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CORE, FUR1 2019 VCAA 15-16 MC

The time series plot below shows the monthly rainfall at a weather station, in millimetres, for each month in 2017.
 


 

Part 1

The median monthly rainfall for 2017 was closest to

  1.  53 mm
  2.  82 mm
  3.  96 mm
  4. 103 mm
  5. 111 mm

 
Part 2

If seven-mean smoothing is used to smooth this time series plot, the number of smoothed data points would be

  1.  3
  2.  5
  3.  6
  4.  8
  5. 10
Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(12 data points)`

`text(Median)` `= text{(6th + 7th)}/2`
  `~~ (97 + 109)/2`
  `~~ 103\ text(mm)`

`=>  D`
 

`text(Part 2)`

`text(January – July)\ =>\ text(First data point)`

`text(February – August)\ =>\ text(S)text(econd data point)`

`qquad vdots`

`text(June – December)\ =>\ text(Sixth data point)`

`=>  C`

CORE, FUR1 2019 VCAA 13-14 MC

The time, in minutes, that Liv ran each day was recorded for nine days.

These times are shown in the table below.
  


  

The time series plot below was generated from this data.
  

Part 1

Both three-median smoothing and five-median smoothing are being considered for this data.

Both of these methods result in the same smoothed value on day number

  1. 3
  2. 4
  3. 5
  4. 6
  5. 7

 
Part 2

A least squares line is to be fitted to the time series plot shown above.

The equation of this least squares line, with day number as the explanatory variable, is closest to

  1. day number = 23.8 + 2.29 × time
  2. day number = 28.5 + 1.77 × time
  3. time = 23.8 + 1.77 × day number
  4. time = 23.8 + 2.29 × day number
  5. time = 28.5 + 1.77 × day number
Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ E`

Show Worked Solution

`text(Part 1)`

`text{Add 3-median (dots) and 5-median (Δ) smoothing to the plot:}`
 

`=>  E`
 

`text(Part 2)`

`text(time) = 28.5 + 1.77 xx text(day number)\ \ \ text{(by CAS)}`

`=>  E`

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 8 MC

Percy conducted a survey of people in his workplace. He constructed a two-way frequency table involving two variables.

One of the variables was attitude towards shorter working days (for, against).

The other variable could have been

  1. age (in years).
  2. sex (male, female).
  3. height (to the nearest centimetre).
  4. income (to the nearest thousand dollars).
  5. time spent travelling to work (in minutes).
Show Answers Only

`B`

Show Worked Solution

`text(A variable is required here that only has two possibilities.)`

`=>  B`

CORE, FUR1 2019 VCAA 7 MC

The volume of a cup of soup served by a machine is normally distributed with a mean of 240 mL and a standard deviation of 5 mL.

A fast-food store used this machine to serve 160 cups of soup.

The number of these cups of soup that are expected to contain less than 230 mL of soup is closest to

  1.     3
  2.     4
  3.     8
  4.   26
  5. 156
Show Answers Only

`B`

Show Worked Solution
`z text(-score)\ (230)` `= (230 – 240)/5`
  `= -2`

 
`:.\ text(Number of cups) < 230\ text(mL)`

`= 2.5/100 xx 160`

`= 4`
 

`=>  B`

CORE, FUR1 2019 VCAA 6 MC

The time taken to travel between two regional cities is approximately normally distributed with a mean of 70 minutes and a standard deviation of 2 minutes.

The percentage of travel times that are between 66 minutes and 72 minutes is closest to

  1. `text(2.5%)`
  2. `text(34%)`
  3. `text(68%)`
  4. `text(81.5%)`
  5. `text(95%)`
Show Answers Only

`D`

Show Worked Solution

`bar x = 70,\ \ s = 2`

`z text(-score)\ (66) = (66 – 70)/2 = -2`

`z text(-score)\ (72) = (72 – 70)/2 = 1`
 


 

`:.\ text(Percentage between 66 – 72)`

`= 47.5 + 34`

`= 81.5\ text(%)`
 

`=>  D`

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`

Statistics, EXT1 S1 SM-Bank 11

Within a particular population, it is known that the percentage of left-handers is 17%.

A research project randomly selects 200 people from this population.

Assuming this sample proportion is normally distributed, what is the probability that the percentage of people that are left-handed in this sample is

  1. greater than 20%   (2 marks)

  2. less than 10%   (2 marks)

Show Answers Only
  1. `0.1292`
  2. `0.0041`
Show Worked Solution

i.   `E(hat p)=p=0.17`

`text(Var)(hat p)` ` = (p(1-p))/n`  
`sigma^2(hat p)` `=(0.17(1-0.17))/200`  
  `=0.0007055…`  
`sigma(hat p)` `=0.02656…`  
     

`hatp\ ~\ N(mu,sigma)\ ~\ N(0.17, 0.02656)`

`ztext{-score (20%)}` `=(x-mu)/sigma`  
  `~~(0.20 – 0.17)/0.02656`  
  `~~1.13`  

 
`text{Using the probability table (attached):}`

`:.P(hat p>20text(%))` `=P(z>1.13)`  
  `=1-0.8708`  
  `=0.1292`  

 

ii.    `ztext{-score (10%)}` `~~(0.10-0.17)/0.02656`
    `~~-2.64`

 
`text{Using the probability table (attached):}`

`:.P(hatp<10text(%))` `=P(z<-2.64)`  
  `=0.0041`  

Functions, EXT1 F2 SM-Bank 5

The polynomial  `P(x) = x^3 + px^2 + qx + r`  has roots  `sqrtk`, `−sqrtk`  and  `alpha`.

  1.  Explain why  `alpha + p = 0`.  (1 mark)

  2.  Show that  `kalpha = r`.  (1 mark)

  3.  Show that  `pq = r`.  (2 marks)

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

i.   `text(Sum of roots:)`

`sqrtk – sqrtk + alpha` `= −b/a`
`alpha` `= −p`
`alpha + p` `= 0`

 

ii.   `text(Product of roots:)`

`sqrtk xx −sqrtk xx alpha` `= −d/a`
`−kalpha` `= −r`
`:.kalpha` `= r`

 

iii.    `sqrtk(−sqrtk) + sqrtk alpha – sqrtk alpha` `= c/d`
  `−k` `= q`

 
`text(Substitute)\ \ k = −q\ \ text(into part (ii)):`

`−qalpha` `= r`
`−q xx – p` `= r\ \ \ text{(using part(i))}`
`:. pq` `= r`

Functions, EXT1 F2 SM-Bank 2

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
  1. `2`
  2. `−24`
  3. `−12`
Show Worked Solution

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

 

ii.   `alphabetagamma = −d/a = −24`

 

iii.   `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, EXT1 F2 SM-Bank 1

The equation  `2x^3 + x^2 - kx + 6 = 0`  has 2 roots which are reciprocals of each other.

