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Calculus, SPEC1-NHT 2019 VCAA 9

  1. Show that  `tan((5pi)/(12)) = sqrt3 + 2`.   (2 marks)

     
         

  2. Hence, find the area bounded by the graph of  `f(x) = (2)/(x^2-4x + 8)`  shown above, the `x`-axis and the lines  `x = 0`  and  `x = 2 sqrt3 +6`.   (4 marks)

Show Answers Only

  1. `text(Proof (See Worked Solution))`
  2. `(2pi)/(3)`

Show Worked Solution

a.    `text(Method 1:)`

`tan \ (5pi)/(12)` `= tan ((pi)/(4) + (pi)/(6))`
  `= (tan \ (pi)/(4) + tan\ (pi)/(6))/(1-tan \ (pi)/(4) · tan \ (pi)/(6))`
  `= (1 + (1)/(sqrt3))/(1-(1)/(sqrt3))`
  `= (sqrt3+1)/(sqrt3-1) xx (sqrt3+1)/(sqrt3+1)`
  `= (3+ 2 sqrt3 + 1)/(3-1)`
  `= sqrt3 +2`

  
`text(Method 2:)`

`tan \ (5pi)/(6)` `= (2tan \ (5pi)/(12))/(1-tan^2 \ (5pi)/(12))`
`- 1/sqrt3` `=(2tan \ (5pi)/(12))/(1-tan^2 \ (5pi)/(12))`
`-2 sqrt3 tan \ (5pi)/(12)` `= 1-tan^2 \ (5pi)/(12)`

 

`tan^2 \ (5pi)/(12)-2 sqrt(3) tan \ (5pi)/(12)-1 = 0`

`tan \ (5pi)/(12)` `= (2 sqrt3 ± sqrt(12 + 4))/(2)`
  `= sqrt3 + 2 \ \ \ (tan theta > 0)`

 

b.   `text(Area)` `= int_0 ^(2 sqrt3 + 6) \ (2)/(x^2-4x + 8)\ dx`
  `= int_0 ^(2 sqrt3 + 6) \ (2)/((x -2)^2 + 2^2)`
  `= [tan^-1 ((x-2)/(2))]_0 ^(2 sqrt3 + 6)`
  `= tan^-1 (sqrt3 + 2)-tan^-1 (-1)`
  `= (5pi)/(12)-(-(pi)/(4))`
  `= (2pi)/(3)`

Calculus, SPEC1-NHT 2019 VCAA 8

Find the length of the arc of the curve defined by  `y = (x^4)/(4) + (1)/(8x^2) + 3`  from  `x = 1`  to  `x = 2`. Give your answer in the form  `(a)/(b)`, where `a` and `b` are positive integers.   (4 marks)

Show Answers Only

`(123)/(32)`

Show Worked Solution

`(dy)/(dx) = x^3 + ((-2)/(8) x^-3) = x^3 – (1)/(4x^3)`

`1 + ((dy)/(dx))^2` `= 1 + (x^6 – (1)/(2) + (1)/(16x^6))`
  `= (16x^6 + 16x^12 – 8x^6 + 1)/(16x^6)`
  `= (16x^12 + 8x^6 + 1)/(16x^6)`
  `= ((4x^6 + 1)^2)/(16x^6)`

 

`text(Arc Length)` `= int_1 ^2 sqrt(((4x^6 + 1)^2)/(16x^6))\ dx`
  `= int_1 ^2 (4x^6 + 1)/(4x^3)\ dx`
  `= int_1 ^2 x^3 + (1)/(4x^3)\ dx`
  `= [(x^4)/(4) – (1)/(8x^2)]_1 ^2`
  `= [4 – (1)/(32) – ((1)/(4) – (1)/(8))]`
  `= (123)/(32)`

Calculus, SPEC1-NHT 2019 VCAA 7

Given that  `3x^2 + 2xy + y^2 = 6`, find  `(d^2 y)/(dx^2)`  at the point  (1, 1).   (5 marks)

Show Answers Only

`- (3)/(2)`

Show Worked Solution
`6x + 2y + 2x · (dy)/(dx) + 2y · (dy)/(dx)` `= 0`
`(2x + 2y) (dy)/(dx)` `= -6x – 2y`
`(dy)/(dx)` `= (-6x – 2y)/(2x + 2y)`

 
`6 + 2 (dy)/(dx) + 2 (dy)/(dx) + 2x *(d^2y)/dx^2 + 2 ((dy)/(dx))^2 + 2y · (d^2 y)/(dx^2) = 0`

`(2x + 2y)  (d^2 y)/(dx^2) + 4 (dy)/(dx) + 2 ((dy)/(dx))^2 + 6 = 0`

`text(At) \ (1, 1):`

`4 (d^2 y)/(dx^2) + 4 ((-6 -2)/(2+2)) + 2 ((-6 -2)/(2 + 2))^2 + 6 ` `= 0`
`4 (d^2 y)/(dx^2) – 8 + 8 + 6` `= 0`
`(d^2 y)/(dx^2)` `= – (3)/(2)`

Calculus, SPEC1-NHT 2019 VCAA 6

Part of the graph of  `y = (2)/(sqrt(x^2-4x+3))`, where  `x > 3`, is shown below.
 


 

Find the volume of the solid of revolution formed when the graph of  `y = (2)/(sqrt(x^2-4x+3))`  from  `x = 4`  to  `x = 6`  is rotated about the `x`-axis. Give your answer in the form  `a log_e(b)`  where `a` and `b` are real numbers.   (5 marks)

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`pi log_e ((9)/(5))`

Show Worked Solution

`V = pi int_4 ^6 (4)/(x^2 – 4x + 3)\ dx`
 

`text(Using partial fractions:)`

`(4)/(x^2 – 4x + 3) = (a)/((x-3)) + (b)/((x-1))`

`a(x -1) + b(x – 3)= 4`
 

`text(When)\ \ x = 1, \ -2b = 4  => \ b = -2`

`text(When)\ \ x = 3, \ 2a = 4  => \ a = 2`

`:. \ V` `= pi int_4 ^6 (2)/(x-3) – (2)/(x-1)\ dx`
  `= 2 pi [log_e(x-3) -log_e(x-1)]_4 ^6`
  `= 2 pi [log_e 3 – log_e 5 – (log_e 1 – log_e3)]`
  `= 2 pi (2log_e 3 – log_e 5)`
  `= 2pi log_e((9)/(5))`

Vectors, EXT2 V1 2019 SPEC2 4

The base of a pyramid is the parallelogram  `ABCD`  with vertices at points  `A(2,−1,3),  B(4,−2,1),  C(a,b,c)`  and  `D(4,3,−1)`. The apex (top) of the pyramid is located at  `P(4,−4,9)`.

  1. Find the values of  `a, b`  and  `c`.  (2 marks)

  2. Find the cosine of the angle between the vectors  `overset(->)(AB)`  and  `overset(->)(AD)`.  (2 marks)

  3. Find the area of the base of the pyramid.  (2 marks)

Show Answers Only
  1. `a = 6, b = 2, c = −3`
  2. `4/9`
  3. `2sqrt65\ text(u²)`
Show Worked Solution
i.    `overset(->)(AB)` `= (4 – 2)underset~i + (−2 + 1)underset~j + (1 – 3)underset~k`
    `= 2underset~i – underset~j – 2underset~k`

 
`text(S)text(ince)\ ABCD\ text(is a parallelogram)\ => \ overset(->)(AB)= overset(->)(DC)`

`overset(->)(DC) = (a – 4)underset~i + (b – 3)underset~j + (c + 1)underset~k`

`a – 4 = 2 \ => \ a = 6`

`b – 3 = −1 \ => \ b = 2`

`c + 1 = −2 \ => \ c = −3`

 

ii.   `overset(->)(AB) = 2underset~i – underset~j – 2underset~k`

`overset(->)(AD) = 2underset~i + 4underset~j – 4underset~k`

`cos angleBAD` `= (overset(->)(AB) · overset(->)(AD))/(|overset(->)(AB)| · |overset(->)(AD)|)`
  `= (4 – 4 + 8)/(sqrt(4 + 1 + 4) · sqrt(4 + 16 + 16))`
  `= 4/9`

 

iii.    `1/2 xx text(Area)_(ABCD)` `= 1/2 ab sin c`
  `text(Area)_(ABCD)` `= |overset(->)(AB)| · |overset(->)(AD)| *sin(cos^(−1)\ 4/9)`

  

  

`:. text(Area)_(ABCD)` `= 3 · 6 · sqrt65/9`
  `= 2sqrt65\ text(u²)`

Vectors, EXT2 V1 2019 SPEC1-N 5

A triangle has vertices  `A(sqrt3 + 1, –2, 4), \ B(1, –2, 3)`  and  `C(2, –2, sqrt3 + 3)`.

  1.  Find angle `ABC`   (3 marks)

  2.  Find the area of the triangle.   (2 marks)

Show Answers Only
  1. `∠ABC = (pi)/(6)`
  2. `1 \ text(u²)`
Show Worked Solution

i.     `overset(->)(BA) = ((sqrt3 + 1), (-2), (4)) – ((1), (-2), (3)) = ((sqrt3), (0), (1))`

`overset(->)(BC) = ((2), (-2), (sqrt3 + 3)) – ((1), (-2), (3)) = ((1), (0), (sqrt3))`

`cos ∠ABC` `= (overset(->)(BA) · overset(->)(BC))/ (|overset(->)(BA)| |overset(->)(BC)|`
  `= (2 sqrt3)/(sqrt4 sqrt4)`
  `= (sqrt3)/(2)`

 
`:. \ ∠ABC = (pi)/(6)`

 

ii.     `text(Area)` `= (1)/(2) · |overset(->)(BA)| |overset(->)(BC)| \ sin ∠ABC`
    `= (1)/(2) xx 2 xx 2 xx sin \ (pi)/(6)`
    `= 1 \ text(u²)`

Vectors, SPEC1-NHT 2019 VCAA 5

A triangle has vertices  `A(sqrt3 + 1, –2, 4), \ B(1, –2, 3)`  and  `C(2, –2, sqrt3 + 3)`.

