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Financial Maths, STD2 F4 2017 HSC 28c

Michelle borrows $100 000. The interest rate charged is 12% per annum compounded monthly. The monthly payment is $1029 and the first repayment is made after one month.

What is the amount outstanding immediately after the SECOND monthly repayment is made?  (3 marks)

Show Answers Only

`$99\ 941.71`

Show Worked Solution
`text(Interest per month)` `= text(12%)/12= 1text(%)`

`text(Let)\ \ A=\ text(amount owing)`

♦ Mean mark 41%.

 
`text(After 1st repayment:)`

`A_1` `= (100\ 000 + text(1%) xx 100\ 000) – 1029`
  `= $99\ 971`

 
`text(After 2nd repayment:)`

`A_2` `= (99\ 971 + text(1%) xx 99\ 971) – 1029`
  `= $99\ 941.71`

Probability, 2UG 2017 HSC 28b

Five people are in a team. Two of them are selected at random to attend a competition.

  1. How many different groups of two can be selected?  (1 mark)
  2. If Mary is one of the five people in the team, what is the probability that she is selected to attend the competition?  (1 mark)

 

Show Answers Only
  1. `10`
  2. `2/5`
Show Worked Solution
(i)   `text(# Groups of 2)` `= (5 xx 4)/(2 xx 1)`
    `=10`
♦♦ Mean mark part (ii) 33%.

 

(ii)   `text(Number of pairs with Mary = 4)`

`:. Ptext{(Mary attends)}` `= 4/10`
  `= 2/5`

Algebra, 2UG 2017 HSC 28a

Temperature can be measured in degrees Celsius (`C`) or degrees Fahrenheit (`F`).

The two temperature scales are related by the equation  `F = (9C)/5 + 32`.

  1. Calculate the temperature in degrees Fahrenheit when it is  −20 degrees Celsius.  (1 mark)
  2. Solve the following equations simultaneously, using either the substitution method or the elimination method.  (2 marks)
    `qquadF = (9C)/5 + 32`

    `qquadF = C`

     

  3. The graphs of  `F = (9C)/5 + 32`  and  `F = C`  are shown below.
  4.  


  5. What does the result from part (ii) mean in the context of the graph?  (1 mark)     

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Show Answers Only
  1. `−4^@F`
  2. `C = −40, F = −40`
  3. `text(It means the two graphs intersect)`
    `text(at)\ (−40,−40).`

 

 

Show Worked Solution
(i)   `F` `= (9(−20))/5 + 32`
    `= −4^@F`

 

(ii)   `F` `= (9C)/5 + 32` `…\ (1)`
  `F` `= C` `…\ (2)`

 

♦♦ Mean mark 31%.
MARKER’S COMMENT: An area that requires attention.

`text(Substitute)\ \ F = C\ \ text{from (2) into (1)}`

`C` `= (9C)/5 + 32`
`(9C)/5 – C` `= −32`
`(4C)/5 – C` `= −32`
`C` `= −32 xx 5/4 = −40`

 

`text{From (2),}`

`F = −40`

`text{(i.e. when}\ C = −40, F = −40)`

♦♦♦ Mean mark 20%.

 

(iii)   `text(It means the two graphs intersect)`

`text{at (−40,−40).}`

Algebra, 2UG 2017 HSC 30b

The cost of a jewellery box varies directly with the cube of its height.

A jewellery box with a height of 10 cm costs $50.

Calculate the cost of a jewellery box with a height of 12 cm.  (2 marks)

Show Answers Only

`$86.40`

Show Worked Solution

`C\ prop\ h^3`

♦ Mean mark 49%.

`C = kh^3`

`text(When)\ h = 10, C = 50,`

`:. k = 50/(10^3) = 0.05`

 

`text(Find)\ \ C\ \ text(when)\ \ h = 12,`

`C` `= 0.05 xx 12^3`
  `= $86.40`

Measurement, 2UG 2017 HSC 27d

Island `A` and island `B` are both on the equator. Island `B` is west of island `A`. The longitude of island `A` is 5°E and the angle at the centre of Earth (`O`), between `A` and `B`, is 30°.
 


 

  1. What is the longitude of island `B`?  (1 mark)
  2. What time is it on island `B` when it is 10 am on island `A`?  (1 mark)
  3. A ship leaves island `A` and travels west along the equator to island `B`. It travels at a constant speed of 40 km/h.
    How long will the ship take to arrive at island `B`? Give your answer in days and hours to the nearest hour.  (3 marks)

 

 

Show Answers Only
  1. `25^@\ text(W)`
  2. `8\ text(am)`
  3. `3\ text(days 12 hours)`
Show Worked Solution
(i)    `text{Longitude (island}\ B)` `= 5 – 30`
    `= −25`
    `= 25^@\ text(W)`

 

(ii)   `text(Time difference) = 30^@ ÷ 15^@ = 2\ text(hours)`

♦ Mean marks of 40% and 45% for parts (i) and (ii) respectively.

`text(S)text(ince)\ B\ text(is west of)\ A,`

`text(Time on island)\ B` `= 10\ text(am less 2 hours)`
  `= 8\ text(am)`

 

(iii)    `text(Arc Distance)` `= theta/360 xx 2pir`
    `= 30/360 xx 2pi xx 6400`
    `= 3351.032…`

♦ Mean mark part (iii) 38%.
`:.\ text(Time to arrive)` `= (3351.032…)/40`
  `= 83.77…\ text(hours)`
  `= 3\ text(days 12 hours)`

Financial Maths, STD2 F5 2017 HSC 27c

A table of future value interest factors for an annuity of $1 is shown.
 


 

An annuity involves contributions of $12 000 per annum for 5 years. The interest rate is 4% per annum, compounded annually.

  1. Calculate the future value of this annuity.  (1 mark)

  2. Calculate the interest earned on this annuity.  (1 mark)

Show Answers Only
  1. `$64\ 995.60`
  2. `$4995.60`
Show Worked Solution

i.   `text(FV factor = 5.4163)`

`:.\ text(FV of Annuity)` `= 12\ 000 xx 5.4163`
  `= $64\ 995.60`
♦♦ Mean mark part (ii) 22%.
COMMENT: A very poorly answered question dealing with a core concept in this area.

 

ii.   `text(Interest earned)` `=\ text(FV − total repayments)`
    `= 64\ 995.60 – (5 xx 12\ 000)`
    `= $4995.60`

FS Comm, 2UG 2017 HSC 27b

How many 20 megabyte files can fit on a 3 terabyte external hard disc?  (2 marks)

Show Answers Only

`157\ 286\ text(files)`

Show Worked Solution
`text(3 TB)` `= 3 xx 2^10 xx 2^10\ text(MB)`
  `= 3\ 145\ 728\ text(MB)`

 

`:.\ text(Number of complete files that can fit)`

♦ Mean mark 49%.

`= (3\ 145\ 728)/20`

`= 157\ 286.4`

`= 157\ 286\ text(files)`

Statistics, STD2 S1 2017 HSC 27a

Jamal surveyed eight households in his street. He asked them how many kilolitres (kL) of water they used in the last year. Here are the results.

`220, 105, 101, 450, 37, 338, 151, 205`

  1. Calculate the mean of this set of data.  (1 mark)

  2. What is the standard deviation of this set of data, correct to one decimal place?  (1 mark)

Show Answers Only
  1. `200.875`
  2. `127.4\ \ text{(1 d.p.)}`
Show Worked Solution
i.   `text(Mean)` `= (220 + 105 + 101 + 450 + 37 + 338 + 151 + 205) ÷ 8`
    `= 200.875`
♦ Mean mark part (ii) 47%.
IMPORTANT: The population standard deviation is required here.

 

ii.   `text(Std Dev)` `= 127.357…\ \ text{(by calc)}`
    `= 127.4\ \ text{(1 d.p.)}`

Measurement, STD2 M1 2017 HSC 25 MC

In the circle, centre `O`, the area of the quadrant is 100 cm².
 


 

What is the arc length `l`, correct to one decimal place?

  1. 8.9 cm
  2. 11.3 cm
  3. 17.7 cm
  4. 25.1 cm
Show Answers Only

`C`

Show Worked Solution

`text(Find)\ r:`

♦ Mean mark 44%.
`text(Area)` `= 1/4 pir^2`
`100` `= 1/4 pir^2`
`r^2` `= 400/pi`
`:. r` `= 11.283…\ text(cm)`

 

`text(Arc length)` `= theta/360 xx 2pir`
  `= 90/360 xx 2pi xx 11.283…`
  `= 17.724…`
  `= 17.7\ text(cm)`

 
`=> C`

Measurement, STD2 M1 2017 HSC 21 MC

The length of a netball court is measured to be 30.50 metres, correct to the nearest centimetre.

What is the lower limit for the length of the netball court?

  1. 30.45 m
  2. 30.49 m
  3. 30.495 m
  4. 30.499 m
Show Answers Only

`C`

Show Worked Solution
`text(A)text(bsolute error)` `= 0.5\ text(cm)`
  `= 0.005\ text(m)`

 

`:.\ text(Lower limit)` `= 30.50-0.005`
  `= 30.495\ text(m)`

 
`=>C`

Algebra, STD2 A2 2017 HSC 20 MC

A pentagon is created using matches.

By adding more matches, a row of two pentagons is formed.

Continuing to add matches, a row of three pentagons can be formed.

