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Vectors, SPEC2 SM-Bank 23

A cube with side length 3 units is pictured below.
 

     
 

  1. Calculate the magnitude of vector `vec(AG)`.  (1 mark)
  2. Find the acute angle between the diagonals `vec(AG)` and `vec(BH)`.  (3 marks)
Show Answers Only
  1. `3 sqrt 3\ text(units)`
  2. `70^@32′`
Show Worked Solution

i.   `A(3, 0 , 0), \ \ G(0, 3, 3)`

  `vec(AG)` `= ((0), (3), (3)) – ((3), (0), (0)) = ((text{−3}), (3), (3))`
  `|\ vec(AG)\ |` `= sqrt (9 + 9 + 9)`
    `= 3 sqrt 3\ text(units)`

 

ii.    `H (3, 3, 3)`
  `vec(BH) = ((3), (3), (3))`
`vec(AG) ⋅ vec(BH)` `= |\ vec(AG)\ | ⋅ |\ vec(BH)\ |\ cos theta`
`((text{−3}), (3), (3)) ⋅ ((3), (3), (3))` `= sqrt (9 + 9 + 9) ⋅ sqrt (9 + 9 + 9) cos theta`
`-9 + 9 + 9` `= 27 cos theta`
`cos theta` `= 1/3`
`theta` `= 70.52…`
  `= 70^@32′`

GRAPHS, FUR1 2020 VCAA 3 MC

The delivery fee for a parcel, in dollars, charged by a courier company is based on the weight of the parcel, in kilograms.

This relationship is shown in the step graph below for parcels that weigh up to 20 kg.
 

Which one of the following statements is not true?

  1. The delivery fee for a 4 kg parcel is $20.
  2. The delivery fee for a 12 kg parcel is $26.
  3. The delivery fee for a 13 kg parcel is the same as the delivery fee for a 20 kg parcel.
  4. The delivery fee for a 10 kg parcel is $14 more than the delivery fee for a 2 kg parcel.
  5. The delivery fee for a 12 kg parcel is $18 more than the delivery fee for a 2 kg parcel.
Show Answers Only

`E`

Show Worked Solution

`text(Consider option)\ E:`

`text(Delivery fee for 12kg parcel) = $26`

`text(Delivery fee for 2kg parcel) = $12\ text{($14 difference)}`

`:.\ text(Statement)\ E\ text(is correct.)`

`=>  E`

GRAPHS, FUR1 2020 VCAA 1 MC

The graph below shows the number of people who attended a market over a seven-hour time period.
 

For how many hours were there at least 600 people at the market?

  1. 2
  2. 3
  3. 4
  4. 6
  5. 7
Show Answers Only

`C`

Show Worked Solution

`text(Hours where graph) >= 600 = 6-2=4`

`=>  C`

GEOMETRY, FUR1 2020 VCAA 3 MC

Two trees stand on horizontal ground.

A 25 m cable connects the two trees at point A and point B, as shown in the diagram below.
 


 

Point A is 45 m above the ground and point B is 30 m above the ground.

The horizontal distance, in metres, between point A and point B is

  1. 10
  2. 15
  3. 20
  4. 30
  5. 35
Show Answers Only

`C`

Show Worked Solution

`CB` `= sqrt(AB^2 – AC^2)`
  `= sqrt(25^2 – 15^2)`
  `= 20`

 
`=>  C`

GEOMETRY, FUR1 2020 VCAA 2 MC

A flag consists of three different coloured sections: red, white and blue.

The flag is 3 m long and 2 m wide, as shown in the diagram below.
 

The blue section is an isosceles triangle that extends to half the length of the flag.

The area of the blue section, in square metres, is

  1.  0.75
  2.  1.5
  3.  2
  4.  3
  5.  6
Show Answers Only

`B`

Show Worked Solution
`text(Area)` `= 1/2 xx 2 xx 1.5`
  `= 1.5\ text(m²)`

 
`=>  B`

NETWORKS, FUR1 2020 VCAA 2 MC

Consider the graph below.
 


 

Which one of the following is not a Hamiltonian cycle for this graph?

  1. `ABCDFEGA`
  2. `BAGEFDCB`
  3. `CDFEGABC`
  4. `DCBAGFED`
  5. `EGABCDFE`
Show Answers Only

`D`

Show Worked Solution

`text(Hamiltonian cycle – start and finish at the same)`

`text(vertex and touch all other vertices once.)`

`text(Option)\ D\ text(is not a valid path as it needs an edge)\ E\ text(to)\ D.`

`=>  D`

MATRICES, FUR1 2020 VCAA 5 MC

The diagram below shows the direct communication links that exist between Sam (S), Tai (T), Umi (U) and Vera (V). For example, the arrow from Umi to Vera indicates that Umi can communicate directly with Vera.
 


 

A communication matrix can be used to convey the same information.

In this matrix:

  • a ‘1’ indicates that a direct communication link exists between a sender and a receiver
  • a ‘0’ indicates that a direct communication link does not exist between a sender and a receiver.

The communication matrix could be

A.  

`{:(qquadqquadqquadqquadqquadqquadquad receiver),(\ quad qquadqquadqquadqquad qquadS\ \ \ Tquad U\ \ V),(sender qquad{:(S),(T),(U),(V):}[(0,1,0,1),(0,0,0,1),(0,1,0,1),(1,0,1,0)]):}`

 

 B.  

`{:(qquadqquadqquadqquadqquadqquadquad receiver),(\ quad qquadqquadqquadqquad qquadS\ \ \ Tquad U\ \ V),(sender qquad{:(S),(T),(U),(V):}[(0,1,0,1),(1,0,0,1),(0,1,0,1),(1,1,1,0)]):}`

 
C.  

`{:(qquadqquadqquadqquadqquadqquadquad receiver),(\ quad qquadqquadqquadqquad qquadS\ \ \ Tquad U\ \ V),(sender qquad{:(S),(T),(U),(V):}[(0,1,0,1),(0,0,0,1),(0,1,0,0),(1,1,1,0)]):}`

 

 D.  

`{:(qquadqquadqquadqquadqquadqquadquad receiver),(\ quad qquadqquadqquadqquad qquadS\ \ \ Tquad U\ \ V),(sender qquad{:(S),(T),(U),(V):}[(0,1,0,1),(0,0,0,1),(0,1,0,1),(1,1,1,0)]):}`

 
E.  

`{:(qquadqquadqquadqquadqquadqquadquad receiver),(\ quad qquadqquadqquadqquad qquadS\ \ \ Tquad U\ \ V),(sender qquad{:(S),(T),(U),(V):}[(0,1,0,2),(0,0,0,2),(0,1,0,2),(2,2,2,0)]):}`

 

    
Show Answers Only

`D`

Show Worked Solution

`text(By Elimination):`

`text(Vera can talk to Tai) \ => \ e_42 = 1`

`:.\ text(Eliminate)\ A and E`
 

`text(Umi can talk to Vera) \ => \ e_34 = 1`

`:.\ text(Eliminate)\ C`
 

`text(Tai cannot talk to Sam) \ => \ e_21 = 0`

`:.\ text(Eliminate)\ B`

`=>  D`

MATRICES, FUR1 2020 VCAA 4 MC

In a particular supermarket, the three top-selling magazines are Angel (A), Bella (B) and Crystal (C).

The transition diagram below shows the way shoppers at this supermarket change their magazine choice from week to week.
 

A transition matrix that provides the same information as the transition diagram is

A.  

`{:(qquadqquadquad this\ week),(qquadquadAqquadquad\ Bqquadquad\ C),([(0.55,0.70,0.35),(0.70,0.60,0.40),(0.35,0.40,0.40)]{:(A),(B),(C):}qquad n\ext\ week):}`

 

 B.  

`{:(qquadqquadquad this\ week),(qquadquadAqquadquad\ Bqquadquad\ C),([(0.55,0.60,0.25),(0.45,0.15,0.35),(0,0.25,0.40)]{:(A),(B),(C):} qquad n\ext\ week):}`

 
C.  

`{:(qquadqquadquad this\ week),(qquadquadAqquadquad\ Bqquadquad\ C),([(0.55,0.25,0.35),(0.45,0.60,0.25),(0,0.15,0.40)]{:(A),(B),(C):} qquad n\ext\ week):}`

 

 D.  

`{:(qquadqquadquad this\ week),(qquadquadAqquadquad\ Bqquadquad\ C),([(0.55,0.25,0.35),(0.45,0.60,0.25),(0.35,0.15,0.40)]{:(A),(B),(C):} qquad n\ext\ week):}`

 
E.  

