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GEOMETRY, FUR2 2020 VCAA 3

Khaleda manufacturers the face cream in Dhaka, Bangladesh.

Dhaka is located at latitude 24° N and longitude 90° E.

Assume that the radius of Earth is 6400 km.

  1. Write a calculation that shows that the radius of the small circle of Earth at latitude 24° N is 5847 km, rounded to the nearest kilometre.   (1 mark)

Khaleda receives an order from Abu Dhabi, United Arab Emirates (24° N, 54° E).

  1. Find the shortest small circle distance between Dhaka and Abu Dhabi.
  2. Round your answer to the nearest kilometre.   (1 mark)

Khaleda sends the order by plane from Dhaka (24° N, 90° E) to Abu Dhabi (24° N, 54° E).

The flight departs Dhaka at 1.00 pm and arrives in Abu Dhabi 11 hours later.

The time difference between Dhaka and Abu Dhabi is two hours.

  1. What time did the flight arrive in Abu Dhabi?   (1 mark)

A helicopter takes the order from the airport to the customer's hotel.

The hotel is 27 km south and 109 km east of the airport.

  1. Show that the bearing of the hotel from the airport is 104°, correct to the nearest degree.   (1 mark)
  2. After the delivery to the hotel, the helicopter returns to its hangar.
  3. The hangar is located due south of the airport.
  4. The helicopter flies directly from the hotel to the hangar on a bearing of 282° .
  5. How far south of the airport is the hangar?
  6. Round your answer to the nearest kilometre.   (1 mark)

Show Answers Only
  1. `text{See Worked Solution}`
  2. `3674 \ text{km}`
  3. `10:00 \ text{pm}`
  4. `text{See Worked Solution}`
  5. `4 \ text{km}`
Show Worked Solution

a.   `text(Let)\ \ r= \ text(radius of small circle)`
 

`cos \ 24^@` `= r/6400`
`r` `=6400 xx cos 24^@`
  `= 5846.69 …`
  `=5847 \ text{(nearest km)}`

 

b.   `text{Small circle through Dhaka and Abu Dhabi has radius of 5847 km.}`

`text{Longitudinal difference}` `= 90 – 54`
  `= 36^@`

 

`:. \ text{Shortest distance}` `= 36/360 xx 2 xx pi xx 5847`
  `= 3673.77 …`
  `= 3674 \ text{km (nearest km)}`

 

c.   `text{Dhaka} \ (24^@ text{N}, 90^@ \ text{E}) \ text{is further east than Abu Dhabi} \  (24^@ text{N}, 54^@ \ text{E})`

`=> \ text{Dhaka is 2 hours ahead.}`

`:. \ text{Flight arrival time (Abu Dhabi time)}`

`= 1:00 \ text{pm} + 11 \ text{hours} – 2 \ text{hours}`

`= 10:00 \ text{pm}`

 

d.

`tan theta` `= 109/27`
`theta` `=tan^(-1) (109/27) = 76.1^@`

  
`:.\ text{Bearing of hotel from airport}`

`=180 – 76.1`

`=104^@\ \ text{(nearest degree)}`

 

e.

`text{Let X = position of hangar}`

`text{Find OX:}`

`tan 12^@` `= text{OX}/109`
`text{OX}` `= 109 xx tan 12^@`
  `= 23.17 \ text{km}`

 

`:. \ text{AX (distance hangar is south of airport)}`

`= 27 – 23.17`

`= 4 \ text{km (nearest km)}`

GEOMETRY, FUR2 2020 VCAA 2

Khaleda has designed a logo for her business.

The logo contains two identical equilateral triangles,

The side length of each triangle is 4.8 cm, shown in the diagram below.
 

  1. Write a calculation to show that the area of one of the triangles, rounded to the nearest centimetre, is 10 cm2.   (1 mark)

In the logo, the two triangles overlap, as shown below. Part of the logo is shaded and part of the logo is not shaded.
 

  1. What is the area of the entire logo?
  2. Round your answer to the nearest square centimetre.   (1 mark)
  3. What is the ratio of the area of the shaded region to the area of the non-shaded region of the logo?   (1 mark)
  4. The logo is enlarged and printed on the boxes for shipping.
  5. The enlarged logo and the original logo are similar shape.
  6. The area of the enlarged logo is four times the area of the original logo.
  7. What is the height, in centimetres, of the enlarged logo?   (1 mark)

Show Answers Only
  1. `10 \ text{cm}^2 \ (text{nearest cm}^2)`
  2. `15 \ text{cm}^2`
  3. `1:2`
  4. `9.6 \ text{cm}`
Show Worked Solution

a.    `text{Triangle is equilateral (all angles = 60}^@)`

`text{Using the sine rule:}`

`A` `= 1/2 a b sin c`
  `= 1/2 xx 4.8 xx 4.8 xx sin 60^@`
  `= 9.976 …`
  `= 10 text{cm}^2 (text{nearest cm}^2)`

♦♦ Mean mark part (b) 32%.

 

b.   `text{L} text{ogo is made up of 2 identical triangles.}`

`text{Divide each triangle into 4 equal smaller triangles.}`

`text{Total shading = 2 small triangles}\ = 1/2 xx \ text{area of 1 triangle}`

`:. \ text{Area of logo}`

`= 2 xx 10 – 1/2 xx 10`

`= 15 \ text{cm}^2`

♦ Mean mark part (c) 48%.

 

c.    `text{Shaded region}` `: \ text{non-shaded}`
  `text{2 triangles}` `: 4 \ text{triangles}`
  `1` `: 2`

 

d.   `text{Area scale factor = 4 (given)}`

♦♦♦ Mean mark part (d) 17%.

`text{Length scale factor} = sqrt4 = 2`
 

`:. \ text{Height of enlarged logo}`

`= 2 xx \ text{height of original logo}`

`= 2 xx 4.8`

`= 9.6 \ text{cm}`

Calculus, MET2 2020 VCAA 2

An area of parkland has a river running through it, as shown below. The river is shown shaded.

The north bank of the river is modelled by the function  `f_(1):[0,200]rarr R, \ f_(1)(x)=20 cos((pi x)/(100))+40`.

The south bank of the river is modelled by the function  `f_(2):[0,200]rarr R, \ f_(2)(x)=20 cos((pi x)/(100))+30`.

The horizontal axis points east and the vertical axis points north.

All distances are measured in metres.
 

A swimmer always starts at point `P`, which has coordinates  (50, 30).

Assume that no movement of water in the river affects the motion or path of the swimmer, which is always a straight line.

  1. The swimmer swims north from point `P`.
  2. Find the distance, in metres, that the swimmer needs to swim to get to get to the north bank of the river.   (1 mark)

  3. The swimmer swims east from point `P`.
  4. Find the distance, in metres, that the swimmer needs to swim to get to the north bank of the river.   (2 marks)

  5. On another occasion, the swimmer swims the minimum distance from point `P` to the north bank of the river.
  6. Find this minimum distance. Give your answer in metres, correct to one decimal place.   (2 marks)

  7. Calculate the surface area of the section of the river shown on the graph in square metres.   (1 mark)

  8. A horizontal line is drawn through point `P`. The section of the river that is south of the line is declared a no "no swimming" zone.
  9. Find the area of the "no swimming" zone, correct to the nearest square metre.   (3 marks)

  10. Scientists observe that the north bank of the river is changing over time. It is moving further north from its current position. They model its predicted new location using the function with rule  `y=kf_(1)(x)`, where `k >= 1`.
  11. Find the values of  `k` for which the distance north across the river, for all parts of the river, is strictly less than 20 m.   (2 marks)

Show Answers Only
  1. `10\ text{m}`
  2. `16 2/3\ text{m}`
  3. `8.5\ text{m}`
  4. `2000\ text{m}^2`
  5. `837\ text{m²}`
  6. `k in [1, 7/6)`
Show Worked Solution

a.   `text{Since swimmer swims due north,}`

`text{Distance}\ = 40-30=10\ text{m}`

 

b.  `text{Solve} \ f_(1)(x)=30 \ text{or} \ x in[50,100]`

`=> x=200/3`

`:.\ text{Distance to swim (east) to reach north bank}`

`=200/3-50`

`=16 2/3\ text{m}`
 

c.   `text{Let swimmer arrive at north bank at the point}\ \ (x,f_(1)(x))`

`text{By Pythagoras,}`

♦ Mean mark part (c) 39%.

`d(x)=sqrt((x-50)^(2)+(f_(1)(x)-30)^(2))`

`text{Solve} \ d/dx(d(x))=0 \ text{for} \ x:`

`x=54.47…`

`:. d_min=8.5\ text{m (to 1 d.p.)}`

 

d.   `text{Shaded Area}`

`=int_(0)^(200)(f_(1)(x)-f_(2)(x))\ dx`

`=2000\ text{m}^2`

 

e.   `text{Find}\ \ f_(1)(x) = 30 \ text{for} x in [50,150]:`

♦♦ Mean mark part (e) 35%.

