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Complex Numbers, EXT2 N2 2017 HSC 6 MC

It is given that  `z = 2 + i`  is a root of  `z^3 + az^2 - 7z + 15 = 0`, where `a` is a real number.

What is the value of `a`?

  1. `−1`
  2. `1`
  3. `7`
  4. `−7`
Show Answers Only

`A`

Show Worked Solution

`text(S)text(ince)\ \ z_1 = 2 + i\ \ text(is a root.)`

`=> z_2 = 2 – i\ \ text(is a root.)`

 

`text(Roots are)\ \ z_1, z_2, alpha.`

`text(Using product of roots:)`

`z_1z_2alpha` `= −15`
`(2 + i)(2 – i)alpha` `= −15`
`5alpha` `= −15`
`a` `= −3`

 

`text(Using sum of roots:)`

`z_1 + z_2 + alpha` `=-a`
`2 + i + 2 – i + −3` `= −a`
`a` `= −1`

`=> A`

Conics, EXT2 2017 HSC 15c

The ellipse with equation  `(x^2)/(a^2) + (y^2)/(b^2) = 1`, where  `a > b`, has eccentricity `e`.

The hyperbola with equation  `(x^2)/(c^2) - (y^2)/(d^2) = 1`, has eccentricity `E`.

The value of `c` is chosen so that the hyperbola and the ellipse meet at  `P(x_1, y_1)`, as shown in the diagram.

  1. Show that  
  2. `(x_1^(\ 2))/(y_1^(\ 2)) = (a^2c^2)/((a^2 - c^2)) xx ((b^2 + d^2))/(b^2d^2)`.  (2 marks)
  4. If the two conics have the same foci, show that their tangents at `P` are perpendicular.  (3 marks)

 

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

(i)   `text(At)\ \ P(x_1,y_1):`

`(x_1^(\ 2))/(a^2) + (y_1^(\ 2))/(b^2)` `= (x_1^(\ 2))/(c^2) – (y_1^(\ 2))/(d^2) = 1`
`(x_1^(\ 2))/(a^2) – (x_1^(\ 2))/(c^2)` `= −(y_1^(\ 2))/(b^2) – (y_1^(\ 2))/(d^2)`
`x_1^(\ 2)(1/(a^2) – 1/(c^2))` `= −y_1^(\ 2)(1/(b^2) + 1/(d^2))`
`x_1^(\ 2)((c^2 – a^2)/(a^2c^2))` `= −y_1^(\ 2)((d^2 + b^2)/(b^2d^2))`
`(x_1^(\ 2))/(y_1^(\ 2))` `= −((a^2c^2)/(c^2 – a^2))((b^2 + d^2)/(b^2d^2))`
  `= (a^2c^2)/((a^2 – c^2)) xx ((b^2 + d^2))/(b^2d^2)\ …\ text(as required).`

 

(ii)   `text(Find gradient of ellipse at)\ \ P(x_1,y_1):`

♦ Mean mark 51%.
`(x^2)/(a^2) + (y^2)/(b^2)` `= 1`
`(2x)/(a^2) + (2y)/(b^2)·(dy)/(dx)` `= 0`
`(dy)/(dx)` `= −x/(a^2) xx (b^2)/y`

`text(At)\ P,\ \ m = (−x_1b^2)/(y_1a^2)`

 

`text(Similarly for gradient of hyperbola at)\ \ P(x_1,y_1):`

`(x^2)/(c^2) – (y^2)/(d^2)` `= 1`
`(2x)/(c^2) – (2y)/(d^2)·(dy)/(dx)` `= 0`
`(dy)/(dx)` `= x/(c^2) xx (d^2)/y`

`text(At)\ P,\ \ m = (x_1d^2)/(y_1c^2)`

`:. m_1m_2` `= −(x_1b^2)/(y_1a^2) xx (x_1d^2)/(y_1c^2)`
  `= −(x_1^2)/(y_1^2) xx (b^2d^2)/(a^2c^2)`

 

`text{Rearranging part (i)}`

`(x_1^(\ 2))/(y_1^(\ 2))` `= ((b^2 + d^2))/((a^2 – c^2)) xx (a^2c^2)/(b^2d^2)`
`(a^2c^2)/(b^2d^2)` `= (x_1^(\ 2))/(y_1^(\ 2))·((a^2 – c^2))/((b^2 + d^2))`
`(b^2d^2)/(a^2c^2)` `= (y_1^(\ 2)(b^2 + d^2))/(x_1^(\ 2)(a^2 – c^2))`
`:. m_1m_2` `= −(x_1^(\ 2))/(y_1^(\ 2)) xx (y_1^(\ 2)(b^2 + d^2))/(x_1^(\ 2)(a^2 – c^2))`
  `= −((b^2 + d^2))/(a^2 – c^2)\ \ …(1)`

 

`text(S)text{ince conics have same foci (given),}`

`ae` `= cE`    
`a^2e^2` `= c^2E^2`    
`a^2 – b^2` `= c^2 + d^2` `\ \ \ text{(using}` `\ \ b^2=a^2(1-e^2) and` 
`a^2 – c^2` `= b^2 + d^2`   `\ \ d^2=c^2(E^2-1) text{)}`

 

`text{Substituting into (1)}`

`m_1m_2` `= −((b^2 + d^2))/(b^2 + d^2)`
  `= −1`

 

`:.\ text(The two tangents are perpendicular.)`

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`

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.  

Volumes, EXT2 2017 HSC 11e

The region bounded by the lines  `y = 3 - x,\ y = 2x`  and the `x`-axis is rotated about the `x`-axis.

Use the method of cylindrical shells to find an integral whose value is the volume of the solid of revolution formed. Do NOT evaluate the integral.  (2 marks)

Show Answers Only

`2 pi int_0^2 y (3 – (3y)/2)\ dy`

Show Worked Solution
`text(Radius)\ (r)` `= y`
`text(Height)\ (h)` `= x_2 – x_1`
  `= (3 – y) – y/2`
  `= 3 – (3y)/2`

 

`V` `= 2 pi int_0^2 r h\ dy`
  `= 2 pi int_0^2 y (3 – (3y)/2)\ dy`

Complex Numbers, EXT2 N2 2017 HSC 3 MC

Which complex number lies in the region  `2 < |z - 1| < 3`?

  1. `1 + sqrt 3 i`
  2. `1 + 3i`
  3. `2 + i`
  4. `3 - i`
Show Answers Only

`D`

Show Worked Solution

`|z – 1| = 2\ \ text(is a circle with centre)\ \ (1, 0),\ text(radius 2)`

`|z – 1| = 3\ \ text(is a circle with centre)\ \ (1, 0),\ text(radius 3)`

`:.\ text(Only)\ \ 3 – i\ \ text(lies within the two circles,)`

`text(i.e. satisfies)\ \ 2 < |z-1| < 3`

`=>  D`

Conics, EXT2 2017 HSC 2 MC

Suppose `ℓ` is a line and `S` is a point NOT on `ℓ`.

The point `P` moves so that the distance from `P` to `S` is half the perpendicular distance from `P` to `ℓ`.

Which conic best describes the locus of `P`?

  1. A circle
  2. An ellipse
  3. A parabola
  4. A hyperbola
Show Answers Only

`B`

Show Worked Solution

`PS = ePN,\ text(where)\ e = 1/2`

`:. P\ text(is an ellipse)`

`=>  B`

MATRICES, FUR1 2017 VCAA 4 MC

A permutation matrix, `P`, can be used to change `[(F),(E),(A),(R),(S)]` into `[(S),(A),(F),(E),(R)]`.

Matrix `P` is

A.

`[(0,0,1,0,1),(0,0,1,1,0),(1,1,0,0,0),(0,1,0,0,1),(1,0,0,1,0)]`

 

 B.

`[(0,0,0,1,0),(0,0,1,0,0),(0,1,0,0,0),(0,0,0,0,1),(1,0,0,0,0)]`

 
C.

`[(0,0,0,0,1),(0,0,1,0,0),(1,0,0,0,0),(0,1,0,0,0),(0,0,0,1,0)]`

 

 D.

