SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Vectors, SPEC2 2020 VCAA 13 MC

The vectors  `underset~a = underset~i + 2underset~j - underset~k, \ underset~b = lambdaunderset~i + 3underset~j + 2underset~k`  and  `underset~c = underset~i + underset~k`  will be linearly dependent when the value of `lambda` is

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

`E`

Show Worked Solution

`text(Linearly dependent:)\ \ underset~b = m underset~a + n underset~c`

`m + n` `= lambda\ …\ (1)`
`2m` `= 3\ …\ (2)`
`−m + n` `= 2\ …\ (3)`

 
`m = 3/2\ \ text{(using (2))}`

`n = 2 + 3/2 = 7/2\ \ text{(using (3))}`

`:. lambda = 3/2 + 7/2 = 5`

`=>E`

Calculus, SPEC2 2020 VCAA 12 MC

If  `(dy)/(dx) = e^(cos(x))`  and  `y_0 = e`  when  `x_0 = 0`, then, using Euler's formula with step size 0.1, `y_3` is equal to

  1. `e + 0.1(1 + e^(cos(0.1)))`
  2. `e + 0.1(1 + e^(cos(0.1)) + e^(cos(0.2)))`
  3. `e + 0.1(e + e^(cos(0.1)) + e^(cos(0.2)))`
  4. `e + 0.1(e^(cos(0.1)) + e^(cos(0.2)) + e^(cos(0.3)))`
  5. `e + 0.1(e + e^(cos(0.1)) + e^(cos(0.2)) + e^(cos(0.3)))`
Show Answers Only

`C`

Show Worked Solution

`text(When)\ \ y_0 = e, x_0 = 0`

`y_1 = e + 0.1 e^(cos(0)) = e + 0.1e`

`y_2` `= e + 0.1e + 0.1e^(cos(0.1))`
  `= e + 0.1(e + e^(cos(0.1)))`
`y_3` `= e + 0.1(e + e^(cos(0.1))) + 0.1e^(cos(0.2))`
  `= e + 0.1(e + e^(cos(0.1)) + e^(cos(0.2)))`

 
`=>C`

Calculus, EXT2 C2 2020 SPEC2 11 MC

With a suitable substitution  `int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) - 3 tan(x) + 1)\ dx`  can be expressed as

  1. `int_1^(1/sqrt3) (1/(u - 1) - 1/(u - 2))\ du`
  2. `int_1^(sqrt3) (1/(u - 2) - 1/(u - 1))\ du`
  3. `int_1^(sqrt3) (1/(u - 1) - 1/(u - 2))\ du`
  4. `int_(pi/4)^(pi/3) (1/(3(u - 1)) - 1/(3(u + 2)))\ du`
Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ u = tan(x)`

`(du)/(dx) = sec^2 (x) \ => \ du = sec^2(x)\ dx`

`text(When)\ `   `x = pi/3,\ u = sqrt3`
    `x = pi/4,\ u = 1`

 

`int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) – 3 tan(x) + 1)\ dx`

`= int_1^(sqrt3)\ 1/(u^2 + 1 – 3u + 1)\ du`
  `= int_1^(sqrt3) 1/(u^2 – 3u + 2)\ du`
  `= int_1^(sqrt3) 1/((u – 2)(u – 1))\ du`
  `= int_1^(sqrt3) 1/(u – 2) – 1/(u – 1)\ du`

`=>B`

Calculus, SPEC2 2020 VCAA 11 MC

With a suitable substitution  `int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) - 3 tan(x) + 1)\ dx`  can be expressed as

  1. `int_1^(1/sqrt3) (1/(u - 1) - 1/(u - 2))\ du`
  2. `int_1^(sqrt3) (1/(3(u - 3)) - 1/(3u))\ du`
  3. `int_1^(sqrt3) (1/(u - 2) - 1/(u - 1))\ du`
  4. `int_1^(sqrt3) (1/(u - 1) - 1/(u - 2))\ du`
  5. `int_(pi/4)^(pi/3) (1/(3(u - 1)) - 1/(3(u + 2)))\ du`
Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ u = tan(x)`

`(du)/(dx) = sec^2 (x) \ => \ du = sec^2(x)\ dx`

`text(When)\ `   `x = pi/3,\ u = sqrt3`
    `x = pi/4,\ u = 1`

 

`int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) – 3 tan(x) + 1)\ dx`

`= int_1^(sqrt3)\ 1/(u^2 + 1 – 3u + 1)\ du`
  `= int_1^(sqrt3) 1/(u^2 – 3u + 2)\ du`
  `= int_1^(sqrt3) 1/((u – 2)(u – 1))\ du`
  `= int_1^(sqrt3) 1/(u – 2) – 1/(u – 1)\ du`

`=>C`

Calculus, EXT1 C3 2020 SPEC2 10

A tank initially contains 300 grams of salt that is dissolved in 50 L of water. A solution containing 15 grams of salt per litre of water is poured into the tank at a rate of 2 L per minute and the mixture in the tank is kept well stirred. At the same time, 5 L of the mixture flows out of the tank per minute.

Find the differential equation,  `(dm)/(dt)`, where  `m` represents the mass, in grams, of salt in the tank at time `t` minutes, for a non-zero volume of mixture.   (2 marks)

Show Answers Only

`(dm)/(dt) = 30 – (5m)/(50 – 3t)`

Show Worked Solution

`V(t) = 50 + 2t – 5t = 50 – 3t`

`(dm)/(dt)\ text(in) = 15 xx 2 = 30\ text(g/min)`

`(dm)/(dt)\ text(out) = 5 xx m/(50 – 3t) = (5m)/(50 – 3t)`

`:. (dm)/(dt) = 30 – (5m)/(50 – 3t)`

Calculus, SPEC2 2020 VCAA 10 MC

A tank initially contains 300 grams of salt that is dissolved in 50 L of water. A solution containing 15 grams of salt per litre of water is poured into the tank at a rate of 2 L per minute and the mixture in the tank is kept well stirred. At the same time, 5 L of the mixture flows out of the tank per minute.

A differential equation representing the mass, `m` grams, of salt in the tank at time `t` minutes, for a non-zero volume of mixture is

  1. `(dm)/(dt) = 0`
  2. `(dm)/(dt) = −(5m)/(50 - 5t)`
  3. `(dm)/(dt) = 30 - m/10`
  4. `(dm)/(dt) = 30 - (5m)/(50 - 3t)`
  5. `(dm)/(dt) = 30 - (5m)/(50 - 5t)`
Show Answers Only

`D`

Show Worked Solution

`V(t) = 50 + 2t – 5t = 50 – 3t`

`(dm)/(dt)\ text(in) = 15 xx 2 = 30\ text(g/min)`

`(dm)/(dt)\ text(out) = 5 xx m/(50 – 3t) = (5m)/(50 – 3t)`

`:. (dm)/(dt) = 30 – (5m)/(50 – 3t)`
 

`=> D`

Complex Numbers, EXT2 N2 2020 SPEC2 5 MC

Given the complex number  `z = a + bi`, where  `a ∈ R text{\}{0}`  and  `b ∈ R, \ (4zbarz)/((z + barz)^2)` is equivalent to

