)]` |
| `=[(1.8), (37.4), (42.55), (38.25)]` |
`:.\ text(43 advanced-level students expected to remain.)`
d. `text(After 4 assessments,)`
`(T_2)^4 R_0\ text(shows 54 students left)`
`text(After 5 assessments,)`
`(T_2)^5 R_0\ text(shows 43 students left.)`
`:.\ text(Numbers drop below 50 after 5 assessments.)`
| e. | `R_1` | `= T_2 R_0 + V` |
| `= [(6), (50), (43), (21)] + [(4), (2), (3), (0)] = [(10), (52), (46), (21)]` | ||
| `R_2` | `= [(0.30, 0, 0, 0), (0.40, 0.70, 0, 0), (0.05, 0.20, 0.75, 0), (0.25, 0.10, 0.25, 1)] [(10), (52), (46), (21)] + [(4), (2), (3), (0)] = [(7), (42.4), (48.4), (40.2)]` | |
| `R_3` | `= [(0.30, 0, 0, 0), (0.40, 0.70, 0, 0), (0.05, 0.20, 0.75, 0), (0.25, 0.10, 0.25, 1)] [(7), (42.4), (48.4), (40.2)] + [(4), (2), (3), (0)] = [(6.1), (34.48), (48.13), (58.29)]` |
`:.\ text(After 3 assessments, the number expected to)`
`text(move from Intermediate to Advanced)`
`= 20text(%) xx 34.48`
`= 6.896`
`= 7\ text{students (nearest whole)}`
When a new industrial site was established at the beginning of 2011, there were 350 staff at the site.
The staff comprised 100 apprentices (`A`), 200 operators (`O`) and 50 professionals (`P`).
At the beginning of each year, staff can choose to stay in the same job, move to a different job at the site or leave the site (`L`).
The number of staff in each category at the beginning of 2011 is given in the matrix
`S_2011 = [(100), (200), (50), (0)]{:(A), (O), (P), (L):}`
The transition diagram below shows the way in which staff are expected to change their jobs at the site each year.
The information in the transition diagram has been used to write the transition matrix `T`.
`{:(qquad qquad qquad qquad qquad qquad text(this year)),((qquad qquad qquad\ A, qquad O, qquad P, qquad L)),(T = [(0.70, 0, 0, 0),(0.10, 0.80, 0, 0), (0, 0.10, 0.90, 0), (0.20, 0.10, 0.10, 1.00)]):} {:(), (), (A), (O), (P), (L):} {:(), (), (qquad text(next year)):}`
If staff at the site continue to change their jobs in this way, the matrix `S_n` will contain the number of apprentices (`A`), operators (`O`), professionals (`P`) and staff who leave the site (`L`) at the beginning of the `n`th year.
Suppose the manager decides to bring 30 new apprentices, 20 new operators and 10 new professionals to the site at the beginning of each year.
The matrix `S_(n + 1)` will then be given by
`S_(n + 1) = T S_n + A` where `S_2011 = [(100), (200), (50), (0)] {:(A), (O), (P), (L):}` and `A = [(30), (20), (10), (0)] {:(A), (O), (P), (L):}`
A company repairs phones and laptops.
Let `x` be the number of phones repaired each day
`y` be the number of laptops repaired each day.
It takes 35 minutes to repair a phone and 50 minutes to repair a laptop.
The constraints on the company are as follows.
Constraint 1 `x ≥ 0`
Constraint 2 `y ≥ 0`
Constraint 3 `35x + 50y ≤ 1750`
Constraint 4 `y ≤ 4/5 x`
Use this constraint to complete the following sentence.
For every ten phones repaired, at most _______ laptops may be repaired. (1 mark)
The line `y = 4/5 x` is drawn on the graph below.
The profit from repairing one phone is $60 and the profit from repairing one laptop is $100.
a. `text(Constraint 3 means that the maximum time)`
`text(available to repair phones and laptops is 1750)`
`text(minutes on any given day.)`
b. `y <= 4/5x`
`text(When)\ x = 10,`
| `y` | `<= 4/5 xx 10` |
| `<= 8` |
`:.\ text(At most, 8 laptops may be repaired)`
| c. |  |
d. `text(From the graph, the highest number)`
`text(of laptops below the intersection)`
`text{(in the feasible region) is 18.}`
e. `text(When)\ y = 9,\ text(constraint 3 requires)`
| `35x + (50 xx 9)` | `<= 1750` |
| `35x` | `<= 1300` |
| `:. x` | `<= 37.14…` |
`:.\ text(The maximum number of phones = 37.)`
f.i. `text(Profit)\ = 60x + 100y`
`text(In the feasible region, maximum)`
`text(profits occur at the intersection.)`
`:. 18\ text(laptops)`
`text(When)\ y = 18,\ text(constraint 3 requires)`
| `35x + (50 xx 18)` | `<= 1750` |
| `35x` | `<= 850` |
| `x` | `<= 24.28…` |
`:. text(Maximum profit occurs at)\ (24,18)`
f.ii. `text(Maximum daily profit)`
`= 60 xx 24 + 100 xx 18`
`= $3240`
Hugo took out a reducing balance loan of $25 000 to compete in road races overseas.
Interest was charged at a rate of 12% per annum compounding quarterly.
His loan is to be repaid fully in four years with equal quarterly payments.
