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Algebra, MET1 2019 VCAA 8

The function  `f: R -> R, \ f(x)`  is a polynomial function of degree 4. Part of the graph of  `f`  is shown below.

The graph of  `f`  touches the `x`-axis at the origin.
 


 

  1. Find the rule of  `f`.   (1 mark) 

Let  `g`  be a function with the same rule as  `f`.

Let  `h: D -> R, \ h(x) = log_e (g(x))-log_e (x^3 + x^2)`, where  `D`  is the maximal domain of  `h`.

  1. State  `D`.   (1 mark)

  2. State the range of  `h`.   (2 marks)

Show Answers Only
  1. `f(x) = -4x^2(x^2-1)`
  2. `x in (-1, 1) text(\{0})`
  3. `h(x) in (-oo, 3 log_e 2) text(\ {)2 log_e 2 text(})`
Show Worked Solution
a.    `y` `= ax^2 (x-1)(x + 1)`
    `= ax^2 (x^2-1)\ …\ (1)`

 
`text(Substitute)\ (1/ sqrt 2, 1)\ text{into  (1):}`

`1 = a ⋅ (1/sqrt 2)^2 ((1/sqrt 2)^2-1)`

`1 = a(1/2)(-1/2)`

`a = -4`

`:. f(x) = -4x^2(x^2-1)`

 

b.    `g(x) > 0` `=> -4x^2 (x^2-1) > 0`
    `=> x in (-1, 1)\  text(\{0})`

`text(and)`

`x^3 + x^2 > 0 => text(true for)\ \ x in (-1 , 1)\ text(\{0})`

`:. D:\ x in (-1, 1)\ text(\{0})`

 

c.    `h(x)` `= log_e ((-4x^2(x^2-1))/(x^3 + x^2))`
    `= log_e ((-4x^2(x + 1)(x-1))/(x^2(x + 1)))`
    `= log_e (4(1-x))\ \ text(where)\ \ x in (-1, 1)\ text(\{0})`

  
`text(As)\ \ x -> -1,\ \ h(x) -> log_e 8 = 3 log_e 2`

`text(As)\ \ x -> 1,\ \ h(x) -> -oo`

`text(As)\ \ x -> 0,\ \ h(x) -> log_e 4 = 2 log_e 2`

`text{(}h(x)\ text(undefined when)\ \ x = 0 text{)}`
 

`:.\ text(Range)\ \ h(x) in (-oo, 3 log_e 2)\ text(\{) 2 log_e 2 text(})`

Measurement, NAP-L4-CA11

This is part of a timetable for the ferry from Apollo Road to West End.
 

 
Skye is catching a ferry from Apollo Road to West End.

She catches the first ferry that leaves Apollo Road after 1 pm.

At what time will Skye arrive at West End?

12:41 pm 1:09 pm 1:12 pm 1:41 pm 2:41 pm
 
 
 
 
 
Show Answers Only

`2:41\ text(pm)`

Show Worked Solution

`text(Skye catches the 13:12 ferry.)`

`=>\ text(She arrives at West End at 14:41 = 2:41 pm)`

Measurement, NAP-L3-CA17

This is part of a timetable for the ferry from Apollo Road to West End.
 

 
Skye is catching a ferry from Apollo Road to West End.

She catches the first ferry that leaves Apollo Road after 1 pm.

At what time will Skye arrive at West End?

12:41 pm 1:09 pm 1:12 pm 1:41 pm 2:41 pm
 
 
 
 
 
Show Answers Only

`2:41\ text(pm)`

Show Worked Solution

`text(Skye catches the 13:12 ferry.)`

`=>\ text(She arrives at West End at 14:41 = 2:41 pm)`

Statistics, NAP-L3-CA15

A survey was conducted that asked students how long it took them to travel to school. This graph show the results.
 


 

What percentage of students had a commute between 15-30 minutes?

 
 less that 25%
 
 between 25% and 50%
 
 between 50% and 75%
 
 more than 75%
Show Answers Only

`text(between 25% and 50%)`

Show Worked Solution

`text(S)text{ection (15-30 minutes) is more than one quarter}`

`text(of the chart and less than half.)`

`:.\ text(between 25% and 50%)`

Number, NAP-L3-CA08

Kelly measures the length of fish she catches for her research.

Which fish has a length closest to 25 cm?

 
 
 
 
Show Answers Only

`25.14\ text(cm)`

Show Worked Solution

`text(Difference of each option:)`

`25.14 – 25` `= 0.14\ text(cm)`
`25 – 24.8` `= 0.20\ text(cm)`
`25 – 24.76` `= 0.24\ text(cm)`
`25.3 – 25` `= 0.30\ text(cm)`

 
`:. 25.14\ text(cm is closest.)`

Complex Numbers, EXT2 N2 2019 HSC 16b

Let  `P(z) = z^4 - 2kz^3 + 2k^2z^2 + mz + 1`, where  `k`  and  `m`  are real numbers.

The roots of  `P(z)`  are  `alpha, bar alpha, beta, bar beta`.

It is given that  `|\ alpha\ | = 1`  and  `|\ beta\ | = 1`.

  1. Show that  `(text{Re} (alpha))^2 + (text{Re} (beta))^2 = 1`.  (3 marks)

  2. The diagram shows the position of  `alpha`.
     


 

On the diagram, accurately show all possible positions of  `beta`.  (2 marks)

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

i.    `P(z) = z^4 – 2kz^3 + 2k^2z^2 + mz + 1,\ \ k, m in RR`

`text(Roots):\ \ alpha, bar alpha, beta, bar beta and |\ alpha\ | = 1, |\ beta\ | = 1`

`text(Show)\ \ (text{Re} (alpha))^2 + (text{Re} (beta))^2 = 1`

♦♦ Mean mark part (i) 26%.

`alpha + bar alpha + beta + bar beta` `= 2k`
`2 text{Re} (alpha) + 2 text{Re} (beta)` `= 2k`
`text{Re} (alpha) + text{Re} (beta)` `= k`

 

`alpha bar alpha + alpha beta + alpha bar beta + bar alpha beta + bar alpha bar beta + beta bar beta` `= 2k^2`
`|\ alpha\ |^2 + alpha(beta + bar beta) + bar alpha(beta + bar beta) + |\ beta\ |^2` `= 2k^2`
`1 + (alpha + bar alpha)(beta + bar beta) + 1` `= 2k^2`
`2 + 2 text{Re} (alpha) ⋅ 2 text{Re} (beta)` `= 2 (text{Re} (alpha) + text{Re} (beta))^2`
`2 + 4 text{Re} (alpha) text{Re} (beta)` `= 2 text{Re} (alpha)^2 + 4 text{Re} (alpha) text{Re} (beta) + 2 text{Re} (beta)^2`
`2` `= 2(text{Re} (alpha)^2 + text{Re} (beta)^2)`
`:. 1` `= text{Re} (alpha)^2 + text{Re} (beta)^2`

 

ii.    `|\ alpha\ | = |\ beta\ |\ \ \ text{(given)}`
  `text{Re}(alpha)^2 + text{Re}(beta)^2 = 1\ \ \ text{(see part (i))}`
  `text{Re}(alpha)^2 + text{Im}(alpha)^2 = 1\ \ \ (|\ alpha\ | = 1)`
  `=> text{Re}(beta)^2 = text{Im} (alpha)^2`
  `\ \ \ \ \ \ text{Re}(beta) = +-text{Im}(alpha)`

 

♦♦♦ Mean mark part (ii) 10%.

Number, NAP-L4-CA07

The table shows the fractions of the Australian workforce in some industries.
 


 

Which of these industries has the least number of employees in the workforce?

 
 Automotive
 
 Finance
 
 Health Care
 
 Telecommunications
Show Answers Only

`text(Finance)`

Show Worked Solution

`text(The smallest fraction is)\ \ 1/30.`

`:.\ text(The finance industry has least number.)`

Number, NAP-L3-CA07

The table shows the fractions of the Australian workforce in some industries.
 


 

Which of these industries has the least number of employees in the workforce?

 
 Automotive
 
 Finance
 
 Health Care
 
 Telecommunications
Show Answers Only

`text(Finance)`

Show Worked Solution

`text(The smallest fraction is)\ \ 1/30.`

`:.\ text(The finance industry has least number.)`

Geometry, NAP-L3-CA06

Blackbeard finds part of a treasure map.
 

The palm tree is at I5.

The skull is at G4.

Where is the treasure chest?

E1 F2 D3 H3
 
 
 
 
Show Answers Only

`text(D3)`

Show Worked Solution

`text(The palm tree and skull are 1 vertical position apart.)`

`=>\ text(numbers must run vertically)`

`=>\ text(letters run horizontally)`

`:.\ text(Treasure is in D3.)`

Combinatorics, EXT1′ S1 2019 HSC 10 MC

An access code consists of 4 digits chosen from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The code will only work if the digits are entered in the correct order.

Some access codes contain exactly two different digits, for example 3377 or 5155.

How many such access codes can be made using exactly two different digits?

  1. 630
  2. 900
  3. 1080
  4. 2160
Show Answers Only

`A`

Show Worked Solution

`text(Code has 1x first digit, 3x other digit:)`

`text(First digit = 10 choices)`

`text(Position of first digit = 4 choices)`

`text(Other digit = 9 choices)`

`text(Combinations)` `= 10 xx 4 xx 9`
  `= 360`

 
`text{Code has 2 × each digit  (say}\ X and Y\text{):}`

`text(Combinations of 2 digits) = \ ^10 C_2`

`text(If)\ X\ text(in position 1) => 3\ text(combinations of last 3 digits)`

`text(If)\ Y\ text(in position 1) => 3\ text(combinations of last 3 digits)`

`text(Combinations)` `= \ ^10 C_2 xx 6`
  `= 270`

 
`:.\ text(Total access codes) = 360 + 270 = 630`

`=>   A`

Calculus, EXT1 C2 2019 HSC 14c

The diagram shows the two curves  `y = sin x`  and  `y = sin(x - alpha) + k`, where  `0 < alpha < pi`  and  `k > 0`. The two curves have a common tangent at `x_0` where  `0 < x_0 < pi/2`.
 


 

  1. Explain why   `cos x_0 = cos (x_0 - alpha)`.  (1 mark)

  2. Show that  `sin x_0 = -sin(x_0 - alpha)`.  (2 marks)

  3. Hence, or otherwise, find  `k`  in terms of  `alpha`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `k = 2 sin\ alpha/2`
Show Worked Solution
i.    `y_1` `= sin x`
  `(dy_1)/(dx)` `= cos x`
  `y_2` `= sin(x – alpha) + k`
  `(dy_2)/(dx)` `= cos (x – alpha)`

 
`text(At)\ \ x = x_0,\ \ text(tangent is common)`

♦ Mean mark part (i) 47%.

`:. cos x_0 = cos(x_0 – alpha)`

 

ii.   `x_0\ \ text{is in 1st quadrant (given)}`

`text{Using part  (i):}`

`cos\ x_0 = cos(x_0 – alpha) >0`

♦♦♦ Mean mark part (ii) 19%.

`=> x_0 – alpha\ \ \ text(is in 4th quadrant)\ \ (0 < alpha < pi)`

`text(S)text(ince sin is positive in 1st quadrant and)`

`text(negative in 4th quadrant)`

`=> sin x_0 = -sin(x_0 – alpha)`

 

iii.   

