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PHYSICS, M7 2021 HSC 11 MC

What is the peak wavelength of electromagnetic radiation emitted by a person with a body temperature of 37°C (310 K)?

  1. `9.3 × 10^(-6)\ `m
  2. `7.8 × 10^(-5)\ `m
  3. `9.3 × 10^(-3)\ `m
  4. `7.8 × 10^(-2)\ `m
Show Answers Only

`A`

Show Worked Solution
`lambda` `=(b)/(T)`  
  `=(2.898 xx10^(-3))/(310)`  
  `=9.3 xx10^(-6)`  

 
`=>A`

PHYSICS, M7 2021 HSC 8 MC

Light from a point source is incident upon a circular metal disc, forming a shadow on a screen as shown. A bright spot is observed in the centre of the shadow.
 

The bright spot is caused by a combination of

  1. interference and refraction.
  2. refraction and polarisation.
  3. polarisation and diffraction.
  4. diffraction and interference.
Show Answers Only

`D`

Show Worked Solution

Light diffracts around metal disc and constructive interference causes the bright spot.

`=>D`

PHYSICS, M7 2021 HSC 5 MC

The spectrum of an object is shown.
 

Which row of the table correctly identifies the most likely source of the spectrum and the features labelled \(Y\)?

\begin{align*}
\begin{array}{l}
\rule{0pt}{1.5ex} \ \rule[-0.5ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{1.5ex}\text{Source of spectrum}\rule[-0.5ex]{0pt}{0pt}& \textit{Features labelled \(Y\)} \\
\hline
\rule{0pt}{2.5ex}\text{Star}\rule[-1ex]{0pt}{0pt}&\text{Absorption lines}\\
\hline
\rule{0pt}{2.5ex}\text{Discharge tube}\rule[-1ex]{0pt}{0pt}& \text{Absorption lines}\\
\hline
\rule{0pt}{2.5ex}\text{Star}\rule[-1ex]{0pt}{0pt}& \text{Emission lines} \\
\hline
\rule{0pt}{2.5ex}\text{Discharge tube}\rule[-1ex]{0pt}{0pt}& \text{Emission lines} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

Overall shape resembles a blackbody curve so the source of the spectrum is a star. The wavelengths of lower intensity are absorption lines.

\(\Rightarrow A\)

PHYSICS, M7 2021 HSC 4 MC

An astronaut is travelling towards Earth in a spaceship at 0.8c. At regular intervals, a radio pulse is sent from the spaceship to an observer on Earth.

Which quantity would the astronaut and the observer measure to be the same?

  1. Length of the spaceship
  2. Speed of the radio pulses
  3. Momentum of the astronaut
  4. Time interval between the radio pulses
Show Answers Only

`B`

Show Worked Solution

Speed of radio waves (light) is equal in all frames of reference. All other quantities depend on the observer.

`=>B`

PHYSICS, M6 2021 HSC 21

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

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

a.   0.04  Nm

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

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

 

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

CHEMISTRY, M7 2021 HSC 7 MC

Methanol undergoes a substitution reaction using hydrogen bromide.

Compared to methanol, the product of this reaction has a

  1. lower boiling point.
  2. lower molecular mass.
  3. greater solubility in water.
  4. different molecular geometry at the carbon atom.
Show Answers Only

`A`

Show Worked Solution

→ The product of the substitution reaction between methanol and hydrogen bromide is bromomethane.

→ Methanol contains an OH functional group and thus can form strong hydrogen bonds.

→ Bromomethane can only form dipole-dipole forces which are weaker than hydrogen bonds. As a result, bromomethane requires less energy to break these intermolecular forces, resulting in a lower boiling point than methanol.

`=> A`

CHEMISTRY, M7 2021 HSC 8 MC

Which diagram shows the expected arrangement of soap anions in an emulsion?
 

Show Answers Only

`D`

Show Worked Solution

→ Soap molecules contain a non-polar hydrophobic hydrocarbon tail and a polar hydrophilic head.

→ The non-polar tail forms dispersion forces with the oil molecule while the polar head forms ion-dipole forces with the polar water molecules.