  1. `1 + ((text(Im)(z))/(text(Re)(z)))^2`
  2. `4[text(Re)(z) xx text(Im)(z)]`
  3. `4([text(Re)(z)]^2 + [text(Im)(z)]^2)`
  4. `(2 xx text(Im)(z))/([text(Re)(z)]^2)`
Show Answers Only

`A`

Show Worked Solution

`z = a + ib, \ barz = a – ib`

`(4zbarz)/((z + barz)^2)` `= (4(a^2 + b^2))/(4a^2)`
  `= 1 + (b/a)^2`
  `= 1 + ((text(Im)(z))/(text(Re)(z)))^2`

`=>A`

Complex Numbers, EXT2 N2 2014 SPEC2 7 MC

The sum of the roots of  `z^3 - 5z^2 + 11z - 7 = 0`,  where  `z ∈ C`,  is

  1. `1 + 2sqrt3i`
  2. `4 - 2sqrt3i`
  3. `2sqrt3i`
  4. `5`
Show Answers Only

`D`

Show Worked Solution

`text(Coefficients are real)`

`=>\ text(Conjugate root theorem applies.)`

`z_1=alpha + beta i,\ \ z_2=alpha – beta i, \ \ z_3=gamma`

`z_1+z_2+z_3` `= alpha + beta i + alpha – beta i + gamma`
  `= 2 alpha + gamma\ \ \ (∈R)`

 

`=> D`

Complex Numbers, SPEC2 2020 VCAA 6 MC

For the complex polynomial  `P(z) = z^3 + az^2 + bz + c`  with real coefficients  `a, b` and `c, P(−2) = 0`  and  `P(3i) = 0`.

The values of  `a, b` and `c`  are respectively

  1. `− 2, 9,− 18`
  2. `3, 4, 12`
  3. `2, 9, 18`
  4. `−3, −4, 12`
  5. `2, −9, −18`
Show Answers Only

`C`

Show Worked Solution

`text(S)text(ince coefficients are real,)`

`P(3i) = 0 => P(−3i) = 0\ \ (text(conjugate root theorem))`

`P(z)` `= (z + 2)(z – 3i)(z + 3i)`
  `= z^3 + 2z^2 + 9z + 18`

 
`:. a = 2, b = 9, c = 18`

`=> C`

Complex Numbers, SPEC2 2020 VCAA 5 MC

Given the complex number \(z=a+b i\), where \(a \in R \backslash\{0\}\) and \(b \in R, \dfrac{4 z \bar{z}}{(z+\bar{z})^2}\) is equivalent to

  1. \(1+\left(\dfrac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right)^2\)
  2. \(4\bigl[\operatorname{Re}(z) \times \operatorname{Im}(z)\bigr]\)
  3. \(4\Bigl([\operatorname{Re}(z)]^2+[\operatorname{Im}(z)]^2\Bigr)\)
  4. \(4\Bigl[1+(\operatorname{Re}(z)+\operatorname{Im}(z))^2\Bigr]\)
  5. \(\dfrac{2 \times \operatorname{Im}(z)}{[\operatorname{Re}(z)]^2}\)
Show Answers Only

\(A\)

Show Worked Solution

\(z=a+i b, \quad \bar{z}=a-i b\)

\(\dfrac{4 z \bar{z}}{(z+\bar{z})^2}\) \(=\dfrac{4\left(a^2+b^2\right)}{4 a^2}\)
  \(=1+\left(\dfrac{b}{a}\right)^2\)
  \(=1+\left(\dfrac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right)^2\)

\(\Rightarrow A\)

Calculus, SPEC2 2020 VCAA 3 MC

A train is travelling from Station A to Station B. The train starts from rest at Station A and travels with constant acceleration for 30 seconds until it reaches a velocity of 10 ms−1. It then travels at this velocity for 200 seconds. The train then slows down, with constant acceleration, and stops at Station B having travelled for 260 seconds in total. Let `v` ms−1 be the velocity of the train at time `t` seconds. The velocity `v` as a function of `t` is given by

  1. `v(t) = {(1/3 t, ),(10, ),(1/3 (260 - t), ):}{:(0 <= t <= 30),(30 < t <= 230),(230 < t <= 260):}`
  2. `v(t) = {(1/3 t, ),(10, ),(1/3 (230 - t), ):}{:(0 <= t <= 30),(30 < t <= 230),(230 < t <= 260):}`
  3. `v(t) = {(3t, ),(10, ),(3 (230 - t), ):}{:(0 <= t <= 30),(30 < t <= 230),(230 < t <= 260):}`
  4. `v(t) = {(3t, ),(10, ),(3 (260 - t), ):}{:(0 <= t <= 30),(30 < t <= 230),(230 < t <= 260):}`
  5. `v(t) = {(1/3 t, ),(10, ),(1/3 (230 - t), ):}{:(0 <= t <= 30),(30 < t <= 200),(200 < t <= 230):}`
Show Answers Only

`A`

Show Worked Solution

`text(When)\ \ t = 30, v(t) = 10\ \ => \ text(Eliminate C and D)`

`v(t) = 10\ \ text(for)\ \ 30 <= t <= 230\ \ =>\ text(Eliminate E)`
 

`text(When)\ \ 230 <= t <= 260:`

`(v – 0)/(t – 260)` `= (10 – 0)/(230 – 260)`
`v` `= 1/3(260 – t)`

`=>A`

Measurement, STD2 M6 FUR1 2011 2


 

The point `Q` on building `B` is visible from the point `P` on building `A`, as shown in the diagram above.

Building `A` is 16 metres taller than building `B`.

The horizontal distance between point `P` and point `Q` is 23 metres.

Calculate the angle of depression of point `Q` from point `P`. Give your answer correct to 1 decimal place.   (2 marks)

Show Answers Only

`34.8^@`

Show Worked Solution

`tan theta` `= 16/23`
`theta` `= 34.8^@`

 

`:.\ text(Angle of depression from)\ P\ text(to)\ Q = 34.8^@`

`text{(alternate,}\ PS\ text(||)\ RQ text{)}`

GRAPHS, FUR1 2020 VCAA 4 MC

A straight line is graphed below.

An equation for this straight line is

  1. `10x + 8y = 40`
  2. `4x + 5y = 40`
  3. `4x - 5y = 40`
  4. `5x + 4y = 40`
  5. `8x + 10y = 40`
Show Answers Only

`B`

Show Worked Solution

`text(Gradient) = -8/10 = -4/5`

`text(Passes through) quad (10, 0)`

`y – y_1` `= m(x – x_1)`
`y` `= -4/5(x – 10)`
`5y` `= -4x + 40`
`4x + 5y` `= 40`

 
`=>  B`

GEOMETRY, FUR1 2020 VCAA 8 MC

A cake is in the shape of a rectangular prism, as shown in the diagram below.
 

The cake is cut in half to create two equal portions.

The cut is made along the diagonal, as represented by the dotted line.

The total surface area, in square centimetres, of one portion of the cake is

  1. 132
  2. 180
  3. 192
  4. 212
  5. 264
Show Answers Only

`C`

Show Worked Solution

`x = sqrt(6^2 + 8^2) = 10`

`:. SA` `= 2 xx (1/2 xx 6 xx 8)  + 6 xx 6 + 6 xx 10 + 6 xx 8`
  `= 192\ text(cm²)`

 
`=>  C`

GEOMETRY, FUR1 2020 VCAA 7 MC

Cape Town has latitude 34° S and longitude 18° E.

