SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Harder Ext1 Topics, EXT2 2018 HSC 11e

The circle centred at `O` with radius `r` has a diameter `AB`. Points `C` and `D` are chosen on the circle as shown in the diagram. The chord `AC` has length `d`.
 

 
Show that  `d = 2r sin D`.  (2 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(Join)\ CB`

`angleB = angleD\ \ (text{angles on circumference on arc}\ AC)`

`angleACB = 90°\ \ (text{angle in a semi-circle})`
 

`text(In)\ DeltaABC,`

`sin B` `= d/(2r)`
`:. d` `= 2r sin B`
  `= 2r sin D\ \ \ text(.. as required)`

Complex Numbers, EXT2 N2 2018 HSC 11d

The points `A`, `B` and `C` on the Argand diagram represent the complex numbers `u`, `v` and `w` respectively.

The points `O`, `A`, `B` and `C` form a square as shown on the diagram.
 

 
It is given that  `u = 5 + 2i`.

  1.  Find  `w`.  (1 mark)

  2.  Find  `v`.  (1 mark)

  3.  Find  `text(arg)(w/v)`.  (1 mark)

Show Answers Only
  1. `−2 + 5i`
  2. `3 + 7i`
  3. `pi/4`
Show Worked Solution
i.    `w` `= iu`
    `= i(5 + 2i)`
    `= −2 + 5i`

 

ii.    `v` `= u + w`
    `= 5 + 2i + (−2 + 5i)`
    `= 3 + 7i`

 

iii.    `text(arg)(w/v)` `= text(arg)(w) – text(arg)(v)`
    `= pi/4\ \ (text(diagonal of square bisects corner))`

Harder Ext1 Topics, EXT2 2018 HSC 10 MC

Consider the functions  `f( x ) = sinx`  and  `g( x ) = x sinx`.

The `x`-coordinate of each stationary point of  `f(x)`  is very close to the `x`-coordinate of a stationary point of `g(x)`.

Suppose  `f(x)`  has a stationary point at  `x = a`  and the stationary point of  `g(x)`  with `x`-coordinate closest to  `x = a`  is at  `x = b`.

Which statement is always true?

  1. `a < b`
  2. `a > b`
  3. `|\ a\ | < |\ b\ |`
  4. `|\ a\ | > |\ b\ |`
Show Answers Only

`text(C)`

Show Worked Solution

`f(x) = sin x\ =>  fprime(x) = cos x`

Mean mark 57%. A challenging question well answered. 

`fprime(x) = 0\ \ text(when)\ \ x = pi/2, (3pi)/2, …`

 
`g(x) = x sin x\ =>  gprime(x) = sin x – x cos x`

`gprime(x) = 0\ \ text(when)\ \ −x = tan x`
 

`text(When)\ x > 0,\ text{intersection (where}\ gprime(x) = 0)\ text(is to)`

`text(the right of)\ \ x = pi/2.`

`text(When)\ x < 0,\ text(intersection is to the left of)\ \ x = – pi/2 .`

`:. |\ a\ | < |\ b\ |`

`=>\ text(C)`

Proof, EXT2 P1 2018 HSC 9 MC

It is given that  `a`, `b` are real and  `p`, `q` are purely imaginary.

Which pair of inequalities must always be true?

  1. `a^2p^2 + b^2q^2 <= 2abpq,qquada^2b^2 + p^2q^2 <= 2abpq`
  2. `a^2p^2 + b^2q^2 <= 2abpq,qquada^2b^2 + p^2q^2 >= 2abpq`
  3. `a^2p^2 + b^2q^2 >= 2abpq,qquada^2b^2 + p^2q^2 <= 2abpq`
  4. `a^2p^2 + b^2q^2 >= 2abpq,qquada^2b^2 + p^2q^2 >= 2abpq`
Show Answers Only

`B`

Show Worked Solution

`a, b ->\ text(real)qquad\ p, q ->\ text(purely imaginary)`

`=> ab, pq, (ab – pq)\ text(are real)`

`=> ap, bq, (ap – bq)\ text(are purely imaginary.)`
 

`(ab – pq)^2` `>= 0`
`a^2b^2 + p^2q^2` `>= 2abpq\ \ (text(Eliminate A and C))`

 

`(ap – bq)^2` `<= 0`
`a^2p^2 + b^2q^2` `<= 2abpq`

 
`=>B`

Complex Numbers, EXT2 N2 2018 HSC 7 MC

Which diagram best represents the solutions to the equation  `text(arg)(z) = text(arg)(z + 1 - i)`?

A. B.
C. D.
Show Answers Only

`text(D)`

Show Worked Solution
`text(arg)(z)` `= text(arg)(z + 1 – i)`
  `=text(arg)(z – (−1 + i))`

 
`=>\ text(arg)(z – (−1 + i))\ \ text(is the argument of)\ \ z\ \ text(from)\ \ (-1+i).`

 
`text(Plot)\ \ (-1 + i)\ \ text(on the argand diagram and then test different)`

`text(positions of)\ \ z\ \ text(along the solutions for each option.)`

`=>\ text(D)`

Complex Numbers, EXT2 N2 2018 HSC 6 MC

Which complex number is a 6th root of `i`?

  1. `−1/sqrt2 + 1/sqrt2i`
  2. `−1/sqrt2 - 1/sqrt2i`
  3. `−sqrt2 + sqrt2i`
  4. `−sqrt2 - sqrt2i`
Show Answers Only

`A`

Show Worked Solution

`text(Consider option A:)`
 

`|−1/sqrt2 + 1/sqrt2i|= sqrt((−1/sqrt2)^2 + (1/sqrt2)^2) = 1`

`text(arg)(z)` `= (3pi)/4`
`z` `= 1(cos\ (3pi)/4 + i sin\ (3pi)/4)`
`z^6` `= cos\ (18pi)/4 + i sin\ (18pi)/4\ \ \ text{(De Moivre)}`
  `= i`

 
`=> A`

Integration, 2UA 2018 HSC 15c

The shaded region is enclosed by the curve  `y = x^3 - 7x`  and the line  `y = 2x`, as shown in the diagram. The line  `y = 2x`  meets the curve  `y = x^3 - 7x`  at  `O(0, 0)`  and  `A(3, 6)`. Do NOT prove this.
 


 

  1. Use integration to find the area of the shaded region.  (2 marks)
  2. Verify that one application of Simpson’s rule gives the exact area of the shaded region.  (2 marks)
     
  3. The point `P` is chosen on the curve  `y = x^3 − 7x`  so that the tangent at `P` is parallel to the line  `y = 2x`  and the `x`-coordinate of `P` is positive.
     
  4. Show that the coordinates of `P` are  `(sqrt 3, -4 sqrt 3)`.  (2 marks)
  5. Find the area of  `Delta OAP`.  (2 marks)
Show Answers Only
  1. `81/4\ text(units²)`

  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `9 sqrt 3\ text(units²)`
Show Worked Solution
(i)   `text(Area)` `= int_0^3 2x – (x^3 – 7x)\ dx`
    `= int_0^3 9x – x^3\ dx`
    `= [9/2 x^2 – 1/4 x^4]_0^3`
    `= [(9/2 xx 3^2 – 1/4 xx 3^4) – 0]`
    `= 81/2 – 81/4`
    `= 81/4\ text(units²)`

 

(ii)  `text(Using Simpson’s rule)`

♦ Mean mark 45%.

