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Statistics, NAP-E2-13

A combined year 5 and year 6 class were asked a question in a survey.

The first seven answers were

`9, 12, 10, 11, 11, 12, 11`

Which one of these questions could have been asked?

 
 Are you left or right handed?
 
 How many centimetres tall are you?
 
 What is your favourite colour?
 
 How many years old are you?
Show Answers Only

`text(How many years old are you?)`

Show Worked Solution

`text(How many years old are you?)`

Number and Algebra, NAP-G2-07

Wayne needs to buy 14 pairs of socks for the players in his soccer team.

The socks are sold in packets of 3.

How many packets does Wayne need to buy?

`4` `5` `17` `42`
 
 
 
 
Show Answers Only

`5`

Show Worked Solution

`text(Socks in 4 packets)`

`=4 xx 3 = 12`

`text(Socks in 5 packets)`

`=5 xx 3= 15`

 

`:.\ text(5 packets are needed.)`

Statistics, NAP-G2-05

The runs scored by Ricky and Steve in 4 games of cricket are recorded in the table below.

In which game did Ricky score 8 more runs than Steve?

`text(Game 1)` `text(Game 2)` `text(Game 3)` `text(Game 4)`
 
 
 
 
Show Answers Only

`text(Game 4)`

Show Worked Solution

`text(Consider Game 4,)`

`38 – 30 = 8`

`:.\ text(Ricky scored 8 more runs than Steve)`

`text(in Game 4.)`

Statistics, NAP-D2-13

Benson asked his friends what their favourite ice cream flavour was.

He recorded their answers in the table below but left off same labels.

Benson remembered that

  • strawberry was the least liked flavour
  • chocolate was the most liked flavour
  • vanilla was more liked than caramel

How many of Benson's friends liked caramel the most?
Show Answers Only

`8`

Show Worked Solution

`text(Using the graph and the information given,)`

`=>\ text(7 preferred strawberry,)`

`=>\ text(13 preferred chocolate.)`

`:.\ text(11 preferred vanilla and 8 liked caramel the most.)`

Number and Algebra, NAP-I2-11

George has no money in his bank account.

He deposits $6 in his account in week 1.

He then deposits twice the amount into his account each week than he did the previous week.

The total amount in his account is?

 
 always odd.
 
 always even.
 
 sometimes odd and sometimes even.
Show Answers Only

`text(always even.)`

Show Worked Solution

`text(In each successive week, an even number will)`

`text(will be added to an existing even number.)`

`:.\ text(His account total will always be even.)`

Number and Algebra, NAP-I2-9

Olivia was the top scorer in her soccer team over two seasons.

In season 1, she scored 25 goals.

In season 2, she scored 35 goals.

Which number sentence could be used to find the total number of goals she scored over both seasons?

 
 `20 + 30 = 50`
 
 `30 + 40 = 70`
 
 `20 + 30 + 5 = 55`
 
 `20 + 30 + 10 = 60`
Show Answers Only

`20 + 30 + 10 = 60`

Show Worked Solution
`25+35` `=20+5+30+5`
  `=20+30+10`
  `=60`

 
`:. 20 + 30 +10=60`

Number and Algebra, NAP-I2-6

The world record for the most push-ups ever completed in one hour is three thousand and ninety-two.
 

 
The number of push-ups can be written as:

`392` `3092` `3920` `30\ 092`
 
 
 
 
Show Answers Only

`3092`

Show Worked Solution

`text(Three thousand and ninety-two contains:)`

`=>\ text(3 “thousands”)`

`=>\ text(0 “hundreds”)`

`=>\ text(9 “tens”)`

`=>\ text(2 “ones”)`

`:. 3092`

Number and Algebra, NAP-I2-01

Bojacks lives in South Australia and receives 10 cents for every glass bottle that he takes to the recycling depot.

Bojacks recycles 28 glass bottles.

How much money will he receive?

`28 ¢` `$2.80` `$28` `$280`
 
 
 
 
Show Answers Only

`$2.80`

Show Worked Solution

`28 xx 10\ text(cents)`

`= 280\ text(cents)`

`= $2.80`

Number and Algebra, NAP-D2-10

A chocolate bar has a mass of 100 grams.

The dotted line shows where Sharon cuts the chocolate bar.
 

 
Sharon takes the larger piece.

About what mass is Sharon's piece of chocolate?

`text(30 grams)` `text(50 grams)` `text(70 grams)` `text(90 grams)`
 
 
 
 
Show Answers Only

`text(70 grams)`

Show Worked Solution

`text(The cut is on a line where the smaller piece is)`

`text{just over one quarter (25 g) of the chocolate bar.}`

`:.\ text(The larger piece is about 70 grams.)`

Measurement, NAP-D2-07

A Summer Camp is split into four colour teams, red, yellow, blue and green.

Which teams has archery lessons first?

`text(Red)` `text(Yellow)` `text(Blue)` `text(Green)`
 
 
 
 
Show Answers Only

`text(Blue)`

Show Worked Solution

`text(Archery is earliest at 11:00 am.)`

`:.\ text(Blue has archery first.)`

Geometry, NAP-D2-06

Kate is going to draw the other half of this picture so that the picture is symmetrical.

In which square should she draw the other dark spot on the butterfly's wings.

`text(A5)` `text(B5)` `text(C5)` `text(D5)`
 
 
 
 
Show Answers Only

`text(B5)`

Show Worked Solution

`text(B5)`

Statistics, NAP-C4-NC02

This diagram shows the proportion of milkshake flavours sold by a corner store over a 1 month period.
 

