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Algebra, NAP-I3-NC02

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(be added to an existing even number.)`

`text(even + even = even)`

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

Geometry, NAP-F4-CA02

The distance between Newcastle and Canberra on the map below is 9 cm.
 

 
The scale of the map is  1 cm = 50 km.

What is the actual distance between Newcastle and Canberra?

`350\ text(km)` `450\ text(km)` `600\ text(km)` `900\ text(km)`
 
 
 
 
Show Answers Only

`450\ text(km)`

Show Worked Solution
`text(Distance)` `= 9 xx 50`
  `= 450\ text(km)`

Statistics, NAP-G4-CA03

10 of the tallest mountains in the United States are listed in the table below:
 

 

How much taller than Utah's tallest mountain is Colorado's tallest mountain?

`225\ text(metres)` `227\ text(metres)` `273\ text(metres)` `330\ text(metres)`
 
 
 
 
Show Answers Only

`273\ text(metres)`

Show Worked Solution

`text(Mount Massive – Kings Peak)`

`= 4398 – 4125`

`= 273\ text(metres)`

Probability, NAP-G4-CA02

Arun flips an unbiased coin 200 times.

Which result is most likely?

`20\ text(tails)` `98\ text(tails)` `108\ text(tails)` `196\ text(tails)`
 
 
 
 
Show Answers Only

`98\ text(tails)`

Show Worked Solution

`text(The expected result is 100 tails.)`

`:.\ text(The most likely result is the one closest to 100 tails.)`

`=> 98\ text(tails)`

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`

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)`

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`

Geometry and Calculus, EXT1 2016 HSC 9 MC

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

ext1-2016-hsc-9-mc

Which of the following is a correct statement?

A.   `f″(1)` `< f(1)` `<quad1` `< f prime (1)`
B.   `f″(1)` `< f prime(1)` `< f(1)` `<quad1`
C.   `f(1)` `<quad1` `< f prime (1)` `< f″(1)`
D.   `f prime(1)` `< f(1)` `<quad1` `< f″(1)`
Show Answers Only

`A`

Show Worked Solution

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

`f(1)` `< 1`
`f prime (1)` `> 1\ \ text{(graph slope}^+\ text{> 45°)}`
`f ″ (1)` `< 1\ \ text{(concave down)}`

`=>   A`

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, EXT1* C3 2016 HSC 15a

The diagram shows two curves  `C_1` and `C_2.` The curve `C_1` is the semicircle  `x^2 + y^2 = 4, \ -2 <= x <= 0.` The curve `C_2` has equation  `x^2/9 + y^2/4 = 1, \ 0 <= x <= 3.`
 

hsc-2016-15a
 

An egg is modelled by rotating the curves about the `x`-axis to form a solid of revolution.

Find the exact value of the volume of the solid of revolution.  (4 marks)

Show Answers Only

`(40 pi)/3\ text(u³)`

Show Worked Solution

`text(Consider)\ \ C_1,`

`V_1` `= pi int_-2^0 y^2\ dx`
  `= pi int_-2^0 4 – x^2\ dx`
  `= pi [4x – x^3/3]_-2^0`
  `= pi [0 – (-8 + 8/3)]`
  `= (16 pi)/3\ u³`

 

`text(Consider)\ \ C_2`

`x^2/9 + y^2/4` `= 1`
`y^2` `= 4 – (4x^2)/9`

 

`V_2` `= pi int_0^3 4 – (4x^2)/9\ dx`
  `= pi [4x – (4x^3)/27]_0^3`
  `= pi [(12 – (4 · 3^3)/27) – 0]`
  `= 8 pi\ text(u³)`

 

`text(Volume)` `= V_1 + V_2`
  `= (16 pi)/3 + 8 pi`
  `= (40 pi)/3\ text(u³)`

L&E, 2ADV E1 2016 HSC 14e

Write  `log 2 + log 4 + log 8 + … + log 512`  in the form  `a log b`  where `a` and `b` are integers greater than `1.`  (2 marks)

Show Answers Only

`45 log 2`

Show Worked Solution

`log 2 + log 4 + log 8 + … + log 512`

`= log 2^1 + log 2^2 + log2^3 + … + log 2^9`

`= log 2 + 2 log 2 + 3 log 2 + … + 9 log 2`

`= 45 log 2`

♦ Mean mark 40%.
 
TIP: Note that `log 2 = log_10 2`

Financial Maths, 2ADV M1 2016 HSC 14d

By summing the geometric series  `1 + x + x^2 + x^3 + x^4`, or otherwise,

find  `lim_(x -> 1) (x^5 - 1)/(x - 1).`  (2 marks)

Show Answers Only

`5`

Show Worked Solution

`1 + x + x^2 + x^3 + x^4`

♦ Mean mark 45%.

`=> text(GP where)\ \ a = 1,\ \ r = x,\ \ n = 5`

`S_n` `= (a(r^n – 1))/(r – 1)`
 `S_5` `= ((x^5 – 1))/(x – 1)`

 
`:. lim_(x -> 1) (x^5 – 1)/(x – 1)`

`= lim_(x -> 1) (1 + x + x^2 + x^3 + x^4)`

`= 5`

Proof, EXT2 P2 2016 HSC 16c

In a group of `n` people, each has one hat, giving a total of `n` different hats. They place their hats on a table. Later, each person picks up a hat, not necessarily their own.

A situation in which none of the `n` people picks up their own hat is called a derangement.

Let  `D(n)`  be the number of possible derangements.