`text(When)\ \ x = x_0,`

`y_1` `=sin x_0`  
`y_2` `=sin(x_0 – alpha) + k`  
`sin x_0` `=sin (x_0 – alpha) + k`  
  `= -sin x_0 + k`  
`k` `== 2\ sin x_0`  

 

♦♦ Mean mark part (iii) 21%.

`text(S)text(ince)\ \ cos x_0` `= cos(x_0 – alpha)`
`x_0` `= -(x_0 – alpha)`
`2x_0` `= alpha`
`x_0` `= alpha/2`

 
 `:. k = 2 sin\ alpha/2`

Proof, EXT2 P2 SM-Bank 5

Use mathematical induction to prove that

`n^5 + n^3 + 2n`

is divisible by 4 for integers  `n >= 1.`  (4 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(If)\ n = 1,`

`1 + 1 + 2 xx 1 = 4\ \ text{(divisible by 4)}`

`:. text(True for)\ n = 1`
 

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

`text(i.e.)\ \ k^5 + k^3 + 2k = 4P\ \ …\ text{(1)}\ \ (text(where)\ P ∈\ text(integer))`
 

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

`text(i.e.)\ \ (k + 1)^5 + (k + 1)^3 + 2(k + 1)\ \ text(is divisible by 4)`

`text(Expanding:)`

`k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 + k^3 + 3k^2 + 3k + 1 + 2k + 2`

`= k^5 + 5k^4 + 11k^3 + 13k^2 + 8k + 4`

`= 4P + 5k^4 + 10k^3 + 13k^2 + 8k + 4\ \ \ \ text{(see (1) above)}`

`= 4P + 4 underbrace{(k^4 + 2k^3 + 3k^2+2k + 1)}_(text(integer)\ Q) + k^4 + 2k^3 + k^2`

`= 4(P + Q) + k^2(k^2 + 2k + 1)`

`= 4(P + Q) + k^2(k + 1)^2`
 

`text(For any integer)\ \ k >= 2, \ k^2(k + 1)^2\ \ text(is  (odd))^2 xx (text(even)^2)`

`text(and  (even))^2 = (2R)^2\ \ text(where)\ \ R ∈\ text(integer)`

`=> k^2(k + 1)^2 = 4R^2 xx (text(odd))^2\ \ \ text{(divisible by 4)}`

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

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

Algebra, STD1 A2 2019 HSC 33

The relationship between British pounds `(p)` and Australian dollars `(d)` on a particular day is shown in the graph.
 


 

  1. Write the direct variation equation relating British pounds to Australian dollars in the form  `p = md`. Leave `m` as a fraction.  (1 mark)

  2. The relationship between Japanese yen `(y)` and Australian dollars `(d)` on the same day is given by the equation  `y = 76d`.

     

    Convert 93 100 Japanese yen to British pounds.  (2 marks)

Show Answers Only
  1. `p = 4/7 d`
  2. `93\ 100\ text(Yen = 700 pounds)`
Show Worked Solution

a.   `m = text(rise)/text(run) = 4/7`

♦♦♦ Mean mark part (a) 8%.

`p = 4/7 d`

 

b.   `text(Yen to Australian dollars:)`

♦♦ Mean mark part (b) 16%.

`y` `=76d`
`93\ 100` `= 76d`
`d` `= (93\ 100)/76`
  `= 1225`

 
`text(Aust dollars to pounds:)`

`p` `= 4/7 xx 1225`
  `= 700\ text(pounds)`

 
`:. 93\ 100\ text(Yen = 700 pounds)`

Financial Maths, STD1 F3 2019 HSC 32

Ashley has a credit card with the following conditions:

  • There is no interest-free period.
  • Interest is charged at the end of each month at 18.25% per annum, compounding daily, from the purchase date (included) to the last day of the month (included).

Ashley's credit card statement for April is shown, with some figures missing.

The minimum payment is calculated as 2% of the closing balance on 30 April.

Calculate the minimum payment.  (3 marks)

Show Answers Only

`$74.40`

Show Worked Solution
`text(Daily interest)` `= 18.25/(100 xx 365)`
  `= 0.0005`

♦♦♦ Mean mark 12%.

 

`text(Closing balance)` `= 3700(1.0005)^11`
  `= 3720.40`

 

`:.\ text(Minimum payment)` `= 3720.40 xx 0.02`
  `= $74.408…`
  `= $74.41\ \ text{(nearest cent)}`

Measurement, STD1 M3 2019 HSC 31

Two right-angled triangles, `ABC` and `ADC`, are shown.
 


 

Calculate the size of angle `theta`, correct to the nearest minute.  (3 marks)

Show Answers Only

`41°4′\ \ text{(nearest minute)}`

Show Worked Solution

`text(Using Pythagoras in)\ DeltaACD:`

♦♦♦ Mean mark 15%.

`AC^2` `= 2.5^2 + 6^2`
  `= 42.25`
`:.AC` `= 6.5\ text(cm)`

 
`text(In)\ DeltaABC:`

`costheta` `= 4.9/6.5`
`theta` `= cos^(−1)\ 4.9/6.5`
  `= 41.075…`
  `= 41°4′31″`
  `= 41°5′\ \ text{(nearest minute)}`

Functions, EXT1 F1 2019 HSC 10 MC

The function  `f(x) = -sqrt(1 + sqrt(1 + x))`  has inverse  `f^(-1) (x)`.

The graph of  `y = f^(-1) (x)`  forms part of the curve  `y = x^4 - 2x^2`.

The diagram shows the curve  `y = x^4 - 2x^2`.
 

 
How many points do the graphs of  `y = f(x)`  and  `y = f^(-1) (x)`  have in common?

A.   1

B.   2

C.   3

D.   4

Show Answers Only

`A`

Show Worked Solution

`f(x) = -sqrt(1 + sqrt(1 + x))`

♦♦♦ Mean mark 29%.

`text(Domain)\ \ f(x)\ \ text(is)\ \ x >= -1`

`text(Range)\ \ f(x)\ \ text(is)\ \ y <= -1\ \ text(since)\ \ y\ \ text(decreases)`

`text(as)\ \ x\ \ text(increases.)`

`f(-1) = -sqrt(1 + sqrt 0) = -1`

`:.\ text{One intersection (only) occurs at}\ \ (-1, -1).`

`=>  A`

Calculus, 2ADV C4 2019 HSC 16c

The diagram shows the region  `R`, bounded by the curve  `y = x^r`, where  `r >= 1`, the `x`-axis and the tangent to the curve at the point  `(1, 1)`.
 

  1. Show that the tangent to the curve at  `(1, 1)`  meets the `x`-axis at
     
         `qquad ((r - 1)/r, 0)`.  (2 marks)

  2. Using the result of part (i), or otherwise, show that the area of the region  `R`  is
     
         `qquad (r - 1)/(2r (r + 1))`.  (2 marks)

  3. Find the exact value of  `r`  for which the area of  `R`  is a maximum.  (3 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `r = 1 + sqrt 2`
Show Worked Solution
i.    `y` `= x^r`
  `(dy)/(dx)` `= r x^(r – 1)`

 
`text(When)\ \ x = 1, \ (dy)/(dx) = r`

♦♦ Mean mark part (i) 31%.

`text(Equation of tangent:)`

`y – 1` `= r(x – 1)`
`y` `= rx – r + 1`

 
`text(When)\ \ y = 0:`

`rx – r + 1` `= 0`
`rx` `= r – 1`
`x` `= (r – 1)/r`

 
`:.\ text(T)text(angent meets)\ x text(-axis at)\ \ ((r – 1)/r, 0)`

 

ii.   `text(Area under curve)`

♦♦♦ Mean mark part (ii) 21%.

`= int_0^1 x^r`

`= [1/(r + 1) ⋅ x^(r + 1)]_0^1`

`= 1/(r + 1) xx 1^(r + 1) – 0`

`= 1/(r + 1)`

 
`text(Area under tangent)`

`= 1/2 xx b xx h`

`= 1/2 (1 – (r – 1)/r) xx 1`

`= 1/2 (1 – (r – 1)/r)`
 

`:. R` `= 1/(r + 1) – 1/2(1 – (r – 1)/r)“
  `= 1/(r + 1) – 1/(2r) [r – (r – 1)]`
  `= 1/(r + 1) – 1/(2r)`
  `= (2r – (r + 1))/(2r(r + 1))`
  `= (r – 1)/(2r(r + 1))`

 

iii.    `R` `= (r – 1)/(2r(r + 1)) = (r – 1)/(2r^2 + 2r)`
  `(dR)/(dr)` `= ((2r^2 + 2r) xx 1 – (r – 1)(4r + 2))/(2r^2 + 2r)^2`
    `= (2r^2 + 2r – 4r^2 – 2r + 4r + 2)/(2r^2 + 2r)^2`
    `= (-2r^2 + 4r + 2)/(2r^2 + 2r)^2`
    `= (-2(r^2 – 2r – 1))/(2r^2 + 2r)^2`

 

`text(Find)\ \ r\ \ text(when)\ \ (dR)/(dr) = 0:`

♦♦ Mean mark part (iii) 23%.

`r^2 – 2r – 1 = 0`

`r` `= (2 +- sqrt((-2)^2 – 4 xx 1 xx (-1)))/2 `
  `= (2 +- sqrt 8)/2`
  `= 1 + sqrt 2\ \ (r >= 1)`

 

  `qquadr qquad` `qquad 1 qquad` `\ \ 1 + sqrt 2\ \ ` `qquad 3 qquad`
  `(dR)/(dr)` `1/4` `0` `-1/144`

 

`:. R_text(max)\ text(occurs when)\ \ r = 1 + sqrt 2`

Financial Maths, 2ADV M1 2019 HSC 16a

A person wins $1 000 000 in a competition and decides to invest this money in an account that earns interest at 6% per annum compounded quarterly. The person decides to withdraw $80 000 from this account at the end of every fourth quarter. Let  `A_n`  be the amount remaining in the account after the `n`th withdrawal.

  1.  Show that the amount remaining in the account after the withdrawal at the end of the eighth quarter is

     

    `qquad A_2 = 1\ 000\ 000 xx 1.015^8 - 80\ 000(1 + 1.015^4)`.  (2 marks)

  2.  For how many years can the full amount of $80 000 be withdrawn?  (3 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `23\ text(years)`
Show Worked Solution

i.  `6 text(% p.a.)= 1.5 text(%)\ text(per quarter)`

  `A_1` `= 1\ 000\ 000 xx 1.015^4 – 80\ 000`
  `A_2` `= A_1 xx 1.015^4 – 80\ 000`
    `= (1\ 000\ 000 xx 1.015^4 – 80\ 000) xx 1.015^4 – 80\ 000`
    `= 1\ 000\ 000 xx 1.015^8 – 80\ 000 xx 1.015^4 – 80\ 000`
    `= 1\ 000\ 000 xx 1.015^8 – 80\ 000 (1 + 1.015^4)`

 

ii.    `A_3` `= 1\ 000\ 000 xx 1.015^12 – 80\ 000 (1 + 1.015^4 + 1.015^8)`
  `vdots`  
  `A_n` `= 1\ 000\ 000 xx 1.015^(4n) – 80\ 000\ underbrace{(1 + 1.015^4 + … + 1.015^(4n – 4))}_{text(GP where)\ a = 1,\ r = 1.015^4}`
    `= 1\ 000\ 000 xx 1.015^(4n) – 80\ 000 [(1(1.015^(4n) – 1))/(1.015^4 – 1)]`
    `= 1\ 000\ 000 xx 1.015^(4n) – 1\ 303\ 706 (1.015^(4n) – 1)`

♦♦♦ Mean mark part (ii) 19%.