→ The resulting orientation has the tail within the oil and the head group at the surface of the oil and water.

`=>D`

CHEMISTRY, M7 2021 HSC 3 MC

The structure of a compound is shown.
 


 

What is the preferred IUPAC name of this compound?

  1. `N`-methylpropanamide
  2. `N`-methylpropanamine
  3. `N`-propanylamine
  4. `N`-propylmethanamide
Show Answers Only

`A`

Show Worked Solution

Compound is a secondary amide.

→ Pre-fix (longest carbon chain) → -propan

→ Suffix (functional group) → -amide

→ Alkyl chain bound to the amide nitrogen is treated as a substituent and as it is bound to the nitrogen atom → `N`-methyl

∴ Compound name is `N`-methylpropanamide

`=>A`

CHEMISTRY, M8 2021 HSC 2 MC

Which ion can be detected using a precipitation reaction with silver nitrate?

  1. \(\ce{Ag+}\)
  2. \(\ce{Cl-}\)
  3. \(\ce{Mg^2+}\)
  4. \(\ce{NO3-}\)
Show Answers Only

`B`

Show Worked Solution

Silver Nitrate →`\ text{Ag}^(+)\ text{and}\ text{NO}_(3)^(\ \ -)`

Solubility rules → all nitrates are soluble

→ `text{Ag}^(+)` is involved in the precipitation reaction

→ `text{Ag}^(+)` will form a precipitate with `text{Cl}^(-)`

`=>B`

BIOLOGY, M7 2021 HSC 8 MC

Howard Florey conducted a breakthrough experiment in the development and use of antibiotics. He infected eight mice with Streptococcus bacteria. Four mice were given penicillin and survived while the four untreated mice died.

What conclusion could be drawn from the data obtained?

  1. The experiment should be repeated with more mice.
  2. The use of penicillin causes antibiotic resistance in mice.
  3. Penicillin may be used on humans safely to treat bacterial infections.
  4. Penicillin may have played a role in the survival of the four treated mice.
Show Answers Only

`D`

Show Worked Solution

Conclusion must be directed to the research problem.

→ Option A is not a conclusion.

→ Options B and C are definitive beyond the supporting data.

`=>D`

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^{prime}(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^{primeprime}-text(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^{prime}(x) = 0`

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

d.i.  `text(Solve)\ {2a^{3}}/{3 a^{2}-1}=1.1\ text(for)\ a:`

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

d.ii.  `text(Solve)\ 1 <= (2a^(3))/(3a^(2)-1) < 1.1\ text(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^{prime}(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 cdot((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^{prime}(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.)}`

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.
    1. Express, in terms of `n`, the probability that one or more deliveries will not arrive on time or earlier.   (1 mark)

    2. 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)
  3. 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`
  4. 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
  5. Let the probability that a delivery is made after 4 pm be `y`.
  6. 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, FUR1 2020 VCAA 9 MC

The graph below shows the volume of water, `V` litres, in a tank at time `t` minutes.
 

The 1100 litre tank took 80 minutes to empty.

The volume of water in the tank initially decreased by 300 litres in 20 minutes.

It then did not change for a period of time.

Finally, the volume of water in the tank decreased at the rate of 32 litres per minute.

The period of time, in minutes, for which the volume of water in the tank did not change is

  1. 20
  2. 25
  3. 35
  4. 55
  5. 65
Show Answers Only

`C`

Show Worked Solution

`text(Find)\ \ V\ \ text(when)\ \ t = 20:`

`V = 1100 – 300 = 800\ text(L)`

`text(Time to empty 800 L)`

`= 800/32`

`= 25\ text(minutes)`
 

`:.\ text(Period where volume did not change)`

`= 80 – 20 – 25`

`= 35`
 

`=>  C`

GRAPHS, FUR1 2020 VCAA 6 MC

Justin makes and sells electrical circuit boards.

He has one fixed cost of $420 each week.

Each circuit board costs $15 to make.

The selling price of each circuit board is $27.

