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Calculus, MET1 2008 VCAA 1b

Let  `f(x) = xe^(3x)`.  Evaluate  `f^{prime}(0)`.   (3 marks)

Show Answers Only

`1`

Show Worked Solution

`text(Using Product Rule:)`

`(gh)^{prime} = g^{prime}h + gh^{prime}`

`f^{prime}(x)` `= x(3e^(3x)) + 1 xx e^(3x)`
`:.f^{prime}(0)` `= 0 + e^0`
  `= 1`

Calculus, 2ADV C3 SM-Bank 13

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

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

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

 vcaa-2011-meth-10a
 

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

  2. Find the length, `L` cm, of the wire in the frame, including length `BD`, in terms of `a` and `theta`.  (1 mark)

  3. Find  `(dL)/(d theta)`,  and hence show that  `(dL)/(d theta) = 0` when  `BD = 2CD`.  (2 marks)

  4. Find the maximum value of `L` if  `a = 3 sqrt 5`.  (1 mark)

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

i.   `text(In)\ \ Delta BCD,`

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

 

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

 

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

♦ Mean mark (Vic) 35%.

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

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

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

 

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

♦♦♦ Mean mark (Vic) 5%.
`- 2 a sin theta+ a cos theta` `= 0`
`sin theta` `=1/2 cos theta`
`tan theta` `=1/2`

 

 vcaa-2011-meth-10ai

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

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

Calculus, 2ADV C4 SM-Bank 12

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

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

 vcaa-2013-meth-10
 

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

  2. Find the maximum area of triangle `OQP` and the value of `x` for which the maximum occurs.  (3 marks)

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

     

 

      vcaa-2013-meth-10i

Show Answers Only
  1. `x e^(-x/5)`
  2. `5/e\ text(u²)`
     
  3. `25/4 log_e (4) – 15/2\ text(u²)`
Show Worked Solution
i.   `text(Area)` `= 1/2 xx b xx h`
    `= 1/2x(2e^(-x/5))`
  `:. A` `= xe^(-x/5)`

 

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

♦ Mean mark (Vic) 35%.
`x(-1/5 e^(-x/5)) + e^(-x/5)` `= 0`
`e^(-x/5)(1 – x/5)` `= 0`
`:. x` `= 5\ \ \ \ (e^(-x/5) >0,\ \ text(for all)\ x)`

 

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

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

 

iii.   `text(Find)\ \ S:\ F(0) = 2`

`=>S(0, 2)`

♦♦ Mean mark (Vic) 32%.

 

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

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

 

vcaa-2013-meth-10ii

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

GRAPHS, FUR2 2017 VCAA 3

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

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

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

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

 Constraint 1  `y >= x/4`
 

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

 
There are three other constraints.

 Constraint 2  `x ≥ 6`

 Constraint 3  `y ≥ 4`

 Constraint 4  `x + y ≥ 12`

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

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

All lifeguards receive a meal allowance per day.

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

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

`C = 15x + 25y`

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

     

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

     

    • Constraint 1 and Constraint 4 are to remain.
     

     

              Constraint 1  `y >= x/4`

     

              Constraint 4  `x + y >= 12`

     


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

     

    `C = 15x + 25y`

     


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

     

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


Show Answers Only

a.   `2`

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

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

c.   `$220`

d.   
Show Worked Solution

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

 

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

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

 

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

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

 

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

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

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

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

`text(⇒ Minimum allowance when)`

GRAPHS, FUR2 2017 VCAA 2

A swimming race is being held in the ocean.

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

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

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

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

     

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

     

    `E = kt,`  where  `0 < t ≤ 12.5`

     

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

     

    Show that `k = 40`.  (1 mark)

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

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

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

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

     

    Sketch the relation representing Zlatko’s swim on the graph below.  (2 marks)

  2. How many minutes after the start of the race were Zlatko and Edgar the same distance from the shore?

     

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

Show Answers Only

a.   `600\ text(metres)`

b.   `text(See Worked Solutions)`

c.   

d.   `10.98\ text(minutes)`

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

 

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

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

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

 

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

♦ Mean mark 47%

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

`text(Zlatko returns to shore when)`

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

 

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

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

GEOMETRY, FUR2 2017 VCAA 3

Some hostel buildings are arranged around a grassed area.

The grassed area is shown shaded in the diagram below.

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

The square has side lengths of 65 m.

The circle has a radius of 50 m.

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

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

     

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

  3. The hostel’s management is planning to build a pathway from point A to point B, as shown on the diagram below.
     

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

     

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

Show Answers Only

a.   `81^@`

b.   `438\ text(m (nearest metre))`

c.   `153\ text(m)`

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

 

♦♦ Mean mark part (b) 34%.

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

 

♦♦ Mean mark part (c) 34%.

c.   

`text(Using Pythagoras,)`

`x = sqrt(50^2 – 32.5^2) = 37.99…`

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

GEOMETRY, FUR2 2017 VCAA 2

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

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

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

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

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

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

    1. What distance will Miki have to walk if she were to walk in a straight line from the Nemuro railway station to the hostel?

       

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

    2. What is the three-figure bearing of the hostel from the Nemuro railway station?

       

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


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

Assume that the radius of Earth is 6400 km.

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

     

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

  2. Find the shortest great circle distance between Melbourne (38° S, 145° E) and Nemuro (43° N, 145° E).

     

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

Show Answers Only
  1. `8\ text(pm (Wed))`
  2. i. `193\ text(m  (nearest metre))`
  3. ii. `345^@`
  4. `4681\ text(km  (nearest km))`
  5. `9048\ text(km  (nearest km))`
Show Worked Solution

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

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

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

 

b.i.  `text(Using Pythagoras:)`

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

 

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

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

 

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

`= 360 – 15`

`= 345^@`

 

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

♦ Mean mark part (c) 43%

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

 

d.   

`text(Shortest distance)`

♦ Mean mark part (d) 38%

`= text(Arc length)\ NM`

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

`= 9047.78…`

`= 9048\ text(km  (nearest km))`

NETWORKS, FUR2 2017 VCAA 4

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

This project involves activities A to O.

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

  1. Write down the two immediate predecessors of activity I.   (1 mark)

  2. The minimum completion time for the project is 19 days.

     

     i.  There are two critical paths. One of the critical paths is A–E–J–L–N.
    Write down the other critical path.   (1 mark)

    ii.  Determine the float time, in days, for activity F.   (1 mark)

  3. The project could finish earlier if some activities were crashed.

     

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

     

    The cost of this crashing is $1000 per activity.

     

     i.  What is the minimum number of days in which the project could now be completed?   (1 mark)

    ii.  What is the minimum cost of completing the project in this time?   (1 mark)

Show Answers Only

a.     `D\ text(and)\ E`

b.i.   `A E I L N`

b.ii.   `6\ text(days)`

c.i.    `17`

c.ii.    `$4000`

Show Worked Solution

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

b.i.   `A  E  I  L  N`

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

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

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

♦♦ Mean mark part (c)(i) 35%.

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

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

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

`text(⇒ Minimum Days = 17)`
  

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

♦♦♦ Mean mark part (c)(ii) 15%.

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

NETWORKS, FUR2 2017 VCAA 3

While on holiday, four friends visit a theme park where there are nine rides.

On the graph below, the positions of the rides are indicated by the vertices.

The numbers on the edges represent the distances, in metres, between rides.
 

  1. Electrical cables are required to power the rides.
    These cables will form a connected graph.
    The shortest total length of cable will be used.
  2. i. Give a mathematical term to describe a graph that represents these cables.   (1 mark)

  3. ii. Draw in the graph that represents these cables on the diagram below.   (1 mark)

     

Show Answers Only

a.i.   `text{Minimum spanning tree (shortest length required).}`

a.ii.   
Show Worked Solution

a.i.   `text{Minimal spanning tree (shortest length required).}`

a.ii.   

NETWORKS, FUR2 2017 VCAA 2

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

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

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

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

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

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


 

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

  2. Determine the minimum total planning time, in minutes, for all four tours.   (1 mark)

Show Answers Only

a.   `text(Colin must plan tour 2.)`

b.   `43\ text(minutes)`

Show Worked Solution

a.   `text(Colin must plan tour 2.)`

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

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

 

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

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

MATRICES, FUR2 2017 VCAA 3

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

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

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

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

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

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

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

  1. How many senior students will not change their elective activity from Term 1 to Term 2?   (1 mark)

  2. Complete `S_2`, the state matrix for Term 2, below.   (1 mark)


             

  3. Of the senior students expected to choose investigation (`I`) in Term 3, what percentage chose service (`S`) in Term 2?   (2 marks)

  4. What is the maximum number of senior students expected in investigation (`I`) at any time during 2018?   (1 mark)

Show Answers Only
  1. `320`
  2.  
    `S_2 = [(250),(250),(300),(200)]`
  3. `text(25%)`
  4. `250`
Show Worked Solution

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

♦ Mean mark 47%.

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

`= 120 + 80 + 60 + 60`

`= 320`

 

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

 

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

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

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

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

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

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

 

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

♦ Mean mark 43%.

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

`(text(During term 2))`

MATRICES, FUR2 2017 VCAA 2

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

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

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

 

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

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

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

  1. 306 junior students are expected to choose sport (`S`) in Term 2.
     
    Complete the calculation below to show this.   (1 mark)


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

Show Answers Only
  1. `text(40%)`
  2.  `300 xx 0.4 + 240 xx 0.6 + 210 xx 0.2 = 306`
  3. `750`
Show Worked Solution

a.   `text(40%)`
 

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

♦♦ Mean mark part (c) 30%.

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


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

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

MATRICES, FUR2 2017 VCAA 1

A school canteen sells pies (`P`), rolls (`R`) and sandwiches (`S`).

The number of each item sold over three school weeks is shown in matrix `M`.