`text(Find)\ \ n\ \ text(when)\ \ A_n =0:`

`0` `= 1\ 000\ 000 xx 1.015^(4n) – 1\ 303\ 706 xx 1.015^(4n) + 1\ 303\ 706`
`-1\ 303\ 706` `> -303\ 706 xx 1.015^(4n)`
`1.015^(4n)` `> (1\ 303\ 706)/(303\ 706)`
`4n` `> (ln((1\ 303\ 706)/(303\ 706)))/(ln 1.015)`
  `> 97.853…`
`n` `> 24.46…`

 
`:.\ text(Full amount can be drawn for 24 years.)`

Networks, STD2 N3 2019 HSC 40

A museum is planning an exhibition using five rooms.

The museum manager draws a network to help plan the exhibition. The vertices `A`, `B`, `C`, `D` and `E` represent the five rooms. The number on the edges represent the maximum number of people per hour who can pass through the security checkpoints between the rooms.
 


 

  1. What is the capacity of the cut shown?  (1 mark)

  2. The museum manager is planning for a maximum of 240 visitors to pass through the exhibition each hour. By using the 'minimum cut-maximum flow' theorem, the manager determines that the plan does not provide sufficient flow capacity.

     

    Draw the minimum cut onto the network below and recommend a change that the manager could make to one or more security checkpoints to increase the flow capacity to 240 visitors per hour.   (2 marks)
     
       

Show Answers Only
  1. `290`
  2.   

Show Worked Solution
a.    `text(Capacity)` `= 130 + 90 + 70`
    `= 290`

♦♦ Mean mark 32%.
COMMENT: In part (a), edge BC flows from the exit to the entry and is therefore not counted.

b.   `text(Maximum flow capacity:)`

 

`text(Minimum cut = 80 + 40 + 65 + 45 = 230)`

♦♦♦ Mean mark 19%.
COMMENT: In part (b), edge BC now flows from entry to exit in the new “minimum” cut and is counted.

`text(If security is improved to increase the flow)`

`text(between Room C and Room B by 10 visitors)`

`text(per hour, the network’s flow capacity increases)`

`text(to 240.)`

Statistics, STD2 S5 2019 HSC 15 MC

The scores on an examination are normally distributed with a mean of 70 and a standard deviation of 6. Michael received a score on the examination between the lower quartile and the upper quartile of the scores.

Which shaded region most accurately represents where Michael's score lies?
 

A. B.
C. D.
Show Answers Only

`A`

Show Worked Solution

`text{68% of marks lie between 64 and 76 (mean ± 1 σ).}`

♦♦♦ Mean mark 17%.

`text(50% of marks lie between)\ Q_1\ text(and)\ Q_3.`

`=> A`

Statistics, STD2 S1 2019 HSC 10 MC

A school collected data related to the reasons given by students for arriving late. The Pareto chart shows the data collected.
 

What percentage of students gave the reason 'Train or bus delay'?

  1. `text(6%)`
  2. `text(15%)`
  3. `text(30%)`
  4. `text(92%)`
Show Answers Only

`A`

Show Worked Solution

`text(Train or bus delay (%))`

♦♦♦ Mean mark 18%.

`= 92 – 86`

`= 6text(%)`

`=> A`

Financial Maths, STD2 F1 2019 HSC 9 MC

What is the interest earned, in dollars, if $800 is invested for `x` months at a simple interest rate of 3% per annum?

  1. `2x`
  2. `24x`
  3. `200x`
  4. `2400x`
Show Answers Only

`A`

Show Worked Solution

♦♦♦ Mean mark 20%!

`text(Interest)` `= 800 xx x/12 xx 3/100`
  `= 2x`

 
`=> A`

Statistics, EXT1 S1 2012 MET2 3

Steve and Jess are two students who have agreed to take part in a psychology experiment. Each has to answer several sets of multiple-choice questions. Each set has the same number of questions, `n`, where `n` is a number greater than 20. For each question there are four possible options A, B, C or D, of which only one is correct.

  1. Steve decides to guess the answer to every question, so that for each question he chooses A, B, C or D at random.

     

    Let the random variable `X` be the number of questions that Steve answers correctly in a particular set.

    1. What is the probability that Steve will answer the first three questions of this set correctly?  (1 mark)

    2. Use the fact that the variance of `X` is `75/16` to show that the value of `n` is 25.  (1 mark)

  1. The probability that Jess will answer any question correctly, independently of her answer to any other question, is  `p\ (p > 0)`. Let the random variable `Y` be the number of questions that Jess answers correctly in any set of 25.

    If   `P(Y > 23) = 6 xx P(Y = 25)`, show that the value of  `p=5/6`.  (2 marks)

Show Answers Only

a.i.  `1/64`

a.ii.  `text(See Worked Solutions)`

b.  `text(See Worked Solutions)`

Show Worked Solution
a.i.    `Ptext{(3 correct in a row)}` `= (1/4)^3`
    `= 1/64`

 

a.ii.    `text(Var)(X)` `= np(1 – p)`
  `75/16` `= n(1/4)(3/4)`
  `75` `= 3n`
  `:. n` `= 25`

 

b.   `Y ∼\ text(Bin)(25,p)`

♦♦♦ Mean mark part (c) 19%.
`P(Y > 23)` `= 6xx P(Y = 25)`
`P(Y = 24) + P(Y = 25)` `= 6xx P(Y = 25)`
`P(Y = 24)` `= 5xx P(Y = 25)`
`((25),(24))p^24(1 – p)^1` `= 5p^25`
`25p^24(1 – p)` `= 5p^25`
`25p^24-25p^25-5p^25` `=0`
`25p^24-30p^25` `=0`
`5p^24(5 – 6p)` `= 0`

 
`:. p = 5/6,\ \ (p>0)\ \ text(… as required)`

Vectors, EXT1 V1 2018 SPEC2 12 MC

If  `|underset ~a + underset ~b| = |underset ~a| + |underset ~b|`  and  `underset ~a, underset ~b != underset ~0`, which one of the following is necessarily true?

A.   `underset ~a\ text(is parallel to)\ underset ~b`

B.   `|underset ~a| = |underset ~b|`

C.   `underset ~a = underset ~b`

D.   `underset ~a\ text(is perpendicular to)\ underset ~b` 

Show Answers Only

`A`

Show Worked Solution
`|underset ~a + underset ~b|^2` `= (|underset ~a| + |underset ~b|)^2\ \ \ text{(given)}`
  `= |underset ~a|^2 + 2|underset ~a||underset ~b|+|underset ~b|^2`
`underset ~a ⋅ underset ~b` `= |underset ~a||underset ~b| cos theta`

 
`=> 2|underset ~a||underset ~b| = (2 underset ~a ⋅ underset ~b)/(cos theta)`

♦♦♦ Mean mark 36%.

`=>|underset ~a + underset ~b|^2 = |underset ~a|^2 + (2 underset ~a ⋅ underset ~b)/(cos theta) + |b|^2`

`(underset ~a + underset ~b) * (underset ~a + underset ~b) = underset ~a ⋅ underset ~a + (2 underset ~a ⋅ underset ~b)/(cos theta) + underset ~b ⋅ underset ~b`

`underset ~a ⋅ underset ~a + 2underset ~a ⋅ underset ~b + underset ~b ⋅ underset ~b = underset ~a ⋅ underset ~a + (2 underset ~a ⋅ underset ~b)/(cos theta) + underset ~b ⋅ underset ~b`

`2 underset ~a ⋅ underset ~b = (2 underset ~a ⋅ underset ~b)/(cos theta)`

`:. cos theta = 1\ \ =>\ \  theta = 0`

`=>  A`

Calculus, MET1 2018 VCAA 9

Consider a part of the graph of  `y = xsin(x)`, as shown below.

  1.  i. Given that  `int(xsin(x))\ dx = sin(x)-xcos(x) + c`, evaluate  `int_(npi)^((n + 1)pi)(xsin(x))\ dx`  when `n` is a positive even integer or 0.
      
    Give your answer in simplest form.   (2 marks)

    ii. Given that  `int(xsin(x))\ dx = sin(x)-xcos(x) + c`, evaluate  `int_(npi)^((n + 1)pi)(xsin(x))\ dx`  when `n` is a positive odd integer.
    Give your answer in simplest form.   (1 mark)

  2.  
  3. Find the equation of the tangent to  `y = xsin(x)`  at the point  `(−(5pi)/2,(5pi)/2)`.   (2 marks)

  4. The translation `T` maps the graph of  `y = xsin(x)`  onto the graph of  `y = (3pi-x)sin(x)`, where
  5. `qquad T: R^2 -> R^2, T([(x),(y)]) = [(x),(y)] + [(a),(0)]`
  6. and `a` is a real constant.
  7. State the value of `a`.   (1 mark)

  8. Let  `f:[0,3pi] -> R, f(x) = (3pi-x)sin(x)`  and  `g:[0,3pi] -> R, g(x) = (x-3pi)sin(x)`.
    The line `l_1` is the tangent to the graph of `f` at the point `(pi/2,(5pi)/2)` and the line `l_2` is the tangent to the graph of `g` at `(pi/2,-(5pi)/2)`, as shown in the diagram below.
     
         
    Find the total area of the shaded regions shown in the diagram above.   (2 marks)

Show Answers Only
  1.  i. `(2n + 1)pi`
    ii. `-(2n + 1)pi`
  2. `:. y =-x`
  3. `-3pi`
  4. `9pi(pi-2)`
Show Worked Solution

a.i.  `text(Given)\ \ n\ \ text(is a positive even integer:)`

♦♦ Mean mark 31%.

`int_(npi)^((n + 1)pi)(xsin(x))dx`

`= [sin(x)-xcos(x)]_(npi)^((n + 1)pi)`

`= [sin((n + 1)pi)-(n + 1)pi · cos((n + 1)pi)]-[sin(npi)-npi · cos(npi)]`

`= [0-(n + 1)pi (−1)]-[0-npi]`

`= (n + 1)pi + npi`

`= (2n + 1)pi`

 

a.ii. `text(Given)\ \ n\ \ text(is a positive odd integer:)`

♦♦♦ Mean mark 19%.

`int_(npi)^((n + 1)pi)(xsin(x))dx`

`= [sin((n + 1)pi)-(n + 1)pi · cos((n + 1)pi)]-[sin(npi)-npi · cos(npi)]`

`= [0-(n + 1)pi(1)]-[0-npi(−1)]`

`= -(n + 1)pi-npi`

`= -(2n + 1)pi`

 

b.    `y` `= xsin(x)`
  `(dy)/(dx)` `= x · cos(x) + sin(x)`

 

♦ Mean mark part (b) 49%.

`(dy)/(dx)` `= -(5pi)/2 · cos(-(5pi)/2) + sin(-(5pi)/2)`
  `= -(5pi)/2 · (0) + (-1)`
  `= -1`

 
`text(T)text(angent has equation)\ \ y =-x + c\ \ text(and passes through)\ \ (-(5pi)/2, (5pi)/2):`

`(5pi)/2` `= +(5pi)/2 + c`
`c` `= 0`

 
`:. y = -x`

 

♦♦ Mean mark part (c) 34%.

c.    `y` `= (3pi-x)sin(x)`
    `=-(x-3pi)sin(x)`

 
`:. a = 3pi`
 

d.   `f(x) = (3pi-x)sin(x)`

♦♦♦ Mean mark 7%.