  1. Find angle `ABC`   (3 marks)

  2. Find the area of the triangle.   (2 marks)

Show Answers Only

  1. `∠ABC = (pi)/(6)`
  2. `1 \ text(u²)`

Show Worked Solution

a.     `overset(->)(BA) = ((sqrt3 + 1), (-2), (4))-((1), (-2), (3)) = ((sqrt3), (0), (1))`

`overset(->)(BC) = ((2), (-2), (sqrt3 + 3))-((1), (-2), (3)) = ((1), (0), (sqrt3))`

`cos ∠ABC` `= (overset(->)(BA) · overset(->)(BC))/ (|overset(->)(BA)| |overset(->)(BC)|`
  `= (2 sqrt3)/(sqrt4 sqrt4)`
  `= (sqrt3)/(2)`

 
`:. \ ∠ABC = (pi)/(6)`

 

b.     `text(Area)` `= (1)/(2) · |overset(->)(BA)| |overset(->)(BC)| \ sin ∠ABC`
    `= (1)/(2) xx 2 xx 2 xx sin \ (pi)/(6)`
    `= 1 \ text(u²)`

Calculus, SPEC1-NHT 2019 VCAA 4

Evaluate  `int_(e^3) ^(e^4) (1)/(x log_e (x))\ dx`.   (3 marks)

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`log_e ((4)/(3))`

Show Worked Solution

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

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

`text(When) \ \ x = e^4 \ => \ u = 4`

`text(When) \ \ x = e^3 \ => \ u = 3`

`int_(e^3) ^(e^4) (1)/(x log_e (x))` `= int_3 ^4 (1)/(u)\ du`
  `= [ log_e u]_3 ^4`
  `= log_e 4 – log_e 3`
  `= log_e ((4)/(3))`

Statistics, SPEC1-NHT 2019 VCAA 3

The number of cars per day making a U-turn at a particular location is known to be normally distributed with a standard deviation of 17.5. In a sample of 25 randomly selected days, a total of 1450 cars were observed making the U-turn.

  1. Based on this sample, calculate an approximate 95% confidence interval for the number of cars making the U-turn each day. Use an integer multiple of the standard deviation in your calculations.   (3 marks)

  2. The average number of U-turns made at the location is actually 60 per day.

     

    Find an approximation, correct to two decimal places, for the probability that on 25 randomly selected days the average number of U-turns is less than 53.   (1 mark)

Show Answers Only
  1. `(51, 65)`
  2. `0 .02`
Show Worked Solution

a.      `barx = (1450)/(25) = 58`

`(σ)/(sqrtn) = (17.5)/(sqrt25) = (35)/(2xx5) = (7)/(2)`

`text(Limit:)\ ` `(58-2 xx (7)/(2) \ , \ 58 + 2 xx (7)/(2)) `
  `= (51, 65)`

 

b.     `μ = 60 \ , \ (σ)/(sqrtn) = (7)/(2)`
 
          `\ barX ∼ N (60, ((7)/(2))^2)`
 
 
         
 

`text(Pr) (barX < 53)` `= text(Pr) (z <–2)`
  `= (0.05)/(2)`
  `= 0.025`
  `≈ 0.02 \ \ text{(round down)}`

 
`text{(0.03 was also accepted as a correct answer)}`

Complex Numbers, EXT2 N2 2019 SPEC1-N 1

A cubic polynomial has the form  `p(z) = z^3 + bz^2 + cz + d, \ z ∈ C`, where  `b, c, d ∈ R`.

Given that a solution of  `p(z) = 0`  is  `z_1 = 3 - 2i`  and that  `p(–2) = 0`, find the values of  `b, c` and `d`.   (4 marks)

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`b =-4 \ , \ c = 1 \ , \ d = 26`

Show Worked Solution

`text(Roots:)\  \ z_1 = 3 – 2i \ , \ z_2 = 3 + 2i \ , \ z_3 = -2`
 

`p(z)` `= (z – 3 + 2i )(z – 3 – 2i )(z + 2)`
  `= ((z-3)^2 – (2i)^2)(z+2)`
  `= (z^2 – 6z + 9 + 4)(z + 2)`
  `= (z^2 – 6z + 13)(z + 2)`
  `= z^3 + 2z^2 – 6z^2 – 12z + 13z + 26`
  `= z^3 – 4z^2 + z + 26`

 

`:. \ b =-4 \ , \ c = 1 \ , \ d = 26`

Complex Numbers, SPEC2-NHT 2019 VCAA 2

A cubic polynomial has the form  `p(z) = z^3 + bz^2 + cz + d, \ z ∈ C`, where  `b, c, d ∈ R`.

Given that a solution of  `p(z) = 0`  is  `z_1 = 3 - 2i`  and that  `p(–2) = 0`, find the values of  `b, c` and `d`.   (4 marks)

Show Answers Only

`b =-4 \ , \ c = 1 \ , \ d = 26`

Show Worked Solution

`text(Roots:)\  \ z_1 = 3 – 2i \ , \ z_2 = 3 + 2i \ , \ z_3 = -2`
 

`p(z)` `= (z – 3 + 2i )(z – 3 – 2i )(z + 2)`
  `= ((z-3)^2 – (2i)^2)(z+2)`
  `= (z^2 – 6z + 9 + 4)(z + 2)`
  `= (z^2 – 6z + 13)(z + 2)`
  `= z^3 + 2z^2 – 6z^2 – 12z + 13z + 26`
  `= z^3 – 4z^2 + z + 26`

 

`:. \ b =-4 \ , \ c = 1 \ , \ d = 26`

Mechanics, SPEC1-NHT 2019 VCAA 1

A 10 kg mass is placed on a rough plane that inclined at 30° to the horizontal, as shown in the diagram below. A force of 40 N is applied to the mass up the slope and parallel to the slope. There is also a frictional resistance force of magnitude `F` that opposes  the motion of the mass.
 


 

  1. Find the magnitude of the frictional resistance force, in newtons, acting up the slope if the force is just sufficient to stop the mass from sliding down the slope.  (2 marks)
     
  2. An additional force of magnitude `P` newtons is applied to the mass  up the slope and parallel to the slope. The sum of the additional force and the frictional resistance force of magnitude `F` that now acts down the slope is such that it is just sufficient to stop the mass from sliding up the slope.  (2 marks)
Show Answers Only
  1. `9 \ text(N)`
  2. `18 \ text(N)`
Show Worked Solution

a.       

`40 + F` `= 10g sin30°`
`F` `= 98 xx 0.5 – 40`
  `= 9\ text(N)`

 

b.   `text(Frictional force)\ F\ text(acts down slope)`

`40 + P` `= 10g sin30° + F`
`P` `= 5g + F – 40`
  `= 49 + 9 – 40`
  `= 18\ text(N)`

Calculus, 2ADV C2 SM-Bank 4

Differentiate  `5^(x^2)5x`.  (2 marks)

Show Answers Only

`5^(x^2 + 1)(ln5*2x^2 + 1)`

Show Worked Solution

COMMENT: See HSC exam reference sheet when differentiating  `5^x`.

`y` `= 5^(x^2) * 5x`
`(dy)/(dx)` `= ln5*2x*5^(x^2)*5x + 5^(x^2)*5`
  `=5^(x^2)(ln5*10x^2 + 5)`
  `=5^(x^2 + 1)(ln5*2x^2 + 1)`

Trigonometry, 2ADV* T1 2011 HSC 24c

A ship sails 6 km from `A` to `B` on a bearing of 121°. It then sails 9 km to `C`.  The
size of angle `ABC` is 114°.
 

2UG 2011 24c

Copy the diagram into your writing booklet and show all the information on it.

  1. What is the bearing of `C` from `B`?   (1 mark)

  2. Find the distance `AC`. Give your answer correct to the nearest kilometre.   (2 marks)

  3. What is the bearing of `A` from `C`? Give your answer correct to the nearest degree.   (3 marks)

Show Answers Only
  1.  `055^@`
  2.  `13\ text(km)`
  3.  `261^@`
Show Worked Solution
STRATEGY: This deserves repeating again: Draw North-South parallel lines through major points to make the angle calculations easier!
i.     2011 HSC 24c

 `text(Let point)\ D\ text(be due North of point)\ B`

`/_ABD` `=180-121\ text{(cointerior with}\ \ /_A text{)}`
  `=59^@`
`/_DBC` `=114-59`
  `=55^@`

   
`:. text(Bearing of)\ \ C\ \ text(from)\ \ B\ \ text(is)\ 055^@`

 

ii.    `text(Using cosine rule:)`

`AC^2` `=AB^2+BC^2-2xxABxxBCxxcos/_ABC`
  `=6^2+9^2-2xx6xx9xxcos114^@`
  `=160.9275…`
`:.AC` `=12.685…\ \ \ text{(Noting}\ AC>0 text{)}`
  `=13\ text(km)\ text{(nearest km)}`

 

iii.    `text(Need to find)\ /_ACB\ \ \ text{(see diagram)}`

MARKER’S COMMENT: The best responses showed clear working on the diagram.
`cos/_ACB` `=(AC^2+BC^2-AB^2)/(2xxACxxBC)`
  `=((12.685…)^2+9^2-6^2)/(2xx(12.685..)xx9)`
  `=0.9018…`
`/_ACB` `=25.6^@\ text{(to 1 d.p.)}`

 

`text(From diagram,)`

`/_BCE=55^@\ text{(alternate angle,}\ DB\ text(||)\ CE text{)}`

`:.\ text(Bearing of)\ A\ text(from)\ C`

  `=180+55+25.6`
  `=260.6`
  `=261^@\ text{(nearest degree)}`

Trigonometry, 2ADV* T1 2005 HSC 27c

2UG-2005-27c
 

The bearing of `C` from `A` is 250° and the distance of `C` from `A` is 36 km.

  1. Explain why  `theta`  is  `110^@`.   (1 mark)

  2. If  `B`  is 15 km due north of  `A`, calculate the distance of  `C`  from  `B`, correct to the nearest kilometre.   (3 marks)

Show Answers Only
  1. `110^@`
  2. `text{43 km (nearest km)}`
Show Worked Solution

i.  `text(There is 360)^@\ text(about point)\ A`

`:.theta + 250^@` `= 360^@`
`theta` `= 110^@`

 

ii.   
`a^2` `= b^2 + c^2 − 2ab\ cos\ A`
`CB^2` `= 36^2 + 15^2 − 2 xx 36 xx 15 xx cos\ 110^@`
  `= 1296 + 225 −(text(−369.38…))`
  `= 1890.38…`
`:.CB` `= 43.47…`
  `= 43\ text{km  (nearest km)}`

Trigonometry, 2ADV* T1 2007 26a

The diagram shows information about the locations of towns  `A`,  `B`  and  `Q`.
 

 
 

  1. It takes Elina 2 hours and 48 minutes to walk directly from Town  `A`  to Town  `Q`.

     

    Calculate her walking speed correct to the nearest km/h.    (1 mark)

  2. Elina decides, instead, to walk to Town  `B`  from Town  `A`  and then to Town  `Q`.

     

    Find the distance from Town  `A`  to Town  `B`. Give your answer to the nearest km.   (2 marks)

  3. Calculate the bearing of Town  `Q`  from Town  `B`.   (1 mark)

Show Answers Only
  1. `5\ text(km/hr)\ text{(nearest km/hr)}`
  2. `18\ text(km)\ text{(nearest km)}`
  3. `236^@`
Show Worked Solution

i.  `text(2 hrs 48 mins) = 168\ text(mins)`

`text(Speed)\ text{(} A\ text(to)\ Q text{)}` `= 15/168`
  `= 0.0892…\ text(km/min)`

 

`text(Speed)\ text{(in km/hr)}` `= 0.0892… xx 60`
  `= 5.357…\ text(km/hr)`
  `= 5\ text(km/hr)\ text{(nearest km/hr)}`

 

ii.  

