SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

PHYSICS, M7 2021 HSC 33

Two experiments are performed with identical light sources having a wavelength of 400 nm.

In experiment `A`, the light is incident on a pair of narrow slits `5.0 xx 10^(-5)` m apart, producing a pattern on a screen located 3.0 m behind the slits.
 

In experiment `B`, the light is incident on different metal samples inside an evacuated tube as shown. The kinetic energy of any emitted photoelectrons can be measured.
 

Some results from experiment `B` are shown.

How do the results from Experiment `A` and Experiment `B` support TWO different models of light? In your answer, include a quantitative analysis of each experiment.   (9 marks)

Show Answers Only

→ Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.

→ When light is incident upon the slits, it diffracts and causes the slit to act as a source of wavefronts. When light from the slits arrives at the screen, bright bands are produced when light waves arrive in phase and undergo constructive interference.

→ Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.

→ The spacing between adjacent bright bands can be calculated:

              `d sin theta` `=m lambda`  
`5xx10^(-5)sin theta` `=1xx400 xx10^(-9)`  
                      `theta` `=0.46^(@)`  
                      `s` `=3 xx tan(0.46) = 0.024` m  

 
→ Experiment B supports Einstein’s particle, or photon model of light. This model can calculate the photon energy of incident light and explain why photons are emitted from calcium but not nickel:

`f=c/lambda =(3.00 xx10^(8))/(400 xx10^(-9))=7.50 xx10^(14)` Hz

`E=hf=6.626 xx10^(-34)xx7.50 xx10^(14)=4.97 xx10^(-19)` J
 

→ This energy is greater than the work function of calcium, explaining why one photon has enough energy to liberate a photoelectron from the calcium sample. However, this energy is less than the work function of nickel, explaining why no photoelectrons were observed from the nickel sample.

→ These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:

`K_(max)=hf-phi=4.97 xx10^(-19)-4.60 xx10^(-19)=3.70 xx10^(-20)\ \text{J}`

Show Worked Solution

→ Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.

→ When light is incident upon the slits, it diffracts and causes the slit to act as a source of wavefronts. When light from the slits arrives at the screen, bright bands are produced when light waves arrive in phase and undergo constructive interference.

→ Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.

→ The spacing between adjacent bright bands can be calculated:

              `d sin theta` `=m lambda`  
`5xx10^(-5)sin theta` `=1xx400 xx10^(-9)`  
                      `theta` `=0.46^(@)`  
                      `s` `=3 xx tan(0.46) = 0.024` m  

 
→ Experiment B supports Einstein’s particle, or photon model of light. This model can calculate the photon energy of incident light and explain why photons are emitted from calcium but not nickel:

`f=c/lambda =(3.00 xx10^(8))/(400 xx10^(-9))=7.50 xx10^(14)` Hz

`E=hf=6.626 xx10^(-34)xx7.50 xx10^(14)=4.97 xx10^(-19)` J
 

→ This energy is greater than the work function of calcium, explaining why one photon has enough energy to liberate a photoelectron from the calcium sample. However, this energy is less than the work function of nickel, explaining why no photoelectrons were observed from the nickel sample.

→ These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:

`K_(max)=hf-phi=4.97 xx10^(-19)-4.60 xx10^(-19)=3.70 xx10^(-20)\ \text{J}`


♦ Mean mark 52%.

PHYSICS, M8 2021 HSC 32

Two students perform an investigation with a piece of elastic laid out straight on a table. The elastic is fixed at one end and has three markings at regular intervals. The distances from each marking to the fixed end, `d_1, d_2` and `d_3`, are measured as shown.
 

A student pulls the elastic to extend it, and the new values of `d_1, d_2` and `d_3` are measured. The student observes that each value has doubled.

How well do the observations from this investigation model the evidence that led to Hubble's discovery of the expansion of the universe? Justify your answer.   (5 marks)

Show Answers Only

→ Hubble observed that galaxies further away from Earth are receding at a faster rate that is proportional to their distance from Earth.

→ From this he discovered that space is expanding.

→ The investigation successfully models this as markings further away from the fixed end moved by a larger amount as the elastic expands.

→ This increase is proportional to a marking’s distance from the fixed end and links to the greater recessional velocity of more distant galaxies.
 

Limitations:

→ The elastic only expands in one direction whereas Hubble’s evidence demonstrated that the universe is expanding in three dimensions.

→ The investigation measures the change in the position of markings whereas Hubble measured speed of galaxies.

→ The markings themselves stretch. Hubble didn’t find that the galaxies themselves were spreading out.

Show Worked Solution

→ Hubble observed that galaxies further away from Earth are receding at a faster rate that is proportional to their distance from Earth.

→ From this he discovered that space is expanding.

→ The investigation successfully models this as markings further away from the fixed end moved by a larger amount as the elastic expands.

→ This increase is proportional to a marking’s distance from the fixed end and links to the greater recessional velocity of more distant galaxies.
 

Limitations:

→ The elastic only expands in one direction whereas Hubble’s evidence demonstrated that the universe is expanding in three dimensions.

→ The investigation measures the change in the position of markings whereas Hubble measured speed of galaxies.

→ The markings themselves stretch. Hubble didn’t find that the galaxies themselves were spreading out.


♦ Mean mark 51%.

PHYSICS, M6 2021 HSC 31

Two identical solenoids are mounted on carts as shown. Each solenoid is connected to a galvanometer, and the solenoid on cart 1 is also connected to an open switch and a battery. The total mass of cart 1 is twice that of cart 2.
 

Explain what would be observed when the switch on cart 1 is closed. In your answer, refer to the current in each galvanometer and the initial movement of the carts.   (7 marks)

Show Answers Only

When the switch on cart one is closed, a direct current will flow through the cart 1 solenoid causing a deflection on `G_1.`

Using the right hand grip rule, this will create a magnetic field with a south pole on the right hand side of this solenoid. As this magnetic field is generated, the solenoid on cart two experiences a change in magnetic flux.

According to Faraday’s law, an induced emf and current will be generated in the cart 2 solenoid.

By Lenz’s law, this current will flow in a direction that opposes the original change in flux, causing a magnetic south pole on the left hand side of this solenoid, and a momentary deflection of `G_2` in the opposite direction to `G_1`.

As the two solenoids produce south poles facing each other, they will repel causing the carts to move away from each other. The law of conservation of momentum dictates that since cart 1 is double the mass of cart 2, it will have half the velocity of cart 2.

Show Worked Solution

When the switch on cart one is closed, a direct current will flow through the cart 1 solenoid causing a deflection on `G_1.`

Using the right hand grip rule, this will create a magnetic field with a south pole on the right hand side of this solenoid. As this magnetic field is generated, the solenoid on cart two experiences a change in magnetic flux.

According to Faraday’s law, an induced emf and current will be generated in the cart 2 solenoid.

By Lenz’s law, this current will flow in a direction that opposes the original change in flux, causing a magnetic south pole on the left hand side of this solenoid, and a momentary deflection of `G_2` in the opposite direction to `G_1`.

As the two solenoids produce south poles facing each other, they will repel causing the carts to move away from each other. The law of conservation of momentum dictates that since cart 1 is double the mass of cart 2, it will have half the velocity of cart 2.


♦ Mean mark 45%.

PHYSICS, M8 2021 HSC 29

Bohr, de Broglie and Schrödinger EACH proposed a model for the structure of the atom.

How does the nature of the electron proposed in each of the three models differ?   (5 marks)

Show Answers Only

Bohr’s model of the atom proposed that the electron was a negatively charged particle which orbits the nucleus in a circular path at discrete, quantised energy levels.

De Broglie’s model proposed that electrons had a dual wave/particle nature existing as stable standing waves around the nucleus.

Schrödinger’s quantum mechanical model described electrons as having a wave nature and existing as orbitals. His equation described a cloud surrounding the nucleus in which electrons had a high probability to be found.

Show Worked Solution

Bohr’s model of the atom proposed that the electron was a negatively charged particle which orbits the nucleus in a circular path at discrete, quantised energy levels.

De Broglie’s model proposed that electrons had a dual wave-particle nature existing as stable standing waves around the nucleus.

Schrödinger’s quantum mechanical model described electrons as having a wave nature and existing as orbitals. His equation described a cloud surrounding the nucleus in which electrons had a high probability to be found.


♦ Mean mark 49%.

PHYSICS, M7 2021 HSC 28

A spaceship travels to a distant star at a constant speed, `v`. When it arrives, 15 years have passed on Earth but 9.4 years have passed for an astronaut on the spaceship.