Continuing this pattern, what is the maximum number of complete pentagons that can be formed if 100 matches in total are available?

A.     `25`

B.     `24`

C.     `21`

D.     `20`

Show Answers Only

`=>\ text(B)`

Show Worked Solution

`text(1 pentagon: 5 matches)`

`text(2 pentagons: 5 + 4 = 9)`

`text(3 pentagons: 5 + 4 × 2 = 13)`

`vdots`

`n\ text(pentagons:)\ 5 + 4(n – 1)`

`5 + 4(n – 1)` `= 100`
`4n – 4` `= 95`
`4n` `= 99`
`n` `= 24.75`

 

`:.\ text(Complete pentagons possible = 24)`

`=>\ text(B)`

Probability, STD2 S2 2017 HSC 15 MC

The faces on a twenty-sided die are labelled  $0.05, $0.10, $0.15, … , $1.00.

The die is rolled once.

What is the probability that the amount showing on the upper face is more than 50 cents but less than 80 cents?

A.     `1/4`

B.     `3/10`

C.     `7/20`

D.     `1/2`

Show Answers Only

`A`

Show Worked Solution

`text(Possible faces that satisfy are:)`

♦ Mean mark 50%.

`55text(c),60text(c),65text(c),70text(c),75text(c)`

`:.\ text(Probability)` `= 5/20`
  `= 1/4`

`=>A`

Statistics, NAP-J2-13

Ms Richards asked her class a question and recorded the results in the table below.
 


 

Which question could Ms Richards have asked?

 
 What type of farm animals do you own?
 
 How many farm animals do you have?
 
 What month was your farm animal born?
 
 How old are your farm animals?
Show Answers Only

`text(What type of farm animals do you own?)`

Show Worked Solution

`text(What type of farm animals do you own?)`

Number and Algebra, NAP-J2-11 SA

A school teacher allocates pieces of cardboard to class groups depending on the number of students in each group.

The table below is used.
 

 
Using the pattern in the table, how many pieces of cardboard should a group of 3 students receive?
Show Answers Only

`9`

Show Worked Solution

`text(The pattern shows that each student receives)`

`text(3 pieces of cardboard.)`

`:.\ text(A group of 3 will be given 9 pieces.)`

Quadratic, 2UA SM-Bank 10

Solve  `3e^t = 5 + 8e^(−t)`  for `t`.  (3 marks)

Show Answers Only

`log_e(8/3)`

Show Worked Solution

`3e^t – 5 – 8e^(−t) = 0`

`text(Multiply both sides by)\ e^t,`

`3e^(2t) – 5e^t – 8 = 0`

`text(Let)\ y = e^t`

`3y^2 – 5y – 8` `= 0`
`(3y – 8)(y + 1)` `= 0`

 

`y` `=8/3\ quadquadquadquadquad text(or)\ \ \ \ ` `y` `=-1\ \ text{(No soln)}`
`e^t` `= 8/3`  
`:. t` `= log_e(8/3)\ \ \ `    

Quadratic, 2UA SM-Bank 05

Solve the equation  `4^x - 15 × 2^x = 16`  for `x.`  (3 marks)

Show Answers Only

`x = 4`

Show Worked Solution
`4^x – 15 xx 2^x – 16` `= 0`
`2^(2x) – 15 xx 2^x – 16` `= 0`

 

COMMENT: Understand why this working is incorrect: if  `y=2^x`, then `2y=4^x`.

`text(Let)\ \ y = 2^x`

`y^2 – 15y – 16` `= 0`
`(y – 16) (y + 1)` `= 0`
`y` `= 16` `\ \ \ or\ \ \ ` `y` `= – 1`
`2^x` `= 16`   `2^x` `= – 1`
`:. x` `= 4`   `text(No solution)`

L&E, 2ADV E1 SM-Bank 6 MC

The expression

`log_c(a) + log_a(b) + log_b(c)`

is equal to

  1. `1/(log_c(a)) + 1/(log_a(b)) + 1/(log_b(c))`
  2. `1/(log_a(c)) + 1/(log_b(a)) + 1/(log_c(b))`
  3. `-1/(log_a(b))-1/(log_b(c))-1/(log_c(a))`
  4. `1/(log_a(a)) + 1/(log_b(b)) + 1/(log_c(c))`
Show Answers Only

`B`

Show Worked Solution

`text(Solution 1)`

`text(Using Change of Base:)`

`log_c(a) + log_a(b) + log_b(c)`

`=(log_a(a))/(log_a(c)) + (log_b(b))/(log_b(a)) + (log_c(c))/(log_c(b))`

`=1/(log_a(c)) + 1/(log_b(a)) + 1/(log_c(b))`

 
`=> B`

 

`text(Solution 2)`

`text(Let)\ \ x` `=log_c(a)`
`c^x` `=a`
`x log_a c` `=log_a a`
`x` `=1/log_a c`

 

`text(Apply similarly for the other terms.)`

`=> B`

GEOMETRY, FUR1 SM-Bank 35 MC

Kim lives in Perth (32°S, 115°E). He wants to watch an ice hockey game being played in Toronto  (44°N, 80°W)  starting at 10.00 pm on Wednesday.

What is the time in Perth when the game starts?

A.   9.00 am on Wednesday

B.   7.40 pm on Wednesday 

C.   9.00 pm on Wednesday

D.   12.20 am on Thursday

E.   11.00 am on Thursday

Show Answers Only

`E`

Show Worked Solution

`text(Perth is East of Toronto) =>\ text(Ahead)`

`text(Longitudinal difference)`

`= 115 + 80`

`= 195^@`
 

`text(Time difference)` `= 195/15`
  `= 13\ text(hours)`

 

`:.\ text(Time in Perth)`

`=\ text{10 pm (Wed) + 13 hours}`

`=\ text(11 am on Thursday)`
 

`=>  E`

GEOMETRY, FUR2 SM-Bank 26

Two cities lie on the same meridian of longitude. One is 40° north of the other.

What is the distance between the two cities, correct to the nearest kilometre?  (2 marks)

Show Answers Only

`4468\ text{km   (nearest km)}`

Show Worked Solution

2UG 2015 23c Answer

`text(Distance between two cities)`

`=\ text(Arc length)\  AB`

`= 40/360 xx 2 xx pi xx r`

`= 1/9 xx 2 xx pi xx 6400`

`= 4468.04…`

`= 4468\ text{km  (nearest km)}`

GEOMETRY, FUR2 SM-Bank 15

Osaka is at  34°N, 135°E, and Denver is at  40°N, 105°W. 

  1. Show that there is a 16-hour time difference between the two cities.
    (Ignore time zones.)    (1 mark)
  2. John lives in Denver and wants to ring a friend in Osaka. In Denver it is 9 pm Monday.     

     

    What time and day is it in Osaka then?   (1 mark)

  3. John’s friend in Osaka sent him a text message which happened to take 14 hours to reach him. It was sent at 10 am Thursday, Osaka time.

     

    What was the time and day in Denver when John received the text?    (1 mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `1\ text(pm Tuesday)`
  3. `8\ text(am Thursday)`
Show Worked Solution
a.   `text(Longitude difference)` `= 135 + 105`
    `= 240^@`

 

`text(Time difference)` `= 240/15`
  `= 16\ text(hours   … as required)`

 

b.  `text(Denver is behind Osaka time because it is further west.)`

`:.\ text(Time in Osaka)` `= 9\ text(pm Monday plus 16 hours)`
  `= 1\ text(pm Tuesday)`

 

c.  `text(Denver is 16 hours behind Osaka)`

`:.\ text(John will receive the text at 10 am Thursday less 16)`

`text{plus 14 hours (i.e. 8 am Thursday.)}`

GEOMETRY, FUR2 SM-Bank 14

Pontianak has a longitude of 109°E, and Jarvis Island has a longitude of 160°W.

Both places lie on the Equator

  1. Find the shortest great circle distance between these two places, to the nearest kilometre. You may assume that the radius of the Earth is 6400 km.    (2 marks)
  2. The position of Rabaul is 4° to the south and 48° to the west of Jarvis Island. What is the latitude and longitude of Rabaul?     (1 mark)
Show Answers Only
  1. `10\ 165\ text(km)\ \ \ text{(nearest km)}`
  2. `152^@ text(E)`
Show Worked Solution
a.   `text(Longitude difference)` `= 109 + 160`
    `= 269^@`

 

`=> text(Shortest distance)\ text{(by degree)}` `= 360 – 269`
  `= 91^@`

 

`:.\ text(Shortest distance)` `= 91/360 xx 2 pi r`
  `= 91/360 xx 2 xx pi xx 6400`
  `= 10\ 164.79…`
  `=10\ 165\ text(km)\ text{(nearest km)}`

 

b.   `text(Latitude)`
  `4^@\ text(South of Jarvis Island)`
  `text(S)text(ince Jarvis Island is on equator)`
  `=>\ text(Latitude is)\ 4^@ text(S)`
   
  `text(Longitude)`
  `text(Jarvis Island is)\ 160^@ text(W)`
  `text(Rubail is)\ 48^@\ text(West of Jarvis Island, or 208° West)`
  `text(which is)\ 28^@\ text{past meridian (180°)}`

 

`=>\ text(Longitude)` `= (180\ -28)^@ text(E)`
  `= 152^@ text(E)`

 

`:.\ text(Position is)\ (4^@text{S}, 152^@text{E})`

Algebra, MET2 2007 VCAA 17 MC

The function  `f`  satisfies the functional equation  `f (f (x)) = x`  for the maximal domain of  `f.`

The rule for the function is

  1. `f(x) = x + 1`
  2. `f(x) = x - 1`
  3. `f(x) = (x - 1)/(x + 1)`
  4. `f(x) = log_e (x)`
  5. `f(x) = (x + 1)/(x - 1)`
Show Answers Only

`E`

Show Worked Solution

`text(Solution 1)`

`text(Define each specific function)`

♦ Mean mark 47%.