 
Which flavour makes up 55% of the milkshakes sold?

vanilla strawberry caramel chocolate
 
 
 
 
Show Answers Only

`text(Chocolate)`

Show Worked Solution

`text(Chocolate)`

Statistics, NAP-D3-CA02

The table below shows the number of car accidents reported in four country towns in June.
 

 
 

Which country town had about 70 reported car accidents in June?

`text(Dubbo)` `text(Warren)` `text(Bourke)` `text(Dunedoo)`
 
 
 
 
Show Answers Only

`text(Warren)`

Show Worked Solution
`3.5\ text(cars)` `= 3.5 xx 20`
  `= 70\ text(accidents)`

 

`:.\ text(Warren had about 70 reported car accidents.)`

Measurement, NAP-E3-NC03

Richard started his jog at 1:25. He finished at 2:08.

How long did Richard jog for?

`text(35 minutes)` `text(43 minutes)` `text(73 minutes)` `text(83 minutes)`
 
 
 
 
Show Answers Only

`text(43 minutes)`

Show Worked Solution

`text(43 minutes)`

Number, NAP-G3-CA01

In Australia, 288 701 children were enrolled in kindergarten in 2013.

Of these children, 150 125 were boys.

How many girls were enrolled in kindergarten in 2013?

`122\ 896` `138\ 576` `288\ 701` `438\ 826`
 
 
 
 
Show Answers Only

`138\ 576`

Show Worked Solution

`text(Number of girls) = 288\ 701  – 150\ 125 = 138\ 576`

Number, NAP-F4-CA01

A return trip from Brodie's house to the beach is 5.78 kilometres. 

How far does Brodie travel if he does this 14 times?

`8.09\ text(km)` `23.12\ text(km)` `42.12\ text(km)` `69.36\ text(km)` `80.92\ text(km)`
 
 
 
 
 
Show Answers Only

`80.92\ text(km)`

Show Worked Solution
`text(Distance)` `= 14 xx 5.78`
  `= 80.92\ text(km)`

Number, NAP-G4-CA01

In Africa, a national park estimated the population of flamingos was 183 409 in 2021.

Of these, 87 396 were male.

How many female flamingos were there?

`96\ 013` `123\ 586` `212\ 786` `270\ 805`
 
 
 
 
Show Answers Only

`96\ 013`

Show Worked Solution

`text(Number of females) = 183\ 409  – 87\ 396 = 96\ 013`

Quadratic, EXT1 2016 HSC 14c

The point  `T(2at,at^2)` lies on the parabola `P_1` with the equation  `x^2=4ay`.

The tangent to the parabola `P_1` at `T` meets the directrix at `D`.

The normal to the parabola `P_1` at `T` meets the vertical line through `D` at the point `R`, as shown in the diagram.

 

ext1-2016-hsc-q14

  1. Show that the point `D` has coordinates `(at - a/t, −a)`.  (1 mark)
  2. Show that the locus of `R` lies on another parabola `P_2`.  (3 marks)
  3. State the focal length of the parabola `P_2`.  (1 mark)

It can be shown that the minimum distance between `R` and `T` occurs when the normal to `P_1` at `T` is also the normal to `P_2` at `R`.  (Do NOT prove this.)

  1. Find the values of `t` so that the distance between `R` and `T` is a minimum.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `a/4`
  4. `+- sqrt 2`
Show Worked Solution

i.     `text(Show)\ \ D (text{at}\ -a/t, -a)`

`text(T)text(angent equation at)\ \ T:`

`y = tx – at^2`

`D\ \ text(occurs when)\ \ y = -a,`

`tx – at^2` `= -a`
`tx` `= at^2 – a`
`x` `= at – a/t`

`:. D\ text(has coordinates)\ \ (text{at}\ -a/t, -a)`

 

ii.    `text(Normal equation at)\ \ T:`

♦♦ Mean mark 32%.
MARKER’S COMMENT: Normal equation is found in the Reference Sheet. Save time by using it!

`x + ty = 2at + at^3`

`R\ text(occurs when)\ \ x = at – a/t`

`at – a/t + ty` `= 2at + at^3`
`ty` `= at + a/t + at^3`
`y` `= a + a/t^2 + at^2`
  `= a(1 + 1/t^2 + t^2)\ \ …\ text{(*)}`

 

`:. R (a (t – 1/t), a (1 + 1/t^2 + t^2))`

`x^2` `= a^2 (t – 1/t)^2`
  `= a^2 (t^2 – 2 + 1/t^2)`
  `= a + (1+1/t^2 + t^2 – 3)`
  `= a^2 (y/a – 3)\ \ text{(see (*) above)}`
  `= ay – 3a^2`

 

`:.\ text(Locus of)\ R\ text(is)\ \ x^2 = ay – 3a^2`

 

iii.   `text(In the form)\ \ x^2 = 4ay,`

♦♦♦ Mean mark 15%.
`x^2` `= a(y – 3a)`
  `= 4 · a/4 (y – 3a)`

 

`:.\ text(Focal length) = a/4`

 

iv.   `text(Equation of)\ \ P_2`

♦♦♦ Mean mark 11%.
`x^2` `= ay – 3a^2`
`y` `= x^2/a + 3a`
`y prime` `= (2x)/a`

`text(At)\ \ R,\ \ x = a (t – 1/t)`

`:.\ text(Gradient of normal at)\ \ R`

`=(-a)/(2x)`

`= (-a)/(2a(t – 1/t)) xx t/t`

`= (-t)/(2(t^2 – 1))`

 

`text(Gradient of normal at)\ \ T:`

`x + ty` `= 2at + at^3`
`y` `= -1/t x + 2a + at^2`
`:. m` `= -1/t`

 

`text(Distance)\ \ RT\ \ text(is a minimum when)`

`(-t)/(2(t^2 – 1))` `= -1/t`
`t^2` `= 2t^2 – 2`
`t^2` `= 2`
`:. t` `= +- sqrt 2`

Binomial, EXT1 2016 HSC 14b

Consider the expansion of  `(1 + x)^n`, where `n` is a positive integer.