Find the value of `k`.  (2 marks)

Show Answers Only

`13`

Show Worked Solution

`text(Let roots be)\ \ alpha, \ 1/alpha, \ beta:`

COMMENT: Substituting  `beta=-3`  into the equation solves this question efficiently.

`alpha · 1/alpha · beta` `= −d/a`
`beta` `= −6/2`
  `= −3`

 

`text(Substitute)\ \ beta = −3\ \ text(into equation:)`

`2(−3)^3 + (−3)^2 – (−3)k + 6` `= 0`
`−54 + 9 + 3k + 6` `= 0`
`3k` `= 39`
`k` `= 13`

Statistics, EXT1 S1 SM-Bank 6

After counting votes in an election, it is known that 40% of people voted for the Katter party.

A sample of 600 voting cards are taken and inspected. Assuming this sample proportion is approximately normally distributed, what is the probability that the percentage of voting cards inspected that chose the Katter party is less than 36%?   (3 marks)

Show Answers Only

`2.5\ text(%)`

Show Worked Solution

`mu = E(hat p)=p=0.4`

`text(Var)(hat p)` ` = (p(1-p))/n`  
`sigma^2(hat p)` `=(0.4(1-0.4))/600`  
  `=0.0004`  
`sigma(hat p)` `=0.02`  
     
`ztext{-score (36%)}` `=(0.36-0.40)/0.02`  
  `=-2`  

 

`:.P(hat p<36text(%))` `=P(z<-2)`  
  `=2.5 text(%)`  

Statistics, EXT1 S1 SM-Bank 4

In an experiment, a pair of dice are rolled 70 times.

A success is recorded if the sum of the dice roll is 5 or less.

  1. What is the mean of this binomial distribution?

     

    Give your answer to one decimal place.  (3 marks)

  2. What is the standard deviation?

     

    Give your answer to one decimal place.  (1 mark)

Show Answers Only
  1. `19.4`
  2. `3.7\ \ (text(to 1 d.p.))`
Show Worked Solution

i.   `text(Array of possible roll totals:)`

STRATEGY: A table (or array) can be a very efficient and error minimising strategy in questions like this.

`P(5\ text(or less)) = 10/36 = 5/18`

`text(Let)\ X =\ text(number of rolls) <= 5`

`X\ ~\ text(Bin)(70, 5/18)`

`E(X)` `= np`
  `= 70 xx 5/18`
  `= 19.4`

 

ii.    `text(Var)(X)` `= np(1 – p)`
  `sigma^2` `= 70 xx 5/18(1 – 5/18)`
    `= 14.043`
  `:. sigma` `= 3.7\ \ (text(to 1 d.p.))`

Trigonometry, EXT1 T3 SM-Bank 2

Show that

`cos3x = 4cos^3 x - 3cosx`.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

COMMENT: High achieving students know the 3 variants of  `cos 2x`  back to front. Here,  `cos2x` `=cos^2x-sin^2x`  breaks the back of this problem.

`text(LHS)` `= cos(2x + x)`
  `= cos2xcosx – sin2xsinx`
  `= (cos^2x – sin^2x)cosx – 2sinxcosxsinx`
  `= cos^3x – sin^2xcosx – 2sin^2xcosx`
  `= cos^3x – (1 – cos^2x)cosx – 2(1 – cos^2x)cosx`
  `= cos^3x – cosx + cos^3x – 2cosx + 2cos^3x`
  `= 4cos^3x – 3cosx`

Trigonometry, EXT1 T1 SM-Bank 3

Determine the exact value of

`sin^(-1)(sqrt3/2)-sin^(-1)(-sqrt3/2)-cos^(-1)(-1/2)`.  (3 marks)

Show Answers Only

`0`

Show Worked Solution

`text(Base angle:)\ \ sin\ pi/3 = sqrt3/2`

`sin =>\ text(– ve in 4th quadrant)`

`sin^(-1)(sqrt3/2)` `= pi/3`
`sin^(-1)(-sqrt3/2)` `= -pi/3`

 
`text(Base angle:)\ \ cos\ pi/3 = 1/2`

`cos =>\ text(– ve in 2nd quadrant)`

`cos^(-1)(-1/2)` `= pi-pi/3= (2pi)/3`

 
`:. sin^(-1)(sqrt3/2)-sin^(-1)(-sqrt3/2)-cos^(-1)(-1/2)`

`= pi/3-(-pi/3)-(2pi)/3`

`= 0`

Trigonometry, EXT1 T3 SM-Bank 13

Find all solutions of  `tan(2theta) = -tan theta`  for  `0<= theta<=2pi`.  (3 marks)

Show Answers Only

`theta =0, \ pi/3, \ (2pi)/3, \ pi, \ (4pi)/3, \ (5pi)/3, \ 2pi`

Show Worked Solution

`(2 tan theta)/(1 – tan^2 theta) = -tan theta`

 
`text(Let)\ \ tan theta = k,`

`(2k)/(1 – k^2)` `= -k`
`2k` `= -k(1 – k^2)`
`2k + k(1 – k^2)` `= 0`
`3k-k^3` `=0`
`k(3 – k^2)` `= 0`

 
`k(3 – k^2) = 0 \ \ =>\ \  k = 0, \ k = +- sqrt3`
 

`text(If)\ \  tan theta = 0 \ => \ theta=0, \ pi, \ 2pi`

`text(If)\ \ tan theta = +- sqrt 3 => \ theta = pi/3, \ (2pi)/3, \ (4pi)/3, \ (5pi)/3`

`:. theta =0, \ pi/3, \ (2pi)/3, \ pi, \ (4pi)/3, \ (5pi)/3, \ 2pi`

Trigonometry, 2ADV T2 SM-Bank 2

Find all solutions of the equation  `2 cos theta = sqrt 3 cot theta`,  for  `0<=theta<=2pi`  (3 marks)

Show Answers Only

`theta=pi/3, pi/2, (2pi)/3, (3pi)/2`

Show Worked Solution

`2 cos theta = sqrt 3 cot theta`

`2 cos theta – sqrt 3 cot theta` `= 0`  
`2 cos theta – sqrt 3 (cos theta)/(sin theta)` `=0`  
`(2 – sqrt 3/sin theta) cos theta` `=0`  

 
`text(If)\ \ cos theta=0,`

`theta=pi/2, (3pi)/2`
  

`text(If)\ \ 2 – sqrt 3/sin theta = 0\ \ =>\ \ sin theta = sqrt 3/2`

`theta = pi/3, (2pi)/3`

Trigonometry, EXT1 T3 SM-Bank 12

 Solve  `sin(2x) = sin x`,  for  `0<= x <=2 pi.`  (3 marks)

Show Answers Only

`x = 0, \ pi/3, \ pi, \ (5pi)/3, \2 pi`

Show Worked Solution

`2sin x cos x = sin x`

MARKER’S COMMENT: Many students expanded `sin(2x)` and then cancelled `sin x` on both sides which lost a set of solutions!