  1. Tom is one of the `n` people. In some derangements Tom finds that he and one other person have each other's hat.

     

    Show that, for  `n > 2`, the number of such derangements is  `(n - 1) D (n - 2).`  (1 mark)

  2. By also considering the remaining possible derangements, show that, for  `n > 2,`

    `qquad qquad D(n) = (n - 1) [D(n - 1) + D(n - 2)].`  (2 marks)

  3. Hence, show that  `D(n) - nD(n - 1) = -[D(n - 1) - (n - 1) D(n - 2)]`,  for  `n > 2.`  (1 mark)

  4. Given  `D(1) = 0`  and  `D(2) = 1`,  deduce that  `D(n) - n D(n - 1) = (-1)^n`,  for  `n > 1.`  (1 mark)

  5. Prove by mathematical induction, or otherwise, that for all integers  `n >= 1,\ D(n) = n! sum_(r = 0)^n (-1)^r/(r!).`  (2 marks)

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

i.   `text(Tom and one other person can have each)`

♦♦♦ Mean mark 22%.

`text(other’s hat in)\ (n – 1)\ text(combinations.)`

`text(There are)\ \ D(n – 2)\ \ text(possibilities for the)`

`text(rest of the people selecting the wrong hats.)`

`:. text(Number of derangements) = (n – 1)D(n – 2)`

 

ii.   `text(Consider all possible derangements:)`

♦♦♦ Mean mark 6%.

`text(Any of)\ n\ text(people choose the wrong hat.)`

`text(Remainder)\ (n – 1)\ text(people can select the)`

`text(wrong hat in)\ D(n – 1)\ text(ways.)`

`:. nD(n – 1)\ text(derangements.)`

`text{This includes part (i) combinations.}`

 

`:.\ text(Remaining possible derangements)`

`= nD(n – 1) – (n – 1)D(n – 2)`

`= nD(n – 1) – D(n – 1)`

`= (n – 1)D(n – 1)`

`:. D(n)` `= (n – 1)D(n – 1) + (n – 1)D(n – 2)`
  `= (n – 1)[D(n – 1) + D(n – 2)]`

 

♦♦ Mean mark part (iii) 37%.
iii.    `D(n)` `= (n – 1)[D(n – 1) + D(n – 2)]`
    `= nD(n – 1) – D(n – 1) + (n – 1)D(n – 2)`

 

`:. Dn – nD(n – 1) = −[D(n – 1) – (n – 1)D(n – 2)]`

 

iv.   `D(1) = 0, D(2) = 1`

♦♦♦ Mean mark 1%!

`D(2) – 2D(1) = 1`

`D(3) – 3D(2) = −[D(2) – 2D(1)] = −1`

`D(4) – 4D(3) = −[D(3) – 3D(2)] = −(−1) = 1`

`D(5) – 5D(4) = −[D(4) – 4D(3)] = −1`

 

`:. D(n) – nD(n – 1) = (−1)^n\ text(for)\ n > 1`

 

v.   `text(Prove)\ D(n) = n! sum_(r = 0)^n ((−1)^r)/(r!)\ text(for)\ n >= 1`

`text(When)\ n = 1,`

`D(1) = 1! sum_(r = 0)^1 ((−1)^r)/(r!) = 1 – 1 = 0`

`text(S)text(ince)\ D(1) = 0\ (text(given)),`

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

♦♦♦ Mean mark 13%.

 

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

`text(i.e.)\ \ D(k) = k! sum_(r = 0)^k ((−1)^r)/(r!)`

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

`text(i.e.)\ \ D(k+1) = (k+1)!sum_(r = 0)^(k+1) ((−1)^r)/(r!)`

`D(k+1)`

`= (k + 1)D(k) + (−1)^(k + 1)qquad(text{from part (iv)})`
`= (k + 1) · k! sum_(r = 0)^k ((−1)^r)/(r!) + (−1)^(k + 1)`
`= (k + 1)!(1 – 1/(1!) + 1/(2!) – 1/(3!) + … + ((−1)^k)/(k!)) + (−1)^(k + 1) · ((k + 1)!)/((k + 1)!)`
`= (k + 1)!(1 – 1/(1!) + 1/(2!) – 1/(3!) + … + ((−1)^k)/(k!) + ((−1)^(k+1))/((k+1)!))`
`= (k + 1)! sum_(r = 0)^(k + 1) ((−1)^r)/(r!)`

 

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

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

Complex Numbers, EXT2 N2 2016 HSC 16b

  1. The complex numbers  `0, \ u`  and  `v`  form the vertices of an equilateral triangle in the Argand diagram.

     

    Show that  `u^2 + v^2 = uv.`  (2 marks)

  2. Give an example of non-zero complex numbers  `u`  and  `v`, so that  `0, \ u`  and  `v`  form the vertices of an equilateral triangle in the Argand diagram.  (1 mark)

Show Answers Only
  1. `text(Show Worked Solutions)`
  2. `u = 1/2 + sqrt3/2 i`
Show Worked Solution

i.   `0,u,v\ text(are vertices of an equilateral triangle.)`

♦♦ Mean mark 23%.

 

ext2-2016-hsc-16b-answer

`=> u` `= v text(cis)(±pi/3)`
`u^3` `= v^3text(cis)(±pi)`
`u^3` `= − v^3`
`u^3 + v^3` `= 0`

 
`(u + v)(u^2 – uv + v^2) = 0`

`text(S)text(ince)\ u != v,`

`u^2 – uv + v^2` `= 0`
`:. u^2 + v^2` `= uv`

 

ii.   `text(Let)\ \ v = 1,`

♦♦ Mean mark 33%.
`:. u` `= text(cis)(pi/3)`
  `= cos\ pi/3 + isin\ pi/3`
  `= 1/2 + sqrt3/2 i`

Mechanics, EXT2 M1 2016 HSC 15b

A particle is initially at rest at the point `B` which is `b` metres to the right of `O.`

The particle then moves in a straight line towards `O.`

For `x != 0,` the acceleration of the particle is given by  `(- mu^2)/x^2,`  where `x` is the distance from `O` and `mu` is a positive constant.