  1. Show that  `2^n = ((n),(0)) + ((n),(1)) + ((n),(2)) + ((n),(3)) + … + ((n),(n))`.  (1 mark)
  2. Show that  `n2^(n - 1) = ((n),(1)) + 2((n),(2)) + 3((n),(3)) + … + n((n),(n))`.  (1 mark)
  3. Hence, or otherwise, show that  `sum_(r = 1)^n ((n),(r))(2r - n) = n`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text[(See Worked Solutions)}`
  2. `text(Proof)\ \ text[(See Worked Solutions)}`
  3. `text(Proof)\ \ text[(See Worked Solutions)}`
Show Worked Solution

i.     `text(Show)`

`2^n = ((n),(0)) + ((n),(1)) + ((n),(2)) + ((n),(3)) + … + ((n),(n))`

`text(Using binomial expansion)`

`(1 + x)^n = ((n), (0)) + ((n), (1))x + ((n),(2)) x^2 + … + ((n), (n)) x^n`

`text(Let)\ \ x = 1,`

`2^n = ((n), (0)) + ((n), (1)) + ((n), (2)) + … + ((n), (n))`

`text(… as required.)`

 

ii.   `text(Differentiate both sides of expansion,)`

`n (1 + x)^(n – 1) = ((n), (1)) + 2 ((n), (2))x + 3 ((n), (3)) x^2 + … + n ((n), (n)) x^(n – 1)`

`text(Let)\ \ x = 1,`

`n2^(n – 1) = ((n), (1)) + 2 ((n), (2)) + 3 ((n), (3)) + … + n ((n), (n))`

`text(… as required.)`

 

iii.  `text{Multiply part (i)} xx n`

♦♦♦ Mean mark 14%.
`n2^n` `= n[((n), (0)) + ((n), (1)) + … + ((n), (n))]`
  `= sum_(r = 0)^n ((n), (r)) n\ \ text{…  (1)}`

 

`text{Multiply part (ii)} xx 2`

`2 xx n2^(n – 1)` `= 2[((n), (1)) + 2 ((n), (2)) + … + n ((n), (n))]`
`n2^n` `= sum_(r = 1)^n ((n), (r)) 2r\ \ text{…  (2)}`

 

 `text(Subtract) qquad (2) – (1)`

`sum_(r = 1)^n ((n), (r)) 2r – sum_(r = 0)^n ((n), (r))n` `= n2^n – n2^n`
`sum_(r = 1)^n ((n), (r)) 2r – sum_(r = 1)^n ((n), (r)) n – n` `= 0`
`sum_(r = 1)^n ((n), (r)) (2r – n)` `= n\ \ text(…  as required)`

Proof, EXT1 P1 2016 HSC 14a

  1. Show that  `4n^3 + 18n^2 + 23n + 9`  can be written as
     

     

    `qquad (n + 1)(4n^2 + 14n + 9)`.  (1 marks)

  2. Using the result in part (i), or otherwise, prove by mathematical induction that, for  `n >= 1`,

     

    `qquad 1 × 3 + 3 × 5 + 5 × 7 + … + (2n - 1)(2n + 1) = 1/3 n(4n^2 + 6n - 1)`.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.   `text(RHS)` `= (n + 1) (4n^2 + 14n + 9)`
    `= 4n^3 + 14n^2 + 9n + 4n^2 + 14n + 9`
    `= 4n^3 + 18n^2 + 23n + 9`

 

ii.   `text(Prove)\ \ 1 xx 3 + 3 xx 5 + 5 xx 7 + … + (2n – 1) (2n + 1)`

`= 1/3n (4n^2 + 6n – 1)`

`text(If)\ \ n = 1,`

`text(LHS) = (1) (3) = 3`

`text(RHS) = 1/3 (1) (4 + 6 – 1) = 3`

`:.\ text(True for)\ \ n = 1`

 

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

`text(i.e.)\ \ 1 xx 3 + 3 xx 5 + … + (2k – 1) (2k + 1)`

`= 1/3 k (4k^2 + 6k – 1)`

`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ \ 1 xx 3 + … + (2k + 1) (2k + 3)`

`= 1/3 (k + 1) [4 (k + 1)^2 + 6(k + 1) – 1]`

`= 1/3 (k + 1) (4k^2 + 14k + 9)`

`= 1/3 (4k^3 + 18k^2 + 23k + 9)`

 

`text(LHS)` `= 1 xx 3 + … + (2k – 1) (2k + 1) + (2k + 1) (2k + 3)`
  `= 1/3k (4k^2 + 6k – 1) + (2k + 1) (2k + 3)`
  `= 1/3 (4k^3 + 6k^2 – k) + (4k^2 + 8k + 3)`
  `= 1/3 (4k^3 + 6k^2 – k + 12k^2 + 24k + 9)`
  `= 1/3 (4k^3 + 18k^2 + 23k + 9)`
  `=\ text(RHS …)`

 

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

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

Plane Geometry, EXT1 2016 HSC 13c

The circle centred at `O` has a diameter `AB`. From the point `M` outside the circle the line segments `MA` and `MB` are drawn meeting the circle at `C` and `D` respectively, as shown in the diagram. The chords `AD` and `BC` meet at `E`. The line segment `ME` produced meets the diameter `AB` at `F`.

ext1-2016-hsc-q13_1

Copy or trace the diagram into your writing booklet.