Stockholm has latitude 59° N and longitude 18° E.

Assume that the radius of Earth is 6400 km.

The shortest distance along the meridian between Cape Town and Stockholm, in kilometres, is closest to

  1.   2011
  2.   3798
  3.   4021
  4.   6590
  5. 10 388
Show Answers Only

`E`

Show Worked Solution


 

`text(Distance)` `= {(59 + 34)}/360 xx 2pi xx 6400`
  `= 10\ 388.19…\ text(km)`

 
`=>  E`

GEOMETRY, FUR1 2020 VCAA 6 MC

The cities of Lhasa (30° N, 91° E), Chengdu (31° N, 104° E) and Shanghai (31° N, 121° E) are all in the same time zone.

Assume that 15° of longitude equates to a one-hour time difference.

Which one of the following statements is true?

  1. The sun rises at the same time in all three cities.
  2. The sun rises in Lhasa approximately one hour before it rises in Chengdu.
  3. The sun rises in Lhasa approximately two hours after it rises in Shanghai.
  4. Sunrise in each of the three cities is approximately 15 minutes apart.
  5. The sun rises in Shanghai and Chengdu at the same time but it rises earlier in Lhasa.
Show Answers Only

`C`

Show Worked Solution

`text(Consider longitudinal differences:)`

`text(Lhasa to Chengdu) = 104 – 91 = 13^@`

`→ text(Time difference) = 13 xx 4 = 42\ text(mins)`

`text(Lhasa to Shanghai) = 121 – 91 = 30^@`

`→ text(Time difference) = 30 xx 4 = 120\ text(mins)`

`:.\ text(Sun rises in Lhasa approx 2 hours after Shanghai)`

`=>  C`

GEOMETRY, FUR1 2020 VCAA 5 MC

The pie chart below displays the results of a survey.
 


 

Eighty per cent of the people surveyed selected ‘agree’.

Twenty per cent of the people surveyed selected ‘disagree’.

The radius of the pie chart is 16 mm.

The area of the sector representing ‘agree’, in square millimetres, is closest to

  1.   80
  2. 161
  3. 483
  4. 643
  5. 804
Show Answers Only

`D`

Show Worked Solution
`text(Area)` `= 0.8 xx pi xx 16^2`
  `= 643.39…\ text(mm²)`

 
`=>  D`

GEOMETRY, FUR1 2020 VCAA 4 MC

A regular hexagon is divided into three equal parts, as shown in the diagram below.
 


 

The size of the angle  `theta`  is

  1.   45°
  2.   60°
  3. 100°
  4. 120°
  5. 180°
Show Answers Only

`B`

Show Worked Solution

`text(3 parts are identical rhombuses.)`

`text(Angle at centre) = 360/3 = 120^@`

`:. theta^@` `= 180 -120\ \ \ text{(cointerior angles)}`
  `= 60^@`

  
`=>  B`

GEOMETRY, FUR1 2020 VCAA 1 MC

Each year begins on 1 January.

The location that begins each year first is

  1. Kathmandu (28° N, 85° E).
  2. Cairo (30° N, 31° E).
  3. Kabul (34° N, 69° E).
  4. Nairobi (1° S, 37° E).
  5. Port Moresby (9° S, 147° E).
Show Answers Only

`E`

Show Worked Solution

`text(Location that is first has the highest “East”)`

`text{longitude (can’t be greater than 180°).}`

`=>  E`

NETWORKS, FUR1 2020 VCAA 8 MC

The adjacency matrix below shows the number of pathway connections between four landmarks: `J, K, L and M`.
 

`{:(qquadqquad J\ \ \ Kquad L\ \ M),({:(J),(K),(L),(M):}[(1,3,0,2),(3,0,1,2),(0,1,0,2),(2,2,2,0)]):}`
 

A network of pathways that could be represented by the adjacency matrix is

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

`E`

Show Worked Solution

`text(Matrix requires:)`

`text(3 paths directly between)\ J and K`

`:.\ text(Eliminate)\ A and D`

`text(2 paths directly between)\ M and J`

`:.\ text(Eliminate)\ C`

`text(1 path directly between)\ J and J`

`:.\ text(Eliminate)\ B`

`=>  E`

Networks, STD2 N2 2020 FUR1 5 MC

The network below shows the distances, in metres, between camp sites at a camping ground that has electricity.

The vertices `A` to `I` represent the camp sites.
 


 

The minimum length of cable required to connect all the camp sites is 53 m.

The value of `x`, in metres, is at least

  1.  6
  2.  8
  3.  9
  4. 11
Show Answers Only

`C`

Show Worked Solution

`text(One strategy – Using Prim’s Algorithm)`

`text(Starting at)\ A`

`text(1st edge) : AH = 6`

`text(2nd edge) : HG = 5`

`text(then …)\ AB = 7,\ GI = 9,\ IE = 6,\ EF = 5`

`DE = 8,\ CD = 7`
 

`text {Total length = 53 m (not including}\ x text{)}`

`text(If)\ \ x < 9, x\ text(would replace)\ GI\ text(and minimum`

`text(length would be less than 53m.)`

`=>  C`

Networks, STD2 N3 FUR1 2020 6

The activity network below shows the sequence of activities required to complete a project.

The number next to each activity in the network is the time it takes to complete that activity, in days.

What is the critical path and minimum completion time for this project.   (2 marks)

Show Answers Only

`BEIJM = 22\ text(days)`

Show Worked Solution

`text(Scanning forward:)`
 

`text(Critical path)` `= BEIJM`
  `= 7 + 4 + 3 + 3 + 5`
  `= 22\ text(days)`

NETWORKS, FUR1 2020 VCAA 6 MC

The activity network below shows the sequence of activities required to complete a project.

The number next to each activity in the network is the time it takes to complete that activity, in days.

 

The minimum completion time for this project, in days, is

  1. 18
  2. 19
  3. 20
  4. 21
  5. 22
Show Answers Only

`E`

Show Worked Solution

`text(Scanning forward:)`
 

`text(Critical path)` `= BEIJM`
  `= 7 + 4 + 3 + 3 + 5`
  `= 22\ text(days)`

 
`=>  E`

NETWORKS, FUR1 2020 VCAA 5 MC

The network below shows the distances, in metres, between camp sites at a camping ground that has electricity.

The vertices `A` to `I` represent the camp sites.
 


 

The minimum length of cable required to connect all the camp sites is 53 m.

The value of `x`, in metres, is at least

  1.  5
  2.  6
  3.  8
  4.  9
  5. 11
Show Answers Only

`D`

Show Worked Solution

`text(One strategy – Using Prim’s Algorith)`

`text(Starting at)\ A`

`text(1st edge) : AH = 6`

`text(2nd edge) : HG = 5`

`text(then …)\ AB = 7,\ GI = 9,\ IE = 6,\ EF = 5`

`DE = 8,\ CD = 7`
 

`text {Total length = 53 m (not including}\ x text{)}`

`text(If)\ \ x < 9, x\ text(would replace)\ GI\ text(and minimum`

`text(length would be less than 53m.)`

`=>  D`

MATRICES, FUR1 2020 VCAA 9 MC

Five competitors, Andy (A), Brie (B), Cleo (C), Della (D) and Eddie (E), participate in a darts tournament.

