A worker has received an annual salary increase of 3% for the past two years.
This year, the worker’s annual salary is $46 500.
Two years ago, her salary was closest to
A. `$42\ 315`
B. `$43\ 750`
C. `$43\ 830`
D. `$45\ 140`
E. `$49\ 330`
Aussie Maths & Science Teachers: Save your time with SmarterEd
A worker has received an annual salary increase of 3% for the past two years.
This year, the worker’s annual salary is $46 500.
Two years ago, her salary was closest to
A. `$42\ 315`
B. `$43\ 750`
C. `$43\ 830`
D. `$45\ 140`
E. `$49\ 330`
`C`
`text(Let)\ x\ text(be the worker’s salary 2 years ago)`
`x xx 1.03 xx 1.03` | `= 46\ 500` |
`x (1.03)^2` | `= 46\ 500` |
`:. x` | `= (46\ 500) / 1.03^2` |
`= 43\ 830.70…` |
`=> C`
Leslie borrowed $35 000 from a bank.
Interest is charged at the rate of 4.75% on the reducing monthly balance.
The loan is to be repaid with 47 monthly payments of $802.00 and a final payment that is to be adjusted so that the loan will be fully repaid after exactly 48 monthly payments.
Correct to the nearest cent, the amount of the final payment will be
A. $0.39
B. $3.57
C. $802.00
D. $802.39
E. $805.57
`E`
`text(Calculate the FV of the loan after 47 payments)`
`text(By TVM Solver:)`
`N` | `= 47` |
`I(%)` | `= 4.75text(%)` |
`PV` | `= -35\ 000` |
`PMT` | `= 802` |
`FV` | `= text(?)` |
`text(P/Y)` | `= text(C/Y) = 12` |
`:.\ text(Balance owing after 47 payments)\ =$802.39…`
`text(Final payment)`
`=\ text(Balance after 47 payments + month’s interest)`
`=802.39… + (802.39… xx 4.75/12 xx 100text{%})`
`=$805.57`
`=> E`
The graph above shows a relationship between `y` and `x^3`.
The graph that shows the same relationship between `y` and `x` is
`C`
`text(The gradient of the linear graph of)`
`y\ text(versus)\ x^3 = 1/3`
`:.\ y = 1/3x^3`
`text(Test if the coordinates in each graph satisfy)`
`text(the equation)\ \ y=1/3x^3.`
`text(Consider)\ C,`
`text(Substitute)\ (1, 1/3)\ text(into equation,)`
`1/3 = 1/3 xx 1^3\ \ \ text{(true)}`
`text(All other options can be shown to not satisfy equation.)`
`=> C`
Russell is a wine producer. He makes both red and white wine.
Let `x` represent the number of bottles of red wine he makes and `y` represent the number of bottles of white wine he makes.
This year he plans to make at least twice as many bottles of red wine as white wine.
An inequality representing this situation is
A. `y <= x + 2`
B. `y <= 2x`
C. `y >= 2x`
D. `x <= 2y`
E. `x >= 2y`
`E`
`x = text(no. red wine bottles)`
`y = text(no. white wine bottles)`
`text(The constraint states that the number of)`
`text(of bottles of red wine will be at least twice)`
`text(the number of bottles of white wine.)`
`x >= 2y`
`=> E`
When shopping, Betty can use either Easypark or Safepark to park her car.
At Easypark, cars can be parked for up to 8 hours per day.
The fee structure is as follows.
`text(Fee)={(text($5.00,), 0 < text(hours) <= 2),(text($8.00,), 2 < text(hours) <= 5),(text($11.00,), 5 < text(hours) <= 8) :}`
Safepark charges fees according to the formula
`text(Fee) = $2.50 xx text(hours)`
Betty wants to park her car for 5 hours on Monday and 3 hours on Tuesday.
The minimum total fee that she can pay for parking for the two days is
A. `$7.50`
B. `$11.00`
C. `$15.50`
D. `$16.00`
E. `$20.00`
`C`
`text(Fee for 5 hours)`
`text(Easypark = $8)`
`text(Safepark = $2.50) xx 5 = $12.50`
`text(Fee for 3 hours)`
`text(Easypark = $8)`
`text(Safepark = $2.50) xx 3 = $7.50`
`:.\ text(The minimum total fee is)`
`$8 + $7.50 = $15.50`
`=> C`
The graph below shows the cost (in dollars) of producing birthday cards.
If the profit from the sale of 150 birthday cards is $175, the selling price of one card is
A. `$0.30`
B. `$1.60`
C. `$3.10`
D. `$3.50`
E. `$4.40`
`D`
`text{Profit = Revenue – Costs = $175 (given)}`
`text{Revenue}\ (R) = 150 xx S,\ \ \ (S= text{selling price})`
`text(Calculating cost of 150 cards,)\ C,`
`text(Fixed cost) = $50\ \ text{(}y\ text(intercept) text{)}`
`text{40 cards cost $130 (from graph)}`
`:.\ text(C)text(ost per card)= (130 − 50)/40=$2`
`:. C=50 + 150 xx 2=350`
`R-C` | `=175` |
`150S-350` | `= 175` |
`150S` | `=525` |
`:. S` | `= $3.50` |
`=> D`
Points `M` and `P` are the same distance from a third point `O`.