`{:(qquadqquadquad this\ week),(qquadquadAqquadquad\ Bqquadquad\ C),([(0.55,0.25,0),(0.45,0.60,0.25),(0,0.15,0.75)]{:(A),(B),(C):} qquad n\ext\ week):}`

 

    
Show Answers Only

`C`

Show Worked Solution

`text(By Elimination):`

`25text(%)\ text(of)\ B\ text(moves to)\ A \ => \ e_12 = 0.25`

`:.\ text(Eliminate)\ A and B`
 

`C\ text(retains 40% from week to week)\ => \ e_33 = 0.4`

`:.\ text(Eliminate)\ E`

 
`0text(%)\ text(of)\ A\ text(moves to)\ C \ => \ e_31 = 0`

`:.\ text(Eliminate)\ D`

`=>  C`

MATRICES, FUR1 2020 VCAA 2 MC

Matrix  `A = [(1, 2), (0, 3), (1, 0), (4, 5)]`  and matrix  `B = [(2, 0, 3, 1), (4, 5, 2, 0)]`.
 

Matrix  `Q = A xx B`.

The element in row  `i`  and column  `j`  of matrix `Q` is  `q_(ij)`.

Element  `q_41` is determined by the calculation

  1. `0 × 0 + 3 × 5`
  2. `1 × 1 + 2 × 0`
  3. `1 × 2 + 2 × 4`
  4. `4 × 1 + 5 × 0`
  5. `4 × 2 + 5 × 4`
Show Answers Only

`E`

Show Worked Solution

`Q = [(1, 2), (0, 3), (1, 0), (4, 5)][(2, 0, 3, 1), (4, 5, 2, 0)]`
 

`q_41 = 4 xx 2 + 5 xx 4`

`=>  E`

MATRICES, FUR1 2020 VCAA 1 MC

The matrix  `[(1, 0, 0), (0, 1, 1), (1, 0, 1)]`  is an example of

  1. a binary matrix.
  2. an identity matrix.
  3. a triangular matrix.
  4. a symmetric matrix.
  5. a permutation matrix.
Show Answers Only

`A`

Show Worked Solution

`text(All elements are 0 or 1 and other definitions)`

`text(don’t apply.)`

`=>  A`

CORE, FUR1 2020 VCAA 24 MC

Manu invests $3000 in an account that pays interest compounding monthly.

The balance of his investment after `n` months, `B_n` , can be determined using the recurrence relation

`B_0 = 3000, qquad B_(n+1) = 1.0048 xx B_n`

The total interest earned by Manu’s investment after the first five months is closest to

  1. $57.60
  2. $58.02
  3. $72.00
  4. $72.69
  5. $87.44
Show Answers Only

`D`

Show Worked Solution
`text(Total interest)` `= 3000 xx 1.0048^5 – 3000`
  `~~ $72.69`

`=>  D`

CORE, FUR1 2020 VCAA 23 MC

Consider the following four recurrence relations representing the value of an asset after `n` years, `V_n`.
 

  • `V_0 = 20\ 000, qquad V_(n+1) = V_n + 2500`
  • `V_0 = 20\ 000, qquad V_(n+1) = V_n - 2500`
  • `V_0 = 20\ 000, qquad V_(n+1) = 0.875 V_n`
  • `V_0 = 20\ 000, qquad V_(n+1) = 1.125V_n - 2500`

How many of these recurrence relations indicate that the value of an asset is depreciating?

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show Answers Only

`C`

Show Worked Solution
`V_(n+1)` `= V_n + 2500\ => \ text(appreciating)`
`V_(n-1)` `= V_n – 2500\ => \ text(depreciating)`
`V_(n+1)` `= 0.875 V_n\ => \ text(depreciating)`

 
`text(When)\ \ V_(n+1) = 1.125 V_n – 2500`

`V_1` `= 1.125 xx 20\ 000 – 2500`
  `= 20\ 000\ text{(same price)}`

 
`=>  C`

CORE, FUR1 2020 VCAA 17-18 MC

Table 4 below shows the monthly rainfall for 2019, in millimetres, recorded at a weather station, and the associated long-term seasonal indices for each month of the year.
 


 

Part 1

The deseasonalised rainfall for May 2019 is closest to

  1.   71.3 mm
  2.   75.8 mm
  3.   86.1 mm
  4.   88.1 mm
  5. 113.0 mm

 
Part 2

The six-mean smoothed monthly rainfall with centring for August 2019 is closest to

  1. 67.8 mm
  2. 75.9 mm
  3. 81.3 mm
  4. 83.4 mm
  5. 86.4 mm
Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ C`

Show Worked Solution

Part 1

`text(Deseasonalised rainfall for May)`

`= 92.6/1.222`

`= 75.8\ text(mm)`
 

`=> B`
 

Part 2

`text{Six-mean smoothed average (Aug)}`

`=[(92.6 + 77.2 + 80 + 86.8 + 93.8 + 55.2) ÷6 +`

`(77.2 + 80 + 86.8 + 93.8 + 55.2 + 97.3) ÷ 6] ÷ 2`

`~~ (80.93 + 81.72) ÷ 2`

`~~ 81.3\ text(mm)`
 

`=> C`

CORE, FUR1 2020 VCAA 1-3 MC

The times between successive nerve impulses (time), in milliseconds, were recorded.

Table 1 shows the mean and the five-number summary calculated using 800 recorded data values.
 


 

Part 1

The difference, in milliseconds, between the mean time and the median time is

  1.  10
  2.  70
  3.  150
  4.  220
  5.  230

 
Part 2

Of these 800 times, the number of times that are longer than 300 milliseconds is closest to

  1. 20
  2. 25
  3. 75
  4. 200
  5. 400

 
Part 3

The shape of the distribution of these 800 times is best described as

  1. approximately symmetric.
  2. positively skewed.
  3. positively skewed with one or more outliers.
  4. negatively skewed.
  5. negatively skewed with one or more outliers.
Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ D`

`text(Part 3:)\ C`

Show Worked Solution

Part 1

`text(Difference)` `= 220 -150`
  `= 70`

`=> B`
 

Part 2

`Q_3 = 300`

`:.\ text(Impulses longer than 300 milliseconds)`

`= 25text(%) xx 800`

`= 200`

`=> D`
 

Part 3

`text(Distribution has a long tail to the right)`

♦ Mean mark 50%.

`:.\ text(Positively skewed)`

`text(Upper fence)` `= Q_3 + 1.5 xx IQR`
  `= 300 + 1.5 (300 – 70)`
  `= 645`

 
`=> C`

Calculus, EXT1 C2 2020 SPEC1 6

Let  `f(x) = tan^(-1) (3x - 6) + pi`.

  1. Show that  `f^{prime}(x) = 3/(9x^2 - 36x + 37)`.  (1 mark)

  2. Hence, show that the graph of  `f`  has a point of inflection at  `x = 2`.  (2 marks)

  3. Sketch the graph of  `y = f(x)`  on the axes provided below. Label any asymptotes with their equations and the point of inflection with its coordinates.   (2 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`
Show Worked Solution
a.    `f^{\prime}(x)` `= (d/(dx) (3x – 6))/(1 + (3x – 6)^2)`
    `= 3/(9x^2 – 36x + 37)`

 

b.   `f^{\prime\prime}(x) = (3(18x – 36))/(9x^2 – 36x + 37)^2`

`f^{\prime\prime}(x) = 0\ \ text(when)\ \ 18x – 36 = 0 \ => \ x = 2`

`text(If)\ \ x < 2, 18x – 36 < 0 \ => \ f^{\prime\prime}(x) < 0`

`text(If)\ \ x > 2, 18x – 36 > 0 \ => \ f^{\prime\prime}(x) > 0`

`text(S) text(ince)\ \ f^{\prime\prime}(x)\ \ text(changes sign about)\ \ x = 2,`

`text(a POI exists at)\ \ x = 2`

 

c.   

Calculus, SPEC1 2020 VCAA 6

Let  `f(x) = arctan (3x - 6) + pi`.

  1. Show that  `f^{\prime}(x) = 3/(9x^2 - 36x + 37)`.  (1 mark)
  2. Hence, show that the graph of  `f`  has a point of inflection at  `x = 2`.  (2 marks)
  3. Sketch the graph of  `y = f(x)`  on the axes provided below. Label any asymptotes with their equations and the point of inflection with its coordinates.   (2 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`
Show Worked Solution
a.    `f^{\prime}(x)` `= (d/(dx) (3x – 6))/(1 + (3x – 6)^2)`
    `= 3/(9x^2 – 36x + 37)`

 

b.   `f^{\prime\prime}(x) = (3(18x – 36))/(9x^2 – 36x + 37)^2`

♦ Mean mark part (b) 42%.

`f^{\prime\prime}(x) = 0\ \ text(when)\ \ 18x – 36 = 0 \ => \ x = 2`

`text(If)\ \ x < 2, 18x – 36 < 0 \ => \ f^{\prime\prime}(x) < 0`

`text(If)\ \ x > 2, 18x – 36 > 0 \ => \ f^{\prime\prime}(x) > 0`

`text(S) text(ince)\ \ f^{\prime\prime}(x)\ \ text(changes sign about)\ \ x = 2,`

`text(a POI exists at)\ \ x = 2`

 

c.   