`{:(qquadqquadqquadquadPqquadRqquadS),(M = [(35,24,60),(28,32,43),(32,30,56)]{:(text(week 1)),(text(week 2)),(text(week 3)):}):}` 

  1. In total, how many sandwiches were sold in these three weeks?   (1 mark)

  2. The element in row `i` and column `j` of matrix `M` is `m_(ij)`.
  3. What does the element `m_12` indicate?  (1 mark)

  4. Consider the matrix equation

    `[(35,24,60),(28,32,43),(32,30,56)] xx [(a),(b),(c)] = [(491.55),(428.00),(487.60)]`

    where `a` = cost of one pie, `b` = cost of one roll and `c` = cost of one sandwich.
  5.  i. What is the cost of one sandwich?   (1 mark)

The matrix equation below shows that the total value of all rolls and sandwiches sold in these three weeks is $915.60

`L xx [(491.55),(428.00),(487.60)] = [915.60]`

Matrix `L` in this equation is of order `1 × 3`.

  1. ii. Write down matrix `L`.   (1 mark)

Show Answers Only
  1. `159`
  2. `text(It represents the number of rolls sold in week 1.)`
    1. `$3.80`
    2. `text(Matrix)\ L = [0,1,1]`
Show Worked Solution
a.    `text(Total sandwiches)` `= 60 + 43 + 56`
    `= 159`

 
b.   `m_12 = 24`

`text(It represents the number of rolls sold in week 1.)`
 

c.i.    `[(a),(b),(c)]` `= [(35,24,60),(28,32,43),(32,30,56)]^(−1)[(491.55),(428.00),(487.60)]`
    `= [(4.65),(4.20),(3.80)]`

 
`:.\ text(C)text(ost of 1 sandwich = $3.80)`
 

c.ii.   `text(Matrix)\ L = [0,1,1]`

CORE, FUR2 2017 VCAA 5

Alex is a mobile mechanic.

He uses a van to travel to his customers to repair their cars.

The value of Alex’s van is depreciated using the flat rate method of depreciation.

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

`V_0 = 75\ 000 qquad V_(n + 1) = V_n - 3375`

  1. Recursion can be used to calculate the value of the van after two years.

     

    Complete the calculations below by writing the appropriate numbers in the boxes provided.   (2 marks)


    1. By how many dollars is the value of the van depreciated each year?   (1 mark)
    2. Calculate the annual flat rate of depreciation in the value of the van.
    3. Write your answer as a percentage.   (1 mark)
  3. The value of Alex’s van could also be depreciated using the reducing balance method of depreciation.
  4. The value of the van, in dollars, after `n` years, `R_n`, can be modelled by the recurrence relation shown below.

     

            `R_0 = 75\ 000 qquad R_(n + 1) = 0.943R_n`

    At what annual percentage rate is the value of the van depreciated each year?   (1 mark)

Show Answers Only

a.

b.i.  `$3375`

b.ii. `4.5text(%)`

c.  `5.7text(%)`

Show Worked Solution

a.   

  
b.i.   `$3375`
  

b.ii.    `text(Annual Rate)` `= 3375/(75\ 000) xx 100`
    `= 4.5text(%)`

 

c.    `text(Annual Rate)` `= (1-0.943) xx 100text(%)`
    `= 5.7text(%)`

CORE, FUR2 2017 VCAA 6

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

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

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

The interest on this amount will compound monthly.

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

     

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

     

    How much did Marcus pay?   (1 mark)

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

Lily did not pay the bill by the due date.

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

  1. Write down a recurrence relation, in terms of `A_0`, `A_(n + 1)` and `A_n`, that models the amount of the bill.   (2 marks)

  2. Lily paid the full amount of her bill four months after the due date.

     

    How much interest was Lily charged?

     

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

Show Answers Only
  1. `$203`
  2. `A_o = 428,qquadA_(n + 1) = 1.015A_n`
  3. `$26.26\ \ (text(nearest cent))`
Show Worked Solution
a.    `text(Amount paid)` `= 200 + 200 xx 1.5text(%)`
    `= 1.015 xx 200`
    `= $203`

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

 

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

 

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

♦♦ Mean mark part (c) 29%.

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

CORE, FUR2 2017 VCAA 4

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

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

The data used to generate this plot is also given.
 

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

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

  1. When the equation of the least squares line that can be used to predict log10 (area) from year is determined, the slope of this line is approximately 0.0085385
  2. Round this value to three significant figures.   (1 mark)
  3. Perform the log10 transformation to the variable area and determine the equation of the least squares line that can be used to predict log10 (area) from year.
  4. Write the values of the intercept and slope of this least squares line in the appropriate boxes provided below.
  5. Round your answers to three significant figures.  (2 marks)

The least squares line predicts that the log10 (area) of forest eaten by the caterpillars by the year 2020 will be approximately 2.85

  1. Using this value of 2.85, calculate the expected area of forest that will be eaten by the caterpillars by the year 2020.
  2.  i. Round your answer to the nearest hectare.   (1 mark)

  3. ii. Give a reason why this prediction may have limited reliability.   (1 mark)

Show Answers Only

a.  `0.00854\ (text(3 sig fig))`

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

c.i.  `708\ text(hectares)`

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

Show Worked Solution

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

♦ Mean marks of part (a) and (b) 44%.

 

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

 

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

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

 

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

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

Calculus, 2ADV C3 SM-Bank 7

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


 

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

  2. For `x` in the interval `(0, 1)`, show that the gradient of the tangent to the graph of  `f(x)`  is  `(1 - 3x)/(2 sqrt x)`.  (1 mark)

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

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


 

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

Show Answers Only
  1. `4/15\ text(units)^2`
  2. `text(Proof)\ \ text{(See Workes Solutions)}`
  3. `y = -x + 1`
Show Worked Solution
i.   `text(Area)` `= int_0^1 (sqrt x – x sqrt x)\ dx`
    `= int_0^1 (x^(1/2) – x^(3/2))\ dx`
    `= [2/3 x^(3/2) – 2/5 x^(5/2)]_0^1`
    `= (2/3 – 2/5) – (0 – 0)`
    `= 10/15 – 6/15`
    `= 4/15\ text(units)^2`

 

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

 

iii.  `m_(AC) = tan 45^@=1`

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

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

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

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

 
`text(Let)\ \ a=sqrtx,`

`3a^2-2a-1` `=0`
`(3a+1)(a-1)` `=0`
`a=1 or -1/3`   
`:. sqrt x` `=1` `or`   `sqrt x=- 1/3\ \ text{(no solution)}`
`x` `=1`    

 
`f(1)=sqrt1(1-1)=0\ \ =>B(1,0)`
 

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

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

Trigonometry, 2ADV T2 SM-Bank 8

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

  1. State all possible values of  `tan theta`.  (1 mark)
  2. Hence, find all possible solutions for  `(tan theta - 1) (sin^2 theta - 3 cos^2 theta) = 0`,
    where  `0 <= theta <= pi`.  (2 marks)
Show Answers Only
  1. `tan theta = 1 or tan theta = +- sqrt 3`
  2. `theta = pi/4, pi/3 or (2 pi)/3`
Show Worked Solution

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

`=> tan theta = 1`

♦ Mean mark (Vic) 42%.
`=>sin theta – sqrt 3 cos theta` `=0`
`sin theta` `=sqrt3 cos theta`
`tan theta` `=sqrt3`

 

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

 

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

 

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

`text{Using part (i):}`

♦ Mean mark (Vic) 42%.

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

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

 

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

CORE, FUR2 2017 VCAA 3

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

  1. Determine the equation of the least squares line that can be used to predict the egg density in the forest from the number of male moths caught in the trap.
  2. Write the values of the intercept and slope of this least squares line in the appropriate boxes provided below.
  3. Round your answers to one decimal place.  (2 marks)

     

         

  4. The number of female moths caught in a trap set in a forest and the egg density (eggs per square metre) in the forest can also be examined.

     

    A scatterplot of the data is shown below.
     


     
    The equation of the least squares line is

     

                  egg density = 191 + 31.3 × number of female moths

    1. Draw the graph of this least squares line on the scatterplot (provided above).   (1 mark)

    2. Interpret the slope of the regression line in terms of the variables egg density and number of female moths caught in the trap.   (1 mark)

    3. The egg density is 1500 when the number of female moths caught is 55.
    4. Determine the residual value if the least squares line is used to predict the egg density for this number of female moths.   (1 mark)
    5. The correlation coefficient is  `r = 0.862`
    6. Determine the percentage of the variation in egg density in the forest explained by the variation in the number of female moths caught in the trap.
    7. Round your answer to one decimal place.   (1 mark)
Show Answers Only

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

b.i.  

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

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

`text(caught in the trap.)`

b.iii.  `−412.5`

b.iv.  `text(74.3%)`

Show Worked Solution

a.   `text(By calculator)`

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

 

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

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

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

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

 

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

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

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

 `text(caught in the trap.)`

 

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

♦ Mean mark 48%.

`= 191 + 31.3 xx 55`

`= 1912.5`

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

 

b.iv.   `r = 0.862`

♦ Mean mark 47%.

