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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)

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  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)

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  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)

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  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)

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

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)

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  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)

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  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}`
Show Worked Solution

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

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))`

Show Worked Solution

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

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)

Show Answers Only
  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)

Show Answers Only

`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)

Show Answers Only

`p= ± 3`

Show Worked Solution

`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)

Show Answers Only

`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)

Show Answers Only

`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
Show Answers Only

`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`
Show Answers Only

`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`
Show Answers Only

`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)

Show Answers Only

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%.

Statistics, STD1 S1 2020 HSC 6 MC

When blood pressure is measured, two numbers are recorded: systolic pressure and diastolic pressure. If the measurements recorded are 140 systolic and 90 diastolic, then the blood pressure is written as 140/90 mmHg.
 
Blood pressure measurements are categorized as shown in the diagram.
 


 

Amy has a blood pressure of 97/76 mmHg and Betty has a blood pressure of 125/75 mmHg.

Which row of the table describes the blood pressure for Amy and Betty?
  

Show Answers Only

`D`

Show Worked Solution

`text{Amy: 97/76 is Ideal}`

`text{Betty: 125/75 is Pre-high}`

`=> \ D`

Functions, EXT1 F2 2020 HSC 11a

Let  `P(x) = x^3 + 3x^2-13x + 6`.

  1. Show that  `P(2) = 0`.  (1 mark)

  2. Hence, factor the polynomial  `P(x)`  as  `A(x)B(x)`, where  `B(x)`  is a quadratic polynomial.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `P(x) = (x-2)(x^2 + 5x – 3)`
Show Worked Solution
i.    `P(2)` `= 8 + 12-26 + 6`
    `= 0`

 

ii.   

`:. P(x) = (x-2)(x^2 + 5x – 3)`

Measurement, STD2 M6 2020 HSC 31

Mr Ali, Ms Brown and a group of students were camping at the site located at `P`. Mr Ali walked with some of the students on a bearing of 035° for 7 km to location `A`. Ms Brown, with the rest of the students, walked on a bearing of 100° for 9 km to location `B`.
 


 

  1. Show that the angle `APB` is 65°.  (1 mark)

  2. Find the distance `AB`.  (2 marks)

  3. Find the bearing of Ms Brown's group from Mr Ali's group. Give your answer correct to the nearest degree.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `8.76\ text{km  (to 2 d.p.)}`
  3. `146^@`
Show Worked Solution
a.    `angle APB` `= 100 – 35`
    `= 65^@`

 

b.   `text(Using cosine rule:)`

Mean mark 53%.
`AB^2` `= AP^2 + PB^2 – 2 xx AP xx PB cos 65^@`
  `= 49 + 81 – 2 xx 7 xx 9 cos 65^@`
  `= 76.750…`
`:.AB` `= 8.760…`
  `= 8.76\ text{km  (to 2 d.p.)}`

 
c.

 
`anglePAC = 35^@\ (text(alternate))`

♦♦ Mean mark 22%.

`text(Using cosine rule, find)\ anglePAB:`

`cos anglePAB` `= (7^2 + 8.76 – 9^2)/(2 xx 7 xx 8.76)`  
  `= 0.3647…`  
`:. angle PAB` `= 68.61…^@`  
  `= 69^@\ \ (text(nearest degree))`  

 

`:. text(Bearing of)\ B\ text(from)\ A\ (theta)` 

`= 180 – (69 – 35)`

`= 146^@`

Statistics, EXT1 S1 2020 HSC 12b

When a particular biased coin is tossed, the probability of obtaining a head is `3/5`.

This coin is tossed 100 times.

Let `X` be the random variable representing the number of heads obtained. This random variable will have a binomial distribution.

  1. Find the expected value, `E(X)`.  (1 mark)

  2. By finding the variance, `text(Var)(X)`, show that the standard deviation of `X` is approximately 5.  (1 mark)

  3. By using a normal approximation, find the approximate probability that `X` is between 55 and 65.  (1 mark)

Show Answers Only
  1. `60`
  2. `text(See Worked Solutions)`
  3. `68text(%)`
Show Worked Solution

i.   `X = text(number of heads)`

`X\ ~\ text(Bin) (n, p)\ ~\ text(Bin) (100, 3/5)`

`E(X)` `= np`
  `= 100 xx 3/5`
  `= 60`

 

ii.    `text(Var)(X)` `= np(1 – p)`
    `= 60 xx 2/5`
    `= 24`

 

`sigma(x)` `= sqrt24`
  `~~ 5`

 

iii.    `P(55 <= x <=65)` `~~ P(−1 <= z <= 1)`
    `~~ 68text(%)`

Proof, EXT1 P1 2020 HSC 12a

Use the principle of mathematical induction to show that for all integers  `n >= 1`,

`1 xx 2 + 2 xx 5 + 3 xx 8 + … + n(3n-1) = n^2(n + 1)`.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(Prove)`