  1. Show that `CMDE` is a cyclic quadrilateral.  (2 marks)
  2. Hence, or otherwise, prove that `MF` is perpendicular to `AB`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.  
`/_ ACB` `= /_ ADB = 90^@ qquad text{(angle in semi-circle)}`
`/_ MCE` `= 90^@ qquad (/_ MCA\ text{is a straight angle)}`
`/_ MDA` `= 90^@ qquad (/_ MDB\ text{is a straight angle)}`

 
`:. CMDE\ text(is a cyclic quad)`

`qquad text{(opposite angles are supplementary)}`
 

ii.   `text(Consider)\ \ Delta MAB,`

♦♦♦ Mean mark 14%.

`CB\ \ text(is an altitude.)`

`AD\ \ text(is an altitude.)`

`text(S) text(ince the altitudes of a triangle are)`

`text(concurrent)\ \ text{(in}\ Delta MAB, text{at}\ Etext{),}`

`=> MF\ \ text(must be an altitude.)`

`:. MF _|_ AB`

Mechanics, EXT2* M1 2016 HSC 13b

The trajectory of a projectile fired with speed  `u\ text(ms)^-1`  at an angle  `theta`  to the horizontal is represented by the parametric equations

`x = utcostheta`   and   `y = utsintheta - 5t^2`,

where `t` is the time in seconds.

  1. Prove that the greatest height reached by the projectile is  `(u^2 sin^2 theta)/20`.  (2 marks)

A ball is thrown from a point `20\ text(m)` above the horizontal ground. It is thrown with speed `30\ text(ms)^-1` at an angle of `30^@` to the horizontal. At its highest point the ball hits a wall, as shown in the diagram.
 

     ext1-2016-hsc-q13
 

  1. Show that the ball hits the wall at a height of `125/4\ text(m)` above the ground.  (2 marks)

The ball then rebounds horizontally from the wall with speed `10\ text(ms)^-1`. You may assume that the acceleration due to gravity is `10\ text(ms)^-2`.

  1. How long does it take the ball to reach the ground after it rebounds from the wall?  (2 marks)

  2. How far from the wall is the ball when it hits the ground?  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `2.5\ text(seconds)`
  4. `25\ text(m)`
Show Worked Solution
i.    `y` `= u t sin theta – 5t^2`
  `y prime` `= u sin theta – 10t`

 

`text(Maximum height when)\ \ y prime = 0`

`10 t` `= u sin theta`
`t` `= (u sin theta)/10`

 

`:.\ text(Maximum height)`

`= u ((u sin theta)/10) · sin theta – 5 ((u sin theta)/10)^2`

`= (u^2 sin^2 theta)/10 – (u^2 sin^2 theta)/20`

`= (u^2 sin^2 theta)/20\ text(… as required)`

 

ii.   `text{Using part (i)},`

`text(Height that ball hits wall)`

`= (30^2 · (sin 30)^2)/20 + 20`

`= (30^2 · (1/2)^2)/20 + 20`

`= 11 1/4 + 20`

`= 125/4\ text(m … as required)`

 

♦♦ Mean mark part (iii) 35%.
iii.   ext1-hsc-2016-13bi
`y ″` `= -10`
`y prime` `= -10 t`
`y` `= 125/4 – 5t^2`

 

`text(Ball hits ground when)\ \ y = 0,`

MARKER’S COMMENT: Many students struggled to solve: `5t^2=125/4`.
`5t^2` `= 125/4`
`t^2` `= 25/4`
`:. t` `= 5/2,\ \ t > 0`

 

`:.\ text(It takes the ball 2.5 seconds to hit the ground.)`

 

iv.   `text(Distance from wall)`

♦ Mean mark 43%.

`= 2.5 xx 10`

`= 25\ text(m)`

Mechanics, EXT2* M1 2016 HSC 13a

The tide can be modelled using simple harmonic motion.

At a particular location, the high tide is 9 metres and the low tide is 1 metre.

At this location the tide completes 2 full periods every 25 hours.

Let `t` be the time in hours after the first high tide today.

  1. Explain why the tide can be modelled by the function  `x = 5 + 4cos ((4pi)/25 t)`.  (2 marks)

  2. The first high tide tomorrow is at 2 am.

     

    What is the earliest time tomorrow at which the tide is increasing at the fastest rate?  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `11:22:30\ text(am)`
Show Worked Solution
i.    `text(High tide)` `= 9\ text(m)`
  `text(Low tide)` `= 1\ text(m)`

 
`:. A = (9 – 1)/2 = 4\ text(m)`

`T = 25/2`

`:. (2 pi)/n` `= 25/2`
`n` `= (4 pi)/25`

 

`text(Centre of motion) = 5`

 