`=>x=200/3, 400/3`

`text{Find}\ \ f_(2)(x) = 30 \ text{for} \ x in [50,150]:`

`=>x=50, 150`

`:.\ text{Area}` `=int_(50)^(150)(30-f_(2)(x))\ dx-int_((200)/(3))^((400)/(3))(30-f_(1)(x))\ dx`  
  `=837\ text{m² (to nearest m²)}`  

♦♦♦Mean mark part (f) 15%.

 

f.   `text{Let}\ \ D(x)=\ text{vertical distance between banks}`

`D(x)` `=kf_(1)(x)-f_(2)(x)`  
  `=20k cos((pi x)/(100))+40k-(20 cos((pi x)/(100))+30)`   
  `=(20k-20)cos((pi x)/(100)) +40k-30`  

  
`text{Given}\ \ D(x)<20 \ text{for}\ x in[0,200]`

`text{Maximum} \ cos((pi x)/(100)) = 1\  text{when}\ \ x=0, 200`

`text{Solve} \ 20k-20+40k-30<20\ \ text{for}\ k:`

`=> k<7/6`

`:. k in [1,7/6)`

GEOMETRY, FUR2 2020 VCAA 1

Khaleda manufactures a face cream. The cream comes in a cylindrical container.

The area of the circular base is 43 cm2. The container has a height of 7 cm, as shown in the diagram below.
 

  1. What is the volume of the container, in cubic centimetres?   (1 mark)
  2. Write a calculation that shows that the radius of the cylindrical container, rounded to one decimal place, is 3.7 cm.   (1 mark)
  3. What is the total surface area of the container, in square centimetres, including the base and the lid?
  4. Round your answer to the nearest square centimetre.   (1 mark)

The diagram below shows the dimensions of a shelf that will display the containers.

 
               

  1. What is the perimeter of the shelf, in centimetres?   (1 mark)
  2. The shelf will display the containers in a single layer. Each container will stand upright on the shelf.
  3. What is the maximum number of containers that can fit on the shelf?   (1 mark)
  4. The shelf sits against a wall at a 90° angle.
  5. The shelf is supported by a 26 cm bracket that forms a 30° angle with the wall, as shown in the diagram below.

 
                 
 

  1. Find the value of  `h`, the distance between the edge of the shelf and the bracket, in centimetres.   (1 mark)

Show Answers Only
  1. `301 \ text{cm}^3`
  2. `3.7 \ text{cm (to d.p)}`
  3. `249 \ text{cm}^2`
  4. `296 \ text{cm}^2`
  5. `75`
  6. `24 \ text{cm}`
Show Worked Solution
a.    `V` `= text{Area of base} xx text{height}`
    `= 43 xx 7`
    `= 301 \ text{cm}^3`

 

b.   `text{Area of base} = 43 \ text{cm}^2`

`pi r^2` `= 43`
`r^2` `= 43/pi`
`r` `= sqrt{43/pi} = 3.699 … = 3.7 \ text{cm (to 1 d.p.)}`

 

c.    `text{S.A.}` `= 2 xx text{base + 2 pi r h}`
    `= 2 xx 43 + 2 xx pi xx 3.7 xx 7`
    `= 248.73 …`
    `= 249 \ text{cm}^2 \ (text{nearest cm}^2)`

 

d.    `text{Perimeter}` `= 2 xx 74 + 4 xx 37`
    `=296 \ text{cm}`

 

e.   `text{Diameter of container} = 2 xx 3.7 = 7.4 \ text{cm}`

♦♦ Mean mark part (e) 27%.

`text{Divide area into 2 sections}`
 


 

`text{S} text{ection 1 dimensions (in containers)}`

`text{Width} = 37/7.4 = 5 \ text{containers}`

`text{Height} = 74/7.4 = 10 \ text{containers}`
 

`text{S} text{ection 2 dimensions}`

`text{Width} = 5`

`text{Height} = 5`
 

`:. \ text{Maximum containers}`

`= 5 xx 10 + 5 xx 5`

`= 75`

 

f.    `sin 30^@` `= x/26`
  `x` `= 13 \ text{cm}`

 

`:. h` `= 37 – x`
  `= 37 – 13`
  `= 24 \ text{cm}`

GRAPHS, FUR2 2020 VCAA 4

Another section of Kyla's business services and details cars and trucks.

Every vehicle is both serviced and detailed.

Each car takes two hours to service and one hour to detail.

Each truck takes three hours to service and three hours to detail.

Let `x` represent the number of cars that are serviced and detailed each day.

Let `y` represent the number of trucks that are serviced and detailed each day.

Past records suggest there are constraints on the servicing and detailing of vehicles each day.

These constraints are represented by Inequalities 1 to 4 below.

`text{Inequality 1}`   `x >= 16`    
`text{Inequality 2}`   `y >= 10`    
`text{Inequality 3}`   `2x + 3y <= 96`   `text{(servicing department)}`
`text{Inequality 4}`   `x + 3y <= 72`   `text{(detailing department)}`
     
  1. Explain the meaning of Inequality 1 in the context of this problem.   (1 mark)
  2. Each employee at the business works eight hours per day.
  3. What is the maximum number of employees who can work in the servicing department each day?   (1 mark)

The graph below shows the feasible region (shaded) that satisfies Inequalities 1 to 4 .
 


 

  1. On a day when 20 cars are serviced and detailed, what is the maximum number of trucks that can be serviced and detailed?   (1 mark)
  2. When servicing and detailing, the business makes a profit of $150 per car and $225 per truck.
  3. List the points within the feasible region that will result in a maximum profit for the day.   (2 marks)

Show Answers Only
  1. `text{At least 16 cars are serviced and detailed each day.}`
  2. `12`
  3. `17 \ text{trucks}`
  4. `(24, 16),(27, 14),(30, 12), \ text{and} (33, 10)`
Show Worked Solution

a.   `text{At least 16 cars are serviced and detailed each day.}`

♦♦ Mean mark part (b) 34%.

 

b.   `text{Consider Inequality 3:} \ 2x + 3y ≤ 96`

`text{Total time} \ ≤ 96 \ text{hours and each employee works 8 hours.}`

`:. \ text{Maximum employees in servicing}`

`=96/8`

`=12`
 

c.    `text{Find} \ y_text{max} \ text{when} \ x= 20`

♦ Mean mark part (c) 42%.

`text{With reference to the feasible region,}`

`y_text{max} \ text{lies on the line of Inequality 4}`

`x + 3y` `= 72`
`20 + 3y` `= 72`
`3y`  `= 52`
`y` `= 17.33`

 
`:. y_text{max} = 17 \ text{trucks (highest integer within the feasible region)}`

 

d.    `text{Profits: $150 per car, $225 per truck}`

♦♦♦ Mean mark part (d) 19%.

`P = 150x + 225y`

`=> \ m_P = – 150/225 = – 2/3`

`text{The objective function} \ (P) \ text{is parallel to}\ \ 2x+3y=96\ \ text{(Inequality 3)}`
 


 

`=> text{Maximum profit occurs at integer co-ordinates on line} \ AB.`

`A\ text(occurs at intersection of:)`

`2x+3y=96 and x+3y=72\ \ =>\ \ x=24`

`B\ text(occurs at intersection of:)`

`2x+3y=96 and y=10\ \ =>\ \ x=33`

`text(Max profit requires)\ \ x in [24,33]`

`text{Test each integer}\ xtext{-value for an integer}\ ytext{-value:}`

`:.\ text{Maximum profit occurs at}`

`text{(24, 16), (27, 14), (30, 12) and (33, 10)}`

GRAPHS, FUR2 2020 VCAA 3

Kyla's business also manufactures car seat covers.

The monthly revenue, `R`, in dollars, from selling `n` seat covers is given by

`R = 80 n`

This relationship is shown on the graph below.
 

The monthly cost, `C`, in dollars, of making `n` seat covers is given by

`C = 36n + 5000`

  1. On the graph above, sketch the monthly cost, `C`, of making `n` seat covers.   (1 mark) 
  2. Find the least number of seat covers that need to be sold in order to make a profit.   (1 mark)

Last month, 180 seat covers were sold and the profit was $2920.

Kyla believes that if she lowers the selling price of the seat covers she will sell 60 more seat covers this month.

  1. If Kyla maintains the same profit of $2920 this month, find the new selling price of the seat covers.   (1 mark)
  2. Kyla finds that the monthly cost, `C`, of making `n` seat covers changes when more than 300 seat covers are made.
  3. The monthly cost, `C`, of making `n` seat covers is now given by
     
              `C={[36 n+5000,0 <= n <= 300],[mn+p,n > 300]:}`
     
    This is represented on the graph below.

