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PHYSICS, M6 2019 HSC 24

A step-up transformer is constructed using a solid iron core. The coils are made using copper wires of different thicknesses as shown.
 

The table shows electrical data for this transformer.
 

  1. Explain how the operation of this transformer remains consistent with the law of conservation of energy. Include a relevant calculation in your answer.   (3 marks)
  1. Explain how TWO modifications to this transformer would improve its efficiency.   (4 marks)
Show Answers Only

a.    The energy input into this transformer is 500 J `text{s}^(-1)`

The energy output is given by:

`P` `=V_(s)I_(s)`  
  `=50xx 9`  
  `=450 ` J `text{s}^(-1)`  

 
This is consistent with the law of conservation of energy as 500 J `text{s}^(-1)` of energy is converted into other energy forms such as heat.
   

b.    → Laminating the iron core prevents large eddy currents from being induced in it.

→ This reduces energy loss in the form of heat, increasing efficiency.

→ Increasing the thickness of the wire in the primary coil reduces its resistance, increasing the transformer’s efficiency.

Show Worked Solution

a.   The energy input into this transformer is  500  J `text{s}^(-1)`

The energy output is given by:

`P` `=V_(s)I_(s)`  
  `=50xx 9`  
  `=450 ` J `text{s}^(-1)`  

 
This is consistent with the law of conservation of energy as 500 J `text{s}^(-1)` of energy is converted into other energy forms such as heat.
   

b.     Modification 1:

→ Laminating the iron core prevents large eddy currents from being induced in it.

→ This reduces energy loss in the form of heat, increasing efficiency.

Modification 2:

→ Increasing the thickness of the wire in the primary coil.

→ This reduces its resistance, increasing the transformer’s efficiency.


♦ Mean mark part 48%.

PHYSICS, M7 2019 HSC 2 MC

Two stars were observed from Earth. Their spectra are shown with the wavelength in nanometres.
 

Using these spectra, what can be concluded about the motion of the stars relative to Earth and their chemical compositions?
 

Show Answers Only

`B`

Show Worked Solution

Both stars have the same set of spectral lines → Chemical composition is the same.

Spectral lines at different wavelengths → Different doppler shift → Different motion relative to the earth.

`=>B`

PHYSICS, M7 2020 HSC 30b

  1. Calculate the wavelength of a proton travelling at 0.1\(c\).   (2 marks)
  1. Explain the relativistic effect on the wavelength of a proton travelling at 0.95\(c\).   (2 marks)
Show Answers Only

i.   `lambda=1xx10^(-14)` m

ii.  See Worked Solutions

Show Worked Solution
i.     `lambda` `=(h)/(mv)`
    `=(6.626 xx10^(-34))/(1.673 xx10^(-27)xx0.1 xx3xx10^(8))`
    `=1xx10^(-14)\ text{m}`

   

ii.    Due to momentum dilation:

→ The momentum of the proton is greater than that predicted by classical mechanics. 

→ Since  `lambda=(h)/(mv)\ \ =>\ \ lambda prop 1/(mv)`

→ The wavelength of the proton is shorter than would be predicted by classical mechanics.


♦ Mean mark (ii) 45%.

PHYSICS, M6 2020 HSC 23

The graph shows data for a motor connected to a 240 V power supply.
 

The equation for the torque, `tau`, produced by the motor is  `tau=frac{V I \eta}{\omega}`

where    `tau\ ` `=` torque (N m)
  `V \` `=`  voltage (V)
  `I\ ` `=` current (A)
  `eta\ ` `=` efficiency `=` 0.3
  `omega\ ` `=` angular velocity (rad `text{s}^(-1)`)

 
A circuit breaker cuts the current to the motor if the current exceeds 5  A.

Determine what will happen when the motor produces a torque of 2.95 N m. Show relevant calculations.   (3 marks)

Show Answers Only

`I` = 4.1 A

Current is below 5 A, so the circuit breaker will not cut the current and the motor will continue to run.

Show Worked Solution

From the graph, when `tau =` 2.95, `omega ~~ 100`

`tau` `=(VI eta)/(omega)`  
`I` `=tau(omega)/(Veta)`  
  `=2.95 xx(100)/(240 xx0.3)`  
  `=4.1` A  

 Current is below 5 A, so the circuit breaker will not cut the current and the motor will continue to run.

PHYSICS, M8 2020 HSC 8 MC

A uranium isotope, `U`, undergoes four successive decays to produce `Q`.
 

Which row of the table correctly shows the decay process `R` and product `Q` ?
 

Show Answers Only

`C`

Show Worked Solution

Decay process `R`  reduces the number of both protons and neutrons by two → alpha decay.

Thorium has atomic number 90.

`=>C`

PHYSICS, M8 2020 HSC 4 MC

The graph shows the mass of a radioactive isotope as a function of time.
 

What is the decay constant, in `text{years}^(-1)`, for this isotope?

  1. 0.0030
  2. 0.0069
  3. 2.0
  4. 100
Show Answers Only

`B`

Show Worked Solution
`t_((1)/(2))` `~~100` years  
`lambda` `=(ln 2)/(t_((1)/(2)))`  
  `=(ln 2)/(100)`  
  `=0.0069  text{years}^(-1)`  

 
`=>B`

PHYSICS, M7 2020 HSC 3 MC

What was the basis for Maxwell's prediction of the velocity of electromagnetic waves?

  1. Experiments using magnetic fields to accelerate particles
  2. Experiments using light and mirrors to establish the finite speed of light
  3. Equations showing how oscillating electric and magnetic fields propagate
  4. Equations showing how electromagnetic waves are affected by gravitational fields
Show Answers Only

`C`

Show Worked Solution

Maxwell used equations to show interactions between electric and magnetic fields.

`=>C`

CHEMISTRY, M7 2021 HSC 4 MC

The structure of ethyl pentanoate is shown.
 

Which pair of chemicals would produce ethyl pentanoate by esterification?

  1. Ethene and pentan-1-ol
  2. Ethane and pentanoic acid
  3. Ethanol and pentanoic acid
  4. Ethanoic acid and pentan-1-ol
Show Answers Only

`C`

Show Worked Solution

→ Esterification involves a reaction between a carboxylic aciand an alcohol.

→ Creating ethyl pentanoate requires ethanol (which is an alcohol) and pentanoic acid (which is a carboxylic acid).

→ Ethanol + pentanoic acid  ↔ ethyl pentanoate + water

`=> C`

PHYSICS, M7 2021 HSC 26

A student performs an experiment to measure Planck's constant, `h`, using a device that emits specific frequencies of light when specific voltages are applied.

The voltage, `V`, needed to produce each frequency, `f`, is given by

`V=(hf)/q_e`

where `q_e` is the charge on an electron.

Data from the experiment is shown.
 

