SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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 2

The neck size, in centimetres, of 250 men was recorded and displayed in the dot plot below.
 

  1. Write down the modal neck size, in centimetres, for these 250 men.   (1 mark)

  2. Assume that this sample of 250 men has been drawn at random from a population of men whose neck size is normally distributed with a mean of 38 cm and a standard deviation of 2.3 cm.
  3.  i. How many of these 250 men are expected to have a neck size that is more than three standard deviations above or below the mean? Round your answer to the nearest whole number.   (1 mark)

  4. ii. How many of these 250 men actually have a neck size that is more than three standard deviations above or below the mean?   (1 mark)

  5. The five-number summary for this sample of neck sizes, in centimetres, is given below.
     

    Use the five-number summary to construct a boxplot, showing any outliers if appropriate, on the grid below.   (2 marks)

Show Answers Only

a.    `text(Mode) = 38\ text(cm)`

b.i.   `1`

b.ii. `1`

c.    `text(See Worked Solutions)`

Show Worked Solution

a.   `text(Mode) = 38\ text(cm)`

 

♦ Mean mark part b.i. 40%.
b.i.   `text(Expected number of men)` `= (1-0.997) xx 250`
    `= 0.75`
    `= 1\ text{(nearest whole)}`

 

♦ Mean mark part b.ii. 42%.
b.ii.   `text(When)\ \ z = +- 3`
  `text(Neck size limits)` `= 38 +- (2.3 xx 3)`
    `= 44.9 or 31.1`

 
`:.\ text(1 man has neck size outside 3 s.d.)`
 

c.   `IQR = 39-36=3`

`text(Upper fence)\ =Q_3 + 1.5 xx 3=39 + 4.5=43.5`

`text(Lower fence)\ =Q_1-1.5 xx 3=36-4.5=31.5`

 

GRAPHS, FUR1 2020 VCAA 2 MC

At a school concert, the entry fee for adults was different from the entry fee for children.

The entry fee for three adults and four children was $67.00

The entry fee for two adults and five children was $57.50

Let  `x`  be the entry fee for an adult.

Let  `y`  be the entry fee for a child.

A pair of simultaneous equations that could be used to represent the situation above is

A.   `3x + 2y = 57.5` B.   `3x + 2y = 67`
 

`4x + 5y = 67`

 

  

`4x + 5y = 57.5`

 
C.   `3x + 4y = 57.5` D.   `3x + 4y = 67`
 

`2x + 5y = 67`

 

  

`2x + 5y = 57.5`

 
E.   `4x + 3y = 67`    
  `5x + 2y = 57.5`

 

    
Show Answers Only

`D`

Show Worked Solution
`3x + 4y` `= 67`
`2x + 5y` `= 57.5`

 
`=>  D`

CORE, FUR1 2020 VCAA 10-12 MC

The data in Table 2 was collected in a study of the association between the variables frequency of nightmares (low, high) and snores (no, yes).
 


 

Part 1

The variables in this study, frequency of nightmares (low, high) and snores (no, yes), are

  1. ordinal and nominal respectively.
  2. nominal and ordinal respectively.
  3. both numerical.
  4. both ordinal.
  5. both nominal.

 
Part 2

The percentage of participants in the study who did not snore is closest to

  1. 42.0%
  2. 43.5%
  3. 49.7%
  4. 52.2%
  5. 56.5%

 
Part 3

Of those people in the study who did snore, the percentage who have a high frequency of nightmares is closest to

  1.   7.5%
  2. 17.1%
  3. 47.8%
  4. 52.2%
  5. 58.0%
Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ E`

`text(Part 3:)\ B`

Show Worked Solution

Part 1

`text{frequency of nightmares (low, high) is ordinal.}`

`text{snores (no, yes) is nominal.}`

`=> A`
 

Part 2

`text(Percentage)` `= text(not snore)/text(total participants) xx 100`
  `= 91/161 xx 100`
  `= 56.5%`

`=> E`
 

Part 3

`text(Percentage)` `= text(High frequency and snore)/text(total who snore) xx 100`
  `= 12/70 xx 100`
  `= 17.1%`

`=> B`

CORE, FUR1 2020 VCAA 1-3 MC

The times between successive nerve impulses (time), in milliseconds, were recorded.

Table 1 shows the mean and the five-number summary calculated using 800 recorded data values.
 


 

Part 1

The difference, in milliseconds, between the mean time and the median time is

  1.  10
  2.  70
  3.  150
  4.  220
  5.  230

 
Part 2

Of these 800 times, the number of times that are longer than 300 milliseconds is closest to

  1. 20
  2. 25
  3. 75
  4. 200
  5. 400

 
Part 3

The shape of the distribution of these 800 times is best described as

  1. approximately symmetric.
  2. positively skewed.
  3. positively skewed with one or more outliers.
  4. negatively skewed.
  5. negatively skewed with one or more outliers.
Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ D`

`text(Part 3:)\ C`

Show Worked Solution

Part 1

`text(Difference)` `= 220 -150`
  `= 70`

`=> B`
 

Part 2

`Q_3 = 300`

`:.\ text(Impulses longer than 300 milliseconds)`

`= 25text(%) xx 800`

`= 200`

`=> D`
 

Part 3

`text(Distribution has a long tail to the right)`

♦ Mean mark 50%.

`:.\ text(Positively skewed)`

`text(Upper fence)` `= Q_3 + 1.5 xx IQR`
  `= 300 + 1.5 (300 – 70)`
  `= 645`

 
`=> C`

Complex Numbers, EXT2 N1 2004 HSC 2b

Let  `alpha = 1 + i sqrt3`  and  `beta = 1 + i`.

  1. Find  `frac{alpha}{beta}`, in the form  `x + i y`.   (1 mark)

  2. Express `alpha` in modulus-argument form.   (3 marks)

  3. Given that `beta` has the modulus-argument form
     
         `beta = sqrt2 (cos frac{pi}{4} + i sin frac{pi}{4})`.
     
    find the modulus-argument form of  `frac{alpha}{beta}`.   (1 mark)

  4. Hence find the exact value of  `sin frac{pi}{12}`   (1 mark)

Show Answers Only
  1. `frac{1+sqrt3}{2} + i (frac{sqrt3 – 1}{2})`
  2. `2 \ text{cis} (frac{pi}{3})`
  3. `sqrt2 \ text{cis} (frac{pi}{12})`
  4. `frac{sqrt6-sqrt2}{4}`
Show Worked Solution
i.     `frac{alpha}{beta}` `= frac{1 + i sqrt3}{1 + i} xx frac{1 – i}{1 – i}`
    `= frac{(1 + i sqrt3)(1 – i)}{1^2 – i^2}`
    `= frac{1 – i + i sqrt3 – i^2 sqrt3}{2}`
    `= frac{1+sqrt3}{2} + i (frac{sqrt3 – 1}{2})`

 

ii.   `alpha` `= 1 + i sqrt3`
  `| alpha |` `= sqrt(1^2 + (sqrt3)^2) = 2`

`text{arg} \ (alpha) = tan^-1 (frac{sqrt3}{1}) = frac{pi}{3}`

`therefore \ alpha = 2 text{cis} (frac{pi}{3})`

 

iii.   `beta` `= sqrt2 text{cis} (frac{pi}{4})`
  `frac{alpha}{beta}` `= frac{2}{sqrt2} \ text{cis} (frac{pi}{3} – frac{pi}{4})`
    `= sqrt2 text{cis}  (frac{pi}{12})`

 

iv.  `text{Equating imaginary parts of i and ii:}`

`sqrt2 \ sin \ (frac{pi}{12})` `= frac{sqrt3 – 1}{2}`
`sin (frac{pi}{12})` `= frac{sqrt3 – 1}{2 sqrt2} xx frac{sqrt2}{sqrt2}`
  `= frac{sqrt6 – sqrt2}{4}`

Complex Numbers, EXT1 N1 SM-Bank 5

Let  `z = 1 + 2 i`  and  `w = 3 - i`.

