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NETWORKS, FUR1 2020 VCAA 7 MC

Four friends go to an ice-cream shop.

Akiro chooses chocolate and strawberry ice cream.

Doris chooses chocolate and vanilla ice cream.

Gohar chooses vanilla ice cream.

Imani chooses vanilla and lemon ice cream.

This information could be presented as a graph.

Consider the following four statements:

  • The graph would be connected.
  • The graph would be bipartite.
  • The graph would be planar.
  • The graph would be a tree.

How many of these four statements are true?

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show Answers Only

`E`

Show Worked Solution

`text(True statements :)`

♦♦♦ Mean mark 11%.

`text{The graph is connected (a path exists between all vertices).}`

`text{The graph is bipartite (vertices can be divided into two groups).}`

`text{The graph is planar (no edges cross if strawberry is moved to the top).}`

`text{The graph forms a tree (no cycles).}`

`=>  E`

Networks, STD2 N3 SM-Bank 50

Roadworks planned by the local council require 13 activities to be completed.

The network below shows these 13 activities and their completion times in weeks.
 

  1. What is the earliest start time, in weeks, of activity `K`?  (1 mark)

  2. How many of these activities have zero float time?  (2 marks)

  3. It is possible to reduce the completion time for activities `A, E, F, L` and `K`.
  4. The reduction in completion time for each of these five activities will incur an additional cost.
  5. The table below shows the five activities that can have their completion time reduced and the associated weekly cost, in dollars.
     
       
  6. The completion time for each these five activities can be reduced by a maximum of two weeks.
  7. The overall completion time for the roadworks can be reduced to 16 weeks.
  8. What is the minimum cost, in dollars, of this change in completion time?  (3 marks)

Show Answers Only
  1. `14 \ text{weeks}`
  2. `7`
  3. `$ 380 \ 000`
Show Worked Solution

a.    `text{Scan forwards:}`

`EST \ (text{activity} \ K)`

`= A \ E \ J`

`= 6 + 5 + 3`

`= 14 \ text{weeks}`
 

b.     `text{Scan backwards:}`

`text{Critical paths:}\ ADGLM\ text(and)\ AEHLM`

`:. \ text{7 activities have no float time:} \ ADEGHLM`
 

c.    `text{There are 9 possible paths}`

`ADGLM\ (19), AEHLM\ (19) , AEIM\ (14)`

`AEJK\ (17), BCEIM \ (13), BCEJK\ (16)`

`BCEHLM\ (18), BCDGLM \ (18), BFK (11)`

`text{Consider the 5 paths with completions over 16 weeks}`

` to \ text{all contain}\ A\ text{or}\ L \ text{or both.}`
  

`text{Consider} \ ADGLM, AEHLM \ (text{contains both} \ A\ text{and} \ L )`

`to \ text{reduce} \ L xx 2 \ ,  A xx 1 \ text{to reach 16 weeks}`

`to \ text{cheaper than} \ L xx 1 \ , \ A xx 2`
  

`text{Consider} \ BCEHLM , BCDGLM\ (text{both contain} \ L \ text{only})`

`to \ text{reduce} \ L xx 2 \ text{to reach 16 weeks}`
 

`text{Consider} \ AEJK \ ( text{contains} \ A \ text{only} )`

`to \ text{reduce} \ A xx 1 \ text{reach 16 weeks}.`
 

`:. \ text{Minimum cost to reduce time to 16 weeks}`

`= 2 xx 120\ 000 + 1 xx  140\ 000`

`= $ 380\ 000`

GRAPHS, FUR2 2021 VCAA 4

Health and training sessions are held each day at the new community centre.

  • Let `x` be the number of sessions for children each day.
  • Let `y` be the number of sessions for adults each day.
  • The total number of all sessions each day must be at least 10.
  • The number of sessions for adults must not be less than the number of sessions for children.
  • It takes 30 minutes for each session for children and 40 minutes for each session for adults.

The constraints on the health and training sessions each day can be represented by the following five inequalities.

Inequality 1      `x >= 0`

Inequality 2      `y >= 0`

Inequality 3      `x + y >= 10`

Inequality 4       `y >= x`

Inequality 5        `30x + 40y <= 600`

  1. Explain the meaning of Inequality 5 in the context of this situation.  (1 mark)

The graph below shows the feasible region (shaded) that satisfies Inequality 1 to 5.

  1. What is the maximum number of sessions for children each day?  (1 mark)
  2. The community profits from these sessions.
  3. Each session for children makes a profit of $45 and each session for adults makes a profit of $60.
  4.  i.  Determine the maximum profit per day from all health and training sessions. (1 mark)
  5. ii. List all the points within the feasible region that result in this maximum profit. (1 mark)
Show Answers Only
  1. `text{The time of all adult and children sessions each day, in total,}`
    `text{must be less than 600 minutes (10 hours).}`
  2. `8`
  3.  i.  `$900`
    ii. `(0, 15), (4, 12) and (8, 9)`
Show Worked Solution

a.    `text{The time of all adult and children sessions each day, in total,}`

`text{must be less than 600 minutes (10 hours).}`
 

b.    `text{Maximum children sessions = 8}`

`text{(largest integer value of} \ x \ text{in the feasible region)}`
 

c.i.  `P = 45 x + 60 y`

`y = – 3/4 x + P/60`

`text{Objective function’s gradient is the same as} \ 30x + 40y = 600`

`=> \ text{maximum profit occurs at integer points on graph of}`

 `30x+40y=600\ \ text{in feasible region.}`
 

`text{Using (0, 15):}`

`P_text{max}` `= 45 xx 0 + 60 xx 15`
  `= $900`

 

c.ii.

 

`P_text{max} \ text{occurs at (0, 15), (4, 12) and (8, 9)}`

Statistics, SPEC2 2021 VCAA 6

The maximum load of a lift in a chocolate company's office building is 1000 kg. The masses of the employees who use the lift are normally distributed with a mean of 75 kg and a standard deviation of 8 kg. On a particular morning there are `n` employees about to use the lift.

  1. What is the maximum possible value of `n` for there to be less than a 1% chance of the lift exceeding the maximum load?   (2 marks)

Clare, who is one of the employees, likes to have a hot drink after she exits the lift. The time taken for the drink machine to dispense a hot drink is normally distributed with a mean of 2 minutes and a standard deviation of 0.5 minutes. Times taken to dispense successive hot drinks are independent.

  1. Clare has a meeting at 9.00 am and at 8.52 am she is fourth in the queue for a hot drink. Assume that the waiting time between hot drinks dispensed is negligible and that it takes Clare 0.5 minutes to get from the drink machine to the meeting room.
  2. What is the probability, correct to four decimal places, that Clare will get to her meeting on time?   (2 marks)

Clare is a statistician for the chocolate company. The number of chocolate bars sold daily is normally distributed with a mean of 60 000 and a standard deviation of 5000. To increase sales, the company decides to run an advertising campaign. After the campaign, the mean daily sales from 14 randomly selected days was found to be 63 500.

Clare has been asked to investigate whether the advertising campaign was effective, so she decides to perform a one-sided statistical test at the 1% level of significance.

  1.   i. Write down suitable null and alternative hypotheses for this test.   (1 mark)

  2.  ii. Determine the `p` value, correct for decimal places, for this test.   (1 mark)

  3. iii. Giving a reason, state whether there is any evidence for the success of the advertising campaign.   (1 mark)

  4. Find the range of values for the mean daily sales of another 14 randomly selected days that would lead to the null hypothesis being rejected when tested at the 1% level of significance. Give your answer correct to the nearest integer.   (1 mark)

  5. The advertising campaign has been successful to the extent that the mean daily sales is now 63 000.
  6. A statistical test is applied at the 5% level of significance.
  7. Find the probability that the null hypothesis would be incorrectly accepted, based on the sales of another 14 randomly selected days and assuming a standard deviation of 5000. Give your answer correct to three decimal places.   (2 marks)

Show Answers Only
  1. `12`
  2. `0.3085`
  3.   i. `H_0: \ mu = 60 \ 000, \ H_1: \ mu > 60 \ 000`
  4.  ii.  `0.0044`
  5. iii.  `text{Successful as} \ p text{-value is below 0.01.}`
  6. `n ≥ 63\ 109`
  7. `0.274`
Show Worked Solution

a.    `text{Method 1}`

♦♦ Mean mark 22%.

`text{Let} \ \ M_i\ ~\ N (75, 8^2) \ \text{for} \ \ i = 1, 2, 3, … , n`

`W_n = W_1 + W_2 + … + W_n`

`E (W_n) = E(M_1 + … + M_n) = 75n`

`text(s.d.) (W_n) = text{s.d.} (M_1 + … M_n) = 8 sqrtn`

`W_n\ ~\ N (75n, 8^2 n)`

`text{Using} \ Z\ ~\ N (0, 1)`

`text(Pr) (W_n > 1000) = 0.01`

`text(Pr) (Z > {1000-75 n}/{8 sqrtn}) = 0.01`
 
`n = 12.5`

`:. \ text{Largest} \ n = 12`
 

`text{Method 2}`

`text{By trial and error}`
 
`text(Pr) (M_1 + … + M_11 > 1000) ≈ 0`

`text(Pr) (M_1 + … + M_12 > 1000) ≈ 0.0002`

`text(Pr) (M_1 + … + M_13 > 1000) ≈ 0.193`

`:. \ text{Largest} \ n = 12`

 

b.   `T_i\ ~\ N (2, 0.5^2) \ \ text{for} \ \ i = 1, 2, …`

Mean mark part (b) 51%.

`text{Wait time} \ (T) = T_1 + T_2 + T_3 + T_4`

`E(T) = 4 xx 2 = 8`

`text{s.d.}(T) = text{s.d.}(T_1 + T_2 + T_3 + T_4) = sqrt4 xx 0.5 = 1`
 
`T\ ~\ N (8, 1)`

`text(Pr) (T < 7.5) = 0.3085`

`text{By CAS: normCdf} (0, 7.5, 8, 1)`

 

c.i.  `H_0: \ mu = 60 \ 000`

`H_1: \ mu > 60 \ 000`
 

c.ii.  `text(Pr) (barX > 63\ 500 | mu = 60 \ 000) = 0.004407`

  `:. \ p \ text{value} = 0.0044`

 `text{By CAS: norm Cdf} (63\ 500, oo, 60\ 000, 5000/sqrt14)`
 

c.iii.  `text{S} text{ince the} \ p text{-value is below 0.01, there is strong evidence}`

   `text{the advertising was effective (against the null hypothesis).}`

 

d.   `text(Pr) (barX > n | mu = 60 \ 000) < 0.01`

♦♦♦ Mean mark part (d) 16%.

`n ≥ 63\ 109`

`text{By CAS: inv Norm} (0.99, 60 \ 000, 5000/sqrt14)`

 

e.    `text{Similar to part (d):}`

♦♦♦ Mean mark part (e) 10%.

`text(Pr) (barX > n | mu = 60 \  000) < 0.05`

`n ≤ 62198`

`text{By CAS: invNorm} \ (0.95, 60 \ 000, 5000/sqrt14)`
 
`text{If} \ \ mu = 63\ 000, text{find probability null hypothesis incorrectly accepted:}`

`text(Pr) (barX < 62\ 198 | mu = 63\ 000) = 0.274`

`text{By CAS: normCdf} (0, 62\ 198, 63\ 000, 5000/sqrt14)`


NETWORKS, FUR2 2021 VCAA 4

Roadworks planned by the local council require 13 activities to be completed.

The network below shows these 13 activities and their completion times in weeks.
 

  1. What is the earliest start time, in weeks, of activity `K`?   (1 mark)

  2. How many of these activities have zero float time?   (1 mark)

  3. It is possible to reduce the completion time for activities `A, E, F, L` and `K`.
  4. The reduction in completion time for each of these five activities will incur an additional cost.
  5. The table below shows the five activities that can have their completion time reduced and the associated weekly cost, in dollars.
     
       
  6. The completion time for each these five activities can be reduced by a maximum of two weeks.
  7. The overall completion time for the roadworks can be reduced to 16 weeks.
  8. What is the minimum cost, in dollars, of this change in completion time?   (1 mark)

Show Answers Only
  1. `14 \ text{weeks}`
  2. `7`
  3. `$ 380 \ 000`
Show Worked Solution

a.    `text{Scan forwards:}`

`EST \ (text{activity} \ K)`

`= A \ E \ J`

`= 6 + 5 + 3`

`= 14 \ text{weeks}`
 

b.     `text{Scan backwards:}`

`text{Critical paths:}\ ADGLM\ text(and)\ AEHLM`

`:. \ text{7 activities have no float time:} \ ADEGHLM`
 

c.    `text{There are 9 possible paths}`

`ADGLM\ (19), AEHLM\ (19) , AEIM\ (14)`

`AEJK\ (17), BCEIM \ (13), BCEJK\ (16)`

`BCEHLM\ (18), BCDGLM \ (18), BFK (11)`

`text{Consider the 5 paths with completions over 16 weeks}`

` to \ text{all contain}\ A\ text{or}\ L \ text{or both.}`
  

`text{Consider} \ ADGLM, AEHLM \ (text{contains both} \ A\ text{and} \ L )`

`to \ text{reduce} \ L xx 2 \ ,  A xx 1 \ text{to reach 16 weeks}`

`to \ text{cheaper than} \ L xx 1 \ , \ A xx 2`
  

`text{Consider} \ BCEHLM , BCDGLM\ (text{both contain} \ L \ text{only})`

`to \ text{reduce} \ L xx 2 \ text{to reach 16 weeks}`
 

`text{Consider} \ AEJK \ ( text{contains} \ A \ text{only} )`

`to \ text{reduce} \ A xx 1 \ text{reach 16 weeks}.`
 

`:. \ text{Minimum cost to reduce time to 16 weeks}`

`= 2 xx 120\ 000 + 1 xx  140\ 000`

`= $ 380\ 000`

NETWORKS, FUR2 2021 VCAA 3

The network diagram below shows the local road network of Town `M`.

The number on the edges indicate the maximum number of vehicles per hour that can travel along each road in this network.

The arrows represent the permitted direction of travel.

The vertices `A, B, C, D, E` and `F` represent the intersections of the roads.
 

  1. Determine the maximum number of vehicles that can travel from the entrance to the exit per hour.   (1 mark)

  2. The local council plans to increase the number of vehicles per hour that can travel from the entrance to the exit by increasing the capacity of only one road.
  3.  i. Complete the following sentence by filling in the boxes provided.   (1 mark)

        The road that should have its capacity increased is the road from vertex  
     
    		  to 
     
  4. ii. What should be the minimum capacity of this road to maximise the flow of vehicles from the entrance to the exit?   (1 mark)

Show Answers Only
  1. `1330`
  2.  i. `A\ text(to)\ D`
  3. ii. `780`
Show Worked Solution

a.     

