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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`

Calculus, 2ADV C3 2020 HSC 8 MC

The graph of  `y = f(x)`  is shown.
 

Which of the following inequalities is correct?

  1. `f^{″}(1) < 0 < f^{′}(1) < f(1)`
  2. `f^{″}(1) < 0 < f(1) < f^{′}(1)`
  3. `0 < f^{″}(1) < f^{′}(1) < f(1)`
  4. `0 < f^{″}(1) < f(1) < f^{′}(1)`
Show Answers Only

`A`

Show Worked Solution

`y = f(x)\ text(is concave down at)\ \ x = 1`

Mean mark 52%.

`f^{″}(1) < 0`

`=>\ text(Eliminate C and D.)`
 
 

`text(T)text(angent)\ => y=mx+b`

COMMENT: Note the concavity causes any tangent to cut the positive y-axis. No y-axis scale is therefore required.

`text(At)\ \ x = 1, m_text(tang) = f^{′}(1)`

`text(At intersection:)`

`mx+b` `=f(1)`  
`f^{′}(1) xx 1 + b` `=f(1)`  
`:. f^{′}(1)` `<f(1),\ \ \  (b>0)`  

 
`:. f^{″}(1) < 0 < f^{′}(1) < f(1)`

`=>A`

Calculus, 2ADV C4 2020 HSC 7 MC

The diagram show the graph  `y = f(x)`, which is made up of line segments and a semicircle.
 

What is the value of  `int_0^12 f(x)\ dx`?

  1.  `24 + 2pi`
  2.  `24 + 4pi`
  3.  `30 + 2pi`
  4.  `30 + 4pi`
Show Answers Only

`A`

Show Worked Solution

`text(Consider the interval between)\ \ x=8 and x=12:`

`text(Area above and below the)\ xtext(-axis are equal.)`

`int_8^12 f(x) = 0`

♦ Mean mark 48%.
 

`:. int_0^12 f(x)\ dx` `= int_0^8 f(x)\ dx`
  `=\ text(Area of rectangle + area of semi-circle)`
  `= 8 xx 3 + 1/2 xx pi xx 2^2`
  `= 24 + 2pi`

 
`=>A`

Financial Maths, STD2 F4 2020 HSC 11 MC

An asset is depreciated using the declining-balance method with a rate of depreciation of 8% per half year. The asset was bought for $10 000.

What is the salvage value of the asset after 5 years?

  1.  $1749.01
  2.  $4182.12
  3.  $4343.88
  4.  $6590.82
Show Answers Only

`C`

Show Worked Solution

♦ Mean mark 43%.
COMMENT: 8% depreciation is applicable every 6 months here (n=10). Read carefully!

`V_0 = 10\ 000 \ , \ r = 0.08 \ , \ n = 10`

`S` `= V_0 (1 – r)^n`
  `= 10\ 000 (1 – 0.08)^10`
  `= 10\ 000 (0.92)^10`
  `= $ 4343.88`

 
`=> \ C`

Algebra, STD2 A2 2020 HSC 10 MC

A plumber charges a call-out fee of $90 as well as $2 per minute while working.

Suppose the plumber works for `t` hours.

Which equation expresses the amount the plumber charges ($`C`) as a function of time (`t` hours)?

  1.  `C = 2 + 90t`
  2.  `C = 90 + 2t`
  3.  `C = 120 + 90t`
  4.  `C = 90 + 120t`
Show Answers Only

`D`

Show Worked Solution

♦ Mean mark 42%.

`text(Hourly rate)\ = 60 xx 2=$120`

`:. C = 90 + 120t`

`=>D`

Networks, STD2 N2 2020 HSC 9 MC

Team `A` and Team `B` have entered a chess competition.

Team `A` and `B` have three members each. Each member of Team `A` must play each member of Team `B` once.

Which of the following network diagrams could represent the chess games to be played?
 

 

 

  
Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 38%.

`text(Vertices = players)`

`text(Edges = games between 2 players)`

`text(S)text(ince each player plays once against the three players)`

`text(in the other team, each vertex must be degree 3.)`

`=> \ B`

Statistics, STD2 S1 2020 HSC 7 MC

Which histogram best represents a dataset that is positively skewed?

 
 
Show Answers Only

`A`

Show Worked Solution

♦♦ Mean mark 32%.

`text(Positive skew occurs when the tail on the)`

`text{histogram is longer on the right-hand}`

`text{(positive) side.}`

`=> \ A`

Measurement, STD2 M1 2020 HSC 5 MC

A plant stem is measured to be 16.0 cm, correct to one decimal place.

What is the percentage error in this measurement?

  1.  0.3125%
  2.  0.625%
  3.  3.125%
  4.  6.25%
Show Answers Only

`A`

Show Worked Solution

♦ Mean mark 41%.

`text{Absolute error} = 1/2 xx \text{precision} = 1/2 xx 0.1 = 0.05\ text{cm}`

`%  text(error)` `= frac(0.05)(16.0) xx 100`
  `= 0.3125%`

 
`=> \ A`

Measurement, STD2 M1 2020 HSC 2 MC

What is 0.002073 expressed in standard form with two significant figures?

  1. `2.07 xx 10^(-2)`
  2. `2.1 xx 10^(-2)`
  3. `2.07 xx 10^(-3)`
  4. `2.1 xx 10^(-3)`
Show Answers Only

`D`

Show Worked Solution

♦♦ Mean mark 35%.
COMMENT: A notable knowledge gap flagged by this result.
`0.002073` `= 2.073 xx 10^(-3)`  
  `=2.1 xx 10^(-3)\ \ \ text{(to 2 sig fig)}`  

 
`=> D`

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 2

A particle with mass `m` moves horizontally against a resistance force `F`, equal to  `mv(1 + v^2)`  where `v` is the particle's velocity.

Initially, the particle is travelling in a positive direction from the origin at velocity `T`.

  1. Show that the particle's displacement from the origin, `x`, can be expressed as
     
         `x = tan^(-1)((T - v)/(1 + Tv))`  (2 marks)

     

  2. Show that the time, `t`, when the particle is travelling at velocity `v`, is given by
     
         `t = 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`  (4 marks)

     

  3. Express `v^2` as a function of `t`, and hence find the limiting values of `x` and `v`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)“
  3. `v -> 0, \ x -> tan^(−1)T`
Show Worked Solution

i.   `text(Newton’s 2nd law:)`

`F = mddotx = −mv(1 + v^2)`

`v · (dv)/(dx)` `= −v(1 + v^2)`
`(dv)/(dx)` `= −(1 + v^2)`
`(dx)/(dv)` `= −1/(1 + v^2)`
`x` `= −int 1/(1 + v^2)\ dv`
  `= −tan^(−1) v + C`

 

`text(When)\ \ x = 0, v = T:`

`C = tan^(−1)T`

`x = tan^(−1)T – tan^(−1)v`
 

`text(Let)\ \ x = A – B`

`A = tan^(−1) T \ => \ T = tan A`

`B = tan^(−1)v \ => \ v = tan B`

`tan x` `= tan(A – B)`
  `= (tan A – tan B)/(1 + tan A tan B)`
  `= (T – v)/(1 + Tv)`
`:. x` `= tan^(−1)((T – v)/(1 + Tv))`

 

ii.   `text(Using)\ \ ddotx = (dv)/(dt):`

`(dv)/(dt) = −1/(v(1 + v^2))`

`t = −int 1/(v(1 + v^2)) dv`
 

`text(Using Partial Fractions):`

`1/(v(1 + v^2)) = A/v + (Bv + C)/(1 + v^2)`

`A(1 + v^2) + (Bv + C)v = 1`

`A = 1`

`(A + B)v^2` `= 0`   `=> `    `B` `= −1`
`Cv` `= 0`   `=> `    `C` `= 0`

 

`t` `= −int 1/v\ dv + int v/(1 + v^2)\ dv`
  `= −log_e v + 1/2 int(2v)/(1 + v^2)\ dv`
  `= −log_e v + 1/2 log_e (1 + v^2) + C`
  `= −1/2 log_e v^2 + 1/2 log_e (1 + v^2) + C`
  `= 1/2 log_e ((1 + v^2)/(v^2)) + C`

 

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

`C = −1/2 log_e ((1 + T^2)/(T^2))`
 

`:. t` `= 1/2 log_e ((1 + v^2)/(v^2)) – 1/2 log_e((1 + T^2)/(T^2))`
  `= 1/2 log_e (((1 + v^2)/(v^2))/((1 + T^2)/(T^2)))`
  `= 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`

 

iii. `t` `= 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`
  `e^(2t)` `= (T^2(1 + v^2))/(v^2(1 + T^2))`
  `1 + v^2` `= (e^(2t)v^2(1 + T^2))/(T^2)`
  `1` `= v^2((e^(2t)(1 + T^2))/(T^2) – 1)`
  `1` `= v^2((e^(2t)(1 + T^2) – T^2)/(T^2))`
  `:. v^2` `= (T^2)/(e^(2t)(1 + T^2) – T^2)`

 
`text(As)\ \ t -> ∞:`

`v^2 -> 0, \ v -> 0`

`x = tan^(−1)T – tan^(−1)v`

`x -> tan^(−1)T`

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)`

Functions, 2ADV F2 EQ-Bank 13

The curve  `y = kx^2 + c`  is subject to the following transformations

    • Translated 2 units in the positive `x`-direction
    • Dilated in the positive `y`-direction by a factor of 4
    • Reflected in the `y`-axis

The final equation of the curve is  `y = 8x^2 + 32x - 8`.

  1.  Find the equation of the graph after the dilation.  (1 mark)

  2.  Find the values of  `k`  and  `c`.  (2 marks)

Show Answers Only
  1. `y = 4k(x – 2)^2 + 4c`
  2. `k = 2, c = −10`
Show Worked Solution

i.    `y = kx^2 + c`

`text(Translate 2 units in positive)\ xtext(-direction.)`

`y = kx^2 + c \ => \ y = k(x – 2)^2 + c`

`text(Dilate in the positive)\ ytext(-direction by a factor of 4.)`

`y = k(x – 2)^2 + c \ => \ y = 4k(x – 2)^2 + 4c`

 

ii.    `y` `= 4k(x^2 – 4x + 4) + 4c`
    `= 4kx^2 – 16kx + 16k + 4c`

 

 
`text(Reflect in the)\ ytext(-axis.)`

COMMENT: Using “swap” terminology for reflections in the y-axis is simpler and more intelligible for students in our view.