`text(S) text(ince high tide occurs at)\ \ t = 0,`

`x` `= 5 + 4 cos (nt)`
  `= 5 + 4 cos ((4 pi)/25 t)`

 

ii.   `x = 5 + 4 cos ((4 pi)/25 t)`

`(dx)/(dt)` `= -4 · (4 pi)/25 *sin ((4 pi)/25 t)`
  `= -(16 pi)/25 *sin ((4 pi)/25 t)`

 

`text(Tide increases at maximum rate)`

`text(when)\ \ sin ((4 pi)/25 t) = -1,`

`:. (4 pi)/25 t` `= (3 pi)/2`
  `= 75/8`
  `= 9\ text(hours 22.5 minutes)`

 

`:.\ text(Earliest time is)\ 11:22:30\ text(am)`

Polynomials, EXT1 2016 HSC 12c

The graphs of  `y = tan x`  and  `y = cos x`  meet at the point where  `x = α`, as shown.

ext1-2016-hsc-q12_1

  1. Show that the tangents to the curves at  `x = α`  are perpendicular.  (2 marks)
  2. Use one application of Newton’s method with  `x_1 = 1`  to find an approximate value for `α`. Give your answer correct to two decimal places.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `0.76\ text{(2 d.p.)}`
Show Worked Solution
i.   `text(Let)\ \ \ y_1` `= tan x`
  `(dy_1)/(dx)` `= sec^2 x`

`text(At)\ \ x = alpha,`

`m_1 = sec^2 alpha`

 

`text(Let)\ \ \ y_2` `= cos x`
`(dy_2)/(dx)` `= -sin x`
`m_2` `= -sin alpha`

 

`text(Intersection occurs when:)`

`tan alpha` `= cos alpha`
`(sin alpha)/(cos alpha)` `= cos alpha`
`:. sin alpha` `= cos^2 alpha`

 

`m_1 m_2` `= sec^2 alpha · -sin alpha`
  `= 1/(cos^2 alpha) · -sin alpha`
  `= -(sin alpha)/(sin alpha)`
  `= -1`

 

`:.\ text(T)text(angents at)\ \ x = alpha\ \ text(are perpendicular)`

 

ii.   `f(x)` `= tan x – cos x`
  `f prime (x)` `= sec^2 x + sin x`

 

`text(Let)\ \ x_1 = 1,`

`alpha` `~~ x_1 – (f(1))/(f prime (1))`
  `~~ 1 – (tan 1 – cos 1)/(1/(cos^2 1) + sin 1)`
  `~~ 1 – 0.238…`
  `~~ 0.761…`
  `~~ 0.76\ \ text{(2 d.p.)}`

Calculus, EXT1 C1 2016 HSC 12b

In a chemical reaction, a compound `X` is formed from a compound `Y`. The mass in grams of `X` and `Y` are `x(t)` and `y(t)` respectively, where `t` is the time in seconds after the start of the chemical reaction.

Throughout the reaction the sum of the two masses is 500 g. At any time `t`, the rate at which the mass of compound `X` is increasing is proportional to the mass of compound `Y`.

At the start of the chemical reaction, `x = 0`  and  `(dx)/(dt) = 2`.

  1.  Show that  `(dx)/(dt) = 0.004(500 - x)`.  (3 marks)

  2.  Show that  ` x = 500 - Ae^(−0.004t)`  satisfies the equation in part (i), and find the value of `A`.  (2 marks)

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

i.   `x + y = 500\ \ text{(given)}`

♦ Mean mark (part i) 43%.

`(dx)/(dt)` `= ky`
  `= k(500 – x)`

 
`text(When)\ \ t = 0,\ \ x = 0,\ \ (dx)/(dt) = 2`

`2` `= k (500 – 0)`
`:. k` `= 0.004`

 
`:. (dx)/(dt) = 0.004 (500 – x)\ text(… as required)`

 

ii.    `x` `= 500 – Ae^(-0.004t)`
  `Ae^(-0.004t)` `= 500 – x`

 

`(dx)/(dt)` `= 0.004 Ae^(-0.004t)`
  `= 0.004 (500 – x)`

 
`text(When)\ \ t = 0,\ \ x = 0,`

`0` `= 500 – Ae^0`
`:. A` `= 500`

Calculus, EXT1 C1 2016 HSC 12a

The diagram shows a conical soap dispenser of radius 5 cm and height 20 cm.
 

     ext1-2016-hsc-q12
 

At any time `t` seconds, the top surface of the soap in the container is a circle of radius `r` cm and its height is `h` cm.

The volume of the soap is given by  `v = 1/3 pir^2h`.

  1.  Explain why  `r = h/4`.  (1 mark)

  2.  Show that  `(dv)/(dh) = pi/16 h^2`.  (1 mark)

The dispenser has a leak which causes soap to drip from the container. The area of the circle formed by the top surface of the soap is decreasing at a constant rate of  `0.04\ text(cm² s)^-1`.
 

