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Networks, STD2 N2 2010 FUR1 5 MC

vcaa-networks-fur1-2010-5
 

For the network above, the length of the minimal spanning tree is

A.   `30`

B.   `31`

C.   `35`

D.   `39`

Show Answers Only

`C`

Show Worked Solution

`text(Using Kruskal’s algorithm:)`

`text{Edge 1: 2 (least weight)}`

`text(Edge 2: 3)`

`text(Edge 3: 4)`

`text(Edges 4-5: 5)`

`text{Edges 6-7: 8 (unused edges with weights <8 create circuits and are ignored)}`
 

vcaa-networks-fur1-2010-5i

 
`:.\ text(Minimal spanning tree)`

`= 2+3+4+5+5+8+8`

`= 35`

`=>  C`

Networks, STD2 N2 2017 FUR1 2 MC

Two graphs, labelled Graph 1 and Graph 2, are shown below.
 

 
The sum of the degrees of the vertices of Graph 1 is

  1. two less than the sum of the degrees of the vertices of Graph 2.
  2. one less than the sum of the degrees of the vertices of Graph 2.
  3. equal to the sum of the degrees of the vertices of Graph 2.
  4. two more than the sum of the degrees of the vertices of Graph 2.
Show Answers Only

`C`

Show Worked Solution

`text(Graph 1)`

`∑\ text(degrees)\ = 3 + 3 + 3 + 3 = 12`

`text(Graph 2)`

`∑\ text(degrees)\ = 2 + 2 + 2 + 2 + 2 + 2 = 12`

`=> C`

Networks, FUR1 2017 VCE 1 MC

Which one of the following graphs contains a loop?

A. B.
C. D.
Show Answers Only

`B`

Show Worked Solution

`text(A loop occurs when an edge connects a vertex to)`

`text{itself (without going through an other node).}`

`=> B`

Networks, STD2 N2 SM-Bank 3 MC

A store manager is directly in charge of five department managers.

Each department manager is directly in charge of six sales people in their department.

This staffing structure could be represented graphically by

A.   a tree.

B.   a path.

C.   a cycle.

D.   a weighted graph.

Show Answers Only

`A`

Show Worked Solution

`=>  A`

Calculus, MET1 2011 ADV 4d

  1. Differentiate  `y=sqrt(9-x^2)`  with respect to  `x`.   (2 marks)

Show Answers Only
  1. `- x/sqrt(9-x^2)`
  2. `-6 sqrt(9-x^2) + c`
Show Worked Solution
IMPORTANT: Some students might find calculations easier by rewriting the equation as `y=(9-x^2)^(1/2)`.
a.    `y` `= sqrt(9-x^2)`
    `= (9-x^2)^(1/2)`

 
`text{Using the function of a function rule (or “chain” rule)}`

`dy/dx` `=1/2 xx (9-x^2)^(-1/2) xx d/dx (9-x^2)`
  `= 1/2 xx (9-x^2)^(-1/2) xx -2x`
  `=-x/sqrt(9-x^2)`

 

b.    `int (6x)/sqrt(9-x^2)\ dx` `= -6 int (-x)/sqrt(9-x^2)\ dx`
    `= -6 sqrt(9-x^2) + c`

Calculus, MET1 2015 ADV 12c

Find  `f^{′}(x)`, where  `f(x) = (x^2 + 3)/(x-1).`   (2 marks)

Show Answers Only

`((x-3) (x + 1))/(x-1)^2`

Show Worked Solution

`f(x) = (x^2 + 3)/(x-1)`

`text(Using the quotient rule)`

`u` `= x^2 + 3` `\ \ \ \ \ \ v` `= x-1`
`u^{′}` `= 2x` `\ \ \ \ \ \ v^{′}` `= 1`
`f′(x)` `= (u^{′} v-uv^{′})/v^2`
  `= (2x (x-1)-(x^2 + 3) xx 1)/(x-1)^2`
  `= (2x^2-2x-x^2-3)/(x-1)^2`
  `= (x^2-2x-3)/(x-1)^2`
  `= ((x-3) (x + 1))/(x-1)^2`

Calculus, MET1 2015 VCAA 1b

Let  `f(x) = (log_e(x))/(x^2)`.

  1. Find  `f^{prime}(x)`.   (2 marks)

  2. Evaluate  `f^{prime}(1)`.   (1 mark)

Show Answers Only
  1. `(1 – 2log_e(x))/(x^3)`
  2. `1`
Show Worked Solution

i.   `text(Using Quotient Rule:)`

`(h/g)^{prime}` `= (h^{prime}g-hg^{prime})/(g^2)`
`f^{prime}(x)` `= ((1/x)x^2-log_e(x)*2x)/(x^4)`
  `= (1-2log_e(x))/(x^3)`

 

ii.    `f^{prime}(1)` `= (1-2log_e(1))/(1^3)`
    `= 1`

GRAPHS, FUR2 2017 VCAA 3

Lifeguards are required to ensure the safety of swimmers at the beach.

Let `x` be the number of junior lifeguards required.

Let `y` be the number of senior lifeguards required.

The inequality below represents the constraint on the relationship between the number of senior lifeguards required and the number of junior lifeguards required.

 Constraint 1  `y >= x/4`
 

  1. If eight junior lifeguards are required, what is the minimum number of senior lifeguards required?  (1 mark)

 
There are three other constraints.

 Constraint 2  `x ≥ 6`

 Constraint 3  `y ≥ 4`

 Constraint 4  `x + y ≥ 12`

  1. Interpret Constraint 4 in terms of the number of junior lifeguards and senior lifeguards required.  (1 mark)

 
The shaded region of the graph below contains the points that satisfy Constraints 1 to 4.

All lifeguards receive a meal allowance per day.

Junior lifeguards receive $15 per day and senior lifeguards receive $25 per day.

The total meal allowance cost per day, `$C`, for the lifeguards is given by

`C = 15x + 25y`

  1. Determine the minimum total meal allowance cost per day for the lifeguards.  (2 marks)
  2. On rainy days there will be no set minimum number of junior lifeguards or senior lifeguards required, therefore:

     

    • Constraint 2  `(x ≥ 6)`  and Constraint 3  `(y ≥ 4)`  are removed

     

    • Constraint 1 and Constraint 4 are to remain.
     

     

              Constraint 1  `y >= x/4`

     

              Constraint 4  `x + y >= 12`

     


    The total meal allowance cost per day,     `$C`, for the lifeguards remains as

     

    `C = 15x + 25y`

     


    How many junior lifeguards and senior lifeguards work on a rainy day if the total meal allowance cost     is to be a minimum?

     

    Write your answers in the boxes provided below.  (1 mark)


Show Answers Only

a.   `2`

b.   `text(The total combined number of junior and)`

`text(senior lifeguards must be at least 12.)`

c.   `$220`

d.   
Show Worked Solution

a.   `text(Minimum senior lifeguards) = 8/4 = 2`

 

b.   `text(The total combined number of junior and)`

`text(senior lifeguards must be at least 12.)`

 

c.   `text(Minimum cost occurs at (8, 4))`

`:. C_text(min)` `= 15 xx 8 + 25 xx 4`
  `= $220`

 

d.   `text(Consider the graph without the restrictions)`

♦♦ Mean mark 24%.
MARKER’S COMMENT: A common incorrect answer was 10 and 3.

`x >= 6quadtext(and)quady >= 4:`

`text(By inspection, intersection around (9.5, 2.4))`

`text(⇒ Minimum allowance when)`

GRAPHS, FUR2 2017 VCAA 1

The graph below shows the depth of water in a sea bath from 6.00 am to 8.00 pm.
 

 

  1. What was the maximum depth, in metres, of water in the sea bath?   (1 mark)

  2. The sea bath was open to the public when the depth of water was above 1.5 m.
  3. Between which times was the sea bath open?   (1 mark)
Show Answers Only

a.   `text(2 metres)`

b.   `text(8 am – 6 pm)`

Show Worked Solution

a.   `text(2 metres)`

b.   `text(8 am – 6 pm)`

GEOMETRY, FUR2 2017 VCAA 1

Miki is planning a gap year in Japan.

She will store some of her belongings in a small storage box while she is away.

This small storage box is in the shape of a rectangular prism.

The diagram below shows that the dimensions of the small storage box are 40 cm × 19 cm × 32 cm.
 

The lid of the small storage box is labelled on the diagram above.