`(text{i.e. define}\ \ f(x) = (x + 1)/(x – 1))`

`f(f(x)) = x\ \ text(when)\ \ f(x) = (x + 1)/(x – 1)`

`=>   E`

 

`text(Solution 2)`

`text(Consider)\ \ f(x) = (x + 1)/(x – 1)`

`f(f(x))` `=((x + 1)/(x – 1) +1)/((x + 1)/(x – 1) -1)`
  `=(x+1+x-1)/(x+1-x+1)`
  `=x`

`=>E`

Algebra, MET2 2007 VCAA 5 MC

The simultaneous linear equations

`mx + 12y = 24`

`3x + my = m`

have a unique solution only for

  1. `m = 6 or m = – 6`
  2. `m = 12 or m = 3`
  3. `m in R\ text(\){– 6, 6}`
  4. `m = 2 or m = 1`
  5. `m in R\ text(\){– 12, – 3}`
Show Answers Only

`C`

Show Worked Solution
`mx + 12y` `=24`
`y` `=-m/12 x +2\ \ …\ (1)`
`3x + my` `=m`
`y` `= -3/m x+1\ \ …\ (2)`

 

`text(Unique solution occurs when)\ \ m_1!= m_2,`

♦♦ Mean mark 36%.
`-m/12` `!=-3/m`
`m^2` `!= 36`
`:. m` `!= +- 6`

`=>   C`

Graphs, MET2 2008 VCAA 20 MC

The function  `f: B -> R`  with rule  `f(x) = 4x^3 + 3x^2 + 1`  will have an inverse function for

  1. `B = R`
  2. `B = (1/2, oo)`
  3. `B = (text{−∞}, 1/2]`
  4. `B = (text{−∞}, 1/2)`
  5. `B = [−1/2, oo)`
Show Answers Only

`B`

Show Worked Solution

`text(Inverse exists if)\ \ f(x)\ \ text(is)\ \ 1 – 1:`

 

vcaa-2008-20i

`text(Option B’s domain ensures)\ \ f(x)\ \ text(is)\ \ 1- 1`

`=>   B`

Calculus, MET2 2008 VCAA 19 MC

The graph of a function  `f` is shown below.

VCAA 2008 19mc

The graph of an antiderivative of  `f` could be

VCAA 2008 19mci

VCAA 2008 19mcii 

VCAA 2008 19mciii

Show Answers Only

`B`

Show Worked Solution

`text(S)text(ince)\ \ f(x)>0\ \ text(for all)\ x,\ text(the)`

♦ Mean mark 50%.

`text(antiderivative function has no stationary)`

`text(points.)`

`=>\ text(Only B or E are possibilities.)`

 

`text(Also, the antiderivative function must have)`

`text(an increasing gradient for all)\ x.`

`=>   B`

Algebra, MET2 2008 VCAA 12 MC

Let  `f: R -> R,\ f(x) = e^x + e^(–x).`

For all  `u in R,\ f(2u)`  is equal to

  1. `f(u) + f(-u)`
  2. `2 f(u)`
  3. `(f(u))^2 - 2`
  4. `(f(u))^2`
  5. `(f(u))^2 + 2`
Show Answers Only

`C`

Show Worked Solution

`text(Solution 1)`

♦ Mean mark 44%.

`text(Define)\ \ f(x) = e^x + e^-x`

`text(Enter each functional equation)`

`[text(i.e.)\ \ f(2u) = (f(u))^2 – 2]`

`text(until CAS output is “true”)`

`=>   C`

 

`text(Solution 2)`

`f(2u)` `=e^(2u) + e^(-2u)`
`(f(u))^2` `=(e^u + e^(-u))^2`
  `=e^(2u) + 2 + e^(-2u)`
   

`:. f(2u) = (f(u))^2-2`

`=>C`

Graphs, MET2 2008 VCAA 9 MC

The transformation  `T: R^2 -> R^2` with rule

`T([(x), (y)]) = [(4, 0), (0, -2)] [(x), (y)] + [(1), (3)]`

maps the curve with equation  `y = x^3`  to the curve with equation

  1. `y = (-(x - 1)^3)/32 + 3`
  2. `y = (-(4x + 1)^3 + 3)/2`
  3. `y = (-(x + 1)^3)/32 - 3`
  4. `y = ((1 - x)^3)/64 - 3`
  5. `y = ((4x - 1)^3 + 3)/2`
Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ (x′,y′)\ \ text(be the transformation of)\ \ (x,y).`

♦ Mean mark 38%.
`x′` `= 4x + 1`
`x` `=(x′ – 1)/4`
`y′` `= – 2y + 3`
`y` `=(y′ – 3)/(- 2)`

 

`text(Substitute)\ \ x, y\ \ text(into)\ \ y = x^3:`

`(y′ – 3)/(- 2)` `= ((x′ – 1)/4)^3`
`y′ – 3` `= (- 2)/4^3 (x′ – 1)^3`
`y` `= (- (x – 1)^3)/32 + 3`

`=>   A`

Algebra, MET2 2008 VCAA 6 MC

The simultaneous linear equations

`ax + 3y = 0`

`2x + (a + 1)y = 0`

where `a` is a real constant, have infinitely many solutions for

  1. `a in R`
  2. `a in\ text({−3, 2})`
  3. `a in R\ text(\ {−3, 2})`
  4. `a in\ text({−2, 3})`
  5. `a in R\ text(\ {−2, 3})`
Show Answers Only

`B`

Show Worked Solution

`text(Infinite solution:)`

♦ Mean mark 45%.
`m_1` `= m_2` `and` `c_1` `= c_2`
`a/2` `= 3/(a + 1)`   `text(always true as both)`
`a` `= – 3, 2`   `text(lines have)\ y text(-int) = 0`

 
`=>   B`

Calculus, MET2 2008 VCAA 4 MC

If  `int_1^3 f(x)\ dx = 5`, then  `int_1^3 (2f(x) - 3)\ dx`  is equal to

A.   `4`

B.   `5`

C.   `7`

D.   `10`

E.   `16`

Show Answers Only

`A`

Show Worked Solution

`int_1^3 (2 f(x) – 3)\ dx`

♦ Mean mark 49%.
MARKER’S COMMENT: Over one third of students did not integrate –3.

`= 2 int_1^3 f(x)\ dx + int_1^3 (– 3)\ dx`

`=2(5) +[-3x]_1^3`

`= 10 + (– 6)`

`= 4`

 
`=>   A`

Calculus, MET1 SM-Bank 28

The function  `f`  has the rule  `f(x) = 1 + 2 cos x`.

  1. Show that the graph of  `y = f(x)`  cuts the `x`-axis at  `x = (2 pi)/3`.   (1 mark)

  2. Sketch the graph  `y = f(x)`  for  `x  in [-pi,pi]`  showing where the graph cuts each of the axes.   (2 marks)

  3. Find the area under the curve  `y = f(x)`  between  `x = -pi/2`  and  `x = (2 pi)/3`.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}` 
  2.  
       
  3. `((7 pi)/6 + sqrt 3 + 1)\ text(u²)`
Show Worked Solution

a.   `f(x) = 1 + 2 cos x`

`f(x)\ text(cuts the)\ x text(-axis when)\ f(x) = 0`

`1 + 2 cos x` `= 0`
`2 cos x` `=-1`
 `cos x` `= -1/2`

 
`:.  x = (2 pi)/3\ …\ text(as required)`

 

b.   2UA HSC 2006 7b

 

c.  `text(Area)` `= int_(-pi/2)^((2 pi)/3) 1 + 2 cos x\ \ dx`
  `= [x + 2 sin x]_(-pi/2)^((2 pi)/3)`
  `= [((2 pi)/3 + 2 sin­ (2 pi)/3)-((-pi)/2 + 2 sin­ (-pi)/2)]`
  `= ((2 pi)/3 + 2 xx sqrt 3/2)-((-pi)/2 +2(- 1))`
  `= (2 pi)/3 + sqrt(3) + pi/2 + 2`
  `= ((7 pi)/6 + sqrt(3) + 2)\ text(u²)`

Algebra, MET1 SM-Bank 24

The rule for  `f`  is  `f(x) = e^x-e^(-x)`.