       
     
    If 350 seat covers cost $17 100 to make, find the value of `m`.   (1 mark)

Show Answers Only

  1.  
  2. `text{114 seat covers}`
  3. `$69`
  4. `26`
Show Worked Solution

a.   `”Draw graph through endpoints (0, 5000) and (250, 14 000).”`
 

 

b.    `C=36 n+5000`

`R=80 n`

`text{Profit occurs when} \ \ R > C`

`text{Solve for} \ n:`

`80 n` `> 36 n+5000`  
`44n` `>5000`  
`n` `> (5000)/(44)`  
`n` `>113.6`  

 
`:.\ text(114 seat covers is the least to make a profit.)`

♦♦ Mean mark part (c) 31%.

 

c.    `n` `= 240\ \ (text{60 more than last month})`
  `C` `= 36 xx 240 + 5000`
    `= $13\ 640`

 
`text(Let)\ \ x=\ text(selling price where profit is $2920)`

`2920` `= 240x – 13\ 640`
`240x` `= 16\ 560`
`:. x` `= (16\ 560)/240`
  `= $69`
♦♦ Mean mark part (d) 28%.

 

d.   `m \ text{is the gradient of}\ \ C = mn + p \ \ text{when}\ \ p>300.`

`C = mn + p \ \ text{passes through} \ (300, 15\ 800) \ text{and} \ (350, 17\ 100)`

`m` `= {17\ 100 – 15\ 800}/{350 – 300}`
  `= 26`

GRAPHS, FUR2 2020 VCAA 2

Kyla organises fundraiser car shows at the business.

  1. Fifteen stallholders pay a set-up fee of $120 each.
  2. Twenty-six competitors pay $30 each to enter their cars to win prizes.
  3. How much money in total is raised from stallholders and competitors at each car show?   (1 mark)
  4. Admission fees for the show are $5 per adult and $2 per child.
  5. $1644 was raised from the 537 people who attended the most recent car show.
  6. How many children attended this car show?   (1 mark)
  7. The table below displays the speed of the car, `s`, in kilometres per hour, and the fuel consumption, `F`, in kilometres per litre, of one of the cars at two particular speeds.
     
       

  8. The equation `F = k/s`, where `k` is the constant of proportionality, is used to model the relationship between fuel consumption and speed of the car.
  9. Determine the fuel consumption of the car, in kilometres per litre, when it is travelling at 80 km/h.   (1 mark)

Show Answers Only
  1. `$2580`
  2. `347`
  3. `7.5 \ text{km/litre}`
Show Worked Solution
a.    `text{Total raised}` `= 15 xx 120 + 26 xx 30`
    `= $2580`

 

b.   `text{Let} \ \ A = text{number of adults}`

`text{Let} \ \ C = text{number of children}`

`A + C = 537 \ …\ (1)`

`5A + 2C = 1644 \ …\ (2)`
 
`text{Multiply} \ (1) xx 2`

`2A + 2C = 1074 \ …\ (3)`

`”Subtract “(2) – (3)`

`3A = 570\ \ =>\ \ A = 190`

`text{Substitute}\ \ A = 190 \ \ text{into (1)}`

`:. \ C = 537 – 190 = 347`

♦ Mean mark part (c) 48%.

 

c.    `F` `= k/s`
  `10` `= k/60`
  `k` `= 600`

 
`text{Find} \ F \ text{when} \ S = 80:`

`F` `= 600/80`
  `= 7.5 \text{km/litre}`

Calculus, MET2 2020 VCAA 1

Let  `f:R rarr R, \ f(x)=a(x+2)^(2)(x-2)^(2)`, where `a in R`. Part of the graph of `f` is shown below. 
 

  1. Show that  `a = 1/4`.    (1 mark)

  2. Express  `f(x)=(1)/(4)(x+2)^(2)(x-2)^(2)`  in the form  `f(x)=(1)/(4)x^(4)+bx^(2)+c`  where `b` and `c` are integers.   (1 mark)

Part of the graph of the derivative function  `f^{′}` is shown below.
 
     
 

  1.  i. Write the rule for `f^{′}` in terms of `x`.   (1 mark)

  2. ii. Find the minimum value of the graph of `f^{′}` on the interval  `x in (0, 2)`.   (2 marks)  

Let  `h:R rarr R, \ h(x)=-(1)/(4)(x+2)^(2)(x-2)^(2)+2`. Parts of the graph of `f` and `h` are shown below.
 
 
       
 

  1. Write a sequence of two transformations that map the graph of `f` onto the graph of `h`.   (1 mark)

     
       
     

  2.   i. State the values of `x` for which the graphs of `f`and `h` intersect.   (1 mark)

  3.  ii. Write down a definite integral that will give the total area of the shaded regions in the graph above.   (1 mark)

  4. iii. Find the total area of the shaded regions in the graph above. Give your answer correct to two decimal places.   (1 mark)

  5. Let `D` be the vertical distance between the graphs of `f`and`h`.
  6. Find all values of `x` for which `D` is at most 2 units. Give your answers correct to two decimal places.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `f(x)=(1)/(4)x^(4)-2x^(2)+4`
  3.  i. `f^{′}(x)=x(x-2)(x+2)=x^(3)-4x`
  4. ii. `-(16sqrt3)/(9)`
  5. `text{Reflect in the x-axis then translate 2 units up, or translate 2 units down then reflect in the x-axis.}`
  6.   i. `x in{-sqrt6,-sqrt2,sqrt2,sqrt6}`
  7.  ii. `2int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx” or “int_(-sqrt6)^(-sqrt2)(h(x)-f(x))\ dx+int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx`
  8. iii. `text(2.72 u²)`
  9. `-2.61 <= x <= -1.08,1.08 <= x <= 2.61`
Show Worked Solution

a.   `text(S)text(ince)\ f(x)\ text{passes through (0, 4):}`

`4` `=a(2)^2(-2)^2`  
`4` `=16a`  
`:.a` `=1/4\ \ text(… as required)`  

 

b.  `f(x)` `=(1)/(4)(x+2)^(2)(x-2)^(2)`  
  `=(1)/(4)x^4-2x^2+4`  

 
c.i.  `f^{′}(x)=x(x-2)(x+2)=x^(3)-4x`
 

c.ii.  `text{Solve}\ \ f^{″}(x)=0\ \ \text{for}\ x:`

`3x^2-4=0`

`=> \ x=(2sqrt3)/(3)`

`f^{′}((2sqrt3)/(3))=-(16sqrt3)/(9)`
 

d.  `text{Reflect in the x-axis then translate 2 units up, or}`

`text{translate 2 units down then reflect in the x-axis.}`
 

e.i.  `text{By CAS, solve}\ \ f(x)=h(x)\ \ text{for}\ x:}`

`x in{-sqrt6,-sqrt2,sqrt2,sqrt6}`
 

e.ii.  `2int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx” or”`

`int_(-sqrt6)^(-sqrt2)(h(x)-f(x))\ dx+int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx`
 

e.iii.  `text(Area = 2.72 u²)`

♦♦ Mean mark part (f) 26%.

 

f.   `text{By CAS, solve}\ -2 <= h(x)-f(x) <= 2\ \ \text{for}\ x:`

`-2.61 <= x <= -1.08, \ 1.08 <= x <= 2.61,” or”`

`-sqrt(4+2sqrt2) <= x <= -sqrt(4-2sqrt2), \ sqrt(4-2sqrt2) <= x <= sqrt(4+2sqrt2)`

GRAPHS, FUR2 2020 VCAA 1

Kyla owns and manages a truck and car care business.

After a major repair on a truck, one of the mechanics took it on a long test drive.

The test drive started at 12 noon.

After four hours, the mechanic stopped to rest for one hour and then returned to the workshop.

The graph below shows the truck's distance from the workshop, `d`, in kilometres, and the number of hours of test driving, `n`, after 12 noon. 
 

  1. At what time of the day did the mechanic arrive back at the workshop?  (1 mark)
  2. Find the average speed, in kilometres per hour, of the truck during the first four hours of the test drive.   (1 mark)
  3. On the return trip, the truck travelled at an average speed of 80 km/h.
  4. The equation of the line segment `VW` that represents this part of the test drive is of the form
  5.     `d = k - 80n`
  6. Find the value of `k`.   (1 mark)

Show Answers Only
  1. `8 \ text{pm}`
  2. `60 \ text{km/h}`
  3. `k = 640`
Show Worked Solution

a.    `n = 0 \ text{at} \ 12 \ text{noon}`

`text{Mechanic arrives back at}\ \ n = 8\ \ text{which is} \ 8 text{pm.}`
 

b.    `text{Average speed}` `= text{distance}/text{time}`
    `= 240/4`
    `= 60\ text{km/h}`

 

c.    `text{Substitute (8, 0) into}\ \ d = k – 80n:`

♦ Mean mark 43%.

`0 = k – 80 xx 8`

`k = 640`

GRAPHS, FUR1 2020 VCAA 8 MC

A vet recommends that dogs eat no more than two units of meat for every five units of dry food.

