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Networks, STD2 N2 2011 FUR2 1

Aden, Bredon, Carrie, Dunlop, Enwin and Farnham are six towns.

The network shows the road connections and distances between these towns in kilometres.

 

  1. In kilometers, what is the shortest distance between Farnham and Carrie?  (1 mark)

  2. How many different ways are there to travel from Farnham to Carrie without passing through any town more than once?  (1 mark)

Show Answers Only
  1. `200\ text(km)`
  2. `6`
Show Worked Solution

a.   `text{Farnham to Carrie (shortest)}`

`= 60 + 140`

`= 200\ text(km)`

 

b.   `text(Different paths are)`

`FDC, FEDC, FEBC,`

`FEABC, FDEBC,`

`FDEABC`

 
`:. 6\ text(different ways)`

Probability, STD2 S2 2011 HSC 26a

The two spinners shown are used in a game.

2UG 2011 26a1

Each arrow is spun once. The score is the total of the two numbers shown by the arrows.
A table is drawn up to show all scores that can be obtained in this game.

2UG 2011 26a2

  1. What is the value of `X` in the table?   (1 mark)

  2. What is the probability of obtaining a score less than 4?   (1 mark)

  3. On Spinner `B`, a 2 is obtained. What is the probability of obtaining a score of 3?   (1 mark)

Show Answers Only
  1. `5`
  2. `1/2`
  3. `2/3`
Show Worked Solution

i.   `X=3+2=5`
 

ii.   `P(text{score}<4)=6/12=1/2`
 

iii.   `P(3)=2/3`

Vectors, SPEC2 2017 VCAA 13 MC

Given the vectors  `underset~a = 3underset~i - 4underset~j + 12underset~k`  and  `underset~b = 2underset~i + 2underset~j - underset~k`, the vector resolute of  `underset~a` in the direction of  `underset~b`  is

  1.  `−14/3`
  2.  `−14/3(2underset~i - 2underset~j - underset~k)`
  3.  `−14/9(2underset~i + 2underset~j - underset~k)`
  4.  `−14/13`
  5.  `−14/169(3underset~i - 4underset~j + 12underset~k)`
Show Answers Only

`C`

Show Worked Solution
`(underset~a · overset^(underset~b))overset^(underset~b)` `= (3 xx 2 + −4 xx 2 + 12 xx −1)/sqrt(2^2 + 2^2 +(−1)^2) xx (2underset~i + 2underset~j – underset~k)/sqrt(2^2 + 2^2 + (−1)^2)`
  `= – 14/9 (2underset~i + 2underset~j – underset~k)`

`=> C`

Vectors, SPEC2 2017 VCAA 11 MC

The vectors  `underset~a = 2underset~i + 3underset~j + d underset~k, \ underset~b = underset~i + underset~j - 4underset~k`  and  `underset~c = 2underset~i + underset~j - 2underset~k`  where `d` is a real constant, are linearly dependent if

  1. `d = −10`
  2. `d ∈ R\ text(\)\ {−14}`
  3. `d = −14`
  4. `d = R\ text(\)\ {−10}`
  5. `d ∈ R`
Show Answers Only

`C`

Show Worked Solution

`alphaunderset~a + betaunderset~b = underset~c`

`2alpha + beta = 2\ \ …\ (1)`

`3alpha + beta = 1\ \ …\ (2)`

`dalpha + 4beta = −2\ \ …\ (3)`

 
`(2) – (1):`

`alpha = −1`

`=> beta = 2+2=4`
 

`text(Substitute)\ \ alpha = −1, beta = 4\ \ text{into (3):}`

`-d – 16` `= −2`
`d` `= −14`

 
`=> C`

Networks, STD2 N3 2018 FUR2 1

The graph below shows the possible number of postal deliveries each day between the Central Mail Depot and the Zenith Post Office.

The unmarked vertices represent other depots in the region.

The weighting of each edge represents the maximum number of deliveries that can be made each day.
 


 

  1.  Cut A, shown on the graph, has a capacity of 10.

     

     Two other cuts are labelled as Cut B and Cut C.

    1.  Write down the capacity of Cut B.  (1 mark)

    2.  Write down the capacity of Cut C. (1 mark)

  2.  Determine the maximum number of deliveries that can be made each day from the Central Mail  Depot to the Zenith Post Office.  (1 mark)

Show Answers Only
    1. `9`
    2. `13`
  2.  `7`
Show Worked Solution
a.i.    `text{Capacity (Cut B)}` `= 3 + 2 + 4`
    `= 9`

 

a.ii.    `text{Capacity (Cut C)}` `= 3 + 6 + 4`
    `= 13`

♦ Mean mark part (b) 32%.
COMMENT: Review carefully! Most common incorrect answer was 9.

 

b.  `text(Minimum cut) = 2 + 2 + 3 = 7`

`:.\ text(Maximum deliveries) = 7`

Probability, MET1 2018 VCAA 4

Let `X` be a normally distributed random variable with a mean of 6 and a variance of 4. Let `Z` be a random variable with the standard normal distribution.

  1.  Find  `text(Pr)(X > 6)`.  (1 mark)
  2.  Find  `b`  such that  `text(Pr)(X > 7) = text(Pr)(Z < b)`.  (1 mark)
Show Answers Only
  1. `0.5`
  2. `−1/2`
Show Worked Solution

a.   `text(Mean)\ (X) = 6`

`:. text(Pr)(X > 6) = 0.5`
 

♦ Mean mark part (b) 41%.

b.    `text(Pr)(X > 7)` `= text(Pr)(X < 5)`
    `= text(Pr)(Z < (5 – 6)/2)`
    `= text(Pr)(Z < −1/2)`

 
`:. b = −1/2`

Graphs, MET1 2018 VCAA 3

Let  `f:[0,2pi] -> R, \ f(x) = 2cos(x) + 1`.

  1. Solve the equation  `2cos(x) + 1 = 0`  for  `0 <= x <= 2pi`.  (2 marks)
  2. Sketch the graph of the function  `f` on the axes below. Label the endpoints and local minimum point with their coordinates.  (3 marks)

 

 

Show Answers Only
  1. `(2pi)/3, (4pi)/3`
  2.  
Show Worked Solution
a. `2cos(x) + 1` `= 0`
  `cos(x)` `= −1/2`

`=> cos\ pi/3 = 1/2\ text(and cos is negative)`

`text(in 2nd/3rd quadrant)`

`:.x` `= pi – pi/3, pi + pi/3`
  `= (2pi)/3, (4pi)/3`

 

b.   

Algebra, MET2 2018 VCAA 7 MC

Let  `f: R^+ -> R,\ f(x) = k log_2(x),\ k in R`.

Given that  `f^(-1)(1) = 8`, the value of `k` is

  1. `0`
  2. `1/3`
  3. `3`
  4. `8`
  5. `12`
Show Answers Only

`B`

Show Worked Solution

`y = k log_2(x)`

`text(Inverse: swap)\ x ↔ y`

`x` `= k log_2 (y)`
`log_2(y)` `= x/k`
`y` `= 2^(x/k)`

 
`text(Given)\ \ f^(-1)(1) = 8,`

`8` `= 2^(1/k)`
`1/k` `= 3`
`:. k` `= 1/3`

 
`=>   B`

Graphs, MET2 2018 VCAA 2 MC

The maximal domain of the function  `f`  is  `R text(\{1})`.

A possible rule for  `f` is

  1. `f(x) = (x^2 - 5)/(x - 1)`
  2. `f(x) = (x + 4)/(x - 5)`
  3. `f(x) = (x^2 + x + 4)/(x^2 + 1)`
  4. `f(x) = (5 - x^2)/(1 + x)`
  5. `f(x) = sqrt (x - 1)`
Show Answers Only

`A`

Show Worked Solution

`text(Maximal domain of)\ \ f -> R text(\{1})`

`->\ text(all real)\ x,\ x != 1`

`:. f(x) = (x^2 – 5)/(x – 1)`

`=>   A`

GRAPHS, FUR1 2018 VCAA 01 MC

The graph below shows a line intersecting the `x`-axis at `(4, 0)` and the `y`-axis at `(0, 2)`.