  1.  Show that  `(dh)/(dt) = (−0.32)/(pih)`.  (2 marks)

  2.  What is the rate of change of the volume of the soap, with respect to time, when `h = 10`?  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `-0.2\ text(cm³ s)^-1`
Show Worked Solution
i.   ext1-hsc-2016-12a

`text(Using similar triangles,)`

`r/h` `= 5/20`
 `:. r` `= h/4\ text(… as required)`

 

ii.   `v` `= 1/3 pi r^2 h`
    `= 1/3 pi · (h/4)^2 h`
    `= (pi h^3)/48`
  `:. (dv)/(dh)` `= 3 xx (pi h^2)/48`
    `= (pi h^2)/16\ text(… as required.)`

 

iii.   `(dA)/(dt)` `= -0.04\ text(cm² s)^-1`
  `A` `= pi r^2`
    `= (pi h^2)/16`
  `:. (dA)/(dh)` `= (pi h)/8`

 

`(dA)/(dt)` `= (dA)/(dh) xx (dh)/(dt)`
`-0.04` `= (pi h)/8 xx (dh)/(dt)`
`:. (dh)/(dt)` `= (-0.32)/(pi h)\ text(… as required.)`

 

iv.   `(dv)/(dt)` `= (dv)/(dh) · (dh)/(dt)`
    `= (pi h^2)/16 · (-0.32)/(pi h)`
    `= (-0.32 h)/16`

 

`text(When)\ \ h =10,`

`(dv)/(dt)` `= (-0.32 xx 10)/16`
  `= -0.2\ text(cm³ s)^-1`

Statistics, EXT1 S1 2016 HSC 11f

A darts player calculates that when she aims for the bullseye the probability of her hitting the bullseye is  `3/5`  with each throw.

  1. Find the probability that she hits the bullseye with exactly one of her first three throws.  (1 mark)

  2. Find the probability that she hits the bullseye with at least two of her first six throws.  (2 marks)

Show Answers Only
  1. `36/125`
  2. `2997/3125`
Show Worked Solution

i.   `P text{(exactly 1 bullseye)}`

`=\ ^3C_1 · (3/5)^1 (2/5)^2`

`= 3 · (3/5) · (4/25)`

`= 36/125`

 

ii.   `P text{(at least 2 from 6 throws)}`

`= 1 – [P(0) + P(1)]`

`= 1 – [(2/5)^6 + \ ^6C_1 · (3/5)^1· (2/5)^5]`

`= 1 – [128/3125]`

`= 2997/3125`

Functions, EXT1 F1 2016 HSC 11e

Solve  `3/(2x + 5) - x > 0`.  (3 marks)

Show Answers Only

`x < -3,\ \ – 5/2 < x < 1/2`

Show Worked Solution

`3/(2x + 5) – x > 0`

`3 (2x + 5) > x (2x + 5)^2,\ \ x != – 5/2`

`3 (2x + 5)-x(2x+5)^2` `> 0`
`(2x + 5) [3- x(2x + 5)]` `> 0`
`(2x + 5) (-2x^2 – 5x + 3)` `> 0`
`(2x + 5) (1-2x) (x + 3)` `> 0`

 

ext1-hsc-2016-11ei

`x < -3,\ \ \ – 5/2 < x < 1/2`

Functions, EXT1 F2 2016 HSC 10 MC

Consider the polynomial  `p(x) = ax^3 + bx^2 + cx - 6`  with `a` and `b` positive.

Which graph could represent  `p(x)`?

 

ext1-2016-hsc-10-mc-ab

ext1-2016-hsc-10-mc-cd

Show Answers Only

`A`

Show Worked Solution
`p(x)` `= ax^3 + bx^2 + cx – 6`
`p prime (x)` `= 3ax^2 + 2bx + c`
`p ″ (x)` `= 6ax + 2b`

 

`text(S) text(ince)\ \ a, b > 0,`

`text(As)\ \ x -> oo,\ \ p(x) -> oo`

`:.\ text(Eliminate)\ \ C and D.`

 

`text(POI occurs when)\ \ p ″ (x) = 0,`

`6 ax + 2b` `= 0`
 `x` `= -b/(3a) < 0,\ \ \ (a, b > 0)`

 

`:.\ text(Eliminate)\ \ B`

`=>   A`

Combinatorics, EXT1 A1 2016 HSC 8 MC

A team of 11 students is to be formed from a group of 18 students. Among the 18 students are 3 students who are left-handed.

What is the number of possible teams containing at least 1 student who is left-handed?

  1. `19\ 448`
  2. `30\ 459`
  3. `31\ 824`
  4. `58\ 344`
Show Answers Only

`B`

Show Worked Solution

`text(Teams with at least 1 left-hander)`

`=\ ^18C_11 -\ ^15C_11`

`= 30\ 459`
 

`=>   B`

Mechanics, EXT2* M1 2016 HSC 7 MC

The displacement `x` of a particle at time `t` is given by

`x = 5 sin 4t + 12 cos 4t`.

What is the maximum velocity of the particle?

  1. `13`
  2. `28`
  3. `52`
  4. `68`
Show Answers Only

`C`

Show Worked Solution

`x = 5 sin 4t + 12 cos 4t`

`(dx)/(dt) = 20 cos 4t – 48 sin 4t`

 `=>\ text(Can be written in the form:)`

`A cos (4t + alpha),\ text(where)`

`A` `= sqrt (20^2 + 48^2)`
  `= 52`

 
`:. text(Max)\ v = 52\ text(ms)^-1`

`=>   C`

Plane Geometry, EXT1 2016 HSC 4 MC

In the diagram, `O` is the centre of the circle `ABC`, `D` is the midpoint of `BC`, `AT` is the tangent at `A` and  `∠ATB = 40^@`.
 

What is the size of the reflex angle `DOA`?

