SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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}`

Vectors, EXT1 V1 2020 HSC 4 MC

Maria starts at the origin and walks along all of the vector  `2underset~i + 3underset~j`, then walks along all of the vector  `3underset~i - 2underset~j`  and finally along all of the vector  `4underset~i - 3underset~j`.

How far from the origin is she?

  1. `sqrt77`
  2. `sqrt85`
  3. `2sqrt13 + sqrt5`
  4. `sqrt5 + sqrt7 + sqrt13`
Show Answers Only

`B`

Show Worked Solution
`underset~v` `= ((2),(3)) + ((3),(−2)) + ((4),(−3))`
  `= ((9),(−2))`
`|underset~v|` `= sqrt(9^2 + (−2)^2)`
  `= sqrt85`

 
`=>B`

Calculus, EXT1 C2 2020 HSC 3 MC

Which of the following is an anti-derivative of  `1/(4x^2 + 1)`?

  1. `2tan^(−1)(x/2) + c`
  2. `1/2tan^(−1)(x/2) + c`
  3. `2tan^(−1)(2x) + c`
  4. `1/2tan^(−1)(2x) + c`
Show Answers Only

`D`

Show Worked Solution
`int 1/(4x^2 + 1)\ dx` `= 1/2 int 2/(1+ (2x)^2)\ dx`
  `= 1/2 tan^(−1) (2x) + c`

 
`=>D`

Financial Maths, 2ADV M1 2020 HSC 26

Tina inherits $60 000 and invests it in an account earning interest at a rate of 0.5% per month. Each month, immediately after the interest has been paid, Tina withdraws $800.

The amount in the account immediately after the `n`th withdrawal can be determined using the recurrence relation

`A_n = A_(n - 1)(1.005) - 800`,

where `n = 1, 2, 3, …`  and  `A_0 = 60\ 000`

  1. Use the recurrence relation to find the amount of money in the account immediately after the third withdrawal.  (2 marks)

  2. Calculate the amount of interest earned in the first three months.  (2 marks)

  3. Calculate the amount of money in the account immediately after the 94th withdrawal.  (3 marks) 

Show Answers Only
  1. `$58\ 492.49`
  2. `$892.49`
  3. `$187.85`
Show Worked Solution
a.    `A_1` `= 60\ 000(1.005) – 800 = $59\ 500`
  `A_2` `= 59\ 500(1.005) – 800 = $58\ 997.50`
  `A_3` `= 58\ 997.50(1.005) – 800 = $58\ 492.49`

 

b.   `text{Amount (not interest)}`

`= 60\ 000 – (3 xx 800)`

`= $57\ 600`
 

`:.\ text(Interest earned in 3 months)`

`= A_3 – 57\ 600`

`= 58\ 492.49 – 57\ 600`

`= $892.49`
 

c.   `A_1 = 60\ 000(1.005) – 800`

`A_2` `= [60\ 000(1.005) – 800](1.005) – 800`
  `= 60\ 000(1.005)^n – 800(1.005 + 1)`
`vdots`  
`A_n` `= 60\ 000(1.005)^n – 800(1 + 1.005 + … + 1.005^(n – 1))`
`A_94` `= 60\ 000(1.005)^94 – 800\ underbrace((1 + 1.005 + … + 1.005^93))_(text(GP where)\ a = 1,\ r = 1.005,\ n = 94)`
  `= 60\ 000(1.005)^94 – 800 ((1(1.005^94 – 1))/(1.005 – 1))`
  `= 60\ 000(1.005)^94 – 160\ 000(1.005^94 – 1)`
  `= $187.85`

Probability, 2ADV S1 2020 HSC 14

History and Geography are two of the subjects students may decide to study. For a group of 40 students, the following is known.

    • 7 students study neither History nor Geography
    • 20 students study History
    • 18 students study Geography
  1. A student is chosen at random. By a using a Venn diagram, or otherwise, find the probability that the student studies both History and Geography.  (2 marks)

  2. A students is chosen at random. Given that the student studies Geography, what is the probability that the student does NOT study History?  (1 mark)

  3. Two different students are chosen at random, one after the other. What is the probability that the first student studies History and the second student does NOT study History?  (2 marks)

Show Answers Only
  1.  
  2. `13/18`
  3. `10/39`
Show Worked Solution
a.   
`P(text(H and G))` `= 5/40`
  `= 1/8`

 

♦ Mean mark (b) 49%.
b.    `P(bartext(H) | text(G))` `= (P(bartext(H) ∩ text(G)))/(Ptext{(G)})`
    `= 13/18`

 

c.    `P(text(H), bartext(H))` `= 20/40 xx 20/39`
    `= 10/39`

Calculus, 2ADV C3 2020 HSC 21

Hot tea is poured into a cup. The temperature of tea can be modelled by  `T = 25 + 70(1.5)^(−0.4t)`, where `T` is the temperature of the tea, in degrees Celsius, `t` minutes after it is poured.

