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MATRICES, FUR2 2019 VCAA 1

The car park at a theme park has three areas, `A, B` and `C`.

The number of empty `(E)` and full `(F)` parking spaces in each of the three areas at 1 pm on Friday are shown in matrix `Q`  below.
 

`{:(qquad qquad qquad \ E qquad F),(Q = [(70, 50),(30, 20),(40, 40)]{:(A),(B),(C):}quad text(area)):}`
 

  1. What is the order of matrix `Q`?   (1 mark)

  2. Write down a calculation to show that 110 parking spaces are full at 1 pm.   (1 mark)

Drivers must pay a parking fee for each hour of parking.

Matrix `P`, below, shows the hourly fee, in dollars, for a car parked in each of the three areas.
 

`{:(qquad qquad qquad qquad qquad text{area}), (qquad qquad qquad A qquad quad quad B qquad qquad C), (P = [(1.30, 3.50, 1.80)]):}`
 

  1. The total parking fee, in dollars, collected from these 110 parked cars if they were parked for one hour is calculated as follows.  

     

     

    `qquad qquad qquad P xx L = [207.00]`

     

    where matrix  `L`  is a  `3 xx 1`  matrix.

     

    Write down matrix  `L`.   (1 mark)

The number of whole hours that each of the 110 cars had been parked was recorded at 1 pm. Matrix `R`, below, shows the number of cars parked for one, two, three or four hours in each of the areas `A, B` and `C`.

`{:(qquadqquadqquadqquadquadtext(area)),(quad qquadqquadquad \ A qquad B qquad C),(R = [(3, 1, 1),(6, 10, 3),(22, 7,10),(19, 2, 26)]{:(1),(2),(3),(4):}\ text(hours)):}`
 

  1. Matrix  `R^T`  is the transpose of matrix  `R`.

      

    Complete the matrix  `R^T`  below.   (1 mark)

      

    `qquad R^T = [( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , )]`
     

  2. Explain what the element in row 3, column 2 of matrix  `R^T`  represents.   (1 mark)

Show Answers Only
  1. `3 xx 2`
  2. `50 + 20 + 40 = 110`
  3. `L = [(50), (20), (40)]`
  4. `R^T = [(3 ,6 , 22, 19), (1, 10, 7, 2), (1, 3, 10, 26)]`
  5. `text(Number of cars parked in area)\ C\ text(for 2 hours).`
Show Worked Solution

a.  `text(Order) : 3 xx 2`
 

b.  `text(Add 2nd column): \ 50 + 20 + 40 = 110`
 

c.  `L = [(50), (20), (40)]`
 

d.  `R^T = [(3 ,6 , 22, 19), (1, 10, 7, 2), (1, 3, 10, 26)]`
 

e.   `e_32\ text(in)\ R^T =>` `text(number of cars parked in area)\ C`
    `text(for 2 hours.)`

Proof, EXT2 P1 SM-Bank 10

Prove that  `sqrt11 - sqrt5 < sqrt2`  by contradiction.   (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ sqrt11 – sqrt5 >= sqrt2`

`( sqrt11 – sqrt5)^2` `>=  2`
`11  –  2 sqrt55  +  5` `>= 2`
`2 sqrt55` `<= 14`
`sqrt55` `<= 7`
`55` `<= 49 \ \ text{(which is incorrect)}`

 
`:.\ text(By contradiction,) \ sqrt11 – sqrt5 < sqrt2`

Financial Maths, GEN2 2019 NHT 8

Phil invests $200 000 in an annuity from which he receives a regular monthly payment.

The balance of the annuity, in dollars, after `n` months, `A_n`, can be modelled by the recurrence relation

`A_0 = 200\ 000, qquad A_(n + 1) = 1.0035\ A_n - 3700`

  1. What monthly payment does Phil receive?   (1 mark)

  2. Show that the annual percentage compound interest rate for this annuity is 4.2%.   (1 mark)

At some point in the future, the annuity will have a balance that is lower than the monthly payment amount.

  1. What is the balance of the annuity when it first falls below the monthly payment amount?

     

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

  2. If the payment received each month by Phil had been a different amount, the investment would act as a simple perpetuity.

     

    What monthly payment could Phil have received from this perpetuity?   (1 mark)

Show Answers Only
  1. `$3700`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `$92.15`
  4. `$700`
Show Worked Solution

a.  `$3700`

b.   `text(Monthly rate)` `= 0.0035 = 0.35%`
  `text(Annual rate)` `= 12 xx 0.35 = 4.2%`

  
c.  `text(Find)\ N\ text(when)\ FV = 0\ \ text{(by TVM solver)}:`

`N` `= ?`
`I(%)` `= 4.2`
`PV` `= 200\ 000`
`PMT` `= 3700`
`FV` `= 0`
`text(P/Y)` `= 12`
`text(C/Y)` `= 12`

 
`=> N = 60.024951`

 
`text(Find)\ \ FV\ \ text(when)\ \ N = 60.024951\ \ text{(by TVM solver):}`

`=>FV = $92.15`  

d.  `text(Perpetuity) => text(monthly payment) = text(monthly interest)`

`:.\ text(Perpetuity payment)` `= 200\ 000 xx 4.2/(12 xx 100)`
  `= $700`

Financial Maths, 2ADV M1 SM-Bank 15

Phil is a builder who has purchased a large set of tools.

The value of Phil’s tools is depreciated using the reducing balance method.

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

`V_0 = 60\ 000, qquad qquad  V_(n + 1) = 0.9 V_n`
 

  1. Use recursion to show that the value of the tools after two years.   (1 mark)

  2. Phil plans to replace these tools when their value first falls below $20 000.

     

    After how many years will Phil replace these tools?  (1 mark)

  3. Phil has another option for depreciation. He depreciates the value of the tools by a flat rate of 8% of the purchase price per annum.

     

    Let `V_n` be the value of the tools after `n` years, in dollars.

     

    Write down a recurrence relation, in terms of `V_0, V_(n + 1)` and `V_n`, that could be used to model the value of the tools using this flat rate depreciation.  (1 mark)

Show Answers Only
  1. `$48\ 600`
  2. `11\ text(years)`
  3. `V_(n + 1) = V_n – 0.08 V_0`
Show Worked Solution
a.   `V_0` `= 60\ 000`
  `V_1` `= 0.9 xx 60\ 000 = 54\ 000`
  `V_2` `= 0.9 xx 54\ 000 = $48\ 600`

 

b.  `text(Find)\ \ n\ \ text(such that:)`

COMMENT: Dividing by  `log_e 0.9`  is dividing by a negative number!

`60\ 000 xx 0.9^n` `< 20\ 000`  
`log_e 0.9^n` `<log_e (1/3)`  
`n` `> log_e (1/3) / log_e 0.9`  
  `> 10.427`  

 
`:.\ text(Phil will replace in the 11th year.)`

 

c.  `text(Annual depreciation) = 0.08 xx V_0`

`:. V_(n + 1) = V_n – 0.08V_0`

CORE, FUR2 2019 VCAA 7

Phil is a builder who has purchased a large set of tools.

The value of Phil’s tools is depreciated using the reducing balance method.

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

`V_0 = 60\ 000, qquad V_(n + 1) = 0.9 V_n` 

  1. Use recursion to show that the value of the tools after two years, `V_2` , is $48 600.   (1 mark)

  2. What is the annual percentage rate of depreciation used by Phil?   (1 mark)

  3. Phil plans to replace these tools when their value first falls below $20 000.

     

    After how many years will Phil replace these tools?   (1 mark)

  4. Phil has another option for depreciation. He depreciates the value of the tools by a flat rate of 8% of the purchase price per annum.

     

    Let `V_n` be the value of the tools after `n` years, in dollars.

     

    Write down a recurrence relation, in terms of `V_0, V_(n + 1)` and `V_n`, that could be used to model the value of the tools using this flat rate depreciation.   (1 mark)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(10%)`
  3. `11\ text(years)`
  4. `V_0=60\ 000,\ \ \ V_(n + 1) = V_n-4800`
Show Worked Solution
a.   `V_0` `= 60\ 000`
  `V_1` `= 0.9 xx 60\ 000 = 54\ 000`
  `V_2` `= 0.9 xx 54\ 000 = $48\ 600`

  
b.  `text(Depreciation rate) = 0.1 = 10%`

 
c.  `text(Find)\ \ n\ \ text(such that)`

`60\ 000 xx 0.9^n = 20\ 000`

`=> n = 10.427\ \ \ text{(by CAS)}`

`:.\ text(Phil will replace in the 11th year.)`

 
d.  `text(Annual depreciation) = 0.08 xx V_0 = 4800`

`:.\ text(Recurrence relation is:)`

`V_0=60\ 000,\ \ \ V_(n + 1) = V_n-4800`

CORE, FUR2 2019 VCAA 5

The scatterplot below shows the atmospheric pressure, in hectopascals (hPa), at 3 pm (pressure 3 pm) plotted against the atmospheric pressure, in hectopascals, at 9 am (pressure 9 am) for 23 days in November 2017 at a particular weather station.
 

A least squares line has been fitted to the scatterplot as shown.