Let `x` be the number of units of meat eaten.

Let `y` be the number of units of dry food eaten.

An inequality representing this situation is

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

`A`

Show Worked Solution

`text(Consider a ratio of 2 units of meat to 5 units of dry food.)`

♦ Mean mark 43%.

`x:y = 2:5`

`text(Given)\ x\ text(can be no more than 2 units:)`

`x/y` `<=2/5`
`x` `<= 2/5 y`
`y` `>= 5/2x`

 
`=>  A`

GRAPHS, FUR1 2020 VCAA 7 MC

An energy company charges a monthly service fee of $38.70 plus a supply charge of 2.5 cents per megajoule of energy used.

Caitlin’s energy bill for the 30 days of the month of June was $169.90

On average, the number of megajoules of energy that Caitlin used per day in June is closest to

  1.       6
  2.     52
  3.   175
  4.   227
  5. 5248
Show Answers Only

`C`

Show Worked Solution

`text(C)text(ost of energy usage)\ = 169.90 -38.70 = $131.20`

♦♦ Mean mark 38%.

`text(Total megajoules used)`

`= 131.20/0.025`

`= 5248`
 

`:.\ text(Megajoules used per day)`

`= 5248/30`

`= 174.93…`

`=>  C`

Graphs, MET2 2020 VCAA 18 MC

Let `a \in(0, \infty)` and `b \in R`.

Consider the function  `h:[-a, 0) \cup(0, a] \rightarrow R, h(x)=\frac{a}{x}+b`.

The range of  `h`  is

  1. `[b-1,b+1]`
  2. `(b-1,b+1)`
  3. `(-oo,b-1)uu(b+1,oo)`
  4. `(-oo,b-1]uu[b+1,oo)`
  5. `[b-1,oo)`
Show Answers Only

`D`

Show Worked Solution

`h(x)=(a)/(x)+b`

♦ Mean mark 43%.
MARKER’S COMMENT: Use  `y=2/x-3`  to illustrate an example of endpoints etc..

`text(Graph will be in the shape of a hyperbola)`

`text(Endpoint coordinates:)`

`(-a,-1+b) and (a, 1+b)`

`:.\ “Range”=(-oo,-1+b]uu[b+1,oo)`

`=>D`

Graphs, MET2 2020 VCAA 17 MC

Let  `f(x)=-\log _{e}(x+2)`.
 
A tangent to the graph of  `f`  has a vertical axis intercept at `(0, c)`.

The maximum value of `c` is

  1. `-1`
  2. `-1+log_(e)(2)`
  3. `-log_(e)(2)`
  4. `-1-log_(e)(2)`
  5. `log_(e)(2)`
Show Answers Only

`C`

Show Worked Solution

`text(Graph)\ \ y=-\log _{e}(x+2)\ \ text(by CAS:)`

♦ Mean mark 42%.

`”The maximum value of”\ c\ “occurs when the tangent “`

`”occurs at “x=0`

`f(0) = – log_e(2)`

`=>C`

Calculus, MET2 2020 VCAA 15 MC

Part of the graph of a function  `f`, where `a>0`, is shown below.
 

The average value of  the function  `f` over the interval  `[2a, a]` is

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

`B`

Show Worked Solution

♦♦ Mean mark 32%.

`text(Average Value)`

`=(1)/(a-(-2a))int_(-2a)^(a)f(x)\ dx`

`=(1)/(3a)(int_(-2a)^(0)(-(3)/(2)x-a)\ dx+int_(0)^(a)(2x-a)\ dx)`

`=(a)/(3)`

`=>B`

Statistics, MET2 2020 VCAA 14 MC

The random variable `X` is normally distributed.

The mean of `X` is twice the standard deviation of `X`.

If  `text{Pr}(X>5.2)=0.9`, then the standard deviation of `X` is closest to

  1.   `7.238`
  2. `14.476`
  3.   `3.327`
  4.   `1.585`
  5.   `3.169`
Show Answers Only

`A`

Show Worked Solution

♦ Mean mark 44%.

`X∼N(2sigma,sigma^(2))`

`text(Pr)(X > 5.2)=0.9`

`text(Pr)(Z < (5.2-2sigma)/(sigma))=0.1`

`”Solve (by CAS): “(5.2-2sigma)/(sigma)=-1.281…`

`sigma=7.238″ (to 3 d.p.)”`

`=>A`

Functions, MET2 2020 VCAA 12 MC

A clock has a minute hand that is 10 cm long and a clock face with a radius of 15 cm, as shown below.
 


 

At 12.00 noon, both hands of the clock point vertically upwards and the tip of the minute hand is at its maximum distance above the base of the clock face.

The height, `h` centimetres, of the tip of the minute hand above the base of the clock face `t` minutes after 12.00 noon is given by

  1. `h(t)=15+10 sin((pi t)/(30))`
  2. `h(t)=15-10 sin((pi t)/(30))`
  3. `h(t)=15+10 sin((pi t)/(60))`
  4. `h(t)=15+10 cos((pi t)/(60))`
  5. `h(t)=15+10 cos((pi t)/(30))`
Show Answers Only

`E`

Show Worked Solution

`text(By elimination)`

♦ Mean mark 45%.

`text(At)\ \ t=0,\ text(height = 25 cm)`

`=>\ text(Eliminate)\ A, B, C`
 

`text(Period = 60)`

`(2pi)/n` `=60`  
`60n` `=2pi`  
`n` `=pi/30`  

 
`=>\ text(Eliminate)\ D`

`=>E`

Calculus, MET2 2020 VCAA 9 MC

If   `int_(4)^(8)f(x)\ dx=5`, then  `int_(0)^(2)f(2(x+2))\ dx`  is equal to

  1. `12`
  2. `10`
  3. `8`
  4. `1/2`
  5. `5/2`
Show Answers Only

`E`

Show Worked Solution

`int_(4)^(8)f(x)\ dx=5`

♦♦ Mean mark 35%.

`text{Dilate by a factor of} \ (1)/(2)\ text{from the} \ y text{-axis}`

`int_(2)^(4)f(2x)\ dx=(5)/(2)`

`text{Translating 2 units to the left does not change the area}`

`int_(0)^(2)f(2(x+2))\ dx=(5)/(2)`

`=>E`

Probability, MET2 2020 VCAA 8 MC

Items are packed in boxes of 25 and the mean number of defective items per box is 1.4

Assuming that the probability of an item being defective is binomially distributed, the probability that a box contains more than three defective items, correct to three decimal places, is

  1. `0.037`
  2. `0.048`
  3. `0.056`
  4. `0.114`
  5. `0.162`
Show Answers Only

`B`

Show Worked Solution

`X\ ~\ text(Bi)(25, p)`

♦ Mean mark 50%.
`E(X)` `=np = 1.4`  
`25p` `=1.4`  
`p` `=1.4/25 = 0.056`  

 

`text(Pr)(X>3)` `=\ text(Pr) (X>=4)`  
  `=0.048\ \ text{(to 3 d.p.)}`  

 
`=>B`

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`

Combinatorics, EXT1 A1 SM-Bank 21

Eight points `P_1, P_2, ..., P_8`, are arranged in order around a circle, as shown below.
 

  1. How many triangles can be drawn using these points as vertices?   (1 mark)

  2. How many pairs of triangles can be drawn, where the vertices of each triangle are distinct points?   (2 marks)

Show Answers Only
  1. `56`
  2. `280`
Show Worked Solution
i.    `text{Total triangles}` `= \ ^8 C_3`
    `= 56`
COMMENT: In part (ii), divide by 2 to account for duplicate pairs.

 

ii.   `text{Total pairs}` `= (\ ^8 C_3 xx \ ^5 C_3)/{2}`
    `= 280`

Combinatorics, EXT1 A1 SM-Bank 20

How many rectangles, including all squares, can be found in the 4 × 5 grid below, in total?   (2 marks)
 

Show Answers Only

`150`

Show Worked Solution

`text{Consider the 5 horizontal lines}`

`text{Height combinations of rectangle} =\ ^5C_2`

`text{Consider the 6 vertical lines}`

`text{Width combinations of rectangle} =\ ^6C_2`
  

`:. \ text{Total rectangles in grid}`

`= \ ^5C_2 xx \ ^6C_2`

`= 150`

Trigonometry, EXT1 T2 SM-Bank 8

Let  `cos (x) = 3/5`  and  `sin^2(y) = 25/169`, where  `x ∈ [{3pi}/{2} , 2 pi]`  and  `y ∈ [{3pi}/{2} , 2 pi]`.

