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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`

Algebra, SPEC1 2020 VCAA 4

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

Show Answers Only

`x ∈ (– oo, (7 – sqrt 5)/2)`

Show Worked Solution

`3 – x > 1/|x – 4|`

♦ Mean mark 46%.

`|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)`

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`

Calculus, MET1 2020 VCAA 6

Let  `f:[0,2] -> R`, where  `f(x) = 1/sqrt2 sqrtx`.

  1. Find the domain and the rule for  `f^(-1)`, the inverse function of  `f`.   (2 marks)

The graph of  `y = f(x)`, where  `x ∈ [0, 2]`, is shown on the axes below.
 

     
 

  1. On the axes above, sketch the graph of  `f^(-1)`  over its domain. Label the endpoints and point(s) of intersection with the function  `f`, giving their coordinates.   (2 marks)

  2. Find the total area of the two regions: one region bounded by the functions  `f` and `f^(-1)`, and the other region bounded by  `f, f^(-1)`  and the line  `x = 1`. Give your answer in the form  `(a-bsqrtb)/6`, where  `a, b ∈ ZZ^+`.   (4 marks)

Show Answers Only
  1. `text(Domain) = [0, 1]`

     

    `f^(-1)(x) = 2x^2`

  2.  
       
  3.  `(5 + 2sqrt2)/6\ \ text(u²)`
Show Worked Solution
a.    `text(Domain)\ \ f^(-1)(x)` `= text(Range)\ \ f(x)=[0,1]`

 
`y = 1/sqrt2 x`

`text(Inverse: swap)\ \ x ↔ y`

`x` `= 1/sqrt2 sqrty`
`sqrty` `= sqrt2 x`
`y` `= 2x^2`

 
`:. f^(-1)(x) = 2x^2`

 

b.     

 

c.     
`A` `= int_0^(1/2) 1/sqrt2 sqrtx-2x^2 dx + int_(1/2)^1 2x^2-1/sqrt2 sqrtx\ dx`
  `= [sqrt2/3 x^(3/2)-2/3 x^3]_0^(1/2) + [2/3 x^3-sqrt2/3 x^(3/2)]_(1/2)^1`
  `= [sqrt2/3 (1/sqrt2)^3-2/3(1/2)^3] + [(2/3-sqrt2/3)-(2/24-sqrt2/3 · 1/(2sqrt2))]`
  `= (1/6-1/12) + 2/3-sqrt2/3-(1/12-1/6)`
  `= 1/12 + 2/3-sqrt2/3 + 1/12`
  `= (1 + 8-4sqrt2 + 1)/12`
  `= (5-2sqrt2)/6\ \ text(u²)`

L&E, 2ADV E1 2020 MET1 4

Solve the equation  `2 log_2(x + 5)-log_2(x + 9) = 1`.  (3 marks)

Show Answers Only

`x = text{−1}`

Show Worked Solution
`2 log_2(x + 5)-log_2(x + 9)` `= 1`
`log_2(x + 5)^2-log_2(x + 9)` `= 1`
`log_2(((x + 5)^2)/(x + 9))` `= 1`
`((x + 5)^2)/(x + 9)` `= 2`
`x^2 + 10x + 25` `= 2x + 18`
`x^2 + 8x + 7` `= 0`
`(x + 7)(x + 1)` `= 0`

 
`:. x = -1\ \ \ \ (x != text{−7}\ \ text(as)\ \ x > text{−5})`

Statistics, STD2 S1 2006 HSC 23c*

Vicki wants to investigate the number of hours spent on homework by students at her high school.

She asks each student how many hours (to the nearest hour) they usually spend on homework during one week. 

The responses are shown in the frequency table.

\begin{array} {|c|c|}
\hline
\textit{  Number of hours spent  } & \ \ \ \ \ \textit{Frequency}\ \ \ \ \  \\ \textit{  on homework in a week  } & \\
\hline
\rule{0pt}{2.5ex} \text{0 to 4} \rule[-1ex]{0pt}{0pt} & 69 \\
\hline
\rule{0pt}{2.5ex} \text{5 to 9} \rule[-1ex]{0pt}{0pt} & 72 \\
\hline
\rule{0pt}{2.5ex} \text{10 to 14} \rule[-1ex]{0pt}{0pt} & 38 \\
\hline
\rule{0pt}{2.5ex} \text{15 to 19} \rule[-1ex]{0pt}{0pt} & 21 \\
\hline
\end{array}

What is the mean amount of time spent on homework?   (2 marks)

Show Answers Only

`text(7.275 hours)`

Show Worked Solution

MARKER’S COMMENT: This “routine” exercise of finding a mean from grouped data was incorrectly answered by most students! The best responses copied the table and inserted a class-centre column (see solution).
      2UG-2006-23c Answer

 

`text(Mean)` `= text(Sum of Scores) / text(Total scores)`
  `= 1455/200`
  `= 7.275\ text(hours)`

Complex Numbers, EXT2 N1 2008 HSC 2b

  1. Write  `frac{1 + i sqrt3}{1 + i}`  in the form  `x + iy`, where `x` and `y` are real.  (2 marks)

  2. By expressing both  `1 + i sqrt3`  and  `1 + i`  in  modulus-argument form, write  `frac{1 + i sqrt3}{1 + i}`  in modulus-argument form.   (3 marks)

  3. Hence find  `cos frac{pi}{12}`  in surd form.  (1 mark)

  4. By using the result of part (ii), or otherwise, calculate  `(frac{1 + i sqrt3}{1 + i})^12`.   (1 mark)

Show Answers Only
  1. `frac{1 + sqrt3}{2} – i ( frac{1 – sqrt3}{2} )`
  2. `sqrt2 (cos (frac{pi}{12}) + i sin (frac{pi}{12}))`
  3. `frac{sqrt2 + sqrt6}{4}`
  4. `-64`
Show Worked Solution
i.      `frac{1 + i sqrt3}{1 + i} xx frac{1 – i}{1 – i}` `= frac{(1 + i sqrt3)(1 – i)}{1 – i^2}`
    `= frac{1 – i + i sqrt3 – sqrt3 i^2}{2}`
    `= frac{1 + sqrt3}{2} – i ( frac{1 – sqrt3}{2} )`