  `-> l_1\ text(is the tangent)\ \ y =-x\ \ (text{using part (b)})`

`-> g(x)\ text(is)\ \ y=xsin(x)\ \ text(translated 3π to the right.)`

`-> f(x)\ text(is)\ \ g(x)\ \ text(reflected in the)\ \ x text(-axis.)`

 
`text(Area between)\ f(x)\ text(and)\ xtext(-axis)`

`= int_0^pi xsin(x)\ dx + | int_pi^(2pi) xsin(x)\ dx | + int_(2pi)^(3pi) xsin(x)\ dx`

`= (2 xx 0 + 1)pi + |-1(2 xx 1 + 1)pi | + (2 xx 2 + 1)pi\ \ (text{using part (a)})`

`= pi + 3pi + 5pi`

`= 9pi`
 

`:.\ text(Shaded Area)`

`= 2 xx (1/2 xx 3pi xx 3pi)-2 xx 9pi`

`= 9pi^2-18pi`

`= 9pi(pi-2)`

Statistics, 2ADV S3 SM-Bank 4

The continuous random variable \(X\) has a probability density function given  by
 

\(f(x)= \begin{cases}
\cos(2x)& \text {if}\quad \dfrac{3 \pi}{4}<x<\dfrac{5 \pi}{4} \\
\ \\
0 & \text{elsewhere}
\end{cases}\)
 

Find the value of  \(a\) such that  \(P(X < a) = 0.25\). Give your answer correct to 2 decimal places.   (3 marks)

Show Answers Only

\(2.8\)

Show Worked Solution

\(\displaystyle \int_{\frac{3 x}{4}}^a \cos (2 x) d x=0.25\)

\begin{aligned}
\frac{1}{2}[\sin (2 x)]_{\frac{3 \pi}{4}}^a & =0.25 \\
{\left[\sin (2 a)-\sin \left(\frac{3 \pi}{2}\right)\right] } & =0.5 \\
\sin (2 a)+1 & =0.5 \\
2 a & =\sin ^{-1}(-0.5) \\
& =-0.5235
\end{aligned}

\(\text {Since sin is negative in 3rd/4th quadrants: }\)

\begin{aligned}
2 a & =\pi+0.5234 \\
a &=1.832 \ldots \quad \text { (not in range) } \\
&\text { or } \\
2a & =2 \pi-0.5234 \\
 a &=2.87989 \ldots \\
&=2.88 \quad\left(\text{in range: } \frac{3 \pi}{4}<x<\frac{5 \pi}{4}\right)
\end{aligned}

Calculus, SPEC2 2011 VCAA 16 MC

The gradient of the the perpendicular line to a curve at any point  `P(x,y)`  is twice the gradient of the line joining `P` and the point  `Q(1,1)`.

The coordinate of points on the curve satisfy the differential equation

  1. `(dy)/(dx) + (x - 1)/(2(y - 1)) = 0`
  2. `(dy)/(dx) - (x - 1)/(2(y - 1)) = 0`
  3. `(dy)/(dx) + (2(y - 1))/(x - 1) = 0`
  4. `(dy)/(dx) + (2(x - 1))/(y - 1) = 0`
  5. `(dy)/(dx) - (2(y - 1))/(x - 1) = 0`
Show Answers Only

`A`

Show Worked Solution

`m_T = dy/dx`

♦♦♦ Mean mark 25%.
COMMENT: Easily the lowest mean mark in the 2011 MC section.

`m_⊥ = −1/(m_T) = – (dx)/(dy)`

`m_(PQ) = (y-1)/(x-1)`

`- (dx)/(dy)` `= 2m_(PQ)`
`-(dx)/(dy)` `= (2(y-1))/(x-1)`
`- (dy)/(dx)` `= (x – 1)/(2(y-1))`
`0` `=dy/dx + (x – 1)/(2(y-1))`

 
`=> A`

Graphs, SPEC1 2012 VCAA 10

Consider the functions with rules  `f(x) = arcsin (x/2) + 3/sqrt (25 x^2-1)`  and  `g(x) = arcsin (3x)-3/sqrt (25x^2-1).`

    1. Find the maximal domain of  `f_1(x) = arcsin (x/2).`   (1 mark)

    2. Find the maximal domain of  `f_2(x) = 3/sqrt (25x^2-1).`   (1 mark)

    3. Find the largest set of values of  `x in R`  for which  `f(x)`  is defined.   (1 mark)

  2. Given that  `h(x) = f(x) + g(x)`  and that  `theta = h(1/4)`, evaluate  `sin (theta).`

     

    Give your answer in the form  `(a sqrt b)/c, \ a, b, c in Z.`   (3 marks)

Show Answers Only
    1. `-2 <= x <= 2`
    2. `x < -1/5 uu x > 1/5`
    3. `-2 <= x < -1/5 uu 1/5 < x <= 2`
  2. `(5 sqrt 7)/16`
Show Worked Solution

a.i.   `text(Maximal Domain occurs when:)`

  `-1 <= x/2 <= 1`
  `{x: -2 <= x <= 2}`

 
a.ii.   `text(Maximal Domain occurs when:)`

♦♦ Mean mark part (a)(ii) 33%.

  `25x^2-1 > 0`
  `x^2 > 1/25`
  `{x: x < -1/5 uu x > 1/5}`

 

a.iii.  `text(Max domain for which)\ \ f(x)\ \ text(is defined:)`

♦♦ Mean mark part (a)(iii) 26%.

  `(-2 <= x <= 2) nn (x < -1/5 uu x > 1/5)`
  `{x: -2 <= x < -1/5 \ uu \ 1/5 < x <= 2}`

 

b.    `h(x)` `= sin^(-1) (x/2) + 3/sqrt(25x^2-1) + sin^(-1)(3x)-3/sqrt(25x^2-1)`
    `= sin^(-1)(x/2) + sin^(-1)(3x)`

 
`text(When)\ \ x=1/4,\ \ h(x)=theta`

♦♦♦ Mean mark part (b) 20%.

`theta=sin^(-1) (1/8) + sin^(-1) (3/4)`

`text(Let)\ \ theta_1 = sin^(-1) (1/8),\ \ theta_2 = sin^(-1) (3/4)`

`text(Using)\ \ sin(theta_1 + theta_2)=sin theta_1 cos theta_2 + cos theta_1 sin theta_2:`

`sin(theta)` `= 1/8 * sqrt7/4 + sqrt63/8 * 3/4`
  `=sqrt7/32 + (9sqrt7)/32`
  `=(5 sqrt7)/16`

Mechanics, SPEC2 2015 VCAA 16 MC

SPEC2 2015 VCAA 16 MC

The diagram above shows a mass suspended in equilibrium by two light strings that make angles of `60^@` and `30^@` with a ceiling. The tensions in the strings are `T_1` and `T_2`, and the weight force acting on the mass is `underset~W`. The correct statement relating the given forces is

A.   `underset~T_1 + underset~T_2 + underset~W = underset~0`

B.   `underset~T_1 + underset~T_2 - underset~W = underset~0`

C.   `underset~T_1 xx 1/2 + underset~T_2 xx sqrt3/2 = underset~0`

D.   `underset~T_1 xx sqrt3/2 + underset~T_2 xx 1/2 = underset~W`

E.   `underset~T_1 xx 1/2 + underset~T_2 xx sqrt3/2 = underset~W`

Show Answers Only

`A`

Show Worked Solution

`text(S)text(ince equilibrium exists:)`

♦♦♦ Mean mark 26%.

`sumunderset~F = underset~(T_1) + underset~(T_2) + underset~W = underset~0`

 
`=> A`

Vectors, SPEC2 2017 VCAA 5

On a particular morning, the position vectors of a boat and a jet ski on a lake  `t`  minutes after they have started moving are given by  `underset~r_B(t) = (1-2cos(t)) underset~i + (3 + sin(t))underset~j`  and  `underset~r_J(t) = (1-sin(t)) underset~i + (2-cos(t))underset~j`  respectively for  `t >= 0`, where distances are measured in kilometres. The boat and the jet ski start moving at the same time. The graphs of their paths are shown below.

  1. On the diagram above, mark the initial positions of the boat and the jet ski, clearly identifying each of them. Use arrows to show the directions in which they move.   (2 marks)

    1. Find the first time for  `t > 0`  when the speeds of the boat and the jet ski are the same.   (2 marks)

    2. State the coordinates of the boat at this time.   (1 mark)

    1. Write down an expression for the distance between the jet ski and the boat at any time `t`.   (1 mark)

    2. Find the minimum distance separating the boat and the jet ski. Give your answer in kilometres, correct to two decimal places.   (1 mark)

  4. On another morning, the boat’s position vector remained the same but the jet skier considered starting from a different location with a new position vector given by  `underset~r(t) = (1-2sin(t)) underset~i + (a-cos(t))underset~j, \ t >= 0`, where `a` is a real constant. Both vessels are to start at the same time.
    Assuming the vessels would collide shortly after starting, find the time of the collision and the value of `a`.   (3 marks)

Show Answers Only
  1.  
  2.   
    1. `t = 0, t_2 = pi`
    2. `(3,3)`
  3.   
    1. `sqrt((sin(t)-2cos(t))^2 + (1 + sin(t) + cos (t))^2)`
    2. `d_text(min) ~~ 0.33`
  4. `a = 3 + 3/sqrt5, t = tan^(−1)(2)`
Show Worked Solution
a.   

`text(Initial positions occur at)\ \ t=0.`

`text(Consider the graph)\ \ underset~r_B(t)\ \ text(at)\ \ t=pi/4:`

`(1-2cos0) < (1-2cos(pi/4)) =>\ text(moves higher)`
  

`text(Consider the graph)\ \ underset~r_J(t)\ \ text(at)\ \ t=pi/4:`

`(1-sin(0)) > (1-sin(pi/4)) =>\ text(moves left)`

 

b.i.    `underset~dotr_B(t)` `= 2sin(t)underset~i + cos(t)underset~j`
  `|underset~dotr_B(t)|` `= sqrt(4sin^2(t) + cos^2(t))`
`underset~dotr_J(t)` `= −cos(t)underset~i + sin(t)underset~j`
`|underset~dotr_J(t)|` `= sqrt(cos^2(t) + sin(t))=1`
   

`text(Find)\ \ t\ \ text(when)\ \ sqrt(4sin^2(t) + cos^2(t))=1:`

`t = pi\ text(seconds)\ \ \ (t!=0)`

 

♦ Mean mark (b)(ii) 48%.

b.ii.    `(underset~r)_B(pi)` `= (1-2cos(pi))underset~i + (3 + sin(pi))underset~j`
    `= 3underset~i + 3underset~j`

`:.\ text(Boat coordinates):\  (3,3)`

 

c.i.    `underset~r_B-underset~r_J` `=(sin(t)-2 cos(t))i +(1+sin(t) + cos(t))`
  `:. d` `= |underset~r_B-underset~r_J|`
    `= sqrt((sin(t)-2cos(t))^2 + (1 + sin(t) + cos (t))^2)`

 

c.ii.  `d_text(min) ~~ 0.33\ \ \ text{(by CAS)}`

♦♦♦ Mean mark part (c)(ii) 15%.