Each competitor plays each of the other competitors once only, and each match results in a winner and a loser.

The matrix below shows the results of this darts tournament.

There are still two matches that need to be played.
 

`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquad loser),(quadqquadqquadqquadqquadqquad \ A qquad\ B qquad \ C qquad\ D qquad E),(wi\n\n\er qquad{:(A),(B),(C),(D),(E):}[(0,…,0,1,0),(…,0,1,0,1),(1,0,0,…,1),(0,1,…,0,0),(1,0,0,1,0)]):}`
 

A ‘1’ in the matrix shows that the competitor named in that row defeated the competitor named in that column.

For example, the ‘1’ in row 2, column 3 shows that Brie defeated Cleo.

A ‘…’ in the matrix shows that the competitor named in that row has not yet played the competitor named in that column.

The winner of this darts tournament is the competitor with the highest sum of their one-step and two-step dominances.

Which player, by winning their remaining match, will ensure that they are ranked first by the sum of their one-step and two-step dominances?

  1. Andie
  2. Brie
  3. Cleo
  4. Della
  5. Eddie
Show Answers Only

`B`

Show Worked Solution

`text(Case 1): e_12 = 1,\ e_21 = 0,\ e_34 = 1,\ e_43 = 0`
 

`[(0,1,0,1,0),(0,0,1,0,1),(1,0,0,1,1),(0,1,0,0,0),(1,0,0,1,0)]+[(0,1,0,1,0),(0,0,1,0,1),(1,0,0,1,1),(0,1,0,0,0),(1,0,0,1,0)]^2=[(0,2,1,1,1),(2,0,1,2,2),(2,2,0,3,1),(0,1,1,0,1),(1,2,0,2,0)]{:(5),(7),(8),(3),(5):}`
 

`text(Case 2): e_12 = 0,\ e_21 = 1,\ e_34 = 1,\ e_43 = 0`
 

`[(0,0,0,1,0),(1,0,1,0,1),(1,0,0,1,1),(0,1,0,0,0),(1,0,0,1,0)]+[(0,0,0,1,0),(1,0,1,0,1),(1,0,0,1,1),(0,1,0,0,0),(1,0,0,1,0)]^2=[(0,1,0,1,0),(3,0,1,3,2),(2,1,0,3,1),(1,1,1,0,1),(1,1,0,2,0)]{:(2),(9),(7),(4),(4):}`
 

`text(Similarly, totals can be calculated for:)`

`text(Case 3):\ e_34 = 0, e_43 = 1, and e_12=0, e_21=1`

`text(Case 4):\ e_34 = 0, e_43 = 1, and e_12=1, e_21=0`

`text(Only Brie can ensure she is top ranked by winning.)`

`=>  B`

MATRICES, FUR1 2020 VCAA 8 MC

The table below shows information about three matrices: `A, B`  and  `C`.

 

  `qquad qquad text(Matrix) qquad qquad` `quad qquad qquad text(Order) quad qquad qquad`
  `A` `2 xx 4`
  `B` `2 xx 3`
  `C` `3 xx 4`

 

The transpose of matrix `A`, for example, is written as `A^T`.

What is the order of the product  `C^T xx (A^T xx B)^T`?

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

`E`

Show Worked Solution

`C^T xx (A^T xx B)^T`

`= (3 xx 4)^T xx ((2 xx 4)^T xx (2 xx 3))^T`

`= (4 xx 3) xx ((4 xx 2) xx (2 xx 3))^T`

`= (4 xx 3) xx (4 xx 3)^T`

`= (4 xx 3) xx (3 xx 4)`

`= 4 xx 4`

`=>  E`

MATRICES, FUR1 2020 VCAA 6 MC

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

`M`  is a  3 × 3  matrix. It is constructed using the rule  `m_(ij) = 3i + 2j`.

`M`  is

A.  

`[(5,7,9),(7,9,11),(11,13,15)]`

 

 B.  

`[(5,7,9),(8,10,12),(11,13,15)]`

 
C.  

`[(5,7,10),(8,10,13),(11,13,16)]`

 

 D.  

`[(5,8,11),(7,10,13),(9,12,15)]`

 
E.  

`[(5,8,11),(8,11,14),(11,14,17)]`

 

    
Show Answers Only

`B`

Show Worked Solution

`text(By Elimination):`

`m_13 = 3 xx 1 + 2 xx 3 = 9`

`:.\ text(Eliminate)\ C, D and E`
 

`m_21 = 3 xx 2 + 2 xx 1 = 8`

`:.\ text(Eliminate)\ A`

`=>  B`

MATRICES, FUR1 2020 VCAA 3 MC

Matrices `P` and `W` are defined below.
 

`P = [(0,0,1,0,0),(0,0,0,0,1),(0,1,0,0,0),(0,0,0,1,0),(1,0,0,0,0)] qquad qquad W = [(A),(S),(T),(O),(R)]`
 

If  `P^n xx W = [(A),(S),(T),(O),(R)]`, the value of  `n`  could be

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

`D`

Show Worked Solution

`text(Find)\ n\ text(where)\ P^n\ text(is a  5 × 5  identity matrix:)`
 

`P^4 = [(1,0,0,0,0),(0,1,0,0,0),(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1)]`

 

`=>  D`

CORE, FUR1 2020 VCAA 29 MC

The value of a van purchased for $45 000 is depreciated by `k`% per annum using the reducing balance method.

After three years of this depreciation, it is then depreciated in the fourth year under the unit cost method at the rate of 15 cents per kilometre.

The value of the van after it travels 30 000 km in this fourth year is $26 166.24

The value of `k` is

  1.   9
  2. 12
  3. 14
  4. 16
  5. 18
Show Answers Only

`B`

Show Worked Solution

`text(Depreciation rate) = 1 – k/100`

`S = 45\ 000 (1 – k/100)^3 – 30\ 000 xx 0.15`

`26\ 166.24` `= 45\ 000 (1 – k/100)^3 + 4500`
`30\ 666.24` `= 45\ 000 (1 – k/100)^3`

 
`text(Testing each option:)`

`k = 12`

`=>  B`

CORE, FUR1 2020 VCAA 28 MC

The nominal interest rate for a loan is 8% per annum.

When rounded to two decimal places, the effective interest rate for this loan is not

  1. 8.33% per annum when interest is charged daily.
  2. 8.32% per annum when interest is charged weekly.
  3. 8.31% per annum when interest is charged fortnightly.
  4. 8.30% per annum when interest is charged monthly.
  5. 8.24% per annum when interest is charged quarterly.
Show Answers Only

`C`

Show Worked Solution

`text(Consider option C:)`

`r_text(effective)` `= [(1 + 8/(100 xx 26))^26 – 1] xx 100`
  `~~ 8.3154`
  `~~ 8.32%`

 
 `=>  C`

CORE, FUR1 2020 VCAA 27 MC

Gen invests $10 000 at an interest rate of 5.5% per annum, compounding annually.

After how many years will her investment first be more than double its original value?