 

ii.   `z_1 = 1 +  i sqrt3`

`| z_1 | = sqrt(1 + ( sqrt3)^2) = 2`

`text{arg} (z_1) = tan^-1 (sqrt3) = frac{pi}{3}`
  

`z_1 = 2 (cos frac{pi}{3} + i sin frac{pi}{3})`
 
`z_2 = 1 + i`

`| z_2 | = sqrt(1^2 + 1^2) = sqrt2`

`text{arg} (z_2) = tan^-1 (1) = frac{pi}{4}`

`z_2 = sqrt2 (cos frac{pi}{4} + i sin frac{pi}{4})`
 

`frac{1 + i sqrt3}{1 + i}` `= frac{z_1}{z_2}`
  `= frac{2}{sqrt2} ( cos ( frac{pi}{3} – frac{pi}{4} ) + i sin ( frac{pi}{3} – frac{pi}{4} ) )`
  `= sqrt2 ( cos (frac{pi}{12}) + i sin (frac{pi}{12}) )`

 

iii.  `text{Equating real parts of i and ii:}`

`sqrt2 cos (frac{pi}{12})` `= frac{1 + sqrt3}{2}`
`cos(frac{pi}{12})` `= frac{1 + sqrt3}{2 sqrt2} xx frac{sqrt2}{sqrt2}`
  `= frac{sqrt2 + sqrt6}{4}`

 

iv.     `(frac{1 + i sqrt2}{1 + i})^12` `= (sqrt2)^12 (cos (frac{pi}{12} xx 12) + i sin (frac{pi}{12} xx 12))`
    `= 64 (cos pi + i sin pi)`
    `= – 64`

Calculus, EXT2 C1 2020 HSC 10 MC

Which of the following is equal to  `int_0^(2a) f(x)\ dx`?

  1. `int_0^a f(x) - f(2a - x)\ dx`
  2. `int_0^a f(x) + f(2a -x)\ dx`
  3. `2 int_0^a f(x - a)\ dx`
  4. `int_0^a frac{1}{2} f(2x)\ dx`
Show Answers Only

`B`

Show Worked Solution

`int_0^(2a) f(x)\ dx – int_0^a f(x)\ dx + int_a^(2a) f(x)\ dx`

♦ Mean mark 54%.

`text(Let)\ \ x = 2a – u\ \ => \ u = 2a – x`

`frac{du}{dx} = -1 \ \ => \ du = -dx`

 

`text{When}`   `x = a,` `\ u = a`
  `x= 2a,`  `\ u = 0`

 

`int_0^(2a) f(x)\ dx= int_0^a f(x)\ dx – int_a^0 f(2a – u)\ du`

`text(Use substitution for)\ \ int_a^0 f(2a – u)\ du`

`text(Let)\ \ 2a – u = 2a-x \ \ => \ x=u`

`dx/(du) = 1 \ => \ du=dx`

`text{When}`   `u = a,` `\ x = a`
  `u= 0,`  `\ x = 0`

 

`:. int_0^(2a) f(x)\ dx` `= int_0^a f(x)\ dx – int_a^0 f(2a – x)\ dx`
  `= int_0^a f(x)\ dx + int_0^a f(2a -x)\ dx`
  `= int_0^a f(x) + f(2a – x)\ dx`

 
`=> \ B`

Proof, EXT2 P1 2020 HSC 8 MC

Consider the statement:

'If `n` is even, then if `n` is a multiple of 3, then `n` is a multiple of 6'.

Which of the following is the negation of this statement?

  1. `n` is odd and `n` is not a multiple of 3 or 6.
  2. `n` is even and `n` is a multiple of 3 but not a multiple of 6.
  3. If `n` is even, then `n` is not a multiple of 3 and `n` is not a multiple of 6.
  4. If `n` is odd, then if `n` is not a multiple of 3 then `n` is not a multiple of 6.
Show Answers Only

`B`

Show Worked Solution

`text{Proposition: If} \ \ X =>Y`

♦♦ Mean mark part 33%.

`X \ text{is a compound statement}`

`text{“If} \ n \ text{is even and a multiple of 3.”}`

`Y \ text{states “} n \ text{is a multiple of 6.”}`
 
`text{Negation if} \ X \ text{but} \ ¬ \ Y.`
 
`=> \ B`

Calculus, EXT2 C1 2020 HSC 16b

Let  `I_n = int_0^(frac{pi}{2}) sin^(2n + 1)(2theta)\ d theta, \ n = 0, 1, ...`

  1. Prove that  `I_n = frac{2n}{2n + 1} I_(n-1) , \ n ≥ 1`.  (3 marks)

  2. Deduce that  `I_n = frac{2^(2n)(n!)^2}{(2n +1)!}`.  (3 marks)

Let  `J_n = int_0^1 x^n (1 - x)^n\ dx , \ n = 0, 1, 2,...`

  1. Using the result of part (ii), or otherwise, show that  `J_n = frac{(n!)^2}{(2n + 1)!}`.  (3 marks)

  2. Prove that  `(2^n n!)^2 ≤ (2n + 1)!`.  (2 marks)

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

i.    `text{Prove} \ \ I_n = frac{2n}{2n + 1} I_(n-1) , \ n ≥ 1`

♦ Mean mark (i) 48%.