The gradient of this line is

  1.  `−4`
  2.  `−2`
  3.  `−1/2`
  4.  `1/2`
  5.  `4`
Show Answers Only

`C`

Show Worked Solution
`text(Gradient)` `= (y_2 – y_1)/(x_2 – x_1)`
  `= (0 – 2)/(4 – 0)`
  `= −1/2`

`=> C`

GRAPHS, FUR2 2018 VCAA 3

Robert wants to hire a geologist to help him find potential gold locations.

One geologist, Jennifer, charges a flat fee of $600 plus 25% commission on the value of gold found.

The following graph displays Jennifer’s total fee in dollars.
 


 

Another geologist, Kevin, charges a total fee of $3400 for the same task.

  1. Draw a graph of the line representing Kevin’s fee on the axes above.  (1 mark)

     

    `qquad qquad`(answer on the axes above.)

  2. For what value of gold found will Kevin and Jennifer charge the same amount for their work?  (1 mark)
  3. A third geologist, Bella, has offered to assist Robert.
  4. Below is the relation that describes Bella’s fee, in dollars, for the value of gold found.

  5. `qquad  text{fee (dollars)} = {(quad 500),(1000),(2600),(4000):}qquad qquad quad{:(qquad quad 0 <),(2000 <=),(6000 <=),(quad):}{:(text(value of gold found) < 2000),(text(value of gold found) < 6000),(text(value of gold found) < 10\ 000),(text(value of gold found) >= 10\ 000):}`


    The step graph below representing this relation is incomplete.

     

    Complete the step graph by sketching the missing information.  (2 marks)
     

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `$11\ 200`
  3. `text(See Worked Solutions)`
Show Worked Solution
a.  

 

b.    `text(Let)\ \ x = text(value of gold)`

♦ Mean mark 41%.

`text(Jennifer’s total fee) = 600 + 0.25x`

`text(Equating fees:)`

`600 + 0.25x` `= 3400`
`0.25x` `= 2800`
`:.x` `= 2800/0.25`
  `= $11\ 200`

 

c.  

Networks, STD2 N2 2018 FUR1 4 MC

Consider the graph below.
 


 

Which one of the following is not a path for this graph?

  1. `PRQTS`
  2. `PRTSQ`
  3. `PTQSR`
  4. `PTRQS`
Show Answers Only

`C`

Show Worked Solution

`text(By trial and error:)`

`text(Consider option)\ C,`

`PTQSR\ text(is not a path because)\ S\ text(to)\ R`

`text(must go through another vertex.)`

`=> C`

GRAPHS, FUR2 2018 VCAA 2

The weight of gold can be recorded in either grams or ounces.

The following graph shows the relationship between weight in grams and weight in ounces.
 


 

The relationship between weight measured in grams and weight measured in ounces is shown in the equation

     weight in grams = `M` × weight in ounces
 

  1. Show that  `M = 28.35`  (1 mark)
  2. Robert found a gold nugget weighing 0.2 ounces.

     

    Using the equation above, calculate the weight, in grams, of this gold nugget.  (1 mark)

  3. Last year Robert sold gold to a buyer at $55 per gram.
    The buyer paid Robert a total of $12 474.

     

    Using the equation above, calculate the weight, in ounces, of this gold.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `5.67\ text(g)`
  3. `8\ text(ounces)`
Show Worked Solution

a.    `text{Substitute (20, 567) into equation:}`

Mean mark 53%.
MARKER’S COMMENT: Giving the equation and writing “solve” did not receive a mark in part (a).

`567` `= M xx 20`
`:. M` `= 567/20`
  `= 28.35\ text(… as required.)`

 

b.   `text{Weight (grams)}` `= 28.35 xx 0.2`
    `= 5.67\ text(g)`

 

c.   `text(Total grams sold)` `= 12474/55`
    `= 226.8\ text(g)`

 

`226.8` `= 28.35 xx W_text(ounces)`
`W_text(ounces)` `= 226.8/28.35`
  `= 8\ text(ounces)`

GRAPHS, FUR2 2018 VCAA 1

The following chart displays the daily gold prices (dollars per gram) for the month of November 2017.
 

  1. Which day in November had the lowest gold price?  (1 mark)
  2. Between which two consecutive days did the greatest increase in gold price occur?  (1 mark)
Show Answers Only
  1. `text(Day 2)`
  2. `text(between days 16 and 17)`
Show Worked Solution

a.   `text(Day 2)`
 

b.   `text(Between days 16 and 17)`

GEOMETRY, FUR1 2018 VCAA 02 MC

A triangle  `PQR`  is shown in the diagram below.


 

The length of the side  `QR`  is 18 cm.

The length of the side  `PR`  is 26 cm.

The angle  `QRP`  is 30°.

The area of triangle `PQR`, in square centimetres, is closest to

  1.  117
  2.  162
  3.  171
  4.  234
  5.  468
Show Answers Only

`A`

Show Worked Solution

`text(Using sine rule:)`

`A` `= 1/2 ab sinc`
  `= 1/2 xx 18 xx 26 xx sin30°`
  `= 117\ text(cm²)`

 
`=> A`

NETWORKS, FUR1 2018 VCAA 4 MC

Consider the graph below.
 


 

Which one of the following is not a path for this graph?

  1. `PRQTS`
  2. `PQRTS`
  3. `PRTSQ`
  4. `PTQSR`
  5. `PTRQS`
Show Answers Only

`D`

Show Worked Solution

`text(By trial and error:)`

`text(Consider option)\ D,`

`PTQSR\ text(is not a path because)\ S\ text(to)\ R`

`text(must go through another vertex.)`

`=> D`

NETWORKS, FUR1 2018 VCAA 3 MC

A planar graph has five faces.

This graph could have

  1.  eight vertices and eight edges.
  2.  six vertices and eight edges.
  3.  eight vertices and five edges.
  4.  eight vertices and six edges.
  5.  five vertices and eight edges.
Show Answers Only

`E`

Show Worked Solution

`text(Using Euler’s formula:)`

`v + f` `= e + 2`
`v + 5` `= e + 2`
`v + 3` `= e`

 
`:. 5\ text(vertices and 8 edges ⇒ Euler holds)`

`=> E`

GEOMETRY, FUR2 2018 VCAA 1

Tennis balls are packaged in cylindrical containers.

Frank purchases a container of tennis balls that holds three standard tennis balls, stacked one on top of the other.

This container has a radius of 3.4 cm and a height of 20.4 cm, as shown in the diagram below.
 

  1. What is the diameter, in centimetres, of this container?   (1 mark)
  2. What is the total outside surface area of this container, including both ends?

     

    Write your answer in square centimetres, rounded to one decimal place.  (1 mark)

A standard tennis ball is spherical in shape with a radius of 3.4 cm.

    1. Write a calculation that shows that the volume, rounded to one decimal place, of one standard tennis ball is 164.6 cm³.  (1 mark)
    2. Write a calculation that shows that the volume, rounded to one decimal place, of the cylindrical container that can hold three standard tennis balls is 740.9 cm³. (1 mark)
    3. How much unused volume, in cubic centimetres, surrounds the tennis balls in this container?