    1. What is the surface area of the lid, in square centimetres?  (1 mark)
    2. What is the total outside surface area of this storage box, including the lid and base, in square centimetres?  (1 mark)
  2. Miki has a large storage box that is also a rectangular prism.

      

    The large storage box and the small storage box are similar in shape.

      

    The volume of the large storage box is eight times the volume of the small storage box.

      

    The length of the small storage box is 40 cm.

    What is the length of the large storage box, in centimetres?  (1 mark)

Show Answers Only

a.i.   `760\ text(cm²)`

a.ii.  `5296\ text(cm²)`

b.     `80\ text(cm)`

Show Worked Solution
a.i.    `text{Area (lid)}` `= 40 xx 19`
    `= 760\ text(cm²)`

 

a.ii.    `text(Total S.A.)` `= 2 xx (32 xx 19) + 2 xx (40 xx 32) + 2 xx 760`
    `= 5296\ text(cm²)`

 

b.   `text(If volume scale factor = 8,)`

♦♦ Mean mark 32%.
MARKER’S COMMENT: A lack of understanding between linear and volume scale factors was again notable.

`text(⇒ Linear scale factor) = root(3)8 = 2`

`:.\ text(Length of large storage box)`

`= 2 xx 40`

`= 80\ text(cm)`

NETWORKS, FUR2 2017 VCAA 1

Bus routes connect six towns.

The towns are Northend (`N`), Opera (`O`), Palmer (`P`), Quigley (`Q`), Rosebush (`R`) and Seatown (`S`).

The graph below gives the cost, in dollars, of bus travel along these routes.

Bai lives in Northend (`N`) and he will travel by bus to take a holiday in Seatown (`S`).
 


 

  1. Bai considers travelling by bus along the route Northend (`N`) – Opera (`O`) – Seatown (`S`).

     

    How much would Bai have to pay?   (1 mark)

  2. If Bai takes the cheapest route from Northend (`N`) to Seatown (`S`), which other town(s) will he pass through?   (1 mark)

  3. Euler’s formula, `v + f = e + 2`, holds for this graph.

    Complete the formula by writing the appropriate numbers in the boxes provided below.   (1 mark)

     

Show Answers Only

a.   `$120`

b.   `text(Quigley and Rosebush.)`

c. 

       

Show Worked Solution
a.    `text(C)text(ost)` `= 15 + 105`
    `= $120`

 

b.   `text(Cheapest route is)\ N – Q – R – S`

`:.\ text(Other towns are Quigley and Rosebush.)`

 

c.   

MATRICES, FUR2 2017 VCAA 1

A school canteen sells pies (`P`), rolls (`R`) and sandwiches (`S`).

The number of each item sold over three school weeks is shown in matrix `M`.

`{:(qquadqquadqquadquadPqquadRqquadS),(M = [(35,24,60),(28,32,43),(32,30,56)]{:(text(week 1)),(text(week 2)),(text(week 3)):}):}` 

  1. In total, how many sandwiches were sold in these three weeks?   (1 mark)

  2. The element in row `i` and column `j` of matrix `M` is `m_(ij)`.
  3. What does the element `m_12` indicate?  (1 mark)

  4. Consider the matrix equation

    `[(35,24,60),(28,32,43),(32,30,56)] xx [(a),(b),(c)] = [(491.55),(428.00),(487.60)]`

    where `a` = cost of one pie, `b` = cost of one roll and `c` = cost of one sandwich.
  5.  i. What is the cost of one sandwich?   (1 mark)

The matrix equation below shows that the total value of all rolls and sandwiches sold in these three weeks is $915.60

`L xx [(491.55),(428.00),(487.60)] = [915.60]`

Matrix `L` in this equation is of order `1 × 3`.

  1. ii. Write down matrix `L`.   (1 mark)

Show Answers Only
  1. `159`
  2. `text(It represents the number of rolls sold in week 1.)`
    1. `$3.80`
    2. `text(Matrix)\ L = [0,1,1]`
Show Worked Solution
a.    `text(Total sandwiches)` `= 60 + 43 + 56`
    `= 159`

 
b.   `m_12 = 24`

`text(It represents the number of rolls sold in week 1.)`
 

c.i.    `[(a),(b),(c)]` `= [(35,24,60),(28,32,43),(32,30,56)]^(−1)[(491.55),(428.00),(487.60)]`
    `= [(4.65),(4.20),(3.80)]`

 
`:.\ text(C)text(ost of 1 sandwich = $3.80)`
 

c.ii.   `text(Matrix)\ L = [0,1,1]`

CORE, FUR2 2017 VCAA 5

Alex is a mobile mechanic.

He uses a van to travel to his customers to repair their cars.

The value of Alex’s van is depreciated using the flat rate method of depreciation.

The value of the van, in dollars, after `n` years, `V_n`, can be modelled by the recurrence relation shown below.

`V_0 = 75\ 000 qquad V_(n + 1) = V_n - 3375`

  1. Recursion can be used to calculate the value of the van after two years.

     

    Complete the calculations below by writing the appropriate numbers in the boxes provided.   (2 marks)


    1. By how many dollars is the value of the van depreciated each year?   (1 mark)
    2. Calculate the annual flat rate of depreciation in the value of the van.
    3. Write your answer as a percentage.   (1 mark)
  3. The value of Alex’s van could also be depreciated using the reducing balance method of depreciation.
  4. The value of the van, in dollars, after `n` years, `R_n`, can be modelled by the recurrence relation shown below.

     

            `R_0 = 75\ 000 qquad R_(n + 1) = 0.943R_n`

    At what annual percentage rate is the value of the van depreciated each year?   (1 mark)

Show Answers Only

a.

b.i.  `$3375`

b.ii. `4.5text(%)`

c.  `5.7text(%)`

Show Worked Solution

a.   

  
b.i.   `$3375`
  

b.ii.    `text(Annual Rate)` `= 3375/(75\ 000) xx 100`
    `= 4.5text(%)`

 

c.    `text(Annual Rate)` `= (1-0.943) xx 100text(%)`
    `= 5.7text(%)`

Algebra, MET2 2017 VCAA 2

Sammy visits a giant Ferris wheel. Sammy enters a capsule on the Ferris wheel from a platform above the ground. The Ferris wheel is rotating anticlockwise. The capsule is attached to the Ferris wheel at point `P`. The height of `P` above the ground, `h`, is modelled by  `h(t) = 65-55cos((pit)/15)`, where `t` is the time in minutes after Sammy enters the capsule and `h` is measured in metres.

Sammy exits the capsule after one complete rotation of the Ferris wheel.
 


 

  1. State the minimum and maximum heights of `P` above the ground.   (1 mark)

  2. For how much time is Sammy in the capsule?   (1 mark)

  3. Find the rate of change of `h` with respect to `t` and, hence, state the value of `t` at which the rate of change of `h` is at its maximum.   (2 marks)

As the Ferris wheel rotates, a stationary boat at `B`, on a nearby river, first becomes visible at point `P_1`. `B` is 500 m horizontally from the vertical axis through the centre `C` of the Ferris wheel and angle `CBO = theta`, as shown below.
 

   
 

  1. Find `theta` in degrees, correct to two decimal places.   (1 mark)

Part of the path of `P` is given by  `y = sqrt(3025-x^2) + 65, x ∈ [-55,55]`, where `x` and `y` are in metres.

  1. Find `(dy)/(dx)`.   (1 mark)

As the Ferris wheel continues to rotate, the boat at `B` is no longer visible from the point  `P_2(u, v)` onwards. The line through `B` and `P_2` is tangent to the path of `P`, where angle `OBP_2 = alpha`.
 

   
 

  1. Find the gradient of the line segment `P_2B` in terms of `u` and, hence, find the coordinates of `P_2`, correct to two decimal places.   (3 marks)

  2. Find `alpha` in degrees, correct to two decimal places.   (1 mark)

  3. Hence or otherwise, find the length of time, to the nearest minute, during which the boat at `B` is visible.   (2 marks)

Show Answers Only
  1. `h_text(min) = 10\ text(m), h_text(max) = 120\ text(m)`
  2. `30\ text(min)`
  3. `t = 7.5`
  4. `7.41^@`
  5. `(-x)/(sqrt(3025-x^2))`
  6. `P_2(13.00, 118.44)`
  7. `13.67^@`
  8. `7\ text(min)`
Show Worked Solution
a.    `h_text(min)` `= 65-55` `h_text(max)` `= 65 + 55`
    `= 10\ text(m)`   `= 120\ text(m)`

 

b.   `text(Period) = (2pi)/(pi/15) = 30\ text(min)`

 

c.   `h^{prime}(t) = (11pi)/3\ sin(pi/15 t)`

♦ Mean mark 50%.
MARKER’S COMMENT: A number of commons errors here – 2 answers given, calc not in radian mode, etc …

 

`text(Solve)\ h^{primeprime}(t) = 0, t ∈ (0,30)`

`t = 15/2\ \ text{(max)}`   `text(or)`   `t = 45/2\ \ text{(min – descending)}`

`:. t = 7.5`

 

d.   