`I_n = int_0^(frac{pi}{2}) sin^(2n) (2 theta) * sin (2 theta)\ d theta`

`text{Integrating by parts:}`

`u = sin^(2n) (2 theta)`   `u^(′) = 2n sin^(2n -1) (2 theta)  xx -frac(1)(2) cos (2 theta)`
`v = -frac{1}{2} cos (2 theta)`   `v^(′) = sin 2 theta`

 

`I_n` `= [ sin^(2n) (2 theta) * -frac{1}{2} cos (2 theta)]_0^(frac{pi}{2}) -2n int_0^(frac{pi}{2}) sin^(2n -1) (2 theta) * 2 cos (2 theta) * -frac{1}{2} cos (2 theta)\ d theta`
`I_n` `= 0 + 2n int_0^(frac{pi}{2}) sin^(2n-1) (2 theta) * cos^2 (2 theta)\ d theta`
`I_n` `= 2 n int_0^(frac{pi}{2}) sin^(2n-1) (2 theta) (1 – sin^2 (2 theta))\ d theta`
`I_n` `= 2 n int_0^(frac{pi}{2}) sin^(2n-1) ( 2 theta) – sin^(2n+1) (2 theta)\ d theta`
`I_n` `= 2n (I_(n-1) – I_n)`
`I_n + 2 n  I_n` `= 2 n I_(n-1)`
`I_n (2n + 1)` `= 2 n I_(n-1)`
`therefore I_n` `= frac{2n}{2n +1} I_(n-1)`
♦ Mean mark (ii) 36%.

 

ii.     `I_0` `= int_0^(frac{pi}{2}) sin (2 theta)\ d theta`
    `= [ -frac(1)(2) cos (2 theta) ]_0^(frac{pi}{2}`
    `=( -frac{1}{2} cos pi + frac{1}{2} cos 0 )`
    `= 1`
     
`I_n` `= frac{2n}{2n + 1} I_(n-1)`
`I_(n-1)` `= frac{2(n -1)}{2n -1} I_(n-2)`
  `vdots`
`I_1` `= frac{2}{3} I_0`

 

`I_n` `= frac{2n}{2n + 1} xx frac{2(n-1)}{2n-1} xx frac{2(n-2)}{2n-3} xx … xx frac{2}{3} xx 1`
  `= frac{2n}{2n+1} xx frac{2n}{2n} xx frac{2(n-1)}{2n-1} xx frac{2(n-1)}{2n-2} xx … xx frac{2}{3} xx frac{2}{2} xx 1`
  `= frac{2^n (n xx (n-1) xx .. xx 1) xx 2^n (n xx (n – 1) xx … xx 1)}{(2n + 1)!}`
  `= frac{2^(2n) (n!)^2}{(2n + 1)!}`
♦♦♦ Mean mark (iii) 16%.

 

iii.   `J_n = int_0^1 x^n (1-x)^n\ dx ,  \ n = 0, 1, 2, …`

`text{Let} \ \ x` `= sin^2 theta`
`frac{dx}{d theta}` `= 2 sin theta \ cos theta \ => \ dx = 2 sin theta \ cos theta \ d theta`

 

`text{When}`    `x = 0 \ ,` ` \ theta = 0`
  `x = 1  \ ,` ` \ theta = frac{pi}{2}`

 

`J_n` `= int_0^(frac{pi}{2}) (sin^2 theta)^n (1 – sin^2 theta)^n * 2 sin theta \ cos theta \ d theta`
  `= int_0^(frac{pi}{2}) sin^(2n) theta \ cos^(2n) theta * sin (2 theta)\ d theta`
  `= frac{1}{2^(2n)} int_0^(frac{pi}{2}) 2^(2n) sin^(2n) theta \ cos^(2n) theta * sin (2 theta)\ d theta`
  `= frac{1}{2^(2n)} int_0^(frac{pi}{2}) sin^(2n) (2 theta) * sin (2 theta)\ d theta`
  `= frac{1}{2^(2n)} int_0^(frac{pi}{2}) sin^(2n+1) (2 theta)\ d theta`
  `= frac{1}{2^(2n)} * frac{2^(2n) (n!)^2}{(2n+1)!}\ \ \ text{(using part (ii))}`
  `= frac{(n!)^2}{(2n + 1)!}`
♦♦♦ Mean mark (iv) 10%.

 

iv.   `text{If} \ \ I_n ≤ 1,`

`2^(2n) (n!)^2` ` ≤ (2n + 1)!`
`(2^n n!)^2` `≤ (2n + 1)!`

  
`text{Show} \ \ I_n ≤ 1 :`

`text{Consider the graphs}`

`y = sin(2 theta) \ \ text{and}\ \ y = sin^(2n + 1) (2 theta) \ \ text{for} \ \ 0 ≤ theta ≤ frac{pi}{2}`
 

`int_0^(frac{pi}{2}) sin(2 theta)` `= [ – frac{1}{2} cos (2 theta) ]_0^(frac{pi}{2})`
  `= – frac{1}{2} cos \ pi + frac{1}{2} cos \ 0`
  `= 1`

 
`y = sin(2 theta) \ => \ text{Range} \ [0, 1] \ \ text{for}\ \ theta ∈ [0, frac{pi}{2}]`

`sin^(2n+1) (2 theta)` `≤ sin (2 theta) \ \ text{for}\ \ theta ∈ [0, frac{pi}{2}]`
`sin^(2n+1) (2 theta)` `≤ 1`
`I_n` `≤ 1`
`therefore (2^n n!)^2` `≤ (2n + 1) !`

Mechanics, EXT2 M1 2020 HSC 16a

Two masses, `2m` kg and `4m` kg, are attached by a light string. The string is placed over a smooth pulley as shown.

The two masses are at rest before being released and `v` is the velocity of the larger mass at time `t` seconds after they are released.
 


 

The force due to air resistance on each mass has magnitude `kv`, where `k` is a positive constant.

  1. Show that  `frac{dv}{dt} = frac{gm-kv}{3m}`.   (2 marks)

  2. Given that  `v < frac{gm}{k}`, show that when  `t = frac{3m}{k} ln 2`, the velocity of the larger mass is  `frac{gm}{2k}`.  (3 marks)

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

i. 

♦ Mean mark part (i) 37%.