       

      Round your answer to the nearest whole number.  (1 mark)

Show Answers Only
  1. `6.8\ text(cm)`
  2. `508.4\ text(cm)^2\ text{(to 1 d.p.)}`
    1. `164.6\ text(cm)^3\ text{(to 1 d.p.)}`
    2. `740.9\ text(cm)^3\ text{(to 1 d.p.)}`
    3. `247\ text(cm)^3\ text{(nearest cm}^3 text{)}`
Show Worked Solution
a.    `text(Diameter)` `= 2 xx text(radius)`
    `= 2 xx 3.4`
    `= 6.8\ text(cm)`

 

b.    `text(S.A.)` `= 2 pi r^2 + 2 pi rh`
    `= 2 xx pi xx 3.4^2 + 2 xx pi xx 3.4 xx 20.4`
    `= 508.43…`
    `= 508.4\ text(cm)^2\ text{(to 1 d.p.)}`

 

c.i.    `text(Volume)` `= 4/3 pi r^3`
    `= 4/3 xx pi xx 3.4^3`
    `= 164.6\ text(cm)^3\ text{(to 1 d.p.)}`

 

c.ii.    `text(Volume)` `= Ah`
    `= pi xx 3.4^2 xx 20.4`
    `= 740.86…`
    `= 740.9\ text(cm)^3\ text{(to 1 d.p.)}`

 

c.iii.    `text(Unused volume)` `= text(cylinder volume) – text(volume of balls)`
    `= 740.9 – 3 xx 164.6`
    `= 247.1`
    `= 247\ text(cm)^3\ text{(nearest cm}^3 text{)}`

NETWORKS, FUR2 2018 VCAA 3

At the Zenith Post Office all computer systems are to be upgraded.

This project involves 10 activities, `A` to `J`.

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

  1. Determine the earliest starting time, in hours, for activity `I`.  (1 mark)
  2. The minimum completion time for the project is 15 hours.

     

    Write down the critical path.  (1 mark)

  3. Two of the activities have a float time of two hours.

     

    Write down these two activities.  (1 mark)

  4. For the next upgrade, the same project will be repeated but one extra activity will be added.
    This activity has a duration of one hour, an earliest starting time of five hours and a latest starting time of 12 hours.

     

    Complete the following sentence by filling in the boxes provided.  (1 mark)

     

    The extra activity could be represented on the network above by a directed edge from the

      end of activity   
 
 to the start of activity   
 
Show Answers Only
  1. `10\ text(hours)`
  2. `B – E – G – H – J`
  3.  `text(Activity)\ A\ text(and)\ C`
  4. `text(end of activity)\ E\ text(to the start of activity)\ J`
Show Worked Solution

a.  `text(Longest path to)\ I:`

`B -> E -> G`

`:.\ text(EST for)\ \ I` `= 2 + 3 + 5`
  `= 10\ text(hours)`

 
b.  `B – E – G – H – J`
 

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

♦ Mean mark 45%.

 


 

`:.\ text(Activity)\ A\ text(and)\ C\ text(have a 2 hour float time.)`
 

d.   `text(end of activity)\ E\ text(to the start of activity)\ J`

♦♦ Mean mark 25%.
 

`text(By inspection of forward and backward scanning:)`

`text(EST of 5 hours is possible after activity)\ E.`

`text(LST of 12 hours after activity)\ E -> text(edge has weight)`

`text(of 1 and connects to)\ J`

NETWORKS, FUR2 2018 VCAA 1

The graph below shows the possible number of postal deliveries each day between the Central Mail Depot and the Zenith Post Office.

The unmarked vertices represent other depots in the region.

The weighting of each edge represents the maximum number of deliveries that can be made each day.
 


 

  1.  Cut A, shown on the graph, has a capacity of 10.

     

     Two other cuts are labelled as Cut B and Cut C.

    1.  Write down the capacity of Cut B.  (1 mark)
    2.  Write down the capacity of Cut C. (1 mark)
  2.  Determine the maximum number of deliveries that can be made each day from the Central Mail   Depot to the Zenith Post Office.  (1 mark)
Show Answers Only
    1. `9`
    2. `13`
  2.  `7`
Show Worked Solution
a.i.    `text{Capacity (Cut B)}` `= 3 + 2 + 4`
    `= 9`

 

a.ii.    `text{Capacity (Cut C)}` `= 3 + 6 + 4`
    `= 13`

♦ Mean mark part (b) 32%.
COMMENT: Review carefully! Most common incorrect answer was 9.

 

b.  `text(Minimum cut) = 2 + 2 + 3 = 7`

`:.\ text(Maximum deliveries) = 7`

MATRICES, FUR1 2018 VCAA 6 MC

A transition matrix, `V`, is shown below.

`{:(),(),(),(),(V =),(),():}{:(qquadqquadtext(this month)),(qquadLqquadquadTqquadquadFqquad\ M),([(0.6,0.6,0.6,0.0),(0.1,0.2,0.0,0.1),(0.3,0.0,0.8,0.4),(0.0,0.2,0.0,0.5)]):}{:(),(),(L),(T),(F),(M):}{:\ text(next month):}`

The transition diagram below has been constructed from the transition matrix `V`.

The labelling in the transition diagram is not yet complete.
 

The proportion for one of the transitions is labelled `x`.

The value of `x` is

  1. 0.2
  2. 0.5
  3. 0.6
  4. 0.7
  5. 0.8
Show Answers Only

`C`

Show Worked Solution

`x = 0.6`

`=> C`

MATRICES, FUR2 2018 VCAA 3

The Hiroads company has a contract to maintain and improve 2700 km of highway.

Each year sections of highway must be graded `(G)`, resurfaced `(R)` or sealed `(S)`.

The remaining highway will need no maintenance `(N)` that year.

Let `S_n` be the state matrix that shows the highway maintenance schedule for the `n`th year after 2018.

The maintenance schedule for 2018 is shown in matrix `S_0` below.

 
`S_0 = [(700),(400),(200),(1400)]{:(G),(R),(S),(N):}`
 

The type of maintenance in sections of highway varies from year to year, as shown in the transition matrix `T`, below.
 

`{:(qquad qquad qquad qquad qquad quad text(this year)),(qquad qquad quad quad G qquad quad R qquad quad S quad quad \ N),(T = [(0.2,0.1,0.0,0.2),(0.1,0.1,0.0,0.2),(0.2,0.1,0.2,0.1),(0.5, 0.7,0.8,0.5)]{:(G),(R),(S),(N):} \ text (next year)):}`
 

  1. Of the length of highway that was graded `(G)` in 2018, how many kilometres are expected to be resurfaced `(R)` the following year?  (1 mark)
  2. Show that the length of highway that is to be graded `(G)` in 2019 is 460 km by writing the appropriate numbers in the boxes below.  (1 mark)
     
 
 `× 700 +`
 
 `× 400 +`
 
 `× 200 +`
 
 `× 1400 = 460`

 

The state matrix describing the highway maintenance schedule for the nth year after 2018 is given by

`S_(n + 1) = TS_n`
 

  1. Complete the state matrix, `S_1`, below for the highway maintenance schedule for 2019 (one year after 2018).  (1 mark)

     


    `qquad qquad S_1 = [(460),(text{____}),(text{____}),(1490)]{:(G),(R),(S),(N):}`
     

  2. In 2020, 1536 km of highway is expected to require no maintenance `(N)`

     

    Of these kilometres, what percentage is expected to have had no maintenance `(N)` in 2019?

     

    Round your answer to one decimal place.  (1 mark)

  3. In the long term, what percentage of highway each year is expected to have no maintenance `(N)`?

     

    Round your answer to one decimal place.  (1 mark)

Show Answers Only
  1. `70\ text(km)`
  2. `G_2019 = 0.2 xx 700 + 0.1 xx 400 + 0 xx 200 + 0.2 xx 1400 = 460`
  3. `[(460),(390),(360),(1490)]`
  4. `48.5 text{%  (to 1 d.p.)}`
  5. `56.7 text{%  (to 1 d.p.)}`
Show Worked Solution
a.    `G -> R` `= 0.1 xx 700`
    `= 70\ text(km)`

♦ Mean mark part (a) 48%.