Find the value of  `sin(x) + cos(y)`.   (2 marks)

Show Answers Only

`8/65`

Show Worked Solution

`text{Both angles are in 4th quadrant (given)}`

`cos(x) = 3/5`
 

`sin(x)` `= – 4/5\ \ text{(4th quadrant)}`
`sin^2(y)` `= 25/169`
`sin(y)` `= – 5/13\ \ text{(4th quadrant)}`

 

`cos(y) = 12/13`
 

`:. \ sin(x) + cos(y)` `= – 4/5 + 12/13`
  `= 8/65`

Statistics, 2ADV S3 SM-Bank 20

A random variable \(X\) has the probability density function \(f(x)\) given by

\(f(x)= \begin{cases}
\dfrac{k}{x^2} & 1 \leq x \leq 2 \\
\ \\
0 & \text {elsewhere }
\end{cases}\)

where \(k\) is a positive real number.

Show that  \(k = 2\).  (2 marks)

Show Answers Only

\(\text{See Worked Solution}\)

Show Worked Solution

♦ Mean mark 48%.

\(\begin{aligned} \int_1^2 \dfrac{k}{x^2} d x & =1 \\
k\left[-\dfrac{1}{x}\right]_1^2 & =1 \\
k\left(-\dfrac{1}{2}+1\right) & =1 \\
\dfrac{k}{2} & =1 \\
\therefore k & =2 \quad \ldots \text{ as required }
\end{aligned}\)

 

Functions, 2ADV F2 SM-Bank 20

  1. Sketch the graph of  `y = 1 - 2/(x - 2)`  on the axes below. Label asymptotes with their equations and axis intercepts with their coordinates.  (3 marks)
     

     
  2. Find the values of  `x`  for which  `1 - 2/(x - 2) >= 3`.  (1 mark)

Show Answers Only

a. 

b. `x in [1, 2)`

Show Worked Solution

a.   `text(Asymptotes:)`

`x=2`

`text(As)\ \ x→ +-oo, \ y→1\ \ =>\ text(Asymptote at)\ \ y=1`

`ytext(-intercept at)\ (0,2)`

`xtext(-intercept at)\ (4,0)`


 

b.   `text(By inspection of the graph:)`

`1-2/(x-2) >=3\ \ text(for)\ \ x in [1, 2)`

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)`


GRAPHS, FUR2 2021 VCAA 3

Lam is a builder constructing a community centre at the new housing estate.

The cost, `C`, in dollars, for Lam to work onsite for `n` weeks is given by

`C = 10 \ 000 + k xx n` 

  1. The cost of 15 weeks of onsite work is $92 500.
  2. Show that the value of `k` is 5500.   (1 mark)

Lam's revenue for this job is $6500 per week

  1. Determine the number of weeks that Lam needs to work onsite to break even.   (1 mark)
  2. An equation for the profit, `P`, in dollars, made by Lam in terms of the number of weeks worked onsite, `n`, can be determined.
  3. Complete this equation by filling in the boxes below.   (1 mark)
`P =` 
 
  `xx n +` 
 
Show Answers Only
  1. `5500`
  2. `10 \ text{weeks}`
  3. `1000 n – 10 \ 000`
Show Worked Solution
a. `C` `= 10 \ 000 + k n`
     `92\ 500` `= 10 \ 000 + 15 k`
  `15 k` `= 82 \ 500`
  `:. k` `= (82\ 500)/15`
    `= 5500`

 

b.   `text{Breakdown occurs when} \ \ R = C`

`6500n` `= 10 \ 000 + 5500 n`
`1000n` `= 10 \ 000`
`:. n` `= {10\ 000}/{1000}`
  `= 10 \ text{weeks}`

 

c.   `P` `= R – C`
    `= 6500n – (10 \ 000 + 5500n)`
    `= 1000n – 10 \ 000`

 

GEOMETRY, FUR2 2021 VCAA 3

Players will travel from around the world to a squash competition in New York City (41° N, 74° W).

The top three female players are from three different cities:

  • Wei-Yi is from Shanghai (31° N, 121° E) in China.
  • Camilla is from Durban (30° S, 31° E) in South Africa.
  • Ozlem is from Istanbul (41° N, 29° E) in Turkey.
  1. Which players are from the Northern Hemisphere?   (1 mark)

The diagram below shows the location of Shanghai, Durban and New York City.


           
 

  1. The city Istanbul (41° N, 29° E) is missing from the diagram.
  2. On the diagram above, mark with an `X` the location of Istanbul.   (1 mark)
  3. The flight from Istanbul (41° N, 29° E) to New York City (41° N, 74° W) travels along a small circle.
  4. Assume that the radius of Earth is 6400 km.
  5.  i. Show that the radius of this small circle, rounded to the nearest kilometres, is 4830.  (1 mark)
  6. ii. Calculate the distance that the plane travels between Istanbul (41° N, 29° E) and New York City (41° N, 74° W).
  7.     Round your answer to the nearest kilometre.   (1 mark)
  8. Wei-Yi and Camilla both arrive in New York City on Monday 11 January at 10.00 pm, local time.
  9. Wei-Yi's flight left Shanghai on Monday 11 January at 8.00 pm, local time.
  10. Camilla's flight left Durban on Monday 11 January at 4.00 am, local time.
  11. Assume that the time difference between Shanghai and New York City is 13 hours, and that the time difference between Durban and New York City is seven hours.
  12. How much longer, in hours, was Camilla's flight compared to Wei-Yi's flight?   (1 mark)
Show Answers Only
  1. `text{Wei-Yi and Ozlem}`
  2.  
     

     
  3.  i.  `text(See Worked Solutions)`
  4. ii. `8683 \ text{km (nearesrt km)}`
  5. `10 \ text{hours}`
Show Worked Solution

a.    `text{Northern Hemisphere} \ to \ text{latitude is °N}`

`text{Wei-Yi and Ozlem}`
 

b.    `text{Istanbul} \ to \ text{same latitude as New York}`

`text{Longitude 29° E is just left of Durban (31° E)}`
 

 

c.i.

`cos 41^@` `= r/6400`
`r` `= 6400 xx cos 41^@`
  `= 4830.14 …`
  `= 4830 \ text{km (nearest km)}`

 

c.ii.

    `text{Arc Length}` `= 103/360 xx 2 xx pi xx 4830`
    `= 8682.83`
    `= 8683 \ text{km (nearest km)}`

 

d.    `text{In New York time}`

`text{Wei-Yi departure = 8:00 pm less 13 hours = 7:00 am Monday (NYT)}`

`text{Wei-Yi’s flight time = 7:00 am to 10:00 pm = 15 hours}`

`text{Camilla’s departure = 4:00 am less 7 hours = 9:00 pm Saturday (NYT)}`

`text{Camilla’s flight time =  9:00 pm (Sunday) to 10:00 pm (Monday) = 25 hours}`

`:. \ text{Extra flight time}` `= 25 – 15`
  `= 10 \ text{hours}`

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 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 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`

CORE, FUR2 2021 VCAA 7

Sienna owns a coffee shop.

A coffee machine, purchased for $12 000, is depreciated in value using the unit cost method.

The rate of depreciation is $0.05 per cup of coffee made.

The recurrence relation that models the year-to-year value, in dollars, of the coffee machine is

`M_0 = 12 \ 000,`     `M_{n+1} = M_n - 1440`

  1. Calculate the number of cups of coffee that the machine produces per year.   (1 mark)

  2. The recurrence relation above could also represent the value of the coffee machine depreciating at a flat rate.
  3. What annual flat rate percentage of depreciation is represented?   (1 mark)

  4. Complete the rule below that gives the value of the coffee machine, `M_n`, in dollars, after `n` cups have been produced. Write your answers in the boxes provided.   (1 mark)

     
       `M_n =`
     
    		  `+`  
     
    		 `xx n`  

Show Answers Only

  1. `2880`
  2. `12text{% p.a.}`
  3. `12 \ 000 + (-0.05) xx n`

Show Worked Solution

a.   `text{Cups of coffee}` `= 1440/0.05`
    `= 28 \ 800`

 

b.   `text{Flat rate (%)}` `= {1440}/{12 \000}`
    `= 12text{% p.a.}`

   
c.  `M_n = 12 \ 000 + (-0.05) xx n`

CORE, FUR2 2021 VCAA 5

A method for predicting future time differences in the 100 m freestyle swim is to use the formula

difference = winning time women – winning time men

A resulting data and time series plot are shown below. The plot is clearly non-linear.
      

  1. Apply a reciprocal transformation to the variable difference to linearise the data. Fit a least squares line to the transformed data and write its equation below.
  2. Round the values of the intercept and the slope to four significant figures.   (2 marks)

  3. Use the equation from part a. to predict, in seconds, the difference the women's and men's winning times in the year 2032.
  4. Round your answer to one decimal place.   (1 mark)

Show Answers Only
  1. `1/text{difference} = -2.234 + 0.001209 xx text{year}`
  2. `4.5 \ text{seconds}`
Show Worked Solution

a.  `text{By CAS}: \ x text{-variable = year}, \ \ y text{-variable} = 1/text{difference}`

`:.  \ text{Equation of least squares line:}`

`1/text{difference} = -2.234 + 0.001209 xx text{year}`
 

b.  `1/text{difference}\ = -2.234 + 0.001209 xx 2032= 0.222688`

`:. \ text{difference}\ = 4.490 …= 4.5 \ text{seconds (to 1 d.p.)}`

CORE, FUR2 2021 VCAA 4

The time series plot below shows that the winning time for both men and women in the 100 m freestyle swim in the Olympic Games has been decreasing during the period 1912 to 2016.
 