The equation of this line is

pressure 3 pm = 111.4 + 0.8894 × pressure 9 am

  1. Interpret the slope of this least squares line in terms of the atmospheric pressure at this weather station at 9 am and at 3 pm.   (1 mark)

  2. Use the equation of the least squares line to predict the atmospheric pressure at 3 pm when the atmospheric pressure at 9 am is 1025 hPa.
  3. Round your answer to the nearest whole number.   (1 mark)

  4. Is the prediction made in part b. an example of extrapolation or interpolation?   (1 mark)

  5. Determine the residual when the atmospheric pressure at 9 am is 1013 hPa.
  6. Round your answer to the nearest whole number.   (1 mark)

  7. The mean and the standard deviation of pressure 9 am and pressure 3 pm for these 23 days are shown in Table 4 below.

    1. Use the equation of the least squares line and the information in Table 4 to show that the correlation coefficient for this data, rounded to three decimal places, is  `r` = 0.966   (1 mark)

    2. What percentage of the variation in pressure 3 pm is explained by the variation in pressure 9 am?
    3. Round your answer to one decimal place.   (1 mark)

  1. The residual plot associated with the least squares line is shown below.
     

    1. The residual plot above can be used to test one of the assumptions about the nature of the association between the atmospheric pressure at 3 pm and the atmospheric pressure at 9 am.
    2. What is this assumption?   (1 mark)

    3. The residual plot above does not support this assumption.
    4. Explain why.   (1 mark)

Show Answers Only
  1. `text(An increase in 1hPa of pressure at 9 am is associated)`
    `text(with an increase of 0.8894 hPa of pressure at 3 pm.)`
  2. `1023\ text(hPa)`
  3. `text(Interpolation)`
  4. `3\ text(hPa)`
    1. `0.966`
    2. `93.3%`
    1. `text(The assumption is that a linear relationship)`
      `text(exists between the pressure at 9 am and the)`
      `text(pressure at 3 pm.)`
    2. `text(The residual plot does not appear to be random.)`
Show Worked Solution

a.    `text(An increase in 1hPa of pressure at 9 am is associated)`

`text(with an increase of 0.8894 hPa of pressure at 3 pm.)`

 

b.   `text(pressure 3 pm)` `= 111.4 + 0.8894 xx 1025`
    `= 1023\ text(hPa)`

 

c.  `text{Interpolation (1025 is within the given data range)}`

 

d.   `text(Residual)` `= text(actual) – text(predicted)`
    `= 1015 – (111.4 + 0.8894 xx 1013)`
    `= 1015 – 1012.36`
    `= 2.63…`
    `~~ 3\ text(hPa)`

 

e.i.   `r= b (s_x)/(s_y)`

    `= 0.8894 xx 4.5477/4.1884`
    `= 0.96569…`
    `= 0.966`

 

e.ii.   `r` `= 0.966`
  `r^2` `= 0.9331`
    `= 93.3%`

 

f.i.   `text(The assumption is that a linear relationship)`
 

`text(exists between the pressure at 9 am and the)`

`text(pressure at 3 pm.)`

 

f.ii.  `text(The residual plot does not appear to be random.)`

CORE, FUR2 2019 VCAA 4

The relative humidity (%) at 9 am and 3 pm on 14 days in November 2017 is shown in Table 3 below.

A least squares line is to be fitted to the data with the aim of predicting the relative humidity at 3 pm (humidity 3 pm) from the relative humidity at 9 am (humidity 9 am).

  1. Name the explanatory variable.   (1 mark)

  2. Determine the values of the intercept and the slope of this least squares line.
  3. Round both values to three significant figures and write them in the appropriate boxes provided.   (1 mark)

humidity 3 pm = 
 
   +  
 
   × humidity 9 am  (1 mark)
  1. Determine the value of the correlation coefficient for this data set.
  2. Round your answer to three decimal places.   (1 mark)

Show Answers Only
  1. `text(humidity 9 am)`
  2. `text(humidity 3 pm) = -1.26 + 0.765 xx text(humidity 9 am)`
  3. `r = 0.871`
Show Worked Solution

a.  `text(humidity 9 am)`
 

b.  `text(Input all data points into CAS:)`

`text(humidity 3 pm) = -1.26 + 0.765 xx text(humidity 9 am)`
 

c.    `r = 0.871\ \ text{(3 d.p.)}`

Calculus, EXT2 C1 2005 HSC 1a

Find  `int(cos theta)/(sin^5 theta)  d theta`   (2 marks)

Show Answers Only

`(-1)/(4sin^4 theta ) + C`

Show Worked Solution
`text(Let) \ u` `= sin theta `
`(du)/(d theta)` `=cos theta =>  d u = cos theta \ d theta`

 

`int(cos theta)/(sin^5 theta)  d theta` `= int u^-5 d u`
  `=(-1)/(4) u ^-4 + C`
  `=(-1)/(4sin^4 theta ) + C`

Proof, EXT2 P1 SM-Bank 1 MC

Consider the following statement.

"If you have no treasure, I have no kingdom."

Which of the following is logically equivalent to this statement?

  1.  If I have no kingdom then you have no treasure.
  2.  If you have treasure then I have a kingdom.
  3.  If you have no kingdom then I have no treasure.
  4.  If I have a kingdom then you have treasure.
Show Answers Only

`D`

Show Worked Solution

`text(The statement is conditional.)`

`text(If)\ X\ text(then)\ Y\ \ text(or)\  \ X=>Y`

`text(The contrapositive statement is logically equivalent.)`

`text{(i.e.)}\ \ ¬Y => ¬X`

`text(If I have a kingdom then you have treasure.)`

`=>  D`

Proof, EXT2 P1 SM-Bank 3

Prove that  `1/sqrt2`  is irrational.   (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ 1/sqrt2\ \ text(is rational.)`

`:. 1/sqrt2 = p/q\ \ \ text(where)\ \ p,q in ZZ\ \ text(with no common factor except 1)`

`1/2` `=p^2/q^2`  
`q^2` `=2p^2\ …\ (1)`  

 
`q^2\ \ text(is even) \ =>\ q\ text(is even)`

`text{(i.e.)}\ \ ∃ k,\ \ k in ZZ\ \ text(such that)\ \ q=2k`
 

`text{Substitute}\ \ q=2k\ \ text{into (1)}`

`(2k)^2` `=2p^2`  
`2k^2` `=p^2`  

 

`p^2\ \ text(is even) \ =>\ p\ text(is even)`

`:. p and q\ \ text(have a common factor of 2)`

`=> \ text(Contradiction)`

`:.\ 1/sqrt2\ \ \ text(is irrational.)`

CORE, FUR2 2019 VCAA 2

 

The parallel boxplots below show the maximum daily temperature and minimum daily temperature, in degrees Celsius, for 30 days in November 2017.
 

  1. Use the information in the boxplots to complete the following sentences.
  2. For November 2017
  3.    i. the interquartile range for the minimum daily temperature was _____ °C   (1 mark)

  4.  ii. the median value for maximum daily temperature was _____ °C higher than the median value for minimum daily temperature   (1 mark)

  5. iii. the number of days on which the maximum daily temperature was less than the median value for minimum daily temperature was _____    (1 mark)

  1. The temperature difference between the minimum daily temperature and the maximum daily temperature in November 2017 at this location is approximately normally distributed with a mean of 9.4 °C and a standard deviation of 3.2 °C.
  2. Determine the number of days in November 2017 for which this temperature difference is expected to be greater than 9.4 °C.  (1 mark)

Show Answers Only

    1. `5text(°C)`
    2. `10text(°C)`
    3. `1\ text(day)`
  2. `15\ text(days)`

Show Worked Solution

a.i.  `text(IQR)\ = 17-12= 5text(°C)`
 

a.ii.    `text{Median (maximum temperature)}` `= 25`
  `text{Median (minimum temperature)}` `= 15`

 
`:.\ text(Maximum is 10°C higher)`
 

a.iii.  `text{Median (minimum temperature)} = 15text(°C)`

   `text(1 day) => text(maximum temperature is below)\ 15text(°C)`
 

b.    `text(Number of days)` `= 0.50 xx 30`
    `= 15\ text(days)`

CORE, FUR2 2019 VCAA 1

Table 1 shows the day number and the minimum temperature, in degrees Celsius, for 15 consecutive days in May 2017.
 

  1. Which of the two variables in this data set is an ordinal variable?   (1 mark)

The incomplete ordered stem plot below has been constructed using the data values for days 1 to 10.

  1. Complete the stem plot above by adding the data values for days 11 to 15.   (1 mark)

  2. The ordered stem plot below shows the maximum temperature, in degrees Celsius, for the same 15 days.

     

     

    Use this stem plot to determine

  3.  i. the value of the first quartile `(Q_1)`   (1 mark)

  4. ii. the percentage of days with a maximum temperature higher than 15.3 °C.   (1 mark)

Show Answers Only
  1. `text(Day number)`
  2. `text(See Worked Solutions)`
    1. `12.2`
    2. `text(20%)`
Show Worked Solution

a.   `text(Day number)`

b.  

 
c.i.   `15\ text(data points)`

 `Q_1 = 12.2\ text{(4th data point)}`
 

c.ii.    `text(% Days)` `= 3/15 xx 100`
    `= 20%`

GRAPHS, FUR1 2019 VCAA 2 MC

The cost, `$C`, of using `K` kilowatt hours of electricity can be calculated using the equation below.
 