`text{Minimum cut}` `= 680 + 650`
  `= 1330`

 
`:. \ text{Maximum number of vehicles} = 1330`

 

b.i.  `text{Vehicles from} \ A \ text{to} \ B \ text{restricted to 700 per hour.}`

`:. \ text{Road to increase capacity is from} \ A \ text{to} \ D.`
 

b.ii. `text{Minimum cut} = 1330 \ text{(partial)}`

 `text{Next lowest cut} = 620 + 840 = 1460`
 


 

`text{Extra capacity} = 1460 – 1330 = 130`

`:. \ text{Minimum capacity of road} \ A \ text{to}\ D \ text{should be}`

`= 650 + 130`

`= 780`

MATRICES, FUR2 2021 VCAA 4

Five staff members in Elena's office played a round-robin video game tournament, where each employee played each of the other employees once. In each game there was a winner and a loser.

A table of their one-step and two-step dominances was prepared to summarise the results.
 

   

Consider the results matrix shown below.

A '1' in this matrix shows that the player named in that row defeated the player named in that column.

A '0' in this matrix shows that the player named in that row lost to the player named in that column.

Use all of the information provided to complete the results matrix.   (2 marks)

`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquad loser),(quadqquadqquadqquadqquadqquad \ \ I qquad\ J qquad \ K qquad\ L qquad M),(wi\n\n\er qquad{:(I),(J),(K),(L),(M):}[(0,…,…,…,…),(…,0,…,…,…),(0,0,0,1,0),(…,…,…,0,…),(…,…,…,…,0)]):}`

Show Answers Only

`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquad ),(quadqquadqquad \ \ \ I \ \ \ J  \ \ K \ \ L \ \ \ M),(qquad{:(I),(J),(K),(L),(M):}[(0,1,1,1,0),(0,0,1,1,1),(0,0,0,1,0),(0,0,0,0,1),(1,0,1,0,0)]):}`

Show Worked Solution

`text{Katie only beat Leslie} \ => \ text{everyone else beat Katie}`

`{:(qquadqquadqquadqquadqquad),(quadqquadqquadqquad I qquad\ J qquad \ K qquad\ L qquad M),(qquad{:(I),(J),(K),(L),(M):}[(0,…,…,…,…),(…,0,…,…,…),(0,0,0,1,0),(…,…,…,0,…),(…,…,…,…,0)]):} \ to \ {:(qquadqquadqquadqquadqquadqquadqquadqquadqquad),(quadqquadqquadqquad I qquad\ J qquad \ K qquad\ L qquad M),(qquad{:(I),(J),(K),(L),(M):}[(0,…,1,…,…),(…,0,1,…,…),(0,0,0,1,0),(…,…,0,0,…),(…,…,1,…,0)]):}`
 

`text{Katie only has one 2-step dominances. She only beat Leslie.}`

`=>\ text{Leslie only won 1 game (note she has two 2-step dominances)}`

`=>\ text{Leslie must have beaten Mikki (she is the only person with two 1-step dominances)}`

`to \ {:(qquadqquadqquadqquadqquad),(quadqquadqquadqquad I quad\ J qquad \ K qquad\ L qquad M),(qquad{:(I),(J),(K),(L),(M):}[(0,…,1,1,…),(…,0,1,1,…),(0,0,0,1,0),(0,0,0,0,1),(…,…,1,1,0)]):} `
 

`text{Ike beat either Jolene or Mikki}`

`=> \ text{Jolene has more wins than Mikki and since}`

`  qquadqquad text{Ike has the most two-step dominances}`

`=> \ text{Ike beat Jolene and Jolene beat Mikki}`

`to \ {:(qquadqquadqquadqquadqquadqquadqquadqquadqquad),(quadqquadqquad \ \ \ I \ \ J  \ \ K \ \ L \ \  M),(qquad{:(I),(J),(K),(L),(M):}[(0,1,1,1,0),(0,0,1,1,1),(0,0,0,1,0),(0,0,0,0,1),(1,0,1,0,0)]):}`

MATRICES, FUR2 2021 VCAA 3

A market research study of shoppers showed that the buying preferences for the three olive oils, Carmani (`C`), Linelli (`L`) and Ohana (`O`), change from month to month according to the transition matrix  `T` below.

`qquadqquadqquadqquad \ text(this month)`

`T = {:(qquad\ C quadquadqquad \ L quadquad \ O ),([(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:(C),(L),(O):} qquad text(next month)):}`
 

The initial state matrix `S_0` below shows the number of shoppers who bought each brand of olive oil in July 2021.

`S_0 = {:[(3200),(2000),(2800)]{:(C),(L),(O):} :}`

Let `S_n` represent the state matrix describing the number of shoppers buying each brand `n` months after July 2021.

  1. How many of these 8000 shoppers bought a different brand of olive oil in August 2021 from the brand bought in July 2021?   (1 mark)

  2. Using the rule `S_(n+1) = T xx S_(n)`, complete the matrix `S_1` below.   (1 mark)

`S_1 = {:[(3060),(text{_____}),(text{_____})]{:(C),(L),(O):} :}`

  1. Consider the shoppers who were expected to buy Carmani olive oil in August 2021.
  2. What percentage of these shoppers also bought Carmani olive oil in July 2021?
  3. Round your answer to the nearest percentage.   (1 mark)

  4. Write a calculation that shows Ohana olive oil is the brand bought by 50% of these shoppers in the long run.   (1 mark)

  5. Further research suggests more shoppers will buy olive oil in the coming months.
  6. A rule to model this situation is  `R_(n+1) = T xx R_n + B`, where `R_n` represents the state matrix describing the number of shoppers `n` months after July 2021.

`qquadqquadqquadqquad \ text(this month)`
    `T = {:(qquad\ C quadquadqquad \ L quadquad \ O ),([(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:(C),(L),(O):} ):} qquad text(next month) \ , \ B = {:[(200),(100),(k)]{:(C),(L),(O):} :}, \ R_0 = {:[(3200),(2000),(2800)]{:(C),(L),(O):} :}`

  1. `k` represents the extra number of shoppers expected to buy Ohana olive oil each month.
  2. If  `R_2 = {:[(3333),(2025),(3642)]:}`, what is the value of `k`?   (1 mark)

Show Answers Only
  1. `1160`
  2. `L = 1900 \ , \ O = 3040`
  3. `89text(%)`
  4. `text(See Worked Solutions)`
  5. `200`
Show Worked Solution

a.    `text{Consider matrix} \ T`

`text{Carmani – 15% bought new brand}`

`text{Linelli – 20%, Ohana – 10%}`

`:.  text{Shoppers}` `= 0.15 xx 3200 xx + 0.2 xx 2000 + 0.1 xx 2800`
  `= 1160`

 
b.    `S_1 = [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:[(3200),(2000),(2800)]:} = {:[(3060),(1900),(3040)]:}`
 

`:. \ L = 1900 \ , \ O = 3040`
 

c.    `text{Carmani purchasers in August)} \ = 3060 \ text{(see part b)}`

`text{Carmani purchasers in July} = 3200`

`text{Carmani purchasers in both July and August}`

`= 0.85 xx 3200`

`= 2720`
 

`:.\ text{% of August purchasers who bought in July}`

`= 2720/3060 xx 100`

`= 88.88 …`

`=89text(%)`
 

d.    `S = T^50 xx S_0 = {:[(2400),(1600),(4000)]:}`

`text{Total shoppers} = 8000`

`:.\ text{Ohana purchasers (in long run)}`

`= 4000/8000 xx 100`

`= 50text(%)`
 

e.     `R_1` `= [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)] [(3200),(2000),(2800)] + [(200),(180),(k)]` 
  `R_2` `= [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)] [(3260),(2000),(k+3040)] + [(200),(100),(k)]`
    `= [(3171 + 0.05 (k + 3040)),(text{not required}),(text{not required})]`

 
`text{Equating matrices, solve for} \ k:`

`3171 + 0.05 (k + 3040) = 3333`

`:. k = 200`

CORE, FUR2 2021 VCAA 9

Sienna invests $152 431 into an annuity from which she will receive a regular monthly payment of $900 for 25 years. The interest rate for this annuity is 5.1% per annum, compounding monthly.

  1. Let `V` be the balance of the annuity after `n` monthly payments. A recurrence relation written in terms of `V_0 , V_{n + 1}` and `V_n` can model the value of this annuity from month to month.
  2. Showing recursive calculations, determine the value of the annuity after two months.
  3. Round your answer to the nearest cent.   (2 marks)

  4. After two years, the interest rate for this annuity will fail to 4.6%.
  5. To ensure that she will still receive the same number of $900 monthly payments, Sienna will add an extra one-pff amount into the annuity at this time.
  6. Determine the value of this extra amount that Sienna will add.
  7. Round your answer to the nearest cent.   (1 mark)

Show Answers Only
  1. `$151 \ 925.59`
  2. `$7039.20`
Show Worked Solution

a.   `text{Payments are monthly} \ => \ r = 5.1/12 = 0.425text(%) \ text{per month}`

`R` `= 1 + r/100 = 1.00425`
`V_1` `= RV_0-text{payment}`
  `= 1.00425 xx 152 \ 431-900`
  `= $152\ 178.83`
`:.V_2` `= 1.00425 xx 152\ 178.83-900`
  `= $ 151 \ 925.59`

 

b.      `text{Find} \ V_24 \ text{(annuity value after 2 years) by TVM Solver:}`

`N` `= 24`
`I text{(%)}` `= 5.1`
`PV` `= -152 \ 431`
`PMT` `= 900`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

 
`=> FV = 146 \ 073.7405`
 

`text{Find}\ PV\ text{of annuity needed (by TVM solver):}`

`N` `= 243 xx 12 = 276`
`Itext{(%)}` `= 4.6`
`PV` `= ?`
`PMT` `= 900`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 
`=> PV = -153\ 112.9399`
 

`:. \ text{Amount to add}` `= 153 \ 112.94-146 \ 073.74`
  `= $ 7039.20`

CORE, FUR2 2021 VCAA 8

For renovations to the coffee shop, Sienna took out a reducing balance loan of $570 000 with interest calculated fortnightly.

The balance of the loan, in dollars, after `n` fortnights, `S_n` can be modelled by the recurrence relation
 

`S_0 = 570 \ 000,`                  `S_{n+1} = 1.001 S_n - 1193`
 

  1. Calculate the balance of this loan after the first fortnightly repayment is made.   (1 mark)

  2. Show that the compound interest rate for this loan is 2.6% per annum.   (1 mark)

  3. For the loan to be fully repaid, to the nearest cent, Sienna's final repayment will be a larger amount.
  4. Determine this final repayment amount.
  5. Round your answer to the nearest cent.   (1 mark)

Show Answers Only
  1. `$ 569 \ 377`
  2. `2.6text(%)`
  3. `$ 1198.56`
Show Worked Solution
a.   `S_1` `= 1.001 xx 570 \ 000-1193`
    `= $ 569 \ 377`


b.  `text{Fortnights in 1 year} = 26`

`text{Rate per fortnight} = (1.001-1) xx 100text(%) = 0.1text(%)`
 

`:.\ text(Annual compound rate)` `= 26 xx 0.1text(%)`
  `= 2.6text(%)`

 

c.  `text{Find}\ N \ text{by TVM Solver:}`

`N` `= ?`
`I text{(%)}` `= 2.6`
`PV` `= 570 \ 000`
`PMT` `= -1193`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 26`

 
`=> N = 650.0046 …`
 

`text{Find} \ FV \ text{after exactly 650 payments:}`

`N` `= 650`
`Itext{(%)}` `= 2.6`
`PV` `= 570 \ 000`
`PMT` `= -1193`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 26`

 
`=> FV = -5.59`
 

`:. \ text{Final repayment}` `= 1193 + 5.59`
  `= $ 1198.59`

Mechanics, SPEC2 2021 VCAA 5

A mass of `m_1` kilograms is placed on a plane inclined at 30° to the horizontal. It is connected by a light inextensible string to a second mass of `m_2` kilograms that hangs below a frictionless pulley situated at the top end of the incline, over which the string passes.
 


 

  1. Given that the inclined plane is smooth, find the relationship between `m_1` and `m_2` if the mass `m_1` moves down the plane at constant speed.  (2 marks)

The masses are now placed on a rough plane inclined at 30°, with the light inextensible string passing over a frictionless pulley in the same way, as shown in the diagram above. Let `N` be the magnitude of the normal force exerted on the mass `m_1` by the plane. A resistance force of magnitude `lambdaN` acts on and opposes the motion of the mass `m_1`.

  1. The mass `m_1` moves up the plane.
  2.   i. Mark and label all forces acting on this mass on the diagram above.  (1 mark)
  3.  ii. Taking the direction up the plane as positive, find the acceleration of the mass `m_1` in terms of `m_1`, `m_2` and `lambda`.  (2 marks)

Some time after the masses have begun to move, the mass `m_2` hits the ground at 4.5 ms`\ ^(-1)` and the string becomes slack. At this instant, the mass `m_1` is at the point `P` on the plane, which is 2 m from the pulley. Take the value of `lambda` to be 0.1

  1. How far from point `P` does the mass `m_1` travel before it starts to slide back down the plane?
  2. Give your answer in metres, correct to two decimal places.  (2 marks)
  3. Find the time taken, from when the string becomes slack, for the mass `m_1` to return to point `P`.
  4. Give your answer correct to the nearest tenth of a second.  (3 marks)
Show Answers Only
  1. `2m_2`
  2. i. 
       
     
  3. ii. `a = ((2m_2 – m_1 – lambda m_1 sqrt3)g)/(m_1 + m_2)`
  4. `s = 1.76\ text{m}`
  5. `1.7\ text{seconds}`
Show Worked Solution

a.   `m_1g sin30 – m_2g = (m_1 + m_2)a`

`text(S)text(ince)\ m_1\ text(moves at constant speed,)\ a = 0`

`m_1 g · 1/2 – m_2 g` `= 0`
`m_1` `= 2m_2`

 
b.i.   

 

b.ii.   `text(S)text(ince)\ m_1\ text(is moving up the slope)`

♦ Mean mark part (b)(ii) 43%.

