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Geometry, NAPX-p100504v01

Alexa was walking on Ruby Street in the direction of the arrow shown below.
 


 

In what direction is Alexa heading?

 
  North-East
 
  North-West
 
  South-East
 
  South-West

Show Answers Only

`text(South-East)`

Show Worked Solution

`text{Alexa was heading in the South-East direction.}`

Measurement, NAPX-p167365v02

Which shape has an area of exactly 12 square units?

 
   A
 
   B
 
   C
 
   D

Show Answers Only

`B`

Show Worked Solution

`text{Area of triangle} = text{b} xx text{h}`

`text{Check each option:}`

`text{A.} \ \ 5 xx 3 = 15 `

`text{B.} \ \ 2 xx 6 = 12 \ text{(Correct)}`

`text{C.} \ \ 8 xx 1 = 8`

`text{D.} \ \ 9 xx 2 = 18`

`therefore \ text{square B has an area of exactly 12 square units.}`

Measurement, NAPX-p167365v01

What shape has an area greater than 11 square units?

 
   A
 
   B
 
   C
 
   D

Show Answers Only

`A`

Show Worked Solution

`text{Area of triangle} = frac{1}{2} xx text{b} xx text{h}`

`text{Check each option:}`

`text{A.} \ frac{1}{2} xx 6 xx 4 = 12 \ text{(Correct)}`

`text{B.} \ frac{1}{2} xx 3 xx 3 = 4.5`

`text{C} \ frac{1}{2} xx 6 xx 3 = 9`

`text{D} \ frac{1}{2} xx 4 xx 4 = 8`

`therefore \ text{triangle A has an area greater than 11 square units.}`

Measurement, NAPX-p124624v04

Barnsy owned a carp that was one quarter of a metre long.

How long was Barnsy's carp in millimetres?

 
   4000
 
   2000
 
   400
 
   250
 
   50
Show Answers Only

`250`

Show Worked Solution

`1 \ text{metre} = 1000 \ text{millimetres}`

`frac{1}{4} \ text{metre}` ` = frac{1}{4} xx 1000`
  `= 250 \ text{millimetres}`

Measurement, NAPX-p116865v04

Scott expects his wine to be delivered on the 28th March.

He is told by the distributor that its delivery will be delayed by 5 days.
 

 
What day of the week will the wine now be delivered?

Monday Tuesday Wednesday Saturday
 
 
 
 
Show Answers Only

`text{Tuesday}`

Show Worked Solution

`text{28th March is a Thursday.}`

`therefore \ text{5 days later is a Tuesday.}`

Measurement, NAPX-p116865v03

Marty expects his gold mining drench to be delivered on the 28th of June.

Delays at customs mean it will be delivered 6 days later.
 


 

Which day of the week should the drench now be delivered?

Saturday Sunday Monday Tuesday
 
 
 
 
Show Answers Only

`text{Monday}`

Show Worked Solution

`text{28th June is a Tuesday.}`

`therefore \ text{6 days later is a Monday.}`

Measurement, NAPX-p116876v04

Flanno builds the shape below using identical small blocks.
 


 
What is the smallest number of blocks Flanno needs to build a cube?
  10   11  12   13
 
 
 
 
Show Answers Only

`10`

Show Worked Solution

`text{The smallest cube will be} \ 3 xx 3 xx 3`

`text{Bottom row = 1}`

`text{2nd row = 5}`

`text{Top row = 7}`

`therefore \ text{13 blocks needed}`

Measurement, NAPX-p116876v03

Ella builds the shape below using identical small blocks.
 


 
What is the smallest number of blocks Ella needs to build a cube?
  8   9   10   11
 
 
 
 
Show Answers Only

`10`

Show Worked Solution

`text{The smallest cube will be} \ 3 xx 3 xx 3`

`text{Bottom row = 2}`

`text{2nd row = 3}`

`text{Top row = 5}`

`therefore \ text{10 blocks needed}`

Algebra, MET1 2015 VCAA 7b

Solve  `3e^t = 5 + 8e^(−t)`  for `t`.  (3 marks)

Show Answers Only

`log_e(8/3)`

Show Worked Solution

`3e^t – 5 – 8e^(−t) = 0`

♦ Part (b) mean mark 44%.
MARKER’S COMMENT: Many students could not create a quadratic, and others who did made mistakes by trying to solve using the quadratic formula rather than simple factorising.

`text(Multiply both sides by)\ e^t:`

`3e^(2t) – 5e^t – 8 = 0`

`text(Let)\ \ y = e^t`

`log_2((6 – x)/(4 – x))` `= 2`
`3y^2 – 5y – 8` `= 0`
`(3y – 8)(y + 1)` `= 0`
`y` `=8/3\ quadquadquadquadquad text(or)\ \ \ \ ` `y` `=-1\ \ text{(No solution)}`
`e^t` `= 8/3`  
`:. t` `= log_e(8/3)\ \ \ `    

Mechanics, SPEC2 2020 VCAA 5

Two objects, each of mass `m` kilograms, are connected by a light inextensible strings that passes over a smooth pulley, as shown below. The object on the platform is initially at point A and, when it is released, it moves towards point C. The distance from point A to point C is 10 m. The platform has a rough surface and, when it moves along the platform, the object experiences a horizontal force opposing the motion of magnitude `F_1` newtons in the section AB and a horizontal force opposing the motion of magnitude `F_2` newtons when it moves in the section BC.
 

  1. On the diagram above, mark all forces that act on each object once the object on the platform has been released and the system is in motion.  (2 marks)

The force `F_1` is given by  `F_1 = kmg, \ k ∈ R^+`.