  1. What is the temperature of the tea 4 minutes after it has been poured?  (1 mark)

  2. At what rate is the tea cooling 4 minutes after it has been poured?  (2 marks)

  3. How long after the tea is poured will it take for its temperature to reach 55°C?  (3 marks)

Show Answers Only
  1. `61.6\ \ (text(to 1 d.p.))`
  2. `−5.9^@text(C/min)`
  3. `5.2\ text(minutes  (to 1 d.p.))`
Show Worked Solution
a.    `T` `= 25 + 70(1.5)^(−0.4 xx 4)`
    `= 61.58…`
    `= 61.6\ \ (text(to 1 d.p.))`

 

b.    `(dT)/(dt)` `= 70 log_e(1.5) xx −0.4(1.5)^(−0.4t)`
    `= −28log_e(1.5)(1.5)^(−0.4t)`

 

`text(When)\ \ t = 4,`

`(dT)/(dt)` `= −28log_e(1.5)(1.5)^(−1.6)`
  `= −5.934…`
  `= −5.9^@text(C/min  (to 1 d.p.))`

 

c.   `text(Find)\ \ t\ \ text(when)\ \ T = 55:`

♦ Mean mark part (c) 44%.
`55` `= 25 + 70(1.5)^(−0.4t)`
`30` `= 70(1.5)^(0.4t)`
`(1.5)^(−0.4t)` `= 30/70`
`−0.4t log_e(1.5)` `= log_e\ 3/7`
`−0.4t` `= (log_e\ 3/7)/(log_e (1.5))`
`:. t` `= (−2.08969)/(−0.4)`
  `= 5.224…`
  `= 5.2\ text(minutes  (to 1 d.p.))`

Trigonometry, 2ADV T1 2020 HSC 15

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:)`

`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))`

`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^@`

Measurement, STD2 M6 2020 HSC 16

Consider the triangle shown.
 


 

  1. Find the value of `theta`, correct to the nearest degree.   (2 marks)

  2. Find the value of `x`, correct to one decimal place.   (2 marks)

Show Answers Only
  1. `39^@`
  2. `12.1 \ text{(to 1 d.p.)}`
Show Worked Solution
a.      `tan theta` `= frac{8}{10}`
  `theta` `= tan ^(-1) frac{8}{10}`
    `= 38.659…`
    `= 39^@ \ text{(nearest degree)}`

 

b.     `text{Using Pythagoras:}`

`x` `= sqrt{8^2 + 10^2}`
  `= 12.806…`
  `= 12.8 \ \ text{(to 1 d.p.)}`

Calculus, 2ADV C4 2020 HSC 18

  1. Differentiate  `e^(2x) (2x + 1)`.  (2 marks)

  2. Hence, find  `int(x + 1)e^(2x)\ dx`.  (1 marks)

Show Answers Only
  1. `4e^(2x)(x + 1)`
  2. `1/4 e^(2x)(2x + 1) + c`
Show Worked Solution
a.    `y` `= e^(2x) (2x + 1)`
  `(dy)/(dx)` `= 2e^(2x)(2x + 1) + 2e^(2x)`
    `= 2e^(2x)(2x + 2)`
    `= 4e^(2x)(x + 1)`

♦ Mean mark part (b) 40%.

 

b.    `int(x + 1)e^(2x)dx` `= 1/4 int 4e^(2x)(x + 1)`
    `= 1/4 e^(2x)(2x + 1) + c`

Financial Maths, 2ADV M1 2020 HSC 12

Calculate the sum of the arithmetic series  `4 + 10 + 16 + … + 1354`.  (3 marks)

Show Answers Only

`153\ 454`

Show Worked Solution

`a = 4, \ l = 1354, \ d = 10 – 4 = 6`

`text(Find)\ n:`

`T_n` `= a + (n + 1)d`
`1354` `= 4 + (n – 1)6`
`1354` `= 6n – 2`
`n` `= 1356/6`
  `= 226`

 

`:. S_226` `= n/2 (a + l)`
  `= 226/2(4 + 1354)`
  `= 153\ 454`

Algebra, STD2 A1 2020 HSC 3 MC

The distance between Bricktown and Koala Creek is 75 km. A person travels from Bricktown to Koala Creek at an average speed of 50 km/h.

How long does it take the person to complete the journey?

  1.  40 minutes
  2.  1 hour 25 minutes
  3.  1 hour 30 minutes
  4.  1 hour 50 minutes
Show Answers Only

`C`

Show Worked Solution
`text(Time)` `= frac(text(Distance))(text(Speed))`
  `= frac(75)(50)`
  `=1.5 \ text(hours)`
  `= 1 \ text(hour) \ 30 \ text(minutes)`

 
`=> \ C`

Functions, 2ADV F2 2020 HSC 2 MC

The function  `f(x) = x^3`  is transformed to  `g(x) = (x - 2)^3 + 5`  by a horizontal translation of 2 units followed by a vertical translation of 5 units.

Which row of the table shows the directions of the translations?
 

Show Answers Only

`B`

Show Worked Solution

`text(Horizontal translation: 2 units to the right)`

`x^3 -> (x – 2)^3`

`text(Vertical translation: 5 units up`

`(x – 2)^3 -> (x – 2)^3 + 5`

`=>\ B`

Mechanics, EXT2 M1 EQ-Bank 4

A torpedo with a mass of 80 kilograms has a propeller system that delivers a force of `F` on the torpedo, at maximum power. The water exerts a resistance on the torpedo proportional to the square of the torpedo's velocity `v`.