After two years, how much of the $25 000 will Hugo have repaid?
Write your answer, correct to the nearest dollar. (2 marks)
Hugo paid $7500 for a second bike under a hire-purchase agreement.
A flat interest rate of 8% per annum was charged.
He will fully repay the principal and the interest in 24 equal monthly instalments.
Write your answer in dollars, correct to the nearest cent. (2 marks)
Write your answer as a percentage, correct to two decimal places. (1 mark)
One year after it was purchased, this bike was valued at $6375.
Determine the value of the bike five years after it was purchased.
Write your answer, correct to the nearest dollar. (2 marks)
The school group may hire two types of camp sites: powered sites and unpowered sites.
Let `x` be the number of powered camp sites hired
`y` be the number of unpowered camp sites hired.
Inequality 1 and inequality 2 give some restrictions on `x` and `y`.
inequality 1 `x ≤ 5`
inequality 2 `y ≤ 10`
There are 48 students to accommodate in total.
A powered camp site can accommodate up to six students and an unpowered camp site can accommodate up to four students.
Inequality 3 gives the restrictions on `x` and `y` based on the maximum number of students who can be accommodated at each type of camp site.
inequality 3 `ax + by ≥ 48`
School groups must hire at least two unpowered camp sites for every powered camp site they hire.
The graph below shows the three lines that represent the boundaries of inequalities 1, 3 and 4.
School regulations require boys and girls to be accommodated separately.
The girls must all use one type of camp site and the boys must all use the other type of camp site.
a. `a = 6`
`b = 4`
b. `text(Inequality 4)`
`y ≥ 2x`
| c. | |
d. `text(From the graph, minimum number)`
`text(of camps is 11, which occurs at)\ (2,9)`
`text(and)\ (3,8).`
e.i. `text(Test)\ (2,9)\ text(and)\ (3,8)\ text(for minimum cost.)`
`text(At)\ (2,9),`
`text(C)text(ost) = 60 xx 2 + 30 xx 9 = $390`
`text(At)\ (3,8),`
`text(C)text(ost) = 60 xx 3 + 30 xx 8 = $420`
`:. text(Minimum cost per day) = $390.`
e.ii. `text(If equal boys and girls at each site.)`
`text(Powered sites) = 24/6 = 4`
`text(Unpowered sites) = 24/4 = 6`
`:. x >= 4\ text(and)\ y = 8`
| `:.\ text(Minimum cost)` | `= 60 xx 4 + 30 xx 8` |
| `= $480` |
Daniel threw a javelin a distance of 68.32 m on a bearing of 057° on his first throw.
On his second throw from the same point, he threw the javelin a distance of 72.51 m.
The second throw landed at a point on a bearing of 125°, measured from the point where the first throw landed.
Determine the distance between the point where Daniel’s first throw landed and the point where his second throw landed.
Write your answer in metres, correct to one decimal place. (2 marks)
`text(Using the sine rule,)`
| `(sin∠ABT)/68.32` | `= (sin∠TAB)/72.51` |
| `sin∠ABT` | `= (68.32 xx sin112^@)/72.51` |
| `= 0.873…` | |
| `:. ∠ABT` | `= 60.88…^@` |
| `:. ∠ATB` | `= 180 – (112 + 60.88…)` |
| `= 7.11…^@` |
`text(Using the cosine rule,)`
| `AB^2` | `= 68.32^2 + 72.51^2 – 2 xx 68.32 xx 72.51 xx cos7.11…^@` |
| `= 93.74…` | |
| `:. AB` | `= 9.68…` |
| `= 9.7\ text{m (1 d.p.)}` |
A chicken escapes into a neighbouring field through an open gate.
The chicken’s owner is 50 m due north of the gate, searching for the chicken.
The chicken is 40 m from the gate on a bearing of 295°.
What is the bearing of the chicken from its owner?
Write your answer, correct to the nearest degree. (2 marks)
`text(Using the cosine rule,)`
| `OC` | `= sqrt(40^2 + 50^2 − 2 xx 40 xx 50 xx cos65^@)` |
| `=49.08…\ text(m)` |
`text(Using the sine rule,)`
| `(sin theta)/40` | `= (sin65^@)/49` |
| `sin theta` | `=(40 xx sin65^@)/49` |
|
`=0.739…` |
|
| `:.theta` | `=47.7…^@` |
| `=48^@\ \ text{(nearest degree)` |
`:.\ text(The bearing of)\ C\ text(from)\ O`
`=180^@ + 48^@`
`= 228^@`
Competitors in the intermediate division of the discus use a smaller discus than the one used in the senior division, but of a similar shape. The total surface area of each discus is given below.
By what value can the volume of the intermediate discus be multiplied to give the volume of the senior discus? (2 marks)
The cricket club borrowed $400 000 to build a clubhouse.
Interest is calculated at the rate of 4.5% per annum, compounding monthly.
The cricket club will make monthly repayments of $2500.
After a number of monthly repayments, the balance of the loan will be reduced to $143 585.33.
What percentage of the next monthly repayment will reduce the balance of the loan?
Write your answer, correct to the nearest percentage. (2 marks)
A shop owner bought 100 kg of Arthur’s tomatoes to sell in her shop.
She bought the tomatoes for $3.50 per kilogram.