Find, in the form  `x + i y`,

  1.  `zw`   (1 mark)

  2.  `overset_((frac{10}{z}))`.   (1 mark)

Show Answers Only
  1. `5 + 5 i`
  2. `2+4 i`
Show Worked Solution
i.    `zw` `= (1 + 2 i)(3 – i)`
    `= 3 – i + 6 i – 2 i^2`
    `= 5 + 5 i`

 

ii.    `frac{10}{z}` `= frac{10}{1 + 2 i} xx frac{1-2 i}{1-2 i}`
    `= frac{10-20 i}{1^2 – (2 i)^2`
    `= frac{10-20 i}{1+4}`
    `= 2-4i`

 
`therefore \ overset_(frac{10}{z})= 2+4 i`

Complex Numbers, EXT2 N2 EQ-Bank 1

`z = sqrt2 e^((ipi)/15)`  is a root of the equation  `z^5 = alpha(1 + isqrt3), \ alpha ∈ R`.

  1. Express  `1 + isqrt3`  in exponential form.  (2 marks)

  2. Find the value of `alpha`.  (1 mark)

  3. Find the other 4 roots of the equation in exponential form.  (3 marks)

Show Answers Only
  1. `2e^((ipi)/3)`
  2. `alpha = 2sqrt2`
  3. `e^((i11pi)/15), e^(-(ipi)/3), e^((i13pi)/15), e^(−(i11pi)/15)`
Show Worked Solution

i.   `beta = 1 + isqrt3`

`|beta| = sqrt(1 + (sqrt3)^2) = 2`

`text(arg)(beta) = tan^(−1) (sqrt3/1) = pi/3`

`beta = 2e^((ipi)/3)`

 

ii.    `z` `= sqrt2 e^((ipi)/15)`
  `z^5` `= (sqrt2 e^((ipi)/15))^5`
    `= (sqrt2)^5 e^((ipi)/15 xx 5)`
    `= 4sqrt2 e^((ipi)/3)`

 
`:. alpha = 2sqrt2`

 

iii.   `text(arg)(z^5) = pi/3 + 2kpi, \ \ k = 0, ±1, ±2, …`

`text(arg)(z) = pi/15 + (2kpi)/5`

`k = 1:\ text(arg)(z) = pi/15 + (2pi)/5 = (11pi)/15`

`k = text(−1):\ text(arg)(z) = pi/15 – (2pi)/5 = −pi/3`

`k = 2:\ text(arg)(z) = pi/15 + (4pi)/5 = (13pi)/15`

`k =text(−2):\ text(arg)(z) = pi/15 – (4pi)/5 = −(11pi)/15`
 

`:. 4\ text(other roots are:)`

`e^((i11pi)/15), e^(−(ipi)/3), e^((i13pi)/15), e^(−(i11pi)/15)` 

Complex Numbers, EXT2 N1 2005 HSC 2b

Let  `beta = 1-i sqrt3`.

  1. Express  `beta`  in modulus-argument form.   (2 marks)

  2. Express  `beta^5`  in modulus-argument form.   (2 marks)

  3. Hence express  `beta^5`  in the form  `x+iy`.   (1 mark)

Show Answers Only
  1. `2 \ text{cis} (-frac{pi}{3})`
  2. `32 \ text{cis} (frac{pi}{3})`
  3. `16 + i 16 sqrt3`
Show Worked Solution

i.    `beta = 1 – i sqrt3`
 

 
`| beta | = sqrt(1^2 + (sqrt3)^2) = 2`

`tan theta` `= frac{sqrt3}{1} = sqrt3`
`theta` `= frac{pi}{3}`
`text{arg} (beta)` `= -frac{pi}{3}`

`therefore \ beta = 2 \ text{cis} (-frac{pi}{3})`

 

ii.   `beta^5` `= 2^5 \ text{cis} (-frac{pi}{3} xx5)`
    `= 32 \ text{cis} (-frac{5pi}{3} + 2 pi)`
    `= 32 \ text{cis} (frac{pi}{3})`

 

iii.   `beta^5` `= 32 ( cos (frac{pi}{3}) + i sin (frac{pi}{3}) )`
    `= 32 ( frac{1}{2} + i  frac{sqrt3}{2})`
    `= 16 + i 16 sqrt3`

Complex Numbers, EXT2 N1 2005 HSC 2a

Let  `z = 3 + i`  and  `w=1-i`.  Find, in the form  `x+iy`,

  1.  `2z+iw`   (1 mark)

  2.  `bar z w`   (1 mark)

  3.  `6/w`   (1 mark)

Show Answers Only
  1.  `7+3i`
  2.  `2-4i`
  3.  `3+3i`
Show Worked Solution

i.    `z = 3 + i \ , \ w = 1 – i`

`2z + i w` `=2 (3 + i) + i(1 – i)`
  `= 6 + 2i + i – i^2`
  `= 7 + 3 i`

 

ii.    `overset_z w` `= (3 – i)(1 – i)`
    `= 3 – 3i –  i + i^2`
    `= 2 – 4 i`

 

iii.   `frac{6}{w}` `= frac{6}{1 – i} xx frac{1 + i}{1 + i}`
    `= frac{6 + 6i}{1^2 – i^2}`
    `= frac{6 + 6i}{2}`
    `= 3 + 3i`

Complex Numbers, EXT2 N1 2020 HSC 13d

  1. Show that for any integer `n`,  `e^(i n theta) + e^(-i n theta) = 2 cos (n theta)`.   (1 mark)

  2. By expanding  `(e^(i theta) + e^(-i theta))^4`  show that
     
       `cos^4 theta = frac{1}{8} ( cos (4 theta) + 4 cos (2 theta) + 3 )`.   (3 marks)

  3. Hence, or otherwise, find  `int_0^(frac{pi}{2}) cos^4 theta\ d theta`.   (2 marks)

Show Answers Only
  1. `text{See Worked Solution}`
  2. `text{See Worked Solution}`
  3. `text{See Worked Solution}`
Show Worked Solution
i.      `e^(i n theta) + e^(-i n theta)` `= cos(n theta) + i sin(n theta) + cos(-n theta) + i sin(-n theta)`
    `= cos(n theta) + i sin(n theta) + cos(n theta) – i sin(n theta)`
    `= 2 cos (n theta)`

 

ii.    `(e^{i theta} + e^{-i theta})^4` `= (2 cos theta)^4`
    `= 16 cos^4 theta`

 
`text{Expand} \ (e^{i theta} + e^{-i theta})^4 :`

`e^(i 4 theta) + 4 e^(i 3 theta) e^(-i theta) + 6 e^(i 2 theta) e^(-i 2 theta) + 4 e^(i theta) e^(-i 3 theta) + e^(-i 4 theta)`

`= e^(i 4 theta) + 4e^(i 2 theta) + 6 + 4^(-i 2 theta) + e^(-i 4 theta)`

`= e^(i 4 theta) + e^(i 4 theta) + 4 (e^{i 2 theta} + e^{-i 2 theta}) + 6`

`= 2 cos (4 theta) + 8 cos (2 theta) + 6`
 

`therefore \ 16 cos^4 theta` `= 2 cos (4 theta) + 8 cos (2 theta) + 6`
`cos^4 theta` `= frac{1}{8} cos(4 theta) + 1/2 cos(2 theta) + 3/8`
`cos^4 theta` `= frac{1}{8} (cos(4 theta) + 4 cos(2 theta) + 3)`

 

iii.    `int_0^(frac{pi}{2}) cos^4 theta\ d theta` `= frac{1}{8} int_0^(frac{pi}{2}) cos(4 theta) + 4 cos(2 theta) + 3\ d theta`
    `= frac{1}{8} [ frac{1}{4} sin(4 theta) + 2 sin (2 theta) + 3 theta ]_0^(frac{pi}{2}`
    `= frac{1}{8} [( frac{1}{4} sin (2 pi) + 2 sin pi  + frac{3 pi}{2}) – 0 ]`
    `= frac{1}{8} ( frac{3 pi}{2})`
    `= frac{3 pi}{16}`

Complex Numbers, EXT2 N1 2020 HSC 11a

Consider the complex numbers  `w = -1 + 4i`  and  `z = 2 -i`.