  1. Explain why  `(dv)/(dt) = 1/80 (F - kv^2)`
     
    where `k` is a positive constant.  (1 mark)

  2. If the torpedo increases its velocity from  `text(10 ms)\ ^(−1)`  to  `text(20 ms)\ ^(−1)`, show that the distance it travels in this time, `d`, is given by
     
         `d = 40/k log_e((F - 100k)/(F - 400k))`  (3 marks)

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

i.   `R ∝ v^2`

`R = −kv^2\ \ (k\ text{is positive constant})`

`text(Newton’s 2nd Law:)`

`text(Net Force)= mddotx` `= F – R`
`80ddotx` `= F – kv^2`
`ddotx` `= 1/80 (F – kv^2)`
`(dv)/(dt)` `= 1/80 (F – kv^2)`

 

ii.    `v · (dv)/(dx)` `= 1/80 (F – kv^2)`
  `(dv)/(dx)` `= (F – kv^2)/(80v)`
  `(dx)/(dv)` `= (80v)/(F – kv^2)`
  `x` `= −40 int (−2v)/(F – kv^2)\ dv`
    `= −40/k log_e (F – kv^2) + C`

 

`text(When)\ \ v = 10:`

`x_1 = −40/k log_e (F – 100k) + C`
 

`text(When)\ \ v = 20:`

`x_2 = −40/k log_e (F – 400k) + C`
 

`d` `= x_2 – x_1`
  `= −40/k log_e (F – 400k) + 40/k log_e (F – 100k)`
  `= 40/k log_e ((F – 100k)/(F – 400k))`

Mechanics, EXT2 M1 EQ-Bank 1

A canon ball of mass 9 kilograms is dropped from the top of a castle at a height of `h` metres above the ground.

The canon ball experiences a resistance force due to air resistance equivalent to  `(v^2)/500`, where `v` is the speed of the canon ball in metres per second. Let  `g=9.8\ text(ms)^-2`  and the displacement, `x` metres at time `t` seconds, be measured in a downward direction.

  1. Show the equation of motion is given by
     
         `ddotx = g - (v^2)/4500`  (1 mark)

  2. Show, by integrating using partial fractions, that
     
         `v = 210((e^(7/75 t) - 1)/(e^(7/75 t) + 1))`  (5 marks)

     

  3. If the canon hits the ground after 4 seconds, calculate the height of the castle, to the nearest metre.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `78\ text(metres)`
Show Worked Solution

i.   `text(Newton’s 2nd Law:)`

`text(Net Force)` `= mddotx`
`mddotx` `= mg – (v^2)/500`
`9ddotx` `= 9g – (v^2)/500`
`ddotx` `= g – (v^2)/4500`

 

ii.    `(dv)/(dt)` `= g – (v^2)/4500`
    `= (44\ 100 – v^2)/4500`

 
`(dt)/(dv) = 4500/(44\ 100 – v^2)`
 

`text(Using Partial Fractions:)`

`1/(44\ 100 – v^2) = A/(210- v) + B/(210 – v)`

`A(210 – v) + B(210 + v) = 1`
 

`text(If)\ \ v = 210,`

`420B = 1 \ => \ B = 1/420`
 

`text(If)\ \ v = −210,`

`420A = 1 \ => \ A = 1/420`
 

`t` `= int 4500/(210^2 – v^2)\ dv`
  `= 4500/420 int 1/(210 + v) + 1/(210 – v)\ dv`
  `= 75/7 [ln(210 + v) – ln(210 – v)] + c`
  `= 75/7 ln((210 + v)/(210 – v)) + c`

 

`text(When)\ \ t = 0, v = 0:`

`0` `= 75/7 ln(210/210) + c`
`:. c` `= 0`

 

` t` `= 75/7 ln((210 + v)/(210 – v))`
`7/75 t` `= ln((210 + v)/(210 – v))`
`e^(7/75 t)` `= (210 + v)/(210 – v)`
`e^(7/75 t) (210 – v)` `= 210 + v`
`210e^(7/75 t) – 210` `= ve^(7/75 t) + v`
`210(e^(7/75 t) – 1)` `= v(e^(7/75 t) + 1)`
`:. v` `= 210((e^(7/75 t) – 1)/(e^(7/75 t) + 1))`

 

iii.    `v · (dv)/(dx)` `= (44\ 100 – v^2)/4500`
  `(dx)/(dv)` `= (4500v)/(44\ 100 – v^2)`
  `int (dx)/(dv)\ dv` `= −4500/2 int (−2v)/(44\ 100 – v^2)\ dv`
  `x` `= −2550 log_e(44\ 100 – v^2) + c`

 
`text(When)\ \ x = 0, v = 0:`

`0` `= −2250 log_e(44\ 100) + c`
`c` `= 2250 log_e(44\ 100)`
`x` `= −2250 log_e(44\ 100 – v^2) + 2250 log_e44\ 100`
  `= 2250 log_e((44\ 100)/(44\ 100 – v^2))`

 
`text(When)\ \ t = 4:`

`v` `= 210((e^(28/75) – 1)/(e^(28/75) + 1))`
  `= 38.7509…`

 

`:. h` `= 2250 log_e((44\ 100)/(44\ 100 – 38.751^2))`
  `= 77.94…`
  `= 78\ text(metres)`

Complex Numbers, EXT2 N1 SM-Bank 4

  1. Express `z` in the form  `a + bi`, given  `z = sqrt(−3 - 4i)`  (2 marks)

  2. Hence, using the quadratic formula, solve
     
         `z^2 - 7z + 13 + i = 0`  (1 mark)

Show Answers Only
  1. `z_1 = −1 + 2i`
  2. `z = 3 + i\ \ text(or)\ \ 4 – i`
Show Worked Solution
i.    `z` `= sqrt(−3 – 4i)`
  `z^2` `= −3 – 4i`
`−3 – 4i` `= (x + iy)^2`
  `= x^2 – y^2 + 2xyi`
   
`x^2 – y^2` `= −3\ …\ \ (1)`
`2xy` `= −4`
`xy` `= −2\ …\ \ (2)`

 
`x=-1,\ \ y=2`

`x=1,\ \ y=-2`
 

`:. z_1` `= −1 + 2i`
`z_2` `= 1 – 2i`


ii.   `z^2 – 7z + 13 + i = 0`

 
`text(Using general formula:)`

`z` `= (−b ± sqrt(b^2 – 4ac))/(2a)`
  `= (7 ± sqrt(49 – 4 · 1(13 + i)))/2`
  `= (7 ± sqrt(−3 – 4i))/2`

 
`text(Using)\ \ z_1 = −1 + 2i,`

COMMENT: Since  `z_1 = – z_2`, the general formula only produces 2 distinct solutions.