Show that the inverse function is given by

    `f^(-1)(x) = log_e((x + sqrt(x^2 + 4))/2)`  (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions.)}`

Show Worked Solution

`y = e^x-e^(-x)`

`text(Inverse: swap)\  x harr y`

`x` `= e^y-1/(e^y)`
`xe^y` `= e^(2y)-1`
`e^(2y)-xe^y-1` `= 0`

 

`text(Let)\ \ A = e^y`

`:.A^2-xA-1 = 0`
 

`text(Using the quadratic formula)`

`A` `=(x ± sqrt((-x)^2-4 · 1 · (-1)))/(2 · 1)`
  `=(x ± sqrt(x^2 + 4))/2`

 

`text(S)text(ince)\ \ (x-sqrt(x^2 + 4))/2<0\ \ text(and)\ \ e^y>0,`

`:.e^y` `= (x + sqrt(x^2 + 4))/2`
`log_e e^y` ` = log_e((x + sqrt(x^2 + 4))/2)`
`y` `= log_e((x + sqrt(x^2 + 4))/2)`
`:.f^(-1)(x)` `= log_e((x + sqrt(x^2 + 4))/2)\ \  …\ text(as required)`

Algebra, MET1 SM-Bank 23

The function  `f: [0,oo) → R`  with rule  `f(x) = 1/(1 + x^2)`  is drawn below.

Inverse Functions, EXT1 2004 HSC 5b

  1. Copy or trace this diagram into your writing booklet.
  2. On the same set of axes, sketch  `y=f^(-1)(x)`  where  `f^(-1)` is the inverse function of  `f(x)`.   (1 mark)

  3. Find the domain of the inverse  `f^(-1)`.   (1 mark)

  4. Find an expression for  `y = f^(-1)(x)`  in terms of  `x`.   (2 marks)

  5. The graphs of  `y = f(x)`  and  `y = f^(-1)(x)`  meet at exactly one point  `P`.Let  `α`  be the `x`-coordinate of  `P`. Explain why  `α`  is a root of the equation
  6.      `x^3 + x-1 = 0`.   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `0 < x ≤ 1`
  3. `y = sqrt((1-x)/x), y > 0`
  4. `text(See Worked Solutions)`
Show Worked Solution
a.  

Inverse Functions, EXT1 2004 HSC 5b Answer

b.   `text(Range of)\ \ f(x):\ (0,1]`

`:.\ text(Domain of)\ \ f^(-1)(x):  (0,1]`

 

c.  `f(x) = 1/(1 + x^2)`

`text(Inverse: swap)\  x harr y`

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

 

`:.y = sqrt((1-x)/x), \ \ y >= 0`

 

d.   `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`

`text(i.e. where)`

`1/(1 + x^2)` `= x`
`1` `= x(1 + x^2)`
`1` `= x + x^3`
`x^3 + x-1` `= 0`

 

`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`

`text(it is a root of)\ \ \ x^3 + x-1 = 0`

Calculus, MET1 SM-Bank 21

The rule for function  `f` is  `f(x) = e^(-x^2)`.  The diagram shows the graph  `y = f(x)`.

 Inverse Functions, EXT1 2010 HSC 3b

The graph has two points of inflection. 

  1. Find the `x` coordinates of these points.   (2 marks)

  2. Explain why the domain of `f(x)` must be restricted if `f(x)` is to have an inverse function.    (1 mark)

  3. Find the rule for the inverse function `f^(-1)` if the domain of `f(x)` is restricted to  `x ≥ 0.`   (2 marks)

  4. Find the domain for `f^(-1)`.    (1 mark)

  5. Sketch the curve  `y = f^(-1) (x)`.   (1 mark)

Show Answers Only
  1. `x = +- 1/sqrt2`
  2. `text(There can only be 1 value of)\ y\ text(for each value of)\ x.`
  3. `f^(-1)x = sqrt(ln(1/x))`
  4. `0 <= x <= 1`
  5. Inverse Functions, EXT1 2010 HSC 3b Answer

Show Worked Solution
a.    `y` `= e^(-x^2)`
  `dy/dx` `= -2x * e^(-x^2)`
  `(d^2y)/(dx^2)` `= -2x (-2x * e^(-x^2)) + e ^(-x^2) (-2)`
    `= 4x^2 e^(-x^2)-2e^(-x^2)`
    `= 2e^(-x^2) (2x^2-1)`

 
`text(P.I. when)\ \ (d^2y)/(dx^2) = 0`

`2e^(-x^2) (2x^2-1)` `= 0` 
 `2x^2-1` `= 0` 
 `x^2` `= 1/2`
 `x` `= +- 1/sqrt2` 
COMMENT: It is also valid to show that `f(x)` is an even function and if a P.I. exists at `x=a`, there must be another P.I. at `x=–a`.
`text(When)\ \ ` `x < 1/sqrt2,` `\ (d^2y)/(dx^2) < 0`
  `x > 1/sqrt2,` `\ (d^2y)/(dx^2) > 0`

 
`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = 1/sqrt2`
 

`text(When)\ \ ` `x <-1/sqrt2,` `\ (d^2y)/(dx^2) > 0`
  `x >-1/sqrt2,` `\ (d^2y)/(dx^2) < 0`

 
`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x =-1/sqrt2`

 

b.   `text(In)\ f(x), text(there are 2 values of)\ y\ text(for)`
  `text(each value of)\ x.`
  `:.\ text(The domain of)\ f(x)\ text(must be restricted)`
  `text(for)\ \ f^(-1) (x)\ text(to exist).`

 

c.  `y = e^(-x^2)`

`text(Inverse: swap)\  x harr y` 

`x` `= e^(-y^2),\ \ \ x >= 0`
`lnx` `= ln e^(-y^2)`
`-y^2` `= lnx`
`y^2` `= -lnx`
  `=ln(1/x)`
`y` `= +- sqrt(ln (1/x))`

 

`text(Restricting)\ \ x>=0,\ \ =>y>=0`

`:.  f^(-1) (x)=sqrt(ln (1/x))`

 

d.   `f(0) = e^0 = 1`

`:.\ text(Range of)\ \ f(x)\ \ text(is)\ \ 0 < y <= 1`

`:.\ text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ 0 < x <= 1`

 

e. 

Inverse Functions, EXT1 2010 HSC 3b Answer

Calculus, MET1 SM-Bank 1

Part of the graph of a function  `f: R→R, \ f(x) = 2/x`  is shown below.

The diagram shows the area under `f(x)` from  `x = 1`  to  `x = d`.

What value of `d` makes the shaded area equal to 2?   (2 marks)

Show Answers Only

`e`

Show Worked Solution

`int_1^d 2/x\ dx = 2`

`[2log_e x]_1^d = 2`

`2log_e d-2log_e 1` `= 2`
`2log_e d` `= 2`
`log_e d` `= 1`
`:. d` `= e`

Algebra, MET1 SM-Bank 10

Solve the equation  `log_e x-3/log_ex=2`  for `x`.   (3 marks)

Show Answers Only

`x=e^3\ \ text(or)\ \ e^-1`

Show Worked Solution
 
IMPORTANT: Students should recognise this equation as a quadratic, and substitute `log_ex` with a variable such as `X`.
`log_e x-3/(log_ex)` `=2`
`(log_ex)^2-3` `=2log_e x`
`(log_ex)^2-2log_ex-3` `=0`
   
`text(Let)\  X=log_ex`  
`:.\ X^2-2X-3` `=0`
`(X-3)(X+1)` `=0`
COMMENT: Must know how to find `x` from the equations  `log_ex=-1`  and  `log_ex=3`.
`X` `=3` `\ \ \ \ \ \ \ \ \ \ ` `X` `=-1`
`log_ex` `=3` `\ \ \ \ \ \ \ \ \ \ ` `log_ex` `=-1`
`x` `=e^3` `\ \ \ \ \ \ \ \ \ \ ` `x` `=e^-1`

 

`:.x=e^3\ \ text(or)\ \ e^-1`

Algebra, MET1 2011 VCAA 2b

Solve the equation  `4^x - 15 × 2^x = 16`  for `x.`  (3 marks)

Show Answers Only

`x = 4`

Show Worked Solution
`4^x – 15 xx 2^x – 16` `= 0`
`2^(2x) – 15 xx 2^x – 16` `= 0`

 

♦♦ Mean mark 33%.
MARKER’S COMMENT: “Poorly answered”. Many students incorrectly stated that if  `y=2^x`, then `2y=4^x`.

`text(Let)\ \ y = 2^x`

`y^2 – 15y – 16` `= 0`
`(y – 16) (y + 1)` `= 0`
`y` `= 16` `\ \ \ or\ \ \ ` `y` `= – 1`
`2^x` `= 16`   `2^x` `= – 1`
`:. x` `= 4`   `text(No solution)`

Probability, MET1 2016 VCAA 8*

Let `X` be a continuous random variable with probability density function

`f(x) = {(−4xlog_e(x),0<x<=1),(0,text(elsewhere)):}`

Part of the graph of  `f` is shown below. The graph has a turning point at  `x = 1/e`.

  1. Show by differentiation that  `(x^k)/(k^2)(k log_e(x)-1)`  is an antiderivative of  `x^(k – 1) log_e(x)`, where `k` is a positive real number.   (2 marks)

  2. Calculate `text(Pr)(X > 1/e)`.   (2 marks)

Show Answers Only

  1. `text(See Worked Solutions)`
  2.  `1-3/(e^2)`

Show Worked Solution

a.   `text(Using Product Rule:)`

♦♦ Mean mark part (a) 28%.
MARKER’S COMMENT: Students who expanded before differentiating tended to score more highly.

`d/(dx) ((x^k)/(k^2)(klog_e(x)-1))`

`=d/(dx)((x^k)/k log_e(x)-(x^k)/(k^2))`

`= x^(k-1) log_e(x) + 1/k x^(k-1)-1/k x^(k-1)`

`= x^(k-1) log_e(x)`

`:. intx^(k-1) log_e(x)\ dx = (x^k)/(k^2)(klog_e(x)-1)`
 

b.   `text(Pr)(x > 1/e)`

♦♦♦ Mean mark 16%.