`text(Using cosine rule)`

`AB^2` `= 15^2 + 10^2 – 2 xx 15 xx 10 xx cos 87^@`
  `= 309.299…`
`AB` `= 17.586…`
  `= 18\ text(km)\ text{(nearest km)}`

 

`:.\ text(The distance from Town)\ A\ text(to Town)\ B\ text(is 18 km.)`

 

iii.  
`/_CAQ` `= 31^@\ \ \ text{(} text(straight angle at)\ A text{)}`
`/_AQD` `= 31^@\ \ \ text{(} text(alternate angle)\ AC\ text(||)\ DQ text{)}`
`/_DQB` `= 87 – 31 = 56^@`
`/_QBE` `= 56^@\ \ \ text{(} text(alternate angle)\ DQ \ text(||)\ BE text{)}`

 

`:.\ text(Bearing of)\ Q\ text(from)\ B`

`= 180 + 56`

`= 236^@`

Vectors, EXT1 V1 EQ-Bank 26

`OABC`  is a quadrilateral.

`P`, `Q`, `R` and `S` divide each side of the quadrilateral in half as shown below.
  


 

Prove, using vectors, that  `PQRS`  is a parallelogram.  (3 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution

`text(Consider diagonal)\ \ overset(->)(OB):`

`overset(->)(OB) = overset(->)(OA) + overset(->)(AB) = overset(->)(OC)+ overset(->)(CB)`

`overset(->)(PQ) = overset(->)(PA) + overset(->)(AQ) = 1/2(overset(->)(OA) + overset(->)(AB)) = 1/2overset(->)(OB)`

`overset(->)(SR) = overset(->)(SC) + overset(->)(CR) = 1/2(overset(->)(OC) + overset(->)(CB)) = 1/2overset(->)(OB)`

`:.overset(->)(PQ) = overset(->)(SR)`

 

`text(Consider diagonal)\ \ overset(->)(AC):`

`overset(->)(AC) = overset(->)(AB) + overset(->)(BC) = overset(->)(AO) + overset(->)(OC)`

`overset(->)(QR) = overset(->)(QB) + overset(->)(BR) = 1/2(overset(->)(AB) + overset(->)(BC)) = 1/2overset(->)(AC)`

`overset(->)(PS) = overset(->)(PO) + overset(->)(OS) = 1/2(overset(->)(AO) + overset(->)(OC)) = 1/2overset(->)(AC)`

`:. overset(->)(QR) = overset(->)(PS)`

 

`text(S)text(ince)\ PQRS\ text(has equal opposite sides,)`

`PQRS\ text(is a parallelogram.)`

Vectors, EXT1 V1 EQ-Bank 24

`PQRS`  is a parallelogram, where  `overset(->)(PQ) = underset~a`  and  `vec(PS) = underset~b`
 

Prove, using vectors, that the sum of the squares of the lengths of the diagonals is equal to the sum of the squares of the lengths of the sides.  (3 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution


 

`overset(->)(PR) = underset~a + underset~b,\ \ overset(->)(SQ) = underset~a-underset~b`

`overset(->)(PQ) = overset(->)(SR) = underset~a`

`overset(->)(PS) = overset(->)(QR) = underset~b`

`text(Prove)\ \ |underset~a + underset~b|^2 + |underset~a-underset~b|^2 = 2|underset~a|^2 + 2|underset~b|^2`

`|underset~a + underset~b|^2 + |underset~a-underset~b|^2` `= (underset~a + underset~b) · (underset~a + underset~b) + (underset~a-underset~b)(underset~a-underset~b)`
  `= underset~a · underset~a + underset~b · underset~b + 2underset~a · underset~b + underset~a · underset~a + underset~b · underset~b-2underset~a · underset~b`
  `= |underset~a|^2 + |underset~b|^2 + |underset~a|^2 + |underset~b|^2`
  `= 2|underset~a|^2 + 2|underset~b|^2\ \ …\ text(as required)`

Vectors, EXT2, V1 SM-Bank 16

 

Let  `OABCD`  be a right square pyramid where  `underset ~a = vec(OA),\ underset ~b = vec(OB),\ underset ~c = vec(OC)`  and  `underset ~d = vec(OD)`.

Show that  `underset~a + underset~c = underset~b + underset~d`.   (3 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution

`text(Let)\ \ A=(p,–p,–k),`

`underset ~a` `= overset(->)(OA) = punderset~i – punderset~j – qunderset~k`
`underset ~b` `= overset(->)(OB) = punderset~i + punderset~j – qunderset~k`
`underset ~c` `= overset(->)(OC) = −punderset~i + punderset~j – qunderset~k`
`underset ~d` `= overset(->)(OA) = −punderset~i – punderset~j – qunderset~k`

 

`underset~a + underset~c` `= −2qunderset~k`
`underset ~b + underset ~d` `= −2qunderset~k`

 
`:. underset~a + underset~c = underset~b + underset~d`

Vectors, EXT2 V1 SM-Bank 23

A cube with side length 3 units is pictured below.
 

     
 

  1. Calculate the magnitude of vector `vec(AG)`.  (1 mark)

  2. Find the acute angle between the diagonals `vec(AG)` and `vec(BH)`.  (3 marks)

Show Answers Only
  1. `3 sqrt 3\ text(units)`
  2. `70^@32′`
Show Worked Solution

i.   `A(3, 0 , 0), \ \ G(0, 3, 3)`

  `vec(AG)` `= ((0), (3), (3)) – ((3), (0), (0)) = ((text{−3}), (3), (3))`
  `|\ vec(AG)\ |` `= sqrt (9 + 9 + 9)`
    `= 3 sqrt 3\ text(units)`

 

ii.    `H (3, 3, 3)`
  `vec(BH) = ((3), (3), (3))`
`vec(AG) ⋅ vec(BH)` `= |\ vec(AG)\ | ⋅ |\ vec(BH)\ |\ cos theta`
`((text{−3}), (3), (3)) ⋅ ((3), (3), (3))` `= sqrt (9 + 9 + 9) ⋅ sqrt (9 + 9 + 9) cos theta`
`-9 + 9 + 9` `= 27 cos theta`
`cos theta` `= 1/3`
`theta` `= 70.52…`
  `= 70^@32′`

Vectors, EXT2 V1 SM-Bank 22

`ABCDEFGH` are the vertices of a rectangular prism.
  


 

  1. Show that the internal diagonals of the prism, `AG` and `DF`, intersect.  (2 marks)

  2. Calculate the acute angle, `theta`, between the diagonals, to the nearest minute.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `83^@37′`
Show Worked Solution
i.    `A(2, text{−2}, 0),`   `G(text{−2}, 2, 2)`
  `D(2, 2, 0),`   `F (text{−2}, text{−2}, 2)`

 

`text(Midpoint)\ AG = ((1/2 (2 – 2)),(1/2 (text{−2} + 2)),(1/2 (0 + 2))) = ((0), (0), (1))`

`text(Midpoint)\ DF = ((1/2 (2 – 2)),(1/2 (2 – 2)),(1/2 (0 + 2))) = ((0), (0), (1))`
 

`text(S) text(ince midpoints are the same), AG and DF\ text(intersect.)`

 

ii.    `vec(AG) = ((text{−2}), (2), (2)) – ((2), (text{−2}), (0)) = ((text{−4}), (4), (2))`
  `vec(DF) = ((text{−2}), (text{−2}), (2)) – ((2), (2), (0)) = ((text{−4}), (text{−4}), (2))`

 

`vec (AG) ⋅ vec (DF) = |\ vec (AG)\ | ⋅ |\ vec(DF)\ |\  cos theta`

`((text{−4}), (4), (2)) ⋅ ((text{−4}), (text{−4}), (2)) = sqrt 36 sqrt 36 cos theta`

`16 – 16 + 4` `= 36 cos theta`
`cos theta` `= 1/9`
`theta` `= 83.62…`
  `= 83^@37′`

Vectors, EXT2 V1 EQ-Bank 15

Point `B` sits on the arc of a semi-circle with diameter  `AC`.
 


 

Using vectors, show `angle ABC`  is a right angle.  (2 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution

`text(Let)\ \ vec (OA) = underset~a, \ text(and)\ \ vec(OC)=underset~c`
 


  

`text(Prove) \ overset(->)(AB) ⊥ overset(->)(BC)`

`|underset~a| = |underset~b| = |underset~c|\ \ \ text{(radii)}`

`underset~c = – underset~a `
 

`overset(->)(AB) ⋅ overset(->)(BC)` `= (underset~b-underset~a) (underset~c-underset~b)`
  `= underset~b · underset~c-|underset~b|^2-underset~a · underset~c + underset~a · underset~b`
  `= underset~b · underset~c-|underset~b|^2 + underset~c · underset~c-underset~c · underset~b`
  `= |underset~c|^2-|underset~b|^2`
  `= 0`

 
`:. \ ∠ABC \ text(is a right angle.)`

Vectors, EXT2 V1 SM-Bank 20

`OABD`  is a trapezium in which  `overset(->)(OA) = underset~a`  and  `overset(->)(OB) = underset~b`.

`OC`  is parallel to  `AB`  and  `DC : CB = 1:2`
 


 

Using vectors, express  `overset(->)(DA)`  in terms of  `underset~a`  and  `underset~b`.  (3 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution

`overset(->)(DA) = overset(->)(DB) + overset(->)(BA)`

`overset(->)(BA) = underset~a – underset~b`
 
`text(S) text(ince) \  OC  ||  AB \ \ text(and) \ \ OA  ||  CB`

`=> OABC \ text(is a parallelogram)`

`overset(->)(OA)` `= overset(->)(CB) = underset~a`
`overset(->)(DC)` `= (1)/(2) \ underset~a `
`overset(->)(DB)` `= overset(->)(DC) + overset(->)(CB)`
  `= (3)/(2) \ underset~a`

 

`:. \ overset(->)(DA)` `= (3)/(2) \ underset~a + (underset~a – underset~b)`
  `= (5)/(2) \ underset~a – underset~b`

Vectors, EXT2 V1 SM-Bank 15

Consider the two vector line equations

`underset~(v_1) = ((1),(4),(−2)) + lambda_1((3),(0),(−1)), qquad underset~(v_2) = ((3),(2),(2)) + lambda_2((4),(2),(−6))`

  1. Show that  `underset~(v_1)`  and  `underset~(v_2)`  intersect and determine the point of intersection . (2 marks)

  2. What is the acute angle between the vector lines, to the nearest minute.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `40°29’\ \ (text(nearest minute))`
Show Worked Solution

i.   `text(Solve simultaneously:)`

`1 + 3lambda_1` `= 3 + 4lambda_2` `\ \ …\ (1)`
`4 + 0lambda_1` `= 2 + 2lambda_2` `\ \ …\ (2)`
`−2 – lambda_1` `= 2 – 6lambda_2` `\ \ …\ (3)`