Calculus, EXT1 C2 2020 SPEC1 2

Evaluate  `int_(-1)^0 (1 + x)/sqrt(1 - x)\ dx`, using the substitution  `u=1-x`.  (3 marks)

Show Answers Only

`(8 sqrt 2)/3 – 10/3`

Show Worked Solution
`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, EXT2 C1 2020 SPEC1 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, MET1 2020 VCAA 7

Consider the function  `f(x) = x^2 + 3x + 5`  and the point  `P(1, 0)`. Part of the graph  `y = f(x)`  is shown below.
 

   
 

  1. Show the point `P` is not on the graph of  `y = f(x)`.  (1 mark)
  2. Consider a point  `Q(a, f(a))`  to be a point on the graph of `f`.

     

    1. Find the slope of the line connecting points `P` and `Q` in terms of `a`.  (1 mark)
    2. Find the slope of the tangent to the graph of `f` at point `Q` in terms of `a`.  (1 mark)
    3. Let the tangent to the graph of `f` at  `x = a`  pass through point `P`.

       

      Find the values of `a`.  (2 marks)

    4. Give the equation of one of the lines passing through point `P` that is tangent to the graph of `f`.  (1 mark)
  3. Find the value of `k`, that gives the shortest possible distance between the graph of the function of  `y = f(x - k)`  and point `P`.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2.  
    1. `(a^2 + 3a + 5)/(a – 1)`
    2. `2a + 3`
    3. `4\ text(or)\ −2`
    4. `y = 11x – 11`
  3. `5/2`
Show Worked Solution

a.   `f(1) = 1 + 3 + 5 = 9`

`text(S)text(ince)\ \ f(1) != 0, P(1, 0)\ text(does not lie on)\ \ y = f(x)`

 

b.i.   `P(1, 0), Q(a, f(a))`

`m_(PQ)` `= (f(a) – 0)/(a – 1)`
  `= (a^2 + 3a + 5)/(a – 1)`

 

b.ii.   `f′(x) = 2x + 3`

`m_Q = f′(a) = 2a + 3`

 

b.iii.   `text(T)text(angent:)\ m = 2a + 3,\ text(passes through)\ (a, a^2 + 3a + 5)`

`y – (a^2 + 3a + 5) = (2a + 3)(x – 1)`

`text(Passes through)\ P(1, 0):`

`0 – (a^2 + 3a + 5)` `= (2a + 3)(1 – a)`
`−(a^2 + 3a + 5)` `= 2a – 2a^2 + 3 – 3a`
`a^2 – 2a – 8` `= 0`
`(a – 4)(1 + 2)` `= 0`

 
`:. a = 4\ text(or)\ −2`

 

b.iv.   `text(When)\ \ a = −2`

`m_text(tang) = 2x – 2 + 3 = −1`

`text(Equation of line)\ \ m = − 1,\ text(through)\ P(1, 0)`

`y – a` `= −1(x – 1)`
`y` `= -x + 1`

 
`text(Similarly, if)\ \ a = 4:`

`y = 11x – 11`

 

c.   `f(x)\ text(is a quadratic with no roots.`

`text(Shortest distance needs S.P. to occur when)\ \ x = 1`

`f′(x) = 2x + 3`

`text(MIN S.P. of)\ \ f(x)\ \ text(occurs when)\ \ f′(x) = 0`

`x = −3/2`

`f(−2/3 – k) = f(1)\ \ text(for shortest distance.)`

`:. k = 5/2`

Calculus, MET1 2006 ADV 2aii

Differentiate with respect to `x`:

Let  `y=sin x/(x + 1)`.  Find  `dy/dx `.   (2 marks)

Show Answers Only

`dy/dx = {cos x (x + 1) – sin x} / (x + 1)^2`

Show Worked Solution

`y = sinx/(x + 1)`

`d/dx (u/v) = (u^{\prime} v – uv^{\prime})/v^2`

`u` `= sin x` `v` `= x + 1`
`u^{\prime}` `= cos x` `\ \ \ v^{\prime}` `= 1`

 

`:.dy/dx = {cos x (x + 1) – sin x} / (x + 1)^2`

Functions, EXT1 F1 2020 MET1 6

`f(x) = 1/sqrt2 sqrtx`, where  `x in [0,2]`

  1. Find  `f^(-1)(x)`, and state its domain.  (2 marks)

The graph of  `y = f(x)`, where  `x ∈ [0, 2]`, is shown on the axes below.
 

     
 

  1. On the axes above, sketch the graph of  `f^(-1)(x)`  over its domain. Label the endpoints and point(s) of intersection with  `f(x)`, giving their coordinates.  (2 marks)

Show Answers Only
  1. `text(Domain) = [0, 1]`

     

    `f^(-1)(x) = 2x^2`

  2.  
       
Show Worked Solution

a.    `text(Domain)\ \ f^(-1)(x)= text(Range)\ \ f(x)=[0,1]`

`y = 1/sqrt2 x`

`text(Inverse: swap)\ \ x ↔ y`

`x` `= 1/sqrt2 sqrty`  
`sqrty` `= sqrt2 x`  
`y` `= 2x^2`  

 
`:. f^(-1)(x) = 2x^2`

 

b.     
  

Probability, 2ADV S1 2012 MET1 2

A car manufacturer is reviewing the performance of its car model X. It is known that at any given six-month service, the probability of model X requiring an oil change is `17/20`, the probability of model X requiring an air filter change is `3/20` and the probability of model X requiring both is `1/20`.

  1. State the probability that at any given six-month service model X will require an air filter change without an oil change.  (1 mark)

  2. The car manufacturer is developing a new model. The production goals are that the probability of model Y requiring an oil change at any given six-month service will be `m/(m + n)`, the probability of model Y requiring an air filter change will be `n/(m + n)` and the probability of model Y requiring both will be `1/(m + n)`, where `m, n ∈ Z^+`.
     
    Determine `m` in terms of `n` if the probability of model Y requiring an air filter change without an oil change at any given six-month service is 0.05.  (2 marks)

Show Answers Only
  1. `1/10`
  2. `m = 19n – 20`
Show Worked Solution
a.   
  `text(Pr)(F ∩ O′)` `= text(Pr)(F) – text(Pr)(F∩ O)`
    `= 3/20 – 1/20`
    `= 1/10`

 

 

b.   
`text(Pr)(F ∩ O′)` `= n/(m + n) – 1/(m + n)`
`1/20` `= (n – 1)/(m + n)`
`m + n` `= 20n – 20`
`m` `= 19n – 20`

Probability, MET1 2020 VCAA 2

A car manufacturer is reviewing the performance of its car model X. It is known that at any given six-month service, the probability of model X requiring an oil change is `17/20`, the probability of model X requiring an air filter change is `3/20` and the probability of model X requiring both is `1/20`.

  1. State the probability that at any given six-month service model X will require an air filter change without an oil change.  (1 mark)
  2. The car manufacturer is developing a new model. The production goals are that the probability of model Y requiring an oil change at any given six-month service will be `m/(m + n)`, the probability of model Y requiring an air filter change will be `n/(m + n)` and the probability of model Y requiring both will be `1/(m + n)`, where `m, n ∈ Z^+`.
     
    Determine `m` in terms of `n` if the probability of model Y requiring an air filter change without an oil change at any given six-month service is 0.05.  (2 marks)
Show Answers Only
  1. `1/10`
  2. `m = 19n – 20`
Show Worked Solution
a.   
  `text(Pr)(F ∩ O′)` `= text(Pr)(F) – text(Pr)(F∩ O)`
    `= 3/20 – 1/20`
    `= 1/10`

 

 

b.   
`text(Pr)(F ∩ O′)` `= n/(m + n) – 1/(m + n)`
`1/20` `= (n – 1)/(m + n)`
`m + n` `= 20n – 20`
`m` `= 19n – 20`

Proof, EXT2 P1 SM-Bank 15

Prove  `sqrt5 + sqrt3 > sqrt14`  by contradiction.   (2 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume)\ \ sqrt5 + sqrt3 <= sqrt14`

`(sqrt5 + sqrt3)^2` `<= (sqrt14)^2`
`5 + 2sqrt15 + 3` `<= 14`
`2sqrt15` `<= 6`
`sqrt15` `<= 3`
`15` `<= 9\ \ \ text{(incorrect)}`

 
`:. text(By contradiction,)\ sqrt5 + sqrt3 > sqrt14`

Complex Numbers, EXT2 N1 SM-Bank 9

Let  `z = sqrt3 - 3 i`

  1. Express `z` in modulus-argument form.   (2 marks)

  2. Find the smallest integer `n`, such that  `z^n + (overset_z)^n = 0`.  (3 marks)

Show Answers Only
  1. `2 sqrt3 text{cis} (frac{-pi}{3})`
  2. `3`
Show Worked Solution
i.    `z` `= sqrt3 – 3 i`
  `|z|` `= sqrt((sqrt3)^2 + 3^2) = 2 sqrt3`