`z = (7 + (−1 + 2i))/2 = 3+i`
 

`text(Using)\ \ z_2 = 1 – 2i,`

`z = (7 + (1 – 2i))/2 = 4 – i`
 

`:. z = 3 + i\ \ text(or)\ \ z=4 – i`

Complex Numbers, EXT2 N1 EQ-Bank 3

Find the values of `z`, in the form  `z = x + iy`, such that

`z = sqrt(-15 + 8i)`  (2 marks)

Show Answers Only

`z= 1 + 4i\ \ text(or)\ \ z= -1 -4i`

Show Worked Solution
`z` `= sqrt(-15 + 8i)`
`z^2` `= -15 + 8i`
`-15 + 8i` `= (x + iy)^2`
  `= x^2-y^2 + 2xyi`

 

`x^2-y^2` `= -15\ …\ (1)`
`2xy` `= 8`
`xy` `= 4\ …\ (2)`

 
`x=1,\ \ y=4`

`x=-1,\ \ y=-4`
 

`:. z_1` `= 1 + 4i`
`z_2` `= -1-4i`

Complex Numbers, EXT2 N1 EQ-Bank 2

Find the values of `z`, in the form  `z = a + ib`, such that

`z = sqrt(7 + 24i)`  (2 marks)

Show Answers Only

`z = 4 + 3i\ \ text(or)\ \ z=-4-3i`

Show Worked Solution
`z` `= sqrt(7 + 24i)`
`z^2` `= 7 + 24i`
`7 + 24i` `= (a + ib)^2`
  `= a^2 – b^2 + 2abi`

 

`a^2 – b^2` `= 7\ …\ (1)`
`2ab` `= 24`
`ab` `= 12\ …\ (2)`

 
`a=4,\ \ b=3`

`a=-4,\ \ b=-3`
 

`:. z_1` `= 4 + 3i`
`z_2` `= −4 – 3i`

Calculus, 2ADV C2 2006 HSC 2aii

Differentiate  `sin x/(x + 1)`  with respect to `x`.  (2 marks)

Show Answers Only

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

Show Worked Solution

`y = sin x / (x + 1)`

`text(Using)\ \ 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`

Algebra, STD2 A2 SM-Bank 11 MC

What is the equation for the line shown?
 

  1. `y = -3/2 x + 2`
  2. `y = 3/2 x + 2`
  3. `y = -2/3 x + 2`
  4. `y = 2/3 x + 2`
Show Answers Only

`=>B`

Show Worked Solution

`text(Graph cuts)\ y text(-axis at 2)`

`text(Gradient) = text(rise)/text(run) = 3/2`

`:. y = 3/2 x + 2`

`=>B`

Algebra, STD2 A1 SM-Bank 15

Fabio drove 300 km in  `4 1/2`  hours.

His average speed for the first 210 km was 70 km per hour.

How long did he take to travel the last 90 km?  (2 marks)

Show Answers Only

`1 1/2`

Show Worked Solution

`text(Time for 1st 210 km)`

`= 210/70`

`= 3\ text(hours)`

 

`:.\ text(Time for last 90 km)`

`= 4 1/2-3`

`= 1 1/2\ text(hours)`

Calculus, EXT1 C3 EQ-Bank 1

  1. Sketch the region bounded by the curve  `y = x^2`  and the lines  `y = 16`  and  `y = 9`.  (1 mark)

  2. Calculate the area of this region.  (3 marks)

Show Answers Only
  1.  
  2. `148/3 \ text(u²)`
Show Worked Solution
i.   

 

ii.   `text(Areas either side of)\ ytext(-axis are equal.)`

`y = x^2\ \ =>\ \ x = sqrty`

`:. A` `= 2 int_9^16 x\ dy`
  `= 2 int_9^16 sqrty\ dy`
  `= 2[2/3 y^(3/2)]_9^16`
  `= 4/3[(sqrt16)^3 – (sqrt9)^3]`
  `= 4/3[64 – 27]`
  `= 148/3 \ text(u²)`

Functions, 2ADV F2 EQ-Bank 2 MC

Which diagram best shows the graph

`y = 1 - 2(x + 1)^2`

A. B.
C. D.
Show Answers Only

`A`

Show Worked Solution

`text(Transforming)\ \ y = x^2 :`

`text(Translate 1 unit left)\ \ => \ y = (x + 1)^2`

`text(Dilate from)\ xtext(-axis by a factor of 2)\ => \ y = 2(x + 1)^2`

`text(Reflect in)\ xtext(-axis)\ \ => \ y= −2(x + 1)^2`

`text(Translate 1 unit up)\ \ => \ y = 1 – 2(x + 1)^2`

`:.\ text(Transformations describe graph)\ A.`

`=>\ A`

Functions, 2ADV F2 EQ-Bank 16

`y = -(x + 2)^4/3`  has been produced by three successive transformations: a translation, a dilation and then a reflection.