`= −4 int_(1/e)^1 (xlog_e(x))\ dx,\ text(where)\ k = 2`

`= −4[(x^2)/4(2log_e(x)-1)]_(1/e)^1`

`= −4[1/4(0 -1)-1/(4e^2)(2log_e(e^(−1))-1)]`

`= −4[−1/4 + 1/(4e^2) + 1/(2e^2)]`

`= 1-3/(e^2)`

Probability, MET1 2016 VCAA 7

A company produces motors for refrigerators. There are two assembly lines, Line A and Line B. 5% of the motors assembled on Line A are faulty and 8% of the motors assembled on Line B are faulty. In one hour, 40 motors are produced from Line A and 50 motors are produced from Line B. At the end of an hour, one motor is selected at random from all the motors that have been produced during that hour.

  1. What is the probability that the selected motor is faulty? Express your answer in the form `1/b`, where `b` is a positive integer.  (2 marks)

  2. The selected motor is found to be faulty.
  3. What is the probability that it was assembled on Line A? Express your answer in the form `1/c`, where `c` is a positive integer.  (1 mark)

Show Answers Only

  1. `1/15`
  2. `1/3`

Show Worked Solution

a.   `text(Construct tree diagram)`

`text(Pr)(AF) + text(Pr)(BF)`

`= 4/9 xx 1/20 + 5/9 xx 2/25`

`= 1/45 + 2/45`

`= 1/15`
 

`:. text(Pr)(F) = 1/15`

 

♦♦ Mean mark 32%.

b.    `text(Pr)(A|F)` `= (text(Pr)(A ∩ F))/(text(Pr)(F))`
    `= (4/9 xx 1/20)/(1/15)`
    `= 1/45 xx 15`
    `= 1/3`

Calculus, MET1 2016 VCAA 6a

Let  `f : [-pi,  pi] → R`, where  `f (x) = 2 sin (2x)-1`.

Calculate the average rate of change of  `f` between  `x = -pi/3`  and  `x = pi/6`.   (2 marks)

Show Answers Only

`(4sqrt3)/pi`

Show Worked Solution

`text(Average rate of change)`

`= (f(pi/6)-f(-pi/3))/(pi/6-(-pi/3))`

`= [(2sin(pi/3)-1)-(2sin(-(2pi)/3)-1)] xx 2/pi`

`= [(2 xx (sqrt3)/2-1)-(2 xx (-sqrt3)/2-1)] xx 2/pi`

♦ Mean mark 48%.
MARKER’S COMMENT: Evaluating  `f(-pi/3)` caused significant problems here.

`= (sqrt3-1 + sqrt3 + 1) xx 2/pi`

`= (4sqrt3)/pi`

 

 

Graphs, MET1 2016 VCAA 5

Let  `f : (0, ∞) → R`, where  `f(x) = log_e(x)`  and  `g: R → R`, where  `g (x) = x^2 + 1`.

  1.   i. Find the rule for `h`, where  `h(x) = f (g(x))`.   (1 mark)

    ii. State the domain and range of `h`.   (2 marks)

  2. iii. Show that  `h(x) + h(-x) = f ((g(x))^2 )`.   (2 marks)

  3. iv. Find the coordinates of the stationary point of `h` and state its nature.   (2 marks)

     

  4. Let  `k: (-∞, 0] → R`  where  `k (x) = log_e(x^2 + 1)`.
  5.  i. Find the rule for  `k^(-1)`.   (2 marks)

  6. ii. State the domain and range of  `k^(-1)`.   (2 marks)

Show Answers Only
  1.   i. `log_e(x^2 + 1)`
  2.  ii. `[0,∞)`
  3. iii. `text(See Worked Solutions)`
  4. iv. `(0, 0)`
  5.  i. `-sqrt(e^x-1)`
  6. ii. `text(Domain)\ (k) = (-∞,0]`
  7.    `text(Range)\ (k) = [0,∞)`
Show Worked Solution
a.i.    `h(x)` `= f(x^2 + 1)`
    `= log_e(x^2 + 1)`

 

a.ii.   `text(Domain)\ (h) =\ text(Domain)\ (g) = R`

♦♦ Mean mark part (a)(ii) 30%.
  `text(For)\ x ∈ R` `-> x^2 + 1 >= 1`
  `-> log_e(x^2 + 1) >= 0`

`:.\ text(Range)\ (h) = [0,∞)`

 

MARKER’S COMMENT: Many students were unsure of how to present their working in this question. Note the layout in the solution.
a.iii.   `text(LHS)` `= h(x) + h(−x)`
    `= log_e(x^2 _ 1) + log_e((-x)^2 + 1)`
    `= log_e(x^2 + 1) + log_e(x^2 + 1)`
    `= 2log_e(x^2 + 1)`

 

`text(RHS)` `= f((x^2 + 1)^2)`
  `= 2log_e(x^2 + 1)`

 

`:. h(x) + h(-x) = f((g(x))^2)\ \ text(… as required)`

 

a.iv.   `text(Stationary points when)\ \ h^{prime}(x) = 0`

♦ Mean mark part (a)(iv) 41%.
MARKER’S COMMENT: Solving a fraction is zero

   `text(Using Chain Rule:)`

`h^{prime}(x)` `= (2x)/(x^2 + 1)`

`:.\ text(S.P. when)\ \ x=0`

 

`text(Find nature using 1st derivative test:)`

`:.\ text{Minimum stationary point at (0, 0)}.`

 

b.i.   `text(Let)\ \ y = k(x)`

♦ Mean mark (b)(i) 49%.
MARKER’s COMMENT: Many students failed to consider the restrictions on the domain in `k(x)` and only select the negative root.

  `text(Inverse: swap)\ x ↔ y`

`x` `= log_e(y^2 + 1)`
`e^x` `= y^2 + 1`
`y^2` `= e^x-1`
`y` `= ±sqrt(e^x-1)`

 

`text(But range)\ \ (k^(-1)) =\ text(domain)\ (k)`

`:.k^(-1)(x) =-sqrt(e^x-1)`

♦ Mean mark part (b)(ii) 44%.

 

b.ii.   `text(Range)\ (k^(-1)) =\ text(Domain)\ (k) = (-∞,0]`

  `text(Domain)\ (k^(-1)) =\ text(Range)\ (k) = [0,∞)`

Calculus, MET1 2016 VCAA 2

Let  `f: (-∞,1/2] -> R`, where  `f(x) = sqrt(1-2x)`.

  1. Find  `f^{prime}(x)`.   (1 mark)

  2. Find the angle `theta` from the positive direction of the `x`-axis to the tangent to the graph of  `f` at  `x =-1`, measured in the anticlockwise direction.   (2 marks)

Show Answers Only
  1. `(-1)/(sqrt(1-2x))`
  2. `(5pi)/6`
Show Worked Solution
a.    `f(x)` `= (1-2x)^(1/2)`
  `f^{prime}(x)` `= 1/2(1-2x)^(-1/2) (-2)qquadtext([Using Chain Rule])`
    `= (-1)/(sqrt(1-2x))`

 

♦ Mean mark 40%.
b.    `tan theta` `= f^{prime}(-1)`
    `= (-1)/(sqrt(1-2(-1)))`
    `= (-1)/(sqrt3)`

 

`text(S)text(ince)\ theta ∈ [0,pi],`

`=> theta = (5pi)/6`

Probability, MET2 2009 VCAA 3

The Bouncy Ball Company (BBC) makes tennis balls whose diameters are normally distributed with mean 67 mm and standard deviation 1 mm. The tennis balls are packed and sold in cylindrical tins that each hold four balls. A tennis ball fits into such a tin if the diameter of the ball is less than 68.5 mm.

  1. What is the probability, correct to four decimal places, that a randomly selected tennis ball produced by BBC fits into a tin?  (2 marks)

BBC management would like each ball produced to have diameter between 65.6 and 68.4 mm.

  1. What is the probability, correct to four decimal places, that the diameter of a randomly selected tennis ball made by BBC is in this range?  (2 marks)

    1. What is the probability, correct to four decimal places, that the diameter of a tennis ball which fits into a tin is between 65.6 and 68.4 mm?  (1 mark)

    2. A tin of four balls is selected at random. What is the probability, correct to four decimal places, that at least one of these balls has diameter outside the desired range of 65.6 to 68.4 mm?  (2 marks)

BBC management wants engineers to change the manufacturing process so that 99% of all balls produced have diameter between 65.6 and 68.4 mm. The mean is to stay at 67 mm but the standard deviation is to be changed.

  1. What should the new standard deviation be (correct to two decimal places)?  (3 marks)

Show Answers Only

a.  `0.9332`

b.  `0.8385`

c.i.  `0.8985`

c.ii.  `0.3482`

d.  `0.54\ text(mm)`

Show Worked Solution

a.   `text(Let)\ \ X = text(diameter),\ \ X ∼ text(N) (67, 1^2)`

`text(Pr) (X < 68.5) = 0.9332\ \ text{(to 4 d.p.)}`

`[text(CAS: normCdf) (– oo, 68.5, 67, 1)]`

 

b.   `text(Pr) (65.6 < X < 68.4)`

`= 0.8385\ \ text{(to 4 d.p.)}`

`[text(CAS: normCdf) (65.6, 68.4, 67, 1)]`

 

c.i.   `text(Pr) (65.6 < X < 68.4 \|\ X < 68.5)`

♦ Mean mark 42%.