  1. `80^@`
  2. `140^@`
  3. `220^@`
  4. `280^@`
Show Answers Only

`C`

Show Worked Solution

 

`/_ ODT` `=90^@\ \ text{(line through centre bisecting chord)}`
`/_OAT` `= 90^@\ \ text{(tangent ⊥ to radius at point of contact)}`
`/_ DOA` `= 360-(90 + 90 + 40)`
  `= 140^@`

 

`:. DOA\ \ text{(reflex)}` `= 360-140`
  `= 220^@`

`=>   C`

Calculus, 2ADV C3 2016 HSC 16b

Some yabbies are introduced into a small dam. The size of the population, `y`, of yabbies can be modelled by the function

`y = 200/(1 + 19e^(-0.5t)),`

where `t` is the time in months after the yabbies are introduced into the dam.

  1. Show that the rate of growth of the size of the population is
  2. `qquad qquad (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2`.  (2 marks)

  3. Find the range of the function `y`, justifying your answer.  (2 marks)

  4. Show that the rate of growth of the size of the population can be written as
  5. `qquad qquad y/400 (200-y)`.  (1 mark)

  6. Hence, find the size of the population when it is growing at its fastest rate.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `10 <= y < 200`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `100`
Show Worked Solution
i.   `y` `= 200/(1 + 19 e^(-0.5t))`
  `(dy)/(dt)` `= 200/(1 + 19 e^(-0.5t))^2 xx d/(dt) (1 + 19 e^(-0.5t))`
    `= (-200)/(1 + 19 e^(-0.5t))^2 xx -0.5 xx 19 e^(-0.5t)`
    `= (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2\ \ text(… as required)`

 

ii.   `text(When)\ \ t = 0,`

♦♦♦ Mean mark (ii) 21%.

`y = 200/(1 + 19) = 10`

`text(As)\ \ t -> oo,\ \ (1 + 19^(-0.5t)) -> 1`

`:. y -> 200`

`:.\ text(Range)\ \ \ 10 <= y < 200`

 

iii.  `(dy)/(dt) = (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2`

♦♦♦ Mean mark (iii) 18%.

`text(S) text(ince)\ \ y = 200/(1 + 19 e^(-0.5t))`

`=> (1 + 19 e^(-0.5t)) = 200/y`

`=> 19 e^(-0.5t) = 200/y-1 = (200-y)/y`

`text(Substituting into)\ \ (dy)/(dt):`

`(dy)/(dt)` `= (100 ((200-y)/y))/(200/y)^2`
  `= 100 ((200-y)/y) xx y^2/200^2`
  `= y/400 (200-y)\ \ text(… as required)`

 

iv.   `(dy)/(dt) = -y^2/400 + y/2`

♦♦♦ Mean mark (iv) 14%.

`text(Sketching the parabola:)`

`(-y^2)/400 + y/2` `= 0`
`-y^2 + 200y` `= 0`
`y (200-y)` `= 0`

 

hsc-2016-16bi

`:.\ text(Maximum)\ \ (dy)/(dt)\ \ text(occurs when)\ \ y = 100.`

Calculus, 2ADV C4 2016 HSC 16a

A particle moves in a straight line. Its velocity `v\ text(ms)^-1` at time `t` seconds is given by

`v = 2 - 4/(t + 1).`

  1. Find the initial velocity.  (1 mark)

  2. Find the acceleration of the particle when the particle is stationary.  (2 marks)

  3. By considering the behaviour of `v` for large `t`, sketch a graph of `v` against `t` for  `t >= 0`, showing any intercepts.  (2 marks)

  4. Find the exact distance travelled by the particle in the first 7 seconds.  (3 marks)

Show Answers Only
  1. `-2\ text(ms)^-1`
  2. `1\ text(ms)^-2`
  3.  
    hsc-2016-16ai
  4. `10 – 4 ln 2\ \ text(metres)`
Show Worked Solution

i.   `text(Initial velocity when)\ \ t = 0`

Mean mark 93%.
COMMENT: Great example of low hanging fruit late in the exam for students who progress efficiently.

`v = 2 – 4/1 = -2\ text(ms)^-1`

 

ii.   `v` `= 2 – 4/(t + 1)`
  `a` `=(dv)/(dt)= 4/(t + 1)^2`

 

`text(Particle is stationary when)\ \ v = 0,`

`2 – 4/(t + 1)` `= 0`
`2 (t + 1)` `= 4`
`t` `= 1`
   

`text(When)\ \ t=1,`

`:.a` `= 4/(1 + 1)^2`
  `= 1\ text(ms)^-2` 

 

iii.  `v = 2 – 4/(t + 1)`

♦ Mean mark 47%.

`text(As)\ \ t -> oo,\ \ \ 4/(t + 1) -> 0`

`:. v -> 2`

 hsc-2016-16ai

 

♦♦ Mean mark 26%.

iv.   `text(Distance travelled in 1st 7 seconds)`

`= |\ int_0^1 (2 – 4/(t + 1))\ dt\ | + int_1^7 (2 – 4/(t + 1))\ dt`

`= -[2t – 4 ln (t + 1)]_0^1 + [2t – 4 ln (t + 1)]_1^7`

`= -[(2 – 4 ln 2) – 0] + [(14 – 4 ln 8) – (2 – 4 ln 2)]`

`= 4 ln 2 – 2 + 12 – 4 ln 2^3 + 4 ln 2`

`= 10 + 8 ln 2 – 12 ln 2`

`= 10 – 4 ln 2\ \ text(metres)`

Plane Geometry, 2UA 2016 HSC 15c

Maryam wishes to estimate the height, `h` metres, of a tower, `ST`, using a square, `ABCD,` with side length `1` metre.