♦ Mean mark 36%.
MARKER’S COMMENT: Choosing degrees vs radians in the correct context was critical here.

`tan(theta)` `= 65/500`
`:. theta` `=7.406…`
  `= 7.41^@`

 

e.    `(dy)/(dx)` `= (-x)/(sqrt(3025-x^2))`

 

f.   

`P_2(u,sqrt(3025-u^2) + 65),\ \ B(500,0)`

`:. m_(P_2B)` `= (sqrt(3025-u^2) + 65)/(u-500)`

 

`text{Using part (e), when}\ \ x=u,`

♦♦♦ Mean mark part (f) 18%.
MARKER’S COMMENT: Many students were unable to use the rise over run information to calculate the second gradient.

`dy/dx=(-u)/(sqrt(3025-u^2))`

 

`text{Solve (by CAS):}`

`(sqrt(3025-u^2) + 65)/(u-500)` `= (-u)/(sqrt(3025-u^2))\ \ text(for)\ u`

 

`u=12.9975…=13.00\ \ text{(2 d.p.)}`

 

`:. v` `= sqrt(3025-(12.9975…)^2) + 65`
  `= 118.4421…`
  `= 118.44\ \ text{(2 d.p.)}`

 

`:.P_2(13.00, 118.44)`

 

♦♦♦ Mean mark part (g) 7%.

g.    `tan alpha` `=v/(500-u)`
    `= (118.442…)/(500-12.9975…)`
  `:. alpha` `= 13.67^@\ \ text{(2 d.p.)}`

 

h.   

♦♦♦ Mean mark 5%.

`text(Find the rotation between)\ P_1 and P_2:`

`text(Rotation to)\ P_1 = 90-7.41=82.59^@`

`text(Rotation to)\ P_2 = 180-13.67=166.33^@`

`text(Rotation)\ \ P_1 → P_2 = 166.33-82.59 = 83.74^@`

 

`:.\ text(Time visible)` `= 83.74/360 xx 30\ text(min)`
  `=6.978…`
  `= 7\ text{min  (nearest degree)}`

Calculus, MET2 2017 VCAA 1

Let  `f : R → R,\  f (x) = x^3-5x`. Part of the graph of `f` is shown below.
 

  1. Find the coordinates of the turning points.   (2 marks)

  2. `A(−1, f (−1))`  and  `B(1, f (1))`  are two points on the graph of `f`.

     

    1. Find the equation of the straight line through `A` and `B`.   (2 marks)

    2. Find the distance `AB`.   (1 mark)

Let  `g : R → R, \ g(x) = x^3-kx, \ k ∈ R^+`.

  1. Let  `C(–1, g(−1))` and `D(1, g(1))` be two points on the graph of `g`.

     

    1. Find the distance `CD` in terms of `k`.   (2 marks)

    2. Find the values of `k` such that the distance `CD` is equal to  `k + 1`.   (1 mark)

  2. The diagram below shows part of the graphs of `g` and  `y = x`. These graphs intersect at the points with the coordinates `(0, 0)` and `(a, a)`.
  3.  
       
  4.  
    1. Find the value of `a` in terms of `k`.   (1 mark)

    2. Find the area of the shaded region in terms of `k`.   (2 marks)

Show Answers Only
  1. `(- sqrt15/3,(10sqrt15)/9)\ text(and)\ (sqrt15/3,-(10sqrt15)/9)`
    1. `y =-4x`
    2. `2sqrt17`
    1. `2sqrt(k^2-2k + 2)`
    2. `k = 1quadtext(or)quadk = 7/3`
    1. `sqrt(k + 1)`
    2. `((k + 1)^2)/4\ text(units)²`
Show Worked Solution
a.   
`text(Solve)\ \ f^{^{′}}(x)` `= 0\ \ text(for)\ x:`
`x` `= ± sqrt15/3`

 
`f(sqrt15/3) = -(10sqrt15)/9`

`f(−sqrt15/3) = (10sqrt15)/9`

`:.\ text(Turning points:)`

`(- sqrt15/3,(10sqrt15)/9)\ text(and)\ (sqrt15/3,-(10sqrt15)/9)`
 

b.i.   `A(-1,4),\ \ B(1,–4)`

`m_(AB) = (4-(−4))/(−1-(1)) = −4`

`text(Equation of)\ AB, m=-4, text(through)\ \ (-1,4):`

` y-4` `= −4(x-(−1))`
`:. y` `= −4x`

 

MARKER’S COMMENT: Students skilled in the use of technology will be much more efficient and minimise errors here.
b.ii.    `d_(text(AB))` `=sqrt((x_2-x_1)^2+(f(x_2)-f(x_1))^2)`
    `= sqrt((1- (- 1))^2 + (f(1)-f(−1))^2)`
    `= 2sqrt17`

 

c.i.    `d_(text(CD))` `=sqrt((x_2-x_1)^2+(g(x_2)-g(x_1))^2)`
    `=sqrt((1-(-1))^2+(g(1)-g(-1))^2)`
    `= 2sqrt(k^2-2k + 2)`

 

c.ii.   `text(Solve:)quad2sqrt(k^2-2k + 2) = k + 1quadtext(for)quadk > 0`

`:. k = 1quadtext(or)quadk = 7/3`
 

d.i.   `text(Solve:)quada^3-ka = a\ \ \ text(for)quad a,\ \ (a > 0)`

`:. a = sqrt(k + 1)`
 

d.ii.    `text(Area)` `= int_0^(sqrt(k + 1))(x-g(x))\ dx`
    `= ((k + 1)^2)/4\ text(units)²`

Calculus, MET2 2017 VCAA 9 MC

The average rate of change of the function with the rule  `f(x) = x^2 - 2x`  over the interval  `[1, a]`, where  `a > 1`, is `8`.

The value of `a` is

  1.  `9`
  2.  `8`
  3.  `7`
  4.  `4`
  5.  `1+ sqrt2`
Show Answers Only

`A`

Show Worked Solution
`(f(a) – f(1))/(a – 1)` `= ((a^2-2a)-(1-2))/(a-1)`
`8` `=(a-1)^2/(a-1),\ \ \ (a>0)`
`8` `=a-1`
`a` `=9`

`=> A`

Probability, MET2 2017 VCAA 3 MC

A box contains five red marbles and three yellow marbles. Two marbles are drawn at random from the box without replacement.

The probability that the marbles are of different colours is

  1. `5/8`
  2. `3/5`
  3. `15/28`
  4. `15/56`
  5. `30/28`
Show Answers Only

`C`

Show Worked Solution

 

`text(Pr)\ (RY) + text(Pr)\ (YR)`

`= 5/8 xx 3/7 +3/8 xx 5/7`

`= 15/28`

`=> C`

Graphs, MET2 2017 VCAA 2 MC

Part of the graph of a cubic polynomial function  `f` and the coordinates of its stationary points are shown below.
 


 

`f′(x) < 0`  for the interval

  1. `(0,3)`
  2. `(−oo,−5) ∪ (0,3)`
  3. `(−oo,−3) ∪ (5/3,oo)`
  4. `(−3,5/3)`
  5. `((−400)/27,36)`
Show Answers Only

`D`

Show Worked Solution

`f′(x) < 0\ \ text(when the gradient of the curve is negative.)`

`=> D`

Probability, MET1 2017 VCAA 5

For Jac to log on to a computer successfully, Jac must type the correct password. Unfortunately, Jac has forgotten the password. If Jac types the wrong password, Jac can make another attempt. The probability of success on any attempt is `2/5`. Assume that the result of each attempt is independent of the result of any other attempt. A maximum of three attempts can be made.