`= (text{Pr} (65.6 < X < 68.4))/(text{Pr} (X < 68.5))`

`= (0.838487…)/(0.933193…)`

`= 0.8985\ \ text{(to 4 d.p.)}`

 

  ii.  `text(Let)\ \ Y = text(Number of balls with diameter outside range)`

♦♦ Mean mark 30%.

`Y ∼ text(Bi)(4, 1 – 0.8985…) -> Y ∼ text(Bi) (4, 0.101486)`

`text(Pr) (Y >= 1) = 0.3482\ \ text{(to 4 d.p.)}`

 

d.   `X = text(diameter),\ \ X ∼ text(N) (67, sigma^2)`

♦♦ Mean mark 35%.

 vcaa-graphs-fur2-2009-3di

`text(Pr) (Z < a)` `= 0.005`
`a` `= – 2.5758`

 

`text(Relate 65.6 to its corresponding)\ \ z text(-score):`

`– 2.5758…` `= (65.6 – 67)/sigma`
`:. sigma` `= 0.54\ text(mm)\ \ text{(to 2 d.p.)}`

Calculus, MET2 2009 VCAA 2

VCAA 2009 2a

A train is travelling at a constant speed of `w` km/h along a straight level track from `M` towards `Q.`

The train will travel along a section of track `MNPQ.`

Section `MN` passes along a bridge over a valley.

Section `NP` passes through a tunnel in a mountain.

Section `PQ` is 6.2 km long.

From `M` to `P`, the curve of the valley and the mountain, directly below and above the train track, is modelled by the graph of
 

`y = 1/200 (ax^3 + bx^2 + c)` where `a, b` and `c` are real numbers.
 

All measurements are in kilometres.

  1. The curve defined from `M` to `P` passes through `N (2, 0)`. The gradient of the curve at `N` is – 0.06 and the curve has a turning point at  `x = 4`.
  2.  i. From this information write down three simultaneous equations in `a`, `b` and `c`.   (3 marks)

  3. ii. Hence show that  `a = 1`, `b = – 6` and `c = 16`.   (2 marks)

  4. Find, giving exact values
  5.   i. the coordinates of `M and P`.   (2 marks)

  6.  ii. the length of the tunnel.   (1 mark)

  7. iii. the maximum depth of the valley below the train track.   (1 mark)

The driver sees a large rock on the track at a point `Q`, 6.2 km from `P`. The driver puts on the brakes at the instant that the front of the train comes out of the tunnel at `P`.

From its initial speed of `w` km/h, the train slows down from point `P` so that its speed `v` km/h is given by

`v = k log_e ({(d + 1)}/7)`,

where `d` km is the distance of the front of the train from `P` and `k` is a real constant.

  1. Find the value of `k` in terms of `w`.   (1 mark)

  2. Find the exact distance from the front of the train to the large rock when the train finally stops.   (2 marks)

Show Answers Only

a.i.    `1/200 (8a + 4b + c) = 0`

a.ii.   `1/200 (12a + 4b) = (– 3)/50`

a.iii.  `1/200 (48a + 8b) = 0`

`text(Proof)\ \ text{(See Worked Solutions)}`

    1. `M (2 + 2 sqrt 3, 0),\ \ P (2-2 sqrt 3, 0)`
    2. `2 sqrt 3\ text(km)`
    3. `2/25\ text(km)`
  2. `(– w)/(log_e (7))`
  3. `120`
  4. `0.2\ text(km)`
Show Worked Solution

 a.i.   `N (2, 0),`

`1/200 (8a + 4b + c) = 0\ \ text{… (1)}`
 

`(dy)/(dx) (x = 2) = (– 3)/50,`

`1/200 (12a + 4b) = (– 3)/50\ \ text{… (2)}`
 

`(dy)/(dx)(x = 4) = 0,`

`1/200 (48a + 8b) = 0\ \ text{… (3)}`
 

  ii.   `text(Using the above equations,)`

♦ Mean mark part (a)(ii) 41%.
`12a + 4b` `=-12` `\ \ \ …\ (2^{′})`
`24 a + 4b` `=0` `\ \ \ …\ (3^{′})`

 
`text{Solve simultaneous equations (By CAS):}`

`a=1, \ b=-6, \ c=16`
 

`\text{Solving manually:}`

`(3^{′})-(2^{′}):`

`12a=12 \ \Rightarrow\ \ a=1`

`text(Substitute)\ \ a = 1\ \ text(into)\ \ (3^{′}):`

`124(1)+4b=0 \ \Rightarrow\ \ b=-1`

`text(Substitute)\ \ a = 1,\ \ b = – 6\ \ text(into)\ \ (1):`

`8(1) + 4 (– 6) + c=0\ \ \Rightarrow\ \ c=16`
 

b.i.   `text(For)\ \ x text(-intercepts),`

`text(Solve:)\ \ x^3 + -6x^2 + 16` `= 0\ \ text(for)\ x,`
 `x= 2-2 sqrt 3, 2, 2 + 2 sqrt 3`  

 
`:. M (2 + 2 sqrt 3, 0),\ \ P (2-2 sqrt 3, 0)`
 

  ii.   `N (2, 0)`

♦ Mean mark part (a) 41%.
`bar (NP)` `=\ text(Tunnel length)`
  `= 2-(2-2 sqrt 3)`
  `= 2 sqrt 3\ text(km)`

 

   iii.  `text(Solve)\ \ frac (dy) (dx) = 0\ \ text(for)\ \ x in (2, 2 + 2 sqrt 3)`

♦ Mean mark part (b) 50%.

`x = 4`

`text(When)\ \ x=4,\ \ y = – 2/25\ text(km) = 80\ text{m (below track)}`

`:.\ text(Max depth below is)\ \ 80\ text(m.)`
 

c.  `text(Solution 1)`

♦ Mean mark part (c) 35%.

`text(Let)\ \ v(d) = k log_e ({(d + 1)}/7)`

`text(S)text(ince)\ \ v=w\ \ text(when)\ \ d=0\ \ text{(given),}`

`k log_e ({(0 + 1)}/7)` `=w`
`k` `=w/log_e (1/7)`
  `=(-w)/log_e7`

 
`text(Solution 2)`

`text(Solve:)\ \ v (0)` `= w\ \ text(for)\ \ k`
`:. k` `= (– w)/(log_e (7))`

 

d.  `v (2.5) = k log_e(1/2)`

♦ Mean mark part (d) 40%.
`text(Solve)\ \ v(2.5)` `= (120 log_e (2))/(log_e (7))\ \ text(for)\ \ k,`
`:. k` `= (– 120)/(log_e (7))`
`(– w)/(log_e (7))` `= (– 120)/(log_e (7))`
`:. w` `= 120`

 

e.   `text(Define)\ \ v (d) = (– 120)/(log_e (7)) log_e ((d + 1)/7)`

♦♦ Mean mark part (e) 32%.
`text(Solve:)\ \ v(d)` `= 0\ \ text(for)\ d,`
`:. d` `= 6\ text(km from)\ \ P`

 
`:.\ text(Distance between train and)\ \ Q`

`= 6.2-6`

`= 0.2\ text(km)`

Calculus, MET2 2009 VCAA 22 MC

Consider the region bounded by the `x`-axis, the `y`-axis, the line with equation  `y = 3`  and the curve with equation  `y = log_e (x - 1).`

The exact value of the area of this region is

A.   `e^-3 - 1`

B.   `16 + 3 log_e (2)`

C.   `3e^3 - e^-3 + 2`

D.   `e^3 + 2`

E.   `3e^2`

Show Answers Only

`D`

Show Worked Solution
♦ Mean mark 45%.

vcaa-2009-22i

`text(Solution 1)`

`text(Area)` `= 3 (e^3 + 1) – int_2^(e^3 + 1) (log_e (x – 1))\ dx`
  `= 3 (e^3 + 1) – (2e^3 + 1)`
  `= e^3 + 2`

`=>   D`

 

`text(Solution 2)`

`text(Inverse of)\ \ y = log_e (x – 1)\ \ text(is)`

`y=e^x+1`

`text(Area)` `=int_0^3 (e^x+1)\ dx`
  `=[e^x+x]_0^3`
  `=[(e^3+3)-e^0]`
  `=e^3 +2\ \ text(u²)`

Calculus, MET2 2009 VCAA 21 MC

A cubic function has the rule  `y = f (x)`. The graph of the derivative function  `f prime` crosses the `x`-axis at  `(2, 0)`  and  `(– 3, 0)`. The maximum value of the derivative function is 10.