  1. What is the distance to the star as measured by an observer on Earth?   (3 marks)
  1. Outline how special relativity imposes a limitation on the maximum velocity of the spaceship.  (2 marks)
Show Answers Only
  1.   12 ly
  2.   According to special relativity, as ` v `→ `c`:
  3.   → the momentum of the spaceship approaches infinity
  4.   → the force required to accelerate the spaceship approaches infinity
  5.   → maximum velocity is limited to the speed of light.
Show Worked Solution

a.    `t_v =t_0/sqrt((1-(v^2)/(c^2)))`

`15=9.4/sqrt((1-(v^2)/(c^2)))`

`1-(v^2)/(c^2)` `=((9.4)/(15))^2`  
`(v^2)/(c^2)` `=1-((9.4)/(15))^2`  
`v^2` `=0.60729c^2`  
`v` `=0.779c`  

  
Distance to star from Earth observer:

`s` `=ut`  
  `=0.779 xx15`  
  `=12\ text{ly}`  

♦ Mean mark part (a) 51%.

b.  According to special relativity, as ` v `→ `c`:

→ the momentum of the spaceship approaches infinity

→ the force required to accelerate the spaceship approaches infinity

→ maximum velocity is limited to the speed of light.

PHYSICS, M8 2021 HSC 19 MC

Rh-106 is a metallic, beta-emitting radioisotope with a half-life of 30 seconds.

A sample of Rh-106 and an electrode are placed inside an evacuated chamber. They are connected to a galvanometer and a variable DC power supply.
 

A student measures the current, `I`, when the power supply is set to zero. They then measure the stopping voltage, `V_s`. The stopping voltage is the minimum voltage needed to prevent current flowing.

A few minutes later, these measurements are repeated.

How do the TWO sets of measurements compare?

  1. Only `I` changes.
  2. Only `V_s` changes.
  3. Both `I` and `V_s` change.
  4. Neither `I` nor `V_s` changes.
Show Answers Only

`A`

Show Worked Solution

Due to the short half life of the isotope, there will be significantly less present when the second reading is made.

→ Lower rate of emission of electrons → Lower recorded current

The energy of emitted electrons will remain the same → Voltage will not change

`=>A`


♦ Mean mark 29%.

PHYSICS, M6 2021 HSC 18 MC

An evacuated chamber contains a pair of parallel plates connected to a power supply and a switch which is initially closed.

A positively charged mass (•) falls within the chamber, under the influence of gravity, from the position shown.
 

When the mass has fallen half the height of the chamber, the switch is opened.

Which of the following correctly shows the trajectory of the mass?
 

 

Show Answers Only

`A`

Show Worked Solution

The mass experiences a constant force to the right due to the electric field and a constant force downwards due to gravity.

This results in a net force pointing diagonally down and to the right. The mass travels in the direction of this net force.

When the switch is opened the mass falls only under the influence of gravity, following a parabolic path.

`=>A`


♦ Mean mark 35%.

PHYSICS, M6 2021 HSC 17 MC

Two long, parallel conductors `X` and `Y` are connected to a light bulb and an AC power supply. The conductors are suspended horizontally from fixed points using sensitive spring balances. `X` is positioned directly below `Y`.
 


 

Which statement correctly compares the forces measured by the spring balances?

  1. The forces measured on `X` and `Y` will always be equal.
  2. The force measured on `Y` will be greater than or equal to that on `X`.
  3. The force measured on `X` will be greater than or equal to that on `Y`.
  4. There will be a continuous reversal of which measured force is greater.
Show Answers Only

`C`

Show Worked Solution

Current in wires in opposite direction → The two conductors will always repel (Ampere’s Law)

Resulting forces:

→ upwards force on Y which partially cancels its weight, decreasing the net downwards force acting on it

→ downwards force on X which increases the net downwards force acting on it.

`=>C`


♦ Mean mark 23%.

PHYSICS, M7 2021 HSC 15 MC

Unpolarised light is incident upon two consecutive polarisers as shown. The second polariser has a fixed transmission axis which cannot be rotated. `I_1` is the intensity of light after the first polariser, and `I_2` is the intensity of light after the second polariser.
 

How would `I_1` and `I_2` be affected if the transmission axis of the first polariser was rotated?

  1. Both would change.
  2. Only `I_1` would change.
  3. Only `I_2` would change.
  4. Neither would change.
Show Answers Only

`C`

Show Worked Solution

`I_(1)`  is constant at 50% of the original intensity.

`I_(2)`  varies depending on the angle between the two polarisers (Malus’ Law).

`=>C`


♦ Mean mark 42%.

PHYSICS, M6 2021 HSC 12 MC

Which graph shows the magnitude of back emf induced in a DC motor rotating continuously at different angular velocities?
 

 

Show Answers Only

`D`

Show Worked Solution

`epsi=-(Delta Phi)/(Delta t)`

`omega=(Delta theta)/(Delta t)`

`∴ epsi=-(Delta Phi)/(Delta theta)omega\ \ \` (linear)

`=>D`


♦ Mean mark 31%.

PHYSICS, M5 2021 HSC 9 MC

A mass, `M`, is positioned at an equal distance from two identical stars as shown.

The mass is then moved to position `X`.
 

Which graph best represents the gravitational potential energy, `U`, of the mass during this movement?
 

Show Answers Only

`C`

Show Worked Solution

Potential energy decreases as the mass moves closer to the stars and then increases as it moves away from them.

`=>C`


♦ Mean mark 43%.

PHYSICS, M6 2021 HSC 7 MC

In a certain ideal transformer, the current in the secondary coil is four times as large as the current in the primary coil.

Which row of the table correctly identifies the type of transformer and the ratio of turns?
 

Show Answers Only

`C`

Show Worked Solution

Larger current in secondary coil → Lower voltage in secondary coil

→ Step down transformer → Primary coil has more turns

`=>C`


♦ Mean mark 42%.

PHYSICS, M7 2021 HSC 16 MC

The Sun has an energy output of `3.85 × 10^{28}\ `W.

By how much does the Sun's mass decrease each minute?

  1. `4.28 × 10^{11}` kg
  2. `2.57 × 10^{13}` kg
  3. `1.28 × 10^{20}` kg
  4. `7.70 × 10^{21}` kg
Show Answers Only

`B`

Show Worked Solution

Energy released = `3.85 xx 10^(28) xx 60 = 2.31 xx 10^(30)\` J/min

`E` `=mc^2`  
`m` `=(E)/(c^(2))`  
  `=(2.31 xx10^(30))/((3xx10^(8))^(2))`  
  `=2.57 xx10^(13)\` kg  

 
`=>B`


♦ Mean mark 45%.

PHYSICS, M6 2021 HSC 2 MC

A positively charged particle is moving at velocity, `v`, in an electric field as shown.
 

What is the direction of the force acting on the particle due to the electric field?

  1. Into the page
  2. Out of the page
  3. Up the page
  4. Down the page
Show Answers Only

`C`

Show Worked Solution

Force of positive charge → direction of electric field

`=>C`


♦ Mean mark 38%.

CHEMISTRY, M6 2021 HSC 5 MC

A student used the following method to titrate an acetic acid solution of unknown concentration with a standardised solution of dilute sodium hydroxide.

    • Rinse burette with deionised water.
    • Fill burette with sodium hydroxide solution.
    • Rinse pipette and conical flask with acetic acid solution.
    • Pipette 25.00 mL of acetic acid solution into conical flask.
    • Add appropriate indicator to the conical flask.
    • Titrate to endpoint and record volume of sodium hydroxide solution used.

Compared to the actual concentration of the acetic acid, the calculated concentration will be

  1. lower.
  2. higher.
  3. the same.
  4. different, but higher or lower cannot be predicted.
Show Answers Only

`B`

Show Worked Solution

Two points to consider in method:

Burette rinsed with water instead of sodium hydroxide.

→ Sodium hydroxide solution diluted requiring more sodium hydroxide to neutralise the acetic acid.

This will result in greater moles of acetic acid, and thus the calculated concentration of acetic acid would be greater than the actual concentration.

 
Conical flask rinsed with acetic acid

→ Results in a greater number of moles of acid. More volume of sodium hydroxide would be added

→ Greater number of moles of acetic acid.

This would result in a greater number of moles, and thus the calculated concentration would be higher than the actual concentration.

In both cases, the calculated concentration would be higher than the actual concentration.