`sinx(2cos x – 1) = 0`

`sinx = 0, cosx = 1/2`

`text(If)\ \ sinx=0 \ => \ x=0, \ pi, \ 2pi`

`text(If)\ \ cos x=1/2 \ => \ x = pi/3, \ (5pi)/3`

`:. x = 0, pi/3, pi, (5pi)/3, 2pi`

Trigonometry, EXT1 T1 SM-Bank 1

Find `sin theta` given that  `theta = arccos (12/13) + arctan (3/4)`, and  `0<=theta<=pi/2`.   (3 marks)

Show Answers Only

`56/65`

Show Worked Solution
`sin theta` `= sin (cos^(-1) (12/13) + tan^(-1)(3/4))`
  `= sin (cos^(-1) (12/13)) cos (tan^(-1) (3/4)) + sin (tan^(-1)(3/4)) cos (cos^(-1)(12/13))`

 
 

 

COMMENT: Syllabus states that students must understand and use the notation `arccos,\ arcsin` etc..

`:. sin theta` `= 5/13 xx 4/5 + 3/5 xx 12/13`
  `= 20/65 + 36/65`
  `= 56/65`

Trigonometry, EXT1 T3 SM-Bank 11

Given that  `cos (theta - phi) = 3/5`  and  `tan theta tan phi = 2`, find  `cos(theta + phi)`.  (3 marks)

Show Answers Only

`-1/5`

Show Worked Solution

`cos(theta+phi)= cos theta cos phi – sin theta sin phi`
 

`cos theta cos phi + sin theta sin phi = 3/5\ \ …\ (1)`

`(sin theta sin phi)/(cos theta cos phi)` `=2`  
`sin theta sin phi` `= 2 cos theta cos phi\ \ …\ (2)`  

  
`text{Substitute (2) into (1):}`

`cos theta cos phi + 2 cos theta cos phi` `= 3/5`
`3 cos theta cos phi` `= 3/5`
`cos theta cos phi` `= 1/5`

 
`text(Substitute)\ \ cos theta cos phi=1/5\ \ text{into (1):}`

`1/5 + sin theta sin phi` `= 3/5`
`sin theta sin phi` `= 2/5`

 

`:. cos(theta + phi)` `= cos theta cos phi – sin theta sin phi`
  `= 1/5 – 2/5`
  `= -1/5`

Trigonometry, EXT1 T3 SM-Bank 8

Solve for `x`, given that

`x sin(x) sec(2x) = 0,\ \ 0<=x<=2pi`  (2 marks)

Show Answers Only

`x = 0, \ pi\ \ text(or)\ \ 2pi`

Show Worked Solution
`xsin(x)sec(2x)` `= (xsin(x))/(cos(2x))=0`

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

`x = 0\ \ text(and)\ \ cos(2x) != 0 \ => \ x=0`

`text(or,)`

`sin(x) = 0\ \ text(and)\ \ cos(2x) != 0 \ => \ x=pi, \ 2pi`

`:. x = 0, \ pi\ \ text(or)\ \ 2pi`

Statistics, SPEC2 2019 VCAA 6

A company produces packets of noodles. It is known from past experience that the mass of a packet of noodles produced by one of the company's machines is normally distributed with a mean of 375 grams and a standard deviation of 15 grams.

To check the operation of the machine after some repairs, the company's quality control employees select two independent random samples of 50 packets and calculate the mean mass of the 50 packets for each random sample.

  1. Assume that the machine is working properly. Find the probability that at least one random sample will have a mean mass between 370 grams and 375 grams. Give your answer correct to three decimal places.   (2 marks)

  2. Assume that the machine is working properly. Find the probability that the means of the two random samples differ by less than 2 grams. Give your answer correct to three decimal places.   (3 marks)

To test whether the machine is working properly after the repairs and is still producing packets with a mean mass of 375 grams, the two random samples are combined and the mean mass of the 100 packets is found to be 372 grams. Assume that the standard deviation of the mass of the packets produced is still 15 grams. A two-tailed test at the 5% level of significance is to be carried out.

  1. Write down suitable hypotheses `H_0` and `H_1` for this test.   (1 mark)

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

  3. Does the mean mass of the sample of 100 packets suggest that the machine is working properly at the 5% level of significance for a two-tailed test? Justify your answer.   (1 mark)

  4. What is the smallest value of the mean mass of the sample of 100 packets for `H_0` to be not rejected? Give your answer correct to one decimal place.   (1 mark)

Show Answers Only
  1. `0.741`
  2. `0.495`
  3. `H_0: mu = 375`
    `H_1: mu != 375`
  4. `0.046`
  5. `text(See Worked Solutions)`
  6. `372.1`
Show Worked Solution

a.   `barX\ ~\ N(375, (15/sqrt50)^2)`

`text(Pr)(370 <= barX <= 375)`

`= 0.490788…`
 

`text(Pr)(text(not in range)) ~~ 1-0.490788 ~~ 0.509212`

`text(Pr)(text(Samples within range) >= 1)`

`= 1-text(Pr)(text(0 samples in range))`

`~~ 1-(0.509212)^2`

`~~ 0.741`
 

b.   `text(Let)\ \ Y = barX_1-barX_2`

`Y\ ~\ N (0, 2 xx (15^2)/50)`

`text(Pr)(−2 < Y < 2)\ \ (text(by CAS))`

`= text(norm cdf)\ (−2,2,0, sqrt(2 xx (15^2)/50))`

`~~ 0.495`
 

c.   `H_0: \ mu = 375`

`H_1: \ mu != 375`
 

d.   `text(By CAS:)\ \ mu_0 = 375, \ barx = 372, \ sigma = 15, \ n = 100`

`p ~~ 0.046`
 

e.   `p < 0.05 \ => \ text(reject)\ H_0`

`=>\ text(The machine is NOT working properly.)`
 

f.   `text(Pr)(barx <= a) = 0.025\ \ (text(2 sided test))`

`a_text(min)` `= text(inv norm)\ (0.025, 375, 15/sqrt100)`
  `= 372.06…`
  `=372.1`

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)`

Calculus, SPEC2 2019 VCAA 3

  1. The growth and decay of a quantity `P` with respect to time `t` is modelled by the differential equation
     
    `qquad qquad(dP)/(dt) = kP`
     
    where `t >= 0`.