`[(1,0,0,0,1),(0,1,1,0,0),(1,0,1,0,0),(0,1,0,1,0),(0,0,0,1,1)]`

 
E.

`[(0,0,0,0,1),(0,0,1,0,0),(0,1,0,0,0),(1,0,0,0,0),(0,0,0,1,0)]`

 

    
Show Answers Only

`C`

Show Worked Solution

`[(0,0,0,0,1),(0,0,1,0,0),(1,0,0,0,0),(0,1,0,0,0),(0,0,0,1,0)][(F),(E),(A),(R),(S)] = [(S),(A),(F),(E),(R)]`

`=> C`

MATRICES, FUR1 2017 VCAA 3 MC

Which one of the following matrix equations has a unique solution?

A.

`[(1,1),(1,1)][(x),(y)] = [(2),(10)]`

 

 B.

`[(6,−6),(−4,4)][(x),(y)] = [(60),(36)]`

 
C.

`[(8,−4),(4,2)][(x),(y)] = [(12),(18)]`

 

 D.

`[(7,0),(5,0)][(x),(y)] = [(14),(15)]`

 
E.

`[(4,−2),(6,−3)][(x),(y)] = [(36),(24)]`

 

    
Show Answers Only

`C`

Show Worked Solution

`text(Consider option)\ C:`

`Delta` `= text(det)[(8,−4),(4,2)]`
  `= 8 xx 2 – (4 xx −4)`
  `= 32`
  `!= 0`

 

`text(In all other systems,)\ Delta = 0`

`=> C`

MATRICES, FUR1 2017 VCAA 2 MC

The matrix below shows how five people, Alan (`A`), Bevan (`B`), Charlie (`C`), Drew (`D`) and Esther (`E`), can communicate with each other.
 

`{:(),(),(text(sender)\ ):}{:(qquadqquadqquad\ text(receiver)),(qquadqquadAquadBquadCquadDquadE),({:(A),(B),(C),(D),(E):}[(0,1,0,1,0),(1,0,0,0,0),(0,0,0,1,1),(1,0,1,0,0),(0,0,1,0,0)]):}`
 

A 1 in the matrix shows that the person named in that row can send a message directly to the person named in that column.

For example, the 1 in row 3 and column 4 shows that Charlie can send a message directly to Drew.

Esther wants to send a message to Bevan.

Which one of the following shows the order of people through which the message is sent?

  1. Esther – Bevan
  2. Esther – Charlie – Bevan
  3. Esther – Charlie – Alan – Bevan
  4. Esther – Charlie – Drew – Bevan
  5. Esther – Charlie – Drew – Alan – Bevan
Show Answers Only

`E`

Show Worked Solution

`text(Esther can only send to Charlie.)`

`text(Charlie can send to Drew.)`

`text(Drew can send to Alan.)`

`text(Alan can send to Bevan.)`

`:.\ text(Esther – Charlie – Drew – Alan – Bevan)`

`=> E`

GRAPHS, FUR1 2017 VCAA 4 MC

The annual fee for membership of a car club, in dollars, based on years of membership of the club is shown in the step graph below.

In the Martin family:

• Hayley has been a member of the club for four years
• John has been a member of the club for 20 years
• Sharon has been a member of the club for 25 years.

What is the total fee for membership of the car club for the Martin family?

  1. `$200`
  2. `$600`
  3. `$720`
  4. `$900`
  5. `$940`
Show Answers Only

`E`

Show Worked Solution
`text(Total Fee)` `= 500 + 240 + 200`
  `= $940`

`=> E`

NETWORKS, FUR1 2017 VCAA 4-5 MC

The directed graph below shows the sequence of activities required to complete a project.

The time to complete each activity, in hours, is also shown.
 


 

Part 1

The earliest starting time, in hours, for activity `N` is

  1. 3
  2. 10
  3. 11
  4. 12
  5. 13

 

Part 2

To complete the project in minimum time, some activities cannot be delayed.

The number of activities that cannot be delayed is

  1. 2
  2. 3
  3. 4
  4. 5
  5. 6
Show Answers Only

`text(Part 1:)\ D`

`text(Part 1:)\ C`

Show Worked Solution

`text(Part 1)`

`text(EST of)\ N:\ CGJ`

`:.\ text(EST of)\ N` `= 4 + 3 +5`
  `= 12\ text(hours)`

`=> D`

 

`text(Part 2)`

`text(Critical path is:)\ CFHM`

`:. 4\ text(activities can’t be delayed.)`

`=> C`

GEOMETRY, FUR1 2017 VCAA 4 MC

A grain storage silo in the shape of a cylinder with a conical top is shown in the diagram below.

The volume of this silo, in cubic metres, is closest to

  1.   `550`
  2. `1304`
  3. `1327`
  4. `1398`
  5. `2560`
Show Answers Only

`B`

Show Worked Solution
`V` `=\ text(volume of cylinder + volume of cone)`
  `= pir^2h + 1/3pir^2h`
  `= pi xx 5^2 xx 16 + 1/3 xx pi xx 5^2 xx 1.8`
  `= 1303.76…\ text(m³)`

`=> B`

GEOMETRY, FUR1 2017 VCAA 3 MC

The locations of four cities are given below.

Adelaide (35° S, 139° E)     Buenos Aires (35° S, 58° W) 
Melilla (35° N, 3° W)      Heraklion (35° N, 25° E)

 

In which order, from first to last, will the sun rise in these cities on New Year’s Day 2018?

  1. Adelaide, Heraklion, Buenos Aires, Melilla
  2. Adelaide, Heraklion, Melilla, Buenos Aires
  3. Heraklion, Adelaide, Melilla, Buenos Aires
  4. Melilla, Adelaide, Heraklion, Buenos Aires
  5. Melilla, Adelaide, Buenos Aires, Heraklion
Show Answers Only

`B`

Show Worked Solution

`text(Sun rises in the East.)`

`text(Adelaide is further east than Heraklion.)`

`text(Melilla is further east (i.e. less west))`

`text(than Buenos Aires.)`

`:.\ text{Order is: Adelaide (139°E) , Heraklion (25°E),}`

    `text{Melilla (3°W), Buenos Aires (58°W)}`

`=> B`

Measurement, STD2 M1 2013 HSC 15a*

The diagram shows the front of a tent supported by three vertical poles. The poles are 1.2 m apart. The height of each outer pole is 1.5 m, and the height of the middle pole is 1.8 m. The roof hangs between the poles.

2013 15a

The front of the tent has area `A\ text(m²)`. 

  1. Use the trapezoidal rule to estimate `A`.    (2 marks)

  2. Explain whether the trapezoidal rule give a greater or smaller estimate of  `A`?  (1 mark)

Show Answers Only
  1. `3.96\ text(m²)`
  2. `text(The trapezoidal rule assumes a straight line between)`

     

    `text(all points and therefore would estimate a greater)`

     

    `text(area than the actual area of the tent front.)`

Show Worked Solution
i.    `A` `~~ h/2 [y_0 + 2y_1 + y_2]`
    `~~ 1.2/2 [1.5 + (2 xx 1.8) + 1.5]`
    `~~ 0.6 [6.6]`
    `~~ 3.96\ text(m²)`

 

ii.  `text(The trapezoidal rule assumes a straight line between)`

`text(all points and therefore would estimate a greater)`

`text(area than the actual area of the tent front.)`

CORE, FUR1 2017 VCAA 19-20 MC

Shirley would like to purchase a new home. She will establish a loan for $225 000 with interest charged at the rate of 3.6% per annum, compounding monthly.

Each month, Shirley will pay only the interest charged for that month.

Part 1

After three years, the amount that Shirley will owe is

  1.    $73 362
  2.  $170 752
  3.  $225 000
  4.  $239 605
  5.  $245 865

 
Part 2

Let `V_n` be the value of Shirley’s loan, in dollars, after `n` months.