The bearing of `M` from `O` is 038° and the bearing of `P` from `O` is 152°.
The bearing of `P` from `M` is
A. between 000° and 090°
B. between 090° and 180°
C. exactly 180°
D. between 180° and 270°
E. between 270° and 360°
`D`
`/_ MOP` | `= 152 − 38` |
`= 114^@` |
`text(Let)\ R\ text(be directly south of)\ O`
`/_POR` | `= 180 − (38 + 114)\ \ text{(straight line)}` |
`= 28^@` |
`text(Given)\ OM = OP`
`text(Diagram shows that)\ M\ text(is further east than)\ P.`
`:.\ text(Bearing of)\ P\ text(from)\ M\ text(just over 180°)`
`=> D`
The cross-section of a water pipe is circular with a radius, `r`, of 50 cm, as shown above.
The surface of the water has a width, `w`, of 80 cm.
The depth of water in the pipe, `d`, could be
A. `20\ text(cm)`
B. `25\ text(cm)`
C. `30\ text(cm)`
D. `40\ text(cm)`
E. `50\ text(cm)`
`A`
A piece of plaster has a uniform cross-section, which has been shaded, and has dimensions as shown.
(i) |
`A` | `= h/3 [y_0 +4y_1 + y_2]\ text(… applied twice)` |
`= 3.6/3 (5 + 4 xx 4.6 + 3.7) + 3.6/3 (3.7 + 4 xx 2.8 + 0)` | |
`= 32.52 + 17.88` | |
`= 50.4\ text(cm²)` |
(ii) `text(Total Area) = 7480.8\ text(cm²)`
`text(Area of Base)` | `= 14.4 xx 200` |
`= 2880\ text(cm²)` |
`text(Area of End)` | `= 5 xx 200` |
`= 1000\ text(cm²)` |
`text(Area of sides)` | `= 2 xx 50.4\ \ \ text{(from (i))}` |
`= 100.8\ text(cm²)` |
`:.\ text(Area of curved surface)`
`= 7480.8 – (2880 + 1000 + 100.8)`
`= 3500\ text(cm²)`
This shape is made up of a right-angled triangle and a regular hexagon.
The area of a regular hexagon can be estimated using the formula `A = 2.598H^2` where `H` is the side-length.
Calculate the total area of the shape using this formula. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
`22.784\ text(cm²)`
`text(Area) = 2.598H^2`
`text(Using Pythagoras)`
`H^2` | `= 2^2 + 2^2` |
`= 8` | |
`H` | `= sqrt 8` |
`:.\ text(Area of hexagon)` | `= 2.598 xx (sqrt 8)^2` |
`= 20.784\ text(cm²)` |
`text(Area of triangle)` | `= 1/2 bh` |
`= 1/2 xx 2 xx 2` | |
`= 2\ text(cm²)` |
`:.\ text(Total Area)` | `= 20.784 + 2` |
`= 22.784\ text(cm²)` |
Two unbiased dice, `A` and `B`, with faces numbered `1`, `2`, `3`, `4`, `5` and `6` are rolled.
The numbers on the uppermost faces are noted. This table shows all the possible outcomes.
A game is played where the difference between the highest number showing and the lowest number showing on the uppermost faces is calculated.
(i) `text(# Outcomes with a difference of 1)`
`= 10`
`:.\ P text{(diff of 1)} = 10/36 = 5/18`
(ii) `P text{(no difference)} = 6/36 = 1/6`
`P text{(2, 3, 4 or 5)}` | `= 1 – [P(0) + P(1)]` |
`= 1 – [5/18 + 1/6]` | |
`= 1 – 8/18` | |
`= 5/9` |
`:.\ text(Financial Expectation)`
`= (1/6 xx 3.50) – (5/18 xx 5) + (5/9 xx 2.80)`
`= $0.75`
(iii) `text(If Jack pays $1 to play, he should expect)`
`text(a loss of $0.25.)`
A solid cylinder has a height of 30 cm and a diameter of 40 cm.
A hemisphere is cut out of the top of the cylinder as shown below.
In square centimetres, the total surface area of the remaining solid (including its base) is closest to
A. `1260`
B. `2510`
C. `6280`
D. `7540`
E. `10\ 050`
`D`
`text(Total surface area of solid)`
`=\ text{S.A. (base)} + text{S.A. (side)} + text{S.A. (hemisphere)}`
`text{S.A. (base)}` | `= pir^2` |
`= pi xx 20^2` | |
`= 1256.63…` |
`text{S.A. (side)}` | `= 2 pi r h` |
`= 2 xx pi xx 20 xx 30` | |
`= 3769.91…` |
`text{S.A. (hemisphere)}` | `= 1/2 xx 4pir^2` |
`= 1/2 xx 4 xx pi xx 20^2` | |
`= 2513.27…` |
`:.\ text(Total S.A.)` | `= 1256.63… + 3769.91… + 2513.27…` |
`= 7539.81…\ text(cm²)` |
`=> D`
A block of land has an area of 4000 m².
When represented on a map, this block of land has an area of 10 cm².