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

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

Algebra, MET2 2017 VCAA 4

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

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

     

    State the values of `c` and `d`.   (2 marks)

  2. Find the rule and domain for  `f^(-1)`, the inverse function of  `f`.   (2 marks)

  3. Find the area bounded by the graphs of  `f` and  `f^(-1)`.   (3 marks)

  4. Part of the graphs of  `f` and  `f^(-1)` are shown below.
     

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

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

  1. Find the value of `k` such that  `g_k(x) = f(x)`.  (1 mark)

  2. Find the rule for the inverse functions  `g_k^(-1)` of  `g_k`, where  `k ∈ R^+`.   (1 mark)

  3. i. Describe the transformation that maps the graph of  `g_1` onto the graph of  `g_k`.   (1 mark)

    ii. Describe the transformation that maps the graph of  `g_1^(-1)` onto the graph of  `g_k^(-1)`.   (1 mark)

  4. The lines `L_1` and `L_2` are the tangents at the origin to the graphs of  `g_k`  and  `g_k^(-1)`  respectively.
  5. Find the value(s) of `k` for which the angle between `L_1` and `L_2` is 30°.   (2 marks)

  6. Let `p` be the value of `k` for which  `g_k(x) = g_k^(−1)(x)`  has only one solution.
  7.  i. Find `p`.   (2 marks)

  8. ii. Let  `A(k)`  be the area bounded by the graphs of  `g_k`  and  `g_k^(-1)`  for all  `k > p`.
  9.     State the smallest value of `b` such that  `A(k) < b`.   (1 mark)

Show Answers Only

a.  `c =-1, \ d =-2`

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

c.  `3-2/(log_e(2))\ text(units)²`

d. `f^{prime}(0)= 2log_e(2) and f^(-1)^{prime}(0)= 1/(2log_e(2))`

e.  `k = log_e(2)`

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

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

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

h.  `k=sqrt3/6\ or\ sqrt3/2`

i.i.  `p = 1/2`

i.ii.  `b=4`

Show Worked Solution

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

`x^{prime}` `=x+c\ \ => x=x^{prime}-c`
`y^{prime}` `=y+d\ \ =>y= y^{prime}-d`
   
`y^{prime}-d` `=2^((x)^{prime}-c)`
`y^{prime}` `= 2^((x)^{prime}-c)+d`

 

`:. c = -1, \ d = -2`

 

b.   `text(Let)\ \ y = f(x)`

`text(Inverse : swap)\ x\ ↔ \ y`

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

 

`text(dom)(f^(-1)) = text(ran)(f)`

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

 

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

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

`x` `= -1, 0`

 

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

 

d.    `f^{prime}(0)` `= 2log_e(2)`
  `f^{(-1)^prime}(0)` `= 1/(2log_e(2))`

 

e.   `g_k(x) = 2e^(kx)-2`

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

`:. k = log_e(2)`

 

f.   `text(Let)\ \ y = g_k(x)`

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

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

 

♦♦ Mean mark part (g)(i) 31%.

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

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

 

♦♦ Mean mark part (g)(ii) 30%.

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

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

 

h.   `text(When)\ \ x=0,`

♦♦♦ Mean mark part (h) 13%.

`g_k^{prime}(0)=2k\ \ => m_(L_1)=2k`

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

 

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

`text(Solve:)\ \ 1/sqrt3` `=+- ((2k-1/(2k)))/2\ \ text(for)\ k`

 
`:. k=sqrt3/6\ or\ sqrt3/2`

 

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

♦♦♦ Mean mark part (i)(i) 4%.

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

 
`:. p = 1/2`

 

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

♦♦♦ Mean mark part (i)(ii) 2%.

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

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

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

 

`:. lim_(k→oo) A(k) = 4`

`:.b=4`

Probability, MET2 2017 VCAA 3

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

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

 

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

 

  1. Sketch the graph of `f` on the axes provided below.  (3 marks)


     

       
     

  2. Find  `text(Pr)(25 ≤ T ≤ 55)`.  (2 marks)

  3. Find  `text(Pr)(T ≤ 25 | T ≤ 55)`.  (2 marks)

  4. Find `a` such that  `text(Pr)(T ≥ a) = 0.7`, correct to four decimal places.  (2 marks)

  5. The probability that Jennifer spends more than 50 minutes on her homework on any given day is `8/25`. Assume that the amount of time spent on her homework on any day is independent of the time spent on her homework on any other day.

     

    1. Find the probability that Jennifer spends more than 50 minutes on her homework on more than three of seven randomly chosen days, correct to four decimal places.  (2 marks)

    2. Find the probability that Jennifer spends more than 50 minutes on her homework on at least two of seven randomly chosen days, given that she spends more than 50 minutes on her homework on at least one of those days, correct to four decimal places.  (2 marks)

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

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

  1. Express `q` as a polynomial in terms of `p`.  (2 marks)

    1. Find the maximum value of `q`, correct to four decimal places, and the value of `p` for which this maximum occurs, correct to four decimal places.  (2 marks)

    2. Find the value of `d` for which the maximum found in part g.i. occurs, correct to the nearest minute.  (2 marks)


Show Answers Only

  1.  

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

Show Worked Solution

a.   

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

 

b.   `text(Pr)(25 <= T <= 55)`

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

`= 4/5`

 

c.   `text(Pr)(T ≤ 25 | T ≤ 55)`

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

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

`= (1/50)/(1 – 9/50)`

`= 1/41`

 

d.   `text(Pr)(T ≥ a) = 0.7`

♦ Mean mark part (d) 36%.

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

`text(Solve:)`

`int_20^a 1/625(t – 20)\ dt` `= 0.3quadtext(for)quada ∈ (20, 45)`

 

`:. a == 39.3649`

 

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

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

`text(Pr)(X >= 4) = 0.1534`

 

e.ii.    `text(Pr)(X >= 2 | X >= 1)` `= (text(Pr)(X >= 2))/(text(Pr)(X >= 1))`
    `= (0.7113…)/(0.9327…)`
    `= 0.7626\ \ text{(to 4 d.p.)}`

 

f.   `text(Let)\ Y =\ text(Number of days Jenn spends more than)\ d\ text(min)`

`Y ~\ text(Bi)(7,p)`

♦ Mean mark part (f) 36%.

`q` `= text(Pr)(Y = 2) + text(Pr)(Y = 3)`
  `= ((7),(2))p^2(1 – p)^5 + ((7),(3))p^3(1 – p)^4`
  `= 21p^2(1 – p)^5 + 35p^3(1 – p)^4`
   `=7p^2(1-p)^4[3(1-p)+5p]`
  `=7p^2(1-p)^4(2p+3)`

 

g.i.   `text(Solve)\ \ q′(p) = 0,`

♦♦ Mean mark part (g)(i) 30%.

`p` `=0.35388…`
  `=0.3539\ \ text{(to 4 d.p.)}`

`:. q_text(max)= 0.5665\ \ text{(to 4 d.p.)}`

 

g.ii.   `text(Pr)(T > d) = p= 0.35388…`

♦♦♦ Mean mark part (g)(ii) 8%.

  `text(Solve:)`

`int_d^70 (1/625(70 – t))dt` `= 0.35388… quadtext(for)quadd ∈ (45,70)`

 

`:. d` `=48.967…`
  `=49\ text(mins)`

Algebra, MET2 2017 VCAA 2

Sammy visits a giant Ferris wheel. Sammy enters a capsule on the Ferris wheel from a platform above the ground. The Ferris wheel is rotating anticlockwise. The capsule is attached to the Ferris wheel at point `P`. The height of `P` above the ground, `h`, is modelled by  `h(t) = 65-55cos((pit)/15)`, where `t` is the time in minutes after Sammy enters the capsule and `h` is measured in metres.

Sammy exits the capsule after one complete rotation of the Ferris wheel.
 


 

  1. State the minimum and maximum heights of `P` above the ground.   (1 mark)

  2. For how much time is Sammy in the capsule?   (1 mark)

  3. Find the rate of change of `h` with respect to `t` and, hence, state the value of `t` at which the rate of change of `h` is at its maximum.   (2 marks)

As the Ferris wheel rotates, a stationary boat at `B`, on a nearby river, first becomes visible at point `P_1`. `B` is 500 m horizontally from the vertical axis through the centre `C` of the Ferris wheel and angle `CBO = theta`, as shown below.
 

   
 

  1. Find `theta` in degrees, correct to two decimal places.   (1 mark)

Part of the path of `P` is given by  `y = sqrt(3025-x^2) + 65, x ∈ [-55,55]`, where `x` and `y` are in metres.

  1. Find `(dy)/(dx)`.   (1 mark)

As the Ferris wheel continues to rotate, the boat at `B` is no longer visible from the point  `P_2(u, v)` onwards. The line through `B` and `P_2` is tangent to the path of `P`, where angle `OBP_2 = alpha`.
 

   
 

  1. Find the gradient of the line segment `P_2B` in terms of `u` and, hence, find the coordinates of `P_2`, correct to two decimal places.   (3 marks)

  2. Find `alpha` in degrees, correct to two decimal places.   (1 mark)

  3. Hence or otherwise, find the length of time, to the nearest minute, during which the boat at `B` is visible.   (2 marks)

Show Answers Only
  1. `h_text(min) = 10\ text(m), h_text(max) = 120\ text(m)`
  2. `30\ text(min)`
  3. `t = 7.5`
  4. `7.41^@`
  5. `(-x)/(sqrt(3025-x^2))`
  6. `P_2(13.00, 118.44)`
  7. `13.67^@`
  8. `7\ text(min)`
Show Worked Solution
a.    `h_text(min)` `= 65-55` `h_text(max)` `= 65 + 55`
    `= 10\ text(m)`   `= 120\ text(m)`

 

b.   `text(Period) = (2pi)/(pi/15) = 30\ text(min)`

 

c.   `h^{prime}(t) = (11pi)/3\ sin(pi/15 t)`

♦ Mean mark 50%.
MARKER’S COMMENT: A number of commons errors here – 2 answers given, calc not in radian mode, etc …

 

`text(Solve)\ h^{primeprime}(t) = 0, t ∈ (0,30)`

`t = 15/2\ \ text{(max)}`   `text(or)`   `t = 45/2\ \ text{(min – descending)}`

`:. t = 7.5`

 

d.   

♦ Mean mark 36%.
MARKER’S COMMENT: Choosing degrees vs radians in the correct context was critical here.

`tan(theta)` `= 65/500`
`:. theta` `=7.406…`
  `= 7.41^@`

 

e.    `(dy)/(dx)` `= (-x)/(sqrt(3025-x^2))`

 

f.   

`P_2(u,sqrt(3025-u^2) + 65),\ \ B(500,0)`

`:. m_(P_2B)` `= (sqrt(3025-u^2) + 65)/(u-500)`

 

`text{Using part (e), when}\ \ x=u,`

♦♦♦ Mean mark part (f) 18%.
MARKER’S COMMENT: Many students were unable to use the rise over run information to calculate the second gradient.