  1. What is the probability that Jac does not log on to the computer successfully?   (1 mark)

  2. Calculate the probability that Jac logs on to the computer successfully. Express your answer in the form `a/b`, where `a` and `b` are positive integers.   (1 mark)

  3. Calculate the probability that Jac logs on to the computer successfully on the second or on the third attempt. Express your answer in the form `c/d`, where `c` and `d` are positive integers.   (2 marks)

Show Answers Only

  1. `27/125`
  2. `98/125`
  3. `48/125`

Show Worked Solution

a.  `text(Pr)\ (S^{′} S^{′} S^{′})`

`= (3/5)^3`

`= 27/125`

 

b.  `1-text(Pr)\ (S^{′} S^{′} S^{′})`

`= 1-27/125`

`= 98/125`

 

c.  `text(Pr)\ (S^{′} S) + text(Pr)\ (S^{′} S^{′} S)`

`= 3/5 xx 2/5 + (3/5)^2 xx 2/5`

`= 48/125`

Calculus, MET1 2017 VCAA 3

Let  `f: [-3, 0] -> R,\ f(x) = (x + 2)^2 (x-1).`

  1. Show that  `(x + 2)^2 (x-1) = x^3 + 3x^2-4`.   (1 mark)

  2. Sketch the graph of  `f` on the axes below. Label the axis intercepts and any stationary points with their coordinates.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2.  
Show Worked Solution

a.  `text(Expand LHS:)`

`(x^2 + 4x + 4) (x-1)`

`=x^3-x^2 + 4x^2-4x + 4x-4`

`=x^3 + 3x^2-4\ \ text(… as required.)`

 

b.  

`x text(-intercepts:)\ \ (–2, 0), (1, 0)\ \ text(but)\ \ x ∈ [–3, 0]`

`:. x text(-int occurs at)\ (–2, 0)`
 

`text(Endpoints:)\ \ (–3, –4),\ \ (0, –4)`

`text(Stationary points occur when)\ \ f prime (x)=0`

`3x^2 + 6x` `= 0`
`3x(x + 2)` `= 0`
`:. x=0 or -2`  

Calculus, MET1 2017 VCAA 1a

Let  `f: (-2, oo) -> R,\ f(x) = x/(x + 2)`.

Differentiate  `f` with respect to `x`.  (2 marks)

Show Answers Only

`f prime(x) = 2/(x + 2)^2`

Show Worked Solution

`text(Using Quotient Rule:)`

`(h/g)′` `= (h′ g – h g′)/g^2`
`:. f prime (x)` `= (1 xx (x + 2) – x xx 1)/(x + 2)^2`
  `= 2/(x + 2)^2`

Harder Ext1 Topics, EXT2 2017 HSC 15b

Consider the curve  `sqrt x + sqrt y = sqrt a`, for  `x >= 0`  and  `y >= 0`, where `a` is a positive constant.

  1. Show that the equation of the tangent to the curve at the point  `P (c, d)`  is given by  `y sqrt c + x sqrt d = d sqrt c + c sqrt d`.  (2 marks)
  2. The tangent to the curve at the point `P` meets the `x` and `y` axes at  `A` and `B` respectively. Show that  `OA + OB = a`, where `O` is the origin.  (3 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)    `x^(1/2) + y^(1/2) = sqrt a`

`text(Using implicit differentiation:)`

`1/2 x^(-1/2) + 1/2 y^(-1/2) · (dy)/(dx)` `= 0`
`:. (dy)/(dx)` `= -(x^(-1/2))/(y^(-1/2))`
  `= -(y^(1/2))/(x^(1/2))`

 

`text(At)\  \P(c, d):`

`m_text(tang) = – sqrt d/sqrt c`

`text(Equation of tangent:)`

`y – d` `= – sqrt d/ sqrt c (x – c)`
`y sqrt c – d sqrt c` `= -x sqrt d + c sqrt d`
`y sqrt c + x sqrt d` `= d sqrt c + c sqrt d\ text(… as required.)`

 

(ii)  `text(At)\ \ A, y = 0`

`x sqrt d` `= d sqrt c + c sqrt d`
`x` `= (d sqrt c + c sqrt d)/sqrt d`

 

`text(At)\ \ B, x = 0`

`y sqrt c` `= d sqrt c + c sqrt d`
`y`  `=  (d sqrt c + c sqrt d)/sqrt c`

 

`AO + OB` `= x + y`
  `= (d sqrt c + c sqrt d)/sqrt d xx sqrt d/sqrt d + (d sqrt c + c sqrt d)/sqrt c xx sqrt c/sqrt c`
  `= sqrt d sqrt c + c + d + sqrt c sqrt d`
  `= c + 2 sqrt c sqrt d + d`
  `= (sqrt c + sqrt d)^2`
  `= (sqrt a)^2 qquad text{(S} text(ince)\ \ P(c, d)\ \ text(lies on)\ sqrt x + sqrt y = sqrt a text{)}`
  `= a\ text(… as required)`

Financial Maths, STD2 F1 SM-Bank 12

A golf shop is having a Boxing Day sale.

  1. What is the percentage discount on a putter which is reduced from $120.00 to $102.00?  (1 mark)

  2. The same discount applies storewide. What discount amount is applicable to a box of golf balls whose original price was $25.00?  (1 mark)

  3. What is the sale price of a golf bag which originally cost $160.00  (1 mark)

Show Answers Only
  1. `15 text(%)`
  2. `$3.75`
  3. `$136.00`
Show Worked Solution
a.     `text(Percentage Discount)` `=(120-102)/120 xx 100`
    `=18/120 xx 100`
    `=15 text(%)`

 

b.    `text(Discounted Amount)` `=15 text(%) xx 25`
    `=$3.75`

 

c.    `text(Sale Price)` `= 160 – 15 text(%) xx 160`
    `=0.85 xx 160`
    `=$136.00`

Financial Maths, STD2 F1 SM-Bank 6

Michelle intends to keep a car purchased for $17 000 for 15 years. At the end of this time its value will be $3500.

  1. By what amount, in dollars, would the car’s value depreciate annually if Michelle used the flat rate method of depreciation?  (1 mark)

  2. Determine the annual flat rate of depreciation correct to one decimal place.  (1 mark) 

Show Answers Only
  1. `$900`
  2. `5.3text{%  (1 d.p.)}`
Show Worked Solution
i.    `text(Depreciation)` `= 17\ 000 – 3500`
    `= $13\ 500`

 

`:.\ text(Annual depreciation)`

`= (13\ 500)/15`

`= $900`

 

ii.   `:.\ text(Flat rate of depreciation )`

`= 900/(17\ 000) xx 100text(%)`

`= 5.29…`

`= 5.3text{%  (1 d.p.)}`

Financial Maths, STD2 F1 SM-Bank 5

Khan paid $900 for a printer.

This price includes 10% GST (goods and services tax).

  1. Determine the price of the printer before GST was added.

     

    Write your answer correct to the nearest cent.  (2 marks)

  2. Khan is able to depreciate the full $900 purchase price of his printer for taxation purposes.

     

    Under flat rate depreciation the printer will be valued at $300 after five years.

     

    Calculate the annual depreciation in dollars.  (1 mark)

Show Answers Only
  1. `$818.18`
  2. `$120`
Show Worked Solution

i.   `text(Let)\ \ $P = text(price ex-GST)`

COMMENT: Reverse GST questions regularly cause problems for many students.
`:. P + 10text(%) xx P` `= 900`
`1.1P` `= 900`
`P` `= 900/1.1`
  `= 818.181…`
  `= $818.18\ \ text(nearest cent)`

 

ii.   `text(Annual depreciation)`

`= ((900 – 300))/5`

`= $120`

Financial Maths, STD2 F1 SM-Bank 3

A company purchased a machine for $60 000.

For taxation purposes the machine is depreciated over time using the straight line depreciation method.

The machine is depreciated at a flat rate of 10% of the purchase price each year.

  1. By how many dollars will the machine depreciate annually?  (1 mark)

  2. Calculate the value of the machine after three years.  (1 mark)

  3. After how many years will the machine be $12 000 in value?  (1 mark)

Show Answers Only
  1. `$6000`
  2. `$42\ 000`
  3. `8\ text(years)`
Show Worked Solution
i.    `text(Annual depreciation)` `= 10text(%) xx 60\ 000`
    `= $6000`

 

ii.   `text(After 3 years,)`

`text(Value)` `=V_0 – Dn`
  `= 60\ 000 – (3 xx 6000)`
  `= $42\ 000`

 

iii.   `text(Find)\ n\ text(when value = $12 000)`

`12\ 000` `= 60\ 000 – 6000 xx n`
`6000n` `= 48\ 000`
`:.n` `=(48\ 000)/6000`
  `= 8\ text(years)`

Financial Maths, STD2 F1 SM-Bank 10

Hugo is a professional bike rider.