 

b.    `G_2019` `= 0.2 xx 700 + 0.1 xx 400 + 0 xx 200 + 0.2 xx 1400`
    `= 460`

 

c.    `S_1` `= TS_0`
    `= [(0.2,0.1,0.0,0.2),(0.1,0.1,0.0,0.2),(0.2,0.1,0.2,0.1),(0.5,0.7,0.8,0.5)][(700),(400),(200),(1400)]=[(460),(390),(360),(1490)]`

 

d.   `N_2019 = 1490`

`:.\ text(Percentage)` `= (0.5 xx 1490)/1536 xx 100`
  `= 48.502…`
  `= 48.5 text{%  (to 1 d.p.)}`

 

e.  `text(Raise)\ \ T\ \ text(to a high power)\ (n = 50):`

`T^50 = [(0.160,0.160,0.160,0.160),(0.144,0.144,0.144,0.144),(0.129,0.129,0.129,0.129),(0.567,0.567,0.567,0.567)]`

`:.\ %N` `= 0.567`
  `= 56.7 text{%  (to 1 d.p.)}`

CORE, FUR2 2018 VCAA 4

Julie deposits some money into a savings account that will pay compound interest every month.

The balance of Julie’s account, in dollars, after `n` months, `V_n` , can be modelled by the recurrence relation shown below.

`V_0 = 12\ 000, qquad V_(n + 1) = 1.0062 V_n` 

  1.  How many dollars does Julie initially invest?  (1 mark)
  2.  Recursion can be used to calculate the balance of the account after one month.

    1. Write down a calculation to show that the balance in the account after one month, `V_1`, is  $12 074.40. (1 mark)
    2. After how many months will the balance of Julie’s account first exceed $12 300  (1 mark)
  3.  A rule of the form  `V_n = a xx b^n`  can be used to determine the balance of Julie's account after `n` months.

    1. Complete this rule for Julie’s investment after `n` months by writing the appropriate numbers in the boxes provided below. (1 mark)
       
balance = 
 
  × 
 
  `n`

 

    1. What would be the value of  `n`  if Julie wanted to determine the value of her investment after three years?  (1 mark)
Show Answers Only
  1. `$12\ 000`
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `4\ text(months)`
    1. `text(balance) = 12\ 000 xx 1.0062^n`
    2. `36`
Show Worked Solution

a.   `$12\ 000`
 

b.i.   `V_1` `= 1.0062 xx V_0`
    `= 1.0062 xx 12000`
    `= $12\ 074.40\ text(… as required.)`

 

b.ii.   `V_2` `= 1.0062 xx 12\ 074.40 = 12\ 149.26`
  `V_3` `= 1.0062 xx 12\ 149.26 = 12\ 224.59`
  `V_4` `= 1.0062 xx 12\ 224.59 = 12\ 300.38`

 
`:.\ text(After 4 months)`

 
c.i.  `text(balance) = 12\ 000 xx 1.0062^n`

 
c.ii.  `n = 12 xx 3 = 36`

CORE, FUR2 2018 VCAA 3

Table 3 shows the yearly average traffic congestion levels in two cities, Melbourne and Sydney, during the period 2008 to 2016. Also shown is a time series plot of the same data.

The time series plot for Melbourne is incomplete.
 

  1. Use the data in Table 3 to complete the time series plot above for Melbourne.  (1 mark)

     

                (Answer on the time series plot above.)

  2. A least squares line is used to model the trend in the time series plot for Sydney. The equation is
     
         `text(congestion level = −2280 + 1.15 × year)`
     

    1. Draw this least squares line on the time series plot.  (1 mark)

       

            (Answer on the time series plot above.)

    2. Use the equation of the least squares line to determine the average rate of increase in percentage congestion level for the period 2008 to 2016 in Sydney.

       

      Write your answer in the box provided below.  (1 mark)

 
  % per year
    1. Use the least squares line to predict when the percentage congestion level in Sydney will be 43%.  (1 mark)

 
The yearly average traffic congestion level data for Melbourne is repeated in Table 4 below.
 


 

  1. When a least squares line is used to model the trend in the data for Melbourne, the intercept of this line is approximately –1514.75556

     

    Round this value to four significant figures.  (1 mark)

  2. Use the data in Table 4 to determine the equation of the least squares line that can be used to model the trend in the data for Melbourne. The variable year is the explanatory variable.

     

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

     

    Round both values to four significant figures.  (2 marks)
     

congestion level = 
 
  + 
 
  × year

 

  1. Since 2008, the equations of the least squares lines for Sydney and Melbourne have predicted that future traffic congestion levels in Sydney will always exceed future traffic congestion levels in Melbourne.

     

    Explain why, quoting the values of appropriate statistics.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
    1. `text(See Worked Solutions)`
    2. `1.15 text(%)`
    3. `2020`
  2. `-1515`
  3. `text(congestion level) = -1515 + 0.7667 xx text(year)`
  4. `text(See Worked Solutions)`
Show Worked Solution
a.   

 

b.i.   

 

b.ii.  `text(The least squares line is 1.15% higher each year.)`

♦ Mean mark (b)(ii) 36%.
COMMENT: Major problems caused by part (b)(ii). Review!

  ` :.\ text(Average rate of increase) = 1.15 text(%)`

 

b.iii.    `text(Find year when:)`
  `43` `= -2280 + 1.15 xx text(year)`
  `text(year)` `= 2323/1.15`
    `= 2020`

 

c.  `-1515`

 

d.   `text(congestion level) = -1515 + 0.7667 xx text(year)`

 

e.   `text(Melbourne congestion level in 2008)`

♦♦♦ Mean mark 18%.

`= -1515 + 0.7667 xx 2008`

`= 24.5 text(%)`

 
`text{In 2008 Sydney has higher congestion (29.2 > 24.5)}`

`text(After 2008, Sydney congestion grows at 1.15% per)`

`text(year and Melbourne grows at 0.7667% per year.)`

`:.\ text(Sydney predicted to always exceed Melbourne.)`

CORE, FUR2 2018 VCAA 2

The congestion level in a city can be recorded as the percentage increase in travel time due to traffic congestion in peak periods (compared to non-peak periods).

This is called the percentage congestion level.

The percentage congestion levels for the morning and evening peak periods for 19 large cities are plotted on the scatterplot below.
 


 

  1. Determine the median percentage congestion level for the morning peak period and the evening peak period.

     

    Write your answers in the appropriate boxes provided below.  (2 marks)

Median percentage congestion level for morning peak period
%
Median percentage congestion level for evening peak period
%

 
A least squares line is to be fitted to the data with the aim of predicting evening congestion level from morning congestion level.

The equation of this line is.

evening congestion level = 8.48 + 0.922 × morning congestion level
 

  1. Name the response variable in this equation.  (1 mark)
  2. Use the equation of the least squares line to predict the evening congestion level when the morning congestion level is 60%.  (1 mark)
  3. Determine the residual value when the equation of the least squares line is used to predict the evening congestion level when the morning congestion level is 47%.

     

    Round your answer to one decimal place?  (2 marks)

  4. The value of the correlation coefficient `r` is 0.92

     

    What percentage of the variation in the evening congestion level can be explained by the variation in the morning congestion level?

     

    Round your answer to the nearest whole number.  (1 mark)

Show Answers Only
  1. `52 text(%);\ 56 text(%)`
  2. `text(evening congestion level)`
  3. `63.8 text(%)`
  4. `-1.8 text{% (to 1 d.p.)}`
  5. `85 text(%)`
Show Worked Solution

a.   `19\ text(data points of morning peak)`

Mean mark 56%.
COMMENT: This question was surprisingly poorly answered. Review carefully!