    1. Given that  `P(a) = r`  and  `P(b) = s`, where `P` is a function of `t`,

       

      show that  `k = 1/(a-b)log_e(r/s)`.   (2 marks)

    2. Specify the condition(s) for which  `k >0`.   (2 marks)

  2. The growth of another quantity `Q` with respect to time `t` is modelled by the differential equation
     
       `qquad qquad (dQ)/(dt) = e^(t-Q)`
      
    where  `t >= 0`  and  `Q = 1`  when  `t = 0`.

    1. Express this differential equation in the form  `int f(Q)\ dQ = int h(t)\ dt`.   (1 mark)

    2. Hence, show that  `Q = log_e(e^t + e-1)`.   (2 marks)

    3. Show that the graph of `Q` as a function of `t` does not have a point of inflection.   (2 marks)

Show Answers Only
    1. `text(See Worked Solutions)`
    2. `a-b\ text(and)\ s > r > 0`
    1. `int e^Q\ dQ =  int e^t\ dt`
    2. `text(See Worked Solutions)`
    3. `text(See Worked Solutions)`
Show Worked Solution
a.i.    `(dP)/(dt)` `= kP`
  `(dt)/(dP)` `= 1/(kP)`

`t = 1/k int 1/P\ dP = 1/k log_e P + c`

`P(a) = r`

`a = 1/k · log_e r + c`

`P(b) = r`

`b = 1/k · log_e s + c`

`a-b` `= 1/k(log_e r + c-log_e s + c)`
`:.k` `= 1/(a-b)log_e (r/s)`

 

a.ii.   `text(Conditions for)\ \ k > 0,`

`text(Scenario 1:)`

`a-b > 0\ \ text(and)\ \ log_e(r/s) > 0`

`=> a > b\ \ text(and)\ \ r > s > 0`
 

`text(Scenario 2:)`

`a-b < 0\ \ text(and)\ \ log_e(r/s) < 0`

`=> a < b\ \ text(and)\ \ s > r > 0`
 

b.i.    `(dQ)/(dt)` `= (e^t)/(e^Q)`
  `int e^Q dQ` `= int e^t\ dt`

 

b.ii.    `int e^Q dQ` `= int e^t dt`
  `e^Q` `= e^t + c`

 
`text(When)\ \ Q = 1,  t = 0`

`e` `= e^0 + c`
`c` `= e-1`
`e^Q` `= e^t + e-1`

 
`:. Q = log_e(e^t + e-1)`

 

b.iii.   `Q = log_e(e^t + e-1)`

`(dQ)/(dt)` `= (e^t)/(e^t + e-1)`
`(d^2Q)/(dt^2)` `= (e^t(e-1))/((e^t + e-1)^2)`

 
`(d^2Q)/(dt^2) > 0\ text(for)\ t >= 0`

`:.\ text(No POI)`

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`

Statistics, SPEC2 2019 VCAA 19 MC

`X` and `Y` are independent random variables where each has a mean of 4 and a variance of 9.

If the random variable  `Z = aX + bY`  has a mean of 8 and a variance of 90, possible values of `a` and `b` are

  1. `a = 1, \ b = 1`
  2. `a = 4, \ b = −2`
  3. `a = 3, \ b = −1`
  4. `a = 1, \ b = 3`
  5. `a = −2, \ b = 4`
Show Answers Only

`C`

Show Worked Solution
`E(aX + bY)` `= aE(X) + bE(Y)`
`8` `= 4a + 4b\ \ …\ (1)`
`text(Var)(aX + bY)` `= a^2text(Var)(X) + b^2text(Var)(Y)`
`9a^2 + 9b^2` `= 90\ \ …\ (2)`

 
`text{Solve simultaneous equations (by CAS):}`

`a = 3, b = −1\ \ text(or)\ \ a = −1, b = 3`

`=>C`

Statistics, SPEC2 2019 VCAA 18 MC

The masses of a random sample of 36 track athletes have a mean of 65 kg. The standard deviation of the masses of all track athletes is known to be 4 kg.

A 98% confidence interval for the mean of the masses of all track athletes, correct to one decimal place, would be closest to

  1. (51.0, 79.0)
  2. (63.6, 66.4)
  3. (63.3, 66.7)
  4. (63.4, 66.6)
  5. (64.3, 65.7)
Show Answers Only

`=>D`

Show Worked Solution

`text(Solution 1)`

`mu = 65, \ sigma = 4, \ n = 36`

`text(Confidence level 98%:)`

`(63.4, 66.6)\ \ \ text{(by CAS)}`

 

`text(Solution 2)`

`text(Find)\ z\ text(such that)\ \ text(Pr)(z < 2) = 0.99`

`z ~~ 2.3263,\ \ \ text{(by CAS)}`
 

`:.\ text(Confidence Interval)`

`= (65 – 2.3263 xx 4/sqrt36, 65 + 2.3263 xx 4/sqrt36)`

`= (63.4, 66.6)`

`=>D`

Mechanics, SPEC2 2019 VCAA 17 MC

A particle is held in equilibrium by three coplanar forces of magnitudes  `F_1, F_2`  and  `F_3`.

The angles between these forces are  `alpha, beta`  and  `gamma`  as shown in the diagram below.
 

If  `beta = 2alpha`, then  `(F_1)/(F_2)`  is equal to

  1. `1/2 sin(alpha)`
  2. `2sin(alpha)`
  3. `1/2text(cosec)(alpha)`
  4. `1/2cos(alpha)`
  5. `1/2sec(alpha)`
Show Answers Only

`E`

Show Worked Solution

`text(Using Lami’s theorem:)`

`(F_1)/(sin alpha)` `= (F_2)/(sinbeta)`
`(F_1)/(F_2)` `= (sin alpha)/(sin beta)`
  `= (sin alpha)/(sin 2alpha)`
  `= (sin alpha)/(2sin alphacos alpha)`
  `= 1/2 sec alpha`

 
`=>E`

Calculus, SPEC2 2019 VCAA 16 MC

A variable force acts on a particle, causing it to move in a straight line. At time `t` seconds, where  `t >= 0`, its velocity `v` metres per second and position `x` metres from the origin are such that  `v = e^x sin(x)`.