  1. Evaluate  `|w|`.   (1 mark)

  2. Evaluate  `w overset_ z`.   (2 marks)

Show Answers Only
  1. `text{See Worked Solutions}`
  2. `text{See Worked Solutions}`
Show Worked Solution
i.      `w` ` = -1 + 4 i`
  `|w| ` `= sqrt{(-1)^2 + 4^2} = 17`

 

ii.     `z = 2 – i`

`overset_z = 2 + i`

`w overset_z` `= (-1 + 4i)(2 + i)`
  `= -2 – i +8 i + 4i^2`
  `= -6 + 7i`

Calculus, EXT2 C1 2020 HSC 6 MC

Which expression is equal to `int frac{1}{x^2 + 4x + 10}\ dx`?

  1. `frac{1}{sqrt(6)} tan^-1 (frac{x + 2}{sqrt{6} )) + c`
  2. `tan^-1 (frac{x + 2}{sqrt{6} )) + c`
  3. `frac{1}{2 sqrt(6)} ln | frac{x + 2 - sqrt(6)}{x + 2 + sqrt(6)} | + c`
  4. `ln | frac{x + 2 - sqrt(6)}{x + 2 + sqrt(6)} | + c`
Show Answers Only

`A`

Show Worked Solution
`int frac{1}{x^3 + 4x + 10}\ dx` `= int frac{1}{(x + 2)^2 + (sqrt6)^2}\ dx`
  `= frac{1}{6} tan^-1 (frac{x + 2}{sqrt6}) + c`

`=> \ A`

Functions, EXT1 F2 2020 HSC 11a

Let  `P(x) = x^3 + 3x^2-13x + 6`.

  1. Show that  `P(2) = 0`.  (1 mark)

  2. Hence, factor the polynomial  `P(x)`  as  `A(x)B(x)`, where  `B(x)`  is a quadratic polynomial.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `P(x) = (x-2)(x^2 + 5x – 3)`
Show Worked Solution
i.    `P(2)` `= 8 + 12-26 + 6`
    `= 0`

 

ii.   

`:. P(x) = (x-2)(x^2 + 5x – 3)`

Algebra, STD2 A4 2020 HSC 33

The graph shows the number of bacteria, `y`, at time `n` minutes. Initially (when `n = 0`) the number of bacteria is 1000.
 


 

  1. Find the number of bacteria at 40 minutes.   (1 mark)

  2. The number of bacteria can be modelled by the equation  `y = A xx b^n`, where `A` and `b` are constants.
    Use the guess and check method to find, to two decimal places, an upper and lower estimate for the value of `b`. The upper and lower estimates must differ by 0.01. (2 marks)

Show Answers Only
  1. `4000`
  2. `text{Show Worked Solutions}`
Show Worked Solution

a.    `text{When} \ \ n = 40,`

`text{Number of Bacteria} \ (y) = 4000`
 

b.   `A = 1000 \ => \ y = 1000  b^n`

♦♦♦ Mean mark 14%.

`text{By inspection, graph passes through (40, 4000)}`

`=> \ 4000 = 1000  b^40`
 

`text(Guess and check possible values of)\ b:`

`text{If} \ \ b = 1.03 \ , \ \ y = 1000 xx 1.03^40 = 3262 \ text{(too low)}`

`text{If} \ \ b = 1.04 \ , \ \ y = 1000 xx 1.04^40 = 4801 \ text{(too high)}`

`therefore \ 1.03 < b < 1.04`

Statistics, EXT1 S1 2020 HSC 12b

When a particular biased coin is tossed, the probability of obtaining a head is `3/5`.

This coin is tossed 100 times.

Let `X` be the random variable representing the number of heads obtained. This random variable will have a binomial distribution.

  1. Find the expected value, `E(X)`.  (1 mark)

  2. By finding the variance, `text(Var)(X)`, show that the standard deviation of `X` is approximately 5.  (1 mark)

  3. By using a normal approximation, find the approximate probability that `X` is between 55 and 65.  (1 mark)

Show Answers Only
  1. `60`
  2. `text(See Worked Solutions)`
  3. `68text(%)`
Show Worked Solution

i.   `X = text(number of heads)`

`X\ ~\ text(Bin) (n, p)\ ~\ text(Bin) (100, 3/5)`

`E(X)` `= np`
  `= 100 xx 3/5`
  `= 60`

 

ii.    `text(Var)(X)` `= np(1 – p)`
    `= 60 xx 2/5`
    `= 24`

 

`sigma(x)` `= sqrt24`
  `~~ 5`

 

iii.    `P(55 <= x <=65)` `~~ P(−1 <= z <= 1)`
    `~~ 68text(%)`

Functions, 2ADV F1 2020 HSC 11

There are two tanks on a property, Tank `A` and Tank `B`. Initially, Tank `A` holds 1000 litres of water and Tank B is empty.

  1.  Tank `A` begins to lose water at a constant rate of 20 litres per minute. The volume of water in Tank `A` is modelled by  `V = 1000 - 20t`  where  `V`  is the volume in litres and  `t`  is the time in minutes from when the tank begins to lose water.   (1 mark)
     
    On the grid below, draw the graph of this model and label it as Tank `A`.
     
       

  2. Tank `B` remains empty until  `t=15`  when water is added to it at a constant rate of 30 litres per minute.

     

    By drawing a line on the grid (above), or otherwise, find the value of  `t`  when the two tanks contain the same volume of water.  (2 marks)

  3. Using the graphs drawn, or otherwise, find the value of  `t`  (where  `t > 0`) when the total volume of water in the two tanks is 1000 litres.  (1 mark)

Show Answers Only
  1.  `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}` 
      
  2. `29 \ text{minutes}`
  3. `45 \ text{minutes}`
Show Worked Solution

a.     `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}` 
 


 

b.   `text{T} text{ank} \ B \ text{will pass through (15, 0) and (45, 900)}`  
 

   

`text{By inspection, the two graphs intersect at} \ \ t = 29 \ text{minutes}`

 
c.   `text{Strategy 1}`

`text{By inspection of the graph, consider} \ \ t = 45`

`text{T} text{ank A} = 100 \ text{L} , \ text{T} text{ank B} =900 \ text{L} `

`:.\ text(Total volume = 1000 L when  t = 45)`
  

`text{Strategy 2}`

`text{Total Volume}` `=text{T} text{ank A} + text{T} text{ank B}`
`1000` `= 1000 – 20t + (t – 15) xx 30`
`1000` `= 1000 – 20t + 30t – 450 `
`10t` `= 450`
`t` `= 45 \ text{minutes}`

Calculus, 2ADV C3 2020 HSC 21

Hot tea is poured into a cup. The temperature of tea can be modelled by  `T = 25 + 70(1.5)^(−0.4t)`, where `T` is the temperature of the tea, in degrees Celsius, `t` minutes after it is poured.