 

`tan theta` `= frac{3}{sqrt3}=sqrt3`
`theta` `= frac{pi}{3}`
`text{arg} (z)` `= – frac{pi}{3}`

 
`therefore  z = 2 sqrt3 \ text{cis} (frac{-pi}{3})`

 

ii.   `z^n + (overset_z)^n = 0`

`[2 sqrt3 \ cos (frac{-pi}{3}) + i sin (frac{-pi}{3})]^n + [ 2 sqrt3 \ cos (frac{-pi}{3}) – i sin (frac{-pi}{3}) ]^n = 0`

`(2 sqrt3)^n [cos (frac{-n pi}{3}) + i sin (frac{-n pi}{3}) + cos (frac{-n pi}{3}) – i sin (frac{-n pi}{3}) = 0`

`2 \ cos (frac{-n pi}{3})` `= 0`
`cos (frac{n pi}{3})` `= 0`
`frac{n pi}{3}` `= frac{pi}{2} + k pi \ , \ k = 0, ± 1, ± 2, …`
`frac{n}{3}` `= frac{(2k + 1)}{2}`
`n` `= frac{3 (2k + 1)}{2}`

 
`text{Numerator will always be odd  ⇒  no solution exists}`

Complex Numbers, EXT2 N1 SM-Bank 1 MC

Which of the following is the complex number  \(-\sqrt{3}+3 i ?\)?

  1. \(2 \sqrt{3} e^{-\small{\dfrac{i \pi}{3}}}\)
  2. \(2 \sqrt{3} e^{\small{\dfrac{i 2 \pi}{3}}}\)
  3. \(12 e^{-\small{\dfrac{i \pi}{3}}}\)
  4. \(-12 e^{\small{\dfrac{i 2 \pi}{3}}}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\abs{z}\) \(=\sqrt{(\sqrt{3})^2+3^2}=2 \sqrt{3}\)
\(\tan \theta\) \(=\dfrac{\sqrt{3}}{3}=\dfrac{1}{\sqrt{3}}\)
\(\theta\) \(=\dfrac{\pi}{6}\)

\(\arg (z)=\dfrac{\pi}{2}+\dfrac{\pi}{6}=\dfrac{2 \pi}{3}\)
  

\(\therefore z\) \(=2 \sqrt{3}\left(\cos \left(\dfrac{2 \pi}{3}\right)+i \sin \left(\dfrac{2 \pi}{3}\right)\right.\)
  \(=2 \sqrt{3} e^{\small{\dfrac{i 2 \pi}{3}}}\)

\(\Rightarrow B\)

Complex Numbers, EXT2 N1 2004 HSC 2b

Let  `alpha = 1 + i sqrt3`  and  `beta = 1 + i`.

  1. Find  `frac{alpha}{beta}`, in the form  `x + i y`.   (1 mark)

  2. Express `alpha` in modulus-argument form.   (3 marks)

  3. Given that `beta` has the modulus-argument form
     
         `beta = sqrt2 (cos frac{pi}{4} + i sin frac{pi}{4})`.
     
    find the modulus-argument form of  `frac{alpha}{beta}`.   (1 mark)

  4. Hence find the exact value of  `sin frac{pi}{12}`   (1 mark)

Show Answers Only
  1. `frac{1+sqrt3}{2} + i (frac{sqrt3 – 1}{2})`
  2. `2 \ text{cis} (frac{pi}{3})`
  3. `sqrt2 \ text{cis} (frac{pi}{12})`
  4. `frac{sqrt6-sqrt2}{4}`
Show Worked Solution
i.     `frac{alpha}{beta}` `= frac{1 + i sqrt3}{1 + i} xx frac{1 – i}{1 – i}`
    `= frac{(1 + i sqrt3)(1 – i)}{1^2 – i^2}`
    `= frac{1 – i + i sqrt3 – i^2 sqrt3}{2}`
    `= frac{1+sqrt3}{2} + i (frac{sqrt3 – 1}{2})`

 

ii.   `alpha` `= 1 + i sqrt3`
  `| alpha |` `= sqrt(1^2 + (sqrt3)^2) = 2`

`text{arg} \ (alpha) = tan^-1 (frac{sqrt3}{1}) = frac{pi}{3}`

`therefore \ alpha = 2 text{cis} (frac{pi}{3})`

 

iii.   `beta` `= sqrt2 text{cis} (frac{pi}{4})`
  `frac{alpha}{beta}` `= frac{2}{sqrt2} \ text{cis} (frac{pi}{3} – frac{pi}{4})`
    `= sqrt2 text{cis}  (frac{pi}{12})`

 

iv.  `text{Equating imaginary parts of i and ii:}`

`sqrt2 \ sin \ (frac{pi}{12})` `= frac{sqrt3 – 1}{2}`
`sin (frac{pi}{12})` `= frac{sqrt3 – 1}{2 sqrt2} xx frac{sqrt2}{sqrt2}`
  `= frac{sqrt6 – sqrt2}{4}`

Complex Numbers, EXT2 N2 EQ-Bank 1

`z = sqrt2 e^((ipi)/15)`  is a root of the equation  `z^5 = alpha(1 + isqrt3), \ alpha ∈ R`.

  1. Express  `1 + isqrt3`  in exponential form.  (2 marks)

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

  3. Find the other 4 roots of the equation in exponential form.  (3 marks)

Show Answers Only
  1. `2e^((ipi)/3)`
  2. `alpha = 2sqrt2`
  3. `e^((i11pi)/15), e^(-(ipi)/3), e^((i13pi)/15), e^(−(i11pi)/15)`
Show Worked Solution

i.   `beta = 1 + isqrt3`

`|beta| = sqrt(1 + (sqrt3)^2) = 2`

`text(arg)(beta) = tan^(−1) (sqrt3/1) = pi/3`

`beta = 2e^((ipi)/3)`

 

ii.    `z` `= sqrt2 e^((ipi)/15)`
  `z^5` `= (sqrt2 e^((ipi)/15))^5`
    `= (sqrt2)^5 e^((ipi)/15 xx 5)`
    `= 4sqrt2 e^((ipi)/3)`

 
`:. alpha = 2sqrt2`

 

iii.   `text(arg)(z^5) = pi/3 + 2kpi, \ \ k = 0, ±1, ±2, …`

`text(arg)(z) = pi/15 + (2kpi)/5`

`k = 1:\ text(arg)(z) = pi/15 + (2pi)/5 = (11pi)/15`

`k = text(−1):\ text(arg)(z) = pi/15 – (2pi)/5 = −pi/3`

`k = 2:\ text(arg)(z) = pi/15 + (4pi)/5 = (13pi)/15`

`k =text(−2):\ text(arg)(z) = pi/15 – (4pi)/5 = −(11pi)/15`
 

`:. 4\ text(other roots are:)`

`e^((i11pi)/15), e^(−(ipi)/3), e^((i13pi)/15), e^(−(i11pi)/15)` 

Complex Numbers, EXT2 N1 EQ-Bank 9

Calculate the value of  \(\dfrac{e^{\small{\dfrac{i \pi}{3}}}-e^{-\small{\dfrac{i \pi}{3}}}}{2 i}\).  (2 marks)

Show Answers Only

\(\dfrac{\sqrt{3}}{2}\)

Show Worked Solution
\(\dfrac{e^{\small{\dfrac{i \pi}{3}}}-e^{-\small{\dfrac{i \pi}{3}}}}{2 i}\) \(=\dfrac{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}-\left(\cos \left(-\frac{\pi}{3}\right)+i \sin \left(-\frac{\pi}{3}\right)\right)}{2 i}\)
  \(=\dfrac{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}-\left(\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}\right)}{2 i}\)
  \(=\dfrac{2 i \sin \frac{\pi}{3}}{2 i}\)
  \(=\sin \frac{\pi}{3}\)
  \(=\dfrac{\sqrt{3}}{2}\)

Complex Numbers, EXT2 N1 EQ-Bank 5 MC

In which quadrant of the complex plane is the complex number  \(4e^{\small{\dfrac{i16}{3}}}\)  found?