`=>\ text(Swap:)\ \ x →\ – x`

`y` `= 4k(−x)^2 – 16k(−x) + 16k + 4c`
  `= 4kx^2 + 16kx + 16k + 4c`

 

 
`text(Equating co-efficients:)`

`4k` `=8`  
`:. k` `=2`  

 

`16k + 4c` `= −8`
`4c` `= −40`
`:. c` `=-10`

Trigonometry, 2ADV T3 EQ-Bank 3

By drawing graphs on the number plane, show how many solutions exist for the equation  `cosx = |(x - pi)/4|`  in the domain  `(−∞, ∞)`  (3 marks)

Show Answers Only

`text(4 solutions)`

Show Worked Solution

`text(Sketch:)`

`y` `= cos x`
`y` `= |(x – pi)/4|`

 
`text(Translate)\ pi\ text(units to the right: )`

`y=|x| \ => \ y=|x-pi|`

`text(Multiply by)\ 1/4 :`

`y=|x-pi| \ => \ y= 1/4 |x-pi| = |(x-pi)/4|`

`:.\ text(There are 4 solutions.)`

Calculus, 2ADV C4 EQ-Bank 5

The velocity of a particle moving along the `x`-axis at `v` metres per second at `t` seconds, is shown in the graph below.
 


 

Initially, the displacement `x` is equal to 12 metres.

  1. Write an equation that describes the displacement, `x`, at time `t` seconds.  (2 marks)

  2. Draw a graph that shows the displacement of the particle, `x`  metres from the origin, at a time `t` seconds between  `t= 0`  and  `t = 5`. Label the coordinates of the endpoints of your graph.  (2 marks)

Show Answers Only
  1. `x = 3t – (3)/(10) t ^2 + 12`
  2.  

Show Worked Solution
i.    `m_v` `= -(3)/(5)`
  `v` `= 3 – (3)/(5) t`

 

`x` `= int v \ dt`
  `= int 3 – (3)/(5) t \ dt`
  `= 3t – (3)/(10) t^2 + c`

 
`text(When) \ \ t = 0, x = 12  \ => \ c = 12`

`:. \ x = 3t – (3)/(10) t ^2 + 12`

 

ii.  `text(When) \ \ t = 5:`

`x = 15 – (3)/(10) xx 25 + 12 = 19.5`
 

Statistics, 2ADV S3 EQ-Bank 1

A probability density function can be used to model the lifespan of a termite, `X`, in weeks, is given by
 

`f(x) = {(k(36 - x^2)),(0):}\ \ \ {:(3 <= x <= 6),(text(otherwise)):}`
 

  1. Show that the value of  `k`  is  `1/45`.  (2 marks)

  2. Find the cumulative distribution function.  (2 marks)

  3. Find the probability that a termite's lifespan is greater than 5 weeks.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `F(x) = {(0),(1/135 (108t – t^3  – 297)),(1):}\ \ \ {:(x < 3),(3 <= x <= 6),(x > 6):}`
  3. `17/135`
Show Worked Solution
i.    `k int_3^6 36 – x^2\ dx` `= 1`
  `k[36x – (x^3)/3]_3^6` `= 1`
  `k[(216 – 72)-(108 – 9)]` `= 1`
  `45k` `= 1`
  `k` `= 1/45`

 

ii.    `F(t)` `= int_(-∞)^t f(x)\ dx`
    `= int_3^t f(x)\ dx`
    `= 1/45 int_3^t 36 – x^2\ dx`
    `= 1/45 [36x – (x^3)/3]_3^t`
    `= 1/135[108x – x^3]_3^t`
    `= 1/35[(108t – t^3) – (324 – 27)]`
    `= 1/135(108t – t^3 – 297)`

 
`:. F(x) = {(0),(1/135 (108t – t^3  – 297)),(1):}\ \ \ {:(x < 3),(3 <= x <= 6),(x > 6):}`

 

iii.    `P(X > 5)` `= 1 – F(5)`
    `= 1 – 1/135(108 xx 5 – 5^3 – 297)`
    `= 1 – 118/135`
    `= 17/135`

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`

CORE, FUR1-NHT 2019 VCAA 22-23 MC

Armand borrowed $12 000 to pay for a holiday.

He will be charged interest at the rate of 6.12% per annum, compounding monthly.

This loan will be repaid with monthly repayments of $500.
 

Part 1

After four months, the total interest that Armand will have paid is closest to

  1.   $231
  2.   $245
  3.   $255
  4.   $734
  5. $1796

 
Part 2

After eight repayments, Armand decided to increase the value of his monthly repayments.

He will make a number of monthly repayments of $850 and then one final repayment that will have a smaller value.

This final repayment has a value closest to

  1. $168
  2. $169
  3. $180
  4. $586
  5. $681
Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ B`

Show Worked Solution

`text(Part 1)`

`text(By TVM Solver:)`

`N` `= 4`
`Itext(%)` `= 6.12`
`PV` `= 12\ 000`
`PMT` `= −500`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

 
`=>\ text(FV) = 10\ 231.33`

`:.\ text(Total interest)` `=\ text(total payments − reduction in principle)`
  `= (4 xx 500) – (12\ 000 – 10\ 231.33)`
  `= 231.33`

`=>\ A`

 

`text(Part 2)`

`text(Loan after 8 repayments (by TVM Solver)):`

`N` `= 8`
`I(text(%))` `= 6.12`
`PV` `= 12\ 000`
`PMT` `= −500`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

 

 
`=> FV = 8426.30`

 

`text(Number of repayments required at $850)`

`N` `= ?`
`I(text(%))` `= 6.12`
`PV` `= 8426.30`
`PMT` `= −500`
`FV` `= 0`
`text(P/Y)` `= 12`
`text(C/Y)` `= 12`

 
`=> N = 10.2`

 
`text(Balance after 10 repayments = $168.29)`

`:.\ text(Final repayment)` `= 168.29 xx (1 + 6.12/12)`
  `= 169.15`

`=>\ B`

GEOMETRY, FUR1-NHT 2019 VCAA 7 MC

Two similar right-angled triangles are shown in the diagram below.
 

     

The total perimeter of the two triangles, in metres, is

  1. 21
  2. 30
  3. 36
  4. 42
  5. 45
Show Answers Only

`C`

Show Worked Solution

     

`text(Ratio of similar sides)\ = 3 : 6 = 1 : 2`

`text(Base of 12 m in ratio)\ 1 : 2 = 4 : 8`

 
 

`text(Length of hypotenuses:)`

`h_1 = sqrt(3^2 + 4^2) = 5`

`h_2 = sqrt(6^2 + 8^2) = 10`
 

`:. \ text(Perimeter) ` `= 3 + 4 + 5 + 6 + 8 + 10`
  `= 36 \ text(m)`

 
`=> \ C`

GEOMETRY, FUR1-NHT 2019 VCAA 4 MC

A pizza in the shape of a circle has been cut into 12 equal slices.

The area of the top surfaces of one slice is 9.42 cm2, as shown shaded in the diagram below.
  

     
 

The perimeter of the top surfaces of one slice of pizza, in centimetres, is closest to

  1.   9.14
  2. 15.14
  3. 18.00
  4. 21.99
  5. 40.84
Show Answers Only

`B`

Show Worked Solution
`text(Total Area)` `= 12 xx 9.42`
  `= 113.04 \ text(cm)^2`

 

`pi r^2` `= 113.04`
`r` `= sqrt({113.04}/{pi})`
  `= 5.998 \ …`
  `≈ 6`

 
`:. \ text(Perimeter of 1 slice)`

`= 2 xx 6 + (1)/(12) (2 xx pi xx 6)`

`= 15.14 \ text(cm)`
 

`=> \ B`

MATRICES, FUR2-NHT 2019 VCAA 4

After 5.00 pm, tourists will start to arrive in Gillen and they will stay overnight.

As a result, the number of people in Gillen will increase and the television viewing habits of the tourists will also be monitored.

Assume that 50 tourists arrive every hour.

It is expected that 80% of arriving tourists will watch only `C_2` during the hour that they arrive.

The remaining 20% of arriving tourists will not watch television during the hour that they arrive.

Let `W_m` be the state matrix that shows the number of people in each category `m` hours after 5.00 pm on this day.

The recurrence relation that models the change in the television viewing habits of this increasing number of people in Gillen `m` hours after 5.00 pm on this day is shown below.

`W_(m + 1) = TW_m + V` 

where

`{:(quad qquad qquad qquadqquadqquadquadtext(this hour)),(qquadqquadqquad quad \ C_1 qquad quad C_2 qquad \ C_3 quad \ NoTV),(T = [(quad 0.50, 0.05, 0.10, 0.20 quad),(quad 0.10, 0.60, 0.20, 0.20 quad),(quad 0.25, 0.10, 0.50, 0.10 quad),(quad 0.15, 0.25, 0.20, 0.50 quad)]{:(C_1),(C_2),(C_3),(NoTV):}\ text(next hour,) qquad and qquad  W_0 = [(400), (600), (300),(700)]):}`
 

  1. Write down matrix `V`.   (1 mark)

  2. How many people in Gillen are expected to watch `C_2` at 7.00 pm on this day?   (2 marks)

Show Answers Only
  1.  `V=[(0),(40),(0),(10)]`
  2.  `666`
Show Worked Solution

a.   `V=[(0),(40),(0),(10)]`

b.    `W_1` `=TW_0+V`
    `=[(quad 0.50, 0.05, 0.10, 0.20 quad),(quad 0.10, 0.60, 0.20, 0.20 quad),(quad 0.25, 0.10, 0.50, 0.10 quad),(quad 0.15, 0.25, 0.20, 0.50 quad)] [(400),(600),(300),(700)]+[(0),(40),(0),(10)]=[(400),(640),(380),(630)]`
     
  `W_2` `=TW_1+V=[(396),(666),(417),(621)]`

 
`:.\ text(666 people are expected to watch)\ C_2\ text(at 7 pm.)`

Calculus, MET1-NHT 2018 VCAA 9

Let diagram below shows a trapezium with vertices at  `(0, 0), (0, 2), (3, 2)`  and  `(b, 0)`, where  `b`  is a real number and  `0 < b < 2`

On the same axes as the trapezium, part of the graph of a cubic polynomial function is drawn. It has the rule  `y = ax(x-b)^2`, where  `a`  is a non-zero real number and  `0 ≤ x ≤ b`.

  1. At the local maximum of the graph, `y = b`.
  2. Find `a` in terms of `b`.   (3 marks)

The area between the graph of the function and the `x`-axis is removed from the trapezium, as shown in the diagram.
 