  1.  i. Show that an expression for the acceleration, in `text(ms)^(−2)`, of the object on the platform, in terms of `k`, as it moves from point A to point B is given by  `(g(1 - k))/2`.  (2 marks)
  2. ii. The system will only be in motion for certain values of `k`.
  3.     Find these values of `k`.  (1 mark)

Point B is midway between points A and C.

  1. Find, in terms of `k`, the time taken, is seconds, for the object on the platform to reach point B.  (2 marks)
  2. Express, in terms of `k`, the speed `v_B`, in `text(ms)^(−1)`, of the object on the platform when it reaches point B.  (2 marks)
  3. When the object on the platform is at point B, the string breaks. The velocity of the object at point B is  `v_B = 2.5\ text(ms)^(−1)`. The force that opposes motion from point B to point C is  `F_2 = 0.075 mg + 0.4 mv^2`, where `v` is the velocity of the object when it is a distance of `x` metres from point B. The object on the platform comes to rest before point C.
  4. Find the object's distance from point C when it comes to rest. Give your answer in metres, correct to two decimal places.  (4 marks)
Show Answers Only
  1.  
  2.  i. `text(See Worked Solutions)`
  3. ii. `k ∈ (0, 1), k ∈ R^+`
  4. `2sqrt(5/(g(1 – k)))`
  5. `v_text(B) = sqrt(5g(1 – k))`
  6. `3.15\ text(m)`
Show Worked Solution
a.   

 

b. i.   `text(Horizontally:)`

`ma = T – F_1 = T – kmg\ …\ (1)`

`text(Vertically:)`

`ma = mg – T\ …\ (2)`
 

`text(Add)\ \ (1) + (2) :`

`2ma = mg – kmg`

`:. a` `= (g – kg)/2`
  `= (g(1 – k))/2`
♦ Mean mark (b)(ii) 45%.

 

b. ii.   `text(System in motion when)\ a > 0`

`(g(1 – k))/2 > 0`

`:. k ∈ (0, 1), \ k ∈ R^+`

 

c.   `text(AB) = 5\ (text(given)), u = 0\ (text(given))`

`text(Find)\ t\ text(when)\ s = 5:`

`s = ut + 1/2at^2`

`5 = 0 + 1/2 · (g(1 – k))/2 · t^2`

`t^2` `= 20/(g(1 – k))`
`t` `= sqrt(20/(g(1 – k)))`
  `= 2sqrt(5/(g(1 – k)))`

 

d.   `text(At B,)\ s = 5`

`v_text(B)^2` `= u^2 + 2as`
  `= 0 + 2 · (g(1 – k))/2 · 5`
  `= 5g(1 – k)`

 
`:. v_text(B) = sqrt(5g(1 – k))`

 

e.   `text(Acceleration is against the direction of motion.)`

♦♦ Mean mark (e) 35%.
`a` `= −F/m`
  `= −0.075g – 0.4v^2`
  `= −0.4(0.1875g + v^2)`

 

`d/(dx)(1/2 v^2)` `= −0.4(0.1875g + v^2)`
`d/(dx)(v^2)` `= −0.8(0.1875g + v^2)`
`(dx)/(d(v^2))` `= −1.25(1/(0.1875g + v^2))`
`:. x` `= −1.25 int_(2.5^2)^0 1/(0.1875 + v^2)\ dv^2`
  `= 1.85\ text(m)`

 

`:.\ text(Distance from C)` `= 5 – 1.85`
  `= 3.15\ text(m)`

Complex Numbers, EXT2 N2 2020 SPEC2 2

Two complex numbers, `u` and `v`, are defined as  `u = −2 - i`  and  `v = −4 - 3i`.

  1. Express the relation  `|z - u| = |z - v|`  in the cartesian form  `y = mx + c`, where  `m, c ∈ R`.  (3 marks)

  2. Plot the points that represent `u` and `v` and the relation `|z - u| = |z - v|` on the Argand diagram below.  (2 marks)
     
         
     
  3. State a geometrical interpretation of the graph of  `|z - u| = |z - v|`  in relation to the points that represent `u` and `v`.  (1 mark)

  4.  i. Sketch the ray given by  `text(Arg)(z - u) = pi/4`  on the Argand diagram in part b.  (1 mark)

  5. ii. In Cartesian form, write down the function that describes the ray  `text(Arg)(z - u) = pi/4`.  (1 mark)

Show Answers Only
  1. `y = −x – 5`
  2.  
  3. `|z – u| = |z – v|\ text(is the perpendicular bisector of the line joining)\ u and v.`
  4.  i.
  5. ii. `f: (−2, ∞) ->, f(x) = x + 1`
Show Worked Solution

a.   `text(Let)\ \ z = x + iy`

`z – u = x + 2 + iy + i`

`z – v = x + 4 + iy + 3i`

`|z – u| = |z – v|`

`(x + 2)^2 + (y + 1)^2` `= (x + 4)^2 + (y + 3)^2`
`x^2 + 4x + 4 + y^2 + 2y + 1` `= x^2 + 8x + 16 + y^2 + 6y + 9`
`-4y` `= 4x + 20`
`y` `= −x – 5`

 

b.   

 

c.   `|z – u| = |z – v|\ text(is the graph of the perpendicular bisector of the)`

`text(line joining)\ u and v.`

 

d.i.   

 

d.ii.   `text(Arg)(z – u) = pi/4 =>\ text(gradient) = 1, ytext(-intercept at)\ (0, 1)`

`:. f: (−2, ∞) -> RR, \ f(x) = x + 1`

Complex Numbers, SPEC2 2020 VCAA 2

Two complex numbers, `u` and `v`, are defined as  `u = -2-i`  and  `v = −4-3i`.