On the map 1 cm would represent an actual distance of
A. `10\ text(m)`
B. `20\ text(m)`
C. `40\ text(m)`
D. `400\ text(m)`
E. `4000\ text(m)`
`B`
`text(Area scale factor) = k^2`
`k^2` | `= 4000/10` |
`= 400` | |
`:. k` | `= sqrt(400)` |
`= 20` |
`:.\ text(Linear scale factor) = 20,`
`text(i.e.)\ \ 1\ text(cm):20\ text(m)`
`=> B`
In parallelogram `PQRS`, `/_ QPS = 74°.`
In this parallelogram, `PQ = 18\ text(cm)` and `PS = 25\ text(cm.)`
The length of the longer diagonal of this parallelogram is closest to
A. `26.5\ text(cm)`
B. `30.1\ text(cm)`
C. `30.8\ text(cm)`
D. `34.6\ text(cm)`
E. `39.9\ text(cm)`
`D`
A vertical pole, `TP`, is 4 metres tall and stands on level ground near a vertical wall.
The wall is 6 metres long and 4 metres high.
The base of the pole, `T`, is 5 metres from one end of the wall at `N` and 4 metres from the other end of the wall at `M`.
The pole falls and hits the wall.
The maximum height above ground level at which the pole could hit the wall is closest to
A. `0\ text(m)`
B. `1.5\ text(m)`
C. `2.3\ text(m)`
D. `2.7\ text(m)`
E. `3.3\ text(m)`
`C`
`text(Using cosine rule in)\ Delta MTN\ text(to find)\ theta`
`cos theta` | `= (6^2 + 4^2 − 5^2) / (2 × 6 × 4)` |
`= 0.5625` | |
`theta` | `= 55.77…^@` |
`text(In)\ Delta AMT,` | |
`sin theta` | `= x/4` |
`x` | `= 4 xx sin 55.77…^@` |
`= 3.307…` | |
`text(When the pole falls, max height) = h` |
`text(Using Pythagoras,)`
`4^2` | `= h^2 + 3.307^2` |
`h^2` | `= 4^2 − 3.307^2` |
`= 5.0625` | |
`h` | `= 2.25…\ text(m)` |
`=> C` |
There are 10 checkpoints in a 4500 metre orienteering course.
Checkpoint 1 is the start and checkpoint 10 is the finish.
The distance between successive checkpoints increases by 50 metres as each checkpoint is passed.
The distance, in metres, between checkpoint 2 and checkpoint 3 is
A. `225`
B. `275`
C. `300`
D. `350`
E. `400`
`D`
`text(9 distances exist between 10 checkpoints)`
`text(9 distances form an AP, where)`
`d = 50,\ \ \ S_9 = 4500`
`S_n` | `= n/2[2a + (n − 1)d]` |
`4500` | `= 9/2 [2a + (9 − 1) × 50]` |
`4500` | `= 9/2[2a + 400]` |
`9a` | `= 2700` |
`a` | `= 300` |
`:.\ text(Distance between checkpoint 1 and 2)`
`= a=300\ text(m)`
`:.\ text(Distance between checkpoint 2 and 3)`
`= a+d=350\ text(m)`
`=> D`
The middle section of a cone is shaded, as shown in the diagram below.
The surface area of the unshaded top section of the cone is 180 cm².
The surface area of the middle section of the cone, in square centimetres, is
A. `80`
B. `120`
C. `300`
D. `320`
E. `500`
`D`
`text(Ratio of lengths of)`
`text(Upper Cone : Top Section, is)`
`15:9 = 5:3`
`∴\ text(Ratio of Areas)` | `=5^2:3^2` |
`=25:9` |
`text(Let Area of Upper Cone)\ = x`
`x/180` | `= 25/9` |
`x` | `= 180 xx 25/9` |
`= 500\ \ text(cm²)` |
`∴\ text(Area of shaded middle section)`
`= 500 – 180`
`= 320\ \ text(cm²)`
`=> D`
The solid `OPQR`, as shown below, is one-eighth of a sphere of radius 15 cm.
The point `O` is the centre of the sphere and the points `P, Q` and `R` are on the surface of the sphere.
`∠POQ = ∠QOR = ∠ROP = 90°`
The total surface area of the solid `OPQR`, in cm², is closest to
A. 619
B. 648
C. 706
D. 884
E. 1767
`D`
`text (Surface Area of sphere) = 4pir^2`
`:.\ text (S.A. of)\ \ PQR\ \ text{(on surface)}`
`=1/8 xx 4 pi r^2`
`= 1/8 xx 4 xx pi xx 15^2`
`= 353.43\ text(cm²)`
`text (Surface Area of sector)\ POQ` |
`= 1/4pir^2` |
`= 1/4pi × 15^2` |
`= 176.71\ text(cm²)` |
`:. text (Surface Area of solid)\ OPQR` |
`= 353.43 + 3 × 176.71` |
`= 883.56\ text(cm²)` |
`rArr D`
Chris started to make this pattern of shapes using matchsticks.
If the pattern of shapes is continued, which shape would use exactly 486 matchsticks?
(A) Shape 96
(B) Shape 97
(C) Shape 121
(D) Shape 122
`C`
The seasonal index for heaters in winter is 1.25.