`=>B`


♦ Mean mark 45%.

BIOLOGY, M5 2021 HSC 9 MC

Streptomycin is an antibiotic that kills bacteria by interfering with the function of their ribosomes.

The primary effect of the antibiotic is that it prevents the bacteria from producing 

  1. tRNA.
  2. mRNA.
  3. amino acids.
  4. polypeptides.
Show Answers Only

`D`

Show Worked Solution

→ Ribosomes form polypeptide chains as part of protein synthesis.

→ By interfering with the function of ribosomes, Streptomycin prevents bacteria from producing polypeptides.

`=>D`

♦♦ Mean mark 34%.

BIOLOGY, M6 2021 HSC 6 MC

A mutation involving a DNA deletion is illustrated.
 

Which statement about the mutation is correct?

  1. It will have an effect on many genes.
  2. It will have an effect on only one codon.
  3. It may be the result of an error during translation.
  4. It may be the result of an error during transcription.
Show Answers Only

`A`

Show Worked Solution

→ A single chromosome has thousands of genes.

→ Deleted area will have an effect on many genes.

`=>A`


♦ Mean mark 40%.

BIOLOGY, M8 2021 HSC 5 MC

Glucose levels are maintained by the hormones insulin and glucagon.

Which statement best describes the changes in hormone levels of a healthy human soon after a high glucose meal?

  1. Insulin levels fall and glucagon levels rise.
  2. Insulin levels fall and glucagon levels fall.
  3. Insulin levels rise and glucagon levels fall.
  4. Insulin levels rise and glucagon levels rise.
Show Answers Only

`C`

Show Worked Solution

→ High glucose meal, blood glucose levels rise.

→ Insulin levels rise to return blood glucose levels to normal range causing glucagon levels to fall.

`=>C`


Mean mark 53%.

Calculus, MET2 2020 VCAA 5

Let  `f: R to R, \ f(x)=x^{3}-x`.

Let  `g_{a}: R to R`  be the function representing the tangent to the graph of `f` at  `x=a`, where  `a in R`.

Let `(b, 0)` be the `x`-intercept of the graph of `g_{a}`.

  1. Show that  `b= {2a^{3}}/{3 a^{2}-1}`.   (3 marks)
  2. State the values of `a` for which `b` does not exist.   (1 mark)
  3. State the nature of the graph of `g_a` when `b` does not exist.   (1 mark)
  4. i.  State all values of `a` for which  `b=1.1`. Give your answer correct to four decimal places.   (1 mark)
  5. ii. The graph of `f` has an `x`-intercept at (1, 0).
  6.      State the values of  `a`  for which  `1 <= b <= 1.1`.
  7.      Give your answers correct to three decimal places.   (1 mark)

The coordinate `(b, 0)` is the horizontal axis intercept of `g_a`.

Let `g_b` be the function representing the tangent to the graph of `f` at  `x=b`, as shown in the graph below.
 
 
     
 

  1. Find the values of `a` for which the graphs of `g_a` and `g_b`, where `b` exists, are parallel and where  `b!=a`.   (3 marks)

Let  `p:R rarr R, \ p(x)=x^(3)+wx`, where  `w in R`.

  1. Show that  `p(-x)=-p(x)`  for all  `w in R`.   (1 mark)

A property of the graphs of `p` is that two distinct parallel tangents will always occur at `(t, p(t))` and `(-t,p(-t))` for all  `t!=0`.

  1. Find all values of `w` such that a tangent to the graph of `p` at `(t, p(t))`, for some  `t > 0`, will have an `x`-intercept at `(-t, 0)`.   (1 mark)
  2. Let  `T:R^(2)rarrR^(2),T([[x],[y]])=[[m,0],[0,n]][[x],[y]]+[[h],[k]]`, where  `m,n in R text(\{0})`  and  `h,k in R`.
     
    State any restrictions on the values of `m`, `n`, `h`, and `k`, given that the image of `p` under the transformation `T` always has the property that parallel tangents occur at  `x = -t`  and  `x = t`  for all  `t!=0`.   (1 mark)
Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `a=+-sqrt3/3`
  3. `text(Horizontal line.)`
  4.  i. `a=-0.5052, 0.8084, 1.3468`
  5. ii. `a in (-0.505,-0.500]uu(0.808,1.347)`
  6. `a=+- sqrt5/5`
  7. `text(See Worked Solutions.)`
  8. `w=-5t^2`
  9. `h=0`
Show Worked Solution

a.   `f^(‘)(a) = 3a^2-1`

`g_a(x)\ \ text(has gradient)\ \ 3a^2-1\ \ text(and passes through)\ \ (a, a^3-a)`

`g_a(x)-(a^3-a)` `=(3a^2- 1)(x-a)`  
`g_a(x)` `=(3a^2-1)(x-a)+a^3-a`  

  
`x”-intercept occurs at”\ (b,0):`

`0=(3a^2-1)(b-a) + a^3-a`

`(3a^2-1)(b-a)` `=a-a^3`  
`3a^2b-3a^3-b+a` `=a-a^3`  
`b(3a^2-1)` `=a-a^3+3a^3-a`  
`:.b` `=(2a^3)/(3a^2-1)`  

 
b.   `b\ text{does not exist when:}`

♦ Mean mark part (b) 46%.

`(3a^2-1)=0`

`a=+-sqrt3/3`

♦♦ Mean mark part (c) 23%.
 

c.   `text{If}\ \ a=+-sqrt3/3,\ \ g_a^(‘)(x) = 0`

`=>\ text{the graph is a horizontal line (does not cross the}\ xtext{-axis).}`
 

d.i.  `”Solve “{2a^{3}}/{3 a^{2}-1}=1.1” for “a:`

`a=-0.5052″ or “a=0.8084” or “a=1.3468\ \ text{(to 4 d.p.)}`
  

d.ii.  `” Solve “1 <= (2a^(3))/(3a^(2)-1) < 1.1” for “a:`

♦♦♦ Mean mark part (d)(ii) 13%.

`a in (-0.505,-0.500]uu(0.808,1.347)\ \ text{(to 3 d.p.)}`
 

e.   `f^(‘)(b) = 3b^2-1`

`g_b(x)\ \ text(has gradient)\ \ 3b^2-1\ \ text(and passes through)\ \ (b, b^3-b)`

`g_b(x)-(b^3-b)` `=(3b^2- 1)(x-b)`  
`g_b(x)` `=(3b^2-1)(x-b)+b^3-b`  

 
`g_a(x)\ text{||}\ g_b(x)\ \ text{when}`

♦♦♦ Mean mark part (e) 13%.
`3a^2-1` `=3b^2-1`  
  `=3(2a^3)/(3a^2-1)-1`  

 
`=> a=+-1, +- sqrt5/5, 0`

`text(Test each solution so that)\ \ b!=a :`

`text(When)\ \ a=+-1, 0 \ => \ b=a`

`:. a=+- sqrt5/5`
 

f.    `p(-x)` `=(-x)^3-wx`
    `=-x^3-wx`
    `=-(x^3+wx)`
    `=-p(x)`

 
g.  `p^(‘)(t) = 3t^2+w`

♦♦♦ Mean mark part (g) 3%.

`p(t)\ \ text(has gradient)\ \ 3t^2+w\ \ text(and passes through)\ \ (t, t^3+wt)`

`p(t)-(t^3+wt)` `=(3t^2+w)(x-t)`  
`p(t)` `=(3t^2+w)(x-t) + t^3+wt`  

 
`text{If}\ p(t)\ text{passes through}\ \ (-t, 0):`

`0=(3t^2+w)(-2t) + t^3+wt`

`=>w=-5t^2\ \ (t>0)`
 

h.   `text{Property of parallel tangents is retained under transformation}`

♦♦♦ Mean mark part (h) 2%.

`text{if rotational symmetry remains (odd function).}`

`=>h=0`

`text(No further restrictions apply to)\ m, n\ \ text{or}\ \ k.`

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.)}`

NETWORKS, FUR2 2020 VCAA 5

The Sunny Coast cricket clubroom is undergoing a major works project.

This project involves nine activities: `A` to `I`.

The table below shows the earliest start time (EST) and duration, in months, for each activity.

The immediate predecessor(s) is also shown.

The duration for activity `C` is missing.
 

   

The information in the table above can be used to complete a directed network.

This network will require a dummy activity.