`text{Taking} \ v \ text{downwards as positive.}`

`text{Forces acting on}\ 2m\ text{mass:}`
  
`kv + 2 mg-T = -2m * frac{dv}{dt}\ …\ (1)`
  
`text{Forces acting our} \ 4m \ text{mass:}`
 
`4mg-kv-T = 4m * frac{dv}{dt}\ …\ (2)`

`text{Subtract} \ \ (2)-(1)`

`2 mg-2 kv` `= 6 m * frac{dv}{dt}`
`:. frac{dv}{dt}` `= frac{2mg-2 kv}{6 m}`
  `= frac{gm-kv}{3m}`

 

ii.    `frac{dv}{dt}` `= frac{gm-kv}{3m}`
  `frac{dt}{dv}` `= frac{3m}{gm-kv}`
  `t` `= int frac{3m}{gm-kv}\ dv`
    `= -frac{3m}{k} log_e |gm-kv | + c`

 
`text{When} \ \ t = 0, v = 0:`

`0` `= -frac{3m}{k} log_e \ | gm | + c`
`c` `= frac{3m}{k} log_e \ | gm | `
`t` `= frac{3m}{k} log_e  \ | gm | \-frac{3m}{k} log_e  \ | gm -kv |`
  `= frac{3m}{k} log_e \ | frac{mg}{gm-kv} |`

 
`text{Find} \ \ v\ \ text{when} \ \ t = frac{3m}{k} log_e 2 :`

`frac{3m}{k} log_e 2` `= frac{3m}{k} log_e | frac{gm}{gm-kv} |`
`2` `= frac{gm}{gm-kv}`
`2gm-2kv` `= gm`
`2kv` `= gm`
`therefore \ v` `= frac{gm}{2k}`

 

Vectors, EXT2 V1 2020 HSC 15b

The point `C` divides the interval `AB` so that  `frac{CB}{AC} = frac{m}{n}`. The position vectors of `A` and `B` are `underset~a` and `underset~b` respectively, as shown in the diagram.
 

  1. Show that  `overset->(AC) = frac{n}{m + n} (underset~b - underset~a)`.  (2 marks)

     

  2. Prove that  `overset->(OC) = frac{m}{m + n} underset~a + frac{n}{m + n} underset~b`.  (1 mark)

Let `OPQR` be a parallelogram with  `overset->(OP) = underset~p`  and  `overset->(OR) = underset~r`. The point `S` is the midpoint of `QR` and `T` is the intersection of `PR` and `OS`, as shown in the diagram.
  
 
         
 

  1. Show that  `overset->(OT) = frac{2}{3} underset~r + frac{1}{3} underset~p`.  (3 marks)

  2. Using parts (ii) and (iii), or otherwise, prove that `T` is the point that divides the interval `PR` in the ratio 2 :1.   (1 mark)

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

i.  

`frac{overset->(AC)}{overset->(AB)}` `= frac{n}{m + n}`
`overset->(AC)` `= frac{n}{m + n} * overset->(AB)`
  `= frac{n}{m + n} (underset~b – underset~a)`

 

ii.    `overset->(OC)` `= overset->(OA) + overset->(AC)`
    `= underset~a + frac{n}{m + n} (underset~b – underset~a)`
    `= underset~a – frac{n}{m + n} underset~a + frac{n}{m + n} underset~b`
    `= (1 – frac{n}{m + n}) underset~a + frac{n}{m + n} underset~b`
    `= (frac{m + n – n}{m + n}) underset~a + frac{n}{m + n} underset~b`
    `= frac{m}{m + n} underset~a + frac{n}{m + n} underset~b`

 

iii.  `text{Show} \ \ overset->(OT) = frac{2}{3} underset~r + frac{1}{3} underset~p`

♦ Mean mark part (iii) 50%.
 

 
`text{Consider} \ \ Delta PTO \ \ text{and} \ \ Delta RTS:`

`angle PTO = angle RTS \ (text{vertically opposite})`

`angle OPT = angle SRT \ (text{vertically opposite})`

`therefore \ Delta PTO \ text{|||} \ Delta RTS\ \ text{(equiangular)}`
 

`OT : TS = OP : SR = 2 : 1`

`(text{corresponding sides in the same ratio})`

`frac{overset->(OT)}{overset->(OS)}` `= frac{2}{3}`
`overset->(OT)` `= frac{2}{3} overset->(OS)`
  `= frac{2}{3} ( underset~r + frac{1}{2} underset~p)`
  `= frac{2}{3} underset~r + frac{1}{3} underset~p`
Mean mark part (iv) 51%.

 

iv.   `text{Let} \ \ overset->(OT) \ text{divide} \ PR\ text{so that}\ \ frac{TR}{PT} = frac{m}{n}`

`text{Using part (ii):}`

`overset->(OT)` `= frac{m}{m + n} underset~p + frac{n}{m + n} c`
`overset->(OT)` `= frac{1}{3} underset~p + frac{2}{3} underset~r \ \ \ (text{part (iii)})`
`frac{m}{m + n}` `= frac{1}{3} , frac{n}{m + n} = frac{2}{3}`

 
`=> \ m = 1 \ , \ n = 2`
 

`therefore \ T \ text{divides} \ PR \ text{in ratio  2 : 1}.`

Proof, EXT2 P1 2020 HSC 15a

In the set of integers, let `P` be the proposition:

'If  `k + 1`  is divisible by 3, then  `k^3 + 1`  divisible by 3.'

  1. Prove that the proposition `P` is true.  (2 marks)

  2. Write down the contrapositive of the proposition `P`.  (1 mark)

  3. Write down the converse of the proposition `P` and state, with reasons, whether this converse is true or false.  (3 marks)

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

i.     `text{Let} \ \ k + 1 = 3N, \ N∈ Z`

`=>  k = 3N – 1`

`k^3 + 1` `= (3N -1)^3 + 1`
  `= (3N)^3 + 3(3N)^2 (-1) + 3(3N)(-1)^2 + (-1)^3 + 1`
  `= 27N^3 – 27N^2 + 9N – 1 + 1`
  `= 3 (9N^3 – 9N^2 + 3N)`
  `= 3Q \ , \ Q ∈ Z`

 
`therefore \ text{If} \ \ k+ 1 \ \ text{is divisible by 3}, text{then} \ \ k^3 + 1 \ \ text{is divisible by 3.}`
 

ii.    `text{Contrapositive}`

`text{If} \ \ k^3 + 1 \ \ text{is not divisible by 3, then}\ \ k + 1\ \ text{is not divisible by 3.}`
 

♦♦ Mean mark part (iii) 36%.

iii.   `text{Converse:}`

`text{If} \ \ k^3 + 1\ \ text{is divisible by 3, then}\ \ k + 1\ \ text{is divisible by 3.}`