The shop owner will offer a discount to her customers based on the number of kilograms of tomatoes they buy in one bag.
The revenue, in dollars, that the shop owner receives from selling the tomatoes is given by
`text(revenue)\ {{:(qquadqquad5.4n),(10.8 + 4(n - 2)),(a + 2(n - 10)):}qquadqquad{:(0 < n <= 2),(2 < n <= 10),(10 < n < 100):}`
where `n` is the number of kilograms of tomatoes that a customer buys in one bag.
Jane and Michael borrow $50 000 to expand their business.
Interest on the unpaid balance is charged to the loan account monthly.
The $50 000 is to be fully repaid in equal monthly repayments of $485.60 for 12 years.
Write your answer correct to two decimal places. (1 mark)
Write your answer correct to the nearest dollar. (1 mark)
Assume no other changes are made to their loan conditions.
Determine how much time Jane and Michael will save in repaying their loan.
Give your answer correct to the nearest number of months. (2 marks)
Let `f: R -> R, \ \ f (x) = (x - 3)(x - 1)(x^2 + 3) and g: R-> R, \ \ g (x) = x^4 − 8x.`
a. `text(Solution 1)`
`g(x) = x^4 – 8x`
`= x(x – 2)(x^2 + 2x + 4)\ \ \ text([by CAS])`
`= x(x- 2)((x + 1)^2 + 3)`
`text(Solution 2)`
| `x^4 – 8x` | `=x(x^3-2^3)` |
| `=x(x-2)(x^2 +2x+4)` | |
| `=x(x-2)(x^2 +2x+1+3)` | |
| `=x(x-2)((x+1)^2+3)` |
| b. | `f(x + 1)` | `= ((x + 1) – 3)((x + 1) – 1)((x + 1)^2 + 3)` |
| `= x(x – 2)((x + 1)^2 + 3)` | ||
| `= g(x)` |
`:.\ text(Horizontal translation of 1 unit to the left.)`
c.i. `text(Consider part of the)\ \ f(x)\ text(graph below:)`
`text(For one positive)\ xtext(-axis intercept, translate at least)`
`text(1 unit left, but not more than 3 units left.)`
`:. d ∈ [1,3)`
c.ii. `text(Translate less than 1 unit left, or translate)`
`text(right.)`
`:. d < 1`
d. `text(If)\ \ g(x)=n\ \ text(has one solution, then it)`
`text(will occur when)\ \ g′(x)=0 and x>0.`
| `g′(x)` | `=4x^3-8` |
| `4x^3` | `=8` |
| `x` | `=2^(1/3)` |
| `n` | `=g(2^(1/3))` |
| `=2^(4/3)-8xx2^(1/3)` | |
| `=-6 xx 2^(1/3)` |
e.i. `gprime(u) = mqquadgprime(v) = −m`
| `gprime(u)` | `= −gprime(v)` |
| `4u^3 – 8` | `= −(4v^3 – 8)` |
| `4u^3 + 4v^3` | `= 16` |
| `:. u^3 + v^3` | `= 4` |
e.ii. `u^3 + v^3 = 4\ …\ (1)`
`u + v = 1\ …\ (2)`
`text(Solve simultaneous equation for)\ \ u > 0:`
`:. u = (sqrt5 + 1)/2,quad v = (−sqrt5 + 1)/2`
f.i. `text(Solution 1)`
`text(Using the point-gradient formula,)`
| `y-g(p)` | `=g′(p)(x-p)` |
| `y-(p^4-8p)` | `=(4p^3-8)(x-p)` |
| `y` | `=(4p^3-8)x -3p^4` |
`text(Solution 2)`
`y = 4(p^3 – 2)x – 3p^4`
`text([CAS: tangentLine)\ (g(u),x,p)]`
f.ii. `text(Sub)\ (3/2,−12)\ text(into tangent equation,)`
`text(Solve:)\ \ −12 = 4(p^3 – 2)(3/2) – 3p^4\ \ text(for)\ p,`
`p = 0\ \ text(or)\ \ p = 2`
| `text(When)\ \ p = 0:` | `y` | `= 4(−2)x` |
| `= −8x` |
| `text(When)\ \ p = 2:` | `y` | `= 4(2^3 – 2)x – 3(2)^4` |
| `= 24x – 48` |
`:. text(Equations are:)\ \ y = −8x, \ y = 24x – 48`
Let `V = [0, 5] -> R,\ \ \ V(t) = de^(t/3) + (10 - d)e^(-(2t)/3)`, where `d` is a real number and `d` in `(0, 10)`.