Least squares lines are used to model the trend for both men and women.

The least squares line for the men's winning time has been drawn on the time series plot above.

The equation of the least squares line for men is

winning time men = 356.9 – 0.1544 × year

The equation of the least squares line for women is

winning time women = 538.9 – 0.2430 × year

  1. Draw the least squares line for winning time women on the time series plot above.   (1 mark)

  2. The difference between the women's predicted winning time and the men's predicted winning time can be calculated using the formula.
  3.       difference = winning time women – winning time men
  4. Use the equation of the least squares lines and the formula above to calculate the difference predicted for the 2024 Olympic Games.
  5. Round your answer to one decimal place.   (2 marks)

  6. The Olympic Games are held every four years. The next Olympic Games will be held in 2024, then 2028, 2032 and so on.
  7. In which Olympic year do the two least squares lines predict that the wining time for women will first be faster than the winning time for men in the 100 m freesytle?   (2 marks)

Show Answers Only
  1. `text{See Worked Solutions}`
  2. `2.7 \ text{seconds}`
  3. `2056`
Show Worked Solution

a.   

`text{Find end points for the women’s graph.}`

`1908 \ text{data point} = 538.9-0.2430 xx 1908 = 75.256`

`2020 \ text{data point} = 538.9-0.2430 xx 2020 = 48.04`

`=> \ text{Line passes through} \ (1908, 75.3) \ text{and} \ (2020, 48.0)`
 

b.   `text{difference}` `= (538.9-0.2430 xx 2024)-(356.9-0.1544 xx 2024)`
    `= 2.673 …`
    `= 2.7 \ text{seconds (to 1 d.p.)}`

 
c.    `text{Times will be equal when}`

`538.9-0.2430 xx text{year} = 356.9-0.1544 xx text{year}`

`text{year} = 2054.17 …`

`text{1st Olympic year after} \ 2054.17 = 2056`

CORE, FUR2 2021 VCAA 3

The time series plot below shows the winning time, in seconds, for the women's 100 m freestyle swim plotted against year, for each year that the Olympic Games were held during the period 1956 to 2016.

A least squares line has been fitted to the plot to model the decreasing trend in the winning time over this period.
 

The equation of the least squares line is

winning time = 357.1 – 0.1515 × year

The coefficient of determination is 0.8794

  1. Name the explanatory variable in this time series plot.   (1 mark)

  2. Determine the value of the correlation coefficient (`r`).
  3. Round your answer to three decimal places.   (1 mark)

  4. Write down the average decrease in winning time, in seconds per year, during the period 1956 to 2016.   (1 mark)

  5. The predicted winning time for the women's 100 m freestyle in 2000 was 54.10 seconds.
  6. The actual winning time for the women's 100 m freestyle in 2000 was 53.83 seconds.
  7. Determine the residual value in seconds.   (1 mark)

  8. The following equation can be used to predict the winning time for the women's 100 m freestyle in the future.
  9.      winning time =  357.1 – 0.1515 × year
  10.  i. Show that the predicted winning time for the women's 100 m freestyle in 2032 is 49.252 seconds.   (1 mark)

  11. ii. What assumption is being made when this equation is used to predict the winning time for the women's 100 m freestyle in 2032?   (1 mark)

Show Answers Only
  1. `text{year}`
  2. `- 0.938`
  3. `0.1515 \ text{seconds}`
  4. `-0.27`
  5.  i. `49.252 \ text{seconds}`
  6. ii. `text{The same trend continues when the graph is extended beyond 2016.}`
Show Worked Solution

a.      `text{year}`
 

b. `r^2` `= 0.8794 \ text{(given)}`
  `r` `= ± sqrt{0.8794}`
    `= ± 0.938 \ text{(to 3 d.p.)}`

 
`text{By inspection of graph, correlation is negative}`

`:. \ r = -0.938`
 

c.    `text{Average decrease in winning time = 0.1515 seconds}`

`text{(this is given by the slope of the line.)}`
 

d.    `text{Residual Value}` `= text{actual}-text{predicted}`
    `= 53.83-54.10`
    `= -0.27`

 

e.i.      `text{winning time (2032)}` `= 357.1-0.1515 xx 2032`
      `=49.252 \ text{seconds}`

 

e.ii.  `text{The assumption is that the graph is accurate when it is extended}`

  `text{beyond 2016 (i.e decreasing trend continues).}`

CORE, FUR2 2021 VCAA 2

The two running events in the heptathlon are the 200 m run and the 800 m run. The times taken by the athletes in these two events, times200 and time800, are linearly related.

When a least squares line is fitted to the data, the equation of this line is found to be

                      time800 = 0.03931 + 5.2756 × time200

  1. Round the values for the intercept and the slope to three significant figures. Write your answers in the boxes provided.   (1 mark)
    time800=  
     
    		   +  
     
    		   × time200
  1. The mean and the standard deviation for each variable, time200 and time800, are shown in the table below.

  1. The equation of the least squares line is
  2.     time800 = 0.03931 + 5.2756 × time200
  3. Use this information to calculate the coefficient of determination as a percentage.
  4. Round your answer to the nearest percentage.   (2 marks)

Show Answers Only
  1. `text{See Worked Solutions}`
  2. `38text{%}`
Show Worked Solution

a.    `text{time800} = 0.0393 + 5.28 xx text{time200}`

b.    `b = r xx s_y/s_x`

`text{Solve for} \ r:`

`5.2756` `= r xx 8.2910/0.96956`
`r` `= 0.6169 …`
`r^2` `= (0.6169 …)^2`
  `= 0.3806 …`
  `= 38text{% (nearest %)}`

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)`

Complex Numbers, SPEC2 2021 VCAA 2

The polynomial  `p(z) = z^3 + alpha z^2 + beta z + gamma`, where  `z ∈ C`  and  `alpha, beta, gamma ∈ R`, can also be written as  `p(z) = (z-z_1)(z-z_2)(z-z_3)`, where  `z_1 ∈ R`  and  `z_2, z_3 ∈ C`.

  1.  i. State the relationship between `z_2` and `z_3`.  (1 mark)

  2. ii. Determine the values of `alpha, beta` and `gamma`, given that  `p(2) = -13, |z_2 + z_3| = 0`  and  `|z_2-z_3| = 6`.  (3 marks)

Consider the point  `z_4 = sqrt3 + i`.

  1. Sketch the ray given by  `text(Arg)(z-z_4) = (5pi)/6`  on the Argand diagram below.  (2 marks)
     
  2. The ray  `text(Arg)(z-z_4) = (5pi)/6`  intersects the circle  `|z-3i| = 1`, dividing it into a major and a minor segment.
  3.  i. Sketch the circle  `|z-3i| = 1` on the Argand diagram in part b.  (1 mark)

  4. ii. Find the area of the minor segment.  (2 marks)

Show Answers Only
    1. `z_2 = barz_3`
    2. `alpha = -3, beta = 9, gamma = -27`
  2.  
     

     

c.i.   

c.ii.   `(2pi-3sqrt3)/12\ text(u²)`

Show Worked Solution

a.i.   `text(By conjugate root theory)`

`z_2 = barz_3`

 

a.ii.   `text(Let)\ \ z_1 = a + bi, \ z_2 = a-bi`

♦♦ Mean mark part (a.ii.) 35%.

`|z_2 + z_3| = |2a| = 0 \ => \ a = 0`

`|z_2-z_3| = |2b| = 6 \ => \ b = ±3`
 

`text(Using)\ \ p(2) = -13`

`(2-z_1)(2-3i)(2 + 3i)` `= -13`
`(2-z_1)(4 + 9)` `= -13`
`2-z_1` `= -1`
`z_1` `= 3`

 

`p(z)` `= (z-3)(z-3i)(z + 3i)`
  `= (z-3)(z^2 + 9)`
  `= z^3-3z^2 + 9z-27`

 
`:. alpha = –3, \ beta = 9, \ gamma = –27`

♦ Mean mark part (b) 45%.
MARKER’S COMMENT: Point of emanation is not part of required ray (see open circle).

 

b. 

 

c.i.  

 

c.ii.   `text(Arg)(z-z_4) = (5pi)/6\ \ text(cuts)\ \ ytext(-axis at angle)\ pi/3`

♦♦ Mean mark part (c.ii.) 30%.

`=>\ text(angle at centre of segment) = pi/3\ (text(equilateral triangle))`

`text(Area)` `= (pi/3)/(2pi) xx pi xx 1^2-1/2 xx 1 xx 1 xx sin (pi/3)`
  `= pi/6-sqrt3/4`
  `= (2pi-3sqrt3)/12\ text(u²)`

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 1

A rectangular sheet of cardboard has a width of `h` centimetres. Its length is twice its width.