 

d.   `text(Equating coefficients for collision:)`

  `x:\ \ \ 1-sin(t)` `= 1-2cos(t)\ …\ (1)`
  `sin(t)` `= 2cos(t)`
  `tan(t)` `= 2`
  `t` `=tan^(−1)(2)`

 
`y:\ \ \ a-cos(t) = 3 + sin(t)\ …\ (2)`

♦ Mean mark part (d) 41%.
 

 
`text(Using)\ \ tan(t)=2,`

`=> sin(t) = (2sqrt5)/5,\ \ cos(t) = sqrt5/5`

`text{Substitute into (1):}`

`a-1/sqrt5` `= 3 + 2/sqrt5`  
`:. a` `= 3 + 3/sqrt5`  

 
`:.\ text(Collision occurs when)\ \ t=tan^(-1)2\ \ text(and)\ \ a=3 + 3/sqrt5`

Complex Numbers, SPEC2 2017 VCAA 4

  1. Express  `−2-2sqrt3 i`  in polar form.  (1 mark)

  2. Show that the roots of  `z^2 + 4z + 16 = 0`  are  `z = −2-sqrt3 i`  and  `z = −2 + 2sqrt3 i`.  (1 mark)

  3. Express the roots of  `z^2 + 4z + 16 = 0`  in terms of  `2-2sqrt3 i`.  (1 mark)

  4. Show that the cartesian form of the relation  `|z| = |z-(2-2sqrt3 i)|`  is  `x-sqrt3 y-4 = 0`  (2 marks)

  5. Sketch the line represented by  `x-sqrt3y -4 = 0`  and plot the roots of  `z^2 + 4z + 16 = 0`  on the Argand diagram below.  (2 marks)

     
       

  6. The equation of the line passing through the two roots of  `z^2 + 4z + 16 = 0`  can be expressed as  `|z-a| = |z-b|`, where  `a, b ∈ C`.

     

    Find `b` in terms of `a`.  (1 mark)

  7. Find the area of the major segment bounded by the line passing through the roots of  `z^2 + 4z + 16 = 0`  and the major arc of the circle given by  `|z| = 4`.  (2 marks)

Show Answers Only
  1. `4text(cis)((−2pi)/3)`
  2. `text(See Worked Solutions)`
  3. `−(2-2sqrt3 i) and-bar((2-2sqrt3 i))`
  4. `text(See Worked Solutions)`
  5.  
  6. `−4-bara`
  7. `4sqrt3 + (32pi)/3`
Show Worked Solution
a.    `r` `= sqrt((−2)^2 + (−2sqrt3)^2)=4`

 

`theta` `= −pi + tan^(−1)((2sqrt3)/2)`
  `= −pi + pi/3`
  `=(-2pi)/3`

 
`:. −2-2sqrt3 i = 4text(cis)((−2pi)/3)`
 

b.   `z^2 + 4z + z^2-4 + 16` `=0`
`(z + 2)^2 + 12` `= 0`
`(z + 2)^2` `= -12`
`(z + 2)^2` `= 12i^2`
`z + 2` `= ±sqrt12 i`
`z + 2` `= ±2sqrt3 i`
`:. z` `= -2 ± 2sqrt3 i`

 

♦♦ Mean mark part (c) 31%.

c.    `z_1` `= -2 + 2sqrt3 i = -(2-2sqrt3 i)`
  `z_2` `=-2-2sqrt3 i =-bar((2-2sqrt3 i))`

 

d. `|z|` `= |z-(2-2sqrt3 i)|`
     `x^2 + y^2` `= (x-2)^2 + (y + 2sqrt3)^2`
    `x^2 + y^2` `= x^2-4x + 4+ y^2 + 4sqrt3 y + 12`
  `0` `= −4x + 4sqrt3 y + 16`
  `0` `= −x + sqrt3 y + 4`

 
`:. x-sqrt3 y-4=0`

 

e.   

 

f.   `x = − 2\ \ text(is equidistant from)\ \ z_1 = a\ \ text(and)\ \ z_2 = b`

♦♦♦ Mean mark 1%!

`=> text(Im)(a) = text(Im)(a)`
 

`text(Let)\ \ a = alpha + betai, \ b = gamma + betaj`

`(alpha + gamma)/2` `= −2`
`alpha + gamma` `= −4`
`gamma` `= -4-alpha`

  

`:. b` `= -4-alpha + betaj`
  `= -4-(alpha + betaj)`
  `= -4-bara`

 

♦♦ Mean mark 31%.

g.    `text(Area)\ DeltaOAB` `= 1/2 xx (4sqrt3 xx 2)`
    `= 4sqrt3`

 

`text(Area of sector)\ AOB` `= pi xx 4^2 xx (2 xx pi/3)/(2pi)`
  `= (16pi)/3`

 
`:.\ text(Area of major segment area)`

`=pi(4)^2-((16pi)/3-4sqrt3)`

`= 4sqrt3 + (32pi)/3`

Complex Numbers, SPEC2-NHT 2017 VCAA 2

One root of a quadratic equation with real coefficients is  `sqrt 3 + i`.

    1. Write down the other root of the quadratic equation.   (1 mark)

    2. Hence determine the quadratic equation, writing it in the form  `z^2 + bz + c = 0`.   (2 marks)

  1. Plot and label the roots of   `z^3-2 sqrt 3 z^2 + 4z = 0`  on the Argand diagram below.   (3 marks)

  

 

  1. Find the equation of the line that is the perpendicular bisector of the line segment joining the origin and the point  `sqrt 3 + i`. Express your answer in the form  `y = mx + c`.   (2 marks)

  1. The three roots plotted in part b. lie on a circle.

     

    Find the equation of this circle, expressing it in the form  `|z-alpha| = beta`,  where  `alpha, beta in R`.   (3 marks)

Show Answers Only
    1. `z_2 = sqrt 3-i`
    2. `z^2-2 sqrt 3 z + 4 = 0`
  2. `text(See Worked Solutions)`
  3. `y = -sqrt 3 x + 2`
  4. `|z-2/sqrt 3 | = 2/sqrt 3`
Show Worked Solution

a.i.   `z_1 = sqrt 3 + i`

 `z_2 = bar z_1 = sqrt 3-i\ \ \ text{(conjugate root)}`

 

a.ii.    `(z-(sqrt 3 + i))(z-(sqrt 3-i))` `= 0`
  `((z-sqrt 3)-i)((z-sqrt 3) + i)` `= 0`
  `(z-sqrt 3)^2-i^2` `= 0`
  `z^2-2 sqrt 3 z + 3 + 1` `= 0`
  `z^2-2 sqrt 3 z + 4` `= 0`

 

b.    `z(z^2-2 sqrt 3 z + 4)` `= 0`
  `z(z-(sqrt 3 + i))(z-(sqrt 3-i))` `= 0`

 
`text(Convert to polar form:)`

  `sqrt 3 + i` `= sqrt((sqrt 3)^2 + 1^2) * text(cis) (tan^(-1)(1/sqrt 3))`
    `= 2 text(cis) (pi/6)`
  `=> sqrt 3-i` `= 2 text(cis) (-pi/6)`

 

   

 

c.   `text(Equidistant from)\ (0, 0) and (sqrt 3, 1)`

`text(Midpoint)\ (x_1, y_1)` `= (sqrt 3/2, 1/2)`

 
`m = (1-0)/(sqrt 3-0) = 1/sqrt 3`

`m_(_|_) = (-1)/m = -sqrt 3`
 

`:.\ text(Equation of ⊥ bisector:)`

`y-1/2` `=-sqrt3(x-sqrt3/2)`  
`y` `=-sqrt3 x +3/2+1/2`  
`:.y` `=-sqrt3 x +2`  

 

d.   `text(Let)\ O = (0, 0),\ P = (sqrt 3, 1),\ Q = (sqrt 3, -1)`

`text(⊥ bisector of two points on arc of a circle passes)`

`text(through the centre of the circle.)`
 

`OP = OQ = PQ = 2`

`=> Delta OPQ\ text(is equilateral)`
 

`text(⊥ bisector of)\ PQ\ text(is)\ y=0.`

`text(Centre of circle occurs when:)`

`0 = -sqrt 3 x + 2\ \ text{(using part c)`

`x=2/sqrt3`

`=>\ text(Radius)\ = 2/sqrt3`
 

`:. |z-2/sqrt 3| = 2/sqrt 3`

Calculus, SPEC2-NHT 2018 VCAA 5

A horizontal beam is supported at its endpoints, which are 2 m apart. The deflection `y` metres of the beam measured downwards at a distance `x` metres from the support at the origin `O` is given by the differential equation  `80 (d^2y)/(dx^2) = 3x-4`.
 


 

  1. Given that both the inclination, `(dy)/(dx)`, and the deflection, `y`, of the beam from the horizontal at  `x = 2`  are zero, use the differential equation above to show that  `80 y = 1/2 x^3-2x^2 + 2x`.   (2 marks)

  2. Find the angle of inclination of the beam to the horizontal at the origin `O`. Give your answer as a positive acute angle in degrees, correct to one decimal place.   (2 marks)

  3. Find the value of `x`, in metres, where the maximum deflection occurs, and find the maximum deflection, in metres.   (3 marks)

  4. Find the maximum angle of inclination of the beam to the horizontal in the part of the beam where  `x >= 1`. Give your answer as a positive acute angle in degrees, correct to one decimal place.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1.4^@`
  3. `x = 2/3; quad 1/135\ text(m)`
  4. `0.5^@`
Show Worked Solution
a.   `80 * int (d^2y)/(dx^2)\ dx` `= int 3x-4\ dx`
  `80* (dy)/(dx)` `= (3x^2)/2-4x + c_0`

 
`text(When)\ \ x=2,\ \ dy/dx=0:`

`80 xx 0` `= (3xx2^2)/2-4xx2 + c_0`
`c_0` `= 2`

 

`80* (dy)/(dx)` `= (3x^2)/2-4x + 2`
`80y` `= int (3x^2)/2-4x + 2\ dx`
  `= 1/2 x^3-2x^2+2x +c_1`

 

`text(When)\ \ x=2,\ \ y=0:`

`0` `= 1/2 2^3-2 xx 2^2 + 2 xx 2 + c_1`
`c_1` `= 0`

`:. 80y = 1/2x^3-2x^2 + 2x`

 

b.   `text(Find)\ \ dy/dx\ \ text(when)\ \ x=0:`

  `80 *(dy)/(dx)` `= 0-0 + 2`
  `(dy)/(dx)` `= 1/40`
  `:. theta` `= tan^(-1) (1/40) ~~ 1.4^@`

 

c.   `text(Find)\ \ x\ \ text(when)\ \ dy/dx=0:`

`x=2\ \ text(or)\ \ 2/3`

`=>y_text(max)\ \ text(occurs at)\ \ x=2/3\ \ (y=0\ \ text(at)\ \ x=2)`

 

`80*y_max` `= 1/2(2/3)^3-2(2/3)^2 + 2(2/3)`
`:. y_max` `= 1/80 xx 16/27`
  `= 1/135\ text(metres)`

 

d.   `text(Max inclination when convexity changes)\ => (d^2y)/(dx^2) =0`

`80*(d^2y)/(dx^2) = 3x-4 = 0`

`=> x = 4/3`

`text(Find)\ \ dy/dx\ \ text(when)\ \ x = 4/3 :`

`(dy)/(dx)` `= 1/80 (3/2(4/3)^2-4(4/3) + 2)`
  `= (-1)/120`

 

`theta` `= tan^(-1) (|-1/120|)`
  `~~ 0.5^@`

Statistics, SPEC2 2018 VCAA 6

The heights of mature water buffaloes in northern Australia are known to be normally distributed with a standard deviation of 15 cm. It is claimed that the mean height of the water buffaloes is 150 cm.