  1. Graph this data on the axes provided. Include a line of best fit.   (3 marks)
     
     

       
  2. The student proposes using data point 1 to calculate a value for Planck's constant.
  3. Justify a better method to calculate Planck's constant from the experimental data provided.   (3 marks)
Show Answers Only
  1.  
       
     
  2. The gradient of the line of best fit should be determined using the graph ` V=(hf)/(q_(e))`.
  3. The gradient is equal to `(V)/(f)`, and ` h` can be calculated as `h=`gradient`xxq_(e)`.
  4. This method increases the accuracy by taking more data points into account.
Show Worked Solution
  1.  
       
     
  2. The gradient of the line of best fit should be determined using the graph ` V=(hf)/(q_(e))`.
    The gradient is equal to `(V)/(f)`, and ` h` can be calculated as `h=`gradient`xxq_(e)`.
    This method increases the accuracy by taking more data points into account.

 

PHYSICS, M6 2021 HSC 21

A DC motor is constructed from a single loop of wire with dimensions 0.10 m × 0.07 m. The magnetic field strength is 0.40 T and a current of 14 A flows through the loop.
 

  1. Calculate the magnitude of the maximum torque produced by the motor.   (2 marks)
  1. Describe how the magnitude of the torque changes as the loop moves through half a rotation from the position shown.   (2 marks)
Show Answers Only

a.   0.04  Nm

b.   The torque is initially at a maximum and decreases to zero after a 90º rotation. The torque then increases to a maximum at 180º from the coil’s original position.

Show Worked Solution
a.    `tau_max` `=nIAB sin theta`
    `=1xx14 xx0.1 xx0.07 xx0.40 xx sin(90^(@))`
    `=0.04`  Nm

 

b.  The torque is initially at a maximum and decreases to zero after a 90º rotation. The torque then increases to a maximum at 180º from the coil’s original position.

BIOLOGY, M8 2021 HSC 4 MC

Which is the most effective strategy for treating non-infectious diseases?

  1. Hygiene
  2. Pharmaceuticals
  3. Quarantine
  4. Vaccination
Show Answers Only

`B`

Show Worked Solution

→ Pharmaceuticals is the only option that can be used against non-infectious disease.

→ All other options treat infectious diseases.

`=>B`

BIOLOGY, M6 2021 HSC 3 MC

A scientist transferred male gametes from one plant to another to achieve a desired characteristic in the offspring.

Which genetic technology was the scientist using?

  1. Gene cloning
  2. Artificial pollination
  3. Artificial insemination
  4. Whole organism cloning
Show Answers Only

`B`

Show Worked Solution

The statement describes the manual transferral of pollen from the anther of one plant to the stigma of another which is known as artificial pollination.

`=>B`

BIOLOGY, M5 2021 HSC 2 MC

Which of the following photographs shows an example of sexual reproduction?
 

Show Answers Only

`D`

Show Worked Solution

By Elimination

→ Budding in fungi, binary fission in bacteria and runners in plants are all examples of asexual reproduction (Eliminate A, B and C).

→ Frog spawning requires a female to lay eggs and a male frog to fertilise them with sperm.

`=>D`

Calculus, MET2 2020 VCAA 4

The graph of the function  `f(x)=2xe^((1-x^(2)))`, where  `0 <= x <= 3`, is shown below.
 

  1. Find the slope of the tangent to `f` at  `x=1`.   (1 mark)
  2. Find the obtuse angle that the tangent to `f` at  `x = 1`  makes with the positive direction of the horizontal axis. Give your answer correct to the nearest degree.   (1 mark)
  3. Find the slope of the tangent to `f` at a point  `x =p`. Give your answer in terms of  `p`.   (1 mark)
  4.  i. Find the value of `p` for which the tangent to `f` at  `x=1` and the tangent to `f` at  `x=p`  are perpendicular to each other. Give your answer correct to three decimal places.   (2 marks)
  5. ii. Hence, find the coordinates of the point where the tangents to the graph of `f` at  `x=1`  and  `x=p`  intersect when they are perpendicular. Give your answer correct to two decimal places.   (3 marks)

Two line segments connect the points `(0, f(0))`  and  `(3, f(3))`  to a single point  `Q(n, f(n))`, where  `1 < n < 3`, as shown in the graph below.
 
         
 

  1.   i. The first line segment connects the point `(0, f(0))` and the point `Q(n, f(n))`, where `1 < n < 3`.
  2.      Find the equation of this line segment in terms of  `n`.   (1 mark)
  3.  ii. The second line segment connects the point `Q(n, f(n))` and the point  `(3, f(3))`, where  `1 < n < 3`.
  4.      Find the equation of this line segment in terms of `n`.   (1 mark)
  5. iii. Find the value of `n`, where  `1 < n < 3`, if there are equal areas between the function `f` and each line segment.
  6.      Give your answer correct to three decimal places.   (3 marks)
Show Answers Only
  1. `-2`
  2. `117^@`
  3. `2(1-2p^(2))e^(1-p^(2))” or “(2e-4p^(2)e)e^(-p^(2))`
  4.  i. `0.655`
  5. ii. `(0.80, 2.39)`
  6.   i. `y_1=2e^((1-n^2))x`
  7.  ii. `y_2=(2n e^((1-n^2)) – 6e^(-8))/(n-3) (x-3) + 6e^(-8)`
  8. iii. `n= 1.088`
Show Worked Solution

a.   `f(x)=2xe^((1-x^(2)))`

`f^{′}(1)=-2`

♦ Mean mark part (b) 37%.

 

b.   `text{Solve:}\ tan theta = -2\ \ text{for}\ \ theta in (pi/2, pi)`

`theta = 117^@`
 

c.   `text{Slope of tangent}\ = f^{′}(p)`

`f^{′}(p)=2(1-2p^(2))e^(1-p^(2))\ \ text{or}\ \ (2e-4p^(2)e)e^(-p^(2))`
 

d.i.   `text{If tangents are perpendicular:}`

`f^{′}(p) xx -2 =-1\ \ =>\ \ f^^{′}(p)=1/2`

`text{Solve}\ \ 2(1-2p^(2))e^(1-p^(2))=1/2\ \ text{for}\ p:`

`p=0.655\ \ text{(to 3 d.p.)}`
 

d.ii.  `text{Equation of tangent at}\ \ x=1: \ y=4-2x`

♦ Mean mark part (d)(ii) 41%.

`text{Equation of tangent at}\ \ x=p: \ y= x/2 + 1.991…`

`text{Solve}\ \ 4-2x = x/2 + 1.991…\ \ text{for}\ x:`

`=> x = 0.8035…`

`=> y=4-2(0.8035…) = 2.392…`

`:.\ text{T}text{angents intersect at (0.80, 2.39)}`

♦ Mean mark part (e)(i) 44%.

 
e.i.  `Q (n, 2n e^(1-n^2))`

`m_(OQ) = (2n e^((1-n^2)) – 0)/(n-0) = 2e^((1-n^2))`

`:.\ text{Equation of segment:}\ \ y_1=2e^((1-n^2))x`

♦♦ Mean mark part (e)(ii) 28%.
 

e.ii.  `P(3, f(3)) = (3, 6e^(-8))`

`m_(PQ) = (2n e^((1-n^2)) – 6e^(-8))/(n-3)`

`text{Equation of line segment:}`

`y_2-6e^(-8)` `=(2n e^((1-n^2)) – 6e^(-8))/(n-3) (x-3)`  
`y_2` `=(2n e^((1-n^2)) – 6e^(-8))/(n-3) (x-3) + 6e^(-8)`  
♦♦ Mean mark part (e)(iii) 28%.