To correct for seasonality, the actual heater sales in winter should be
A. reduced by 20%.
B. increased by 20%.
C. reduced by 25%.
D. increased by 25%.
E. reduced by 75%.
`A`
`text(Deseasonalised Sales)`
`=\ text(Actual Sales)/text(Seasonal index)`
`=\ text(Actual Sales)/1.25`
`=80 text(%) xx text(Actual Sales)`
`∴\ text(Winter Sales should be reduced by 20%)`
`=> A`
The seasonal index for headache tablet sales in summer is 0.80.
To correct for seasonality, the headache tablet sales figures for summer should be
A. reduced by 80%
B. reduced by 25%
C. reduced by 20%
D. increased by 20%
E. increased by 25%
`E`
`text(Deseasonalised Sales)`
`= \ \ text(Actual Sales) / text(Seasonal index)` |
`= \ \ text(Actual Sales) / 0.8` |
`= \ \ 1.25 xx text(Actual Sales)` |
`:.\ text(Deseasonalised Sales need Actual Sales to)`
`text(be increased by 25%)`
`=> E`
The `n`th term in a geometric sequence is `t_n`.
The common ratio is greater than one.
A graph that could be used to display the terms of this sequence is
`A`
`text(GP where)\ \ r>1`
`text(General term is)\ \ \ t_n=ar^(n – 1)`
`text(Consider)\ \a>0,`
`text(the graph of)\ \t_n\ \text(will be positive and)`
`text(increase exponentially, but no graph shows this.)`
`text(Consider)\ \a<0,`
`text(the graph will be negative and increase)`
`text(exponentially in a negative direction.)`
`=> A`
A single back-to-back stem plot would be an appropriate graphical tool to investigate the association between a car’s speed, in kilometres per hour, and the
A. driver’s age, in years.
B. car’s colour (white, red, grey, other).
C. car’s fuel consumption, in kilometres per litre.
D. average distance travelled, in kilometres.
E. driver’s sex (female, male).
`E`
`text(In a back-to-back stem plot, the numerical speed)`
`text(data has to be plotted against categorical data)`
`text(with two options.)`
`∴\ text{Driver’s sex (M or F)}`
`=> E`
The following table shows the data collected from a sample of seven drivers who entered a supermarket car park. The variables in the table are:
distance – the distance that each driver travelled to the supermarket from their home
Part 1
The mean, `barx`, and the standard deviation, `s_x`, of the variable, distance, are closest to
A. `barx = 2.5\ \ \ \ \ \ \s_x = 3.3`
B. `barx = 2.8\ \ \ \ \ \ \s_x = 1.7`
C. `barx = 2.8\ \ \ \ \ \ \s_x = 1.8`
D. `barx = 2.9\ \ \ \ \ \ \s_x = 1.7`
E. `barx = 3.3\ \ \ \ \ \ \s_x = 2.5`
Part 2
The number of categorical variables in this data set is
A. `0`
B. `1`
C. `2`
D. `3`
E. `4`
Part 3
The number of female drivers with three children in the car is
A. `0`
B. `1`
C. `2`
D. `3`
E. `4`
`text(Part 1:) \ C`
`text(Part 2:)\ D`
`text(Part 3:)\ B`
`text(Part 1)`
`text(By calculator)`
`text(Distance)\ \ barx` | `=2.8` |
`s_x` | `≈1.822` |
`=>C`
`text(Part 2)`
`text(Categorical variables are sex, type)`
`text(of car, and post code.)`
`=>D`
`text(Part 3)`
`text(1 female driver has 3 children.)`
`=>B`
A tunnel is excavated with a cross-section as shown.
(i) | `A` | `~~ h/3 [y_0 + 4y_1 + y_2] + h/3 [y_0 + 4y_1 + y_2]` |
`~~ h/3 [0 + 4a + b] + h/3 [b + 4a + 0]` | ||
`~~ (2h)/3 (4a + b)` |
(ii) | `text(Given)\ \ A = 600\ text(m²)` |
`text(If 80 m wide) \ => h = 20` |
`A` | `= (2h)/3 (4a + b)` |
`600` | `= ((2 xx 20))/3 (4a + b)` |
`4a + b` | `= (600 xx 3)/40` |
`b` | `= 45 – 4a` |
`:.\ text(If)\ a\ text(increases by 2 m,)\ b\ text(will)`
`text(decrease by 8 m.)`
The following graph indicates `z`-scores of ‘height-for-age’ for girls aged 5 – 19 years.
--- 2 WORK AREA LINES (style=lined) ---
(1) If 2.5% of girls of the same age are taller than Rachel, how tall is she? (1 mark)
--- 1 WORK AREA LINES (style=lined) ---
(2) Rachel does not grow any taller. At age 15 ½, what percentage of girls of the same age will be taller than Rachel? (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
--- 1 WORK AREA LINES (style=lined) ---
For adults (18 years and older), the Body Mass Index is given by
`B = m/h^2` where `m = text(mass)` in kilograms and `h = text(height)` in metres.
The medically accepted healthy range for `B` is `21 <= B <= 25`.
