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Statistics, SPEC1 2021 VCAA 3

A company produces a particular type of light globe called Shiny. The company claims that the lifetime of these globes is normally distributed with a mean of 200 weeks and it is known that the standard deviation of the lifetime of Shiny globes is 10 weeks. Customers have complained, saying Shiny globes were lasting less than the claimed 200 weeks. It was decided to investigate the complaints. A random sample of 36 Shiny globes was tested and it was found that the mean lifetime of the sample was 195 weeks.

Use  `text(Pr)(-1.96 < Z < 1.96) = 0.95`  and  `text(Pr)(-3 < Z < 3) = 0.9973`  to answer the following questions.

  1. Write down the null and alternative hypotheses for the one-tailed test that was conducted to investigate the complaints.  (1 mark)
    1. Determine the `p` value, correct to three places decimal places, for the test.  (2 marks)
    2. What should the company be told if the test was carried out at the 1% level of significance?  (1 mark)
  3. The company decided to produce a new type of light globe called Globeplus.
    Find the approximate 95% confidence interval for the mean lifetime of the new globes if a random sample of 25 Globeplus globes is tested and the sample mean is found to be 250 weeks. Assume that the standard deviation of the population is 10 weeks. Give your answer correct to two decimal places.  (1 mark)
Show Answers Only
  1. `H_0 : mu = 200`
    `H_1 : mu < 200`
    1. `0.001\ \ (text(to 3 d.p.))`
    2. `(246.08, 253.92)`
  3. `text(Reject the null hypothesis.)`
Show Worked Solution

a.   `H_0 : mu = 200`

`H_1 : mu < 200`

 

b.i.   `E(barX) = mu = 200`

♦ Mean mark part (b)(i) 48%.

`sigma(barX) = sigma/sqrtn = 10/sqrt36 = 5/3`

`p` `= text(Pr)(barX< 195 | mu = 200)`
  `= text(Pr)(z < (195 – 200)/(3/5))`
  `= text(Pr)(z < -3)`
  `= 1/2 (1 – 0.9973)`
  `= 0.00135`
  `= 0.001\ \ (text(to 3 d.p.))`

 

b.ii.   `text(1% level) => 0.01`

♦ Mean mark part (b)(ii) 49%.

  `text(S)text(ince)\ \ 0.001 < 0.01\ \ =>\ text(Strong evidence against)\ H_0`

  `:.\ text(Reject the null hypothesis.)`

 

c.   `sigma(barx) = sigma/sqrtn = 10/sqrt25 = 2`

`text(95% C.I.)` `= barx – 1.96 xx 2, barx + 1.96 xx 2`
  `= (250 – 3.92, 250 + 3.92)`
  `= (246.08, 253.92)`

Calculus, EXT2 C1 2021 SPEC2 2

Evaluate  `int_0^1 (2x + 1)/(x^2 + 1)\ dx`.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution
`int_0^1 (2x + 1)/(x^2 + 1)\ dx` `= int_0^1 (2x)/(x^2 + 1)\ dx + int_0^1 1/(x^2 + 1)\ dx`
  `= [log_e(x^2 + 1)]_0^1 + [tan^(-1)(x)]_0^1`
  `= log_e 2 – log_e 1 + tan^(-1)(1) – tan^(-1)(0)`
  `= log_e 2 + pi/4`

Vectors, SPEC1 2021 VCAA 1

The net force acting on a body of mass 10 kg is  `underset~F = 5underset~i + 12underset~j`  newtons.

  1. Find the acceleration of the body in `text(ms)^(-2)`.  (1 mark)
  2. The initial velocity of the body is  `-3underset~j\ \ text(ms)^(-1)`.
    Find the velocity of the body, in `text(ms)^(-1)`, at any time  `t`  seconds.  (2 marks)
  3. Find the momentum of the body, in kg `text(ms)^(-1)`,  when  `t = 2`  seconds.  (1 mark)
Show Answers Only
  1. `1/2 underset~i + 6/5 underset~j`
  2. `1/2 t underset~i + (6/5 t – 3) underset~j`
  3. `10 underset~i – 6underset~j`
Show Worked Solution

a.   `text(Using)\ underset~F = m underset~a:`

`10underset~a` `= 5underset~i + 12underset~j`
`underset~a` `= 1/10 (5underset~i + 12underset~j)`
  `= 1/2 underset~i + 6/5 underset~j`

 

b.    `underset~v(t)` `= int underset~a\ dt`
    `= int 1/2 underset~i + 6/5 underset~j\ dt`
    `= 1/2 t underset~i + 6/5 t underset~j + c`

 
`text(When)\ t = 0, v(t) = -3underset~j`

`=> c = -3underset~j`

`underset~v(t)` `= 1/2 t underset~i + 6/5 t underset~j – 3underset~j`
  `= 1/2 t underset~i + (6/5 t – 3) underset~j`

 

c.    `underset~v(2) = underset~i – 3/5 underset~j`

`underset~p(2)` `= m underset~v(2)`
  `= 10(underset~i – 3/5 underset~j)`
  `= 10 underset~i – 6underset~j`

MATRICES, FUR1 2021 VCAA 1 MC

If matrix  `M = [(3,2), (8,9), (13,7)]` , then its transpose, `M^T` , is

A.  `[(2,3), (9,8), (7,13)]` B.  `[(2,9,7), (3,8,13)]`  
     
C.  `[(7,9,2), (13,8,3)]` D.  `[(3,8,13), (2,9,7)]`  
     
E.  `[(13,8,3), (7,9,2)]`    
Show Answers Only

`D`

Show Worked Solution

`text{Transpose} \ x_{ij} -> x_{ji}`

`[(3,2), (8,9), (13,7)]^T = [(3,8,13), (2,9,7)]`

`=>D`

CORE, FUR1 2021 VCAA 17 MC

The following recurrence relation can generate a sequence of numbers.
 

`L_0 = 37 , \ L_{n+1} = L_n + C`
 

The value of `L_2` is 25.

The value of `C` is

  1. – 6
  2. – 4
  3.    4
  4.    6
  5.  37
Show Answers Only

`A`

Show Worked Solution

`L_{n+1} = L_n + C => text{recurrence with with common difference}`

`L_1` `=37 + C`  
`L_2` `=L_1+C`  
`25` `=37 + C + C`  
`2C` `=-12`  
`:.C` `=-6`  

 
`=> A`

CORE, FUR1 2021 VCAA 1-3 MC

The percentaged segmented bar chart below shows the age (under 55 years, 55 years and over) of visitors at a travel convention, segmented by preferred travel destination (domestic, international).
 

Part 1

The variables age (under 55 years, 55 years and over) and preferred travel destination (domestic, international) are

  1. both categorial variables.
  2. both numerical variables.
  3. a numerical variable and a categorical variable respectively.
  4. a categorical variable and a numerical variable respectively.
  5. a discrete variable and a continuous variable respectively.

 
Part 2

The data displayed in the percentaged segmented bar chart supports the contention that there is an association between preferred travel destination and age because

  1. more visitors favour international travel.
  2. 35% of visitors under 55 years favour international travel.
  3. 45% of visitors 55 years and over favour domestic travel.
  4. 65% of visitors under 55 years favour domestic travel while 45% of visitors 55 years and over favour domestic travel.
  5. the percentage of visitors who prefer domestic travel is greater than the percentage of visitors who prefer international travel.

 
Part 3

The results could also be summarised in a two-way frequency table.

Which one of the following frequency tables could match the pecentaged segmented bar chart?