  1. Prove that  `(dx)/(dt) = -mu sqrt 2 sqrt((b - x)/(bx)).`  (2 marks)

  2. Using the substitution  `x = b cos^2 theta,` show that the time taken to reach a distance `d` metres to the right of `O` is given by
     
         `t = (b sqrt (2b))/mu int_0^(cos^-1 sqrt (d/b)) cos^2 theta\ d theta.`  (3 marks)

It can be shown that   `t = 1/mu sqrt (b/2) (sqrt(bd - d^2) + b cos^-1 sqrt (d/b)).`  (Do NOT prove this.)

  1. What is the limiting time taken for the particle to reach `O?`  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `(pi(sqrtb)^3)/(2sqrt2mu)\ text(seconds)`
Show Worked Solution

i.   `a = d/dx(1/2 v^2) = −(mu^2)/(x^2)`

`:. 1/2 v^2` `= int −(mu^2)/(x^2)\ dx`
  `= (mu^2)/x + c`

 

`text(Initially,)\ v = 0\ text(and)\ x = b`

`:. c = −(mu^2)/b`

`v^2` `= 2mu^2(1/x – 1/b)`
`v` `= −musqrt2 · sqrt(1/x – 1/b)qquad(text(negative since moving to left))`
  `= −musqrt2 · sqrt((b – x)/(bx))\ …\ text(as required.)`

 

ii.    `dx/dt` `= −musqrt2 · sqrt((b – x)/(bx))`
  `dt/dx` `= −1/(musqrt2) · sqrt((bx)/(b – x))`
  `int_0^t dt` `= −1/(musqrt2) · int_b^d sqrt((bx)/(b – x))\ dx`

 

`text(Integration by substitution:)`

`text(Let)\ \ \ x` `= bcos^2theta`
`dx` `= −2bcosthetasintheta\ d theta`

 

`text(When)quadx` `= b,` `theta` `= 0`
`x` `= d,` `theta` `= cos^(−1)sqrt(d/b)`

 

`:. t` `= −1/(musqrt2) · int_0^(cos^(−1)sqrt(d/b))sqrt((b^2cos^2theta)/(b(1 – cos^2theta))) · −2bcosthetasintheta\ d theta`
 

`= (2b)/(musqrt2) · int_0^(cos^(−1)sqrt(d/b))(sqrtb costheta)/(sintheta) · costhetasintheta\ d theta`
 

`= (bsqrt(2b))/mu int_0^(cos^(−1)sqrt(d/b)) cos^2theta\ d theta\ …\ text(as required)`

 

iii.   `t = 1/mu sqrt(b/2)(sqrt(bd – d^2) + bcos^(−1)sqrt(d/b))`

 

`text(As)\ \ d->0,`

♦♦ Mean mark 29%.
`t` `= 1/mu sqrt(b/2) (sqrt0 + bcos^(−1)0)`
  `= 1/mu sqrt(b/2) · b · pi/2`
  `= (pi(sqrtb)^3)/(2sqrt2mu)\ text(seconds)`

Proof, EXT2 P1 2016 HSC 14c

Show that  `x sqrt x + 1 >= x + sqrt x,`  for  `x >= 0.`  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(Show)\ \ xsqrtx + 1 >= x + sqrtx,quadtext(for)\ x >= 0`

♦ Mean mark 39%.
`(text(LHS))^2` `= x^3 + 2xsqrtx + 1`
  `= 2xsqrtx + (x + 1)(x^2 – x + 1)`

 

`(x – 1)^2` `>= 0`
`x^2 – 2x + 1` `>= 0`
`:. x^2 + 1` `>= 2x`

 

`:.\ (text(LHS))^2` `>= 2xsqrtx + (x + 1)(2x – x)`
  `>= 2xsqrtx + x(x + 1)`
  `>= x^2 +2xsqrtx+x`
  `>= (x + sqrtx)^2`
  `>= (text(RHS))^2`

 

`:.\ text(LHS ≥ RHS for)\ \ x >= 0`

Calculus, EXT2 C1 2016 HSC 14b

Let  `I_n = int_0^1 x^n/(x^2 + 1)^2\ dx,`  for   `n = 0, 1, 2, … .`

  1. Using a suitable substitution, show that  `I_0 = pi/8 + 1/4.`  (3 marks)

  2. Show that  `I_0 + I_2 = pi/4.`  (1 mark)

  3. Find  `I_4.`  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `5/4 – (3pi)/8`
Show Worked Solution

i.   `I_n = int_0^1 (x^n)/((x^2 + 1)^2) dx,quadn >= 0`

`text(Let)\ \ \ x` `= tantheta`
`dx` `= sec^2theta\ d theta`

 

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

 

`:. I_0` `= int_0^(pi/4)(sec^2theta)/((tan^2theta + 1)^2)\ d theta`
  `= int_0^(pi/4)(sec^2theta)/(sec^4theta)\ d theta`
  `= int_0^(pi/4)cos^2theta\ d theta`
  `= int_0^(pi/4)1/2(1 + cos2theta)\ d theta`
  `= 1/2[x + 1/2sin2theta]_0^(pi/4)`
  `= 1/2[(pi/4 + 1/2 · sin\ pi/2) – 0]`
  `= pi/8 + 1/4`

 

ii.    `I_0 + I_2` `= int_0^1 1/((x^2 + 1)^2)\ dx + int_0^1 (x^2)/((x^2 + 1)^2)\ dx`
    `= int_0^1 (x^2 + 1)/((x^2 + 1)^2)\ dx`
    `= int_0^1 1/(x^2 + 1)\ dx`
    `= [tan^(−1)x]_0^1`
    `= pi/4`