`dy/dx=(-u)/(sqrt(3025-u^2))`

 

`text{Solve (by CAS):}`

`(sqrt(3025-u^2) + 65)/(u-500)` `= (-u)/(sqrt(3025-u^2))\ \ text(for)\ u`

 

`u=12.9975…=13.00\ \ text{(2 d.p.)}`

 

`:. v` `= sqrt(3025-(12.9975…)^2) + 65`
  `= 118.4421…`
  `= 118.44\ \ text{(2 d.p.)}`

 

`:.P_2(13.00, 118.44)`

 

♦♦♦ Mean mark part (g) 7%.

g.    `tan alpha` `=v/(500-u)`
    `= (118.442…)/(500-12.9975…)`
  `:. alpha` `= 13.67^@\ \ text{(2 d.p.)}`

 

h.   

♦♦♦ Mean mark 5%.

`text(Find the rotation between)\ P_1 and P_2:`

`text(Rotation to)\ P_1 = 90-7.41=82.59^@`

`text(Rotation to)\ P_2 = 180-13.67=166.33^@`

`text(Rotation)\ \ P_1 → P_2 = 166.33-82.59 = 83.74^@`

 

`:.\ text(Time visible)` `= 83.74/360 xx 30\ text(min)`
  `=6.978…`
  `= 7\ text{min  (nearest degree)}`

Calculus, MET2 2017 VCAA 1

Let  `f : R → R,\  f (x) = x^3-5x`. Part of the graph of `f` is shown below.
 

  1. Find the coordinates of the turning points.   (2 marks)

  2. `A(−1, f (−1))`  and  `B(1, f (1))`  are two points on the graph of `f`.

     

    1. Find the equation of the straight line through `A` and `B`.   (2 marks)

    2. Find the distance `AB`.   (1 mark)

Let  `g : R → R, \ g(x) = x^3-kx, \ k ∈ R^+`.

  1. Let  `C(–1, g(−1))` and `D(1, g(1))` be two points on the graph of `g`.

     

    1. Find the distance `CD` in terms of `k`.   (2 marks)

    2. Find the values of `k` such that the distance `CD` is equal to  `k + 1`.   (1 mark)

  2. The diagram below shows part of the graphs of `g` and  `y = x`. These graphs intersect at the points with the coordinates `(0, 0)` and `(a, a)`.
  3.  
       
  4.  
    1. Find the value of `a` in terms of `k`.   (1 mark)

    2. Find the area of the shaded region in terms of `k`.   (2 marks)

Show Answers Only
  1. `(- sqrt15/3,(10sqrt15)/9)\ text(and)\ (sqrt15/3,-(10sqrt15)/9)`
    1. `y =-4x`
    2. `2sqrt17`
    1. `2sqrt(k^2-2k + 2)`
    2. `k = 1quadtext(or)quadk = 7/3`
    1. `sqrt(k + 1)`
    2. `((k + 1)^2)/4\ text(units)²`
Show Worked Solution
a.   
`text(Solve)\ \ f^{^{′}}(x)` `= 0\ \ text(for)\ x:`
`x` `= ± sqrt15/3`

 
`f(sqrt15/3) = -(10sqrt15)/9`

`f(−sqrt15/3) = (10sqrt15)/9`

`:.\ text(Turning points:)`

`(- sqrt15/3,(10sqrt15)/9)\ text(and)\ (sqrt15/3,-(10sqrt15)/9)`
 

b.i.   `A(-1,4),\ \ B(1,–4)`

`m_(AB) = (4-(−4))/(−1-(1)) = −4`

`text(Equation of)\ AB, m=-4, text(through)\ \ (-1,4):`

` y-4` `= −4(x-(−1))`
`:. y` `= −4x`

 

MARKER’S COMMENT: Students skilled in the use of technology will be much more efficient and minimise errors here.
b.ii.    `d_(text(AB))` `=sqrt((x_2-x_1)^2+(f(x_2)-f(x_1))^2)`
    `= sqrt((1- (- 1))^2 + (f(1)-f(−1))^2)`
    `= 2sqrt17`

 

c.i.    `d_(text(CD))` `=sqrt((x_2-x_1)^2+(g(x_2)-g(x_1))^2)`
    `=sqrt((1-(-1))^2+(g(1)-g(-1))^2)`
    `= 2sqrt(k^2-2k + 2)`

 

c.ii.   `text(Solve:)quad2sqrt(k^2-2k + 2) = k + 1quadtext(for)quadk > 0`

`:. k = 1quadtext(or)quadk = 7/3`
 

d.i.   `text(Solve:)quada^3-ka = a\ \ \ text(for)quad a,\ \ (a > 0)`

`:. a = sqrt(k + 1)`
 

d.ii.    `text(Area)` `= int_0^(sqrt(k + 1))(x-g(x))\ dx`
    `= ((k + 1)^2)/4\ text(units)²`

Probability, MET2 2017 VCAA 19 MC

A probability density function  `f` is given by

`f(x) = {{:(cos(x) + 1),(0):}qquad{:(k < x < (k + 1)),(text(elsewhere)):}:}`

where `0 < k < 2`.

The value of `k` is

  1. `1`
  2. `(3pi - 1)/2`
  3. `pi - 1`
  4. `(pi - 1)/2`
  5. `pi/2`
Show Answers Only

`D`

Show Worked Solution
`int_k^(k+1)(cos(x) + 1)\ dx` `=[sinx +x]_k^(k+1)`
`sin(k+1)+(k+1) -sink – k` `=1`
`sin(k+1)-sink` `=0`
`sin(k+1)-sin(pi-k)` `=0`
   
`k+1` `=pi-k`
`2k` `=pi-1`
`:.k` `=(pi-1)/2`

`=> D`

Calculus, MET2 2017 VCAA 15 MC

A rectangle `ABCD` has vertices `A(0, 0)`, `B(u, 0)`, `C(u, v)` and `D(0, v)`, where `(u, v)` lies on the graph of  `y = -x^3 + 8`, as shown below.
 

 

The maximum area of the rectangle is

  1. `root3(2)`
  2. `6 root3(2)`
  3. `16`
  4. `8`
  5. `3root3(2)`
Show Answers Only

`B`

Show Worked Solution

`A` `=u(-u^3+8)`
  `=8u-u^4`
`(dA)/(du)` `=8-4u^3`
   

`(dA)/(du)=0\ \ text(when):`

`4u^3` `=8`
`u^3` `=2`
`u` `=root3(2)`

 

`:.A_text(max)=8root3(2)-(root3(2))^4 = 6 root3(2)`

`=> B`

Probability, MET2 2017 VCAA 14 MC

The random variable `X` has the following probability distribution, where  `0 < p < 1/3`.
 

 
The variance of `X` is

  1. `2p(1 - 3p)`
  2. `1 - 4p`
  3. `(1 - 3p)^2`
  4. `6p - 16p^2`
  5. `p(5 - 9p)`
Show Answers Only

`D`

Show Worked Solution
`text(Var)(X)` `= text(E)(X^2) – [text(E)(X)]^2`
  `= [(−1)^2p + 0^2 xx 2p + 1^2(1 – 3p)] – [−p + 0 + 1 – 3p]^2`
  `= 6p – 16p^2`

`=> D`

Calculus, MET2 2017 VCAA 11 MC

The function  `f : R → R, \ f (x) = x^3 + ax^2 + bx`  has a local maximum at  `x = –1`  and a local minimum at  `x = 3`.

The values of `a` and `b` are respectively

  1. `–2\ text(and)\ \ –3`
  2. `text(2 and 1)`
  3. `text(3 and)\ \ –9`
  4. `–3\ text(and)\ \ –9`
  5. `– 6\ text(and)\ \ –15`
Show Answers Only

`D`

Show Worked Solution

`f′(x) = 3x^2 +2ax +b`

`text(Solve:)quadquadf′(−1)` `=3-2a+b= 0qquadtext(and)`
 `f′(3)` `=27+6a+b = 0qquadtext(for)\ a, b` 

 

`:. a = −3, \ b = − 9`

`=> D`

Algebra, MET2 2017 VCAA 8 MC

If  `y = a^(b - 4x) + 2`, where  `a > 0`, then `x` is equal to

  1. `1/4(b - log_a(y - 2))`
  2. `1/4(b - log_a(y + 2))`
  3. `b - log_a(1/4(y + 2))`
  4. `b/4 - log_a(y - 2)`
  5. `1/4(b + 2 - log_a(y))`
Show Answers Only

`A`

Show Worked Solution

`text(Solving for)\ x:`

`y – 2` `= a^(b – 4x)`
`b – 4x` `= log_a(y – 2)`
`4x` `= b – log_a(y – 2)`
`:. x` `= 1/4(b – log_a(y – 2))`

 
`=> A`

Calculus, MET1 2017 VCAA 9

The graph of  `f: [0, 1] -> R,\ f(x) = sqrt x (1-x)`  is shown below.
 

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

  2. For `x` in the interval `(0, 1)`, show that the gradient of the tangent to the graph of `f` is  `(1-3x)/(2 sqrt x)`.   (1 mark)

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

Let `theta` be the angle that `AC` makes with the positive direction of the horizontal axis, where  `45^@ <= theta < 90^@`.