`1 xx 2 + 2 xx 5 + 3 xx 8 + … + n(3n-1) = n^2(n + 1)`
 

`text(Prove true for)\ \ n = 1:`

`text(LHS) = 1(3 xx 1-1) = 2`

`text(RHS) = 1^2(1 + 1) = 2`
 

`text(Assume true for)\ \ n = k:`

`text(i.e.)\ \ 1 xx 2 + 2 xx 5 + 3 xx 8 + … + k(3k-1) = k^2(k + 1)`
 

`text(Prove true for)\ \ n = k + 1:`

`text(i.e.)\ 1 xx 2 + 2 xx 5 + … + k(3k-1) + (k + 1)(3k + 2) = (k + 1)^2(k + 2)`

`text(LHS)` `= 1 xx 2 + … + k(3k-1) + (k + 1)(3k + 2)`
  `= k^2(k + 1) + (k + 1)(3k + 2)`
  `= (k + 1)(k^2 + 3k + 2)`
  `= (k + 1)(k + 1)(k + 2)`
  `= (k + 1)^2(k + 2)`
  `=\ text(RHS)`

 
`=>\ text(True for)\ n = k + 1`

`:. text(S)text(ince true for)\ n = 1,\ text(by PMI, true for integral)\ n >= 1`.

Functions, 2ADV F1 2020 HSC 11

There are two tanks on a property, Tank `A` and Tank `B`. Initially, Tank `A` holds 1000 litres of water and Tank B is empty.

  1.  Tank `A` begins to lose water at a constant rate of 20 litres per minute. The volume of water in Tank `A` is modelled by  `V = 1000 - 20t`  where  `V`  is the volume in litres and  `t`  is the time in minutes from when the tank begins to lose water.   (1 mark)
     
    On the grid below, draw the graph of this model and label it as Tank `A`.
     
       

  2. Tank `B` remains empty until  `t=15`  when water is added to it at a constant rate of 30 litres per minute.

     

    By drawing a line on the grid (above), or otherwise, find the value of  `t`  when the two tanks contain the same volume of water.  (2 marks)

  3. Using the graphs drawn, or otherwise, find the value of  `t`  (where  `t > 0`) when the total volume of water in the two tanks is 1000 litres.  (1 mark)

Show Answers Only
  1.  `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}` 
      
  2. `29 \ text{minutes}`
  3. `45 \ text{minutes}`
Show Worked Solution

a.     `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}` 
 


 

b.   `text{T} text{ank} \ B \ text{will pass through (15, 0) and (45, 900)}`  
 

   

`text{By inspection, the two graphs intersect at} \ \ t = 29 \ text{minutes}`

 
c.   `text{Strategy 1}`

`text{By inspection of the graph, consider} \ \ t = 45`

`text{T} text{ank A} = 100 \ text{L} , \ text{T} text{ank B} =900 \ text{L} `

`:.\ text(Total volume = 1000 L when  t = 45)`
  

`text{Strategy 2}`

`text{Total Volume}` `=text{T} text{ank A} + text{T} text{ank B}`
`1000` `= 1000 – 20t + (t – 15) xx 30`
`1000` `= 1000 – 20t + 30t – 450 `
`10t` `= 450`
`t` `= 45 \ text{minutes}`

Algebra, STD2 A4 2020 HSC 24

There are two tanks on a property, Tank A and Tank B. Initially, Tank A holds 1000 litres of water and Tank B is empty.

  1. Tank A begins to lose water at a constant rate of 20 litres per minute.

     

    The volume of water in Tank A is modelled by  `V = 1000 - 20t`  where `V` is the volume in litres and  `t`  is the time in minutes from when the tank begins to lose water.

     

    On the grid below, draw the graph of this model and label it as Tank A.   (1 mark)
     

     

  2. Tank B remains empty until  `t=15`  when water is added to it at a constant rate of 30 litres per minute.
     By drawing a line on the grid (above), or otherwise, find the value of  `t`  when the two tanks contain the same volume of water.  (2 marks)

  3. Using the graphs drawn, or otherwise, find the value of  `t`  (where  `t > 0`) when the total volume of water in the two tanks is 1000 litres.  (1 mark)

Show Answers Only
  1.  `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}` 
      
  2. `29 \ text{minutes}`
  3. `45 \ text{minutes}`
Show Worked Solution

a.     `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}` 
 

 

b.   `text{T} text{ank} \ B \ text{will pass through (15, 0) and (45, 900)}`  
 

   

`text{By inspection, the two graphs intersect at} \ \ t = 29 \ text{minutes}`

 
c.   `text{Strategy 1}`

♦♦ Mean mark part (c) 22%.

`text{By inspection of the graph, consider} \ \ t = 45`

`text{T} text{ank A} = 100 \ text{L} , \ text{T} text{ank B} =900 \ text{L} `

`:.\ text(Total volume = 1000 L when  t = 45)`
  

`text{Strategy 2}`

`text{Total Volume}` `=text{T} text{ank A} + text{T} text{ank B}`
`1000` `= 1000 – 20t + (t – 15) xx 30`
`1000` `= 1000 – 20t + 30t – 450 `
`10t` `= 450`
`t` `= 45 \ text{minutes}`