`C = 52.00 + 0.15 xx K`
 

From this equation, it can be concluded that there is

  1. no fixed charge and the electricity used is charged at $0.15 per kilowatt hour.
  2. no fixed charge and the electricity used is charged at $52.00 per kilowatt hour.
  3. a fixed charge of $0.15 and the electricity used is charged at $52.00 per kilowatt hour.
  4. a fixed charge of $52.00 and the electricity used is charged at $0.15 per kilowatt hour.
  5. a fixed charge of $52.00 and the electricity used is charged at $15.00 per kilowatt hour.
Show Answers Only

`D`

Show Worked Solution

`text(Fixed charged) = $52.00`

`text(Electricity used) = $0.15\ text(per kilowatt hour)` 

`=>  D`

Calculus, EXT2 C1 2003 HSC 1d

  1.  Find the real numbers  `a`  and  `b`  such that
     
    `qquad (5x^2-3x+13)/((x-1)(x^2+4)) ≡ a/(x-1) + (bx-1)/(x^2+4)`   (2 marks)

  2.  Hence find  `int (5x^2-3x+13)/((x-1)(x^2+4)) \ dx`   (2 marks)

Show Answers Only
  1. `a = 3, b = 2`
  2. `3 log_e|x-1| + log_e |x^2+4|-1/2 tan^(-1)(x/2) +C`
Show Worked Solution

i.   `(5x^2-3x+13)/((x-1)(x^2+4)) ≡ (a(x^2+4) + (bx-1)(x-1))/((x-1)(x^2+4))`

 
`text(Equating numerators:)`

`5x^2-3x+13` `=ax^2+4a+bx^2-bx-x+1`  
  `=(a+b)x^2+(-b-1)x+4a+1`  

 

`-b-1` `=-3\ \ =>\ \ b=2`  
`a` `=3`  

 

ii.    `int (5x^2-3x+13)/((x-1)(x^2+4)) \ dx` `=int 3/(x-1)\ dx-int (2x)/(x^2+4)\ dx-int 1/(4+x^2)\ dx`
    `=3 log_e|x-1|-log_e |x^2+4|-1/2 tan^(-1)(x/2)+C`

GEOMETRY, FUR1 2019 VCAA 3 MC

An ice cream dessert is in the shape of a hemisphere. The dessert has a radius of 5 cm.
 

The top and the base of the dessert are covered in chocolate.

The total surface area, in square centimetres, that is covered in chocolate is closest to

  1.   52
  2. 157
  3. 236
  4. 314
  5. 942
Show Answers Only

`C`

Show Worked Solution

`text(Total surface area covered)`

`= text(base) + text(dome)`

`=pir^2 + 1/2 xx 4pi r^2`

`= pi xx 5^2 + 1/2 xx 4 pi xx 5^2`

`~~ 236\ text(cm²)`

`=>  C`

Calculus, EXT2 C1 2004 HSC 1d

  1. Find real numbers  `a`  and  `b`  such that
     
    `qquad (x^2-7x+4)/((x+1)(x-1)^2) ≡ a/(x+1) + b/(x-1) - 1/(x-1)^2`   (2 marks)
     
  2. Hence find  `int (x^2-7x+4)/((x+1)(x-1)^2)\ dx`   (2 marks)

 

Show Answers Only
  1. `a = 3, b = -2`
  2. `3 log_e|x+1| – 2log_e |x-1| + 1/(x-1)+C`
Show Worked Solution

i.   `(x^2-7x+4)/((x+1)(x-1)^2) ≡ (a(x-1)^2 +b(x^2-1) -(x+1))/((x+1)(x-1)^2)`

`text(Equating numerators:)`

`x^2-7x+4` `=ax^2-2ax+a+bx^2-b-x-1`  
  `=(a+b)x^2 + (-2a-1)x +a-b-1`  

 

`a+b` `=1\ \ …\ (1)`  
`a-b-1` `=4`  
`a-b` `=5\ \ …\ (2)`  

 
`(1) + (2)`

`2a` `=6`
`a` `= 3`
`b` `=-2`

 

ii.    `int (x^2-7x+4)/((x+1)(x-1)^2)\ dx` `=int 3/(x+1)\ dx – int 2/(x-1)\ dx – int 1/(x-1)^2\ dx`
    `=3 log_e|x+1| – 2log_e |x-1| + 1/(x-1)+C`

Calculus, EXT2 C1 2005 HSC 1b

  1. Find real numbers  `a`  and  `b`  such that
     
    `qquad (5x)/(x^2-x-6) ≡ a/(x-3) + b/(x+2)`   (2 marks)

  2. Hence find  `int (5x)/(x^2-x-6)\ dx`   (1 mark)

Show Answers Only
  1. `a = 3, b = 2`
  2. `3 log_e | x-3 | + 2 log_e | x+2 |+C`
Show Worked Solution

i.   `(5x)/(x^2-x-6) ≡ (a(x+2) +b(x-3))/((x-3)(x+2))`

`text(Equating numerators:)`

`5x` `=ax+2a+bx-3b`  
`5x` `=(a+b)x+2a-3b`  

 

`a+b` `=5\ \ …\ (1)`  
`2a-3b` `=0\ …\ (2)`  

 
`(1) xx 2 – (2)`

`5b` `=10`
`b` `= 2`
`a` `=3`

 

ii.    `int (5x)/(x^2-x-6)\ dx` `=int 3/(x-3)\ dx + int 2/(x+2)\ dx`
    `=3 log_e | x-3 | + 2 log_e | x+2 |+C`

MATRICES, FUR1 2019 VCAA 2 MC

There are two rides called The Big Dipper and The Terror Train at a carnival.

The cost, in dollars, for a child to ride on each ride is shown in the table below.
 

  Ride       Cost ($)      
       The Big Dipper    7
       The Terror Train    8

   
Six children ride once only on The Big Dipper and once only on The Terror Train.

The total cost of the rides, in dollars, for these six children can be determined by which one of the following calculations?

A.   `[6] xx [(7, 8)]` B.   `[6] xx [(7), (8)]`
C.   `[(6, 6)] xx [(7, 8)]` D.   `[(6, 6)] xx [(7), (8)]`
E.   `[(6), (6)] xx [(7, 8)]`    
Show Answers Only

`D`

Show Worked Solution
`text(Total Cost)` `= 6 xx 7 + 6 xx 8`
  `= [(6, 6)] xx [(7), (8)]`

 
`=>  D`

CORE, FUR1 2019 VCAA 17 MC

Consider the recurrence relation shown below.

`A_0 = 3, qquad  A_(n + 1) = 2A_n + 4`

The value of `A_3` in the sequence generated by this recurrence relation is given by

  1. `2 xx 3 + 4`
  2. `2 xx 4 + 4`
  3. `2 xx 10 + 4`
  4. `2 xx 24 + 4`
  5. `2 xx 52 + 4`
Show Answers Only

`D`

Show Worked Solution

`A_1 = 2A_0 + 4 = 2 xx 3 + 4 = 10`

`A_2 = 2 xx 10 + 4 = 24`

`A_3 = 2 xx 24 + 4`

`=>  D`

CORE, FUR1 2019 VCAA 9-10 MC

A least squares line is used to model the relationship between the monthly average temperature and latitude recorded at seven different weather stations. The equation of the least squares line is found to be
 

`quad text(average temperature) = 42.9842 - 0.877447 xx text(latitude)`

 
Part 1

When the numbers in this equation are correctly rounded to three significant figures, the equation will be

  1. `text(average temperature) = 42.984 - 0.877 xx text(latitude)`
  2. `text(average temperature) = 42.984 - 0.878 xx text(latitude)`
  3. `text(average temperature) = 43.0 - 0.878 xx text(latitude)`
  4. `text(average temperature) = 42.9 - 0.878 xx text(latitude)`
  5. `text(average temperature) = 43.0 - 0.877 xx text(latitude)`

 
Part 2

The coefficient of determination was calculated to be 0.893743

The value of the correlation coefficient, rounded to three decimal places, is

  1. − 0.945
  2. − 0.898
  3.  0.806
  4.  0.898
  5.  0.945
Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ A`

Show Worked Solution

`text(Part 1)`

`42.9842 \ => \ 43.0\ text{(3 sig)}`

`0.877447 \ => \ 0.877\ text{(3 sig)}`

`=>  E`
 

`text(Part 2)`

`text(Correlation coefficient)`

`= -sqrt(0.893743)`

`= -0.945`

`=>  A`

CORE, FUR1 2019 VCAA 4-5 MC

The stem plot below shows the distribution of mathematics test scores for a class of 23 students.
 


 

Part 1

For this class, the range of test scores is

  1. 22
  2. 40
  3. 45
  4. 49
  5. 89

 
Part 2

For this class, the interquartile range (IQR) of test scores is

  1. 14.5
  2. 17.5
  3. 18
  4. 24
  5. 49
Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(Range)` `= 89 – 40`
  `= 49`

`=> D`

 
`text(Part 2)`

`text(23 data points)`

`Q_1 =\ text(6th data point) =57`

`Q_3 =\ text(18th data point) = 75`

`text(IQR)` `= 75 – 57`
  `= 18`

 
`=>  C`

CORE, FUR1 2019 VCAA 1-3 MC

The histogram below shows the distribution of the population size of 48 countries in 2018.

Part 1

The number of these countries with a population size between 5 million and 20 million people is

  1.  11
  2.  17
  3.  23
  4.  34
  5.  35

 
Part 2

The shape of this histogram is best described as

  1. positively skewed with no outliers. 
  2. positively skewed with outliers.
  3. approximately symmetric.
  4. negatively skewed with no outliers.
  5. negatively skewed with outliers.

 
Part 3

The histogram below shows the population size for these 48 countries plotted on a `log_10` scale.

Based on this histogram, the number of countries with a population size that is less than `100\ 000` people is

  1. 1
  2. 5
  3. 7
  4. 8
  5. 48
Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ B`

`text(Part 3:)\ D`

Show Worked Solution

Part 1

`text{Countries (5 – 20 million)}`

`= 11 + 4 + 2`

`= 17`

`=> B`
 

Part 2

`text(Histogram is positively skewed with outliers.)`

`=> B`
 

Part 3

`log_10 100\ 000 = 5`

`text{Countries (population < 100 000)}`

`= 1 + 7`

`= 8`

`=> D`

Functions, EXT1 F2 EQ-Bank 16

The polynomial  `P(x) = x^3-2x^2 + kx + 24`  has roots  `alpha, beta, gamma`.