  1. Show that the expression for the area of the shaded region is  `b + 3-(9b^2)/(16)`  square units.   (3 marks)

  2. Find the value of  `b`  for which the area of the shaded region is a maximum and find this maximum area.   (2 marks)

Show Answers Only
  1. `(27)/(4b^2)`
  2. `text(Proof (See Worked Solution))`
  3. `(31)/(9)`
Show Worked Solution
a.     `y` `= ax(x-b)^2`
`y^{′}` `= a(x-b)^2 + 2ax(x-b)`
  `= a(x-b)(x-b + 2x)`
  `= a(x-b)(3x-b)`

 
`text(S) text(olve) \ \ y^{′} = 0 \ \ text(for) \ x:`

`x = b \ \ text(or) \ \ (b)/(3)`
 
`y_text(max) = b \ \ text{(given) and occurs at} \ \ x = (b)/(3)`
 

`b` `= a((b)/(3))((b)/(3)-b)^2`
`b` `= a((b)/(3))((4b^2)/(9))`
`b` `= a((4b^3)/(27))`

 
`:. \ a = (27)/(4b^2)`
 

b.   `text(Let)\ \ A_1 = \ text(area under the graph)`

`A_1` `= int_0^b ax (x-b)^2 dx`
  `= a int_0^b x(x^2-2bx + b^2)\ dx`
  `= (27)/(4b^2) int_0^b x^3-2bx^2 + b^2 x \ dx`
  `= (27)/(4b^2) [(x^4)/(4)-(2bx^3)/(3) + (b^2 x^2)/(2)]_0^b`
  `= (27)/(4b^2) ((b^4)/(4)-(2b^4)/(3) + (b^4)/(2))`
  `= (27)/(4b^2) ((b^4)/(12))`
  `= (9b^2)/(16)`

 
`text(Let) \ \ A_2 = \ text(area of triangle)`

`A_2 = (1)/(2) xx 2 xx (3-b) = 3-b`
 

`:. \ text(Shaded region)` `= 6-(9b^2)/(16)-(3-b)`
  `= b + 3-(9b^2)/(16)`

 

c.   `A = b + 3-(9b^2)/(16)`

`(dA)/(db) = 1-(9b)/(8)`

`(dA^2)/(db^2) =-(9)/(8) <0`
 

`=>\ text{SP (max) when} \ \ (dA)/(db) = 0`

`(9b)/(8) = 1 \ => \ \ b = (8)/(9)`

`:. A_(max)` `= (8)/(9) + 3-{9((8)/(9))^2}/{16}`
  `= (35)/(9)-(64)/(9 xx 16)`
  `= (31)/(9)`

Statistics, MET1-NHT 2018 VCAA 8

Let  `overset^p`  be the random variable that represents the sample proportions of customers who bring their own shopping bags to a large shopping centre.

From a sample consisting of all customers on a particular day, an approximate 95% confidence interval for the proportion  `p`  of customers who bring their own shopping bags to this large shopping centre was determined to be  `((4853)/(50\ 000) , (5147)/(50\ 000))`.

  1. Find the value of  `hatp`  that was used to obtain this approximate 95% confidence interval.   (1 mark)

  2. Use the fact that  `1.96 = (49)/(25)`  to find the size of the sample from which this approximate 95% confidence interval was obtained.   (2 marks)

Show Answers Only

  1. `(1)/(10)`
  2. `40\ 000`

Show Worked Solution

a.    `overset^p – z sqrt((overset^p(1 – overset^p))/(n)) = (4853)/(50\ 000) \ \ , \ \ overset^p + z sqrt((overset^p(1 – overset^p))/(n)) = (5147)/(50\ 000)`

`2 overset^p` `= (4853)/(50\ 000) + (5147)/(50\ 000)`  
`:. \ overset^p` `=1/10`  

 

b.    `(1)/(10) + (49)/(25) sqrt(({1)/{10}(1 – {1}/{10}))/(n)` `= (5147)/(50000)`
`(49)/(25) sqrt((9)/(100n))` `= (147)/(50000)`
`(49)/(25) * (3)/(10) * (1)/(sqrtn)` `= (147)/(25 xx 2000)`
`(147)/(sqrtn)` `= (147)/(200)`
`sqrtn` `= 200`
`:. \ n` `= 40\ 000`

Statistics, MET1-NHT 2018 VCAA 6

The discrete random variable `X` has the probability mass function
 

`text(Pr)(X = x) = {(kx), (k), (0):} qquad {:(x∈{1, 4, 6}), (x = 3), (text(otherwise)):}`
 

  1. Show that  `k = (1)/(12)`.  (2 marks)

  2. Find  `text(E)(X)`.  (1 mark)

  3. Evaluate  `text(Pr)(X ≥ 3 | X ≥ 2)`.  (1 mark)

Show Answers Only

  1. `text(Proof (See Worked Solution))`
  2. `(14)/(3)`
  3. `1`

Show Worked Solution

a.     `qquad X` `qquad 1 qquad` `qquad 3 qquad` `qquad 4 qquad` `qquad 6 qquad`
  `qquad Pr(X = x) qquad` `k` `k` `4k` `6k`

 

`k + k+4k + 6k` `= 1`
`12k` `= 1`
`k` `= (1)/(12)`

 

b.    `E(X)` `= ∑ x text(Pr)(X = x)`
  `= k + 3k + 16k + 36k`
  `= 56k`
  `= (56)/(12)`
  `= (14)/(3)`

 

c.    `text(Pr) (X ≥ 3 | X ≥ 2)` `= (text(Pr) (X ≥ 3 ∩ X ≥ 2))/(text(Pr) (X ≥ 2))`
  `= (text(Pr) (X ≥ 3))/(text(Pr) (X ≥ 2))`
  `= (k + 4k + 6k)/(k + 4k + 6k)`
  `= 1`

CORE, FUR1-NHT 2019 VCAA 21 MC

Yazhen has a reducing balance loan.

Six lines of the amortisation table for Yazhen’s loan are shown below.

The interest rate for Yazhen’s loan increased after one of these six repayments had been made.

The first repayment made at the higher interest rate was repayment number

  1. 15
  2. 16
  3. 17
  4. 18
  5. 19
Show Answers Only

`C`

Show Worked Solution

`text(Interest rate) = text(Interest)/text{Balance (higher row)}`
 

`text(Payment 16:)\ \ \ 343/(34\ 299.50) = 0.0100`

`text(Payment 17:)\ \ \ 403.71/(33\ 642.50) = 0.0120`

`text(Payment 18:)\ \ \ 396.55/(33\ 046.21) = 0.0120`
 

`:.\ text(Payment 17 is the 1st at the higher rate.)`

`=>\ C`

Statistics, EXT1 S1 EQ-Bank 14

It is known that 65% of adults over the age of 60 have been tested for bowel cancer.

A random sample of 140 adults aged over 60 years is surveyed.

Using a normal approximation to the binomial distribution and the probability table attached, calculate the probability that at least 85 of the adults chosen have been tested for bowel cancer.   (3 marks)

Show Answers Only

`0.8554`

Show Worked Solution

`text(Let)\ \ X =\ text(number tested for cancer)`

`X\ ~\ text(Bin)(140, 0.65)`

`text(Find)\ \ P(X>=85):`

`E(hat p)` `=p=0.65`  
`text(Var)(hat p)` ` = (0.65(0.35))/140 = 0.001625`  
`sigma(hat p)` `=0.04031…`  

 
`hatp\ ~\ N(mu,sigma)\ ~\ N(0.65, 0.04031…)`

`text(If)\ \ X=85\ \ => \ hatp=85/140=0.60714…`

`ztext{-score}\ (X=85)` `=(0.60714-0.65)/0.04031`  
  `~~ -1.063`  

 
`text{Using the probability table (attached):}`

`P(z<–1.06) = 0.1446`

`:. P(X>=85)` `=1-0.1446`  
  `=0.8554`  

CORE, FUR2-NHT 2019 VCAA 8

A record producer gave the band $50 000 to write and record an album of songs.

This $50 000 was invested in an annuity that provides a monthly payment to the band.

The annuity pays interest at the rate of 3.12% per annum, compounding monthly.

After six months of writing and recording, the band has $32 667.68 remaining in the annuity.

  1. What is the value, in dollars, of the monthly payment to the band?   (1 mark)

  2. After six months of writing and recording, the band decided that it needs more time to finish the album.

     

    To extend the time that the annuity will last, the band will work for three more months without withdrawing a payment.

     

    After this, the band will receive monthly payments of $3800 for as long as possible.

     

    The annuity will end with one final monthly payment that will be smaller than all of the others.

     

    Calculate the total number of months that this annuity will last.   (2 marks)

Show Answers Only
  1. `$3000`
  2. `18`
Show Worked Solution

a.    `text(By TVM Solver:)`

`N` `= 6`  
`I text(%)` `=3.12`  
`PV` `=50\ 000`  
`PMT` `= ?`  
`FV` `=-32\ 667.68`  
`text(PY)` `= text(CY) = 12`  

 
`=> PMT = -3000.00`

`:. \ text(Monthly payment) = $3000`

 

b.   `text{Value after 3 more months (by TVM Solver):}`

`N` `= 3`  
`I text(%)` `=3.12`  
`PV` `=32\ 667.68`  
`PMT` `= 0`  
`FV` `=?`  
`text(PY)` `= text(CY) = 12`  

 
`=> FV = -32\ 923.15`
 

`text(Find) \ \ N \ text(when) \ FV= 0 \ text{(by TVM Solver):}`

`N` `= ?`  
`I text(%)` `=3.12`  
`PV` `=-32\ 923.15`  
`PMT` `= 3800`  
`FV` `=0`  
`text(PY)` `= text(CY) = 12`  

 
`=> N = 8.77`

`:. \ text(Total months of annuity)` `= 6 + 3 + 9`
  `= 18`

CORE, FUR2-NHT 2019 VCAA 7

Tisha plays drums in the same band as Marlon.

She would like to buy a new drum kit and has saved $2500.

  1. Tisha could invest this money in an account that pays interest compounding monthly.

     

    The balance of this investment after `n` months, `T_n` could be determined using the recurrence relation below
     
          `T_0 = 2500, \ \ \ \ T_(n+1) = 1.0036 xx T_n` 
     
    Calculate the total interest that would be earned by Tisha's investment in the first five months.

     

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

Tisha could invest the $2500 in a different account that pays interest at the rate of 4.08% per annum, compounding monthly. She would make a payment of $150 into this account every month.

  1. Let `V_n` be the value of Tisha's investment after `n` months.

     

    Write down a recurrence relation, in terms of `V_0`, `V_n` and `V_(n + 1)`, that would model the change in the value of this investment.   (1 mark)

  2. Tisha would like to have a balance of $4500, to the nearest dollar, after 12 months.

     

    What annual interest rate would Tisha require?

     

    Round your answer to two decimal places.   (1 mark)

Show Answers Only
  1. `$45.33`
  2. `V_0 = 2500, \ V_(n+1) = 1.0034 xx V_n + 150`
  3. `5.87%`
Show Worked Solution

a.    `T_1 = 1.0036 xx 2500 = 2509`

`T_2 = 1.0036 xx 2509 = 2518.0324`

`vdots`

`T_5 = 2545.33`

`:. \ text(Total interest) ` `= 2545.33-2500`
  `= $45.33`

 

b.    `text(Monthly interest) = (4.08)/(12) = 0.34%`

`:. \ V_0 = 2500, \ V_(n+1) = 1.0034 xx V_n + 150`

 

c.    `text(By TVM Solver:)`

`N` `= 12`  
`I text(%)` `=?`  
`PV` `=-2500`  
`PMT` `=-150`  
`FV` `=4500`  
`text(PY)` `= text(CY)=12`  

 
`=> I = 5.87%`

CORE, FUR2-NHT 2019 VCAA 5

A random sample of 12 mammals drawn from a population of 62 types of mammals was categorized according to two variables.

likelihood of attack (1 = low, 2 = medium, 3 = high)

exposure to attack during sleep (1 = low, 2 = medium, 3 = high)

The data is shown in the following table.
 