The acceleration of the particle, in ms−2, can be expressed as

  1. `e^(2x)(sin^2(x) + 1/2sin(2x))`
  2. `e^x sin(x)(sin(x) + cos(x))`
  3. `e^x(sin(x) + cos(x))`
  4. `1/2 e^(2x) sin^2(x)`
  5. `e^x cos(x)`
Show Answers Only

`A`

Show Worked Solution

`v = e^x sin(x)`

`(dv)/(dx) = e^x cos(x) + e^x sin(x)`

`v · (dv)/(dx)` `= e^x  sin(x)(e^x  cos(x) + e^x  sin(x))`
  `= e^(2x)(sin(x)cos(x) + sin^2(x))`
  `= e^(2x)(1/2 xx 2sin(x)cos(x) + sin^2(x))`
  `= e^(2x)(sin^2(x) + 1/2sin(2x))`

 
`=>A`

Mechanics, SPEC2 2019 VCAA 14 MC

A 4 kg mass is held at rest on a smooth surface. It is connected by a light inextensible string that passes over a smooth pulley to a 2 kg mass, which in turn is connected by the same type of string to a 1 kg mass. This is shown in the diagram below.
 


 

When the 4 kg mass is released, the tension in the string connecting the 1 kg and 2 kg masses is `T` newtons. The value of `T` is

  1. `(4g)/7`
  2. `(3g)/7`
  3. `g/7`
  4. `(6g)/7`
  5. `g`
Show Answers Only

`A`

Show Worked Solution

`text(Considering the whole system:)`

`text(Total mass = 7 kg)`

`text(Net Force ↓ =)\ 3g`

`F` `= ma`
`3g` `= 7a`
`:.a` `= (3g)/7`

 
`text(Consider the forces on the 1 kg mass:)`

`g – T` `= a`
`g – T` `= (3g)/7`
`T` `= g – (3g)/7`
  `= (4g)/7`

 
`=>A`

Calculus, MET2 2019 VCAA 5

Let  `f: R -> R, \ f(x) = 1-x^3`. The tangent to the graph of `f` at  `x = a`, where  `0 < a < 1`, intersects the graph of `f` again at `P` and intersects the horizontal axis at `Q`. The shaded regions shown in the diagram below are bounded by the graph of `f`, its tangent at  `x = a`  and the horizontal axis.
 

  1. Find the equation of the tangent to the graph of `f` at  `x = a`, in terms of `a`.   (1 mark)

  2. Find the `x`-coordinate of `Q`, in terms of `a`.   (1 mark)

  3. Find the `x`-coordinate of `P`, in terms of `a`.   (2 marks)

Let  `A`  be the function that determines the total area of the shaded regions.

  1. Find the rule of `A`, in terms of `a`.   (3 marks)

  2. Find the value of `a` for which `A` is a minimum.   (2 marks)

Consider the regions bounded by the graph of `f^(-1)`, the tangent to the graph of `f^(-1)` at  `x = b`, where  `0 < b < 1`, and the vertical axis.

  1. Find the value of `b` for which the total area of these regions is a minimum.   (2 marks)

  2. Find the value of the acute angle between the tangent to the graph of `f` and the tangent to the graph of `f^(-1)` at  `x = 1`.   (1 mark)

Show Answers Only
  1. `y = 2a^3-3a^2x + 1`
  2. `(2a^3 + 1)/(3a^2)`
  3. `-2a`
  4. `(80a^6 + 8a^3-9a^2 + 2)/(12a^2)`
  5. `10^(-1/3)`
  6. `9/10`
  7. `tan^(-1)(1/3)`
Show Worked Solution

a.   `f(x) = 1-x^3,\ \ f^{\prime}(x) = -3x^2`

`m_text(tang) = -3a^2\ \ text(through)\ \ (a, 1-a^3)`

`y = 2a^3-3a^2x + 1`
 

b.   `text(Solve for)\ \x:`

`2a^3-3a^2x + 1 = 0`

`x = (2a^3 + 1)/(3a^2)`
 

c.   `text(Solve for)\ \ x:`

`2a^3-3a^2x + 1 = 1-x^3`

`x=-2a`

`:. x text(-coordinate of)\ P = -2a`
 

d.   `P(-2a, 8a^3 + 1),\ \ Q((2a^3 + 1)/(3a^2), 0)`

`A` `= text(Area of triangle)-int_(-2a)^1 f(x)\ dx`
  `= 1/2((2a^3 + 1)/(3a^2) + 2a)(8a^3 + 1)-int_(-2a)^1 1-x^3\ dx`
  `= (80a^6 + 8a^3-9a^2 + 2)/(12a^2)\ \ text{(by CAS)}`

 

e.   `text(Solve for)\ a: \ (dA)/(da) = 0\ \ text{(by CAS)}`

`a = 10^(-1/3)`
 

f.   `text(Consider)\ f(x):\ \ f(10^(-1/3)) = 9/10`

`A_text(min)\ text(for)\ \ f(x)\ \ text(occurs at)\ \ (10^(-1/3), 9/10)`

`=> A_text(min)\ text(for)\ \ f^(-1)(x)\ \ text(occurs when)\ \ x=9/10`

`:. b = 9/10`
 

g.   `f^{\prime} (1)  = -3`

`text(Gradient of)\ \  f^(-1)(x)\ \ text(at)\ \ x = 1\ \ text(is vertical line.)`

`tan theta` `= 1/3`
`theta` `= tan^(-1)(1/3)`
  `=18.4°\ \ text{(to 1 d.p.)}`

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, EXT1 C3 2019 SPEC2 9 MC

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

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

`B`

Show Worked Solution

`text(By elimination:)`

`text(Along line)\ y = x,\ text(gradient = 1)`

`:.\ text(Eliminate A and D.)`
 

`text{At (1, 0),  0 < gradient < 1}`

`1/(cos(-1)) > 1`

`:.\ text(Eliminate C)`

`=>B`

Calculus, SPEC2 2019 VCAA 9 MC

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

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

`B`

Show Worked Solution

`text(Along line)\ y = x,\ text(gradient = 1)`

`:.\ text(Eliminate A, C, and E.)`
 

`text{At (1, 0),  0 < gradient < 1}`

`1/(cos(-1)) > 1`

`:.\ text(Eliminate D)`

`=>B`

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)`

Calculus, MET2 2019 VCAA 2

An amusement park is planning to build a zip-line above a hill on its property.