--- 4 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
(2) Give ONE reason why this equation is not suitable for predicting heights of girls older than 12. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---
(2) `text(84%)`
`text(grow, on average, 6cm per year)`
(2) `text(Girls eventually stop growing, and the)`
`text(equation doesn’t factor this in.)`
i. | `z text(-score) = 1` |
ii. | (1) | `text(If 2 ½ % are taller than Rachel)` |
`=> z text(-score of +2)` | ||
`:.\ text(She is 155 cm)` | ||
(2) | `text(At age)\ 15\ ½,\ 155\ text(cm has a)\ z text(-score of –1)` | |
`text(68% between)\ z = 1\ text(and)\ –1` | ||
`=> text(34% between)\ z = 0\ text(and)\ –1` | ||
`text(50% have)\ z >= 0` | ||
`:.\ text(% Above)\ z text(-score of –1)` | ||
`= 50 + 34` | ||
`= 8text(4%)` |
`:.\ text(84% of girls would be taller than Rachel at age)\ 15 ½.`
iii. | `text(Average height of 18 year old has)\ z text(-score = 0)` |
`:.\ text(Average height) = 163\ text(cm)` |
iv. | `B = m/h^2` |
`h = 163\ text(cm) = 1.63\ text(m)` |
`text(Given)\ \ 21 <= B <= 25,\ text(minimum healthy)`
`text(weight occurs when)\ B = 21`
`=> 21` | `= m/1.63^2` |
`m` | `= 21 xx 1.63^2` |
`= 55.794…` | |
`= 55.8\ text(kg)\ text{(1 d.p.)}` |
v. | (1) | `text(It indicates that 6-11 year old girls, on average, grow)` |
`text(6 cm per year.)` | ||
(2) | `text(Girls eventually stop growing, and the equation doesn’t)` | |
`text(factor this in.)` |
Pieces of cheese are cut from cylindrical blocks with dimensions as shown.
Twelve pieces are packed in a rectangular box. There are three rows with four pieces of cheese in each row. The curved surface is face down with the pieces touching as shown.
--- 8 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
i. `text(Box height) = 15\ text(cm)`
`text{(radius of the arc)}`
`text(Box width)` | `= 3 xx 7` |
`= 21\ text(cm)` | |
`text(Box length)` | `= 4x` |
`text(Using cosine rule)`
`c^2` | `= a^2 + b^2 – 2ab cos C` |
`x^2` | `= 15^2 + 15^2 – 2 xx 15 xx 15 xx cos 40^@` |
`= 450 – 344.7199…` | |
`= 105.2800…` | |
`x` | `= 10.2606…` |
`text(Box length)` | `= 4 xx 10.2606…` |
`= 41.04…` |
`:.\ text(Dimensions are)\ \ 41\ text(cm) xx 21\ text(cm) xx 15\ text(cm)`
ii. `text(Volume) = Ah`
`h = 7\ text(cm)`
`text(Find)\ A:`
`A` | `= 1/2 ab sin C` |
`= 1/2 xx 15 xx 15 xx sin 40^@` | |
`= 72.3136…` |
`:. V` | `= 72.3136… xx 7` |
`= 506.195…` | |
`= 506\ text(cm³)\ \ text{(nearest whole)}` |
Three-digit numbers are formed from five cards labelled 1, 2, 3, 4 and 5.
--- 1 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
i. `text(# Different numbers)`
`= 5 xx 4 xx 3`
`= 60`
ii. `text(The last digit must be one of the)`
`text(5 numbers, of which 3 are odd)`
`:.\ text{P(odd)} = 3/5`
iii. `text{P(even)} = 1- text{P(odd)} = 2/5`
`:.\ text(Number of even numbers)`
`= 2/5 xx 60`
`= 24`
iv. `text(The numbers that satisfy the criteria:)`
`432, 431, 421, 321`
`:.\ text{P(selection)} = 4/60 = 1/15`
The height of a particular termite mound is directly proportional to the square root of the number of termites.
The height of this mound is 35 cm when the number of termites is 2000.
What is the height of this mound, in centimetres, when there are 10 000 termites?
(A) `16`
(B) `78`
(C) `175`
(D) `875`
`B`
`text(Height)\ (h)` | `prop sqrt( text(Termites)\ (T)` |
`=>h` | `= k sqrt T` |
`text(Given)\ \ h = 35\ \ text(when)\ \ T = 2000`
`35` | `= k xx sqrt 2000` |
`k` | `= 35/ sqrt 2000` |
`= 0.7826…` |
`text(Find)\ \ h\ \ text(when)\ \ T = 10\ 000`
`h` | `= 0.7826… xx sqrt (10\ 000)` |
`= 78.26…\ text(cm)` |
`=> B`
A beam is supported at `(-b, 0)` and `(b, 0)` as shown in the diagram.