  1. What is the temperature of the tea 4 minutes after it has been poured?  (1 mark)

  2. At what rate is the tea cooling 4 minutes after it has been poured?  (2 marks)

  3. How long after the tea is poured will it take for its temperature to reach 55°C?  (3 marks)

Show Answers Only
  1. `61.6\ \ (text(to 1 d.p.))`
  2. `−5.9^@text(C/min)`
  3. `5.2\ text(minutes  (to 1 d.p.))`
Show Worked Solution
a.    `T` `= 25 + 70(1.5)^(−0.4 xx 4)`
    `= 61.58…`
    `= 61.6\ \ (text(to 1 d.p.))`

 

b.    `(dT)/(dt)` `= 70 log_e(1.5) xx −0.4(1.5)^(−0.4t)`
    `= −28log_e(1.5)(1.5)^(−0.4t)`

 

`text(When)\ \ t = 4,`

`(dT)/(dt)` `= −28log_e(1.5)(1.5)^(−1.6)`
  `= −5.934…`
  `= −5.9^@text(C/min  (to 1 d.p.))`

 

c.   `text(Find)\ \ t\ \ text(when)\ \ T = 55:`

♦ Mean mark part (c) 44%.
`55` `= 25 + 70(1.5)^(−0.4t)`
`30` `= 70(1.5)^(0.4t)`
`(1.5)^(−0.4t)` `= 30/70`
`−0.4t log_e(1.5)` `= log_e\ 3/7`
`−0.4t` `= (log_e\ 3/7)/(log_e (1.5))`
`:. t` `= (−2.08969)/(−0.4)`
  `= 5.224…`
  `= 5.2\ text(minutes  (to 1 d.p.))`

Trigonometry, 2ADV T1 2020 HSC 15

Mr Ali, Ms Brown and a group of students were camping at the site located at `P`. Mr Ali walked with some of the students on a bearing of 035° for 7 km to location `A`. Ms Brown, with the rest of the students, walked on a bearing of 100° for 9 km to location `B`.
 


 

  1. Show that the angle `APB` is 65°.  (1 mark)

  2. Find the distance `AB`.  (2 marks)

  3. Find the bearing of Ms Brown's group from Mr Ali's group. Give your answer correct to the nearest degree.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `8.76\ text{km  (to 2 d.p.)}`
  3. `146^@`
Show Worked Solution
a.    `angle APB` `= 100-35`
    `= 65^@`

 

b.   `text(Using cosine rule:)`

`AB^2` `= AP^2 + PB^2-2 xx AP xx PB cos 65^@`
  `= 49 + 81-2 xx 7 xx 9 cos 65^@`
  `= 76.750…`
`:.AB` `= 8.760…`
  `= 8.76\ text{km  (to 2 d.p.)}`

 
c.

`anglePAC = 35^@\ (text(alternate))`

`text(Using cosine rule, find)\ anglePAB:`

`cos anglePAB` `= (7^2 + 8.76-9^2)/(2 xx 7 xx 8.76)`  
  `= 0.3647…`  
`:. angle PAB` `= 68.61…^@`  
  `= 69^@\ \ (text(nearest degree))`  

 

`:. text(Bearing of)\ B\ text(from)\ A\ (theta)` 

`= 180-(69-35)`

`= 146^@`

NETWORKS, FUR2-NHT 2019 VCAA 1

A zoo has an entrance, a cafe and nine animal exhibits: bears `(B)`, elephants `(E)`, giraffes `(G)`, lions `(L)`, monkeys `(M)`, penguins `(P)`, seals `(S)`, tigers `(T)` and zebras `(Z)`.

The edges on the graph below represent the paths between the entrance, the cafe and the animal exhibits. The numbers on each edge represent the length, in metres, along that path. Visitors to the zoo can use only these paths to travel around the zoo.
  

 
 

  1. What is the shortest distance, in metres, between the entrance and the seal exhibit `(S)`?   (1 mark)

  2. Freddy is a visitor to the zoo. He wishes to visit the cafe and each animal exhibit just once, starting and ending at the entrance.
  3. i. What is the mathematical term used to describe this route?   (1 mark)

  4. ii. Draw one possible route that Freddy may take on the graph below.   (1 mark)


 

A reptile exhibit `(R)` will be added to the zoo.

A new path of length 20 m will be built between the reptile exhibit `(R)` and the giraffe exhibit `(G)`.

A second new path, of length 35 m, will be built between the reptile exhibit `(R)` and the cafe.

  1. Complete the graph below with the new reptile exhibit and the two new paths added. Label the new vertex `R` and write the distances on the new edges.   (1 mark)

     


 

  1. The new paths reduce the minimum distance that visitors have to walk between the giraffe exhibit `(G)` and the cafe.

     

    By how many metres will these new paths reduce the minimum distance between the giraffe exhibit `(G)` and the cafe?   (1 mark)

Show Answers Only
  1.  `45\ text(metres)`
  2. i.  `text(Hamilton cycle.)`
    ii. `text(See Worked Solutions)`
  3.  `text(See Worked Solutions)`
  4.  `85\ text(metres)`
Show Worked Solution

a.   `45\ text(metres)`
 

b.i.   `text(Hamilton cycle.)`
 

b.ii.  

`text(Possible route:)`

`text(entrance)\ – LGTMCEBSZP\ –\ text(entrance)`

 

c.  

 

d.  `text{Minimum distance (before new exhibit)}`

`= GLTMC`

`= 15 + 35 + 40 + 50`

`= 140\ text(m)`
 

`:.\ text(Reduction in minimum distance)`

`= 140 – (20 + 35)`

`= 85\ text(m)`

MATRICES, FUR2-NHT 2019 VCAA 1

A total of six residents from two towns will be competing at the International Games.

Matrix `A`, shown below, contains the number of male `(M)` and the number of female `(F)` athletes competing from the towns of Gillen `(G)` and Haldaw `(H)`.

`{:(qquad qquad quad \ M quad F), (A = [(2, 2), (1, 1)]{:(G),(H):}):}` 

  1. How many of these athletes are residents of Haldaw?   (1 mark)

Each of the six athletes will compete in one event: table tennis, running or basketball.

Matrices `T` and `R`, shown below, contain the number of male and female athletes from each town who will compete in table tennis and running respectively.
 

            Table tennis                        Running             
 

`{:(qquad qquad quad \ M quad F), (T = [(0, 1), (1, 0)]{:(G),(H):}):}`

`{:(qquad qquad quad \ M quad F), (R = [(1, 1), (0, 0)]{:(G),(H):}):}`

 

  1. Matrix `B` contains the number of male and female athletes from each town who will compete in basketball.

     

    Complete matrix `B` below.   (1 mark)

`{:(qquad qquad qquad \ M qquad quad F), (B = [(\ text{___}, text{___}\ ), (\ text{___}, text{___}\)]{:(G),(H):}):}`

Matrix `C` contains the cost of one uniform, in dollars, for each of the three events: table tennis `(T)`, running `(R)` and basketball `(B)`.

`C = [(515), (550), (580)]{:(T), (R), (B):}`

    1. For which event will the total cost of uniforms for the athletes be $1030?   (1 mark)

    2. Write a matrix calculation, that includes matrix `C`, to show that the total cost of uniforms for the event named in part c.i. is contained in the matrix answer of [1030].   (1 mark)

  1. Matrix `V` and matrix `Q` are two new matrices where  `V = Q xx C`  and:
  • matrix `Q` is a  `4 xx 3`  matrix
  • element `v_11 =` total cost of uniforms for all female athletes from Gillen
  • element `v_21 =` total cost of uniforms for all female athletes from Haldaw
  • element `v_31 =` total cost of uniforms for all male athletes from Gillen
  • element `v_41 =` total cost of uniforms for all male athletes from Haldaw
     
  • `C = [(515), (550), (580)]{:(T), (R), (B):}`
  1. Complete matrix `Q` with the missing values.   (1 mark)

`Q = [(1, text{___}, text{___}\ ), (0, 0, 1), (0, 1, 1), (\ text{___}, text{___}, 0)]`

Show Answers Only
  1.  `2`
  2.  `B = [(1, 0), (0, 1)]`
  3. i.  `text(Table tennis)`
    ii. `[2\ \ \ 0\ \ \ 0] xx [(515), (550), (580)] = [1030]`
  4.  `Q = [(1, \ 1, \ 0),(0, \ 0, \ 1),(0, \ 1, \ 1),(1, \ 0, \ 0)]`
Show Worked Solution

a.  `2`
 

b.  `B = [(1, 0), (0, 1)]`
 

c.i.  `text(Table tennis)`
 

c.ii.  `[2\ \ \ 0\ \ \ 0] xx [(515), (550), (580)] = [1030]`
 

d.  `Q = [(1, \ 1, \ 0),(0, \ 0, \ 1),(0, \ 1, \ 1),(1, \ 0, \ 0)]`

Financial Maths, GEN2 2019 NHT 6

Marlon plays guitar in a band.