  1. Find the value of  `alpha + beta + gamma`.   (1 mark)

  2. Find the value of  `alphabetagamma`.   (1 mark)

  3. It is known that two of the roots are equal in magnitude but opposite in sign.

     

    Find the third root and hence find the value of `k`.   (2 marks)

Show Answers Only

a.    `2`

b.    `−24`

c.    `−12`

Show Worked Solution

a.    `alpha + beta + gamma = −b/a = 2`
 

b.    `alphabetagamma = −d/a = −24`
 

c.    `text(Let roots be)\ \ alpha, −alpha, \ beta:`

`alpha-alpha + beta` `= 2`
`beta` `= 2`

 
`text(Substitute)\ \ beta = 2\ \ text(into equation:)`

`2^3-2 ·2^2 + 2k + 24` `= 0`
`2k` `= -24`
`k` `= -12`

Functions, 2ADV F1 SM-Bank 43

 Solve  `|\ 3x -1\ | = 2`   (2 marks)

Show Answers Only

 ` x= -1/3 or 1`

Show Worked Solution
 MARKER’S COMMENT: Note that both conditions must be satisfied! Dealing with negative signs and division for inequalities produced many errors.

`|\ 3x -1\ | = 2`

`3x -1` `=2`  `\ \ \ \ \-(3x -1)` `= 2`
`3x`  `=3` `-3x + 1` `= 2`
`x` `= 1`  `3x` `= -1`
    `x` `= -1/3`

`:. x= -1/3 or 1`

Functions, 2ADV F1 EQ-Bank 16

Find the values of `x` for which  `|x-3| = 1`.   (2 marks)

Show Answers Only

`x=2 or 4`

Show Worked Solution

`|x-3| = 1`

`text(Method 1)`

`(x-3)^2` `=1`  
`x^2-6x +8` `=0`  
`(x-4)(x-2)` `=0`  

 
`:. x=2 or 4`
 

`text(Method 2)`

`(x-3)` `=1` `-(x-3)` ` = 1`
`x` `=4` `-x +3` `=1`
    `x` `=2`

Trigonometry, EXT1 T3 EQ-Bank 12

Given that  `cot(2x) + 1/2 tan(x) = a cot(x)`, calculate `a`.   (3 marks)

Show Answers Only

`a = 1/2`

Show Worked Solution

`1/(tan(2x)) + (tan(x))/2 = a/(tan(x)),\ \ tan(x) != 0`

`(1-tan^2(x))/(2 tan(x)) + (tan(x))/2 = a/(tan(x))`

 
`text(If)\ \ tan(x) != 0 :`

`(1-tan^2(x)+tan^2(x))/(2tan(x))` `=a/(tan(x))`
`1` `=2a`
`:. a` `=1/2`

Trigonometry, EXT1 T2 EQ-Bank 3 MC

If  \(\sin (\theta+\phi)=a\)  and  \(\sin (\theta-\phi)=b\),  then  \(\sin (\theta) \cos (\phi)\)  is equal to

  1. \(\sqrt{a^2+b^2}\)
  2. \(\sqrt{a b}\)
  3. \(\sqrt{a^2-b^2}\)
  4. \(\dfrac{a+b}{2}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Solution 1}\)

\(\sin \theta \cos \phi=\dfrac{1}{2} [(\sin (\theta+\phi)+\sin (\theta-\phi)]=\dfrac{1}{2}(a+b)\)

\(\Rightarrow D\)
 

\(\text{Solution 2}\)

\(\sin (\theta+\phi)=a\)

\(\sin \theta \cos \phi+\sin \phi \cos \theta\) \(=a \ldots(1)\)
\(\sin (\theta-\phi)=b\)  
\(\sin \theta \cos \phi-\sin \phi \cos \theta\) \(=b \ldots(2)\)

 
\((1)+(2):\)

\(2 \sin \theta \cos \phi\) \(=a+b\)
\(\therefore \sin \theta \cos \phi\) \(=\dfrac{a+b}{2}\)

 

Mechanics, SPEC2 2019 VCAA 5

A mass of  `m_1`  kilograms is initially held at rest near the bottom of a smooth plane inclined at `theta` degrees to the horizontal. It is connected to a mass of  `m_2`  kilograms by a light inextensible string parallel to the plane, which passes over a smooth pulley at the end of the plane. The mass  `m_2`  is 2 m above the horizontal floor.

The situation is shown in the diagram below.
 


 

  1. After the mass  `m_1`  is released, the following forces, measured in newtons, act on the system:

     

    • weight forces  `W_1`  and  `W_2`
    •  the normal reaction force  `N`
    •  the tension in the string  `T`

     

    On the diagram above, show and clearly label the forces acting on each of the masses.  (1 mark) 

  2. If the system remains in equilibrium after the mass  `m_1`  is released, show that  `sin(theta) = (m_2)/(m_1)`.  (1 mark)
  3. After the mass  `m_1`  is released, the mass  `m_2`  falls to the floor.

     

    1. For what values of  `theta`  will this occur? Express your answer as an inequality in terms of  `m_1`  and  `m_2`.  (1 mark)
    2. Find the magnitude of acceleration, in ms−2, of the system after the mass  `m_1`  is released and before the mass  `m_2`  hits the floor. Express your answer in terms of  `m_1, \ m_2`  and  `theta`.  (2 marks)
  4. After the mass  `m_1`  is released, it moves up the plane.
    Find the maximum distance, in metres, that the mass  `m_1`  will move up the plane if  `m_1 = 2m_2`  and  `sin(theta) = 1/4`.  (5 marks)
Show Answers Only
  1.   
  2. `text(See Worked Solutions)`
    1. `theta < sin^(−1)\ ((m_2)/(m_1)), theta ∈ (0, pi/2)`
    2. `a = (g(m_2 – m_1 · sin(theta)))/(m_1 + m_2)`
  3. `10/3 \ text(m)`
Show Worked Solution
a.   

 

b.   `T = m_2g\ \ … \ (1)`

`T = m_1sin(theta)\ \ … \ (2)`

`text{Solve:  (1) = (2)}`

`m_1g sin(theta)` `= m_2g`
`sin(theta)` `= (m_2)/(m_1)`

 

c.i.   `m_2g > m_1gsin(theta)`

`sin(theta) < (m_2)/(m_1)`

`theta < sin^(−1)\ ((m_2)/(m_1)), \ \ theta ∈ (0, pi/2)`

 

c.ii.   `text(Net force)\ (F) = m_2g – m_1gsin(theta)`

`(m_1 + m_2) · a` `= g(m_2 – m_1 sin(theta))`
`:. a` `= (g(m_2 – m_1 sin(theta)))/(m_1 + m_2)`

 

d.   `text(Motion:)\ m_1\ text(will accelerate up the plane for 2 m.)`

 `m_1\ text(will then decelerate up plane until)\ v = 0.`

`text(Find)\ v_(m_1)\ \ text(given)\ \ s_(m_1) = 2, \ u = 0, \ m_1=2m_2`

`a = (g(m_2 – m_1  sintheta))/(m_1 + m_2) = (g(m_2 – 2m_2 · 1/4))/(2m_2 + m_2) = g/6`
 

`text(Using)\ \ v^2 = u^2 + 2as,`

`v_(m_1)^2 = 0 + 2 · g/6 · 2 = (2g)/3`

`text(Find distance)\ (s_2)\ text(for)\ m_1\ text(to decelerate until)\ v = 0:`

`a = −gsintheta = −g/4, \ u = sqrt((2g)/3)`

`0` `= (2g)/3 – 2 · g/4 · s_2`
`s_2` `= (2g)/3 xx 2/g`
  `= 4/3`

 

`:.\ text(Maximum distance)` `= 2 + 4/3`
  `= 10/3\ text(m)`

Vectors, SPEC2 2019 VCAA 4

The base of a pyramid is the parallelogram  `ABCD`  with vertices at points  `A(2,−1,3),  B(4,−2,1),  C(a,b,c)`  and  `D(4,3,−1)`. The apex (top) of the pyramid is located at  `P(4,−4,9)`.

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

  2. Find the cosine of the angle between the vectors  `overset(->)(AB)`  and  `overset(->)(AD)`.  (2 marks)

  3. Find the area of the base of the pyramid.  (2 marks)

  4. Show that  `6underset~i + 2underset~j + 5underset~k`  is perpendicular to both  `overset(->)(AB)`  and  `overset(->)(AD)`, and hence find a unit vector that is perpendicular to the base of the pyramid.  (3 marks)

  5. Find the volume of the pyramid.  (2 marks)

Show Answers Only

  1. `a = 6, b = 2, c = −3`
  2. `4/9`
  3. `2sqrt65 u^2`
  4. `text(See Worked Solutions)`
  5. `24\ text(u³)`

Show Worked Solution

a.    `overset(->)(AB)` `= (4-2)underset~i + (−2 + 1)underset~j + (1-3)underset~k`
    `= 2underset~i-underset~j-2underset~k`

 
`text(S)text(ince)\ ABCD\ text(is a parallelogram)\ => \ overset(->)(AB)= overset(->)(DC)`

`overset(->)(DC) = (a-4)underset~i + (b-3)underset~j + (c + 1)underset~k`

`a-4 = 2 \ => \ a = 6`

`b-3 = −1 \ => \ b = 2`

`c + 1 = −2 \ => \ c = −3`

 

b.   `overset(->)(AB) = 2underset~i-underset~j-2underset~k`

`overset(->)(AD) = 2underset~i + 4underset~j-4underset~k`

`cos angleBAD` `= (overset(->)(AB) · overset(->)(AD))/(|overset(->)(AB)| · |overset(->)(AD)|)`
  `= (4-4 + 8)/(sqrt(4 + 1 + 4) · sqrt(4 + 16 + 16))`
  `= 4/9`