 

Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ D `

`text(Part 3:)\ A`

Show Worked Solution

`text(Part 1)`

`text{Preferred travel destination → categorical (nominal) variable}`

`text{Age → categorical (ordinal) variable}`

`=> A`
 

`text(Part 2)`

`text(Only option)\ D\ text(highlights a change in preference for domestic travel)`

`text(between the two age categories.)`

`=>D` 

 
`text(Part 3)`

♦ Mean mark 50%.

`text(Converting the frequency table data into percentages,)`

`text(consider option)\ A:`

`91/140 xx 100 = 65text(%),\ \ 49/140 xx 100 = 35text(%)`

`90/200 xx 100 = 45text(%), \ \ 110/200 xx 100 = 55text(%)`

`=> A`

Functions, MET1 2021 VCAA 3

Consider the function  `g: R -> R, \ g(x) = 2sin(2x).`

  1. State the range of `g`.  (1 mark)
  2. State the period of `g`.  (1 mark)
  3. Solve  `2 sin(2x) = sqrt3`  for  `x ∈ R`.  (3 marks)
Show Answers Only
  1. `[–2,2]`
  2. `pi`
  3. `x= pi/6 + npi, pi/3 + npi\ \ \ (n in ZZ)`
Show Worked Solution

a.   `text(S)text(ince)  -1<sin(2x)<1,`

`text(Range)\  g(x) = [–2,2]`
 

b.   `text(Period) = (2pi)/n = (2pi)/2 = pi`
 

c.    `2sin(2x)` `=sqrt3`
  `sin(2x)` `=sqrt3/2`
  `2x` `=pi/3, (2pi)/3, pi/3 + 2pi, (2pi)/3 + 2pi, …`
  `x` `=pi/6, pi/3, pi/6+pi, pi/3+pi, …`

 
`:.\ text(General solution)`

`= pi/6 + npi, pi/3 + npi\ \ \ (n in ZZ)`

Calculus, MET1 2021 VCAA 2

Let  `f′(x) = x^3 + x`.

Find  `f(x)`  given that  `f(1) = 2`.  (2 marks)

Show Answers Only

`f(x) =1/4 x^4 + 1/2 x^2 +5/4`

Show Worked Solution
`f′(x)` `= x^3 + x`  
`f(x)` `=int x^3 + x\ dx`  
  `=1/4 x^4 + 1/2 x^2 +c`  

 
`text(Given)\ f(1) = 2:`

`2` `= 1/4 + 1/2 + c`  
`c` `= 5/4`  

 
`:. f(x) =1/4 x^4 + 1/2 x^2 +5/4`

Mechanics, EXT2 M1 2021 HSC 14b

An object of mass 5 kg is on a slope that is inclined at an angle of 60° to the horizontal. The acceleration due to gravity is `g\ text(m s)^(-2)`  and the velocity of the object down the slope is `v\ text(m s)^(-1)`.

As well as the force due to gravity, the object is acted on by two forces, one of magnitude `2v` newtons and one of magnitude `2v^2` newtons, both acting up the slope.

  1. Show that the resultant force down the slope is  `(5sqrt3)/2 g - 2v - 2v^2`  newtons.  (2 marks)

  2. There is one value of `v` such that the object will slide down the slope at a constant speed.
  3. Find this value of `v` in `text(m s)^(-1)`, correct to 1 decimal place, given that  `g = 10`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `4.2\ \ text(ms)^-1`
Show Worked Solution

i.   

`F_s` `=\ text(force down slope)`
  `= 5g cos30 – 2v^2 – 2v`
  `= 5g · sqrt3/2 – 2v^2 – 2v`
  `= (5sqrt3)/2 g – 2v^2 – 2v`

 

ii.   `text(Constant speed occurs if)\ \ a = 0`

`F_s = ma = 0`

`2v^2 + 2v – (5sqrt3)/2 xx 10` `= 0`
`2v^2 + 2v – 25sqrt3` `= 0`

 

`v` `= (-2 + sqrt(2^2 + 4 · 2 · 25sqrt3))/(2 xx 2)`
  `= (-2 + sqrt(4 + 200sqrt3))/4`
  `= (-1 + sqrt(1 + 50 sqrt3))/2`
  `= 4.179…`
  `= 4.2\ \ text(ms)^-1\ text{(1 d.p.)}`

Complex Numbers, EXT2 N2 2021 HSC 14c

  1. Using de Moivre’s theorem and the binomial expansion of `(cos theta + i sin theta)^5`, or otherwise, show that
  2.    `cos5theta = 16cos^5theta - 20cos^3 theta + 5cos theta`.  (3 marks)

  3. By using part (i), or otherwise, show that  `text(Re)(e^((ipi)/10)) = sqrt((5 + sqrt5)/8)`.  (3 marks)

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

i.   `(cos theta + i sin theta)^5 = cos5theta + i sin 5theta\ \ text{(by De Moivre)}`

`text(Using binomial expansion:)`

`(cos theta + i sin theta)^5`

`= cos^5theta + 5cos^4theta · isin theta + 10cos^3theta · i^2sin^2theta + 10 cos^2theta · i^3sin^3theta`

`+ 5costheta · i^4sin^4theta + i^5sin^5theta`

`= cos^5theta – 10cos^3thetasin^2theta + 5costhetasin^4theta + i\ \ text{(imaginary part)}`
 

`text(Equating real parts:)`

`cos5theta` `= cos^5theta – 10cos^3thetasin^2theta + 5costhetasin^4theta`
  `= cos^5theta – 10cos^3theta(1 – cos^2theta) + 5costheta(1 – cos^2theta)sin^2theta`
  `= cos^5theta – 10cos^3theta + 10cos^5theta + (5costheta – 5cos^3theta)(1 – cos^2theta)`
  `= 11cos^5theta – 10cos^3theta + 5costheta – 5cos^3theta – 5cos^3theta + 5cos^5theta`
  `= 16cos^5theta – 20cos^3theta + 5costheta`

 

ii.   `text(Re)(e^((ipi)/10)) = cos\ pi/10`

♦♦ Mean mark 31%.

`text(Using part a:)`

`16 cos^5(pi/10) – 20cos^3(pi/10) + 5cos(pi/10) = cos((5pi)/10) = cos(pi/2) = 0`

`text(Solutions to)\ cos(5theta) = 0`

`5theta` `= pi/2, (3pi)/2, (5pi)/2, …`
`theta` `= pi/10, (3pi)/10, (5pi)/10, …`

  
`text(Let)\ cos(pi/10) = x`

`16x^5 – 20x^3 + 5x` `= 0`
`16x^4 – 20x^2 + 5` `= 0`
`x^2` `= (20 ± sqrt(20^2 – 4 · 16 · 5))/(2 xx 16)`
  `= (20 ± sqrt80)/32`
  `= (5 ±sqrt5)/8`
`x` `= ±sqrt((5 ± sqrt5)/8)`

 
`cos\ pi/10 > 0 \ => \ x = sqrt((5 ± sqrt5)/8)`

`1 > cos\ pi/10 > cos\ (3pi)/10 > 0`

`:.\ text(Re)\ (e^((ipi)/10)) = cos\ pi/10 = sqrt((5 + sqrt5)/8)`

Calculus, EXT2 C1 2021 HSC 13c

  1. The integral  `I_n`  is defined by  `I_n = int_1^e (ln x)^n\ dx`  for integers  `n >= 0`.
    Show that  `I_n = e - nI_(n - 1)`  for   `n >= 1`.  (2 marks)

  2. The diagram shows two regions.

Region `A` is bounded by  `y = 1`  and  `x^2 + y^2 = 1`  between  `x = 0`  and  `x = 1`.