 

♦ Mean mark part (iii) 49%.
iii.    `I_4` `= int_0^1 (x^4)/(x^2 + 1)\ dx`
    `= int_0^1 (x^4 – 1)/((x^2 + 1)^2)\ dx + int_0^1 1/((x^2 + 1)^2)`
    `= int_0^1 ((x^2 + 1)(x^2 – 1))/((x^2 + 1)^2)\ dx + I_0`
    `= int_0^1 (x^2 – 1)/(x^2 + 1)\ dx + I_0`
    `= int_0^1 (1 – 2/(x^2 + 1))\ dx + I_0`
    `= [x – 2tan^(−1)x]_0^1 + pi/8 + 1/4`
    `= [(1 – 2 · pi/4) – 0] + pi/8 + 1/4`
    `= 5/4 – (3pi)/8`

Statistics, STD2 S5 2016 HSC 30d

The formula to calculate  `z`-scores can be rearranged to give

`mu = x - σz`

 

where    `mu`  is the mean
  `x`  is the score
  `σ`  is the standard deviation
  `z`  is the `z`-score
   
  1. In an examination, Aaron achieved a score of 88, which corresponds to a  `z`-score of 2.4.

     

    Substitute these values into the rearranged formula above to form an equation.  (1 mark)

  2. In the same examination, Brock achieved a score of 52, which corresponds to a  `z`-score of  –1.2.

     

    Using this information, form another equation and solve it simultaneously with the equation from part (i) to find the values of  `mu`  and  `σ`.  (3 marks)

Show Answers Only
  1. `mu = 88 – 2.4σ`
  2. `64`
Show Worked Solution

i.   `mu = 88 – 2.4σ`
 

ii.   `mu = 52 + 1.2σ\ …\ (1)`

♦♦ Mean mark part (ii) 32%.
COMMENT: Recognise this is a simultaneous equation problem (now outside Std2 syllabus) and not a normal distribution one.

`mu = 88 – 2.4σ\ …\ (2)`
 

`text(Subtract)\ \ (2) – (1):`

`0` `= 36 – 3.6σ`
`3.6σ` `= 36`
`:. σ` `= 10`

 
`text(Substitute)\ \ σ = 10\ \ text(into)\ (1):`

`:. mu` `= 52 + 1.2 xx 10`
  `= 64`

FS Comm, 2UG 2016 HSC 30b

Michael was transferring some video files from his computer onto a USB stick. At some point during the transfer, he observed the information shown below. 

2ug-2016-hsc-q30_1

  1. Show that, at that time, approximately `3072` MB of data remained to be transferred.  (1 mark)
  2. Calculate the speed required to transfer `3072` MB in `7` minutes. Give your answer in megabits per second (Mbps), correct to the nearest whole number. (Note that `1` megabit = `1\ 000\ 000` bits.)  (3 marks)

 

Show Answers Only
  1. `3072\ text(MB)`
  2. `61\ text{Mbps  (nearest whole)}`
Show Worked Solution

(i)   `text(Data to be transferred)`

`= 6.44 – 3.44`

`= 3\ text(GB)`

`= 3 xx 2^10\ text(MB)`

`= 3072\ text(MB)`

 

(ii)   `text(7 min = 7 × 60 = 420 seconds)`

♦ Mean mark part (ii) 35%.

`text(Convert MB to Megabits:)`

`3072\ text(MB)` `= 3072 xx 2^20 xx 8`
  `= 2.576… xx 10^10\ text(bits)`
  `= 25\ 769.8…\ text(Mb)`

 

`:.\ text(Download speed required)`

`= (25\ 769.8…)/420`

`= 61.35…`

`= 61\ text{Mbps  (nearest whole)}`

Measurement, STD2 M1 2016 HSC 30a

The area of a roof is 30 m². Any rain that falls on the roof flows directly onto a garden.

Calculate how many litres of water flow onto the garden when 20 mm of rain falls on the roof.  (2 marks)

Show Answers Only

`600\ text(L)`

Show Worked Solution
♦♦ Mean Mark 27%.
STRATEGY: Converting 20 mm into metres as a 1st step is an efficient approach here.
`text(Volume)` `= Ah`
  `= 30 xx 20/1000`
  `= 0.6\ text(m³)`
  `= 600\ text(L)`

Algebra, STD2 A2 2016 HSC 29e

The graph shows the life expectancy of people born between 1900 and 2000.
 


  1. According to the graph, what is the life expectancy of a person born in 1932?  (1 mark)

  2. With reference to the value of the gradient, explain the meaning of the gradient in this context.  (2 marks)

Show Answers Only
  1. `text(68 years)`
  2. `text(After 1900, life expectancy increases 0.25 years for each later year someone is born.)`
Show Worked Solution

i.    \(\text{68 years}\)

ii.    \(\text{Using (1900,60), (1980,80):}\)

\(\text{Gradient}\) \(= \dfrac{y_2-y_1}{x_2-x_1}\)
  \(= \dfrac{80-60}{1980-1900}\)
  \(= 0.25\)

 
\(\text{After 1900, life expectancy increases by 0.25 years for}\)

\(\text{each year later that someone is born.}\)

♦♦ Mean mark (ii) 33%.

FS Health, 2UG 2016 HSC 29d

Five students sat both a Physics and a Chemistry exam. Their results are shown in the table. The mean and standard deviation of each exam are also shown.
 

2ug-2016-hsc-q29_3
 

The correlation coefficient for this data set is approximately 0.9.