  1. Describe each transformation and state the equation of the graph after each transformation.  (2 marks)

  2. Sketch the graph.  (1 mark)

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

i.   `text(Transformation 1:)`

`text(Translate)\ \ y = x^4\ \ 2\ text(units to the left.)`

`y = x^4 \ => \ y = (x + 2)^4`
  

`text(Transformation 2:)`

`text(Dilate)\ \ y = (x + 2)^4\ \ text(by a factor of)\ 1/3\ text(from the)\ xtext(-axis)`

`y = (x + 2)^4 \ => \ y = ((x + 2)^4)/3`
 

`text(Transformation 3:)`

`text(Reflect)\ \ y = ((x + 2)^4)/3\ \ text(in the)\ xtext(-axis).`

`y = ((x + 2)^4)/3 \ => \ y = −(x + 2)^4/3`

 

ii.   

Calculus, EXT1 C1 EQ-Bank 1

A particle is moving along the `x`-axis with a velocity, `dotx`, in metres per second at  `t`  seconds, is given by the function

`dotx = sqrt(5t+4t^2-t^3)`

Find the acceleration of the particle when  `t=3`.  (2 marks)

Show Answers Only

`sqrt6/12\ \ text(ms)^(-2)`

Show Worked Solution
`dotx` `=(5t+4t^2-t^3)^(1/2)`  
`ddot x` `=d/(dt)(dotx)`  
  `=1/2(5t+4t^2-t^3)^(-1/2) xx (5+8t-3t^2)`  

 
`text(When)\ \ t=3,`

`ddotx` `=1/2 (15+36-27)^(-1/2) xx (5 + 24-27)`  
  `=2/(2sqrt24)`  
  `=sqrt6/12\ \ text(ms)^(-2)`  

Statistics, 2ADV S3 EQ-Bank 2

Let  `X` denote a normal random variable with mean 0 and standard deviation 1.

The random variable `X` has the probability density function

`f(x) = 1/sqrt(2pi) e^((−x^2)/2)`   where  `x ∈ (−∞, ∞)`

The diagram shows the graph of  `y = f(x)`.
 

 
 

  1. Complete the table of values for the given function, correct to four decimal places.  (1 mark)
     

     
  2. Use the trapezoidal rule and 5 function values in the table in part i. to estimate

     

         `int_(−2)^2 f(x)\ dx`   (2 marks)

  3. The weights of Rhodesian ridgebacks are normally distributed with a mean of 48 kilograms and a standard deviation of 6 kilograms.

     

    Using the result from part ii., calculate the probability of a randomly selected Rhodesian ridgeback weighing less than 36 kilograms.  (2 marks)

Show Answers Only
i.   

ii.  `0.9369`

iii.  `3.155text(%)`

Show Worked Solution
i.   

 

ii.   

 

`int_(−2)^2 f(x)` `~~ 1/2[0.0540 + 2(0.2420 + 0.3989 + 0.2420) + 0.0540]`
  `~~ 1/2 xx 1.8738`
  `~~ 0.9369`

 

iii.   `mu = 48, sigma = 6`

`ztext(-score (36))` `= (x – mu)/sigma= (32 – 48)/6=-2`

`text(Shaded area = 93.69%)`

`text(By symmetry,)`

`P(X < 36\ text(kgs))` `= P(z < −2)`
  `= (100 – 93.69)/2`
  `= 3.155text(%)`

Trigonometry, 2ADV T3 EQ-Bank 5

The function  `f(x) = sin x`  is transformed into the function  `g(x) = (sin(4x))/3`.

Describe in words how the amplitude and period have changed in this transformation.  (2 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`g(x) = 1/3 sin (4x)`

`=>\ text(The new amplitude is one third of the original amplitude.)`

`text(Period)\ = (2pi)/n \ => \ \ n=1/4`

`=>\ text(The new period is one quarter of the original period.)`

Statistics, 2ADV S2 EQ-Bank 3

The table below lists the average life span (in years) and average sleeping time (in hours/day) of 9 animal species.
 


 

  1. Using sleeping time as the independent variable, calculate the least squares regression line. (1 mark)

  2. A wallaby species sleeps for 4.5 hours, on average, each day.

     

    Use your equation from part i to predict its expected life span, to the nearest year.   (1 mark)

Show Answers Only
  1. `text(life span) = 42.89 – 2.85 xx text(sleeping time)`
  2. `30\ text(years)`
Show Worked Solution

i.    `text(By calculator:)`

COMMENT: Issues here? YouTube has short and excellent help videos – search your calculator model and topic – eg. “fx-82 regression line” .

`text(life span) = 42.89 – 2.85 xx text(sleeping time)`
 

ii.   `text(Predicted life span of wallaby)`

`= 42.89 – 2.85 xx 4.5` 

`= 30.06…`

`= 30\ text(years)`

Statistics, 2ADV S2 EQ-Bank 2

The table below lists the average body weight (in kilograms) and average brain weight (in grams) of nine animal species.
 


 

A least squares regression line is fitted to the data using body weight as the independent variable.

  1. Calculate the equation of the least squares regression line. (1 mark)

  2. If dingos have an average body weight of 22.3 kilograms, calculate the predicted average brain weight of a dingo using your answer to part i.   (1 mark)

Show Answers Only
  1. `text(brain weight) = 49.4 + 2.68 xx text(body weight)`
  2. `109\ text(grams)`
Show Worked Solution

i.   `text(By calculator:)`

COMMENT: Know this critical calculator skill!.