 

c.    `1/2 xx text(Area)_(ABCD)` `= 1/2 ab sin c`
  `text(Area)_(ABCD)` `= |overset(->)(AB)| · |overset(->)(AD)| *sin(cos^(−1)\ 4/9)`

 


 

`:. text(Area)_(ABCD)` `= 3 · 6 · sqrt65/9`
  `= 2sqrt65\ text(u²)`

 

d.   `text(Let)\ \ underset~r = 6underset~i + 2underset~j + 5underset~k`

`underset~r · overset(->)(AB)` `= 6 xx 2 + 2 xx −1 + 5 xx −2 = 0`
`underset~r · overset(->)(AD)` `= 6 xx 2 + 2 xx 4 + 5 xx −2 = 0`

 
`:. underset~r\ \ text(is ⊥ to)\ \ overset(->)(AB)\ \ text(and)\ \ overset(->)(AD)`

`text(Let)\ \ hatr\ = text(unit vector ⊥ to pyramid base)`

`underset~overset^r = 1/sqrt(6^2 + 2^2 + 5^2) *underset~r = 1/sqrt65(6underset~i + 2underset~j + 5underset~k)`

 

e.   `text(Find height)\ (h)\ text(of pyramid:)`

`overset(->)(AP)` `= (4-2)underset~i + (−4 + 1)underset~j + (9-3)underset~k`
  `= 2underset~i-3underset~j + 6underset~k`

 

`h` `= overset(->)(AP) · underset~overset^r`
  `= (2 xx 6/sqrt65)-(3 xx 2/sqrt65) + (6 xx 5/sqrt65)`
  `= 36/sqrt65`

 

`:. V` `= 1/3 b xx h`
  `= 1/3 xx 2sqrt65 xx 36/sqrt65`
  `= 24\ text(u³)`

Complex Numbers, SPEC2 2019 VCAA 2

  1. Show that the solutions of  `2z^2 + 4z + 5 = 0`, where  `z ∈ C`, are  `z = −1 ± sqrt6/2 i`.   (1 mark)

  2. Plot the solutions of  `2z^2 + 4z + 5 = 0`  on the Argand diagram below.   (1 mark)

     

     

Let  `|z + m| = n`, where  `m, n ∈ R`, represent the circle of minimum radius that passes through the solutions of  `2z^2 + 4z + 5 = 0`.

    1. Find  `m`  and  `n`.   (2 marks)

    2. Find the cartesian equation of the circle  `|z + m| = n`.   (1 mark)

    3. Sketch the circle on the Argand diagram in part a.ii. Intercepts with the coordinate axes do not need to be calculated or labelled.   (1 mark)

  2. Find all values of  `d`, where  `d ∈ R`, for which the solutions of  `2z^2 + 4z + d = 0`  satisfy the relation  `|z + m| <= n`.   (2 marks)

  3. All complex solutions of  `az^2 + bz + c = 0`  have non-zero real and imaginary parts.

     

    Let  `|z + p| = q`  represent the circle of minimum radius in the complex plane that passes through these solutions, where  `a, b, c, p, q ∈ R`.

     

    Find  `p`  and  `q`  in terms of  `a, b`  and  `c`.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2.   
  3. `m = 1, n = sqrt6/2`
  4. `(x + 1)^2 + y^2 = 3/2`
  5. `−1 <= d <= 5\ \ (text(by CAS))`
  6. `p = b/(2a), \ q = |sqrt(b^2-4ac)/(2a)|`
Show Worked Solution
a.i.    `z` `= (−b ± sqrt(b^2-4ac))/(2a)`
    `= (−4 ± sqrt(16-4 · 2 · 5))/(4)`
    `= (−4 ± 2sqrt6 i)/(4)`
    `= −1 ± sqrt6/2 i\ \ …\ text(as required)`

 

a.ii.   

 

b.i.   `text(Radius of circle = )sqrt6/2`

 `text(Centre) = (0, −1)`

`:. m = 1, \ n = sqrt6/2`

 

b.ii.    `|z + 1|` `= sqrt6/2`
  `|x + iy + 1|` `= sqrt6/2`
  `(x + 1)^2 + y^2` `= 3/2`

 

b.iii.   

 

c.   `text(Solve:)\ 2z^2 + 4z + d = 0`

`z = −1 ± sqrt(4-2d)/2 = −1 ± sqrt((2-d)/2)`

`z + 1 = ± sqrt((2-d)/2)`
 

`text(Solve for)\ d\ text(such that:)`

`|sqrt((2-d)/2)| <= sqrt6/2`

`−1 <= d <= 5\ \ (text(by CAS))`

 

d.   `z = (−b ± sqrt(b^2-4ac))/(2a) = (−b)/(2a) ± sqrt(b^2-4ac)/(2a)`

`z + b/(2a)` `= ± sqrt(b^2-4ac)/(2a)`
`|z + b/(2a)|` `= |sqrt(b^2-4ac)/(2a)|`

 
`:. p = b/(2a), \ q = |sqrt(b^2-4ac)/(2a)|`

Calculus, SPEC2 2019 VCAA 1

A curve is defined parametrically by  `x = sec(t) + 1, \ y = tan(t)`, where  `t ∈ [0, pi/2)`.

  1. Show that the curve can be represent in cartesian form by the rule  `y = sqrt(x^2-2x)`.   (2 marks)

  2. State the domain and range of the relation given by  `y = sqrt(x^2-2x)`.  (2 marks)

  3.  i. Express  `(dy)/(dx)`  in terms of  `sin(t)`.   (2 marks)

  4. ii. State the limiting value of  `(dy)/(dx)`  as  `t`  approaches  `pi/2`.   (1 mark)

  1. Sketch the curve  `y = sqrt(x^2-2x)`  on the axes below for  `x ∈ [2, 4]`, labelling the endpoints with their coordinates.   (2 marks)

     

     

  2. The portion of the curve given by  `y = sqrt(x^2-2x)`  for  `x ∈ [2, 4]`  is rotated about the `y`-axis to form a solid of revolution.
  3. Write down, but do not evaluate, a definite integral in terms of  `t`  that gives the volume of the solid formed.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Domain:)\ x ∈ [2, ∞)`

     

    `text(Range:)\ y ∈ [0, ∞)`

    1. `(dy)/(dx) = ((dy)/(dt))/((dt)/(dx)) = (sec^2(t))/(sin(t)sec^2(t)) = 1/(sin(t))`
    2. `(dy)/(dx) -> 1`
  4.  
  5. ` V = pi int_0^(tan^(−1)(2sqrt2)) (sec(t) + 1)^2sec^2(t)dt`
Show Worked Solution

a.   `x = sec(t) + 1 \ => \ sec(t) = x-1`

`y = tan(t)`

`text(Using)\ \ tan^2(t) + 1 = sec^2(t):`

`y^2 + 1` `= (x-1)^2`
`y^2 + 1` `= x^2-2x + 1`
`y^2` `= x^2-2x`
`y` `= sqrt(x^2-2x), \ y >= 0\ \ text(as)\ \ t ∈ [0, pi/2)`

 

b.   `text(Sketch:)\ \ x = sec(t) + 1, \ y = tan(t)\ \ text(for)\ \ t ∈ [0, pi/2)`

`text(Domain:)\ \ x ∈ [2, ∞)`

`text(Range:)\ \ y ∈ [0, ∞)`

 

c.i.   `(dy)/(dt) = sec^2(t), \ (dx)/(dt) = sin(t)sec^2(t)\ \ \ (text(by CAS))`

`(dy)/(dx) = ((dy)/(dt))/((dx)/(dt)) = (sec^2(t))/(sin(t)sec^2(t)) = 1/(sin(t))`

 

c.ii.   `text(As)\ \ t -> pi/2:`

`(dy)/(dx) -> 1`

 

d.   

 

e.   `V = pi int_0^(2sqrt2) x^2\ dy`

`x^2 = (sec(t) + 1)^2`

`(dy)/(dt) = sec^2(t) \ => \ dy = sec^2(t)\ dt`

`text(When)\ y = 0, t = 0`

`text(When)\ y = 2sqrt2, t = tan^(−1)(2sqrt2)`
 

`:. V = pi int_0^(tan^(−1)(2sqrt2)) (sec(t) + 1)^2sec^2(t)\ dt`

Vectors, SPEC2 2019 VCAA 11 MC

Let point `M` have coordinates  `(a, 1,-2)`  and let point `N` have coordinates  `(-3, b,-1)`.

If the coordinates of the midpoint of  `bar(MN)`  are  `(-5, 3/2, c)`  and `a, b` and `c` are real constants, the the values of `a, b` and `c` are respectively

  1. `−13, 2 and −1/2`
  2. `−2, 1/2 and −3`
  3. `−7, −2 and −3/2`
  4. `−2, −1/2 and −3`
  5. `−7, 2 and −3/2`
Show Answers Only

`E`

Show Worked Solution

`M = 1/2 ([(a),(1),(−2)] + [(−3),(b),(−1)]) = 1/2 [(a-3),(1 + b),(−3)]`

`1/2(a-3)` `= −5`
`a-3` `= −10`
`a` `= −7`
`1/2(1 + b)` `= 3/2`
`1 + b` `= 3`
`b` `= 2`
`c` `= −3/2`

 
`=>E`

Statistics, MET2 2019 VCAA 4

The Lorenz birdwing is the largest butterfly in Town A.