He paid $3264 for a new guitar.

The value of Marlon's guitar will be depreciated by a fixed amount for each concert that he plays.

After 25 concerts, the value of the guitar will have decreased by $200.

  1. What will be the value of Marlon's guitar after 25 concerts?   (1 mark)

  2. Write a calculation that shows that the value of Marlon's guitar will depreciate by $8 per concert.   (1 mark)

  3. The value of Marlon's guitar after `n` concerts, `G_n`, can be determined using a rule.

     

    Complete the rule below by writing the appropriate numbers in the boxes provided.   (1 mark)

     
              `G_n =`   – × `n`
     

  4. The value of the guitar continues to be depreciated by $8 per concert.

     

    After how many concerts will the value of Marlon's guitar first fall below $2500?   (2 marks)

Show Answers Only
  1. `$3064`
  2. `text{Proof (See Worked Solution)}`
  3. `G_n = 3264-8 xx n`
  4. `96 \ text(concerts)`
Show Worked Solution
a.    `text(Value)` `= 3264-200`
  `= $3064`

 

b.    `text(Depreciation per concert)` `= (200)/(25)`
  `= $8 \ text(per concert)`

c.    `G_n = 3264 – 8 xx n`
 

d.    `text(Find) \ n \ text(when) \ \ G_n = 2500:`

`2500` `= 3264 – 8n`
`n` `= (3264-2500)/8`
  `= 95.5`

 
`:. \ text(After) \ 96 \ text(concerts, value first falls below $2500)`

Data Analysis, GEN2 2019 NHT 1

The table below displays the average sleep time, in hours, for a sample of 19 types of mammals.
 

  1. Which of the two variables, type of mammal or average sleep time, is a nominal variable?   (1 mark)

  2. Determine the mean and standard deviation of the variable average sleep time for this sample of mammals.
  3. Round your answer to one decimal place.   (1 mark)

  4. The average sleep time for a human is eight hours.
  5. What percentage of this sample of mammals has an average sleep time that is less than the average sleep time for a human.
  6. Round your answer to one decimal place.   (1 mark)

  7. The sample is increase in size by adding in the average sleep time of the little brown bat.
  8. Its average sleep time is 19.9 hours.
  9. By how many many hours will the range for average sleep time increase when the average sleep time for the little brown bat is added to the sample?   (1 mark)

Show Answers Only
  1. `text(type of mammal)`
  2. `text(mean)= 9.2 \ text(hours)`

     

    `sigma = 4.2 \ text(hours)`

  3. `31.6text(%)`
  4. `5.4 \ text(hours)`
Show Worked Solution

a.    `text(type of mammal is nominal)`

 
b.    `text(mean)= 9.2 \ text(hours) \ \ text{(by CAS)}`

`sigma = 4.2 \ text(hours) \ \ text{(by CAS)}`
 

c.    `text(Percentage)` `= (6)/(19) xx 100`
  `= 0.3157 …`
  `= 31.6text(%)`

 

d.    `text(Range increase)` `= 19.9-14.5`
  `= 5.4 \ text(hours)`

GRAPHS, FUR1-NHT 2019 VCAA 1 MC

The graph below shows the temperature, in degrees Celsius, over 24 hours.
 


 

The difference between the highest and lowest temperatures on this day, in degrees Celsius, is closest to

  1. 16
  2. 20
  3. 24
  4. 28
  5. 32
Show Answers Only

`D`

Show Worked Solution

`text(High) ~~ 32`

`text(Low) ~~ 4`

`:.\ text(Range)` `~~ 32 -4`
  `~~ 28`

`=>  D`

Functions, 2ADV F1 SM-Bank 53

  1. If  `1/(root3(7+pi)) = (7+pi)^x`, find  `x`.  (1 mark)

  2. Calculate the value of  `1/(root3(7+pi))`  to 3 significant figures.   (1 mark)

Show Answers Only
  1. `-1/3`
  2. `0.462`
Show Worked Solution

i.    `1/(root3(7+pi)) = (7+pi)^(-1/3)`

ii.     `1/(root3(7+pi))` `=0.4619…`
    `=0.462\ \ text{(to 3 sig. fig.)}`

Vectors, EXT2 V1 SM-Bank 9

  1. Find the equation of line vector  `underset ~r`, given it passes through  `(1, 3, –2)`  and  `(2, –1, 2)`.   (2 marks)

  2. Determine if  `underset ~r`  passes through  `(4, –9, 10)`.   (1 mark)

Show Answers Only
  1. `underset ~r = ((1), (3), (-2)) + lambda ((1), (-4), (4)) or`

       
      

    `underset ~r = ((2), (-1), (2)) + lambda ((-1), (4), (-4))`

  2. `text(See Worked Solutions)`
Show Worked Solution

i.     `text(Method 1)`

`text(Let)\ \ A(1, 3, –2) and B(2, –1, 2)`

`vec (AB)` `= ((2), (-1), (2))-((1), (3), (-2)) = ((1), (-4), (4))`
`underset ~r` `= ((1), (3), (-2)) + lambda ((1), (-4), (4))`

 

`text (Method 2)`

`vec (BA)` `= ((1), (3), (-2))-((2), (-1), (2)) = ((-1), (4), (-4))`
`underset ~r` `= ((2), (-1), (2)) + lambda ((-1), (4), (-4))`

 

ii.   `text(If)\ \ (4, –9, 10)\ \ text(lies on the vector line,)`

`∃ lambda\ \ text(that satisfies:)`

`1 + lambda` `= 4\ \ text{… (1)}`
`3-4 lambda` `= -9\ \ text{… (2)}`
`-2 + 4 lambda` `= 10\ \ text{… (3)}`

 

`lambda = 3\ \ text(satisfies all equations)`

`:. (4, –9, 10)\ \ text(lies on the line.)`

GRAPHS, FUR2 2019 VCAA 1

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


 

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

a.  `60`

 

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

 

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

GEOMETRY, FUR2 2019 VCAA 1

The following diagram shows a cargo ship viewed from above.
 

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

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

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

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

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

      

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

      

    Each barrel must remain upright in the shipping container
     
     

      

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

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

 

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

 

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

 

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

 

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

NETWORKS, FUR2 2019 VCAA 1

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


 

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

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

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

 

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

a.  `text(Office)`
 

b.i.   `text(Hamiltonian cycle)`

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

  `text{at the same vertex.)}`
 

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

MATRICES, FUR2 2019 VCAA 1

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

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

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

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

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

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

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

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

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

     

     

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

     

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

     

    Write down matrix  `L`.   (1 mark)

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

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

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

      

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

      

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

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

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

a.  `text(Order) : 3 xx 2`
 

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

c.  `L = [(50), (20), (40)]`
 

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

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

CORE, FUR2 2019 VCAA 2

 

The parallel boxplots below show the maximum daily temperature and minimum daily temperature, in degrees Celsius, for 30 days in November 2017.
 