  1. I
  2. II
  3. III
  4. IV
Show Answers Only

\(\D\)

Show Worked Solution

\(z=4 e^{\small{\dfrac{in6}{3}}}\)

\(\arg (z)=\dfrac{16}{3} \approx 5.3\)

\(\dfrac{3 \pi}{2}<5.3<2 \pi\)

\(\therefore \ \text{Quadrant IV}\)

\(\Rightarrow D\)

Calculus, EXT2 C1 2005 HSC 1e

Let  `t=tan(theta/2).`

  1. Show that  `(dt)/(d theta) = 1/2(1+t^2)`   (1 mark)

  2. Show that  `sin theta = (2t)/(1+t^2).`   (2 marks)

  3. Use the substitution  `t=tan(theta/2)`  to find  `int text(cosec)\ theta\ d theta.`   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `log_e | tan frac{theta}{2} | + c`
Show Worked Solution
i.     `t` `= tan frac{theta}{2}`
  `frac{dt}{d theta}` `= frac{1}{2} text{sec}^2 frac{theta}{2}`
    `= frac{1}{2} (1 + tan^2 frac{theta}{2})`
    `= frac{1}{2} (1 + t^2)`

 

ii.   `text{Show} \ \ sin theta = frac{2t}{1 + t^2} :`
 

`sin theta` `= 2 \ sin frac{theta}{2} cos frac{theta}{2}`
  `= 2 * frac{t}{sqrt(1 + t^2)} * frac{1}{sqrt(1 + t^2)}`
  `= frac{2t}{1 + t^2}`

 

iii.   `int \ text{cosec} \ theta \ d theta`

`t = tan frac {theta}{2}`

`frac{dt}{d theta} = frac{1}{2} text{sec}^2 frac{theta}{2} \ , \ d theta = frac{2dt}{sec^2 frac{theta}{2}} = frac{2}{1 + t^2}  dt`

`int \ text{cosec} \ theta\ d theta` `= int frac{1 + t^2}{2t} xx frac{2}{1 + t^2} dt`
  `= int frac{1}{t}\ dt`
  `= log_e | t | + c`
  `= log_e | tan frac{theta}{2} | + c`

Complex Numbers, EXT2 N1 2005 HSC 2b

Let  `beta = 1-i sqrt3`.

  1. Express  `beta`  in modulus-argument form.   (2 marks)

  2. Express  `beta^5`  in modulus-argument form.   (2 marks)

  3. Hence express  `beta^5`  in the form  `x+iy`.   (1 mark)

Show Answers Only
  1. `2 \ text{cis} (-frac{pi}{3})`
  2. `32 \ text{cis} (frac{pi}{3})`
  3. `16 + i 16 sqrt3`
Show Worked Solution

i.    `beta = 1 – i sqrt3`
 

 
`| beta | = sqrt(1^2 + (sqrt3)^2) = 2`

`tan theta` `= frac{sqrt3}{1} = sqrt3`
`theta` `= frac{pi}{3}`
`text{arg} (beta)` `= -frac{pi}{3}`

`therefore \ beta = 2 \ text{cis} (-frac{pi}{3})`

 

ii.   `beta^5` `= 2^5 \ text{cis} (-frac{pi}{3} xx5)`
    `= 32 \ text{cis} (-frac{5pi}{3} + 2 pi)`
    `= 32 \ text{cis} (frac{pi}{3})`

 

iii.   `beta^5` `= 32 ( cos (frac{pi}{3}) + i sin (frac{pi}{3}) )`
    `= 32 ( frac{1}{2} + i  frac{sqrt3}{2})`
    `= 16 + i 16 sqrt3`

Complex Numbers, EXT2 N1 2008 HSC 2b

  1. Write  `frac{1 + i sqrt3}{1 + i}`  in the form  `x + iy`, where `x` and `y` are real.  (2 marks)

  2. By expressing both  `1 + i sqrt3`  and  `1 + i`  in  modulus-argument form, write  `frac{1 + i sqrt3}{1 + i}`  in modulus-argument form.   (3 marks)

  3. Hence find  `cos frac{pi}{12}`  in surd form.  (1 mark)

  4. By using the result of part (ii), or otherwise, calculate  `(frac{1 + i sqrt3}{1 + i})^12`.   (1 mark)

Show Answers Only
  1. `frac{1 + sqrt3}{2} – i ( frac{1 – sqrt3}{2} )`
  2. `sqrt2 (cos (frac{pi}{12}) + i sin (frac{pi}{12}))`
  3. `frac{sqrt2 + sqrt6}{4}`
  4. `-64`
Show Worked Solution
i.      `frac{1 + i sqrt3}{1 + i} xx frac{1 – i}{1 – i}` `= frac{(1 + i sqrt3)(1 – i)}{1 – i^2}`
    `= frac{1 – i + i sqrt3 – sqrt3 i^2}{2}`
    `= frac{1 + sqrt3}{2} – i ( frac{1 – sqrt3}{2} )`

 

ii.   `z_1 = 1 +  i sqrt3`

`| z_1 | = sqrt(1 + ( sqrt3)^2) = 2`

`text{arg} (z_1) = tan^-1 (sqrt3) = frac{pi}{3}`
  

`z_1 = 2 (cos frac{pi}{3} + i sin frac{pi}{3})`
 
`z_2 = 1 + i`

`| z_2 | = sqrt(1^2 + 1^2) = sqrt2`

`text{arg} (z_2) = tan^-1 (1) = frac{pi}{4}`

`z_2 = sqrt2 (cos frac{pi}{4} + i sin frac{pi}{4})`
 

`frac{1 + i sqrt3}{1 + i}` `= frac{z_1}{z_2}`
  `= frac{2}{sqrt2} ( cos ( frac{pi}{3} – frac{pi}{4} ) + i sin ( frac{pi}{3} – frac{pi}{4} ) )`
  `= sqrt2 ( cos (frac{pi}{12}) + i sin (frac{pi}{12}) )`

 

iii.  `text{Equating real parts of i and ii:}`

`sqrt2 cos (frac{pi}{12})` `= frac{1 + sqrt3}{2}`
`cos(frac{pi}{12})` `= frac{1 + sqrt3}{2 sqrt2} xx frac{sqrt2}{sqrt2}`
  `= frac{sqrt2 + sqrt6}{4}`

 

iv.     `(frac{1 + i sqrt2}{1 + i})^12` `= (sqrt2)^12 (cos (frac{pi}{12} xx 12) + i sin (frac{pi}{12} xx 12))`
    `= 64 (cos pi + i sin pi)`
    `= – 64`

Complex Numbers, EXT2 N1 EQ-Bank 2

Express the complex number  \(z=-2 \sqrt{2}-2 \sqrt{6} i\)  in the exponential form.  (2 marks)

Show Answers Only

\(4 \sqrt{2} e^{-i \small{\dfrac{2 \pi}{3}}}\)

Show Worked Solution
\(\abs{z}\) \(=\sqrt{(2 \sqrt{2})^2+(2 \sqrt{6})^2}\)
  \(=\sqrt{8+24}\)
  \(=4 \sqrt{2}\)

 

\(\tan \theta\) \(=\dfrac{2 \sqrt{2}}{2 \sqrt{6}}=\dfrac{1}{\sqrt{3}}\)
\(\theta\) \(=\dfrac{\pi}{6}\)

 

\(\operatorname{Arg}(z)=-\left(\dfrac{\pi}{2}+\dfrac{\pi}{6}\right)=-\dfrac{2 \pi}{3}\)

\(\therefore z\) \(=4 \sqrt{2} \operatorname{cis}\left(\dfrac{-2 \pi}{3}\right)\)
  \(=4 \sqrt{2} e^{-i \small{\dfrac{2 \pi}{3}}}\)

Complex Numbers, EXT2 N2 EQ-Bank 1 MC

The Argand diagram shows the complex number `e^(i theta)`.
 

Which of the following could be the complex number `i e^(-i theta)`?

    

 

  

 
Show Answers Only

`B`

Show Worked Solution

`e^(i theta) = cos theta + i sin theta`

`e^(-i theta) = cos (-theta) + i sin (-theta) = cos theta-i sin theta\ \ \ text{(conjugate)}`

`e^(i theta) \ => \ P^{′}`

`ie^(-i theta) \ => \ text(Rotate)\ e^(-i theta)\ (P^{′})\ text(90° anticlockwise.)`

`=> B`

Vectors, EXT2 V1 2020 HSC 15b

The point `C` divides the interval `AB` so that  `frac{CB}{AC} = frac{m}{n}`. The position vectors of `A` and `B` are `underset~a` and `underset~b` respectively, as shown in the diagram.
 

  1. Show that  `overset->(AC) = frac{n}{m + n} (underset~b - underset~a)`.  (2 marks)

     

  2. Prove that  `overset->(OC) = frac{m}{m + n} underset~a + frac{n}{m + n} underset~b`.  (1 mark)

Let `OPQR` be a parallelogram with  `overset->(OP) = underset~p`  and  `overset->(OR) = underset~r`. The point `S` is the midpoint of `QR` and `T` is the intersection of `PR` and `OS`, as shown in the diagram.
  
 
         
 

  1. Show that  `overset->(OT) = frac{2}{3} underset~r + frac{1}{3} underset~p`.  (3 marks)

  2. Using parts (ii) and (iii), or otherwise, prove that `T` is the point that divides the interval `PR` in the ratio 2 :1.   (1 mark)

Show Answers Only
  1. `text{See Worked Solutions}`
  2. `text{See Worked Solutions}`
  3. `text{See Worked Solutions}`
  4. `text{See Worked Solutions}`
Show Worked Solution

i.  