 

e.iii.  `text{Find}\ n\ text{where shaded areas are equal.}`

`text{Solve}\ int_(0)^(n)(f(x)-y_(1))\ dx=int_(n)^(3)(y_(2)-f(x))\ dx\ \ text{for} n:`

`=> n= 1.088\ \ text{(to 3 d.p.)}`

GRAPHS, FUR2 2020 VCAA 2

Kyla organises fundraiser car shows at the business.

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

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

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

 

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

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

`A + C = 537 \ …\ (1)`

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

`2A + 2C = 1074 \ …\ (3)`

`”Subtract “(2) – (3)`

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

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

`:. \ C = 537 – 190 = 347`

♦ Mean mark part (c) 48%.

 

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

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

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

Calculus, MET2 2020 VCAA 1

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

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

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

  1.  i. Write the rule for `f^{′}` in terms of `x`.   (1 mark)
  2. ii. Find the minimum value of the graph of `f^{′}` on the interval  `x in (0, 2)`.   (2 marks)  

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

  1. Write a sequence of two transformations that map the graph of `f` onto the graph of `h`.   (1 mark)
     
       
     
  2.   i. State the values of `x` for which the graphs of `f`and `h` intersect.   (1 mark)
  3.  ii. Write down a definite integral that will give the total area of the shaded regions in the graph above.   (1 mark)
  4. iii. Find the total area of the shaded regions in the graph above. Give your answer correct to two decimal places.   ( 1 mark)
  5. Let `D` be the vertical distance between the graphs of `f`and`h`.
  6. Find all values of `x` for which `D` is at most 2 units. Give your answers correct to two decimal places.   (2 marks)

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

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

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

 

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

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

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

`3x^2-4=0`

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

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

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

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

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

`x in{-sqrt6,-sqrt2,sqrt2,sqrt6}`
 

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

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

e.iii.  `text(Area = 2.72 u²)`

♦♦ Mean mark part (f) 26%.

 

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

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

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

GRAPHS, FUR2 2020 VCAA 1

Kyla owns and manages a truck and car care business.

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

The test drive started at 12 noon.

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

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

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

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

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

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

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

 

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

♦ Mean mark 43%.

`0 = k – 80 xx 8`

`k = 640`

Graphs, MET2 2020 VCAA 5 MC

The graph of the function  `f:D rarr R,f(x)=(3x+2)/(5-x)`, where `D` is the maximal domain, has asymptotes

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

`E`

Show Worked Solution
`f(x)` `=(3x+2)/(5-x)`  
  `=(-(15-3x)+17)/(5-x)`  
  `=-3+17/(5-x)`  

 
`text(Vertical asymptote:)\ \ x=5`

`text(Horizontal asymptote:)\ \ y=-3`

`=>E`

Calculus, MET2 2020 VCAA 3 MC

 Let  `f^(′)(x)=(2)/(sqrt(2x-3))`. 

If  `f(6)=4`, then
 

  1. `f(x)=2sqrt(2x-3)`
  2. `f(x)=sqrt(2x-3)-2`
  3. `f(x)=2sqrt(2x-3)-2`
  4. `f(x)=sqrt(2x-3)+2`
  5. `f(x)=sqrt(2x-3)`
Show Answers Only

`=>C`

Show Worked Solution
`f^(′)(x)` `=2/(sqrt(2x-3))`  
`f(x)` `=2 int(2x-3)^{- 1/2}`  
  `=2*1/2*2(2x-3)^{1/2}+c`  
  `=2sqrt(2x-3)+c`  

 
`text(When)\ \ x=6, \ f(x)=4:`

`4=2sqrt(12-3) + c \ => \ c=-2`

`:. f(x) = 2sqrt(2x-3) – 2`

`=>C`

Statistics, 2ADV S3 SM-Bank 8 MC

The heights of females living in a small country town are normally distributed:

    • 16% of the females are more than 160 cm tall.
    • 2.5% of the females are less than 115 cm tall.

The mean and the standard deviation of this female population, in centimetres, are closest to

  1. mean = 135               standard deviation = 15
  2. mean = 135               standard deviation = 25
  3. mean = 145               standard deviation = 15
  4. mean = 145               standard deviation = 20
Show Answers Only

`C`

Show Worked Solution

`160 -> ztext{-score} = 1`

`115 -> ztext{-score} = -2`

`1` `= {160 – mu}/sigma`
`mu + sigma` `= 160\ …\ (1)`
`-2` `= {115-mu}/sigma`
`mu-2 sigma` `= 115\ …\ (2)`

 
`(1)-(2)`

`3sigma` `=45`  
`sigma` `=15`  

 
`text{Substitute}\ \ sigma = 15\ \ text{into (1)}`

`mu = 145` 

`=> C`

Statistics, 2ADV S2 SM-Bank 5 MC

 The stem plot below shows the height, in centimetres, of 20 players in a junior football team.
 

A player with a height of 179 cm is considered an outlier because 179 cm is greater than

  1. 162 cm
  2. 169 cm
  3. 173 cm
  4. 175.5 cm
Show Answers Only

`D`

Show Worked Solution

`Q_1 = (148 + 148)/2 = 148`

`Q_3 = (158 + 160)/2 = 159`

`IQR = 159 – 148 = 11`

`text{Upper fence}` `= Q_3 + 1.5 xx IQR`
  `= 159 + 1.5 xx 11`
  `= 175.5`

 
`=> D`

Calculus, 2ADV C4 SM-Bank 5

Let  `f^(′)(x) = x^3 + x`.

Find  `f(x)`  given that  `f(1) = 2`.  (2 marks)

Show Answers Only

`f(x) =1/4 x^4 + 1/2 x^2 +5/4`

Show Worked Solution
`f^(′)(x)` `= x^3 + x`  
`f(x)` `=int x^3 + x\ dx`  
  `=1/4 x^4 + 1/2 x^2 +c`  

 
`text(Given)\ f(1) = 2:`

`2` `= 1/4 + 1/2 + c`  
`c` `= 5/4`  

 
`:. f(x) =1/4 x^4 + 1/2 x^2 +5/4`

GRAPHS, FUR2 2021 VCAA 2

z

Christy sells blocks of land at the housing estate.

Her pay is based on the number of blocks that she sells each week.

The maximum number of blocks sold each week is 20.