 

  1. Use this data to complete the two-way frequency table below.   (1 mark)

      
         
     

The following two-way frequency table was formed from the data generated when the entire population of 62 types of mammals was similarly categorized.
 
     

    1. How many of these 62 mammals had both a high likelihood of attack and a high exposure to attack during sleep?   (1 mark)

    2. Of those mammals that had a medium likelihood of attack, what percentage also had a low exposure to attack during sleep?   (1 mark)

    3. Does the information in the table above support the contention that likelihood of attack is associated with exposure to attack during sleep? justify your answer by quoting appropriate percentages. It is sufficient to consider only one category of likelihood of attack when justifying your answer.   (2 marks)

Show Answers Only

  1.  
    1. `15`
    2. `text(Percentage) = (2)/(4) xx 100`
                           `= 50 %`
    3. `text(The data supports the contention that animals with a low likelihood)`
      `text(of attack is associated with low exposure to attack during sleep.)`
       
      `text(- 91%) \ ({31}/{34}) \ text(of animals with low exposure to attack)`
         `text(during sleep, have a low likelihood of attack.)`
       
      `text(- Similarly, 89% of animals with a medium exposure to attack during)`
         `text(sleep have a low likelihood of attack.)`
       
      `text(- 11% of animals with a high exposure to attack during sleep have)`
          `text(a low likelihood of attack)`

Show Worked Solution

a.     

 

b.    i. `15`

ii.   `text(Percentage)` `= (2)/(4) xx 100`
  `= 50%`

 

iii. `text(The data supports the contention that animals with a low likelihood)`

`text(of attack is associated with low exposure to attack during sleep.)`

`text(- 91%) \ ({31}/{34}) \ text(of animals with low exposure to attack)`

`text(during sleep, have a low likelihood of attack.)`

`text(- Similarly, 89% of animals with a medium exposure to attack during)`

`text(sleep have a low likelihood of attack.)`

`text(- 11% of animals with a high exposure to attack during sleep have)`

`text(a low likelihood of attack)`

CORE, FUR2-NHT 2019 VCAA 4

The scatterplot below plots the variable life span, in years, against the variable sleep time, in hours, for a sample of 19 types of mammals.
 

On the assumption that the association between sleep time and life span is linear, a least squares line is fitted to this data with sleep time as the explanatory variable.

The equation of this least squares line is

life span = 42.1 – 1.90 × sleep time

The coefficient of determination is 0.416

  1. Draw the graph of the least squares line on the scatterplot above.   (1 mark)

  2. Describe the linear association between life span and sleep time in terms of strength and direction.   (2 marks)

  3. Interpret the slope of the least squares line in terms of life span and sleep time.   (2 marks)

  4. Interpret the coefficient of determination in terms of life span and sleep time.   (1 mark)

  5. The life of the mammal with a sleep time of 12 hours is 39.2 years.
  6. Show that, when the least squares line is used to predict the life span of this mammal, the residual is 19.9 years.   (2 marks)

Show Answers Only
  1.  

  2. `text(Strength is moderate.)`

     

    `text(Direction is negative.)`

  3. `text(The gradient of –1.9 means that life span decreases by)` 

     

    `text(1.9 years for each additional hour of sleep time.)`

  4. `text(41.6% of the variation in life span can be explained by the)`

     

    `text(variation in sleep time.)`

  5. `text(Proof(See Worked Solution))`
Show Worked Solution

a.    `text{Graph endpoints (0, 42.1) and (18, 7.9)}`
 


 

b.   `text(Strength is moderate.)`

`text(Direction is negative.)`
 

c.    `text(The gradient of –1.9 means that life span decreases by)`

`text(1.9 years for each additional hour of sleep time.)`

 

d.    `text(41.6% of the variation in life span can be explained by the )`

`text(variation in sleep time.)`
 

e.    `text(Predicted value)` `= 42.1 – 1.9 xx 12`
  `= 19.3 \ text(years)`

 

`text(Residual)` `= text(actual) – text(predicted)`
  `= 39.2 – 19.3`
  `= 19.9 \ text(years)`

MATRICES, FUR1-NHT 2019 VCAA 7-8 MC

A farm contains four water sources, `P`, `Q`, `R` and `S`.

Part 1

Cows on the farm are free to move between the four water sources.

The change in the number of cows at each of these water sources from week to week is shown in the transition diagram below.
 

Let `C_n` be the state matrix for the location of the cows in week `n` of 2019.

The state matrix for the location of the cows in week 23 of 2019 is `C_23 = [(180),(200),(240),(180)]{:(P),(Q),(R),(S):}`
 

The state matrix for the location of the cows in week 24 of 2019 is `C_24 = [(160),(222),(203),(215)]{:(P),(Q),(R),(S):}`

Of the cows expected to be at `Q` in week 24 of 2019, the percentage of these cows at `R` in week 23 of 2019 is closest to

  1.   8%
  2.   9%
  3. 20%
  4. 22%
  5. 25%

 

Part 2

Sheep on the farm are also free to move between the four water sources.

The change in the number of sheep at each water source from week to week is shown in matrix `T` below.
 

`{:(),(),(T=):}{:(qquadqquadqquadtext(this week)),((qquadP,quadQ,quadR,quadS)),([(0.4,0.3,0.2,0.1),(0.2,0.1,0.5,0.3),(0.1,0.3,0.1,0.2),(0.3,0.3,0.2,0.4)]):}{:(),(),({:(P),(Q),(R),(S):}):}{:(),(),(text(next week)):}`
 

In the long term, 635 sheep are expected to be at `S` each week.

In the long term, the number of sheep expected to be at `Q` each week is closest to

  1. 371
  2. 493
  3. 527
  4. 607
  5. 635
Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(In week 23, 240 cows are at)\ R.`

`text(In week 24, number of cows moving from)\ R\ text(to)\ Q`

`=20text(%) xx 240`

`= 48\ text(cows)`
 

`text(Total cows at)\ Q = 222`
 

`:.\ text(Percentage)` `= 48/222`
  `= 0.2162`
  `= 22text(%)`

`=>\ D`

 

`text(Part 2)`

`T^50 = [(0.2434, 0.2434, 0.2434, 0.2434),(0.2603, 0.2603, 0.2603, 0.2603),(0.1834, 0.1834, 0.1834, 0.1834),(0.3130, 0.3130, 0.3130, 0.3130)]`

 
`text(S)text(ince 635 sheep are expected long term at)\ S,`

`text(Total sheep)` `= 635/0.3130`
  `= 2029`

 
`:. text(Long term expected at)\ Q`

`~~ 0.2603 xx 2029`

`~~ 528`

`=>\ C`

GRAPHS, FUR1-NHT 2019 VCAA 7 MC

The shaded area in the graph below represents the feasible region for a linear programming problem.
 


 

The maximum value of the objective function  `Z = -2x - 2y` occurs at

  1. point `C` only.
  2. any point along line segment  `BC`.
  3. any point along line segment  `AD`.
  4. any point along line segment  `AB`.
  5. any point along line segment  `DC`.
Show Answers Only

`C`

Show Worked Solution

`text(Equation of)\ \ BC \ => \ x + y = 10`

`text(Equation of)\ \ AD \ => \ x + y = 5`

`Z` `= -2x – 2y`
  `= -2(x + y)`

 
`text(Along)\ \ AD,\ Z = -10`

`text(Along)\ \ BC,\ Z = -20`

`:.\ text(Max value of)\ \ Z = -10`
 

`=>  C`

GRAPHS, FUR1-NHT 2019 VCAA 4 MC

A farm has `x` cows and `y` sheep.

On this farm there are always at least twice as many sheep as cows.

The relationship between the number of cows and the number of sheep on this farm can be represented by the inequality

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

`A`

Show Worked Solution

`text(If twice as many sheep as cows)`

`y = 2x`

`text(If at least twice as many sheep as cows)`

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

 
`=>  A`

MATRICES, FUR1-NHT 2019 VCAA 5 MC

A population of birds feeds at two different locations, `A` and `B`, on an island.

The change in the percentage of the birds at each location from year to year can be determined from the transition matrix `T` shown below.
 

`{:(),(),(T=):}{:(qquadtext(this year)),((qquadA,\ B)),([(0.8,0.4),(0.2,0.6)]):}{:(),(),({:(A),(B):}):}{:(),(),(text(next year)):}`
 

In 2018, 55% of the birds fed at location `B`.

In 2019, the percentage of the birds that are expected to feed at location `A` is

  1. 32%
  2. 42%
  3. 48%
  4. 58%
  5. 62%
Show Answers Only

`D`

Show Worked Solution

`[(A_2019),(B_2019)] = [(0.8,0.4),(0.2,0.6)][(0.45),(0.55)] = [(0.58),(0.42)]`

`=>\ D`

NETWORKS, FUR1-NHT 2019 VCAA 7 MC

A graph has five vertices, `A, B, C, D` and `E`.

The adjacency matrix for this graph is shown below.
 

`{:(qquad qquad A quad B quad C quad D quad E), ({:(A), (B), (C), (D), (E):} [(0, 1, 0, 1, 2),(1, 0, 1, 0, 1),(0, 1, 1, 0, 1),(1, 0, 0, 0, 1),(2, 1, 1, 1, 0)]):}`
 

Which one of the following statements about this graph is not true?

  1. The graph is connected.
  2. The graph contains an Eulerian trail.
  3. The graph contains an Eulerian circuit.
  4. The graph contains a Hamiltonian cycle.
  5. The graph contains a loop and multiple edges.
Show Answers Only

`C`

Show Worked Solution

`text(Consider the degree of each vertex:)`

`A – 4, B – 3, C – 3, D – 2, E – 5`

`text{Graph cannot be an Eulerian circuit because it}`

`text{has odd degree vertices.}`

`text{Note that an Eulerian circuit cannot have an odd degree}`

`text{vertex as it uses all edges exactly once and begins and ends}`

`text{at the same vertex.}`

`=>  C`

NETWORKS, FUR1-NHT 2019 VCAA 3 MC

Four students, Alice, Brad, Charli and Dexter, are working together on a school project.

This project has four parts.

Each of the students will complete only one part of the project.

The table below shows the time it would take each student to complete each part of the project, in minutes.
 

           Part 1               Part 2               Part 3               Part 4       
    Alice 5 5 3 5
    Brad 3 3 6 4
    Charli 6 5 4 3
    Dexter     4 5 6 5

  
The parts of this project must be completed one after the other.

Which allocation of student to part must occur for this project to be completed in the minimum time possible?