  1. Use the information in the boxplots to complete the following sentences.
  2. For November 2017
  3.    i. the interquartile range for the minimum daily temperature was _____ °C   (1 mark)

  4.  ii. the median value for maximum daily temperature was _____ °C higher than the median value for minimum daily temperature   (1 mark)

  5. iii. the number of days on which the maximum daily temperature was less than the median value for minimum daily temperature was _____    (1 mark)

  1. The temperature difference between the minimum daily temperature and the maximum daily temperature in November 2017 at this location is approximately normally distributed with a mean of 9.4 °C and a standard deviation of 3.2 °C.
  2. Determine the number of days in November 2017 for which this temperature difference is expected to be greater than 9.4 °C.  (1 mark)

Show Answers Only

    1. `5text(°C)`
    2. `10text(°C)`
    3. `1\ text(day)`
  2. `15\ text(days)`

Show Worked Solution

a.i.  `text(IQR)\ = 17-12= 5text(°C)`
 

a.ii.    `text{Median (maximum temperature)}` `= 25`
  `text{Median (minimum temperature)}` `= 15`

 
`:.\ text(Maximum is 10°C higher)`
 

a.iii.  `text{Median (minimum temperature)} = 15text(°C)`

   `text(1 day) => text(maximum temperature is below)\ 15text(°C)`
 

b.    `text(Number of days)` `= 0.50 xx 30`
    `= 15\ text(days)`

GRAPHS, FUR1 2019 VCAA 1 MC

The graph below shows the water level in a creek, in metres, over a 12-hour period.
 

  
Compared to the water level at 8 am, the water level at 12 noon was closest to

  1. 1.0 m higher.
  2. 1.4 m higher
  3. 2.4 m higher.
  4. 0.9 m lower.
  5. 1.3 m lower.
Show Answers Only

`B`

Show Worked Solution

`text(Level at 8am) ~~ 1.0\ text(m)`

`text(Level at noon) ~~ 2.4\ text(m)`

`:.\ text(Noon level) ~~ 1.4\ text(m higher)`

`=>  B`

GEOMETRY, FUR1 2019 VCAA 1 MC

The four bases of a baseball field form four corners of a square of side length 27.43 m, as shown in the diagram below.
 


 

A player ran from home base to first base, then to second base, then to third base and finally back to home base.

The minimum distance, in metres, that the player ran is

  1.   27.43
  2.   54.86
  3.   82.29
  4. 109.72
  5. 164.58
Show Answers Only

`D`

Show Worked Solution
`text(Minimum distance)` `= 4 xx 27.43`
  `= 109.72`

`=>  D`

MATRICES, FUR1 2019 VCAA 1 MC

Consider the following four matrix expressions.
 

`[(8), (12)]+[(4), (2)] qquad qquad qquad qquad qquad qquad quad [(8), (12)]+[(4, 0),(0, 2)]`
 

`[(8, 0),(12, 0)] + [(4), (2)] qquad qquad qquad qquad qquad \ [(8, 0),(12, 0)] + [(4, 0),(0, 2)]`
 

How many of these four matrix expressions are defined?

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show Answers Only

`C`

Show Worked Solution

`[(8), (12)]+[(4), (2)] = [(12), (4)]`

`[(8), (12)]+[(4, 0),(0, 2)] = text(not defined)`

`[(8, 0),(12, 0)] + [(4), (2)] = text(not defined)`

`[(8, 0),(12, 0)] + [(4, 0),(0, 2)] = [(12, 0), (12, 2)]`

 

`=>  C`

Calculus, SPEC1 2019 VCAA 5

The graph of  `f(x) = cos^2(x) + cos(x) + 1`  over the domain  `0 <= x <= 2pi`  is shown below.

  1.  i.  Find `f^{′}(x)`.  (1 mark)

  2. ii. Hence, find the coordinates of the turning points of the graph in the interval  `(0, 2pi)`.  (2 marks)

  3. Sketch the graph of  `y = 1/(f(x))`  on the set of axes above. Clearly label the turning points and endpoints of this graph with their coordinates.  (3 marks)
Show Answers Only
  1. i.  `−sin(x)(2cos(x) + 1)`
  2. ii. `(pi, 1), ((2pi)/3,3/4), ((4pi)/3, 3/4)`
  3.  

Show Worked Solution
a.i.    `f(x)` `= cos^2(x) + cos(x) + 1`
  `f^{′}(x)` `= -2sin(x)cos(x)-sin(x)`
    `= -sin(x)(2cos(x) + 1)`

 

a.ii.   `text(SP when)\ \ sin(x) = 0\ \ text(or)\ \ 2cos(x) + 1 = 0`

`sin(x) = 0 \ => \ x = pi\ \ (x = 0\ \ text{not in domain})`

`2cos(x) + 1` `= 0`
`cos(x)` `= -1/2`
`x` `= (2pi)/3, (4pi)/3`

 
`text(When)\ \ cos(x) = −1/2 \ => \ f(x) = 1/4-1/2 + 1 = 3/4`

`:.\ text(Turning Points:)\ (pi, 1), ((2pi)/3,3/4), ((4pi)/3, 3/4)`

 

b.   

Complex Numbers, EXT2 N1 2019 HSC 11e

Let  `z = -1 + i sqrt 3`.

  1. Write  `z`  in modulus-argument form.  (2 marks)

  2. Find  `z^3`, giving your answer in the form  `x + iy`, where  `x`  and  `y`  are real numbers.  (2 marks)

Show Answers Only
  1. `z = 2 text(cis) (2 pi)/3`
  2. `8 + 0i`
Show Worked Solution
i.   `|\ z\ |` `= -1 + i sqrt 3`
    `= sqrt((-1)^2 + (sqrt 3)^2)`
    `= 2`

 

  `tan theta` `= -sqrt 3`
  `text(arg)(z)` `= (2 pi)/3`
  `:. z` `= 2 text(cis) (2 pi)/3`

 

ii.   `z^3 = 2^3 [cos(3 xx (2 pi)/3) + i sin (3 xx (2 pi)/3)]\ \ \ text{(by De Moivre)}`

`= 8(cos 2 pi + i sin 2 pi)`

`= 8(1 + 0i)`

`= 8 + 0i`

Calculus, EXT2 C1 2019 HSC 11d

Find  `int 6/(x^2-9) dx`.  (3 marks)

Show Answers Only

`ln |\ (x-3)/(x + 3)\ |+ c`

Show Worked Solution

`text(Using partial fractions):`

`6/(x^2-9)` `= 6/((x + 3)(x-3))`
  `= A/(x + 3) + B/(x-3)`

 
`:. A(x-3) + B(x + 3) = 6`
 

`text(When)\ \ x = 3,\ 6B = 6 \ => \ B = 1`

`text(When)\ \ x = -3,\ -6A = 6 \ => \ A = -1`

`int 6/(x^2-9)\ dx` `= int (-1)/(x + 3) + 1/(x-3)\ dx`
  `= -ln|\ x + 3\ | + ln |\ x-3\ | + C`
  `= ln |\ (x-3)/(x + 3)\ | + C`

Complex Numbers, EXT2 N1 2019 HSC 11a

Let  `z = 1 + 3i`  and  `w = 2 - i`.

  1. Find  `z + bar w`.  (1 mark)

  2. Express  `z/w`  in the form  `x + iy`, where  `x` and  `y`  are real numbers.  (2 marks)

Show Answers Only
  1. `3 + 4i`
  2. `-1/5 + 7/5 i`
Show Worked Solution

i.   `z = 1 + 3i`

`w = 2 – i \ => \ bar w = 2 + i`

`z + bar w` `= 1 + 3i + 2 + i`
  `= 3 + 4i`

 

ii.    `z/w` `= (1 + 3i)/(2 – i) xx (2 + i)/(2 + i)`
    `= ((1 + 3i)(2 + i))/(2^2 – i^2)`
    `= (2 + i + 6i + 3i^2)/5`
    `= (-1 + 7i)/5`
    `= -1/5 + 7/5 i`

Calculus, EXT1* C1 2019 HSC 12c

The number of leaves, `L(t)`, on a tree `t` days after the start of autumn can be modelled by