The weekly profit if Justin makes and sells 200 circuit boards per week is

  1. $1980
  2. $2400
  3. $2820
  4. $4980
  5. $5400
Show Answers Only

`A`

Show Worked Solution

`text(Revenue) = 200 xx 27 = $5400`

`text(C)text(ost) = 420 + 200 xx 15 = $3420`

`:.\ text(Profit)` `= 5400 – 3420`
  `= $1980`

`=>  A`

GRAPHS, FUR1 2020 VCAA 5 MC

The graphs shown below represent the same relationship, `y = kx^n`.

The gradient of the straight line in Graph 2 is

  1. `15/2`
  2. `15/8`
  3. `450/8`
  4. `2/15`
  5. `8/15`
Show Answers Only

`B`

Show Worked Solution

`(4, 30) quad text(on graph 1 corresponds to)`

`(4^2, 30) = (16, 30) quad text(on graph 2)`

`text(Gradient)` `= (30 – 0)/(16 – 0)`
  `= 15/8`

`=>  B`

Calculus, MET2 2020 VCAA 16 MC

A right-angled triangle, `OBC`, is formed using the horizontal axis and the point  `C(m, 9 - m^2)`, where  `m ∈ (0, 3)`, on the parabola  `y = 9 - x^2`, as shown below.
 

The maximum area of the triangle  `OBC`  is

  1. `sqrt3/3`
  2. `(2sqrt3)/3`
  3. `sqrt3`
  4. `3sqrt3`
  5. `9sqrt3`
Show Answers Only

`D`

Show Worked Solution
`A` `=1/2 m (9-m^2)`  
  `=9/2m – m^3/2`  

  
`text(Solve:)\ (dA)/(dm)=0\ \ text(for)\ \ m`

`m=sqrt3`
 

`:. A_max` `=(9sqrt3)/2 – (3sqrt3)/2`  
  `=3sqrt3`  

 
`=>D`

Calculus, SPEC1 2020 VCAA 2

Evaluate  `int_(-1)^0 (1 + x)/sqrt(1 - x)\ dx`.  (3 marks)

Show Answers Only

`(8 sqrt 2)/3 – 10/3`

Show Worked Solution
`text(Let)\ \ u` `= 1 – x \ => \ x = 1 – u`
`(du)/(dx)` `= -1 \ => \ dx = -du`

 

`text(When)\ \ x` `= 0,\ u = 1`
`x` `= -1,\ u = 2`

 

`int_(-1)^0 (1 + x)/sqrt(1 – x)\ dx` `= -int_2^1 (2 – u)/sqrt u\ du`
  `= int_1^2 2u^(-1/2) – u^(1/2)\ du`
  `= [4u^(1/2) – 2/3u^(3/2)]_1^2`
  `= 4 sqrt 2 – (4 sqrt 2)/3 – (4 – 2/3)`
  `= (8 sqrt 2)/3 – 10/3`

Calculus, 2ADV C2 SM-Bank 1 MC

If  `f(x)=e^(g(x^(2)))`, where `g` is a differentiable function, then  `f^(′)(x)`  is equal to

  1. `2xe^(g(x^(2)))`
  2. `2xg(x^(2))e^(g(x^(2)))`
  3. `2xg^(′)(x^(2))e^(g(x^(2))`
  4. `2xg^(′)(2x)e^(g(x^(2)))`
Show Answers Only

`C`

Show Worked Solution

`f(x)=e^(g(x^2))`

`text{Using the chain rule (twice):}`

`f^(′)(x)` `=d/dx[g(x^2)] * e^(g(x^2))`  
  `=2x*g^(′)(x^2)*e^(g(x^2))`  

 
`=> C`

Statistics, MET2 2020 VCAA 11 MC

The lengths of plastic pipes that are cut by a particular machine are a normally distributed random variable, `X`, with a mean of 250 mm.

`Z` is the standard normal random variable.