The graph below shows Christy's weekly pay, in dollars, for the number of blocks sold.

  1. What is the minimum number of blocks of land that Chirsty must sell to receive $6000 in one week?   (1 mark)

John also sells blocks of land. He is paid $1000 for each block that he sells.

  1. Write down all values for the number of blocks of land sold for which Christy and John will receive the same weekly pay.  (1 mark)
  2. John sells no more than three blocks of land for every five blocks of land sold by Christy.  (1 mark)
  3. Let  `x`  be the number of blocks that Christy sells.
  4. Let  `y`  be the number of blocks that John sells.
  5. Write the inequality, written as  `y` in terms of  `x`, that represents this situation.
Show Answers Only
  1. `6 \ text{blocks}`
  2. `text{Blocks sold for equal pay: 3, 6, 11, 15, 20}`
  3. `y <= 3/5 x`
Show Worked Solution

a.    `6 \ text{blocks}`
 

b.    `text{Blocks sold for equal pay: 3, 6, 11, 15, 20}`
 

c.     `y <= 3/5 x`

GRAPHS, FUR2 2021 VCAA 3

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

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

`C = 10 \ 000 + k xx n` 

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

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

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

 

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

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

 

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

 

GRAPHS, FUR2 2021 VCAA 1

The graph below shows the height, in metres, of a drone flying above a new housing estate over a six-minute period of time.
 

  1. For what length of time, in minutes, was the height of the drone a least 50 m?   (1 mark)
  2. What was the average rate of change in the height of the drone, in metres per minute, in the first two minutes?  (1 mark)
Show Answers Only
  1. `4 \ text{minutes}`
  2. `100 \ text{metres per minute}`
Show Worked Solution

a.   `4 \ text{minutes}`
 

b.   `text{At} \ t = 0 , \ text{height} = 0 \ text{m}`

`text{At} \ t = 2 , \ text{height} = 200 \ text{m}`
  
`:. \ text{Average rate of change}`

`= {200 – 0}/{2}`

`= 100 \ text{metres per minute}`

GEOMETRY, FUR2 2021 VCAA 3

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

The top three female players are from three different cities:

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

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


           
 

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

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

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

`text{Wei-Yi and Ozlem}`
 

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

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

 

c.i.

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

 

c.ii.

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

 

d.    `text{In New York time}`

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

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

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

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

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

GEOMETRY, FUR2 2021 VCAA 2

The game of squash is played indoors on a court with a front wall, a back wall and two side walls, as shown in the image below.
 

 
Each side wall has the following dimensions.
 

The shaded region in the diagram above is considered part of the playing area.

  1. Calculate the area, in square metres, of the shaded region in the diagram above.
  2. Round your answer to two decimal places.   (1 mark)
  3. Calculate the perimeter, in metres, of the shaded region in the diagram.
  4. Round your answer to one decimal place.   (1 mark)
  5. Two squash players, Wei-Yi and Bao, are playing a match.
  6. The diagram below shows the lines on the court floor.

 
             

  1. Wei-Yi is serving from point `A` and and Bao is standing at point `B`.
  2. Point `A` is 2.7 m from Point `C`.
  3. Point `B` is 3.1 m from Point `C`.
  4. The angle `ACB` is 119°.
  5. Show that the distance between Wei-Yi and Bao is 5 m, rounded to the nearest metre.   (1 mark)
Show Answers Only
  1. `32.66 \ text{m}^2`
  2. `26.5 \ text{m}`
  3. `5 \ text{m}`
Show Worked Solution
a.   `text{Shaded Area}` `= 1/2 xx 9.75 xx (4.57 + 2.13)`
    `= 32.6625`
    `= 32.66 \ text{m}^2 \ text{(to 2 d.p.)}`

 

b.

`x = 4.57 – 2.13 = 2.44 \ text{m}`
  

`text{Using Pythagoras}`

`y` `= sqrt{2.44^2 + 9.75^2}`
  `= 10.050 …\ text{m}`

 

`:. \ text{Perimeter}` `= 4.57 + 9.75 + 2.13 + 10.05`
  `= 26.5 \ text{m}`

 
c. 

`text{Using cosine rule:}`

`AB^2` `=  AC^2 + CB^2 – 2 xx AC xx CB xx cos 119^@`
  `= 2.7^2 + 3.1^2 – 2 xx 2.7 xx 3.1 xx cos119^@`
  `= 25.0157`
`:.AB` `= 5 \ text{m (nearest metre)}`

GEOMETRY, FUR2 2021 VCAA 1

The game of squash is played with a special ball that has a radius of 2 cm.

  1. Show that the volume of one squash ball, rounded to two decimal places, is 33.51 cm3.  (1 mark)

Squash balls may be sold in cube-shaped boxes.

Each box contains one ball and has a side length of 4.1 cm, as shown in the diagram below.
 

  1. Calculate the empty space, in cubic centimetres, that surrounds the ball in the box.
  2. Round your answer to two decimal places.   (1 mark)
  3. Calculate the total surface area, in square centimetres, of one box.   (1 mark)
  4. Retail shops store the cube-shaped boxes in a space within a display unit.
  5. The space has a length of 17.0 cm and a width of 12.5 cm. Due to the presence of a shelf above, there is a maximum height of 8.5 cm available. This is shown in the diagram below.
     

     

  1. Calculate the maximum number of cube-shaped boxes that can fit into the space within the display unit.   (1 mark)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `35.41 \ text{cm}^3`
  3. `100.86 \ text{cm}^2`
  4. `text(24 boxes)`
Show Worked Solution
a.   `V` `= 4/3 pi r^3`
    `= 4/3 xx pi xx 2^3`
    `= 33.5103 …`
    `= 33.51 \ text{cm}^3 \ text{(to 2 d.p.)}`

 

b.  `text{Volume of cube} = 4.1^3 = 68.921 \ text{cm}^3`

     `:. \ text{Empty space}` `= 68.921 – 33.51`
  `= 35.41 \ text{cm}^3 \ text{(to 2 d.p.)}`

 

c.   `text(S.A.)` `= 6 xx 4.1 xx 4.1`
    `= 100.86 \ text{cm}^2`

 

d.  `text{Length} = 17 div 4.1 = 4.146 … \ \ to 4 \ text{boxes}`

`text{Width} = 12.5 div 4.1 = 3.048 … \ to 3 \ text{boxes}`

`text{Height} = 8.5 div 4.1 = 2.073 … \ \ to 2\ text{boxes}`
 

`:. \ text{Max boxes}` `= 4 xx 3 xx 2`
  `= 24`

NETWORKS, FUR2 2021 VCAA 2

George lives in Town `G` and Maggie lives in Town `M`.

The diagram below shows the network of main roads between Town `G` and Town `M`.

The vertices `G, H, I, J, K, L, M, N` and `O` represent towns.

The edges represent the main roads. The number on the edges indicate the distance, in kilometres, between adjacent towns.
 