`L(t) = 200\ 000e^(-0.14t)`

  1. What is the number of leaves on the tree when  `t = 31`?  (1 mark)

  2. What is the rate of change of the number of leaves on the tree when  `t = 31`?  (2 marks)

  3. For what value of `t` are there 100 leaves on the tree?  (2 marks)

Show Answers Only
  1. `2607\ text(leaves)`
  2. `-365.02…`
  3. `54.3\ text{(1 d.p.)}`
Show Worked Solution

i.  `text(When)\ \ t = 31`

`L(t)` `= 200\ 000 xx e^(-0.14(31))`
  `=2607.305…`
  `= 2607\ text(leaves)`

 

ii.    `L` `= 2000\ 000^(-0.14t)`
  `(dL)/(dt)` `= -0.14 xx 200\ 000 e^(0.14t`
    `= -28\ 000e^(-0.14t)`

 
`text(When)\ \ t = 31,`

`(dL)/(dt)` `= -28\ 000 xx e^(-0.14(31))`
  `= -365.02…`

 
`:.\ text(365 leaves fall per day.)`

 

iii.   `text(Find)\ t\ text(when)\ \ L = 100:`

`100` `= 200\ 000 e^(-0.14t)`
`e^(-0.14t)` `= 0.0005`
`e^(-0.14t)` `= ln 0.0005`
`t` `= (ln 0.0005)/(-0.14)`
  `= 54.292…`
  `= 54.3\ text{(1 d.p.)}`

Calculus, SPEC1 2011 VCAA 3

  1. Show that  `f(x) = (2x^2 + 3)/(x^2 + 1)`  can be written in the form  `f(x) = 2 + 1/(x^2 + 1).`  (1 mark)

  2. Sketch the graph of the relation  `y = (2x^2 + 3)/(x^2 + 1)`  on the axes below.
  3. Label any asymptotes with their equations and label any intercepts with the axes, writing them as coordinates.  (3 marks)

     
    VCAA 2011 spec 3b
     

  4. Find the area enclosed by the graph of the relation  `y = (2x^2 + 3)/(x^2 + 1)`, the `x`-axis, and the lines  `x = -1`  and  `x = 1.`  (3 marks)

Show Answers Only

  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. “
    VCAA 2011 spec 3bi
  3. `4 + pi/2`

Show Worked Solution

a.    `f(x)` `= (2x^2 + 3)/(x^2 + 1)`
    `= (2(x^2 + 1) + 1)/(x^2 + 1)`
    `= 2 + 1/(x^2 + 1)`

 

b.   `underset (x→oo) (lim y) = 2`

`text(S)text(ince)\ \ x^2>0\ \ text(for all)\ x,`

`=> f(x)_text(max)\ \ text(occurs when)\ \ x=0\ \ text(at)\ \ (0,3)`
 

VCAA 2011 spec 3bi

 

c.   `f(x)\ \ text(is an even function.)`

`:.\ text(Area)` `= 2 int_0^1 2 + 1/(x^2 + 1)\ dx`  
  `= 2[2x + tan^(−1)(x)]_0^1`  
  `=2[(2+pi/4)-0]`  
  `= 4 + pi/2`  

Calculus, SPEC2 2017 VCAA 2

A helicopter is hovering at a constant height above a fixed location. A skydiver falls from rest for two seconds from the helicopter. The skydiver is subject only to gravitational acceleration and air resistance is negligible for the first two seconds. Let downward displacement be positive.

  1. Find the distance, in metres, fallen in the first two seconds.   (2 marks)

  2. Show that the speed of the skydiver after two seconds is 19.6 ms–1.   (1 mark)

After two seconds, air resistance is significant and the acceleration of the skydiver is given by  `a = g -0.01v^2`.

  1. Find the limiting (terminal) velocity, in ms–1, that the skydiver would reach.   (1 mark)

  2. i.  Write down an expression involving a definite integral that gives the time taken for the skydiver to reach a speed of 30 ms–1.   (2 marks)

  3. ii. Hence, find the time, in seconds, taken to reach a speed of 30 ms–1, correct to the nearest tenth of a second.   (1 mark)

  4. Write down an expression involving a definite integral that gives the distance through which the skydiver falls to reach a speed of 30 ms–1. Find this distance, giving your answer in metres, correct to the nearest metre.   (3 marks)

Show Answers Only
  1. `19.6\ text(m)`
  2. `text(See Worked Solutions)`
  3. `v_t = 10sqrtg\ text(ms)^(−1)\ \ text(downwards)`
  4.  i.  `t(30) = int_19.6^30 100/(100g-v^2)dv + 2`
  5. ii. `5.8\ text(s)`
  6. `120\ text(m)`
Show Worked Solution

a.   `u = 0, \ t = 2, \ a = g = 9.8`

`x` `= ut + 1/2 at^2`
  `=0 xx 2 + 1/2 xx 9.8 xx 2^2`
  `= 19.6\ text(m)`

 

b.    `v` `= u+ at`
  `v(2)` `= 9.8 xx 2`
    `= 19.6\ text(ms)^(−1)\ \ \ text{(downward → positive)}`

 

c.   `text(Terminal velocity), v_t, text(occurs when)\ \ a = 0,`

♦ Mean mark 48%.

`g -0.01v_t^2` `=0`
`v_t^2` `= 100 xx 9.8`
`:. v_t` `= 14 sqrt5\ text(ms)^(−1)\ \ \ (v_t > 0\ \ text{as}\ \ v_t\ \ text{is downwards})`

 

d.i.    `(dv)/(dt)` `= g-0.01v^2`
  `(dt)/(dv)` `= 1/(g-0.01v^2)`
    `= 100/(100g-v^2)`
  `t` `= int 100/(100g-v^2)`

 
`text(Time taken until)\ \ v=19.6\ \ text(is 2 seconds.)`

♦ Mean mark part (d)(i) 39%.

 
`:.\ text(Time taken until)\ \ v=30`

`=int_19.6^30 100/(100g-v^2)\ dv + 2`

 

d.ii.   `5.8\ text(seconds)\ \ \ text{(by CAS)}`

♦♦ Mean mark part (d)(ii) 25%.

 

e.    `v *(dv)/(dx)` `= g-0.01 v^2`
  `(dv)/(dx)` `= g/v-(v^2)/(100v)`
    `= (100g-v^2)/(100v)`
  `(dx)/(dv)` `= (100v)/(100g-v^2)`
  `x` `= int (100v)/(100g-v^2)\ dv`

 
`text(Distance fallen in 1st 2 seconds)\ = 19.6\ text(m)`

♦♦ Mean mark part (e) 29%.

 
`:.\ text(Total distance fallen until)\ \ v=30`

`= int_19.6^30 (100v)/(100g-v^2)\ dv + 19.6`

`~~ 120\ text(m)`

Calculus, SPEC2 2018 VCAA 3

Part of the graph of  `y = 1/2 sqrt(4x^2-1)`  is shown below.
 


 

The curve shown is rotated about the `y`-axis to form a volume of revolution that is to model a fountain, where length units are in metres.

  1. Show that the volume, `V` cubic metres, of water in the fountain when it is filled to a depth of `h` metres is given by  `V = pi/4(4/3h^3 + h)`.   (2 marks)

  2. Find the depth `h` when the fountain is filled to half's its volume. Give your answer in metres, correct to two decimal places.   (2 marks)

The fountain is initially empty. A vertical jet of water in the centre fills the fountain at a rate of 0.04 cubic metres per second and, at the same time, water flows out from the bottom of the fountain at a rate of  `0.05 sqrt h`  cubic metres per second when the depth is `h` metres.