  1. Find the equation of the line through `B` and `C` in the form  `y = mx + c`, for  `theta = 45^@`.   (2 marks)

  2. Find the coordinates of `C` when  `theta = 45^@`.   (4 marks)

Show Answers Only
  1. `4/15\ text(units)^2`
  2. `text(Proof)\ \ text{(See Workes Solutions)}`
  3. `y = -x + 1`
  4. `C (11/27, 16/27)`
Show Worked Solution
a.   `text(Area)` `= int_0^1 (sqrt x-x sqrt x)\ dx`
    `= int_0^1 (x^(1/2)-x^(3/2))\ dx`
    `= [2/3 x^(3/2)-2/5 x^(5/2)]_0^1`
    `= (2/3-2/5)-(0-0)`
    `= 10/15-6/15`
    `= 4/15\ text(units)^2`

 

♦♦ Mean mark part (b) 35%.
MARKER’S COMMENT: Establishing the common denominator in the working was required!
b.   `f (x)` `= x^(1/2)-x^(3/2)`
  `f^{′}(x)` `= 1/2 x^(-1/2)-3/2 x^(1/2)`
    `= 1/(2 sqrt x)-(3 sqrt x)/2`
    `= (1-3x)/(2 sqrt x)\ \ text(.. as required.)`

 

c.  `m_(AC) = tan 45^@=1`

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

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

 

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

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

 

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

`3a^2-2a-1` `=0`
`(3a+1)(a-1)` `=0`
`a=1 or -1/3`   
`:. sqrt x` `=1` `or`   `sqrt x=- 1/3\ \ text{(no solution)}`
`x` `=1`    

 
`f(1)=sqrt1(1-1)=0\ \ =>B(1,0)`
 

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

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

 

d.  `text(Find Equation)\ AC:`

♦♦♦ Mean mark part (d) 17%.

`m_(AC) =1`

`text(At point of tangency of)\ AC,\  f^{prime}(x) = 1`

`(1-3x)/(2 sqrt x)` `=1`
`1-3x` `=2sqrtx`
`3x+2sqrt x-1` `=0`
`(3 sqrtx-1)(sqrtx+1)` `=0`
   
`:. sqrt x` `=1/3` `or`   `sqrt x=-1\ \ text{(no solution)}`
`x` `=1/9`    

 
`f(1/9)=sqrt(1/9)(1-1/9)=1/3 xx 8/9 = 8/27\ \ =>P(1/9,8/27)`
 

`:.\ text(Equation of)\ AC, m=1, text(through)\ \ P\ \ text(is):`

`y-8/27` `= 1 (x-1/9)`
`y` `= x + 5/27`

 
`C\ text(is at intersection of)\ AC and CB:`

`-x + 1` `= x + 5/27`
`2x` `= 22/27`
`:. x` `= 11/27`
`y` `= -11/27 + 1 = 16/27`

 
`:. C (11/27, 16/27)`

Probability, MET1 2017 VCAA 8

For events `A` and `B` from a sample space, `text(Pr)(A text(|)B) = 1/5` and `text(Pr)(B text(|)A) = 1/4`.  Let  `text(Pr)(A nn B) = p`.

  1. Find  `text(Pr)(A)` in terms of `p`.  (1 mark)

  2. Find  `text(Pr)(A^{′} nn B^{′})` in terms of `p`.  (2 marks)

  3. Given that  `text(Pr)(A uu B) <= 1/5`, state the largest possible interval for `p`.  (2 marks)

Show Answers Only

  1. `text(Pr) (A) = 4p,\ \ p > 0`
  2. `1-8p`
  3. `0 < p <= 1/40`

Show Worked Solution

a.    `text(Pr)(A)` `=(text(Pr)(A nn B))/(text(Pr)(B text(|) A))`
    `=p/(1/4)`
    `=4p`

 

b.  `text(Consider the Venn diagram:)`

♦ Mean mark 40%.
MARKER’S COMMENT: The most successful answers used a Venn diagram or table.

`text(Pr)(A^{′} nn B^{′}) = 1-8p`

 

c.  `text(Given  Pr)(A uu B) = 8p`

♦ Mean mark 37%.

`=> 0 < 8p <= 1/5`

`:. 0 < p <= 1/40`

Functions, MET1 2017 VCAA 7

Let  `f: [0, oo) -> R,\ f(x) = sqrt(x + 1)`.

  1.  State the range of `f`.   (1 mark)

  2.  Let  `g: (-oo, c] -> R,\ \ g(x) = x^2 + 4x + 3`.
    1. Find the largest possible value of `c` such that the range of `g` is a subset of the domain of `f`.   (2 marks)

    2. For the value of `c` found in part b.i., state the range of `f(g(x))`.   (1 mark) 

  3. Let  `h: R -> R,\ \ h(x) = x^2 + 3`.
  4. State the range of `f(h(x))`.   (1 mark)

Show Answers Only
  1. `[1, oo)`
    1. `-3`
    2. `[1, oo)`
  2. `[2, oo)`
Show Worked Solution

a.  `text(Sketch of)\ \ f(x):`
 

`:.\ text(Range)\ \ (f) = [1, oo)`

 

b.i.  `text(Sketch)\ \ g(x) = (x + 1) (x + 3)`

♦ Mean mark 38%.
 


 

`text(Domain of)\ \ f(x)=[0,oo)`

`text(Find domain of)\ \ g(x)\ \ text(such that Range)\ (g) = [0, oo)`

`text(Graphically, this occurs when)\ \ g(x)\ \ text(has domain:)`

`=> x ∈ (–oo, –3] ∪ [–1,oo)`

`:. c = -3`
 

b.ii.  `text(Range)\ g(x) = [0, oo) = text(Domain)\ \ f(x)`

♦♦♦ Mean mark 20%.

`:.\ text(Range)\ \ f(g(x)) = [1, oo)`
 

c.  `text(Range)\ h(x) = [3, oo)`

♦♦ Mean mark 30%.
`f(3)` `= sqrt (3 + 1)`
  `= sqrt 4`
  `= 2`

 

`:.\ text(Range)\ \ f(h(x)) = [2, oo)`

Probability, MET1 2017 VCAA 5

For Jac to log on to a computer successfully, Jac must type the correct password. Unfortunately, Jac has forgotten the password. If Jac types the wrong password, Jac can make another attempt. The probability of success on any attempt is `2/5`. Assume that the result of each attempt is independent of the result of any other attempt. A maximum of three attempts can be made.

  1. What is the probability that Jac does not log on to the computer successfully?   (1 mark)

  2. Calculate the probability that Jac logs on to the computer successfully. Express your answer in the form `a/b`, where `a` and `b` are positive integers.   (1 mark)

  3. Calculate the probability that Jac logs on to the computer successfully on the second or on the third attempt. Express your answer in the form `c/d`, where `c` and `d` are positive integers.   (2 marks)

Show Answers Only

  1. `27/125`
  2. `98/125`
  3. `48/125`

Show Worked Solution

a.  `text(Pr)\ (S^{′} S^{′} S^{′})`

`= (3/5)^3`

`= 27/125`

 

b.  `1-text(Pr)\ (S^{′} S^{′} S^{′})`

`= 1-27/125`

`= 98/125`

 

c.  `text(Pr)\ (S^{′} S) + text(Pr)\ (S^{′} S^{′} S)`

`= 3/5 xx 2/5 + (3/5)^2 xx 2/5`

`= 48/125`

Probability, MET1 2017 VCAA 4

In a large population of fish, the proportion of angel fish is `1/4`.

Let `hat P` be the random variable that represents the sample proportion of angel fish for samples of size `n` drawn from the population.

Find the smallest integer value of `n` such that the standard deviation of `hat P` is less than or equal to `1/100`.  (2 marks)

Show Answers Only

`n_text(min) = 1875`

Show Worked Solution

`text(Let)\ X =\ text(number of angel fish),\ X ~ text(Bi) (n, 1/4)`

`text(sd)\ (hat P)` `= sqrt((p(1 – p))/n)`
  `= sqrt((1/4 xx 3/4)/n)`
  `= sqrt(3/(16n))`

 

`text(Solve:)\ \ \ sqrt(3/(16n))` `<= 1/100`
`3/(16 n)` `<= 1/(10\ 000)`
`(30\ 000)/16` `<= n`
`:. n` `>= 1875`

 

`:. n_text(min) = 1875`

Calculus, MET1 2017 VCAA 3

Let  `f: [-3, 0] -> R,\ f(x) = (x + 2)^2 (x-1).`

  1. Show that  `(x + 2)^2 (x-1) = x^3 + 3x^2-4`.   (1 mark)

  2. Sketch the graph of  `f` on the axes below. Label the axis intercepts and any stationary points with their coordinates.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2.  
Show Worked Solution

a.  `text(Expand LHS:)`

`(x^2 + 4x + 4) (x-1)`

`=x^3-x^2 + 4x^2-4x + 4x-4`

`=x^3 + 3x^2-4\ \ text(… as required.)`

 

b.  

`x text(-intercepts:)\ \ (–2, 0), (1, 0)\ \ text(but)\ \ x ∈ [–3, 0]`

`:. x text(-int occurs at)\ (–2, 0)`
 

`text(Endpoints:)\ \ (–3, –4),\ \ (0, –4)`

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

`3x^2 + 6x` `= 0`
`3x(x + 2)` `= 0`
`:. x=0 or -2`  

Calculus, MET1 2017 VCAA 2

Let  `y = x log_e(3x).`

  1. Find  `(dy)/(dx)`.   (2 marks)

  2. Hence, calculate  `int_1^2 (log_e(3x) + 1) dx`. Express your answer in the form  `log_e(a)`, where `a` is a positive integer.   (2 marks)

Show Answers Only
  1. `(dy)/(dx) = log_e (3x) + 1`
  2. `log_e(12)`
Show Worked Solution

a.  `text(Using Product Rule:)`

`(hg)^{prime}` `= h^{prime} g + h g^{prime}`
`:. (dy)/(dx)` `= 1 xx log_e (3x) + x (3/(3x))`
  `= log_e (3x) + 1`

 

b.  `text(Using Integration by Recognition:)`

MARKER’S COMMENT: Too many students ignored the hence instruction and were not able to form an integral.

`int (log_e(3x) + 1) dx = x log_e (3x)`

`:. int_1^2 (log_e (3x) + 1) dx`

`= [x log_e (3x)]_1^2`

`= (2 log_e (6))-(log_e(3))`

`= log_e(6^2)-log_e(3)`

`= log_e (36/3)`

`= log_e(12)`

Financial Maths, STD2 F1 SM-Bank 5

Alex is buying a used car which has a sale price of  $13 380. In addition to the sale price there are the following costs:

2014 27a1

  1. Stamp Duty for this car is calculated at $3 for every $100, or part thereof, of the sale price.  
    Calculate the Stamp Duty payable.   (1 mark)

  2. Alex wishes to take out comprehensive insurance for the car for 12 months.

     

    The cost of comprehensive insurance is calculated using the following:
     
          2014 27a2
    Find the total amount that Alex will need to pay for comprehensive insurance.   (3 marks)

  3. Alex has decided he will take out the comprehensive car insurance rather than the less expensive non-compulsory third-party car insurance.