The hill is modelled by  `y = (3x(x-30)^2)/2000,  x in [0, 30]`, where  `x`  is the horizontal distance, in metres, from an origin and  `y`  is the height, in metres, above this origin, as shown in the graph below.
  

  1. Find  `(dy)/(dx)`.   (1 mark)

  2. State the set of values for which the gradient of the hill is strictly decreasing.   (1 mark)

The cable for the zip-line is connected to a pole at the origin at a height of 10 m and is straight for  `0 <= x <= a`, where  `10 <= a <= 20`. The straight section joins the curved section at  `A(a, b)`. The cable is then exactly 3 m vertically above the hill from  `a <= x <= 30`, as shown in the graph below.

  1. State the rule, in terms of `x`, for the height of the cable above the horizontal axis for  `x in [a, 30]`.   (1 mark)

  2. Find the values of `x` for which the gradient of the cable is equal to the average gradient of the hill for  `x in [10, 30]`.   (3 marks)

The gradients of the straight and curved sections of the cable approach the same value at  `x = a`, so there is a continuous and smooth join at `A`.

    1. State the gradient of the cable at `A`, in terms of `a`.   (1 mark)

    2. Find the coordinates of `A`, with each value correct to two decimal places.   (3 marks)

    3. Find the value of the gradient at `A`, correct to one decimal place.   (1 mark)

Show Answers Only
  1. `(9(x-10)(x-30))/2000`
  2. `x in [0, 20]`
  3. `h(x) = 3 + (3x(x-30)^2)/2000`
  4. `x = (60 +- 10 sqrt 3)/3`
    1. `(9(a-10)(a-30))/2000`
    2. `A(11.12, 8.95)`
    3. `-0.1`
Show Worked Solution

a.   `d/(dx)((3x(x-30)^2)/2000) = (9(x-10)(x-30))/2000\ \ text{(by CAS)}`
 

b.   `text(Sketch)\ \ dy/dx\ \ text{(by CAS)}:`

`(dy)/(dx)\ \ text(strictly decreasing for)\ \ x in [0, 20]`
 

c.   `text(Cable height is a vertical shift of +3 of)\ \ y:`

`h(x) = 3 + (3x(x-30)^2)/2000\ \ text(for)\ \ x in [a, 30]`
 

d.   `text(Let)\ \ f(x) = (9(x-10)(x-30))/2000`

`text(Average gradient) = (f(30)-f(10))/(30-10) = -3/10\ \ text{(by CAS)}`

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

`(9(x-10)(x-30))/2000 = -3/10`

`x = (60 +- 10 sqrt 3)/3`
 

e.i.   `text(S)text(ince)\ \ h(x)\ \ text(is a vertical shift of)\ \ f(x),`

`=> h^{′}(a) = f^{′}(a)`

`h^{′}(a) = (9(a-10)(a-30))/2000`
 

e.ii.   `h(a) = (3a(a-30)^2)/2000 + 3`

`:. A(a, (3a(a-30)^2)/2000 + 3),\ B(0, 10)`
 

`m_(AB) = ((3a(a-30)^2)/2000 + 3-10)/a = (3a(a-30)^2-14000)/(2000 a)`

 
`text(Equating gradients:)`

`text(Solve): \ f^{′}(a) = m_(AB)\ \ text{(by CAS)}`

`(9(a-10)(a-30))/2000 = (3a(a-30)^2-14000)/(2000 a)`

`a ~~ 11.12, quad b ~~ 8.95`

`:. A(11.12, 8.95)`

 

e.iii.    `M_A` `= h^{′}(11.12)`
    `~~ -0.1\ text{(by CAS)}`

Calculus, EXT1 C2 SM-Bank 1 MC

With a suitable substitution, `int_1^5(2x - 1)sqrt(2x + 1)\ dx` can be expressed as

  1. `1/2 int_1^5 (u^(3/2) + u^(1/2))\ du`
  2. `2 int_3^11 (u^(3/2) + u^(1/2))\ du`
  3. `2 int_1^5 (u^(3/2) - 2u^(1/2))\ du`
  4. `1/2 int_3^11 (u^(3/2) - 2u^(1/2))\ du`
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ u = 2x + 1 \ => \ u – 2 = 2x – 1`

`(du)/(dx) = 2 \ => \ dx = 1/2 du`

`text(When)\ \ x = 5, \ u = 11`

`text(When)\ \ x = 1, \ u = 3`

`:. int_1^5 (2x – 1)sqrt(2x + 1)\ dx`

`= 1/2 int_3^11 (u – 2)u^(1/2)\ du`

`= 1/2 int_3^11 u^(3/2) – 2u^(1/2)\ du`

 
`=>D`

Calculus, SPEC2 2019 VCAA 8 MC

With a suitable substitution, `int_1^5(2x - 1)sqrt(2x + 1)\ dx` can be expressed as

  1. `1/2 int_1^5 (u^(3/2) + u^(1/2))\ du`
  2. `2 int_3^11 (u^(3/2) + u^(1/2))\ du`
  3. `2 int_1^5 (u^(3/2) - 2u^(1/2))\ du`
  4. `2 int_3^11 (u^(3/2) - 2u^(1/2))\ du`
  5. `1/2 int_3^11 (u^(3/2) - 2u^(1/2))\ du`
Show Answers Only

`E`

Show Worked Solution

`text(Let)\ \ u = 2x + 1 \ => \ u – 2 = 2x – 1`

`(du)/(dx) = 2 \ => \ dx = 1/2 du`

`text(When)\ \ x = 5, \ u = 11`

`text(When)\ \ x = 1, \ u = 3`

`:. int_1^5 (2x – 1)sqrt(2x + 1)\ dx`

`= 1/2 int_3^11 (u – 2)u^(1/2)\ du`

`= 1/2 int_3^11 u^(3/2) – 2u^(1/2)\ du`

 
`=>E`

Calculus, SPEC2 2019 VCAA 7 MC

The length of the curve defined by the parametric equations  `x = 3sin(t)`  and  `y = 4cos(t)`  for  `0 <= t <= pi`  is given by