  1. Express the relation  `|z-u| = |z-v|`  in the cartesian form  `y = mx + c`, where  `m, c ∈ R`.   (3 marks)

  2. Plot the points that represent `u` and `v` and the relation `|z-u| = |z-v|` on the Argand diagram below.   (2 marks)

     
         
     

  3. State a geometrical interpretation of the graph of  `|z-u| = |z-v|`  in relation to the points that represent `u` and `v`.   (1 mark)

  4.  i. Sketch the ray given by  `text(Arg)(z-u) = pi/4`  on the Argand diagram in part b.   (1 mark)

  5. ii. Write down the function that describes the ray  `text(Arg)(z-u) = pi/4`, giving the rule in cartesian form.   (1 mark)

  6. The points representing `u` and `v` and  `−5i`  lie on the circle given by  `|z-z_c| = r`, where `z_c` is the centre of the circle and `r` is the radius.
  7. Find `z_c` in the form  `a + ib`, where  `a, b ∈ R`, and find the radius `r`.   (3 marks)

Show Answers Only
  1. `y = −x-5`
  2.  
  3. `|z-u| = |z-v|\ text(is the perpendicular bisector of the line joining)\ u and v.`
  4.  i.
  5. ii. `f: (−2, ∞) ->, f(x) = x + 1`
  6. `z_c = −5/3-10/3 i and r = (5sqrt2)/3`
Show Worked Solution

a.   `text(Let)\ \ z = x + iy`

`z-u = x + 2 + iy + i`

`z-v = x + 4 + iy + 3i`

`|z-u| = |z-v|`

`(x + 2)^2 + (y + 1)^2` `= (x + 4)^2 + (y + 3)^2`
`x^2 + 4x + 4 + y^2 + 2y + 1` `= x^2 + 8x + 16 + y^2 + 6y + 9`
`-4y` `= 4x + 20`
`y` `= −x-5`

 

b.   

 

c.   `|z-u| = |z-v|\ text(is the graph of the perpendicular bisector of the)`

`text(line joining)\ u and v.`

 

d.i.   

♦♦ Mean mark (d.ii.) 25%.

 

d.ii.   `text(Arg)(z-u) = pi/4 =>\ text(gradient) = 1, ytext(-intercept at)\ (0, 1)`

`:. f: (−2, ∞) -> RR, \ f(x) = x + 1`

 

e.   `|z_c-u| = |z_c-v| = |z_c-(−5i)|`

♦ Mean mark (e) 40%.
`z_c-u` `= (a + 2) + (b + 1)i`
`z_c-v` `= (a + 4) + (b + 3)i`
`z_c +5i` `= a + (b + 5)i`

 
`a^2 + (b + 5)^2 = (a + 2)^2 + (b + 1)^2\ …\ (1)`

`a^2 + (b + 5)^2 = (a + 4)^2 + (b + 3)^2\ …\ (2)`

`a = −5/3, b = −10/3\ \ text{(by CAS)}`
 

`:.z_c = −5/3-10/3 i`

`:.r` `= |z_c-(−5i)|`
  `= sqrt((−5/3)^2 + (−10/3 + 5)^2)`
  `= (5sqrt2)/3`

Vectors, SPEC2 2020 VCAA 1

A particle moves in the  `x\ – y`  plane such that its position in terms of `x` and `y` metres at `t` seconds is given by the parametric equations

`x = 2sin(2t)`

`y = 3cos(t)`

where `t >= 0`

  1. Find the distance, in metres, of the particle from the origin when  `t = pi/6`.   (2 marks)

  2.   i. Express  `(dy)/(dx)`  in terms of `t` and, hence, find the equation of the tangent to the path of the particle at  `t = pi`  seconds.   (3 marks)

  3.  ii. Find the velocity, `underset ~ v`, in `text(ms)^(−1)`, of the particle when  `t = pi`.   (2 marks)

  4. iii. Find the magnitude of the acceleration, in `text(ms)^(−2)`, when  `t = pi`.   (2 marks)

  5. Find the time, in seconds, when the particle first passes through the origin.   (1 mark)

  6. Express the distance, `d` metres, travelled by the particle from  `t = 0`  to  `t = pi/6`  as a definite integral and find this distance correct to three decimal places.   (2 marks)

Show Answers Only
  1. `sqrt39/2\ text(metres)`
  2.   i. `y = −3`
  3.  ii. `4underset~i`
  4. iii. `3\ text(ms)^(−2)`
  5. `pi/2\ \ (text(1st time))`
  6. `1.804\ \ (text(to 3 d.p.))`
Show Worked Solution

a.   `text(At)\ \ t = pi/6,`

`x = 2sin\ pi/3 = sqrt3`

`y = 3cos\ pi/6 = (3sqrt3)/2`

`:.\ text(Distance)` `= sqrt((sqrt3)^2 + ((3sqrt3)/2)^2)`
  `= sqrt39/2\ text(metres)`

 

b.i.   `(dx)/(dt) = 4cos(2t),\ \ (dy)/(dt) = −3sin(t)`

`(dy)/(dx)` `= (dy)/(dt) · (dt)/(dx)`
  `= (−3sin(t))/(4cos(2t))`

 
`text(When)\ t = pi :`

`(dy)/(dx)` `= (−3sin(pi))/(4cos(2pi))=0`

 
`text(Equation of tangent where)\ \ m = 0,\ text(through)\ (0, −3):`

`y = −3`
  

b.ii.   `underset~r(t)` `= 2sin(2t)underset~i + 3cos(t)underset~j`
  `underset~v(t)` `= 4cos(2t)underset~i-3sin(t)underset~j`
  `underset~v(pi)` `= 4cos(2pi)underset~i-3sin(pi)underset~i`
    `= 4underset~i`

 

b.iii.  `underset~a(t) = −8sin(2t)underset~i-3cos(t)underset~j`

♦ Mean mark (b)(iii) 48%.
`underset~a(pi)` `= −8sin(2pi)underset~i-3cos(pi)underset~j`
  `= 3underset~j`
`|underset~a(pi)|` `= sqrt(0^2 + 3^2)`
  `= 3\ text(ms)^(−2)`

 

c.   `text(Find)\ t\ text(when)\ \ x = 2sin(2t) = 0\ \ text(and)\ \ y = 3cos(t) = 0:`

`t = pi/2\ \ (text(1st time))`

 

d.    `text(Distance)` `= int_0^(pi/6) sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2)\ dt`
    `= int_0^(pi/6) sqrt((4cos(2t))^2 + (−3sin(t))^2)\ dt`
    `~~ 1.804\ \ (text(to 3 d.p.))`

Number, NAPX-p116764v04

Pringle has 55 twenty-cent coins in his money pouch.