The value of `x` for which the graph of  `y = f(x)`  has a local maximum is

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

`B`

Show Worked Solution

`text(If)\ \ f(x)\ \ text(is cubic) -> f prime (x)\ \ text(is quadratic)`

vcaa-2009-21aii

`f′(x) >0\ \ text(for)\ \ x in (-3,2), and f′(x) =0\ \ text(at)\ \ x=2.`

`f′(x) <0\ \ text(for)\ \ x>2.`

`:.\ text(Local max occurs when)\ \ x = 2`

`=>   B`

Calculus, MET2 2009 VCAA 9 MC

The tangent at the point (1, 5) on the graph of the curve  `y = f (x)`  has equation  `y = 3 + 2x.`

The tangent at the point (3, 8) on the curve  `y = f (x - 2) + 3`  has equation

A.   `y = 2x - 4`

B.   `y = x + 5`

C.   `y = -2x + 14`

D.   `y = 2x + 4`

E.   `y = 2x + 2`

Show Answers Only

`E`

Show Worked Solution

`f(x)\ \ text(translated right 2, up 3)`

`P (1, 5)\ overset (x + 2,\ \ y + 3) rightarrow\ P prime (3, 8)`

 

`text(T) text(angent equation at)\ \ P prime (3, 8),\ text(has the)`

`text(same gradient as)\ \ y = 3 + 2x.`

`:.\ text(Equation of the second tangent,)`

`y – 8` `= 2 (x – 3)`
`y` `= 2x + 2`

 
`=>   E`

Algebra, MET2 2009 VCAA 1 MC

The simultaneous linear equations

`kx - 3y = 0`

`5x - (k + 2)y = 0`

where `k` is a real constant, have a unique solution provided

  1. `k in {– 5, 3}`
  2. `k in R\ text(\){– 5, 3}`
  3. `k in {– 3, 5}`
  4. `k in R\ text(\){– 3, 5}`
  5. `k in R\ text(\){0}`
Show Answers Only

`B`

Show Worked Solution

`text(Unique solution occurs when:)`

♦ Mean mark 49%.
`m_1` `!= m_2`
`k/5` `!= (– 3)/(– (k + 2))`
`k^2+2k-15` `!=0`
`(k+5)(k-3)` `!=0`
`:. k` `!= – 5, 3`

`=>   B`

Calculus, MET2 2011 VCAA 4

Deep in the South American jungle, Tasmania Jones has been working to help the Quetzacotl tribe to get drinking water from the very salty water of the Parabolic River. The river follows the curve with equation  `y = x^2-1`,  `x >= 0` as shown below. All lengths are measured in kilometres.

Tasmania has his camp site at `(0, 0)` and the Quetzacotl tribe’s village is at `(0, 1)`. Tasmania builds a desalination plant, which is connected to the village by a straight pipeline.
 

met2-2011-vcaa-q4

  1. If the desalination plant is at the point `(m, n)` show that the length, `L` kilometres, of the straight pipeline that carries the water from the desalination plant to the village is given by
  2.    `L = sqrt(m^4-3m^2 + 4)`.   (3 marks)

  3. If the desalination plant is built at the point on the river that is closest to the village
    1. find `(dL)/(dm)` and hence find the coordinates of the desalination plant.   (3 marks)

    2. find the length, in kilometres, of the pipeline from the desalination plant to the village.   (2 marks)

The desalination plant is actually built at `(sqrt7/2, 3/4)`.

If the desalination plant stops working, Tasmania needs to get to the plant in the minimum time.

Tasmania runs in a straight line from his camp to a point `(x,y)` on the river bank where  `x <= sqrt7/2`. He then swims up the river to the desalination plant.

Tasmania runs from his camp to the river at 2 km per hour. The time that he takes to swim to the desalination plant is proportional to the difference between the `y`-coordinates of the desalination plant and the point where he enters the river.

  1. Show that the total time taken to get to the desalination plant is given by

     

    `qquadT = 1/2 sqrt(x^4-x^2 + 1) + 1/4k(7-4x^2)` hours where `k` is a positive constant of proportionality.   (3 marks)

The value of `k` varies from day to day depending on the weather conditions.

  1. If  `k = 1/(2sqrt13)`
    1. find `(dT)/(dx)`   (1 mark)

    2. hence find the coordinates of the point where Tasmania should reach the river if he is to get to the desalination plant in the minimum time.   (2 marks)

  2. On one particular day, the value of `k` is such that Tasmania should run directly from his camp to the point `(1,0)` on the river to get to the desalination plant in the minimum time. Find the value of `k` on that particular day.   (2 marks)

  3. Find the values of `k` for which Tasmania should run directly from his camp towards the desalination plant to reach it in the minimum time.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2.  i. `(sqrt6/2, 1/2)`
  3. ii. `sqrt7/2\ text(km)`
  4. `text(See Worked Solutions)`
  5.  i. `(dT)/(dx) = (x(2x^2-1))/(2sqrt(x^4-x^2 + 1))-(sqrt13 x)/13`
  6. ii. `(sqrt3/2, −1/4)`
  7. `1/4`
  8. `(5sqrt37)/74`
Show Worked Solution

a.   `text(S)text(ince)\ \ (m,n)\ \ text(lies on)\ \ y=x^2-1,`

♦ Mean mark (a) 47%.

`=> n=m^2-1`

`V(0,1), D(m,m^2-1)`

`L` `= sqrt((m-0)^2 + ((m^2-1)-1)^2)`
  `= sqrt(m^2 + m^4-4m^2 + 4)`
  `= sqrt(m^4-3m^2 + 4)\ \ text(… as required)`

 

b.i.   `(dL)/(dm) = (2m^2-3m)/(sqrt(m^4-3m^2 + 4))`

`text(Solve:)\ \ (dL)/(dm) = 0quadtext(for)quadm >= 0`

`\Rightarrow m = sqrt6/2`

`text(Substitute into:)\ \ D(m, m^2-1),`

`:. text(Desalination plant at)\ \ (sqrt6/2, 1/2)`
 

♦ Mean mark part (b)(ii) 41%.
b.ii.   `L(sqrt6/2)` `= sqrt(m^4-3m^2 + 4)`
    `=sqrt(36/16-3xx6/4+4`
    `=sqrt7/2`

 

c.   `text(Let)\ \ P(x,x^2-1)\ text(be run point on bank)`

♦♦♦ Mean mark (c) 16%.

`text(Let)\ \ D(sqrt7/2, 3/4)\ text(be desalination location)`

`T` `=\ text(run time + swim time)`
  `= (sqrt((x-0)^2 + ((x^2-1)-0)^2))/2 + k(3/4-(x^2-1))`
  `= (sqrt(x^2 + x^4-2x^2 + 1))/2 + k/4(3-4(x^2-1))`
`:. T` `= (sqrt(x^4-x^2 + 1))/2 + 1/4k(7-4x^2)`

 

d.i.   `(dT)/(dx) = (x(2x^2-1))/(2sqrt(x^4-x^2 + 1))-(sqrt13 x)/13`

 

d.ii.   `text(Solve:)\ \ (dT)/(dx) = 0`

♦♦ Mean mark (d.ii.) 33%.

`x = sqrt3/2`

`y=x^2-1=-1/4`

`:. T_(text(min)) \ text(when point is)\ \ (sqrt3/2, −1/4)`

 

e.  `(dT)/(dx) = (x(2x^2-1))/(2sqrt(x^4-x^2 + 1))-2kx`

♦♦ Mean mark (e) 39%.

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

`text(Solve:)\ \ (dT)/(dx)` `=0\ \ text(for)\ k,`
`1/2 -2k` `=0`
`:.k` `=1/4`

 

f.   `text(Require)\ T_text(min)\ text(to occur at right-hand endpoint)\ \ x = sqrt7/2.`

`text(This can occur in 2 situations:)`

♦♦♦ Mean mark (f) 13%.

`text(Firstly,)\ \ T\ text(has a local min at)\ \ x = sqrt7/2,`

`text(Solve:)\ \ (x(2x^2-1))/(2sqrt(x^4-x^2 + 1))-2kx=0| x = sqrt7/2,\ \ text(for)\ k,`

`:.k = (5sqrt37)/74`
 

`text(S)text(econdly,)\ \ T\ text(is decreasing function over)\ x ∈ (0, sqrt7/2),`

`text(Solve:)\ \ (dT)/(dx) <= 0 | x = sqrt7/2,\ text(for)\ k,`

`:. k > (5sqrt37)/74`

`:. k >= (5sqrt37)/74`

Calculus, MET2 2011 VCAA 3

  1. Consider the function  `f: R -> R, f(x) = 4x^3 + 5x-9`.

     

    1. Find  `f^{prime}(x).`   (1 mark)

    2. Explain why  `f^{prime}(x) >= 5` for all `x`.   (1 mark)

  2. The cubic function `p` is defined by  `p: R -> R, p(x) = ax^3 + bx^2 + cx + k`, where `a`, `b`, `c` and `k` are real numbers.

     

    1. If `p` has `m` stationary points, what possible values can `m` have?   (1 mark)

    2. If `p` has an inverse function, what possible values can `m` have?   (1 mark)

  3. The cubic function `q` is defined by  `q:R -> R, q(x) = 3-2x^3`.

     

    1. Write down a expression for  `q^(-1)(x)`.   (2 marks)

    2. Determine the coordinates of the point(s) of intersection of the graphs of  `y = q(x)`  and  `y = q^(-1)(x)`.   (2 marks)

  4. The cubic function `g` is defined by  `g: R -> R, g(x) = x^3 + 2x^2 + cx + k`, where `c` and `k` are real numbers.

     

    1. If `g` has exactly one stationary point, find the value of `c`.   (3 marks)

    2. If this stationary point occurs at a point of intersection of  `y = g(x)`  and  `g^(−1)(x)`, find the value of `k`.   (3 marks)

Show Answers Only
    1. `f^{prime}(x) = 12x^2 + 5`
    2. `text(See Worked Solutions)`
    1. `m = 0, 1, 2`
    2. `m = 0, 1`
    1. `q^(-1)(x) = root(3)((3-x)/2), x ∈ R`
    2. `(1, 1)`
    1. `4/3`
    2. `-10/27`
Show Worked Solution

a.i.   `f^{prime}(x) = 12x^2 + 5`
  

a.ii.  `text(S)text(ince)\ \ x^2>=0\ \ text(for all)\ x,`

♦ Mean mark 47%.
` 12x^2` `>= 0`
`12x^2 + 5` `>=  5`
`f^{prime}(x)` `>=  5\ \ text(for all)\ x`

 

b.i.   `p(x) = text(is a cubic)`

♦♦♦ Mean mark part (b)(i) 9%, and part (b)(ii) 20%.
MARKER’S COMMENT: Good exam strategy should point students to investigate earlier parts for direction. Here, part (a) clearly sheds light on a solution!