  1. What is the shortest distance, in kilometres, between Town `G` and Town `M`?   (1 mark)
  2. George plans to travel to Maggie's house. He will pass through all the towns shown above.
  3. George plans to take the shortest route possible.
  4. Which town will George pass through twice?    (1 mark)
Show Answers Only
  1. `86 \ text{km}`
  2. `text{Passes through town} \ K \ text{twice.}`
Show Worked Solution
a.  `text{Shortest route}` `= G \ O \ N \ M`
    `= 28 + 42 + 16`
    `= 86 \ text{km}`

 
b.  `text{Shortest route through all towns} = G \ H \ I \ K \ L \ K \ J\ O \ N \ M`

`:. \ text{Passes through town} \ K \ text{twice}`

NETWORKS, FUR2 2021 VCAA 1

Maggie's house has five rooms, `A, B, C, D` and `E`, and eight doors.

The floor plan of these rooms and doors is shown below. The outside area, `F`, is shown shaded on the floor plan.
 

The floor plan is represented by the graph below.

On this graph, vertices represent the rooms and the outside area. Edges represent direct access to the rooms through the doors.

One edge is missing from the graph.
 

  1. On the graph above, draw the missing edge.  (1 mark)
  2. What is the degree of vertex `E`?  (1 mark)
  3. Maggie hires a cleaner to clean the house.
  4. It is possible for the cleaner to enter the house from the outside area, `F`, and walk through each room only once, cleaning each room as he goes and finishing in the outside area, `F`.
  5.  i. Complete the following to show one possible route that the cleaner could take.  (1 mark)
        
         
    ii. What is the mathematical term for such a journey?  (1 mark)
Show Answers Only
  1.  
  2. `2`
  3.  i. `FABEDCF\ text(or)\ FCDEBAF`
  4. ii. `text{Hamiltonian cycle}`
Show Worked Solution

a.

b.  `text{Degree} = 2`
 

c.i.  `FABEDCF\ text(or)\ FCDEBAF`

c.ii.  `text{Hamiltonian cycle}`

MATRICES, FUR2 2021 VCAA 2

The main computer system in Elena's office has broken down.

The five staff members, Alex (`A`), Brie (`B`), Chai (`C`), Dex (`D`) and Elena (`E`), are having problems sending information to each other.

Matrix `M` below shows the available communication links between the staff members.

`qquadqquadqquadqquadqquadqquadqquadqquadqquad text(receiver)`

`qquadqquadqquadqquadqquadqquadqquad \ \ \ A \ \ B \ \ C \ \ D \ \ E`

`M= \ text{sender} \ \ {:(A),(B),(C),(D),(E):} [(0,1,0,0,1),(0,0,1,1,0),(1,0,0,1,0),(0,1,0,0,0),(0,0,0,1,0)] `

In this matrix:

  • the '1' in row `A`, column `B` indicates that Alex can send information to Brie
  • the '0' in row `D`, column `C` indicates that Dex cannot send information to Chai.
  1. Which two staff members can send information directly to each other?   (1 mark)
  2. Elena needs to send documents to Chai.
  3. What is the sequence of communication links that will successfully get the information from Elena to Chai?  (1 mark)
  4. Matrix  `M^2` below is the square of `M` and shows the number of two-step communication links between each pair of staff members.
     

    `qquadqquadqquadqquadqquadqquadqquadqquadqquad text(receiver)`

    `qquadqquadqquadqquadqquadqquadqquad \ \ \ A \ \ B \ \ C \ \ D \ \ E`

    `M= \ text{sender} \ \ {:(A),(B),(C),(D),(E):} [(0,0,1,2,0),(0,1,0,1,0),(0,1,0,0,0),(0,0,1,1,0),(0,1,0,0,0)] `


    Only one pair of individuals has two different two-step communication links.

    List each two-step communication link for this pair.  (1 mark)

Show Answers Only
  1. `text{B and D}`
  2. `text{Elena} to text{Dex} to text{Brie} to text{Chai}`
  3. `text{Alex} to text{Brie} to text{Dex}`
    `text{Alex} to text{Elena} to text{Dex}`
Show Worked Solution

a.   `text{B (sender) to D (receiver)} => 1`

`text{D (sender) to B (receiver)} => 1`

`:. \ text{B and D can send information to each other}`
 

b.   `text{Elena} to text{Dex} to text{Brie} to text{Chai}`
 

c.   `text{The two 2-step links are from Alex to Dex.}`

`text{These are:}`

`text{Alex} to text{Brie} to text{Dex}`

`text{Alex} to text{Elena} to text{Dex}`

MATRICES, FUR2 2021 VCAA 1

Elena imports three brands of olive oil: Carmani (`C`) Linelli (`L`) and Ohana (`O`).

The number of 1 litre bottles of these oils sold in January 2021 is shown in matrix `J` below.

`J = {:[(2800),(1700),(2400)]:} {:(C),(L),(O):}`

  1. What is the order of matrix `J`?  (1 mark)
  2. Elena expected that in February 2021 the sales of all three brands of olive oil would increase by 5%.
  3. She multiplied matrix `J` by a scalar value, `k` , to determine the expected volume of sales for February.
  4. Write down the value of the scalar `k`.  (1 mark)
       `k =`
     
Show Answers Only
  1. `3 xx 1`
  2. `1.05`
Show Worked Solution

a.      `3 xx 1`

 

b.     `text{All brands} \ ↑ 5%`

 `:. k = 1.05`

CORE, FUR2 2021 VCAA 7

Sienna owns a coffee shop.

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

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

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

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

  1. Calculate the number of cups of coffee that the machine produces per year.  (1 mark)
  2. The recurrence relation above could also represent the value of the coffee machine depreciating at a flat rate.
  3. What annual flat rate percentage of depreciation is represented?  (1 mark)
  4. Complete the rule below that gives the value of the coffee machine, `M_n`, in dollars, after `n` cups have been produced. Write your answers in the boxes provided.  (1 mark)
     

       `M_n =`
     
    		  `+`  
     
    		 `xx n`  
Show Answers Only
  1. `2880`
  2. `12text{% p.a.}`
  3. `12 \ 000 + (-0.05) xx n`
Show Worked Solution
a.   `text{Cups of coffee}` `= 1440/0.05`
    `= 28 \ 800`

 