`y = 9x – x^3`
 

`text(A)` `~~ h/3 (y_0 + 4y_1 + y_2)`
  `= 3/6 (0 + 4 xx 81/8 + 0)`
  `= 1/2 (81/2)`
  `= 81/4\ text( … as required)`

 

(iii)   `y = x^3 – 7x`

`(dy)/(dx) = 3x^2 – 7`

`text(Find)\ \ x\ \ text(such that)\ \ (dy)/(dx) = 2`

`3x^2 – 7` `= 2`
`3x^2` `= 9`
`x^2` `= 3`
`x` `= sqrt 3 qquad (x > 0)`

 

`y` `= (sqrt 3)^3 – 7 sqrt 3`
  `= 3 sqrt 3 – 7 sqrt 3`
  `= -4 sqrt 3`

 
`:. P\ \ text(has coordinates)\ (sqrt 3, -4 sqrt 3)`

 

(iv)  

 

`text(dist)\ OA` `= sqrt((3 – 0)^2 + (6 – 0)^2)`
  `= sqrt 45`
  `= 3 sqrt 5`

 
`text(Find)\ _|_\ text(distance of)\ \ P\ \ text(from)\ \ OA`

♦♦ Mean mark 27%.

`P(sqrt 3, -4 sqrt 3),\ \ 2x – y=0`

`_|_\ text(dist)` `= |(2 sqrt 3 + 4 sqrt 3)/sqrt (3 + 2)|`
  `= (6 sqrt 3)/sqrt 5`
   
`:.\ text(Area)` `= 1/2 xx 3 sqrt 5 xx (6 sqrt 3)/sqrt 5`
  `= 9 sqrt 3\ text(units²)`

Calculus, 2ADV C3 2018 HSC 16a

A sector with radius 10 cm and angle  `theta`  is used to form the curved surface of a cone with base radius `x` cm, as shown in the diagram.
 


 

The volume of a cone of radius `r` and height `h` is given by  `V = 1/3 pi r^2 h`.

  1. Show that the volume, `V` cm³, of the cone described above is given by
     
          `V = 1/3 pi x^2 sqrt(100 - x^2)`.  (1 mark)

  2. Show that  `(dV)/(dx) = (pi x (200 - 3x^2))/(3 sqrt(100 - x^2))`.  (2 marks)

  3. Find the exact value of  `theta`  for which `V` is a maximum.  (3 marks)

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

i.   `V = 1/3 pi r^2 h`

`text(Using Pythagoras,)`

`h = sqrt(100 – x^2)`

`r = x`

`:.\ text(Volume) = 1/3 pi x^2 sqrt(100 – x^2)`
 

♦ Mean mark 45%.

ii.   `V` `= 1/3 pi x^2 sqrt(100 – x^2)`
  `(dV)/(dh)` `= 1/3 pi [2x ⋅ sqrt(100 – x^2) – 2x ⋅ 1/2 (100 – x^2)^(-1/2) ⋅ x^2]`
    `= 1/3 pi [(2x (100 – x^2) – x^3)/sqrt(100 – x^2)]`
    `= 1/3 pi [(200 x – 2x^3 – x^3)/sqrt(100 – x^2)]`
    `= (pi x(200 – 3x^2))/(3 sqrt (100 – x^2))\ \ text{.. as required}`

 

iii.  `text(Find)\ \ x\ \ text(when)\ \ (dV)/(dx) = 0`

♦♦ Mean mark 23%.

`200 – 3x^2` `= 0`
`x` `= sqrt(200/3)`

 
`text(When)\ \ x < sqrt(200/3),\ (200 – 3x^2) > 0`

`=> (dV)/(dx) > 0`

`text(When)\ \ x > sqrt(200/3),\ (200 – 3x^2) < 0`

`=> (dV)/(dx) < 0`

`:.\ text(MAX when)\ \ x = sqrt(200/3)`
 

`text(Equating the arc length of the section)`

`text(to the circumference of the cone:)`

`2 pi r ⋅ theta/(2 pi)` `= 2 pi ⋅ x`
`10 theta` `= 2 pi sqrt (200/3)`
`:. theta` `= (2 pi ⋅ 10 sqrt 2)/(10 ⋅ sqrt 3)`
  `= (2 sqrt 2 pi)/sqrt 3`

Calculus, 2ADV C4 2018 HSC 15b

The diagram shows the region bounded by the curve  `y = 1/(x + 3)`  and the lines  `x = 0`,  `x = 45`  and  `y = 0`. The region is divided into two parts of equal area by the line  `x = k`, where `k` is a positive integer. 
 

 
What is the value of the integer `k`, given that the two parts have equal areas?  (3 marks)

Show Answers Only

`9`

Show Worked Solution
`text(Total Area)` `= int_0^45 1/(x + 3)`
  `= [ln (x + 3)]_0^45`
  `= ln 48 – ln 3`
  `= ln 16`

 

`=> int_0^k 1/(x + 3)` `= 1/2 xx ln 16`
`[ln (x + 3)]_0^k` `= ln 16^(1/2)`
`ln (k + 3) – ln 3` `= ln 4`
`ln ((k + 3)/3)` `= ln 4`
`(k + 3)/3` `= 4`
`:. k` `= 4 xx 3 – 3`
  `= 9`

Trigonometry, 2ADV T3 2018 HSC 15a

The length of daylight, `L(t)`, is defined as the number of hours from sunrise to sunset, and can be modelled by the equation

`L(t) = 12 + 2 cos ((2 pi t)/366)`,

where `t` is the number of days after 21 December 2015, for  `0 ≤ t ≤ 366`.

  1. Find the length of daylight on 21 December 2015.   (1 mark)

  2. What is the shortest length of daylight?  (1 mark)

  3. What are the two values of  `t`  for which the length of daylight is 11?  (2 marks)

Show Answers Only
  1. `14\ text(hours)`
  2. `10\ text(hours)`
  3. `t = 122 or 244`
Show Worked Solution

i.   `L(t) = 12 + 2 cos ((2 pi t)/366)`

`text(On 21 Dec 2015) => t = 0`

`:. L(0)` `= 12 + 2 cos 0`
  `= 14\ text(hours)`

 

ii.   `text(Shortest length of daylight occurs when)`

♦ Mean mark 43%.

`cos ((2 pi t)/366) = -1`
 

`:.\ text(Shortest length)` `= 12 + 2 (-1)`
  `= 10\ text(hours)`

 

iii.   `text(Find)\ \ t\ \ text(such that)\ \ L(t) = 11:`

`11 = 12 + 2 cos ((2 pi t)/366)`

`cos ((2 pi t)/366) = -1/2`
 

`(2 pi t)/366` `= (2 pi)/3` `qquad\ \ text(or)`     `(2 pi t)/366` `= (4 pi)/3`
`t` `= 366/3`   `t` `= (366 xx 2)/3`
  `= 122`     `= 244`

 
`:. t = 122 or 244`

Probability, 2ADV S1 2018 HSC 14e

Two machines, `A` and `B`, produce pens. It is known that 10% of the pens produced by machine `A` are faulty and that 5% of the pens produced by machine `B` are faulty.

  1. One pen is chosen at random from each machine.

     

    What is the probability that at least one of the pens is faulty?  (1 mark)

  2. A coin is tossed to select one of the two machines. Two pens are chosen at random from the selected machine.

     

    What is the probability that neither pen is faulty?  (2 marks)

Show Answers Only
  1. `0.145`
  2. `0.85625`
Show Worked Solution
i.   `text{P(at least 1 faulty)}` `= 1 –  text{P(both faulty)}`
    `= 1 – 0.9 xx 0.95`
    `= 1 – 0.855`
    `= 0.145`

 

ii.   `text{P(2 non-faulty pens})`

♦ Mean mark 48%.