The value of his bike will be depreciated over time using the flat rate method of depreciation.

The graph below shows his bike’s initial purchase price and its value at the end of each year for a period of three years.
 

  1. What was the initial purchase price of the bike?  (1 mark)

  2. Use calculations to show that the bike depreciates in value by $1500 each year.  (1 mark)

  3. Assume that the bike’s value continues to depreciate by $1500 each year. Determine its value five years after it was purchased.  (1 mark)

Show Answers Only
  1. `$8000`
  2. `text(See Worked Solutions)`
  3. `$500`
Show Worked Solution

i.   `$8000`
 

ii.   `text(Value after 1 year) = $6500\ \ \ text{(from graph)}`

`:.\ text(Annual depreciation)` `= 8000-6500`
  `= $1500`

 

iii.  `text(After 5 years:)`

`S` `=V_0-Dn`
  `=8000-5 xx 1500`
  `=$500`

Mechanics, EXT2 2017 HSC 14c

A smooth double cone with semi-vertical angle  `theta < pi/2`  is rotating about its axis with constant angular velocity `w`.

Two particles, each of mass `m`, are sitting on the cone as it rotates, as shown in the diagram.

Particle 1 is inside the cone at vertical distance `h` above the apex, `A`, and moves in a horizontal circle of radius `r`.

Particle 2 is attached to the apex `A` by a light inextensible string so that it sits on the cone at vertical distance `h` below the apex. Particle 2 also moves in a horizontal circle of radius `r`.

The acceleration due to gravity is `g`.

  1. The normal reaction force on Particle 1 is `R`.
    By resolving `R` into vertical and horizontal components, or otherwise, show that  `w^2 = (gh)/r^2`.  (2 marks)
  2. The normal reaction force on Particle 2 is `N` and the tension in the string is  `T`.

  3. By considering horizontal and vertical forces, or otherwise, show that

  4. `qquad qquad N = mg (sin theta - h/r cos theta)`.  (2 marks)
  6. Show that  `theta >= pi/4`.  (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)  

`text(Resolving forces horizontally:)`

`R cos theta = mrw^2`

`text(Resolving forces vertically:)`

`R sin theta` `= mg`
`(R sin theta)/(R cos theta)` `= (mg)/(mrw^2)`
`tan theta` `= g/(rw^2)`
`:. r/h` `= g/(rw^2)`
`r^2w^2` `= gh`
`w^2` `= (gh)/r^2\ \ text(… as required.)`

 

(ii)  

`text(Resolving forces horizontally:)`

`T sin theta – N cos theta = mrw^2\ text{… (1)}`

`text(Resolving forces vertically:)`

`T cos theta + N sin theta = mg\ text{… (2)}`

 

`text{Multiply (1)} xx cos theta`

`T sin theta cos theta – N cos^2 theta = mrw^2 cos theta\ text{… (3)}`

`text{Multiply (2) by}\ sin theta`

`T sin theta cos theta + N sin^2 theta = mg sin theta\ text{… (4)}`

 

`text{Subtract  (4) – (3)}`

`N (sin^2 theta + cos^2 theta)` `=mg sin theta – mrw^2 cos theta`
`:. N` `= mg sin theta – mr · (gh)/r^2 · cos theta`
  `= mg (sin theta – h/r cos theta)`

 

(iii)  `text(Movement of particle 2 requires)\ N >= 0`

♦♦ Mean mark 34%.
`mg (sin theta – h/r cos theta)` `>= 0`
`sin theta – 1/(tan theta) · cos theta` `>= 0`
`sin theta` `>= (cos theta)/(tan theta)`
`tan theta` `>= 1/(tan theta)`
`tan^2 theta` `>= 1`
`tan theta` `>= 1 qquad text{(}theta\ \ text{is in 1st quadrant)}`
`:. theta` `>= pi/4`

Calculus, EXT2 C1 2017 HSC 14a

It is given that  `x^4 + 4 = (x^2 + 2x + 2) (x^2-2x + 2)`.

  1. Find `A` and `B` so that  `16/(x^4 + 4) = (A + 2x)/(x^2 + 2x + 2) + (B-2x)/(x^2-2x + 2)`.  (1 mark)

  2. Hence, or otherwise, show that for any real number `m`
     
         `int_0^m 16/(x^4 + 4)\ dx = ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`.  (2 marks)

  3. Find the limiting value as  `m -> oo`  of
     
         `int_0^m 16/(x^4 + 4)\ dx`.  (1 mark)

Show Answers Only
  1. `A = 4 and B = 4`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `2 pi`
Show Worked Solution

i.  `16/(x^4 + 4) = (A + 2x)/(x^2 + 2x + 2) + (B-2x)/(x^2-2x + 2)`

`text(Multiply each side by)\ \ x^4 + 4`

`16 = (A + 2x) (x^2-2x + 2) + (B-2x) (x^2 + 2x + 2)`

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

`2A + 2B` `=16`
`A + B` `= 8\ text{… (1)}`

 
`text(When)\ \ x = 1`

`(A + 2) (1) + (B-2) (5)` `=16`
`A + 2 + 5B-10` `=16`
`A + 5B` `= 24\ text{… (2)}`

 

`text{Subtract  (2) – (1)}`

`4B=16\ \ =>\ \ B=4`

`A=4`
 

ii.  `int_0^m 16/(x^4 + 4)`

`= int_0^m (4 + 2x)/(x^2 + 2x + 2) dx + int_0^m (4-2x)/(x^2-2x + 2) dx`

`= int_0^m (2x + 2)/(x^2 + 2x + 2) + 2/(x^2 + 2x + 2) dx + int_0^m 2/(x^2-2x + 2)-(2x-2)/(x^2-2x + 2) dx`

`= [ln(x^2 + 2x + 2)]_0^m + int_0^m 2/(1 + (x + 1)^2) dx + int_0^m 2/(1 + (x-1)^2) dx-[ln (x^2-2x + 2)]_0^m`

`= [ln(m^2 + 2m + 2)-ln 2] + [2 tan^(-1) (x + 1)]_0^m + [2 tan^(-1) (x-1)]_0^m-[ln (m^2-2m + 2)-ln 2]`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 [tan^(-1) (m + 1)-tan^(-1) (1)] + 2 [tan^(-1) (m-1)-tan^(-1) (1)]`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1)-2 · pi/4 + 2 tan ^(-1) (m-1) + 2 · pi/4`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`

 

iii.  `I = int_0^m 16/(x^4 + 4) = ln ((1 + 2/m + 2/m^2)/(1-2/m + 2/m^2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`

`lim_(m -> oo) I` `= ln\ 1 + 2 · pi/2 + 2 · pi/2`
  `= 0 + pi + pi`
  `= 2 pi`

Measurement, STD2 M1 SM-Bank 8

A cannon ball is made out of steel and has a diameter of 23 cm.

  1. Find the volume of the sphere in cubic centimetres (correct to 1 decimal place).  (2 marks)

  2. It is known that the mass of the steel used is 8.2 tonnes/m³. Use this information to find the mass of the cannon ball to the nearest gram.  (2 marks)

Show Answers Only
  1. `6370.6\ text{cm³  (to 1 d.p.)}`
  2. `52\ 239\ text(grams)`
Show Worked Solution

a.   `text(Radius)= 23/2 = 11.5\ text(cm)`

`text(Volume)` `= 4/3pir^3`
  `= 4/3 xx pi xx 11.5^3`
  `= 6370.626…`
  `= 6370.6\ text{cm³  (to 1 d.p.)}`

 

b.   `text(Convert m³ to cm³:)`

`text(1 m³)` `= 100\ text(cm × 100 cm × 100 cm)`
  `= 1\ 000\ 000\ text(cm³)`

 

`text(Convert 8.2 tonnes to grams:)`

`text(8.2 tonnes)` `= 8200\ text(kg)`
  `= 8\ 200\ 000\ text(g)`

 

`:.\ text(Weight of cannon ball)`

`= 6370.6 xx (8\ 200\ 000)/(1\ 000\ 000)`

`= 52\ 238.92`

`= 52\ 239\ text(grams)`

Measurement, STD2 M1 SM-Bank 3

A 250-watt television is turned on for an average of 4 hours per day during off-peak periods for a week.