  1. Complete the following sentence by filling in the boxes provided.   (1 mark)
        This dummy activity could be drawn as a directed edge from the end of activity 
 
  to
        the start of activity 
 
  .
  1. What is the duration, in months, of activity  `C`?  (1 mark)
  2. Name the four activities that have a float time.   (1 mark)
  3. The project is to be crashed by reducing the completion time of one activity only.

        What is the minimum time, in months, that the project can be completed in?   (1 mark)

Show Answers Only
  1. `B\ text{to the start of activity}\ C.`
  2. `text{2 months}`
  3. `A, E, F, H`
  4. `text{17 months}`
Show Worked Solution

a.   `B\ text{to the start of activity}\ C.`

♦♦ Mean mark part (a) 26%.

 

b.   `text{Sketch network diagram.}`

♦ Mean mark part (b) 49%.
 

`text{Duration of Activity C = 2 months}`

 

c.   `text{Critical path:}\   BCDGI`

♦♦ Mean mark part (c) 23%.

`text{Activities with a float time are activities}`

`text{not on critical path.}`

`:. \ text{Four activities are:}\ \ A, E, F, H`

 

d.   `text{Completion time of}\ BCDGI = 20\ text{months}`

♦♦♦ Mean mark part (d) 12%.

`text{Reduce the completion of}\ B\ text{by 3 months to create}`

`text{a new minimum completion time of 17 months.}`

Statistics, MET2 2020 VCAA 3

A transport company has detailed records of all its deliveries. The number of minutes a delivery is made before or after its schedule delivery time can be modelled as a normally distributed random variable, `T`, with a mean of zero and a standard deviation of four minutes. A graph of the probability distribution of `T` is shown below.
 

  1. If  `"Pr"(T <= a)=0.6`, find `a` to the nearest minute.   (1 mark)
  2. Find the probability, correct to three decimal places, of a delivery being no later than three minutes after its scheduled delivery time, given that it arrives after its scheduled delivery time.   (2 marks)
  3. Using the model described above, the transport company can make 46.48% of its deliveries over the interval  `-3 <= t <= 2`.
  4. It has an improved delivery model with a mean of `k` and a standard deviation of four minutes.
  5. Find the values of `k`, correct to one decimal place, so that 46.48% of the transport company's deliveries can be made over the interval  `-4.5 <= t <= 0.5`   (3 marks)

A rival transport company claims that there is a 0.85 probability that each delivery it makes will arrive on time or earlier.

Assume that whether each delivery is on time or earlier is independent of other deliveries.

  1. Assuming that the rival company's claim is true, find the probability that on a day in which the rival company makes eight deliveries, fewer than half of them arrive on time or earlier. Give your answer correct to three decimal places.   (2 marks)
  2. Assuming that the rival company's claim is true, consider a day in which it makes `n` deliveries.
  3.  i. Express, in terms of `n`, the probability that one or more deliveries will not arrive on time or earlier.   (1 mark)
  4. ii. Hence, or otherwise, find the minimum value of `n` such that there is at least a 0.95 probability that one or more deliveries will not arrive on time or earlier.   (1 mark)
  5. An analyst from a government department believes the rival transport company's claim is only true for deliveries made before 4 pm. For deliveries made after 4 pm, the analyst believes the probability of a delivery arriving on time or earlier is `x`, where  `0.3 <=x <= 0.7`
  6. After observing a large number of the rival transport company's deliveries, the analyst believes that the overall probability that a delivery arrives on time or earlier is actually 0.75
  7. Let the probability that a delivery is made after 4 pm be `y`.
  8. Assuming that the analyst's belief are true, find the minimum and maximum values of `y`.   (2 marks)
Show Answers Only
  1. `a= 1\ text(minute)`
  2. `0.547`
  3. `k=-1.5, -2.5`
  4. `0.003`
  5.  i. `1-0.85^n`
  6. ii. `19`
  7. `2/3`
Show Worked Solution

a.   `T\ ~\ N(0, 4^2)`

`text(Solve (by CAS): Pr)(T<=a) = 0.6`

`:. a= 1\ text(minute)`
 

b.    `text{Pr}(T <= 3∣T > 0)` `=(text{Pr}(0 < T <= 3))/(text{Pr}(T > 0))`
    `=(0.27337 dots)/(0.5)`
    `=0.547\ \ text{(to 3 d.p.)}`

 

c.   `text(Given)\ \ text{Pr}(-3 <= T <= 2) = 0.4648`

`sigma = 4 text{minutes}`

`=> \ text{Pr}(-4.5 <= T – 1.5 <= 0.5) = 0.4648`

`=> k=-1.5`

`text(By symmetry of the normal distribution)`

`text{Pr}(-2 <= T <= 3) = text{Pr}(-3 <= T <= 2) = 0.4648`

`=> \ text{Pr}(-4.5 <= T – 2.5 <= 0.5) = 0.4648`

`=> k=-2.5`

`:. k=-1.5, -2.5`

 

d.   `text{Let}\ \ X\ ~\ text{Bi}(8, 0.85)`

`text(Solve (by CAS):)`

`text{Pr}(X<=3) = 0.003\ \ text{(to 3 d.p.)}`
 

e.i.   `text{Pr(at least 1 delivery is late)}`

`= 1-\ text{Pr(all deliveries are on time)}`

`=1-0.85^n`
 

e.ii.   `text{Solve for}\ n:`

`1-0.85^n` `<0.95`  
`n` `>18.43…`  

 
`:.n_min=19`
 

f.   `text{Pr(delivery made after 4pm)} = y`

`=>\ text{Pr(delivery made before 4pm)} = 1-y`

`0.85(1-y)+xy` `=0.75`  
`y` `=-(0.1)/(x-0.85)`  
  `=(2)/(17-20 x)`  

 
`text(Given ) 0.3<=x<=0.7:`

`y_min = (2)/(17-20 xx 0.3) = 2/11`

`y_max = (2)/(17-20 xx 0.7) = 2/3`

GEOMETRY, FUR2 2020 VCAA 3

Khaleda manufacturers the face cream in Dhaka, Bangladesh.

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

Assume that the radius of Earth is 6400 km.

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

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

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

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

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

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

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

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

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

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

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

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

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

 

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

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

 

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

 

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

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

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

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

`= 10:00 \ text{pm}`

 

d.

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

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

`=180 – 76.1`

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

 

e.

`text{Let X = position of hangar}`

`text{Find OX:}`

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

 

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

`= 27 – 23.17`

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

GEOMETRY, FUR2 2020 VCAA 2

Khaleda has designed a logo for her business.

The logo contains two identical equilateral triangles,

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

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

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

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

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

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

`text{Using the sine rule:}`

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

♦♦ Mean mark part (b) 32%.

 

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

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

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

`:. \ text{Area of logo}`

`= 2 xx 10 – 1/2 xx 10`

`= 15 \ text{cm}^2`

♦ Mean mark part (c) 48%.

 

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

 

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

♦♦♦ Mean mark part (d) 17%.

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

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

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

`= 2 xx 4.8`

`= 9.6 \ text{cm}`

Calculus, MET2 2020 VCAA 2

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

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

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

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

All distances are measured in metres.
 

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

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

  1. The swimmer swims north from point `P`.
  2. Find the distance, in metres, that the swimmer needs to swim to get to get to the north bank of the river.   (1 mark)
  3. The swimmer swims east from point `P`.
  4. Find the distance, in metres, that the swimmer needs to swim to get to the north bank of the river.   (2 marks)
  5. On another occasion, the swimmer swims the minimum distance from point `P` to the north bank of the river.
  6. Find this minimum distance. Give your answer in metres, correct to one decimal place.   (2 marks)
  7. Calculate the surface area of the section of the river shown on the graph in square metres.   (1 mark)
  8. A horizontal line is drawn through point `P`. The section of the river that is south of the line is declared a no "no swimming" zone.
  9. Find the area of the "no swimming" zone, correct to the nearest square metre.   (3 marks)
  10. Scientists observe that the north bank of the river is changing over time. It is moving further north from its current position. They model its predicted new location using the function with rule  `y=kf_(1)(x)`, where `k >= 1`.
  11. Find the values of  `k` for which the distance north across the river, for all parts of the river, is strictly less than 20 m.   (2 marks)
Show Answers Only
  1. `10\ text{m}`
  2. `16 2/3\ text{m}`
  3. `8.5\ text{m}`
  4. `2000\ text{m}^2`
  5. `837\ text{m²}`
  6. `k in [1, 7/6)`
Show Worked Solution

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

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

 

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

`=> x=200/3`

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

`=200/3-50`

`=16 2/3\ text{m}`
 

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

`text{By Pythagoras,}`

♦ Mean mark part (c) 39%.