Region `B` is bounded by  `y = 1`  and  `y = ln x`  between  `x = 1`  and  `x = e`.
 
   
 
The volume of the solid created when the region between the curve  `y = f(x)`  and the  `x`-axis, between  `x = a`  and  `x = b`, is rotated about the `x`-axis is given by  `V = pi int_a^b [f(x)]^2\ dx`.

The volume of the solid of revolution formed when region `A` is rotated about the `x`-axis is `V_A`.

The volume of the solid of revolution formed when region `B` is rotated about the `x`-axis is `V_B`.

Using part (i), or otherwise, show that the ratio  `V_A : V_B` is `1:3`.  (4 marks)

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

i.   `I_n = int_1^e (ln x)^n\ dx\ \ text(for)\ \ n >= 0`

`text(Show)\ \ I_n = e – nI_(n – 1)\ \ text(for)\ \ n >= 1`

`text(Integrating by parts:)`

`u` `= (ln x)^n` `u′` `= n/x (ln x)^(n – 1)`
`v` `= x` `v′` `= 1`
`I_n` `= [x(ln x)^n]_1^e – n int(ln x)^(n – 1)\ dx`
  `=[e(lne)^n-1(ln1)^n]- n I_(n – 1)`
  `= e – n I_(n – 1)`

 

ii.    `V_A` `=\ text(cylinder volume – hemisphere volume)`
    `= pi int_0^1 1\ dx – 1/2 xx 4/3 xx pir^3`
    `= pi [x]_0^1 – (2pi)/3`
    `= pi(1 – 0) – (2pi)/3`
    `= pi/3`

 

`V_B` `= pi int_1^e 1\ dx – pi int_1^e (ln x)^2\ dx`
  `= pi [x]_1^e – pi I_2`
  `= pi(e – 1) – pi(e – 2I_1)`
  `= pi(e – 1) – pi[e – 2(e – I_0)]`
  `= pi(e – 1) – pi(2I_0 – e)`
  `= pi(e – 1) – pi[2(e – 1) – e]`
  `= pi(e – 1) – pi(e – 2)`
  `= pie – pi – pie + 2pi`
  `= pi`
`:. V_A : V_B` `= pi/3 : pi`
  `= 1 : 3`

Vectors, EXT2 V1 2021 HSC 12e

The diagram shows the pyramid  `ABCDS`  where  `ABCD`  is a square. The diagonals of the square bisect each other at `H`.
 

  1. Show that  `overset->{HA} + overset->{HB} + overset->{HC} + overset->{HD} = underset~0`   (1 mark)

    Let `G` be the point such that  `overset->{GA} + overset->{GB} + overset->{GC} + overset->{GD} + overset->{GS}  =  underset~0`.

  1. Using part (i), or otherwise, show that  `4 overset->{GH} + overset->{GS}  =  underset~0`.   (2 marks)

  2. Find the value of  `λ`  such that  `overset->{HG} = λ overset->{HS}`   (1 mark)

Show Answers Only
  1. `text{See Worked Solution}`
  2. `text{See Worked Solution}`
  3. `1/5`
Show Worked Solution

i.    `text{S} text{ince diagonal} \ overset->{AC} \ text{is bisected by H:}`

`overset->{HA} =- overset->{HC}`

`text{Similarly for diagonal} \ overset->{BD}`

`overset->{HB} = – overset->{HD}`
 

`:. \ overset->{HA} + overset->{HB} + overset->{HC} + overset->{HD}` `= – overset->{HC} – overset->{HD} +  overset->{HC} + overset->{HD}`
  `= underset~0`
 
 
ii.    `overset->{GA} = overset->{GH} + overset->{HA} \ , \ overset->{GB} = overset->{GH} + overset->{HB}`
`overset->{GC} = overset->{GH} + overset->{HC} \ , \ overset->{GD} = overset->{GH} + overset->{HD}`
 
`overset->{GA} + overset->{GB} + overset->{GC} + overset->{GD} + overset->{GS}`
`= overset->{GH} + overset->{HA} + overset->{GH} + overset->{HB} + overset->{GH} + overset->{HC} + overset->{GH} + overset->{HD} + overset->{GS}`
`= 4 overset->{GH} + (overset->{HA} + overset->{HB} + overset->{HC} + overset->{HD} + overset->{GS})`
`= 4 overset->{GH} + overset->{GS}`
 
`overset->{GA} + overset->{GB} + overset->{GC} + overset->{GD} + overset->{GS} = underset~0\ \ \ text{(given)}`
`:. 4 overset->{GH} + overset->{GS} = underset~0`
 
iii.  `overset->{HS} = overset->{HG} + overset->{GS}`
`overset->{GS} = overset->{HS} + overset->{GH}`
♦ Mean mark 48%.

 

`text{Using part (ii):}`
`4 overset->{GH} + overset->{HS} + overset->{GH}` `= underset~0`
`5 overset->{GH}` `= overset->{SH}`
`overset->{GH}` `= 1/5 overset->{SH}`
`overset->{HG}` `= 1/5 overset->{HS}`

`:. \ λ = 1/5`

Vectors, EXT2 V1 2021 HSC 12c

Two lines are given by  `text(r)_1 = ((-2),(1),(3)) + lambda((1),(0),(2))`  and  `text(r)_2 = ((4),(-2),(q)) + mu ((p),(3),(-1))` , where `p` and `q` are real numbers. These lines intersect and are perpendicular.

Find the values of `p` and `q`.  (3 marks)

Show Answers Only

`p = 2 \ , \ q = 20`

Show Worked Solution

`text{S} text{ince vectors are perpendicular}`

`((1),(0),(2)) ((p),(3),(-1))` `= 0`
`p – 2` `= 0`
`p` `= 2`
 

`text{S} text{ince lines intersect, equate}\ y text{-coordinates:} `

`1 + lambda 0` `= -2 + 3 mu`
`mu` `= 1`

 

`text{Find} \ lambda \ text{by equating}\ xtext{-coordinates:}`

`-2 + lambda` `= 4 + 1 xx 2`
`lambda` `= 8`

 
`text{Equating}\ ztext{-coordinates:}`

`3 + 8 xx 2` `= q – 1 xx 1`
`q` `= 20`

Proof, EXT2 P1 2021 HSC 12b

Consider Statement A.

Statement A: ‘If `n^2` is even, then `n` is even.’