If the television is not running at any other time and electricity is charged at $0.36/kWh during off-peak, how much does it cost to run the television for a week?  (2 marks)

Show Answers Only

`$2.52`

Show Worked Solution

`text(Total electricity used)`

`= 7\ text(days × 4 hours × 250)`

`= 7000\ text(watt hours)`

`= 7.0\ text(kWh)`
 

`:.\  text(Running Cost)` `= 7.0 xx 0.36`
  `= $2.52`

Measurement, STD2 M1 SM-Bank 2

A farmer wants to estimate the area of an irregular shaped paddock.
 

 
What is the estimated area of the land using the Trapezoidal Rule?  (2 marks)

Show Answers Only

`370\ text(m²)`

Show Worked Solution

`text(Solution 1)`

`text(Height) = 20 div 4 = 5\ text(m)`

`text(Area)` `~~ 5/2 (28 + 18) + 5/2 (18 + 17) + 5/2 (17 + 16) + 5/2 (16 + 18)`
  `~~ 370\ text(m²)`

 

`text(Solution 2)`

  `x` `0` `5` `10` `15` `20`
  `text(height)` `28` `18` `17` `16` `18`
  `text(weight)` `1` `2` `2` `2` `1`
`text(Area)` `~~ h/2 [28 + 2 (18 + 17 + 16) + 18]`
  `~~ 5/2 xx 148`
  `~~ 370\ text(m²)`

Mechanics, EXT2 M1 2017 HSC 13c

A particle is projected upwards from ground level with initial velocity  `1/2 sqrt(g/k)\ text(ms)^(-1)`, where `g` is the acceleration due to gravity and `k` is a positive constant. The particle moves through the air with speed  `v\ text(ms)^(-1)`  and experiences a resistive force.

The acceleration of the particle is given by  `ddot x = -g - kv^2\ text(ms)^(-2)`. Do NOT prove this.

The particle reaches a maximum height, `H`, before returning to the ground.

Using  `ddot x = v (dv)/(dx)`, or otherwise, show that  `H = 1/(2k) log_e (5/4)`  metres.  (4 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`ddot x = v · (dv)/(dx) = -g – kv^2`

`(dv)/(dx)` `= – (g + kv^2)/v`
`(dx)/(dv)` `= – v/(g + kv^2)`
`:. x` `= – int v/(g + kv^2)\ dv`
  `= -1/(2k) log_e (g + kv^2) + c`

 
`text(When)\ \ x = 0,\ \ v = 1/2 sqrt(g/k)`

`0` `= -1/(2k) log_e (g + k · g/(4k)) + c`
`:. c` `= 1/(2k) log_e ((5g)/4)`

 
`:. x = 1/(2k) log_e ((5g)/4) – 1/(2k) log_e (g + kv^2)`

 
`text{Max height}\ H\ text(occurs when)\ \ v = 0:`

`H` `= 1/(2k) log_e ((5g)/4) – 1/(2k) log_e g`
  `= 1/(2k) log_e ((5g)/(4g))`
  `= 1/(2k) log_e (5/4)\ text(… as required.)`

Functions, EXT1′ F2 2017 HSC 5 MC

The polynomial  `p(x) = x^3 - 2x + 2`  has roots `alpha`, `beta` and `gamma`.

What is the value of  `alpha^3 + beta^3 + gamma^3`?

  1. `−10`
  2. `−6`
  3. `−2`
  4. `0`
Show Answers Only

`B`

Show Worked Solution

`text(S)text(ince)\ \ alpha, beta, gamma\ text(are roots,)`

`p(alpha)` `= alpha^3 – 2alpha + 2 = 0` `…\ (1)`
`p(beta)` `= beta^3 – 2beta + 2 = 0` `…\ (2)`
`p(gamma)` `= gamma^3 – 2gamma + 2 = 0` `…\ (3)`

 
`text(Add:)\ (1) + (2) + (3)`

`alpha^3 + beta^3 + gamma^3 – 2(alpha + beta + gamma) + 6 = 0`

`:. alpha^3 + beta^3 + gamma^3=-6\ \ \ \ text{(note:}\ \ alpha + beta + gamma =- b/a=0 text{)}`

`=> B`

Harder Ext1 Topics, EXT2 2017 HSC 16a

  Let  `alpha = costheta + i sintheta`, where  `0 < theta < 2pi`.

  1. Show that  `alpha^k + alpha^(−k) = 2 cos ktheta`, for any integer `k`.  (1 mark)
  3. Let  `C = alpha^(−n) + … + alpha^(−1) + 1 + alpha + … + alpha^n`, where `n` is a positive integer.
  4. By summing the series, prove that  

    `C = (alpha^n + alpha^(−n) - (a^(n + 1) + alpha^(−(n + 1))))/((1 - alpha)(1 - baralpha))`.  (3 marks)

  6. Deduce, from parts (i) and (ii), that`1 + 2(costheta + cos2theta + … + cosntheta) = (cosntheta - cos(n + 1)theta)/(1 - costheta)`.  (2 marks)

  7. Show that  
    `cos\ pi/n + cos\ (2pi)/n + …  + cos\ (npi)/n`  is independent of `n`.  (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)   `alpha = costheta – isintheta`

`text(Using de Moivre:)`

`alpha^k` `= cos(ktheta) + isin(ktheta)`
`alpha^(−k)` `= cos(−ktheta) + isin(−ktheta)`
  `= cos(ktheta) – isin(ktheta)`
`:. alpha^k + alpha^(−k)` `= cos(ktheta) + isin(ktheta) + cos(ktheta) – isin(ktheta)`
  `= 2cos(ktheta)\ …\ \ text(as required.)`

 

(ii)   `C = alpha^(−n) + … + alpha^(−1) + 1 + alpha + … + alpha^n`

♦♦ Mean mark 34%.

`=> text(GP where)\ \ a = alpha^(−n), r = alpha`

`=> text(Number of terms) = 2n + 1`

`:. C` `= (a(1 – r^n))/(1 – r)`
  `= (alpha^(−n)(1 – alpha^(2n + 1)))/(1 – alpha)`
  `= (alpha^(−n)(1 – alpha^(2n + 1)))/(1 – alpha) xx (1 – baralpha)/(1 – baralpha)\ \ \ \ text{(where}\ baralpha = alpha^(−1) text{)}`
  `= ((alpha^(−n) – alpha^(n + 1))(1 – alpha^(−1)))/((1 – alpha)(1 – baralpha))`
  `= (alpha^(−n) – alpha^(−n – 1) – alpha^(n + 1) + alpha^n)/((1 – alpha)(1 – baralpha))`
  `= (alpha^n + alpha^(−n) – (alpha^(n + 1) + alpha^(−(n + 1))))/((1 – alpha)(1 – baralpha))\ …\ \ text(as required.)`
♦ Mean mark part (iii) 48%.

 

(iii)    `C` `= alpha^(−n) + a^(−n + 1) … + alpha^(−1) + 1 + alpha + … + alpha^n `
    `= 1 + (alpha + alpha^(−1)) + (alpha^2 + alpha^(−2)) + … + (alpha^n + alpha^(−n))`
    `= 1 + 2costheta + 2cos2theta + … + 2cosntheta\ \ \ \ (text{using part (i)})`
    `= 1 + 2(costheta + 2cos2theta + … + cosntheta)`

 

`text{Also, using part (ii)}`

`C` `= (alpha^n + alpha^(−n) – (alpha^(n + 1) + alpha^(−(n + 1))))/((1 – alpha)(1 – baralpha))`
  `= (2cosntheta – 2cos(n + 1)theta)/(1 – baralpha – alpha + 1)\ \ (text{part (i)})`
  `= (2cosntheta – 2cos(n + 1)theta)/(2 – (alpha + alpha^(−1)))`
  `= (2cosntheta – 2cos(n + 1)theta)/(2 – 2costheta)`
  `= (cosntheta – cos(n + 1)theta)/(1 – costheta)`

`:. 1 + 2(costheta, cos2theta + … + cosntheta) = (cosntheta – cos(n + 1)theta)/(1 – costheta)`

♦♦♦ Mean mark part (iv) 28%.

 

(iv)   `text{Rearrange the result from (iii):}`

`costheta + cos2theta + … + cosntheta = 1/2((cosntheta – cos(n + 1)theta)/(1 – costheta) – 1)`

`text(Let)\ \ theta = pi/n`

`:. cos\ pi/n + cos\ (2pi)/n + … + cos\ (npi)/n`

`= 1/2((cos((npi)/n) – cos(((n + 1)pi)/n))/(1 – cos(pi/n)) – 1)`
`= 1/2((cospi – cos(pi + pi/n))/(1 – cos(pi/n)) – 1)`
`= 1/2((−(1 – cos(pi/n)))/(1 – cos(pi/n)) – 1)`
`= 1/2 (−1 – 1)`
`= −1`

 

`=> text(which is independent of)\ n.`

Functions, EXT1′ F2 2017 HSC 13b

Let `a, b` and `c` be real numbers. Suppose that  `P(x) = x^4 + ax^3 + bx^2 + cx + 1`  has roots  `alpha, 1/alpha, beta, 1/beta,`  where  `alpha > 0 and beta > 0`.