 
`=> lambda_2 = 1\ \ \ text{(from (2))}`

`=>lambda_1 = 2\ \ \ text{(from (1) and (3))}`

`:.\ text(vector lines intersect)`
  

`text(P.O.I.) = ((1),(4),(−2)) + 2((3),(0), (−1)) = ((7),(4),(−4))`

 

ii.   `underset~(v_1) = underset~(a_1) + lambda_1*underset~(b_1)`

`underset~(v_2) = underset~(a_2) + lambda_2*underset~(b_2)`

`costheta` `= (underset~(b_1) · underset~(b_2))/(|underset~b_1||underset~b_2|)`
  `= (12 + 0 + 6)/(sqrt10 sqrt56)`
  `= 0.7606…`

 

`theta` `= 40.479…`
  `= 40°29’\ \ (text(nearest minute))`

Vectors, EXT2 V1 SM-Bank 2

  1. Find values of  `a`, `b`, `c`  and  `d`  such that  `underset~v = ((a),(b)) + 2((c),(d))`  is a vector equation of a line that passes through  `((3),(1))`  and  `((−3),(−3))`.  (2 marks)

  2. Determine whether  `underset~u = ((4),(6)) + lambda((−2),(3))`  is perpendicular to  `underset~v`.  (1 mark)

  3. Express  `underset~u`  in Cartessian form.  (1 mark)

Show Answers Only
  1. `a = 3, b = 1, c = −3, d = −2`, or

     

    `a = −3, b = −3, c = 3, d = 2`

  2. `text(See worked solutions)`
  3. `y = −3/2x + 12`
Show Worked Solution

i.   `text(Method 1)`

`overset(->)(OA) = underset~a = ((3),(1)),\ \ overset(->)(OB) = underset~b = ((−3),(−3))`

`overset(->)(AB)` `= overset(->)(OB) – overset(->)(OA)`
  `= ((−3),(−3)) – ((3),(1))`
  `= ((−6),(−4))`

 

`underset~v` `= underset~a + lambdaunderset~b`
  `= ((3),(1)) + lambda((−6),(−4))`
  `= ((3),(1)) + 2((−3),(−2))`

 
`:. a = 3, b = 1, c = −3, d = −2`

 

`text(Method 2)`

`overset(->)(BA)` `= overset(->)(OA) – overset(->)(OB)`
  `= ((3),(1)) – ((−3),(−3))`
  `= ((6),(4))`

 
`underset~v = ((−3),(−3)) + 2((3),(2))`
  

`:. a = −3, b = −3, c = 3, d = 2`
 

ii.   `underset~u = ((4),(6)) + lambda((−2),(3))`

`underset~v = ((3),(1)) + 2((−3),(−2))`

`((−2),(3)) · ((−3),(−2)) = 6 – 6 = 0`

`:. underset~u ⊥ underset~v`
 

iii.   `((x),(y))= ((4),(6)) + lambda((−2),(3))`

`x = 4 – 2lambda\ \ \ …\ (1)`

`y = 6 + 3lambda\ \ \ …\ (2)`

`text(Substitute)\ \ lambda = (4 – x)/2\ \ text{from (1) into (2):}`

`y` `= 6 + 3((4 – x)/2)`
`y` `= 6 + 6 – (3x)/2`
`y` `= −3/2x + 12`

Vectors, EXT1 V1 SM-Bank 11 MC

Which pair of line segments intersect at exactly one point

A.    `{(underset ~u = ((3), (2)) + lambda ((text{−1}),(2)) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((2), (1)) + lambda ((2), (text{−4})) text{,} quad qquad 0 <= lambda <= 1):}`
   
B.    `{(underset ~u = ((4), (1)) + lambda ((3), (text{−1})) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((3), (2)) + lambda ((2), (2)) text{,} quad qquad 0 <= lambda <= 1):}`
   
C.    `{(underset ~u = ((4), (0)) + lambda ((text{−3}),(6)) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((0), (1)) + lambda ((1), (text{−2})) text{,} quad qquad 0 <= lambda <= 1):}`
   
D.    `{(underset ~u = ((0), (2)) + lambda ((3), (text{−2})) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((0), (1)) + lambda ((1), (1)) text{,} quad qquad 0 <= lambda <= 1):}`
Show Answers Only

`D`

Show Worked Solution

`text(S) text(ince)\ ((2), (text{−4})) = -2((text{−1}), (2)) and ((text{−3}), (6)) = -3((text{−1}), (2))`

`=> A and C\ text(are parallel lines.)`
 

`text(Consider)\ D:`

`3 lambda_1` `= lambda_2` `\ text{… (1)}`
`2 – 2 lambda_1` `= 1 + lambda_2` `\ text{… (2)}`

 
`text(Substitute)\ text{(1) into (2)}`

`2 – 2 lambda_1` `= 1 + 3 lambda_1`
`lambda_1` `= 1/5`
`lambda_2` `= 3/5`

 
`text(Similarly,)\ lambda_1, lambda_2\ text(in)\ B\ \ text(can be calculated)`

`text(and found to be outside)\ \ 0 <= lambda <= 1.`

`=> D`

Vectors, EXT2 V1 SM-Bank 10

  1. Determine the point of intersection of  `underset ~a`  and  `underset ~b`  given.

`qquad underset ~a = ((3), (5), (1)) + lambda ((1), (3), (text{−2})),`  and
 

`qquad underset ~b = ((text{−2}), (2), (text{−1})) + mu ((1), (text{−1}), (2))`  (2 marks)

 
  1. Determine if the point  `(2, text{−2}, 5)`  lies on  `underset ~b`.  (1 mark)

Show Answers Only
  1. `((1), (text{−1}), (5))`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.     `text(At point of intersection:)`

`3 + lambda` `= -2 + mu\ \ text{… (1)}`
`5 + 3 lambda` `= 2 – mu\ \ text{… (2)}`
`1 – 2 lambda` `= -1 + 2 mu\ \ text{… (3)}`

 
`(1) + (2)`

`8 + 4 lambda` `= 0`
`lambda` `= -2,\ \ mu = 3`

 
`text{Intersection (using}\ lambda = –2 text{)}:`
 

`((x), (y), (z)) = ((3 – 2 xx 1), (5 – 2 xx 3), (1 – 2 xx text{−2})) = ((1), (text{−1}), (5))`

 

ii.   `text(If)\ \ (2, text{−2}, text{−10})\ \ text(lies on)\ underset ~b, ∃ mu\ \ text(that satisfies:)`

`-2 + mu` `= 2\ \ text{… (1)}\ => \ mu = 4`
`2 – mu` `= ­text{−2}\ \ text{… (2)}\ => \ mu = 4`
`-1 + 2 mu` `= 5\ \ text{… (3)}\ => \ mu = 3`

 
`=>\  text(No solution)`

`:. (2, text{−2}, 5)\ \ text(does not lie on)\ underset ~b.`

Vectors, EXT2 V1 SM-Bank 8

Use the vector form of the linear equations

`3x - 2y = 4`  and  `3y + 2x - 6 = 0`

to show they are perpendicular.  (3 marks)

Show Answers Only

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

Show Worked Solution
`3x-2y` `= 4`
`3x` `= 2y + 4`
`3/2 x` `=y+2`
`x/(2/3)` `= y + 2`

 
`underset ~(v_1) = ((0), (-2)) + lambda ((2/3), (1))`

 

`3y + 2x-6` `= 0`
`2x` `= -3y + 6`
`-2/3 x` `= y-2`
`x/(-3/2)` `= y-2`

 
`underset ~(v_2) = ((0), (2)) + lambda ((-3/2), (1))`

`((2/3), (1)) ((-3/2), (1)) = -1 + 1 = 0`

`:. underset ~(v_1) _|_ underset ~(v_2)`

Vectors, EXT2 V1 SM-Bank 7

Find the value of `x` and `y`, given

`underset ~r = ((5), (-1), (2)) + lambda ((x), (y), (-3))`

and  `underset ~r`  is perpendicular to both `underset ~v` and `underset ~w`, where

`underset ~v = ((1), (2), (1)) + mu_1 ((3), (-3), (-1))`  and  `underset ~w = ((-3), (1), (1)) + mu_2 ((-4), (-1), (-2))`.  (2 marks)

Show Answers Only

`x = 1, y = 2`

Show Worked Solution

`text(S) text(ince)\ underset ~r\ text(is perpendicular to)\ underset ~v and underset ~w:`
 

`((x), (y), (-3))((3), (-3), (-1)) = 0`

`3x – 3y + 3` `= 0`
`x – y` `= -1\ text{… (1)}`

 
`((x), (y), (-3))((-4), (-1), (-2)) = 0`

`-4x – y + 6` `= 0`
`4x + y` `= 6\ text{… (2)}`

 
`(1) + (2)`

`5x` `= 5`
`:.x` `= 1`
`:. y` `= 2`

Functions, EXT1 F1 SM-Bank 16

Given  `f(x) = 2 sinx`  and  `g(x) = x`, sketch  `y = f(x) + g(x)`  for  `0 <= x <= 2pi`.

Clearly label endpoints and all local maximum and minimums.   (3 marks)

Show Answers Only

Show Worked Solution
`y` `= 2sinx + x`
`(dy)/(dx)` `= 2cosx + 1`
`(dy)/(dx)` `= 0 \ => \ cosx = −1/2 => x = (2pi)/3, (4pi)/3`

 
`text(MAX at)\ ((2pi)/3, 3.83)`

`text(MIN at)\ ((4pi)/3, 2.46)`
 

Functions, EXT1 F1 EQ-Bank 15

  1.  Given  `f(x) = sin x`  and  `g(x) = cos x`, sketch  `y = f(x) + g(x)`  for  `−pi <= x <= pi`.   (2 marks)

  2.  State the range of  `y`.  (2 marks)

Show Answers Only
  1.  
  2. `text(Range)\ {−sqrt2 <= y <= sqrt2}`
Show Worked Solution
i.    `y` `= sinx + cosx`
  `(dy)/(dx)` `= cosx – sinx`
  `(dy)/(dx)` `= 0\ \ text(when)\ \ sinx = cosx`

COMMENT: Question style consistent with NESA exemplar questions in official Topic Guidance.

`:. text(SP’s when)\ \ x = pi/4, −(3pi)/4`
 


 

ii.   `text(Max/min when)\ \ (dy)/(dx) = 0`

`x = pi/4, −(3pi)/4`

`y_(text(max))` `= sin\ pi/4 + cos\ pi/4`
  `= 2 xx 1/sqrt2`
  `= sqrt2`

 
`:.\ text(By symmetry, range)\ {−sqrt2 <= y <= sqrt2}`

Functions, EXT1′ F1 2008 HSC 3a

The following diagram shows the graph of  `y = g(x)`.
 

 
Draw separate one-third page sketches of the graphs of the following:

  1.  `y = |g(x)|`  (1 mark)

  2.  `y = 1/(g(x))`  (2 marks)

Show Answers Only
  1.  
  2.  
Show Worked Solution
i.   

 

ii.   