A.           Part 1               Part 2               Part 3               Part 4       
  Brad Dexter Alice Charli
         
B.    Part 1 Part 2 Part 3 Part 4
  Brad Dexter Charli Alice
         
C.    Part 1 Part 2 Part 3 Part 4
  Dexter Alice Charli Brad
         
D.    Part 1 Part 2 Part 3 Part 4
  Dexter Brad Alice Charli
         
E.    Part 1 Part 2 Part 3 Part 4
  Dexter Brad Charli Alice

 

Show Answers Only

`D`

Show Worked Solution

`text(After row and column reduction:)`

 

 
`text{4 lines (minimum) required to cover all zero’s.}`

`:.\ text(Alice)` `\ text(− Part 3)`
`text(Charli)` `\ text(− Part 4)`
`text(Dexter)` `\ text(− Part 1)`
`text(Brad)` `\ text(− Part 2)`

 
`=>  D`

Calculus, 2ADV C4 2019 MET-N 15 MC

The area bounded by the graph of  `y = f(x)`, the line  `x = 2`, the line  `x = 8`  and the `x`-axis, as shaded in the diagram below, is  `3log_e(13)`

The value of  `int_4^10 3 f(x - 2)\ dx`  is

  1.  `3log_e(13)`
  2.  `9log_e(13)`
  3.  `6log_e(39)`
  4.  `9log_e(11)`
Show Answers Only

`B`

Show Worked Solution

`A_1 = int_2^8 f(x)\ dx = 3 log_e 13`

`f(x – 2) = f(x) \ text(shifted 2 units to right.)`

`:. \ int_2^8 f(x)\ dx = int_4^10 f(x – 2)\ dx`

`:. \ 3 int_4^10 f(x – 2)\ dx` `= 3 xx 3log_e 13`
  `= 9 log_e 13`

 
`=> \ B`

Functions, EXT1 F1 2019 MET2-N 11 MC

The function  `f(x) = 5x^3 + 10x^2 + 1`  will have an inverse function for the domain

  1.  `D = (–2, ∞)`
  2.  `D = (–∞ , (1)/(2)]`
  3.  `D = (–∞ , –1]`
  4.  `D = [0 , ∞)`
Show Answers Only

`D`

Show Worked Solution

`y = 5x^3 + 10x^2 + 1`

`y^{′} = 15x^2 + 20x`

`y^{″}=30x+20`

`text(SP’s when)\ \ y^{′}=0\ \ =>\ \ x= – 4/3\ text{(max)}, \ 0\ text{(min)}`

`text(Sketch graph:)`
 

`=> \ D`

Calculus, MET2-NHT 2019 VCAA 5

Let  `f: R → R, \ f(x) = e^((x/2))`  and  `g: R → R, \ g(x) = 2log_e(x)`.

  1. Find  `g^-1 (x)`.   (1 mark)

  2. Find the coordinates of point  `A`, where the tangent to the graph of  `f` at  `A` is parallel to the graph of  `y = x`.   (2 marks)

  3. Show that the equation of the line that is perpendicular to the graph of  `y = x`  and goes through point  `A` is  `y = -x + 2log_e(2) + 2`.   (1 mark)

Let `B` be the point of intersection of the graphs of `g` and  `y =-x + 2log_e(2) + 2`, as shown in the diagram below.
 

               
 

  1. Determine the coordinates of point `B`.   (1 mark)

  2. The shaded region below is enclosed by the axes, the graphs of  `f` and `g`, and the line  `y =-x + 2log_e(2) + 2`.
     
     
               
     
    Find the area of the shaded region.   (2 marks)

Let  `p : R→ R, \ p(x) = e^(kx)`  and  `q : R→ R, \ q(x) = (1)/(k) log_e(x)`.

  1. The graphs of `p`, `q` and  `y = x`  are shown in the diagram below. The graphs of `p` and `q` touch but do not cross.
     
     
               
     

     Find the value of  `k`.   (2 marks)

  2. Find the value of  `k, k > 0`, for which the tangent to the graph of `p` at its `y`-intercept and the tangent to the graph of `q` at its `x`-intercept are parallel.   (1 mark)

Show Answers Only
  1. `e^{(x)/(2)}`
  2. `(2log_e 2, 2)`
  3. `text(Proof(See Worked Solution))`
  4. `(2, 2log_e 2)`
  5. `6-2(log_e 2)^2-4 log_e 2`
  6. `(1)/(e)`
  7. `k =1`
Show Worked Solution

a.    `g(x) = 2log_e x`

`text(Inverse: swap) \ x ↔ y`

`x` `= 2log_e y`
`log_e y` `= (x)/(2)`
`y` `= e^{(x)/(2)}`

 

b.    `f(x) = e^{(x)/(2)}`

`f′(x) = (1)/(2) e^{(x)/(2)}`
 
`text(S) text(olve) \ \ f′(x) = 1 \ text(for) \ x:`

`x = 2log_e 2`

`y = e^(log_e 2) = 2`
 
`:. \ A\ text(has coordinates)\  (2log_e 2, 2)`

 

c.    `m_(⊥) = -1`

`text(Equation of line) \ \ m = -1 \ \ text(through)\ \ (2log_e 2, 2) :`

`y-2` `= -(x-2log_e 2)`
`y` `= -x +  2log_e 2 + 2`

 

d.    `text(Method 1)`

`text(S) text(olve for) \ x :`

`-x + 2log_e 2 + 2 = 2log_e x`

`=> x = 2 , \ y = 2log_e 2`

`:. B ≡ (2, 2log_e 2)`
 

`text(Method 2)`

`text(S) text(ince) \ \ f(x) = g^-1 (x)`

`B \ text(is the reflection of) \ \ A(2log_e 2, 2) \ \ text(in the) \ \ y=x \ \ text(axis)`

`:. \ B ≡ (2, 2log_e 2)`

 

e.
             
 

`y = g(x) \ \ text(intersects) \ x text(-axis at) \ \ x = 1`

`text(Dividing shaded area into 3 sections:)`

`A` `= int_0^1 f(x)\ dx \ + \ int_1^(2log_e 2) f(x)-g(x)\ dx`  
  ` \ + \ int_(2log_e 2)^2 (-x + 2log_e 2 + 2)-g(x)\ dx`  
  `= 6-2(log_e 2)^2-4 log_e 2`  

 

f.   `p(x) = e^(kx) \ , \ q(x) = (1)/(k) log_e x`

`p′(x) = k e^(kx) \ , \ q′(x) = (1)/(kx)`
 
`text(S) text(ince graphs touch on) \ y = x`

`k e^(kx) = 1\ …\ (1)`

`(1)/(kx) = 1\ …\ (2)`

`text(Substitute) \ x = (1)/(k) \ text{from (2) into (1)}`

`k e^(k xx 1/k)` `= 1`
`:. k` `= (1)/(e)`

 

g.   `text(Consider)\ \ p(x):`

`text(When) \ \ x = 0 , \ p(0) = 1 , \ p′(0) = k`
 

 `text(Consider) \ \ q(x):`

`text(When) \ \ y= 0, \ (1)/(k) log_e x = 0 \ => \ x = 1`

`q′(1) = (1)/(k)`

`text(If lines are parallel), \ k = (1)/(k)`
 
`:. \ k = 1`

Calculus, MET2-NHT 2019 VCAA 4

A mining company has found deposits of gold between two points, `A` and `B`, that are located on a straight fence line that separates Ms Pot's property and Mr Neg's property. The distance between `A` and `B` is 4 units.

The mining company believes that the gold could be found on both Ms Pot's property and Mr Neg's property.

The mining company initially models he boundary of its proposed mining area using the fence line and the graph of 

`f : [0, 4] → R, \ f(x) = x(x-2)(x-4)`

where `x` is the number of units from point `A` in the direction of point `B` and `y` is the number of units perpendicular to the fence line, with the positive direction towards Ms Pot's property. The mining company will only mine from the boundary curve to the fence line, as indicated by the shaded area below.
 

  1. Determine the total number of square units that will be mined according to this model.   (2 marks)

The mining company offers to pay Mr Neg $100 000 per square unit of his land mined and Ms Pot $120 000 per square unit of her land mined.

  1. Determine the total amount of money that the mining company offers to pay.   (1 mark)

The mining company reviews its model to use the fence line and the graph of

     `p : [0, 4] → R, \ p(x) = x(x-4 + (4)/(1 + a)) (x-4)`  where `a > 0`.

  1. Find the value of  `a`  for which  `p(x) = f(x)`  for all `x`.   (1 mark)

  2. Solve  `p^{′}(x) = 0`  for `x` in terms of `a`.   (2 marks)

Mr Neg does not want his property to be mined further than 4 units measured perpendicular from the fence line.

  1. Find the smallest value of `a`, correct to three decimal places, for this condition to be met.   (2 marks)

  2. Find the value of `a` for which the total area of land mined is a minimum.   (3 marks)

  3. The mining company offers to pay Ms Pot $120 000 per square unit of her land mined and Mr Neg $100 000 per square unit of his land mined.
  4. Determine the value of `a` that will minimize the total cost of the land purchase for the mining company. Give your answer correct to three decimal places.   (2 marks)

Show Answers Only
  1. `8`
  2. `$880\ 000`
  3. `1`
  4. `x = ((8a + 4) ± 4 sqrt(a^2 + a + 1))/(3a + 3)`
  5. `0.716`
  6. `1`
  7. `0.886`
Show Worked Solution
a.    `A` `= int_0^2 x(x-2)(x-4)\ dx – int_2^4 x(x-2)(x-4)\ dx`
  `= 4-(-4)`
  `= 8`

 

b.    `text(Total payment)` `= 4 xx 100 000 + 4 xx 120 000`
  `= $880\ 000`

 
c.    `text(If) \ \ p(x) = f(x)`

`x-2` `= x-4 + (4)/(1 + a)`  
`(4)/(1 + a)` `=2`  
`:. a` `=1`  

 
d.    `p^{′}(x) = (3(a + 1) x^2-8(2a + 1) x + 16a)/(a + 1) \ \ \ text{(by CAS)}`

`text(S) text(olve) \ \ p^{′}(x) = 0 \ \ text(for) \ \ x :`

`x = ((8a + 4) ± 4 sqrt(a^2 + a + 1))/(3a +3) \ , \ a > 0` 
 

e.    `text(If no mining further than 4 units,) \ \ p(x) ≥-4`

`text(Max distance from Mr Neg’s fence occurs when) \ \ 2< x <4 .`

`text((i.e. the higher) \ x text(-value when) \ \ p^{′}(x) = 0)`
 

`text(At) \ \ x = ((8a + 4) + 4 sqrt(a^2 + a + 1))/(3a + 3) \ ,`
 
`text(S) text(olve for) \ \ a \ \ text(such that) \ \ p(x) = -4`

`=> a = 0.716 \ \ text((to 3 d.p.))`

 

f.    `p(x) \ text(intersects) \ x text(-axis at) \ \ x = 4-(4)/(1 + a) = (4a)/(1 + a)`

`A` `= int_0^{(4a)/(1 +a)} p(x)\ dx-int_{(4a)/(1 +a)}^4 p(x)\ dx`
  `= (64(1 + 2a + 2a^3 + a^4))/(3(1 + a)^4)`

 
`text(S) text(olve) \ \ A^{′}(a) = 0 \ \ text(for) \ \ a :`

`a = 1`

 

g.    `C(a) = 120\ 000 int_0^{(4a)/(1 +a)} p(x)\ dx-100\ 000 int_{(4a)/(1 +a)}^4 p(x)\ dx`

`text(S) text(olve) \ \ C^{′}(a) = 0 \ \ text(for) \ \ a :`

`a = 0.886 \ \ text((to 3 d.p.))`

Statistics, MET2-NHT 2019 VCAA 3

Concerts at the Mathsland Concert Hall begin `L` minutes after the scheduled starting time. `L` is a random variable that is normally distributed with a mean of 10 minutes and a standard deviation of four minutes.

