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CORE, FUR2 2021 VCAA 9

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

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

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

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

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

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

 

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

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

 
`=> FV = 146 \ 073.7405`
 

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

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

 
`=> PV = -153\ 112.9399`
 

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

CORE, FUR2 2021 VCAA 8

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

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

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

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

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

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

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


b.  `text{Fortnights in 1 year} = 26`

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

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

 

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

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

 
`=> N = 650.0046 …`
 

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

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

 
`=> FV = -5.59`
 

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

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 6

Sienna invests $420 000 in a perpetuity from which she will receive a regular monthly payment of $1890.

The perpetuity earns interest at the rate of 5.4% per annum.

  1. Determine the total amount, in dollars, that Sienna will receive after one year of monthly payments.   (1 mark)

  2. Write down the value of the perpetuity after Sienna has received one year of monthly payments.   (1 mark)

  3. Let `S_n` be the value of Sienna's perpetuity after `n` months.
  4. Complete the recurrence relation, in terms of `S_0`, `S_{n + 1}` and `S_n`, that would model the value of this perpetuity over time. Write your answers in the boxes provided.   (1 mark)

    `S_n =`
     
    		  ,          `S_{n+1} =`
     
    		 `xx S_n-1890`
Show Answers Only
  1. `$ 22 \ 680`
  2. `$ 420 \ 000`
  3. `S_n = 420 \ 000, \ \ S_{n+1} = 1.0045 xx S_n-1890`
Show Worked Solution
a.   `text{Total amount}` `= 12 xx 1890`
    `= $ 22 \ 680`

 

b.  `text{Value} = $ 420 \ 00 \ text{(balance remains the same after each payment).}`

 

c.  `S_0 = 420 \ 000 \ => \ S_n = 420 \ 000`

`S_{n+1} = RS_n-1890 \ \ text{where}\ \ R = 1 + r/100`

`r = 5.4/12 = 0.45`

`:. \ S_{n+1} = 1.0045 xx S_n-1890`

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 %)}`

CORE, FUR2 2021 VCAA 1

In the sport of heptathlon, athletes compete in seven events.

These events are the 100 m hurdles, high jump, shot-put, javelin, 200 m run, 800 m run and long jump.

Fifteen female athletes competed to qualify for the heptathlon at the Olympic Games.

Their results for three of the heptathlon events – high jump, shot-put and javelin – are shown in Table 1

  1. Write down the number of numerical variables in Table 1.   (1 mark)

  2. Complete Table 2 below by calculating the mean height jumped for the high jump, in metres, by the 15 athletes. Write your answer in the space provided in the table.   (1 mark)

  3. In shot-put, athletes throw a heavy spherical ball (a shot) as far as they can. Athlete number six, Jamilia, threw the shot 14.50 m.
  4. Calculate Jamilia's standardised score (`z`). Round your answer to one decimal place.   (1 mark)

  5. In the qualifying competition, the heights jumped in the high jump are expected to be approximately normally distributed.
  6. Chara's jump in this competition would give her a standardised score of  `z = –1.0`
  7. Use the 68–95–99.7% rule to calculate the percentage of athletes who would be expected to jump higher than Chara in the qualifying competition.   (1 mark)

  8. The boxplot below was constructed to show the distribution of high jump heights for all 15 athletes in the qualifying competition.

 

  1. Explain why the boxplot has no whisker at its upper end.   (1 mark)

  2. For the javelin qualifying competition (refer to Table 1), another boxplot is used to display the distribution of athlete's results.
  3. An athlete whose result is displayed as an outlier at the upper end of the plot is considered to be a potential medal winner in the event.
  4. What is the minimum distance that an athlete needs to throw the javelin to be considered a potential medal winner?   (2 marks)

Show Answers Only

  1. `3`
  2. `1.81`
  3. `0.5 \ text{(to d.p.)}`
  4. `84text(%)`
  5. `text{See Worked Solutions}`
  6. `46.89 \ text{m}`

Show Worked Solution

a.    `3 \ text{High jump, shot-put and javelin}`

 `text{Athlete number is not a numerical variable}`
  

b.     `text{High jump mean}`

`= (1.76 + 1.79 + 1.83 + 1.82 + 1.87 + 1.73 + 1.68 + 1.82 +`

`1.83 + 1.87 + 1.87 + 1.80 + 1.83 + 1.87 + 1.78) ÷ 15`

`= 1.81`
 

c.   `z text{-score} (14.50)` `= {14.50-13.74}/{1.43}`
    `= 0.531 …`
    `= 0.5 \ text{(to 1 d.p.)}`

 
d.  `P (z text{-score} > -1 ) = 84text(%)`
 

e.  `text{If the} \ Q_3 \ text{value is also the highest value in the data set,}`

`text{there is no whisker at the upper end of a boxplot.}`
 

f.  `text{Javelin (ascending):}`

`38.12, 39.22, 40.62, 40.88, 41.22, 41.32, 42.33, 42.41, `

`42.51, 42.65, 42.75, 42.88, 45.64, 45.68, 46.53`

`Q_1 = 40.88 \ \ , \ Q_3 = 42.88 \ \ , \ \ IQR = 42.88-40.88 = 2`

`text{Upper Fence}` `= Q_3 + 1.5  xx IQR`
  `= 42.88 + 1.5 xx 2`
  `= 45.88`

 
`:. \ text{Minimum distance = 45.89 m  (longer than upper fence value)}`

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`

Functions, EXT1 F1 SM-Bank 17

Solve the inequality  `x^2<=|\ 2x-1\ |`  for `x`.