If  `text{Pr}(X<259)=1-text{Pr}(Z>1.5)`, then the standard deviation of the lengths of plastic pipes, in millimetres, is

  1. 1.5
  2. 3
  3. 6
  4. 9
  5. 12
Show Answers Only

`C`

Show Worked Solution

`X\ ~\ N(250, sigma^2)`

`text(Pr)(X < 259)=text(Pr)(Z < (259-250)/(sigma))=text(Pr)(Z < 1.5)`

`(259-250)/sigma` `=1.5`  
`1.5 sigma` `=9`  
`sigma` `=6`  

 
`=>C`

Algebra, MET2 2020 VCAA 10 MC

Given that  `log_(2)(n+1)=x`, the values of  `n`  for which  `x`  is a positive integer are

  1. `n=2^(k),k inZ^(+)`
  2. `n=2^(k)-1,k inZ^(+)`
  3. `n=2^(k-1),k inZ^(+)`
  4. `n=2k-1,k inZ^(+)`
  5. `n=2k,k inZ^(+)`
Show Answers Only

`B`

Show Worked Solution

`text(If)\ \ x in ZZ^+ ,`

`log_2(n+1)` `in ZZ^+`  
`n+1` `=2^k\ \ text(for)\ \ k in ZZ^+`  
`n` `=2^k -1`  

 
`=>B`

Calculus, MET2 2020 VCAA 7 MC

If  `f(x)=e^(g(x^(2)))`, where `g` is a differentiable function, then  `f^(')(x)`  is equal to

  1. `2xe^(g(x^(2)))`
  2. `2xg(x^(2))e^(g(x^(2)))`
  3. `2xg^(')(x^(2))e^(g(x^(2))`
  4. `2xg^(')(2x)e^(g(x^(2)))`
  5. `2xg^(')(x^(2))e^(g(2x))`
Show Answers Only

`C`

Show Worked Solution

`f(x)=e^(g(x^2))`

`text{Using the chain rule (twice):}`

`f^{‘}(x)` `=d/dx[g(x^2)] * e^(g(x^2))`  
  `=2x*g^{‘}(x^2)*e^(g(x^2))`  

 
`=> C`

Calculus, MET2 2020 VCAA 6 MC

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

 The corresponding part of the graph of  `y=f(x)`  is best represented by 
 

Show Answers Only

`B`

Show Worked Solution

`text(By elimination:)`

`text(Two stationary points exist where)\ \ f^{‘}(x) =0\ \ text(and)\ \ x<0.`

`=>\ text(Eliminate)\ \ E, C, D`

`text{No other stationary points exist}\ →\ f^{‘}(x) !=0\ \ text(for)\ \ x>0.`

`=>\ text(Eliminate)\ \ E`

`=> B`

Calculus, 2ADV C4 SM-Bank 4 MC

 Let  `f^(')(x)=(2)/(sqrt(2x-3))`. 

If  `f(6)=4`, then
 

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

`=>C`

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

 
`text(When)\ \ x=6, \ f(x)=4:`

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

`:. f(x) = 2sqrt(2x-3) – 2`

`=>C`

Algebra, MET2 2020 VCAA 4 MC

The solutions of the equation  `2cos(2x-(pi)/(3))+1=0`  are

  1. `x=(pi(6k-2))/(6)" or "x=(pi(6k-3))/(6)," for "k in Z`
  2. `x=(pi(6k-2))/(6)" or "x=(pi(6k+5))/(6)," for "k in Z`
  3. `x=(pi(6k-1))/(6)" or "x=(pi(6k+2))/(6)," for "k in Z`
  4. `x=(pi(6k-1))/(6)" or "x=(pi(6k+3))/(6)," for "k in Z`
  5. `x=pi" or "x=(pi(6k+2))/(6)," for "k in Z`
Show Answers Only

`D`

Show Worked Solution
`2cos(2x-(pi)/(3))+1` `=0`  
`cos(2x-(pi)/(3))` `=- 1/2`  
`2x-(pi)/(3)` `=(2pi)/3\ \ text(or)\ \ -(2pi)/3`  

 
`text(General Solution:)`

`2x-(pi)/(3)` `=2kpi+(2pi)/3`  
`2x` `=2kpi+pi`  
`x` `=kpi+pi/2`  
  `=pi/6(6k+3)`  

 

`2x-(pi)/(3)` `=2kpi-(2pi)/3`  
`2x` `=2kpi-pi/3`  
`x` `=kpi-pi/6`  
  `=pi/6(6k-1)`  

`=>D`

Algebra, MET2 2020 VCAA 2 MC

Let  `p(x)=x^{3}-2 a x^{2}+x-1`, where  `a \in R`. When `p` is divided by  `x+2`, the remainder is 5.