`= text{(choose A, NF, NF)} + P text{(choose B, NF, NF)}`

`= 1/2 xx 0.9 xx 0.9 + 1/2 xx 0.95 xx 0.95`

`= 0.405 + 0.45125`

`=0.85625`

Calculus, EXT1* C3 2018 HSC 14b

The shaded region shown in the diagram is bounded by the curve  `y = x^4 + 1`, the `y`-axis and the line  `y = 10`.
  


 

Find the volume of the solid of revolution formed when the shaded region is rotated about the `y`-axis.  (3 marks)

Show Answers Only

`18 pi\ text(units²)`

Show Worked Solution

`y = x^4 + 1`

`x^4 = y – 1`

`x^2 = +- (y – 1)^(1/2)`

`text(When)\ \ x = 0,\ \ y = 1`

`=> x^2 = (y – 1)^(1/2)`
 

`:.\ text(Volume)` `= pi int_1^10 x^2\ dy`
  `= pi int_1^10 (y – 1)^(1/2)\ dy`
  `= pi xx 2/3 [(y – 1)^(3/2)]_1^10`
  `= (2 pi)/3 [9^(3/2) – 0]`
  `= (2 pi)/3 (27)`
  `= 18 pi\ \ text(units²)`

Trigonometry, 2ADV T1 2018 HSC 14a

In  `Delta KLM, KL`  has length 3, `LM` has length 6 and `/_KLM` is 60°. The point `N` is chosen on side  `KM`  so that  `LN`  bisects `/_KLM`. The length  `LN`  is `x`.
 


 

  1. Find the exact value of the area of  `Delta KLM`.  (1 mark)

  2. Hence, or otherwise, find the exact value of `x`.  (2 marks)

Show Answers Only
  1. `(9 sqrt 3)/2`
  2. `2 sqrt 3`
Show Worked Solution

i.   `text(Using sine rule:)`

`text(Area)\ \ Delta KLM` `= 1/2 xx 3 xx 6 xx sin 60^@`
  `= (9 sqrt 3)/2\ \ text(u²)`

 

ii.  `text(Area)\ \ Delta KLN + text(Area)\ \ Delta NLM = text(Area)\ \ Delta KLM`

`1/2 xx 3 xx x xx sin 30^@ + 1/2 xx x xx 6 xx sin 30^@ = (9 sqrt 3)/2`

♦ Mean mark 37%.

`3/4 x + 3/2 x` `= (9 sqrt 3)/2`
`9/4 x` `= (9 sqrt 3)/2`
`:. x` `= (9 sqrt 3)/2 xx 4/9`
  `= 2 sqrt 3`

Plane Geometry, 2UA 2018 HSC 13b

In `Delta ABC`, sides `AB` and `AC` have length 3, and `BC` has length 2. The point `D` is chosen on `AB` so that `DC` has length 2.
 

  1. Prove that `Delta ABC` and `Delta CBD` are similar.  (2 marks)

  2. Find the length `AD`.  (2 marks)

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

i.    `text(Prove)\ \ Delta ABC\ text(|||)\ Delta CBD`

`Delta ABC\ text{is isosceles:}`

`/_ ABC = /_ ACB qquad text{(angles opposite equal sides)}`

`Delta CBD\ text{is isosceles:}`

`/_ CBD = /_ CDB qquad text{(angles opposite equal sides)}`

 
`text{Since}\ \ /_ ABC =  /_ CBD`

`:. Delta ABC\ text(|||)\ Delta CDB qquad text{(equiangular)}`
 

ii.   `text(Using ratios of similar triangles)`

`(DB)/(CB)` `= (BC)/(AC)`
`{(3-AD)}/2` `= 2/3`
`3-AD` `= 4/3`
`:. AD` `= 5/3`

 

Calculus, 2ADV C3 2018 HSC 13a

Consider the curve  `y = 6x^2 - x^3`.

  1. Find the stationary points and determine their nature.  (3 marks)

  2. Given that the point  (2,16)  lies on the curve, show that it is a point of inflection.  (2 marks)

  3. Sketch the curve, showing the stationary points, the point of inflection and the `x` and `y` intercepts.  (2 marks)

Show Answers Only
  1. `text(MIN at)\ (0, 0);\ text(MAX at)\ (4, 32)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`
Show Worked Solution

i.    `y = 6x^2 – x^3`

`(dy)/(dx) = 12x – 3x^2`

`(d^2y)/(dx^2) = 12 – 6x`
 

`text(S.P.s occur when)\ \ (dy)/(dx) = 0`

`12x – 3x^2 = 0`

`3x(4 – x) = 0`

`x = 0 or 4`
 

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

`:.\ text(MIN at)\ (0, 0)`
 

`text(When)\ \ x = 4,\ (d^2y)/(dx^2) < 0`

`:.\ text(MAX at)\ (4, 32)`

 

ii.  `text(P.I. occur when)\ \ (d^2y)/(dx^2) = 0,`

`12 – 6x` `= 0`
`x` `= 2`

 
`text(When)\ \ x = 2,\ y = 16`

 
`text(S)text(ince the concavity changes)`

`=>\ text(P.I. occurs at)\ \ (2, 16)`

 

iii.  

Plane Geometry, 2UA 2018 HSC 12c

The diagram shows the square `ABCD`. The point `E` is chosen on `BC` and the point `F` is chosen on `CD` so that  `EC = FC`.
 

  1. Prove that `Delta ADF` is congruent to `Delta ABE`.   (2 marks)

  2. The side length of the square is 14 cm and `EC` has length 4 cm. Find the area of  `AECF`.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `56\ text(cm)^2`
Show Worked Solution

i.    `AB = AD\ \ text{(sides of a square)}`

`DF = DC-CF`

`BE = BC-CE`

`text{Since}\ CE = CF\ \ text{(given), and}\ DC = BC\ \ text{(sides of a square)}`

`=> DF = BE`

`=> /_ ADF = /_ ABE = 90^@`

`:. Delta ADF \equiv Delta ABE\ \ text{(SAS)}`

 

ii.   `text(Area of)\ Delta ABE` `= 1/2 xx 14 xx 10`
    `= 70\ text(cm)^2`

 
`:.\ text(Area of)\ AECF`

`= text(Area of)\ ABCD-(2 xx 70)`

`= (14 xx 14)-140`

`= 56\ text(cm)^2`

Calculus, 2ADV C3 2018 HSC 12b

Find the equation of the tangent to the curve  `y = cos 2x`  at  `x = pi/6`.  (3 marks)

Show Answers Only

`y = -sqrt 3 x + (sqrt 3 pi + 3)/6`

Show Worked Solution
`y` `= cos 2x`
`(dy)/(dx)` `= -2 sin 2x`

 
`text(When)\ \ x = pi/6:`

`y` `= cos  pi/3 = 1/2`
`(dy)/(dx)` `= -sin  pi/3 = -sqrt 3`

 
`text(Equation of tangent)\ \ m = -sqrt 3,\ text(through)\ (pi/6, 1/2):`

`y – y_1` `= m(x – x_1)`
`y – 1/2` `= -sqrt 3 (x – pi/6)`
`y` `= -sqrt 3 x + (sqrt 3 pi)/6 + 1/2`
`:. y` `= -sqrt 3 x + (sqrt 3 pi + 3)/6`

Probability, 2UG 2018 HSC 30d

A game consists of two tokens being drawn at random from a barrel containing 20 tokens. There are 17 tokens labelled 10 cents and 3 tokens labelled $2. The player wins the total value of the two tokens drawn.