`m_2g – m_1 g · 1/2 – lambdam_1 g · cos 30` `= (m_1 + m_2)a`
`m_2g – m_1 g · 1/2 – lambdam_1 g · cos sqrt3/2` `= (m_1 + m_2)a`

`:. a = ((2m_2 – m_1 – lambda m_1 sqrt3)g)/(2(m_1 + m_2))`

 

c.   `text(After)\ m_2\ text(hits the ground)`

♦♦♦ Mean mark part (c) 17%.
`m_1a` `=-m_1g*1/2 – lambda m_1 g sqrt3/2`
`a` `= -g/2(1 + lambda sqrt3)`
  `= -g/2(1 + 0.1 xx sqrt3)`

 
`text(By CAS, solve)\ \ v^2 = u^2 + 2as,\ text(for)\ \ s:`

`0 = 4.5^2 – g(1 + 0.1 xx sqrt3)s`

`s = 1.76\ text{m (to 2 d.p.)}`

 

d.   `u = 4.5\ \ text(ms)^(-1), s = 1.76\ text(m),\ a = -g/2(1 + 0.1sqrt3)`

♦♦♦ Mean mark part (d) 7%.

`text(Let)\ \ t_1 = text(Time travelling up slope until stopping)`

`s = ut_1 + 1/2at^2`

`1.76 = 4.5t_1 – 1/2 · g/2(1 + 0.1sqrt3)t_1^2`

`t_1 = 0.78\ text(seconds)`
 

`text(Let)\ t_2 =\ text(time travelling down the slope)`

`=>\ text(friction is reversed)`

`a` `= g/2(1 – 0.1sqrt3)`
`1.76` `= 0 xx t_2 + 1/2 · g/2(1 – 0.1sqrt3)  t_2^2`
`t_2` `= 0.93\ text(seconds)`

 

`:.\ text(Total time)` `= t_1 + t_2`
  `= 0.78 + 0.93`
  `= 1.7\ text{seconds (to 1 d.p.)}`

Vectors, SPEC2 2021 VCAA 4

A car that performs stunts moves along a track, as shown in the diagram below. The car accelerates from rest at point `A`, is launched into the air by the ramp `BO` and lands on a second section of track at or beyond point `C`.  This second section of track is inclined at `10^@` to the horizontal.

Due to tailwind, the effect of air resistance is negligible. Point `O` is taken as the origin of a cartesian coordinate system and all displacements are measured in metres. Point `C` has the coordinates `(16, 4)`.

At point `O`, the speed of the car is `u` ms`\ ^(-1)` and it takes off at an angle of `theta` to the horizontal direction. After the car passes point `O`, it follows a trajectory where the position of the car's rear wheels relative to point `O`, is given by

`underset~r(t) = ut cos(theta) underset~i + (ut sin(theta)-1/2 g t^2)underset~j`  until the car lands on the second section of track that starts at point `C`.
 

  1. Show that the path of the rear wheels of the car, while in the air, is given in cartesian form by
  2.    `y = x tan(theta)-(4.9x^2)/(u^2cos^2(theta))`.   (1 mark)

  3. If  `theta = 30^@`, find the minimum speed, in ms`\ ^(-1)`, that the car must reach at point `O` for the rear wheels to land on the second section of track at or beyond point `C`. Give your answer correct to two decimal places.   (2 marks)

  4. The ramp  `BO`  is constructed so that the angle `theta` can be varied.
  5. For what values of `theta` and `u` will the path of the rear wheels of the car join up smoothly with the beginning of the second section of track at point `C`? Give your answer for `theta` in degrees, correct to the nearest degree, and give your answer for `u` in ms`\ ^(-1)`, correct to one decimal place.   (3 marks)

The car accelerates from rest along the horizontal section of track `AB`, where its acceleration, `a`  ms`\ ^(-2)`, after it has travelled `s` metres from point `A`, is given by  `a = 60/v`, where `v` is its speed at `s` metres.

  1. Show that `v` in terms of `s` given by  `v = (180x)^(1/3)`.   (2 marks)

  2. After the car leaves point `A`, it accelerates to reach a speed of 20 ms`\ ^(-1)` at point `B`. However, if the stunt is called off, the car immediately brakes and reduces its speed at a rate of 9 ms`\ ^(-2)`.  It is only safe to call off the stunt if the car can come to rest at or before point `B`.  Point `W` is the furthest point along the section `AB` at which the stunt can be called off.
  3. How far is point `W` from point `B`?  Give your answer in metres, correct to one decimal place.   (3 marks)

Show Answers Only
  1. `text(Show Worked Solutions)`
  2. `17.87\ text(ms)^-1`
  3. `u = 16.4\ text(ms)^-1, \ theta = 34.1^@`
  4. `text(Show Worked Solutions)`
  5. `16.4\ text{m}`
Show Worked Solution

a.   `x = ut costheta\ …\ (1)`

`y=ut sin theta-1/2 g t^2\ …\ (2)`

`text{Using (1):}`

`t = x/(u costheta)\ …\ (3)`

`text{Substitute (3) into (2):}`

`y` `= u · x/(u costheta) sintheta-4.9 · (x^2)/(u^2cos^2theta)`
  `= x tan theta-(4.9 x^2)/(u^2 cos^2 theta)`

 

b.   `text(When)\ \ theta = 30^@`

♦ Mean mark part (b) 45%.

`y = x\ tan 30^@-(4.9 x^2)/(u^2 cos^2 30^@)`

`y = x/sqrt3-(4.9x^2)/(u^2) · 4/3`

`text{Substitute (16, 4) into equation and solve for}\ u:`

`4 = 16/sqrt3-(4.9 xx 16^2)/(u^2) · 4/3`

`u = 17.87\ text(ms)^-1\ \ text{(to 2 d.p.)}`

 

c.   `text(Smooth join will occur when)`

♦♦♦ Mean mark part (c) 12%.

`y(16) = 4\ …\ (1), \ \ text{and}`

`text(Gradient of motion equals the gradient of landing ramp)`

`(dy)/(dx)|_(x = 16) = -tan10^@\ …\ (2)`

`text{Solve (1) and (2) by CAS for}\ \ u, theta:`

`u = 16.4\ text(ms)^-1`

`theta = 34.1^@`

 

d.   `a = 60/v`

♦ Mean mark part (d) 40%.

`v · (dv)/(ds) = 60/v`

`int v^2 dv = int 60\ ds`

`1/3 v^3 = 60s + c`

`text(When)\ \ s = 0, v = 0 \ => \ c = 0`

`1/3 v^3` `= 60s`
`v^3` `= 180s`
`v` `= (180s)^(1/3)`

 

e.   `text(Find)\ s\ text{at point}\ B\ text{(by CAS):}`

♦♦♦ Mean mark part (e) 8%.
`(180s)^(1/3)` `= 20`
`s` `= (400)/9\ text(m)`

 
`text(Car decelerates at 9 ms)^(-1)`

`d/(ds)(1/2v^2)` `= -9`
`1/2 v^2` `= -9s + c`

 
`text(When)\ \ s = 400/9, v = 0 \ => \ c = 400`

`1/2 v^2` `= 400-9s`
`v` `= sqrt(800-18s)`

 
`text{Solve for}\ s\ text{(by CAS):}`

`(180s)^(1/3)` `= sqrt(800-18s)`
`s` `= 28.08`

 
`text(Distance of)\ W\ text(from)\ B`

`= 400/9-28.08`

`= 16.4\ text{m (to 1 d.p.)}`

Calculus, SPEC2 2021 VCAA 3

A thin-walled vessel is produced by rotating the graph of  `y = x^3-8`  about the `y`-axis for  `0 <= y <= H`.

All lengths are measured in centimetres.

    1. Write down a definite integral in terms of `y` and `H` for the volume of the vessel in cubic centimetres.   (1 mark)

    2. Hence, find an expression for the volume of the vessel in terms of `H`.   (1 mark)

Water is poured into the vessel. However, due to a crack in the base, water leaks out at a rate proportional to the square root of the depth `h` of water in the vessel, that is  `(dV)/(dt) = -4sqrth`, where `V` is the volume of water remaining in the vessel, in cubic centimetres, after `t` minutes.

    1. Show that  `(dh)/(dt) = (-4sqrth)/(pi(h + 8)^(2/3))`.   (2 marks)

    2. Find the maximum rate, in centimetres per minute, at which the depth of water in the vessel decreases, correct to two decimal places, and find the corresponding depth in centimetres.   (2 marks)

    3. Let  `H = 50`  for a particular vessel. The vessel is initially full and water continues to leak out at a rate of  `4 sqrth`  cm³ min`\ ^(-1)`.
    4. Find the maximum rate at which water can be added, in cubic centimetres per minute, without the vessel overflowing.   (1 mark)

  2. The vessel is initially full where  `H = 50`  and water leaks out at a rate of  `4sqrth`  cm³ min`\ ^(-1)`. When the depth of the water drops to 25 cm, extra water is poured in at a rate of  `40sqrt2`  cm³ min`\ ^(-1)`.
  3. Find how long it takes for the vessel to refill completely from a depht of 25 cm. Give your answer in minutes, correct to one decimal place.   (3 marks)

Show Answers Only
    1. `V = pi int_0^H (y + 8)^(2/3)\ dy`
    2. `V = (3pi)/5 [(H + 8)^(5/3)-32]`
    1. `text(See Worked Solutions)`
    2. `0.62\ text(cm/min when)\ \ h = 24.`
    3. `4sqrt50\ \ text(cm³/min)`
  3. `31.4\ text(mins)`
Show Worked Solution

a.i.   `y = x^3-8 \ => \ x = root3(y + 8) \ => \ x^2 = (y + 8)^(2/3)`

`:. V = pi int_0^H (y + 8)^(2/3)\ dy`

 

a.ii.   `V = (3pi)/5 [(H + 8)^(5/3)-32]`

 

b.i.   `(dV)/(dt) = -4sqrth`

`(dV)/(dh) = pi(h + 8)^(2/3) \ => \ (dh)/(dV) =1/(pi(h + 8)^(2/3))`

`(dh)/(dt)` `= (dV)/(dt) * (dh)/(dV)`
  `= (-4sqrth)/(pi(h + 8)^(2/3))`

 

b.ii.   `text(Solve)\ (d^2h)/(dt^2) = 0\ \ text(for)\ \ h\ \ text{(by CAS):}`

♦ Mean mark part (b)(ii) 42%.

`(d^2h)/(dt^2) = (2(h-24))/(3sqrth pi (h + 8)^(5/3))=0`

`=>h = 24`

`text(At)\ \ h=24, \ (dh)/(dt) = -0.62\ text(cm/min)`

`:.\ text(Max rate at which depth decreases is)`

`0.62\ text(cm/min when)\ \ h = 24.`

♦♦ Mean mark part (b)(iii) 24%.

 

b.iii.   `text(At)\ \ H = 50, text(vessel is full and losing water at)\ \ 4sqrt50\ \ text(cm³/min)`

`:. text(Water can be added at a max-rate of)\ \ 4sqrt50\ \ text(cm³/min and)`

`text(vessel will not overflow.)`

 

c.   `(dV)/(dt) = 40sqrt2-4sqrth`

♦♦♦ Mean mark part (c) 16%.

`(dV)/(dh) · (dh)/(dt) = 40sqrt2-4sqrth`

`pi(h + 8)^(2/3) · (dh)/(dt)` `= 40sqrt2-4sqrth`
`(dh)/(dt)` `= (40sqrt2-4sqrth)/(pi(h + 8)^(2/3)`
`(dt)/(dh)` `= (pi(h + 8)^(2/3))/(40sqrt2-4sqrth)`
`t` `= int (pi(h + 8)^(2/3))/(40sqrt2-4sqrth)\ dh`

 
`text(Time of vessel to refill from)\ \ h = 25\ \ text(to)\ \ h = 50:`

`t` `= int_25^50 (pi(h + 8)^(2/3))/(40sqrt2-4sqrth)\ dh`
  `~~ 31.4\ text(mins)`

Calculus, SPEC2 2021 VCAA 1

Let  `f(x) = ((2x-3)(x + 5))/((x-1)(x + 2))`.

  1. Express `f(x)` in the form  `A + (Bx + C)/((x-1)(x + 2))`, where `A`, `B` an `C` are real constants.   (1 mark)

  2. State the equation of the asymptotes of the graph of `f`.   (2 marks)

  3. Sketch the graph of `f` on the set of axes below. Label the asymptotes with their equations, and label the maximum turning point and the point of inflection with their coordinates, correct to two decimal places. Label the intercepts with the coordinate axes.   (3 marks)

     

     

  4. Let  `g_k(x) = ((2x-3)(x + 5))/((x-k)(x + 2))`, where `k` is a real constant.
  5.  i. For what values of `k` will the graph of `g_k`, have two asymptotes?   (2 marks)

  6. ii. Given that the graph of `g_k` has more than two asymptotes, for what values of `k` will the graph of `g_k` have no stationary points?   (2 marks)

Show Answers Only
  1. `f(x) = 2 + (5x-11)/((x-1)(x + 2))`
  2. `text(Horizontal asymptote:)\ y = 2`

  3.  
  4.  i. `k = 3/2 => g_k(x) = 2 + 6/(x + 2)`
  5. ii. `k , -5\ \ text(or)\ \ k > 3/2`
Show Worked Solution

a.   `text{By CAS  (prop Frac}\ f(x)):`

`f(x) = 2 + (5x-11)/((x-1)(x + 2))`
 

b.   `text(Vertical asymptotes:)\ x = 1, x = –2`

`text(As)\ \ x -> ∞, y -> 2`

`text(Horizontal asymptote:)\ y = 2`

♦ Mean mark part (c) 48%.

 
c.   

 

d.i.   `text(Two asymptotes only when:)`

♦♦ Mean mark part (d)(i) 24%.

`k = -2 \ => \ g_k(x) = 2-(23 + x)/((x + 2)^2)`

`k = -5 \ => \ g_k(x) = 2-7/(x + 2)`

`k = 3/2 \ => \ g_k(x) = 2 + 6/(x + 2)`

 

d.ii.   `text(By CAS, solve)\ \ d/(dx)(g_k(x)) = 0\ \ text(for)\ \ x:`

♦♦♦ Mean mark part (d)(ii) 16%.

`x = (-4k + 15 ± sqrt(-21(2k^2 + 7k-15)))/(2k + 3)`
 

`text(No solutions occur when:)`

`k = -3/2\ \ text(or)`

`2k^2 + 7k-15 < 0`

`=> k < -5\ \ text(or)\ \ k > 3/2`

Calculus, MET2 2021 VCAA 5

Part of the graph of  `f: R to R , \ f(x) = sin (x/2) + cos(2x)`  is shown below.
 