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

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

`x=54.47…`

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

 

d.   `text{Shaded Area}`

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

`=2000\ text{m}^2`

 

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

♦♦ Mean mark part (e) 35%.

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

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

`=>x=50, 150`

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

♦♦♦Mean mark part (f) 15%.

 

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

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

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

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

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

`=> k<7/6`

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

GEOMETRY, FUR2 2020 VCAA 1

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

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

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

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

 
               

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

 
                 
 

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

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

 

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

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

 

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

 

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

 

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

♦♦ Mean mark part (e) 27%.

`text{Divide area into 2 sections}`
 


 

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

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

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

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

`text{Width} = 5`

`text{Height} = 5`
 

`:. \ text{Maximum containers}`

`= 5 xx 10 + 5 xx 5`

`= 75`

 

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

 

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

GRAPHS, FUR2 2020 VCAA 4

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

Every vehicle is both serviced and detailed.

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

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

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

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

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

These constraints are represented by Inequalities 1 to 4 below.

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

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


 

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

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

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

♦♦ Mean mark part (b) 34%.

 

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

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

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

`=96/8`

`=12`
 

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

♦ Mean mark part (c) 42%.

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

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

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

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

 

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

♦♦♦ Mean mark part (d) 19%.

`P = 150x + 225y`

`=> \ m_P = – 150/225 = – 2/3`

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


 

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

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

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

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

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

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

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

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

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

GRAPHS, FUR2 2020 VCAA 3

Kyla's business also manufactures car seat covers.

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

`R = 80 n`

This relationship is shown on the graph below.
 

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

`C = 36n + 5000`

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

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

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

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

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

Show Answers Only

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

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

 

b.    `C=36 n+5000`

`R=80 n`

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

`text{Solve for} \ n:`

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

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

♦♦ Mean mark part (c) 31%.

 

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

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

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

 

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

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

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

GRAPHS, FUR2 2020 VCAA 2

Kyla organises fundraiser car shows at the business.

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

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

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

 

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

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

`A + C = 537 \ …\ (1)`

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

`2A + 2C = 1074 \ …\ (3)`

`”Subtract “(2) – (3)`

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

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

`:. \ C = 537 – 190 = 347`

♦ Mean mark part (c) 48%.

 

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

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

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

Calculus, MET2 2020 VCAA 1

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

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

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

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

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

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

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

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

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

 

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

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

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

`3x^2-4=0`

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

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

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

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

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

`x in{-sqrt6,-sqrt2,sqrt2,sqrt6}`
 

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

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

e.iii.  `text(Area = 2.72 u²)`

♦♦ Mean mark part (f) 26%.

 

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

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

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

GRAPHS, FUR2 2020 VCAA 1

Kyla owns and manages a truck and car care business.

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

The test drive started at 12 noon.

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

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

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

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

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

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

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

 

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

♦ Mean mark 43%.

`0 = k – 80 xx 8`

`k = 640`

GRAPHS, FUR1 2020 VCAA 8 MC

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

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

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

An inequality representing this situation is

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

`A`

Show Worked Solution

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

♦ Mean mark 43%.

`x:y = 2:5`

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

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

 
`=>  A`

GRAPHS, FUR1 2020 VCAA 7 MC

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

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

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

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

`C`

Show Worked Solution

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

♦♦ Mean mark 38%.

`text(Total megajoules used)`

`= 131.20/0.025`

`= 5248`
 

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

`= 5248/30`

`= 174.93…`

`=>  C`

Graphs, MET2 2020 VCAA 18 MC

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

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

The range of  `h`  is

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

`D`

Show Worked Solution

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

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

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

`text(Endpoint coordinates:)`

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

`:.\ “Range”=(-oo,-1+b]uu[b+1,oo)`

`=>D`

Graphs, MET2 2020 VCAA 17 MC

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

The maximum value of `c` is

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

`C`

Show Worked Solution

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

♦ Mean mark 42%.

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

`”occurs at “x=0`

`f(0) = – log_e(2)`

`=>C`

Calculus, MET2 2020 VCAA 15 MC

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

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

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

`B`

Show Worked Solution

♦♦ Mean mark 32%.

`text(Average Value)`

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

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

`=(a)/(3)`

`=>B`

Statistics, MET2 2020 VCAA 14 MC

The random variable `X` is normally distributed.

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

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

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

`A`

Show Worked Solution

♦ Mean mark 44%.

`X∼N(2sigma,sigma^(2))`

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

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

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

`sigma=7.238″ (to 3 d.p.)”`

`=>A`

Functions, MET2 2020 VCAA 12 MC

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


 

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

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

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

`E`

Show Worked Solution

`text(By elimination)`

♦ Mean mark 45%.

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

`=>\ text(Eliminate)\ A, B, C`
 

`text(Period = 60)`

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

 
`=>\ text(Eliminate)\ D`

`=>E`

Calculus, MET2 2020 VCAA 9 MC

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

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

`E`

Show Worked Solution

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

♦♦ Mean mark 35%.

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

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

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

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

`=>E`

Probability, MET2 2020 VCAA 8 MC

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

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

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

`B`

Show Worked Solution

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

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

 

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

 
`=>B`

Networks, STD2 N3 SM-Bank 50

Roadworks planned by the local council require 13 activities to be completed.

The network below shows these 13 activities and their completion times in weeks.
 

  1. What is the earliest start time, in weeks, of activity `K`?  (1 mark)

  2. How many of these activities have zero float time?  (2 marks)

  3. It is possible to reduce the completion time for activities `A, E, F, L` and `K`.
  4. The reduction in completion time for each of these five activities will incur an additional cost.
  5. The table below shows the five activities that can have their completion time reduced and the associated weekly cost, in dollars.
     
       
  6. The completion time for each these five activities can be reduced by a maximum of two weeks.
  7. The overall completion time for the roadworks can be reduced to 16 weeks.
  8. What is the minimum cost, in dollars, of this change in completion time?  (3 marks)

Show Answers Only
  1. `14 \ text{weeks}`
  2. `7`
  3. `$ 380 \ 000`
Show Worked Solution

a.    `text{Scan forwards:}`

`EST \ (text{activity} \ K)`

`= A \ E \ J`

`= 6 + 5 + 3`

`= 14 \ text{weeks}`
 

b.     `text{Scan backwards:}`

`text{Critical paths:}\ ADGLM\ text(and)\ AEHLM`

`:. \ text{7 activities have no float time:} \ ADEGHLM`
 

c.    `text{There are 9 possible paths}`

`ADGLM\ (19), AEHLM\ (19) , AEIM\ (14)`

`AEJK\ (17), BCEIM \ (13), BCEJK\ (16)`

`BCEHLM\ (18), BCDGLM \ (18), BFK (11)`

`text{Consider the 5 paths with completions over 16 weeks}`

` to \ text{all contain}\ A\ text{or}\ L \ text{or both.}`
  

`text{Consider} \ ADGLM, AEHLM \ (text{contains both} \ A\ text{and} \ L )`

`to \ text{reduce} \ L xx 2 \ ,  A xx 1 \ text{to reach 16 weeks}`

`to \ text{cheaper than} \ L xx 1 \ , \ A xx 2`
  

`text{Consider} \ BCEHLM , BCDGLM\ (text{both contain} \ L \ text{only})`

`to \ text{reduce} \ L xx 2 \ text{to reach 16 weeks}`
 

`text{Consider} \ AEJK \ ( text{contains} \ A \ text{only} )`

`to \ text{reduce} \ A xx 1 \ text{reach 16 weeks}.`
 

`:. \ text{Minimum cost to reduce time to 16 weeks}`

`= 2 xx 120\ 000 + 1 xx  140\ 000`

`= $ 380\ 000`

Combinatorics, EXT1 A1 SM-Bank 21

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

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

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

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

 

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

Combinatorics, EXT1 A1 SM-Bank 20

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

Show Answers Only

`150`

Show Worked Solution

`text{Consider the 5 horizontal lines}`

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

`text{Consider the 6 vertical lines}`

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

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

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

`= 150`

Trigonometry, EXT1 T2 SM-Bank 8

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

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

Show Answers Only

`8/65`

Show Worked Solution

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

`cos(x) = 3/5`
 

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

 

`cos(y) = 12/13`
 

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

Statistics, 2ADV S3 SM-Bank 20

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

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

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

Show that  \(k = 2\).  (2 marks)

Show Answers Only

\(\text{See Worked Solution}\)

Show Worked Solution

♦ Mean mark 48%.