  4.  

    What extra cover is provided by the comprehensive car insurance?   (1 mark)

Show Answers Only
  1. `$402`
  2. `$985.74`
  3. `text(Comprehensive insurance covers Alex)`
  4. `text(for damage done to his own car as well.)`
Show Worked Solution
♦♦♦ Mean mark 12%
IMPORTANT: “or part thereof ..” in the question requires students to round up to 134 to get the right multiple of $3 for their calculation.
i.    `($13\ 380)/100 = 133.8`
`:.\ text(Stamp duty)` `= 134 xx $3`
  `= $402`

 

ii.   `text(Base rate) = $845`

 
`text(FSL) =\ text(1%) xx 845 = $8.45`
 

`text(Stamp)` `=\ text(5.5%) xx(845 + 8.45)`
  `= 46.9397…`
  `= $46.94\ text{(nearest cent)}`

 

`text(GST)` `= 10 text(%) xx(845 + 8.45)`
  `= 85.345`
  `= $85.35`

 

`:.\ text(Total cost)` `= 845 + 8.45 + 46.94 + 85.35`
  `= $985.74`

 

♦ Mean mark 34%.
iii.   `text(Comprehensive insurance covers Alex)`
  `text(for damage done to his own car as well.)`

Harder Ext1 Topics, EXT2 2017 HSC 15b

Consider the curve  `sqrt x + sqrt y = sqrt a`, for  `x >= 0`  and  `y >= 0`, where `a` is a positive constant.

  1. Show that the equation of the tangent to the curve at the point  `P (c, d)`  is given by  `y sqrt c + x sqrt d = d sqrt c + c sqrt d`.  (2 marks)
  2. The tangent to the curve at the point `P` meets the `x` and `y` axes at  `A` and `B` respectively. Show that  `OA + OB = a`, where `O` is the origin.  (3 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)    `x^(1/2) + y^(1/2) = sqrt a`

`text(Using implicit differentiation:)`

`1/2 x^(-1/2) + 1/2 y^(-1/2) · (dy)/(dx)` `= 0`
`:. (dy)/(dx)` `= -(x^(-1/2))/(y^(-1/2))`
  `= -(y^(1/2))/(x^(1/2))`

 

`text(At)\  \P(c, d):`

`m_text(tang) = – sqrt d/sqrt c`

`text(Equation of tangent:)`

`y – d` `= – sqrt d/ sqrt c (x – c)`
`y sqrt c – d sqrt c` `= -x sqrt d + c sqrt d`
`y sqrt c + x sqrt d` `= d sqrt c + c sqrt d\ text(… as required.)`

 

(ii)  `text(At)\ \ A, y = 0`

`x sqrt d` `= d sqrt c + c sqrt d`
`x` `= (d sqrt c + c sqrt d)/sqrt d`

 

`text(At)\ \ B, x = 0`

`y sqrt c` `= d sqrt c + c sqrt d`
`y`  `=  (d sqrt c + c sqrt d)/sqrt c`

 

`AO + OB` `= x + y`
  `= (d sqrt c + c sqrt d)/sqrt d xx sqrt d/sqrt d + (d sqrt c + c sqrt d)/sqrt c xx sqrt c/sqrt c`
  `= sqrt d sqrt c + c + d + sqrt c sqrt d`
  `= c + 2 sqrt c sqrt d + d`
  `= (sqrt c + sqrt d)^2`
  `= (sqrt a)^2 qquad text{(S} text(ince)\ \ P(c, d)\ \ text(lies on)\ sqrt x + sqrt y = sqrt a text{)}`
  `= a\ text(… as required)`

Financial Maths, STD2 F1 SM-Bank 7

A dress was on sale with 25% discount.

As a regular customer, Kate received a further 10% on the already discounted price.

What was the overall percentage discount Kate received?  (2 marks)

Show Answers Only

`text(32.5%)`

Show Worked Solution

`text{Solution 1 (efficient method)}`

`text(Overall discount)` `=1-(0.75 xx 0.90)`
  `=1-0.675`
  `=0.325`
  `=32.5text(%)`

 

`text(Solution 2)`

`text(Let the dress cost) = $100`

`text(C) text(ost after 25% discount) = $75`

 

`text(C) text(ost after another 10% discount)`

`= 75 – (10 text{%} xx 75)`

`= $67.50`
 

`:.\ text(Overall discount) = 32.5text(%)`

Financial Maths, STD2 F1 SM-Bank 12

A golf shop is having a Boxing Day sale.

  1. What is the percentage discount on a putter which is reduced from $120.00 to $102.00?  (1 mark)

  2. The same discount applies storewide. What discount amount is applicable to a box of golf balls whose original price was $25.00?  (1 mark)

  3. What is the sale price of a golf bag which originally cost $160.00  (1 mark)

Show Answers Only
  1. `15 text(%)`
  2. `$3.75`
  3. `$136.00`
Show Worked Solution
i.     `text(Percentage Discount)` `=(120-102)/120 xx 100`
    `=18/120 xx 100`
    `=15 text(%)`

 

ii.    `text(Discounted Amount)` `=15 text(%) xx 25`
    `=$3.75`

 

iii.    `text(Sale Price)` `= 160 – 15 text(%) xx 160`
    `=0.85 xx 160`
    `=$136.00`

Financial Maths, STD2 F1 SM-Bank 2

George makes a single deposit of $9000 into an account that pays simple interest.

After 4 years, George's account has a balance of $10 350.

What simple interest rate did George receive on his investment?  (2 marks)

Show Answers Only

`3.75text(%)`

Show Worked Solution
`text(Interest earned)` `= 10\ 350 – 9000`
  `= $1350`

 

`text(Using)\ \ I = Prn,`

`1350` `= 9000 xx r xx 4`
`:. r` `= 1350/(4 xx 9000)`
  `= 0.0375`
  `= 3.75text(%)`

Financial Maths, STD2 F1 SM-Bank 6

Michelle intends to keep a car purchased for $17 000 for 15 years. At the end of this time its value will be $3500.

  1. By what amount, in dollars, would the car’s value depreciate annually if Michelle used the flat rate method of depreciation?  (1 mark)

  2. Determine the annual flat rate of depreciation correct to one decimal place.  (1 mark) 

Show Answers Only
  1. `$900`
  2. `5.3text{%  (1 d.p.)}`
Show Worked Solution
i.    `text(Depreciation)` `= 17\ 000 – 3500`
    `= $13\ 500`

 

`:.\ text(Annual depreciation)`

`= (13\ 500)/15`

`= $900`

 

ii.   `:.\ text(Flat rate of depreciation )`

`= 900/(17\ 000) xx 100text(%)`

`= 5.29…`

`= 5.3text{%  (1 d.p.)}`

Financial Maths, STD2 F1 SM-Bank 5

Khan paid $900 for a printer.

This price includes 10% GST (goods and services tax).

  1. Determine the price of the printer before GST was added.

     

    Write your answer correct to the nearest cent.  (2 marks)

  2. Khan is able to depreciate the full $900 purchase price of his printer for taxation purposes.

     

    Under flat rate depreciation the printer will be valued at $300 after five years.

     

    Calculate the annual depreciation in dollars.  (1 mark)

Show Answers Only
  1. `$818.18`
  2. `$120`
Show Worked Solution

i.   `text(Let)\ \ $P = text(price ex-GST)`

COMMENT: Reverse GST questions regularly cause problems for many students.
`:. P + 10text(%) xx P` `= 900`
`1.1P` `= 900`
`P` `= 900/1.1`
  `= 818.181…`
  `= $818.18\ \ text(nearest cent)`

 

ii.   `text(Annual depreciation)`

`= ((900 – 300))/5`

`= $120`

Financial Maths, STD2 F1 SM-Bank 3

A company purchased a machine for $60 000.

For taxation purposes the machine is depreciated over time using the straight line depreciation method.

The machine is depreciated at a flat rate of 10% of the purchase price each year.

  1. By how many dollars will the machine depreciate annually?  (1 mark)

  2. Calculate the value of the machine after three years.  (1 mark)

  3. After how many years will the machine be $12 000 in value?  (1 mark)

Show Answers Only
  1. `$6000`
  2. `$42\ 000`
  3. `8\ text(years)`
Show Worked Solution
i.    `text(Annual depreciation)` `= 10text(%) xx 60\ 000`
    `= $6000`

 

ii.   `text(After 3 years,)`

`text(Value)` `=V_0 – Dn`
  `= 60\ 000 – (3 xx 6000)`
  `= $42\ 000`

 

iii.   `text(Find)\ n\ text(when value = $12 000)`

`12\ 000` `= 60\ 000 – 6000 xx n`
`6000n` `= 48\ 000`
`:.n` `=(48\ 000)/6000`
  `= 8\ text(years)`

Financial Maths, STD2 F1 SM-Bank 10

Hugo is a professional bike rider.

The value of his bike will be depreciated over time using the flat rate method of depreciation.

The graph below shows his bike’s initial purchase price and its value at the end of each year for a period of three years.
 

  1. What was the initial purchase price of the bike?  (1 mark)

  2. Use calculations to show that the bike depreciates in value by $1500 each year.  (1 mark)

  3. Assume that the bike’s value continues to depreciate by $1500 each year. Determine its value five years after it was purchased.  (1 mark)

Show Answers Only
  1. `$8000`
  2. `text(See Worked Solutions)`
  3. `$500`
Show Worked Solution

i.   `$8000`
 

ii.   `text(Value after 1 year) = $6500\ \ \ text{(from graph)}`

`:.\ text(Annual depreciation)` `= 8000-6500`
  `= $1500`

 

iii.  `text(After 5 years:)`

`S` `=V_0-Dn`
  `=8000-5 xx 1500`
  `=$500`

Financial Maths, STD2 F1 SM-Bank 1 MC

Rae paid  $40 000  for new office equipment at the start of the 2013 financial year.