Express your answer in interval notation.  (3 marks)

Show Answers Only

`:. x in [-1-sqrt2, -1 + sqrt2]\ \ ∪\ \ x in [1]`

Show Worked Solution

`text(Case 1:)`

`x^2` `<=2x-1`
`x^2-2x+1` `<=0`
`(x-1)^2` `<=0`

 
`x=1`
 

`text(Case 2:)`

`x^2` `<=-(2x-1)`
`x^2+2x-1` `<=0`

 

`x` `=(-2+-sqrt(4+4*1*1))/2`  
  `=-1+-sqrt2`  

 
`text(Test)\ \ x=0\ \ =>\ \ 0<=-(-1)\ \ text{(correct)}`

`:. x in [-1-sqrt2, -1 + sqrt2]\ \ ∪\ \ x in [1]`

Calculus, MET2 2021 VCAA 5

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

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

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

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

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

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

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

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

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

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

`text{Period}\ = 4 pi`
 

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

`f_min = -1.722`

♦ Mean mark part (c) 21%.
 

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

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

`:. h = 2 pi`
 

d.    `f(x) = sin(x/2) + cos(2x)`

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

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

e.i.  `int g_a (x)\ dx`

♦ Mean mark part (e)(i) 50%.

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

e.ii.   `int_0^{2a pi} g_a(x)\ dx`

♦ Mean mark part (e)(ii) 29%.

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

`= 0 \ \ (a ∈ ZZ^+)`
 

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

f.    `g_a (x) = sin(x/a) + cos (ax)`

♦ Mean mark part (f) 13%.

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

`:. -2 <= g_a (x) <= 2`
 

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

♦ Mean mark part (g) 2%.

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

`g_a(x)_min = – sqrt2`

Probability, MET2 2021 VCAA 4

A teacher coaches their school's table tennis team.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

a.   `W\ ~\ N (10,0.8^2)`

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

`text(Pr) (W >= 11) = 0.106`
 

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

c.    `E(overset^P) =  0.08 = 2/25`

♦ Mean mark part (c) 45%.

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

d.    `X\ ~\ text(Bi) (25, 0.08)`

♦ Mean mark part (d) 49%.

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

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

 

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

♦♦ Mean mark part (e) 21%.

 

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

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

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

Calculus, MET2 2021 VCAA 3

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

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

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

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

Let  `p(x) = e^{-2x}-2e^-x + 1.`

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

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

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

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

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

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

a.    `text(Method 1)`

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

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

`text(Method 2)`

`text{Sketch graph by CAS}`

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

`text{Domain:} \ x ∈ (-∞, -1)`

`text{Range:} \ y ∈ R`
 
 

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

`y = -x-2`
 

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

`y = x + 2`
  

c.    `text{Sketch graph by CAS.}`

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

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

d.   `p^{′}(x) = -2e^{-2x} + 2e^-x`

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

e. 

 
`text{Case 1}`

`text{By CAS, solve:}`

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

`a = -0.11`
 

`text{Case 2}`

`text{Case 1}`

`text{By CAS, solve:}`

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

`a = -0.67`
 

f.     `text{At intersection,}`

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

`x = -0.750`
 

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

Calculus, MET2 2021 VCAA 2

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

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

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

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

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

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

a.   `0.25`

 

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

 

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

 

d.

♦♦♦ Mean mark part (d) 16%.

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

 

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

 
f.   `text{Case 1:} \ a ≤ 1`

♦♦♦ Mean mark part (f) 18%.

`int_0^a (sqrtx – ax^2) dx = 1/3`

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

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

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

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

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

`:. a = 1.13`

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}`

Mechanics, EXT2 M1 2021 SPEC2 15

The diagram below shows a stationary body being acted on by four forces whose magnitudes are in newtons. The force of magnitude `F_1` newtons acts in the opposite direction to the force of magnitude 8 N.
 

Calculate the value of `F_1` in newtons.   (3 marks)

Show Answers Only

`8 – 2sqrt3\ \ text(N)`

Show Worked Solution

`text(Resolve forces vertically:)`

`F_2sin60^@` `= 6sin30^@`
`F_2 sqrt3/2` `= 6 xx 1/2`
`F_2` `= 6/sqrt3`
  `= 2sqrt3`

 
`text(Resolve forces horizontally:)`

`F_1 + 6cos30^@` `= 2sqrt3 cos60^@ + 8`
`F_1 + 3sqrt3` `= sqrt3 + 8`
`F_1` `= 8 – 2sqrt3\ \ text(N)`

Vectors, EXT1 V1 2021 SPEC2 12 MC

Consider the vectors  `underset~a = x underset~i + underset~j, \ underset~b = underset~i - underset~j`  and  `underset~c = underset~i + x underset~j`.

Given that  `theta`  is the angle between  `underset~a`  and  `underset~b`,  and  `phi`  is the angle between  `underset~b`  and  `underset~c, cos(theta) cos (phi)`  is

  1. `(2(1 + x^2))/(1 - x^2)`
  2. `(sqrt2(1 - x^2))/(1 + x^2)`
  3. `-((x + 1)^2)/(2(1 + x^2))`
  4. `-((x - 1)^2)/(2(1 + x^2))`
Show Answers Only

`D`

Show Worked Solution

`underset~a = x underset~i – underset~j, \ underset~b = underset~i – underset~j, \ underset~c = underset~i + x underset~j`

`underset~a · underset~b = |underset~a||underset~b|costheta`

`costheta = (x – 1)/(sqrt(x^2 + 1)sqrt2)`

`cos phi = (underset~b · underset~c)/(|underset~b||underset~c|) = (1 – x)/(sqrt2 sqrt(1 + x^2))`

`costheta · cos phi` `= ((x – 1)(1 – x))/(2(1 + x^2))`
  `= -((x – 1)^2)/(2(1 + x^2))`

 
`=>\ D`

Vectors, EXT1 V1 2021 SPEC2 11

Let `underset~i` be a unit vector pointing east and let `underset~j` be a unit vector pointing north.