  1. 12
  2. 13
  3. 14
  4. 15
  5. 16
Show Answers Only

`B`

Show Worked Solution

`text(Find)\ n\ text(such that:)`

`10\ 000 xx 1.055^n` `> 20\ 000`
`1.055^n` `> 2`

 
`text(Testing some answer options:)`

`1.055^12 = 1.90`

`1.055^13 = 2.005`

`=>  B`

CORE, FUR1 2020 VCAA 25 MC

The graph below represents the value of an annuity investment, `A_n`, in dollars, after `n` time periods.
 


 

A recurrence relation that could match this graphical representation is

  1. `A_0 = 200\ 000, qquad A_(n+1) = 1.015A_n - 2500`
  2. `A_0 = 200\ 000, qquad A_(n+1) = 1.025A_n - 5000`
  3. `A_0 = 200\ 000, qquad A_(n+1) = 1.03A_n - 5500`
  4. `A_0 = 200\ 000, qquad A_(n+1) = 1.04A_n - 6000`
  5. `A_0 = 200\ 000, qquad A_(n+1) = 1.05A_n - 8000`
Show Answers Only

`B`

Show Worked Solution

`text(The value doesn’t change.)`

`text(Consider option B:)`

`A_1` `= 1.025 xx 200\ 000 – 5000`
  `= 200\ 000`
  `= A_0`

 
`=>  B`

CORE, FUR1 2020 VCAA 22 MC

An asset is purchased for $2480.

The value of this asset after `n` time periods, `V_n` , can be determined using the rule

`V_n = 2480 + 45n`

A recurrence relation that also models the value of this asset after `n` time periods is

  1. `V_0 = 2480, qquad V_(n + 1) = V_n + 45n`
  2. `V_n = 2480, qquad V_(n + 1) = V_n + 45n`
  3. `V_0 = 2480, qquad V_(n + 1) = V_n + 45`
  4. `V_1 = 2480, qquad V_(n + 1) = V_n + 45`
  5. `V_n = 2480, qquad V_(n + 1) = V_n + 45`
Show Answers Only

`C`

Show Worked Solution

`V_0 = 2480`

`V_(n+1)` `= 2480 + 45(n+1)`
  `= 2480 + 45n + 45`
  `= V_n + 45`

 
`=>  C`

CORE, FUR1 2020 VCAA 19-20 MC

The time series plot below displays the number of airline passengers, in thousands, each month during the period January to December 1960.
 


 

Part 1

During 1960, the median number of monthly airline passengers was closest to

  1. 461 000
  2. 465 000
  3. 471 000
  4. 573 000
  5. 621 000

 
Part 2

During the period January to May 1960, the total number of airline passengers was 2 160 000.

The five-mean smoothed number of passengers for March 1960 is

  1. 419 000
  2. 424 000
  3. 430 000
  4. 432 000
  5. 434 000
Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ D`

Show Worked Solution

Part 1

♦ Mean mark part (1) 45%.

`text(12 data points)`

`text(Median)` `= {text(6th + 7th data point)}/2`
  `~~ (460\ 000 + 460\ 000)/2`
  `~~ 460\ 000`

`=> A`
 

Part 2

`text(Five-mean smoothed number)`

`= (2\ 160\ 000)/5`

`= 432\ 000`
 

`=> D`

CORE, FUR1 2020 VCAA 17-18 MC

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


 

Part 1

The deseasonalised rainfall for May 2019 is closest to

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

 
Part 2

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

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

`text(Part 1:)\ B`

`text(Part 2:)\ C`

Show Worked Solution

Part 1

`text(Deseasonalised rainfall for May)`

`= 92.6/1.222`

`= 75.8\ text(mm)`
 

`=> B`
 

Part 2

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

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

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

`~~ (80.93 + 81.72) ÷ 2`

`~~ 81.3\ text(mm)`
 

`=> C`

CORE, FUR1 2020 VCAA 15-16 MC

Table 3 below shows the long-term mean rainfall, in millimetres, recorded at a weather station, and the associated long-term seasonal indices for each month of the year.

The long-term mean rainfall for December is missing.
 


 

Part 1

To correct the rainfall in March for seasonality, the actual rainfall should be, to the nearest per cent

  1. decreased by 26%
  2. increased by 26%
  3. decreased by 35%
  4. increased by 35%
  5. increased by 74%

 
Part 2

The long-term mean rainfall for December is closest to

  1. 64.7 mm
  2. 65.1 mm
  3. 71.3 mm
  4. 76.4 mm
  5. 82.0 mm
Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ D`

Show Worked Solution

Part 1

`text(Deseasonalised rainfall for March)`

`= 52.8/0.741`

`= 71.255`

`:.\ text(Percentage increase)` `= (71.255 – 52.8)/() xx 100`
  `~~ 35%`

`=> D`
 

Part 2

`text(Mean) = 51.9/0.728 = 71.3`

`text(Actual December)` `= 71.3 xx 1.072`
  `~~ 76.4\ text(mm)`

`=> D`

CORE, FUR1 2020 VCAA 14 MC

In a study, the association between the number of tasks completed on a test and the time allowed for the test, in hours, was found to be non-linear.

The data can be linearised using a log10 transformation applied to the variable number of tasks.

The equation of the least squares line for the transformed data is

log10 (number of tasks) = 1.160 + 0.03617 × time

This equation predicts that the number of tasks completed when the time allowed for the test is three hours is closest to

  1. 13
  2. 16
  3. 19
  4. 25
  5. 26
Show Answers Only

`C`

Show Worked Solution
`log_10 text{(number of tasks)}` `= 1.160 + 0.03617 xx 3`
  `= 1.26851`

 

`:.\ text(Number of tasks)` `= 10^(1.26851)`
  `= 18.557…`
  `~~ 19`

`=>  C`

CORE, FUR1 2020 VCAA 10-12 MC

The data in Table 2 was collected in a study of the association between the variables frequency of nightmares (low, high) and snores (no, yes).
 


 

Part 1

The variables in this study, frequency of nightmares (low, high) and snores (no, yes), are

  1. ordinal and nominal respectively.
  2. nominal and ordinal respectively.
  3. both numerical.
  4. both ordinal.
  5. both nominal.

 
Part 2

The percentage of participants in the study who did not snore is closest to

  1. 42.0%
  2. 43.5%
  3. 49.7%
  4. 52.2%
  5. 56.5%

 
Part 3

Of those people in the study who did snore, the percentage who have a high frequency of nightmares is closest to

  1.   7.5%
  2. 17.1%
  3. 47.8%
  4. 52.2%
  5. 58.0%
Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ E`

`text(Part 3:)\ B`

Show Worked Solution

Part 1

`text{frequency of nightmares (low, high) is ordinal.}`

`text{snores (no, yes) is nominal.}`

`=> A`
 

Part 2

`text(Percentage)` `= text(not snore)/text(total participants) xx 100`
  `= 91/161 xx 100`
  `= 56.5%`

`=> E`
 

Part 3

`text(Percentage)` `= text(High frequency and snore)/text(total who snore) xx 100`
  `= 12/70 xx 100`
  `= 17.1%`

`=> B`

CORE, FUR1 2020 VCAA 9 MC

The lifetime of a certain brand of light globe, in hours, is approximately normally distributed.

It is known that 16% of the light globes have a lifetime of less than 655 hours and 50% of the light globes have a lifetime that is greater than 670 hours.