| a.i. | `S(0)` | `=2e^0+8e^0` |
| `=10` | ||
| `S(5)` | `=2e^(5/3)+8e^(-10/3)` |
a.ii. `text(When)\ \ Sprime(t)=0,`
| `t` | `= 3ln(2)` |
| `= log_e(2^3)` |
`:. c = 8`
`:. S_text(min) = S(log_e(8)) = 6`
| a.iii. | |
| a.iv. | `text(Average ROC)` | `= (S(log_e(8)) – S(0))/(log_e(8) – 0)` |
| `=(6-10)/(log_(8))` | ||
| `= (−4)/(log_e(8))` |
b. `V(t) = de^(t/3) + (10 – d)e^(-(2t)/3)`
`text(Solve:)\ \ Vprime(log_e(9)) = 0\ text(for)\ d,`
`:. d = 20/11`
c.i. `text(Solve:)\ \ Vprime(0) = 0\ \ text(for)\ \ d,`
`d=20/3,`
`:. 20/3 <= d <10`
c.ii. `text(Solve:)\ \ Vprime(5) = 0\ \ text(for)\ \ d,`
`d=20/(2+e^5),`
`:. 0< d <=20/(2 + e^5)`
| d. | `text(Local min when)\ Vprime(a)` | `= 0` |
`text(Solve)\ \ Vprime(a)=0\ \ text(for)\ a`
`a= log_e(20/d -2)`
`text(Solve)\ \ V(log_e(20/d-2))=k/2 d^(3/2)(10-d)^(1/3)\ \ text(for)\ k,`
`k= 3 xx 2^(1/3)`
An electronics company is designing a new logo, based initially on the graphs of the functions
`f(x) = 2 sin (x) and g(x) = 1/2 sin (2x),\ text(for)\ 0 <= x <= 2 pi`
These graphs are shown in the diagram below, in which the measurements in the `x` and `y` directions are in metres.
The logo is to be painted onto a large sign, with the area enclosed by the graphs of the two functions (shaded in the diagram) to be painted red.
What is the value of `a`? (1 mark)
The electronics company considers changing the circular functions used in the design of the logo.
Its next attempt uses the graphs of the functions `f(x) = 2 sin(x) and h(x) = 1/3 sin (3x),\ text(for)\ \ 0 <= x <= 2 pi`.
On the same axes, draw the graph of `y = h(x)`. (2 marks)
The electronics company now considers using the graphs of the functions `k(x) = m sin(x) and q (x) = 1/n sin (nx)`, where `m` and `n` are positive integers with `m >= 2 and 0<= x <= 2pi`.
| a. | `text(Area)` | `= 2 xx int_0^pi (2 sin (x))\ dx` |
| `= 4 xx int_0^pi sin (x)\ dx` |
`:. a = 4`
| b. | |
c. `text(Find sequence that takes)\ f(x) -> h(x)`
`text(A dilation by factor of)\ 1/6\ text(from the)\ xtext(-axis.)`
`text(A dilation by factor of)\ 1/3\ text(from the)\ ytext(-axis.)`
d.i. `text(Area when)\ n\ text(is even)`
| `= 2 int_0^pi (msin(x) – 1/n sin(nx))\ dx` |
| `=2 [-mcos x + 1/n^2 cos (nx)]_0^pi` |
| `=2[m+1/n^2 cos (npi) – (-m+1/n^2)]` |
| `= 2(2m+(cos(npi)-1)/n^2)` |
| `= 4m + 0/(n^2)` |
d.ii. `text(Area when)\ n\ text(is even)`
| `= 2 int_0^pi (msin(x) – 1/n sin(nx))\ dx` |
| `=2 [-mcos x + 1/n^2 cos (nx)]_0^pi` |
| `=2[m+1/n^2 cos (npi) – (-m+1/n^2)]` |
| `= 2(2m+(cos(npi)-1)/n^2)` |
| `= 4m + (-4)/(n^2)` |
Harry offers dog washing and dog clipping services.
Let `x` be the number of dogs washed in one day
`y` be the number of dogs clipped in one day.
It takes 20 minutes to wash a dog and 25 minutes to clip a dog.
There are 200 minutes available each day to wash and clip dogs.
This information can be written as Inequalities 1 to 3.
Inequality 1: `x ≥ 0`
Inequality 2: `y ≥ 0`
Inequality 3: `20x + 25y ≤ 200`
In any one day the number of dogs clipped is at least twice the number of dogs washed.
The profit from washing one dog is $40 and the profit from clipping one dog is $30.
Let `P` be the total profit obtained in one day from washing and clipping dogs.
| a. | |
b. `text(Inequality 4:)\ \ y >= 2x`
| c.i. | |
c.ii. `text(When)\ y =5,\ text(the maximum value of)\ x`
`text{(whole number) in the feasible region is 2.)`
`:. 2\ text(dogs can be washed.)`
d. `P = 40x + 30y`
e.i. `text(Test points for maximum profit)`
`text(At)\ (0, 8),`
`text(Profit) = 40 xx 0 + 30 xx 8 = $240`
`text(At)\ (1, 7),`
`text(Profit) = 40 xx 1 + 30 xx 7 = $250`
`text(At)\ (2, 6),`
`text(Profit) = 40 xx 2 + 30 xx 6 = $260`
`:.\ text(2 washes and 6 clips produce maximum profit.)`
e.ii. `text(Maximum profit)`
`= 40 xx 2 + 30 xx 6`
`= $260`
A farmer owns a flat allotment of land in the shape of triangle `ABC` shown below.
Boundary `AB` is 251 metres.
Boundary `AC` is 142 metres.
Angle `BAC` is 45°.
A straight track, `XY`, runs perpendicular to the boundary `AC`.
Point `Y` is 55 m from `A` along the boundary `AC`.
Write your answer, in metres, correct to one decimal place. (1 mark)
Determine the bearing of `B` from `A`. (1 mark)
Write your answer, in metres, correct to one decimal place. (2 marks)
The length of the boundary `BC` is 181 metres (correct to the nearest metre).