To decide whether the claim about the mean height is true, rangers selected a random sample of 50 mature water buffaloes. The mean height of this sample was found to be 145 cm.

A one-tailed statistical test is to be carried out to see if the sample mean height of 145 cm differs significantly from the claimed population mean of 150 cm.

Let `bar X` denote the mean height of a random sample of 50 mature water buffaloes.

  1. State suitable hypotheses `H_0` and `H_1` for the statistical test.   (1 mark)

  2. Find the standard deviation of `bar X`.   (1 mark)

  3. Write down an expression for the `p` value of the statistical test and evaluate your answer to four decimal places.   (2 marks)

  4. State with a reason whether `H_0` should be rejected at the 5% level of significance.   (1 mark)

  5. What is the smallest value of the sample mean height that could be observed for `H_0` to be not rejected? Give your answer in centimetres, correct to two decimal places.   (1 mark)

  6. If the true mean height of all mature water buffaloes in northern Australia is in fact 145 cm, what is the probability that `H_0` will be accepted at the 5% level of significance? Give your answer correct to two decimal places.   (1 mark)

  7. Using the observed sample mean of 145 cm, find a 99% confidence interval for the mean height of all mature water buffaloes in northern Australia. Express the values in your confidence interval in centimetres, correct to one decimal place.   (1 mark)

Show Answers Only
  1.  `H_0: mu = 150; qquad H_1: mu < 150`
  2. `3/sqrt 2`
  3. `p = text(Pr)(bar X < 145); qquad p~~ 0.0092`
  4. `text(Yes, he should be rejected as)`
    `p~~ 0.0092 < 0.05`
  5. `bar X_min = 146.52`
  6. `0.24`
  7. `(139.5, 150.5)`
Show Worked Solution
a.    `H_0: mu =150`
  `H_1: mu < 150`

 

b.    `sigma_bar X` `= (sigma_X)/sqrt n`
    `= 15/sqrt 50`
    `= (3sqrt 2)/2`

 

c.   `p = text(Pr)(bar X < 145), qquad bar X\ ~\ N (150, 9/2)`

`p ~~ 0.0092`

 

d.   `text(Yes, he should be rejected as)`

`p~~ 0.0092 < 0.05`

 

e.  `text(Pr)(bar X < x) = 0.05`

♦ Mean mark part (e) 48%.

`x ~~ 146.511`

`text(NOT rejected:) quad bar X_min = 146.52`

 

f.   `bar X_2\ ~\ N (145, 9/2)`

♦♦♦ Mean mark part (f) 11%.

`text(Pr)(bar X_2 > x)` `= text(Pr)(bar X_2 > 146.51074)`  
  `~~ 0.24`  

 

g.   `(145-(Z_99 xx 3)/sqrt 2, 145 + (Z_99 xx 3)/sqrt 2)`

`text(Pr)(Z < Z_99) = 0.995, \ Z\ ~\ N(0, 1)`

`=>Z_99 ~~ 2.57583`

`:. 99%\ text(C.I:)\ (139.5, 150.5)`

Calculus, SPEC1 2015 VCAA 9

Consider the curve represented by  `x^2 - xy + 3/2 y^2 = 9.`

  1. Find the gradient of the curve at any point  `(x, y).`  (2 marks)
  2. Find the equation of the tangent to the curve at the point  `(3, 0)`  and find the equation of the tangent to the curve at the point `(0, sqrt 6).`

     

    Write each equation in the form  `y = ax + b.`  (2 marks)

  3. Find the acute angle between the tangent to the curve at the point  `(3, 0)`  and the tangent to the curve at the point  `(0, sqrt 6).`

     

    Give your answer in the form  `k pi`, where `k` is a real constant  (2 marks)

Show Answers Only
  1. `(dy)/(dx) = (2x – y)/(x – 3y)`
  2. `y = 2(x – 3);\ \ \ y = 1/3 x + sqrt 6`
  3. `theta = pi/4`
Show Worked Solution
a.    `d/(dx)(x^2) – d/(dx)(xy) + 3/2* d/(dx) (y^2)` `= 0`
  `2x – x*(dy)/(dx) – y + 3/2(2y)*(dy)/(dx)` `= 0`
  `(dy)/(dx)(−x + 3y)` `= y – 2x`
  `:. (dy)/(dx)` `= (y – 2x)/(3y – x)`

 

b.   `m_{(3,0)} = (0 – 6)/(0 – 3) = 2`

`:.\ text{Equation of tangent at (3, 0):}`

`y = 2(x – 3)`

  `= 2x – 6`

 

`m_{(0,sqrt6)} = (sqrt6 – 0)/(3sqrt6 – 0) = 1/3`

`:.\ text{Equation of tangent at}\ (0,sqrt6):`

`y -sqrt6` `= 1/3(x – 0)`  
`y` `=1/3 x + sqrt6`  

 

c.   `m_1 = 2 = tan(theta_1), \ \ m_2 = 1/3 = tan(theta_2)`

`alpha` `= theta_1 – theta_2`
  `= tan^(−1)(2) – tan^(−1)(1/3)`
`tan(alpha)` `= tan(tan^(−1)(2) – tan^(−1)(1/3))`
  `= (tan(tan^(−1)(2)) – tan(tan^(−1)(1/3)))/(1 + tan(tan^(−1)(2)tan(tan^(−1)(1/3))))`
  `= (2 – 1/3)/(1 + 2/3)`
  `= 1`

 
`:. alpha = pi/4\ \ \ (alpha ∈ (0, pi/2))`

Calculus, SPEC1 2014 VCAA 7

Consider  `f(x) = 3x arctan (2x)`.

  1.  Write down the range of  `f`.  (1 mark)

  2.  Show that  `f prime(x) = 3 arctan (2x) + (6x)/(1 + 4x^2)`.  (1 mark)

  3.  Hence evaluate the area enclosed by the graph of  `g(x) = arctan (2x)`, the `x`-axis and the lines  `x = 1/2`  and  `x = sqrt 3/2`.  (3 marks)

Show Answers Only

  1. `f(x) in [0, oo)`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `(pi sqrt 3)/6-pi/8-1/4 ln2`

Show Worked Solution

a.  `f(x) = 3x arctan (2x)`

`f(0) = 0`

♦♦♦ Mean mark 4%.

`text(When)\ \ x < 0, \ \ 3x<0,\ \ text(and)\ \ y = tan^(-1) (2x) < 0`

`text(When)\ \ x>0,\ \ 3x>0,\ \ text(and)\ \ y = tan^(-1) (2x) > 0`
 


 
 `:. f(x) in [0, oo)`

 

b.  `u = 3x, qquad v = tan^(-1)(2x)`

`u prime = 3, qquad v prime = 2/(1 + 4x^2)`

`:. f prime (x)` `= 3(arctan (2x)) + (2/(1 + 4x^2))(3x)`
  `= 3 arctan (2x) + (6x)/(1 + 4x^2)`

 

c.    `A` `= int_(1/2)^(sqrt 3/2) arctan (2x)\ dx`
    `= 1/3 int_(1/2)^(sqrt 3/2) 3arctan (2x)\ dx`
    `=1/3 int_(1/2)^(sqrt 3/2) (3 arctan (2x) + (6x)/(1 + 4x^2)-(6x)/(1 + 4x^2))\ dx`
    `= 1/3 int_(1/2)^(sqrt 3/2) 3 arctan (2x) + (6x)/(1 + 4x^2) dx-int_(1/2)^(sqrt 3/2) (2x)/(1 + 4x^2)\ dx`
    `= 1/3 [3x arctan (2x)]_(1/2)^(sqrt 3/2)-1/4 int_(1/2)^(sqrt 3/2) (8x)/(1 + 4x^2)\ dx`
    `= [sqrt 3/2 arctan (sqrt 3)-1/2 arctan (1)]-1/4[ln (1 + 4x^2)]_(1/2)^(sqrt 3/2)`
    `= sqrt 3/2 (pi/3)-1/2 (pi/4)-1/4 [ln (1 + 4(3/4))-ln(1 + 4 (1/4)]`
    `= (pi sqrt 3)/6-pi/8-1/4 (ln4-ln 2)`
    `= (pi sqrt 3)/6-pi/8-1/4 ln2`

Calculus, MET1 2018 VCAA 8

Let  `f: R -> R, \ f(x) = x^2e^(kx)`, where `k` is a positive real constant.

  1.  Show that  `f^{′}(x) = xe^(kx)(kx + 2)`.   (1 mark)

  2. Find the value of `k` for which the graphs of  `y = f(x)`  and  `y = f^{′}(x)`  have exactly one point of intersection.   (3 marks)

Let  `g(x) = −(2xe^(kx))/k`. The diagram below shows sections of the graphs of `f` and `g` for  `x >= 0`.
 


 

Let `A` be the area of the region bounded by the curves  `y = f(x), \ y = g(x)` and the line  `x = 2`.

  1. Write down a definite integral that gives the value of `A`.   (1 mark)

  2. Using your result from part a., or otherwise, find the value of `k` such that  `A = 16/k`.   (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(One point of intersection)\ x = 0\ text(when)\ k = 1.`
  3. `int_0^2 (x^2e^(kx) + (2xe^(kx))/k)dx`
  4. `1/2 ln 4\ \ text(or)\ \ ln2`
Show Worked Solution

a.   `f(x) = x^2e^(kx)`

`f^{′}(x)` `= 2x · e^(kx) + x^2 · k · e^(kx)`
  `= xe^(kx)(kx + 2)\ \ …text(as required)`

 

b.  `text(Intersection occurs when:)`

♦♦♦ Mean mark 7%.
MARKER’S COMMENT: Most students found the correct quadratic equation but solved for `k`.

`x^2e^(kx)` `= xe^(kx)(kx + 2)`
`x^2-x(kx + 2)` `= 0,\ \ e^(kx) != 0`
`x^2-kx^2-2x` `= 0`
`x^2(1-k)-2x` `= 0`
`x[x(1-k)-2]` `= 0`

 

`:.x = 0\ \ text(or)\ \ x(1-k)-2` `= 0`
`x` `= 2/(1-k)`

 
`text(S)text(ince)\ \ x = 2/(1-k)\ \ text(is undefined when)\ \ k = 1,`

`=>\ text(One point of intersection only at)\ \ x = 0\ \ text(when)\ \ k = 1.`

 

c.    `A` `= int_0^2 f(x)\ dx-int_0^2 g(x)\ dx`
    `= int_0^2 x^2e^(kx)\ dx-int_0^2 −(2xe^(kx))/k \ dx`
    `= int_0^2 (x^2e^(kx) + (2xe^(kx))/k)dx`

♦♦ Mean mark 29%.