  1. What is the converse of Statement A?.  (1 mark)

  2. Show that the converse of Statement A is true. (1 mark)

Show Answers Only
  1. `text{See Worked Solution}`
  2. `text{See Worked Solution}`
Show Worked Solution

i.     `text{Converse}`

`text{If} \ n \ text{is even, then} \ n^2 \ text{is even.}`
 

ii.    `text{If} \ n \ text{is even:}`

`n` `= 2p, \ p  ∈  ZZ`
`n^2` `= (2 p)^2`
  `=4 p^2`
  `= 2 (2p^2)`
  `=2q, \ q  ∈  ZZ`

 

`:. \ text{If} \ n \ text{is even, then} \ n^2 \ text{is even.}`

Calculus, EXT2 C1 2021 HSC 12a

Find  `int {2x + 3}/{x^2 + 2x + 2} dx`.  (3 marks)

Show Answers Only

`ln (x^2 + 2x + 2) + tan^{-1} (x + 1) + c`

Show Worked Solution
`int {2x + 3}/{x^2 + 2x + 2} dx` `= int {2x + 2}/{x^2 + 2x + 2} dx + int {1}/{(x – 1)^2 + 1} dx`
  `= ln (x^2 + 2x + 2) + tan^{-1} (x + 1) + c`

Calculus, EXT2 C1 2021 HSC 11f

Express  `{3x^2-5}/{(x-2)(x^2 + x + 1)}`  as a sum of partial fractions over `RR`.  (3 marks)

Show Answers Only

`{3x^2-5}/{(x-2)(x^2 + x + 1)} = {1}/{x-2} + {2x + 3}/{x^2 + x + 1}`

Show Worked Solution

`{3x^2-5}/{(x-2)(x^2 + x + 1)} = {A}/{(x-2)} + {B x + C}/{(x^2 + x + 1)}`

`A (x^2 + x + 1) + (Bx + C)(x-2) ≡ 3x^2-5`

`text{If} \ \ x = 2,`

`7A = 7 \ => \ A = 1`

`text{If} \ \ x = 0,`

`1-2C=-5 \ => \ C = 3`

`text(Equating coefficients:)`

`x^2 + x + 1 + Bx^2-2Bx + 3x -6 \ ≡ \ 3x^2 -5`

`(B + 1) x^2 + (4 -2B) x -5 \ ≡ \ 3x^2-5`

`=> B = 2`
 

`:. \ {3x^2-5}/{(x-2)(x^2 + x + 1)} = {1}/{x-2} + {2x + 3}/{x^2 + x + 1}`

Complex Numbers, EXT2 N1 2021 HSC 11e

The complex numbers  `z = 5 + i`  and  `w = 2 − 4 i`  are given.

Find  `bar z/{w}`, giving your answer in Cartesian form.  (2 marks)

Show Answers Only

`7/10 + 9/10 i`

Show Worked Solution

`z = 5 + i \ => \ bar z= 5 – i`

`w = 2- 4 i`

`bar z/w` `= {5 – i}/{2 – 4 i} xx {2 + 4 i}/{2 + 4 i}`
  `= {(5 – i)(2 +  4i)}/{4 + 16}`
  `= {10 + 20 i – 2i + 4}/{20}`
  `= 7/10 + 9/10 i`

Complex Numbers, EXT2 N2 2021 HSC 11d

  1. Find the two square roots of  `−i` , giving the answers in the form  `x + iy`, where `x` and `y` are real numbers.  (2 marks)

  2. Hence, or otherwise, solve  `z^2 + 2z + 1 + i = 0`  giving your solutions in the form  `a + i b`  where `a` and `b` are real numbers. (2 marks)

Show Answers Only
  1. `z = 1/sqrt2 – 1/sqrt2 i \ \ text{or} \ \ z = – 1/sqrt2+ 1/sqrt2 i`
  2. `z = -1 + 1/sqrt2 – 1/sqrt2 i \ \ text{or} \ \ z = -1 – 1/sqrt2+ 1/sqrt2 i`
Show Worked Solution

i.     `text{Solution 1}`

`z = x + iy \ , \ z^2 = -i`

`z^2 = x^2 – y^2 + 2x yi = -i`

`x^2 – y^2 = 0 \ …\ (1)`

`2xy = -1 \ …\ (2)`

`x= ± y \ …\ (1)′`
 

`text{Substitute} \ \ x = y \ \ text{into} \ (2)`

`2x^2 = -1 \ -> \ text{no real solutions}`

`text{Substitute} \ \ x= -y \ \ text{into} \ (2)`

`-2x^2` `= -1`  
`x` `= ± 1/sqrt2`  

 
`:. \ z = 1/sqrt2 – 1/sqrt2 i \ \ text{or} \ \ z = – 1/sqrt2+ 1/sqrt2 i`
 

`text{Solution 2}`

`z = re^{i theta} \ , \ z^2 = -i`

`r^2 e^{i2 theta} = e^{i {3pi}/{2}} \ \ text{or} \ \ e^{-i pi/2}`

`=> \ r = 1 \ , \  theta = {3pi}/{4} \ \ text{or} \ \ – pi/4`
 

`z= text{cos} {3pi}/{4} + i \ text{sin} {3pi}/{4} = – 1/sqrt2 + 1/sqrt2 i`

`z= text{cos} (- pi/4) + i \ text{sin} (- pi/4) = 1/sqrt2 – 1/sqrt2 i`

 

ii.    `z^2 + 2z + 1 + i = 0`

`z` `= {2 ± sqrt{4 – 4 * 1 * (1 + i)}}/{2}`  
  `= {-2 ± sqrt{4 – 4 – 4 \ i}}/{2}`  
  `= -1 ± sqrt(-i)`  

 
`:. \ z = -1 + 1/sqrt2 – 1/sqrt2 i \ \ text{or} \ \ z = -1 – 1/sqrt2 + 1/sqrt2 i`

Proof, EXT2 P1 2021 HSC 4 MC

Consider the statement:

‘For all integers `n`, if `n` is a multiple of 6, then `n` is a multiple of 2’.

Which of the following is the contrapositive of the statement?

  1. There exists an integer `n` such that `n` is a multiple of 6 and not a multiple of 2.
  2. There exists an integer `n` such that `n` is a multiple of 2 and not a multiple of 6.
  3. For all integers `n`, if `n` is not a multiple of 2, then `n` is not a multiple of 6.
  4. For all integers `n`, if `n` is not a multiple of 6, then `n` is not a multiple of 2.
Show Answers Only

`C`

Show Worked Solution

`text{L} text{ogically equivalent statements:}`

`text{If} \ n \ text{is multiple of 6} \ => \ n \ text{is a multiple of 2}`

`¬\ n\ text{is a multiple of 2}\ => \ ¬\ n \ text{is a multiple of 6}`

`text{i.e. if} \ n \ text{is not multiple of 2} \ => \ n \ text{is not a multiple of 6}`

`=>\ C`

Calculus, EXT1 C2 2021 HSC 11c

Use the substitution  `u = x + 1`  to find  `int xsqrt(x + 1)\ dx`.  (3 marks)

Show Answers Only

`2/5 (x + 1)^(5/2) – 2/3(x + 1)^(3/2) + c`

Show Worked Solution

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

`int x sqrt(x – 1)\ dx` `= int (u – 1) sqrtu\ du`
  `= int u^(3/2) – u^(1/2)\ du`
  `= 2/5 u^(5/2) – 2/3 u^(3/2) + c`
  `= 2/5 (x + 1)^(5/2) – 2/3(x + 1)^(3/2) + c`

Combinatorics, EXT1 A1 2021 HSC 11b

Expand and simplify  `(2a - b)^4`.  (2 marks)

Show Answers Only

`16a^4 – 32a^3b + 24a^2b^2 – 8ab^3 + b^4`

Show Worked Solution

`(2a – b)^4`

`= (2a)^4 + \ ^4C_1(2a)^3(-b) + \ ^4C_2(2a)^2(-b)^2 + \ ^4C_3(2a)(-b)^3 + (-b)^4`

`= 16a^4 – 4 · 8a^3b + 6 · 4a^2b^2 – 4 · 2ab^3 + b^4`

`= 16a^4 – 32a^3b + 24a^2b^2 – 8ab^3 + b^4`

Calculus, EXT1 C3 2021 HSC 4 MC

Consider the differential equation  `(dy)/(dx) = x/y`.