A group of hikers travels 5 km in the direction south 30° west and then north for 10 km before stopping for lunch at position vector  `(a underset~i, b underset~j)`.

Find the values of `a` and `b`.   (2 marks)

Show Answers Only

`a=-5/2, b=10-(5sqrt3)/2`

Show Worked Solution

`sin30 = x/5 \ => \ x = 5/2`

`cos30 = y/5 \ => \ y = (5sqrt3)/2`

`P_1\ text(vector:)\  (-5/2 underset~i, -(5sqrt3)/2 underset~j)`

`P_2\ text(vector:)\  (-5/2 underset~i, (10-(5sqrt3)/2)underset~j)`

`:. a=-5/2, b=10-(5sqrt3)/2`

Calculus, EXT1 C3 2021 SPEC2 10 MC

The differential equation that has the diagram above as its direction field is

  1. `(dy)/(dx) = y + 2x`
  2. `(dy)/(dx) = 2x - y`
  3. `(dy)/(dx) = x+2y`
  4. `(dy)/(dx) = y - 2x`
Show Answers Only

`D`

Show Worked Solution

`text(By elimination:)`

`text(At)\ (1, 2), m = 0`

`->\ text(Eliminate)\ A, C`

`text(At)\ (0, 1),\ m\ text(is positive)`

`->\ text(Eliminate)\ B`

`=>\ D`

Functions, EXT1 F1 2021 SPEC2 7

A function is defined parametrically by

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

If  `A(6, –1)`  and  `B(1, 4)`  are two points that lie on the graph of the function, then find the shortest distance along the graph from `A` to `B`.   (2 marks)

Show Answers Only

`(5pi)/2`

Show Worked Solution

`x = 5cos(2t) + 1 \ => \ cos(2t) = (x-1)/5`

`y = 5sin(2t)-1 \ => \ sin(2t) = (y + 1)/5`

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

 

`text(Distance)` `= 1/4 xx 2 xx pi xx r`
  `= (5pi)/2`

Complex Numbers, EXT2 N2 2021 SPEC2 5

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


 

For points on this circle, find the maximum value of  `|z|`.   (2 marks)

Show Answers Only

`sqrt7+1`

Show Worked Solution

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

`text(Radius = 1)`

`text(By Pythagoras, line from origin to the centre of the circle)`

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

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

Complex Numbers, EXT2 N1 2021 SPEC2 4 MC

For  the complex number `z `, if  `text(Im)(z) > 0`, then  `text(Arg)((zbarz)/(z - barz))` is

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

`A`

Show Worked Solution

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

`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`

Probability, MET2 2021 VCAA 17 MC

A discrete random variable `X` has a binomial distribution with a probability of success of  `p = 0.1`  for `n` trials, where  `n > 2`.

If the probability of obtaining at least two successes after `n` trials is at least 0.5, then the smallest possible value of `n` is

  1. 15
  2. 16
  3. 17
  4. 18
  5. 19
Show Answers Only

`C`

Show Worked Solution

`text(Pr) (X ≥ 2) ≥ 0.5`

`text(Pr) (X = 0)` `=\ ^n C_0 xx 0.1^0 xx 0.9^n=0.9^n`
`text(Pr) (X = 1)` `=\ ^n C_1 xx 0.1^1 xx 0.9^{n-1}=nxx0.1xx0.9^(n-1)`
`text(Pr) (X ≥ 2)` `= 1 – (text(Pr) (X=0) + text(Pr) (X=1))`

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

`1 – (0.9^n + n xx 0.1 xx 0.9^{n-1}) ≥ 0.5`

`n = 17`

`=> \ C`

Calculus, MET2 2021 VCAA 14 MC

A value of `k` for which the average value of  `y = cos (kx - pi/2)`  over the interval  `[0, pi]`  is equal to the average value of  `y = sin(x)` over the same interval is

  1. `1/6`
  2. `1/5`
  3. `1/4`
  4. `1/3`
  5. `1/2`
Show Answers Only

`E`

Show Worked Solution

`text{By CAS, solve for} \ k:`

`1/pi int_0^pi cos (kx – pi/2) dx = 1/pi int_0^pi sin (x)\ dx`

`k = 1/2`

`=> E`

Statistics, MET2 2021 VCAA 12 MC

For a certain species of bird, the proportion of birds with a crest is known to be `3/5`.

Let `overset^P` be the random variable representing the proportion of birds with a crest in samples of size `n` for this specific bird.

The smallest sample size for which the standard deviation of `overset^P` is less than 0.08 is

  1.   7
  2. 27
  3. 37
  4. 38
  5. 43
Show Answers Only

`D`

Show Worked Solution

`text{Solve by CAS:}`

`sqrt{{0.6 xx 0.4}/{n}} < 0.08`

`n > 37.5`

`=> D`

Functions, MET2 2021 VCAA 10 MC

Consider the functions  `f(x) = sqrt{x+2}`  and  `g(x) = sqrt{1-2x}`, defined over their maximal domains.

The maximal domain of the function  `h = f + g`  is.

  1. `(–2, 1/2)`
  2. `[–2,∞)`
  3. `(–∞, –2) ∪ (1/2, ∞)`
  4. `[–2, 1/2]`
  5. `[–2, 1]`
Show Answers Only

`D`

Show Worked Solution

`f(x) \ = \ sqrt{x + 2} \ => \ text{domain} \ x ≥ -2`

`g(x) \ = \ sqrt{1-2x} \ => \ text{domain} \ x ≤ 1/2`

`text{Intersection of domains = domain} \ h(x)`

`:. \ h(x) ∈ [-2, 1/2]`

`=> D`

Functions, MET2 2021 VCAA 9 MC

Let  `g(x) = x + 2`  and  `f(x) = x^2 - 4`

If `h` is the composite function given by  `h : [–5, –1) to R, h(x) = f(g(x))`, then the range of `h` is

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

`E`

Show Worked Solution
`h(x)` `= (x + 2)^2 – 4`
  `= x^2 + 4x`

 
`text{By CAS, graph} \ \ y = x^2 + 4x , \ x∈ [–5, –1)`

`text{Min at} \ (–2, –4)`

`text{Max at} \ (–5, 5)`

`:. \ text{Range of} \ h(x) ∈ [–4, 5]`

`=> E`

Calculus, MET2 2021 VCAA 7 MC

The tangent to the graph of  `y = x^3 - ax^2 + 1`  at  `x = 1` passes through the origin.