  1.  i. Show that  `(dh)/(dt) = (4-5sqrt h)/(25 pi (4h^2 + 1))`.   (2 marks)

  2. ii. Find the rate, in metres per second, correct to four decimal places, at which the depth is increasing when the depth is 0.25 m.   (1 mark)

  3. Express the time taken for the depth to reach 0.25 m as a definite integral and evaluate this integral correct to the nearest tenth of a second.   (2 marks)

  4. After 25 seconds the depth has risen to 0.4 m.
    Using Euler's method with a step size of five seconds, find an estimate of the depth 30 seconds after the fountain began to fill. Give your answer in metres, correct to two decimal places.   (2 marks)

  5. How far from the top of the fountain does the water level ultimately stabilise? Give your answer in metres, correct to two decimal places.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `h~~ 0.59\ text(m)`
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `0.0153\ text(ms)^(-1)`
  1. `9.8\ text(seconds)`
  2. `0.43\ text(m)`
  3. `0.23\ text(m)`
Show Worked Solution

a.   `V= pi int_0^h x^2\ dy`

`y` `=1/2 sqrt(4x^2-1)`
`2y` `=sqrt(4x^2-1)`
`4y^2` `= 4x^2-1`
`4x^2` `= 4y^2 + 1`
`x^2` `= y^2 + 1/4`

 

`:. V` `= pi int_0^h y^2 + 1/4\ dy`
  `= pi[y^3/3 + y/4]_0^h`
  `= pi(h^3/3 + h/4-0)`
  `= pi(1/4((4h^3)/3 + h))`
  `= pi/4((4h^3)/3 + h)\ \ …\ text(as required)`

 

b.    `V_text(max)` `= pi/4 (4/3 xx (sqrt 3/2)^3 + sqrt 3/2)`
    `= pi/4 (sqrt 3/2 + sqrt 3/2)`
    `= (pi sqrt 3)/4`

 

`1/2 V_text(max)` `= (pi sqrt 3)/8`
`(pi sqrt 3)/8` `= pi/4 (4/3 h^3 + h)`
`sqrt 3/2` `= 4/3 h^3 + h`
`:. h` `~~0.59\ text(m)`

  

c.i.   `((dV)/(dt))_text(in)` `= 0.04`
  `((dV)/(dt))_text(out)` `= 0.05 sqrt h`
  `(dV)/(dt)` `= 0.04-0.05 sqrt h`
    `= (4-5 sqrt h)/100`
     
  `(dV)/(dh)` `= pi/4(4h^2 + 1)`
  `:. (dh)/(dt)` `= (dh)/(dV) ⋅ (dV)/(dt)`
    `= 4/(pi(4h^2 + 1)) xx (4-5 sqrt h)/100`
    `= (4-5 sqrt h)/(25 pi (4h^2 + 1))`

 

c.ii.   `(dh)/(dt)|_(h = 0.25)` `= (4-5 sqrt(0.25))/(25 pi (4(0.25)^2 + 1))`
    `~~ 0.0153\ text(ms)^(-1)`

 

d.   `(dt)/(dh) = (25 pi (4h^2 + 1))/(4-5 sqrt h)`

`:. t(0.25)` `= int_0^0.25 (25 pi (4h^2 + 1))/(4-5 sqrt h)\ dh`
  `~~9.8\ text(seconds)`

 

e.   `text(When)\ \ t=25,\ \ h=0.4\ \ text{(given)}`

♦♦ Mean mark part (e) 30%.

`:. h(30)` `~~ h(25) + 5 xx (dh)/(dt)|_(h = 0.4)`
  `~~ 0.4 + 5 xx ((4-5 sqrt 0.4)/(25 pi (4(0.4)^2 + 1)))`
  `~~ 0.43\ text(m)`

 

f.    `(dV)/(dt) = 0`

♦♦ Mean mark part (f) 32%.

`0.04-0.05 sqrt h` `= 0`
`0.04` `= 0.05 sqrt h“
`sqrt h` `= 4/5`
`h` `= 16/25`

  

`d` `= h_max-16/25`
  `= sqrt 3/2-16/25`
  `~~ 0.23\ text(m)`

Mechanics, SPEC1 2016 VCAA 1

A taut rope of length `1 2/3` m suspends a mass of 20 kg from a fixed point `O`. A horizontal force of `P` newtons displaces the mass by 1 m horizontally so that the taut rope is then at an angle of  `theta`  to the vertical.

  1. Show all the forces acting on the mass on the diagram below.  (1 mark)

 

 

  1. Show that  `sin (theta) = 3/5`.  (1 mark)
  2. Find the magnitude of the tension force in the rope in newtons.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `T = 245\ text(N)`
Show Worked Solution
a.   

 

b.    `sin(theta)` `= 1/(1 2/3)`
    `= 1/((5/3)`
    `= 3/5`

 

c.   

`cos theta = (20g)/T`

`sin theta = 3/5\ \ text{(using part b)}`

`underbrace{(400g^2)/T^2 + 9/25}_(cos^2 theta + sin^2 theta = 1)` `= 1`
`(400g^2)/T^2` `= 16/25`
`(20g)/T` `=4/5`
`T/(20g)` `= 5/4`
`T` `= 25g`
  `=245\ text(N)`

Networks, STD2 N2 2011 FUR2 1

Aden, Bredon, Carrie, Dunlop, Enwin and Farnham are six towns.

The network shows the road connections and distances between these towns in kilometres.

 

  1. In kilometers, what is the shortest distance between Farnham and Carrie?  (1 mark)

  2. How many different ways are there to travel from Farnham to Carrie without passing through any town more than once?  (1 mark)

Show Answers Only
  1. `200\ text(km)`
  2. `6`
Show Worked Solution

a.   `text{Farnham to Carrie (shortest)}`

`= 60 + 140`

`= 200\ text(km)`

 

b.   `text(Different paths are)`

`FDC, FEDC, FEBC,`

`FEABC, FDEBC,`

`FDEABC`

 
`:. 6\ text(different ways)`

GRAPHS, FUR2 2018 VCAA 1

The following chart displays the daily gold prices (dollars per gram) for the month of November 2017.
 

  1. Which day in November had the lowest gold price?  (1 mark)
  2. Between which two consecutive days did the greatest increase in gold price occur?  (1 mark)
Show Answers Only
  1. `text(Day 2)`
  2. `text(between days 16 and 17)`
Show Worked Solution

a.   `text(Day 2)`
 

b.   `text(Between days 16 and 17)`

GEOMETRY, FUR1 2018 VCAA 1 MC

Henry flies a kite attached to a long string, as shown in the diagram below.
 

The horizontal distance of the kite to Henry’s hand is 8 m.

The vertical distance of the kite above Henry’s hand is 15 m.

The length of the string, in metres, is

  1.  13
  2.  17
  3.  23
  4.  161
  5.  289
Show Answers Only

`B`

Show Worked Solution

`text(Using Pythagoras)`

`s^2` `= 8^2 + 15^2`
  `= 289`
`:. s` `= 17\ text(metres)`

  
`=> B`

NETWORKS, FUR1 2018 VCAA 1 MC

Consider the graph with five isolated vertices shown below.
 


 

To form a tree, the minimum number of edges that must be added to the graph is

  1.   1
  2.   4
  3.   5
  4.   6
  5. 10
Show Answers Only

`B`

Show Worked Solution

`text(5 vertices.)`

`text(Minimum edges for a tree = 4)`

`=> B`

NETWORKS, FUR2 2018 VCAA 2

In one area of the town of Zenith, a postal worker delivers mail to 10 houses labelled as vertices `A` to `J` on the graph below.
 

  1. Which one of the vertices on the graph has degree 4?   (1 mark)

For this graph, an Eulerian trail does not currently exist.

  1. For an Eulerian trail to exist, what is the minimum number of extra edges that the graph would require.   (1 mark)

  2. The postal worker has delivered the mail at `F` and will continue her deliveries by following a Hamiltonian path from `F`.

     

    Draw in a possible Hamiltonian path for the postal worker on the diagram below.   (1 mark)

 

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

a.   `text(Vertex)\ F`
 

b. `text(Eulerian trail)\ =>\ text(all edges used exactly once.)`

`text(6 vertices are odd)`

`=> 2\ text(extra edges could create graph with only 2)`

`text(odd vertices)`

`:.\ text(Minimum of 2 extra edges.)`
 

c.  `text{One example (of a number) beginning at)\ F:}`
 


 

`text{(Note: path should not return to}\ F text{)}`