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

 

c.  `M_n = 12 \ 000 + (-0.05) xx n`

CORE, FUR2 2021 VCAA 3

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

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

The equation of the least squares line is

winning time = 357.1 – 0.1515 × year

The coefficient of determination is 0.8794

  1. Name the explanatory variable in this time series plot.  (1 mark)
  2. Determine the value of the correlation coefficient (`r`).
  3. Round your answer to three decimal places.  (1 mark)
  4. Write down the average decrease in winning time, in seconds per year, during the period 1956 to 2016.  (1 mark)
  5. The predicted winning time for the women's 100 m freestyle in 2000 was 54.10 seconds.
  6. The actual winning time for the women's 100 m freestyle in 2000 was 53.83 seconds.
  7. Determine the residual value in seconds.  (1 mark)
  8. The following equation can be used to predict the winning time for the women's 100 m freestyle in the future.
  9.      winning time =  357.1 – 0.1515 × year
  10.  i. Show that the predicted winning time for the women's 100 m freestyle in 2032 is 49.252 seconds.  (1 mark)
  11. ii. What assumption is being made when this equation is used to predict the winning time for the women's 100 m freestyle in 2032?   (1 mark)
Show Answers Only
  1. `text{year}`
  2. `- 0.938`
  3. `0.1515 \ text{seconds}`
  4. `-0.27`
  5.  i. `49.252 \ text{seconds}`
  6. ii. `text{The same trend continues when the graph is extended beyond 2016.}`
Show Worked Solution

a.      `text{year}`
 

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

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

`:. \ r = -0.938`
 

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

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

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

 

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

 

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

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

CORE, FUR2 2021 VCAA 1

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

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

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

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

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

     
  3. In shot-put, athletes throw a heavy spherical ball (a shot) as far as they can.
    Athlete number six, Jamilia, threw the shot 14.50 m.
    Calculate Jamilia's standardised score (`z`).
    Round your answer to one decimal place.  (1 mark)
  4. In the qualifying competition, the heights jumped in the high jump are expected to be approximately normally distributed.
    Chara's jump in this competition would give her a standardised score of  `z = –1.0`
    Use the 68–95–99.7% rule to calculate the percentage of athletes who would be expected to jump higher than Chara in the qualifying competition.  (1 mark)
     
  5. The boxplot below was constructed to show the distribution of high jump heights for all 15 athletes in the qualifying competition.
     

     Explain why the boxplot has no whisker at its upper end.  (1 mark)
  6. For the javelin qualifying competition (refer to Table 1), another boxplot is used to display the distribution of athlete's results.
     An athlete whose result is displayed as an outlier at the upper end of the plot is considered to be a potential medal winner in the event.
    What is the minimum distance that an athlete needs to throw the javelin to be considered a potential medal winner?  (2 marks)
Show Answers Only
  1. `3`
  2. `1.81`
  3. `0.5 \ text{(to d.p.)}`
  4. `84text(%)`
  5. `text{See Worked Solutions}`
  6. `46.89 \ text{m}`
Show Worked Solution

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

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

b.     `text{High jump mean}`

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

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

`= 1.81`
 

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

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

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

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

f.  `text{Javelin (ascending):}`

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

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

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

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

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

Complex Numbers, SPEC2 2021 VCAA 2

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

  1.  i. State the relationship between `z_2` and `z_3`.  (1 mark)
  2. ii. Determine the values of  `alpha, beta` and `gamma`, given that  `p(2) = -13, |z_2 + z_3| = 0`  and  `|z_2 - z_3| = 6`.  (3 marks)

Consider the point  `z_4 = sqrt3 + 1`.

  1. Sketch the ray given by  `text(Arg)(z - z_4) = (5pi)/6`  on the Argand diagram below.  (2 marks)
     
  2. The ray  `text(Arg)(z - z_4) = (5pi)/6`  intersects the circle  `|z - 3i| = 1`, dividing it into a major and a minor segment.
  3.  i. Sketch the circle  `|z - 3i| = 1` on the Argand diagram in part b.  (1 mark)
  4. ii. Find the area of the minor segment.  (2 marks)
Show Answers Only
    1. `z_2 = barz_3`
    2. `alpha = -3, beta = 9, gamma = -27`
  2.  
     

     
    1.  

       
    2. `(2pi – 3sqrt3)/12\ text(u²)`
Show Worked Solution

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

`z_2 = barz_3`

 

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

♦♦ Mean mark part (a)(ii) 35%.

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

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

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

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

 

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

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

♦ Mean mark part (b) 45%.

 

b. 

 

c.i.  

 

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

♦♦ Mean mark part (c)(ii) 30%.

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

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

Calculus, MET2 2021 VCAA 2

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

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

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

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

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

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

a.   `0.25`

 

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

 

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

 

d.

♦♦♦ Mean mark part (d) 16%.

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

 

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

 
f.   `text{Case 1:} \ a ≤ 1`

♦♦♦ Mean mark part (f) 18%.

`int_0^a (sqrtx – ax^2) dx = 1/3`

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

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

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

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

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

`:. a = 1.13`

Calculus, MET2 2021 VCAA 1

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

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

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

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

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

  1. Show that the volume, `V_text{box} > 0`, in cubic centimetres, is given by  `V_text{box} (x) = 2x (25-2x)(25-x)`.  (1 mark)
  2. State the domain of  `V_text{box}`.  (1 mark)
  3. Find the derivative of  `V_text{box}`  with respect to `x`.  (1 mark)
  4. Calculate the maximum possible volume of the box and for which value of `x` this occurs.  (3 marks)
  5. Waste minimisation is a goal when making cardboard boxes.
  6. Percentage wasted is based on the area of the sheet of cardboard that is cut out before the box is made.
  7. Find the percentage of the sheet of cardboard that is wasted when `x = 5`.  (2 marks)

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

  1.  i. Let  `V_text{box}` be the function that gives the volume of the box.
  2.     State the domain `V_text{box}` in terms of `h`.  (1 mark)
  3. ii. Find the maximum volume for any such rectangular box, `V_text{box}`, in terms of  `h`.  (3 marks)
  4. Now consider making a box from a square sheet of cardboard with side lengths of `h` centimetres.
  5. Show that the maximum volume of the box occurs when  `x = h/6`.  (2 marks)
Show Answers Only
  1. `text{Show Worked Solutions}`
  2. `(0, 12.5)`
  3. `text{Show Worked Solutions}`
  4. `{15625 sqrt3}/{9}  \ text{or} \ {15625}/{3 sqrt3}`
  5. `8%`
  6. i.  `text{Domain} \ (V_text{box}) = (0, h/2)`
  7. (sqrt3 h^3)/9`
  8. `text(See Worked Solutions)`
Show Worked Solution
a.   `V` `= x (h-2x)(2h-2x)`
    `= x (25-2x)(50-2x)`
    `= 2x (25-2x)(25-x)`

 

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

♦ Mean mark part (b) 42%.

`text{Domain is} \ (0, 12.5)`
 

c.    `(dV)/dx = 12x^2-300x + 1250`
 

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

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

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

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

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

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

♦ Mean mark part (f)(i) 33%.

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

f.ii.  `V = x (h-2x)(2h-2x)`

♦ Mean mark part (f)(ii) 45%.

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

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

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

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

 

g.    `V = x(h-2x)(h-2x)`

♦ Mean mark part (g) 40%.