`text(Median is 10th data point moving left to right.)`

`:.\ text{Median (morning peak) = 52%}`

 

`text(Median of evening peak is 10th data point)`

`text(moving bottom to top.)`

`:.\ text{Median (afternoon peak) = 56%}`
 

b.   `text(Response variable is evening congestion level.)`
 

c.    `text(evening congestion level)` `= 8.48 + 0.922 xx 60`
    `= 63.8 text(%)`

 
d.  `text(When morning level = 47%, Actual = 50%)`

Mean mark part (d) 53%.
COMMENT: Many students had problems at a number of stages in this part.

`text(Residual)` `=\ text(Actual evening congestion − predicted)`
  `= 50 – (8.48 + 0.922 xx 47)`
  `= -1.814`
  `= -1.8 text{% (to 1 d.p.)}`

 

e.    `r` `= 0.92`
  `r^2` `= 0.8464`
    `= 85 text{% (nearest whole)}`

 
`:. 85 text(% of the variations is explained.)`

CORE, FUR2 2018 VCAA 1

The data in Table 1 relates to the impact of traffic congestion in 2016 on travel times in 23 cities in the United Kingdom (UK).

The four variables in this data set are:

  • city — name of city
  • congestion level — traffic congestion level (high, medium, low)
  • size — size of city (large, small)
  • increase in travel time — increase in travel time due to traffic congestion (minutes per day).
  1. How many variables in this data set are categorical variables?  (1 mark)
  2. How many variables in this data set are ordinal variables  (1 mark)
  3. Name the large UK cities with a medium level of traffic congestion.  (1 mark)
  4. Use the data in Table 1 to complete the following two-way frequency table, Table 2.  (2 marks)
     

     


     

  5. What percentage of the small cities have a high level of traffic congestion?  (1 mark)

Traffic congestion can lead to an increase in travel times in cities. The dot plot and boxplot below both show the increase in travel time due to traffic congestion, in minutes per day, for the 23 UK cities.
 


 

  1. Describe the shape of the distribution of the increase in travel time for the 23 cities.  (1 mark)
  2. The data value 52 is below the upper fence and is not an outlier.

     

    Determine the value of the upper fence.  (1 mark)

Show Answers Only
  1. `3\ text(city, congestion level, size)`
  2. `2\ text(congestion level, size)`
  3. `text(Newcastle-Sunderland and Liverpool)`
  4. `text(See Worked Solutions)`
  5. `25 text(%)`
  6. `text(Positively skewed)`
  7. `52.5`
Show Worked Solution

a.   `3\ – text(city, congestion level, size)`
 

b.   `2\ – text(congestion level, size)`
 

c.   `text(Newcastle-Sunderland and Liverpool)`
 

d.   

 

e.    `text(Percentage)` `= text(Number of small cities high congestion)/text(Number of small cities) xx 100`
    `= 4/16 xx 100`
    `= 25 text(%)`

 
f.   `text(Positively skewed)`
 

g.   `IQR = 39 – 30 = 9`
 

`text(Calculate the Upper Fence:)`

`Q_3 + 1.5 xx IQR` `= 39 + 1.5 xx 9`
  `= 52.5`

MATRICES, FUR1 2018 VCAA 3 MC

Five people, India (`I`), Jackson (`J`), Krishna (`K`), Leanne (`L`) and Mustafa (`M`), competed in a table tennis tournament.

Each competitor played every other competitor once only.

Each match resulted in a winner and a loser.

The matrix below shows the tournament results.
 

`{:(),(),(),(),(text(winner)),(),():}{:(),(),(I),(J),(K),(L),(M):}{:(qquadqquadqquadtext(loser)),(qquadIquadJquadKquadLquadM),([(0,1,0,1,0),(0,0,1,0,1),(1,0,0,1,1),(0,1,0,0,0),(0,0,0,1,0)]):}`
 

A 1 in the matrix shows that the competitor named in that row defeated the competitor named in that column.

For example, the 1 in the fourth row shows that Leanne defeated Jackson.

A 0 in the matrix shows that the competitor named in that row lost to the competitor named in that column.

There is an error in the matrix. The winner of one of the matches has been incorrectly recorded as a 0.

This match was between

  1. India and Mustafa.
  2. India and Krishna.
  3. Krishna and Leanne.
  4. Leanne and Mustafa.
  5. Jackson and Mustafa.
Show Answers Only

`A`

Show Worked Solution

`text(S)text(ince each match has a winner and a loser,)`

`text(If)\ \ x_(ij) = 1 \ => \ x_(ji) = 0`

`text(By trial and error,)`

`text(Consider option)\ A:`

`x_15 = 0\ (text(India loses to Mustafa))`

`x_51 = 0\ (text(Mustafa loses to India))`

`:.\ text(Both cannot be correct.)`

`=> A`

MATRICES, FUR1 2018 VCAA 1 MC

Which one of the following matrices has a determinant of zero?

A. `[(0,1),(1,0)]` B. `[(1,0),(0,1)]` C. `[(1,2),(−3,6)]`
           
D. `[(3,6),(2,4)]` E. `[(4,0),(0,−2)]`    
Show Answers Only

`D`

Show Worked Solution

`text(By trial and error:)`

`text(Consider option)\ D,`

`text(det)[(3,6),(2,4)]` `= 3 xx 4 – 6 xx 2`
  `= 0`

`=> D`

CORE, FUR1 2018 VCAA 17-18 MC

The value of an annuity investment, in dollars, after `n` years, `V_n` , can be modelled by the recurrence relation shown below.

`V_0 = 46\ 000, quadqquadV_(n + 1) = 1.0034V_n + 500`

 
Part 1

What is the value of the regular payment added to the principal of this annuity investment?

  1.   $34.00
  2. $156.40
  3. $466.00
  4. $500.00
  5. $656.40

 
Part 2

Between the second and third years, the increase in the value of this investment is closest to

  1.   $656
  2.   $658
  3.   $661
  4. $1315
  5. $1975
Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(Regular payment = $500)`

`=> D`
 

`text(Part 2)`

`V_1` `= 1.0034 xx 46\ 000 + 500 = $46\ 656.40`
`V_2` `= 1.0034 xx 46\ 656.40 + 500 = $47\ 315.03`
`V_3` `= 1.0034 xx 47\ 315.03 + 500 = $47\ 975.90`

 

`:.\ text(Increase)` `= 47\ 975.90 – 47\ 315.03`
  `= $660.87`

`=> C`

CORE, FUR1 2018 VCAA 15 MC

The table below shows the monthly profit, in dollars, of a new coffee shop for the first nine months of 2018.

Using four-mean smoothing with centring, the smoothed profit for May is closest to

  1. $2502
  2. $3294
  3. $3503
  4. $3804
  5. $4651
Show Answers Only

`B`

Show Worked Solution

`text(4 point mean:)`

`text(Mar-Jun)` `= (2402 + 2456 + 4651 + 3456)/4 = 3241.25`
`text(Apr-Jul)` `= (2456 + 4651 + 3456 + 2823)/4 = 3346.5`

 

`:.\ text(Centred Mean)` `= (3241.25 + 3346.5)/2`
  `= 3293.875`

`=> B`

CORE, FUR1 2018 VCAA 3-5 MC

The pulse rates of a population of Year 12 students are approximately normally distributed with a mean of 69 beats per minute and a standard deviation of 4 beats per minute.
 

Part 1

A student selected at random from this population has a standardised pulse rate of  `z = –2.5`

This student’s actual pulse rate is

  1.  59 beats per minute.
  2.  63 beats per minute.
  3.  65 beats per minute.
  4.  73 beats per minute.
  5.  79 beats per minute.
     