  1. What proportion of concerts begin before the scheduled starting time, correct to four decimal places?   (1 mark)

  2. Find the probability that a concert begins more than 15 minutes after the scheduled starting time, correct to four decimal places.   (1 mark)

If a concert begins more than 15 minutes after the scheduled starting time, the cleaner is given an extra payment of $200. If a concert begins up to 15 minutes after the scheduled starting time, the cleaner is given an extra payment of $100. If a concert begins at or before the scheduled starting time, there is no extra payment for the cleaner.

Let `C` be the random variable that represents the extra payment for the cleaner, in dollars.

    1. Using your responses from part a. and part b., complete the following table, correct to three decimal places.   (1 mark)

       

    2. Calculate the expected value of the extra payment for the cleaner, to the nearest dollar.   (1 mark)

    3. Calculate the standard deviation of `C`, correct to the nearest dollar.   (1 mark)

The owners of the Mathsland Concert Hall decide to review their operation. They study information from 1000 concerts at other similar venues, collected as a simple random sample. The sample value for the number of concerts that start more than 15 minutes after the scheduled starting time is 43.

    1. Find the 95% confidence interval for the proportion of the concerts that begin more than 15 minutes after the scheduled starting time. Give values correct to three decimal places.   (1 mark)

    2. Explain why this confidence interval suggests that the proportion of concerts that begin more than 15 minutes after the scheduled starting time at the Mathsland Concert Hall is different from the proportion at the venue in the sample.   (1 mark)

The owners of the Mathsland Concert Hall decide that concerts must not begin before the scheduled starting time. They also make changes to reduce the number of concerts that begin after the scheduled starting time. Following these changes, `M` is the random variable that represents the number of minutes after the scheduled starting time that concerts begin. The probability density function for `M` is
 

`qquad qquad f(x) = {(8/(x + 2)^3), (0):} qquad {:(x ≥ 0), (x < 0):}`
 

where `x` is the the time, in minutes, after the scheduled starting time.

  1. Calculate the expected value of `M`.   (2 marks) 
    1. Find the probability that a concert now begins more than 15 minutes after the scheduled starting time.   (1 mark)

    2. Find the probability that each of the next nine concerts begins more than 15 minutes after the scheduled starting time and the 10th concert begins more than 15 minutes after the scheduled starting time. Give your answer correct to four decimal places.   (2 marks)

    3. Find the probability that a concert begins up to 20 minutes after the scheduled starting time, given that it begins more than 15 minutes after the scheduled starting time. Give your answer correct to three decimal places.   (2 marks)

Show Answers Only

  1. `0.0062`
  2. `0.1056`
  3. i.   

     ii.   `$110 \ \ text((nearest dollar))`
     iii.  `$32 \ \ text(nearest dollar))`
  4. i.   `(0.030, 0.056)`
    ii.   `text(The proportion of concerts that begin more than 15 minutes)` 

     

    `qquad text(late is not within the sample 95% confidence interval.)`

  5.  `2`
  6. i.   `(4)/(289)`
    ii.   `0.0122 \ \ text((to 4 decimal places))`
    iii.  `0.403 \ \ text((to 3 decimal places))`

Show Worked Solution

a.    `L\ ~\ N (10, 4^2)`

`text(Pr) (L < 0)` `= P(z < –2.5)`
  `= 0.0062`

 

b.    `text(Pr) (L > 15)` `= text(Pr) ( z > 1.25)`
  `= 0.1056`

 

c.i.

ii.  `E(C)` `= 0.8882 xx 100 + 0.1056 xx 200`
  `= 109.94`
  `= $110 \ \ text((nearest dollar))`

 

iii.  `E(C^2)` `= 100^2 xx 0.8882 + 200^2 xx 0.1056`
  `= 13\ 106`

 

`text(s.d.) (C)` `= sqrt(E(C^2) – [E(C)]^2)`
  `= sqrt(13\ 106 – (109.94)^2)`
  `= sqrt(1019.1 …)`
  `= 31.92 …`
  `= $32 \ \ text((nearest dollar))`

 

 
d.i.  `overset^p = (43)/(1000) \ \ , \ n = 1000`

`95% \ text(C. I.)` `= overset^p ± 1.96 sqrt((overset^p(1 – overset^p))/(n))`
  `= (0.030, 0.056)`

 
ii.   `text(The proportion of concerts that begin more than 15 minutes)`

`text(late is not within the sample 95% confidence interval.)`

 

e.    `E(M)` `= int_0^∞ x ((8)/((x +2)^3))\ dx`
  `= 2`

 

f.i.    `text(Pr)(M > 15)` `=int_0^∞ x ((8)/((x +2)^3))\ dx`
  `= (4)/(289)`

 
ii.  `text(Pr)(M > 15) = (4)/(289) \ \ , \ \ text(Pr)(M ≤ 15) = (285)/(289)`
 

`:. \ text(Pr) text{(9 concerts}\ \ M ≤ 15 ,\ text{10th concert}\ \ M > 15)`

`= ((205)/(289))^9 xx (4)/(289)`

`= 0.0122 \ \ text((to 4 decimal places))`

 
iii.  `text(Pr)(15 < M < 20)= int_15^20 (8)/((x + 2)^3)\ dx = (195)/(34\ 969)`

 `text(Pr)(M > 15)= int_15^oo (8)/((x + 2)^3)\ dx = (4)/(289)`
 

`text(Pr)(M < 20 | M>15)` `= (text(Pr)(15 < M < 20))/(text(Pr)(M > 15))`
  `= ((195)/(34\ 969))/((4)/(289))`
  `= (195)/(484)`
  `= 0.403 \ \ text((to 3 decimal places))`

Trigonometry, EXT1 T3 EQ-Bank 2

θθ

A particular energy wave can be modelled by the function

`f(t) = sqrt5 sin 0.2t + 2 cos 0.2t, \ \ t ∈ [0, 50]`

  1. Express this function in the form  `f(t) = Rsin(nt - alpha), \ \ alpha ∈ [0, 2pi]`.  (2 marks)

  2. Find the time the wave first attains its maximum value. Give your answer to one decimal place.  (2 marks)

Show Answers Only
  1. `f(t) = 3sin(0.2t – 5.553)`
  2. `t = 4.2`
Show Worked Solution

i.   `f(t) = sqrt5 sin 0.2t + 2cos 0.2t`

`Rsin(nt – alpha)` `= Rsin(0.2t – alpha)`
  `= Rsin 0.2t cosalpha – Rcos 0.2t sinalpha`

 

`=> Rcosalpha = sqrt5,\ \ R sinalpha = −2`

`R^2` `= (sqrt5)^2 + (−2)^2 = 9`
`R` `= 3`

 
`cosalpha = sqrt5/3, sinalpha = −2/3`

`=> alpha\ text(is in 4th quadrant)`
 

`text(Base angle) = cos^(−1)(sqrt5/3) = 0.7297`

`:.alpha` `= 2pi – 0.7297`
  `= 5.553…`

 
`:. f(t) = 3sin(0.2t – 5.553)`

 

ii.   `text(Max value occurs when)\ sin(0.2t – 5.553) = 1`

`0.2t – 5.553` `= pi/2`
`0.2t` `= 7.124…`
`t` `= 35.62…`

 
`text(Test if)\ \ t > 0\ \ text(for:)`

`0.2t – 5.553` `= -(3pi)/2`
`0.2t` `= 0.8406`
`t` `= 4.20…`

 
`:. text(Time of 1st maximum:)\ \ t = 4.2`

Trigonometry, MET2-NHT 2019 VCAA 2

The wind speed at a weather monitoring station varies according to the function

`v(t) = 20 + 16sin ((pi t)/(14))`

where `v` is the speed of the wind, in kilometres per hour (km/h), and  `t`  is the time, in minutes, after 9 am.

  1. What is the amplitude and the period of  `v(t)`?   (2 marks)

  2. What are the maximum and minimum wind speeds at the weather monitoring station?   (1 mark)

  3. Find  `v(60)`, correct to four decimal places.   (1 mark)

  4. Find the average value of  `v(t)`  for the first 60 minutes, correct to two decimal places.   (2 marks)

A sudden wind change occurs at 10 am. From that point in time, the wind speed varies according to the new function

                    `v_1(t) = 28 + 18 sin((pi(t-k))/(7))`

where  `v_1`  is the speed of the wind, in kilometres per hour, `t` is the time, in minutes, after 9 am and  `k ∈ R^+`. The wind speed after 9 am is shown in the diagram below.
 

  1. Find the smallest value of `k`, correct to four decimal places, such that  `v(t)`  and  `v_1(t)`  are equal and are both increasing at 10 am.   (2 marks)

  2. Another possible value of `k` was found to be 31.4358

     

    Using this value of `k`, the weather monitoring station sends a signal when the wind speed is greater than 38 km/h.

     

    i.  Find the value of `t` at which a signal is first sent, correct to two decimal places.   (1 mark)

     

    ii. Find the proportion of one cycle, to the nearest whole percent, for which  `v_1 > 38`.   (2 marks)

  3. Let  `f(x) = 20 + 16 sin ((pi x)/(14))`  and  `g(x) = 28 + 18 sin ((pi(x-k))/(7))`.
     
    The transformation  `T([(x),(y)]) = [(a \ \ \ \ 0),(0 \ \ \ \ b)][(x),(y)] + [(c),(d)]`  maps the graph of  `f`  onto the graph of  `g`.