The probability density function that describes its life span, `X`, in weeks, is given by
 

`f(x) = {(4/625 (5x^3-x^4), quad 0 <= x <= 5),(0, quad text(elsewhere)):}`
 

  1. Find the mean life span of the Lorenz birdwing butterfly.  (2 marks)

  2. In a sample of 80 Lorenz birdwing butterflies, how many butterflies are expected to live longer than two weeks, correct to the nearest integer?  (2 marks)

  3. What is the probability that a Lorenz birdwing butterfly lives for at least four weeks, given that it lives for at least two weeks, correct to four decimal places?  (2 marks)

The wingspans of Lorenz birdwing butterflies in Town A are normally distributed with a mean of 14.1 cm and a standard deviation of 2.1 cm.

  1. Find the probability that a randomly selected Lorenz birdwing butterfly in Town A has a wingspan between 16 cm and 18 cm, correct to four decimal places.  (1 mark)

  2. A Lorenz birdwing butterfly is considered to be very small if its wingspan is in the smallest 5% of all the Lorenz birdwing butterflies in Town A.

     

    Find the greatest possible wingspan, in centimetres, for a very small Lorenz birdwing butterfly in Town A, correct to one decimal place.  (1 mark)

Each year, a detailed study is conducted on a random sample of 36 Lorenz birdwing butterflies in Town A.

A Lorenz birdwing butterfly is considered to be very large if its wingspan is greater than 17.5 cm. The probability that the wingspan of any Lorenz birdwing butterfly in Town A is greater than 17.5 cm is 0.0527, correct to four decimal places.

    1. Find the probability that three or more of the butterflies, in a random sample of 36 Lorenz birdwing butterflies from Town A, are very large, correct to four decimal places.  (1 mark)

    2. The probability that `n` or more butterflies, in a random sample of 36 Lorenz birdwing butterflies from Town A, are very large is less than 1%.

       

      Find the smallest value of `n`, where `n` is an integer.  (2 marks)

    3. For random samples of 36 Lorenz birdwing butterflies in Town A, `hat p` is the random variable that represents the proportion of butterflies that are very large.
    4. Find the expected value and the standard deviation of `hat p`, correct to four decimal places.  (2 marks)

    5. What is the probability that a sample proportion of butterflies that are very large lies within one standard deviation of 0.0527, correct to four decimal places? Do not use a normal approximation.  (2 marks)

  2. The Lorenz birdwing butterfly also lives in Town B.

     

    In a particular sample of Lorenz birdwing butterflies from Town B, an approximate 95% confidence interval for the proportion of butterflies that are very large was calculated to be (0.0234, 0.0866), correct to four decimal places.

     

    Determine the sample size used in the calculation of this confidence interval.  (2 marks)

Show Answers Only

  1. `10/3`
  2. `73`
  3. `0.2878`
  4. `0.1512`
  5. `10.6\ text(cm)`
    1. `0.2947`
    2. `7`
    3. `0.0372`
    4. `0.7380`
  7. `200`

Show Worked Solution

a.    `mu` `= 4/625 int_0^5 x(5x^3-x^4)\ dx`
    `= 4/625[x^5-1/6 x^6]_0^5`
    `= 10/3\ \ \ text{(by CAS)}`

 

b.    `text(Pr)(X > 2)` `= 4/625 int_2^5 5x^3-x^4\ dx`
    `= 4/625[5/4x^4-x^5/5]_2^5`
    `= 0.9129…`

 

`:.\ text(Expected number)` `= 80 xx 0.9129…`
  `~~ 73.03`
  `~~ 73`

 

c.    `text(Pr)(X = 4|X> 2)` `= (text(Pr)(X >= 4))/(text(Pr)(X >= 2))`
    `= 0.26272/0.91296`
    `= 0.2878`

 

d.   `W\ ~\ N (14.1, 2.1^2)`

`text(Pr)(16 < W < 18) = 0.1512\ \ \ text{(by CAS)}`

 

e.   `text(Solution 1:)`

`text(Pr)(W < w) = 0.05`

`text(Pr)(Z < z) = 0.05\ \ =>\ \ z = -1.6449\ \ text{(by CAS)}`

`(w-14.1)/2.1` `= -1.6449`
`w` `= 10.6\ text(cm)`

 
`text(Solution 2:)`

`text(invNorm)`

`text(Tail setting: left)`

`text(prob: 0.05)`

`sigma: 2.1`

`mu: 14.1`

`=> 10.6\ \ \ text{cm (by CAS)}`

 

f.i.   `L\ ~\ text(Bi)(n, p)\ ~\ text(Bi) (36, 0.0527)`

`text(Pr)(L >= 3) ~~ 0.2947`

 

f.ii.    `text(Pr)(L >= n) < 0.01`
 

`text(CAS: binomialCdf) (x, 36, 36, 0.0527)`

`text(Pr)(L >= 0) = 0.011 > 0.01`

`text(Pr)(L >= 7) = 0.002 < 0.01`

`:.\ text(Smallest)\ n = 7`

 

f.iii.    `E(hat p)` `= p = 0.0527`
  `sigma(hat p)` `= sqrt((p(1-p))/n) = sqrt((0.0527(1-0.0527))/36) ~~ 0.0372`

 

f.iv.        `hat p +- 1 sigma: (0.0527-0.0372, 0.0527 + 0.0372) = (0.0155, 0.0899)`

`text(Pr)(0.0155 < hat p < 0.0899)`

`= text(Pr)(36 xx 0.0155 < L < 36 xx 0.0899)`

`= text(Pr)(0.56 < L < 3.24)`

`= text(Pr)(1 <= L <= 3)`

`~~ 0.7380`

 

g.   `0.0234 = hat p-1.96 sqrt((hat p(1-hat p))/n) qquad text{… (1)}`

`0.0866 = hat p + 1.96 sqrt((hat p(1-hat p))/n) qquad text{… (2)}`

`text(Solve simultaneous equations:)`

`hat p ~~ 0.055, quad n ~~ 199.96`

`:.\ text(Sample size) = 200`

Calculus, MET2 2019 VCAA 3

During a telephone call, a phone uses a dual-tone frequency electrical signal to communicate with the telephone exchange.

The strength, `f`, of a simple dual-tone frequency signal is given by the function  `f(t) = sin((pi t)/3) + sin ((pi t)/6)`, where  `t`  is a measure of time and  `t >= 0`.

Part of the graph of `y = f(t)`  is shown below

  1. State the period of the function.   (1 mark)

  2. Find the values of  `t`  where  `f(t) = 0`  for the interval  `t in [0, 6]`.   (1 mark)

  3. Find the maximum strength of the dual-tone frequency signal, correct to two decimal places.   (1 mark)

  4. Find the area between the graph of  `f`  and the horizontal axis for  `t in [0, 6]`.   (2 marks)

Let  `g`  be the function obtained by applying the transformation  `T`  to the function  `f`, where
 

`T([(x), (y)]) = [(a, 0), (0, b)] [(x), (y)] + [(c), (d)]`
 

and `a, b, c` and `d` are real numbers.

  1. Find the values of `a, b, c` and `d` given that  `int_2^0 g(t)\ dt + int_2^6 g(t)\ dt`  has the same area calculated in part d.   (2 marks)

  2. The rectangle bounded by the line  `y = k, \ k in R^+`, the horizontal axis, and the lines  `x = 0`  and  `x = 12`  has the same area as the area between the graph of  `f`  and the horizontal axis for one period of the dual-tone frequency signal.

     

    Find the value of  `k`.   (2 marks)

Show Answers Only
  1. `12`
  2. `0, 4, 6`
  3. `1.760`
  4. `15/pi\ text(u²)`
  5. `a = 1,\ b =-1,\ c =-6,\ d = 0`
  6. `5/(2pi)`
Show Worked Solution

a.   `text(Period) = 12`
  

b.   `t = 0, 4, 6`
  

c.   `f(t) = sin ((pi t)/3) + sin ((pi t)/6)`

`f(t)_max ~~ 1.76\ \ text{(by CAS)}`

 

d.    `text(Area)` `= int_0^4 sin((pi t)/3) + sin ((pi t)/6) dt-int_4^6 sin ((pi t)/3) + sin ((pi t)/6) dt`
    `= 15/pi\ text(u²)`

 

e.   `text(Same area) => f(t)\ text(is reflected in the)\ x text(-axis and)`

`text(translated 6 units to the left.)`

`x′=ax+c`

`y′=by+d`

`text(Reflection in)\ xtext(-axis) \ => \ b=-1, \ d=0`

`text(Translate 6 units to the left) \ => \ a=1, \ c=-6`

`:. a = 1,\ b = -1,\ c = -6,\ d = 0`
  

f.    `text(Area of rectangle)` `= 2 xx text(Area between)\ f(t) and x text(-axis)\ \ t in [0, 6]`
  `12k` `= 2 xx 15/pi`
  `:. k` `=5/(2pi)`