Part 2

Another student selected at random from this population has a standardised pulse rate of  `z=–1.`

The percentage of students in this population with a pulse rate greater than this student is closest to

  1.   2.5%
  2.   5%
  3.  16%
  4.  68%
  5.  84%
     

Part 3

A sample of 200 students was selected at random from this population.

The number of these students with a pulse rate of less than 61 beats per minute or greater than 73 beats per minute is closest to

  1.  19
  2.  37
  3.  64
  4.  95
  5. 190
Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ E`

`text(Part 3:)\ B`

Show Worked Solution

`text(Part 1)`

`ztext(-score)` `= (x – barx)/s`
`−2.5` `= (x – 69)/4`
`x – 69` `= −10`
`:. x` `= 59`

 
`=> A`
 

`text(Part 2)`

`ztext(-score) = −1`
 

`text(% above)` `= 34 + 50`
  `= 84 text(%)`

`=> E`
 

`text(Part 3)`

`z-text(score)\ (61)` `= (61 – 69)/4 = −2`
`z-text(score)\ (73)` `= (73 – 69)/4 = 1`

 


 

`text(Percentage)` `= 2.5 + 16`
  `= 18.5text(%)`

 

`:.\ text(Number of students)` `= 18.5 text(%) xx 200`
  `= 37`

`=> B`

CORE, FUR1 2018 VCAA 1-2 MC

The dot plot below displays the difference in travel time between the morning peak and the evening peak travel times for the same journey on 25 days.
 


 

Part 1

The percentage of days when there was five minutes difference in travel time between the morning peak and the evening peak travel times is

  1.    0%
  2.    5%
  3.  20%
  4.  25%
  5.  28%
     

Part 2

The median difference in travel time is

  1.  3.0 minutes.
  2.  3.5 minutes.
  3.  4.0 minutes.
  4.  4.5 minutes.
  5.  5.0 minutes.
Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ A`

Show Worked Solution

`text(Part 1)`

`text(Percentage)` `= text(days with 5 m difference)/text(total days) xx 100`
  `= 5/25 xx 100`
  `= 20\ text(%)`

`=> C`

 

`text(Part 2)`

`text(Median)` `= 13text(th data point)`
  `= 3\ text(minutes)`

`=> A`

Networks, STD2 N3 SM-Bank 31

Murray is building a new garage. The project involves activities `A` to `L`.
 

 

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

  1. Which TWO activities immediately precede activity `G`?  (1 mark)

  2. By completing the diagram shown, calculate the minimum time required to build the new garage.  (2 marks)

  3. Hence, what is the float time for activity `E`?  (1 mark)

Show Answers Only
  1. `text(Activity)\ C\ text(and)\ D.`
  2. `30\ text(days)`
  3. `6\ text(days)`
Show Worked Solution

a.   `text(Activity)\ C\ text(and)\ D.`

 

b.   

`text(Critical path)\ \ A – D – G – L`

`= 1 + 10 + 13 + 6`

`= 30\ text(days)`

 

c.   `text(Float time of Activity)\ E`

`= 14 – 8`

`= 6\ text(days)`

Networks, STD2 N2 SM-Bank 28

In central Queensland, there are four petrol stations `A`, `B`, `C` and `D`. The table shows the length, in kilometres, of roads connecting these petrol stations.
 


 

  1. Construct a network diagram to represent the information in the table.  (2 marks)

  2. A petrol tanker needs to refill each station. It starts at Station `A` and visits each station.

     

    Calculate the shortest distance that can be travelled by the petrol tanker. In your answer, include the order the petrol stations are refilled.  (2 marks)

Show Answers Only
  1.  
  2. `380\ text(km)`
Show Worked Solution
a.   

 

b.   `text(Shortest Path from)\ A\ (text(visiting all stations))`

`A – B – D – C`

`text(Distance)` `= 170 + 90 + 120`
  `= 380\ text(km)`

Networks, STD2 N2 SM-Bank 27 MC

This diagram shows the possible paths (in km) for laying gas pipes between various locations.
 

 
Gas is be supplied from one location. Any one of the locations can be the source of the supply.

What is the minimum total length of length of the pipes required to provide gas to all the locations?

  1.  46 km
  2.  48 km
  3.  50 km
  4.  52 km
Show Answers Only

`text(A)`

Show Worked Solution

`text(Using Kruskul’s Theorem:)`

`=>\ text(5 vertices – 4 edges needed)`

`text{Edge 1: The Hill → Carnie (9)}`

`text{Edge 2: Carnie → Bally (10)}`

`text{Edge 3: Bally → Eden (13)}`

`text{Edge 4: Carnie → Shallow End (14)}`
  


  

`:.\ text(Maximum length)` `= 9 + 10 + 13 + 14`
  `= 46\ text(km)`

 
`=>\ text(A)`

Measurement, STD2 M2 SM-Bank 2

Bert is in Moscow, which is three hours behind of Coordinated Universal Time (UTC).

Karen is in Sydney, which is eleven hours ahead of UTC.

  1. Bert is going to ring Karen at 9 pm on Tuesday, Moscow time. What day and time will it be in Sydney when he rings?  (1 mark)

  2. Karen is going to fly from Sydney to Moscow. Her flight will leave on Wednesday at 8 am, Sydney time, and will take 15 hours. What day and time will it be in Moscow when she arrives?  (2 marks)

Show Answers Only
  1. `11\ text(am Wednesday)`
  2. `9\ text(am Wednesday)`
Show Worked Solution

a.   `text(Bert is 14 hours behind Karen.)`

`text(At 9 pm in Moscow:)`

`text(Time in Sydney)` `= 9\ text(pm + 14 hours)`
  `= 11\ text(am Wednesday)`

 

b.    `text(Arrival time)` `= 8\ text(am + 15 hours)`
    `= 11\ text{pm Wednesday (Sydney time)}`

 

`:.\ text(Time in Moscow)` `= 11\ text(pm less 14 hours)`
  `= 9\ text(am Wednesday)`

Measurement, STD2 M7 SM-Bank 4

Blood pressure is measured using two numbers: systolic pressure and diastolic pressure. If the measurement shows 120 systolic and 80 diastolic, it is written as 120/80.

The bars on the graph show the normal range of blood pressure for people of various ages.

  1. What is the normal range of blood pressure for a 53-year-old?  (2 marks)

  2. Ralph, aged 53, had a blood pressure reading of 173 over 120. A doctor prescribed Ralph a medication to reduce his blood pressure to be within the normal range. To check that the medication was being effective, the doctor measured Ralph's blood pressure for 10 weeks and recorded the following results.  
     

     
    With reference to the data provided, comment on the effectiveness of the medication during the 10-week period in returning Ralph’s blood pressure to the normal range.  (3 marks)

Show Answers Only

a.   `text{Normal range (systolic): 117 – 145}`

       `text{Normal range (diastolic): 81 – 89}`

b.   `text(Effectiveness of medication)`

`text(• systolic and diastolic pressure reduced below original)`

   `text(readings from week 3 onwards.)`

`text(• systolic pressure in normal range from week 5 onwards.)`

`text(• diastolic pressure in normal range from week 7 onwards.)`

`text(• blood pressure only in the normal range from week 7.)`

Show Worked Solution

a.   `text{Normal range (systolic): 117 – 145}`

`text{Normal range (diastolic): 81 – 89}`

 

b.   `text(Effectiveness of medication)`

`text(• systolic and diastolic pressure reduced below original)`

     `text(readings from week 3 onwards.)`

`text(• systolic pressure in normal range from week 5 onwards.)`

`text(• diastolic pressure in normal range from week 7 onwards.)`

`text(• blood pressure only in the normal range from week 7.)`

Measurement, STD2 M2 SM-Bank 27 MC

Part of a train timetable is shown.

Kalyn arrives at New Castle station at 1.25 pm and needs to get to Gosford as quickly as possible.