How much money does Pringle have?

 
    $9.50
 
   $11.00
 
   $20.50
 
   $55.00
Show Answers Only

`$11.00`

Show Worked Solution

`text(5 twenty-cent coins = $1.00)`

`:.\ text(55 twenty-cent coins)` `=11 xx $1.00`  
  `=$11.00`  

Number, NAPX-p116764v03

Sheena has 26 twenty-cent coins in her piggy bank.

How much money does Sheena have?

 
   $26.20
 
   $5.20
 
   $5.00
 
   $4.40
Show Answers Only

$5.20

Show Worked Solution

`text(5 twenty-cent coins = $1.00)`

`:.\ text(26 twenty-cent coins)` `=5 xx $1.00 + 20\ text(cents)`  
  `=$5.20`  

Measurement, NAPX-p116903v03

Wilde compared the weight of four different solid figures by using a balancing scale.
 

          

 
Which of the four objects is the heaviest?

 
 
 
 
Show Answers Only

Show Worked Solution

`text{By inspection:}`

`text(Triangle → heavier than the hexagon.)`

`text(Triangle → same weight as the oval.)`

`text(Star → heavier than the oval.)`
 

`:.\ text(Star is the heaviest.)`

Measurement, NAPX-p116903v02

Banjo compared the weight of four different solid figures by using a balancing scale.
 

      
 

Which of the four objects is the lightest?

 
 
 
 
Show Answers Only

Show Worked Solution

`text{By inspection:}`

`text(Sphere → lighter than the cube.)`

`text(Sphere → same weight as the cylinder.)`

`text(Cylinder → heavier than the pentagonal prism.)`
 

`:.\ text(Pentagonal prism is the lightest.)`

Measurement, NAPX-p116893v03

Mince decided to paint some triangles red in the four figures below.

All triangles are the same size.

Which of these figures has the smallest area that is painted red.

 
 
 
 

Show Answers Only

Show Worked Solution

`text{Smallest area painted red = lowest number of triangles painted red.}`

`text{Check each option:}`

`text{1st Figure  – 9 triangles}`

`text{2nd Figure – 9 triangles}`

`text{3rd Figure – 10 triangles}`

`text{4th Figure – 7 triangles}`

`therefore \ text{4th Figure has the smallest area painted red}`

Measurement, NAPX-p116893v02

Jane painted some hexagons in the four figures pictured below.

All hexagons are the same size.

Which of these figures has the largest area that is painted green.

 
 
 
 
Show Answers Only

Show Worked Solution

`text{Largest area painted = figure with the most hexagons painted green.}`

`text{Check each option:}`

`text{1st Figure – 11 hexagons}`

`text{2nd Figure – 9 hexagons}`

`text{3rd Figure – 13 hexagons}`

`text{4th Figure – 9 hexagons}`

`therefore \ text{3rd Figure has the largest area painted green.}`

Number, NAPX-p116838v04

In a bakery Michelle bought one slice of cake, one donut and a croissant.
 


 

She gave the cashier $10.00.

How much change should Michelle receive?

$5.10 $5.60 $5.80 $6.00
 
 
 
 
Show Answers Only

`$6.00`

Show Worked Solution
`text{Total cost}` `= 1.35 + 1.20 + 1.45 `
  `= $4.00`

 

`text{Change}` `= 10 – 4`
  `= $6.00`

Number, NAPX-p116838v03

Alexander went shopping and bought the following items.
 


 

He gave the cashier $50.

How much change should Alexander receive?

$17.00 $18.20 $31.80 $33.00
 
 
 
 
Show Answers Only

`$17.00`

Show Worked Solution
`text{Total cost}` `= 9.50 + 12.10 + 11.40 `
  `= $33.00`

 

`text{Change}` `= 50 – 33.00 `
  `= $17.00`

Measurement, NAPX p124557v03

Timothy started with 53 millilitres of oil in a beaker.

He saw that there was some unused oil in a test tube and returned it to the beaker.


 

How much of the chemical solution was poured int0 the test tube?

 
  13 millilitres
 
  23 millilitres
 
  27 millilitres
 
  37 millilitres
Show Answers Only

`27 \ text{millilitres}`

Show Worked Solution
`text{Oil in the test tube }` `= text{New volume} – text{Original volume}`
  `= 80 – 53`
  `= 27 \ text{millilitres}`

Measurement, NAPX p124557v02

A chemist started with 685 millilitres of a solution.

He then poured some of the solution into a test tube.
 


 

How much of the chemical solution was poured into the test tube?

 
  90 millilitres
 
  135 millilitres
 
  145 millilitres
 
  185 millilitres
Show Answers Only

`185 \ text{millilitres}`

Show Worked Solution
`text{Solution in test tube }` `= text{New volume} – text{Original volume}`
  `= 685 – 500`
  `= 185 \ text{millilitres}`

Statistics, NAPX-p116802v04

In a neighborhood 10 villagers were asked if how many pets they own.