It is known that the shape formed by the beam has equation `y = f(x)`, where `f(x)` satisfies
`f^{″}(x)` | `= k (b^2-x^2),\ \ \ \ \ `(`k` is a positive constant) | |
and | `f^{′}(-b)` | `= -f'(b)`. |
--- 5 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. | `text(Show)\ \ f^{′}(x) = k (b^2x-x^3/3)` |
`f^{″}(x)` | `= k (b^2 – x^2)` |
`f^{′}(x)` | `= int k (b^2-x^2)\ dx` |
`= k int b^2-x^2\ dx` | |
`= k (b^2x-x^3/3) + c` |
`text(S)text(ince S.P. exists at)\ \ x = 0`
`=> f^{′}(x)` | `= 0\ \ text(when)\ \ x = 0` |
`0` | `= k (b^2 * 0-0) + c` |
`c` | `= 0` |
`:.\ f^{′}(x) = k (b^2x-x^3/3)\ \ \ text(… as required)`
ii. | `f(x)` | `= int f^{′}(x)\ dx` |
`= k int b^2x-x^3/3\ dx` | ||
`= k ((b^2x^2)/2-x^4/12) + c` |
`text(We know)\ \ f(x) = 0\ \ text(when)\ \ x = b`
`=> 0` | `= k ( (b^2*b^2)/2-b^4/12) + c` |
`c` | `= -k ( (6b^4)/12-b^4/12)` |
`= -k ( (5b^4)/12 )` | |
`= -(5kb^4)/12` |
`:.\ text(When)\ \ x = 0, text(the beam is)\ \ (5kb^4)/12\ \ text(units)`
`text(below the)\ x text(-axis.)`
Two players `A` and `B` play a game that consists of taking turns until a winner is determined. Each turn consists of spinning the arrow on a spinner once. The spinner has three sectors `P`, `Q` and `R`. The probabilities that the arrow stops in sectors `P`, `Q` and `R` are `p`, `q` and `r` respectively.
The rules of the game are as follows:
• If the arrow stops in sector `P`, then the player having the turn wins.
• If the arrow stops in sector `Q`, then the player having the turn loses and the other player wins.
• If the arrow stops in sector `R`, then the other player takes a turn.
Player `A` takes the first turn.
--- 6 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
i. `text(Show)\ P text{(}A\ text(wins)\ T_1\ text(or)\ T_2 text{)} = (1 – r)(p + r)`
`P (A\ text(wins on)\ T_1) = p`
`P (A\ text(wins on)\ T_2)`
`= P text{(} A\ text(lands on)\ R,\ B\ text(lands on)\ Q text{)}`
`= rq`
`:.\ P text{(}A\ text(wins)\ T_1\ text(or)\ T_2 text{)}`
`= p + rq`
`text(S)text(ince)\ q = 1 – (p + r),`
`P text{(} A\ text(wins)\ T_1\ text(or)\ T_2 text{)}`
`= p + r [1 – (p + r)]`
`= p + r – rp – r^2`
`= (1 – r)(p + r)\ \ \ text(… as required.)`
ii. `text(Show)\ P text{(}A\ text(wins eventually) text{)} = (p + r)/(1 + r)`
`P text{(} A\ text(wins)\ T_1\ text(or)\ T_2 text{)} = (1 – r)(p + r)`
`P text{(} text(No result)\ T_1\ text(or)\ T_2 text{)} = r * r = r^2`
`:.\ P text{(} A\ text(wins eventually) text{)}`
`= underbrace{(1 – r)(p + r) + r^2 (1 – r)(p + r) + r^2*r^2 (1 – r)(p + r) + …}_{text(GP where)\ \ a = (1 – r)(p + r),\ \ r = r^2}`
`text(S)text(ince)\ \ 0 < r < 1 \ \ =>\ \ 0 < r^2 < 1`
`:.\ S_oo` | `= a/(1 – r)` |
`= ((1 – r)(p + r))/(1 – r^2)` | |
`= ((1 – r)(p + r))/((1 – r)(1 + r))` | |
`= (p + r)/(1 + r)\ \ \ text(… as required)` |
Consider the billboard below. There is a unique circle that passes through the top and bottom of the billboard (points `Q` and `R` respectively) and is tangent to the street at `T`.
Let `phi` be the angle subtended by the billboard at `S`, the point where `PQ` intersects the circle.
Copy the diagram into your writing booklet.
(i) |
`text(Show)\ \ theta < phi\ \ text(when)\ \ P != T`
`text(Let)\ \ /_PRS = alpha`
`phi` | `= theta + alpha\ \ text{(exterior angle of}\ Delta PSR text{)}` |
`:.\ theta < phi,\ alpha != 0`
`text(When)\ P\ text(and)\ T\ text(are the same point,)`
`S,\ P, and T\ \ text(are the same point.)`
`text(i.e. when)\ \ /_PRS = alpha = 0`
`:. theta\ text(will be a maximum)\ (= phi)`
`text(when)\ P = T.`
(ii) `text(Square of tangent)\ =\ text(product of)`
`text(intercepts from an external point.)`
`text(i.e.)\ \ OT^2` | `= OR xx QR` |
`= h (a + h)` | |
`:.\ OT` | `= sqrt (h (a + h))` |
A billboard of height `a` metres is mounted on the side of a building, with its bottom edge `h` metres above street level. The billboard subtends an angle `theta` at the point `P`, `x` metres from the building.