`text(Contrapositive of converse:)`

`text{If}\ \ k + 1\ \ text{is not divisible by 3, then}\ \ k^3 + 1\ \ text{is not divisible by 3.}`
 
`text(i.e.)\ \ k + 1 \ \ text{is not divisible by 3 when}\ \ k + 1 = 3Q + 1\ \ text{or}\ \ k + 1 = 3Q + 2, text{where}\ Q ∈ Z`
 

`text{If} \ \ k + 1` `= 3Q + 1\ \ => \ k=3Q`
`k^3 + 1` `= (3Q)^3 + 1`
  `= 27Q^3 + 1`
  `= 3(9Q^3) + 1`
  `= 3M + 1 \ \ (text{not divisible by 3,}\ M ∈ Z)`

 

`text{If} \ \ k + 1` `= 3Q + 2\ \ => \ k=3Q+1`
`k^3 + 1` `= (3Q + 1)^3 + 1`
  `= (3Q)^3 + 3(3Q)^2 + 3(3Q) + 1 + 1`
  `= 27Q^3 + 27Q^2 + 9Q + 2`
  `= 3(9Q^3 + 9Q^2 + 3Q) + 2`
  `= 3M + 2 \ (text{not divisible by 3,}\ M ∈ Z) `

 

`therefore \ text{By contrapositive, if}\ \ k^3 + 1\ \ text {is divisible by 3, k + 1 is divisible by 3.}`

Statistics, STD1 S1 2020 HSC 24

  1. The ages in years, of ten people at the local cinema last Saturday afternoon are shown.

\(38 \ \ 25 \ \ 38 \ \ 46 \ \ 55 \ \ 68 \ \ 72 \ \ 55 \ \ 36 \ \ 38\)

  1. The mean of this dataset is 47.1 years.
  2. How many of the ten people were aged between the mean age and the median age?  (2 marks)

  3. On Wednesday, ten people all aged 70 went to this same cinema.
  4. Would the standard deviation of the age dataset from Wednesday be larger than, smaller than or equal to the standard deviation of the age dataset given in part (a)? Briefly explain your answer without performing any calculations.  (2 marks)

Show Answers Only

a.     \(1\)

b.    \(\text{Standard deviation is a measure of how much the}\)

\(\text{ages of individuals differ from the mean age of the group.}\)
 

\(\Rightarrow\ \text{Standard deviation of Wednesday’s group would be}\)

\(\text{less as the mean is 70 and everyone’s age is 70.}\)

Show Worked Solution

a.     \(\text{Reorder ages in ascending order:}\)

    \(25, 36, 38, 38, 38, 46, 55 , 55, 68, 72\)

\(\text{Median} = \dfrac{\text{5th + 6th}}{2} = \dfrac{38 + 46}{2} = 42\)

\(\therefore\ \text{People with age between 42 − 47.1 = 1}\)

♦ Mean mark (a) 39%.

 
b.    \(\text{Standard deviation is a measure of how much the}\)

\(\text{ages of individuals differ from the mean age of the group.}\)
 

\(\Rightarrow\ \text{Standard deviation of Wednesday’s group would be}\)

\(\text{less as the mean is 70 and everyone’s age is 70.}\)

♦♦♦ Mean mark (b) 20%.

Financial Maths, STD1 F3 2020 HSC 30

Colin takes out a 5-year reducing balance loan of $19 000 with interest charged at 6% per annum. He uses this money to buy a car valued at $19 000.

The table shows some of the output from a spreadsheet used to model the reducing balance loan.
 


 

Colin's car is depreciated using the declining-balance method, with a depreciation rate of 20% per annum.

At the end of 3 years, after making the third repayment on the loan, Colin sells the car at its salvage value. He uses the money from the sale of the car to repay the amount owing on the loan at the end of the third year.

How much money will he have left over?  (4 marks)

Show Answers Only

`$1458.43`

Show Worked Solution

`V_0 = 19\ 000 \ , \ r = 20text(%) \ , \ n=3`

♦ Mean mark 23%.
`S` `= V_0 (1 – r)^n`
  `= 19\ 000 (1 – 0.2)^3`
  `= $9728`

 

`text{Find the amount owing on the loan after 3 years:}`

`text(Using the table,)`

`text{Interest (year 3)} = 0.06 xx 12\ 056.70 = $ 723.40`

`text{Amount owing (end of year 3)`

`= 12\ 056.70 + 723.40 – 4510.53`

`= $ 8269.57`

 

`therefore \ text{Money left over}`

`= 9728 – 8269.57`

`= $1458.43`

Probability, STD1 S2 2020 HSC 26

Barbara plays a game of chance, in which two unbiased six-sided dice are rolled. The score for the game is obtained by finding the difference between the two numbers rolled. For example, if Barbara rolls a 2 and a 5, the score is 3.

The table shows some of the scores.
 


 

  1. Complete the six missing values in the table to show all possible scores for the game.   (1 mark)
  2. What is the probability that the score for a game is NOT 0?  (2 marks)

Show Answers Only
  1.  

     
  2. `frac{5}{6}`
Show Worked Solution

a.     

♦ Mean mark part (b) 47%.
b.       `Ptext{(not zero)}` `= frac{text(numbers) ≠ 0}{text(total numbers)}`
    `= frac{30}{36}`
    `= frac{5}{6}`

 
\(\text{Alternate solution (b)}\)

b.       `Ptext{(not zero)}` `= 1 – Ptext{(zero)}`
    `= 1 – frac{6}{36}`
    `= frac{5}{6}`

Financial Maths, STD1 F2 2020 HSC 25

Tom is offered two different investment options.

Option A: 10% per annum simple interest.

Option B: 9% per annum interest, compounded annually.

Tom has $1000 to invest. The graph shows the future values over time of $1000 invested using Option B.
 


 

Tom wants to find the difference between the future values after 8 years using these two investment options.

By first drawing, on the grid above, the graph of the future values of $1000 invested using Option A, estimate the difference between the future values after 8 years.   (3 marks)

Show Answers Only

`$200`

Show Worked Solution

♦♦ Mean mark 30%.