Squares of side length `x` centimetres, where  `x > 0`  are cut from each of the corners, as shown in the diagram below.
 

The sides  of this sheet of cardboard are then folded up to make a rectangular box with an open top, as shown in the diagram below.

Assume that the thickness of the cardboard is negligible and that  `V_text{box} > 0`.
 

A box is to be made from a sheet of cardboard with  `h` = 25 cm.

  1. Show that the volume, `V_text{box} > 0`, in cubic centimetres, is given by  `V_text{box} (x) = 2x (25-2x)(25-x)`.   (1 mark)

  2. State the domain of  `V_text{box}`.   (1 mark)

  3. Find the derivative of  `V_text{box}`  with respect to `x`.   (1 mark)

  4. Calculate the maximum possible volume of the box and for which value of `x` this occurs.   (3 marks)

  5. Waste minimisation is a goal when making cardboard boxes.
  6. Percentage wasted is based on the area of the sheet of cardboard that is cut out before the box is made.
  7. Find the percentage of the sheet of cardboard that is wasted when `x = 5`.   (2 marks)

Now consider a box made from a rectangular sheet of cardboard where  `h>0` and the box's length is still twice its width.

    1. Let  `V_text{box}` be the function that gives the volume of the box.
    2. State the domain `V_text{box}` in terms of `h`.   (1 mark)

    3. Find the maximum volume for any such rectangular box, `V_text{box}`, in terms of  `h`.   (3 marks)

  2. Now consider making a box from a square sheet of cardboard with side lengths of `h` centimetres.
  3. Show that the maximum volume of the box occurs when  `x = h/6`.   (2 marks)

Show Answers Only
  1. `text{Show Worked Solutions}`
  2. `(0, 12.5)`
  3. `text{Show Worked Solutions}`
  4. `{15625 sqrt3}/{9}  \ text{or} \ {15625}/{3 sqrt3}`
  5. `8%`
  6. i.  `text{Domain} \ (V_text{box}) = (0, h/2)`
  7. (sqrt3 h^3)/9`
  8. `text(See Worked Solutions)`
Show Worked Solution
a.   `V` `= x (h-2x)(2h-2x)`
    `= x (25-2x)(50-2x)`
    `= 2x (25-2x)(25-x)`

 

b.   `text{Sketch} \ y = 2x (25-2x)(25-x) \ text{by CAS:} `

♦ Mean mark (b) 42%.

`text{Domain is} \ (0, 12.5)`
 

c.    `(dV)/dx = 12x^2-300x + 1250`
 

d.    `text{Solve} \ V^{′}(x) = 0 \ text{for} \ x :`

`x = {-25 (sqrt3-3)}/6 \ or \ x = {25(sqrt3 + 3)}/6`

`:. \ x =  {-25 (sqrt3-3)}/6 \ , \ \ x ∈ (0, 12.5)`

`:. \ V_text{max} = {15\ 625 sqrt3}/9  \ text{or} \  \ {15\ 625}/{3 sqrt3}`
 

e.   `text{Area cut out} = 4 xx 5^2 = 100\ text(cm)^2`

`text{% Wasted}` `= {100}/{25 xx 50} xx 100`
  `= 8text(%)`

♦ Mean mark (f.i.) 33%.

 
f.i.  `text{Domain} \ (V_text{box}) = (0, h/2), \ \ text{(using part b)}`
 

f.ii.  `V = x (h-2x)(2h-2x)`

♦ Mean mark (f.ii.) 45%.

`text{Solve} \ V^{′}(x) = 0 \ text{for} \ x:`

`x = {-h (sqrt3-3)}/{6} \ \ text{or} \ \ x = {h(sqrt{3} + 3)}/{6}`

`V_text{max} \ text{occurs when} \ \ x = {-h(sqrt3-3)}/{6} \ \ text{(see part a)}`

`V_text{max} = (sqrt3 h^3)/9`

 

g.    `V = x(h-2x)(h-2x)`

♦ Mean mark part (g) 40%.

`text{Solve} \ V^{′}(x) = 0 \ text{for} \ x:`

`x = h/2 \ \ text{or} \ \ x =  h/6`

`V (h/6) = {2h^3}/{27} \ , \ V(h/2) = 0`

`:. V_text{max} \ text{occurs when} \ \ x = {h}/{6}`

Calculus, MET2 2021 VCAA 19 MC

Which one of the following functions is differentiable for all real values of `x`?

  1. `f(x) = {{:(qquadx),(–x):} {:(qquadx<0),(qquadx≥0):}`
  2.  `f(x) = {{:(qquadx),(–x):} {:(qquadx<0),(qquadx>0):}`
  3.  `f(x) = {{:(8x + 4),((2x + 1)^2):} {:(qquadx<0),(qquadx≥0):}`
  4.  `f(x) = {{:(2x + 1),((2x + 1)^2):} {:(qquadx<0),(qquadx≥0):}`
  5.  `f(x) = {{:(4x + 1),((2x + 1)^2):} {:(qquadx<0),(qquadx≥0):}`
Show Answers Only

`E`

Show Worked Solution

`text{Differentiable → function must be continuous and smooth}`

♦♦ Mean mark 35%.

`text{Consider}\ E:`
 

`f(x) = {{:(4x + 1),((2x + 1)^2):} {:(qquadx<0),(qquadx≥0):}  \ => \ f′(x) = {{:(4),(4(2x + 1)):} {:(qquadx<0),(qquadx≥0):}`
 

`text{As} \ x to 0^- \ , \ f(x) to \ 1 \ , \ f′(x) = 4`

`text{As} \ x to 0^+ \ , \ f(x) to \ 1 \ , \ lim f′(x) = 4`

`:. \ text{Function is continuous and smooth}`

`=> E`

Graphs, MET2 2021 VCAA 18 MC

Let  `f : R to R, \ f(x) = (2x - 1)(2x + 1)(3x - 1)`  and  `g : (–∞, 0) to R, \ g(x) = x log_e(–x)`.

The maximum number of solutions for the equation  `f(x - k) = g(x)`, where  `k ∈ R`, is

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

`D`

Show Worked Solution

`text{By CAS, graph}`

♦ Mean mark 39%.

`f(x) = (2x – 1)(2x + 1)(3x – 1) , text{and}`

`g(x) = x log_e (-x) , x < 0`
 


 

`text{If} \ f(x) \ text{is translated to the left, by inspection,}`

`text{a maximum of 3 intersections can occur.}`
 

`=> \ D`

Algebra, MET2 2021 VCAA 16 MC

Let  `cos (x) = 3/5`  and  `sin^2(y) = 25/169`, where  `x ∈ [{3pi}/{2} , 2 pi]`  and  `y ∈ [{3pi}/{2} , 2 pi]`.

The value of  `sin(x) + cos(y)`  is 

  1. `8/65`
  2. `– 112/65`
  3. `112/65`
  4. `–8/65`
  5. `64/65`
Show Answers Only

`A`

Show Worked Solution

`text{Both angles are in 4th quadrant (given)}`

♦♦ Mean mark 31%.

`cos(x) = 3/5`
 

`sin(x)` `= – 4/5\ \ text{(4th quadrant)}`
`sin^2(y)` `= 25/169`
`sin(y)` `= – 5/13\ \ text{(4th quadrant)}`

 

`cos(y) = 12/13`
 

`:. \ sin(x) + cos(y)` `= – 4/5 + 12/13`
  `= 8/65`

`=> A`

Probability, MET2 2021 VCAA 15 MC

Four fair coins are tossed at the same time.

The outcome for each coin is independent of the outcome for any other coin.

The probability that there is an equal number of heads and tails, given that there is at least one head; is

  1. `1/2`
  2. `1/3`
  3. `3/4`
  4. `2/5`
  5. `4/7`
Show Answers Only

`D`

Show Worked Solution

`X\ ~\ text(Bi) (4, 1/2)`

♦ Mean mark 48%.

`text(By CAS):`

`text(Pr) (X = 2 | X ≥ 1)` `= {text(Pr) (X = 2)}/{text(Pr) (X ≥ 1)}`
  `= 0.375/0.9375`
  `= 2/5`

`=> D`

Calculus, MET2 2021 VCAA 8 MC

The graph of the function  `f`  is shown below.
 

The graph corresponding to  `f^{′}` is
 


 


 

Show Answers Only

`E`

Show Worked Solution

`text(By elimination:)`

♦ Mean mark 40%.