The mean and the standard deviation of this normal distribution are closest to

A.     mean = 655 hours standard deviation = 10 hours
B.     mean = 655 hours standard deviation = 15 hours
C.     mean = 670 hours standard deviation = 10 hours
D.     mean = 670 hours standard deviation = 15 hours
E.     mean = 670 hours standard deviation = 20 hours
Show Answers Only

`D`

Show Worked Solution

`text(mean) = 670`

`sigma = 670 – 655 = 15`

`=>  D`

CORE, FUR1 2020 VCAA 8 MC

The wing length of a species of bird is approximately normally distributed with a mean of 61 mm and a standard deviation of 2 mm.

Using the 68–95–99.7% rule, for a random sample of 10 000 of these birds, the number of these birds with a wing length of less than 57 mm is closest to

  1.   50
  2. 160
  3. 230
  4. 250
  5. 500
Show Answers Only

`D`

Show Worked Solution

`z text(-score)\ (57) = (x – bar x)/s = (57 – 61)/2 = -2`
 


 

`:.\ text(Number less than 57)`

`= 2.5text(%) xx 10\ 000`

`= 250`

`=>  D`

CORE, FUR1 2020 VCAA 7 MC

Data relating to the following five variables was collected from insects that were caught overnight in a trap:

    • colour
    • name of species
    • number of wings
    • body length (in millimetres)
    • body weight (in milligrams)

The number of these variables that are discrete numerical variables is

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

`A`

Show Worked Solution

`text(Discrete numeral variables:)`

`text(number of wings only)`

`=>  A`

CORE, FUR1 2020 VCAA 6 MC

A percentaged segmented bar chart would be an appropriate graphical tool to display the association between month of the year (January, February, March, etc.) and the

  1. monthly average rainfall (in millimetres).
  2. monthly mean temperature (in degrees Celsius).
  3. annual median wind speed (in kilometres per hour).
  4. monthly average rainfall (below average, average, above average).
  5. annual average temperature (in degrees Celsius).
Show Answers Only

`D`

Show Worked Solution

`text(The segments could be used to show below average,)`

`text(average and above average.)`

`=>  D`

CORE, FUR1 2020 VCAA 5 MC

The histogram below shows the distribution of weight, in grams, for a sample of 20 animal species. The histogram has been plotted on a `log_10` scale.
 


 

The percentage of these animal species with a weight of less than 10 000 g is

  1. 17%
  2. 70%
  3. 75%
  4. 80%
  5. 85%
Show Answers Only

`E`

Show Worked Solution

`log_10 10\ 000 = 4`

`text(Animals)\ (x <= 4) = 3 + 2 + 12 = 17`

`:.\ text(Percentage)` `= 17/20`
  `= 85%`

`=>  E`

CORE, FUR1 2020 VCAA 1-3 MC

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

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


 

Part 1

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

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

 
Part 2

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

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

 
Part 3

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

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

`text(Part 1:)\ B`

`text(Part 2:)\ D`

`text(Part 3:)\ C`

Show Worked Solution

Part 1

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

`=> B`
 

Part 2

`Q_3 = 300`

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

`= 25text(%) xx 800`

`= 200`

`=> D`
 

Part 3

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

♦ Mean mark 50%.

`:.\ text(Positively skewed)`

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

 
`=> C`

Calculus, SPEC1 2020 VCAA 9

Consider the curve defined parametrically by

`x = arcsin (t)`

`y = log_e(1 + t) + 1/4 log_e (1-t)`

where `t in [0, 1)`.

  1. `((dy)/(dt))^2` can be written in the form  `1/(a(1 + t)^2) + 1/(b(1-t^2)) + 1/(c(1-t)^2)` where  `a, b`  and  `c`  are real numbers.
  2. Show that  `a = 1, b = – 2`  and  `c = 16`.  (2 marks)

  3. Find the arc length, `s`, of the curve from  `t = 0`  to  `t = 1/2`. Give your answer in the form  `s = log_e(m) + n log_e (p)`, where  `m, n, p in Q`.  (3 marks)

Show Answers Only

  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `log_e(3/2)-1/4 log_e(1/2)`

Show Worked Solution

a.   `y` `= log_e(1 + t) + 1/4 log_e (1-t)`
  `(dy)/(dt)` `= 1/(1 + t)-1/(4(1-t))`
  `((dy)/(dt))^2` `= 1/(1 + t)^2-2 ⋅ 1/(1 + t) ⋅ 1/(4(1-t)) + 1/(16(1-t)^2)`
    `= 1/(1 + t)^2-1/(2(1-t^2)) + 1/(16(1-t)^2)`

 
`:. a = 1, b = – 2, c = 16`

♦♦ Mean mark part (b) 31%.

 

b.   `s` `= int_0^(1/2) sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2)\ dt`
    `= int_0^(1/2) sqrt(1/(1-t^2) + 1/(1 + t^2)-1/(2(1-t^2)) + 1/(16(1-t)^2))\ dt`
    `= int_0^(1/2) sqrt(1/(1 + t^2) + 1/(2(1-t^2)) + 1/(16(1-t)^2))\ dt`
    `= int_0^(1/2) sqrt((1/(1 + t) + 1/(4(1-t)))^2)\ dt`
    `= int_0^(1/2) 1/(1 + t) + 1/(4(1-t))\ dt`
    `= [log_e(1 + t)-1/4 log_e(1-t)]_0^(1/2)`
    `= log_e(3/2)-1/4 log_e(1/2)`

Calculus, SPEC1 2020 VCAA 7

Consider the function defined by

`f(x) = {({:mx + n,:}, x < 1), ({: frac{4}{(1 + x^2)},:}, x >= 1):}`

where `m` and `n` are real numbers.

  1. Given that  `f(x)`  and  `f^{′}(x)`  are continuous over  `R`, show that  `m = -2`  and  `n = 4`.  (2 marks)

  2. Find the area enclosed by the graph of the function, the x-axis and the lines  `x = 0`  and  `x = sqrt 3`.  (3 marks)

Show Answers Only

  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `3 + pi/3\ text(u²)`

Show Worked Solution

a.  `f(x)\ \ text(continuous when:)`

`m(1) + n` `= 4/(1 + 12)`
`m + n` `= 2`

 
`f^{′} (x) = {({:m,:}, x < 1), ({:(-8x)/(1 + x^2),:}, x >= 1):}`
 

`f(x)\ \ text(continuous when:)`

`m = (-8(1))/(1+1^2)^2 = -2`

`text(When)\ \ m =-2,\ n = 4`

 

b.   `f(x) = {({:4-2x,:}, x < 1), ({:4/(1 + x^2),:}, x >= 1):}`

`f(x) > 0\ \ text(for)\ \ x ∈ [0, sqrt 3]`

`A` `= int_1^sqrt 3 4/(1 + x^2)\ dx + int_0^1 4-2x\ dx`
  `= [4 tan^(-1)(x)]_1^sqrt 3 + [4x-x^2]_0^1`
  `= (4 pi)/3-(4 pi)/4 + 4-1`
  `= 3 + pi/3\ text(u²)`

Calculus, EXT1 C2 2020 SPEC1 6

Let  `f(x) = tan^(-1) (3x - 6) + pi`.

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

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

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

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

 

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

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

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

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

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

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

 

c.   

Calculus, SPEC1 2020 VCAA 6

Let  `f(x) = arctan (3x - 6) + pi`.