The value of `a` is

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

`B`

Show Worked Solution
`y` `= x^3 – ax^2 + 1`
`dy/dx` `= 3x^2 – 2ax`

 
`text{At} \ \ x = 1 \ => \  y = 2-a, \ dy/dx = 3-2a`
 

`text{T} text{angent passes through} \ (1,2-a) \ text{and} \ (0,0)`

`m_text{tang} = 2 – a`

`text{Equating gradients:}`

`3-2a` `= 2-a`
`:. a` `= 1`

 
`=> B`

Functions, MET2 2021 VCAA 5 MC

Consider the following four functional relations.

  `f(x) = f(-x)\ \ \ \ \ -f(x) = f(-x)\ \ \ \ \ f(x) = -f(x)\ \ \ \ \ (f(x))^2 = f(x^2)`

The number of these functional relations that are satisfied by the function  `f : R -> R, \ f(x) = x`  is

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

`C`

Show Worked Solution

`text{Consider each relation}`

`f(x) = x \ ,` `f(–x) = –x \ \ ✘`
`–f(x) = –x  \ ,` ` f(-x) = –x \ \ ✓`
`f(x) = x \ ,` `–f(x) = –x \ \ ✘`
`(f(x))^2 = x^2 \ ,` `f(x^2) = x^2 \ \ ✓`

 

`=> C`

Statistics, MET2 2021 VCAA 3 MC

A box contains many coloured glass beads.

A random sample of 48 beads is selected and it is found that the proportion of blue-coloured beads in this sample is 0.125

Based on this sample, a 95% confidence interval for the proportion of blue-coloured glass beads is

  1. (0.0314, 0.2186)
  2. (0.0465, 0.2035)
  3. (0.0018, 0.2482)
  4. (0.0896, 0.1604)
  5. (0.0264, 0.2136)
Show Answers Only

`A`

Show Worked Solution

`text{Method 1}`

`overset^p = 0.125 , \ n= 48`

`text{Successes} = 0.125 xx 48 = 6`
  

`text{By CAS:}`

`text{95% C.I.} = (0.0314, 0.2186)`
 

`text{Method 2}`

`text(sd)(overset^p)` `= sqrt({0.125 (1-0.125)}/{48})`
  `= 0.04773`

 

`:.\ text{95% C.I.}` `= (0.125 – 1.96 xx 0.04773 , 0.125 + 1.96 xx 0.04773)`
  `= (0.0314 , 0.2186)`

`=> A`

Statistics, SPEC2 2021 VCAA 18 MC

A scientist investigates the distribution of the masses of fish in a particular river. A 95% confidence interval for the mean mass of a fish, in grams, calculated from a random sample of 100 fish is (70.2, 75.8).

The sample mean divided by the population standard deviation is closest to

  1.   1.3
  2.   2.6
  3.   5.1
  4. 10.2
  5. 13.0
Show Answers Only

`C`

Show Worked Solution

`barx = (70.2 + 75.8)/2 = 73`

`1.96 xx s/sqrt100` `= 73 – 70.2`
`s` `= (2.8 xx 10)/1.96`
  `= 14.29`

 
`:. barx/s = 73/14.29 ~~ 5.1`

`=>\ C`

Mechanics, SPEC2 2021 VCAA 15 MC

The diagram below shows a stationary body being acted on by four forces whose magnitudes are in newtons. The force of magnitude `F_1` newtons acts in the opposite direction to the force of magnitude 8 N.
 

The value of `F_1` is

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

`A`

Show Worked Solution

`text(Resolve forces vertically:)`

`F_2sin60^@` `= 6sin30^@`
`F_2 sqrt3/2` `= 6 xx 1/2`
`F_2` `= 6/sqrt3`
  `= 2sqrt3`

 
`text(Resolve forces horizontally:)`

`F_1 + 6cos30^@` `= 2sqrt3 cos60^@ + 8`
`F_1 + 3sqrt3` `= sqrt3 + 8`
`F_1` `= 8 – 2sqrt3`

`=>\ A`

Mechanics, SPEC2 2021 VCAA 14 MC

A body of mass 5 kg is acted on by a net force of magnitude `F` newtons. This force causes the body to move so that its velocity,  `v\ text(ms)^(-1)`, along a straight line of motion is given by  `v = 3 + 2x`, where `x` metres is the position of the body at time `t` seconds.

When  `x = 2, F` is equal to

  1. 10
  2. 14
  3. 35
  4. 70
  5. 175
Show Answers Only

`D`

Show Worked Solution
`a` `= v · (dv)/(dx)`
  `= (3 + 2x) · 2`

 
`text(Find)\ F\ text(when)\ x = 2:`

`F` `= ma`
  `= 5 xx 2(3 + 2 xx 2)`
  `= 70\ text(N)`

 
`=>\ D`

Vectors, SPEC2 2021 VCAA 12 MC

Consider the vectors  `underset~a = x underset~i + underset~j, \ underset~b = underset~i - underset~j`  and  `underset~c = underset~i + x underset~j`.