Statistics, STD2 S1 EQ-Bank 4

A high school conducted a survey asking students what their favourite Summer sport was.

The Pareto chart shows the data collected.
 


 

  1. What percentage of students chose Hockey as their favourite Summer sport?  (1 mark)

  2. What percentage of students chose Touch Football as their favourite Summer sport?  (1 mark)

Show Answers Only
  1. `text(2%)`
  2. `text(22%)`
Show Worked Solution

i.   `text(Method 1)`

`text(Percentage who chose hockey)`

`=\ text{cum freq (at hockey column) – cum freq (at tennis column)}` 

`=100-98`

`=2 text(%)`
 

`text(Method 2)`

`text(Column 1 → 180 people = 30%)`

`text(Total surveyed) = 180 -: 0.3 = 600`

`:.\ text(Hockey %)` `=12/600 xx 100`  
  `=2\text(%)`  

 

ii.   `text(Percentage who chose touch football)`

`=\ text{cum freq (at touch football column) – cum freq (at cricket column)}` 

`=76-54`

`=22 text(%)`

GRAPHS, FUR2 2019 VCAA 3

Members of the association will travel to a conference in cars and minibuses:

  • Let `x` be the number of cars used for travel.
  • Let `y` be the number of minibuses used for travel.
  • A maximum of eight cars and minibuses in total can be used.
  • At least three cars must be used.
  • At least two minibuses must be used.

The constraints above can be represented by the following three inequalities.
 

`text(Inequality 1) qquad qquad x + y <= 8`

`text(Inequality 2) qquad qquad x >= 3`

`text(Inequality 3) qquad qquad y >= 2`
 

  1. Each car can carry a total of five people and each minibus can carry a total of 10 people.

      

    A maximum of 60 people can attend the conference.

      

    Use this information to write Inequality 4.   (1 mark)

The graph below shows the four lines representing Inequalities 1 to 4.

Also shown on this graph are four of the integer points that satisfy Inequalities 1 to 4. Each of these integer points is marked with a cross (✖).
 


 

  1. On the graph above, mark clearly, with a circle (o), the remaining integer points that satisfy Inequalities 1 to 4.  (1 mark)

Each car will cost $70 to hire and each minibus will cost $100 to hire.

  1. What is the cost for 60 members to travel to the conference?  (1 mark)
  2. What is the minimum cost for 55 members to travel to the conference?  (1 mark)
  3. Just before the cars were booked, the cost of hiring each car increased.

      

    The cost of hiring each minibus remained $100.

      

    All original constraints apply.

      

    If the increase in the cost of hiring each car is more than `k` dollars, then the maximum cost of transporting members to this conference can only occur when using six cars and two minibuses.

      

    Determine the value of  `k`.  (1 mark)

Show Answers Only
  1. `5x + 10y <= 60`
  2. `text(See Worked Solutions)`
  3. `$680`
  4. `$610`
  5. `30`
Show Worked Solution

a.  `5x + 10y <= 60`

 

b.    

 

c.    `text(Consider the line)\ \ 5x + 10y = 60\ \ text(on the graph)`

`text(Touches the feasible region at)\ (4, 4)\ text(only)`

`:.\ text (C) text(ost of 60 members)`

`= 4 xx 70 + 4 xx 100`

`= $680`

 

d.    `text(Coordinates that allow 55 to travel) \ => \ (5, 3) and (3, 4)`

`text(C) text(ost)\ (5, 3) = 5 xx 70 + 3 xx 100 = $650`

`text(C) text(ost)\ (3, 4) = 3 xx 70 + 4 xx 100 = $610`

`:.\ text(Minimum cost is $610)`

 

e.    `text(Objective function): \ C = ax + 100y`

`text(Max cost occurs at)\ \ (6, 2)\ \ text(when)\ \ a > 100`

`text{(i.e. graphically when the slope of}\ \ C = ax + 100y`

   `text(is steeper than)\ \ x + y= 8 text{)}`

`:. k + 70` `= 100`
`k` `= 30`

GRAPHS, FUR2 2019 VCAA 2

Each branch within the association pays an annual fee based on the number of members it has.

To encourage each branch to find new members, two new annual fee systems have been proposed.

Proposal 1 is shown in the graph below, where the proposed annual fee per member, in dollars, is displayed for branches with up to 25 members.
 


 

  1. What is the smallest number of members that a branch may have?  (1 mark)
  2. The incomplete inequality below shows the number of members required for an annual fee per member of $10.

      

    Complete the inequality by writing the appropriate symbol and number in the box provided.   (1 mark)
     

3 ≤ number of members  
 

 

Proposal 2 is modelled by the following equation.

annual fee per member = – 0.25 × number of members + 12.25

  1. Sketch this equation on the graph for Proposal 1, shown below.  (1 mark)

 

 

  1. Proposal 1 and Proposal 2 have the same annual fee per member for some values of the number of members.

      

    Write down all values of the number of members for which this is the case.  (1 mark)

Show Answers Only
  1. `3`
  2. `3 <= text(number of members) <= 10`
  3. `text(See Worked Solutions)`
  4. `9, 17, 25`
Show Worked Solution

a.  `3`
 

b.  `3 <= text(number of members) <= 10`
 

c.  

 

d.    `text(Same annual fee occurs when graphs intersect.)`

`text(Member numbers): 9, 17, 25`

GRAPHS, FUR2 2019 VCAA 1

The graph below shows the membership numbers of the Wombatong Rural Women’s Association each year for the years 2008–2018.
 


 

  1. How many members were there in 2009?  (1 mark)
    1. Show that the average rate of change of membership numbers from 2013 to 2018 was − 6 members per year.  (1 mark)
    2. If the change in membership numbers continues at this rate, how many members will there be in 2021?  (1 mark)
Show Answers Only
  1. `60`
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `14`
Show Worked Solution

a.  `60`

 

b.i.   `text(Members in 2013)` `= 62`
  `text(Members in 2018)` `= 32`
  `text(Average ROC)` `= (32 – 62)/5`
    `= -6\ text(members per year.)`

 

b.ii.   `text(Members in 2021)` `= 32 – 3 xx 6`
    `= 14`

GEOMETRY, FUR2 2019 VCAA 3

The following diagram shows a crane that is used to transfer shipping containers between the port and the cargo ship.
 


 

The length of the boom, `BC`, is 25 m. The length of the hoist, `AB`, is 15 m.

    1. Write a calculation to show that the distance `AC` is 20 m.  (1 mark)
    2. Find the angle `ACB`.

        

      Round your answer to the nearest degree.  (1 mark)

  1. The diagram below shows a cargo ship next to a port. The base of a crane is shown at point `Q`.
     

      

      

    `qquad`
     

      

    The base of the crane (`Q`) is 20 m from a shipping container at point `R`. The shipping container will be moved to point `P`, 38 m from `Q`. The crane rotates 120° as it moves the shipping container anticlockwise from `R` to `P`.

      

    What is the distance `RP`, in metres?

      

    Round your answer to the nearest metre.  (1 mark)

  2. A shipping container is a rectangular prism.

      

    Four chains connect the shipping container to a hoist at point `M`, as shown in the diagram below.
     

      

    `qquad` 
     

      

    The shipping container has a height of 2.6 m, a width of 2.4 m and a length of 6 m.

     

    Each chain on the hoist is 4.4 m in length.

      

    What is the vertical distance, in metres, between point `M` and the top of the shipping container?

      

    Round your answer to the nearest metre.  (2 marks)

Show Answers Only
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `37^@`
  1. `51\ text(m)`
  2. `3\ text(m)`
Show Worked Solution
a.i.   `AC` `= sqrt(BC^2 – AB^2)`
    `= sqrt(25^2 – 15^2)`
    `= sqrt 400`
    `= 20`

 

a.ii.   `tan\  /_ ACB` `= 15/20`
  `/_ ACB` `= tan^(-1) (3/4)`
    `= 36.86…`
    `~~ 37^@`

 

b.  `text(Using cosine rule:)`

`RP^2` `= 20^2 + 38^2 – 2 xx 20 xx 38 xx cos 120^@`
  `= 2604`
`:. RP` `= 51.029…`
  `~~ 51\ text(metres)`

 

c.  

`text(Find)\ h => text(need to find)\ x`

`text(Consider the top of the container)`
 


 

`x` `= sqrt(3^2 + 1.2^2)`
  `~~ 3.2311`

 

`:. h` `= sqrt(4.4^2 – 3.2311^2)`
  `= 2.98…`
  `~~ 3\ text(metres)`

GEOMETRY, FUR2 2019 VCAA 2

A cargo ship travels from Magadan (60° N, 151° E) to Sydney (34° S, 151° E).

  1. Explain, with reference to the information provided, how we know that Sydney is closer to the equator than Magadan.   (1 mark)
  2. Assume that the radius of Earth is 6400 km.

      

    Find the shortest great circle distance between Magadan and Sydney.

      

    Round your answer to the nearest kilometre.  (1 mark)

  3. The cargo ship left Sydney (34° S, 151° E) at 6 am on 1 June and arrived in Perth (32° S, 116° E) at 10 am on 11 June.

      

    There is a two-hour time difference between Sydney and Perth at that time of year.

      

    How many hours did it take the cargo ship to travel from Sydney to Perth?  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `10\ 500\ text(km)`
  3. `246\ text(hours)`
Show Worked Solution
a.  

`text{Sydney’s angle with the equator (34°S) is smaller than Magadan’s}`

`text{(60°N) resulting in a shorter distance to the equator.)`

 

b.   `text(Distance)` `= (34 + 60)/360 xx 2pi xx 6400`
    `~~ 10\ 500\ text(km)`

 

c.    `text(Perth is 2 hours behind Sydney)`

`text(6 am Sydney = 4 am Perth)`

`:.\ text(Travel time)` `= 4\ text{am (1 Jun)} – 10\ text{am (11 Jun)}`
  `= 6\ text(hours) + 10\ text(days)`
  `= 6 + 10 xx 24`
  `= 246\ text(hours)`

GEOMETRY, FUR2 2019 VCAA 1

The following diagram shows a cargo ship viewed from above.
 

 
The shaded region illustrates the part of the deck on which shipping containers are stored.

  1. What is the area, in square metres, of the shaded region?  (1 mark)

Each shipping container is in the shape of a rectangular prism.

Each shipping container has a height of 2.6 m, a width of 2.4 m and a length of 6 m, as shown in the diagram below.
 

  1. What is the volume, in cubic metres, of one shipping container?  (1 mark)
  2. What is the total surface area, in square metres, of the outside of one shipping container?  (1 mark)
  3. One shipping container is used to carry barrels. Each barrel is in the shape of a cylinder.

      

    Each barrel is 1.25 m high and has a diameter of 0.73 m, as shown in the diagram below.