`text(brain weight) = 49.4 + 2.68 xx text(body weight)`

 

ii.   `text(Predicted brain weight of a dingo)`

`= 49.4 + 2.68 xx 22.3` 

`=109.164`

`= 109\ text(grams)`

Statistics, 2ADV S2 EQ-Bank 1

The arm spans (in cm) and heights (in cm) for a group of 13 boys have been measured. The results are displayed in the table below.
 

CORE, FUR2 2008 VCAA 4

The aim is to find a linear equation that allows arm span to be predicted from height.

  1. What will be the independent variable in the equation?  (1 mark)

  2. Assuming a linear association, determine the equation of the least squares regression line that enables arm span to be predicted from height. Write this equation in terms of the variables arm span and height. Give the coefficients correct to two decimal places.  (2 marks)

  3. Using the equation that you have determined in part b., interpret the slope of the least squares regression line in terms of the variables height and arm span.  (1 mark)

Show Answers Only
  1. `text(Height)`
  2. `text(Arm span)\ = 1.09 xx text(height) – 15.63`
  3. `text(On average, arm span increases by 1.09 cm)`
    `text(for each 1 cm increase in height.)`
Show Worked Solution

a.   `text(Height)`

COMMENT: Calculator skills for finding the least squares regression line were required in NESA sample exam – know this critical skill well!

 

b.   `text(By calculator,)`

`text(Arm span)\ = 1.09 xx text(height) – 15.63`

 

c.   `text(On average, arm span increases by 1.09 cm)`

`text(for each 1 cm increase in height.)`

Calculus, 2ADV C1 EQ-Bank 2

When differentiating  `f(x) = 3-2x-x^2`  from first principles, a student began the solution as shown below.

Complete the solution.  (2 marks)

   `f^{′}(x) = lim_(h->0) (f(x + h)-f(x))/h`

Show Answers Only

`f^{′}(x) = -2x-2`

Show Worked Solution
   `f^{′}(x)` `= lim_(h->0) (f(x + h)-f(x))/h`
    `= lim_(h->0) (3-2(x + h) – (x+h)^2-(3-2x-x^2))/h`
    `= lim_(h->0) (3-2x-2h-x^2-2hx-h^2 – 3 + 2x + x^2)/h`
    `= lim_(h->0) (-2h-2hx-h^2)/h`
    `= lim_(h->0) (h(-2x-2-h))/h`

 
`:.\ f^{′}(x) = -2x-2`

Calculus, 2ADV C2 EQ-Bank 2

Differentiate with respect to  `x`: 

`e^(tan(2x))`   (2 marks)

Show Answers Only

 `2 sec^2(2x)* e^(tan(2x))`

Show Worked Solution
`y` `=e^(tan(2x))`
`dy/dx` `= d/(dx)tan(2x) xx e^(tan(2x))`
  `= 2 sec^2(2x)* e^(tan(2x))`

NETWORKS, FUR2-NHT 2019 VCAA 3

The zoo’s management requests quotes for parts of the new building works.

Four businesses each submit quotes for four different tasks.

Each business will be offered only one task.

The quoted cost, in $100 000, of providing the work is shown in Table 1 below.
  


 

The zoo’s management wants to complete the new building works at minimum cost.

The Hungarian algorithm is used to determine the allocation of tasks to businesses.

The first step of the Hungarian algorithm involves row reduction; that is, subtracting the smallest element in each row of Table 1 from each of the elements in that row.

The result of the first step is shown in Table 2 below.
 


 

The second step of the Hungarian algorithm involves column reduction; that is, subtracting the smallest element in each column of Table 2 from each of the elements in that column.

The results of the second step of the Hungarian algorithm are shown in Table 3 below. The values of Task 1 are given as `A, B, C` and `D`.
 


 

  1. Write down the values of `A, B, C` and `D`.   (1 mark)

  2. The next step of the Hungarian algorithm involves covering all the zero elements with horizontal or vertical lines. The minimum number of lines required to cover the zeros is three.

     

    Draw these three lines on Table 3 above.   (1 mark)

  3. An allocation for minimum cost is not yet possible.

     

    When all steps of the Hungarian algorithm are complete, a bipartite graph can show the allocation for minimum cost.

     

    Complete the bipartite graph below to show this allocation for minimum cost.   (1 mark)

     

  1. Business 4 has changed its quote for the construction of the pathways. The new cost is $1 000 000. The overall minimum cost of the building works is now reduced by reallocating the tasks.

     

    How much is this reduction?   (1 mark)

Show Answers Only
  1. `A = 2, B = 1, C = 1, D = 0`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(See Worked Solutions)`
Show Worked Solution

a.   `A = 2, \ B = 1, \ C = 1, \ D = 0`

 

b.  

 

c.  `text(After next step:)`
 

`text(Allocation): `
 

 

d.  `text{Hungarian Algorithm table (complete):}`
 

`text(Allocation): B2 -> T3,\ B1 -> T4,\ B3 -> T2,\ B4 -> T1`
 

`text(C)text(ost) = 4 + 7 + 8 + 10 = 29`

`text(Previous cost) = 5 + 10 + 8 + 8 = 31`
 

`:.\ text(Reduction)` `= 2 xx 100\ 000`
  `= $200\ 000`

GEOMETRY, FUR1-NHT 2019 VCAA 1 MC

A piece of cardboard is shown in the diagram below.
 