A recurrence relation that models the value of `V_n` is

  1. `V_0 = 225\ 000,qquadV_(n + 1) = 1.003 V_n`
  2. `V_0 = 225\ 000,qquadV_(n + 1) = 1.036 V_n`
  3. `V_0 = 225\ 000,qquadV_(n + 1) = 1.003 V_n - 8100`
  4. `V_0 = 225\ 000,qquadV_(n + 1) = 1.003 V_n - 675`
  5. `V_0 = 225\ 000,qquadV_(n + 1) = 1.036 V_n - 675`
Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`text(If the loan payments are interest only,)`

`text(the principal outstanding after 3 years)`

`text(remains $225 000.)`

`=> C`
 

`text(Part 2)`

`text(Monthly interest rate)`

`= 3.6/12 = 0.3text(%) = 0.003`
 

`text(Monthly payment)`

`= 225\ 000 xx 0.3text(%)`

`= $675`
 

`:.\ text(Recurrence Relation is)`

`V_(n + 1) = 1.003V_n – 675`

`=> D`

CORE, FUR1 2017 VCAA 18 MC

The first five terms of a sequence are 2, 6, 22, 86, 342 …

The recurrence relation that generates this sequence could be

  1. `P_0 = 2,qquadP_(n + 1) = P_n + 4`
  2. `P_0 = 2,qquadP_(n + 1) = 2 P_n + 2`
  3. `P_0 = 2,qquadP_(n + 1) = 3 P_n`
  4. `P_0 = 2,qquadP_(n + 1) = 4 P_n – 2`
  5. `P_0 = 2,qquadP_(n + 1) = 5 P_n – 4`
Show Answers Only

`D`

Show Worked Solution

`text(Using trial and error,)`

`text(Consider option)\ D:`

`P_1 = 4 xx 2 – 2 = 6`

`P_2 = 4 xx 6 – 2 = 22`

`P_3 = 4 xx 22 – 2 = 86\ \ text(etc…)`

`=> D`

CORE, FUR1 2017 VCAA 17 MC

The value of a reducing balance loan, in dollars, after `n` months, `V_n`, can be modelled by the recurrence relation shown below.

`V_0 = 26\ 000,qquadV_(n + 1) = 1.003 V_n - 400`

What is the value of this loan after five months?

  1. `$24\ 380.31`
  2. `$24\ 706.19`
  3. `$25\ 031.10`
  4. `$25\ 355.03`
  5. `$25\ 678.00`
Show Answers Only

`=> A`

Show Worked Solution

`V_1 = 1.003 xx 26\ 000 – 400 = 25\ 678`

`V_2 = 1.003 xx 25\ 678 – 400 = 25\ 355.03`

`V_3 = 1.003 xx 25\ 355.03 – 400 = 25\ 031.10`

`V_4 = 1.003 xx 25\ 031.10 – 400 = 24\ 706.19`

`V_5 = 1.003 xx 24\ 706.19 – 400 = 24\ 380.31`

`=> A`

CORE, FUR1 2017 VCAA 13-15 MC

The wind speed at a city location is measured throughout the day.

The time series plot below shows the daily maximum wind speed, in kilometres per hour, over a three-week period.
  

 

Part 1

The time series is best described as having

  1. seasonality only.
  2. irregular fluctuations only.
  3. seasonality with irregular fluctuations.
  4. a decreasing trend with irregular fluctuations.
  5. an increasing trend with irregular fluctuations.

 

Part 2

The seven-median smoothed maximum wind speed, in kilometres per hour, for day 4 is closest to

  1. `22`
  2. `26`
  3. `27`
  4. `30`
  5. `32`

 

Part 3

The table below shows the daily maximum wind speed, in kilometres per hour, for the days in week 2.

A four-point moving mean with centring is used to smooth the time series data above.

The smoothed maximum wind speed, in kilometres per hour, for day 11 is closest to

  1. `22`
  2. `24`
  3. `26`
  4. `28`
  5. `30`
Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ D`

`text(Part 3:)\ D`

Show Worked Solution

`text(Part 1)`

`text(The time series plot shows no obvious trend and)`

`text(is over too short a period to show seasonality.)`

`=> B`

 

`text(Part 2)`

`text(Consider the 7 values where day 4 is the middle)`

`text(data point.)`

`text(By inspection of the graph, the 4th highest point = 30.)`

`=> D`

 

`text(Part 3)`

`text(Mean for Day 9 – 12)`

`= (22 + 19 + 22 + 43)/4 = 26.5`

 

`text(Mean for Day 10 – 13)`

`= (19 + 22 + 43 + 37)/4 = 30.25`

 

`:. 4text(-point moving mean with centring)`

`= (26.5 + 30.25)/2`

`= 28.375`

`=> D`

`

CORE, FUR1 2017 VCAA 12 MC

Data collected over a period of 10 years indicated a strong, positive association between the number of stray cats and the number of stray dogs reported each year (`r = 0.87`) in a large, regional city.

A positive association was also found between the population of the city and both the number of stray cats (`r = 0.61`) and the number of stray dogs (`r = 0.72`).

During the time that the data was collected, the population of the city grew from 34 564 to 51 055.

From this information, we can conclude that

  1. if cat owners paid more attention to keeping dogs off their property, the number of stray cats reported would decrease.
  2. the association between the number of stray cats and stray dogs reported cannot be causal because only a correlation of +1 or –1 shows causal relationships.
  3. there is no logical explanation for the association between the number of stray cats and stray dogs reported in the city so it must be a chance occurrence.
  4. because larger populations tend to have both a larger number of stray cats and stray dogs, the association between the number of stray cats and the number of stray dogs can be explained by a common response to a third variable, which is the increasing population size of the city.
  5. more stray cats were reported because people are no longer as careful about keeping their cats properly contained on their property as they were in the past.
Show Answers Only

`D`

Show Worked Solution

`text(Because larger populations tend to have both a larger)`

`text(number of stray cats and stray dogs, the association)`

`text(between the number of stray cats and the number of stray)`

`text(dogs can be explained by a common response to a third)`

`text(variable, which is the increasing population size of the city.)`

`=> D`

CORE, FUR1 2017 VCAA 11 MC

Which one of the following statistics can never be negative?

  1. the maximum value in a data set
  2. the value of a Pearson correlation coefficient
  3. the value of a moving mean in a smoothed time series
  4. the value of a seasonal index
  5. the value of the slope of a least squares line fitted to a scatterplot
Show Answers Only

`D`

Show Worked Solution

`text(The value of a seasonal index cannot be negative.)`

`=> D`

CORE, FUR1 2017 VCAA 8-10 MC

The scatterplot below shows the wrist circumference and ankle circumference, both in centimetres, of 13 people. A least squares line has been fitted to the scatterplot with ankle circumference as the explanatory variable.
 

Part 1

The equation of the least squares line is closest to

  1. ankle = 10.2 + 0.342 × wrist
  2. wrist = 10.2 + 0.342 × ankle
  3. ankle = 17.4 + 0.342 × wrist
  4. wrist = 17.4 + 0.342 × ankle
  5. wrist = 17.4 + 0.731 × ankle

 

Part 2

When the least squares line on the scatterplot is used to predict the wrist circumference of the person with an ankle circumference of 24 cm, the residual will be closest to

  1. `–0.7`
  2. `–0.4`
  3. `–0.1`
  4.    `0.4`
  5.    `0.7`

 

Part 3

The residuals for this least squares line have a mean of 0.02 cm and a standard deviation of 0.4 cm.

The value of the residual for one of the data points is found to be  – 0.3 cm.

The standardised value of this residual is

  1. `–0.8`
  2. `–0.7`
  3. `–0.3`
  4.    `0.7`
  5.    `0.8`
Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ A`

`text(Part 3:)\ A`

Show Worked Solution

`text(Part 1)`

`ytext(-intercept = 10.2)`

`(text(note)\ xtext(-axis in graph begins at 21 cm))`

`:. text(wrist) = 10.2 + 0.342 xx text(ankle)`

`=> B`

 

`text(Part 2)`

`text(Predicted wrist)` `= 10.2 + 0.342 xx 24`
  `= 18.4`

 

`text(Residual)` `=\ text(actual − predicted)`
  `~~ 17.7 – 18.4`
  `~~ −0.7`

 
`=> A`

 
`text(Part 3)`

`barx = 0.02,qquad s_x = 0.4`

`text(If)\ \ x = −0.3,`

`z` `= (x – barx)/s_x`
  `= (−0.3 – 0.02)/0.4`
  `= −0.8`

 
`=> A`

CORE, FUR1 2017 VCAA 5-7 MC

A study was conducted to investigate the association between the number of moths caught in a moth trap (less than 250, 250–500, more than 500) and the trap type (sugar, scent, light). The results are summarised in the percentaged segmented bar chart below.
 