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

`x = h/2 \ \ text{or} \ \ x =  h/6`

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

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

Calculus, MET2 2021 VCAA 13 MC

The value of an investment, in dollars, after `n` months can be modelled by the function

`f(n) =  2500 xx (1.004)^n`

where  `n ∈ {0, 1, 2, ...}`.

The average rate of change of the value of the investment over the first 12 months is closest to

  1. $10.00 per month.
  2. $10.20 per month.
  3. $10.50 per month.
  4. $125.00 per month.
  5. $127.00 per month.
Show Answers Only

`B`

Show Worked Solution

`text(By CAS):`

`text{Average ROC}` `={f(12) – f(0)}/{12 – 0}`
  `= 10.2229 …`

`=> B`

Probability, MET2 2021 VCAA 6 MC

The probability of winning a game is 0.25

The probability of winning a game is independent of winning any other game.

If Ben plays 10 games, the probability that he will win exactly four rimes is closest to

  1. 0.1460
  2. 0.2241
  3. 0.9219
  4. 0.0781
  5. 0.7759
Show Answers Only

`A`

Show Worked Solution

`X\ ~\ text(Bi)(10, 0.25)`

`text{By CAS (binomPdf(10, 0.25, 4)):}`

`text(Pr) (X = 4) = 0.145998 …`

`=> A`

Complex Numbers, EXT2 N2 2021 SPEC1 8

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

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

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

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

 

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

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

 

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

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

 
`text(When)\ \ 2xy – 2y = 0`

`2y(x – 1)` `= 0`
`x` `= 1`

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

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

 
`:. z = 1 ± sqrt5 i`

Statistics, SPEC1 2021 VCAA 3

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

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

  1. Write down the null and alternative hypotheses for the one-tailed test that was conducted to investigate the complaints.  (1 mark)
    1. Determine the `p` value, correct to three places decimal places, for the test.  (2 marks)
    2. What should the company be told if the test was carried out at the 1% level of significance?  (1 mark)
  3. The company decided to produce a new type of light globe called Globeplus.
    Find the approximate 95% confidence interval for the mean lifetime of the new globes if a random sample of 25 Globeplus globes is tested and the sample mean is found to be 250 weeks. Assume that the standard deviation of the population is 10 weeks. Give your answer correct to two decimal places.  (1 mark)
Show Answers Only
  1. `H_0 : mu = 200`
    `H_1 : mu < 200`
    1. `0.001\ \ (text(to 3 d.p.))`
    2. `(246.08, 253.92)`
  3. `text(Reject the null hypothesis.)`
Show Worked Solution

a.   `H_0 : mu = 200`

`H_1 : mu < 200`

 

b.i.   `E(barX) = mu = 200`

♦ Mean mark part (b)(i) 48%.

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

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

 

b.ii.   `text(1% level) => 0.01`

♦ Mean mark part (b)(ii) 49%.

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

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

 

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

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

Calculus, EXT2 C1 2021 SPEC2 2

Evaluate  `int_0^1 (2x + 1)/(x^2 + 1)\ dx`.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution
`int_0^1 (2x + 1)/(x^2 + 1)\ dx` `= int_0^1 (2x)/(x^2 + 1)\ dx + int_0^1 1/(x^2 + 1)\ dx`
  `= [log_e(x^2 + 1)]_0^1 + [tan^(-1)(x)]_0^1`
  `= log_e 2 – log_e 1 + tan^(-1)(1) – tan^(-1)(0)`
  `= log_e 2 + pi/4`

Vectors, SPEC1 2021 VCAA 1

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

  1. Find the acceleration of the body in `text(ms)^(-2)`.  (1 mark)
  2. The initial velocity of the body is  `-3underset~j\ \ text(ms)^(-1)`.
    Find the velocity of the body, in `text(ms)^(-1)`, at any time  `t`  seconds.  (2 marks)
  3. Find the momentum of the body, in kg `text(ms)^(-1)`,  when  `t = 2`  seconds.  (1 mark)
Show Answers Only
  1. `1/2 underset~i + 6/5 underset~j`
  2. `1/2 t underset~i + (6/5 t – 3) underset~j`
  3. `10 underset~i – 6underset~j`
Show Worked Solution

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

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

 

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

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

`=> c = -3underset~j`

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

 

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

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

MATRICES, FUR1 2021 VCAA 1 MC

If matrix  `M = [(3,2), (8,9), (13,7)]` , then its transpose, `M^T` , is

A.  `[(2,3), (9,8), (7,13)]` B.  `[(2,9,7), (3,8,13)]`  
     
C.  `[(7,9,2), (13,8,3)]` D.  `[(3,8,13), (2,9,7)]`  
     
E.  `[(13,8,3), (7,9,2)]`    
Show Answers Only

`D`

Show Worked Solution

`text{Transpose} \ x_{ij} -> x_{ji}`

`[(3,2), (8,9), (13,7)]^T = [(3,8,13), (2,9,7)]`

`=>D`

CORE, FUR1 2021 VCAA 17 MC

The following recurrence relation can generate a sequence of numbers.
 

`L_0 = 37 , \ L_{n+1} = L_n + C`
 

The value of `L_2` is 25.

The value of `C` is

  1. – 6
  2. – 4
  3.    4
  4.    6
  5.  37
Show Answers Only

`A`

Show Worked Solution

`L_{n+1} = L_n + C => text{recurrence with with common difference}`

`L_1` `=37 + C`  
`L_2` `=L_1+C`  
`25` `=37 + C + C`  
`2C` `=-12`  
`:.C` `=-6`  

 
`=> A`

CORE, FUR1 2021 VCAA 1-3 MC

The percentaged segmented bar chart below shows the age (under 55 years, 55 years and over) of visitors at a travel convention, segmented by preferred travel destination (domestic, international).
 

Part 1

The variables age (under 55 years, 55 years and over) and preferred travel destination (domestic, international) are

  1. both categorial variables.
  2. both numerical variables.
  3. a numerical variable and a categorical variable respectively.
  4. a categorical variable and a numerical variable respectively.
  5. a discrete variable and a continuous variable respectively.

 
Part 2

The data displayed in the percentaged segmented bar chart supports the contention that there is an association between preferred travel destination and age because

  1. more visitors favour international travel.
  2. 35% of visitors under 55 years favour international travel.
  3. 45% of visitors 55 years and over favour domestic travel.
  4. 65% of visitors under 55 years favour domestic travel while 45% of visitors 55 years and over favour domestic travel.
  5. the percentage of visitors who prefer domestic travel is greater than the percentage of visitors who prefer international travel.

 
Part 3

The results could also be summarised in a two-way frequency table.

Which one of the following frequency tables could match the pecentaged segmented bar chart?

 

Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ D `

`text(Part 3:)\ A`

Show Worked Solution

`text(Part 1)`

`text{Preferred travel destination → categorical (nominal) variable}`

`text{Age → categorical (ordinal) variable}`

`=> A`
 

`text(Part 2)`

`text(Only option)\ D\ text(highlights a change in preference for domestic travel)`

`text(between the two age categories.)`

`=>D` 

 
`text(Part 3)`

♦ Mean mark 50%.