  1. Complete the probability tree by writing the missing probabilities in the boxes.  (2 marks)
     

     
  2. Considering that the game costs $1 to play, calculate the financial expectation of the game.  (3 marks)
Show Answers Only
  1.  
  2. `−$0.23\ \ (text(or $0.23 loss))`
Show Worked Solution
i.   

 

ii.   `text(Expectated Gain/Loss)`

♦♦ Mean mark part (ii) 27%.
COMMENT: Financial expectation included here (one example only in database) as it is possible to argue that it is an application of expectation.

`= [17/20 xx 16/19 xx 0.20 + 17/20 xx 3/19 xx 2.1`

`+ 3/20 + 17/19 xx 2.1 + 3/20 xx 2/19 xx 4] – 1`

`= 0.77 – 1`

`= −$0.23\ \ (text(or $0.23 loss))`

Financial Maths, STD2 F1 2018 HSC 30b

Last year, Luke’s taxable income was `$87\ 000` and the tax payable on this income was `$19\ 822`. This year, Luke’s taxable income has increased by `$16\ 800`.

  1. Use the table to calculate the tax payable by Luke this year.  (2 marks)
     

  2. How much extra money will Luke have this year, after paying tax, as a result of the increase in his taxable income? Ignore the Medicare levy.  (2 marks)

Show Answers Only
  1. `$26\ 038`
  2. `$10\ 584`
Show Worked Solution

i.   `text(Taxable income) = 87\ 000 + 16\ 800 = $103\ 800`

`:.\ text(Tax payable)` `= 19\ 822 + 0.37 (103\ 800-87\ 000)`
  `= $26\ 038`

 

ii.   `text(Net income from last year)`

`= 87\ 000-19\ 822`

`= $67\ 178`
 

`text(Net income in current year)`

`= 103\ 800-26\ 038`

`= $77\ 762`
 

`:.\ text(Extra money)` `= 77\ 762-67\ 178`
  `= $10\ 584`

Functions, 2ADV F1 2018 HSC 11c

Simplify  `(8x^3 - 27y^3)/(2x - 3y)`.  (2 marks)

Show Answers Only

`4x^2 + 6xy + 9y^2`

Show Worked Solution
`(8x^3 – 27y^3)/(2x – 3y)` `= ((2x)^3 – (3y)^3)/(2x – 3y)`
  `= ((2x – 3y)(4x^2 + 6xy + 9y^2))/(2x – 3y)`
  `= 4x^2 + 6xy + 9y^2`

Financial Maths, STD2 F4 2018 HSC 29e

Andrew borrowed $20 000 to be repaid in equal monthly repayments of $243 over 10 years. Having made this monthly repayment for 4 years, he increased his monthly repayment to $281. As a result, Andrew paid off the loan one year earlier.

How much less did he repay altogether by making this change?  (2 marks)

Show Answers Only

`$636`

Show Worked Solution
`text(Total original repayments)` `= 10 xx 12 xx 243`
  `= $29\ 160`
   
`text(Actual repayments)` `= 4 xx 12 xx 243\ +\ 5 xx 12 xx 281`
  `= $28\ 524`
   
`:.\ text(Savings)` `= 29\ 160 – 28\ 524`
  `= $636`

Measurement, STD2 M2 2018 HSC 29a

The time in Brisbane is 4.5 hours ahead of the time in New Delhi. John flew from New Delhi to Brisbane via Singapore. His plane left New Delhi at 11.30 am (New Delhi time), stopped for 3 hours in Singapore, and arrived in Brisbane at 9.00 am the following day (Brisbane time).

What was the plane’s total flying time?  (3 marks)

Show Answers Only

`14\ text(hours)`

Show Worked Solution

`text(11:30 am in New Delhi = 4 pm in Brisbane)`

`text(Total travel time)` `= 4\ text{pm → 9 am (next day)}`
  `= 17\ text(hours)`

 

`:.\ text(Flying time)` `= 17-3`
  `= 14\ text(hours)`

Measurement, STD2 M7 2018 HSC 28c

Every day, a 1200-watt microwave oven is used for 45 minutes at 40% power. Electricity is charged at $0.25 per kWh.

What is the cost of running this microwave oven for 180 days?  (3 marks)

Show Answers Only

`$16.20`

Show Worked Solution
`text(Daily usage)` `= 1200 xx 45/60 xx 40text(%)`
  `= 360\ text(watts)`

 

`text(180 day usage)` `= 180 xx 360`
  `= 64\ 800\ text(watts)`
  `= 64.8\ text(kW)`

 

`:.\ text(C)text(ost over 180 days)` `= 64.8 xx 0.25`
  `= $16.20`

Algebra, STD2 A4 SM-Bank 5

`y` `= x + 5`
`y + 2x` `= 2`

 
Draw these two linear graphs on the number plane below and determine their intersection.  (3 marks)
 

 

 
Show Answers Only

`(−1,4)`

Show Worked Solution

`text(Table of coordinates:)\ \ y = x + 5`

 
`text(Table of coordinates:)\ \ y + 2x = 2 \ => \ y = −2x + 2`

 
`text{From graph (and table), intersection occurs}`

`text(at)\ \ (−1,4).`

Functions, 2ADV F1 2018 HSC 4 MC

The line  `3x-4y + 3 = 0`  is a tangent to a circle with centre (3, – 2).

What is the equation of the circle?

  1. `(x + 3)^2 + (y-2)^2 = 4`
  2. `(x-3)^2 + (y + 2)^2 = 4`
  3. `(x + 3)^2 + (y-2)^2 = 16`
  4. `(x-3)^2 + (y + 2)^2 = 16`
Show Answers Only

`D`

Show Worked Solution

`text(Radius) = _|_\ text{distance of  (3, −2)  from}\ \ 3x-4y + 3 = 0`

`_|_\ text(dist)` `= |(ax_1 + by_1 + c)/sqrt(a^2 + b^2)|`
  `= |(3(3)-4(-2) + 3)/sqrt(3^2 + (-4)^2)|`
  `= |(9 + 8 + 3)/sqrt 25|`
  `= 4`

 
`text{Equation of circle, centre (3, –2) with radius 4}`

`(x-3)^2 + (y + 2)^2 = 16`

`=>  D`

Statistics, STD2 S5 2018 HSC 27e

Joanna sits a Physics test and a Biology test.

  1. Joanna’s mark in the Physics test is 70. The mean mark for this test is 58 and the standard deviation is 8.

     

    Calculate the `z`-score for Joanna’s mark in this test.  (1 mark)

  2. In the Biology test, the mean mark is 64 and the standard deviation is 10.

     

    Joanna’s `z`-score is the same in both the Physics test and the Biology test.

     

    What is her mark in the Biology test?  (2 marks)

Show Answers Only
  1. `1.5`
  2. `79`
Show Worked Solution

i.   `x = 70, \ mu = 58, \ sigma = 8`

`:. ztext(-score)` `= (x – mu)/sigma`
  `= (70 – 58)/8`
  `= 1.5`

 

ii.    `1.5` `= (x – 64)/10`
  `x – 64` `= 15`
  `:. x` `= 79`

Algebra, STD2 A4 2018 HSC 27d

The graph displays the cost (`$c`) charged by two companies for the hire of a minibus for `x` hours.
 