`text{When}\ \ t = 8 \ text{years:}`

`text{Option} \ A = $1800 \ , \ text{Option} \ B = $2000`

`therefore \ text{Estimate of difference}` `= 2000 – 1800`
  `= $ 200`

Measurement, STD1 M5 2020 HSC 23

The diagram shows a scale drawing of the floor of a room.
 


 

Carpet is to be laid to cover the entire floor. The cost of the carpet is $100 per square metre.

Find the total cost of the carpet required.  (3 marks)

Show Answers Only

`$9000`

Show Worked Solution


 

`text{Ratio of sides}` `= \ text{base : height}`
  `=8 : 5`

 

`text{Find actual} \ h :`

COMMENT: Note the image dimensions on the actual HSC paper were exactly 8 cm × 5 cm, making this ratio easy to calculate.
`frac{h}{12}` `= frac{5}{8}`
`h` `= frac{5 xx 12}{8} = 7.5`

 

♦ Mean mark 41%.

`text{Area (actual)} = 12 xx 7.5 = 90 \ text{m}^2`

`therefore \ text{C} text{ost}` `= 90 xx 100`
  `= $9000`

Statistics, STD1 S3 2020 HSC 22

A group of students sat a test at the end of term. The number of lessons each student missed during the term and their score on the test are shown on the scatterplot.
 


 

  1. Describe the strength and direction of the linear association observed in this dataset.  (2 marks)

  2. Calculate the range of the test scores for the students who missed no lessons.  (1 mark)

  3. Draw a line of the best fit in the scatterplot above.  (1 mark)
  4. Meg did not sit the test. She missed five lessons.

     

    Use the line of the best fit drawn in part (c) to estimate Meg's score on this test. (1 mark)

  5. John also did not sit the test and he missed 16 lessons.

     

    Is it appropriate to use the line of the best fit to estimate his score on the test? Briefly explain your answer. (1 mark)

Show Answers Only

a.    \(\text{Strength : strong}\)

\(\text{Direction : negative} \)

b.    \(\text{Range}\ = \text{high}-\text{low}\ = 100-80=20\)
 

c.   

d. 


 

e.    \(\text{John’s missed days are too extreme and the LOBF is not}\)

\(\text{appropriate. The model would estimate a negative score for}\)

\(\text{John which is impossible.}\)

Show Worked Solution

a.    \(\text{Strength : strong}\)

\(\text{Direction : negative} \)

♦ Mean mark (a) 45%.
♦♦ Mean mark (b) 31%.

b.    \(\text{Range}\ = \text{high}-\text{low}\ = 100-80=20\)
 

c.   

d. 


 
\(\therefore\ \text{Meg’s estimated score = 40}\)
 

e.    \(\text{John’s missed days are too extreme and the LOBF is not}\)

\(\text{appropriate. The model would estimate a negative score for}\)

\(\text{John which is impossible.}\)

♦ Mean mark (e) 38%.

Algebra, STD1 A2 2020 HSC 20

The weight of a bundle of A4 paper (`W` kg) varies directly with the number of sheets (`N`) of A4 paper that the bundle contains.

This relationship is modelled by the formula  `W = kN`, where  `k`  is a constant.

The weight of a bundle containing 500 sheets of A4 paper is 2.5 kilograms.

  1. Show that the value of  `k`  is 0.005.   (1 mark)

  2. A bundle of A4 paper has a weight of 1.2 kilograms. Calculate the number of sheets of A4 paper in the bundle.   (2 marks)

Show Answers Only
  1. `text{Show Worked Solutions}`
  2. `240 \ text{sheets}`
Show Worked Solution

a.     `W = 2.5\ text{kg when} \  N = 500:`

`2.5` `= k xx 500`
`therefore \ k` `= frac{2.5}{500}`
  `= 0.005`

 

b.     `text{Find}\ \ N \ \ text{when} \ \ W = 1.2\ text{kg:}`

♦ Mean mark 50%.
`1.2` `= 0.005 xx N`
`therefore N` `= frac{1.2}{0.005}`
  `= 240 \ text{sheets}`

Algebra, STD1 A3 2020 HSC 19

Each year the number of fish in a pond is three times that of the year before.

  1. The table shows the number of fish in the pond for four years.
    \begin{array} {|l|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
    \hline
    \rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & & & 2700\\
    \hline
    \end{array}

    Complete the table above showing the number of fish in 2021 and 2022.   (2 marks)
     

  2. Plot the points from the  table in part (a) on the grid.   (2 marks)
     
  3. Which model is more suitable for this dataset: linear or exponential? Briefly explain your answer.   (2 marks)

Show Answers Only

a.   

\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & 300 & 900 & 2700\\
\hline
\end{array}

b.   
       

c.     The more suitable model is exponential.

A linear dataset would graph a straight line which is not the case here.

An exponential curve can be used to graph populations that grow at an increasing rate, such as this example.

Show Worked Solution

a.        

\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & 300 & 900 & 2700\\
\hline
\end{array}

b.  

c.     The more suitable model is exponential.

A linear dataset would graph a straight line which is not the case here.

An exponential curve can be used to graph populations that grow at an increasing rate, such as this example.

♦ Mean mark (c) 31%.

Algebra, STD1 A1 2020 HSC 18

The distance, `d` metres, travelled by a car slowing down from `u` km/h to `v` km/h can be obtained using the formula

`v^2 = u^2-100 d`

What distance does a car travel while slowing down from 70 km/h to 40 km/h?   (2 marks)

Show Answers Only

`33 \ text{metres}`

Show Worked Solution

`u = 70 \ , \ v = 40`

♦♦ Mean mark 21%.
`v^2` `= u^2-100d`
`40^2` `= 70^2-100d`
`100d` `= 70^2-40^2`
`:. d` `= frac{70^2-40^2}{100}`
  `= 33 \ text{metres}`

Measurement, STD1 M2 2020 HSC 15

The time in Melbourne is 11 hours ahead of Coordinated Universal Time (UTC). The time in Honolulu is 10 hours behind UTC. A plane departs from Melbourne at 7 pm on Tuesday and lands in Honolulu 9 hours later.