Which of the following equations best represents this relationship between `x` and `y`?

  1. `y^2 = x^2 + c`
  2. `y^2 = (x^2)/2 + c`
  3. `y = x ln\ | y | + c`
  4. `y = (x^2)/2 ln\ |y| + c`
Show Answers Only

`A`

Show Worked Solution
`(dy)/(dx)` `= x/y`
`int y\ dy` `= int x\ dx`
`1/2 y^2` `= 1/2 x^2 + c`
`y^2` `= x^2 + c_1`

 
`=> A`

Statistics, STD1 S3 2021 HSC 18

People are placed into groups to complete a puzzle. There are 9 different groups.

The table shows the number of people in each group and the amount of time, in minutes, for each group to complete the puzzle.

\begin{array} {|l|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Number of people} \rule[-1ex]{0pt}{0pt} & 2 & 2 & 3 & 5 & 5 & 7 & 7 & 7 & 8 \\
\hline
\rule{0pt}{2.5ex} \textit{Time taken (min)} \rule[-1ex]{0pt}{0pt} & 28 & 30 & 26 & 19 & 21 & 12 & 13 & 11 & 8 \\
\hline
\end{array}

  1. Complete the scatterplot by adding the last four points from the table.  (2 marks)
     
       
  2. Add a line of best fit by eye to the graph in part (a).  (1 mark)
  3. The graph in part (a) shows the association between the time to complete the puzzle and the number of people in the group.
  4. Identify the form (linear or non-linear), the direction and the strength of the association.  (3 marks)

  5. Calculate the mean of the time taken to complete the puzzle for the three groups of size 7 observed in the dataset.  (1 mark)

Show Answers Only
  1.  
       
  2.  
       
  3. `text(The association is linear, negative and strong.)`
  4. `12\ text(minutes)`
Show Worked Solution

a.

b.


 

c.    `text(Form: linear)`

♦ Mean mark (c) 50%.

`text{Direction: negative}`

`text{Strength: strong}`
 

d. `text{Mean time (7 people)}` `= (12 + 13 + 11)/3`
    `= 12\ text(minutes)`

Statistics, 2ADV S2 2021 HSC 17

For a sample of 17 inland towns in Australia, the height above sea level, `x` (metres), and the average maximum daily temperature, `y` (°C), were recorded.

The graph shows the data as well as a regression line.
 

     
 

The equation of the regression line is  `y = 29.2 − 0.011x`.

The correlation coefficient is  `r = –0.494`.

  1. i.  By using the equation of the regression line, predict the average maximum daily temperature, in degrees Celsius, for a town that is 540 m above sea level. Give your answer correct to one decimal place.  (1 mark)

  2. ii. The gradient of the regression line is −0.011. Interpret the value of this gradient in the given context.  (2 marks)

  3. The graph below shows the relationship between the latitude, `x` (degrees south), and the average maximum daily temperature, `y` (°C), for the same 17 towns, as well as a regression line.
     
     
         
     
    The equation of the regression line is  `y = 45.6 − 0.683x`.
  4. The correlation coefficient is  `r = − 0.897`.
  5. Another inland town in Australia is 540 m above sea level. Its latitude is 28 degrees south.
  6. Which measurement, height above sea level or latitude, would be better to use to predict this town’s average maximum daily temperature? Give a reason for your answer.  (1 mark)

Show Answers Only
  1. i.  `23.3°text(C)`
  2. ii. `text(See Worked Solutions)`
  3. `text(Latitude. Correlation coefficient shows a stronger relationship.)`
Show Worked Solution
a.i.    `y` `=29.2 – 0.011(540)`
    `=23.26`
    `=23.3°text{C  (1 d.p.)`
 

a.ii.  `text(On average, the average maximum daily temperature of)`

`text(inland towns drops by 0.011 of a degree for every metre)`

`text(above sea level the town is situated.)`
 

b.  `text(The correlation co-efficient of the regression line using)`

`text(latitude is significantly stronger than the equivalent)`

`text(co-efficient for the regression line using height above sea)`

`text(level.)`

`:.\ text(The equation using latitude is preferred.)`

Trigonometry, 2ADV T1 2021 HSC 12

A right-angled triangle  `XYZ`  is cut out from a semicircle with centre `O`. The length of the diameter  `XZ`  is 16 cm and  `angle YXZ`  = 30°, as shown on the diagram.
 


 

  1. Find the length of  `XY`  in centimetres, correct to two decimal places.  (2 marks)

  2. Hence, find the area of the shaded region in square centimetres, correct to one decimal place.  (3 marks)

Show Answers Only
  1. `13.86 \ text{cm}`
  2. `45.1 \ text{cm}^2`
Show Worked Solution

 

a.    `cos 30^@` `=(XY)/16`
  `XY` `= 16 \ cos 30^@`
    `= 13.8564`
    `= 13.86 \ text{cm (2 d.p.)}`

 

b.    `text{Area of semi-circle}` `= 1/2 times pi r^2`
    `= 1/2 pi times 8^2`
    `= 100.531 \ text{cm}^2`

 

`text{Area of} \ Δ XYZ` `= 1/2 a b sin C`  
  `= 1/2 xx 16 xx 13.856 xx sin 30^@`  
  `= 55.42 \ text{cm}^2`  

 

`:. \ text{Shaded Area}` `= 100.531 – 55.42`  
  `= 45.111`  
  `= 45.1 \ text{cm}^2 \ text{(1 d.p.)}`  

Algebra, STD2 A4 2021 HSC 24

A population,  `P`,  is to be modelled using the function  `P = 2000 (1.2)^t`,  where `t` is the time in years.

  1. What is the initial population?  (1 mark)

  2. Find the population after 5 years.  (1 mark)

  3. On the axes below, draw the graph of the population against time, showing the points at  `t = 0`  and at  `t = 5`.  (2 marks)
      

Show Answers Only
  1. `2000`
  2. `4977`
  3. `text{See Worked Solutions}`
Show Worked Solution
a.    `text{Initial population occurs when}\ \  t = 0.`
`P` `= 2000 (1.2)^0`  
  `= 2000`  

 

b.    `text{Find} \ P \ text{when} \ \ t = 5: `

`P` `= 2000 (1.2)^5`  
  `= 4976.64`  
  `= 4977 \ text{(nearest whole)}`  

 

♦ Mean mark (c) 48%.

c. 

Financial Maths, 2ADV M1 2021 HSC 14

The first term of an arithmetic sequence is 5. The sum of the first 43 terms is 2021.

What is the common difference of the sequence?   (2 marks)

Show Answers Only

`2`

Show Worked Solution

`T_1 = a = 5`

`S_43` `= n/2 [2a + (n – 1)d]`
`2021` `= 43/2 (10 + 42d)`
`2021` `= 215 + 903d`
`903d` `= 1806`
`:. d` `= 2`

Statistics, 2ADV S2 2021 HSC 4 MC

The number of downloads of a song on each of twenty consecutive days is shown in the following graph.
 

Which of the following graphs best shows the cumulative number of downloads up to and including each day?

Show Answers Only

`C`

Show Worked Solution

`text{The gradient of the cumulative frequency histogram}`

`text{will increase gradually, be steepest at day 10 then}`

`text{decrease gradually.}`

`=> C`

Statistics, STD2 S1 2021 HSC 3 MC

The stem-and-leaf plot shows the number of goals scored by a team in each of ten netball games.
  

What is the mode of this dataset?