     
 

The area of the cardboard, in square centimetres, is

  1.   4
  2.   5
  3. 21
  4. 25
  5. 29
Show Answers Only

`C`

Show Worked Solution
`text(Area)` `= (5 xx 5) – 4 xx (1 xx 1)`
  `= 21 \ text(cm)^2`

 
`=> \ C`

Networks, STD2 N3 2019 FUR2-N 2

The construction of the new reptile exhibit is a project involving nine activities, `A` to `I`.

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


 

  1. Which activities have more than one immediate predecessor?  (1 mark)

  2. Write down the critical path for this project.  (1 mark)

  3. What is the latest start time, in weeks, for activity `B`?  (1 mark)

Show Answers Only
  1. `D, G and I`
  2. `text(See Worked Solutions)`
  3. `2\ text(weeks)`
Show Worked Solution

a.   `D, G and I`

 

b.   `text(Scanning forwards and backwards:)`
 

​
 

`text(Critical Path:)\ ACDFGI`

 

c.   `text{LST (activity}\ B text{)}` `= 7 – 5`
    `= 2\ text(weeks)`

NETWORKS, FUR2-NHT 2019 VCAA 1

A zoo has an entrance, a cafe and nine animal exhibits: bears `(B)`, elephants `(E)`, giraffes `(G)`, lions `(L)`, monkeys `(M)`, penguins `(P)`, seals `(S)`, tigers `(T)` and zebras `(Z)`.

The edges on the graph below represent the paths between the entrance, the cafe and the animal exhibits. The numbers on each edge represent the length, in metres, along that path. Visitors to the zoo can use only these paths to travel around the zoo.
  

 
 

  1. What is the shortest distance, in metres, between the entrance and the seal exhibit `(S)`?   (1 mark)

  2. Freddy is a visitor to the zoo. He wishes to visit the cafe and each animal exhibit just once, starting and ending at the entrance.
  3. i. What is the mathematical term used to describe this route?   (1 mark)

  4. ii. Draw one possible route that Freddy may take on the graph below.   (1 mark)


 

A reptile exhibit `(R)` will be added to the zoo.

A new path of length 20 m will be built between the reptile exhibit `(R)` and the giraffe exhibit `(G)`.

A second new path, of length 35 m, will be built between the reptile exhibit `(R)` and the cafe.

  1. Complete the graph below with the new reptile exhibit and the two new paths added. Label the new vertex `R` and write the distances on the new edges.   (1 mark)

     


 

  1. The new paths reduce the minimum distance that visitors have to walk between the giraffe exhibit `(G)` and the cafe.

     

    By how many metres will these new paths reduce the minimum distance between the giraffe exhibit `(G)` and the cafe?   (1 mark)

Show Answers Only
  1.  `45\ text(metres)`
  2. i.  `text(Hamilton cycle.)`
    ii. `text(See Worked Solutions)`
  3.  `text(See Worked Solutions)`
  4.  `85\ text(metres)`
Show Worked Solution

a.   `45\ text(metres)`
 

b.i.   `text(Hamilton cycle.)`
 

b.ii.  

`text(Possible route:)`

`text(entrance)\ – LGTMCEBSZP\ –\ text(entrance)`

 

c.  

 

d.  `text{Minimum distance (before new exhibit)}`

`= GLTMC`

`= 15 + 35 + 40 + 50`

`= 140\ text(m)`
 

`:.\ text(Reduction in minimum distance)`

`= 140 – (20 + 35)`

`= 85\ text(m)`

MATRICES, FUR2-NHT 2019 VCAA 1

A total of six residents from two towns will be competing at the International Games.

Matrix `A`, shown below, contains the number of male `(M)` and the number of female `(F)` athletes competing from the towns of Gillen `(G)` and Haldaw `(H)`.

`{:(qquad qquad quad \ M quad F), (A = [(2, 2), (1, 1)]{:(G),(H):}):}` 

  1. How many of these athletes are residents of Haldaw?   (1 mark)

Each of the six athletes will compete in one event: table tennis, running or basketball.

Matrices `T` and `R`, shown below, contain the number of male and female athletes from each town who will compete in table tennis and running respectively.
 

            Table tennis                        Running             
 

`{:(qquad qquad quad \ M quad F), (T = [(0, 1), (1, 0)]{:(G),(H):}):}`

`{:(qquad qquad quad \ M quad F), (R = [(1, 1), (0, 0)]{:(G),(H):}):}`

 

  1. Matrix `B` contains the number of male and female athletes from each town who will compete in basketball.

     

    Complete matrix `B` below.   (1 mark)

`{:(qquad qquad qquad \ M qquad quad F), (B = [(\ text{___}, text{___}\ ), (\ text{___}, text{___}\)]{:(G),(H):}):}`

Matrix `C` contains the cost of one uniform, in dollars, for each of the three events: table tennis `(T)`, running `(R)` and basketball `(B)`.

`C = [(515), (550), (580)]{:(T), (R), (B):}`

    1. For which event will the total cost of uniforms for the athletes be $1030?   (1 mark)

    2. Write a matrix calculation, that includes matrix `C`, to show that the total cost of uniforms for the event named in part c.i. is contained in the matrix answer of [1030].   (1 mark)

  1. Matrix `V` and matrix `Q` are two new matrices where  `V = Q xx C`  and:
  • matrix `Q` is a  `4 xx 3`  matrix
  • element `v_11 =` total cost of uniforms for all female athletes from Gillen
  • element `v_21 =` total cost of uniforms for all female athletes from Haldaw
  • element `v_31 =` total cost of uniforms for all male athletes from Gillen
  • element `v_41 =` total cost of uniforms for all male athletes from Haldaw
     