  

Both companies charge $360 for the hire of a minibus for 3 hours.

  1. What is the hourly rate charged by Company A?  (1 mark)

  2. Company B charges an initial booking fee of $75.

     

    Write a formula, in the form of  `c = mx + b`, for the cost of hiring a minibus from Company B for `x` hours.  (2 marks)

  3. A minibus is hired for 5 hours from Company B.

     

    Calculate how much cheaper this is than hiring from Company A.  (2 marks)

Show Answers Only
  1. `$120`
  2. `c = 95x + 75`
  3. `$50`
Show Worked Solution
i.    `text(Hourly rate)\ (A)` `= 360 ÷ 3`
    `= $120`

 

ii.   `m = text(hourly rate)`

`text(Find)\ m,\ text(given)\ c = 360,\ text(when)\ \ x = 3\ \ text(and)\ \ b = 75`

`360` `= m xx 3 + 75`
`3m` `= 285`
`m` `= 95`

 
`:. c = 95x + 75`
 

iii.    `text(C)text(ost)\ (A)` `= 120 xx 5 = $600`
  `text(C)text(ost)\ (B)` `= 95 xx 5 + 75 = $550`

 
`:.\ text(Company)\ B’text(s hiring cost is $50 cheaper.)`

Measurement, STD2 M1 2018 HSC 27c

A shade shelter is to be constructed in the shape of half a cylinder with open ends. The diameter is 3.8 m and the length is 10 m.
 

 
The curved roof is to be made of plastic sheeting.

What area of plastic sheeting is required, to the nearest m²?  (2 marks)

Show Answers Only

`60\ text(m²  (nearest m²))`

Show Worked Solution

`text(Flatten out the half cylinder,)`

`text(Width)` `= 1/2 xx text(circumference)`
  `= 1/2 xx pi xx 3.8`
  `= 5.969…`

 

`:.\ text(Sheeting required)` `= 10 xx 5.969…`
  `= 59.69…`
  `= 60\ text(m²  (nearest m²))`

Measurement, STD2 M7 2018 HSC 26g

A field diagram of a block of land has been drawn to scale. The shaded region `ABFG` is covered in grass.
 

 
The actual length of `AG` is 24 m.

  1. If the length of `AG` on the field diagram is 8 cm, what is the scale of the diagram?  (1 mark)

  2. How much fertiliser would be needed to fertilise the grassed area  `ABFG`  at the rate of 26.5 g /m²?  (3 marks)

Show Answers Only
  1. `text(1 : 300)`
  2. `5008.5\ text(grams)`
Show Worked Solution

♦♦ Mean mark 32%.

i.    `text(Scale    8 cm)` `\ :\ text(24 m)`
  `text(1 cm)` `\ :\ text(3 m)`
  `1` `\ :\ 300`

 

ii.   `text(Area of rectangle)\ ABFE`

`= 6\ text(cm × 3 cm)`

`= 18\ text(m × 9 m)`

`= 162\ text(m²)`
 

`text(Area of)\ DeltaEFG`

`= 1/2 xx 3\ text(cm) xx 2\ text(cm)`

`= 1/2 xx 9 xx 6`

`= 27\ text(m²)`
 

`:.\ text(Fertiliser needed)` `= (162 + 27) xx 26.5`
  `= 5008.5\ text(grams)`

Measurement, STD2 M1 2018 HSC 22 MC

A shape consisting of a quadrant and a right-angled triangle is shown.
 

 
What is the perimeter of this shape, correct to one decimal place?

  1. 28.6 cm
  2. 36.6 cm
  3. 66.3 cm
  4. 74.3 cm
Show Answers Only

`text(B)`

Show Worked Solution

`text(Using Pythagoras to find radius)\ (r):`

`r` `= sqrt(10^2 – 6^2)`
  `= sqrt64`
  `= 8\ text(cm)`

 

`text(Arc length)` `= 1/4 xx 2 pi r`
  `= 1/4 xx 2 xx pi xx 8`
  `= 12.56…\ text(cm)`

 

`:.\ text(Perimeter)` `= 8 + 6 + 10 + 12.56…`
  `= 36.57…`

`=>\ text(B)`

Financial Maths, STD2 F1 2018 HSC 21 MC

David earns a gross income of $890 per week. Each week, 25% of this income is deducted in taxation. David budgets to save 20% of his net income.

How much does he budget to save each week?

  1. $44.50
  2. $133.50
  3. $489.50
  4. $534.00
Show Answers Only

`text(B)`

Show Worked Solution
`text(Net Income)` `= 75text(%) xx 890`
  `= $667.50`

 

`text(Savings)` `= 20text(%) xx 667.50`
  `= $133.50`

`=>\ text(B)`

Probability, STD2 S2 2018 HSC 20 MC

During a year, the maximum temperature each day was recorded. The results are shown in the table.
  


  

From the days with a maximum temperature less than 25°C, one day is selected at random.

What is the probability, to the nearest percentage, that the selected day occurred during winter?

  1. 19%
  2. 25%
  3. 32%
  4. 77%
Show Answers Only

`text(C)`

Show Worked Solution
`text{P(winter day)}` `= (text(winter days < 25))/text(total days < 25) xx 100`
  `= 71/223 xx 100`
  `= 31.8…%`

`=>\ text(C)`

Measurement, STD2 M1 2018 HSC 18 MC

The length of a window is measured as 2.4 m.

Which calculation will give the percentage error for this measurement?

  1. `0.05/2.4 xx 100`
  2. `0.05/100 xx 2.4`
  3. `0.5/2.4 xx 100`
  4. `0.5/100 xx 2.4`
Show Answers Only

`A`

Show Worked Solution

`text{Absolute error}\ =1/2 xx text{precision}\ = 1/2 xx 0.1 = 0.05\ text{m}`

`text{% error}` `=\ frac{text{absolute error}}{text{measurement}} xx 100%`  
  `=0.05/2.4 xx 100%`  

 
`=>A`

Measurement, STD2 M1 2018 HSC 13 MC

A rectangular pyramid has base side lengths `3x` and `4x`. The perpendicular height of the pyramid is `2x`. All measurements are in metres.
 

What is the volume of the pyramid in cubic metres?

  1. `8x^3`
  2. `9x^3`
  3. `12x^3`
  4. `24x^3`
Show Answers Only

`A`

Show Worked Solution
`text(Volume)` `= 1/3Ah`
  `= 1/3(4x xx 3x xx 2x)`
  `= 8x^3`

 
`=>A`

Measurement, STD2 M6 2018 HSC 12 MC

The diagram shows a triangle with side lengths 8 m, 9 m and 10m.
 


 

What is the value of `theta`, marked on the diagram, to the nearest degree?

  1. 49°
  2. 51°
  3. 59°
  4. 72°
Show Answers Only

`text(D)`

Show Worked Solution

`text(Using the cosine rule:)`

`costheta` `= (8^2 + 9^2 – 10^2)/(2 xx 8 xx 9)`
  `= 0.3125`
`:.theta` `= cos^(−1)(0.3125)`
  `= 71.790…^@`

 
`=>D`

Statistics, STD2 S1 2018 HSC 11 MC

A set of data is summarised in this frequency distribution table.
 