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

 

Functions, 2ADV F2 SM-Bank 20

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

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

Show Answers Only

a. 

b. `x in [1, 2)`

Show Worked Solution

a.   `text(Asymptotes:)`

`x=2`

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

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

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


 

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

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

GRAPHS, FUR2 2021 VCAA 4

Health and training sessions are held each day at the new community centre.

  • Let `x` be the number of sessions for children each day.
  • Let `y` be the number of sessions for adults each day.
  • The total number of all sessions each day must be at least 10.
  • The number of sessions for adults must not be less than the number of sessions for children.
  • It takes 30 minutes for each session for children and 40 minutes for each session for adults.

The constraints on the health and training sessions each day can be represented by the following five inequalities.

Inequality 1      `x >= 0`

Inequality 2      `y >= 0`

Inequality 3      `x + y >= 10`

Inequality 4       `y >= x`

Inequality 5        `30x + 40y <= 600`

  1. Explain the meaning of Inequality 5 in the context of this situation.  (1 mark)

The graph below shows the feasible region (shaded) that satisfies Inequality 1 to 5.

  1. What is the maximum number of sessions for children each day?  (1 mark)
  2. The community profits from these sessions.
  3. Each session for children makes a profit of $45 and each session for adults makes a profit of $60.
  4.  i.  Determine the maximum profit per day from all health and training sessions. (1 mark)
  5. ii. List all the points within the feasible region that result in this maximum profit. (1 mark)
Show Answers Only
  1. `text{The time of all adult and children sessions each day, in total,}`
    `text{must be less than 600 minutes (10 hours).}`
  2. `8`
  3.  i.  `$900`
    ii. `(0, 15), (4, 12) and (8, 9)`
Show Worked Solution

a.    `text{The time of all adult and children sessions each day, in total,}`

`text{must be less than 600 minutes (10 hours).}`
 

b.    `text{Maximum children sessions = 8}`

`text{(largest integer value of} \ x \ text{in the feasible region)}`
 

c.i.  `P = 45 x + 60 y`

`y = – 3/4 x + P/60`

`text{Objective function’s gradient is the same as} \ 30x + 40y = 600`

`=> \ text{maximum profit occurs at integer points on graph of}`

 `30x+40y=600\ \ text{in feasible region.}`
 

`text{Using (0, 15):}`

`P_text{max}` `= 45 xx 0 + 60 xx 15`
  `= $900`

 

c.ii.

 

`P_text{max} \ text{occurs at (0, 15), (4, 12) and (8, 9)}`

Statistics, SPEC2 2021 VCAA 6

The maximum load of a lift in a chocolate company's office building is 1000 kg. The masses of the employees who use the lift are normally distributed with a mean of 75 kg and a standard deviation of 8 kg. On a particular morning there are `n` employees about to use the lift.

  1. What is the maximum possible value of `n` for there to be less than a 1% chance of the lift exceeding the maximum load?   (2 marks)

Clare, who is one of the employees, likes to have a hot drink after she exits the lift. The time taken for the drink machine to dispense a hot drink is normally distributed with a mean of 2 minutes and a standard deviation of 0.5 minutes. Times taken to dispense successive hot drinks are independent.

  1. Clare has a meeting at 9.00 am and at 8.52 am she is fourth in the queue for a hot drink. Assume that the waiting time between hot drinks dispensed is negligible and that it takes Clare 0.5 minutes to get from the drink machine to the meeting room.
  2. What is the probability, correct to four decimal places, that Clare will get to her meeting on time?   (2 marks)

Clare is a statistician for the chocolate company. The number of chocolate bars sold daily is normally distributed with a mean of 60 000 and a standard deviation of 5000. To increase sales, the company decides to run an advertising campaign. After the campaign, the mean daily sales from 14 randomly selected days was found to be 63 500.

Clare has been asked to investigate whether the advertising campaign was effective, so she decides to perform a one-sided statistical test at the 1% level of significance.

  1.   i. Write down suitable null and alternative hypotheses for this test.  (1 mark)
  2.  ii. Determine the `p` value, correct for decimal places, for this test.  (1 mark)
  3. iii. Giving a reason, state whether there is any evidence for the success of the advertising campaign.  (1 mark)
  4. Find the range of values for the mean daily sales of another 14 randomly selected days that would lead to the null hypothesis being rejected when tested at the 1% level of significance. Give your answer correct to the nearest integer.  (1 mark)
  5. The advertising campaign has been successful to the extent that the mean daily sales is now 63 000.
  6. A statistical test is applied at the 5% level of significance.
  7. Find the probability that the null hypothesis would be incorrectly accepted, based on the sales of another 14 randomly selected days and assuming a standard deviation of 5000. Give your answer correct to three decimal places.  (2 marks)
Show Answers Only
  1. `12`
  2. `0.3085`
  3.   i. `H_0: \ mu = 60 \ 000, \ H_1: \ mu > 60 \ 000`
  4.  ii.  `0.0044`
  5. iii.  `text{Successful as} \ p text{-value is below 0.01.}`
  6. `n ≥ 63\ 109`
  7. `0.274`
Show Worked Solution

a.    `text{Method 1}`

♦♦ Mean mark 22%.

`text{Let} \ \ M_i\ ~\ N (75, 8^2) \ \text{for} \ \ i = 1, 2, 3, … , n`

`W_n = W_1 + W_2 + … + W_n`

`E (W_n) = E(M_1 + … + M_n) = 75n`

`text(s.d.) (W_n) = text{s.d.} (M_1 + … M_n) = 8 sqrtn`

`W_n\ ~\ N (75n, 8^2 n)`

`text{Using} \ Z\ ~\ N (0, 1)`

`text(Pr) (W_n > 1000) = 0.01`

`text(Pr) (Z > {1000 – 75 n}/{8 sqrtn}) = 0.01`
 
`n = 12.5`

`:. \ text{Largest} \ n = 12`
 

`text{Method 2}`

`text{By trial and error}`
 
`text(Pr) (M_1 + … + M_11 > 1000) ≈ 0`

`text(Pr) (M_1 + … + M_12 > 1000) ≈ 0.0002`

`text(Pr) (M_1 + … + M_13 > 1000) ≈ 0.193`

`:. \ text{Largest} \ n = 12`

 

b.   `T_i\ ~\ N (2, 0.5^2) \ \ text{for} \ \ i = 1, 2, …`

Mean mark part (b) 51%.

`text{Wait time} \ (T) = T_1 + T_2 + T_3 + T_4`

`E(T) = 4 xx 2 = 8`

`text{s.d.}(T) = text{s.d.}(T_1 + T_2 + T_3 + T_4) = sqrt4 xx 0.5 = 1`
 
`T\ ~\ N (8, 1)`

`text(Pr) (T < 7.5) = 0.3085`

`text{By CAS: normCdf} (0, 7.5, 8, 1)`

 

c.i.  `H_0: \ mu = 60 \ 000`

`H_1: \ mu > 60 \ 000`
 

c.ii.  `text(Pr) (barX > 63\ 500 | mu = 60 \ 000) = 0.004407`

  `:. \ p \ text{value} = 0.0044`

 `text{By CAS: norm Cdf} (63\ 500, oo, 60\ 000, 5000/sqrt14)`
 

c.iii.  `text{S} text{ince the} \ p text{-value is below 0.01, there is strong evidence}`

   `text{the advertising was effective (against the null hypothesis).}`

 

d.   `text(Pr) (barX > n | mu = 60 \ 000) < 0.01`

♦♦♦ Mean mark part (d) 16%.

`n ≥ 63\ 109`

`text{By CAS: inv Norm} (0.99, 60 \ 000, 5000/sqrt14)`

 

e.    `text{Similar to part (d):}`

♦♦♦ Mean mark part (e) 10%.

`text(Pr) (barX > n | mu = 60 \  000) < 0.05`

`n ≤ 62198`

`text{By CAS: invNorm} \ (0.95, 60 \ 000, 5000/sqrt14)`
 
`text{If} \ \ mu = 63\ 000, text{find probability null hypothesis incorrectly accepted:}`

`text(Pr) (barX < 62\ 198 | mu = 63\ 000) = 0.274`

`text{By CAS: normCdf} (0, 62\ 198, 63\ 000, 5000/sqrt14)`


GRAPHS, FUR2 2021 VCAA 3

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

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

`C = 10 \ 000 + k xx n` 

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

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

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

 

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

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

 

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

 

GEOMETRY, FUR2 2021 VCAA 3

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

The top three female players are from three different cities:

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

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


           
 

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

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

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

`text{Wei-Yi and Ozlem}`
 

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

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

 

c.i.