     

    State the values of  `a`, `b`, `c` and `d`, in terms of `k` where appropriate.   (3 marks)

Show Answers Only
  1. `28`
  2. `v_text(max) \ = 36 \ text(km/h)`

     

    `v_text(min) \ = 4 \ text(km/h)`

  3. `32.5093 \ text(km/h)`
  4. `20.45 \ text(km/h)`
  5. `3.4358`
  6. i. `60.75 \ text(m)`

     

    ii. `31text(%)`

  7. `a = (1)/(2) \ , \ b = (9)/(8) \ , \ c = k \ , \ d = (11)/(2)`
Show Worked Solution

a.    `text(Amplitude) = 16`

`text{Find Period (n):}`

`(2 pi)/(n)` `= (pi)/(14)`
`n` `= 28`

 

b.    `v_text(max) = 20 + 16 = 36 \ text(km/h)`

`v_text(min) = 20-16 = 4 \ text(km/h)`

 

c.    `v(60)` `= 20 + 16 sin ((60 pi)/(14))`
  `= 32.5093 \ \ text(km/h)`

 

d.    `v(t)\ \ text(is always positive.)`

`s(t) = int_0^60 v(t) \ dt`

`v(t)_(avg)` `= (1)/(60) int_0^60 20 + 16 sin ((pi t)/(14))\ dt`
  `= 20.447`
  `= 20.45 \ text(km/h) \ \ text((to 2 d.p.))`

 

e.    `text(S) text{olve (for}\ k text{):} \ \ v(60) = v_1(60)`

`k = 3.4358 \ \ text((to 4 d.p.))`

 

f.i.  `text(S) text(olve for) \ t , \ text(given) \ \ v_1(t) = 38 \ \ text(and) \ \ k = 31.4358`

`=> t = 60.75 \ text(minutes)`
 

f.ii.  `text(S) text(olving for) \ \ v_1(t) = 38 \ , \ k = 31.4358`

`t_1 = 60.75 \ text{(part i)}, \ t_2 = 65.123`

`text(Period of) \ \ v_1 = (2 pi)/(n) = (pi)/(7)\ \ => \ n = 14`

`:. \ text(Proportion of cycle)` `= (65.123-60.75)/(14)`
  `= 0.312`
  `= 31 text{%  (nearest %)}`

 

g.    `f(x) → g(x)`

`y^{prime} = 28 + 18 sin ({pi(x^{prime}-k)}/{7})`

`x^{prime} = ax + c` `\ \ \ \ \ \ y^{prime} = by + d`

 

`text(Using) \ \ y^{prime} = by + d`

`28 + 18 sin ({pi(x^{prime}-k)}/{7}) = b (20 + 16 sin ({pi x}/{14})) + d`
 

`text(Equating coefficients of) \ \ sin theta :`

`16b = 18 \ \ \ => \ b = (9)/(8)`
 

`text(Equating constants:)`

`20 xx (9)/(8) + d = 28 \ \ \ => \ \ d = (11)/(2)`

`(x^{prime}-k)/(7)` `= (x)/(14)`
`x^{prime}` `= (x)/(2) + k \ \ => \ \ a = (1)/(2) \ , \ c = k`

 

`a = (1)/(2) \ , \ b = (9)/(8) \ , \ c = k \ , \ d = (11)/(2)`

Graphs, MET2-NHT 2019 VCAA 1

Parts of the graphs of  `f(x) = (x-1)^3(x + 2)^3`  and  `g(x) = (x-1)^2(x + 2)^3`  are shown on the axes below.
 


 

The two graphs intersect at three points,  (–2, 0),  (1, 0)  and  (`c`, `d`). The point  (`c`, `d`)  is not shown in the diagram above.

  1. Find the values of `c` and `d`.   (2 marks)

  2. Find the values of `x` such that  `f(x) > g(x)`.   (1 mark)

  3. State the values of `x` for which
    1. `f^{'}(x) > 0`   (1 mark)

    2. `g^{'}(x) > 0`   (1 mark)

  4. Show that  `f(1 + m) = f(–2-m)`  for all  `m`.   (1 mark)

  5. Find the values of `h` such that  `g(x + h) = 0`  has exactly one negative solution.   (2 marks)

  6. Find the values of `k` such that  `f(x) + k = 0`  has no solutions.   (1 mark)

Show Answers Only
  1. `c = 2 \ , \ d = 64`
  2. `(–∞, –2) \ ∪ \ (2, ∞)`
  3. i.  `(–(1)/(2), 1) \ ∪ \ (1, ∞)`
    ii. `(–∞, –2) \ ∪ \ (–2, –(1)/(5)) \ ∪ \ (1, ∞)`
  4. `text{Proof (Show Worked Solution)}`
  5.  `-2< h <=1`
  6. `(729)/(64)`
Show Worked Solution

a.    `text(Solve:) \ \ f(x) = g(x)`

`x = 1 , \ –2 \ text(and) \ 2`
 
`f(2) = 1^3 xx 4^3 = 64`
 
`text(Intersection at) \ (2, 64)`

`:. \ c = 2 \ , \ d = 64`

 

b.    `text(Using the graph and intersection at) \ (2, 64):`

`f(x) > g(x) \ \ text(for) \ \ (–∞, –2) \ ∪ \ (2, ∞)`

 

c.i.  `f'(x) > 0 \ \ text(for) \ \ (–(1)/(2), 1) \ ∪ \ (1, ∞)`

c.ii.  `g'(x) > 0 \ \ text(for) \ \ (–∞, –2) \ ∪ \ (–2, –(1)/(5)) \ ∪ \ (1, ∞)`

 

d.     `f(1 + m)` `= (1 + m-1)^3 (1 + m + 2)^3`
  `= m^3 (m + 3)^3`

 

`f(–2-m)` `= (–2-m -1)^3 (-2-m + 2)^3`
  `= (-m-3)^3 (-m)^3`
  `= (–1)^3 (m + 3)^3 (–1)^3 m^3`
  `= m^3 (m + 3)^3`

 

e.     `g(x + h)` `= (x + h-1)^2(x + h + 2)^3`
  `= underbrace{(x-(1 -h))^2}_{text(+ve solution)} * underbrace{(x-(h-2))^3}_{text(–ve solution)}`

 
`1-h ≥ 0 \ \ => \ \ h ≤ 1`

`-h-2 < 0 \ \ =>\ \ h > -2`

`:. -2< h <=1`

 

f.    `f(x) \ \ text(minimum S.P. when) \ \ f ′(x) = 0 \ =>  \ x =-(1)/(2)`

`text(S.P. at) \ \ (-(1)/(2) \ , \ -(729)/(64))`

`:. \ text(No solution if) \ \ k > (729)/(64)`

Trigonometry, EXT1 T3 EQ-Bank 1

  1. Show that `sinx + sin3x = 2sin2xcosx`.  (2 marks)

  2. Hence or otherwise, find all values of `x` that satisfy
     
    `qquad sinx + sin2x + sin3x = 0,\ \ \ x in [0,2pi]`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `x = 0, pi/2, pi, (3pi)/2, 2pi, (2pi)/3, (4pi)/3`
Show Worked Solution
i.   `sinx + sin3x` `= sinx + sin2xcosx + cos2xsinx`
  `= sinx + 2sinxcos^2x + (cos^2x – sin^2x)sinx`
  `= sinx + 2sinxcos^2x + cos^2xsinx – sin^3x`
  `= sinx + 3sinx(1 – sin^2x) – sin^3x`
  `= sinx + 3sinx – 3sin^3x- sin^3x`
  `= 4sinx(1 – sin^2x)`
  `= 4sinxcos^2x`
  `= 2sin2xcosx`
  `=\ text(RHS)`

 

ii.    `sinx + sin2x + sin3x` `= 0`
  `sin2x + 2sin2xcosx` `= 0`
  `sin2x(1 + 2cosx)` `= 0`

 

`text(If)\ sin2x` `= 0:`
`2x` `= 0, pi, 2pi, 3pi, 4pi`
`:.x` `= 0, pi/2, pi, (3pi)/2, 2pi`
`text(If)\ \ cosx` `= -1/2:`
`:.x` `= (2pi)/3, (4pi)/3`

CORE, FUR1-NHT 2019 VCAA 12 MC

Which one of the following statements could be true when written as part of the results of a statistical investigation?

  1. The correlation coefficient between height (in centimetres) and foot length (in centimetres) was found to be  `r = 1.24`
  2. The correlation coefficient between height (below average, average, above average) and arm span (in centimetres) was found to be  `r = 0.64`
  3. The correlation coefficient between blood pressure (low, normal, high) and age (under 25, 25–49, over 50) was found to be  `r = 0.74`
  4. The correlation coefficient between the height of students of the same age (in centimetres) and the money they spent on snack food (in dollars) was found to be  `r = 0.22`
  5. The correlation coefficient between height of wheat (in centimetres) and grain yield (in tonnes) was found to be  `r = –0.40`  and the coefficient of determination was found to be  `r^2 = –0.16`
Show Answers Only

`D`

Show Worked Solution

`text(Consider each option,)`

`A: \ r <= 1\ (text(incorrect))`

`B: \ text(height categories are ordinal) :. r\ text(not applicable)`

`C: \ text(both variables are ordinal) :. r\ text(not applicable)`

`D: \ text(possibly true)`

`E: \ text(Coefficient of determination cannot be negative (incorrect))`

`=>\ D`

CORE, FUR1-NHT 2019 VCAA 11 MC

The Human Development Index (HDI) and the mean number of children per woman for 13 countries are related.

This relationship is non-linear.

To linearise the data, a  `log_10`  transformation is applied to the response variable children.

A least squares line is then fitted to the linearised data.

The equation of this least squares line is

`log_10 (text(children)) = 1.06 - 0.00674 xx HDI`

Using this equation, the mean number of children per woman for a country with a HDI of 95 is predicted to be closest to

  1.  0.4
  2.  1.5
  3.  2.6
  4.  2.9
  5.  3.1
Show Answers Only

`C`

Show Worked Solution
`log_10(text(children))` `= 1.06 – 0.00674 xx 95`
  `= 0.4197`
`:.\ text(children)` `= 10^(0.4197)`
  `= 2.62…`

`=>\ C`

CORE, FUR1-NHT 2019 VCAA 5-7 MC

The birth weights of a large population of babies are approximately normally distributed with a mean of 3300 g and a standard deviation of 550 g.

Part 1

A baby selected at random from this population has a standardised weight of  `z = – 0.75`

Which one of the following calculations will result in the actual birth weight of this baby?
 

  1. `text(actual birth weight)\ = 550 - 0.75 × 3300`
  2. `text(actual birth weight)\ = 550 + 0.75 × 3300`
  3. `text(actual birth weight)\ = 3300 - 0.75 × 550`
  4. `text(actual birth weight)\ = 3300 + 0.75/550`
  5. `text(actual birth weight)\ = 3300 - 0.75/550`

 

Part 2

Using the 68–95–99.7% rule, the percentage of babies with a birth weight of less than 1650 g is closest to

  1. 0.14%
  2. 0.15%
  3. 0.17%
  4. 0.3%
  5. 2.5%

 

Part 3

A sample of 600 babies was drawn at random from this population.

Using the 68–95–99.7% rule, the number of these babies with a birth weight between 2200 g and 3850 g is closest to

  1. 111
  2. 113
  3. 185
  4. 408
  5. 489
Show Answers Only

`text(Part 1:)\ \ C`

`text(Part 2:)\ \ B`

`text(Part 3:)\ \ E`

Show Worked Solution

`text(Part 1)`

`text(Actual weight)` `= text(mean) + z xx text(std dev)`
  `= 3300 – 0.75 xx 550`

`=> C`

 

`text(Part 2)`

`z-text(score)` `= (x – barx)/5`
  `= (1650 – 3300)/550`
  `= −3`

 

`:. P(x < 1650)` `= P(z < −3)`
  `= 0.3/2`
  `= 0.15\ text(%)`

`=>B`
 

`text(Part 3)`

`ztext(-score)\ (2200) = (2200 – 3300)/550 = −2`

`ztext(-score)\ (3850) = (3850 – 3300)/550 = 1`
 


 

`text(Percentage)` `= (47.5 + 34)text(%) xx 600`
  `= 81.5text(%) xx 600`
  `= 48text(%)`

`=>\ E`

Probability, MET2-NHT 2019 VCAA 19 MC

A random sample of computer users was surveyed about whether the users had played a particular computer game. An approximate 95% confidence interval for the proportion of computer users who had played this game was calculated from this random sample to be (0.6668, 0.8147).