 

d.    `int_0^2(x^2e^(kx) + (2xe^(kx))/k)\ dx` `= 16/k`
  `1/k int_0^2(kx^2e^(kx) + 2xe^(kx))\ dx` `= 16/k`
  `1/k int_0^2(xe^(kx)(kx + 2))\ dx` `= 16/k`
  `1/k[x^2e^(kx)]_0^2` `= 16/k\ \ \ text{(using part a)}`
  `[2^2 · e^(2k)-0]` `= 16`
  `e^(2k)` `= 4`
  `2k` `= ln 4`
  `k` `= 1/2 ln 4\ \ text(or)\ \ ln2`

Algebra, MET2 2018 VCAA 20 MC

The differentiable function  `f : R -> R`  is a probability density function. It is known that the median of the probability density function  `f`  is at  `x = 0`  and  `f^{′} (0) = 4`.

The transformation  `T : R^2 -> R^2`  maps the graph of  `f`  to the graph of  `g`, where  `g : R -> R`  is a probability density function with a median at  `x = 0`  and  `g^{′} (0) = -1`.

The transformation `T` could be given by

  1. `T([(x), (y)]) = [(-2, 0), (0, 1/2)][(x), (y)]`
  2. `T([(x), (y)]) = [(2, 0), (0, -1/2)][(x), (y)]`
  3. `T([(x), (y)]) = [(2, 0), (0, 1/2)][(x), (y)]`
  4. `T([(x), (y)]) = [(-1/2, 0), (0, 2)][(x), (y)]`
  5. `T([(x), (y)]) = [(1/2, 0), (0, -2)][(x), (y)]`
Show Answers Only

`A`

Show Worked Solution

`m_f = 4`

♦♦♦ Mean mark 20%.

`text(Reflect in)\ y text(-axis): m = -4`
 

`text(Dilate by a factor of 2 from)\ y text(-axis):`

`m = (-4)/2 = -2`
 

`text(Dilate by a factor of)\ 1/2\ text{from}\ x text(-axis):`

`m = -1`

`[(-2, 0),(0, 1/2)]\ text(describes this transformation)`

`=>   A`

Algebra, MET2 2018 VCAA 18 MC

Consider the functions  `f:R^+ -> R,\ f(x) = x^(p/q)`  and  `g: R^+ -> R,\ g(x) = x^(m/n)`, where `p, q, m` and `n` are positive integers, and `p/q` and `m/n` are fractions in simplest form.

If  `{x: f(x) > g(x)} = (0, 1)`  and  `{x: g(x) > f(x)} = (1, oo)`, which of the following must be false?

A.   `q > n and p = m`

B.   `m > p and q = n`

C.   `pn < qm`

D.   `f prime (c) = g prime(c)\ \ text(for some)\ \ c in (0, 1)`

E.   `f prime (d) = g prime(d)\ \ text(for some)\ \ d in (1, oo)`

Show Answers Only

`E`

Show Worked Solution

`text(False statement:)`

♦♦♦ Mean mark 14%.

`f prime(d) = g prime (d)\ \ text(for some)\ \ d in (1, oo)`

`text(By testing with substituted values, options)`

`A\ text(to)\ D\ text(can be shown to be true.)`

`=>   E`

Financial Maths, STD2 2014 HSC 27a

Alex is buying a used car which has a sale price of  $13 380. In addition to the sale price there are the following costs:

2014 27a1

  1. Stamp Duty for this car is calculated at $3 for every $100, or part thereof, of the sale price.  
  2. Calculate the Stamp Duty payable.   (1 mark)
  3.  
  4. Alex wishes to take out comprehensive insurance for the car for 12 months. The cost of comprehensive insurance is calculated using the following: 
    1.  
    2. 2014 27a2
  5. Find the total amount that Alex will need to pay for comprehensive insurance.   (3 marks)
  6.  
  7. Alex has decided he will take out the comprehensive car insurance rather than the less expensive non-compulsory third-party car insurance.
  8. What extra cover is provided by the comprehensive car insurance?   (1 mark)

 

Show Answers Only
  1. `$402`
  2. `$985.74`
  3. `text(Comprehensive insurance covers Alex)`
  4. `text(for damage done to his own car as well.)`
  5.  
Show Worked Solution
♦♦♦ Mean mark 12%
IMPORTANT: “or part thereof ..” in the question requires students to round up to 134 to get the right multiple of $3 for their calculation.
(i)    `($13\ 380)/100 = 133.8`
`:.\ text(Stamp duty)` `= 134 xx $3`
  `= $402`

 

(ii)   `text(Base rate) = $845`

`text(FSL) =\ text(1%) xx 845 = $8.45`
 

`text(Stamp)` `=\ text(5.5%) xx(845 + 8.45)`
  `= 46.9397…`
  `= $46.94\ text{(nearest cent)}`

 

`text(GST)` `= 10 text(%) xx(845 + 8.45)`
  `= 85.345`
  `= $85.35`

 

`:.\ text(Total cost)` `= 845 + 8.45 + 46.94 + 85.35`
  `= $985.74`

 

♦ Mean mark 34%.
(iii)   `text(Comprehensive insurance covers Alex)`
  `text(for damage done to his own car as well.)`

Graphs, MET2 2018 VCAA 11 MC

The graph of  `y = tan(ax)`, where  `a ∈ R^+`, has a vertical asymptote  `x = 3 pi`  and has exactly one `x`-intercept in the region  `(0, 3 pi)`.

The value of `a` is

  1. `1/6`
  2. `1/3`
  3. `1/2`
  4. `1`
  5. `2`
Show Answers Only

`C`

Show Worked Solution

`y = tan(ax)`

`tan x ->\ text(period of)\ pi,\ text(asymptotes at)\ \ x = pi/2, (3 pi)/2`

`tan(x/2) ->\ text(period of)\ 2 pi,\ text(asymptotes at)\ \ x=pi, 3 pi`

`tan(x/2) -> text(has one)\ x text(-intercept of)\ 2 pi\ \ {x: (0, 3 pi)}`

`:. a = 1/2`
 

`=>   C`

GEOMETRY, FUR2 2018 VCAA 3

Frank owns a tennis court.

A diagram of his tennis court is shown below

Assume that all intersecting lines meet at right angles.

Frank stands at point `A`. Another point on the court is labelled point `B`.
 

  1. What is the straight-line distance, in metres, between point `A` and point `B`?

     

    Round your answer to one decimal place.   (1 mark)

  2. Frank hits a ball when it is at a height of 2.5 m directly above point `A`.

     

    Assume that the ball travels in a straight line to the ground at point `B`.

     

    What is the straight-line distance, in metres, that the ball travels?

     

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

Frank hits two balls from point `A`.

For Frank’s first hit, the ball strikes the ground at point `P`, 20.7 m from point `A`.

For Frank’s second hit, the ball strikes the ground at point `Q`.

Point `Q` is `x` metres from point `A`.

Point `Q` is 10.4 m from point `P`.

The angle, `PAQ`, formed is 23.5°.
 


 

    1. Determine two possible values for angle `AQP`.

       

      Round your answers to one decimal place.   (1 mark)

    2. If point `Q` is within the boundary of the court, what is the value of `x`?

       

      Round your answer to the nearest metre.  (1 mark)

Show Answers Only
  1. `18.8\ text{m (to 1 d.p.)}`
  2. `19\ text(m)\ text{(nearest m)}`
    1. `52.5^@,\ 127.5^@`
    2. `13\ text(m)\ text{(nearest m)}`
Show Worked Solution
a.    `text(Using Pythagoras,)`
  `AB` `= sqrt(4.1^2 + (6.4 + 6.4 + 5.5)^2`
    `= 18.75…`
    `= 18.8\ text{m (to 1 d.p.)}`

 

b. `text(Let)\ \ d = text(distance travelled)`
  


 

`d` `= sqrt(2.5^2 + 18.8^2)`
  `= 18.96…`
  `= 19\ text{m (nearest m)}`

 

c.i.  

`/_AQP ->\ text(2 possibilities)`

♦♦ Mean mark 25%.

`text(Using Sine rule,)`

`(sin/_AQP)/20.7` `= (sin 23.5^@)/10.4`
`sin /_AQP` `= (20.7 xx sin 23.5^@)/10.4`
  `= 0.7936…`
`:. /_AQP` `= 52.5^@ or 127.5^@\ \ \ text{(to 1 d.p.)}`

 

♦♦♦ Mean mark 18%.

c.ii.    `Q\ text(is within court when)\ \ /_AQP = 127.5^@`
  `/_APQ = 180 – (127.5 + 23.5) = 29^@`
   
  `text(Using sine rule,)`
  `x/(sin 29^@)` `= 10.4/(sin 23.5^@)`
  `:. x` `= (10.4 xx sin 29^@)/(sin 23.5^@)`
    `= 12.64…`
    `= 13\ text{m (nearest m)}`

NETWORKS, FUR1 2018 VCAA 8 MC

Annie, Buddhi, Chuck and Dorothy work in a factory.

Today each worker will complete one of four tasks, 1, 2, 3 and 4.

The usual completion times for Annie, Chuck and Dorothy are shown in the table below.
 


 

Buddhi takes 3 minutes for Task 3.

He takes `k` minutes for each other task.

Today the factory supervisor allocates the tasks as follows

    • Task 1 to Dorothy
    • Task 2 to Annie
    • Task 3 to Buddhi
    • Task 4 to Chuck

This allocation will achieve the minimum total completion time if the value of `k` is at least

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

`C`

Show Worked Solution

`text(Completion time of given allocation)`

`= 4 + 3 + 3 + 2`

`= 12\ text(hours)`
 

`text(Using Hungarian algorithm:)`

`text(If)\ \ k = 0\ ->\ text(Min completion time = 10)`

`text(If)\ \ k = 1\ ->\ text(Min completion time = 11)`

`text(If)\ \ k = 2\ ->\ text(Min completion = 12)`
 

`:. k\ text(must be at least 2 minutes.)`

`=> C`

NETWORKS, FUR2 2018 VCAA 4

Parcel deliveries are made between five nearby towns, `P` to `T`.

The roads connecting these five towns are shown on the graph below. The distances, in kilometres, are also shown.
 

A road inspector will leave from town `P` to check all the roads and return to town `P` when the inspection is complete. He will travel the minimum distance possible.

  1.  How many roads will the inspector have to travel on more than once?   (1 mark)

  2.  Determine the minimum distance, in kilometres, that the inspector will travel.   (1 mark)

Show Answers Only
  1. `2\ text(roads)`
  2. `108\ text(km)`
Show Worked Solution

a.   `text(Minimum distance if Eulerian circuit exists.)`

♦♦ Mean mark 31%.

`->\ text(no Eulerian circuit possible since 4 vertices are odd)`

`->\ text(if 2 edges added, Eulerian circuit exists)`

`:.\ text(Inspector will travel on 2 roads more than once.)`

 

b.   `text(By inspection, an extra edge added to)\ PQ\ text{(10)}`

♦♦♦ Mean mark 14%.

`text(and)\ ST\ text{(12) creates an Eulerian circuit with}`

`text(minimum distance.)`
 

`:.\ text(Min distance)`

`= (10 xx 2) + (12 xx 2) + 14 + 20 + 6 + 7 + 8 + 9`

`= 108\ text(km)`

CORE, FUR2 2018 VCAA 6

 

Julie has retired from work and has received a superannuation payment of $492 800.

She has two options for investing her money.

Option 1

Julie could invest the $492 800 in a perpetuity. She would then receive $887.04 each fortnight for the rest of her life.

  1. At what annual percentage rate is interest earned by this perpetuity?  (1 mark)

Option 2

Julie could invest the $492 800 in an annuity, instead of a perpetuity.