Assuming all trains run to schedule, what is the EARLIEST time that Kalyn can arrive at Gosford station?

  1. 2.32 pm
  2. 2.36 pm
  3. 2.51 pm
  4. 2.58 pm
Show Answers Only

`text(C)`

Show Worked Solution

`text(1st train – Kalyn arrives at Newcastle late)`

`text(2nd train – doesn’t stop at Gosford)`

`text(3rd train – arrives 2.51 pm)`

`text(4th train – arrives 2.58 pm)`

`=>\ text(C)`

Graphs, SPEC2 2018 VCAA 1MC

Part of the graph of  `y = 1/2 tan^(-1)(x)`  is shown below.
 

 
The equations of its asymptotes are

A.  `y = +- 1/2`

B.  `y = +- 3/4`

C.  `y = +- 1`

D.  `y = +- pi/2`

E.  `y = +- pi/4` 

Show Answers Only

`E`

Show Worked Solution

`y = tan(x): qquad x in (-pi/2, pi/2)`

`-> y = tan^(-1)(x): qquad y in (-pi/2, pi/2)`

`-> y = 1/2 tan^(-1)(x): qquad y in (-pi/4, pi/4)`

 
`=>  E`

Measurement, STD2 M1 SM-Bank 25 MC

A cockroach is measured in a school science experiment and its length is recorded as 5.2 cm.

What is the upper limit of accuracy of this measurement?

  1.  5.21 cm
  2.  5.25 cm
  3.  5.5 cm
  4.  5.9 cm
Show Answers Only

`=>\ B`

Show Worked Solution

`text(A)text(bsolute error = 0.05 cm)`

`text(Upper limit)` `= 5.2 + 0.05`
  `= 5.25\ text(cm)`

 
`=>\ B`

L&E, 2ADV E1 SM-Bank 2

The population of Indian Myna birds in a suburb can be described by the exponential function

`N = 35e^(0.07t)`

where `t` is the time in months.

  1.  What will be the population after 2 years?  (1 mark)

  2.  Draw a graph of the population.  (2 marks)

Show Answers Only
  1. `188\ text(birds)`

  2.  

Show Worked Solution

i.   `N = 35e^(0.07t)`

`text(Find)\ N\ text(when)\ \ t = 24:`

`N` `= 35e^(0.07 xx 24)`
  `= 35e^(1.68)`
  `= 187.79…`
  `= 188\ text(birds)`

 

ii.

Calculus, 2ADV C1 SM-Bank 3

The displacement `x` metres from the origin at time `t` seconds of a particle travelling in a straight line is given by

`x = 2t^3 - t^2 - 3t + 11`     when   `t >= 0`

  1.  Calculate the velocity when  `t = 2`.  (1 mark)

  2.  When is the particle stationary?  (2 marks)

Show Answers Only
  1.  `17\ text(ms)^(−1)`
  2.  `(1 + sqrt19)/6`
Show Worked Solution

i.   `x =2t^3 – t^2 – 3t + 11` 

`v = (dx)/(dt) = 6t^2 – 2t – 3`

 
`text(When)\ t = 2,`

`v` `= 6 xx 2^2 – 2 · 2 – 3`
  `= 17\ text(ms)^(−1)`

 

ii.   `text(Particle is stationary when)\ \ v = 0`

`6t^2 – 2t – 3` `= 0`
`:. t` `= (2 ±sqrt((−2)^2 – 4 · 6 · (−3)))/12`
  `= (2 ± sqrt76)/12`
  `= (1 ± sqrt19)/6`
  `= (1 + sqrt19)/6 qquad(t >= 0)`

Calculus, 2ADV C1 SM-Bank 2

  1.  Find the equations of the tangents to the curve  `y = x^2 - 3x`  at the points where the curve cuts the `x`-axis.  (2 marks)

  2.  Where do the tangents intersect?  (2 marks)

Show Answers Only
  1. `y = −3x`
    `y = 3x – 9`
  2. `(3/2, −9/4)`
Show Worked Solution
i.   `y` `= x^2 – 3x`
  `= x(x – 3)`

 
`text(Cuts)\ xtext(-axis at)\ \ x = 0\ \ text(or)\ \ x = 3`
 

`(dy)/(dx) = 2x – 3`

 
`text(At)\ \ x = 0 \ => \ (dy)/(dx) = -3`

`T_1\ text(has)\ \ m = −3,\ text{through (0, 0)}`

`y – 0` `= -3(x – 0)`
`y` `= -3x`

  

`text(At)\ \ x = 3 \ => \ (dy)/(dx) = 3`

`T_2\ text(has)\ \ m = 3,\ text{through (3, 0)}`

`y – 0` `= 3(x – 3)`
`y` `= 3x – 9`

 

ii.   `text(Intersection occurs when:)`

`3x – 9` `= -3x`
`6x` `= 9`
`x` `= 3/2`

  

`y = -3 xx 3/2 = −9/2`

`:.\ text(Intersection at)\ \ (3/2, −9/2)`

Calculus, 2ADV C1 EQ-Bank 3

  1.  Use differentiation by first principles to find `y^{′}`, given  `y = 2x^2 + 5x`.   (2 marks)

  2.  Find the equation of the tangent to the curve when  `x = 1`.   (1 mark)

Show Answers Only
  1.  `y^{′} = 4x + 5`
  2.  `y = 9x-2`
Show Worked Solution
i.    `f(x)` `= 2x^2 + 5x`
  `f^{′}(x)` `= lim_(h->0) (f(x + h)-f(x))/h`
    `= lim_(h->0) ((2(x + h)^2 + 5(x + h))-(2x^2 + 5x))/h`
    `= lim_(h->0)(2x^2 + 4xh + 2h^2 + 5x + 5h-2x^2-5x)/h`
    `= lim_(h->0)(4xh + 2h^2 + 5h)/h`
    `= lim_(h->0)(h(4x + 5 + 2h))/h`

 
`:.\ y^{′} = 4x + 5`
 

ii.   `text(When)\ \ x = 1, y = 7`

`y^{′} = 4 + 5 = 9`
 

`:. y-7` `= 9(x-1)`
`y` `= 9x-2`

Trigonometry, 2ADV T2 SM-Bank 31

Given  `cottheta = −24/7`  for  `−pi/2 < theta < pi/2`, find the exact value of

  1.  `sectheta`  (2 marks)

  2.  `sintheta`  (1 mark)

Show Answers Only
  1. `25/24`
  2. `-7/25`
Show Worked Solution

i.   `cot theta = −24/7\ \ =>\ \ tan theta=– 7/24`

`text(Graphically, given)\ −pi/2 < theta < pi/2:`

`x= sqrt(24^2 + 7^2)=25`
 

`sectheta` `= 1/(costheta)`
  `= 1/(24/25)`
  `= 25/24`

 

ii.   `sintheta = −7/25`

Functions, 2ADV F1 SM-Bank 25

Damon owns a swim school and purchased a new pool pump for $3250.

He writes down the value of the pool pump by 8% of the original price each year.

  1.  Construct a function to represent the value of the pool pump after `t` years.   (1 mark)

  2.  Draw the graph of the function and state its domain and range.   (2 marks)

Show Answers Only
  1. `text(Value) = 3259 – 260t`
  2. `text(Domain)\ {t: 0 <= t <= 12.5}`
    `text(Range)\ {y: 0 <= y <= 3250}`
Show Worked Solution
i.   `text(Depreciation each year)` `= 8text(%) xx 3250`
  `= $260`

 
`:.\ text(Value) = 3250 – 260t`
 

ii.   

`text(Find)\ \ t\ \ text(when value = 0)`

`3250 – 260t` `= 0`
`t` `= 3250/260`
  `= 12.5\ text(years)`

 
`text(Domain)\ \ {t: 0 <= t <= 12.5}`

`text(Range)\ \ {y: 0 <= y <= 3250}`

Functions, 2ADV F1 SM-Bank 24

Ita publishes and sells calendars for $25 each. The cost of producing the calendars is $8 each plus a set up cost of $5950.