The results were: 3, 2, 1, 1, 1, 4, 2, 4, 5, 2

Select the dot plot that displays the data recorded

 
 
 
 
Show Answers Only

Show Worked Solution

`text{A dot represents the number of persons that owns N pets}`

`text{From the data above:}`

`text{Owns one pet = 3}`

`text{Owns two pets = 3}`

`text{Owns three pets = 1}`

`text{Owns four pets = 2}`

`text{Owns five pets = 1}`

`therefore \ text{the correct dot plot is:}`
 

Statistics, NAPX-p116802v03

10 people were asked how often they went to the doctor in the last 12 months.

Their responses were: 2, 3, 3, 4, 1, 2, 5, 1, 1, 2

Choose the dot plot which correctly displays the data.

 
    
 
    
       
 
    
 
    
Show Answers Only

Show Worked Solution

`text{From the data above:}`

`text{1 visit =  3 people}`

`text{2 visits =  3 people}`

`text{3 visits =  2 people}`

`text{4 visits =  1 person}`

`text{5 visits =  1 person}`
 

   

Geometry, NAPX-p116732v03

A bridge is drawn below.
 

   
 

The angle `y°` is marked from the foot of the bridge to the diagonal support.

The value of angle `y°`  is

 
  Less than 90°
 
  More than 180°
 
  Equal to 90°
 
  More than 90° but less than 180°
Show Answers Only

`text{Less than} \ 90^@`

Show Worked Solution

`text{The angle from the foot to the diagonal support is less than} \ 90^@`

Statistics, NAPX-p110558v02

A charity event was held for three weeks. On the first week $5000 was raised, on the second week $3500 was raised and on the last week $2000 was raised.

In the tables below, O = $500

Which table correctly shows the amount of donations raised in the 3 weeks?

 
  
 
  
 
  
 
    
Show Answers Only

Show Worked Solution

`text{O represents $500 in donations.}`

`text{1st week:} \ 5000 \div 500 = 10`

`therefore \ text{Ten O symbols needed.}`
 

`text{2nd week:} \ 3500 \div 500 = 7`

`therefore \ text{Seven O symbols needed.}`
 

`text{3rd week:} \ 2000 \div 500 = 4`

`therefore \ text{Four O symbols needed.}`

Statistics, NAPX-p110558v01

In a certain school, the number of Year 7 students is 120, Year 8 students is 80, and Year 9 students is 100.

In the tables below, V = 20 students

Which table correctly shows the number of students per year level?

 
    
 
    
 
    
 
    
Show Answers Only

Show Worked Solution

`text{V represents 20 students.}`

`text{Year 7 students:} \ 120 \div 20 = 6`

`therefore \ text{Six V symbols needed}`
 

`text{Year 8 students:} \ 80 \div 20 = 4`

`therefore \ text{Four V symbols needed}`
 

`text{Year 9 students:} \ 100 \div 20 = 5`

`therefore \ text{Five V symbols needed}`

Geometry, NAPX-p116721v03

Sarah drew a straight line on a shape.

The line divided the shape into two rectangles.

Which of these could have been Sarah’s shape?

 
 
 
 
Show Answers Only

Show Worked Solution

`text{Drawing any horizontal straight line on this shape divides it}`

`text{into two rectangles.}`

Algebra, NAPX-p116672v03

Rudy uses the number sentence  24 – 6 = 18

Which of the following problems can he solve with this number sentence?

 
   Rudy shared 24 slices of cake equally between 6 of his friends.
  How many slices does each of his friends get?
 
   Rudy has 24 slices of cake and gave 6 slices to his friends.
  How many slices does Rudy have left?
 
   Rudy took 6 slices of cake and shared them between 24 of his friends.
  How many slices does each friend receive?
 
   Rudy has 24 slices of cake and gave 18 of them to his friends.
  How many slices does Rudy have left?
Show Answers Only

`text{Rudy has 24 slices of cake and gave 6 slices to his friends.}`

`text{How many slices does Rudy have left?}`

Show Worked Solution

`text{Rudy has 24 slices of cake and gave 6 slices to his friends.}`

`text(Expressing this as an equation:)`

`text(Slices Rudy has left)` `=24 -6`  
`18` `=20 – 6`  

Algebra, NAPX-p116672v02

Shawn uses the number sentence  20 × 6 = 120

Which of the following problems can he solve with this number sentence?

 
   Shawn buys 20 toys and gave 6 of them to his nephew.
  How many toys does Shaun's nephew have now?
 
   Shawn buys 20 toys and received another 6 toys from friends.
  How many toys does Shaun have now?
 
   Shawn buys 20 toys and divides them up between his 6 nephews.
  How many toys does each nephew get?
 
   Shawn has 20 nephews and gives 6 toys to each one.
  Altogether, how many toys does Shaun give to his nephews?
Show Answers Only

`text{Shawn has 20 nephews and gives 6 toys to each one.}`

Show Worked Solution

`text{Shawn has 20 nephews and gives 6 toys to each one.}`

`text(Expressing this as an equation:)`

`text(Total toys to nephews)` `=20 xx 6`  
`120` `=20 xx 6`  

Measurement, NAPX-p167325v01

Sam painted grids in the larger square areas pictured below.

All of the grids are the same size.

Which of the following squares has the largest area painted?

 
 
 
 

Show Answers Only

Show Worked Solution

`text(The largest area painted is the figure with the most squares painted grey.)`

`text(Checking each option:)`

`text(Option 1 – 9 square grids painted)`

`text(Option 2 – 13 square grids painted)`

`text{Option 3 – 16 square grids painted (Correct)}`

`text(Option 4 – 14 square grids painted)`

Mechanics, EXT2 M1 2017 SPEC2 17 MC

The acceleration,  `a\ text(ms)^(-2)`, of a particle moving in a straight line is given by  `a = v^2 + 1`,  where  `v`  is the velocity of the particle at any time `t`. The initial velocity of the particle when at origin O is  `2\ text(ms)^(-1)`.