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

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

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

Show Answers Only

  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`

Show Worked Solution

a.    `f^{\prime}(x)` `= (d/(dx) (3x – 6))/(1 + (3x – 6)^2)`
    `= 3/(9x^2 – 36x + 37)`

 

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

♦ Mean mark part (b) 42%.

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

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

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

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

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

 

c.   

Vectors, EXT2 V1 2020 SPEC1 5

Let  `underset ~ a = 2 underset ~i - 3 underset ~j + underset ~k`  and  `underset ~b = underset ~i + m underset ~j - underset ~k`, where `m` is an integer.

The projection of  `underset ~a`  onto  `underset ~b`  is  `-11/18 (underset ~i + m underset ~j - underset ~k)`.

  1. Find the value of `m`.  (3 marks)

  2. Find the component of  `underset ~a`  that is perpendicular to  `underset ~b`.  (1 mark)

Show Answers Only
  1. `m = 4`
  2. `47/18 underset ~i – 5/9 underset ~j + 7/18 underset ~k`
Show Worked Solution
a.    `underset ~b ⋅ underset ~a` `= ((1), (m), (-1))((2), (-3), (1)) = 2 – 3m – 1 = 1 – 3m`
  `|underset ~b|^2` `= 1^2 + m^2 + (-1)^2 = m^2 + 2`
`(underset ~b ⋅ underset ~a)/|underset ~b|^2 ⋅ underset ~b` `= -11/18 underset ~b`
`(1 -3m)/(m^2 + 2)` `= -11/18`
`18 – 54m` `= -11m^2 – 22`
`11m^2 – 54m + 40` `=0`  
`(11m – 10)(m – 4)` `=0`  

 
`:. m= 4\ \ (m != 10/11, \ m ∈ Z)`
 

b.    `underset ~a_(⊥ underset ~b)` `= underset ~a + 11/18 (underset ~i + 4 underset ~j – underset ~k)`
    `= 2 underset ~i + 11/18 underset ~i – 3 underset ~j + 44/18 underset ~j + underset ~k – 11/18 underset ~k`
    `= 47/18 underset ~i – 5/9 underset ~j + 7/18 underset ~k`

Vectors, SPEC1 2020 VCAA 5

Let  `underset ~ a = 2 underset ~i-3 underset ~j + underset ~k`  and  `underset ~b = underset ~i + m underset ~j-underset ~k`, where `m` is an integer.

The vector resolute of  `underset ~a`  in the direction of  `underset ~b`  is  `-11/18 (underset ~i + m underset ~j-underset ~k)`.

  1. Find the value of `m`.   (3 marks)

  2. Find the component of  `underset ~a`  that is perpendicular to  `underset ~b`.   (1 mark)

Show Answers Only

  1. `m = 4`
  2. `47/18 underset ~i-5/9 underset ~j + 7/18 underset ~k`

Show Worked Solution

a.    `underset ~b ⋅ underset ~a` `= ((1), (m), (-1))((2), (-3), (1)) = 2-3m-1 = 1-3m`
  `|underset ~b|^2` `= 1^2 + m^2 + (-1)^2 = m^2 + 2`
`(underset ~b ⋅ underset ~a)/|underset ~b|^2 ⋅ underset ~b` `= -11/18 underset ~b`
`(1 -3m)/(m^2 + 2)` `= -11/18`
`18-54m` `= -11m^2-22`
`11m^2-54m + 40` `=0`  
`(11m-10)(m-4)` `=0`  

 
`:. m= 4\ \ (m != 10/11, \ m ∈ Z)`

♦♦ Mean mark part (b) 28%.

 

b.    `underset ~a_(⊥ underset ~b)` `= underset ~a + 11/18 (underset ~i + 4 underset ~j-underset ~k)`
    `= 2 underset ~i + 11/18 underset ~i-3 underset ~j + 44/18 underset ~j + underset ~k-11/18 underset ~k`
    `= 47/18 underset ~i-5/9 underset ~j + 7/18 underset ~k`

Functions, EXT1 F1 2020 SPEC1 4

Solve the inequality  `3 - x > 1/|x - 4|`  for `x`, expressing your answer in interval notation.  (3 marks)

Show Answers Only

`x ∈ (– oo, (7 – sqrt 5)/2)`

Show Worked Solution

`3 – x > 1/|x – 4|`

`|x – 4| (3 – x) > 1`
 

`text(If)\ \ x – 4 > 0, x > 4`

`(x – 4) (3 – x)` `> 1`
`3x – x^2 – 12 + 4x` `> 1`
`-x^2 + 7x – 13` `> 0`

 
`Delta = 7^2 – 4 ⋅ 1 ⋅ 13 = -3 < 0`

`=>\ text(No Solutions)`
 

`text(If)\ \ x – 4 < 0, x < 4`

`-(x – 4) (3 – x)` `> 1`
`x^2 – 7x + 12` `> 1`
`x^2 – 7x + 11` `> 0`
`x` `= (7 +- sqrt(7^2 – 4 ⋅ 1 ⋅ 11))/2`
  `= (7 +- sqrt 5)/2`

`text(Combining solutions)`

`(x < (7 – sqrt 5)/2  ∪ x > (7 + sqrt 5)/2)  nn x < 4`

`x ∈ (– oo, (7 – sqrt 5)/2)`

Complex Numbers, EXT2 N2 2020 SPEC1 3

Find the cube roots of  `1/sqrt 2 - 1/sqrt 2 i`. Express your answers in modulus-argument form.  (3 marks)

Show Answers Only

`z_1 = text(cis)((7 pi)/12)`

`z_2 = text(cis)(-pi/12)`

`z_3 = text(cis)((-3 pi)/4)`

Show Worked Solution

`z^3 = 1/sqrt 2 – 1/sqrt 2 i = text(cis)(-pi/4) = text(cis)(-pi/4 + 2 pi k), k ∈ Z`
 

`text(By De Moivre):`

`z = text(cis)(-pi/12 + (2 pi k)/3)`

`text(If)\ \ k = 1, \ z_1 = text(cis)(-pi/12 + (2 pi)/3) = text(cis)((7 pi)/12)`

`text(If)\ \ k = 0, \ z_2 = text(cis)(-pi/12)`

`text(If)\ \ k = – 1, \ z_3= text(cis)(-pi/12 – (2pi)/3) = text(cis)((-3 pi)/4)`

Complex Numbers, SPEC1 2020 VCAA 3

Find the cube roots of  `1/sqrt 2 - 1/sqrt 2 i`. Express your answers in polar form using principal values of the argument.  (3 marks)

Show Answers Only

`z_1 = text(cis)((7 pi)/12)`

`z_2 = text(cis)(-pi/12)`

`z_3 = text(cis)((-3 pi)/4)`

Show Worked Solution

`z^3 = 1/sqrt 2 – 1/sqrt 2 i = text(cis)(-pi/4) = text(cis)(-pi/4 + 2 pi k), k ∈ Z`
 

`text(By De Moivre):`

`z = text(cis)(-pi/12 + (2 pi k)/3)`

`text(If)\ \ k = 1, \ z_1 = text(cis)(-pi/12 + (2 pi)/3) = text(cis)((7 pi)/12)`

`text(If)\ \ k = 0, \ z_2 = text(cis)(-pi/12)`

`text(If)\ \ k = – 1, \ z_3= text(cis)(-pi/12 – (2pi)/3) = text(cis)((-3 pi)/4)`

Mechanics, SPEC1 2020 VCAA 1

A 2 kg mass is initially at rest on a smooth horizontal surface. The mass is then acted on by two constant forces that cause the mass to move horizontally. One force has magnitude 10 N and acts in a direction 60° upwards from the horizontal, and the other force has magnitude 5 N and acts in a direction 30° upwards from the horizontal, as shown in the diagram below.
 