  1.  5
  2.  18
  3.  25
  4.  29
Show Answers Only

`C`

Show Worked Solution

`\text{Mode}  -> \text{data point with highest frequency}`

`\text{Mode}  = 25`

`=> C`

Financial Maths, STD2 F1 2021 HSC 19

Adam purchased some office furniture five years ago. It depreciated by $2300 each year based on the straight-line method of depreciation. The salvage value of the furniture is now $7500.

Find the initial value of the office furniture.  (2 marks)

Show Answers Only

`$19\ 000`

Show Worked Solution

`text{Find initial value}\ (V_0):`

`S` `=V_0-Dn`  
`7500` `=V_0-2300 xx 5`  
`V_0` `=7500 + 11\ 500`  
  `=$19\ 000`  

Algebra, STD2 A2 2021 HSC 18

The fuel consumption for a car is 6.7 litres/100 km. On a road trip, the car travels a distance of 1560 km and the fuel cost is $1.45 per litre.

What is the total fuel cost for the trip?  (2 marks)

Show Answers Only

`$151.55`

Show Worked Solution
`text(Total fuel used)` `=6.7 xx 1560/100`  
  `=104.52\ text(litres)`  

 

`text(Total fuel cost)` `=104.52 xx 1.45`  
  `=$151.55`  

Financial Maths, STD2 F4 2021 HSC 4 MC

Three years ago an appliance was valued at $2467. Its value has depreciated by 15% each year, based on the declining-balance method.

What is the salvage value today, to the nearest dollar?

  1. $952
  2. $1110
  3. $1357
  4. $1515
Show Answers Only

`D`

Show Worked Solution
`S` `= V_0 (1-r)^n`
  `= 2467 (1-0.15)^3`
  `= 2467 (0.85)^3`
  `= $1515`

 
`=>  D`

MATRICES, FUR2 2020 VCAA 3

An offer to buy the Westmall shopping centre was made by a competitor.

One market research project suggested that if the Westmall shopping centre were sold, each of the three centres (Westmall, Grandmall and Eastmall) would continue to have regular shoppers but would attract and lose shoppers on a weekly basis.

Let  `S_n`  be the state matrix that shows the expected number of shoppers at each of the three centres  `n`  weeks after Westmall is sold.

A matrix recurrence relation that generates values of  `S_n`  is

`S_(n+1) = T xx S_n`

`{:(quad qquad qquad qquad qquad qquad qquad qquad text(this week)),(qquad qquad qquad qquad qquad qquad quad \ W qquad quad G qquad quad \ E),(text(where)\ T = [(quad 0.80, 0.09, 0.10),(quad 0.12, 0.79, 0.10),(quad 0.08, 0.12, 0.80)]{:(W),(G),(E):}\ text(next week,) qquad qquad  S_0 = [(250\ 000), (230\ 000), (200\ 000)]{:(W),(G),(E):}):}`
 

  1. Calculate the state matrix, `S_1`, to show the expected number of shoppers at each of the three centres one week after Westmall is sold.  (1 mark)

Using values from the recurrence relation above, the graph below shows the expected number of shoppers at Westmall, Grandmall and Eastmall for each of the 10 weeks after Westmall is sold.
 


 

  1. What is the difference in the expected weekly number of shoppers at Westmall from the time Westmall is sold to 10 weeks after Westmall is sold?

      

    Give your answer correct to the nearest thousand.  (1 mark)

  2. Grandmall is expected to achieve its maximum number of shoppers sometime between the fourth and the tenth week after Westmall is sold.

      

    Write down the week number in which this is expected to occur.  (1 mark)

  3. In the long term, what is the expected weekly number of shoppers at Westmall?

      

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

Show Answers Only
  1. `S_1 =[(240\ 700),(231\ 700),(207\ 600)]`
  2. `30\ 000`
  3. `S_6 = T^6S_0 =  [(text(__)), (233\ 708), (text(__))]`
  4. `218\ 884`
Show Worked Solution
a.   `S_1` `= TS_0`
    `= [(0.80, 0.09, 0.10),(0.12, 0.79, 0.10),(0.08, 0.12, 0.80)][(250\ 000),(230\ 000),(200\ 000)]=[(240\ 700),(231\ 700),(207\ 600)]`

 

b.   `text(Using the graph)`
  `text(Difference)` `= 250\ 000 – 220\ 000`
    `= 30\ 000`

 

♦♦ Mean mark part (c) 27%.

c.  `text(Testing options:)`

`S_6 = T^6S_0 = [(0.80, 0.09, 0.10),(0.12, 0.79, 0.10),(0.08, 0.12, 0.80)]^6[(250\ 000),(230\ 000),(200\ 000)] = [(text(__)), (233\ 708), (text(__))]`
 

`:.\ text(Maximum shoppers in Grandmall expected in week 6.)`


♦ Mean mark part (d) 39%.

d.  `text(Test with high integer)\ n:`

`S_50 = T^50S_0 -> text(Westmall) = 218\ 884`

MATRICES, FUR2 2020 VCAA 1

The three major shopping centres in a large city, Eastmall `(E)`, Grandmall `(G)` and Westmall `(W)`, are owned by the same company.

The total number of shoppers at each of the centres at 1.00 pm on a typical day is shown in matrix `V`.
 

`qquad qquad qquad {:(qquad qquad qquad \ E qquad qquad G qquad qquad \  W),(V = [(2300,2700,2200)]):}`

 

  1. Write down the order of matrix `V`.  (1 mark)

Each of these centres has three major shopping areas: food `(F)`, clothing `(C)` and merchandise `(M)`.

The proportion of shoppers in each of these three areas at 1.00 pm on a typical day is the same at all three centres and is given in matrix `P` below.
 

`qquad qquad qquad P = [(0.48), (0.27), (0.25)] {:(F),(C),(M):}`
 

  1. Grandmall’s management would like to see 700 shoppers in its merchandise area at 1.00 pm.

     

    If this were to happen, how many shoppers, in total, would be at Grandmall at this time?  (1 mark)

  2. The matrix  `Q = P xx V`  is shown below. Two of the elements of this matrix are missing.

     

     

     

    `{:(quad qquad qquad qquad \ E qquad qquad G qquad qquad W), (Q = [(1104, \ text{___}, 1056 ), (621,\ text{___}, 594), (575, 675, 550)]{:(F),(C), (M):}):}`
     

    1. Complete matrix `Q` above by filling in the missing elements. (1 mark)
    2. The element in row `i` and column `j` of matrix `Q` is `q_(ij)`.

       

      What does the element `q_23` represent?  (1 mark)

The average daily amount spent, in dollars, by each shopper in each of the three areas at Grandmall in 2019 is shown in matrix  `A_2019`  below.
 

`qquad qquad A_2019 = [(21.30), (34.00), (14.70)] {:(F),(C),(M):}`
 

On one particular day, 135 shoppers spent the average daily amount on food, 143 shoppers spent the average daily amount on clothing and 131 shoppers spent the average daily amount on merchandise.

  1. Write a matrix calculation, using matrix  `A_2019`, showing that the total amount spent by all these shoppers is $9663.20  (1 mark)
  2. In 2020, the average daily amount spent by each shopper was expected to change by the percentage shown in the table below.
     