  1. `int_0^pi sqrt(9cos^2(t) - 16sin^2(t))\ dt`
  2. `int_0^pi sqrt(9 + 7sin^2(t))\ dt`
  3. `int_0^pi sqrt(1 + 16sin^2(t))\ dt`
  4. `int_0^pi (3cos(t) - 4sin(t))\ dt`
  5. `int_0^pi sqrt(3cos^2(t) + 4sin^2(t))\ dt`
Show Answers Only

`B`

Show Worked Solution

`x = 3sin(t) \ => \ (dx)/(dt) = 3cos(t)`

`y = 4cos(t) \ => \ (dy)/(dt) = −4sin(t)`

`text(Length)` `= int_0^pi sqrt((3cos(t))^2 + (−4sin(t))^2)\ dt`
  `= int_0^pi sqrt(9cos^2(t) + 16sin^2(t))\ dt`
  `= int_0^pi sqrt(9cos^2(t) + 9sin^2(t) + 7sin^2(t))\ dt`
  `= int_0^pi sqrt(9 + 7sin^2(t))\ dt`

 
`=>B`

Complex Numbers, SPEC2 2019 VCAA 6 MC

Let  `z, w ∈ C`, where  `text(Arg)(z) = pi/2`  and  `text(Arg)(w) = pi/4`.

The value of  `text(Arg)((z^5)/(w^4))`  is

  1. `−pi/2`
  2. `pi/2`
  3. `pi`
  4. `(5pi)/2`
  5. `(7pi)/2`
Show Answers Only

`A`

Show Worked Solution
`text(Arg)((z^5)/(w^4))` `= text(Arg)(z^5) – text(Arg)(w^4)`
  `= 5text(Arg)(z) – 4text(Arg)(w)`
  `= (5pi)/2 – (4pi)/4`
  `= (3pi)/2`

 
`:. text(Arg)((z^5)/(w^4)) = −pi/2`

`=>A`

Complex Numbers, SPEC2 2019 VCAA 5 MC

Let  `z = x + yi`, where  `x, y ∈ R`. The rays  `text(Arg)(z - 2)=pi/4`  and  `text(Arg)(z - (5 + i)) = (5pi)/6`, where  `z ∈ C`, intersect on the complex plane at a  point  `(a,b)`.

The value of `b` is

  1. `−sqrt3`
  2. `2 - sqrt3`
  3. `0`
  4. `sqrt3`
  5. `2 + sqrt3`
Show Answers Only

`D`

Show Worked Solution

`text(Convert to cartesian equations)`

`text(Arg)(z – 2) = pi/4`

`y = x – 2\ …\ (1)`

`text(Arg)(z – (5 + i)) = (5pi)/6`

`y – 1 = −1/sqrt3 (x – 5)\ …\ (2)`

`text{Solve (1) and (2) simultaneously:}`

`text(Intersection)\ (a,b) :\ \ x = 2+sqrt3, \ y = sqrt3`

`:. b = sqrt3`

`=>D`

Complex Numbers, EXT2 N1 SM-Bank 7

Calculate  `i^(1!) + i^(2!) + i^(3!) + …+ i^(100!)`

giving your answer in the form  `x+iy`  (3 marks)

Show Answers Only

`95+i`

Show Worked Solution

`i^(1!) =i^1 =  i`

`i^(2!) =i^2= −1`

`i^(3!) = i^6=-1`

`i^(4!) = i^24 = 1`

`i^(5!) = i^(24 xx 5)= 1^5 = 1`

`=> i^(n!) = 1\ \ text(for)\ \ n >= 4`

`:.\ text(Sum)` `= i – 1 – 1 + 97`
  `= 95 + i`

Graphs, SPEC2 2019 VCAA 3 MC

The implied domain of the function with rule  `f(x) = 1 - sec(x + pi/4)`  is

  1. `R`
  2. `[0,2]`
  3. `R\ text(\)\ {((4n - 1)pi)/4}, n ∈ ZZ`
  4. `R\ text(\)\ {((4n + 1)pi)/4}, n ∈ ZZ`
  5. `R\ text(\)\ {((2n - 1)pi)/4}, n ∈ ZZ`
Show Answers Only

`D`

Show Worked Solution

`y = sec (x)=1/cos(x)\ \ text(has asymptotes when)`

`x = −pi/2, pi/2, (3pi)/2, …`

`=> y = sec (x + pi/4)\ \ text(has asymptotes at when)`

`x = (−3pi)/4, pi/4, (5pi)/4, …`

`:.\ text(Domain:)\ R\ text(\)\ {((4n + 1)pi)/4}, n ∈ ZZ`

`=>D`

Mechanics, SPEC1 2019 VCAA 9

  1. A light inextensible string is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs in equilibrium from a smooth ring on the string, as shown in the diagram below. The string makes an angle `alpha` with the ceiling.
      
    `qquad qquad`
     
    Express the tension, `T` newtons, in the string in terms of `m`, `g` and `alpha`.  (1 mark)
  2. A different light inextensible sting is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs from a smooth ring on the string. A horizontal force of `F` newtons is applied to the ring. The tension in the sting has a constant magnitude and the system is in equilibrium. At one end the string makes an angle `beta` with the ceiling and at the other end the string makes an angle `2beta` with the ceiling, as shown in the diagram below.
     

     
    Show that  `F = mg((1 - cos(beta))/(sin(beta)))`.  (3 marks)
Show Answers Only
  1. `T = (mg)/(2sinalpha)`
  2. `text(See Worked Solutions)`
Show Worked Solution
a.   
`2 xx Tsinalpha` `= mg`
`:.T` `= (mg)/(2sinalpha)`

 

b.   

 
`text(Resolving forces vertically:)`

`Tsin(beta) + Tsin(2beta)` `= mg`
`T` `= (mg)/(sin(beta) + sin(2beta))`

 
`text(Resolving forces horizontally:)`

`F + Tcos(2beta)` `= Tcos(beta)`
`F` `= Tcos(beta) – Tcos(2beta)`
  `= T(cos(beta) – cos(2beta))`
  `= T[cos(beta) – (2cos^2beta – 1)]`
  `= T(−2cos^2(beta) + cos(beta) + 1)`
  `= T(−2cos(beta) – 1)(cos(beta) – 1)`
  `= (mg(1 -cos(beta))(2cosbeta + 1))/(sin(beta) + 2sin(beta)cos(beta))`
  `= (mg(1 – cos(beta))(2cos(beta) + 1))/(sin(beta)(1+2cos(beta)))`
  `= mg((1 – cos(beta))/(sin(beta)))`