Part 1

There were 300 sugar traps.

The number of sugar traps that caught less than 250 moths is closest to

  1. 30
  2. 90
  3. 250
  4. 300
  5. 500

 
Part 2

The data displayed in the percentaged segmented bar chart supports the contention that there is an association between the number of moths caught in a moth trap and the trap type because

  1. most of the light traps contained less than 250 moths.
  2. 15% of the scent traps contained 500 or more moths.
  3. the percentage of sugar traps containing more than 500 moths is greater than the percentage of scent traps containing less than 500 moths.
  4. 20% of sugar traps contained more than 500 moths while 50% of light traps contained less than 250 moths.
  5. 20% of sugar traps contained more than 500 moths while 10% of light traps contained more than 500 moths.

 
Part 3

The variables number of moths (less than 250, 250–500, more than 500) and trap type (sugar, scent, light) are

  1. both nominal variables.
  2. both ordinal variables.
  3. a numerical variable and a categorical variable respectively.
  4. a nominal variable and an ordinal variable respectively.
  5. an ordinal variable and a nominal variable respectively.
Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ E`

`text(Part 3:)\ E`

Show Worked Solution

`text(Part 1)`

`text(Sugar traps that caught < 250)`

`= 30text(%) xx 300`

`= 90`

`=> B`

 

`text(Part 2)`

`text(An association should compare different)`

`text(trap types against the same value of)`

`text(number of moths caught.)`

`=> E`

 

`text(Part 3)`

`text{Number of moths (grouped) – ordinal variable}`

`text(Trap type – nominal variable)`

`=> E`

CORE, FUR1 2017 VCAA 4 MC

The histogram below shows the distribution of the log10 (area), with area in square kilometres, of 17 islands.
 

The median area of these islands, in square kilometres, is between

  1. 2 and 3
  2. 3 and 4
  3. 10 and 100
  4. 1000 and 10 000
  5. 10 000 and 100 000
Show Answers Only

`D`

Show Worked Solution

`text(17 data points)\ =>\ text(median is 9th)`

`3 < text(log)_10(text(area)) < 4`

`1000 < text(area) < 10\ 000`

`=> D`

Geometry, NAP-A3-NC01

This map shows the location of different animal enclosures at a zoo.
 

 
Which animal enclosure is West of the Giraffes and North of the Monkeys?

Reptiles Bears Tigers Crocodiles
 
 
 
 
Show Answers Only

`text(Crocodiles)`

Show Worked Solution

`text(Crocodiles)`

Quadratic, EXT1 2017 HSC 14b

Let  `P(2p, p^2)`  be a point on the parabola  `x^2 = 4y`.

The tangent to the parabola at `P` meets the parabola  `x^2 = −4ay`, `a > 0`, at `Q` and `R`. Let `M` be the midpoint of `QR`.

  1. Show that the `x` coordinates of `R` and `Q` satisfy
  2. `qquadx^2 + 4apx - 4ap^2 = 0`.  (2 marks)

  3. Show that the coordination of `M` are  `(−2ap, −p^2(2a + 1))`.  (2 marks)
  4. Find the value of `a` so that the point `M` always lies on the parabola  `x^2 = −4y`.  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `1 + sqrt2, a > 0`
Show Worked Solution

(i)   `x^2 = 4y,\ \ =>y = (x^2)/4`

`(dy)/(dx) = x/2`

`text(At)\ P(2p, p^2),`

`(dy)/(dx) = p`

`text(Equation of tangent:)`

`y = px – p^2`

 

`R and Q\ text(at intersection)`

`y` `= px – p^2\ …\ (1)`
`y` `= −(x^2)/(4a)\ …\ (2)`

 

`text(Subtract)\ (1) – (2)`

`px – p^2 + (x^2)/(4a)` `= 0`
`x^2 + 4apx – 4ap^2` `= 0\ …\ text(as required)`

 

(ii)   `x^2 + 4apx – 4ap^2 = 0`

♦♦ Mean mark 30%.
`x=` `\ (−4ap ± sqrt((4ap)^2 – 4 · 1 · (−4ap^2)))/2`
`=`  `\ (−4ap ± sqrt(16ap^2(a + 1)))/2`
`=`  `\ −2ap ± 2psqrt(a(a + 1))`

 

`=> xtext(-coordinate of)\ M\ text(is)\ −2ap.`

 

`M\ text(lies on tangent)\ \ y = px – p^2`

`:.y` `= p(−2ap) – p^2`
  `= −2ap^2 – p^2`
  `= −p^2(2a + 1)`

`:. M\ text(has coordinates)\ \ (−2ap, −p^2(2a + 1))`

 

(iii)   `text(If)\ M\ text(always lies on)\ \ x^2 = −4y`

♦♦ Mean mark 23%.
`(−2ap)^2` `= −4(−p^2(2a + 1))`
`4a^2p^2` `= 4p^2(2a + 1)`
`a^2` `= 2a + 1`
`a^2 – 2a – 1` `= 0`
`:. a` `= (+2 ± sqrt(4 + 4 · 1· 1))/2`
  `= 1 ± sqrt2`
  `= 1 + sqrt2, \ a > 0`

Proof, EXT1 P1 2017 HSC 14a

Prove by mathematical induction that  `8^(2n + 1) + 6^(2n − 1)`  is divisible by 7, for any integer  `n ≥ 1`.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(Prove)\ \ 8^(2n + 1) + 6^(2n − 1)\ \ text(is divisible)`

`text(by 7 for integers)\ \ n >= 1`

`text(If)\ n = 1,`

`8^3 + 6^1 = 518 = 74 xx 7`

`:. text(True for)\ n = 1`
 

`text(Assume true for)\ n = k`

`text(i.e.)\ \ 8^(2k + 1) + 6^(2k – 1)` `= 7P\ \ (text(where)\ P\ text(is an integer))`
`8^(2k + 1)` `= 7P – 6^(2k – 1)`

 
`text(Prove true for)\ n = k + 1`

`8^(2k + 3) + 6^(2k − 1)` `= 64 · 8^(2k + 1) + 36*6^(2k – 1)`
  `= 64(7P – 6^(2k – 1)) + 36 · 6^(2k – 1)`
  `= 64 · 7P – 64 · 6^(2k – 1) + 36 · 6^(2k – 1)`
  `= 64 · 7P – 28 · 6^(2k – 1)`
  `= 7(64P – 4 · 6^(2k – 1))`

 
`…\ text(which is divisible by 7.)`
 

`=> text(True for)\ \ n = k + 1`

`:.\ text(S) text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ \ n >= 1.`

Mechanics, EXT2* M1 2017 HSC 13c

A golfer hits a golf ball with initial speed `V\ text(ms)^(−1)` at an angle `theta` to the horizontal. The golf ball is hit from one side of a lake and must have a horizontal range of 100 m or more to avoid landing in the lake.
 

     

Neglecting the effects of air resistance, the equations describing the motion of the ball are

`x = Vt costheta`

`y = Vt sintheta - 1/2 g t^2`,

where `t` is the time in seconds after the ball is hit and `g` is the acceleration due to gravity in `text(ms)^(−2)`. Do NOT prove these equations.

  1. Show that the horizontal range of the golf ball is
     
         `(V^2sin 2theta)/g` metres.  (2 marks)

  2. Show that if  `V^2 < 100 g`  then the horizontal range of the ball is less than 100 m.  (1 mark)

It is now given that  `V^2 = 200 g`  and that the horizontal range of the ball is 100 m or more.