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

  2. State the minimum value of `f`, correct to three decimal places.   (1 mark)

  3. Find the smallest positive value of `h` for which  `f(h-x) = f(x)`.   (1 mark)
Consider the set of functions of the form  `g_a : R to R, \ g_a (x) = sin(x/a) + cos(ax)`, where `a` is a positive integer.
  1. State the value of `a` such that  `g_a (x) = f(x)`  for all `x`.   (1 mark)

  2. i.  Find an antiderivative of `g_a` in terms of `a`.   (1 mark)

  3. ii. Use a definite integral to show that the area bounded by `g_a` and the `x`-axis over the interval  `[0, 2a pi]`  is equal above and below the `x`-axis for all values of `a`.  (3 marks)

  4. Explain why the maximum value of `g_a` cannot be greater than 2 for all values of `a` and why the minimum value of `g_a` cannot be less than –2 for all values of `a`.   (1 mark)

  5. Find the greatest possible minimum value of `g_a`.   (1 mark)

Show Answers Only
  1. `4 pi`
  2. `-1.722`
  3. `2 pi`
  4. `text{See Worked Solutions}`
  5. i.  `text{See Worked Solutions}`
    ii. `text{See Worked Solutions}`
  6. `text{See Worked Solutions}`
  7. `-sqrt2`
Show Worked Solution

a.    `text{By inspection, graph begins to repeat after 4pi.}`

`text{Period}\ = 4 pi`
 

b.    `text{By CAS: Sketch}\ \ f(x) = sin (x/2) + cos(2x)`

`f_min = -1.722`

♦ Mean mark part (c) 21%.
 

c.     `text{If} \ \ f(x)\ \ text{is reflected in the} \ y text{-axis and translated} \ 2 pi \ text{to the right} => text{same graph}`

`f(x) = f(-x + h) = f(2 pi-x)`

`:. h = 2 pi`
 

d.    `f(x) = sin(x/2) + cos(2x)`

`g_a(x) = sin(x/a) + cos(2a)`

`g_a(x) = f(x) \ \ text{when} \ \ a = 2`
 

e.i.  `int g_a (x)\ dx`

♦ Mean mark part (e)(i) 50%.

`= -a cos (x/a) + {sin (ax)}/{a} , \ (c = 0)`
 

e.ii.   `int_0^{2a pi} g_a(x)\ dx`

♦ Mean mark part (e)(ii) 29%.

`= {sin (2a^2 pi)}/{a}`

`= 0 \ \ (a ∈ ZZ^+)`
 

`text{When integral = 0, areas above and below the} \ x text{-axis are equal.}`
 

f.    `g_a (x) = sin(x/a) + cos (ax)`

♦ Mean mark part (f) 13%.

`-1 <= sin(x/a) <= 1 \ \ text{and}\ \ -1 <= cos(ax) <= 1`

`:. -2 <= g_a (x) <= 2`
 

g.    `text{Sketch}\ \ g_a (x) \ \ text{by CAS}`

♦ Mean mark part (g) 2%.

`text{Minimum access at}\ \ a = 1`

`g_a(x)_min = – sqrt2`

Probability, MET2 2021 VCAA 4

A teacher coaches their school's table tennis team.

The teacher has an adjustable ball machine that they use to help the players practise.

The speed, measured in metres per second, of the balls shot by the ball machine is a normally distributed random variable `W`.

The teacher sets the ball machine with a mean speed of 10 metres per second and standard deviation of 0.8 metres per second.

  1. Determine  `text(Pr) (W ≥11)`, correct to three decimal places.   (1 mark)

  2. Find the value of `k`, in metres per second, which 80% of ball speeds are below. Give your answer in metres per second, correct to one decimal place.   (1 mark) 

The teacher adjusts the height setting for the ball machine. The machine now shoots balls high above the table tennis table.

Unfortunately, with the new height setting, 8% of balls do not land on the table.

Let  `overset^P`  be the random variable representing the sample proportion of the balls that do not land on the table in random samples of 25 balls.

  1. Find the mean and the standard deviation of  `overset^P`.   (2 marks)

  2. Use the binomial distribution to find  `text(Pr) (overset^P > 0.1)`, correct to three decimal places.   (2 marks) 

The teacher can also adjust the spin setting on the ball machine.

The spin, measured in revolutions per second, is a continuous random variable  `X` with the probability density function
 

       `f(x) = {(x/500, 0 <= x < 20), ({50-x}/{750}, 20 <= x <= 50), (\ 0, text(elsewhere)):}`
 

  1. Find the maximum possible spin applied by the ball machine, in revolutions per second.   (1 mark)

Show Answers Only

  1. `0.106`
  2. `0.8`
  3. `sqrt46/125`
  4. `0.323`
  5. `50 \ text{revolutions per second}`
  6. This content is no longer in the Study Design.
  7. This content is no longer in the Study Design.
  8. This content is no longer in the Study Design.

Show Worked Solution

a.   `W\ ~\ N (10,0.8^2)`

`text{By CAS: norm Cdf} \ (11, ∞, 10, 0.8)`

`text(Pr) (W >= 11) = 0.106`
 

b.    `text(Pr) (W < k) = 0.8 \ \ \ text{By CAS: inv Norm} \ (0.8, 10, 0.8)`
 

c.    `E(overset^P) =  0.08 = 2/25`

♦ Mean mark part (c) 45%.

`text(s.d.) (overset^P) = sqrt{{0.08 xx 0.92}/{25}} = sqrt46/125`
 

d.    `X\ ~\ text(Bi) (25, 0.08)`

♦ Mean mark part (d) 49%.

`text{By CAS: binomCdf} \ (25, 0.08, 3, 25)`

`text(Pr) (overset^P > 0.1)` `= text(Pr) (X > 0.1 xx 25)`
  `= text(Pr) (X > 2.5)`
  `= text(Pr) (X >= 3)`
  `= 0.323`

 

e.    `text{Maximum spin = 50 revolutions per second}`

♦♦ Mean mark part (e) 21%.

 

f.    This content is no longer in the Study Design.

g.    This content is no longer in the Study Design.

h.    This content is no longer in the Study Design.

Calculus, MET2 2021 VCAA 3

Let  `q(x) = log_e (x^2-1)-log_e (1-x)`.

  1. State the maximal domain and the range of `q`.   (2 marks)

  2.  i. Find the equation of the tangent to the graph of `q` when  `x =-2`.   (1 mark)

  3. ii. Find the equation of the line that is perpendicular to the graph of `q` when  `x =-2`  and passes through the point  (-2, 0).   (1 mark)

Let  `p(x) = e^{-2x}-2e^-x + 1.`

  1. Explain why `p` is not a one-to one function.   (1 mark)

  2. Find the gradient of the tangent to the graph of `p` at  `x = a`.   (1 mark)

The diagram below shows parts of the graph of `p` and the line  `y = x + 2`.
 
 
                     
 
The line  `y = x + 2`  and the tangent to the graph of `p` at  `x = a`  intersect with an acute angle of `theta` between them.

  1. Find the value(s) of `a` for which  `theta = 60^@`. Give your answer(s) correct to two decimal places.   (3 marks)

  2. Find the `x`-coordinate of the point of intersection between the line  `y = x + 2` and the graph of `p`, and hence find the area bounded by  `y = x + 2`, the graph of `p` and the `x`-axis, both correct to three decimal places.   (3 marks)

Show Answers Only
  1. `text{Domain:} \ x ∈ (-∞, -1)`
    `text{Range:} \ y ∈ R`
  2. i.  `-x-2`
    ii. `x + 2`
  3. `text{Not a one-to-one function as it fails the horizontal line test.}`
  4. `-2e^{-2a} + 2e^{-a}`
  5. `-0.67`
  6. `1.038\ text(u)^2`
Show Worked Solution

a.    `text(Method 1)`

`x^2-1 > 0 \ \ => \ \ x > 1 \ \ ∪ \ \ x < -1`

`1-x > 0 \ \ => \ \ x < 1`
 
`:. \ x ∈ (-∞, -1)`
 

`text(Method 2)`

`text{Sketch graph by CAS}`

`text{Asymptote at} \ x = -1`

`text{Domain:} \ x ∈ (-∞, -1)`

`text{Range:} \ y ∈ R`
 
 

b.     i.   `text{By CAS (tanLine} \ (q (x), x, -2)):`

`y = -x-2`
 

ii.   `text{By CAS (normal} \ (q (x), x, -2)):`

`y = x + 2`
  

c.    `text{Sketch graph by CAS.}`

`p(x) \ text{is not a one-to-one function as it fails the horizontal line test}`

`text{(i.e. it is a many-to-one function)}`
 

d.   `p^{′}(x) = -2e^{-2x} + 2e^-x`

`p^{prime}(a) = -2e^{-2a} + 2e^{-a}`
 
 

e. 

 
`text{Case 1}`

`text{By CAS, solve:}`

`2e^{-a} -2e^{-2a} =-tan (15^@) \ \ text{for}\ a:`

`a = -0.11`
 

`text{Case 2}`

`text{Case 1}`

`text{By CAS, solve:}`

`2e^{-a}-2e^{-2a} = -tan 75^@\ \ text{for}\ a:`

`a = -0.67`
 

f.     `text{At intersection,}`

`x + 2 = e^{-2x} -2e^{-x} + 1`

`x = -0.750`
 

`text{Area}` `= int_{-2}^{-0.750} x + 2\ dx + int_{-0.750}^0 e^{-2x}-2e^{-x} + 1\ dx`
  `= 1.038 \ text(u)^2`

Calculus, MET2 2021 VCAA 2

Four rectangles of equal width are drawn and used to approximate the area under the parabola  `y = x^2`  from  `x = 0`  to  `x = 1`.

The heights of the rectangles are the values of the graph of  `y = x^2` at the right endpoint of each rectangle, as shown in the graph below.
 

  1. State the width of each of the rectangles shown above.  (1 mark)
  2. Find the total area of the four rectangles shown above.  (1 mark)
  3. Find the area between the graph of  `y = x^2`, the `x`-axis and the line  `x=1`.  (2 marks)
  4. The graph of `f` is shown below.
     
         

    Approximate  `int_(-2)^2 f(x)\ dx`  using four rectangles of equal width and the right endpoint of each rectangle.  (1 mark)

Parts of the graphs of  `y = x^2`  and  `y = sqrtx`  are shown below.
 
     

  1. Find the area of the shaded region.  (1 mark)
  2. The graph of  `y=x^2` is transformed to the graph of  `y = ax^2`, where  `a ∈ (0, 2]`.
  3. Find the values of `a` such that the area defined by region(s) bounded by the graphs of  `y = ax^2`  and  `y = sqrtx`  and the lines  `x = 0`  and  `x = a`  is equal to `1/3`. Give your answer correct to two decimal places.  (4 marks)
Show Answers Only
  1. `0.25`
  2. `15/32\ text(u)^2`
  3. `1/3\ text(u)^2`
  4. `-2`
  5. `1/3\ text(u)^2`
  6. `1.13`
Show Worked Solution

a.   `0.25`

 

b.    `text{Area}` `= 1/4 (1/16 + 1/4 + 9/16 + 1)`
    `= 15/32\ text(u)^2`

 

c.    `text{Area}` `= int_0^1 x^2 dx`
    `= 1/3\ text(u)^2`

 

d.

♦♦♦ Mean mark part (d) 16%.

`int_-2^2 f(x)\ dx` `≈ 6 + 2 – 4 – 6`
  `≈ -2`

 

e.   `text{Area}` `= int_0^1 (sqrtx – x^2) dx`
    `= 1/3\ text(u)^2`

 
f.   `text{Case 1:} \ a ≤ 1`

♦♦♦ Mean mark part (f) 18%.

`int_0^a (sqrtx – ax^2) dx = 1/3`

`a =  0.77 \ \ text{or} \ \ a = 1.00`
  

`text{Case 2:} \ a > 1`

`sqrtx = ax^2 \ => \ x = a^{- 3/2}`

`text{Solve for}\ a\ text{(by CAS)}:`

`int_0^{a^{- 3/2}} (sqrtx – ax^2) dx + int_{a^{- 3/2}}^a (sqrtx – ax^2) dx = 1/3`

`:. a = 1.13`

Probability, MET2 2021 VCAA 20 MC

Let `A` and `B` be two independent events from a sample space.

If  `text{Pr} (A) = p , \ text{Pr}(B) = p^2`  and  `text{Pr} (A) + text{Pr} (B) = 1`, then  `text{Pr}(A′ ∪ B)`  is equal to

  1. `1-p-p^2`
  2. `p^2-p^3`
  3. `p-p^3`
  4. `1-p + p^3`
  5. `1-p-p^2 + p^3`
Show Answers Only

`D`

Show Worked Solution

`text{Pr} (A) = p , \ text{Pr}(B) = p^2, text{Pr} (A) + text{Pr} (B) = 1`

♦ Mean mark 39%.
`text{Pr}(A ∩ B)` `= p xx p^2 = p^3 \ \ (A, B\ text{are independent)}`
`text{Pr}(A ∩ B′)` `= p (1-p^2) = p-p^3`
`text{Pr}(A′ ∪ B)` `= 1-text{Pr} (A ∩ B′)`
  `= 1-p + p^3`

 
`=> D`

Complex Numbers, SPEC2 2021 VCAA 6 MC

If  `z ∈ C, z != 0`  and  `z^2 ∈ R`, then the possible values of  `text(arg)(z)`  are

  1. `(kpi)/2, k ∈ Z`
  2. `kpi, k ∈ Z`
  3. `((2k + 1)pi)/2, k ∈ Z`
  4. `((4k + 1)pi)/2, k ∈ Z`
  5. `((4k - 1)pi)/2, k ∈ Z`
Show Answers Only

`A`

Show Worked Solution
♦♦♦ Mean mark 23%.
`text(Let)\ \ z` `= r text(cis) theta`
`z^2` `= r^2 text(cis)(2theta)`

 
`text(If)\ \ z^2 ∈ R,`

`2theta` `= kpi`
`theta` `= (kpi)/2`
`text(arg)(z)` `= (kpi)/2`

 
`=>\ A`

Vectors, SPEC1 2021 VCAA 9

Let  `underset~r(t) = (-1 + 4cos(t))underset~i + 2/sqrt3\ sin(t)underset~j`  and  `underset~s(t) = (3 sec(t)-1)underset~i + tan(t)underset~j`  be the position vectors relative to a fixed point `O` of particle `A` and particle `B` respectively for  `0 <= 1 <= c`, where `c` is a positive real constant.

    1. Show that the cartesian equation of the path of particle `A` is  `((x + 1)^2)/16 + (3y^2)/4 = 1`.   (1 mark)

    2. Show that the cartesian equation of the path of particle `A` in the first quadrant can be written as  `y = sqrt3/6 sqrt(-x^2-2x + 15)`.   (1 mark)

    1. Show that the particles `A` and `B` will collide.   (1 mark)

    2. Hence, find the coordinates of the point of collision of the two particles.   (1 mark)

    1. Show that  `d/(dx)(8arcsin ((x + 1)/4) + ((x + 1)sqrt(-x^2 -2x + 15))/2) = sqrt(-x^2-2x + 15)`.   (2 marks)


    2.    