 
Which of the following is true about the data?

  1. Mode = 7, median = 5.5
  2. Mode = 7, median = 6
  3. Mode = 9, median = 5.5
  4. Mode = 9, median = 6
Show Answers Only

`text(B)`

Show Worked Solution

`text{Mode = 7  (highest frequency of 9)}`

`text(Median = average of 15th and 16th data points.)`

`:.\ text(Median = 6)`

`=>\ text(B)`

Measurement, STD2 M6 2018 HSC 7 MC

The diagram shows the positions of towns `A`, `B` and `C`.

Town `A` is due north of town `B` and `angleCAB = 34°`
  


 

What is the bearing of town `C` from town `A`?

  1. 034°
  2. 146°
  3. 214°
  4. 326°
Show Answers Only

`C`

Show Worked Solution

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

`text(Bearing)` `= 180 + 34`
  `= 214^@`

 
`=>C`

Statistics, STD2 S1 2018 HSC 3 MC

A survey asked the following question.

'How many brothers do you have?'

How would the responses be classified?

  1. Categorical, ordinal
  2. Categorical, nominal
  3. Numerical, discrete
  4. Numerical, continuous
Show Answers Only

`text(C)`

Show Worked Solution

`text(The number of brothers a person has is)`

`text(an exact whole number.)`

`:.\ text(Classification is numerical, discrete.)`

`=>\ text(C)`

Mechanics, EXT2* M1 2018 HSC 13c

An object is projected from the origin with an initial velocity of  `V` at an angle  `theta`  to the horizontal. The equations of motion of the object are

`x(t)` `= Vt cos theta`
`y(t)` `= Vt sin theta - (g t^2)/2.`  (Do NOT prove this.)

 

  1. Show that when the object is projected at an angle  `theta`, the horizontal range is

     

         `V^2/g sin 2 theta`  (2 marks)

  2. Show that when the object is projected at an angle  `pi/2 - theta`, the horizontal range is also 

     

         `V^2/g sin 2 theta`.  (1 mark)

  3. The object is projected with initial velocity `V` to reach a horizontal distance `d`, which is less than the maximum possible horizontal range. There are two angles at which the object can be projected in order to travel that horizontal distance before landing.

     

    Let these angles be `alpha`  and  `beta`, where  `beta = pi/2 - alpha.`

     

    Let  `h_alpha`  be the maximum height reached by the object when projected at the angle `alpha` to the horizontal.

     

    Let  `h_beta`  be the maximum height reached by the object when projected at the angle `beta` to the horizontal.
     
         
     
    Show that the average of the two heights, `(h_alpha + h_beta)/2`, depends only on `V` and `g`.  (3 marks)

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

i.    `text(Horizontal range occurs when)\ \ y = 0`

`Vt sin theta – (g t^2)/2` `= 0`
`V sin theta – (g t)/2` `= 0`
`t` `= (2V sin theta)/g`

 
`text(Find)\ \ x\ \ text(when)\ \ t=(2V sin theta)/g :`

`x` `= Vt cos theta`
  `= V((2V sin theta)/g) cos theta`
  `= (V^2 * 2sin theta cos theta)/g`
  `= V^2/g sin 2 theta\ \ \ text(.. as required)`

 

ii.  `text(Find)\ \ x\ \ text(when)\ \ theta = (pi/2 – theta):`

`x` `= V^2/g sin 2 (pi/2 – theta)`
  `= V^2/g underbrace {sin (pi – 2 theta)}_{text(Using)\ \ sin (pi-theta) = sin theta}`
  `= V^2/g sin 2 theta\ \ \ text(.. as required)`

 

iii.  `text(Highest point → half way through the flight.)`

`=> h_alpha\ \ text(occurs when)\ \ t=(V sin alpha)/g\ \ text{(by symmetry)}`
  

`:. h_alpha` `= V((V sin alpha)/g) sin alpha – g/2 ((V sin alpha)/g)^2`
  `= (V^2 sin^2 alpha)/g – g/2 ((V^2 sin^2 alpha)/g^2)`
  `= (V^2 sin^2 alpha)/g – (V^2 sin^2 alpha)/(2g)`
  `= (V^2 sin^2 alpha)/(2g)`

 

`text(Similarly,)\ \ h_beta = (V^2 sin^2 beta)/(2g)`

♦ Mean mark 44%.
 

`:. (h_alpha + h_beta)/2` `= 1/2 ((V^2 sin^2 alpha)/(2g) + (V^2 sin^2 beta)/(2g))`
  `= V^2/(4g) (sin^2 alpha + underbrace{sin^2 beta}_{text(Using)\ \ beta= pi/2 -alpha})`
  `= V^2/(4g) (sin^2 alpha + sin^2(pi/2 – alpha)) `
  `= V^2/(4g) (sin^2 alpha + cos^2 alpha)`
  `= V^2/(4g)`

 
`:. text(The average height depends only on)\ V\ text(or)\ g.`

Quadratic, EXT1 2018 HSC 12e

The points  `P(2ap, ap^2)`  and  `Q(2aq, aq^2)`  lie on the parabola  `x^2 = 4ay`. The focus of the parabola is  `S(0, a)`  and the tangents at  `P`  and  `Q`  intersect at  `T(a(p + q), apq)`. (Do NOT prove this.)

The tangents at  `P`  and  `Q`  meet the `x`-axis at  `A`  and  `B`  respectively, as shown.
 

 

  1. Show that  `/_ PAS = 90^@`.  (2 marks)
  2. Explain why  `S, B, A, T`  are concyclic points.  (1 mark)
  3. Show that the diameter of the circle through  `S, B, A`  and  `T`  has length
     
    `qquad qquad a sqrt((p^2 + 1)(q^2 + 1))`.  (2 marks)

 

 

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

(i)    `y=x^2/(4a)\ \ => dy/dx=x/(2a)`

`text(At)\ \ x=2ap,\ \ m_text(tang) = p`

`text(Equation of tangent)\ \ m=p, text(through)\ (2ap, ap^2):`

`y-ap^2` `=p(x-2ap)`  
`y` `=px-ap^2`  

 
`text(When)\ \ y=0, x=ap`

`:. A(ap,0)`
 

`m_(AP)` `= (ap^2 – 0)/(2ap-ap)`
  `= p`
`m_(AS)` `= (0 – a)/(ap – 0)`
  `= -1/p`

  
`m_(AP) xx m_(AS) = -1`

`:. /_PAS = 90°\ \ text(… as required)`

 

(ii)   `text{Similarly to part (i):}`

♦ Mean mark 37%.

`=> B(aq,0)` 

`=> m_(BQ) xx m_(BS) = -1`

`=> /_QBS = 90°`
 

`text(S)text(ince)\ \ T\ \ text(is the intersection of both tangents,)`

`/_ PAS = /_ TAS = 90^@`

`/_QBS = /_ TBS = 90^@`
 

`=> ST\ text(can be considered the extreme points of a circle diameter)`

`text(that is subtending 2 right angles on the circle circumference.)`   

`:. SBAT\ text(are concyclic points.)`

 

(iii)  `text(Diameter = distance)\ ST:`

♦ Mean mark 44%.