The number of computer users in the sample is closest to

  1.      5
  2.    33
  3.  135
  4.  150
  5.  180
Show Answers Only

`C`

Show Worked Solution

`95% text(C.I.) \ => \ z = 1.96`

`overset^p = (0.6668 + 0.8147)/(2) = 0.74075`

`text(S) text(olve for) \ \n\ \ text{(by CAS):}`

`2 xx 1.96 xx sqrt((0.74075(1 – 0.74075))/(n)) = \ 0.8147 – 0.6668`

`n ≈ 134.9`
 
`=> \ C`

Calculus, MET2-NHT 2019 VCAA 18 MC

Part of the graph of the function `f`, where  `f(x) = 8 - 2^(x-1)`, is shown below. It intersects the axes at the points `A` and `B`. The line segment joining `A` and `B` is also shown on the graph.
 


 

The area of the shaded region is

  1.  `16 - (15)/(log_e (2))`
  2.  `17 - (15)/(2log_e (2))`
  3.  `(7)/(log_e (2)) - (159)/(16)`
  4.  `16 - (15)/(2log_e (2))`
  5.  `(17)/(2log_e (2)) - 15`
Show Answers Only

`B`

Show Worked Solution

`text(When) \ \ x = 0 \ => \ y = 8 – (1)/(2) = 15/2`

`text(When) \ \ y = 0,\ => \ 8 – 2^(x-1)=0 \ => x=4`

`text(Area)_(AOB)` `= (1)/(2) xx 4 xx (15)/(2)`
  `= 15 \ text(u²)`

 

`text(Shaded Area)` `= int_0^4 8 – 2^(x -1)\ dx – (15)/(2)`
  `= 17 – (15)/(2 log_e 2)`

 
`=> \ B`

Functions, MET2-NHT 2019 VCAA 17 MC

The graph of the function `g` is obtained from the graph of the function `f` with rule  `f(x) = cos(x) - (3)/(8)`  by a dilation of factor  `(4)/(pi)`  from the `y`-axis, a dilation of factor  `(4)/(3)`  from the `x`-axis, a reflection in the  `y`-axis and a translation of  `(3)/(2)`  units in the positive `y` direction, in that order.

The range and period of `g` are respectively

  1. `[–(1)/(3) , (7)/(3)] \ text(and) \ 2`
  2. `[–(1)/(3) , (7)/(3)] \ text(and) \ 8`
  3. `[–(7)/(3) , (1)/(3)] \ text(and) \ 2`
  4. `[–(7)/(3) , (1)/(3)] \ text(and) \ 8`
  5. `[–(4)/(3) , (4)/(3)] \ text(and) \ (pi^2)/(2)`
Show Answers Only

`B`

Show Worked Solution

`text(Dilation of) \ \ (4)/(pi) \ \ text(from) \ y text(-axis:)`

`cos x -(3)/(8) \ => \ cos ((pi x)/(4)) – (3)/(8)`
 
`text(Dilation of) \ \ (4)/(3) \ \ text(from) \ x text(-axis:)`

`cos((pi x)/(4)) – (3)/(8) \ => \ (4)/(3) (cos({pi x}/{4}) -(3)/(8)) = (4)/(3) \ cos((pi x)/(4)) – (1)/(2)`
 
`text(Reflection in) \ y text(-axis and translation up) \ (3)/(2) :`

`(4)/(3) cos((pi x)/(4)) – (1)/(2) \ => \ (4)/(3) \ cos((–pi x)/(4)) + 1`

`:. y = (4)/(3) \ cos ((–pi x)/(4)) + 1`
 
`text(Range:) \ [1 – (4)/(3) , 1 + (4)/(3)] = [–(1)/(3) , (7)/(3)]`
 
`text(Period:) \ (2 pi)/(n) = (pi)/(4) \ => \ n = 8`

`=> \ B`

Graphs, MET2-NHT 2019 VCAA 16 MC

Parts of the graphs of  `y = f(x)`  and  `y = g(x)`  are shown below.
 


 

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

A.     B.  
     
C.     D.  
     
E.        
       
Show Answers Only

`E`

Show Worked Solution

`text(Possible graphs are:)`
 
`g(x) = -(1)/(x)`

`f(x) = (x – 1)^2 + 1`
 
`text{Graph (by CAS):}`

`g(f(x)) = -(1)/((x – 1)^2 + 1)`
 
`=>  E`

Calculus, MET2-NHT 2019 VCAA 15 MC

The area bounded by the graph of  `y = f(x)`, the line  `x = 2`, the line  `x = 8`  and the `x`-axis, as shaded in the diagram below, is  `3log_e(13)`

The value of  `int_4^10 3 f(x - 2)\ dx`  is

  1.  `3log_e(13)`
  2.  `9log_e(13)`
  3.  `6log_e(39)`
  4.  `log_e(13)`
  5.  `9log_e(11)`
Show Answers Only

`B`

Show Worked Solution

`A_1 = int_2^8 f(x)\ dx = 3 log_e 13`

`f(x – 2) = f(x) \ text(shifted 2 units to right.)`

`:. \ int_2^8 f(x)\ dx = int_4^10 f(x – 2)\ dx`

`:. \ 3 int_4^10 f(x – 2)\ dx` `= 3 xx 3log_e 13`
  `= 9 log_e 13`

 
`=> \ B`

Graphs, MET2-NHT 2019 VCAA 12 MC

The transformation  `T : R^2 → R^2`,  which maps the graph of  `y = -sqrt(2x + 1)-3`  onto the graph of  `y = sqrtx`, has rule

  1. `T ([(x),(y)]) = [((1)/(2) \ \ \ \ \ 0),(\ \ 0\ \ \ \–1)] [(x),(y)] + [(–1),(–3)]`
  2. `T ([(x),(y)]) = [((1)/(2) \ \ \ \ \ 0),(\ \ 0\ \ \ \–1)] [(x),(y)] + [(–1),(3)]`
  3. `T ([(x),(y)]) = [((1)/(2) \ \ \ \ \ 0),(\ \ 0 \ \ \ \ –1)] [(x),(y)] + [(1),(–3)]`
  4. `T ([(x),(y)]) = [(2 \ \ \ \ \ 0),(\ \ 0 \ \ \ –1)] [(x),(y)] + [(1),(–3)]`
  5. `T ([(x),(y)]) = [(2 \ \ \ \ \ 0),(\ \ 0 \ \ \ –1)] [(x),(y)] + [(–1),(3)]`
Show Answers Only

`D`

Show Worked Solution
`y` `= -sqrt(2x + 1)-3`
`y + 3` `= -sqrt(2x + 1)`
`-y-3` `= -sqrt(2x + 1)`

 
`text(S) text(ince) \ \ y^{′} = sqrt(x^{′})`

`=> x^{′} = 2x + 1 \ , \ \ y^{′} = -y-3` 
 

`[(x^{′}),(y^{′})] = [(2 \ \ \ 0),(0 \ \–1)] [(x),(y)] + [(1),(–3)]` 
 

`=> D`

Algebra, MET2-NHT 2019 VCAA 11 MC

The function  `f : D → R, \ f(x) = 5x^3 + 10x^2 + 1`  will have an inverse function for

  1.  `D = R`
  2.  `D = (–2, ∞)`
  3.  `D = (–∞ , (1)/(2)]`
  4.  `D = (–∞ , –1]`
  5.  `D = [0 , ∞)`
Show Answers Only

`E`

Show Worked Solution

`text(Graph:) \ \ y = 5x^3 + 10x^2 + 1 \ \ text{(by CAS)}`
 


 

`text(Max when) \ \ x = -(4)/(3)`

`text(Min when) \ \ x = 0`

`=> \ E`

Algebra, MET2-NHT 2019 VCAA 7 MC

If  `m = int_1^3 (2)/(x)\ dx`, then the value of  `e^m`  is

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

`D`

Show Worked Solution
`int_1^3 (2)/(x)\ dx` `= [2 log_e x]_1^3`
`m` `= 2 log_e 3 – 2 log_e 1`
`m` `= 2 log_e 3`
`e^(2 log_e 3)` `= e^(log_e 9)`
  `= 9`

 
`=>D`

Combinatorics, EXT1 A1 2019 MET1 8

A fair standard die is rolled 50 times. Let `W` be a random variable with binomial distribution that represents the number of times the face with a six on it appears uppermost.

  1. Write down the expression for  `P(W = k)`, where  `k in {0, 1, 2, …, 50}`.  (1 mark)

  2. Show that  `(P(W = k + 1))/(P(W = k)) = (50 - k)/(5(k + 1))`.  (2 marks)

Show Answers Only
  1. `\ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50 – k)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

a.  `P(W = k) = \ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50 – k)`

 

b.   `(P(W = k + 1))/(P(W = k))` `= (\ ^50C_(k+1) ⋅ (1/6)^(k+1) ⋅ (5/6)^(49 – k))/(\ ^ 50C_k ⋅ (1/6)^k ⋅ (5/6)^(50-k))`
    `= ((50!)/((49 – k)!(k + 1)!) ⋅ (1/6))/((50!)/((50 – k)! k!) ⋅ (5/6))`
    `= ((50 – k)!k!)/(5(49 – k)!(k + 1)!)`
    `= (50 – k)/(5(k + 1))`

Statistics, EXT1 S1 2019 MET1 6

Jacinta tosses a coin five times.

  1. Assuming that the coin is fair and given that Jacinta observes a head on the first two tosses, find the probability that she observes a total of either four or five heads.  (2 marks)

  2. Albin suspects that a coin is not actually a fair coin and he tosses it 18 times.

     

    Albin observes a total of 12 heads from the 18 tosses.

  3. Let  `X` = probability of obtaining a head.
  4. Find the range of `X` in which 95% of observations are expected to lie within.  (2 marks)

Show Answers Only
  1.  `1/2`
  2.  `(4/9, 8/9)`
Show Worked Solution

a.   `text(After 2 tosses, 2 heads.)`

`Ptext{(4 or 5}\ H)` `= HHT + HTH + THH + HHH`
  `= (1/2)^3 xx 4`
  `= 1/2`

 

b.    `E(hat p) = p=12/18 = 2/3`

`text(Var)(hatp) = (2/3(1-2/3))/18=1/81`

`sigma(hatp)=1/9`

`text(95% interval)\  (z= +-2)`

`=(hatp-z xx sigma(hatp), \ hatp+z xx sigma(hatp))`

`= (2/3 – 2 xx 1/9, 2/3 + 2 xx 1/9)`

`= (4/9, 8/9)`