The displacement of the particle from O when its velocity is  `3\ text(ms)^(-1)`  is

  1. `log_e(2)`
  2. `1/2 log_e(10/3)`
  3. `1/2 log_e(2)`
  4. `1/2 log_e(5/2)`
Show Answers Only

`C`

Show Worked Solution
`v* (dv)/(dx)` `= v^2 + 1`
`(dv)/(dx)` `= (v^2 + 1)/v`
`(dx)/(dv)` `= v/(v^2 + 1)`
`x` `= int (v/(v^2 + 1))\ dv`
  `=1/2 ln(v^2 + 1)+c`

 
`text(When)\ \ x=0, \ v=2:`

`c=-1/2 ln5`

 
`text(Find)\ \ x\ \ text(when)\ \ v=3:`

`x` `= 1/2ln(9 + 1) – 1/2 ln5`
  `= 1/2 ln (10/5)`
  `= 1/2 ln 2`

 
`=>   C`

Mechanics, SPEC2 2020 VCAA 20 MC

An object of mass 2 kg is suspended from a spring balance that is inside a lift travelling downwards.

If the reading on the spring balance is 30 N, the acceleration of the lift is

  1. `text(5.2 ms)^(−2)` upwards.
  2. `text(5.2 ms)^(−2)` downwards.
  3. `text(9.8 ms)^(−2)` downwards.
  4. `text(10.4 ms)^(−2)` upwards.
  5. `text(10.4 ms)^(−2)` downwards.
Show Answers Only

`A`

Show Worked Solution

`text(Assuming up is positive:)`

♦ Mean mark 43%.
`ma` `= 30 – 2g`
`2a` `= 10.4`
`:.a` `= 5.2\ text(ms)^(−2)`

 
`text{(i.e. acceleration is against the direction of travel.)}`

`=>A`

Vectors, SPEC2 2020 VCAA 15 MC

Two forces, `underset~F_(text(A)) = 4 underset~i - 2 underset~j`  and  `underset~F_(text(B)) = 2 underset~i - 5 underset~j`, act on a particle of mass 3 kg. The particle is initially at rest at position  `underset~i + underset~j`. All force components are measured in newtons and displacements are measured in metres.

The cartesian equation of the path of the particle is

  1. `y = x/2`
  2. `y = x/2 - 1/2`
  3. `y = ((x + 1)^2)/2 + 1`
  4. `y = ((x - 1)^2)/1 + 1`
  5. `y = x/2 + 1/2`
Show Answers Only

`E`

Show Worked Solution

`text(Net Force)\ (underset~F) = 4underset~i – 2underset~j +  2underset~i + 5underset~j = 6underset~i + 3underset~j`

`underset~a = underset~F/m = 2underset~i + underset~j`

♦ Mean mark 38%.

`underset~v = int_0^t 2underset~i + underset~j\ dt = 2tunderset~i + tunderset~j`

`underset~r` `= int_0^t 2tunderset~i + tunderset~j\ dt + (underset~i + underset~j)`
  `= t^2 underset~i + (t^2)/2 underset~j + underset~i + underset~j`
  `= (t^2 + 1)underset~i + ((t^2)/2 + 1)underset~j`

 
`x = t^2 + 1 \ => \ t^2 = x – 1`

`y` `= (t^2)/2 + 1`
  `= ((x – 1))/2 + 1`
  `= x/2 + 1/2`

 
`=>E`

Calculus, EXT1 C3 2020 SPEC2 9 MC

`P(x, y)`  is a point on a curve. The `x`-intercept of a tangent to point  `P(x, y)`  is equal to the `y`-value at `P`.

Which one of the following slope fields best represents this curve?

A.   B.
C. D.
Show Answers Only

`B`

Show Worked Solution

`text(The tangent to the curve passes through)`

`(x, y)\ and\ (y, 0)`

`(dy)/(dx) = (0 – y)/(y – x) = y/(x – y)`
 

`text(When)\ \ x = 0:`

`(dy)/(dx) = y/(−y) = −1`

`=>B`

Calculus, SPEC2 2020 VCAA 9 MC

`P(x, y)`  is a point on a curve. The `x`-intercept of a tangent to point  `P(x, y)`  is equal to the `y`-value at `P`.

Which one of the following slope fields best represents this curve?

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

`B`

Show Worked Solution

`text(The tangent to the curve passes through)`

♦♦ Mean mark 35%.

`(x, y)\ and\ (y, 0)`

`(dy)/(dx) = (0 – y)/(y – x) = y/(x – y)`
 

`text(When)\ \ x = 0:`

`(dy)/(dx) = y/(−y) = −1`

`=>B`

Complex Numbers, SPEC2 2020 VCAA 8 MC

Given that  `(x + iy)^14 = a + ib`, where  `x, y, a, b ∈ R, \ (y - ix)^14`  for all values of `x` and `y` is equal to

  1. `−a - ib`
  2. `b - ia`
  3. `−b + ia`
  4. `−a + ib`
  5. `b + ia`
Show Answers Only

`A`

Show Worked Solution

♦♦ Mean mark 34%.
`(y – ix)^14` `= (−i(x + iy))^14`
  `= −i^14(x + iy)^14`
  `= −(x + iy)^14`
  `= −a – ib`

 
`=>A`

Trigonometry, 2ADV T3 2011 SPEC1 8

Find the coordinates of the points of intersection of the graph of the relation

`y = text(cosec)^2 ((pi x)/6)`  with the line  `y = 4/3`, for  `0 < x < 12.`  (3 marks)

Show Answers Only

`(2, 4/3),(4, 4/3),(8, 4/3), (10, 4/3)`

Show Worked Solution

`text(Intersection occurs when:)`

`text(cosec)^2((pix)/6)` `=4/3`
`text(cosec)((pix)/6)` `= ±2/sqrt3`
`sin((pix)/6)` `= ±sqrt3/2`