  1. Verify the value of the correlation coefficient, using your calculator, and give your value correct to three decimal places.  (1 mark)
  2. By using the appropriate formulae from the Formulae and Data Sheet, and the given information, determine the equation of the least-squares line of best fit.  (3 marks)

 

Show Answers Only
  1. `0.907`
  2. `y = 1.376x – 29.494`
Show Worked Solution

(i)   `r ~~ 0.907`

♦♦ Mean mark part (i) 31%

 

(ii)    `text(gradient)` `= r xx (text(std dev)(y))/(text(std dev)(x))`
    `= 0.907 xx 13.73/9.05`
    `= 1.376\ \ (text(3 d.p.))`
♦ Mean mark part (ii) 48%

 

`ytext(-intercept)` `= bary – (text(gradient) xx barx)`
  `= 66 – 1.376 xx 69.4`
  `= −29.494\ \ (text(3 d.p.))`

 

`:.\ text(Equation:)\ \ y = 1.376x – 29.494`

Statistics, STD2 S1 2016 HSC 29c

The ages of members of a dance class are shown in the back-to-back stem-and-leaf plot.
 

2ug-2016-hsc-q29_2
 

Pat claims that the women who attend the dance class are generally older than the men.

Is Pat correct? Justify your answer by referring to the median and skewness of the two sets of data.  (3 marks)

Show Answers Only

`text(Women:)`

`text(The median is 55 in a data)`

`text(set that is negatively skewed.)`

`text(Men:)`

`text(The median is 45 in a data)`

`text(set that is positively skewed.)`

`:.\ text(Pat is correct.)`

Show Worked Solution

`text(Women:)`

♦ Mean mark 44%.

`text(The median is 55 in a data)`

`text(set that is negatively skewed.)`

`text(Men:)`

`text(The median is 45 in a data)`

`text(set that is positively skewed.)`

`:.\ text(Pat is correct.)`

Functions, EXT1′ F2 2016 HSC 13d

Suppose  `p(x) = ax^3 + bx^2 + cx + d`  with `a, b, c` and `d` real, `a != 0.`

  1. Deduce that if  `b^2 - 3ac < 0`  then `p(x)`  cuts the `x`-axis only once.  (2 marks)

  2. If  `b^2 - 3ac = 0 and p(-b/(3a)) = 0`, what is the multiplicity of the root  `x = -b/(3a)?`  (2 marks)

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

i.     `p(x) = ax^3 + bx^2 + cx + d`

`pprime(x) = 3ax^2 + 2bx + c`

`=> p(x)\ text(will cut the)\ xtext(-axis once)`

`text(only if)\ \ Delta(pprime(x)) < 0`

`(2b)^2 – 4(3a)c` `< 0`
`4b^2 – 12ac` `< 0`
`b^2 – 3ac` `< 0`

 

ii.   `p(−b/(3a)) = 0`

`pprime(−b/(3a))` `=3a(- b/(3a))^2 + 2b (- b/(3a))+c`
  `=- b^2/(3a)+c`
  `=0\ \ \ (text{given}\ \ b^2 – 3ac = 0)`

 
`:.\ text(Multiplicity at least 2.)`

♦♦ Mean mark 32%.

 
`p″(x) = 6ax + 2b`

`p″(−b/(3a))` `= 6a(−b/(3a)) + 2b=0`

`:. text(Multiplicity of)\ \ x=− b/(3a)\ \ text(is 3.)`

Mechanics, EXT2 2016 HSC 13c

The ends of a string are attached to points `A` and `B`, with `A` directly above `B.` The points `A` and `B` are  `0.4` m apart.

An object of mass `M` kg is fixed to the string at `C.` The object moves in a horizontal circle with centre `B` and radius  `0.3` m, as shown in the diagram.

ext2-hsc-2016-13c

The tensions in the string from the object to points `A` and `B` are `T_1` and `T_2` respectively. The object rotates with constant angular velocity `omega.` You may assume that the acceleration due to gravity is  `g = 10\ text(ms)^-2.`

  1. Show that  `T_2 = 0.3M (omega^2 - 25).`  (3 marks)
  2. For what range of values of  `omega` is  `T_2 > T_1?`  (1 mark)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `8.16\ text{rad/sec}`
Show Worked Solution
i.    ext2-hsc-2016-13c-answer1

`tantheta = 4/3, costheta = 3/5, sintheta = 4/5`

`text(Resolving the forces at)\ C\ text(vertically:)`

`sintheta · T_1 – 10M` `= 0`
`4/5 · T_1` `= 10M`
`T_1` `= (50M)/4\ …\ (1)`

 

`text(Resolving the forces at)\ C\ text(horizontally:)`

`costheta · T_1 + T_2` `= Mr omega^2`
`3/5T_1 + T_2` `= 0.3M omega^2`

 

`text(Substitute)\ \ T_1 = (50M)/4\ \ text(from)\ (1)`

`T_2` `= 0.3Momega^2 – 3/5 · (50M)/4`
  `= 0.3Momega^2 – (30M)/4`
  `= 0.3M(omega^2 – 25)\ …\ text(as required.)`

 

ii.   `text(Find)\ omega,\ text(such that)\ T_2 > T_1,`

♦ Mean mark 41%.
`0.3M(omega^2 – 25)` `> (50M)/4`
`omega^2 – 25` `> 125/3`
`omega^2` `> 200/3`
`omega` `> 8.16\ text{rad/sec  (2 d.p.)}`

Harder Ext1 Topics, EXT2 2016 HSC 13a

The function `f(x) = x^x` is defined and positive for all `x > 0.`

By differentiating `ln(f(x))`, find the value of `x` at which `f(x)` has a minimum.  (3 marks)

Show Answers Only

`1/e`

Show Worked Solution
`f(x)` `= x^x`
`ln f(x)` `= ln x^x`
  `= x ln x`

 

`:. d/dx (x ln x) = ln x + 1`

♦ Mean mark 41%.