The annuity earns interest at the rate of 4.32% per annum, compounding monthly.

The balance of Julie’s annuity at the end of the first year of investment would be $480 242.25

    1. What monthly payment, in dollars, would Julie receive?   (1 mark)
    2. How much interest would Julie’s annuity earn in the second year of investment?
    3. Round your answer to the nearest cent.    (1 mark)

Show Answers Only

  1. `4.68 text(%)`
    1. `$2800`
    2. `$20\ 488.89`

Show Worked Solution

a.    `text(Annual interest)` `= 26 xx 887.04`
    `= $23\ 063.04`

♦ Mean mark 41%.
 

`:.\ text(Annual percentage rate)` `= (23\ 063.04)/(492\ 800)`
  `= 4.68 text(%)`

 

b.i.   `text(Find the monthly payment by TVM Solver:)`

♦ Mean mark 48%.

`N` `= 12`
`I(%)` `= 4.32`
`PV` `= -492\ 800`
`PMT` `= ?`
`FV` `= 480\ 242.25`
`text(P/Y)` `= text(C/Y) = 12`

 
`=> PMT = $2800.00`
 

♦♦♦ Mean mark 16%.

b.ii.   `text(Year 2 start balance)` `= $480\ 242.25`
  `text(Year 2 end balance)` `= $467\ 131.14`
  `text(Balance reduction)` `= 480\ 242.25-467\ 131.14`
    `= 13\ 111.11`

 
`text(Year 2 total payment) = 12 xx 2800 = 33\ 600`

`:.\ text(Interest)` `= 33\ 600-13\ 111.11`
  `= $20\ 488.89`

CORE, FUR2 2018 VCAA 3

Table 3 shows the yearly average traffic congestion levels in two cities, Melbourne and Sydney, during the period 2008 to 2016. Also shown is a time series plot of the same data.

The time series plot for Melbourne is incomplete.

  1. Use the data in Table 3 to complete the time series plot above for Melbourne.   (1 mark)

  2. A least squares line is used to model the trend in the time series plot for Sydney. The equation is

       `text(congestion level = −2280 + 1.15 × year)`

  1.   i. Draw this least squares line on the time series plot.   (1 mark)

  2.  ii. Use the equation of the least squares line to determine the average rate of increase in percentage congestion level for the period 2008 to 2016 in Sydney.   (1 mark)

    iii. Use the least squares line to predict when the percentage congestion level in Sydney will be 43%.   (1 mark)

The yearly average traffic congestion level data for Melbourne is repeated in Table 4 below.

  1. When a least squares line is used to model the trend in the data for Melbourne, the intercept of this line is approximately –1514.75556
  2. Round this value to four significant figures.   (1 mark)

  3. Use the data in Table 4 to determine the equation of the least squares line that can be used to model the trend in the data for Melbourne. The variable year is the explanatory variable.
  4. Write the values of the intercept and the slope of this least squares line in the appropriate boxes provided below.
  5. Round both values to four significant figures.   (2 marks)

congestion level = 
 
  + 
 
  × year
  1. Since 2008, the equations of the least squares lines for Sydney and Melbourne have predicted that future traffic congestion levels in Sydney will always exceed future traffic congestion levels in Melbourne.

     

    Explain why, quoting the values of appropriate statistics.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
    1. `text(See Worked Solutions)`
    2. `1.15 text(%)`
    3. `2020`
  2. `-1515`
  3. `text(congestion level) = -1515 + 0.7667 xx text(year)`
  4. `text(See Worked Solutions)`
Show Worked Solution
a.   

 

b.i.   

 

b.ii.  `text(The least squares line is 1.15% higher each year.)`

♦ Mean mark (b)(ii) 36%.
COMMENT: Major problems caused by part (b)(ii). Review!

  ` :.\ text(Average rate of increase) = 1.15 text(%)`

 

b.iii.    `text(Find year when:)`
  `43` `= -2280 + 1.15 xx text(year)`
  `text(year)` `= 2323/1.15`
    `= 2020`

 

c.  `-1515`

 

d.   `text(congestion level) = -1515 + 0.7667 xx text(year)`

 

e.   `text(Melbourne congestion level in 2008)`

♦♦♦ Mean mark 18%.

`= -1515 + 0.7667 xx 2008`

`= 24.5 text(%)`

 
`text{In 2008 Sydney has higher congestion (29.2 > 24.5)}`

`text(After 2008, Sydney congestion grows at 1.15% per)`

`text(year and Melbourne grows at 0.7667% per year.)`

`:.\ text(Sydney predicted to always exceed Melbourne.)`

Trigonometry, 2ADV T1 SM-Bank 1

A tower is built on flat ground.

Three tourists, `A`, `B` and `C` are observing the tower from ground level.

`A` is due north of the tower, `C` is due east and `B` is on the line of sight from `A` and `C` and between them.

The angles of elevation to the top of the tower from `A`, `B` and `C` are 26°, 28° and 30°, respectively.

What is the bearing of `B` from the tower?  (4 marks)

Show Answers Only

`005°`

Show Worked Solution

`text(Let)\ \ h =\ text(height of tower)`

`text(In)\ DeltaOAT:`

`tan26^@` `= h/(d_A)`
`d_A` `= h/(tan26^@)`

 
`text(Similarly,)`

`d_B` `= h/(tan28^@)`
`d_C` `= h/(tan30^@)`

   
`text(In)\ DeltaOAC:`

`tan angleOAC` `= (d_C)/(d_A)`
  `= (h/(tan30^@))/(h/(tan26^@))`
  `= (tan26^@)/(tan30^@)`
  `= 0.8447…`
`angleOAC` `= 40.19^@`

 
`text(Using sine rule in)\ DeltaOAB:`

`(sinangleABO)/(d_A)` `= (sinangleOAC)/(d_B)`
`sinangleABO` `= sin40.2^@ xx (tan28^@)/(tan26^@)`
  `= 0.7035…`
`angleABO` `= 44.71^@\ text(or)\ 135.29^@`

 

`text(S)text(ince)\ \ angleOCA` `= tan^(−1)((tan30)/(tan26))`
  `= 49.8^@`

 
`=> angleOBC = 44.71°`

`(text(otherwise angle sum)\ DeltaOBC > 180°)`
 

`angleAOB` `= 180 – (40.19 + 135.29)`
  `= 4.52`

 
`:.\ text(Bearing of)\ B\ text(from tower is)\ 005°.`

Calculus, SPEC2 2017 VCAA 10 MC

A function  `f`, its derivative  `fprime` and its second derivative  `f″` are defined for  `x ∈ R`  with the following properties.

`f(a) = 1, f(−a) = −1`

`f(b) = −1, f(−b) = 1`

and  `f″(x) = ((x + a)^2(x - b))/(g(x))`, where  `g(x) < 0`

The coordinates of any points of inflection of   `|\ f(x)\ |`  are

  1.  `(−a,1) and (b,1)`
  2.  `(b,−1)`
  3.  `(−a,−1) and (b,−1)`
  4.  `(−a,1)`
  5.  `(b,1)`
Show Answers Only

`B`

Show Worked Solution

`text(P.O.I. requires:)`

♦♦♦ Mean mark 6%.

`f″(x) =0\ \ text(and)\ \ f″(x)\ \ text(to change sign)`

`f″(x)\ \ text(does not change sign around)\ \ x=a`

`(b, −1)\ text(is the only P.O.I. of)\ \ f(x)`

`:. (b, 1)\ text(is the only P.O.I. of)\ \ |\ f(x)\ |`

`=>B`

Functions, 2ADV F1 SM-Bank 11

Given  `f(x) = sqrt (x^2 - 9)`  and  `g(x) = x + 5`

  1.  Find integers `c` and `d` such that  `f(g(x)) = sqrt {(x + c) (x + d)}`   (2 marks)

  2.  State the domain for which  `f(g(x))`  is defined.   (2 marks)

Show Answers Only
  1. `c = 2, d = 8 or c = 8, d = 2`
  2. `x in (– oo, – 8] uu [– 2, oo)`
Show Worked Solution
a.   `f(g(x))` `= sqrt {(x + 5)^2 – 9}`
    `= sqrt (x^2 + 10x + 16)`
    `= sqrt {(x + 2) (x + 8)}`

 
 `:. c = 2, d = 8 or c = 8, d = 2`

 

b.   `text(Find)\ x\ text(such that:)`

♦♦♦ Mean mark 13%.
MARKER’S COMMENT: “Very poorly answered” with a common response of  `-3 <=x<=3`  that ignored the information from part (a).

`(x+2)(x+8) >= 0`
 

 vcaa-2011-meth-4ii

`(x + 2) (x + 8) >= 0\ \ text(when)`

`x <= -8 or x >= -2`
 

`:.\ text(Domain:)\ \ x<=-8\ \ and\ \  x>=-2`

Calculus, 2ADV C4 2007* HSC 10a

An object is moving on the `x`-axis. The graph shows the velocity, `(dx)/(dt)`, of the object, as a function of time, `t`. The coordinates of the points shown on the graph are  `A (2, 1), B (4, 5), C (5, 0) and D (6, –5)`. The velocity is constant for  `t >= 6`.
 


 

  1. The object is initially at the origin. During which time(s) is the displacement of the object decreasing?  (1 mark)

  2. If the object travels 7 units in the first 4 seconds, estimate the time at which the object returns to the origin. Justify your answer.  (2 marks)

  3. Sketch the displacement, `x`, as a function of time.  (2 marks)

Show Answers Only
  1. `t > 5\ \ text(seconds)`
  2. `7.2\ \ text(seconds)`
  3.    
Show Worked Solution

i.  `text(Displacement is reducing when the velocity is negative.)`

`:. t > 5\ \ text(seconds)`

 

ii.  `text(At)\ B,\ text(the displacement) = 7\ text(units)`

`text(Considering displacement from)\ B\ text(to)\ D:`

`text(S)text(ince the area below the graph from)`

`B\ text(to)\ C\ text(equals the area above the)`

`text(graph from)\ C\ text(to)\ D,\ text(there is no change)`

`text(in displacement from)\ B\ text(to)\ D.`

 

`text(Considering)\ t >= 6`

`text(Time required to return to origin)`

`t` `= d/v`
  `= 7/5`
  `= 1.4\ \ text(seconds)`

 

`:.\ text(The particle returns to the origin after 7.4 seconds.)`
   

iii.   

Calculus, 2ADV C4 2013* HSC 15a

The diagram shows the front of a tent supported by three vertical poles. The poles are 1.2 m apart. The height of each outer pole is 1.5 m, and the height of the middle pole is 1.8 m. The roof hangs between the poles.

2013 15a

The front of the tent has area  `A\ text(m²)`. 

  1. Use the trapezoidal rule to estimate  `A`.    (1 mark)

  2. Does the Trapezoidal rule give a higher or lower estimate of the actual area? Justify your answer.  (1 mark)

Show Answers Only
  1. `3.96\ text(m²)`
  2. `text(See Worked Solutions)`
Show Worked Solution
i.    `A` `~~ h/2 [y_0 + 2y_1 + y_2]`
    `~~ 1.2/2 [1.5 + (2 xx 1.8) + 1.5]`
    `~~ 0.6 [6.6]`
    `~~ 3.96\ text(m²)`

 

ii.       

`text(S)text(ince the tent roof is concave up, the)`

`text(Trapezoidal rule uses straight lines and)`

`text(will estimate a higher area.)`