A farmer plans to build a fence, `MN`, perpendicular to the boundary `AC`.
The land enclosed by triangle `AMN` will have an area of 3200 m².
| a. | `/_ AXY` | `= 180 – (90 + 45)` |
| `= 45^@` |
b. `text(In)\ Delta AXY,`
| `cos 45^@` | `= 55/(AX)` |
| `:. AX` | `= 55/(cos 45^@` |
| `= 77.78…` | |
| `= 77.8\ text{m (1 d.p.)}` |
| c. |  |
`text(Bearing of)\ B\ text(from)\ A`
`= 78 – 45`
`= 033^@`
| d. |  |
`text(Using the cosine rule,)`
| `XC^2` | `= 77.8^2 + 142^2 – 2 xx 77.8 xx 142 xx cos 45^@` |
| `= 10\ 593.17…` | |
| `:. XC` | `=sqrt (10\ 593.17…)` |
| `= 102.92…` | |
| `= 102.9\ text{m (1 d.p.)}` |
e. `text(Using sine rule,)`
| `text(Area)\ Delta ABC` | `= 1/2 bc sin A` |
| `= 1/2 xx 142 xx 251 xx sin 45^@` | |
| `= 12601.34…` | |
| `= 12\ 601\ text{m² (nearest m²)}` |
f.i. `text(Using the cosine rule,)`
| `BC^2` | `= 142^2 + 251^2 – 2 xx 142 xx 251 xx cos 45^@` |
| `= 32\ 759.60…` | |
| `:. BC` | `= 180.99…` |
| `= 181\ text{m (nearest m) … as required}` |
f.ii. `text(Using the cosine rule,)`
| `cos /_ ABC` | `= (251^2 + 181^2 – 142^2)/(2 xx 251 xx 181)` |
| `= 0.832…` | |
| `:. /_ ABC` | `= 33.69…` |
| `= 33.7^@\ text{(1 d.p.)}` |
g. `/_ AMN = 180 – (90 + 45) = 45^@`
`:. Delta AMN\ text(is isosceles.)`
| `text(Area)\ Delta AMN` | `= 1/2 xx AN xx MN` |
| `3200` | `= 1/2 xx MN xx MN\ (AN = MN)` |
| `MN^2` | `= 6400` |
| `:. MN` | `= 80\ text(m)` |
Tessa has a task that involves removing the top 24 cm of the height of her right regular pyramid below.
The shape remaining is shown in Figure 5 below. The top surface, `JKLM`, is parallel to the base, `ABCD`.
By the end of each academic year, students at the university will have either passed, failed or deferred the year. Experience has shown that
Twelve hundred and thirty students began a business degree in 2007.
By the end of the 2007 academic year, 880 students had passed, 230 had failed, while 120 had deferred the year.
No students have dropped out of the business degree permanently.
Use this information to predict the number of business students who will defer the 2009 academic year. (2 marks)
A tree, 12 m tall, is growing at point `T` near a shed.
The distance, `CT`, from corner `C` of the shed to the centre base of the tree is 13 m.
Write your answer, in degrees, correct to one decimal place. (1 mark)
`N` and `C` are two corners at the base of the shed. `N` is due north of `C`.
The angle, `TCN`, is 65°.
Explain your answer showing appropriate calculations. (2 marks)
| a. |  |
| `text(Let)\ \ theta` | `=\ text(angle of elevation)` |
| `= 12/13` | |
| `:. theta` | `= 42.709…` |
| `= 42.7^@\ text{(1 d.p.)}` |
| b. |  |
`text(Using the cosine rule:)`
| `NT^2` | `= 10^2 + 13^2 – 2 xx 10 xx 13 xx cos65^@` |
| `= 159.11…` | |
| `:. NT` | `= sqrt(159.11…)` |
| `= 12.61…` | |
| `= 12.6\ text{m (1 d.p.) … as required}` |
c. `text(Using the sine rule:)`
| `(sin /_ CNT)/13` | `= (sin 65^@)/12.6` |
| `sin /_ CNT` | `= (13 xx sin 65^@)/12.6` |
| `= 0.935…` | |
| `:. /_ CNT` | `= 69.24…` |
| `= 69^@\ text{(nearest degree)}` |
| d. |  |
`text(Bearing of)\ T\ text(from)\ N`
`= 180 – 69`
`= 111^@`
| e. |  |
`ST\ text(is the shortest distance between the)`
`text(tree and the shed.)`
`text(In)\ \ Delta NST,`
| `sin 69^@` | `= (ST)/12.6` |
| `:. ST` | `= 12.6 xx sin 69^@` |
| `= 11.76…\ text(m)` |
`:.\ text(S) text(ince the tree is 12 m, and 12 m) > 11.76\ text(m),`
`text(it could hit the shed if it falls.)`
The ferry has two fuel filters, `A` and `B`.
Filter `A` has a hemispherical base with radius 12 cm.
A cylinder of height 30 cm sits on top of this base.
Filter `B` is a right cone with height 50 cm.
If the height of the oil in the cone is now 20 cm, what percentage of the original volume of oil was removed? (2 marks)
Part of the graph of a function `g: R -> R, \ g (x) = (16 - x^2)/4` is shown below.