  1. Show that  `pi/12 <= theta <= (5pi)/12`.  (2 marks)

  2. Find the greatest height the ball can achieve.  (2 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `text(See Worked Solution)`
  4. `25(2 – sqrt 3)\ text(metres)`
Show Worked Solution

i.   `text(Find)\ \ t\ \ text(when)\ \ y = 0:`

`1/2 g t^2` `= Vtsintheta`
`1/2 g t` `= Vsintheta`
`t` `= (2Vsintheta)/g`

 

`text(Horizontal range)\ (x)\ text(when)\ \ t = (2Vsintheta)/g :`

`x` `= V · (2Vsintheta)/g costheta`
  `= (V^2 2sintheta costheta)/g`
  `= (V^2sin2theta)/g\ … text(as required)`

 

ii.   `text(If)\ \ V^2 < 100 g`

♦ Mean mark 44%.
`x` `< (100 g sin2theta)/g`
`x` `< 100 sin2theta`

 

`text(S)text(ince)\ −1 <= 2theta <= 1,`

`x < 100\ text(metres)`

 

iii.   `V^2 = 200g,\ \ x >= 100`

`(200 g · sin2theta)/g` `>= 100`
`sin2theta` `>= 1/2`

`:. pi/6 <= 2theta <= (5pi)/6`

`:. pi/12 <= theta <= (5pi)/12\ …\ text(as required)`

 

iv.   `text(Max height occurs when)`

♦♦ Mean mark 35%.
`t` `= 1/2 xx text(time of flight)`
  `= (Vsintheta)/g`

 
`text(Find)\ \ y\ \ text(when)\ \ t = (Vsintheta)/g`

`y` `= V · (Vsintheta)/g · sintheta – 1/2 g ((Vsintheta)/g)^2`
  `= (V^2 sin^2theta)/g – 1/2 · (V^2 sin^2 theta)/g`
  `= (V^2 sin^2theta)/(2g)`

 

`text(Max height when)\ theta = (5pi)/12\ (text(steepest angle)), V^2 = 200 g\ (text(given))`

`y_text(max)` `= (200 g · sin^2 ((5pi)/12))/(2g)`
  `= 100 sin^2 ((5pi)/12)`
  `= 50(1 – cos((5pi)/6))`
  `= 50(1 + sqrt3/2)`
  `= 25(2 + sqrt 3)\ text(metres)`

Binomial, EXT1 2017 HSC 13b

Let `n` be a positive EVEN integer.

  1. Show that
  2. `(1 + x)^n + (1 - x)^n = 2[((n),(0)) + ((n),(2))x^2 + … + ((n),(n))x^n]`.  (2 marks)

  3. Hence show that
  4. `n[(1 + n)^(n - 1) - (1 - x)^(n - 1)] = 2[2((n),(2))x + 4((n),(4))x^3 + … + n((n),(n))x^(n - 1)]`.  (1 mark)

  5. Hence show that
  6. `((n),(2)) + 2((n),(4)) + 3((n),(6)) + … + n/2((n),(n)) = n2^(n - 3)`.  (2 marks)

 

 

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

(i)   `text(Show)\ (1 + x)^n + (1 – x)^n = 2[\ ^nC_0 + \ ^nC_2x^2 + … + \ ^nC_nx^n]`

`text(Using binomial expansion)`

`(1 + x)^n + (1 – x)^n`

`= \ ^nC_0 + \ ^nC_1x + \ ^nC_2x^2 + … + \ ^nC_nx^n`

`+ \ ^nC_0 – \ ^nC_1x + \ ^nC_2x^2 + … + \ ^nC_nx^n`

`= 2[\ ^nC_0 + \ ^nC_2x^2 + \ ^nC_4x^4 + … + \ ^nC_nx^n]`

`…\ text(as required)`

 

(ii)   `text{Differentiate both sides of part (i):}`

`n(1 + x)^(n-1) – n(1 – x)^(n-1)`

`= 2[2\ ^nC_2x + 4\ ^nC_4x^3 + … + n \ ^nC_nx^(n – 1)]`

`n[(1 + x)^(n – 1) + (1 – x)^(n – 1)]`

`= 2[2\ ^nC_2x + 4\ ^nC_4x^3 + … + n \ ^nC_nx^(n – 1)]`

`…\ text(as required.)`

♦ Mean mark part (iii) 48%.

 

(iii)   `text(Let)\ \ x = 1\ \ text{in part (ii)}`

`n(2^(n – 1) + 0) = 2[2\ ^nC_2 + 4\ ^nC_4 + 6\ ^nC_6 + … + n\ ^nC_n]`

`text(Divide both sides by)\ 2^2:`

`(n2^(n – 1))/(2^2)` `= 2/(2^2)[2\ ^nC_2 + 4\ ^nC_4 + 6\ ^nC_6 + … + n\ ^nC_n]`
`n2^(n – 3)`

`= \ ^nC_2 + 2\ ^nC_4 + 3\ ^nC_6 + … + n/2\ ^nC_ n`

`…\ text(as required)`

Mechanics, EXT2* M1 2017 HSC 12d

At time `t` the displacement, `x`, of a particle satisfies  `t=4-e^(-2x)`.

Find the acceleration of the particle as a function of `x`.  (3 marks)

Show Answers Only

`(e^(4x))/2`

Show Worked Solution
`t` `= 4 – e^(−2x)`
`(dt)/(dx)` `= 2e^(−2x)`
`(dx)/(dt)` `= (e^(2x))/2`

 

`a` `= {:d/(dx):}^(1/2v^2)`
  `= d/(dx)(1/2 · ((e^(2x))/2)^2)`
  `= d/(dx)((e^(4x))/8)`
  `= 4 · (e^(4x))/8`
  `= (e^(4x))/2`

Integration, EXT1 2017 HSC 12c

The region enclosed by the semicircle  `y = sqrt(1 - x^2)`  and the `x`-axis is to be divided into two pieces by the line  `x = h`, when  `0 <= h <1`.
 


 

The two pieces are rotated about the `x`-axis to form solids of revolution. The value of `h` is chosen so that the volumes of the solids are in the ratio `2 : 1`.

  1. Show that `h` satisfies the equation  `3h^3 - 9h + 2 = 0`.  (3 marks)
  2. Given  `h_1 = 0`  as the first approximation for `h`, use one application of Newton’s method to find a second approximation for `h`.  (1 mark)

 

Show Answers Only
  1. `text(Show Worked Solution)`
  2. `2/9`
Show Worked Solution

(i)   `text(Volume of smaller solid)`

`= pi int_h^1 (sqrt(1 – x^2))^2\ dx`

`= pi int_h^1 1 – x^2\ dx`

`= pi[x – (x^3)/3]_h^1`

`= pi[(1 – 1/3) – (h – (h^3)/3)]`

`= pi(2/3 – h + (h^3)/3)`

 
`text(S)text(ince smaller solid is)\ 1/3\ text(volume of sphere,)`

`pi(2/3 – h + (h^3)/3)` `= 1/3 xx 4/3 · pi · 1^3`
`(h^3)/3 – h + 2/3` `= 4/9`
`3h^3 – 9h + 6` `= 4`
`:. 3h^3 – 9h + 2` `= 0\ \ text(… as required)`

 

(ii)    `f(h)` `= 3h^3 – 9h + 2`
  `f′(h)` `= 6h^2 – 9`
  `f(0)` `= 2`
  `f′(0)` `= −9`

 

`h_2` `= h_1 – (f(0))/(f′(0))`
  `= 0 + 2/9`
  `= 2/9`

Functions, EXT1 F1 2017 HSC 12b

  1. Carefully sketch the graphs of  `y = |\ x + 1\ |`  and  `y = 3 - |\ x - 2\ |`  on the same axes, showing all intercepts.  (3 marks)

  2. Using the graphs from part (i), or otherwise, find the range of values of `x` for which
     
    `qquad qquad |\ x + 1\ | + |\ x - 2\ | = 3`.  (1 mark)

Show Answers Only
  1.  
  2. `−1 <= x <= 2`
Show Worked Solution
i.   

 

ii.    `|\ x + 1\ | + |\ x – 2\ |` `= 3`
  `|\ x + 1\ |` `= 3 – |\ x – 2\ |`

 

`:.\ text(Solution is intersection of the graphs)`

`:. −1 <= x <= 2`

Calculus, 2ADV C3 2017 HSC 16a

John’s home is at point `A` and his school is at point `B`. A straight river runs nearby.