At the start of each following financial year, she used flat rate depreciation to revalue her equipment.

At the start of the 2016 financial year she revalued her equipment at  $22 000.

The annual flat rate of depreciation she used, as a percentage of the purchase price, was

  1. 11.25%
  2. 15%
  3. 17.5%
  4. 35%
Show Answers Only

`B`

Show Worked Solution

`text(Depreciation over 3 years)`

♦ Mean mark 50%.

`=40\ 000 – 22\ 000`

`=$18\ 000`

`:.\ text(Annual depreciation) = (18\ 000)/3 = $6000`

`:.\ text(Depreciation rate) = 6000/(40\ 000) = 0.15 = 15text(%)`

`=> B`

Calculus, EXT2 C1 2017 HSC 15a

Let  `I_n = int_0^1 x^n sqrt (1 - x^2)\ dx`, for  `n = 0, 1, 2, …`

  1. Find the value of `I_1`.  (1 mark)

  2. Using integration by parts, or otherwise, show that for  `n >= 2`
     
         `I_n = ((n - 1)/(n + 2)) I_(n-2)`.  (3 marks)

  3. Find the value of  `I_5`.  (1 mark)

Show Answers Only
  1. `1/3`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `8/105`
Show Worked Solution
i.   `I_1` `= int_0^1 x^1 sqrt (1-x^2)\ dx`
    `= -1/2 int_0^1-2x sqrt (1-x^2)\ dx`
    `= -1/2 [2/3 (1-x^2)^(3/2)]_0^1`
    `= -1/2 [0-2/3 (1)]`
    `= 1/3`

 

ii.  `I_n = int_0^1 x^n sqrt (1-x^2)\ dx`

♦ Mean mark 48%.

`text(Integrating by parts:)`

`u` `= x^(n-1)`    `dv` `= x sqrt (1-x^2)\ dx`
`du` `= (n-1) x^(n – 2)\ dx` `v` `= -1/3 (1-x^2)^(3/2)`
`I_n` `= [x^(n-1) xx -1/3 (1-x^2)^(3/2)]_0^1 + 1/3 int_0^1 (1-x^2)^(3/2)(n-1) x^(n-2)\ dx`
  `= 0 + (n-1)/3 int_0^1 x^(n-2) (1-x^2) · sqrt(1-x^2)\ dx`
  `= (n-1)/3 int_0^1 x^(n-2) sqrt(1-x^2)\ dx-(n-1)/3 int_0^1 x^n sqrt (1-x^2)\ dx`
  `= (n-1)/3 · I_(n-2)-(n-1)/3 · I_n`

 

`:. I_n + (n-1)/3 · I_n` `= (n-1)/3 · I_(n-2)`
`I_n (1 + (n-1)/3)` `= (n-1)/3 · I_(n-2)`
`I_n ((n + 2)/3)` `= (n-1)/3 · I_(n-2)`
`:. I_n` `= ((n-1)/(n + 2)) · I_(n-2)\ text(… as required.)`

 

iii.   `I_5` `= 4/7 xx I_3`
    `= 4/7 xx 2/5 xx I_1`
    `= 4/7 xx 2/5 xx 1/3\ text{(part (i))}`
    `= 8/105`

Mechanics, EXT2 2017 HSC 14c

A smooth double cone with semi-vertical angle  `theta < pi/2`  is rotating about its axis with constant angular velocity `w`.

Two particles, each of mass `m`, are sitting on the cone as it rotates, as shown in the diagram.

Particle 1 is inside the cone at vertical distance `h` above the apex, `A`, and moves in a horizontal circle of radius `r`.

Particle 2 is attached to the apex `A` by a light inextensible string so that it sits on the cone at vertical distance `h` below the apex. Particle 2 also moves in a horizontal circle of radius `r`.

The acceleration due to gravity is `g`.

  1. The normal reaction force on Particle 1 is `R`.
    By resolving `R` into vertical and horizontal components, or otherwise, show that  `w^2 = (gh)/r^2`.  (2 marks)
  2. The normal reaction force on Particle 2 is `N` and the tension in the string is  `T`.

  3. By considering horizontal and vertical forces, or otherwise, show that

  4. `qquad qquad N = mg (sin theta - h/r cos theta)`.  (2 marks)
  6. Show that  `theta >= pi/4`.  (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)  

`text(Resolving forces horizontally:)`

`R cos theta = mrw^2`

`text(Resolving forces vertically:)`

`R sin theta` `= mg`
`(R sin theta)/(R cos theta)` `= (mg)/(mrw^2)`
`tan theta` `= g/(rw^2)`
`:. r/h` `= g/(rw^2)`
`r^2w^2` `= gh`
`w^2` `= (gh)/r^2\ \ text(… as required.)`

 

(ii)  

`text(Resolving forces horizontally:)`

`T sin theta – N cos theta = mrw^2\ text{… (1)}`

`text(Resolving forces vertically:)`

`T cos theta + N sin theta = mg\ text{… (2)}`

 

`text{Multiply (1)} xx cos theta`

`T sin theta cos theta – N cos^2 theta = mrw^2 cos theta\ text{… (3)}`

`text{Multiply (2) by}\ sin theta`

`T sin theta cos theta + N sin^2 theta = mg sin theta\ text{… (4)}`

 

`text{Subtract  (4) – (3)}`

`N (sin^2 theta + cos^2 theta)` `=mg sin theta – mrw^2 cos theta`
`:. N` `= mg sin theta – mr · (gh)/r^2 · cos theta`
  `= mg (sin theta – h/r cos theta)`

 

(iii)  `text(Movement of particle 2 requires)\ N >= 0`

♦♦ Mean mark 34%.
`mg (sin theta – h/r cos theta)` `>= 0`
`sin theta – 1/(tan theta) · cos theta` `>= 0`
`sin theta` `>= (cos theta)/(tan theta)`
`tan theta` `>= 1/(tan theta)`
`tan^2 theta` `>= 1`
`tan theta` `>= 1 qquad text{(}theta\ \ text{is in 1st quadrant)}`
`:. theta` `>= pi/4`

Calculus, EXT2 C1 2017 HSC 14a

It is given that  `x^4 + 4 = (x^2 + 2x + 2) (x^2-2x + 2)`.

  1. Find `A` and `B` so that  `16/(x^4 + 4) = (A + 2x)/(x^2 + 2x + 2) + (B-2x)/(x^2-2x + 2)`.  (1 mark)

  2. Hence, or otherwise, show that for any real number `m`
     
         `int_0^m 16/(x^4 + 4)\ dx = ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`.  (2 marks)

  3. Find the limiting value as  `m -> oo`  of
     
         `int_0^m 16/(x^4 + 4)\ dx`.  (1 mark)

Show Answers Only
  1. `A = 4 and B = 4`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `2 pi`
Show Worked Solution

i.  `16/(x^4 + 4) = (A + 2x)/(x^2 + 2x + 2) + (B-2x)/(x^2-2x + 2)`

`text(Multiply each side by)\ \ x^4 + 4`

`16 = (A + 2x) (x^2-2x + 2) + (B-2x) (x^2 + 2x + 2)`

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

`2A + 2B` `=16`
`A + B` `= 8\ text{… (1)}`

 
`text(When)\ \ x = 1`

`(A + 2) (1) + (B-2) (5)` `=16`
`A + 2 + 5B-10` `=16`
`A + 5B` `= 24\ text{… (2)}`

 

`text{Subtract  (2) – (1)}`

`4B=16\ \ =>\ \ B=4`

`A=4`
 

ii.  `int_0^m 16/(x^4 + 4)`

`= int_0^m (4 + 2x)/(x^2 + 2x + 2) dx + int_0^m (4-2x)/(x^2-2x + 2) dx`

`= int_0^m (2x + 2)/(x^2 + 2x + 2) + 2/(x^2 + 2x + 2) dx + int_0^m 2/(x^2-2x + 2)-(2x-2)/(x^2-2x + 2) dx`

`= [ln(x^2 + 2x + 2)]_0^m + int_0^m 2/(1 + (x + 1)^2) dx + int_0^m 2/(1 + (x-1)^2) dx-[ln (x^2-2x + 2)]_0^m`

`= [ln(m^2 + 2m + 2)-ln 2] + [2 tan^(-1) (x + 1)]_0^m + [2 tan^(-1) (x-1)]_0^m-[ln (m^2-2m + 2)-ln 2]`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 [tan^(-1) (m + 1)-tan^(-1) (1)] + 2 [tan^(-1) (m-1)-tan^(-1) (1)]`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1)-2 · pi/4 + 2 tan ^(-1) (m-1) + 2 · pi/4`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`

 

iii.  `I = int_0^m 16/(x^4 + 4) = ln ((1 + 2/m + 2/m^2)/(1-2/m + 2/m^2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`

`lim_(m -> oo) I` `= ln\ 1 + 2 · pi/2 + 2 · pi/2`
  `= 0 + pi + pi`
  `= 2 pi`

Harder Ext1 Topics, EXT2 2017 HSC 14b

Two circles, `cc"C"_1` and `cc"C"_2`, intersect at the points `A` and `B`. Point `C` is chosen on the arc `AB` of `cc"C"_2` as shown in the diagram.

The line segment `AC` produced meets `cc"C"_1` at `D`.

The line segment `BC` produced meets `cc"C"_1` at `E`.

The line segment `EA` produced meets `cc"C"_2` at `F`.

The line segment `FC` produced meets the line segment `ED` at `G`.

Copy or trace the diagram into your writing booklet.