The tangent to the graph of `g` at a point `P` has a negative gradient and intersects the `y`-axis at point `D(0, k)`, where `5 <= k <= 8.`
a.i. `B(4,0), C(0,4), A(a,(16 – a^2)/4)`
`m_(BC) = m_text(tan) = (4 – 0)/(0 – 4) = −1`
| `g(x)` | `= (16 – x^2)/4` |
| `g′(x)` | `=- x/2` |
`text(S)text(ince)\ \ g′(a) = −1,`
`=>a = 2`
`text(T)text(angent passes through)\ \ (2,3),`
| `y-3` | `=-(x-2)` |
| `y` | `=-x + 5` |
`text(Alternatively, using technology:)`
`[text(CAS: tangent Line)]\ (g(x),x,2)`
a.ii. `text(Solution 1)`
`D(5,0), E(0,5)`
| `text(Area)` | `= DeltaEOD – int_0^4 g(x)\ dx` |
| `= 1/2 xx 5 xx 5 – 32/3` | |
| `= 11/6\ text(u²)` |
`text(Solution 2)`
| `text(Area)` | `= int_0^5 (-x+5)\ dx – int_0^4 g(x)\ dx` |
| `= 11/6\ text(u²)` |
b. `Q(x, (16 – x^2)/4), O(0,0)`
| `z` | `= OQ` |
| `= sqrt(x^2 + ((16 – x^2)/4)^2), x > 0` |
`text(Max or min when)\ \ (dz)/(dx)=0,`
`text(Solve:)\ (dz)/dx = 0quadtext(for)quadx > 0`
`=> x = 2sqrt2`
`=>z(2sqrt2) = 2sqrt3`
`:. text(Min distance of)\ \ 2sqrt3\ \ text(units when)\ \ x = 2sqrt2`
c. `text(Let)\ \ P(p, (16-p^2)/4)`
`m_text(tan)\ text(at)\ P = – p/2`
`m_(PD) = ((16-p^2)/4 – k)/(p-0)`
`text(Equating gradients,)`
`text(Solve:)\ \ ((16-p^2)/4 – k)/p=-p/2\ \ text(for)\ p,`
`=> p=2sqrt(k – 4)`
`:.\ text(Gradient of tangent:)`
`gprime(2sqrt(k – 4)) = −sqrt(k – 4)`
d.i. `text(Equation of tangent:)`
`y = −sqrt(k – 4)x + k`
`text(CAS: tangent Line)\ (g(x),x,2sqrt(k – 4))`
`xtext(-int of tangent:)\ x = k/(sqrt(k – 4))`
| `:. A(k)` | `= 1/2 xx (k/(sqrt(k – 4))) xx k – int_0^4 g(x)\ dx` |
| `= (k^2)/(2sqrt(k – 4)) – 32/3, \ \ k ∈ [5,8]` |
d.ii. `text(Solve:)\ \ A(k)=0quadtext(for)quadk ∈ [5,8]`
`=> k=16/3`
`text(Sketch the graph of)\ \ A(k)\ \ text(for)\ \ 5<=k<=8`
`:. A_text(max) = 16/3quadtext(when)quadk = 8`
d.iii. `A_text(min)\ text(occurs at the turning point)\ (k=16/3).`
| `A_text(min)` | `=A(16/3)` |
| `=(64sqrt3)/9 – 32/3` |
Tasmania Jones is in Switzerland. He is working as a construction engineer and he is developing a thrilling train ride in the mountains. He chooses a region of a mountain landscape, the cross-section of which is shown in the diagram below.
The cross-section of the mountain and the valley shown in the diagram (including a lake bed) is modelled by the function with rule
`f(x) = (3x^3)/64 - (7x^2)/32 + 1/2.`
Tasmania knows that `A (0, 1/2)` is the highest point on the mountain and that `C(2, 0)` and `B(4, 0)` are the points at the edge of the lake, situated in the valley. All distances are measured in kilometres.
Tasmania’s train ride is made by constructing a straight railway line `AB` from the top of the mountain, `A`, to the edge of the lake, `B`. The section of the railway line from `A` to `D` passes through a tunnel in the mountain.
In order to ensure that the section of the railway line from `D` to `B` remains stable, Tasmania constructs vertical columns from the lake bed to the railway line. The column `EF` is the longest of all possible columns. (Refer to the diagram above.)
Tasmania’s train travels down the railway line from `A` to `B`. The speed, in km/h, of the train as it moves down the railway line is described by the function.
`V: [0, 4] -> R, \ V(x) = k sqrt x - mx^2,`
where `x` is the `x`-coordinate of a point on the front of the train as it moves down the railway line, and `k` and `m` are positive real constants.
The train begins its journey at `A (0, 1/2)`. It increases its speed as it travels down the railway line.
The train then slows to a stop at `B(4, 0)`, that is `V(4) = 0.`
Tasmania is able to change the value of `m` on any particular day. As `m` changes, the relationship between `k` and `m` remains the same.
Trigg the gardener is working in a temperature-controlled greenhouse. During a particular 24-hour time interval, the temperature `(Ttext{°C})` is given by `T(t) = 25 + 2 cos ((pi t)/8), \ 0 <= t <= 24`, where `t` is the time in hours from the beginning of the 24-hour time interval.