     

       Area food clothing merchandise
       Expected change       increase by 5%       decrease by 15%       decrease by 1%   

     

      


    The average daily amount, in dollars, expected to be spent in each area in 2020 can be determined by forming the matrix product
     
    `qquad qquad A_2020 = K xx A_2019`
     

     

    Write down matrix `K`.  (1 mark)

Show Answers Only
  1. `1 xx 3`
  2. `2800`
  3.  i.   `{:(quad qquad qquad qquad \ E qquad qquad G qquad qquad W), (Q = [(1104, 1296, 1056 ), (621, 729, 594), (575, 675, 550)]{:(F),(C), (M):}):}`
     
  4. ii. `q_23\ text(represents the number of people)`
  5.    `text(in the clothing area of Westmall.)`
  6. `text(Total spent) = [(135, 143, 131)] = [(21.30), (34.00), (14.70)] = [9663.20]`
  7. `K = [(1.05, 0, 0),(0, 0.85, 0),(0, 0, 0.99)]`
Show Worked Solution

a.  `1 xx 3`

♦ Mean mark part (b) 45%.
b.   `0.25 xx G\ text(shoppers in)\ M` `= 700`
  `:. G\ text(shoppers in)\ M` `= 700/0.25`
    `= 2800`

 

c.i.   `{:(quad qquad qquad qquad \ E qquad qquad G qquad qquad W), (Q = [(1104, 1296, 1056 ), (621, 729, 594), (575, 675, 550)]{:(F),(C), (M):}):}`

 

c.ii.   `q_23\ text(represents the number of people)`
  `text(in the clothing area of Westmall.)`

 

d.  `text(Total spent) = [(135, 143, 131)] [(21.30), (34.00), (14.70)] = [9663.20]`

♦♦ Mean mark part (e) 24%.
 

e.   `A_2020` `= K xx [(21.30), (34.00), (14.70)]`
  `:. K` `= [(1.05, 0, 0),(0, 0.85, 0),(0, 0, 0.99)]`

CORE, FUR2 2020 VCAA 8

Samuel has a reducing balance loan.

The first five lines of the amortisation table for Samuel’s loan are shown below.
 


 

Interest is calculated monthly and Samuel makes monthly payments of $1600.

Interest is charged on this loan at the rate of 3.6% per annum.

  1. Using the values in the amortisation table
  2.  i. calculate the principal reduction associated with payment number 3  (1 mark)
  3. ii. calculate the balance of the loan after payment number 4 is made.
  4.     Round your answer to the nearest cent.  (1 mark)
  5. Let `S_n` be the balance of Samuel’s loan after `n` months.
  6. Write down a recurrence relation, in terms of `S_0, S_(n+1)`  and  `S_n`, that could be used to model the month-to-month balance of the loan.  (1 mark)
Show Answers Only
  1.  i. `$643.85`
  2. ii. `$317\ 428.45`
  3. `text(See Worked Solutions)`
Show Worked Solution
a.i.   `text(Principal reduction)` `=\ text(Payment – interest)`
    `= 1600 – 956.15`
    `= $643.85`

 

♦ Mean mark part a.ii. 39%.
a.ii.   `text(Interest)` `= 318\ 074.23 xx (0.036/12)`
    `= $954.22`

 
`:.\ text(Balance after payment 4)`

`= 318\ 074.23 – 1600 + 954.22`

`= $317\ 428.45`

 

♦♦ Mean mark part b. 30%.

b.   `S_0 = 320\ 000,\ S_(n+1)` `= S_n(1 + 0.036/12) – 1600`
  `S_(n+1)` `= 1.003 S_n – 1600`

CORE, FUR2 2020 VCAA 6

The table below shows the mean age, in years, and the mean height, in centimetres, of 648 women from seven different age groups.
 


 

  1. What was the difference, in centimetres, between the mean height of the women in their twenties and the mean height of the women in their eighties?  (1 mark)

A scatterplot displaying this data shows an association between the mean height and the mean age of these women. In an initial analysis of the data, a line is fitted to the data by eye, as shown.
 

 

  1. Describe this association in terms of strength and direction.  (1 mark)

  2. The line on the scatterplot passes through the points (20,168) and (85,157).

     

    Using these two points, determine the equation of this line. Write the values of the intercept and the slope in the appropriate boxes below.

     

    Round your answers to three significant figures.  (1 mark)

mean height = 
 
   +  
 
   × mean age

 

  1. In a further analysis of the data, a least squares line was fitted.

     

    The associated residual plot that was generated is shown below.

     
     

          

     

    The residual plot indicates that the association between the mean height and the mean age of women is non-linear.

     

    The data presented in the table in part a is repeated below. It can be linearised by applying an appropriate transformation to the variable mean age.

     

      

     

    Apply an appropriate transformation to the variable mean age to linearise the data. Fit a least squares line to the transformed data and write its equation below.

     

    Round the values of the intercept and the slope to four significant figures.  (2 marks)


Show Answers Only
  1. `10.4\ text(cm)`
  2. `text(Strong and negative)`
  3. `text(mean height) = 171 – 0.169 xx text(mean age)`
  4. `text(mean height) = 167.9 – 0.001621 xx text{(mean age)}^2`
Show Worked Solution
a.   `text(Difference)` `= 167.1 – 156.7`
    `= 10.4\ text(cm)`

 

Mean mark part b. 51%.

b.  `text(Strong and negative.)`

 

♦♦ Mean mark part c. 23%.

c.   `text(Slope) = (157 – 168)/(85 – 20) = -0.169`

`text(Equation of line)`

`y – 168` `= -0.1692 (x – 20)`
`y` `= -0.169x + 171`

 
`:.\ text(mean height) = 171 – 0.169 xx text(mean age)`

 

D.    `text(By CAS)`

`text(mean height) = 167.9 – 0.001621 xx text{(mean age)}^2`

CORE, FUR2 2020 VCAA 4

The age, in years, body density, in kilograms per litre, and weight, in kilograms, of a sample of 12 men aged 23 to 25 years are shown in the table below.
 

          Age       
(years)

        Body density        
(kg/litre)

        Weight        
(kg)
  23 1.07 70.1
  23 1.07 90.4
  23 1.08 73.2
  23 1.08 85.0
  24 1.03 84.3
  24 1.05 95.6
  24 1.07 71.7
  24 1.06 95.0
  25 1.07 80.2
  25 1.09 87.4
  25 1.02 94.9
  25 1.09 65.3

 

  1. For these 12 men, determine
  2.  i. their median age, in years  (1 mark)
  3. ii. the mean of their body density, in kilograms per litre.  (1 mark)
  4. A least squares line is to be fitted to the data with the aim of predicting body density from weight.
  5.  i. Name the explanatory variable for this least squares line.  (1 mark)
  6. ii. Determine the slope of this least squares line.
  7.     Round your answer to three significant figures.  (1 mark)
  8. What percentage of the variation in body density can be explained by the variation in weight?
  9. Round your answer to the nearest percentage.  (1 mark)
Show Answers Only
  1.  i. `24`
  2. ii. `1.065\ text(kg/litre)`
  3. i. `text(Weight)`
  4. ii. `text(Slope) = -0.00112\ text{(by CAS)}`
  5. `29 text(%)`
Show Worked Solution
a.i.   `n = 12`  
  `text(Median)` `= (text{6th + 7th})/2`
    `= (24 + 24)/2`
    `= 24`

 

a.ii.   `text(Mean)` `= (∑\ text{body density})/12`
    `= 1.065\ text(kg/litre)`

 

b.i.   `text(Weight)`

♦ Mean mark b.ii. 29%.
MARKER’S COMMENT: Most students did not round correctly.

b.ii.   `text(Slope) = -0.00112\ text{(by CAS)}`

 

c.   `r` `= -0.53847\ text{(by CAS)}`
  `r^2` `= 0.289…`

 
`:. 29 text(%)`

CORE, FUR2 2020 VCAA 3

In a study of the association between BMI and neck size, 250 men were grouped by neck size (below average, average and above average) and their BMI recorded.