What is the time and day in Honolulu when the plane lands?   (2 marks)

Show Answers Only

`7\ text{am (Tuesday)}`

Show Worked Solution

`text{Melbourne} \ to \ text{UTC}\ + 11`

♦♦ Mean mark 28%.

`text{Honolulu} \ to \ text{UTC}\ – 10`

`=>\ text{Honolulu is 21 hours behind Melbourne}`
 

`therefore \ text{Plane landing time}`

STRATEGY: 21 hours behind = 1 day behind + 3 hours.

`= 7\ text{pm} + 9 \ text{hours}`

`= 4\ text{am (Wednesday – Melb)}`

`= 7\ text{am (Tuesday – Honolulu)}`

Algebra, STD1 A3 2020 HSC 14

Adam travels on a straight road away from his home. His journey is shown in the distance – time graph.

  1. Describe the journey in the first 4 minutes by referring to change in speed and distance travelled.   (2 marks)

  2. After the 4 minutes shown on the graph. Adam rests for 2 minutes and then return home by travelling on the same road at a constant speed. Adam is away from home for a total of 10 minutes.

     

    On the above, complete the distance-time using the information provided.   (2 marks)

Show Answers Only

  1. `text{Speed: Adam increases speed until approximately}\ \ t=2,`

    `text{and then decreases speed until he stops when}\ \ t=4.`

    `text{Distance travelled: Adam’s distance from home increases}`

    `text{at an increasing rate until}\ \ t=2, \ text{and then continues}`

    `text{to increase but at a decreasing rate until}\ \ t=4, \ text(when the)`

    `text(distance from home remains the same.)`

  2.  
Show Worked Solution

a.    `text{Speed: Adam increases speed until approximately}\ \ t=2,`

♦ Mean mark part (a) 35%.

`text{and then decreases speed until he stops when}\ \ t=4.`

`text{Distance travelled: Adam’s distance from home increases}`

`text{at an increasing rate until}\ \ t=2, \ text{and then continues}`

`text{to increase but at a decreasing rate until}\ \ t=4, \ text(when the)`

`text(distance from home remains the same.)`

Mean mark part (b) 51%.

 

b.  `text(S)text(ince Adam travels home at a constant speed, the graph is)`

`text{is a straight line and ends at (10, 0).}`
 

Financial Maths, STD1 F2 2020 HSC 13

Taro needs $1000 in 5 years time. He is going to invest some money today in an account earning 3% per annum compounded annually. He will make no further deposits or withdrawals.

How much money does he need to invest today?   (3 marks)

Show Answers Only

`$862.61`

Show Worked Solution

♦♦ Mean mark 29%.
`FV` `= PV (1 + r)^n`
`1000` `= PV (1 + frac{3}{100})^5`
`1000` `= PV (1.0.3)^5`
`:. PV` `= frac{1000}{(1.03)^5}`
  `= $862.61`

Algebra, STD1 A3 2020 HSC 29

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

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

     

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

     

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

     

  2. Tank B remains empty until  `t=15`  when water is added to it at a constant rate of 30 litres per minute.
     By drawing a line on the grid (above), or otherwise, find the value of  `t`  when the two tanks contain the same volume of water.  (2 marks)
  3. Using the graphs drawn, or otherwise, find the value of  `t`  (where  `t > 0`) when the total volume of water in the two tanks is 1000 litres.  (1 mark)

Show Answers Only
  1.  `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}` 
      

     
  2. `29 \ text{minutes}`
  3. `45 \ text{minutes}`
Show Worked Solution

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

♦ Mean mark part (a) 45%.
 


 

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

♦♦ Mean mark part (b) 24%.
 

   

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

 
c.   `text{Strategy 1}`

♦♦♦ Mean mark part (c) 5%.

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

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

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

`text{Strategy 2}`

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

Financial Maths, STD1 F1 2020 HSC 27

The table shows the income tax rates for the 2019 – 2020 financial year.

For the 2019 – 2020 financial year, Wally had a taxable income of $122 680. During the year, he paid $3000 per month in Pay As You Go (PAYG) tax.

Calculate Wally's tax refund, ignoring the Medicare levy.   (3 marks)

Show Answers Only

`$3111.40`

Show Worked Solution
`text(Tax paid)` `=12 xx 3000`
  `=$36\ 000`

 
`text(Tax payable on $122 680)`

♦ Mean mark 35%.

`=20\ 797 + 0.37(122\ 680 – 90\ 000)`

`=20\ 797 + 0.37(32\ 680)`

`=$32\ 888.60`
 

`:.\ text(Tax refund)` `=36\ 000 – 32\ 888.60`  
  `=$3111.40`  

Networks, STD1 N1 2020 HSC 21

The diagram represents a network with weighted edges.
 


 

  1. Draw a minimum spanning tree for this network and determine its length.   (3 marks)

  2. The network is revised by adding another vertex, `K`. Edges `AK` and `CK` have weights of 12 and 10 respectively, as shown.
     

   
 

What is the length of the minimum spanning tree for this revised network?   (1 mark)

Show Answers Only
  1.  `text(Length = 14)`

     

     `text(One of many possibilities:)`
     

     

  2. `24`
Show Worked Solution

a.      `text{Using Kruskal’s Algorithm (one of many possibilities):}`

♦♦ Mean mark part (a) 32%.

`text{Edge 1 :}\ GH\ (1)`
`text{Edge 2 :}\ FH\ (2)`
`text{Edge 3 :}\ CF\ (2)`
`text{Edge 4 :}\ FD\ (2)`
`text{Edge 5 :}\ DE\ (2)`
`text{Edge 6 :}\ BC\ (3)`
`text{Edge 7 :}\ AB\ (2)`
 


 

`text{Minimum length of spanning tree}` `= 1 + 2 + 2 + 2 +2 + 3 +2`
  `= 14`
♦ Mean mark part (b) 45%.

 

b.     `text{Add}\ CK \ text{to the minimum spanning tree in (a).}`

`therefore \ text(Revised length)` `= 14 + 10`
  `= 24`

Measurement, STD1 M3 2020 HSC 11

Consider the triangle shown.
 