Trigg is designing a garden that is to be built on flat ground. In his initial plans, he draws the graph of `y = sin(x)` for `0 <= x <= 2 pi` and decides that the garden beds will have the shape of the shaded regions shown in the diagram below. He includes a garden path, which is shown as line segment `PC.`
The line through points `P((2 pi)/3, sqrt 3/2)` and `C (c, 0)` is a tangent to the graph of `y = sin (x)` at point `P.`
In further planning for the garden, Trigg uses a transformation of the plane defined as a dilation of factor `k` from the `x`-axis and a dilation of factor `m` from the `y`-axis, where `k` and `m` are positive real numbers.
a. `T_text(max)\ text(occurs when)\ \ cos((pit)/8) = 1,`
`T_text(max)= 25 + 2 = 27^@C`
`text(Max occurs when)\ \ t = 0, or 16\ text(h)`
| b. | `text(Period)` | `= (2pi)/(pi/8)` |
| `= 16\ text(hours)` |
c. `text(Solve:)\ \ 25 + 2 cos ((pi t)/8)=26\ \ text(for)\ t,`
| `t` | `= 8/3,40/3,56/3\ \ text(for)\ t ∈ [0,24]` |
| `t_text(min)` | `= 8/3` |
d. `text(Consider the graph:)`
| `text(Time above)\ 26 text(°C)` | `= 8/3 + (56/3 – 40/3)` |
| `= 8\ text(hours)` |
e.i. `(dy)/(dx) = cos(x)`
`text(At)\ x = (2pi)/3,`
| `(dy)/(dx)` | `= cos((2pi)/3)= −1/2` |
e.ii. `text(Solution 1)`
`text(Equation of)\ \ PC,`
| `y-sqrt3/2` | `=-1/2(x-(2pi)/3)` |
| `y` | `=-1/2 x +pi/3 +sqrt3/2` |
`PC\ \ text(passes through)\ \ (c,0),`
| `0` | `=-1/2 c +pi/3 + sqrt3/2` |
| `c` | `=sqrt3 + (2 pi)/3\ …\ text(as required)` |
`text(Solution 2)`
`text(Equating gradients:)`
| `- 1/2` | `= (sqrt3/2 – 0)/((2pi)/3 – c)` |
| `-1` | `= sqrt3/((2pi – 3c)/3)` |
| `3c – 2pi` | `= 3sqrt3` |
| `3c` | `= 3 sqrt3 + 2pi` |
| `:. c` | `= sqrt3 + (2pi)/3\ …\ text(as required)` |
f.i. `Xprime ((2pi)/3 m,0)qquadPprime((2pi)/3 m, sqrt3/2 k)qquadCprime ((sqrt3 + (2pi)/3)m, 0)`
| `XprimePprime` | `= 10` |
| `sqrt3/2 k` | `= 10` |
| `:. k` | `= 20/sqrt3` |
| `=(20sqrt3)/3` |
`XprimeCprime=30`
| `((sqrt3 + (2pi)/3)m) – (2pi)/3 m` | `= 30` |
| `:. m` | `= 30/sqrt3` |
| `=10sqrt3` |
| f.ii. | `Pprime((2pi)/3 m, sqrt3/2 k)` | `= Pprime((2pi)/3 xx 10sqrt3, sqrt3/2 xx 20/sqrt3)` |
| `= Pprime((20pisqrt3)/3,10)` |
The shaded region in the diagram below is the plan of a mine site for the Black Possum mining company.
All distances are in kilometres.
Two of the boundaries of the mine site are in the shape of the graphs of the functions
`f: R -> R,\ f(x) = e^x and g: R^+ -> R,\ g(x) = log_e (x).`
Victoria decides that the graph of `k` will be such that the `x`-coordinate of the midpoint of `AB` is `sqrt 2`.
Find the value of `a` in this case. (2 marks)
Tasmania Jones is in the jungle, searching for the Quetzalotl tribe’s valuable emerald that has been stolen and hidden by a neighbouring tribe. Tasmania has heard that the emerald has been hidden in a tank shaped like an inverted cone, with a height of 10 metres and a diameter of 4 metres (as shown below).
The emerald is on a shelf. The tank has a poisonous liquid in it.
The tank has a tap at its base that allows the liquid to run out of it. The tank is initially full. When the tap is turned on, the liquid flows out of the tank at such a rate that the depth, `h` metres, of the liquid in the tank is given by
`h = 10 + 1/1600 (t^3 - 1200t)`,
where `t` minutes is the length of time after the tap is turned on until the tank is empty.
From the moment the liquid starts to flow from the tank, find how long, in minutes, it takes until `h = 2`.
(Give your answer correct to one decimal place.) (2 marks)
How long, in minutes, after the tank is first empty will the liquid once again reach a depth of 2 metres? (2 marks)
Find the length of time, in minutes, correct to one decimal place, that Tasmania Jones has from the time he enters the tank to the time he leaves the tank. (1 mark)
Steve, Katerina and Jess are three students who have agreed to take part in a psychology experiment. Each student is to answer several sets of multiple-choice questions. Each set has the same number of questions, `n`, where `n` is a number greater than 20. For each question there are four possible options A, B, C or D, of which only one is correct.
Part (b) is no longer in the syllabus and is removed.