Calculus, EXT1 C3 SM-Bank 3

Find the volume of the solid of revolution formed when the graph of  `y = sqrt((1 + 2x)/(1 + x^2))`  is rotated about the `x`-axis over the interval  `[0,1]`.  (3 marks)

Show Answers Only

`pi(pi/4 + ln2)\ \ text(u³)`

Show Worked Solution
`V` `= pi int_0^1 (1 + 2x)/(1 + x^2)\ dx`
  `= pi int_0^1 1/(1 + x^2)\ dx + pi int_0^1 (2x)/(1 + x^2)\ dx`
  `= pi [tan^(−1)(x)]_0^1 + pi [ln(1 + x^2)]_0^1`
  `= pi(tan^(−1)1 – tan^(−1)0) + pi(ln2 – ln1)`
  `= pi(pi/4) + pi(ln2)`
  `= pi(pi/4 + ln2)\ \ text(u³)`

Calculus, SPEC1 2019 VCAA 8

Find the volume of the solid of revolution formed when the graph of  `y = sqrt((1 + 2x)/(1 + x^2))`  is rotated about the `x`-axis over the interval  `[0,1]`.  (4 marks)

Show Answers Only

`pi(pi/4 + ln2)\ \ text(u³)`

Show Worked Solution
`V` `= pi int_0^1 (1 + 2x)/(1 + x^2)\ dx`
  `= pi int_0^1 1/(1 + x^2)\ dx + pi int_0^1 (2x)/(1 + x^2)\ dx`
  `= pi [tan^(−1)(x)]_0^1 + pi [ln(1 + x^2)]_0^1`
  `= pi(tan^(−1)1 – tan^(−1)0) + pi(ln2 – ln1)`
  `= pi(pi/4) + pi(ln2)`
  `= pi(pi/4 + ln2)\ \ text(u³)`

Graphs, SPEC2 2019 VCAA 2 MC

The asymptote(s) of the graph of  `f(x) = (x^2 + 1)/(2x - 8)`  has equation(s)

  1. `x = 4`
  2. `x = 4 and y = x/2`
  3. `x = 4 and y = x/2 + 2`
  4. `x = 8 and y = x/2`
  5. `x = 8 and y = 2x + 2`
Show Answers Only

`C`

Show Worked Solution
`(x^2 + 1)/(2x – 8)` `= ((x – 4)(x + 4) + 17)/(2(x – 4))`
  `= (x + 4)/2 + 17/(2x – 8)`
  `= x/2 + 2 + 17/(2x – 8)`

 
`:.\ text(Asymptotes at)\ \ x = 4 and y = x/2 + 2`

`=>C`

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)`

Vectors, SPEC1 2019 VCAA 6

Find value of  `d`  for which the vectors  `underset~a = 2underset~i - 3underset~j + 4underset~k`,  `underset~b = −2underset~i + 4underset~j - 8underset~k`  and  `underset~c = −6underset~i + 2underset~j + dunderset~k`  are linearly dependent.  (3 marks)

Show Answers Only

`16`

Show Worked Solution

`text(Linear dependent:)\ \ m underset~a + n underset~b = underset~c`

`2m – 2n` `= −6\ \ …\ (1)`
`−3m + 4n` `= −2\ \ …\ (2)`
`4m – 8n` `= d\ \ …\ (3)`

 
`(1) xx 2`

`4m – 4n = −12\ \ …\ (1′)`

`(2) + (1′)`

`m = −10`

`text(Substitute)\ \ m = −10\ \ text(into (1))`

`−20 – 2n` `= −6`
`n` `= −7`

 

`:.d` `= 4 xx −10 – 8 xx −7`
  `= 16`

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)}`

Vectors, SPEC1 2019 VCAA 4

The position vectors of two particles  `A`  and  `B`  at time  `t` seconds after they have started moving are given by  `underset~r_A(t) = (t^2-1)underset~i + (a + t/3)underset~j`  and  `underset~r_B(t) = (t^3-t)underset~i + (text(arccos)(t/2))underset~j`  respectively, where  `a`  is a real constant and  `0 <= t <= 2`.

Find the value of  `a`  if the particles collide after they have started moving.   (3 marks)

Show Answers Only

`(pi-1)/3`

Show Worked Solution

`text(Collision occurs when)\ underset~i\ text(and)\ underset~j\ text(components are equal).`

`text(Equating)\ underset~i\ text(components:)`

`t^2-1` `= t^3-t`
`t^3-t^2-t + 1` `= 0`
`t^2(t-1)-(t-1)` `= 0`
`(t^2-1)(t-1)` `= 0`

 
`:. t = 1\ \ (t > 0)`
 

`text(Equating)\ underset~j\ text{components (when}\ t = 1text{):}`

`a + 1/3` `= cos^(−1)(1/2)`
`:. a` `= (pi-1)/3`

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 F1 SM-Bank 35

  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
  1.   
  2.   
Show Worked Solution

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

 

ii.   `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).`

 

Calculus, SPEC1 2019 VCAA 1

Solve the differential equation  `(dy)/(dx) = (2ye^(2x))/(1 + e^(2x))`  given that  `y(0) = pi`.  (4 marks)

Show Answers Only

`y = (pi(1 + e^(2x)))/2`

Show Worked Solution
`int 1/y\ dy` `= (2e^(2x))/(1 + e^(2x))\ dx`
`log_e |y|` `= log_e |1 + e^(2x)| + c`

 
`text(When)\ \ x=0, \ y= pi:`

`log_e pi` `= log_e |1 + e^0| + c`
`c` `= log_e pi – log_e 2`
  `= log_e\ pi/2`
`log_e |y|` `= log_e(1 + e^(2x)) + log_e\ pi/2`
`log_e |y|` `= log_e\ (pi(1 + e^(2x)))/2`
`:. y` `= (pi(1 + e^(2x)))/2`

Functions, 2ADV F1 SM-Bank 36

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

i.

ii.   

Show Worked Solution

i.   `text(Sketch)\ \ y = 1/(x + 2)`

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

 

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

Calculus, MET2 2019 VCAA 16 MC

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

The corresponding part of the graph of  `y = f^{\prime}(x)`  is best represented by

A.    B.   
C.    D.   
E.       
Show Answers Only

`A`

Show Worked Solution

`text(By Elimination):`

`text(S.P. at)\ \ x = 5`

`=> f^{\prime}(x) = 0\ \ text(when)\ \ x = 0`

`text(Eliminate  C, E)`

`text(Gradient is negative when)\ \ x < 5,\ \ text(and positive when)\ \ x >5`

`text(Eliminate  B, D)`
 

`=>   A`