Given that  `theta`  is the angle between  `underset~a`  and  `underset~b`,  and  `phi`  is the angle between  `underset~b`  and  `underset~c, cos(theta) cos (phi)`  is

  1. `(2(1 + x^2))/(1 - x^2)`
  2. `(sqrt2(1 - x^2))/(1 + x^2)`
  3. `-((x + 1)^2)/(2(1 + x^2))`
  4. `-((x - 1)^2)/(2(1 + x^2))`
  5. `(sqrt2(1 + x^2))/(1 - x^2)`
Show Answers Only

`D`

Show Worked Solution

`underset~a = x underset~i – underset~j, \ underset~b = underset~i – underset~j, \ underset~c = underset~i + x underset~j`

`underset~a · underset~b = |underset~a||underset~b|costheta`

`costheta = (x – 1)/(sqrt(x^2 + 1)sqrt2)`

`cos phi = (underset~b · underset~c)/(|underset~b||underset~c|) = (1 – x)/(sqrt2 sqrt(1 – x^2))`

`costheta · cos phi` `= ((x – 1)(1 – x))/(2(1 + x^2))`
  `= -((x – 1)^2)/(2(1 + x^2))`

 
`=>\ D`

Vectors, SPEC2 2021 VCAA 11 MC

Let `underset~i` be a unit vector pointing east and let `underset~j` be a unit vector pointing north.

A group of hikers travels 5 km in the direction south 30° west and then north for 10 km.

The position vector `underset~a` of the group of hikers with respect to the starting point is

  1. `underset~a = -5/2 underset~i - (5sqrt3)/2 underset~j`
  2. `underset~a = -5/2 underset~i + (10 - (5sqrt3)/2)underset~j`
  3. `underset~a = -5/2 underset~i + 10underset~j`
  4. `underset~a = -(5sqrt3)/2 underset~i + 15/2 underset~j`
  5. `underset~a = 5/2 underset~i + (10 + (5sqrt3)/2)underset~j`
Show Answers Only

`B`

Show Worked Solution

`sin30 = x/5 \ => \ x = 5/2`

`cos30 = y/5 \ => \ y = (5sqrt3)/2`

`:. P_1\ text(vector:)\  (-5/2 underset~i, -(5sqrt3)/2 underset~j)`

`:. P_2\ text(vector:)\  (-5/2 underset~i, (10 – (5sqrt3)/2)underset~j)`

`=>\ B`

Calculus, SPEC2 2021 VCAA 10 MC

The differential equation that has the diagram above as its direction field is

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

`D`

Show Worked Solution

`text(By elimination:)`

`text(At)\ (1, 2), m = 0`

`->\ text(Eliminate)\ A, C, E`

`text(At)\ (0, 1),\ m\ text(is positive)`

`->\ text(Eliminate)\ B`

`=>\ D`

Calculus, SPEC2 2021 VCAA 8 MC

Euler's method, with a step size of 0.1, is used to approximate the solution of the differential equation  `(dy)/(dx) = ysin(x)`.

Given that  `y = 2`  when  `x = 1`, the value of `y`, correct to three decimal places, when  `x = 1.2` is

  1. 2.168
  2. 2.178
  3. 2.362
  4. 2.370
  5. 2.381
Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ y_0 = 2, \ x_0 = 1, \ h = 0.1:`

`y_1` `= y_0 + h(y_0sinx_0)`
  `= 2 + 0.1(2sin1)`
  `= 2.1683`

 

`y_2` `= 2.1683 + 0.1(2.1683 xx sin 1.1)`
  `= 2.362`

 
`=>\ C`

Trigonometry, SPEC2 2021 VCAA 3 MC

The coordinates of the local maxima of the graph of  `y = 1/((cos(ax) + 1)^2 + 3)`, where  `a ∈ Rtext(\){0}`, are

  1. `((2pik)/a , 1/7), k ∈ Z`
  2. `((2pik)/a , 1/3), k ∈ Z`
  3. `(((1 + 2k)pi)/(2a) , 1/4), k ∈ Z`
  4. `((pi(1 + 2k))/a , 1/4), k ∈ Z`
  5. `((pi(1 + 2k))/a , 1/3), k ∈ Z`
Show Answers Only

`E`

Show Worked Solution

`y_text(max)\ \ text(occurs when)`

Mean mark 51%.

`(cos(ax))^2 + 3\ \ text(is a minimum)`
 

`text(Find)\ x\ text(such that:)`

`cos(ax)` `= -1`
`ax` `= pi + 2kpi, k ∈ Z`
`x` `= (pi(1 + 2k))/a`

 
`:.\ text(Local maxima at)\ ((pi(1 + 2k))/a, 1/3)`

`=>\ E`

Graphs, SPEC2 2021 VCAA 2 MC

The implied domain of the function with rule  `f(x) = cos^(-1)(log_e(bx), b > 0)`  is

  1. `(0, 1]`
  2. `[1, e]`
  3. `[1/b, e/b]`
  4. `[1/b, (e^pi)/b]`
  5. `[1/(be), e/b]`
Show Answers Only

`E`

Show Worked Solution

`f(x)\ text(is defined when)`

`-1 <= log_e(bx) <= 1`

`1/e <= bx <= e`

`1/(be) <= x <= e/b`

`=>\ E`

Vectors, SPEC1 2021 VCAA 9

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

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

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

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

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

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


    2.    

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

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

a.i.   `text(Particle A)`

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

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

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

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

 

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

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

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

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

 

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

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

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

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

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

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

 
`:.\ text(Particles collide.)`

 

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

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

 

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

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

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

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

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

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

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

♦ Mean mark part (c)(ii) 45%.