 

  1. Find the normal reaction force, in newtons, that the surface exerts on the mass.  (2 marks)
  2. Find the acceleration of the mass, in ms−2, after it begins to move.  (2 marks)
  3. Find how far the mass travels, in metres, during the first four seconds of motion.   (1 mark)
Show Answers Only
  1. `R = 2g – 5/2 – 5 sqrt 3\ text(N)`
  2. `a = 5/2 – (5 sqrt 3)/2\ text(ms)^(-2)`
  3. `20 – 10 sqrt 3\ text(m)`
Show Worked Solution
a.  

`text(Resolving forces vertically:)`

`2g` `= 5 sin 30 + 10 sin 60 + R`
`2g` `= 5/2 + 5 sqrt 3 + R`
`R` `= 2g – 5/2 – 5 sqrt 3\ text(N)`

 

b.    `2a` `= 10 cos 60 – 5 cos 30`
  `2a` `= 5 – (5 sqrt 3)/2`
  `:.a` `= 5/2 – (5 sqrt 3)/4\ text(ms)^(-2)`
Mean mark part (c) 51%.

 

c.    `text(Distance)` `= ut + 1/2 at^2`
    `= 0+ 1/2 (5/2 – (5 sqrt 3)/4) × 4^2`
    `= 20 – 10 sqrt 3\ text(m)`

Calculus, MET1 2020 VCAA 8

Part of the graph of  `y = f(x)`, where  `f:(0, ∞) -> R, \ f(x) = xlog_e(x)`, is shown below.
 


 

The graph of `f` has a minimum at the point `Q(a, f(a))`, as shown above.

  1. Find the coordinates of the point `Q`.   (2 marks)

  2. Using  `(d(x^2log_e(x)))/(dx) = 2x log_e(x) + x`, show that  `xlog_e(x)`  has an antiderivative  `(x^2log_e(x))/2-(x^2)/4`.   (1 mark)

  3. Find the area of the region that is bounded by `f`, the lines  `x = a`  and the horizontal axis for  `x ∈ [a, b]`, where `b` is the `x`-intercept of `f`.   (2 marks)

  4. Let  `g: (a, ∞) -> R, \ g(x) = f(x) + k`  for  `k ∈ R`.

     

    i. Find the value of `k` for which  `y = 2x`  is a tangent to the graph of `g`.   (1 mark)

    ii. Find all values of `k` for which the graphs of `g` and `g^(-1)` do not intersect.   (2 marks)

Show Answers Only
  1. `Q(1/e, -1/e)`
  2. `text(See Worked Solutions)`
  3. `1/4-3/(4e^2)\ text(u)^2`
  4. i.  `e`
  5. ii.  `k ∈ (1, ∞)`
Show Worked Solution

a.   `y = xlog_e x`

`(dy)/(dx)` `= x · 1/x + log_e x`
  `= 1 + log_e x`

 
`text(Find)\ x\ text(when)\ (dy)/(dx) = 0:`

`1 + log_e x` `= 0`
`log_e x` `= -1`
`x` `= 1/e`
`y` `= 1/e log_e (e^(-1))`
  `= -1/e`

 
`:. Q(1/e, -1/e)`

 

b.    `int 2x log_e(x) + x\ dx` `= x^2 log_e (x) + c`
  `2 int x log_e(x)\ dx` `= x^2 log_e (x)-intx\ dx + c`
  `:. int x log_e(x)\ dx` `= (x^2 log_e (x))/2-(x^2)/4 \ \ (c = 0)`

 

c.   

`text(When)\ \ x log_e x = 0 \ => \ x = 1`

`=> b = 1`

`:.\ text(Area)` `= −int_(1/e)^1 x log_e(x)\ dx`
  `= [(x^2)/4-(x^2 log_e(x))/2]_(1/e)^1`
  `= (1/4-0)-(1/(4e^2)-(log_e(e^(-1)))/(2e^2))`
  `= 1/4-(1/(4e^2) + 1/(2e^2))`
  `= 1/4-3/(4e^2) \ text(u)^2`

 

d.i.   `text(When)\ \ f^{prime}(x) = m_text(tang) = 2,`

`1 + log_e(x)` `= 2`
`x` `= e`

 
`text(T)text(angent meets)\ \ g(x)\ \ text(at)\ \ (e, 2e)`

`g(e)` `= f(e) + k`
`2e` `= e log_e e + k`
`:.k` `= e`

 

d.ii. `text(Find the value of)\ k\ text(when)\ \ y = x\ \ text(is a tangent to)\ g(x):`

`text(When)\ \ f^{prime}(x) = 1,`

`1 + log_e(x)` `= 1`
`x` `= 1`

 
`text(T)text(angent occurs at)\ (1, 1)`

`g(1) = f(1) + k \ => \ k = 1`
 

`:.\ text(Graphs don’t intersect for)\ k ∈ (1, ∞)`

Calculus, MET1 2020 VCAA 7

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

   
 

  1. Show the point `P` is not on the graph of  `y = f(x)`.   (1 mark)

  2. Consider a point  `Q(a, f(a))`  to be a point on the graph of `f`.

     

      i. Find the slope of the line connecting points `P` and `Q` in terms of `a`.   (1 mark)

     ii. Find the slope of the tangent to the graph of `f` at point `Q` in terms of `a`.   (1 mark)

    iii. Let the tangent to the graph of `f` at  `x = a`  pass through point `P`.

     

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

     iv. Give the equation of one of the lines passing through point `P` that is tangent to the graph of `f`.   (1 mark)

  3. Find the value of `k`, that gives the shortest possible distance between the graph of the function of  `y = f(x-k)`  and point `P`.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2.   i.  `(a^2 + 3a + 5)/(a-1)`
     ii.  `2a + 3`
    iii.  `4\ text(or)\ -2`
    iv.  `y = 11x-11`
  3. `5/2`
Show Worked Solution

a.   `f(1) = 1 + 3 + 5 = 9`

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

 

b.i.   `P(1, 0), Q(a, f(a))`

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

 

b.ii.   `f^{prime}(x) = 2x + 3`

`m_Q = f^{prime}(a) = 2a + 3`
  

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

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

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

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

 
`:. a = 4\ text(or)\ -2`
  

b.iv.   `text(When)\ \ a = -2`

`m_text(tang) = 2x-2 + 3 = -1`

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

`y-a` `=-1(x-1)`
`y` `= -x + 1`

 
`text(Similarly, if)\ \ a = 4:`

`y = 11x-11`

 

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

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

`f^{prime}(x) = 2x + 3`

`text(MIN S.P. of)\ \ f(x)\ \ text(occurs when)\ \ f^{prime}(x) = 0`

`x =-3/2`

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

`:. k = 5/2`