      Hence, find the area bounded by the graph of  `y = sqrt3/6 sqrt(-x^2-2x + 15)`,  the `x`-axis and the lines  `x = 1`  and  `x = 2sqrt3-1`,  as shown in the diagram above. Give your answer in the form  `(asqrt3pi)/b`, where `a` and `b` are positive integers.   (2 marks)

Show Answers Only
    1. `text(See Worked Solutions)`
    2. `text(See Worked Solutions)`
    1. `text(See Worked Solutions)`
    2. `-1 + 2sqrt3, 1/sqrt3`
    1. `text(See Worked Solutions)`
    2. `(2sqrt3pi)/9`
Show Worked Solution

a.i.   `text(Particle A)`

`underset~r(t) = (-1 + 4cos(t))underset~i + 2/sqrt3\ sin(t)underset~j`

`x` `= -1 + 4cos(t)`
`x + 1` `= 4cos(t)`
`cos(t)` `= (x + 1)/4`
`y` `= 2/sqrt3\ sin(t)`
`sin(t)` `= (sqrt3 y)/2`

 
`text(Using)\ \ cos^2(t) + sin^2(t) = 1`

`((x + 1)^2)/16 + (3y^2)/4 = 1`

 

a.ii.  `((x + 1)^2)/16 + (3y^2)/4 = 1`

♦ Mean mark part (a)(ii) 41%.
`(x + 1)^2 + 12y^2` `= 16`
`12y^2` `= 16-x^2-2x-1`
`y^2` `= 1/12(15-x^2-2x)`
`y` `= ±sqrt(1/12 (-x^2-2x + 15))`

 
`text(In the 1st quadrant,)\ \ y > 0`

`:. y` `= 1/sqrt12 sqrt(-x^2-2x + 15)`
  `= 1/(2sqrt3) xx sqrt3/sqrt3 sqrt(-x^2-2x + 15)`
  `= sqrt3/6 sqrt(-x^2-2x + 15)`

 

b.i.   `text(If particles collide, find)\ \t\ text(that satisfies)`

`-1 + 4cos(t)` `= 3sec(t)-1\ \ text(and)`
`2/sqrt3 sin(t)` `= tan(t)`

 
`text(Equate)\ underset~j\ text(components:)`

`2/sqrt3 sin(t)` `= (sin(t))/(cos(t))`
`cos(t)` `= sqrt3/2`
`t` `= pi/6`

 
`text(Check)\ underset~i\ text(components at)\ \ t= pi/6 :`

`-1 + 4cos(pi/6)` `= 3 sec(pi/6)-1`
`-1 + 4 · sqrt3/2` `= 3· 2/sqrt3-1`
`2sqrt3` `= 2sqrt3`

 
`:.\ text(Particles collide.)`

 

b.ii.   `text(Collision occurs at)\ \ r(pi/6)`

`r(pi/6)` `= (-1 + 4cos\ pi/6, 2/sqrt3 sin\ pi/6)`
  `= (-1 + 2sqrt3, 1/sqrt3)`
♦ Mean mark part (c)(i) 37%.

 

c.i.   `d/dx (8sin^(-1) ((x + 1)/4) + ((x + 1)sqrt(-x^2 -2x + 15))/2)`

`= 8/(sqrt(1-((x + 1)/4)^2)) xx 1/4 + ((x + 1))/2 xx (-2x-2)/(2sqrt(-x^2-2x + 15)) + sqrt(-x^2-2x + 15) xx 1/2`

`= 8/(sqrt(16-(x + 1)^2))-((x + 1)^2)/(2sqrt(-x^2-2x + 15)) + sqrt(-x^2 -2x + 15)/2`

`= 16/(2sqrt(-x^2-2x + 15))-((x + 1)^2)/(2sqrt(-x^2-2x + 15))+ (-x^2-2x + 15)/(2sqrt(-x^2-2x + 15))`

`= (16-x^2-2x-1-x^2-2x + 15)/(2sqrt(-x^2-2x + 15))`

`= (2(-x^2-2x + 15))/(2sqrt(-x^2-2x + 15))`

`= sqrt(-x^2-2x + 15)`

♦ Mean mark part (c)(ii) 45%.

 

c.ii.    `text(Area)` `= int_1^(2sqrt3-1) sqrt3/6 sqrt(-x^2-2x + 15)\ dx`
    `= sqrt3/6 int_1^(2sqrt3-1) sqrt(-x^2-2x + 15)\ dx`
    `= sqrt3/6 [8sin^(-1)((x + 1)/4) + ((x + 1)sqrt(-x^2-2x + 15))/2]_1^(2sqrt3-1)`
    `= sqrt3/6 [(8sin^(-1)(sqrt3/2) + 2sqrt3 sqrt(-(2sqrt3 -1)^2-2(2sqrt3-1) + 15)/2)-(8sin^(-1)(1/2) + (2sqrt(-1-2 + 15))/2)]`
    `= sqrt3/6 [(8pi)/3 + sqrt3 sqrt(-12 + 4sqrt3-1-4sqrt3 + 2 + 15)-((8pi)/6 + sqrt12)]`
    `= sqrt3/6 ((8pi)/3 + 2sqrt3-(8pi)/6-2sqrt3)`
    `= sqrt3/6 ((8pi)/6)`
    `= (2sqrt3pi)/9`

GRAPHS, FUR1 2021 VCAA 8 MC

A company produces two types of tennis racquets: the Smash and the Volley.

Let  `x`  be the number of Smash tennis racquets that are produced each week.

Let  `y` be the number of Volley tennis racquets that are produced each week.

The constraints on the production of these tennis racquets are given by the following set of inequalities.

`x` `≥ 0`
`y` `≥ 0`
`x + 2y` `<= 100`
`x + y` `≥ 20`
`x` `<= 60`

 

The graph below shows the lines that represent the boundaries of these inequalities.
 

The profit, `P`, that the company can make each week from the sale of these tennis racquets is in the form  `P = ax + by`.

The maximum profit will occur when 20 Smash tennis racquets and 40 Volley tennis racquets are sold each week.

If  `a = 15`, the maximum profit is

  1.   600
  2.   900
  3. 1000
  4. 1200
  5. 1500
Show Answers Only

`E`

Show Worked Solution

`text{Max profit at (20, 40) lies on}`

♦♦♦ Mean mark 29%.

`x + 2y = 100 \  \ text{with gradient} = -1/2`

`=> P = ax + by \ \ text{has gradient} = -1/2`

`P` `= 15 x + by`
`by` `=P – 15 x`
`y` `= P/b – 15/b x`

 
`text(Equating gradients:)`

`-15/b = -1/2\ \ =>\ \ b=30` 
 

`:. P_text{max}` `= 15 xx 20 + 30 xx 40`
  `= 1500`

  
`=> E`

NETWORKS, FUR1 2021 VCAA 8 MC

A network of roads connecting towns in an alpine region is shown below.

The distances between neighbouring towns, represented by vertices, are given in kilometres.
 

The region receives a large snowfall, leaving all roads between the towns closed to traffic.

To ensure each town is accessible by car from every other town, some roads will be cleared.

The minimal total length of road, in kilometres, that needs to be cleared is

  1. 361 if  `x` = 50 and  `y` = 55
  2. 361 if  `x` = 50 and  `y` = 60
  3. 366 if  `x` = 55 and  `y` = 55
  4. 366 if  `x` = 55 and  `y` = 60
  5. 371 if  `x` = 55 and  `y` = 65
Show Answers Only

`B`

Show Worked Solution

`text{A partial minimal spanning tree can be drawn:}`
 

`text{Consider each option:}`

♦ Mean mark 48%.

`A:\ text{If} \ x=50 \ text{(include),} \ y = 55 \ text{(include)}`

   `-> \ text{Total length} = 251 + 50 + 55 != 361 \ text{(incorrect)}`

`B:\ text{If} \ x=50 \ text{(include),} \ y = 60 \ text{(include)}`

   `-> \ text{Total length} = 251 + 50 + 55 = 356 \ text{(correct)}`

`text{Similarly, options} \ C, D, E \ text{can be shown to be incorrect.}`

`=> B`

NETWORKS, FUR1 2021 VCAA 5 MC

Consider the following five statements about the graph above:

    • The graph is planar.
    • The graph contains a cycle.
    • The graph contains a bridge.
    • The graph contains an Eulerian trail.
    • The graph contains a Hamiltonian path.

How many of these statements are true?

  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
Show Answers Only

`D`

Show Worked Solution

`text{Consider each statement}`

♦♦♦ Mean mark 17%.

`text{Graph is planar} \ -> \ text{edges only meet at vertices (TRUE)}` 

`text{Graph contains a cycle} \ -> \ text{multiple cycles exist (TRUE)}`

`text{Graph contains a bridge} \ -> \ text{the far right edge, if removed,}`

`text{would disconnect the graph (TRUE)}`

`text{Graph contains an Eulerian trail} \ -> \ text{requires two (only)}`

`text{vertices of an odd degree (FALSE)}`

`text{Graph contains a Hamiltonian path} \ -> \ text{a path that touches}`

`text{each vertex once only (TRUE)}`
  

`=> D`

MATRICES, FUR1 2021 VCAA 8 MC

A new colony of endangered marsupials is established on a remote island.

For one week, the marsupials can feed from only one of three feeding stations: `A`, `B` or `C`.

On Monday, 50% of the marsupials were observed feeding at station `A` and 50% were observed feeding at station `B`. No marsupials were observed feeding `C`.

The marsupials are expected to change their feeding stations each day this week according to the transition matrix `T`.

`qquadqquadqquadqquad \ text(this day)`

`P = {:(qquad\ A quadquadqquad \ B quadquad \ C ),([(0.4,0.1, 0.2),(0.2,0.5,0.2),(0.4,0.4,0.6)]{:(A),(B),(C):} qquad text(next day)):}`
 

Let  `S_n` represent the state matrix showing the percentage of marsupials observed feeding at each feeding station `n`  days after Monday of this week.

The matrix recurrence rule  `S_{n+1} = TS_n`  is used to model this situation.

From Tuesday to Wednesday, the percentage of marsupials who are not expected to change their feeding location is

  1. 44.5%
  2. 45%
  3. 50%
  4. 51.5%
  5. 52
Show Answers Only

`D`

Show Worked Solution

`text{Let} \ S_0 = text{Monday}`

♦♦♦ Mean mark 29%.

`S_1 = TS_0 = [(0.4,0.1,0.2),(0.2,0.5,0.2),(0.4,0.4,0.6)] [(50),(50),(0)] = [(25),(35),(40)]`
  

`S_2 = TS_1 = [(0.4,0.1,0.2),(0.2,0.5,0.2),(0.4,0.4,0.6)] [(25),(35),(40)]`
 

`text{Percentage not expected to change}`

`= 0.4 xx 25 + 0.5 xx 35 + 0.6 xx 40`

`= 51.5`

`=> D`

Functions, MET1 2021 VCAA 9

Consider the unit circle  `x^2 + y^2 = 1`  and the tangent to the circle at the point `P`, shown in the diagram below.
  

  1. Show that the equation of the line that passes through the points `A` and `P` is given by  `y = -x/sqrt3 + 2/sqrt3`.   (2 marks)

Let  `T : R^2 -> R^2, T ([(x),(y)]) = [(1, 0),(0, q)][(x),(y)]`,  where  `q ∈ R text{\}{0}`, and let the graph of the function `h` be the transformation of the line that passes through the points `A` and `P` under `T`.

    1. Find the values of `q` for which the graph of `h` intersects with the unit circle at least once.   (1 mark)

    2. Let the graph of `h` intersect the unit circle twice.
    3. Find the values of `q` for which the coordinates of the points of intersection have only positive values.   (1 mark)

  2. For  `0 < q <= 1`, let  `P^{′}` be the point of intersection of the graph of `h` with the unit circle, where  `P^{′}` is always the point of intersection that is closest to `A`, as shown in the diagram below.
     
         

Let `g` be the function that gives the area of triangle `OAP^{′}` in terms of `theta`.

    1. Define the function `g`.   (2 marks)

    2. Determine the maximum possible area of triangle `OAP^{′}`.   (2 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2.  i. `q in [–1,1]text{\}{0}`
  3. ii. `q in (sqrt3/2, 1)`
  4. i. `g(theta) = sin theta,\ \ theta in (0, pi/3]`
  5. ii. `sqrt3/2\ text(u²)`
Show Worked Solution

a.   `text(In)\ ΔOAP, text(by Pythagoras:)`

♦♦♦ Mean mark part (a) 18%.

`AP=sqrt(2^2-1^2) = sqrt3`

`tan ∠OAP = 1/sqrt3`

`=> m_(AP) =-1/sqrt3`

`text{Find equation of line}\ \ m=-1/sqrt3\ \ text{through (2, 0):}`

`y-0` `=- 1/sqrt3 (x-2)`  
`y` `=- x/sqrt3 + 2/sqrt3\ \ text{… as required}`  

 

♦♦♦ Mean mark part (b)(i) 6%.
bi.  `text{Transformation matrix → dilates line by a factor of}\ q\ text{from}\ xtext{-axis}`

`:. h(x)\ text{intersects circle for}\ \ q in [–1,1]text{\}{0}`

 

b.ii. `text(Positive coordinates → line can only intersect circle in top right quadrant.)`

 `text(Line)\ AP\ text{can move down until it cuts the circle at (0, 1)`

♦♦♦ Mean mark part (b)(ii) 4%.

`=>\ text(S)text(ince)\ AP\ text(cuts)\ ytext{-axis at}\ 2/sqrt3,`

`q=sqrt3/2\ \ text{dilates}\ AP\ text{where it cuts at (0, 1)`

`:. q in (sqrt3/2, 1)`
 

c.i.   `text(If)\ \ q=1, \ P^{′} = P and theta=pi/3`

♦♦♦ Mean mark part (c)(i) 10%.
`g(theta)` `=1/2 xx 1 xx 2 xx sin theta`  
  `=sin theta, \ \ theta in (0, pi/3]`  

 
c.ii.  `text{S}text{ince}\ \ g(theta) = sin theta\ \ text(for)\ \ theta in (0, pi/3],`

♦♦♦ Mean mark part (c)(ii) 12%.

`text(Maximum)\ g(theta)\ text(occurs when)\ \ theta = pi/3`

`:.g(theta)_text(max)=sin (pi/3)=sqrt3/2\ \ text(u²)`

Probability, MET1 2021 VCAA 6

An online shopping site sells boxes of doughnuts.