`S(0,a),\ \ T((a(p + q), apq)`

`d^2` `= ST^2`
  `=(x_2-x_1)^2 + (y_2-y_1)^2`
  `= (a(p + q) – 0)^2 + (apq – a)^2`
  `= a^2 (p + q)^2 + a^2 (pq – 1)^2`
  `= a^2(p^2 + 2pq + q^2 + p^2q^2 – 2pq + 1)`
  `= a^2(p^2 + q^2 + p^2q^2 + 1)`
  `= a^2(p^2 + 1) (q^2 + 1)`
`:.d` `= a sqrt((p^2 + 1)(q^2 + 1))\ \ text(… as required)`

Statistics, EXT1 S1 2018 HSC 12d

A group of 12 people sets off on a trek. The probability that a person finishes the trek within 8 hours is 0.75.

Find an expression for the probability that at least 10 people from the group complete the trek within 8 hours.  (2 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text{P(finishes within 8 hours)} = 0.75`

`text{P(not finish within 8 hours)} = 0.25`

`text(Let)\ \ X = text(number who finish below 8 hours)`
 

`:.\ text{P(at least 10 finish within 8 hours)}`

`=\ text{P(X=10) + P(X=11) + P(X=12)}`

`= ((12), (10)) (0.75)^10 (0.25)^2 + ((12), (11)) (0.75)^11 (0.25)^1 + ((12), (12)) (0.75)^12`

Calculus, EXT1 C1 2018 HSC 12b

A ferris wheel has a radius of 20 metres and is rotating at a rate of 1.5 radians per minute. The top of a carriage is `h` metres above the horizontal diameter of the ferris wheel. The angle of elevation of the top of the carriage from the centre of the ferris wheel is `theta`.
 


 

  1. Show that  `(dh)/(d theta) = 20 cos theta`.  (1 mark)

  2. At what speed is the top of the carriage rising when it is 15 metres higher than the horizontal diameter of the ferris wheel? Give your answer correct to one decimal place.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solution)}`
  2. `19.8\ text{metres per minute  (1 d.p)}`
Show Worked Solution

i.   `text(From the diagram,)`

`sin theta` `= h/20`
`h` `= 20 sin theta`
`:. (dh)/(d theta)` `= 20 cos theta`

 

ii.   `text(Find)\ \ (dh)/(dt)\ text(when)\ h = 15:`

`(dh)/(dt)` `= (dh)/(d theta) xx (d theta)/(dt)`
  `= (20 cos theta) xx 1.5`
  `= 30 cos theta`

 
`text(Find)\ cos theta\ \ text(when)\ h = 15`

`text(Using Pythagoras,)`

`cos theta` `=sqrt(20^2 – 15^2)/20`  
  `=sqrt7/4`  

 

`:. (dh)/(dt)` `= 30 cos theta`
  `= (15 sqrt 7)/2`
  `~~ 19.8\ text{metres per minute  (1 d.p)}`

Real Functions, EXT1 2018 HSC 11e

Consider the function  `f(x) = 1/(4x - 1)`.
 

  1. Find the domain of  `f(x)`.  (1 mark)
  2. For what values of  `x`  is  `f(x) < 1`?  (2 marks)
Show Answers Only
  1. `{text(all real)\ x,  x!=1/4}`
  2. `x > 1/2 or x < 1/4`
Show Worked Solution

(i)   `text(Domain is)\ {text(all real)\ x,  x!=1/4}`

 

(ii)   `1/(4x – 1)` `< 1`
  `(4x – 1)` `< (4 x – 1)^2`
  `(4x – 1)^2 – (4x – 1)` `> 0`
  `(4x – 1)[4x – 1- 1]` `> 0`
  `2 (4x – 1) (2x – 1)` `> 0`

 
`text(Sketching the parabola:)`

`x > 1/2 or x < 1/4.`

`text(From the graph,)`

`y > 1\ \ text(when)\ \ x < 1/4 or x > 1/2`

Plane Geometry, EXT1 2018 HSC 11d

Two secants from the point `C` intersect a circle as shown in the diagram.
 

 
What is the value of `x`?  (2 marks)

Show Answers Only

`4`

Show Worked Solution

`text(Using formula for intercepts of intersecting secants:)`

`x (x + 2)` `= 3 (3 + 5)`
`x^2 + 2x` `= 24`
`x^2 + 2x – 24` `= 0`
`(x + 6) (x – 4)` `= 0`
`:. x` `= 4 \ \ \ (x > 0)`

Mechanics, EXT2* M1 2018 HSC 7 MC

The velocity of a particle, in metres per second, is given by  `v = x^2 + 2`  where `x` is its displacement in metres from the origin.

What is the acceleration of the particle at  `x = 1`?

A.     `2\ text(m s)^(-2)`

B.     `3\ text(m s)^(-2)`

C.     `6\ text(m s)^(-2)`

D.     `12\ text(m s)^(-2)`

Show Answers Only

`C`

Show Worked Solution
`a` `= d/dx (1/2  v^2)`
  `= d/dx (1/2 (x^2 + 2)^2)`
  `=1/2 xx 2 xx 2x (x^2+2)`
  `= 2x (x^2 + 2)`

 
`text(When)\ \ x = 1,`

`a = 6\ text(m s)^(-2)`
 

`⇒  C`

Polynomials, EXT1 2018 HSC 6 MC

The diagram shows the graph of  `y = f(x)`. The equation  `f(x) = 0`  has a solution at  `x = w`.
 


 

Newton’s method can be used to give an approximation close to the solution  `x = w`.

Which initial approximation, `x_1`, will give the second approximation that is closest to the solution  `x = w`?

A.     `x_1 = a`

B.     `x_1 = b`

C.     `x_1 = c`

D.     `x_1 = d`

Show Answers Only

`C`

Show Worked Solution

`text(Consider where the tangents at each point will cross the)`

`xtext(-axis).`

`x_1 = a\ text(will approximate the negative root.)`

`x_1 = c\ text(will cross the)\ x text(-axis closest to) \ w.`

`⇒  C`

Calculus, EXT1* C1 2018 HSC 5 MC

The diagram shows the number of penguins, `P(t)`, on an island at time `t`.
  


  

Which equation best represents this graph?

A.     `P(t) = 1500 + 1500e^(-kt)`

B.     `P(t) = 3000 - 1500e^(-kt)`

C.     `P(t) = 3000 + 1500e^(-kt)`

D.     `P(t) = 4500 - 1500e^(-kt)`

Show Answers Only

`A`

Show Worked Solution

`P(0) = 3000\ text{(from graph) → Eliminate B and C}`
 

`text(As),\ t -> oo,\ 1500e^(-kt) -> 0,`

`:. 1500 + 1500e^(-kt) => 1500`

`=>  A`

Geometry and Calculus, EXT1 2018 HSC 4 MC

The diagram shows the graph of  `y = a(x + b) (x + c) (x + d)^2`.
 


 

What are possible values of  `a, b, c` and `d`?

A.     `a = -6,\ \ b = -2,\ \ c = -1,\ \ d = 1`

B.     `a = -6,\ \ b = 2,\ \ c = 1,\ \ d = -1`

C.     `a = -3,\ \ b = -2,\ \ c = -1,\ \ d = 1`

D.     `a = -3,\ \ b = 2,\ \ c = 1,\ \ d = -1`

Show Answers Only

`D`

Show Worked Solution

`text(S)text(ingle roots at:)\ \ x = -2, -1`

`text(Double root at:)\ \ x = 1`

`:. b = 2,\ \ c = 1,\ \ d = -1`
 

`y text(-intercept occurs when)\ x=0,`

`=> abcd^2 = -6`

`a(2)(1)(1^2) = -6`

`:.a = -3`
  

`⇒  D`

Trig Calculus, EXT1 2018 HSC 3 MC

What is the value of  `lim_(x -> 0) (sin 3x cos 3x)/(12x)`?