How many calendars does Ita need to sell to breakeven?  (2 marks)

Show Answers Only

`350`

Show Worked Solution

`text(Let)\ \ x =\ text(number of calendars sold)`

`text(C)text(ost) = 5950 + 8x`

`text(Sales revenue) = 25x`
  

`text(Breakeven occurs when:)`

`25x` `= 5950 + 8x`
`17x` `= 5950`
`:. x` `= 350`

Functions, 2ADV F1 EQ-Bank 22

Worker A picks a bucket of blueberries in `a` hours. Worker B picks a bucket of blueberries in `b` hours.

  1.  Write an algebraic expression for the fraction of a bucket of blueberries that could be picked in one hour if A and B worked together.  (2 marks)

  2.  What does the reciprocal of this fraction represent?  (1 mark)

Show Answers Only
  1. `(a + b)/(ab)`
  2. `text(The reciprocal represents the number of hours it would)`
  3. `text(take to fill one bucket, with A and B working together.)`
Show Worked Solution

i.    `text(In one hour:)`

COMMENT: Note that the question asks for “a fraction”.

`text(Worker A picks)\ 1/a\ text(bucket.)`

`text(Worker B picks)\ 1/b\ text(bucket.)`
 

`:.\ text(Fraction picked in 1 hour working together)`

`= 1/a + 1/b`

`= (a + b)/(ab)`
 

ii.   `text(The reciprocal represents the number of hours it would)`

`text(take to fill one bucket, with A and B working together.)`

Calculus, EXT1 C1 EQ-Bank 12

A tank is initially full. It is drained so that at time `t` seconds the volume of water, `V`, in litres, is given by

`V = 50(1 - t/80)^2`  for  `0 <= t <= 100`

  1.  How much water was initially in the tank?  (1 mark)

  2.  After how many seconds was the tank one-quarter full?  (1 mark)

  3.  At what rate was the water draining out the tank when it was one-quarter full?  (2 marks)

Show Answers Only
  1.  `50\ text(L)`
  2.  `40\ text(seconds)`
  3.  `5/8\ text(litres per second)`
Show Worked Solution

i.   `text(Initially,)\ \ t = 0.`

`V` `= 50(1 – 0)^2`
  `= 50\ text(L)`

 

ii.   `text(Find)\ t\ text(when tank is)\ \ 1/4\ \ text(full:)`

`50/4` `= 50(1 – t/80)^2`
`(1 – 1/80)^2` `= 1/4`
`1 – t/80` `= 1/2`
`t/80` `= 1/2`
`t` `= 40\ text(seconds)`

 

iii.   `(dv)/(dt)` `= 2 · 50(1 – t/80)(−1/80)`
  `= −5/4(1 – t/80)`

 
`text(When)\ \ t = 40,`

`(dv)/(dt)` `= −5/4(1 – 40/80)`
  `= −5/8`

 
`:. text(Water is draining out at)\ \ 5/8\ \ text(litres per second.)`

Trigonometry, EXT1 T2 SM-Bank 4

If  `costheta = −3/4`  and  `0 < theta < pi`,

determine the exact value of  `tantheta`.  (2 marks)

Show Answers Only

`tan theta = – sqrt7/3`

Show Worked Solution

`text(Consider the angle graphically:)`

 
`text(S)text(ince)\ \ costheta\ \ text(is negative  ⇒  2nd quadrant.)`

`text(Using Pythagoras,)`

`x^2` `= 4^2 – 3^2`
`x` `= sqrt7`

 
`:. tan theta = – sqrt7/3`

Trigonometry, EXT1 T2 EQ-Bank 1

Find  `a`  and  `b`  such that

`tan75^@ = a + bsqrt3`   (2 marks)

Show Answers Only

`a = 2, b = 1`

Show Worked Solution
`tan75^@` `= tan(45 + 30)^@`
  `= (tan45^@ + tan30^@)/(1 – tan45^@tan30^@)`
  `= (1 + 1/sqrt3)/(1 – 1 · 1/sqrt3) xx sqrt3/sqrt3`
  `= (sqrt3 + 1)/(sqrt3 – 1) xx (sqrt3 + 1)/(sqrt3 + 1)`
  `= (3 + 2sqrt3 + 1)/((sqrt3)^2 – 1^2)`
  `= 2 + sqrt3`

 
`:. a = 2, \ b = 1`

Functions, EXT1 F1 EQ-Bank 7

The parametric equations of a graph are

`x = 1 - 1/t`

`y = 1 + 1/t`
 

Sketch the graph.  (2 marks)

Show Answers Only

Show Worked Solution
`x` `= 1 – 1/t\ \ …\ (1)`
`tx` `= t – 1`
`t(1 – x)` `= 1`
`t` `= 1/(1 – x)\ \ …\ (1′)`
`y` `= 1 + 1/t\ \ …\ (2)`

 
`text(Substitute)\ \ t = 1/(1 – x)\ \ text{from (1′) into (2)}:`

`y` `= 1 + 1/(1/(1 – x))`
`y` `= 1 + 1 – x`
`y` `= 2 – x`

 

Functions, EXT1 F1 EQ-Bank 6

The parametric equations of a graph are

`x = t^2`

`y = 1/t`  for  `t > 0`
 

  1. Find the Cartesian equation for the graph.  (1 mark)

  2. Sketch the graph.  (1 mark)

Show Answers Only

i.   `y = sqrt(1/x)`

ii.

Show Worked Solution

i.     `x = t^2\ …\ (1)`

`y = 1/t\ …\ (2)`
 

`text(Substitute)\ \ t = 1/y\ \ text{from (1) into (1)}`

`x` `= (1/y)^2`
`y^2` `= 1/x`
`y` `= sqrt(1/x)\ \ \ (y > 0\ \ text(as)\ \ t > 0)`

 

ii.

Functions, EXT1 F1 SM-Bank 4

Sketch the curve described by these two parametric equations

`x = 3cost + 2`

`y = 3sint - 3`   for   `0 <= t < 2pi`.  (3 marks)

Show Answers Only

Show Worked Solution

`(x – 2) = 3cost`

`(y + 3) = 3sint`

`(x – 2)^2 + (y + 3)^2` `= (3cost)^2 + (3sint)^2`
  `= 9cos^2t + 9sin^2t`
  `= 9(cos^2t + sin^2t)`
  `= 9`

 
`text(Sketch:) \ (x – 2)^2 + (y + 3)^2 = 3^2`

Functions, EXT1 F1 SM-Bank 2

Solve  `3/(|\ x - 3\ |) < 3`.  (3 marks)

Show Answers Only

`x < 2\ ∪\ x > 4`

Show Worked Solution

`text(Solution 1)`

`3/(|\ x – 3\ |) < 3`

`|\ x – 3\ |` `> 1`
`(x^2 – 6x + 9)` `> 1^2`
`x^2 – 6x + 8` `> 0`
`(x – 4)(x – 2)` `> 0`

 

 

`:. {x: \ x < 2\ ∪\ x > 4}`

 

`text(Solution 2)`

`|\ x – 3\ | > 1`

`text(If)\ \ (x – 3)` `> 0,\ text(i.e.)\ x >3`
`x – 3` `> 1`
`x` `> 4`

 
`=> x > 4\ (text(satisfies both))`
 

`text(If)\ \ (x – 3)` `< 0,\ text(i.e.)\ x <3`
`−(x – 3)` `> 1`
`−x + 3` `> 1`
`x` `< 2`

 
`=> x < 2\ (text(satisfies both))`

`:. {x: \ x < 2\ ∪\ x > 4}`