 

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

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

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

♦ Mean mark 44% and 39% for part (a) and (b) respectively.

 

b.     `text{Using Pythagoras:}`

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

Networks, STD1 N1 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

`text(Vertices = players)`

♦♦ Mean mark 34%.

`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`

Financial Maths, STD1 F2 2020 HSC 8 MC

Joan invests $200. She earns interest at 3% per annum, compounded monthly.

What is the future value of Joan's investment after 1.5 years?

  1. $209.07
  2. $209.19
  3. $279.51
  4. $311.93
Show Answers Only

`B`

Show Worked Solution

`text(Monthly interest rate) \ = frac(0.03)(12)`

♦♦ Mean mark 33%.

`n \ = \ 1.5 xx 12 = 18`
  

`FV` `= PV(1 + r)^n`
  `= 200 (1 + frac(0.03)(12))^18`
  `= $209.19`

 
`=> \ B`

Calculus, EXT1 C2 2020 HSC 13c

Suppose  `f(x) = tan(cos^(-1)(x))`  and  `g(x) = (sqrt(1-x^2))/x`.

The graph of  `y = g(x)`  is given.
 

  1. Show that  `f^(′)(x) = g^(′)(x)`.  (4 marks)

  2. Using part (i), or otherwise, show that  `f(x) = g(x)`.  (3 marks) 

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

i.   `f(x) = tan(cos^(-1)(x))`

♦ Mean mark (i) 50%.
`f^(′)(x)` `= -1/sqrt(1-x^2) · sec^2(cos^(-1)(x))`
  `= -1/sqrt(1-x^2) · 1/(cos^2(cos^(-1)(x)))`
  `= -1/(x^2sqrt(1-x^2))`

 
`g(x) = (1-x^2)^(1/2) · x^(-1)`

`g^(′)(x)` `= 1/2 · -2x(1-x^2)^(-1/2) · x^(-1)-(1-x^2)^(1/2) · x^(-2)`
  `= (-x)/(x sqrt(1-x^2))-sqrt(1-x^2)/(x^2)`
  `= (-x^2-sqrt(1-x^2) sqrt(1-x^2))/(x^2 sqrt(1-x^2))`
  `= (-x^2-(1-x^2))/(x^2sqrt(1-x^2))`
  `= -1/(x^2sqrt(1-x^2))`
  `=f^(′)(x)`

 

ii.   `f^(′)(x) = g^(′)(x)`

♦♦♦ Mean mark (ii) 15%.

`=> f(x) = g(x) + c`
 

`text(Find)\ c:`

`f(1)` `= tan(cos^(-1) 1)`
  `= tan 0`
  `= 0`

`g(1) = sqrt(1-1)/0 = 0`

`f(1) = g(1) + c`

`:. c = 0`

`:. f(x) = g(x)`

Statistics, STD2 S5 2020 HSC 35

The intelligence Quotient (IQ) scores for adults in City A are normally distributed with a mean of 108 and a standard deviation of 10.

The IQ score for adults in City B are normally distributed with a mean of 112 and a standard deviation of 16.

  1. Yin is an adult who lives in City A and has an IQ score of 128.
    What percentage of the adults in this city have an IQ score higher than Yin's?   (2 marks) 

  2. There are 1 000 000 adults living in City B.
    Calculate the number of adults in City B that would be expected to have an IQ score lower than Yin's.  (2 marks)

  3. Simon, an adult who lives in City A, moves to City B. The  `z` -score corresponding to his IQ score in City A is the same as the `z`-score corresponding to his IQ score in City B.
    By first forming an equation, calculate Simon's IQ score. Give your answer correct to one decimal place.   (3 marks)

Show Answers Only
  1. `text(2.5%)`
  2. `840 \ 000`
  3. `101.3`
Show Worked Solution

a.    `text{In City A:}`

♦ Mean mark 50%.

`z text{-score}\ (128) = frac {x -mu}{sigma} = frac{128 – 108}{10} = 2`
 


 

`therefore\ text{2.5% have a higher IQ in City} \ A.`
 

b.    `text{In City B:}`

♦ Mean mark 44%.

`z text{-score}\ (128) = frac{128 – 112}{16} = 1`
 


 

`therefore \ text{Adults in City} \ B \ text{with an IQ}\ <  128`

`= 84text(%) xx 1 \ 000 \ 000`

`= 840 \ 000`
 

c.    `z text{-score in City A}\ = z text{-score in City B}`

♦♦♦ Mean mark 17%.
`frac{x – 108}{10} ` `= frac{x -112}{16} \ \ text{(multiply b.s.} \ xx 160 text{)}`
`16 (x – 108)` `= 10 (x – 112)`
`16 x – 1728` `= 10 x – 1120`
`6x` `= 608`
`x` `= 101.3`

  
`therefore \ text{Simon’s IQ} = 101.3 \ text{(to 1 d.p.)}`

Measurement, STD2 M6 2020 HSC 32

The diagram shows a regular decagon (ten-sided shape with all sides equal and all interior angles equal). The decagon has centre `O`.
 

The perimeter of the shape is 80 cm.

By considering triangle `OAB`, calculate the area of the ten-sided shape. Give your answer in square centimetres correct to one decimal place.  (4 marks)

Show Answers Only

`492.4\ text(cm²)`

Show Worked Solution

`angle AOB = 360/10 = 36^@`

♦♦ Mean mark 33%.

`AB = 80/10 = 8\ text(cm)`

`DeltaAOB\ text(is made up of 2 identical right-angled triangles)`
 

`tan 18^@` `= 4/x`
`x` `= 4/(tan 18^@)`

 

`:.\ text(Area of decagon)` `= 20 xx 1/2 xx 4/(tan 18^@) xx 4`
  `= 492.429…`
  `= 492.4\ text(cm²  (to 1 d.p.))`

Measurement, STD2 M6 2020 HSC 31

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


 

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

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

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

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

 

b.   `text(Using cosine rule:)`

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

 
c.

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

♦♦ Mean mark 22%.

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

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

 

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

`= 180 – (69 – 35)`

`= 146^@`