The value of `a` is

  1. `\ \ \ \ 2`
  2. `- 7/4`
  3. `\ \ \ 1/2`
  4. `- 3/2`
  5. `-2`
Show Answers Only

`E`

Show Worked Solution
`P(-2)` `=5`  
`5` `=(-2)^3-2a(-2)^2-2-1`  
`5` `=-8-8a-2-1`  
`8a` `=-16`  
`:.a` `=-2`  

 
`=>E`

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`

Statistics, STD2 S5 SM-Bank 7

800 participants auditioned for a stage musical. Each participant was required to complete a series of ability tests for which they received an overall score.

The overall scores were approximately normally distributed with a mean score of 69.5 points and a standard deviation of 6.5 points.

Only the participants who scored at least 76.0 points in the audition were considered successful.

How many of the participants were considered unsuccessful?   (2 marks)

Show Answers Only

`672`

Show Worked Solution

`mu = 69.5 \ , \ sigma= 6.5`

`ztext{-score} \ (76) = {76 – 69.5}/6.5 = 1`

`:.\  text{Unsuccessful}` `=84text(%) xx 800`
  `= 672`

Statistics, STD2 S1 SM-Bank 5 MC

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

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

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

`D`

Show Worked Solution

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

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

`IQR = 159 – 148 = 11`

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

 
`=> D`

Complex Numbers, EXT2 N2 SM-Bank 10

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

  1. State the relationship between `z_2` and `z_3`.  (1 mark)

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

Show Answers Only
  1. `z_2 = barz_3`
  2. `alpha = -3, beta = 9, gamma = -27`
Show Worked Solution

i.   `text(By conjugate root theory)`

`z_2 = barz_3`

 

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

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

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

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

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

 

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

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

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`

Statistics, EXT1 S1 SM-Bank 16

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

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

Find the smallest sample size for which the standard deviation of `overset^p` is less than 0.08.  (2 marks)

Show Answers Only

`38`

Show Worked Solution

`E(overset^p) = p = 0.6`

`text(Var)(overset^p) = (0.6(1 – 0.6))/n = 0.24/n`

`sigma(overset^p) = sqrt(0.24/n)`
 

`text(If)\ \ sigma(overset^p) < 0.08,`

`sqrt(0.24/n)` `< 0.08`
`0.0064n` `> 0.024`
`n` `> 0.24/0.0064`
`n` `> 37.5`

 

`:. text(Smallest)\ \ n = 38`

Calculus, 2ADV C3 SM-Bank 1 MC

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

The value of `a` is

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

`B`

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

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

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

`m_text{tang} = 2 – a`

`text{Equating gradients:}`

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

 
`=> B`

Functions, 2ADV F2 SM-Bank 16

Let  `f(x) = x^2 - 4`

Let the graph of `g(x)` be a transformation of the graph of `f(x)` where the transformations have been applied in the following order:
• dilation by a factor of  `1/2`  from the vertical axis (parallel to the horizontal axis)
• translation by two units to the right (in the direction of the positive horizontal axis

Find `g(x)` and the coordinates of the horizontal axis intercepts of the graph of `g(x)`.  (3 marks)

Show Answers Only

`(1,0) and (3,0)`

Show Worked Solution

`text(1st transformation)`

`text(Dilation by a factor of)\ 1/2\ text(from the)\ ytext(-axis:)`

`x^2 – 4 \ => \ (x/(1/2))^2 -4 = 4x^2-4`
 

`text(2nd transformation)`

`text(Translation by 2 units to the right:)`

`4x^2-4 \ => \ g(x) = 4(x-2)^2 – 4`
 

`xtext(-axis intercept of)\ g(x):`

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

 
`x-2=1 \ => \ x=3`

`x-2=-1 \ => \ x=1`

 
`:.\ text(Horizontal axis intercepts occur at)\ (1,0) and (3,0).`

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)}`

GRAPHS, FUR2 2021 VCAA 2

z

Christy sells blocks of land at the housing estate.