`text(Max or min when)`

`ln x + 1` `= 0`
`ln x` `= −1`
`x` `= 1/e`

 

ext2-2016-hsc-13a-answer

`:. f(x)\ text(has a minimum at)\ \ x = 1/e.`

Complex Numbers, EXT2 N2 2016 HSC 10 MC

Suppose that  `x + 1/x = -1.`

What is the value of  `x^2016 + 1/x^2016?`

  1. `1`
  2. `2`
  3. `(2 pi)/3`
  4. `(4 pi)/3`
Show Answers Only

`=> B`

Show Worked Solution

`x + 1/x = −1`

♦ Mean mark 43%.

`x^2 + x + 1 = 0`

`:. x` `= (−1 ± sqrt(1 – 4))/2`
  `= −1/2 ± sqrt3/2 i`

 

`|\ x\ | = sqrt((−1/2)^2 + (sqrt3/2)^2) = 1`

`:. x` `= cos\ (2pi)/3 + isin\ (2pi)/3`
  `= text(cis)(2pi)/3`

`text(or)`

`x` `= cos\ (2pi)/3 – isin\ (2pi)/3`
  `= text(cis)(−(2pi)/3)`

 

`x^2016` `= text(cis)(±1344pi)quad(text(De Moivre))`
  `= text(cis)0`
  `= 1`

 

`:. x^2016 + 1/(x^2016) = 2`

`=> B`

Conics, EXT2 2016 HSC 7 MC

The hyperbola with equation  `xy = 8`  is the hyperbola  `x^2 - y^2 = k`  referred to different axes.

What is the value of `k?`

  1. `2`
  2. `4`
  3. `8`
  4. `16`
Show Answers Only

`=> D`

Show Worked Solution

met1-2008-vcaa-q2-answer

`text(Distance)\ OA = 4.`

♦♦ Mean mark 35%.

`text(Rotate)\ xy = 8quad\ text(clockwise 45°)`

`text{Intersection is (4, 0)}`

`4^2 – 0^2` `= k`
`:.k` `= 16`

`=> D`

Algebra, STD2 A4 2016 HSC 29b

The mass `M` kg of a baby pig at age `x` days is given by  `M = A(1.1)^x`  where `A` is a constant. The graph of this equation is shown.
 

2ug-2016-hsc-q29_1

  1. What is the value of `A`?   (1 mark)

  2. What is the daily growth rate of the pig’s mass? Write your answer as a percentage.   (1 mark)

Show Answers Only
  1. `1.5\ text(kg)`
  2. `10text(%)`
Show Worked Solution

i.   `text(When)\ x = 0,`

♦ Mean mark (i) 48%.
♦♦♦ Mean mark part (ii) 6%!
`1.5` `= A(1.1)^0`
`:. A` `= 1.5\ text(kg)`

 
ii.   `text(Daily growth rate)`

`= 0.1`

`= 10text(%)`

Probability, 2UG 2016 HSC 29a

Two unbiased coins are tossed.

  1. What is the probability that one coin shows heads and the other shows tails?  (1 mark)
  2. A game is played in which one player tosses the two coins. The rules are as follows:
  3. • If both coins show heads, the player wins `$40`
    • If both coins show tails, the player wins `$20`
  4. • If one coin shows heads and the other shows tails, the player loses `$30`.

  5.  

    What is the financial expectation of this game?  (2 marks)

 

Show Answers Only
  1. `1/2`

  2. `text(Expectation is breakeven)`
Show Worked Solution

(i)   `P(H\ text(and)\ T)`

`= P(H,T) + P(T,H)`

`= 1/2 xx 1/2 + 1/2 xx 1/2`

`= 1/2`

 

(ii)   `P(H,H) = 1/2 xx 1/2 = 1/4`

♦ Mean mark part (ii) 43%.

`P(T,T) = 1/2 xx 1/2 = 1/4`

 

`:.\ text(Financial expectation)`

`= 1/4 xx 40 + 1/4 xx 20 – 1/2 xx 30`

`=10+5-15`

`= 0\ \ \ text{(i.e. breakeven.)}`

Measurement, STD2 M1 2016 HSC 28e

A company makes large marshmallows. They are in the shape of a cylinder with diameter 5 cm and height 3 cm, as shown in the diagram.

2ug-2016-hsc-q28_4

  1. Find the volume of one of these large marshmallows, correct to one decimal place.  (2 marks)

A cake is to be made by stacking 24 of these large marshmallows and filling the gaps between them with chocolate. The diagrams show the cake and its top view. The shading shows the gaps to be filled with chocolate.
 

2ug-2016-hsc-q28_5

  1. What volume of chocolate will be required? Give your answer correct to the nearest whole number.  (3 marks)

Show Answers Only
  1. `58.9\ text{cm}^3`
  2. `193\ text{cm}^3`
Show Worked Solution
i.    `V` `= pir^2h`
    `= pi xx 2.5^2 xx 3`
    `= 58.904…`
    `= 58.9\ text{cm³  (1 d.p.)}`

 

ii.    2ug-2016-hsc-q28-answer1

`text(Volume of rectangle)`

♦ Mean mark part (ii) 35%.

`= 15 xx 10 xx 6`

`= 900\ text(cm)^3`
 

`text(Volume of marshmallows in rectangle)`

`= 6 xx 2 xx 58.9`

`= 706.8\ text(cm)^3`

 

`:.\ text(Volume of chocolate)`

`= 900-706.8`

`= 193.2`

`= 193\ text{cm}^3 \ text{(nearest cm}^3 text{)}`

Financial Maths, STD2 F5 2016 HSC 28d

The table gives the contribution per period for an annuity with a future value of $1 at different interest rates and different periods of time. 
 