The point on the river closest to `A` is point `C`, which is 5 km from `A`.

The point on the river closest to `B` is point `D`, which is 7 km from `B`.

The distance from `C` to `D` is 9 km.

To get some exercise, John cycles from home directly to point `E` on the river, `x` km from `C`, before cycling directly to school at `B`, as shown in the diagram.
 


  

The total distance John cycles from home to school is `L` km.

  1. Show that  `L = sqrt (x^2 + 25) + sqrt (49 + (9 - x)^2)`.   (1 mark)

  2. Show that if  `(dL)/(dx) = 0`, then  `sin alpha = sin beta`.   (3 marks)

  3. Find the value of `x` that makes  `sin alpha = sin beta`.   (2 marks)

  4. Explain why this value of `x` gives a minimum for `L`.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `45/12`
  4. `text(See Worked Solutions)`
Show Worked Solution
i.  

`text(Using Pythagoras:)`

`L` `= AE + EB`
  `= sqrt (5^2 + x^2) + sqrt (7^2 + (9 – x)^2)`
  `= sqrt(25 + x^2) + sqrt (49 + (9 – x)^2)\ text(… as required)`

 

ii.  `text(From diagram):`

♦ Mean mark 41%.

`sin alpha = x/sqrt(25 + x^2) and sin beta = (9 – x)/sqrt(49 + (9 – x)^2)`

`L` `= sqrt(25 + x^2) + sqrt (49 + (9 – x)^2)`
`(dL)/(dx)` `= (2x)/sqrt(25 + x^2) – (2(9 – x))/sqrt(49 + (9 – x)^2)`

 

`text(If)\ \ (dL)/(dx) = 0,`

`=> (2x)/sqrt(25 + x^2)` `= (2(9 – x))/sqrt(49 + (9 – x)^2)`
`x/sqrt(25 + x^2)` `= (9 – x)/sqrt(49 + (9 – x)^2)`
`sin alpha` `= sin beta\ text(… as required)`

 

iii.  `text(If)\ sin alpha = sin beta,\ text(then)\ alpha = beta and`

♦♦ Mean mark 29%.

`Delta ACE\ text(|||)\ Delta BDE`

`text(Using corresponding sides of similar triangles:)`

`x/5` `= (9 – x)/7`
`7x` `= 45 – 5x`
`12x` `= 45`
`:. x` `= 45/12\ text(km)`
♦♦♦ Mean mark 9%.

 

iv.  

`text(If point)\ B\ text(is reflected across the)`

`text(river),\ AEB\ text(will be a straight line.)`

`text(If any other point is chosen,)\ AEB`

`text(would not be straight and the distance)`

`text(would be longer.)`

Calculus, EXT1* C1 2017 HSC 15c

Two particles move along the `x`-axis.

When  `t = 0`, particle `P_1` is at the origin and moving with velocity 3.

For  `t >= 0`, particle `P_1` has acceleration given by  `a_1 = 6t + e^(-t)`.

  1. Show that the velocity of particle `P_1` is given by  `v_1 = 3t^2 + 4-e^(-t)`  (2 marks)

When  `t = 0`, particle `P_2` is also at the origin.

For  `t >= 0`, particle `P_2` has velocity given by  `v_2 = 6t + 1-e^(-t)`.

  1. When do the two particles have the same velocity?  (2 marks)

  2. Show that the two particles do not meet for  `t > 0`.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `t = 1`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.   `a_1` `= 6t + e^(-t)`
  `v_1` `= int a_1\ dt`
    `= int 6t + e^(-t)\ dt`
    `= 3t^2-e^(-t) + c`

 

`text(When)\ t = 0,\ v_1 = 3`

`3` `= 0-1 + c`
`c` `= 4`
`:. v_1` `= 3t^2 + 4-e^(-t) …\ text(as required)`

 

ii.  `v_2 = 6t + 1-e^(-t)`

`text(Find)\ \ t\ \ text(when)\ \ v_1 = v_2`

`3t^2 + 4-e^(-t)` `= 6t + 1-e^(-t)`
`3t^2-6t + 3` `= 0`
`t^2-2t + 1` `= 0`
`(t-1)^2` `= 0`
`:. t` `=1`

 

iii.   `x_1` `= int v_1\ dt`
    `= int 3t^2 + 4-e^(-t)\ dt`
    `= t^3 + 4t + e^(-t) + c`

 

`text(When)\ \ t = 0,\ \ x_1 = 0`

♦ Mean mark (iii) 39%.
`0` `= 0 + 0 + 1 + c`
`c` `= -1`
`:. x_1` `= t^3 + 4t + e^(-t)-1`

 

`x_2` `= int 6t + 1-e^(-t)\ dt`
  `= 3t^2 + t + e^(-t) + c`

 
`text(When)\ \ t = 0,\ \ x_2 = 0`

`0` `= 0 + 0 + 1 + c`
`c` `= -1`
`:. x_2` `= 3t^2 + t + e^(-t)-1`

 

`text(Find)\ \ t\ \ text(when)\ \ x_1 = x_2`

`t^3 + 4t + e^(-t)-1` `= 3t^2 + t + e^(-t)-1`
`t^3-3t^2 + 3t` `= 0`
`t(t^2-3t + 3)` `= 0`

 

`text(S)text(ince)\ \ Delta < 0\ \ text(for)\ \ t^2-3t + 3`

`=>\ text(No real solution)`

 

`:.\ text(The particles do not meet)`

`(x_1 != x_2)\ \ text(for)\ \ t > 0.`

Financial Maths, 2ADV M1 2017 HSC 15b

Anita opens a savings account. At the start of each month she deposits `$X` into the savings account. At the end of each month, after interest is added into the savings account, the bank withdraws $2500 from the savings account as a loan repayment. Let `M_n` be the amount in the savings account after the `n`th  withdrawal.

The savings account pays interest of 4.2% per annum compounded monthly.

  1. Show that after the second withdrawal the amount in the savings account is given by
    `qquad qquad M_2 = X(1.0035^2 + 1.0035) - 2500 (1.0035 + 1)`.  (2 marks)

  2. Find the value of  `X`  so that the amount in the savings account is $80 000 after the last withdrawal of the fourth year.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `$ 4019.42\ text{(nearest cent)}`
Show Worked Solution
i.   `M_1` `= X (1 + 4.2/(12 xx 100)) – 2500`
    `= X (1.0035) – 2500`
  `M_2` `= X (1.0035) + M_1 (1.0035) – 2500`
    `=X(1.0035)+(1.0035)[X(1.0035)-2500]-2500`
    `= X (1.0035) + X (1.0035^2) – 2500 (1.0035) – 2500`
    `= X (1.0035^2 + 1.0035) – 2500 (1.0035 + 1)`
♦ Mean mark part (ii) 48%.

 

ii.  

`M_3 = X (1.0035^3 + … + 1.0035) – 2500 (1.0035^2 + 1.0035 + 1)`

`vdots`

`M_n = X (1.0035^n + … + 1.0035) – 2500 (1.0035^(n – 1) + … + 1)`

`text(Find)\ X\ text(such that)\ \ M_n = 80\ 000\ \ text(when)\ n = 48`

`80\ 000 = X (1.0035^48 + …\ 1.0035) – 2500 (1.0035^47 + … + 1)`

`text(Using)\ S_n = (a(r^n – 1))/(r – 1)`

`80\ 000` `= X [(1.0035(1.0035^48 – 1))/(1.0035 – 1)] – 2500 [(1(1.0035^48 – 1))/(1.0035 – 1)]`
`80\ 000` `= X (52.351…) – 130\ 421.20…`
   
`:. X` `= (80\ 000 + 130\ 421.20…)/(52.351…)`
  `= $4019.42\ text{(nearest cent)}`

Plane Geometry, 2UA 2017 HSC 15a

The triangle `ABC` is isosceles with  `AB = AC`  and the size of `/_BAC` is `x^@`.

Points `D` and `E` are chosen so that `Delta ABC, Delta ACD` and `Delta ADE` are congruent, as shown in the diagram.
 