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

 

c.ii.

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

 

d.    `text{In New York time}`

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

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

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

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

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

NETWORKS, FUR2 2021 VCAA 4

Roadworks planned by the local council require 13 activities to be completed.

The network below shows these 13 activities and their completion times in weeks.
 

  1. What is the earliest start time, in weeks, of activity `K`?  (1 mark)
  2. How many of these activities have zero float time?  (1 mark)
  3. It is possible to reduce the completion time for activities `A, E, F, L` and `K`.
  4. The reduction in completion time for each of these five activities will incur an additional cost.
  5. The table below shows the five activities that can have their completion time reduced and the associated weekly cost, in dollars.
     
       
  6. The completion time for each these five activities can be reduced by a maximum of two weeks.
  7. The overall completion time for the roadworks can be reduced to 16 weeks.
  8. What is the minimum cost, in dollars, of this change in completion time?  (1 mark)
Show Answers Only
  1. `14 \ text{weeks}`
  2. `7`
  3. `$ 380 \ 000`
Show Worked Solution

a.    `text{Scan forwards:}`

`EST \ (text{activity} \ K)`

`= A \ E \ J`

`= 6 + 5 + 3`

`= 14 \ text{weeks}`
 

b.     `text{Scan backwards:}`

`text{Critical paths:}\ ADGLM\ text(and)\ AEHLM`

`:. \ text{7 activities have no float time:} \ ADEGHLM`
 

c.    `text{There are 9 possible paths}`

`ADGLM\ (19), AEHLM\ (19) , AEIM\ (14)`

`AEJK\ (17), BCEIM \ (13), BCEJK\ (16)`

`BCEHLM\ (18), BCDGLM \ (18), BFK (11)`

`text{Consider the 5 paths with completions over 16 weeks}`

` to \ text{all contain}\ A\ text{or}\ L \ text{or both.}`
  

`text{Consider} \ ADGLM, AEHLM \ (text{contains both} \ A\ text{and} \ L )`

`to \ text{reduce} \ L xx 2 \ ,  A xx 1 \ text{to reach 16 weeks}`

`to \ text{cheaper than} \ L xx 1 \ , \ A xx 2`
  

`text{Consider} \ BCEHLM , BCDGLM\ (text{both contain} \ L \ text{only})`

`to \ text{reduce} \ L xx 2 \ text{to reach 16 weeks}`
 

`text{Consider} \ AEJK \ ( text{contains} \ A \ text{only} )`

`to \ text{reduce} \ A xx 1 \ text{reach 16 weeks}.`
 

`:. \ text{Minimum cost to reduce time to 16 weeks}`

`= 2 xx 120\ 000 + 1 xx  140\ 000`

`= $ 380\ 000`

NETWORKS, FUR2 2021 VCAA 3

The network diagram below shows the local road network of Town `M`.

The number on the edges indicate the maximum number of vehicles per hour that can travel along each road in this network.

The arrows represent the permitted direction of travel.

The vertices `A, B, C, D, E` and `F` represent the intersections of the roads.
 

  1. Determine the maximum number of vehicles that can travel from the entrance to the exit per hour.  (1 mark)
  2. The local council plans to increase the number of vehicles per hour that can travel from the entrance to the exit by increasing the capacity of only one road.
  3.  i. Complete the following sentence by filling in the boxes provided.
        The road that should have its capacity increased is the road from vertex  
     
    		  to 
     
  4. ii. What should be the minimum capacity of this road to maximise the flow of vehicles from the entrance to the exit?  (1 mark)
Show Answers Only
  1. `1330`
  2.  i. `A\ text(to)\ D`
  3. ii. `780`
Show Worked Solution

a.     

`text{Minimum cut}` `= 680 + 650`
  `= 1330`

 
`:. \ text{Maximum number of vehicles} = 1330`

 

b.i.  `text{Vehicles from} \ A \ text{to} \ B \ text{restricted to 700 per hour.}`

`:. \ text{Road to increase capacity is from} \ A \ text{to} \ D.`
 

b.ii. `text{Minimum cut} = 1330 \ text{(partial)}`

 `text{Next lowest cut} = 620 + 840 = 1460`
 


 

`text{Extra capacity} = 1460 – 1330 = 130`

`:. \ text{Minimum capacity of road} \ A \ text{to}\ D \ text{should be}`

`= 650 + 130`

`= 780`

MATRICES, FUR2 2021 VCAA 3

A market research study of shoppers showed that the buying preferences for the three olive oils, Carmani (`C`), Linelli (`L`) and Ohana (`O`), change from month to month according to the transition matrix  `T` below.
 

`qquadqquadqquadqquad \ text(this month)`

`T = {:(qquad\ C quadquadqquad \ L quadquad \ O ),([(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:(C),(L),(O):} qquad text(next month)):}`
 

The initial state matrix `S_0` below shows the number of shoppers who bought each brand of olive oil in July 2021.
 

`S_0 = {:[(3200),(2000),(2800)]{:(C),(L),(O):} :}`
 

Let `S_n` represent the state matrix describing the number of shoppers buying each brand `n` months after July 2021.

  1. How many of these 8000 shoppers bought a different brand of olive oil in August 2021 from the brand bought in July 2021?  (1 mark)
  2. Using the rule `S_(n+1) = T xx S_(n)`, complete the matrix `S_1` below.  (1 mark)
     
                `S_1 = {:[(3060),(text{_____}),(text{_____})]{:(C),(L),(O):} :}`
     
  3. Consider the shoppers who were expected to buy Carmani olive oil in August 2021.
  4. What percentage of these shoppers also bought Carmani olive oil in July 2021?
  5. Round your answer to the nearest percentage.  (1 mark)
  6. Write a calculation that shows Ohana olive oil is the brand bought by 50% of these shoppers in the long run.   (1 mark)
  7. Further research suggests more shoppers will buy olive oil in the coming months.
  8. A rule to model this situation is  `R_(n+1) = T xx R_n + B`, where `R_n` represents the state matrix describing the number of shoppers `n` months after July 2021.
     

        `qquadqquadqquadqquad \ text(this month)`
        `T = {:(qquad\ C quadquadqquad \ L quadquad \ O ),([(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:(C),(L),(O):} ):} qquad text(next month) \ , \ B = {:[(200),(100),(k)]{:(C),(L),(O):} :}, \ R_0 = {:[(3200),(2000),(2800)]{:(C),(L),(O):} :}`
      
    `k` represents the extra number of shoppers expected to buy Ohana olive oil each month.
     
    If  `R_2 = {:[(3333),(2025),(3642)]:}`, what is the value of `k`?   (1 mark)

Show Answers Only
  1. `1160`
  2. `L = 1900 \ , \ O = 3040`
  3. `89text(%)`
  4. `text(See Worked Solutions)`
  5. `200`
Show Worked Solution

a.    `text{Consider matrix} \ T`

`text{Carmani – 15% bought new brand}`

`text{Linelli – 20%, Ohana – 10%}`

`:.  text{Shoppers}` `= 0.15 xx 3200 xx + 0.2 xx 2000 + 0.1 xx 2800`
  `= 1160`

 
b.    `S_1 = [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:[(3200),(2000),(2800)]:} = {:[(3060),(1900),(3040)]:}`
 

`:. \ L = 1900 \ , \ O = 3040`
 

c.    `text{Carmani purchasers in August)} \ = 3060 \ text{(see part b)}`

`text{Carmani purchasers in July} = 3200`

`text{Carmani purchasers in both July and August}`

`= 0.85 xx 3200`

`= 2720`
 

`:.\ text{% of August purchasers who bought in July}`

`= 2720/3060 xx 100`

`= 88.88 …`

`=89text(%)`
 

d.    `S = T^50 xx S_0 = {:[(2400),(1600),(4000)]:}`

`text{Total shoppers} = 8000`

`:.\ text{Ohana purchasers (in long run)}`

`= 4000/8000 xx 100`

`= 50text(%)`
 

e.     `R_1` `= [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)] [(3200),(2000),(2800)] + [(200),(180),(k)]` 
  `R_2` `= [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)] [(3260),(2000),(k+3040)] + [(200),(100),(k)]`
    `= [(3171 + 0.05 (k + 3040)),(text{not required}),(text{not required})]`

 
`text{Equating matrices, solve for} \ k:`

`3171 + 0.05 (k + 3040) = 3333`

`:. k = 200`

CORE, FUR2 2021 VCAA 8

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

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

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

  1. Calculate the balance of this loan after the first fortnightly repayment is made.  (1 mark)
  2. Show that the compound interest rate for this loan is 2.6% per annum.  (1 mark)
  3. For the loan to be fully repaid, to the nearest cent, Sienna's final repayment will be a larger amount.
  4. Determine this final repayment amount.
  5. Round your answer to the nearest cent.  (1 mark)
Show Answers Only
  1. `$ 569 \ 377`
  2. `2.6text(%)`
  3. `$ 1198.56`
Show Worked Solution
a.   `S_1` `= 1.001 xx 570 \ 000 – 1193`
    `= $ 569 \ 377`


b.  `text{Fortnights in 1 year} = 26`

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

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

 

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

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

 
`=> N = 650.0046 …`
 

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

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

 
`=> FV = -5.59`
 

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

CORE, FUR2 2021 VCAA 7

Sienna owns a coffee shop.