`frac{overset->(AC)}{overset->(AB)}` `= frac{n}{m + n}`
`overset->(AC)` `= frac{n}{m + n} * overset->(AB)`
  `= frac{n}{m + n} (underset~b – underset~a)`

 

ii.    `overset->(OC)` `= overset->(OA) + overset->(AC)`
    `= underset~a + frac{n}{m + n} (underset~b – underset~a)`
    `= underset~a – frac{n}{m + n} underset~a + frac{n}{m + n} underset~b`
    `= (1 – frac{n}{m + n}) underset~a + frac{n}{m + n} underset~b`
    `= (frac{m + n – n}{m + n}) underset~a + frac{n}{m + n} underset~b`
    `= frac{m}{m + n} underset~a + frac{n}{m + n} underset~b`

 

iii.  `text{Show} \ \ overset->(OT) = frac{2}{3} underset~r + frac{1}{3} underset~p`

♦ Mean mark part (iii) 50%.
 

 
`text{Consider} \ \ Delta PTO \ \ text{and} \ \ Delta RTS:`

`angle PTO = angle RTS \ (text{vertically opposite})`

`angle OPT = angle SRT \ (text{vertically opposite})`

`therefore \ Delta PTO \ text{|||} \ Delta RTS\ \ text{(equiangular)}`
 

`OT : TS = OP : SR = 2 : 1`

`(text{corresponding sides in the same ratio})`

`frac{overset->(OT)}{overset->(OS)}` `= frac{2}{3}`
`overset->(OT)` `= frac{2}{3} overset->(OS)`
  `= frac{2}{3} ( underset~r + frac{1}{2} underset~p)`
  `= frac{2}{3} underset~r + frac{1}{3} underset~p`
Mean mark part (iv) 51%.

 

iv.   `text{Let} \ \ overset->(OT) \ text{divide} \ PR\ text{so that}\ \ frac{TR}{PT} = frac{m}{n}`

`text{Using part (ii):}`

`overset->(OT)` `= frac{m}{m + n} underset~p + frac{n}{m + n} c`
`overset->(OT)` `= frac{1}{3} underset~p + frac{2}{3} underset~r \ \ \ (text{part (iii)})`
`frac{m}{m + n}` `= frac{1}{3} , frac{n}{m + n} = frac{2}{3}`

 
`=> \ m = 1 \ , \ n = 2`
 

`therefore \ T \ text{divides} \ PR \ text{in ratio  2 : 1}.`

Proof, EXT2 P1 2020 HSC 15a

In the set of integers, let `P` be the proposition:

'If  `k + 1`  is divisible by 3, then  `k^3 + 1`  divisible by 3.'

  1. Prove that the proposition `P` is true.  (2 marks)

  2. Write down the contrapositive of the proposition `P`.  (1 mark)

  3. Write down the converse of the proposition `P` and state, with reasons, whether this converse is true or false.  (3 marks)

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

i.     `text{Let} \ \ k + 1 = 3N, \ N∈ Z`

`=>  k = 3N – 1`

`k^3 + 1` `= (3N -1)^3 + 1`
  `= (3N)^3 + 3(3N)^2 (-1) + 3(3N)(-1)^2 + (-1)^3 + 1`
  `= 27N^3 – 27N^2 + 9N – 1 + 1`
  `= 3 (9N^3 – 9N^2 + 3N)`
  `= 3Q \ , \ Q ∈ Z`

 
`therefore \ text{If} \ \ k+ 1 \ \ text{is divisible by 3}, text{then} \ \ k^3 + 1 \ \ text{is divisible by 3.}`
 

ii.    `text{Contrapositive}`

`text{If} \ \ k^3 + 1 \ \ text{is not divisible by 3, then}\ \ k + 1\ \ text{is not divisible by 3.}`
 

♦♦ Mean mark part (iii) 36%.

iii.   `text{Converse:}`

`text{If} \ \ k^3 + 1\ \ text{is divisible by 3, then}\ \ k + 1\ \ text{is divisible by 3.}`

`text(Contrapositive of converse:)`

`text{If}\ \ k + 1\ \ text{is not divisible by 3, then}\ \ k^3 + 1\ \ text{is not divisible by 3.}`
 
`text(i.e.)\ \ k + 1 \ \ text{is not divisible by 3 when}\ \ k + 1 = 3Q + 1\ \ text{or}\ \ k + 1 = 3Q + 2, text{where}\ Q ∈ Z`
 

`text{If} \ \ k + 1` `= 3Q + 1\ \ => \ k=3Q`
`k^3 + 1` `= (3Q)^3 + 1`
  `= 27Q^3 + 1`
  `= 3(9Q^3) + 1`
  `= 3M + 1 \ \ (text{not divisible by 3,}\ M ∈ Z)`

 

`text{If} \ \ k + 1` `= 3Q + 2\ \ => \ k=3Q+1`
`k^3 + 1` `= (3Q + 1)^3 + 1`
  `= (3Q)^3 + 3(3Q)^2 + 3(3Q) + 1 + 1`
  `= 27Q^3 + 27Q^2 + 9Q + 2`
  `= 3(9Q^3 + 9Q^2 + 3Q) + 2`
  `= 3M + 2 \ (text{not divisible by 3,}\ M ∈ Z) `

 

`therefore \ text{By contrapositive, if}\ \ k^3 + 1\ \ text {is divisible by 3, k + 1 is divisible by 3.}`

Complex Numbers, EXT2 N2 2020 HSC 14a

Let `z_1` be a complex number and let  `z_2 = e^(frac{i pi}{3}) z_1`

The diagram shows points `A` and `B` which represent `z_1` and `z_2`, respectively, in the Argand plane.
 


 

  1. Explain why triangle  `OAB`  is an equilateral triangle.   (2 marks)

  2. Prove that  `z_1 ^2 + z_2^2 = z_1 z_2`.  (3 marks)

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  1. `text{See Worked Solutions}`
  2. `text{See Worked Solutions}`
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i.

 

`text{Let}`     `z_1` `= r(cos theta + i sin theta)`
  `z_2` `= e^(i frac{pi}{3}) z_1`
    `= r (cos ( theta + frac{pi}{3} ) + i sin (theta + frac{pi}{3}))`

 

`| z_1 | = | z_2 | => OA = OB`

` angle AOB = frac{pi}{3} \ ( z_2 \ text{is a} \ frac{pi}{3} \ text{anti-clockwise rotation of} \ z_1 )`

`=> angle OBA = angle BAO = pi/3\ \ \ text{(angles opposite equal sides)}`

`therefore \ OAB \ text{is equilateral}`

 

ii.     `z_1` `= z_2 e^(i frac{pi}{3})`
  `frac{z_1}{z_2}` `= e^(i frac{pi}{3})`
  `(frac{z_1}{z_2})^3` `= e^((3 xx i frac{pi}{3})) \ \ (text{by De Moivre})`
  `frac{z_1^3}{z_2^3}` `= e^(i pi)`
  `z_1^3` `= -z_2^3`
  `z_1^3 + z_2^3` `= 0`

COMMENT: The identity to prove suggests the addition of 2 cubes is a possible strategy.
`(z_1 + z_2)(z_1^2 – z_1 z_2 + z_2^2)` `= 0`
`z_1^2 – z_1 z_2 + z_2^2` `= 0`
`z_1^2 + z_2^2` `= z_1 z_2`

Complex Numbers, EXT2 N1 2020 HSC 13d

  1. Show that for any integer `n`,  `e^(i n theta) + e^(-i n theta) = 2 cos (n theta)`.   (1 mark)

  2. By expanding  `(e^(i theta) + e^(-i theta))^4`  show that
     
       `cos^4 theta = frac{1}{8} ( cos (4 theta) + 4 cos (2 theta) + 3 )`.   (3 marks)

  3. Hence, or otherwise, find  `int_0^(frac{pi}{2}) cos^4 theta\ d theta`.   (2 marks)

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i.      `e^(i n theta) + e^(-i n theta)` `= cos(n theta) + i sin(n theta) + cos(-n theta) + i sin(-n theta)`
    `= cos(n theta) + i sin(n theta) + cos(n theta) – i sin(n theta)`
    `= 2 cos (n theta)`

 

ii.    `(e^{i theta} + e^{-i theta})^4` `= (2 cos theta)^4`
    `= 16 cos^4 theta`

 
`text{Expand} \ (e^{i theta} + e^{-i theta})^4 :`

`e^(i 4 theta) + 4 e^(i 3 theta) e^(-i theta) + 6 e^(i 2 theta) e^(-i 2 theta) + 4 e^(i theta) e^(-i 3 theta) + e^(-i 4 theta)`

`= e^(i 4 theta) + 4e^(i 2 theta) + 6 + 4^(-i 2 theta) + e^(-i 4 theta)`

`= e^(i 4 theta) + e^(i 4 theta) + 4 (e^{i 2 theta} + e^{-i 2 theta}) + 6`

`= 2 cos (4 theta) + 8 cos (2 theta) + 6`
 

`therefore \ 16 cos^4 theta` `= 2 cos (4 theta) + 8 cos (2 theta) + 6`
`cos^4 theta` `= frac{1}{8} cos(4 theta) + 1/2 cos(2 theta) + 3/8`
`cos^4 theta` `= frac{1}{8} (cos(4 theta) + 4 cos(2 theta) + 3)`