Find the value of `x` for which `AB` is parallel to `ED`, giving reasons.  (3 marks)

Show Answers Only

`x = 36`

Show Worked Solution

`text(All base angles) = 1/2(180-x)\ \ text{(Angle sum of}\ Delta text{)}`

`text(If)\ \ AB\ text(||)\ ED,`

`/_ BAD` `= /_ ADE\ \ \ text{(alternate angles)}`
`2x` `= 1/2 (180-x)`
`4x` `= 180-x`
`5x` `= 180`
`x` `= 36^{\circ}`

Calculus, 2ADV C4 2017 HSC 14d

The shaded region shown is enclosed by two parabolas, each with `x`-intercepts at  `x = –1`  and  `x = 1`.

The parabolas have equations  `y = 2k (x^2 - 1)`  and  `y = k(1 - x^2)`, where  `k > 0`.
  


 

Given that the area of the shaded region is 8, find the value of `k`.  (3 marks)

Show Answers Only

`k = 2`

Show Worked Solution
`text(Area)` `= int_(-1)^1 k (1 – x^2) dx – int_(-1)^1 2k (x^2 – 1) dx`
`8` `= 2 int_0^1 k – kx^2 – 2kx^2 + 2k\ dx`
`8` `= 2 int_0^1 3k – 3kx^2\ dx`
`8` `= 2 [3kx – kx^3]_0^1`
`8` `= 2 [(3k – k) – 0]`
`8` `= 4k`
`:. k` `= 2`

Calculus, EXT1* C1 2017 HSC 14c

Carbon-14 is a radioactive substance that decays over time. The amount of carbon-14 present in a kangaroo bone is given by

`C(t) = Ae^(kt),`

where `A` and `k` are constants, and `t` is the number of years since the kangaroo died.

  1. Show that `C(t)` satisfies  `(dC)/(dt) = kC`.  (1 mark)

  2. After 5730 years, half of the original amount of carbon-14 is present.

     

    Show that the value of `k`, correct to 2 significant figures, is – 0.00012.  (2 marks)

  3. The amount of carbon-14 now present in a kangaroo bone is 90% of the original amount.

     

    Find the number of years since the kangaroo died. Give your answer correct to 2 significant  figures.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `870\ text{years (2 sig. fig.)}`
Show Worked Solution
i.   `C` `= Ae^(kt)`
   `(dC)/(dt)` `= k * Ae^(kt)`
    `= kC …\ text(as required)`

 

ii.  `text(When)\ \ t = 5730, qquad A = 0.5 A_0`

`0.5 A_0` `= A_0 * e^(5730 k)`
`e^(5730 k)` `= 0.5`
`text(ln)\ e^(5730 k)` `= text(ln)\ 0.5`
`5730 k` `= text(ln)\ 0.5`
`k` `= {text(ln)\ 0.5}/5730`
  `= -0.0001209…`
  `= -0.00012\ text{(2 sig fig) … as required}`

 

iii.  `text(Find)\ t\ text(when)\ A = 0.9 A_0`

`0.9 A_0` `= A_0 e^(kt)`
`e^(kt)` `= 0.9`
`kt` `= text(ln)\ 0.9`
`t` `= (text(ln)\ 0.9)/k`
  `= (5730 xx text(ln)\ 0.9)/(text(ln)\ 0.5)`
  `= 870.97…`
  `= 870\ text{years (2 sig.fig.)}`

Combinatorics, EXT1 A1 2017 HSC 10 MC

Three squares are chosen at random from the 3 × 3 grid below, and a cross is placed in each chosen square.
 


 

What is the probability that all three crosses lie in the same row, column or diagonal?

A.     `1/28`

B.     `2/21`

C.     `1/3`

D.     `8/9`

Show Answers Only

`B`

Show Worked Solution
`P` `= text(favourable events)/text(total possible events)`
  `= (3 + 3 + 2)/(\ ^9C_3)`
  `= 2/21`

 
`=>B`

Combinatorics, EXT1 A1 2017 HSC 9 MC

When expanded, which expression has a non-zero constant term?

A.     `(x + 1/(x^2))^7`

B.     `(x^2 + 1/(x^3))^7`

C.     `(x^3 + 1/(x^4))^7`

D.     `(x^4 + 1/(x^5))^7`

Show Answers Only

`C`

Show Worked Solution

`text(Consider the general term for option)\ A:`

`T_k` `= \ ^7C_k · x^(7 – k) · x^(−2k)`
  `= \ ^7C_k · x^(7 – 3k)`

 
`text(Non zero constant term occurs when)`

`7 – 3k` `= 0`
`k` `= 7/3 => text(no terms exists)\ (k\ text{not integer)}`

 
`text(Consider option)\ C:`

`T_k` `= \ ^7C_k · x^(3(7 – k)) · x^(−4k)`
  `= \ ^7C_k · x^(21 – 7k)`
`21 – 7k` `= 0`
`k` `= 3`

 
`:.\ text(Non-zero constant term exists)`

`text(since)\ k\ text(is an integer)`

`⇒C`

Calculus, 2ADV C4 2017 HSC 14b

  1. Find the exact value of
     
  2. `qquad int_0^(pi/3) cos x\ dx`.  (1 mark)

  3. Using Simpson’s rule with one application, find an approximation to the integral
  4.  
    `qquad int_0^(pi/3) cos x\ dx,`
     
  5. leaving your answer in terms of `pi` and `sqrt 3`.  (2 marks)
     

  6. Using parts (i) and (ii), show that
     
  7. `qquad pi ~~ (18 sqrt 3)/(3 + 4 sqrt 3)`.  (1 mark)

 

 

Show Answers Only
(i)   `sqrt 3/2`
(ii)   `((4 sqrt 3 + 3)pi)/36`
(iii)   `text{Proof (See Worked Solutions)}`
Show Worked Solution
(i)   `int_0^(pi/3) cos x\ dx` `= [sin x]_0^(pi/3)`
    `= sin\ pi/3 – 0`
    `= sqrt 3/2`

 

(ii)  
    `x`     `0`    `overset(pi) underset(6) _`     `overset(pi) underset(3) _`  
    `y`     `1`     `overset(sqrt 3) underset(2) _`     `overset(1) underset(2) _`  
      `y_0`     `y_1`     `y_2`  
`int_0^(pi/3) cos x\ dx` `~~ h/3 [y_0 + 4y_1 + y_2]`
  `~~ pi/6 ⋅ 1/3 [1 + 4 ⋅ sqrt 3/2 + 1/2]`
  `~~ pi/18 ((4 sqrt 3 + 3)/2)`
  `~~ ((4 sqrt 3 + 3) pi)/36`

 

(iii)  `text{Using parts (i) and (ii)}`

♦ Mean mark 49%.
`((4 sqrt 3 + 3) pi)/36` `~~ sqrt 3/2`
`:. pi` `~~ (36 sqrt 3)/(2(3 + 4 sqrt 3))`
  `~~ (18 sqrt 3)/(3 + 4 sqrt 3) … text( as required)`

Quadratic, EXT1 2017 HSC 6 MC

The point  `P(2/p, 1/(p^2))`, where  `p != 0`  lies on the parabola  `x^2 = 4y`.

What is the equation of the normal at `P`?

A.     `py - x = −p`

B.     `p^2y + px = −1`

C.     `p^2y - p^3x = 1 − 2p^2`

D.     `p^2y + p^3x = 1 + 2p^2`

Show Answers Only

`D`

Show Worked Solution

`x^2 = 4y,\ \ => a = 1`

`text(Normal equation at)\ (2t,t^2):`

`=> x + ty = t^3 + 2t`

`text(When)\ (2/p, 1/(p^2)) ≡ (2t,t^2),`

`t = 1/p`

`:.\ text(Equation of normal:)`

`x + y/p` `= 1/(p^3) + 2/p`
`p^3x + p^2y` `= 1 + 2p^2`

 

`⇒D`

Trigonometry, EXT1 T3 2017 HSC 4 MC

What is the value of  `tan alpha`  when the expression  `2sinx - cosx`  is written in the form  `sqrt5 sin(x - alpha)`?