(i) `text(Consider angles)\ \ A and B\ \ text(on the graph:)`
`text(Show)\ \ theta = tan^(-1) [(ax)/(x^2 + h(a + h))]`
`tan A` | `= (a + h)/x` |
`tan B` | `= h/x` |
`tan (A – B)` | `= ((a + h)/x – h/x)/(1 + ((a + h)/x)(h/x)) xx (x^2)/(x^2)` |
`= (x(a + h) – xh)/(x^2 + h(a + h))` | |
`= (ax)/(x^2 + h(a + h)` |
`text(S)text(ince)\ \ theta` | `= A – B` |
`theta` | `= tan^(-1) [(ax)/(x^2 + h(a + h))]\ \ \ text(… as required.)` |
(ii) `text(Max when)\ \ (d theta)/(dx) = 0\ \ text(and)\ \ x > 0`
`text(Let)\ \ u` | `= (ax)/(x^2 + h(a + h))` |
`theta` | `= tan^(-1) u` |
`(du)/(dx)` | `=(a[x^2 + h (a + h)] – ax * 2x)/([x^2 + h (a + h)]^2)` |
`=(-ax^2 + ah (a + h))/([x^2 + h (a + h)]^2)` |
`(d theta)/(dx)` | `=(d theta)/(du) * (du)/(dx)` |
`= 1/(1 + u^2) * (du)/(dx)` | |
`= 1/(1 + [(ax)/(x^2 + h(a + h))]^2) xx (-ax^2 + ah (a + h))/[x^2 + h (a + h)]^2` | |
`=(-ax^2 + ah(a + h))/([x^2 + h (a + h)]^2 + a^2 x^2)` |
`text(Note that)\ \ (d theta)/(dx) = 0\ \ text(when)`
`-ax^2 + ah (a + h)` | `= 0` |
`ax^2` | `= ah (a + h)` |
`x^2` | `= h (a + h)` |
`x` | `= sqrt (h (a + h))\ \ \ \ (x > 0)` |
(i) `(1 + x)^r + (1 + x)^(r + 1) + … + (1 + x)^n`
`=> text(GP where)\ \ a` | `= (1 + x)^r` |
`r` | `= (1 + x)` |
`#\ text(Terms)` | `= n – r + 1` |
`text(Sum)` | `= (a (r^n – 1))/(r – 1)` |
`= ((1 + x)^r [(1 + x)^(n\ – r + 1) – 1])/((1 + x) – 1)` | |
`= ((1 + x)^(n\ – r + 1 + r) – (1 + x)^r)/x` | |
`= ((1 + x)^(n + 1) – (1 + x)^r)/x` |
`text(Show)`
`((r),(r)) + ((r + 1),(r)) + … + ((n),(r)) = ((n + 1),(r + 1))`
`text(Consider series)\ (1 + x)^r + (1 + x)^(r + 1) + … + (1 + x)^n`
`text(Co-efficient of)\ \ x^r = ((r),(r)) + ((r + 1),(r)) + … + ((n),(r))`
`text(Consider sum of series)\ \ \ ((1 + x)^(n + 1)\ – (1 + x)^r)/x`
`text(Co-efficient of)\ \ x^r` | `=\ text(co-efficient of)\ \ x^(r + 1)\ \ text(in numerator)` |
`= ((n +1),(r + 1))` |
`:.\ text(S)text(ince co-efficients are equal)`
`((r),(r)) + ((r + 1),(r)) + … + ((n),(r)) = ((n + 1),(r + 1))`
`text(… as required.)`
(ii)(1) | `text(All intervals start and finish at different points.)` |
`text(Any)\ n xx x\ text(grid has)\ n\ text(points on the)\ y = x\ text(diagonal.)` | |
`:.\ text(Possible intervals) = ((n),(2))` | |
(2) | `text(The longest diagonal of)\ n xx n\ text(grid is)` |
`text(the)\ y = x\ text(diagonal with)\ ((n),(2))\ text(intervals.)` | |
`text(Each side of this, there is one less point)` | |
`text(on the diagonals, with)\ ((n – 1),(2))\ text(intervals.)` | |
`text(This continues until there are only 2 points)` | |
`text(on the diagonal with)\ ((2),(2))\ text(intervals.)` |
`:.\ S_n` | `= ((2),(2)) + ((3),(2)) + … + ((n – 1),(2)) + ((n),(2))` |
`+ ((n – 1),(2)) + … + ((2),(2))` | |
`text(… as required.)` |
(iii) | `text(Show)\ \ S_n = (n(n – 1)(2n – 1))/6` |
`S_n` | `= { ((2),(2)) + ((3),(2)) + … + ((n – 1),(2)) + ((n),(2))}` |
`+ {((n – 1),(2)) + … + ((2),(2))}` | |
`= ((n + 1),(3)) + ((n),(3))\ \ \ text{(from part (i))}` | |
`= ((n + 1)*n*(n – 1))/(3 * 2 * 1) + (n* (n – 1)*(n – 2))/(3 * 2 * 1)` | |
`= 1/6 n (n – 1) (n + 1 + n – 2)` | |
`= (n(n – 1)(2n – 1))/6\ \ \ text(… as required.)` |
The cross-section of a 10 metre long tank is an isosceles triangle, as shown in the diagram. The top of the tank is horizontal.
When the tank is full, the depth of water is 3 m. The depth of water at time `t` days is `h` metres.