Complex Numbers, SPEC1 2019 VCAA 7

  1. Show that  `3-sqrt3 i = 2sqrt3 text(cis)(-pi/6)`.   (1 mark)

  2. Find  `(3-sqrt3 i)^3`, expressing your answer in the form  `x + iy`, where  `x`,  `y ∈ R`.   (2 marks)

  3. Find the integer values of  `n`  for which  `(3-sqrt3 i)^n`  is real.   (1 mark)

  4. Find the integer values of  `n`  for which  `(3-sqrt3 i)^n = ai`, where  `a`  is a real number.   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `0-i 24sqrt3`
  3. `n = 6k\ \ (k ∈ ZZ)`
  4. `n = 3 + 6k\ \ (k ∈ ZZ)`
Show Worked Solution

a.  `|3-sqrt3 i|= sqrt(3^2 + (-sqrt3)^2)= sqrt12= 2sqrt3`

`text(Arg)(3-sqrt3 i)` `= tan^(-1)(-(sqrt3)/3)= -pi/6`
   

`:. 3-sqrt3 i = 2sqrt3\ text(cis)(-pi/6)`

b.    `(3-sqrt3 i)^3` `= (2sqrt3)^3\ text(cis)(3 xx-pi/6)`
    `= 24sqrt3\ text(cis)(-pi/2)`
    `= 24sqrt3(cos(-pi/2) + isin(-pi/2))`
    `= 0-i 24sqrt3`

 

c.   `(3-sqrt3 i)^n = (2sqrt3)^n\ text(cis)((-npi)/6)`

`text(Real when)\ \ sin(-(npi)/6) = -sin((npi)/6) = 0`

`(npi)/6 = 0, pi, 2pi, …, kpi\ \ (k ∈ ZZ)`

`:. n = 6k\ \ (k ∈ ZZ)`

 

d.  `(3-sqrt3 i)^n = ai\ \ text(when)\ \ cos(-(npi)/6) = cos((npi)/6) = 0`

`(npi)/6 = pi/2, (3pi)/2, …, pi/2 + kpi\ \ (k ∈ ZZ)`

`:. n = 3 + 6k\ \ (k ∈ ZZ)`

Calculus, SPEC1 2019 VCAA 5

The graph of  `f(x) = cos^2(x) + cos(x) + 1`  over the domain  `0 <= x <= 2pi`  is shown below.

  1.  i.  Find `f^{′}(x)`.  (1 mark)

  2. ii. Hence, find the coordinates of the turning points of the graph in the interval  `(0, 2pi)`.  (2 marks)

  3. Sketch the graph of  `y = 1/(f(x))`  on the set of axes above. Clearly label the turning points and endpoints of this graph with their coordinates.  (3 marks)
Show Answers Only
  1. i.  `−sin(x)(2cos(x) + 1)`
  2. ii. `(pi, 1), ((2pi)/3,3/4), ((4pi)/3, 3/4)`
  3.  

Show Worked Solution
a.i.    `f(x)` `= cos^2(x) + cos(x) + 1`
  `f^{′}(x)` `= -2sin(x)cos(x)-sin(x)`
    `= -sin(x)(2cos(x) + 1)`

 

a.ii.   `text(SP when)\ \ sin(x) = 0\ \ text(or)\ \ 2cos(x) + 1 = 0`

`sin(x) = 0 \ => \ x = pi\ \ (x = 0\ \ text{not in domain})`

`2cos(x) + 1` `= 0`
`cos(x)` `= -1/2`
`x` `= (2pi)/3, (4pi)/3`

 
`text(When)\ \ cos(x) = −1/2 \ => \ f(x) = 1/4-1/2 + 1 = 3/4`

`:.\ text(Turning Points:)\ (pi, 1), ((2pi)/3,3/4), ((4pi)/3, 3/4)`

 

b.   

Calculus, MET2 2019 VCAA 1

Let  `f: R -> R,\ \ f(x) = x^2e^(-x^2)`.

  1. Find `f^{\prime}(x)`.  (1 mark)

  2. i.  State the nature of the stationary point on the graph of  `f`  at the origin.  (1 mark)

  3. ii.  Find the maximum value of the function  `f`  and the values of  `x`  for which the maximum occurs.  (2 marks)

  4. iii. Find the values of  `d in R`  for which  `f(x) + d`  is always negative.  (1 mark)

  5. i.  Find the equation of the tangent to the graph of  `f`  at  `x = –1`.  (1 mark)

  6. ii. Find the area enclosed by the graph of  `f`  and the tangent to the graph of  `f`  at  `x = –1`, correct to four decimal places.  (2 marks)

  7. Let  `M(m, n)`  be a point on the graph of  `f`, where  `m in [0, 1]`.
  8. Find the minimum distance between  `M`  and the point  `(0, e)`, and the value of  `m`  for which this occurs, correct to three decimal places.  (3 marks)

Show Answers Only

  1. `2x e^(-x^2)(1-x^2)`
  2. i.  `text(Local minimum)`
  3. ii. `f(x)_max= 1/e\ \ text(when) \ x = -1 and 1`
  4. iii. `d<1/e`
  5. i.  `y = 1/e` 
  6. ii. `0.3568\ text{(to 4 d.p.)}`
  7. `D_(min) = 2.511\ \ text(when)\ \ m ~~ 0.783`

Show Worked Solution

a.    `f^{\prime}(x)` `= x^2 ⋅ -2x ⋅ e^(-x^2) + e^(-x^2) ⋅ 2x`
    `= 2x e^(-x^2) (1-x^2)`

​

b.i.    `f ^{″}(0) = 2 > 0\ \ \ text{(by CAS)}`
  `:.\ text(Local minimum)`

 

b.ii.    `text(SP’s occur when)\ \ x = –1, 0, 1`
  `f(x)_max = 1/e\ \ text(when)\ \ x = –1  and  1`

 

b.iii.    `f(x)_max + d` `< 0`
  `d` `< 1/e`

 

c.i.    `text(At)\ \ x = –1, \ f(x)\ text(has a max turning point).`
  `:.\ text(T)text(angent:)\ \ y = 1/e`

 

c.ii.    `text(Area)` `= int_(_1)^1 1/e-x^2e^(-x^2) dx`
    `~~ 0.3568\ text{(to 4 d.p.)}`

 

d.   `text(When)\ x = m,\ \ n = m^2 e^(-m^2)`

`text(Find distance between)\ \ M(m, m^2 e^(-m^2)) and P(0, e):`

`D = sqrt((m-0)^2 + (m^2 e^(-m^2)-e)^2)`

`=>\ text(MIN distance when)\ \ (dD)/(dm) = 0`

`D_(min) = 2.511\ \ text(when)\ \ m ~~ 0.783\ \ \ text{(by CAS)}`

Statistics, SPEC1 2019 VCAA 3

A machine produces chocolate in the form of a continuous cylinder of radius 0.5 cm. Smaller cylindrical pieces are cut parallel to its end, as shown in the diagram below.

The lengths of the pieces vary with a mean of 3 cm and a standard deviation of 0.1 cm.
 


 

  1. Find the expected volume of a piece of chocolate in cm³.   (1 mark)

  2. Find the variance of the volume of a piece of chocolate in cm6.   (1 mark)

  3. Find the expected surface area of a piece of chocolate in cm².   (1 mark)

Show Answers Only
  1. `0.75pi\ text(cm³)`
  2. `0.000625pi^2\ text(cm)^6`
  3. `3.5pi\ text(cm²)`
Show Worked Solution
a.    `E(V)` `= pir^2 xx E(h)`
    `= pi xx 0.5^2 xx 3`
    `= 0.75pi\ text(cm³)`

 

b.    `text(Var)(V)` `= text(Var)(pir^2h)`
    `= pi^2 xx 0.5^4 xx text(Var)(h)`
    `= 0.0625pi^2 xx 0.1^2`
    `= 0.000625pi^2\ text(cm)^6`

 

c.    `E text{(Surface Area)}` `= 2pir^2 + 2pir xx E(h)`
    `= 2pir(r +E(h))`
    `= pi(0.5 + 3)`
    `= 3.5pi\ text(cm²)`

Functions, 2ADV F2 EQ-Bank 13

  1. Sketch the function  `y = f(x)`  where  `f(x) = (x-1)^3`  on a number plane, labelling all intercepts.   (1 mark)

  2. On the same graph, sketch  `y = -f(-x)`. Label all intercepts.   (2 marks)

Show Answers Only

a.    

b.    

Show Worked Solution

a.   `y = (x-1)^3 => y = x^3\ text(shifted 1 unit to the right.)`
 

 
b.   `y = -f(x) \ => \ text(reflect)\ \ y = (x-1)^3\ \ text(in)\ xtext(-axis).`

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

 

Functions, 2ADV F1 EQ-Bank 19

Find all values of `x` for which  `| x-4 | = x/2 + 7`.   (3 marks)

Show Answers Only

`x = 22\ \ text(or)\ -2`

Show Worked Solution
`x-4` `= x/2 + 7` `text(or)` `-(x-4)` `= x/2 + 7`
`2x-8` `= x + 14`   `-2x + 8` `= x + 14`
`x` `= 22`   `3x` `= -6`
      `x` `= -2`

 
`:. x = 22\ \ text(or)\ -2`

Functions, 2ADV F1 EQ-Bank 15

Consider the function  `f(x) = 1/(x + 2)`.
 

  1. Sketch the graph  `y = f(-x)`.   (2 marks)

  2. On the same graph, sketch  `y = -f(x)`.   (2 marks)

Show Answers Only

a.    

b.    

Show Worked Solution

a.   `text(Sketch)\ \ y = 1/(x + 2)`

`y = f(-x) \ =>\ text(reflect)\ \ y = 1/(x + 2)\ \ text(in the)\ ytext(-axis).`
 

 

b.   `y = -f(x) \ =>\ text(reflect)\ \ y = 1/(x + 2)\ \ text(in the)\ xtext(-axis).`
 

Probability, MET2 2019 VCAA 7 MC

The discrete random variable `X` has the following probability distribution.
 