Prove that  `a = c`.   (2 marks)

 

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text(Sum of roots) = -b/a:`

`-a = alpha + 1/alpha + beta + 1/beta`
 

`text(Sum of products of 3 roots) = -d/a:`

`- c` `= alpha · 1/alpha · beta + alpha · 1/alpha · 1/beta + alpha · beta · 1/beta + 1/alpha · beta · 1/beta`
  `= beta + 1/beta + alpha + 1/alpha`
  `= -a`

 
`:. a = c\ text(… as required.)`

Functions, EXT1′ F2 2017 HSC 12d

Let `P(x)` be a polynomial.

  1. Given that  `(x - alpha)^2`  is a factor of `P(x)`, show that
     
    `qquad qquad P(alpha) = P prime (alpha) = 0`.  (2 marks)

  2. Given that the polynomial  `P(x) = x^4 - 3x^3 + x^2 + 4`  has a factor  `(x - alpha)^2`, find the value of `alpha`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `2`
Show Worked Solution
i.   `P(x)` `= (x – alpha)^2 · Q(x)`
  `P prime (x)` `= 2 (x – alpha) · Q (x) + (x – alpha)^2 *Q prime (x)`
    `= (x – alpha) [2 Q (x) + (x – alpha) *Q prime (x)]`

 

`P (alpha)` `= 0 xx Q(x) = 0`
`P prime (alpha)` `= 0 [2Q (x) + 0 xx Q prime (x)] = 0`

 
`:. P(alpha) = P prime (alpha) = 0\ text(… as required.)`

 

ii.   `P(x)` `= x^4 – 3x^3 + x^2 + 4`
  `P prime(x)` `= 4x^3 – 9x^2 + 2x`
    `= x (4x^2 – 9x + 2)`
    `= x (4x – 1) (x – 2)`

 
`:. P prime(x) = 0\ \ text(when)\ \ x = 0, 1/4 or 2`

`=>\ text(Multiple roots may exist at)\ \ x=0, 1/4 or 2.`

`text(Test each root in)\ \ P(x):`

`P(0)` `= 0 – 0 + 0 + 4 = 4`
`P(1/4)` `= (1/4)^4 – 3(1/4)^3 + (1/4)^2 + 4= 4 5/256`
`P(2)` `= 16 – 3(8) + 4 + 4 = 0`

 
`:. (x – 2)^2\ \ text(is a factor of)\ \ P(x)`

`:. alpha = 2`

Calculus, EXT2 C1 2017 HSC 12c

Find  `int x tan^(-1) x\ dx`.  (3 marks)

Show Answers Only

`1/2(x^2 tan^(-1)x – x + tan^(-1) x) + c`

Show Worked Solution

`I = int x tan^(-1) x\ dx`

`text(Let)\ \ \ u` `= tan^(-1) x`   `v prime` `= x`
`u prime` `= 1/(1 + x^2)`   `v` `= x^2/2`

 

`I` `= uv – int u prime v\ dx`
  `= tan^(-1) x · x^2/2 – int 1/(1 + x^2) · x^2/2\ dx`
  `= x^2/2  tan^(-1) x – 1/2 int x^2/(1 + x^2)\ dx`
  `= x^2/2  tan^(-1) x – 1/2 int (1 + x^2 – 1)/(1 + x^2)\ dx`
  `= x^2/2  tan^(-1) x – 1/2 int 1 – 1/(1 + x^2)\ dx`
  `= x^2/2  tan^(-1) x – 1/2 [x – tan^(-1) x] + c`
  `= x^2/2  tan^(-1) x – 1/2 x + 1/2 tan^(-1) x + c`
  `= 1/2(x^2 tan^(-1) x – x + tan^(-1) x) + c`

Complex Numbers, EXT2 N2 2017 HSC 12b

Solve the quadratic equation  `z^2 + (2 + 3i)z + (1 + 3i) = 0`, giving your answers in the form  `a + bi`, where `a` and `b` are real numbers.  (3 marks)

Show Answers Only

`-1 or -1 – 3i`

Show Worked Solution

`z^2 + (2 + 3i)z + (1 + 3i) = 0`

 

`z`

 `= (-(2 + 3i) +- sqrt((2 + 3i)^2 – 4 · 1(1 + 3i)))/2`
  `= (-(2 + 3i) +- sqrt (4 + 12i + 9i^2 – 4 – 12i))/2`
  `= (-(2 + 3i) +- sqrt(-9))/2`
  `= ((-2 – 3i) +- 3i)/2`
   
`:. z` `= (-2)/2 or (-2 – 6i)/2`
  `= -1 or -1 – 3i`

CORE, FUR2 2017 VCAA 2

The back-to-back stem plot below displays the wingspan, in millimetres, of 32 moths and their place of capture (forest or grassland).

 

  1. Which variable, wingspan or place of capture, is a categorical variable?  (1 mark)

  2. Write down the modal wingspan, in millimetres, of the moths captured in the forest.  (1 mark)

  3. Use the information in the back-to-back stem plot to complete the table below.  (2 marks)

     

     

  4. Show that the moth captured in the forest that had a wingspan of 52 mm is an outlier.  (2 marks)

  5. The back-to-back stem plot suggests that wingspan is associated with place of capture.
  6. Explain why, quoting the values of an appropriate statistic.  (2 marks)

Show Answers Only

  1. `text(Place of Capture)`
  2. `20\ text(mm)`
  3.    

     

  4. `text(See Worked Solution)`
  5. `text(See Worked Solution)`

Show Worked Solution

a.   `text(Place of Capture is categorical.)`

 

b.   `text(Modal wingspan in forest = 20 mm)`

 

c.   `Q_3\ text(in grassland: 19 data points)`

`:. Q_3\ text{is the 15th data point (lowest to highest) = 36}`

 

 

d.    `Q_1\ (text(forest))` `= (text(3rd + 4th))/2 = (20 + 20)/2 = 20`
  `Q_3\ (text(forest))` `= (text(10th + 11th))/2 = (30 + 34)/2 = 32`

 

`=> IQR = 32 – 20 = 12`

`Q_3 + 1.5 xx IQR` `= 32 + 1.5 xx 12`
  `= 50\ text(mm)`

 

`:.\ text(S)text(ince 52 mm > 50 mm, 52 min is an outlier.)`

 

e.   `text(Comparing the median wingspan of both places:)`

`M_text(forest) = 21,\ \ M_text(grassland) = 30`

`text(The higher median of grassland suggests that)`

`text(wingspan is associated with place of capture.)`

CORE, FUR2 2017 VCAA 1

The number of eggs counted in a sample of 12 clusters of moth eggs is recorded in the table below.
     

  1. From the information given, determine
  2.  i. the range   (1 mark)

  3. ii. the percentage of clusters in this sample that contain more than 170 eggs.   (1 mark)

In a large population of moths, the number of eggs per cluster is approximately normally distributed with a mean of 165 eggs and a standard deviation of 25 eggs.

  1. Using the 68–95–99.7% rule, determine
  2.  i. the percentage of clusters expected to contain more than 140 eggs.   (1 mark)

  3. ii. the number of clusters expected to have less than 215 eggs in a sample of 1000 clusters.   (1 mark)

  4. The standardised number of eggs in one cluster is given by  `z = –2.4`
  5. Determine the actual number of eggs in this cluster.   (1 mark)

Show Answers Only

a.    `72`

b.i.  `25text(%)`

b.ii.  `-1`

c.  `105\ text(eggs)`

Show Worked Solution

a.i.  `text(Range)\ = 197-125= 72`

a.ii.  `text(3 clusters > 170 eggs)`

`:.\ text(Percentage)` `= 3/12 xx 100text(%)`
  `= 25text(%)`

 

b.i.    `ztext{-score (140)}` `= (140-165)/25`
    `= −1`

`:.\ text(Percentage over 140)`

`= 68 + 16`

`= 84text(%)`

 

b.ii.    `ztext{-score (215)}` `= (215-165)/25`
    `= 2`

 

`:.\ text(Percentage less than 215)`

`= 97.5text(%) xx 1000`

`= 975`

 

c.    `text(Using)\ \ \ z` `= (x-barx)/s`
  `−2.4` `= (x-165)/25`
  `x` `= (−2.4 xx 25) + 165`
    `= 105\ text(eggs)`

Functions, EXT1′ F1 2017 HSC 12a

Consider the function  `f(x) = (e^x - 1)/(e^x + 1)`.