A box contains 20 doughnuts. There are only four types of doughnuts in the box. They are:

    • glazed, with custard
    • glazed, with no custard
    • not glazed, with custard
    • not glazed, with no custard

It is known that, in the box:

    • `1/2`  of the doughnuts are with custard
    • `7/10`  of the doughnuts are not glazed
    • `1/10`  of the doughnuts are glazed, with custard
  1. A doughnut is chosen at random from the box.
  2. Find the probability that it is not glazed, with custard.  (1 mark)

  3. The 20 doughnuts in the box are randomly allocated to two new boxes, Box A and Box B.
  4. Each new box contains 10 doughnuts.
  5. One of the two new boxes is chosen at random and then a doughnut from that box is chosen at random.
  6. Let `g` be the number of glazed doughnuts in Box A.
  7. Find the probability, in terms of `g`, that the doughnut comes from box B given that it is glazed.  (2 marks)

  8. The online shopping site has over one million visitors per day.
  9. It is know that half of these visitors are less than 25 years old.
  10. Let  `overset^P`  be the random variable representing the proportion of visitors who are less than 25 years old in a random sample of five visitors.
  11. Find  `text{Pr}(overset^P >= 0.8)`. Do not use a normal approximation.  (3 marks)

Show Answers Only

  1. `2/5`
  2. `(6-g)/6`
  3. `3/16`

Show Worked Solution

a.   `text(Create a 2-way table:)`

`text{Pr(not glazed with custard)}` `=8/20`  
  `=2/5`  

 
b.   `A_text{glazed} = g \ => \ B_text{glazed} = 6-g`

♦♦♦ Mean mark part (b) 20%.

`text{Pr(glazed)} = 6/20`

`text{Pr}(B | text{glazed})` `= (text{Pr}(B ∩ text{glazed}))/text{Pr(glazed)}`  
  `=(1/2 xx (6-g)/10)/(6/20)`  
  `=(6-g)/20 xx 20/6`  
  `=(6-g)/6`  

 
c.   `text(Let)\ \ X=\ text(number of visitors < 25 years)`

♦ Mean mark part (c) 40%.

`X\ ~\ text{Bi}(5, 0.5)`

`text{Pr}(hatP>=0.8)` `= text{Pr}(X>=4)`  
  `=\ ^5C_4 * (1/2)^4(1/2) + (1/2)^5`  
  `=5/32 + 1/32`  
  `=3/16`  

Complex Numbers, EXT2 N2 2021 HSC 10 MC

Consider the two non-zero complex numbers  `z`  and  `w`  as vectors.

Which of the following expressions is the projection of  `z`  onto  `w` ? 

  1.  `{text{Re} (zw)}/{|w|} w`
  2.  `|z/w| w`
  3.  `text{Re} (z/w) w`
  4. `{text{Re}(z)}/{|w|} w`
Show Answers Only

`C`

Show Worked Solution

`text{Let} \ \ z = a + i b \ \ =>\ underset~z = ((a),(b))`

♦♦♦ Mean mark 30%.

`text{Let} \ \ w = c + i d \ \ =>\ underset~w = ((c),(d))`

`underset~z * underset~w = ac + bd`

`|underset~w|^2 = c^2 + d^2`

`text{proj}_(underset~w) underset~z = (underset~z*underset~w)/|underset~w|^2 *underset~w= (ac+bd)/(c^2+d^2)*underset~w`

`z/w` `=(a+ib)/(c+id) xx (c-id)/(c-id)`  
  `=(ac+bd + i(bc-ad))/(c^2+d^2)`  

 
`text{Re}(z/w) =(ac+bd)/(c^2+d^2)`

`:.\ text{proj}_(underset~w) underset~z = text{Re} (z/w) w`
 

`=> C`

Complex Numbers, EXT2 N2 2021 HSC 16c

Sketch the region of the complex plane defined by  `text{Re}(z) ≥ text{Arg}(z)`  where  `text{Arg}(z)`  is the principal argument of `z`.   (3 marks)

Show Answers Only

Show Worked Solution

`text{Find region where} \ \ text{Re}(z) ≥ text{Arg}(z).`

♦♦♦ Mean mark 15%.
`text{In quadrant 1:} \ x ≥ 0 \ , \ y ≥ 0 \ , \ 0 ≤ text{Arg}(z)≤ pi/2`
`x` `≥ tan^-1 (y/x)`  
`x tan (x)` `≥ y`  

  
`text(If)\ \ x>=pi/2, \ x>=tan^-1(y/x)\ \ text(as)\ \ tan^-1(y/x)<pi/2` 

`text{Arg}(0)\ text(is not defined.)`
 

`text{In quadrant 2:} \ \ x ≤ 0 \ , \  y ≤ 0 \ , \ pi/2 ≤ text{Arg}(z ) ≤ pi`

`text{Re}(z) < text{Arg}(z ) \ => \ text{no points satisfy}`
 

`text{In quadrant 3:} \ \ x ≤ 0 \ , \  y ≤ 0 \ , \ -pi ≤ text{Arg}(z) ≤ -pi/2`

`text{Arg}(z)` `= -pi + tan^-1 (y/x)`
`x` `≥ -pi + tan^-1 (y/x)`
`tan^-1 (y/x)` `≤ pi + x`
`y/x` `≤ tan (pi + x)`
`y` `≥ xtan (pi + x)`

 

`text{In the domain} \ \ -pi/2 <= x <= 0,`

`text{Arg}(z) = -pi+ tan^-1(y/x) < -pi/2\ \ \ (text{s}text{ince}\ \ tan^-1(y/x)<pi/2)`

`=>\ text(all points satisfy for)\ \ \ -pi/2 <= x <= 0`
 

`text{In quadrant 4:} \ \ x ≥ 0 \ , \  y ≤ 0 \ , \ -pi/2 ≤ text{Arg}(z) ≤ 0`

`text{Re}(z) > text{Arg}(z) \ => \ text{all points satisfy}`

 

Vectors, EXT2 V1 2021 HSC 16a

  1. The point  `P(x, y, z)`  lies on the sphere of radius 1 centred at the origin `O`.
  2. Using the position vector of `P, overset->{OP} = x underset~i + y underset~j + z underset~k` , and the triangle inequality, or otherwise, show that `| x | + | y | + |z | ≥ 1`.  (2 marks)

  3. Given the vectors  `underset~a = ((a_1),(a_2),(a_3))`  and  `underset~b = ((b_1),(b_2),(b_3))` , show that 
  4.    `|a_1 b_1 + a_2 b_2 + a_3 b_3 | ≤ sqrt{a_1^2 + a_2^2 + a_3^2 }\ sqrt{b_1^2 + b_2^2 + b_3^2}`.  (3 marks)

  5. As in part (i), the point  `P (x, y, z)`  lies on the sphere of radius 1 centred at the origin `O`.
  6. Using part (ii), or otherwise, show that  `| x | + | y | + | z | ≤ sqrt3`.  (2 marks)

Show Answers Only
  1. `text{See Worked Solution}`
  2. `text{See Worked Solution}`
  3. `text{See Worked Solution}`
Show Worked Solution
i.      `text{Triangle inequality:} \ |x| + |y| ≥ |x + y|`
♦ Mean mark (i) 50%.
`|x| + |y| + |z|` ` = |x_underset~i| + |y_underset~j| + |z_underset~k|`
  `≥ |x_underset~i + y_underset~j| + |z_underset~k|`
  `≥ |x_underset~i + y_underset~j + z_underset~k|`
  `≥ 1\ \ (|overset->{OP}| = | x_underset~i + y_underset~j + z_underset~k | = 1)`

 

ii.   `text{Using the dot product:}`

♦ Mean mark (ii) 42%.

`underset~a * underset~b = a_1 b_1 + a_2 b_2 + a_3 b_3`

`underset~a * underset~b = |underset~a| |underset~b| cos theta`

 

`a_1 b_1 + a_2 b_2 + a_3 b_3` `= sqrt{a_1^2 + a_2^2 + a_3^2} * sqrt{b_1^2 + b_2^2 + b_3^2} * cos theta`
`|a_1 b_1 + a_2 b_2 + a_3 b_3|` `= sqrt{a_1^2 + a_2^2 + a_3^2} * sqrt{b_1^2 + b_2^2 + b_3^2} * |cos theta|`

 

`text{S} text{ince} \ -1 ≤ cos theta ≤ 1 \ => \ |cos theta| ≤ 1`

`:. \ | a_1 b_1 + a_2 b_2 + a_3 b_3 | ≤ sqrt{a_1^2 + a_2^3 + a_3^2} * sqrt{b_1^2 + b_2^2 + b_3^2}`

 

iii.  `text{Using part (ii) with vectors:}`

♦♦♦ Mean mark (iii) 14%.

`underset~a = ((1),(1),(1)) \ , \ underset~b = (( | x| ),( |y| ),( |z| ))`

`|\ |x| + |y| + |z|\ |` `≤ sqrt{1^2 + 1^2 + 1^2} * sqrt{ x^2 + y^2 + z^2}`
`|x| + |y| + |z|` `≤ sqrt3`

Proof, EXT2 P1 2021 HSC 15b

For integers  `n ≥ 1`, the triangular numbers  `t_n`  are defined by  `t_n = (n(n + 1))/2`,  giving  `t_1 = 1, t_2 = 3, t_3 = 6, t_4 = 10`  and so on.

For integers  `n >= 1`,  the hexagonal numbers  `h_n`  are defined by  `h_n = 2n^2-n`,  giving  `h_1 = 1, h_2 = 6, h_3 = 15, h_4 = 28`  and so on.

  1. Show that the triangular numbers  `t_1, t_3 , t_5`, and so on, are also hexagonal numbers.  (2 marks)

  2. Show that the triangular numbers  `t_2, t_4 , t_6`, and so on, are not hexagonal numbers.  (1 mark)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
Show Worked Solution

i.   `t_n = (n(n + 1))/2`

♦ Mean mark part (i) 50%.

`text(Odd triangular numbers:)`

`→ \ text(Let)\ \ n = 2k-1\ \ text(for integers)\ \ k >= 1`

`t_(2k – 1)` `= ((2k-1)(2k-1 + 1))/2`
  `= (2k(2k-1))/2`
  `= 2k^2-k\ \ text{(definition of hexagonal numbers)}`

 
`:. t_1, t_3\ …\ text(are also hexagonal numbers.)`

 

ii.   `t_2, t_4, t_6, …`

♦ Mean mark part (ii) 1%!

`→\ text(Let)\ \ n = 2k\ \ text(for integers)\ \ k > = 1`

`t_(2k)` `= (2k(2k + 1))/2`
  `= 2k^2 + k`

 
`text(Find values for)\ \ k, n\ \ text(that satisfy:)`

`2k^2 + k` `= 2n^2 – n`
`2k^2-2n^2 + k + n` `= 0`
`2(k-n)(k + n) + (k + n)` `= 0`
`(k + n)(2k-2n + 1)` `= 0`

 
`k = -n =>\ text(no solution)\ (k, n >= 1)`

`2k = 2n-1 =>\ text(no solution)\ \ (2k\ \ text(is even,)\ \ 2n-1\ \ text{is odd)}`

`:. t_2, t_4, t_6, …\ text(are not hexagonal numbers.)`

Proof, EXT2 P1 2021 HSC 15a

For all non-negative real numbers `x` and `y, \ sqrt(xy) <= (x + y)/2`.  (Do NOT prove this.)

  1. Using this fact, show that for all non-negative real numbers `a`, `b` and `c`,
  2.     `sqrt(abc) <= (a^2 + b^2 + 2c)/4`.  (2 marks)

  3. Using part (i), or otherwise, show that for all non-negative real numbers `a`, `b` and `c`,  
  4.     `sqrt(abc) <= (a^2 + b^2 + c^2 + a + b + c)/6`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
Show Worked Solution

i.   `text(Show)\ \ sqrt(abc) <= (a^2 + b^2 + 2c)/4`

♦ Mean mark part (i) 50%.

`sqrt(abc) = sqrt((ab)c) <= (ab + c)/2\ …\ (1)`
 

`text(Let)\ \ x = a^2\ \ text(and)\ \ y = b^2:`

`sqrt(xy) = sqrt(a^2b^2) ` `<= (a^2 + b^2)/2`
`ab` `<= (a^2 + b^2)/2\ …\ (2)`

 
`text{Substitute (2) into (1):}`

`sqrt(abc) <= ((a^2 + b^2)/2 + c)/2`

`sqrt(abc) <= (a^2 + b^2 + 2c)/4`

 

ii.   `text(Show)\ sqrt(abc) <= (a^2 + b^2 + c^2 + a + b + c)/6`

♦♦♦ Mean mark part (ii) 26%.

`text{Similarly (to part a):}`

`text(If)\ \ x = b^2\ \ text(and)\ \ y = c^2`

  `sqrt(abc) <= (b^2 + c^2 + 2a)/4`

`text(If)\ \ x = c^2\ \ text(and)\ \ y = a^2`

  `sqrt(abc) <= (c^2 + a^2 + 2b)/4`

`:.3sqrt(abc)` `<= (a^2 + b^2 + 2c + b^2 + c^2 + 2a + c^2 + a^2 + 2b)/4`
`3sqrt(abc)` `<= (2(a^2 + b^2 + c^2 + a + b + c))/4`
`sqrt(abc)` `<= (a^2 + b^2 + c^2 + a + b + c)/6`

Mechanics, EXT2 M1 2021 HSC 13d

An object is moving in simple harmonic motion along the `x`-axis. The acceleration of the object is given by  `overset¨x = – 4 (x - 3)`  where `x` is its displacement from the origin, measured in metres, after `t` seconds.

Initially, the object is 5.5 metres to the right of the origin and moving towards the origin. The object has a speed of 8 m s`\ ^(-1)` as it passes through the origin.

  1. Between which two values of `x` is the particle oscillating?  (2 marks)

  2. Find the first value of `t` for which  `x = 0`,  giving the answer correct to 2 decimal places.  (2 marks)

Show Answers Only
  1. `x = -2\ \ text(and)\ \ x = 8`
  2. `0.58\ text(seconds)`
Show Worked Solution

i.   `overset¨x = -4 (x – 3)`

♦ Mean mark 47%.
`d/(dx)(1/2 v^2)` `= -4x + 12`
`1/2 v^2` `= -2x^2 + 12x + c`

 
`v = 8\ \ text(when)\ \ x = 0:`

`1/2 xx 8^2 = c \ => \ c = 32`

`1/2 v^2 = -2x^2 + 12x + 32`
 

`text(Find)\ \ x\ \ text(when)\ \ v = 0:`

`-2x^2 + 12x + 32` `= 0`
`x^2 – 6x – 16` `= 0`
`(x – 8)(x + 2)` `= 0`

 

`:.\ text(Particle oscillates between)\ \ x = -2\ \ text(and)\ \ x = 8`

 

ii.   `overset¨x = -4 (x – 3) \ => \ n = 2`

♦♦ Mean mark 24%.