  `qquad x` `qquad 0 qquad` `qquad 1 qquad` `qquad 2 qquad` `qquad 3 qquad`
  `qquad Pr(X = x) qquad` `a` `3a` `5a` `7a`

 
The mean of `X` is

  1. `1/16`
  2. `1`
  3. `35/16`
  4. `17/8`
  5. `2`
Show Answers Only

`D`

Show Worked Solution

`16a = 1 qquad => qquad a = 1/16`

`text(E)(X)` `= 3/16 xx 1 + 5/16 xx 2 + 7/16 xx 3`
  `=34/16`
  `= 17/8`

 
`=>   D`

Calculus, MET2 2019 VCAA 5 MC

Let  `f prime(x) = 3x^2 - 2x`  such that  `f(4) = 0`.

The rule of  `f`  is

A.   `f(x) = x^3 - x^2`

B.   `f(x) = x^3 - x^2 + 48`

C.   `f(x) = x^3 - x^2 - 48`

D.   `f(x) = 6x - 2`

E.   `f(x) = 6x - 24`

Show Answers Only

`C`

Show Worked Solution
`f(x)` `= int 3x^2 – 2x\ dx`
  `= x^3 – x^2 + c`

 

`text(When)\ \ x = 4, \ f(x) = 0`

`0 = 4^3 – 4^2 + c`

`c = -48`
 

`:. f(x) = x^3 – x^2 – 48`

`=>   C`

Statistics, 2ADV S2 SM-Bank 14

A probability density function  `f(x)`  is given by
 

`f(x) = {(px(3 - x), \ text(if)\ \ 0 <= x <= 3),(0, \ text(if)\ \ x < 0\ \ text(or if)\ \ x > 3):}`
 

where  `p`  is a positive constant.

  1. Find the value of  `p`.  (2 marks)

  2. Find the mode of  `f(x)`.  (2 marks)

Show Answers Only
  1. `2/9`
  2. `3/2`
Show Worked Solution

i.   `text(Total Area under curve = 1)`

`p int_0^3 3x-x^2\ dx` `= 1`
`p[3/2 x^2-(x^3)/3]_0^3` `= 1`
`p[27/2-27/3-0]` `= 1`
`(9p)/2` `= 1`
`p` `= 2/9`

 

ii.   `f(x) = 2/9(3x-x^2)`

`f^{′}(x) = 2/9(3-2x)`

`text(S.P. when)\ \ f^{′}(x) = 0:`

`3-2x` `= 0`
`x` `= 3/2`

 
`f^{″}(x) = -4/9 < 0 \ =>\ text(MAX)`

`:.\ text(Mode) = 3/2`

Statistics, 2ADV S3 EQ-Bank 17

The diastolic measurement for blood pressure in 35-year-old people is normally distributed, with a mean of 75 and a standard deviation of 12.

  1. A person is considered to have low blood pressure if their diastolic measurement is 63 or less.What percentage of 35-year-olds have low blood pressure?  (1 mark)

  2. Calculate the \(z\)-score for a diastolic measurement of 57.  (1 mark)

  3. The probability that a 35-year-old person has a diastolic measurement for blood pressure between 57 and 63 can be found by evaluating
     
    \(\int_a^b f(x) d x\)
     
    where \(a\) and \(b\) are constants and where
     
    \(f(x)=\dfrac{1}{\sqrt{2 \pi}} e^{\small{\dfrac{-x^2}{2}}}\)

    is the normal probability density function with mean 0 and standard deviation 1.

     

    By first finding the values \(a\) and \(b\), calculate an approximate value for this probability by using the trapezoidal rule with 3 function values.  (3 marks)

  4. Hence, find the approximate probability that a 35-year-old person chosen at random has a diastolic measurement of 57 or less.  (1 mark)

Show Answers Only
  1. \(16 \% \ \text{have low blood pressure}\)
  2. \(-1.5\)
  3. \(9.2 \%\)
  4. \(6.8 \%\)
Show Worked Solution
i.    \(z \text{-score}(63)\) \(=\dfrac{x-\mu}{\sigma}\)
    \(=\dfrac{63-75}{12}\)
    \(=-1\)

 

\(\therefore \ 16 \% \ \text {have low blood pressure}\)

 

ii.    \(z \text{-score}\) \(=\dfrac{57-75}{12}\)
    \(=-1.5\)

 

iii. \(y=\dfrac{1}{\sqrt{2 \pi}} e^{\small{\dfrac{-x^2}{2}}}\)

 
\(\text{Area}\) \(=\dfrac{h}{2}\left(y_0+2 y_1+y_2\right)\)
  \(\approx \dfrac{0.25}{2}(0.1295+2 \times 0.1826+0.2420)\)
  \(\approx 0.0920\)
  \(\approx 9.2 \%\)

 

iv.   

 

\(P(\text{blood pressure}\ \leq 57)\) \(=16-9.2\)
  \(\approx 6.8 \%\)

Algebra, MET2 2019 VCAA 2 MC

The set of values of  `k`  for which  `x^2 + 2x-k = 0`  has two real solutions is

  1. `{-1, 1}`
  2. `(-1, oo)`
  3. `(-oo, -1)`
  4. `{-1}`
  5. `[-1, oo)`
Show Answers Only

`B`

Show Worked Solution

`text(Two real solutions):`

`b^2-4ac` `> 0`
`4-4 ⋅ 1 ⋅ (-k)` `> 0`
`4k` `> -4`
`k` `> -1`

 
`k in (-1, oo)`

`=>   B`

Graphs, MET2 2019 VCAA 1 MC

Let  `f: R -> R,\ \ f(x) = 3 sin ((2x)/5) - 2`.

The period and range of  `f`  are respectively

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

`B`

Show Worked Solution
`text(Period)` `= (2pi)/n`
  `= (2 pi)/(2/5)`
  `= 5 pi`
   
`text(Range)` `= [-2 -3, -2 + 3]`
  `= [-5, 1]`

 
`=>   B`

Calculus, MET1 2019 VCAA 7

The graph of the relation  `y = sqrt (1-x^2)`  is shown on the axes below. `P` is a point on the graph of this relation, `A` is the point `(-1, 0)` and `B` is the point `(x, 0)`.
 

  1. Find an expression for the length `PB` in terms of `x` only.   (1 mark) 

  2. Find the maximum area of the triangle `ABP`.  (3 marks)

Show Answers Only
  1. `PB = sqrt(1-x^2)`
  2. `(3 sqrt 3)/8`
Show Worked Solution

a.    `PB = sqrt(1-x^2)`

 

b.    `A` `= 1/2 ⋅ AB ⋅ PB`
    `= 1/2 (x + 1) ⋅ (1-x^2)^(1/2)`
  `(dA)/(dx)` `= 1/2[(x + 1) ⋅ 1/2 ⋅ -2x ⋅ 1/sqrt(1-x^2) + sqrt(1-x^2)]`
    `= 1/2((-x^2-x + 1-x^2)/sqrt(1-x^2))`
    `= (-2x^2-x + 1)/(2 sqrt(1-x^2))`

 
`text(Find max when)\ \ (dA)/(dx) = 0`

`2x^2 + x-1 = 0`

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

`x = 1/2 qquad (x =\ text{–1 is a min)}`

`:. A_max` `= 1/2 (3/2)(1-1/4)^(1/2)`
  `= 3/4 ⋅ sqrt(3/4)`
  `= (3 sqrt 3)/8`

Probability, MET1 2019 VCAA 6

Fred owns a company that produces thousands of pegs each day. He randomly selects 41 pegs that are produced on one day and finds eight faulty pegs.

  1. What is the proportion of faulty pegs in this sample?  (1 mark) 

  2. Pegs are packed each day in boxes. Each box holds 12 pegs. Let  `hat P`  be the random variable that represents the proportion of faulty pegs in a box.
  3. The actual proportion of faulty pegs produced by the company each day is `1/6`.
  4. Find  `text(Pr)(hat P < 1/6)`. Express your answer in the form  `a(b)^n`, where  `a`  and  `b`  are positive rational numbers and  `n`  is a positive integer.  (2 marks)

Show Answers Only

  1. `8/41`
  2. `(17/6) ⋅ (5/6)^11`

Show Worked Solution

a.    `text(Proportion of faulty pegs) = 8/41`
 

b.   `hat P = x/n = 1/6`

`text(Given)\ \ n = 12`

`1/6 = X/12 \ => \ X = 2`

`X\ ~\ text(Bi) (12, 1/6)`

`text(Pr)(hat P < 1/6)` `= text(Pr)(X < 2)`
  `= text(Pr)(X = 0) + text(Pr)(X = 1)`
  `= \ ^12 C_0 * (5/6)^12 + \ ^12 C_1 ⋅ (1/6)(5/6)^11`
  `= (5/6)^11 (5/6 + 12/6)`
  `= (17/6) ⋅ (5/6)^11`

Calculus, MET1 2019 VCAA 5

Let  `f: R\ text(\{1}) -> R, \ f(x) = 2/(x-1)^2 + 1`.

    1. Evaluate  `f(-1)`.   (1 mark)

    2. Sketch the graph of `f` on the axes below, labelling all asymptotes with their equations.   (2 marks)

        

  2. Find the area bounded by the graph of `f`, the `x`-axis, the line  `x = -1`  and the line  `x = 0`.   (2 marks)

Show Answers Only
    1. `3/2`
    2. `text(See Worked Solutions)`
  2. `2`
Show Worked Solution
a.i.    `f(-1)` `= 2/(-1-1)^2 + 1`
    `= 3/2`

​

a.ii.     

 

b.    `text(Area)` `= int_(-1)^0 2/(x-1)^2 + 1\ dx`
    `= int_(-1)^0 2(x-1)^(-2) + 1\ dx`
    `= [-2(x-1)^(-1) + x]_(-1)^0`
    `= [((-2)/-1 + 0)-((-2)/-2-1)]`
    `= 2`