  1.  Show that  `f(x)`  is increasing for all `x`.  (1 mark)

  2.  Show that  `f(x)`  is an odd function.  (1 mark)

  3.  Describe the behaviour of  `f(x)`  for large positive values of `x`.  (1 mark)

  4.  Hence sketch the graph of  `f(x) = (e^x - 1)/(e^x + 1)`.  (1 mark)

  5.  Hence, or otherwise, sketch the graph of  `y = 1/(f(x))`.  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text{(See worked Solutions)}`
  4.  
  5.  

Show Worked Solution

i.  `text(Solution 1)`

`f(x) = (e^x – 1)/(e^x + 1)`

`u = e^x – 1, quadquad u prime = e^x`

`v = e^x + 1, quadquad v prime = e^x`
 

`f′(x)` `= {e^x (e^x + 1) – e^x (e^x – 1)}/(e^x + 1)^2`
  `= (e^(2x) + e^x – e^(2x) + e^x)/(e^x + 1)^2`
  `= (2e^x)/(e^x + 1)^2`

 
`text(S) text(ince)\ \ e^x > 0\ \ text(for all)\ \ x,`

`=>f prime (x) > 0\ text(for all)\ x`

`:. f(x)\ text(is increasing for all)\ x.`
 

`text(Solution 2)`

`f(x)` `= (e^x – 1)/(e^x + 1)`
  `= (e^x + 1 – 2)/(e^x + 1)`
  `= 1 – 2/(e^x + 1)`
`f′(x)` `= (2e^x)/(e^x + 1)^2`

 
`text(See Solution 1 for remainder.)`
 

ii.   `f(-x)` `= (e^(-x) – 1)/(e^(-x) + 1) xx e^x/e^x`
    `= (1 – e^x)/(1 + e^x)` 
    `= – (e^x – 1)/(e^x + 1`
    `= -f(x)`

 
`:. f(x)\ text(is odd.)`

 

iii.  `lim_(x -> oo) (e^x – 1)/(e^x + 1) = 1`

`text(i.e.  As)\ \ x -> oo,\ f(x) -> 1^-\ \ text{(i.e. from lower side)}`

 

iv.  

 

v.  

Calculus, EXT2 C1 2017 HSC 11f

Using the substitution  `x = sin^2 theta`, or otherwise, evaluate  `int_0^(1/2) sqrt(x/(1 - x))\ dx`.  (3 marks)

Show Answers Only

`pi/4 – 1/2`

Show Worked Solution

`x = sin^2 theta`

`dx = 2 sin theta cos theta\ d theta`

`text(If)\ \ x = 1/2, sin theta = 1/sqrt 2, \ theta = pi/4`

`text(If)\ \ x = 0, \ theta = 0`

`int_0^(1/2) sqrt(x/(1 – x))\ dx` `= int_0^(pi/4) sqrt ((sin^2 theta)/(1 – sin^2 theta)) xx 2 sin theta cos theta\ d theta`
  `= int_0^(pi/4) (sin theta)/(cos theta) xx 2 sin theta cos theta\ d theta`
  `= int_0^(pi/4) 2 sin^2 theta\ d theta`
  `= int_0^(pi/4) 1 – cos 2 theta\ d theta`
  `= [theta – 1/2 sin 2 theta]_0^(pi/4)`
  `= (pi/4 – 1/2 sin {:pi/2) – (0 – 0)`
  `= pi/4 – 1/2`

Calculus, EXT2 C1 2017 HSC 11d

Using the substitution  `t = tan {:theta/2:}`, or otherwise, evaluate

`int_0^((2 pi)/3) 1/(1 + cos theta)\ d theta`.  (3 marks)

Show Answers Only

`sqrt 3`

Show Worked Solution

`text(Let)\ \ t = tan {:theta/2:}, \ cos theta = (1 – t^2)/(1 + t^2), \ d theta = (2 dt)/(1 + t^2)`

 

`text(When)\ theta = 0, t = tan 0 = 0`

`text(When)\ theta = (2 pi)/3, t = tan {:pi/3:} = sqrt 3`

`int_0^((2pi)/3) 1/(1 + cos theta)\ d theta` `= int_0^(sqrt 3) ((2dt)/(1 + t^2))/((1 + t^2)/(1 + t^2) + (1 – t^2)/(1 + t^2))`
  `= int_0^(sqrt 3) (2/(1 + t^2))/(2/(1 + t^2))\ dt`
  `= int_0^(sqrt 3) 1\ dt`
  `= [t]_0^ (sqrt 3)`
  `= sqrt 3 – 0`
  `= sqrt 3`

Conics, EXT2 2017 HSC 11b

An asymptote to the hyperbola  `x^2/12 - y^2/4 = 1`  makes an angle `alpha` with the positive `x`-axis, as shown.

Find the value of `alpha`.  (2 marks)

Show Answers Only

`pi/6`

Show Worked Solution

`x^2/12 – y^2/4 = 1`

`a^2 = 12\ \ =>\ \ a = 2 sqrt 3`

`b^2 = 4\ \ =>\ \ b = 2`

`text(Slope of asymptote) = b/a`

`=> tan alpha` `= b/a`
  `= 2/(2 sqrt 3)`
  `= 1/sqrt 3`
`:. alpha` `= pi/6`

Complex Numbers, EXT2 N1 2017 HSC 11a

Let  `z = 1 - sqrt 3 i`  and  `w = 1 + i`.

  1. Find the exact value of the argument of `z`.  (1 mark)
  2. Find the exact value of the argument of  `z/w`.  (2 marks)  
Show Answers Only
  1. `-pi/3`
  2. `-(7 pi)/12`
Show Worked Solution

i.  `z = 1 – i sqrt 3`

`text(arg)\ z = – pi/3`

 

ii.  `w = 1 + i`

`text(arg)\ w = pi/4`

`text(arg)\ (z/w)` `= text(arg)\ (z) – text(arg)\ (w)`
  `= – pi/3 – pi/4`
  `= -(7 pi)/12`

Complex Numbers, EXT2 N2 2017 HSC 1 MC

The complex number `z` is chosen so that  `1, z, …, z^7`  form the vertices of the regular polygon shown.

Which polynomial equation has all of these complex numbers as roots?

  1. `x^7 - 1 = 0`
  2. `x^7 + 1 = 0`
  3. `x^8 - 1 = 0`
  4. `x^8 + 1 = 0`
Show Answers Only

`C`

Show Worked Solution

`P(x)\ \ text(has 8 separate roots.)`

`:.\ text(Must be of degree at least 8.)`

`text(S) text(ince 1 is also a root,)`

`=>  C`

MATRICES, FUR1 2017 VCAA 1 MC

Kai has a part-time job.

Each week, he earns money and saves some of this money.

The matrix below shows the amounts earned (`E`) and saved (`S`), in dollars, in each of three weeks.
 

`{:(qquadqquadqquadqquadquadEquadqquadS),({:(text(week 1)),(text(week 2)),(text(week 3)):}[(300,100),(270,90),(240,80)]):}`

 

How much did Kai save in week 2?

  1.   `$80`
  2.   `$90`
  3. `$100`
  4. `$170`
  5. `$270`
Show Answers Only

`B`

Show Worked Solution

`text(Kai saved $90 in week 2.)`

`=> B`

GRAPHS, FUR1 2017 VCAA 2 MC

The graph below shows the volume of water in a water tank between 7 am and 5 pm on one day.

Which one of the following statements is true?

  1. The volume of water in the tank decreases between 8 am and 11 am.
  2. The volume of water in the tank increases at the greatest rate between 4 pm and 5 pm.
  3. The volume of water in the tank is constant between 12 noon and 2 pm.
  4. The tank is filled with water at a constant rate of 100 L per hour.
  5. More water enters the tank during the first five hours than during the last five hours.
Show Answers Only

`C`

Show Worked Solution

`text(Consider option:)\ C`

`text(The volume in the tank is unchanged between 12-2 pm,)`

`text(denoted by a horizontal line in the graph.)`

`=> C`