`text(Amplitude = 5,  Centre of motion at)\ \ x = 3`

`x = 5 cos(2t + alpha) + 3`
 

`text(When)\ \ t = 0, x = 5.5:`

`5.5` `= 5cos alpha + 3`
`cos alpha` `= 1/2`
`alpha` `= pi/3`

  
`:. x = 5 cos(2t + pi/3) + 3`
  

`text(Find)\ \ t\ \ text(when)\ \ x= 0:`

`cos(2t + pi/3)` `= -3/5`
`2t + pi/3` `= 2.214…`
`t` `= 1/2(2.214… – pi/3)`
  `= 0.58\ text(seconds)\ \ text{(2 d.p.)}`

Calculus, EXT1 C2 2021 HSC 14e

The polynomial  `g(x) = x^3 + 4x - 2`  passes through the point (1, 3).

Find the gradient of the tangent to  `f(x) = xg^(-1)(x)`  at the point where  `x = 3`.  (2 marks)

Show Answers Only

`10/7`

Show Worked Solution

`g(x) = x^3 + 4x – 2`

♦♦♦ Mean mark 10%.

`g′(x) = 3x^2 + 4`

`f(x) = x g^(-1)(x)`

`f′(x) = x · d/(dx) g^(-1)(x) + g^(-1)(x)`

COMMENT: The reciprocal relationship of gradients between `g(x)` and `g^(-1)(x)` is critical here.

 

`g(x)\ text(passes through)\ (1, 3)`

`=> g^(-1)(x)\ text(passes through)\ (3, 1)`

`g′(1) = 3 + 4 = 7`

`=> d/(dx) g^(-1)(3) = 1/(d/(dy) g(y)) = 1/(g′(1)) = 1/7`

`:. f′(x)|_(x = 3)` `= 3 · 1/7 + 1`
  `= 10/7`

Vectors, EXT1 V1 2021 HSC 14c

  1. For vector `underset~v`, show that  `underset~v · underset~v = |underset~v|^2`.  (1 mark)

  2. In the trapezium `ABCD`, `BC` is parallel to `AD` and  `|overset(->)(AC)| = |overset(->)(BD)|`.
     
     
         

  3. Let  `underset~a = overset(->)(AB), underset~b = overset(->)(BC)`  and  `overset(->)(AD) = koverset(->)(BC)`,  where  `k > 0`.
  4. Using part (i), or otherwise, show  `2underset~a · underset~b + (1-k)|underset~b|^2 = 0`.  (3 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
Show Worked Solution

i.  `text(Let)\ \ underset~v = ((x),(y))`

`|underset~v| = sqrt(x^2 + y^2)`

`underset~v · underset~v = x^2 + y^2 = (sqrt(x^2 + y^2))^2 = |underset~v|^2`

 

♦♦♦ Mean mark part (ii) 18%.

ii.   `text(Show)\ \ 2underset~a · underset~b + (1-k) |underset~b|^2 = 0`

`|overset(->)(AC)|` `= |overset(->)(BD)|`
`|underset~a + underset~b|` `= |kunderset~b-underset~a|`
`|underset~a + underset~b|^2` `= |kunderset~b-underset~a|^2`
`(underset~a + underset~b)(underset~a + underset~b)` `= (kunderset~b-underset~a)(kunderset~b-underset~a)\ \ text{(see part a.)}`
`underset~a · underset~a + 2underset~a · underset~b + underset~b · underset~b` `= k^2 underset~b · underset~b-2kunderset~a · underset~b + underset~a · underset~a`
`2underset~a · underset~b + 2kunderset~a · underset~b + underset~b · underset~b-k^2 underset~b · underset~b` `= 0`
`2underset~a · underset~b(1 + k) + underset~b · underset~b(1-k^2)` `= 0`
`2underset~a · underset~b(1 + k) + underset~b · underset~b(1 + k)(1-k)` `= 0`
`2underset~a · underset~b +  (1-k)|underset~b|^2` `= 0`

Calculus, EXT1 C3 2021 HSC 12a

The direction field for a differential equation is shown below.

The graph of a particular solution to the differential equation passes through the point `P`.

On the graph, sketch the graph of this particular solution.  (1 mark)

Show Answers Only

Show Worked Solution

♦♦♦ Mean mark 18%!
MARKER COMMENT: A solution curve does not cross any tangent line.

Financial Maths, STD1 F3 2021 HSC 30

Blake opens a new credit card account on 1 May. He uses it, for the first time, on 4 May to buy concert tickets for $850.

He makes no further purchases or repayments during the month of May.

A statement for the credit card is issued on the last day of each month.

The statement for May shows that interest is charged at 19.75% per annum, compounding daily, from 20 May (included) until 31 May (included).

  1. What is the compound interest shown on the statement issued on 31 May?  (3 marks)

  2. The minimum payment is calculated as 3% of the closing balance on 31 May. Calculate the minimum payment.  (1 mark)

Show Answers Only
  1. `$5.54`
  2. `$25.67`
Show Worked Solution

a.   `text(Daily interest rate) = 19.75/365 = 0.05411text(%) = 0.0005411`

`text(Days incurring interest = 12)`

♦♦♦ Mean mark part (a) 14%.
`text{Card balance (31 May)}` `= PV(1+r)^n`
  `= 850 (1 + 0.0005411)^12`
  `= $855.54`

 

`:.\ text(Interest)` `= 855.54 – 850`
  `= $5.54`

 

♦♦♦ Mean mark part (b) 15%.
b.    `text(Minimum payment)` `= 855.54 xx 0.03`
    `= $25.67`

Measurement, STD1 M3 2021 HSC 28

A right-angled triangle `XYZ` is shown. The length of `XZ` is 16 cm and `angleYXZ = 30°`.
 


 

  1. Find the side length, `XY`, of the triangle in centimetres, correct to two decimal places.  (2 marks)

  2. Hence, find the area of triangle `XYZ` in square centimetres, correct to one decimal place.  (3 marks)

Show Answers Only
  1. `13.86\ text(cm)`
  2. `55.4\ text(cm)^2`
Show Worked Solution
♦♦ Mean mark part (a) 24%.
a.    `cos 30^@` `=(XY)/16`
  `XY` `=16 xx cos30^@`
    `=13.856…`
    `=13.86\ text(cm)\ \ text{(2 d.p.)}`

 

b.  `text(Using the sine rule:)`

♦♦♦ Mean mark part (b) 14%.
  `text(Area)\ DeltaXYZ` `= 1/2 ab sinC`
    `= 1/2 xx 16 xx 13.86 xx sin30^@`
    `=55.44`
    `=55.4\ text(cm)^2\ \ text{(1 d.p.)}`

Financial Maths, STD1 F3 2021 HSC 27

Tracy takes out a 30-year reducing balance loan of $680 000 to buy a house. Interest is charged at 0.25% per month. The loan is to be repaid in equal monthly instalments of $2866.91 over a term of 30 years.

Part of a spreadsheet used to model the reducing balance loan is shown.
 

   

  1. Find the amount owing at the end of the second month.  (2 marks)

  2. Suppose that the interest rate reduces to 0.15% per month and the monthly instalments remain as $2866.91.
  3. What will happen to the term of the loan? Explain your answer without using calculations.  (2 marks)

Show Answers Only
  1. `$677\ 663.26`
  2. `text(Loan term will decrease.)`
    `text(The repayment will pay down more of the principal each)`
    `text(month, reducing the term of the loan.)`
Show Worked Solution

a.   `text(In month 2:)`

♦♦ Mean mark part (a) 31%.

`text(Interest) = 678\ 833.09 xx 0.0025 = $1697.08`

`text(Repayment) = $2866.91`

`text(Balance owing)` `= 678\ 833.09 + 1697.08 – 2866.91`
  `= $677\ 663.26`

 

b.   `text(The term of the loan will decrease.)`

♦♦♦ Mean mark part (b) 18%.

`text(If interest rate reduces, the monthly interest amount)`

`text(payable decreases.)`

`text(The repayment will pay down more of the principal each)`

`text(month, reducing the term of the loan.)`

Financial Maths, STD1 F1 2021 HSC 19

Yin purchased a car for $20 000. The value of the car decreases according to a linear model. The graph shows the value of the car, $`V`, against the time, t months, since it was purchased.
 


 

  1. By how much does the value of the car decrease every 10 months?  (1 mark)

  2. Find the value of the car after 5 years.  (1 mark)

  3. Identify ONE problem with using this model to determine the value of Yin’s car over time.  (1 mark)

Show Answers Only
  1. `$2000`
  2. `$8000`
  3. `text(Car will hare a negative value after 100 months)`
Show Worked Solution
a.    `text{Decrease (10 months)}` `= 20\ 000 – 18\ 000`
    `= $2000`

 

b.   `text(5 years) = 5 xx 12 = 60\ text(months)`

`V_(t = 60) = $8000`

♦♦♦ Mean mark part (c) 11%.

 

c.   `text(Car will have a negative value after 100 months.)`

Calculus, 2ADV C4 2021 HSC 28

The region bounded by the graph of the function  `f(x) = 8 - 2^x`  and the coordinate axes is shown
 

  1. Show that the exact area of the shaded region is given by  `24 - 7/ln2`.  (3 marks)

  2. A new function  `g(x)`  is found by taking the graph of   `y = -f(-x)`  and translating it by 5 units to the right.
  3. Sketch the graph of  `y = g(x)`  showing the `x`-intercept and the asymptote.  (2 marks)

  4. Hence, find the exact value of  `int_2^5 g(x)\ dx`.  (1 mark)

Show Answers Only
  1. `text(See Worked Solution)`
  2.  
  3. `7/(ln2) – 24`
Show Worked Solution

a.   `xtext(-intercept occurs when)`

`8 – 2^x = 0 \ => \ x = 3`

`text(Area)` `= int_0^3 8 – 2^x\ dx`
  `= [8x – (2^x)/(ln2)]_0^3`
  `= 24 – 8/(ln 2) – (0 – 1/(ln2))`
  `= 24 – 8/(ln2) + 1/(ln2)`
  `= 24 – 7/(ln2)\ \ text(u²)`

♦ Mean mark part (b) 48%.

b.

`y = f(–x) -> text(reflect)\ \ y = f(x)\ \ text(in the)\ ytext(-axis)`

`y = -f(–x) -> text(reflect)\ \ y = f(–x)\ \ text(in the)\ xtext(-axis)`
 

♦♦♦ Mean mark part (c) 13%.

c.   `int_2^5 g(x)\ dx\ \ text{is the same area as found in part (a)}`

`text(except it is below the)\ xtext(-axis.)`

`:. int_2^5 g(x)\ dx = 7/(ln2) – 24`

Probability, 2ADV S1 2021 HSC 34

A discrete random variable has probability distribution as shown in the table where  `n`  is a finite positive integer.

\begin{array} {|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ r\ \ \  & \ \ \ r^{2}\ \ \  & \ \ \ r^{3}\ \ \  & \ \ \ ...\ \ \  & \ \ \ r^{k}\ \ \ &\ \ \ ...\ \ \ & \ \ \ r^{n}\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & r^{n} & r^{n-1} & r^{n-2} & ... & r^{n-k+1} & ... & r  \\
\hline
\end{array}

Show that  `E(X) = n( 2r-1)`.   (3 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution
♦♦♦ Mean mark 18%.
`E(X)` `= ∑x · P(X = x)`
  `= r · r^n + r^2 · r^(n + 1) + … + r^n · r`
  `= r^(n + 1) + r^(n + 1) + … + r^(n + 1)`
  `= nr^(n + 1) …\ (1)`

 

`text(Sum of probabilities = 1)`

`underset (text(GP where)\ a = r,\ r = r) (underbrace {r + r^2 + r^3 + … + r^(n-1) + r^n}) = 1`

`(r(1-r^n))/(1-r)` `= 1`
`r-r^(n + 1)` `= 1-r`
`r^(n + 1)` `= 2r-1 …\ (2)`

 
`text{Substitute (2) into (1):}`

`E(X) = n(2r-1)`

Statistics, 2ADV S3 2021 HSC 33

People are given a maximum of six hours to complete a puzzle. The time spent on the puzzle, in hours, can be modelled using the continuous random variable \(X\) which has probability density function

\(f(x)= \begin{cases}
\dfrac{A x}{x^2+4} & \text{for } 0 \leq x \leq 6,(\text { where } A>0) \\
\ \\
0 & \text {for all other values of } x
\end{cases}\)

The graph of the probability density function is shown below. The graph has a local maximum.
 

  1. Show that  \(A=\dfrac{2}{\ln 10}\).  (2 marks)

  2. Show that the mode of \(X\) is two hours.  (2 marks)

  3. Show that  \(P(X<2)=\log _{10} 2\).  (2 marks)

  4. The Intelligence Quotient (IQ) scores of people are normally distributed with a mean of 100 and standard deviation of 15.
  5. It has been observed that the puzzle is generally completed more quickly by people with a high IQ.
  6. It is known that 80% of people with an IQ greater than 130 can complete the puzzle in less than two hours.
  7. A person chosen at random can complete the puzzle in less than two hours.
  8. What is the probability that this person has an IQ greater than 130? Give your answer correct to three decimal places.  (2 marks)

Show Answers Only
  1. \(\text{See Worked Solution}\)
  2. \(\text{See Worked Solution}\)
  3. \(\text{See Worked Solution}\)
  4. \(0.066\)
Show Worked Solution

a.  \(\displaystyle\int_0^6 \dfrac{A x}{x^2+4} \, d x=1\)

\(\begin{aligned} \dfrac{A}{2} \int_0^6 \dfrac{2 x}{x^2+4} d x & =1 \\
\dfrac{A}{2}\left[\ln \left(x^2+4\right)\right]_0^6 & =1 \\
\dfrac{A}{2}(\ln 40-\ln 4) & =1 \\
\dfrac{A}{2} \ln \left(\dfrac{40}{4}\right) & =1 \\
\dfrac{A}{2} \ln 10 & =1 \\
A & =\dfrac{2}{\ln 10}\end{aligned}\)

b.  \(\text{Mode \(\rightarrow f(x)\) is a MAX}\)

♦♦♦ Mean mark part (b) 24%.