2ug-2016-hsc-q28_31
 

Margaret needs to save $75 000 over 6 years for a deposit on a new apartment. She makes regular quarterly contributions into an investment account which pays interest at 3% pa.

How much will Margaret need to contribute each quarter to reach her savings goal?  (2 marks)

Show Answers Only

`$2865`

Show Worked Solution

`text(Periods) = 6 xx 4 = 24`

♦ Mean mark 40%.

`text(Interest rate) = 1/4 xx 3 = 0.75text(%)`

`=>\ text(Table factor = 0.0382)`

`(text(i.e. 3.82 cents contributed per)`

  `text(quarter = $1 after 6 years))`

 

`:.\ text(Quarterly contribution)`

`= 75\ 000 xx 0.0382`

`= $2865`

Probability, STD2 S2 2016 HSC 28c

A cricket team is about to play two matches. The probability of the team having a win, a loss or a draw is 0.7, 0.1 and 0.2 respectively in each match. The possible results in the two matches are displayed in the probability tree diagram.
  

2ug-2016-hsc-q28_2

  1. What is the probability of the team having a win and a draw, in any order?  (2 marks)

  2. Paul claims that 1.4 is the probability of the team winning both matches.

     

    Give one reason why this is NOT correct.  (1 mark)

Show Answers Only
  1. `0.28`
  2. `text(Probabilities cannot exceed 1.)`
Show Worked Solution

i.   `P(W\ text(and)\ D)`

`= P(W,D) + P(D,W)`

`= 0.7 xx 0.2 + 0.2 xx 0.7`

`= 0.28`

♦ Mean mark (i) 45%.
♦ Mean mark (ii) 49%.

 
ii.   `text(Probabilities cannot exceed 1.)`

Measurement, STD2 M7 2016 HSC 28a

Jacob has a large jar of silver coins. He adds 20 gold coins into the jar. He then seals the jar and shakes it to ensure that the gold coins are mixed in thoroughly with the silver coins. Jacob then opens the jar and takes a handful of coins. In his hand he has 33 silver coins and 4 gold coins. 

  1. Based on Jacob’s handful, if a coin is selected at random from the jar, what is the probability that it is a gold coin?  (1 mark)

  2. Jacob returns the handful of coins to the jar. Estimate the total number of coins in the jar.  (2 marks)

Show Answers Only
  1. `4/37`
  2. `185`
Show Worked Solution
i.    `P(G)` `= 4/(4 + 33)`
    `= 4/37`
♦ Mean mark part (i) 28%.

 

ii.  `text(Let)\ \ X =\ text(total coins in jar.)`

`20/X` `=4/37`
`:.X` `=(20 xx 37)/4`
  `=185`

Financial Maths, STD2 F4 2016 HSC 27d

Marge borrowed $19 000 to buy a used car. Interest on the loan was charged at 4.8% pa at the end of each month. She made a repayment of $436 at the end of every month. The table below sets out her monthly repayment schedule for the first four months of the loan. 
 

2ug-2016-hsc-q27_1

  1. Some values in the table are missing. Write down the values for `A` and `B`.  (2 marks)

  2. Calculate the value of `X`.  (2 marks)

  3. Marge repaid this loan over four years.

     

    What is the total amount that Marge repaid?  (1 mark)

Show Answers Only
  1. `A = $19\ 000,quadB = $17\ 551.33`
  2. `$74.56`
  3. `$20\ 928`
Show Worked Solution
i.    `A + 76-436` `= 18\ 640`
  `:. A` `= $19\ 000`

 

`17\ 915.67 + 71.66-436 = B`

`:. B = $17\ 551.33`

 

ii.   `18\ 640 + X-436 = 18\ 278.56`

`:. X` `= 18\ 278.56 + 436-18\ 640`
  `= $74.56`
♦♦ Mean mark part (iii) 28%.
COMMENT: Read carefully whether total paid or total interest paid is required.

 

iii.   `text(Total amount repaid)`

`= 48 xx 436`

`= $20\ 928`

Statistics, STD2 S1 2016 HSC 27c

The heights of 400 students were measured. The results are displayed in this cumulative frequency polygon.
 

2ug-2016-hsc-q27_2

 
Use the polygon to estimate the interquartile range.  (2 marks)

Show Answers Only

`36\ text(cm)`

Show Worked Solution

`text(See graph for values:)`

♦ Mean mark 46%.

2ug-2016-hsc-q27c-answer

`IQR` `= Q_3 – Q_1`
  `= 172 – 136`
  `= 36\ text(cm)`

Statistics, STD2 S1 2016 HSC 27b

A small population consists of three students of heights 153 cm, 168 cm and 174 cm. Samples of varying sizes can be taken from this population.

What is the mean of the mean heights of all the possible samples? Justify your answer.  (2 marks)

Show Answers Only

`165\ text(cm)`

Show Worked Solution

`text(If sample is 1 person,)`

♦ Mean mark 40%.

`text{Possible mean(s): 153, 168 or 174.}`
 

`text(If sample is 2 people,)`

`text{Possible mean(s):}quad(153 + 168)/2` `= 160.5`
`(153 + 174)/2` `= 163.5`
`(168 + 174)/2` `= 171`

 

`text(If sample is 3 people,)`

`text(Mean:)quad(153 + 168 + 174)/3 = 165`
 

`:.\ text(Mean of all mean heights)`

`= (153 + 168 + 174 + 160.5 + 163.5 + 171 + 165)/7`

`= 165\ text(cm)`

Financial Maths, STD2 F1 2016 HSC 27a

Alice intends to buy a car and insure it.