A.     `1/4`

B.     `1/2`

C.     `3/4`

D.     `1`

Show Answers Only

`A`

Show Worked Solution
`lim_(x -> 0) (sin 3x cos 3x)/(12x)` `= lim_(x -> 0) (1/2 xx 2 sin 3x cos 3x)/(12x)`
  `= lim_(x ->0) (1/2 xx sin 6x)/(12x)`
  `= 1/4 lim_(x ->0) (sin 6x)/(6x)`
  `= 1/4`

 

`⇒  A`

Networks, STD2 N3 2007 FUR2 4

A community centre is to be built on the new housing estate.

Nine activities have been identified for this building project.

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

 

  1. Determine the minimum time, in weeks, to complete this project.  (1 mark)

  2. Determine the float time, in weeks, for activity `D`.  (2 marks)

The builders of the community centre are able to speed up the project.

Some of the activities can be reduced in time at an additional cost.

The activities that can be reduced in time are `A`, `C`, `E`, `F` and `G`.

  1. Which of these activities, if reduced in time individually, would not result in an earlier completion of the project?  (1 mark)

The owner of the estate is prepared to pay the additional cost to achieve early completion.

The cost of reducing the time of each activity is $5000 per week.

The maximum reduction in time for each one of the five activities, `A`, `C`, `E`, `F`, `G`, is `2` weeks.

  1. Determine the minimum time, in weeks, for the project to be completed now that certain activities can be reduced in time.  (1 mark)

  2. Determine the minimum additional cost of completing the project in this reduced time.  (1 mark)

Show Answers Only
  1. `19\ text(weeks)`
  2. `5\ text(weeks)`
  3. `A, E, G`
  4. `text(15 weeks)`
  5. `$25\ 000`
Show Worked Solution

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

 
`BCFHI\ \ text(is the critical path.)`

♦ Mean mark of all parts (combined) 40%.
`:.\ text(Minimum time)` `= 4 + 3 + 4 + 2 + 6`
  `= 19\ text(weeks)`

 

b.    `text(EST of)\ D` `= 4`
  `text(LST of)\ D` `= 9`
`:.\ text(Float time of)\ D` `= 9 – 4`
  `= 5\ text(weeks)`

 

c.  `A, E,\ text(and)\ G\ text(are not currently on the critical path,)`

`text(therefore reducing their time will not result in an)`

`text(earlier completion time.)`
 

d.  `text(Reduce)\ C\ text(and)\ F\ text(by 2 weeks each.)`

`text(However, a new critical path created:)\ BEHI\ \ text{(16 weeks)}`

`:.\ text(Also reduce)\ E\ text(by 1 week.)`

`:.\ text(Minimum completion time = 15 weeks)`

 

e.    `text(Additional cost)` `= 5 xx $5000`
    `= $25\ 000`

Networks, STD2 N3 2006 FUR2 3

The five musicians are to record an album. This will involve nine activities.

The activities and their immediate predecessors are shown in the following table.

The duration of each activity is not yet known.
 

NETWORKS, FUR2 2006 VCAA 31
 

  1. Use the information in the table above to complete the network below by including activities `G`, `H` and `I`.  (2 marks)
     

NETWORKS, FUR2 2006 VCAA 32
 

There is only one critical path for this project.

  1. How many non-critical activities are there?  (1 mark)

The following table gives the earliest start times (EST) and latest start times (LST) for three of the activities only. All times are in hours.


Networks, FUR2 2006 VCAA 3_3

  1. Write down the critical path for this project.  (1 mark)

The minimum time required for this project to be completed is 19 hours.

  1. What is the duration of activity `I`?  (1 mark)

The duration of activity `C` is 3 hours.

  1. Determine the maximum combined duration of activities `F` and `H`.  (1 mark) 
Show Answers Only
  1.  
    networks-fur2-2006-vcaa-3-answer
  2. `5`
  3. `B E G I`
  4. `text(7 hours)`
  5. `text(8 hours)`
Show Worked Solution
a.    networks-fur2-2006-vcaa-3-answer

 

b.   `text(S)text(ince no time information, possible critical paths are:)`

`ADGI, BEGI\ text(or)\ CFHI\ \ text{(all have 4 activities)}`
 

`:.\ text(Non-critical activities)`

`= 9 – 4 = 5`

 

c.   `text(Critical activities have zero float time.)`

♦ Mean mark of parts (c)-(e) (combined) was 36%.

`=> A\ text(and)\ C\ text(are non-critical.)`

`:. B E G I\ text(is the critical path.)`

 

d.    `text(Duration of)\ \ I` `= 19 – 12`
    `= 7\ text(hours)`

 

e.   `text(Maximum time for)\ F\ text(and)\ H`

♦♦ MARKER’S COMMENT: Many students incorrectly answered 9 hours in part (e).

`=\ text(LST of)\ I – text(duration)\ C – text(slack time of)\ C`

`= 12 – 3 – 1`

`= 8\ text(hours)`

Networks, STD2 N3 2013 FUR2 2

A project will be undertaken in the wildlife park. This project involves the 13 activities shown in the table below. The duration, in hours, and predecessor(s) of each activity are also included in the table.


NETWORKS, FUR2 2013 VCAA 21
 

Activity `G` is missing from the network diagram for this project, which is shown below.

 
NETWORKS, FUR2 2013 VCAA 22
 

  1. Complete the network diagram above by inserting activity `G`.  (1 mark)
  2. Determine the earliest starting time of activity `H`.  (1 mark)

  3. Given that activity `G` is not on the critical path

     

    1. write down the activities that are on the critical path in the order that they are completed  (1 mark)

    2. find the latest starting time for activity `D`.  (1 mark)

  4. Consider the following statement.

     

    ‘If the time to complete just one of the activities in this project is reduced by one hour, then the minimum time to complete the entire project will be reduced by one hour.’

    Explain the circumstances under which this statement will be true for this project.  (1 mark)

  5. Assume activity `F` is reduced by two hours.
    What will be the minimum completion time for the project?  (1 mark)

Show Answers Only

a.

networks-fur2-2013-vcaa-2-answer

b.  `7\ text(hours)`

c.i.  `AFIM`

c.ii. `14\ text(hours)`

d.  `text(The statement will only be true if the crashed activity)`
      `text(is on the critical path)\ \ A F I M.`

e.  `text(36 hours)`

Show Worked Solution
a.    networks-fur2-2013-vcaa-2-answer

 

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

`text(EST for Activity)\ H`

`= 4 + 3`

`= 7\ text(hours)`
 

c.i.   `A F I M`

♦♦ Mean mark of parts (c)-(e) (combined) was 40%.
 

c.ii.  `text(LST of)\ G = 20 – 4 = 16\ text(hours)`

 `text(LST of)\ D = 16 – 2 = 14\ text(hours)`
 

d.   `text(The statement will only be true if the time reduced activity)`

MARKER’S COMMENT: Most students struggled with part (d).

`text(is on the critical path)\ \ A F I M.`
 

e.   `A F I M\ text(is 37 hours.)`

`text(If)\ F\ text(is reduced by 2 hours, the new critical)`

`text(path is)\ \ C E H G I M\ text{(36 hours)}`

`:.\ text(Minimum completion time = 36 hours)`