 
`text(Given:)\ \ 0 < x < 12 \ \ =>\ \ 0 < (pix)/6 < 2pi`
 

`(pix)/6` `= pi/3, pi – pi/3, pi + pi/3, 2pi – pi/3`
  `= pi/3, (2pi)/3, (4pi)/3,(5pi)/3`
`x` `= 2, 4, 8, 10`

  
`=> y = 4/3\ text(for each)`

`:.\ text(Intersection at:)\ \ (2, 4/3),(4, 4/3),(8, 4/3), (10, 4/3)`

Graphs, SPEC2 2020 VCAA 2 MC

A function  `f` has the rule  `f(x) = |b cos^(−1)(x) - a|`, where  `a > 0, b > 0`  and  `a < (bpi)/2`.

The range of  `f` is

  1. `[−a, bpi - a]`
  2. `[0, bpi - a]`
  3. `[a, bpi - a]`
  4. `[0, bpi + a]`
  5. `[a - bpi, a]`
Show Answers Only

`B`

Show Worked Solution

`text(Range:)\ \ cos^(−1)(x) = [0, pi]`

♦ Mean mark 42%.

`text(Range:)\ \ b cos^(−1)(x) – a = [−a, bpi – a]`

`text(Given)\ \ 0 < a < (bpi)/2, text(Range) = [−a, bpi – a]`

`:.\ text(Range:)\ \ |b cos^(−1) (x) – a| = [0, bpi – a]`

`=>B`

GEOMETRY, FUR1 2020 VCAA 9 MC

Shot-put is an athletics field event in which competitors throw a heavy spherical ball (a shot) as far as they can.

The size of the shot for men and the shot for women is different.

The diameter of the shot for men is 1.25 times larger than the diameter of the shot for women.

The ratio of the total surface area of the women’s shot to the total surface area of the men’s shot is

  1.   1:4
  2.   1:25
  3.   4:5
  4.   5:4
  5. 16:25
Show Answers Only

`E`

Show Worked Solution

♦♦ Mean mark 36%.
`SA_f` `: SA_m` 
`4pir^2` `: 4 pi(5/4r)^2`
`1` `: 25/16`
`16` `: 25`

 
`=>  E`

GEOMETRY, FUR1 2020 VCAA 10 MC

An 80 m high lookout tower stands in the centre of town.

Two landmarks, on the same horizontal plane, are visible from the top of the lookout tower.

The direct distance from the top of the lookout tower to the base of Landmark A is 170 m.

The direct distance from the top of the lookout tower to the base of Landmark B is 234 m.
 

The bearing of Landmark B from Landmark A is 105°.

The bearing of Landmark B from the lookout tower is 142°.

The direct distance along the ground, in metres, between Landmark A and Landmark B is closest to

  1. 127
  2. 135
  3. 246
  4. 297
  5. 320
Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ T = text(top of tower),\ \ O = text(base of tower)`

♦ Mean mark 45%.

`text(In)\ \ Delta TOB:`

`OB = sqrt(234^2 – 80^2) ~~ 220`

`text(Similarly),`

`OA = sqrt(170^2 – 80^2) = 150`

`text(Find)\ \ /_AOB:`

`(sin /_OAB)/220` `= (sin 37^@)/150`
`sin /_OAB` `= (220 xx sin 37^@)/150`
`/_OAB` `~~ 62^@`

 
`/_AOB = 180 – (62 + 37) = 81^@`
 

`text(Using cosine rule:)`

`AB^2` `= 150^2 + 220^2 – 2 xx 150 xx 220 xx cos 81^@`
  `= 60\ 575.32…`
`AB` `~~ 246`

 
`=>  C`

NETWORKS, FUR1 2020 VCAA 4 MC

The directed graph below represents a series of one-way streets.

The vertices represent the intersections of these streets.
 


 

The number of vertices that can be reached from `S` is

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

`C`

Show Worked Solution

`S\ text(can reach vertices)\ V, T\ text(directly.)`

`S\ text(can also reach)\ X\ text(indirectly through)\ V.`

♦ Mean mark 39%.

`=>  C`

MATRICES, FUR1 2020 VCAA 10 MC

Consider the matrix recurrence relation below.
  

`S_0 = [(30),(20),(40)], quad S_(n+1) = TS_n qquad text(where)\ T = [(j, 0.3, l),(0.2, m, 0.3),(0.4, 0.2, n)]`
 

Matrix `T` is a regular transition matrix.

Given the information above and that  `S_1 = [(42),(28),(20)]`, which one of the following is true?

  1. `m >l`
  2. `j + l = 0.7`
  3. `j = n`
  4. `j > m`
  5. `l = m + n`
Show Answers Only

`E`

Show Worked Solution

`text(Transition matrix columns sum to 1)`

♦ Mean mark 41%.

`j = 1 – 0.2 – 0.4 = 0.4`

`m = 1 – 0.3 – 0.2 = 0.5`

`l + n = 0.7`
 

`[(0.4, 0.3, l),(0.2, 0.5, 0.3),(0.4, 0.2, n)][(30),(20),(40)]=[(42),(28),(20)]`
 

`42` `= 0.4 xx 30 + 0.3 xx 20 + l xx 40`
`24` `= 40l`
`l` `= 0.6`

 
`:. n = 0.1`

`=>  E`

MATRICES, FUR1 2020 VCAA 7 MC

A small shopping centre has two coffee shops: Fatima’s (F) and Giorgio’s (G).

The percentage of coffee-buyers at each shop changes from day to day, as shown in the transition matrix `T`.
 