 

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

Complex Numbers, EXT2 N2 2021 SPEC1 8

  1. Solve  `z^2 + 2z + 2 = 0`  for `z`, where  `z ∈ C`.  (1 mark)

  2. Solve  `z^2 + 2barz + 2 = 0`  for `z`, where  `z ∈ C`.  (3 marks)

Show Answers Only
  1. `z = -1 – i\ \ text(or)\ \ -1 + i`
  2. `z = 1 ± sqrt5 i`
Show Worked Solution
a.    `z^2 + 2z + 2` `= 0`
  `z^2 + 2z + 1 + 1` `= 0`
  `(z + 1)^2 + 1` `= 0`
  `(z + 1)^2 – i^2` `= 0`
  `(z + 1 + i)(z + 1 – i)` `= 0`

 
`:. z = -1 – i\ \ \ text(or)\ \ -1 + i`

 

b.   `z = x + yi \ => \ barz = x – yi`

`z^2 + 2barz + 2` `= 0`
`(x + yi)^2 + 2(x – yi) + 2` `= 0`
`x^2 + 2xyi – y^2 + 2x – 2yi + 2` `= 0`
`x^2 – y^2 + 2x + 2 + (2xy – 2y)i` `= 0`

 

`text(Find)\ \ x, y\ text(such that)`

`x^2 – y^2 + 2x + 2` `= 0\ …\ (1)`
`2xy – 2y` `= 0\ …\ (2)`

 
`text(When)\ \ 2xy – 2y = 0`

`2y(x – 1)` `= 0`
`x` `= 1`

 
`text(Substitute)\ \ x = 1\ \ text{into (1)}`

`1 – y^2 + 2 + 2` `= 0`
`y^2` `= 5`
`y` `= ±sqrt5`

 
`:. z = 1 ± sqrt5 i`

Complex Numbers, SPEC1 2021 VCAA 8

  1. Solve  `z^2 + 2z + 2 = 0`  for `z`, where  `z ∈ C`.   (1 mark)

  2. Solve  `z^2 + 2barz + 2 = 0`  for `z`, where  `z ∈ C`.   (3 marks)

Show Answers Only
  1. `z = -1-i\ \ text(or)\ \ -1 + i`
  2. `z = 1 ± sqrt5 i`
Show Worked Solution
a.    `z^2 + 2z + 2` `= 0`
  `z^2 + 2z + 1 + 1` `= 0`
  `(z + 1)^2 + 1` `= 0`
  `(z + 1)^2-i^2` `= 0`
  `(z + 1 + i)(z + 1-i)` `= 0`

 
`:. z = -1-i\ \ \ text(or)\ \ -1 + i`

 

b.   `z = x + yi \ => \ barz = x-yi`

♦♦ Mean mark part (b) 28%.
`z^2 + 2barz + 2` `= 0`
`(x + yi)^2 + 2(x-yi) + 2` `= 0`
`x^2 + 2xyi-y^2 + 2x-2yi + 2` `= 0`
`x^2-y^2 + 2x + 2 + (2xy-2y)i` `= 0`

 

`text(Find)\ \ x, y\ text(such that)`

`x^2-y^2 + 2x + 2` `= 0\ …\ (1)`
`2xy-2y` `= 0\ …\ (2)`

 
`text(When)\ \ 2xy-2y = 0`

`2y(x-1)` `= 0`
`x` `= 1`

 
`text(Substitute)\ \ x = 1\ \ text{into (1)}`

`1-y^2 + 2 + 2` `= 0`
`y^2` `= 5`
`y` `= ±sqrt5`

 
`:. z = 1 ± sqrt5 i`

Calculus, EXT1 C3 2021 SPEC1 7

The velocity of a particle satisfies the differential equation  `(dx)/(dt) = xsin(t)`,  where  `x`  centimetres is its displacement relative to a fixed point `O` at time `t` seconds.

Initially, the displacement of the particle is 1 cm.

  1. Find an expression for `x` in terms of `t`.  (3 marks)

  2. Find the maximum displacement of the particle and the times at which this occurs.  (2 marks)

Show Answers Only
  1. `x = e^(1 – cos(t))`
  2. `x_text(max) = e^2\ \ text(when)\ \ t = pi, 3pi, 5pi, …`

     

    `text(or)\ t = (2k + 1)pi\ \ text(for integral)\ \ k =0,1,2,…`

Show Worked Solution
a.    `(dx)/(dt)` `= x sin(t)`
  `int 1/x\ dx` `= int sin(t)\ dt`
  `log_e x` `= -cos(t) + c`

 

`text(When)\ \ t = 0, x = 1`

`log_e 1` `= -cos0 + c`
`c` `= 1`
`log_e x` `= -cos(t) + 1`
`:. x` `= e^(1 – cos(t))`

 

b.    `x` `= e^(1 – cos(t))`
  `(dx)/(dt)` `= sin(t) · e^(1 – cos(t))`

`text(Find)\ \ t\ \ text(when)\ \ (dx)/(dt) = 0:`

`e^(1 – cos(t)) != 0`

`sin(t) = 0\ \ text(when)\ \ t = 0, pi, 2pi, …`

`x_text(max) = e^2\ \ text(when)\ \ t = pi, 3pi, 5pi, …`

`text(or)\ \ t = (2k + 1)pi\ \ text(for integral)\ \ k =0,1,2,…`

Calculus, SPEC1 2021 VCAA 7

The velocity of a particle satisfies the differential equation  `(dx)/(dt) = xsin(t)`,  where  `x`  centimetres is its displacement relative to a fixed point `O` at time `t` seconds.

Initially, the displacement of the particle is 1 cm.