Briefly describe what each of these types of insurance covers: 

    • Compulsory third-party insurance (CTP)

    • Non-compulsory third-party property insurance.  (2 marks)

Show Answers Only

`text(CTP Insurance:)`

`text(This insures a driver against liability)`

`text(if their car injures or kills a person)`

`text(in an accident.)`

`text(Non-compulsory TP Insurance:)`

`text(This insurance covers damage to)`

`text(other people’s property in an accident,)`

`text(but does not cover the driver’s own vehicle.)`

Show Worked Solution

`text(CTP Insurance:)`

♦♦♦ Mean mark 13%.
MARKER’S COMMENT: Again, car insurance causes major issues for students.

`text(This insures a driver against liability)`

`text(if their car injures or kills a person)`

`text(in an accident.)`

`text(Non-compulsory TP Insurance:)`

`text(This insurance covers damage to)`

`text(other people’s property in an accident,)`

`text(but does not cover the driver’s own vehicle.)`

Measurement, STD2 M6 2016 HSC 25 MC

The diagram shows towns `A`, `B` and `C`. Town `B` is 40 km due north of town `A`. The distance from `B` to `C` is 18 km and the bearing of `C` from `A` is 025°. It is known that  `∠BCA`  is obtuse.
 

2ug-2016-hsc-25-mc

 
What is the bearing of `C` from `B`?

  1.    `070°`
  2.    `095°`
  3.    `110°`
  4.    `135°`
Show Answers Only

`=> D`

Show Worked Solution

`text(Using the sine rule,)`

♦ Mean mark 39%.
`(sin∠BCA)/40` `= (sin25^@)/18`
`sin angle BCA` `= (40 xx sin25^@)/18`
  `= 0.939…`
`angle BCA` `= 180 – 69.9quad(angleBCA > 90^@)`
  `= 110.1°`

 

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

`= 110.1 + 25qquad(text(external angle of triangle))`

`= 135.1`

 
`=> D`

Statistics, STD2 S1 2016 HSC 21 MC

A grouped data frequency table is shown.
 

2ug-2016-hsc-21-mc

 
What is the mean for this set of data?

  1.    6.5
  2.    10.5
  3.    11.9
  4.    12.4
Show Answers Only

`=> D`

Show Worked Solution

`text(Using the centre of each class interval:)`

♦ Mean mark 43%.
`text(Mean)` `= (3 xx 3 + 8 xx 6 + 13 xx 8 + 18 xx 9)/(3 + 6 + 8 + 9)`
  `= 12.42…`

`=> D`

Statistics, STD2 S1 2016 HSC 19 MC

A soccer referee wrote down the number of goals scored in 9 different games during the season.

`2,  \ 3,  \ 3,  \ 3,  \ 5,  \ 5,  \ 8,  \ 9,  \ ...`

The last number has been omitted. The range of the data is 10.

What is the five-number summary for this data set?

  1. `2, 3, 5, 8.5, 12`
  2. `2, 3, 5, 8.5, 10`
  3. `2, 3, 5, 8, 12`
  4. `2, 3, 5, 8, 10`
Show Answers Only

`=> A`

Show Worked Solution

`text{Since range is 10} \ => \ text{Last data point = 12}`

`text{Q}_1 = 3`

`text{Q}_3 = (8 + 9)/2 = 8.5`

`text(Median = 5)`

`=> A`

♦ Mean mark 46%.

Financial Maths, STD2 F4 2016 HSC 17 MC

Ariana is charged compound interest at the rate of  0.036%  per day on outstanding credit card balances. She has $780 outstanding for 24 days.

How much compound interest is she charged?

  1.    $6.74
  2.    $6.77
  3.    $786.74
  4.    $786.77
Show Answers Only

`=> B`

Show Worked Solution

`text(Total owing)`

♦ Mean mark 38%.
COMMENT: Credit card problems consistently produce sub-50% mean marks. Important review area.

`= P(1+r)^n`

`= 780(1 + 0.036/100)^24`

`= 786.77`
 

`:.\ text(Interest charged)`

`= 786.77 – 780`

`= $6.77`

 
`=> B`

Measurement, STD2 M7 2016 HSC 15 MC

Calls on a mobile phone plan are charged at the rate of 54 cents per 30 seconds, or part thereof.

What is the cost of a call lasting 2 minutes and 15 seconds?

  1.    $2.16
  2.    $2.32
  3.    $2.43
  4.    $2.70
Show Answers Only

`=> D`

Show Worked Solution

`5 xx 30\ text(second blocks)`

♦ Mean mark 44%.
`:.\ text(C)text(ost)` `= 5 xx 0.54`
  `= $2.70`

`=> D`

Measurement, STD2 M1 2016 HSC 12 MC

A container is in the shape of a triangular prism which has a capacity of 12 litres. The area of the base is 240 cm².
 

What is the distance, `h`, between the two triangular ends of the container?

  1. 5 cm
  2. 20 cm
  3. 25 cm
  4. 50 cm
Show Answers Only

`=> D`

Show Worked Solution
♦♦ Mean mark 35%.

`text{1 mL = 1 cm}^3\ \ =>\ \ text{1 L = 1000 cm}^3`

`text(Volume)` `= Ah`
`12\ 000` `= 240 xx h`
`h` `= (12\ 000)/240`
  `= 50\ text(cm)`

 
`=> D`

Measurement, STD2 M1 2016 HSC 1 MC

What is  208.345  correct to two significant figures?

  1. 208
  2. 210
  3. 208.34
  4. 208.35
Show Answers Only

`B`

Show Worked Solution

`208.345 = 210\ (2\ text(sig. fig.))`

♦♦ Mean mark 36%!!

`=> B`