`{:(quadqquadqquadqquadquad t\oday),(qquadqquadqquadquad F quadqquad G),(T = [(0.85,0.35),(0.15,0.65)]{:(F),(G):}qquad t\omo\rrow):}`
 

On a particular Monday, 40% of coffee-buyers bought their coffees at Fatima’s.

The matrix recursion relation  `S_(n+1) = TS_n`  is used to model this situation.

The percentage of coffee-buyers who are expected to buy their coffee at Giorgio’s on Friday of the same week is closest to

  1. 31%
  2. 32%
  3. 34%
  4. 45%
  5. 68%
Show Answers Only

`B`

Show Worked Solution
♦ Mean mark 38%.
`text(Tuesday)` `= [(0.85, 0.35),(0.15, 0.65)][(0.40),(0.60)]`
`text(Friday)` `= [(0.85, 0.35),(0.15, 0.65)]^4 [(0.40),(0.60)]`
  `~~ [(0.68),(0.32)]`

`=>  B`

CORE, FUR1 2020 VCAA 30 MC

Twenty years ago, Hector invested a sum of money in an account earning interest at the rate of 3.2% per annum, compounding monthly.

After 10 years, he made a one-off extra payment of $10 000 to the account.

For the next 10 years, the account earned interest at the rate of 2.8% per annum, compounding monthly.

The balance of his account today is $686 904.09

The sum of money Hector originally invested is closest to

  1. $355 000
  2. $370 000
  3. $377 000
  4. $384 000
  5. $385 000
Show Answers Only

`B`

Show Worked Solution

`text(Let)\ I = text(original investment)`

♦ Mean mark 41%.

`text(Strategy 1:)`

`text(Balance)` `= [I(1 + 3.2/(12 xx 100))^120 + 10\ 000](1 + 2.8/(12 xx 100))^120`
  `= $686\ 904.09`

 
`text(Test each option in the equation)`

`text(Option)\ B:\ \ I = $370\ 000\ \ text(is correct)`
 

`text{Strategy 2 (By TVM Solver):}`

`N` `=120`  
`Itext{(%)}` `= 2.8`  
`PMT` `=0`  
`PV` `= ?`  
`FV` `= 686\ 904.09`  
`text(PY)` `= text(CY) =12`  

 
`:. PV = 519\ 320.3`
 

`N` `=120`  
`Itext{(%)}` `= 3.2`  
`PMT` `=0`  
`PV` `= ?`  
`FV` `= 509\ 320.3`  
`text(PY)` `= text(CY) =12`  

 
`:. PV = 370\ 000`

`=>  B`

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 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 13 MC

A least squares line of the form  `y = a + bx`  is fitted to a scatterplot.

Which one of the following is always true?

  1. As many of the data points in the scatterplot as possible will lie on the line.
  2. The data points in the scatterplot will be divided so that there are as many data points above the line as there are below the line.
  3. The sum of the squares of the shortest distances from the line to each data point will be a minimum.
  4. The sum of the squares of the horizontal distances from the line to each data point will be a minimum.
  5. The sum of the squares of the vertical distances from the line to each data point will be a minimum.
Show Answers Only

`E`

Show Worked Solution

`text(The sum of the squares of the vertical distances from)`

♦♦ Mean mark 34%.

`text(the line to each data point will be a minimum.)`

`=>  E`

CORE, FUR1 2020 VCAA 4 MC

The histogram below shows the distribution of the forearm circumference, in centimetres, of 252 men.

Assume that the forearm circumference values were all rounded to one decimal place.
 


 

The third quartile `(Q_3)` for this distribution could be

  1. 29.3
  2. 29.8
  3. 30.3
  4. 30.8
  5. 31.3
Show Answers Only

`C`

Show Worked Solution

`text(S) text(ince)\ \ n = 252,`

♦ Mean mark 50%.

`Q_3\ text(occurs at data point 189.)`

`text(Datapoint 189 falls in the class interval 30 – 30.5)` 

`=>  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 8

Find the volume of, `V`, of the solid of revolution formed when the graph of  `y = 2sqrt((x^2 + x + 1)/((x + 1)(x^2 + 1)))`  is rotated about the `x`-axis over the interval  `[0, sqrt 3]`. Give your answer in the form  `V = 2pi(log_e(a) + b)`, where  `a, b in R`.  (5 marks)

Show Answers Only

`V = 2pi (log_e(2 sqrt 3 + 2) + pi/3)\ text(u³)`

Show Worked Solution
`V` `= pi int_0^sqrt 3 y^2\ dx`
  `= 4 pi int_0^sqrt 3 (x^2 + x + 1)/((x + 1)(x^2 + 1))\ dx`

 

`text(Using partial fractions):`

`(x^2 + x + 1)/((x + 1)(x^2 + 1))` `= A/(x + 1) + (Bx + C)/(x^2 + 1)`
`x^2 + x + 1` `= Ax^2 + A + (Bx + C)(x + 1)`
  `=(A+B)x^2 + (B+C)x + A + C`

 
`text(If)\ \ x = – 1,\ 2A = 1 \ => \ A = 1/2`

♦ Mean mark 50%.

`text(Equating coefficients of)\ x^2: A + B = 1 \ => \ B = 1/2`

`text(Equating constants): A + C = 1 \ => \ C = 1/2`

`:. V` `= 2 pi int_0^sqrt 3 1/(x + 1) + x/(x^2 + 1) + 1/(x^2 + 1)\ dx`
  `= 2 pi [log_e |x + 1| + 1/2 log_e |x^2 + 1| + tan^(-1)(x)]_0^sqrt 3`
  `= 2 pi (log_e (sqrt 3 + 1) + 1/2 log_e4 + pi/3)`
  `= 2 pi (log_e (sqrt 3 + 1) + log_e 2 + pi/3)`
  `= 2 pi (log_e (2 sqrt 3 + 2) + pi/3)\ text(u³)`

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`