`:. m = 0, 1, 2`

`text{(Note: part a.ii shows that a cubic may have no SP’s.)}`

 

b.ii.   `text(For)\ p^(−1)(x)\ text(to exist)`

`:. m = 0, 1`

 

c.i.   `text(Let)\ y = q(x)`

`text(Inverse: swap)\ x ↔ y`

`x` `= 3-2y^3`
`y^3` `= (3-x)/2`

`:. q^(-1)(x) = root(3)((3-x)/2), \ x ∈ R`
  

c.ii.  `text(Any function and its inverse intersect on)`

   `text(the line)\ \ y=x.`

`text(Solve:)\ \ 3-2x^3` `= xqquadtext(for)\ x,`
`x` `= 1`

 

`:.\ text{Intersection at (1, 1)}`
  

♦ Mean mark part (d)(i) 44%.
d.i.    `g^{prime}(x)` `= 0`
  `3x^2 + 4x + c` `= 0`
  `Delta` `= 0`
  `16-4(3c)` `= 0`
  `:. c` `= 4/3`

 

d.ii.   `text(Define)\ \ g(x) = x^3 + 2x^2 + 4/3x + k`

♦♦♦ Mean mark part (d)(ii) 14%.

  `text(Stationary point when)\ \ g^{prime}(x)=0`

`g^{prime}(x) = 3x^2+4x+4/3`

`text(Solve:)\ \ g^{prime}(x)=0\ \ text(for)\ x,`

`x = -2/3`

`text(Intersection of)\ g(x)\ text(and)\ g^(-1)(x)\ text(occurs on)\ \ y = x`

`text(Point of intersection is)\  (-2/3, -2/3)`

`text(Find)\ k:`

`g(-2/3)` `= -2/3\ text(for)\ k`
`:. k` ` = -10/27`

Probability, MET2 2011 VCAA 2*

In a chocolate factory the material for making each chocolate is sent to one of two machines, machine A or machine B.

The time, `X` seconds, taken to produce a chocolate by machine A, is normally distributed with mean 3 and standard deviation 0.8.

The time, `Y` seconds, taken to produce a chocolate by machine B, has the following probability density function
 

`f(y) = {{:(0,y < 0),(y/16,0 <= y <= 4),(0.25e^(−0.5(y-4)),y > 4):}`
 

  1. Find correct to four decimal places

    1. `text(Pr)(3 <= X <= 5)`   (1 mark)

    2. `text(Pr)(3 <= Y <= 5)`   (3 marks)

  2. Find the mean of `Y`, correct to three decimal places.   (3 marks)

  3. It can be shown that  `text(Pr)(Y <= 3) = 9/32`. A random sample of 10 chocolates produced by machine B is chosen. Find the probability, correct to four decimal places, that exactly 4 of these 10 chocolate took 3 or less seconds to produce.   (2 marks)

All of the chocolates produced by machine A and machine B are stored in a large bin. There is an equal number of chocolates from each machine in the bin.

It is found that if a chocolate, produced by either machine, takes longer than 3 seconds to produce then it can easily be identified by its darker colour.

  1. A chocolate is selected at random from the bin. It is found to have taken longer than 3 seconds to produce.
  2. Find, correct to four decimal places, the probability that it was produced by machine A.   (3 marks)

Show Answers Only

a.i.  `0.4938`

a.ii. `0.4155`

b.    `4.333`

c.    `0.1812`

d.    `0.4103`

Show Worked Solution

a.i.   `X ∼\ N(3,0.8^2)`

`text(Pr)(3 <= X <= 5) = 0.4938\ \ text{(4 d.p.)}`

 

a.ii.    `text(Pr)(3 <= Y <= 5)` `= text(Pr)(3 <= Y <= 4) + (4 < Y <= 5)`
    `= int_3^4 (y/16)dy + int_4^5(1/4 e^(−1/2(y-4)))dy`
    `= 0.4155\ \ text{(4 d.p.)}`

 

b.    `text(E)(Y)` `= int_0^4 y(y/16)dy + int_4^∞ 0.25ye^(−1/2(y-4))dy`
    `= 4.333\ \ text{(3 d.p.)}`

 

c.   `text(Solution 1)`

`text(Let)\ \ W = #\ text(chocolates from)\ B\ text(that take less)`

`text(than 3 seconds)`

`W ∼\ text(Bi)(10, 9/32)`

`text(Using CAS: binomPdf)(10, 9/32,4)`

`text(Pr)(W = 4) = 0.1812\ \ text{(4 d.p.)}`
 

`text(Solution 2)`

`text(Pr)(W = 4)`  `=((10),(4)) (9/32)^4 (23/32)^6` 
  `=0.1812`

♦♦♦ Mean mark part (e) 19%.
MARKER’S COMMENT: Students who used tree diagrams were the most successful.

 

d.   
`text(Pr)(A | L)` `= (text(Pr)(AL))/(text(Pr)(L))`
  `= (0.5 xx 0.5)/(0.5 xx 0.5 + 0.5 xx 23/32)`
  `= 0.4103\ \ text{(4 d.p.)}`

Calculus, MET2 2011 VCAA 1

Two ships, the Elsa and the Violet, have collided. Fuel immediately starts leaking from the Elsa into the sea.

The captain of the Elsa estimates that at the time of the collision his ship has 6075 litres of fuel on board and he also forecasts that it will leak into the sea at a rate of `(t^2)/5` liters per minute, where `t` is the number of minutes that have elapsed since the collision.

  1. At this rate how long, in minutes, will it take for all the fuel from the Elsa to leak into the sea?   (3 marks)

*Parts (b) - (d) are no longer in the syllabus. 

Show Answers Only
  1. `45\ text(min)`
Show Worked Solution

a.   `text(Let)\ \ f = text(fuel in Elsa at)\ t\ text(min)`

♦♦♦ Mean mark 23%.

`(df)/(dt) = (-t^2)/5\ \ text{(given)}`
  

`text(Find)\ f\ text(in terms of)\ t:`

`f` `= int-1/5t^2 dt`
`f` `= (-t^3)/15 + c`

 

`text(Substitute)\ \ (0,6075):`

`6075` `= 0/15 + c`
`:. c` `= 6075`

 

`text(Solve:)\ \ 0 = -(t^2)/15 + 6075\ \ text(for)\ t`

`:. t = 45\ text(min)`

 

*Parts (b) – (d) are no longer in the syllabus.

CORE, FUR2 SM-Bank 4

Damon runs a swim school.

The value of his pool pump is depreciated over time using flat rate depreciation.

Damon purchased the pool pump for $28 000 and its value in dollars after `n` years, `P_n`, is modelled by the recursion equation below:

`P_0 = 28\ 000,qquad P_(n + 1) = P_n - 3500`

  1. Write down calculations, using the recurrence relation, to find the pool pump's value after 3 years.   (1 mark)

  2. After how many years will the pump's depreciated value reduce to $7000?   (1 mark)

The reducing balance depreciation method can also be used by Damon.

Using this method, the value of the pump is depreciated by 15% each year.

A recursion relation that models its value in dollars after `n` years, `P_n`, is:

`P_0 = 28\ 000, qquad P_(n + 1) = 0.85P_n`

  1. After how many years does the reducing balance method first give the pump a higher valuation than the flat rate method in part (a)?   (2 marks)

Show Answers Only
  1. `$17\ 500`
  2. `6\ text(years)`
  3. `4\ text(years)`
Show Worked Solution
a.    `P_1` `= 28\ 000-3500 = 24\ 500`
  `P_2` `= 24\ 500-3500 = 21\ 000`
  `P_3` `= 21\ 000-3500 = 17\ 500`

  
`:.\ text(After 3 years, the pump’s value is $17 500.)`
  

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

`7000` `= 28\ 000-3500n`
`3500n` `= 21\ 000`
`n` `= (21\ 000)/3500`
  `= 6\ text(years)`

  
c.   `text(Using the reducing balance method)`

`P_1` `= 0.85 xx 28\ 000 = 23\ 800`
`P_2` `= 0.85 xx 23\ 800 = 20\ 230`
`P_3` `= 0.85 xx 20\ 230 = 17\ 195`
`P_4` `= 0.85 xx 17\ 195 = 14\ 615.75`

  
`text{Using the flat rate method (see part (a))}`

`P_4 = 17\ 500-3500 = 14\ 000`

`14\ 615.75 > 14\ 000`
  

`:.\ text(After 4 years, the reducing balance method)`

`text(first values the pump higher.)`