      

    Each barrel must remain upright in the shipping container
     
     

      

    `qquad qquad`
     
    What is the maximum number of barrels that can fit in one shipping container?  (1 mark)

Show Answers Only
  1. `6700\ text(m²)`
  2. `37.44\ text(m³)`
  3. `72.48\ text(m²)`
  4. `48`
Show Worked Solution
a.   `text(Area)` `= 160 xx 40 + 12 xx 25`
    `= 6700\ text(m²)`

 

b.   `text(Volume)` `= 6 xx 2.4 xx 2.6`
    `= 37.44\ text(m³)`

 

c.   `text(S.A.)` `= 2(6 xx 2.6) + 2 (6 xx 2.4) + 2(2.4 xx 2.6)`
    `= 72.48\ text(m²)`

 

d.   `6 ÷ 0.73 = 8.21` `=> \ text(8 barrels along length)`
  `2.4 ÷ 0.73 = 3.28` `=> \ text(3 barrels along width)`
  `2.6 ÷ 1.25 = 2.08` `=> \ text(2 vertical layers)`

 

`:.\ text(Maximum barrels)` `= 2 xx 8 xx 3`
  `= 48`

Networks, STD2 N3 2019 FUR2 3

Fencedale High School is planning to renovate its gymnasium.

This project involves 12 activities, `A` to `L`.

The directed network below shows these activities and their completion times, in weeks.
 


 

The minimum completion time for the project is 35 weeks.

  1. Identify the critical path and state how many activities are on it?  (2 marks)

  2. Determine the latest start time of activity `E`.  (1 mark)

  3. Which activity has the longest float time?  (1 mark)

It is possible to reduce the completion time for activities `C, D, G, H` and `K` by employing more workers.

  1.  The completion time for each of these five activities can be reduced by a maximum of two weeks.

      

    What is the minimum time, in weeks, that the renovation project could take?  (1 mark)

Show Answers Only
  1. `8\ text(activities)`
  2. `12\ text(weeks)`
  3. `text(Activity)\ J`
  4. `29\ text(weeks)`
Show Worked Solution

a.  `text(Scanning forwards and backwards:)`
 

 

 

`text(Critical path:)\ ABDFGIKL`

`:. 8\ text(activities)`
 

b.  `text(LST for activity)\ E = 12\ text{weeks  (i.e. start of 13th week)}`
 

c.   `text(Consider float times of all activities not on critical path.)`

`J-5, H-1, E-1, C-1`

`:.\ text(Activity)\ J\ text(has the largest float time.)`
 

d.   `text(Critical path after reducing)\ CDGHK\ text(by 2 weeks is)`

`ABDFGIKL.`
 

`:.\ text(Minimum time)` `= 2 + 4 + 7 + 1 + 2 + 2 + 5 + 6`
  `= 29\ text(weeks)`

NETWORKS, FUR2 2019 VCAA 1

Fencedale High School has six buildings. The network below shows these buildings represented by vertices. The edges of the network represent the paths between the buildings.
 


 

  1. Which building in the school can be reached directly from all other buildings?   (1 mark)

  2. A school tour is to start and finish at the office, visiting each building only once.
     i.     What is the mathematical term for this route?   (1 mark)

  3. ii. Draw in a possible route for this school tour on the diagram below.   (1 mark)

 

Show Answers Only
  1. `text(Office)`
  2.  i.  `text(Hamiltonian cycle)`
    ii. `text(See Worked Solutions)` 
Show Worked Solution

a.  `text(Office)`
 

b.i.   `text(Hamiltonian cycle)`

`text{(note a Hamiltonian path does not start and finish}`

  `text{at the same vertex.)}`
 

b.ii.   `text(One of many solutions:)`
 

MATRICES, FUR2 2019 VCAA 3

On Sunday, matrix `V` is used when calculating the expected number of visitors at each location every hour after 10 am. It is assumed that the park will be at its capacity of 2000 visitors for all of Sunday.

Let `L_0` be the state matrix that shows the number of visitors at each location at 10 am on Sunday.

The number of visitors expected at each location at 11 am on Sunday can be determined by the matrix product

 
`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad text(this hour)),(qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad qquad qquad  \ A qquad quad F qquad \  G \ quad quad W),({:V xx L_0 qquad text(where) qquad L_0 = [(500), (600), (500), (400)]{:(A),(F),(G),(W):}, qquad text(and):} qquad V = [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]{:(A),(F),(G),(W):}\ text(next hour)):}`
 

  1. Safety restrictions require that all four locations have a maximum of 600 visitors.
  2. Which location is expected to have more than 600 visitors at 11 am on Sunday?   (1 mark) 
  3. Whenever more than 600 visitors are expected to be at a location on Sunday, the first 600 visitors can stay at that location and all others will be moved directly to Ground World `(G)`.
  4. State matrix `R_n` contains the number of visitors at each location `n` hours after 10 am on Sunday, after the safety restrictions have been enforced.
  5. Matrix `R_1` can be determined from the matrix recurrence relation
  6. `qquad qquad qquad R_0 = [(500),(600),(500),(400)]{:(A),(F),(G),(W):}, qquad qquad R_1 = V xx R_0 + B_1`
  7. where matrix `B_1` shows the required movement of visitors at 11 am.
    1. Determine the matrix `B_1`.   (1 mark)

    2. State matrix `R_2` can be determined from the new matrix rule
    3. `qquad qquad R_2 = VR_1 + B_2`
    4. where matrix `B_2` shows the required movement of visitors at 12 noon.
    5. Determine the state matrix `R_2`.   (1 mark)

Show Answers Only
  1. `text(Location)\ A`
  2.  i. `B_1 = [(-210),(0),(210),(0)]`
    ii. `R_2 = [(600),(288),(512),(600)]`
Show Worked Solution
a.   `text{By inspection (higher decimal values in row 1)}`
  `=>\ text(test location)\ A`
  `text(Visitors at)\ A\ text{(11 am)}` `= 0.3 xx 500 + 0.4 xx 600 + 0.6 xx 500 + 0.3 xx 400`
    `= 810`

 
`text(Location)\ A\ text(will have more than 600 visitors.)`

 

b.i.   `VR_0` `= [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)][(500),(600),(500),(400)]=[(810),(300),(310),(580)]`
     
  `R_1` `= V xx R_0 + B_1`
    `= [(810),(300),(310),(580)] + [(-210),(0),(210),(0)]`
  `:. B_1` `= [(-210),(0),(210),(0)]`

 

b.ii.   `R_2` `= [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)][(600),(300),(520),(580)] + B_2`
    `= [(786),(288),(282),(644)] + [(-186),(0),(230),(-44)]= [(600),(288),(512),(600)]`

MATRICES, FUR2 2019 VCAA 2

The theme park has four locations, Air World `(A)`, Food World `(F)`, Ground World `(G)` and Water World `(W)`.

The number of visitors at each of the four locations is counted every hour.

By 10 am on Saturday the park had reached its capacity of 2000 visitors and could take no more visitors.

The park stayed at capacity until the end of the day

The state matrix, `S_0`, below, shows the number of visitors at each location at 10 am on Saturday.
 

`S_0 = [(600), (600), (400), (400)] {:(A),(F),(G),(W):}`
 

  1. What percentage of the park’s visitors were at Water World `(W)` at 10 am on Saturday?   (1 mark)

Let `S_n` be the state matrix that shows the number of visitors expected at each location `n` hours after 10 am on Saturday.

The number of visitors expected at each location `n` hours after 10 am on Saturday can be determined by the matrix recurrence relation below.
 

`{:(qquad qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad text(  this hour)),(qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad qquad qquad quad A qquad quad F qquad \  G \ quad quad W),({:S_0 = [(600), (600), (400), (400)], qquad S_(n+1) = T xx S_n quad quad qquad text(where):}\ T = [(0.1,0.2,0.1,0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]{:(A),(F),(G),(W):}\ text(next hour)):}`
 

  1. Complete the state matrix, `S_1`, below to show the number of visitors expected at each location at 11 am on Saturday.   (1 mark)

 
`S_1 = [(\ text{______}\ ), (\ text{______}\ ), (300),(\ text{______}\ )]{:(A),(F),(G),(W):}`
 

  1. Of the 300 visitors expected at Ground World `(G)` at 11 am, what percentage was at either Air World `(A)` or Food World `(F)` at 10 am?   (1 mark)

  2. The proportion of visitors moving from one location to another each hour on Sunday is different from Saturday.

     

    Matrix `V`, below, shows the proportion of visitors moving from one location to another each hour after 10 am on Sunday.

     

    `qquad qquad {:(qquadqquadqquadqquadqquadtext(this hour)),(qquad qquad qquad \ A qquad quad F qquad \  G \ quad quad W),(V = [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]{:(A),(F),(G),(W):}\ text(next hour)):}`

     

     
    Matrix `V` is similar to matrix `T` but has the first two rows of matrix `T` interchanged.

     

  3. The matrix product that will generate matrix `V` from matrix `T` is
  4. `qquad qquad V = M xx T`
  5. where matrix `M` is a binary matrix.
  6. Write down matrix `M`.   (1 mark)
Show Answers Only
  1. `20%`
  2. `S_1 = [(300 ), (780), (300),(620)]{:(A),(F),(G),(W):}`
  3. `60%`
  4. `M = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)]`
Show Worked Solution

a.   `text(Total visitors)\ =2000`

`text{Percentage}\  (W)` `= 400/2000= 20%`

 

b.   `S_1` `= TS_0`
    `= [(0.1, 0.2, 0.1, 0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)][(600),(600),(400),(400)]=[(300),(780),(300),(620)]`

 

c.  `text(Visitors from)\ A to G= 0.1 xx 600 = 60`

`text(Visitors from)\ F to G= 0.2 xx 600 = 120`

`:.\ text(Percentage of)\ G\ text(at 11 am)= (60 + 120)/300= 60%`
 

d.   `M = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)]`
 

 
`text(Check:)`

`M xx T = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)][(0.1, 0.2, 0.1, 0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)] = [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]`

Calculus, EXT2 C1 2002 HSC 2b

For  `n = 0, 1, 2,`...

let  `I_n = int_0 ^{(pi)/(4)} tan^(n) theta  d theta`.

  1.  Show that  `I _1 = (1)/(2) ln2`.   (1 mark)

  2.  Show that, for  `n >= 2`,
     
         `I_n + I_(n - 2) = (1)/(n-1)`.   (3 marks)

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

i.         `I_n = int_0 ^{(pi)/(4)} tan^n theta \ d theta`

`I_1` `= int_0 ^{(pi)/(4)} tan theta \ d theta`
  `= [ -ln cos x ]_0 ^{(pi)/(4)}`
  `= [- ln cos\ (pi)/(4) + ln cos 0]`
  `= -ln 2^{-(1)/(2)}`
  `= (1)/(2) ln 2`

 

ii. `I_n` `= int_0 ^{(pi)/(4)} tan^(n – 2) theta ⋅ tan^2 theta \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(n-2) theta ⋅ (sec^2 theta – 1) d theta`
    `= int_0 ^{(pi)/(4)} tan^(n – 2) theta ⋅ sec^2 theta \ d theta – int_0 ^{(pi)/(4)} tan^(n – 2) theta \ d theta`

 
`text(Let) \ \ u = tan theta`

`(du)/(d theta) = sec^2 theta \ => \ du = sec^2 theta \ d theta`

`text(When)` `theta` `= (pi)/(4),`   `\ \ \ \u` `= 1`
  `theta` `=0,`   `u` `= 0`

 

`I_n` `= int_0 ^1 u^(n-2) \ du – I_(n – 2)`
  `= [(1)/(n-1) ⋅ u^(n-1)]_0 ^1 – I_(n – 2)`
  `= (1)/(n – 1) – I_(n-2)`

 
`:. \ I_n + I_(n – 2)=(1)/(n – 1)\ \ \ \ (n>=2)`

Calculus, EXT2 C1 2008 HSC 3c

For  `n >= 0`, let  `I_n = int_0 ^{(pi)/(4)} tan^(2n) theta  d theta`.   