 

iii.    `int_0^(frac{pi}{2}) cos^4 theta\ d theta` `= frac{1}{8} int_0^(frac{pi}{2}) cos(4 theta) + 4 cos(2 theta) + 3\ d theta`
    `= frac{1}{8} [ frac{1}{4} sin(4 theta) + 2 sin (2 theta) + 3 theta ]_0^(frac{pi}{2}`
    `= frac{1}{8} [( frac{1}{4} sin (2 pi) + 2 sin pi  + frac{3 pi}{2}) – 0 ]`
    `= frac{1}{8} ( frac{3 pi}{2})`
    `= frac{3 pi}{16}`

Proof, EXT2 P1 2020 HSC 13c

  1. By considering the right-angled triangle below, or otherwise, prove that
     
         `frac{a + b}{2} ≥ sqrt(ab)`, where  `a > b ≥ 0`.  (2 marks)
     
           

  2. Prove that  `p^2 + 4q^2 ≥ 4 pq`.  (1 marks)

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  2. `text{See Worked Solutions}`
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i.   `text{Strategy 1}`

`text{Using Pythagoras:}`

`x` `= sqrt{(a + b)^2 – (a – b)^2}`
  `= sqrt(4ab)`
  `= 2 sqrt(ab)`

  
`a + b \ text{is a hypotenuse}`

`a + b` `≥ x`
`a + b` `≥ 2sqrt(ab)`
`frac{a + b}{2}` `≥ sqrt(ab)`

 

`text{Strategy 2}`

`(sqrta – sqrtb)^2` `≥ 0`
`a – 2 sqrt(a) sqrt(b) + b` `≥ 0`
`a + b` `≥ 2 sqrt(ab)`
`frac{a + b}{2}` `≥ sqrt(ab)`

 

ii.    `text{Let} \ \ a = p  ,  b = 2 q`

`text(Using part i:)`

`frac{p + 2q}{2}` `≥ sqrt(2 pq)`
`p + 2q` `≥ 2 sqrt(2 pq)`
`p^2 + 4pq + 4 q^2` `≥ 8 pq`
`p^2 + 4 q^2` `≥ 4 pq`

Vectors, EXT2 V1 2020 HSC 13b

Consider the two lines in three dimensions given by
 

`underset~r = ((3),(-1),(7)) + λ_1 ((1),(2),(1))`  and  `underset~r = ((3),(-6),(2)) + λ_2 ((-2),(1),(3))`.
 

By equating components, find the point of intersection of the two lines.   (3 marks)

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`((1),(-5),(5))`

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`underset~(r_1) = ((3),(-1),(7)) + λ_1 ((1),(2),(1)) = ((3 + λ_1),(-1 + 2λ_1),(7 + λ_1))`

`underset~(r_2) = ((3),(-6),(2)) + λ_2 ((-2),(1),(3)) = ((3 – 2λ_2),(-6 + λ_2),(2 + 3λ_2))`

 
`text{Intersection occurs when:}`

`3 + λ_1 ` `= 3 – 2λ_2 \ … \ (1)`
`-1 + 2λ_1` `= -6 + λ_2 \ … \ (2)`
`7 + λ_1` `= 2 + 3λ_2 \ … \ (3)`

 
`text{Subtract} \ (3) – (1):`

`4` `= -1 + 5 λ_2`
`λ_2` `=1`

 
`text{Substitute} \ \ λ_2 = 1\ \ text{into} \ (1):`

`3 + λ_1` `= 1`
`λ_1` `= -2`

 

`text{Test that}\  \ λ_1 = -2 \ , \  λ_2 = 1\ \ text{satisfies} \ (2):`

`-1 + 2 xx  – 2` `= -6 + 1`
`-5` `= -5`

 
`∃ \ λ_1,  λ_2, \ text{that satisfy all equations}`

`=> \ text{3 lines intersect at a point}`

`:.\ text{Point of intersection}`

`= ((3),(-6),(2)) + 1 ((-2),(1),(3)) = ((1),(-5),(5))`

Mechanics, EXT2 M1 2020 HSC 12b

A particle is projected from the origin with initial velocity `u` m/s at an angle  `theta`  to the horizontal. The particle lands at  `x = R`  on the `x`-axis. The acceleration vector is given by  `underset~a = ((0),(-g))`, where `g` is the acceleration due to gravity. (Do NOT prove this.)
 

  1. Show that the position vector  `underset~r (t)`  of the particle is given by
     
         `underset~r (t) = ((ut  cos theta),(ut  sin theta - frac{1}{2} g t^2))`.   (3 marks)

  2. Show that the Cartesian equation of the path of flights is given by
     
         `y = frac{-gx^2}{2u^2} (tan^2 theta - frac{2u^2}{gx} tan theta + 1)`.   (3 marks)

  3. Given  `u^2 > gR`, prove that there are 2 distinct values of  `theta`  for which the particle will land at  `x = R`.   (2 marks)

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i.    `underset~a = ((0),(-g))`

`ddotx = 0`

`dotx = int ddotx \ dt = c`
 
`text{When} \ \ t = 0, \ dotx = u  cos theta \ =>  \ c = u  cos theta`

`=> dotx = u  cos theta`

 
`x = int dotx \ dt =u t cos theta + c`

`text{When} \ \ t = 0 , \ x = 0, \ c = 0`

`therefore \ x = ut  cos theta`
 

`ddoty = -g`

`doty = int – g \ dt = -g t + c`

`text{When} \ \ t = 0 , \ doty = u  sin theta`

`=> doty = u  sin theta – g t`
 

`y = int doty \ dt = ut  sin theta – frac {1}{2} g t^2 + c`

`text{When} \ \ t = 0, \ y = 0  \ => \ c = 0`

`therefore  y = ut  sin theta – frac(1)(2) g t^2`

`:. underset~r = ((x),(y)) = ((ut  cos theta),(ut sin theta – frac{1}{2} g t^2))`
 

ii.   `x = ut \ cos theta`

`t = frac{x}{u \ cos theta}`
 
`text{Substitute into} \ y:`

`y` `= u * frac{x}{u \ cos theta}\ sin theta – frac{1}{2} g ( frac{x}{u \ cos theta} )^2`
  `= x tan theta – frac{gx^2}{2 u^2 cos^2 theta}`
  `= frac{-gx^2}{2u^2} ( frac{1}{cos^2 theta} – frac{2u^2}{gx} tan theta )`
  `= frac{-gx^2}{2u^2} ( sec^2 theta – frac{2u^2}{gx} tan theta )`
  `= frac{-gx^2}{2u^2} ( tan^2 theta – frac{2u^2}{gx} tan theta + 1 )`

 

iii.    `text{When} \ \ x = R \ , \ y = 0`

Mean mark part (iii) 52%.
`frac{-gR^2}{2 u^2}` `( tan^2 theta – frac{2u^2}{gR} tan theta + 1 ) = 0`
   `tan^2 theta – frac{2u^2}{gR} tan theta + 1 = 0`
 
`Delta`		  
`= ( frac{-2u^2}{gR} )^2 – 4 * 1 * 1`
  `= frac{4u^2}{g^2 R^2} – 4`

 

  `u^2` `> gR\ \ \ text{(given)}`
  `u^4` `> g^2 R^2`
  `frac{u^4}{g^2 R^2}` `> 1`
  `frac{4u^4}{g^2 R^2}` `> 4`
  `frac{4u^4}{g^2 R^2} – 4` `> 0`
  `Delta` `> 0`

 
`therefore \ 2 \ text{distinct values of} \ \ theta\  \ text{satisfy} \ \ x = R.`

Mechanics, EXT2 M1 2020 HSC 12a

A 50-kilogram box is initially at rest. The box is pulled along the ground with a force of 200 newtons at an angle of 30° to the horizontal. The box experiences a resistive force of  `0.3R`  newtons, where `R` is the normal force, as shown in the diagram.
 
Take the acceleration `g` due to gravity to be 10m/s2.
 

  1. By resolving the forces vertically, show that  `R =400`.   (2 marks)

  2. Show that the net force horizontally is approximately 53.2 newtons.  (2 marks)

  3. Find the velocity of the box after the first three seconds.  (2 marks)

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  1. `text{See Worked Solutions}`
  2. `text{See Worked Solutions}`
  3. `3.19 \ text{ms}^-1`
Show Worked Solution

i.   