She places the point `A` on the horizontal ground and ensures that the point `D` lies on the line joining `A` to the top of the tower `T.` The point `F` is the intersection of the line joining `B` and `T` and the side `CD.` The point `E` is the foot of the perpendicular from `B` to the ground. Let `CF` have length `x` metres and `AE` have length `y` metres.

hsc-2016-15c

Copy or trace the diagram into your writing booklet.

  1. Show that `Delta FCB` and `Delta BAT` are similar.  (2 marks)
  2. Show that `Delta TSA` and `Delta AEB` are similar.  (2 marks)
  3. Find `h` in terms of `x` and `y`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `y/x`
Show Worked Solution
i.  

`/_ BAT = /_ CBA = 90^@`

`text(Let)\ \ /_ TBA = theta`

`:. /_ ATB = 90 – theta\ \ text{(Angle sum of}\ Delta BAT text{)}`

`/_ CBF = 90 – theta\ \ (/_ CBA\ \ text{is a right angle)}`

 

`/_ FCB = /_ BAT = 90^@`

`:. Delta FCB\ text(|||)\ Delta BAT\ \ text{(equiangular)}`

 

ii.   `text(Let)\ \ /_ BAE = alpha`

♦♦ Mean mark 27%.

`text(In)\ \ Delta TSA,`

`/_ TAS` `= 180 – (90 + alpha) qquad (/_ SAE\ \ text{is a straight angle)}`
  `= 90 – alpha`

 

`text(In)\ \ Delta AEB,`

`/_ EBA = 90 – alpha\ \ text{(angle sum of}\ Delta AEB text{)}`

`/_ BEA = /_ TSA = 90^@\ \ text{(given)}`

 

`:. Delta TSA\ text(|||)\ Delta AEB\ \ text{(equiangular)}`

 

iii.   `text(Using)\ \ Delta TSA\ text(|||)\ Delta AEB,`

♦♦ Mean mark 27%.
`h/(TA)` `= y/(AB)` `text{(corresponding sides of}`
  `text{similar triangles)}`
`:. h` `= y · (TA)/(AB)`  

 

 `text(Using)\ \ Delta FCB\ text(|||)\ Delta BAT,`

`1/x` `= (TA)/(AB)` `qquad qquad text{(corresponding sides of}`
   `qquad qquad text{similar triangles)}`
`:. h` `= y/x`  

Probability, 2ADV S1 2016 HSC 15b

An eight- sided die is marked with numbers  1, 2, … , 8. A game is played by rolling the die until an 8 appears on the uppermost face. At this point the game ends.

  1. Using a tree diagram, or otherwise, explain why the probability of the game ending before the fourth roll is

      

    `qquad qquad 1/8 + 7/8 xx 1/8 + (7/8)^2 xx 1/8`.  (2 marks)

  2. What is the smallest value of `n` for which the probability of the game ending before the `n`th roll is more than  `3/4`?  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `12`
Show Worked Solution

i.   `P text{(game ends before 4th roll)}`

`= P (8) + P (text{not}\ 8, 8) + P (text{not}\ 8, text{not}\ 8, 8)`

`= 1/8 + 7/8 · 1/8 + 7/8 · 7/8 · 1/8`

`= 1/8 + 7/8 · 1/8 + (7/8)^2 · 1/8\ \ text(…  as required)`

 

ii.  `1/8 + 7/8 · 1/8 + (7/8)^2 · 1/8 + …`

`=> text(GP where)\ \ a = 1/8,\ \ r = 7/8`

`text(Find)\ \ n\ \ text(such that)\ \ S_(n – 1) > 3/4,`

♦♦ Mean mark 29%.
ALGEBRA: Note that dividing by `ln\ 7/8` reverses the < sign as it is dividing by a negative number.
`S_(n-1)` `= (a (1 – r^(n – 1)))/(1 – r)`
`3/4` `< 1/8 xx {(1 – (7/8)^(n – 1))}/(1 – 7/8)`
`3/4` `< 1 – (7/8)^(n – 1)`
`(7/8)^(n – 1)` `< 1/4`
`(n-1)* ln\ 7/8` `< ln\ 1/4`
`n – 1` `> (ln\ 1/4)/(ln\ 7/8)`
  `> 11.38…`

`:. n = 12`

Calculus, 2ADV C3 2016 HSC 14c

A farmer wishes to make a rectangular enclosure of area 720 m². She uses an existing straight boundary as one side of the enclosure. She uses wire fencing for the remaining three sides and also to divide the enclosure into four equal rectangular areas of width `x` m as shown.
 

hsc-2016-14c
 

  1. Show that the total length, `l` m, of the wire fencing is given by
     
          `l = 5x + 720/x`.  (1 mark)

  2. Find the minimum length of wire fencing required, showing why this is the minimum length.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `120\ text(m)`
Show Worked Solution
i.   `text(Area)` `= xy`
  `720` `= xy`
  `y` `= 720/x`

 

`l` `= 5x + y`
  `= 5x + 720/x\ \ text(… as required)`

 

ii.  `(dl)/(dx)` `= 5 – 720/x^2`
  `(d^2l)/(dx^2)` `= 1440/x^3`

 

`text(Max/Min when)\ \ (dl)/(dx) = 0,`

`5` `= 720/x^2`
`x^2` `= 144`
`x` `= 12,\ \ x > 0`

 

`text(When)\ \ x = 12,\ \ (d^2l)/(dx^2) > 0`

`:.\ text(Minimum occurs when)\ \ x = 12`

`:.\ text(Minimum fencing)`

`= 5 xx 12 + 720/12`

`= 120\ text(m)`