A.     `−2`

B.     `−1/2`

C.     `1/2`

D.     `2`

Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ sqrt5 sin(x – alpha) = sqrt5 sinx cosalpha – sqrt5cosxsinalpha,`

`=>sqrt5 sinx cosalpha – sqrt5 cosx sinalpha = 2sinx – cosx`

`=> sqrt5 cosalpha = 2, sqrt5 sinalpha = 1`

`(sinalpha)/(cosalpha)` `= 1/2`
`:. tan alpha` `= 1/2`

`⇒C`

Quadratic, 2UA 2017 HSC 13c

By letting  `m = t^(1/3)`, or otherwise, solve  `t^(2/3) + t^(1/3) - 6 = 0`.  (2 marks)

Show Answers Only

`t = -27 or 8`

Show Worked Solution

`t^(2/3) + t^(1/3) – 6 = 0`

`text(Let)\ \ m = t^(1/3),`

`m^2 + m – 6 = 0`

`(m + 3) (m – 2) = 0`

`m = -3 or 2`

`t^(1/3)` `= -3` `or` `t^(1/3)` `= 2`
`t` `= (-3)^3`   `t` `= 2^3`
  `= -27`     `= 8`

 

`:. t = -27 or 8`

Calculus, 2ADV C3 2017 HSC 13b

Consider the curve  `y = 2x^3 + 3x^2 - 12x + 7`.

  1. Find the stationary points of the curve and determine their nature.  (4 marks)

  2. Sketch the curve, labelling the stationary points.  (2 marks)

  3. Hence, or otherwise, find the values of `x` for which `(dy)/(dx)` is positive.  (1 mark)

Show Answers Only
  1. `text(maximum at)\ (-2, 27)`

     

    `text(minimum at)\ (1, 0)`

  2.    
  3. `x < -2 and x > 1`
Show Worked Solution
i.   `y` `= 2x^3 + 3x^2 – 12x + 7`
  `(dy)/(dx)` `= 6x^2 + 6x – 12`
  `(d^2y)/(dx^2)` `= 12x + 6`

 

`text(S.P. when)\ (dy)/(dx)` `= 0`
`6x^2 + 6x – 12` `= 0`
`x^2 + x – 2` `= 0`
`(x + 2) (x – 1)` `= 0`

 
`x = -2 or 1`
 

`text(When)\ \ x = –2, (d^2y)/(dx^2) < 0`

`:.\ text(MAX at)\ (–2, 27)`
 

`text(When)\ \ x = 1, (d^2y)/(dx^2) > 0`

`:.\ text(MIN at)\ (1, 0)`

 

ii.  

 

iii.  `text(Solution 1)`

`text(From graph, gradient is positive for)`

`x < –2 and x > 1`

`:. (dy)/(dx) > 0\ \ text(for)\ \ x < –2 and x > 1`

 

`text(Solution 2)`

`(dy)/(dx) > 0`

`6x^2 + 6x – 12` `> 0`
`(x + 2) (x – 1)` `> 0`

 
 
`:. x < –2 and x > 1`

Trigonometry, 2ADV T1 2017 HSC 13a

Using the cosine rule, find the value of `x` in the following diagram.  (3 marks)
 

Show Answers Only

`x = 11`

Show Worked Solution

`text(C)text(osine Rule:)`

`c^2` `= a^2 + b^2 – 2ab cos C`
`13^2` `= (x – 4)^2 + (x + 4)^2 – 2 (x – 4) (x + 4) cos 60^@`
`169` `= x^2 – 8x + 16 + x^2 + 8x + 16 – (x^2 – 16)`
`169` `= x^2 + 48`
`x^2` `= 121`
`:. x` `= 11, qquad (x != –11)`

Statistics, 2ADV 2017 HSC 12e

A spinner is marked with the numbers 1, 2, 3, 4 and 5. When it is spun, each of the five numbers is equally likely to occur.
 

 
The spinner is spun three times.

  1. What is the probability that an even number occurs on the first spin?  (1 mark)
  2. What is the probability that an even number occurs on at least one of the three spins?  (1 mark)
  3. What is the probability that an even number occurs on the first spin and odd numbers occur on the second and third spins?  (1 mark)
  4. What is the probability that an even number occurs on exactly one of the three spins?  (1 mark)
Show Answers Only
  1. `2/5`
  2. `98/125`
  3. `18/125`
  4. `54/125`
Show Worked Solution

i.   `Ptext{(even)} = 2/5`
 

ii.  `Ptext{(at least 1 even)}`

`= 1 – Ptext{(no evens)}`

`= 1 – 3/5 ⋅ 3/5 ⋅ 3/5`

`= 1 – 27/125`

`= 98/125`
 

iii.  `Ptext{(even, odd, odd)}`

`= 2/5 ⋅ 3/5 ⋅ 3/5`

`= 18/125`
 

iv.  `Ptext{(even occurs exactly once)}`

`= Ptext{(e, o, o)} + P text{(o, e, o)} + P text{(o, o, e)}`

`= 2/5 ⋅ 3/5 ⋅ 3/5 + 3/5 ⋅ 2/5 ⋅ 3/5 + 3/5 ⋅ 3/5 ⋅ 2/5`

`= 54/125`

Linear Functions, 2UA 2017 HSC 12d

The points  `A(–4, 0)`  and  `B(1, 5)`  lie on the line  `y = x + 4`.

The length of  `AB`  is  `5 sqrt 2`.

The points  `C(0, –2)`  and  `D(3, 1)`  lie on the line  `x - y - 2 = 0`.

The points `A, B, D, C` form a trapezium as shown.
 


 

  1. Find the perpendicular distance from point `A(–4, 0)` to the line  `x - y - 2 = 0`.  (1 mark)
  2. Calculate the area of the trapezium.  (2 marks)
Show Answers Only

(i)   `3sqrt2\ text(units)`

(ii)  `24\ text(u²)`

Show Worked Solution

(i)   `A(-4, 0), qquad qquad x – y – 2 = 0`

`_|_\ text(dist)` `= |ax_1 + by_1 + c|/sqrt(a^2 + b^2)`
  `= |-4 + 0 – 2|/sqrt (1 + 1)`
  `= 6/sqrt 2 xx sqrt 2/sqrt 2`
  `= 3 sqrt 2\ text(units)`

 

(ii)  `text(Area) = 1/2 ⋅ h ⋅ (AB + CD)`

`AB` `= 5 sqrt 2\ text{(given)}`
`CD` `= sqrt((3 – 0)^2 + (1 + 2)^2)`
  `= sqrt 18`
  `= 3 sqrt 2`

 

`:.\ text(Area)` `= 1/2 ⋅ 3 sqrt 2\ \ (5 sqrt 2 + 3 sqrt 2)`
  `= 1/2 ⋅ 3 sqrt 2 ⋅ 8 sqrt 2`
  `= 24\ text(u)²`

Financial Maths, 2ADV M1 2017 HSC 12c

In an arithmetic series, the fifth term is 200 and the sum of the first four terms is 1200.

Find the value of the tenth term.  (3 marks)

Show Answers Only

`0`

Show Worked Solution

`T_n = a + (n – 1) d`

`=>a + 4d = 200 …\ (1)`

 

`S_n = n/2 [2a + (n – 1) d]`

`=>4a + 6d = 1200 …\ (2)`

 

`text(Multiply)\ (1) xx 4`

`=>4a + 16d = 800 …\ (1 prime)`

 

`text(Subtract)\ \ (1 prime) – (2)`

`10d` `= -400`
`d` `= -40`

 

`text(Substitute)\ \ d = -40\ \ text(into)\ (1)`

`a – 160` `= 200`
`a` `= 360`
`:. T_10` `= 360 – 9(-40)`
  `= 0`

Functions, 2ADV F1 2017 HSC 11h

Find the domain of the function  `f(x) = sqrt (3-x)`.  (2 marks)

Show Answers Only

`x <= 3 or (-oo,3].`

Show Worked Solution

`text(Domain of)\ \ f(x) = sqrt (3-x)`

`3-x` `>= 0`
`x` `<= 3`

 

`text(Note domain can also be expressed as:)\ \ (-oo,3]`