\(\begin{aligned}
f(x) & =\dfrac{A x}{x^2+4} \\
f^{\prime}(x) & =\dfrac{A\left(x^2+4\right)-A x(2 x)}{\left(x^2+4\right)^2} \\
& =\dfrac{A x^2+4 A-2 A x^2}{\left(x^2+4\right)^2} \\
& =\dfrac{A\left(4-x^2\right)}{\left(x^2+4\right)^2}
\end{aligned}\)

\(\text{\(f(x)\) max occurs when \(f^{\prime}(x)=0\) :}\)

\(\begin{aligned}
4-x^2 & =0 \\
x & =2 \quad(x>0)
\end{aligned}\)

♦♦ Mean mark part (c) 30%.
c.    \(P(X<2)\) \(=\displaystyle \int_0^2 \dfrac{A x}{x^2+4} d x\)
    \(=\dfrac{A}{2}\left[\ln \left(x^2+4\right)\right]_0^2\)
    \(=\dfrac{1}{\ln 10}(\ln 8-\ln 4)\)
    \(=\dfrac{1}{\ln 10}\left(\ln \dfrac{8}{4}\right)\)
    \(=\dfrac{1}{\ln 10} \cdot \ln 2\)
    \(=\log _{10} 2\)

 

d.   \(z \text{-score}(130)=\dfrac{x-\mu}{\sigma}=\dfrac{130-100}{15}=2\)

♦♦ Mean mark part (d) 25%.

\(P(z>2)=2.5 \%\)

\begin{aligned}
P(\text { IQ }>130 \mid x<2) & =\dfrac{P(\text { IQ }>130 \cap X<2)}{P(X<2)} \\
& =\dfrac{0.8 \times 0.025}{\log _{10} 2} \\
& =0.0664 \ldots \\
& =0.066
\end{aligned}

Statistics, STD2 S5 2021 HSC 41

In a particular city, the heights of adult females and the heights of adult males are each normally distributed.

Information relating to two females from that city is given in Table 1.
 

The means and standard deviations of adult females and males, in centimetres, are given in Table 2.
 


 

A selected male is taller than 84% of the population of adult males in this city.

By first labelling the normal distribution curve below with the heights of the two females given in Table 1, calculate the height of the selected male, in centimetres, correct to two decimal places.  (4 marks)

 

Show Answers Only

`178.95 \ text{cm}`

Show Worked Solution

 

`z text{-score (175 cm, female)} = 2`

♦♦♦ Mean mark 16%.

`z text{-score (160.6 cm, female)} = -1`
 

`text{Find} \ mu \ text{of female heights:}`

`mu – sigma` `= 160.6`  
`mu + 2sigma` `= 175`  
`3 sigma` `= 175 – 160.6`  
`sigma` `= 14.4/3`  
  `= 4.8 \ text{cm}`  
`:. \ mu` `= 165.4 \ text{cm}`  

 

`text{Selected male’s height has} \ z text{-score} = 1`

`mu text{(male)} = 1.05 times 165.4 = 173.67`

`sigma \ text{(male)} = 1.1 times 4.8 = 5.28`

 

`:. \ text{Actual male height}` `= 173.67 + 5.28`  
  `= 178.95 \ text{cm}`  

Algebra, STD1 A3 2021 HSC 29

In a park the only animals are goannas and emus. Let `x` be the number of goannas and let `y` be the number of emus.

The number of goannas plus the number of emus in the park is 31. Hence  `x + y = 31`.

Each goanna has four legs and each emu has two legs. In total the emus and goannas have 76 legs.

By writing another relevant equation and graphing both equations on the grid on the following page, find the number of goannas and the number of emus in the park.  (4 marks)

 

   
 

Number of goannas = _______________

Number of emus = _________________

Show Answers Only

`text{7 goannas, 24 emus}`

Show Worked Solution

`text{Total goanna legs} = 4x`

♦♦♦ Mean mark 8%.

`text{Total emu legs} = 2y`

`4x + 2y` `= 76`  
`2x + y` `= 38 \ …\ (1)`  
`x + y` `= 31 \ …\ (2)`  

 

`text{Number of goannas} = 7`

`text{Number of emus} = 24`

Algebra, STD2 A4 2021 HSC 34

In a park the only animals are goannas and emus. Let `x` be the number of goannas and let `y` be the number of emus.

The number of goannas plus the number of emus in the park is 31. Hence  `x + y = 31`.

Each goanna has four legs and each emu has two legs. In total the emus and goannas have 76 legs.

By writing another relevant equation and graphing both equations on the grid on the following page, find the number of goannas and the number of emus in the park.  (4 marks)

   
 

Number of goannas = _______________

Number of emus = _________________

Show Answers Only

`text{7 goannas, 24 emus}`

Show Worked Solution

`text{Total goanna legs} = 4x`

♦♦♦ Mean mark 29%.

`text{Total emu legs} = 2y`

`4x + 2y` `= 76`  
`2x + y` `= 38 \ …\ (1)`  
`x + y` `= 31 \ …\ (2)`  

 

`text{Number of goannas} = 7`

`text{Number of emus} = 24`

Statistics, STD2 S4 2021 HSC 33

For a sample of 17 inland towns in Australia, the height above sea level, `x` (metres), and the average maximum daily temperature, `y` (°C), were recorded.

The graph shows the data as well as a regression line.
 

     
 

The equation of the regression line is  `y = 29.2 − 0.011x`.

The correlation coefficient is  `r = –0.494`.

  1. i.  By using the equation of the regression line, predict the average maximum daily temperature, in degrees Celsius, for a town that is 540 m above sea level. Give your answer correct to one decimal place.  (1 mark)

  2. ii. The gradient of the regression line is −0.011. Interpret the value of this gradient in the given context.  (2 marks)

  3. The graph below shows the relationship between the latitude, `x` (degrees south), and the average maximum daily temperature, `y` (°C), for the same 17 towns, as well as a regression line.
     
     
         
     
    The equation of the regression line is  `y = 45.6 − 0.683x`.
  4. The correlation coefficient is  `r = − 0.897`.
  5. Another inland town in Australia is 540 m above sea level. Its latitude is 28 degrees south.
  6. Which measurement, height above sea level or latitude, would be better to use to predict this town’s average maximum daily temperature? Give a reason for your answer.  (1 mark)

Show Answers Only
  1. i.  `23.3°text(C)`
  2. ii. `text(See Worked Solutions)`
  3. `text(Latitude. Correlation coefficient shows a stronger relationship.)`
Show Worked Solution
a.i.    `y` `=29.2 – 0.011(540)`
    `=23.26`
    `=23.3°text{C  (1 d.p.)`
♦♦ Mean mark part a.ii. 28%.

 
a.ii.  `text(On average, the average maximum daily temperature of)`

`text(inland towns drops by 0.011 of a degree for every metre)`

`text(above sea level the town is situated.)`
 

♦♦♦ Mean mark part b 18%.

b.  `text(The correlation co-efficient of the regression line using)`

`text(latitude is significantly stronger than the equivalent)`

`text(co-efficient for the regression line using height above sea)`

`text(level.)`

`:.\ text(The equation using latitude is preferred.)`

Statistics, STD2 S5 2021 HSC 38

A random variable is normally distributed with mean 0 and standard deviation 1. The table gives the probability that this random variable lies between 0 and `z` for different values of `z`.

 

The probability values given in the table for different values of `z` are represented by the shaded area in the following diagram.
 

  1. Using the table, show that the probability that a value from a random variable that is normally distributed with mean 0 and standard deviation 1 is greater than 0.3 is equal to 0.3821.  (1 mark)

  2. Birth weights are normally distributed with a mean of 3300 grams and a standard deviation of 570 grams. By first calculating a `z`-score, find how many babies, out of 1000 born, are expected to have a birth weight greater than 3471 grams.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `382`
Show Worked Solution

a.   `P(z>0)=0.5`

♦♦♦ Mean mark part (a) 1%.
COMMENT: Note the Std2 and Advanced questions varied slightly but used the same table and graph.

`P(0<z<0.3)=0.1179`

`P(z>0.3) = 0.5-0.1179=0.3821`
 

b.   `z text{-score (3471)}` `=(x-mu)/sigma`
    `=(3471-3300)/570`
    `=0.3`

 
`P(z>0.3) = 0.3821\ \ text{(see part (a))}`
  

`:.\ text(Number of babies > 3471 grams)`  
`=1000 xx 0.3821`  
`=382\ \ text{(nearest whole)}`  

Algebra, STD2 A4 2021 HSC 35

A publisher sells a book for $10. At this price, 5000 copies of the book will be sold and the revenue raised will be  `5000 xx 10=$50\ 000`.

The publisher is considering increasing the price of the book. For every dollar the price of the book is increased, the publisher will sell 50 fewer copies of the book.

If the publisher charges `(10+x)` dollars for each book, a quadratic model for the revenue raised, `R`, from selling books is

`R=-50x^2+4500x + 50\ 000`

 


 

  1. By first finding a suitable value of `x`, find the price the publisher should charge for each book to maximise the revenues raised from sales of the book.  (2 marks)

  2. Find the value of the intercept of the parabola with the vertical axis.  (1 mark) 

Show Answers Only
  1. `$55`
  2. `$50\ 000`
Show Worked Solution

a.   `text{Highest revenue}\ (R)\ text(occurs halfway between)\ \ x= –10 and x=100.`

`text{Midpoint}\ =(-10 + 100)/2 = 45`

♦♦♦ Mean mark part (a) 16%.

`:.\ text(Price of book for)\ R_text(max)`

`=45 + 10`

`=$55`
 

♦♦♦ Mean mark part (b) 21%.

b.   `ytext(-intercept → find)\ \ R\ \ text(when)\ \ x=0:`

`R` `= -50(0)^2 + 4500(0) + 50\ 000`
  `=$50\ 000`

Measurement, STD1 M5 2021 HSC 21

A rectangular sportsground has been drawn to scale on a 1-cm grid as shown. The scale used is `1:3000`.
 

Kerry took 12 minutes to walk around the perimeter of this sportsground.

What was Kerry's average speed in kilometres per hour?   (4 marks)

Show Answers Only

`3.9 \ text{km/hr}`

Show Worked Solution
`text{Distance walked on map} \ = \ 2 times  (8 + 5) = 26 \ text{cm}`
♦♦♦ Mean mark 24%.
 
`text{Actual distance}` ` =26 times 3000`  
  `= 78\ 000 \ text{cm}`  
  `=780 \ text{m}`  
  `= 0.78 \ text{km}`  

 
`12 \ text{minutes}\ = 12/60 = 0.2 \ text{hours}`
 

`text{Speed}` `= text{distance}/text{time}`  
  `= 0.78/0.2`  
  `= 3.9 \ text{km/hr}`  

Calculus, 2ADV C3 2021 HSC 10 MC

The line  `y = mx`  is a tangent to the curve  `y = cos x`  at the point where  `x = a`, as shown in the diagram.
 

Which of the following statements is true?

  1. `m < 1/a < 1/(2pi)`
  2. `1/(2pi) < m < 1/a`
  3. `1/(2pi) < 1/a < m`
  4. `m < 1/(2pi) < 1/a`
Show Answers Only

`B`

Show Worked Solution

`text(By Elimination:)`

♦♦♦ Mean mark 27%.

`a < 2pi \ => \ 1/a > 1/(2pi)`

`->\ text(Eliminate A)`
 

`m ~~ 1/(2pi) ~~ 1/6`

`1/a\ text{is much closer to 1 (by inspection)}`

`1/a > m`

`->\ text(Eliminate C)`
 

`text(At)\ \ x = 2pi,`

`y = cos 2pi = 1\ \ text(and)\ \ y = mx > 1`

`m > 1/(2pi)`

`:. 1/(2pi) < m < 1/a`

 
`=>B`

Measurement, STD2 M7 2021 HSC 15 MC

A total of 11 400 people entered a running race. The ratio of professional runners to amateurs was  `3 : 16`. All the professional runners completed the race while 600 of the amateurs did not complete the race.

For those who completed the race, what is the ratio, in its simplest form, of professional runners to amateurs?

  1. `1 : 2`
  2. `1 : 5`
  3. `1 : 8`
  4. `1 : 19`
Show Answers Only

`B`

Show Worked Solution

`3:16 =>\ text(Total parts = 19)`

♦♦ Mean mark 28%.

`text(Number of professionals) = 3/19 xx 11\ 400 = 1800`

`text(Number of amateurs) = 16/19 xx 11\ 400 = 9600`

`text(Amateurs who finished) = 9600 – 600 = 9000`
 

`:.\ text{Ratio of professionals : amateurs (who finished)}`

`= 1800 : 9000`

`=1:5`

 
`=> B`

Number, NAP-B3-CA05v1

Two places are 3.6 cm apart on a map.

On the map 1 cm represents 3 km.

What is the actual distance between the two places?

`1.2\ text(km)` `6.6\ text(km)` `10.8\ text(km)` `32.4\ text(km)`
 
 
 
 
Show Answers Only

`10.8\ text(km)`

Show Worked Solution
`text(Actual distance)` `=3.6 xx 3`
  `=10.8\ text(km)`

Measurement, NAP-E3-NC13v1

A new plane had test run where it took off and landed at the same airport.

The pilot took off at 5:20 am and landed  `15 1/2` hours later.

At what time did the pilot land the plane?

`7:10\ text(pm)` `8:50\ text(pm)` `9:10\ text(pm)` `10:50\ text(pm)`
 
 
 
 
Show Answers Only

`text(8:50 pm)`

Show Worked Solution

`text(One strategy:)`

`5:20\ text(am plus)\ 15 1/2\ text(hours)` `=\ text(5:20 pm plus)\ 3 1/2\ text(hours)`
  `=\ text(8:50 pm)`

Measurement, NAP-F3-CA06v1

Two sides of a triangle measure 3 cm and 5 cm.

Which one of these statements about the length of the third side is true?

 
 Its length is 4 cm.
 
 Its length is less than 3 cm.
 
 Its length is greater than 8 cm.
 
 Its length is greater than 2 cm but less than 8 cm.
Show Answers Only

`text(Its length is greater than 2 cm but less than 8 cm.)`

Show Worked Solution

`=>\ text(One side of a triangle cannot be longer than)`

`text(the sum of the 2 other sides.)`

`=>\ text(One side of a triangle cannot be shorter than)`

`text(the difference between the other 2 sides.)`

`:.\ text(Its length is greater than 2 cm but less)`

`text(than 8 cm)` 

Measurement, NAP-B3-CA10v1

A petrol container has a capacity of 10.25 L.

How many millilitres does the petrol container hold when it is full?

`1025` `10\ 025` `10\ 250` `102\ 500`
 
 
 
 
Show Answers Only

`10\ 250\ text(mL)`

Show Worked Solution

`text(1 litre = 1000 mL:)`

`:. 10.25\ text(L) xx 1000 = 10\ 250\ text(mL)`

Number, NAP-B3-NC08 SA v1

A number is multiplied by itself and then divided by 2.

The answer is 32.

What is the number?
Show Answers Only

`8`

Show Worked Solution

`text(Reversing the order of operations:)`

`32 xx 2 =64`

`8 xx 8 = 64`
 

`text(Check answer:)`

`8 xx 8 -: 2 = 32`

`:.\ text(The number is 8.)`