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

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

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

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

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

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

 

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

 

c.  `M_n = 12 \ 000 + (-0.05) xx n`

CORE, FUR2 2021 VCAA 5

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

difference = winning time women – winning time men

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

  1. Apply a reciprocal transformation to the variable difference to linearise the data. Fit a least squares line to the transformed data and write its equation below.
  2. Round the values of the intercept and the slope to four significant figures.  (2 marks)
  3. Use the equation from part a. to predict, in seconds, the difference the women's and men's winning times in the year 2032.
  4. Round your answer to one decimal place.  (1 mark)
Show Answers Only
  1. `1/text{difference} = -2.234 + 0.001209 xx text{year}`
  2. `4.5 \ text{seconds}`
Show Worked Solution

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

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

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

 

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

 

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

CORE, FUR2 2021 VCAA 4

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

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

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

The equation of the least squares line for men is

winning time men = 356.9 – 0.1544 × year

The equation of the least squares line for women is

winning time women = 538.9 – 0.2430 × year

  1. Draw the least squares line for winning time women on the time series plot above.  (1 mark)
  2. The difference between the women's predicted winning time and the men's predicted winning time can be calculated using the formula.
  3.       difference = winning time women – winning time men
  4. Use the equation of the least squares lines and the formula above to calculate the difference predicted for the 2024 Olympic Games.
  5. Round your answer to one decimal place.  (2 marks)
  6. The Olympic Games are held every four years. The next Olympic Games will be held in 2024, then 2028, 2032 and so on.
  7. In which Olympic year do the two least squares lines predict that the wining time for women will first be faster than the winning time for men in the 100 m freesytle?  (2 marks)
Show Answers Only
  1. `text{See Worked Solutions}`
  2. `2.7 \ text{seconds}`
  3. `2056`
Show Worked Solution

a.   

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

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

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

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

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

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

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

`text{year} = 2054.17 …`

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

CORE, FUR2 2021 VCAA 3

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

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

The equation of the least squares line is

winning time = 357.1 – 0.1515 × year

The coefficient of determination is 0.8794

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

a.      `text{year}`
 

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

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

`:. \ r = -0.938`
 

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

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

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

 

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

 

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

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

CORE, FUR2 2021 VCAA 2

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

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

                      time800 = 0.03931 + 5.2756 × time200

  1. Round the values for the intercept and the slope to three significant figures. Write your answers in the boxes provided.   (1 mark)
     

    time800=  
     
    		   +  
     
    		   × time200
  2. The mean and the standard deviation for each variable, time200 and time800, are shown in the table below.
     

     
    The equation of the least squares line is
     
            time800 = 0.03931 + 5.2756 × time200 
     
    Use this information to calculate the coefficient of determination as a percentage.

    Round your answer to the nearest percentage.  (2 marks)

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

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

b.    `b = r xx s_y/s_x`
 

`text{Solve for} \ r:`

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

Mechanics, SPEC2 2021 VCAA 5

A mass of `m_1` kilograms is placed on a plane inclined at 30° to the horizontal. It is connected by a light inextensible string to a second mass of `m_2` kilograms that hangs below a frictionless pulley situated at the top end of the incline, over which the string passes.
 


 

  1. Given that the inclined plane is smooth, find the relationship between `m_1` and `m_2` if the mass `m_1` moves down the plane at constant speed.  (2 marks)

The masses are now placed on a rough plane inclined at 30°, with the light inextensible string passing over a frictionless pulley in the same way, as shown in the diagram above. Let `N` be the magnitude of the normal force exerted on the mass `m_1` by the plane. A resistance force of magnitude `lambdaN` acts on and opposes the motion of the mass `m_1`.

  1. The mass `m_1` moves up the plane.
  2.   i. Mark and label all forces acting on this mass on the diagram above.  (1 mark)
  3.  ii. Taking the direction up the plane as positive, find the acceleration of the mass `m_1` in terms of `m_1`, `m_2` and `lambda`.  (2 marks)

Some time after the masses have begun to move, the mass `m_2` hits the ground at 4.5 ms`\ ^(-1)` and the string becomes slack. At this instant, the mass `m_1` is at the point `P` on the plane, which is 2 m from the pulley. Take the value of `lambda` to be 0.1

  1. How far from point `P` does the mass `m_1` travel before it starts to slide back down the plane?
  2. Give your answer in metres, correct to two decimal places.  (2 marks)
  3. Find the time taken, from when the string becomes slack, for the mass `m_1` to return to point `P`.
  4. Give your answer correct to the nearest tenth of a second.  (3 marks)
Show Answers Only
  1. `2m_2`
  2. i. 
       
     
  3. ii. `a = ((2m_2 – m_1 – lambda m_1 sqrt3)g)/(m_1 + m_2)`
  4. `s = 1.76\ text{m}`
  5. `1.7\ text{seconds}`
Show Worked Solution

a.   `m_1g sin30 – m_2g = (m_1 + m_2)a`

`text(S)text(ince)\ m_1\ text(moves at constant speed,)\ a = 0`

`m_1 g · 1/2 – m_2 g` `= 0`
`m_1` `= 2m_2`

 
b.i.   

 

b.ii.   `text(S)text(ince)\ m_1\ text(is moving up the slope)`

♦ Mean mark part (b)(ii) 43%.

`m_2g – m_1 g · 1/2 – lambdam_1 g · cos 30` `= (m_1 + m_2)a`
`m_2g – m_1 g · 1/2 – lambdam_1 g · cos sqrt3/2` `= (m_1 + m_2)a`

`:. a = ((2m_2 – m_1 – lambda m_1 sqrt3)g)/(2(m_1 + m_2))`

 

c.   `text(After)\ m_2\ text(hits the ground)`

♦♦♦ Mean mark part (c) 17%.
`m_1a` `=-m_1g*1/2 – lambda m_1 g sqrt3/2`
`a` `= -g/2(1 + lambda sqrt3)`
  `= -g/2(1 + 0.1 xx sqrt3)`

 
`text(By CAS, solve)\ \ v^2 = u^2 + 2as,\ text(for)\ \ s:`

`0 = 4.5^2 – g(1 + 0.1 xx sqrt3)s`

`s = 1.76\ text{m (to 2 d.p.)}`

 

d.   `u = 4.5\ \ text(ms)^(-1), s = 1.76\ text(m),\ a = -g/2(1 + 0.1sqrt3)`

♦♦♦ Mean mark part (d) 7%.

`text(Let)\ \ t_1 = text(Time travelling up slope until stopping)`

`s = ut_1 + 1/2at^2`

`1.76 = 4.5t_1 – 1/2 · g/2(1 + 0.1sqrt3)t_1^2`

`t_1 = 0.78\ text(seconds)`
 

`text(Let)\ t_2 =\ text(time travelling down the slope)`

`=>\ text(friction is reversed)`

`a` `= g/2(1 – 0.1sqrt3)`
`1.76` `= 0 xx t_2 + 1/2 · g/2(1 – 0.1sqrt3)  t_2^2`
`t_2` `= 0.93\ text(seconds)`

 

`:.\ text(Total time)` `= t_1 + t_2`
  `= 0.78 + 0.93`
  `= 1.7\ text{seconds (to 1 d.p.)}`