  1. Show that for `n >= 1`,
     
         `I _n = (1)/(2n - 1) - I_(n-1)`.  (2 marks)

  2. Hence, or otherwise, calculate  `I_3`.   (2 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `(52 – 15 pi)/(60)`
Show Worked Solution
i. `I_n` `= int_0 ^{(pi)/(4)} tan^(2n-2) theta * tan^2 theta \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(2n-2) theta * (sec^2 theta – 1) \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(2n-2) theta * sec^2 theta  d theta – int_0 ^{(pi)/(4)} tan^(2n-2) theta \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(2n-2)theta * sec^2 theta \ d theta – I_(n-1)`

 

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

`(du)/(d theta) = sec^2 theta \ => \ du = \ sec^2 theta \ d theta`

`text(When)`    `theta` `= (pi)/(4)` `,` `\ \ \ \u` `= 1`
  `theta` `= 0` `,` `u` `= 0`

 

`I_n` `= int_0 ^1 u^(2n – 2) \ du – I_(n – 1)`
  `= [1/(2n -1) ⋅ u^(2n – 1)]_0 ^1 – I_(n – 1)`
  `= 1/(2n-1)(1^(2n-1)-0)  – I_(n – 1)`
  `= 1/(2n-1)  – I_(n – 1)`

 

ii.        `I_0 = int_0 ^{(pi)/(4)} d theta = (pi)/(4)`

`I_3` `= (1)/(5) – I_2`
  `= (1)/(5) – ((1)/(3) – I_1)`
  `= (1)/(5) – (1)/(3) + (1 – I_0)`
  `= (13)/(15) – (pi)/(4)`
  `= (52 – 15 pi)/(60)`

Calculus, EXT2 C1 2005 HSC 1c

Use integration by parts to evaluate  `int_1^e x^7 log_e x  dx`.   (3 marks)

Show Answers Only

`(7e^8 – 1)/(64)`

Show Worked Solution
  `u` `= log_e x,` `\ \ \ \ u^(′)` `= (1)/(x)`
  `v^(′)` `= x^7,` `v` `= (1)/(8) x^8`

 

`int_1^e x^7\ log_e x  dx` `= uv-int u^(′) v \ dx`
  `= [(x^8)/(8) ⋅ log_e x]_1 ^e-(1)/(8) int_1 ^e (1)/(x) ⋅ x^8 dx`
  `= ((e^8)/(8) log_e e-(1)/(8) log_e 1)-(1)/(8)[(1)/(8) x^8]_1 ^e`
  `= (e^8)/(8)-(1)/(8) ((e^8)/(8)-(1)/(8))`
  `= (e^8)/(8)-((e^8-1)/(64))`
  `= (7e^8 +1)/(64)`

Proof, EXT2 P1 SM-Bank 12

If  `ab`  is divisible by 3, prove by contrapositive that  `a`  or  `b`  is divisible by 3.   (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Statement)`

`ab \ text(is divisible by 3)\  => a \ text(or) \ b \ text(is divisible by) \ 3`

`text(Contrapositive)`

`a \ text(or) \ b not \ text(divisible by 3)\ => ab not \ text(divisible by 3)`

`text(Let) \ \ a` `= 3x + p, \ text(where) \ \ x ∈ Ζ \ \ text(and) \ \ p= 1 \ text(or) \ \ 2`
`b` `= 3y + q, \ text(where) \ \ y ∈ Ζ \ \ text(and) \ \ q= 1 \ text(or) \ \ 2`

 

`ab` `= (3x + p)(3y + q)`
  `= 9xy + 3qx + 3py + pq`
  `= 3(3xy + qx + py) + pq`

 
`text(Possible values of) \ \ pq = 1, 2, 4`

`=> \ pq \ text(is not divisible by 3)`

`:. \ ab \ text(is not divisible by 3)`

`:. \ text(By contrapositive, statement is true.)`

MATRICES, FUR2 2019 VCAA 1

The car park at a theme park has three areas, `A, B` and `C`.

The number of empty `(E)` and full `(F)` parking spaces in each of the three areas at 1 pm on Friday are shown in matrix `Q`  below.
 

`{:(qquad qquad qquad \ E qquad F),(Q = [(70, 50),(30, 20),(40, 40)]{:(A),(B),(C):}quad text(area)):}`
 

  1. What is the order of matrix `Q`?   (1 mark)

  2. Write down a calculation to show that 110 parking spaces are full at 1 pm.   (1 mark)

Drivers must pay a parking fee for each hour of parking.

Matrix `P`, below, shows the hourly fee, in dollars, for a car parked in each of the three areas.
 

`{:(qquad qquad qquad qquad qquad text{area}), (qquad qquad qquad A qquad quad quad B qquad qquad C), (P = [(1.30, 3.50, 1.80)]):}`
 

  1. The total parking fee, in dollars, collected from these 110 parked cars if they were parked for one hour is calculated as follows.  

     

     

    `qquad qquad qquad P xx L = [207.00]`

     

    where matrix  `L`  is a  `3 xx 1`  matrix.

     

    Write down matrix  `L`.   (1 mark)

The number of whole hours that each of the 110 cars had been parked was recorded at 1 pm. Matrix `R`, below, shows the number of cars parked for one, two, three or four hours in each of the areas `A, B` and `C`.

`{:(qquadqquadqquadqquadquadtext(area)),(quad qquadqquadquad \ A qquad B qquad C),(R = [(3, 1, 1),(6, 10, 3),(22, 7,10),(19, 2, 26)]{:(1),(2),(3),(4):}\ text(hours)):}`
 

  1. Matrix  `R^T`  is the transpose of matrix  `R`.

      

    Complete the matrix  `R^T`  below.   (1 mark)

      

    `qquad R^T = [( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , )]`
     

  2. Explain what the element in row 3, column 2 of matrix  `R^T`  represents.   (1 mark)

Show Answers Only
  1. `3 xx 2`
  2. `50 + 20 + 40 = 110`
  3. `L = [(50), (20), (40)]`
  4. `R^T = [(3 ,6 , 22, 19), (1, 10, 7, 2), (1, 3, 10, 26)]`
  5. `text(Number of cars parked in area)\ C\ text(for 2 hours).`
Show Worked Solution

a.  `text(Order) : 3 xx 2`
 

b.  `text(Add 2nd column): \ 50 + 20 + 40 = 110`
 

c.  `L = [(50), (20), (40)]`
 

d.  `R^T = [(3 ,6 , 22, 19), (1, 10, 7, 2), (1, 3, 10, 26)]`
 

e.   `e_32\ text(in)\ R^T =>` `text(number of cars parked in area)\ C`
    `text(for 2 hours.)`

Proof, EXT2 P1 SM-Bank 11

If  `a^2-4a + 3`  is even, `a ∈ Ζ`,

prove by contrapositive that  `a`  is odd. (3 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contrapositive)`

`a not \ text(odd)` `\ =>\ \ a^2-4x + 3 not \ text(even)`
`text{(i.e)  If} \ a\ text(is even) ` `\ => \ a^2-4x + 3 \ \ text(is odd)`

 

`text(If) \ a \ text(is even) , ∃ \ k , k ∈ Ζ , text(such that) \ \ a = 2k`

`text(Substitute) \ \ 2k \ \ text(into) \ \ a^2-4x + 3`

`(2k)^2-4(2k) + 3` `= 4k^2-8k + 3`
  `= 2(2k^2-4k + 1) + 1`

 
`:. \ text(By contrapositive, if) \ \ a^2-4x + 3 \ \ text(is even) \ =>  \ a \ text(is odd.)`

Proof, EXT2 P1 SM-Bank 7

  1. Given  `a + b = 6`  and  `a, b > 0,` show
     
        `(1)/(a) + (1)/(b) >= (2)/(3)`   (2 marks)
      
  2. If  `a + b = c,` show
     
        `(1)/(a^2) + (1)/(b^2) >= (8)/(c^2)`   (2 marks)
Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.   `text(Prove) \ \ (1)/(a) + (1)/(b) >= 2/3 :`

`(a – b)^2` `>= 0`
`(a + b)^2 – 4 ab` `>= 0`
`(a + b)^2` `>= 4ab`
`6 (a + b)` `>= 4 ab\ \ \ (a+b=6)`
`3 (a + b)` `>= 2 ab`
`(a + b)/(ab)` `>= (2)/(3)\ \ \ (a, b > 0\ => \ ab>0)`

 

ii.   `a + b = c`

`:. \ text(Statement true)`

Proof, EXT2 P1 SM-Bank 6

If  `x,  y,  z  ∈ R`  and  `x  ≠ y ≠ z`, then   

  1.  Prove  `x^2 + y^2 + z^2 > yz + zx + xy`   (2 marks)

  2.  If  `x + y + z = 1`, show  `yz+zx+xy<1/3`   (2 marks)

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

i.   `x^2 + y^2 + z^2 – yz – zx – xy > 0`

`text(Multiply) × 2`

`2x^2 + 2y^2 + 2z^2 – 2yz – 2zx – 2xy > 0`

`(x^2 – 2xy + y^2) + (y^2 – 2yz + z^2) + (z^2 -2xz + x^2) > 0`

`(x – y)^2 + (y – z)^2 + (z – x)^2 > 0`

`text(Square of any rational number) > 0`

`:.\ text(Statement is true.)`

 

ii.    `(x + y + z)^2` `= x^2 + xy + xz + yx + y^2 + yz + zx + zy + z^2`
  `1` `= x^2 + y^2 + z^2 + 2xy + 2yz + 2xz`

 
`text{Consider statement in part (i):}`

`x^2 + y^2 + z^2 – yz – zx – xy > 0`

`=> (x + y + z)^2 – (x^2 + y^2 + z^2 – yz – zx – xy)<1`

`3xy + 3yz + 3zx` `< 1`
`:. \ yz + zx + xy` `< (1)/(3)`

Financial Maths, GEN2 2019 NHT 8

Phil invests $200 000 in an annuity from which he receives a regular monthly payment.