  1. Find an expression for `x` in terms of `t`.   (3 marks)

  2. Find the maximum displacement of the particle and the times at which this occurs.   (2 marks)

Show Answers Only
  1. `x = e^(1-cos(t))`
  2. `x_text(max) = e^2\ \ text(when)\ \ t = pi, 3pi, 5pi, …`

     

    `text(or)\ t = (2k + 1)pi\ \ text(for)\ \ k ∈ Z^+ ∪ {0}`

Show Worked Solution
a.    `(dx)/(dt)` `= x sin(t)`
  `int 1/x\ dx` `= int sin(t)\ dt`
  `log_e x` `= -cos(t) + c`

 

`text(When)\ \ t = 0, x = 1`

`log_e 1` `= -cos0 + c`
`c` `= 1`
`log_e x` `= -cos(t) + 1`
`:. x` `= e^(1-cos(t))`

 

b.    `x` `= e^(1-cos(t))`
  `(dx)/(dt)` `= sin(t) · e^(1-cos(t))`

`text(Find)\ \ t\ \ text(when)\ \ (dx)/(dt) = 0:`

♦♦ Mean mark part (b) 25%.

`e^(1-cos(t)) != 0`

`sin(t) = 0\ \ text(when)\ \ t = 0, pi, 2pi, …`

`x_text(max) = e^2\ \ text(when)\ \ t = pi, 3pi, 5pi, …`

`text(or)\ \ t = (2k + 1)pi\ \ text(for)\ \ k ∈ Z^+ ∪ {0}`

Vectors, SPEC1 2021 VCAA 6

Consider the three vectors  `underset~a = -underset~i + 6underset~j - 3underset~k, underset~b = 2underset~i - 8underset~j + 5underset~k`  and  `underset~c = 3underset~i + 2underset~j + |1 - p^2|underset~k`,  where  `p`  is a real constant.

Find the values of `p` for which the three vectors are linearly independent.  (4 marks)

Show Answers Only

`p ∈ R\ text(\) {±sqrt5}`

Show Worked Solution

`text(Linearly dependent when)\ \ underset~a = m underset~b + n underset~c`

`text(Equating)\ \ underset~i, underset~j\ \ text(and)\ \ underset~k\ \ text(components:)`

`-1` `= 2m + 3n\ \ …\ \ (1)`  
`6` `= -8m + 2n\ \ …\ \ (2)`  
`-3` `= 5m + |1 – p^2|n\ \ …\ \ (3)`  

 
`text(Mult)\ \ (1) xx 4`

`-4 = 8m + 12n\ \ …\ \ (1)′`

`text(Add)\ \  (2) + (1)′`

`2` `= 14n`
`n` `= 1/7`

 
`text(Substitute)\ \ n = 1/7\ \ text{into (1):}`

`-1` `= 2m + 3/7`
`2m` `= -10/7`
`m` `= -5/7`

 
`text(Substitute)\ \ m, n\ \ text{into (3):}`

`-3` `= – 25/7 + |1 – p^2| · 1/7`
`-21` `= -25 + |1 – p^2|`
`|1-p^2|` `= 4`
`1-p^2` `= ±4`
`p^2` `= 5\ \ \ (p^2 != -3)`
`p` `= ±sqrt5`

 
`:.\ text(Vectors are linearly independent for)\ \ p ∈ R\ text(\) {±sqrt5}`

Calculus, SPEC1 2021 VCAA 5

Find the gradient of the curve with equation  `e^x e^(2y) = 2e^4`  at the point  `(2, 1)`.  (3 marks)

Show Answers Only

`-1/10`

Show Worked Solution

`e^x e^(2y) + e^(4y^2) = 2e^4`

`e^x · 2e^(2y) · (dy)/(dx) + e^x · e^(2y) + e^(4y^2) · 8y · (dy)/(dx)` `= 0`
`(dy)/(dx)(e^x ·2e^(2y) + e^(4y^2) · 8y)` `= -e^x · e^(2y)`

 
`(dy)/(dx) = (-e^x · e^(2y))/(e^x · 2e^(2y) + e^(4y^2) ·8y)`

 
`text(At)\ \ (2, 1):`

`(dy)/(dx)` `= (-e^2 · e^2)/(e^2 · 2e^2 + e^4 · 8)`
  `= (-e^4)/(e^4(2 + 8))`
  `= -1/10`

Calculus, EXT1 C3 2021 SPEC1 4

The shaded region in the diagram below is bounded by the graph of  `y = sin(x)`  and the `x`-axis between the first two non-negative `x`-intercepts of the curve, that is interval  `[0, pi]`.  The shaded region is rotated about the `x`-axis to form a solid of revolution.
 
       
 
Find the volume, `V_s` of the solid formed.  (3 marks)

Show Answers Only

`(pi^2)/2\ text(u)³`

Show Worked Solution
  `V_s` `= pi int_0^pi sin^2(x)\ dx`
    `= pi int_0^pi 1/2(1 – cos(2x))\ dx`
    `= pi/2 int_0^pi 1 – cos(2x)\ dx`
    `= pi/2 [x – 1/2 sin(2x)]_0^pi`
    `= pi/2[pi – 1/2 sin(2pi) – (0 – 1/2 sin 0)]`
    `= (pi^2)/2\ text(u)³`

Calculus, SPEC1 2021 VCAA 4

  1. The shaded region in the diagram below is bounded by the graph of  `y = sin(x)`  and the `x`-axis between the first two non-negative `x`-intercepts of the curve, that is interval  `[0, pi]`.  The shaded region is rotated about the `x`-axis to form a solid of revolution.
     