`text(Converting the frequency table data into percentages,)`

`text(consider option)\ A:`

`91/140 xx 100 = 65text(%),\ \ 49/140 xx 100 = 35text(%)`

`90/200 xx 100 = 45text(%), \ \ 110/200 xx 100 = 55text(%)`

`=> A`

Functions, MET1 2021 VCAA 3

Consider the function  `g: R -> R, \ g(x) = 2sin(2x).`

  1. State the range of `g`.  (1 mark)
  2. State the period of `g`.  (1 mark)
  3. Solve  `2 sin(2x) = sqrt3`  for  `x ∈ R`.  (3 marks)
Show Answers Only
  1. `[–2,2]`
  2. `pi`
  3. `x= pi/6 + npi, pi/3 + npi\ \ \ (n in ZZ)`
Show Worked Solution

a.   `text(S)text(ince)  -1<sin(2x)<1,`

`text(Range)\  g(x) = [–2,2]`
 

b.   `text(Period) = (2pi)/n = (2pi)/2 = pi`
 

c.    `2sin(2x)` `=sqrt3`
  `sin(2x)` `=sqrt3/2`
  `2x` `=pi/3, (2pi)/3, pi/3 + 2pi, (2pi)/3 + 2pi, …`
  `x` `=pi/6, pi/3, pi/6+pi, pi/3+pi, …`

 
`:.\ text(General solution)`

`= pi/6 + npi, pi/3 + npi\ \ \ (n in ZZ)`

Calculus, MET1 2021 VCAA 2

Let  `f′(x) = x^3 + x`.

Find  `f(x)`  given that  `f(1) = 2`.  (2 marks)

Show Answers Only

`f(x) =1/4 x^4 + 1/2 x^2 +5/4`

Show Worked Solution
`f′(x)` `= x^3 + x`  
`f(x)` `=int x^3 + x\ dx`  
  `=1/4 x^4 + 1/2 x^2 +c`  

 
`text(Given)\ f(1) = 2:`

`2` `= 1/4 + 1/2 + c`  
`c` `= 5/4`  

 
`:. f(x) =1/4 x^4 + 1/2 x^2 +5/4`

Mechanics, EXT2 M1 2021 HSC 14b

An object of mass 5 kg is on a slope that is inclined at an angle of 60° to the horizontal. The acceleration due to gravity is `g\ text(m s)^(-2)`  and the velocity of the object down the slope is `v\ text(m s)^(-1)`.

As well as the force due to gravity, the object is acted on by two forces, one of magnitude `2v` newtons and one of magnitude `2v^2` newtons, both acting up the slope.

  1. Show that the resultant force down the slope is  `(5sqrt3)/2 g - 2v - 2v^2`  newtons.  (2 marks)

  2. There is one value of `v` such that the object will slide down the slope at a constant speed.
  3. Find this value of `v` in `text(m s)^(-1)`, correct to 1 decimal place, given that  `g = 10`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `4.2\ \ text(ms)^-1`
Show Worked Solution

i.   

`F_s` `=\ text(force down slope)`
  `= 5g cos30 – 2v^2 – 2v`
  `= 5g · sqrt3/2 – 2v^2 – 2v`
  `= (5sqrt3)/2 g – 2v^2 – 2v`

 

ii.   `text(Constant speed occurs if)\ \ a = 0`

`F_s = ma = 0`

`2v^2 + 2v – (5sqrt3)/2 xx 10` `= 0`
`2v^2 + 2v – 25sqrt3` `= 0`

 

`v` `= (-2 + sqrt(2^2 + 4 · 2 · 25sqrt3))/(2 xx 2)`
  `= (-2 + sqrt(4 + 200sqrt3))/4`
  `= (-1 + sqrt(1 + 50 sqrt3))/2`
  `= 4.179…`
  `= 4.2\ \ text(ms)^-1\ text{(1 d.p.)}`

Complex Numbers, EXT2 N2 2021 HSC 14c

  1. Using de Moivre’s theorem and the binomial expansion of `(cos theta + i sin theta)^5`, or otherwise, show that
  2.    `cos5theta = 16cos^5theta - 20cos^3 theta + 5cos theta`.  (3 marks)

  3. By using part (i), or otherwise, show that  `text(Re)(e^((ipi)/10)) = sqrt((5 + sqrt5)/8)`.  (3 marks)

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

i.   `(cos theta + i sin theta)^5 = cos5theta + i sin 5theta\ \ text{(by De Moivre)}`

`text(Using binomial expansion:)`

`(cos theta + i sin theta)^5`

`= cos^5theta + 5cos^4theta · isin theta + 10cos^3theta · i^2sin^2theta + 10 cos^2theta · i^3sin^3theta`

`+ 5costheta · i^4sin^4theta + i^5sin^5theta`

`= cos^5theta – 10cos^3thetasin^2theta + 5costhetasin^4theta + i\ \ text{(imaginary part)}`
 

`text(Equating real parts:)`

`cos5theta` `= cos^5theta – 10cos^3thetasin^2theta + 5costhetasin^4theta`
  `= cos^5theta – 10cos^3theta(1 – cos^2theta) + 5costheta(1 – cos^2theta)sin^2theta`
  `= cos^5theta – 10cos^3theta + 10cos^5theta + (5costheta – 5cos^3theta)(1 – cos^2theta)`
  `= 11cos^5theta – 10cos^3theta + 5costheta – 5cos^3theta – 5cos^3theta + 5cos^5theta`
  `= 16cos^5theta – 20cos^3theta + 5costheta`

 

ii.   `text(Re)(e^((ipi)/10)) = cos\ pi/10`

♦♦ Mean mark 31%.

`text(Using part a:)`

`16 cos^5(pi/10) – 20cos^3(pi/10) + 5cos(pi/10) = cos((5pi)/10) = cos(pi/2) = 0`

`text(Solutions to)\ cos(5theta) = 0`

`5theta` `= pi/2, (3pi)/2, (5pi)/2, …`
`theta` `= pi/10, (3pi)/10, (5pi)/10, …`

  
`text(Let)\ cos(pi/10) = x`

`16x^5 – 20x^3 + 5x` `= 0`
`16x^4 – 20x^2 + 5` `= 0`
`x^2` `= (20 ± sqrt(20^2 – 4 · 16 · 5))/(2 xx 16)`
  `= (20 ± sqrt80)/32`
  `= (5 ±sqrt5)/8`
`x` `= ±sqrt((5 ± sqrt5)/8)`

 
`cos\ pi/10 > 0 \ => \ x = sqrt((5 ± sqrt5)/8)`

`1 > cos\ pi/10 > cos\ (3pi)/10 > 0`

`:.\ text(Re)\ (e^((ipi)/10)) = cos\ pi/10 = sqrt((5 + sqrt5)/8)`