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

The maximum number of blocks sold each week is 20.

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

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

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

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

a.    `6 \ text{blocks}`
 

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

c.     `y <= 3/5 x`

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}`

GEOMETRY, FUR2 2021 VCAA 2

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

 
Each side wall has the following dimensions.
 

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

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

 
             

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

 

b.

`x = 4.57 – 2.13 = 2.44 \ text{m}`
  

`text{Using Pythagoras}`

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

 

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

 
c. 

`text{Using cosine rule:}`

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

GEOMETRY, FUR2 2021 VCAA 1

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

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

Squash balls may be sold in cube-shaped boxes.

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

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

     

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

 

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

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

 

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

 

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

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

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

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

NETWORKS, FUR2 2021 VCAA 2

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

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

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

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

  1. What is the shortest distance, in kilometres, between Town `G` and Town `M`?   (1 mark)

  2. George plans to travel to Maggie's house. He will pass through all the towns shown above.
  3. George plans to take the shortest route possible.
  4. Which town will George pass through twice?    (1 mark)

Show Answers Only
  1. `86 \ text{km}`
  2. `text{Passes through town} \ K \ text{twice.}`
Show Worked Solution
a.  `text{Shortest route}` `= G \ O \ N \ M`
    `= 28 + 42 + 16`
    `= 86 \ text{km}`

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

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

NETWORKS, FUR2 2021 VCAA 1

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

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

The floor plan is represented by the graph below.

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

One edge is missing from the graph.
 

  1. On the graph above, draw the missing edge.   (1 mark)

  2. What is the degree of vertex `E`?   (1 mark)

  3. Maggie hires a cleaner to clean the house.
  4. It is possible for the cleaner to enter the house from the outside area, `F`, and walk through each room only once, cleaning each room as he goes and finishing in the outside area, `F`.
  5.  i. Complete the following to show one possible route that the cleaner could take.   (1 mark)

        
         
    ii. What is the mathematical term for such a journey?   (1 mark)

Show Answers Only
  1.  
  2. `2`
  3.  i. `FABEDCF\ text(or)\ FCDEBAF`
  4. ii. `text{Hamiltonian cycle}`
Show Worked Solution

a.

b.  `text{Degree} = 2`
 

c.i.  `FABEDCF\ text(or)\ FCDEBAF`

c.ii.  `text{Hamiltonian cycle}`

MATRICES, FUR2 2021 VCAA 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):} :}`

  1. Consider the shoppers who were expected to buy Carmani olive oil in August 2021.
  2. What percentage of these shoppers also bought Carmani olive oil in July 2021?
  3. Round your answer to the nearest percentage.   (1 mark)

  4. Write a calculation that shows Ohana olive oil is the brand bought by 50% of these shoppers in the long run.   (1 mark)

  5. Further research suggests more shoppers will buy olive oil in the coming months.
  6. 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):} :}`

  1. `k` represents the extra number of shoppers expected to buy Ohana olive oil each month.
  2. 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`

MATRICES, FUR2 2021 VCAA 2

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

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

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

`qquadqquadqquadqquadqquadqquadqquadqquadqquad text(receiver)`

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

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

In this matrix:

  • the '1' in row `A`, column `B` indicates that Alex can send information to Brie
  • the '0' in row `D`, column `C` indicates that Dex cannot send information to Chai.
  1. Which two staff members can send information directly to each other?   (1 mark)

  2. Elena needs to send documents to Chai.
  3. What is the sequence of communication links that will successfully get the information from Elena to Chai?   (1 mark)

  4. Matrix  `M^2` below is the square of `M` and shows the number of two-step communication links between each pair of staff members.
     

    `qquadqquadqquadqquadqquadqquadqquadqquadqquad text(receiver)`

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

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

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

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

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

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

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

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

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

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

`text{These are:}`

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

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