  • `C = [(515), (550), (580)]{:(T), (R), (B):}`
  1. Complete matrix `Q` with the missing values.   (1 mark)

`Q = [(1, text{___}, text{___}\ ), (0, 0, 1), (0, 1, 1), (\ text{___}, text{___}, 0)]`

Show Answers Only
  1.  `2`
  2.  `B = [(1, 0), (0, 1)]`
  3. i.  `text(Table tennis)`
    ii. `[2\ \ \ 0\ \ \ 0] xx [(515), (550), (580)] = [1030]`
  4.  `Q = [(1, \ 1, \ 0),(0, \ 0, \ 1),(0, \ 1, \ 1),(1, \ 0, \ 0)]`
Show Worked Solution

a.  `2`
 

b.  `B = [(1, 0), (0, 1)]`
 

c.i.  `text(Table tennis)`
 

c.ii.  `[2\ \ \ 0\ \ \ 0] xx [(515), (550), (580)] = [1030]`
 

d.  `Q = [(1, \ 1, \ 0),(0, \ 0, \ 1),(0, \ 1, \ 1),(1, \ 0, \ 0)]`

Functions, EXT1 F1 2019 MET1-N 5

Let  `h(x) = ( 7)/(x + 2) - 3`  for  `x>=0`.

  1.  State the range of  `h`.  (1 mark)

  2.  Find the rule for  `h^-1`.  (2 marks)

Show Answers Only
  1. `(-3, (1)/(2))`
  2. `(7)/(x + 3) – 2`
Show Worked Solution

a.    `y_text(max)\  text(occurs when) \ x = 0`

`y_text(max) = (7)/(2) – 3 = (1)/(2)`

`text(As) \ \ x → ∞ , \ (7)/(x + 2) \ → \ 0^+`

`:. \ text(Range) \ \ h(x) = (-3, (1)/(2))`
 

b.    `y = (7)/(x + 2)`
 

`text(Inverse: swap) \ x ↔ y`

`x` `= (7)/(y + 2) – 3`
`x + 3` `= (7)/(y + 2)`
`y + 2` `= (7)/(x + 3)`
`y` `= (7)/(x + 3) – 2`

  
`:. \ h^-1 = (7)/(x + 3) – 2`

Functions, 2ADV F1 2019 MET1-N 2

Let  `f(x) = -x^2 + x + 4`  and  `g(x) = x^2 - 2`.

  1. Find  `g(f(3))`.  (2 marks)

  2. Express  `f(g(x))`  in the form  `ax^4 + bx^2 + c`, where  `a`, `b`  and  `c`  are non-zero integers.  (2 marks)

Show Answers Only
  1. `2`
  2. `-x^4 + 5x^2 – 2`
Show Worked Solution
a.    `f(3)` `= -3^2 + 3 + 4`
  `= -2`

 

`g(f(3))` `= g(-2)`
  `= (-2)^2 – 2`
  `= 2`

 

b.    `f(g(x))` `= -(x^2 – 2)^2 + (x^2 – 2) + 4`
  `= -(x^4 – 4x^2 + 4) + x^2 + 2`
  `= -x^4 + 5x^2 – 2`

Calculus, 2ADV C2 SM-Bank 7

Let  `f(x) = (e^x)/((x^2 - 3))`.

Find  `f′(x)`.  (2 marks)

Show Answers Only

`{e^x(x^2 – 2x – 3)}/{(x^2 – 3)^2}`

Show Worked Solution

`text(Let) \ \ u = e^x \ \ => \ \ u′ = e^x`

 `v = (x^2 – 3) \ \ => \ \ v′ = 2x`

`f′(x)` `= {e^x(x^2 – 3) – 2x e^x}/{(x^2 – 3)^2}`
  `= {e^x(x^2 – 2x – 3)}/{(x^2 – 3)^2}`

Functions, MET1-NHT 2018 VCAA 5

Let  `h: R^+ ∪ {0} → R, \ h(x) = ( 7)/(x + 2)-3`.

  1.  State the range of  `h`.   (1 mark)

  2.  Find the rule for  `h^-1`.   (2 marks)

Show Answers Only
  1. `(-3, (1)/(2))`
  2. `(7)/(x + 3)-2`
Show Worked Solution

a.    `y_text(max)\  text(occurs when) \ x = 0`

`y_text(max) = (7)/(2)-3 = (1)/(2)`

`text(As) \ \ x → ∞ , \ (7)/(x + 2) \ → \ 0^+`

`:. \ text(Range) \ \ h(x) = (-3, (1)/(2))`
 

b.    `y = (7)/(x + 2)`

`text(Inverse: swap) \ x ↔ y`

`x` `= (7)/(y + 2)-3`
`x + 3` `= (7)/(y + 2)`
`y + 2` `= (7)/(x + 3)`
`y` `= (7)/(x + 3)-2`

  
`:. \ h^-1 = (7)/(x + 3)-2`

Functions, MET1-NHT 2018 VCAA 2

Let  `f(x) = -x^2 + x + 4`  and  `g(x) = x^2-2`.

  1. Find `g(f(3))`.   (2 marks)

Show Answers Only
  1. `2`
  2. `-x^4 + 5x^2-2`
Show Worked Solution
a.    `f(3)` `= -3^2 + 3 + 4`
  `= -2`

 

`g(f(3))` `= g(-2)`
  `= (-2)^2-2`
  `= 2`

 

b.    `f(g(x))` `= -(x^2-2)^2 + (x^2-2) + 4`
  `= -(x^4-4x^2 + 4) + x^2 + 2`
  `= -x^4 + 5x^2-2`