The balance of the annuity, in dollars, after `n` months, `A_n`, can be modelled by the recurrence relation

`A_0 = 200\ 000, qquad A_(n + 1) = 1.0035\ A_n - 3700`

  1. What monthly payment does Phil receive?   (1 mark)

  2. Show that the annual percentage compound interest rate for this annuity is 4.2%.   (1 mark)

At some point in the future, the annuity will have a balance that is lower than the monthly payment amount.

  1. What is the balance of the annuity when it first falls below the monthly payment amount?

     

    Round your answer to the nearest cent.   (1 mark)

  2. If the payment received each month by Phil had been a different amount, the investment would act as a simple perpetuity.

     

    What monthly payment could Phil have received from this perpetuity?   (1 mark)

Show Answers Only
  1. `$3700`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `$92.15`
  4. `$700`
Show Worked Solution

a.  `$3700`

b.   `text(Monthly rate)` `= 0.0035 = 0.35%`
  `text(Annual rate)` `= 12 xx 0.35 = 4.2%`

  
c.  `text(Find)\ N\ text(when)\ FV = 0\ \ text{(by TVM solver)}:`

`N` `= ?`
`I(%)` `= 4.2`
`PV` `= 200\ 000`
`PMT` `= 3700`
`FV` `= 0`
`text(P/Y)` `= 12`
`text(C/Y)` `= 12`

 
`=> N = 60.024951`

 
`text(Find)\ \ FV\ \ text(when)\ \ N = 60.024951\ \ text{(by TVM solver):}`

`=>FV = $92.15`  

d.  `text(Perpetuity) => text(monthly payment) = text(monthly interest)`

`:.\ text(Perpetuity payment)` `= 200\ 000 xx 4.2/(12 xx 100)`
  `= $700`

Proof, EXT2 P1 SM-Bank 5 MC

Four cards are placed on a table with a letter on one face and a shape on the other.
 

     

 
You are given the rule: "if N is on a card then a circle is on the other side."

Which cards need to be turned over to check if this rule holds?

  1. N and G
  2. G and triangle
  3. circle and N
  4. N and triangle
Show Answers Only

`=> \ D`

Show Worked Solution

`text(Solution 1)`

`text(L)text(ogically equivalent statements are:)`

`N` `=> \ text(circle)`  
`not\ text(circle)` `=> not N`  

 
`text(To confirm rule is not broken,)`

`N \ text(must be turned)`

`text{Triangle (not circle) must be turned – only other shape not a circle.}`

`=> D`
 

`text(Solution 2)`

`text(Consider the flip side of each card.)`

`text(If circle has) \ N \ text(on the other side (or not)) – text(tells us nothing.)`

`text(If) \ G \ text(has a circle on the other side (or not)) – text(tells us nothing.)`

`text(If) \ N \ text(doesn’t have a circle on other side) – text(rule broken.)`

`text(If triangle has an) \ N \ text(on other side) – text(rule broken.)`

`:. \ text(Need to turn) \ N \ text(and triangle)`

Financial Maths, 2ADV M1 SM-Bank 15

Phil is a builder who has purchased a large set of tools.

The value of Phil’s tools is depreciated using the reducing balance method.

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

`V_0 = 60\ 000, qquad qquad  V_(n + 1) = 0.9 V_n`
 

  1. Use recursion to show that the value of the tools after two years.   (1 mark)

  2. Phil plans to replace these tools when their value first falls below $20 000.

     

    After how many years will Phil replace these tools?  (1 mark)

  3. Phil has another option for depreciation. He depreciates the value of the tools by a flat rate of 8% of the purchase price per annum.

     

    Let `V_n` be the value of the tools after `n` years, in dollars.

     

    Write down a recurrence relation, in terms of `V_0, V_(n + 1)` and `V_n`, that could be used to model the value of the tools using this flat rate depreciation.  (1 mark)

Show Answers Only
  1. `$48\ 600`
  2. `11\ text(years)`
  3. `V_(n + 1) = V_n – 0.08 V_0`
Show Worked Solution
a.   `V_0` `= 60\ 000`
  `V_1` `= 0.9 xx 60\ 000 = 54\ 000`
  `V_2` `= 0.9 xx 54\ 000 = $48\ 600`

 

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

COMMENT: Dividing by  `log_e 0.9`  is dividing by a negative number!

`60\ 000 xx 0.9^n` `< 20\ 000`  
`log_e 0.9^n` `<log_e (1/3)`  
`n` `> log_e (1/3) / log_e 0.9`  
  `> 10.427`  

 
`:.\ text(Phil will replace in the 11th year.)`

 

c.  `text(Annual depreciation) = 0.08 xx V_0`

`:. V_(n + 1) = V_n – 0.08V_0`

CORE, FUR2 2019 VCAA 7

Phil is a builder who has purchased a large set of tools.

The value of Phil’s tools is depreciated using the reducing balance method.

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

`V_0 = 60\ 000, qquad V_(n + 1) = 0.9 V_n` 

  1. Use recursion to show that the value of the tools after two years, `V_2` , is $48 600.   (1 mark)

  2. What is the annual percentage rate of depreciation used by Phil?   (1 mark)

  3. Phil plans to replace these tools when their value first falls below $20 000.

     

    After how many years will Phil replace these tools?   (1 mark)

  4. Phil has another option for depreciation. He depreciates the value of the tools by a flat rate of 8% of the purchase price per annum.

     

    Let `V_n` be the value of the tools after `n` years, in dollars.

     

    Write down a recurrence relation, in terms of `V_0, V_(n + 1)` and `V_n`, that could be used to model the value of the tools using this flat rate depreciation.   (1 mark)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(10%)`
  3. `11\ text(years)`
  4. `V_0=60\ 000,\ \ \ V_(n + 1) = V_n-4800`
Show Worked Solution
a.   `V_0` `= 60\ 000`
  `V_1` `= 0.9 xx 60\ 000 = 54\ 000`
  `V_2` `= 0.9 xx 54\ 000 = $48\ 600`

  
b.  `text(Depreciation rate) = 0.1 = 10%`

 
c.  `text(Find)\ \ n\ \ text(such that)`

`60\ 000 xx 0.9^n = 20\ 000`

`=> n = 10.427\ \ \ text{(by CAS)}`

`:.\ text(Phil will replace in the 11th year.)`

 
d.  `text(Annual depreciation) = 0.08 xx V_0 = 4800`

`:.\ text(Recurrence relation is:)`

`V_0=60\ 000,\ \ \ V_(n + 1) = V_n-4800`

CORE, FUR2 2019 VCAA 6

The total rainfall, in millimetres, for each of the four seasons in 2015 and 2016 is shown in Table 5 below.

  1. The seasonal index for winter is shown in Table 6 below.

     

    Use the values in Table 5 to find the seasonal indices for summer, autumn and spring.

  2. Write your answers in Table 6, rounded to two decimal places.   (2 marks)

     


  3. The total rainfall for each of the four seasons in 2017 is shown in Table 7 below.

     


    Use the appropriate seasonal index from Table 6 to deseasonalise the total rainfall for winter in 2017.

     

    Round your answer to the nearest whole number.   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `186\ text(mm)`
Show Worked Solution
a.  

`text{Average rainfall (2015)}\ = (142 + 156 + 222 + 120)/4 = 160`

`text{Average rainfall (2016)}\ = (135 + 153 + 216 + 96)/4= 150`

`text{SI (Summer)}` `= 1/2 (142/160 + 135/150)=0.89`
`text{SI (Autumn)}` `= 1/2(156/160 + 153/150)=1.00`
`text{SI (Spring)}` `= 1/2 (120/160 + 96/150)=0.70`

 

b.   `text{Winter (deseasonalised)}` `= 262/1.41`
    `~~ 186\ text(mm)`

CORE, FUR2 2019 VCAA 5

The scatterplot below shows the atmospheric pressure, in hectopascals (hPa), at 3 pm (pressure 3 pm) plotted against the atmospheric pressure, in hectopascals, at 9 am (pressure 9 am) for 23 days in November 2017 at a particular weather station.
 

A least squares line has been fitted to the scatterplot as shown.

The equation of this line is

pressure 3 pm = 111.4 + 0.8894 × pressure 9 am

  1. Interpret the slope of this least squares line in terms of the atmospheric pressure at this weather station at 9 am and at 3 pm.   (1 mark)

  2. Use the equation of the least squares line to predict the atmospheric pressure at 3 pm when the atmospheric pressure at 9 am is 1025 hPa.
  3. Round your answer to the nearest whole number.   (1 mark)

  4. Is the prediction made in part b. an example of extrapolation or interpolation?   (1 mark)

  5. Determine the residual when the atmospheric pressure at 9 am is 1013 hPa.
  6. Round your answer to the nearest whole number.   (1 mark)

  7. The mean and the standard deviation of pressure 9 am and pressure 3 pm for these 23 days are shown in Table 4 below.

    1. Use the equation of the least squares line and the information in Table 4 to show that the correlation coefficient for this data, rounded to three decimal places, is  `r` = 0.966   (1 mark)

    2. What percentage of the variation in pressure 3 pm is explained by the variation in pressure 9 am?
    3. Round your answer to one decimal place.   (1 mark)

  1. The residual plot associated with the least squares line is shown below.
     

    1. The residual plot above can be used to test one of the assumptions about the nature of the association between the atmospheric pressure at 3 pm and the atmospheric pressure at 9 am.
    2. What is this assumption?   (1 mark)

    3. The residual plot above does not support this assumption.
    4. Explain why.   (1 mark)

Show Answers Only
  1. `text(An increase in 1hPa of pressure at 9 am is associated)`
    `text(with an increase of 0.8894 hPa of pressure at 3 pm.)`
  2. `1023\ text(hPa)`
  3. `text(Interpolation)`
  4. `3\ text(hPa)`
    1. `0.966`
    2. `93.3%`
    1. `text(The assumption is that a linear relationship)`
      `text(exists between the pressure at 9 am and the)`
      `text(pressure at 3 pm.)`
    2. `text(The residual plot does not appear to be random.)`
Show Worked Solution

a.    `text(An increase in 1hPa of pressure at 9 am is associated)`

`text(with an increase of 0.8894 hPa of pressure at 3 pm.)`

 

b.   `text(pressure 3 pm)` `= 111.4 + 0.8894 xx 1025`
    `= 1023\ text(hPa)`

 

c.  `text{Interpolation (1025 is within the given data range)}`

 

d.   `text(Residual)` `= text(actual) – text(predicted)`
    `= 1015 – (111.4 + 0.8894 xx 1013)`
    `= 1015 – 1012.36`
    `= 2.63…`
    `~~ 3\ text(hPa)`

 

e.i.   `r= b (s_x)/(s_y)`

    `= 0.8894 xx 4.5477/4.1884`
    `= 0.96569…`
    `= 0.966`

 

e.ii.   `r` `= 0.966`
  `r^2` `= 0.9331`
    `= 93.3%`

 

f.i.   `text(The assumption is that a linear relationship)`
 

`text(exists between the pressure at 9 am and the)`

`text(pressure at 3 pm.)`

 

f.ii.  `text(The residual plot does not appear to be random.)`