If `text(Pr) (Y > 23) = 6 text(Pr) (Y = 25)`, show that the value of `p` is `5/6`. (2 marks)
Calculate the values of `a` and `b`, correct to three decimal places. (4 marks)
Let `f: R text(\{2}) -> R,\ f(x) = 1/(2x-4) + 3.`
a. `text(Asymptotes:)`
`x = 2`
`y = 3`
b.i. `f^{′}(x) = (−2)/((2x-4)^2)`
b.ii. `text(Range) = (−∞,0), or R^-`
| b.iii. | `text(As)\ \ ` | `f^{′}(x) < 0quadtext(for)quadx ∈ R text(\{2})` |
| `f^{′}(x) != 0` |
`:. f\ text(has no stationary points.)`
c. `text(Point of tangency) = P(p,1/(2p-4) + 3)`
| `m_text(tang)` | `= f^{′}(p)` |
| `= (−2)/((2p-4)^2)` |
`text(Equation of tangent using:)`
| `y-y_1` | `= m(x-x_1)` |
| `y-(1/(2p-4) + 3)` | `= (−2)/((2p-4)^2)(x-p)` |
| `y-3` | `= (−2(x-p))/((2p-4)^2) + (2p-4)/((2p-4)^2)` |
| `(2p-4)^2(y-3)` | `= −2x + 2p + 2p-4` |
| `:. (2p-4)^2(y-3)` | `= −2x + 4p-4\ \ text(… as required)` |
d. `text(Substitute)\ \ (−1,7/2)\ text{into tangent (part c),}`
`text(Solve)\ \ (2p-4)^2(7/2-3) = −2(−1) + 4p-4\ \ text(for)\ p:`
`:. p = 1,\ text(or)\ 5`
`text(Substitute)\ \ p = 1\ text(and)\ p = 5\ text(into)\ \ P(p,1/(2p-4) + 3)`
`:. text(Coordinates:)\ (1,5/2)\ text(or)\ (5,19/6)`
e. `text(Determine transformations that that take)\ f -> g:`
`text(Dilate the graph of)\ \ f(x) = 1/(2x-4) + 3\ \ text(by a)`
`text(factor of 2 from the)\ \ ytext(-axis).`
`y = 1/(2(x/2)-4) + 3= 1/(x-4) + 3`
`text(Translate the graph 4 units to the left and 3)`
`text(units down to obtain)\ \ g(x).`
`text(Using the transformation matrix,)`
| `x^{′}` | `=ax+c` |
| `y^{′}` | `=y+d` |
`f -> g:\ \ 1/(2x-4) -> 1/(x^{′})`
`x^{′}=2x-4`
`=> a=2,\ \ c=-4`
`f -> g:\ \ y -> y^{′} + 3`
`y^{′}=y -3`
`=>\ \ d=-3`
Michael began his hike at the national park office and followed a track towards a camping ground, 16 km away.
The distance-time graph below shows Michael's distance from the national park office, `d` kilometres, after `t` hours of hiking.
It took Michael seven hours to complete this hike.
Write your answer correct to one decimal place. (1 mark)
The equation of Michael's distance-time graph from `t = 3` to `t = 7` is
`d = at + b`
Katie hiked along the same track as Michael, but hiked in the opposite direction.
She begun at the camping ground and hiked towards the national park office.
Katie's distance from the national park office, `d` kilometres, after `t` hours of hiking, can be determined from the equation
`d = -3t + 16`
Katie and Michael both starter hiking at the same time.
Katie and Michael both carry radio transmitters that allow them to talk to each other while hiking.
The transmitters will not work if Katie and Michael are more than three kilometres apart.
Write your answer in hours correct to two decimal places. (2 marks)
| a. | `text(Average speed)` | `= text(total kms)/text(total time)` |
| `= 16/7` | ||
| `= 2.285…` | ||
| `= 2.3\ text{km/hr (1 d.p.)}` |
b. `text(When)\ \ t = 3\ \ text{(from graph)},`
`10 = 3a + b\ \ text{… (1)}`
`text(When)\ \ t = 7`
`16 = 7a + b\ \ text{… (2)}`
`text(Substitute)\ \ b = 10 – 3a\ \ text{from (1) into (2)}`
| `16` | `= 7a + 10 – 3a` |
| `6` | `= 4a` |
| `:. a` | `= 1.5` |
`text(Substitute)\ a = 1.5\ \ text{into (1)}`
| `10` | `= 3 xx 1.5 + b` |
| `:. b` | `= 5.5` |
| c. |  |
`text(From the graph, Katie’s distance from)`
`text(the office is the same as Michael’s at)`
`text(the intersection of)`
| `d` | `= -3t + 16\ \ text{… (1)}` |
| `d` | `= 5t\ \ text{… (2)}` |
`text(Intersection when)`
| `5t` | `= -3t + 16` |
| `8t` | `= 16` |
| `t` | `= 2` |
`:.\ text(Katie passes Michael after 2 hours.)`
d. `text(Find the times when they are 3km apart.)`
| `-3t + 16 – 5t` | `= 3` |
| `8t` | `= 13` |
| `t` | `= 1.625\ text(hours)` |
`text(After passing each other, the graph)`
`text(shows the next time they are 3km)`
`text(apart is when)\ \ t = 3\ text(hours.)`
`:.\ text(Time the radios will work)`
`= 3 – 1.625`
`= 1.375`
`= 1.38\ text{hours (2 d.p.)}`