  1. State why `/_ EAD = /_ EBD`.  (1 mark)
  2. Show that `/_ EDA = /_ AFC`.  (1 mark)
  3. Hence, or otherwise, show that `B, C, G` and `D` are concyclic points.  (3 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)  `text(Join)\ DB`

`/_ EAD = /_ EBD` `text{(angles at circumference}`
  `text(sitting on same arc)\ ED\ text(of)\ cc”C”_1 text{)}`

 

(ii)  `text(Join)\ AB`

`/_ EDA = /_ EBA` `text{(angles at circumference}`
  `text(sitting on same arc)\ EA\ text(of)\ cc”C”_1 text{)}`

 

`text(Consider arc)\ AC\ text(in)\ cc”C”_2`

`/_ EBA = /_ CBA = /_ AFC` `text{(angles at circumference}`
  `text(sitting on same arc)\ AC\ text(of)\ cc”C”_2 text{)}`

 

`:. /_ EDA = /_ AFC\ text(… as required.)`

 

(iii)   `text(Let)\ \ ` `/_ EDA = /_ AFC = theta\ \ text{(part (ii))}`
    `/_ EAD = /_ EBD = phi\ \ text{(part (i))}`

 

`text(Consider)\ Delta EAD:`

♦ Mean mark 47%.

`/_ AED = pi – theta – phi`

`text(Consider)\ Delta EGF:`

`/_ EGF` `= pi – (theta + pi – theta – phi)`
  `= phi`

 

`text(In)\ BCGD`

`text(External angle)\ EGF = phi`

`text(Interior opposite angle)\ EBD = phi`

`:. BCGD\ text(is a cyclic quadrilateral.)`

Measurement, STD2 M1 SM-Bank 4

Steel rods are manufactured in the shape of equilateral triangular prisms.
 


 

  1. Find the volume of the prism (answer correct to 1 decimal place).  (2 marks)

  2. The mass of steel is 7850 kg/m³. Use this information to find the mass of the steel rod correct to the nearest gram.  (2 marks)

Show Answers Only
  1. `3464.1\ text(cm³)`
  2. `27\ 193\ text(g)`
Show Worked Solution

i.   `text{Area of triangular face (using sine rule)}`

`=1/2 xx 10 xx 10 xx sin60°`

`=43.301…`

 

`text(Volume)` `=Ah`
  `=43.301… xx 80`
  `=3464.10…`
  `=3464.1\ text{cm³  (to 1 d.p.)}`

 

ii.  `text(Converting kg to g:)`

`7850\ text(kg) = 7\ 850\ 000\ text(g)`

 

`text(Converting m³ to cm³:)`

`1\ text(m³)` `=100\ text(cm) xx100\ text(cm) xx100\ text(cm)` 
  `=1\ 000\ 000\ text(cm³)`

 

`:.\ text(Weight of steel rod)` `=3464.1 xx (7\ 850\ 000)/(1\ 000\ 000)` 
  `=27\ 193.185`
  `=27\ 193\ text{g  (nearest gram)}`

Measurement, STD2 M7 SM-Bank 1

Bikram runs a hot yoga studio.

If it costs 34 cents for 1-kilowatt (1000 watts) for 1 hour, how much does it cost him to run three 3200-watt heaters from 9:00 am to 12:30 pm on a single day? (Give your answer to the nearest cent)  (2 marks)

Show Answers Only

`$11.42`

Show Worked Solution

`text(Total energy usage)`

`= 3 xx 3200 xx 3.5\ text(hours)`

`= 33\ 600\ text(watts)`

 

`:.\ text(C)text(ost)` `= (33\ 600)/1000 xx 0.34`
  `= 11.424`
  `= $11.42\ \ \ text{(nearest cent)}`

Measurement, STD2 M1 SM-Bank 7

Oscar and Nadine select a meal each from the table below.
 

 
Oscar has a hamburger and Nadine has the lamb souvlaki.

After dinner, Oscar goes for a run where he expends energy at 25 kJ/minute.

At the same time, Nadine goes for a brisk walk where she expends 12 kJ/minute.

Who will expend the kilojoule intake from their dinner the quickest, and by how many minutes?  (3 marks)

Show Answers Only

`text(Nadine will expend her dinner’s kJ intake)`

`text(in 46 minutes less than Oscar.)`

Show Worked Solution

`text(Oscar’s kJ intake = 5050)`

`text(Time to work off)` `= 5050/25`
  `= 202\ text(minutes)`

 

`text(Nadine’s intake = 1872)`

`text(Time to work off)` `= 1872/12`
  `= 156\ text(minutes)`

 

`202 – 156 = 46\ text(minutes)`

`:.\ text(Nadine will expend her dinner’s kJ intake)`

`text(in 46 minutes less than Oscar.)`

Measurement, STD2 M1 SM-Bank 8

A cannon ball is made out of steel and has a diameter of 23 cm.

  1. Find the volume of the sphere in cubic centimetres (correct to 1 decimal place).  (2 marks)

  2. It is known that the mass of the steel used is 8.2 tonnes/m³. Use this information to find the mass of the cannon ball to the nearest gram.  (2 marks)

Show Answers Only
  1. `6370.6\ text{cm³  (to 1 d.p.)}`
  2. `52\ 239\ text(grams)`
Show Worked Solution

i.   `text(Radius)= 23/2 = 11.5\ text(cm)`

`text(Volume)` `= 4/3pir^3`
  `= 4/3 xx pi xx 11.5^3`
  `= 6370.626…`
  `= 6370.6\ text{cm³  (to 1 d.p.)}`

 

ii.   `text(Convert m³ to cm³:)`

`text(1 m³)` `= 100\ text(cm × 100 cm × 100 cm)`
  `= 1\ 000\ 000\ text(cm³)`

 

`text(Convert 8.2 tonnes to grams:)`

`text(8.2 tonnes)` `= 8200\ text(kg)`
  `= 8\ 200\ 000\ text(g)`

 

`:.\ text(Weight of cannon ball)`

`= 6370.6 xx (8\ 200\ 000)/(1\ 000\ 000)`

`= 52\ 238.92`

`= 52\ 239\ text(grams)`

Algebra, STD2 A2 SM-Bank 5 MC

The equation of the line drawn in the diagram below is:
 

  1. `5y = 3x`
  2. `y = 5/3x + 5`
  3. `y = 3/5x + 5`
  4. `y = -5/3x + 5`
Show Answers Only

`D`

Show Worked Solution

`y text(-intercept) = +5`

`text(Gradient)` `= text(rise)/text(run)`
  `=- 5/3`

 

`:.\ text(Equation is):\ y = -5/3x + 5`

`=> D`

Algebra, STD2 A4 SM-Bank 4

Penny is a baker and makes meat pies every day.

The cost of making `p` pies, `$C`,  can be calculated using the equation

`C = 675 + 3.5 p`

Penny sells the pies for $5.75 each, and her income is calculated using the equation

`I = 5.75 p`

  1. On the graph, draw the graphs of  `C`  and  `I`.  (2 marks)

  2. On the graph, label the breakeven point and the loss zone.  (2 marks)

Show Answers Only

(i) and (ii)

Show Worked Solution
i.   

 

ii.  `text(Loss zone occurs when)\ C > I,\ text(which is shaded)`

`text(in the diagram above.)`

Volumes, EXT2 2017 HSC 13d

The trapezium whose vertices are  `A(0, –1), B(0, 1), C(4, 3)`  and  `D(4, –3)`  forms the base of a solid.

Each cross-section perpendicular to the `x`-axis is an isosceles triangle. The height of the isosceles triangle `ABE` with base `AB`, is 1. The height of the isosceles triangle `DCF` with base `DC`, is 2.  The cross-section through the point `(x, 0)` is the isosceles triangle `PQR`, where `R` lies on the line `EF`, as shown in the diagram.

Find the volume of the solid.  (4 marks)

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`38/3\ text(u³)`

Show Worked Solution

`m_(BC) = (3 – 1)/(4 – 0) = 1/2`

♦ Mean mark 51%.

`text(Equation of)\ BC:`

`y = 1/2 x + 1`

`QP_text(dist)` `= 2 xx (1/2 x + 1)`
  `= x + 2`

 

`text(Find height of)\ Delta RQP:`

`text(Using similar triangles),`

`h/x = 1/4`

`h = x/4`

`=>\ text(Height of)\ Delta RQP = 1 + x/4`

 

`:.\ text(Area)\ Delta RQP` `= 1/2 · QP · (1 + x/4)`
  `= 1/2 · (x + 2) (x/4 + 1)`
  `= x^2/8 + (3x)/4 + 1`

 

`:.\ text(Volume)` `= 1/8 int_0^4 x^2 + 6x + 8\ dx`
  `= 1/8 [x^3/3 + 3x^2 + 8x]_0^4`
  `= 1/8 [(64/3 + 48 + 32) – 0]`
  `= 38/3\ text(u³)`

Functions, EXT1′ F1 2017 HSC 8 MC

Suppose that  `f(x)`  is a non-zero odd function.

Which of the functions below is also odd?

  1. `f(x^2)cosx`
  2. `f(f(x))`
  3. `f(x^3)sinx`
  4. `f(x^2) - f(x)`
Show Answers Only

`B`

Show Worked Solution

`text(By trial and error, consider)\ B:`

`f(f(−x))` `= f(−f(x))`
  `= −f(f(x))`

 
`:. f(f(x))\ text(is odd.)`

`=> B`

Calculus, EXT2 C1 2017 HSC 7 MC

It is given that  `f(x)`  is a non-zero even function and  `g(x)`  is a non-zero odd function.

Which expression is equal to  `int_(−a)^a f(x) + g(x)\ dx`?

  1. `2 int_0^a f(x)\ dx`
  2. `2 int_0^a g(x)\ dx`
  3. `int_(−a)^a g(x)\ dx`
  4. `2int_0^a f(x) + g(x)\ dx`
Show Answers Only

`A`

Show Worked Solution
`int_(−a)^a f(x) + g(x)\ dx` `= int_(−a)^a f(x)\ dx + int_(−a)^a g(x)\ dx`
  `= 2 int_0^a f(x)\ dx + 0`

`=> A`