`text{As} \ \ x to a^+ \ , \ f^{′}(x) to ∞`

`:. \ text{Eliminate}\ A, B, C`

`f(x) \ text{exists for} \ x ∈ (a,∞)`

`f^{′}(x) \ text{only exists for} \ \ x ∈ (a, ∞)`

`:. \ text{Eliminate}\ D`

`=> E`

Statistics, SPEC2 2021 VCAA 20 MC

An office has two coffee machines that operate independently of each other. The time taken for each machine to produce a cup of coffee is normally distributed with a mean of 30 seconds and a standard deviation of 5 seconds. On a particular morning, a cup is produced from each machine.

The probability that the time taken by each coffee machine to produce one cup of coffee will differ by less than 3 seconds is closest to

  1. 0.164
  2. 0.236
  3. 0.329
  4. 0.451
  5. 0.671
Show Answers Only

`C`

Show Worked Solution

`T_n~N(30, 5^2), n = 1, 2\ …`

♦ Mean mark 43%.

`E(T_2-T_1) = 0`

`text(Var)(T_2-T_1)` `= text(Var)(T_2) + text(Var)(T_1)`
  `= 2 xx 5^2`
  `= 50`

 
`:. T_2-T_1~N(0, (sqrt50)^2)`

`text(Pr)(-3 < T_2-T_1 < 3) = 0.329`

`text{By CAS: normCdf}(–3,3,0,sqrt50)`

`=>\ C`

Statistics, SPEC2 2021 VCAA 19 MC

The mean unscaled score for a certain assessment task is 25 and the variance is 36. The scores scaled so that the mean score is 30 and the variance is 49. Let `S` be the scaled scores, to the nearest integer, and let `X` be the unscaled scores.

If the scaling function takes the form  `S = mX+n`, where  `m ∈ R^+` and `n ∈ R`, then a score of 32 would be scaled to

  1. 22
  2. 34
  3. 36
  4. 38
  5. 40
Show Answers Only

`D`

Show Worked Solution
`E(S)` `= E(mX + n)`
  `= mE(X) + n`
  `= 25m + n`

 

♦ Mean mark 51%.
`text(Var)(S)` `= text(Var)(mX + n)`
  `= m^2text(Var)(X)`
  `= 36m^2`

 
`text{Solve (by CAS)}:`

`25m + n` `= 30\ …\ (1)`
`36m^2` `= 49\ …\ (2)`

 
`m = 7/6, n = 5/6`
 

`:.\ text(Adjusted score of 32)`

`= 7/6 xx 32 + 5/6`

`~~ 38`

`=>\ D`

Statistics, SPEC2 2021 VCAA 17 MC

Bottles of a particular brand of soft drink are labelled as having a volume of 1.25 L. The machines filling the bottles deliver a volume that is normally distributed with a mean of 1.26 L and a standard deviation of 0.01 L.

The probability that six bottles have a mean volume that is at least the labelled volume of 1.25 L is closest to

  1. 0.5968
  2. 0.8413
  3. 0.9750
  4. 0.9772
  5. 0.9928
Show Answers Only

`E`

Show Worked Solution

`V~N (1.26, 0.01^2)`

♦ Mean mark 50%.

`E(barV) = mu = 1.26`

`text(s.d.)(barV) = 0.01/sqrt6`

`text(Pr)(barV>= 1.25) ≈ 0.9928`

`text{By CAS: normCdf}(1.25,∞,1.26,0.01/sqrt6)`

`=> E`

Mechanics, SPEC2 2021 VCAA 16 MC

An object of mass `m` kilograms slides down a smooth slope that is inclined at an angle of `theta^@` to the horizontal, where  `0^@ < theta^@ < 45^@`. The acceleration of the object down the slope is  `a\ text(ms)^(-2), a > 0`.

If the angle of inclination of the slope is doubled to `2theta^@`, then the acceleration of the object down the slope, in `text(ms)^(-2)`, is

  1. `2a`
  2. `(2a)/gsqrt(g^2 - a^2)`
  3. `(2a^2 - g^2)/g`
  4. `a/g sqrt(g^2 - a^2)`
  5. `2asqrt(g^2 - a^2)`
Show Answers Only

`B`

Show Worked Solution

`ma = mgsintheta`

♦ Mean mark 48%.
`sintheta` `= a/g`
`cos^2theta` `= 1 – (a^2)/(g^2)`
`costheta` `= sqrt(1 – (a^2)/(g^2))`

 
`text(If incline angle) = 2theta`

`ma` `= mgsin(2theta)`
`a` `= g*2sinthetacostheta`
  `= g *2* a/g sqrt(1 – (a^2)/(g^2))`
  `= (2a)/g sqrt(g^2 – a^2)`

 
`=>\ B`

Vectors, SPEC2 2021 VCAA 13 MC

The scalar resolute of vector  `underset~a`  in the direction of vector  `underset~b`  is  `-4`.

If  `underset~b = – sqrt3 underset~i`,  the vector resolute of  `underset~a`  in the direction of  `underset~b`  is

  1. `-4underset~i`
  2. `-3underset~i`
  3. `1/sqrt3 underset~i`
  4. `3underset~i`
  5. `4underset~i`
Show Answers Only

`E`

Show Worked Solution

`text(Scalar resolute of)\ underset~a\ text(is direction of)\ underset~b = -4`

♦ Mean mark 48%.

`underset~a · overset^underset~b = -4`

`underset~b = -sqrt3 underset~i \ => \ overset^underset~b = -underset~i`

`text(Vector resolute of)\ underset~a\ text(in direction of)\ underset~b`

`(underset~a · overset^underset~b)overset^underset~b = 4underset~i`

`=>\ E`

Calculus, SPEC2 2021 VCAA 9 MC

Which one of the following derivatives corresponds to a graph of  `f`  that has no points of inflection?

  1. `f′(x) = 2(x - 3)^2 + 5`
  2. `f′(x) = 2(x - 3)^3 + 5`
  3. `f′(x) = 5/2(x - 3)^2`
  4. `f′(x) = 1/2(x - 3)^2 - 5`
  5. `f′(x) = (x - 3)^3 - 12x`
Show Answers Only

`B`

Show Worked Solution

`text(If no POI,)\ \ f″(x)\ text(does not change sign).`

♦ Mean mark 38%.

`text(Consider option B)`

`f″(x) = 6(x – 3)^2 \ ->\ text(doesn’t change sign)`

`=>\ B`

Trigonometry, SPEC2 2021 VCAA 7 MC

A relation is defined parametrically by

\(x(t)=5 \cos (2 t)+1, \quad y(t)=5 \sin (2 t)-1\)

If  \(A(6,-1)\)  and  \(B(1,4)\)  are two points that lie on the graph of the relation, then the shortest distance along the graph from \(A\) to \(B\) is

  1. \(\dfrac{\pi}{4}\)
  2. \(\dfrac{\pi}{2}\)
  3. \(\pi\)
  4. \(\dfrac{5 \pi}{4}\)
  5. \(\dfrac{5 \pi}{2}\)
Show Answers Only

\(E\)

Show Worked Solution

\(x=5 \cos (2 t)+1 \Rightarrow \cos (2 t)=\dfrac{x-1}{5}\)

♦ Mean mark 39%.

\(y=5 \sin (2 t)-1 \Rightarrow \sin (2 t)=\dfrac{y+1}{5}\)

\((\cos (2 t))^2+(\sin (2 t))^2\) \(=1\)  
\((x-1)^2+(y+1)^2\) \(=25\)  

 

\(\text {Distance}\ =\dfrac{1}{4} \times 2 \times \pi \times r =\dfrac{5 \pi}{2}\)

\(\Rightarrow E\)

Complex Numbers, SPEC2 2021 VCAA 5 MC

The graph of the circle given by  `|z - 2 - sqrt3i| = 1`, where  `z ∈ C`, is shown below.
 


 

For points on this circle, the maximum value of  `|z|`  is

  1. `sqrt3 + 1`
  2. `3`
  3. `sqrt13`
  4. `sqrt7 + 1`
  5. `8`
Show Answers Only

`D`

Show Worked Solution
♦ Mean mark part 42%.

`text(Centre of circle at)\ (2, sqrt3)`

`text(Radius = 1)`

`text(By Pythagoras, line from)`

`text(origin to centre of circle)`

`d = sqrt(2^2 + sqrt(3)^2) = sqrt7`

`:. |z|\ text(max) = sqrt7 + 1`

`=>\ D`

Complex Numbers, SPEC2 2021 VCAA 4 MC

For  `z ∈ C`, if  `text(Im)(z) > 0`, then  `text(Arg)((zbarz)/(z - barz))` is

  1. `-pi/2`
  2. `0`
  3. `pi/4`
  4. `pi/2`
  5. `pi`
Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ z=x+iy \ => \ barz=x-iy`

♦♦ Mean mark 35%.
`text(Arg)((zbarz)/(z – barz))` `= text(Arg)(zbarz) – text(Arg)(z – barz)`
  `= text(Arg)(x^2 + y^2) – text(Arg)(2yi)`
  `= 0 – text(Arg)(2yi),\ \ text(where)\ y > 0`
  `= -pi/2`

`=>\ A`