Five-number summaries describing the distribution of BMI for each group are displayed in the table below along with the group size.

The associated boxplots are shown below the table.
 

 

  1. What percentage of these 250 men are classified as having a below average neck size?   (1 mark)
  2. What is the interquartile range (IQR) of BMI for the men with an average neck size?   (1 mark)
  3. People with a BMI of 30 or more are classified as being obese.
  4. Using this criterion, how many of these 250 men would be classified as obese? Assume that the BMI values were all rounded to one decimal place.  (1 mark)
  5. Do the boxplots support the contention that BMI is associated with neck size? Refer to the values of an appropriate statistic in your response.  (2 marks)
Show Answers Only
  1. `20 text(%)`
  2. `2.6`
  3. `23`
  4. `text(See Worked Solutions)`
Show Worked Solution
a.    `text(Percentage)` `= 50/250 xx 100`
    `= 20text(%)`

 

b.    `text(IQR)` `= 26.0 – 23.4`
    `= 2.6`

 

c.   `text{Outliers in average neck size}\ (text(BMI) >= 30) = 4`

♦♦ Mean mark part c. 22%.
COMMENT: Many students incorrectly counted the two “above average” outliers twice.

`:.\ text(Number classified as obese)`

`= 4 + 1/4 xx 76`

`= 23`

 

d.   `text(The boxplots support a strong association between)`

♦ Mean mark 49%.
MARKER’S COMMENT: General statement of change = 1 mark. Median or IQR values need to be quoted directly for the second mark.

`text(BMI and neck size as median BMI values increase)`

`text(as neck size increases.)`

`text(Below average neck sizes have a BMI of 21.6, which)`

`text(increases to 24.6 for average neck sizes and increases)`

`text(further to 28.1 for above average neck sizes.)`

CORE, FUR2 2020 VCAA 1

Body mass index (BMI), in kilograms per square metre, was recorded for a sample of 32 men and displayed in the ordered stem plot below.
  

  1. Describe the shape of the distribution.  (1 mark)
  2. Determine the median BMI for this group of men.  (1 mark)
  3. People with a BMI of 25 or over are considered to be overweight.
  4. What percentage of these men would be considered to be overweight?  (1 mark)
Show Answers Only
  1. `text(Positively skewed)`
  2. `24.55`
  3. `37.5 text(%)`
Show Worked Solution

a.   `text(Positively skewed)`
 

b.   `32\ text(data points)`

`text(Median)` `= text(16th + 17th)/2`
  `= (24.5 + 24.6)/2`
  `= 24.55`

 

c.    `text(Percentage)` `= 12/32 xx 100`
    `= 37.5%`

Geometry, NAPX-p124059v02

Nancy is making a rectangular prism using plastic balls and sticks.
  

     
 

How many more sticks does Nancy need to finish the rectangular prism?

  1   2   4   5   6
 
 
 
 
 
Show Answers Only

`5`

Show Worked Solution

`text{5 more sticks are needed.}`
 

Geometry, NAPX-p124059v01

Heath is making a square pyramid using plastic balls and sticks.
 


 

How many more sticks does Heath need to finish the square pyramid?

  1   2   3   4   5
 
 
 
 
 
Show Answers Only

`3`

Show Worked Solution

`text{3 more sticks are needed.}`

Calculus, SPEC2 2020 VCAA 3

Let  `f(x) = x^2e^(−x)`.

  1. Find an expression for  `f′(x)`  and state the coordinates of the stationary points of  `f(x)`.  (2 marks)
  2. State the equation(s) of any asymptotes of  `f(x)`.  (1 mark)
  3. Sketch the graph of  `y = f(x)`  on the axes provided below, labelling the local maximum stationary point and all points of inflection with their coordinates, correct to two decimal places.  (3 marks)
     
         

Let  `g(x) = x^n e^(−x)`, where  `n ∈ Z`.

  1. Write down an expression for  `g″(x)`.  (1 mark)
  2.  i. Find the non-zero values of `x` for which  `g″(x) = 0`.  (1 mark)
  3. ii. Complete the following table by stating the value(s) of `n` for which the graph of  `g(x)`  has the given number of points of inflection.  (2 marks)
     
       
         
Show Answers Only
  1. `f(0) = 0; f(2) = 4e^(−2) text(and SP’s at)\ (0, 0), (2, 4e^(−2))`
  2. `y = 0`
  3.   
  4. `g″(x) = x^(n – 2) e^(−x)(x^2 – 2xn + n^2 – n)`
  5.  i. `x = x ± sqrtn`
  6. ii.
       
Show Worked Solution

a.   `f′(x) = 2xe^(−x) – x^2e^(−x)`

`text(SP’s when)\ \ f′(x) = 0:`

`x^2e^(−x)` `= 2xe^(−x)`
`x` `= 2\ \ text(or)\ \ 0`

 
`f(0) = 0; \ f(2) = 4e^(−2)`

`:. text(SP’s at)\ \ (0, 0) and  (2, 4e^(−2))`

 

b.   `text(As)\ \ x -> ∞, \ f(x) -> 0^+`

`:. text(Horizontal asymptote at)\ \ y = 0`

 

c.   

`text(POI when)\ \ f″(x) = 0`

`:. text(POI’s:)\ (0.59, 0.19), \ (3.41, 0.38)`

 

d.   `g′(x) = x^(n – 1) e^(−x)(n – x)`

`g″(x) = x^(n – 2) e^(−x)(x^2 – 2xn + n^2 – n)`

 

e.i.   `text(Solve:)\ \ x^2 – 2xn + n^2 – n = 0`

`x = n ± sqrtn`

♦♦♦ Mean mark (e)(ii) 9%.

 

e.ii.   

Number, NAPX-p116616v04

Justin owns a collection of action figures which is more than 674 action figures and less than 764.

Which of these numbers could represent the number of action figures that Justin owns?

 792  724  648  772
 
 
 
 
Show Answers Only

`724`

Show Worked Solution

`text{The number of action figures must be between 674 and 764 action figures}`

`text{Check each option:}`

`text{792 is more than 764 ⇒ incorrect}`

`text{724 is more than 674 and less than 764 ⇒ correct}`

`text{648 is less than 674 ⇒ incorrect}`

`text{772 is more than 764 ⇒ incorrect}`

Number, NAPX-p116616v03

Eric owns a book which has more than 258 pages but less than 285 pages

Which of these could represent the number of pages in Eric’s book?

 287  294  249  262
 
 
 
 

Show Answers Only

`262`

Show Worked Solution

`text{Check each option:}`

`text{287 is more than 285 ⇒ incorrect}`

`text{294 is more than 285 ⇒ incorrect}`

`text{249 is less than 258 ⇒ incorrect}`

`text{262 is more than 258 and less than 285 ⇒ correct}`

Number, NAPX-p116602v04

Kiara receives 5 cents for every used soft drink she recycles.

If she recycled 12 of these cans, how much did she earn?

  $0.12  $0.40   $0.60   $1.20
 
 
 
 

Show Answers Only

`$0.60`

Show Worked Solution
`text{Total money earned }` `= 12 \ text{cans} xx 5 \ text{cents}`
  `= 60\ text(cents)`
  `= $0.60`