`text{Resolving forces vertically:}`

`R + 200 \ sin 30^@` `= 50g`
`R + 200 xx frac{1}{2}` `= 50 xx 10`
`R + 100` `= 500`
`therefore \ R` `= 400 \ text(N)`

 

ii.    `text{Resolving forces horizontally:}`

`text{Net Force}` `= 200 \ cos 30^@ – 0.3 R`
  `= 200 xx frac{sqrt3}{2} – 0.3 xx 400`
  `= 100 sqrt3 – 120`
  `= 53.2 \ text{N (to 1 d. p.)}`

 

iii.    `F` `=ma`
  `50 a` `=100 sqrt300 – 120`
  `a` `= frac{100 sqrt3 – 120}{50}\ text(ms)^(-2)`

 
`text{Initially} \ \ u = 0,`

`v` `= u + at`
`v_(t=3)` `= 0 + frac{100 sqrt3 – 120}{50} xx 3`
  `= 3.1923 \ …`
  `= 3.19 \ text{ms}^-1 \ text{(to 2 d.p.)}`

Complex Numbers, EXT2 N2 2020 HSC 11e

Solve  `z^2 + 3 z + (3-i) = 0`, giving your answer(s) in the form  `a + bi`, where  `a`  and  `b`  are real.  (4 marks)

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`z= -1 + i \ \ text{or} \ \ -2-i`

Show Worked Solution

`z^2 + 3z + (3-i) = 0`

`z` `= frac{-3 ± sqrt(9-4 · 1 (3-i))}{2}`
  `= frac{-3 ± sqrt(4i-3)}{2} `

 
`text{Consider} \ \ Delta = sqrt(4i-3) :`

`x + i y` `= sqrt(4i-3)`
`(x + iy)^2` `= 4i-3`
`x^2-y^2 + 2xyi` `= 4i-3`

 
`text{Equating real and imaginary parts:}`

`2 xy` `= 4`
`xy` `= 2\ …\ (1)`
`x^2-y^2` `= -3\ …\ (2)`

 
`=> x = 1 \ , \ y =2`

`=> \ x + iy = 1 + 2 i`
 

`therefore z` `= frac{-3 ± (1 + 2i)}{2}`
`z ` `= -1 + i \ \ text{or} \ \ -2-i`

Vectors, EXT2 V1 2020 HSC 11d

Consider the two vectors  `underset~u = 2 underset~i - underset~j + 3 underset~k`  and  `underset~v = p underset~i +  underset~j + 2 underset~k`.
 
For what values of  `p`  are  `underset~u - underset~v`  and  `underset~u + underset~v`  perpendicular?   (3 marks)

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`p= ± 3`

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`underset~u – underset~v = ((-2),(-1),(3)) – ((p),(1),(2)) = ((-2 – p),(-2),(1))`
 

`underset~u + underset~v = ((-2),(-1),(3)) + ((p),(1),(2)) = ((p – 2),(0),(5))`
 

`⊥ \ text{when} \ \ (underset~u – underset~v) · (underset~u + underset~v ) = 0 :`
 

`((-2 – p),(-2),(1)) · ((p – 2),(0),(5)) = 0`
 

`-(p + 2)(p-2) + 5` `= 0`
`-(p^2 – 4) + 5` `= 0`
`-p^2 + 9` `= 0`
`p^2` `= 9`
`p` `= ± 3`

Mechanics, EXT2 M1 2020 HSC 11c

A particle starts at the origin with velocity 1 and acceleration given by

`a = v^2 + v`,

where `v` is the velocity of the particle.

Find an expression for `x`, the displacement of the particle, in terms of `v`.   (3 marks)

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`x= ln \ | frac{v + 1}{2} |`

Show Worked Solution
`a` `= v^2 + v`
`v · frac{dv}{dx}` `= v^2 + v`
`frac{dv}{dx}` `= v + 1`
`frac{dx}{dv}` `= frac{1}{v + 1}`
`x` `= int frac{1}{v + 1}\ dv`
  `= ln \ | v + 1 | + c`

 
`text{When} \ \ x = 0 , \ v = 1`

`0 = ln \ 2 + c`

`c = -ln \ 2`

`therefore \ x` `= ln \ | v + 1| – ln \ 2`
  `= ln \ | frac{v + 1}{2} |`

Calculus, EXT2 C1 2020 HSC 11b

Use integration by parts to evaluate  `int_1^e x ln x \ dx`.   (3 marks)

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`frac{e^2 + 1}{4}`

Show Worked Solution
`u = ln \ x` `v′ = x`
`u′ = frac{1}{x}` `v = frac{x^2}{2}`

 

`int _1^e x \ ln \ x \ dx` `= [ frac{x^2}{2} · ln \ x ]_1^e – int_1^e frac{x^2}{2} · frac{1}{x}\ dx`
  `= [frac{e^2}{2} ln \ e – frac{1}{2} ln 1]- int_1^e frac{x}{2}\ dx`
  `= frac{e^2}{2} – [ frac{x^2}{4}]_1^e`
  `= frac{e^2}{2} – ( frac{e^2}{4} – frac{1}{4} )`
  `= frac{e^2 + 1}{4}`

Proof, EXT2 P1 2020 HSC 7 MC

Consider the proposition:

'If  `2^n - 1`  is not prime, then `n` is not prime'. 

Given that each of the following statements is true, which statement disproves the proposition?

  1. `2^5 - 1` is prime
  2. `2^6 - 1` is divisible by 9
  3. `2^7 - 1` is prime
  4. `2^11 - 1` is divisible by 23
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`D`

Show Worked Solution

`text(Strategy 1 – Contradiction)`

`text(Consider option)\ D,`

`text(S)text(ince)\ \ 2^11 -1\ \ text(is divisible by 23, it is NOT prime.)`

`text(The proposition states that 11 is not prime which is false.)`

`:. 2^11 – 1\ \ text(is divisible by 23, disproves the proposition.)`
 

`text(Strategy 2 – Contrapositive)`

`text{The proposition is conditional}`

`X => Y`

`text{L}text{ogically equivalent contrapositive statement}`

`not \ Y => not \ X`

`text{i.e. If} \ n \ text{is prime} \ => \ 2^n – 1 \ text{is prime.}`
 

`text{Consider D:}`

`n = 11 \ text{(prime)}`

`2^11 – 1 \ text{is divisible by 23 (not prime)}`

`therefore \ text{Contrapositive statement is false and disproves the proposition.}`
  

`=> \ D`

Vectors, EXT2 V1 2020 HSC 3 MC

 What is the Cartesian equation of the line  `underset~r = ((1),(3)) + lambda ((-2),(4))`?

  1. `2y + x = 7`
  2. `y - 2x = -5`
  3. `y + 2x = 5`
  4. `2y - x = -1`
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`C`

Show Worked Solution

`((x),(y)) = ((1),(3)) + λ ((-2),(4))`
 

`x = 1 – 2λ \ => \ λ = frac{1- x}{2}`

`y = 3 + 4λ \ => \ λ = frac{y- 3}{4}`
 

`frac{y – 3}{4}` `= frac{1 – x}{2}`
`y – 3` `= 2 – 2x`
`y + 2x` `= 5`

  
`=> \ C`

Complex Numbers, EXT2 N2 2020 HSC 2 MC

Given that  `z = 3 + i`  is a root of  `z^2 + pz + q = 0`, where `p` and `q` are real, what are the values of `p` and `q`?

  1.  `p = -6 \ , \ q = sqrt(10)`
  2.  `p = -6 \ , \ q = 10`
  3.  `p = 6 \ , \ q = sqrt(10)`
  4.  `p = 6 \ , \ q = 10`
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`B`

Show Worked Solution

`text{S}text{ince} \ \ z_1 = 3 + i \ \ text{is a root}`

`=> \ z_2 = 3 – i \ \ text{is a root}`
 

`z_1 + z_2` `= frac{-b}{a}`
`(3 + i) + (3 – i)` `= -p`
`therefore \ p` `= -6`

 

`z_1 z_2` `= frac{c}{a}`
`(3 + i) (3 – i)` `= q`
`3^2 – i^2` `= q`
`therefore \ q` `= 10`

 
`=> \ B`

Algebra, STD1 A3 2020 HSC 19

Each year the number of fish in a pond is three times that of the year before.

  1. The table shows the number of fish in the pond for four years.
    \begin{array} {|l|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
    \hline
    \rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & & & 2700\\
    \hline
    \end{array}

    Complete the table above showing the number of fish in 2021 and 2022.   (2 marks)
     

  2. Plot the points from the  table in part (a) on the grid.   (2 marks)
     
  3. Which model is more suitable for this dataset: linear or exponential? Briefly explain your answer.   (2 marks)

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a.   

\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & 300 & 900 & 2700\\
\hline
\end{array}

b.   
       

c.     The more suitable model is exponential.

A linear dataset would graph a straight line which is not the case here.

An exponential curve can be used to graph populations that grow at an increasing rate, such as this example.

Show Worked Solution

a.        

\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & 300 & 900 & 2700\\
\hline
\end{array}

b.  

c.     The more suitable model is exponential.

A linear dataset would graph a straight line which is not the case here.

An exponential curve can be used to graph populations that grow at an increasing rate, such as this example.

♦ Mean mark (c) 31%.