           
     
    Find the volume, `V_s` of the solid formed.   (3 marks)

  2. Now consider the function  `y = sin(kx)`, where `k` is a positive real constant. The region bounded by the graph of the function and the `x`-axis between the first two non-negative `x`-intercepts of the graph is rotated about the `x`-axis to form a solid of revolution.
  3. Find the volume of this solid in term of `V_s`.   (1 mark)

Show Answers Only

  1. `(pi^2)/2\ text(u)³`
  2. `1/k V_s`

Show Worked Solution

a.    `V_s` `= pi int_0^pi sin^2(x)\ dx`
    `= pi int_0^pi 1/2(1-cos(2x))\ dx`
    `= pi/2 int_0^pi 1-cos(2x)\ dx`
    `= pi/2 [x-1/2 sin(2x)]_0^pi`
    `= pi/2[pi-1/2 sin(2pi)-(0-1/2 sin 0)]`
    `= (pi^2)/2\ text(u)³`

♦♦ Mean mark part (b) 30%.

 

b.   `y = sin(kx)\ \ text(is the dilation of)\ \ y = sin(x)\ \ text(by a factor of)`

`k\ \ text(from the)\ \ xtext(-axis)`

`:. V = 1/k V_s`

Statistics, SPEC1 2021 VCAA 3

A company produces a particular type of light globe called Shiny. The company claims that the lifetime of these globes is normally distributed with a mean of 200 weeks and it is known that the standard deviation of the lifetime of Shiny globes is 10 weeks. Customers have complained, saying Shiny globes were lasting less than the claimed 200 weeks. It was decided to investigate the complaints. A random sample of 36 Shiny globes was tested and it was found that the mean lifetime of the sample was 195 weeks.

Use  `text(Pr)(-1.96 < Z < 1.96) = 0.95`  and  `text(Pr)(-3 < Z < 3) = 0.9973`  to answer the following questions.

  1. Write down the null and alternative hypotheses for the one-tailed test that was conducted to investigate the complaints.   (1 mark)

    1. Determine the `p` value, correct to three places decimal places, for the test.   (2 marks)

    2. What should the company be told if the test was carried out at the 1% level of significance?   (1 mark)

  3. The company decided to produce a new type of light globe called Globeplus.
    Find the approximate 95% confidence interval for the mean lifetime of the new globes if a random sample of 25 Globeplus globes is tested and the sample mean is found to be 250 weeks. Assume that the standard deviation of the population is 10 weeks. Give your answer correct to two decimal places.   (1 mark)

Show Answers Only
  1. `H_0 : mu = 200`
    `H_1 : mu < 200`
    1. `0.001\ \ (text(to 3 d.p.))`
    2. `(246.08, 253.92)`
  3. `text(Reject the null hypothesis.)`
Show Worked Solution

a.   `H_0 : mu = 200`

`H_1 : mu < 200`

 

b.i.   `E(barX) = mu = 200`

♦ Mean mark part (b)(i) 48%.

`sigma(barX) = sigma/sqrtn = 10/sqrt36 = 5/3`

`p` `= text(Pr)(barX< 195 | mu = 200)`
  `= text(Pr)(z < (195-200)/(3/5))`
  `= text(Pr)(z < -3)`
  `= 1/2 (1-0.9973)`
  `= 0.00135`
  `= 0.001\ \ (text(to 3 d.p.))`

 

b.ii.   `text(1% level) => 0.01`

♦ Mean mark part (b)(ii) 49%.

  `text(S)text(ince)\ \ 0.001 < 0.01\ \ =>\ text(Strong evidence against)\ H_0`

  `:.\ text(Reject the null hypothesis.)`

 

c.   `sigma(barx) = sigma/sqrtn = 10/sqrt25 = 2`

`text(95% C.I.)` `= barx-1.96 xx 2, barx + 1.96 xx 2`
  `= (250-3.92, 250 + 3.92)`
  `= (246.08, 253.92)`

Vectors, SPEC1 2021 VCAA 1

The net force acting on a body of mass 10 kg is  `underset~F = 5underset~i + 12underset~j`  newtons.

  1. Find the acceleration of the body in `text(ms)^(-2)`.   (1 mark)

  2. The initial velocity of the body is  `-3underset~j\ \ text(ms)^(-1)`.
    Find the velocity of the body, in `text(ms)^(-1)`, at any time  `t`  seconds.   (2 marks)

  3. Find the momentum of the body, in kg `text(ms)^(-1)`, when  `t = 2`  seconds.   (1 mark)

Show Answers Only
  1. `1/2 underset~i + 6/5 underset~j`
  2. `1/2 t underset~i + (6/5 t-3) underset~j`
  3. `10 underset~i-6underset~j`
Show Worked Solution

a.   `text(Using)\ underset~F = m underset~a:`

`10underset~a` `= 5underset~i + 12underset~j`
`underset~a` `= 1/10 (5underset~i + 12underset~j)`
  `= 1/2 underset~i + 6/5 underset~j`

 

b.    `underset~v(t)` `= int underset~a\ dt`
    `= int 1/2 underset~i + 6/5 underset~j\ dt`
    `= 1/2 t underset~i + 6/5 t underset~j + c`

 
`text(When)\ t = 0, v(t) = -3underset~j`

`=> c = -3underset~j`

`underset~v(t)` `= 1/2 t underset~i + 6/5 t underset~j-3underset~j`
  `= 1/2 t underset~i + (6/5 t-3) underset~j`

 

c.    `underset~v(2) = underset~i-3/5 underset~j`

`underset~p(2)` `= m underset~v(2)`
  `= 10(underset~i-3/5 underset~j)`
  `= 10 underset~i-6underset~j`

Functions, 2ADV F2 2021 HSC 19

Without using calculus, sketch the graph of  `y = 2 + 1/(x + 4)`, showing the asymptotes and the `x` and `y` intercepts.  (3 marks)

Show Answers Only

Show Worked Solution

`text(Asymptotes:)\ x = -4`

`text(As)\ \ x -> ∞, y -> 2`

`ytext(-intercept occurs when)\ \ x = 0:`

`y = 2.25`

`xtext(-intercept occurs when)\ \ y = 0:`

`2 + 1/(x + 4) = 0 \ => \ x = -4.5`