--- 6 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
Find the rate at which the depth of water is changing at time `t`. (2 marks)
--- 6 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
(i) | |
`text(Let)\ A = text(area of front)` |
`tan 30^@` | `= h/x` |
`x` | `= h/(tan 30^@)` |
`= sqrt3 h` |
`:.\ A` | `= 2 xx 1/2 xx sqrt 3 h xx h` |
`= sqrt 3 h^2\ \ text(m²)` |
`V` | `= Ah` |
`= sqrt 3 h^2 xx 10` | |
`= 10 sqrt3 h^2\ \ text(m³)` |
(ii) | `text(Area of surface)` |
`= 10 xx 2 sqrt 3 h` | |
`= 20 sqrt 3 h\ \ text(m²)` |
(iii) | `(dV)/(dt)` | `= -kA` |
`= -k\ 20 sqrt3 h` |
`V` | `= 10 sqrt3 h^2` |
`(dV)/(dh)` | `= 20 sqrt 3 h` |
`text(Find)\ (dh)/(dt)`
`(dV)/(dt)` | `= (dV)/(dh) * (dh)/(dt)` |
`(dh)/(dt)` | `= ((dV)/(dt))/((dV)/(dh))` |
`= (-k * 20 sqrt 3 h)/(20 sqrt 3 h)` | |
`= -k` |
`:.\ text(The water depth is changing at a rate)`
`text(of)\ -k\ text(metres per day.)`
(iv) | `text(S)text(ince)\ \ (dh)/(dt)\ \ text(is a constant, each metre)` |
`text(takes the same time.)` |
`:.\ text(It takes 100 days to fall from 2 m to 1 m.)`
In `Delta DEF`, a point `S` is chosen on the side `DE`. The length of `DS` is `x`, and the length of `ES` is `y`. The line through `S` parallel to `DF` meets `EF` at `Q`. The line through `S` parallel to `EF` meets `DF` at `R`.
The area of `Delta DEF` is `A`. The areas of `Delta DSR` and `Delta SEQ` are `A_1` and `A_2` respectively.
(i) `text(Need to show)\ Delta DEF\ text(|||) \ Delta DSR`
`/_FDE\ text(is common)`
`/_DSR = /_DEF = theta\ \ text{(corresponding angles,}\ RS\ text(||)\ FE text{)}`
`:.\ Delta DEF \ text(|||) \ Delta DSR\ \ text{(equiangular)}`
(ii) `(DR)/(DF) = (DS)/(DE) = x/(x + y)`
`text{(Corresponding sides of similar triangles)}`
(iii) `text(Show)\ sqrt((A_1)/A) = x/(x + y)`
`text(Using Area)` | `= 1/2 ab sin C` |
`A_1` | `= 1/2 xx DR xx x xx sin alpha` |
`A` | `= 1/2 xx DF xx (x + y) xx sin alpha` |
`(A_1)/A` | `= (1/2 * DR * x * sin alpha)/(1/2 * DF * (x + y) * sin alpha)` |
`= (DR * x)/(DF * (x + y)` | |
`= (x * x)/((x + y)(x + y))\ \ \ \ text{(using part(ii))}` | |
`= (x^2)/((x + y)^2)` | |
`:.\ sqrt ((A_1)/A)` | `= x/((x + y))\ \ \ text(… as required.)` |
(iv) `text(Consider)\ Delta DFE\ text(and)\ Delta SQE`
`/_FED` | `= theta\ text(is common)` |
`/_FDE` | `= /_QSE = alpha\ \ ` | `text{(corresponding angles,}\ DF\ text(||)\ QS text{)}` |
`:.\ Delta DFE\ text(|||)\ Delta SQE\ \ text{(equiangular)}`
`(QE)/(FE) = (SE)/(DE) = y/(x +y)`
`text{(corresponding sides of similar triangles)}`
`A_2` | `= 1/2 xx QE xx y xx sin theta` |
`A` | `= 1/2 xx FE xx (x + y) xx sin theta` |
`(A_2)/A` | `= (QE * y)/(FE * (x + y))` |
`= (y^2)/((x + y)^2)` | |
`sqrt ((A_2)/A)` | `= y/((x + y))` |
`text(Need to show)\ sqrt A = sqrt (A_1) + sqrt (A_2)`
`sqrt(A_2)/sqrtA` | `= y/((x + y))` |
`sqrt (A_2)` | `= (sqrtA * y)/((x + y))` |
`text(Similarly, from part)\ text{(iii)}`
`sqrt (A_1) = (sqrtA * x)/((x + y))`
`sqrt (A_1) + sqrt (A_2)` | `= (sqrt A * x)/((x + y)) + (sqrt A * y)/((x + y))` |
`= (sqrt A (x + y))/((x + y))` | |
`= sqrt A\ \ \ text(… as required.)` |
How many solutions of the equation `(sin x - 1)(tan x + 2) = 0` lie between `0` and `2 pi`?
(A) `1`
(B) `2`
(C) `3`
(D) `4`
`B`
`text(When)\ (sin x\ – 1)(tan x + 2) = 0`
`(sinx\ – 1) = 0\ \ text(or)\ \ tan x + 2 = 0`
`text(If)\ \ sin x\ – 1` | `= 0` |
`sin x` | `= 1` |
`x` | `= pi/2,\ \ \ 0 < x < 2 pi` |
`text(If)\ \ tan x + 2` | `= 0` |
`tan x` | `= -2` |
`=>\ text(Note that since)\ \ tan\ pi/2\ \ text(is undefined, there)`
`text(are only 2 solutions when)\ \ tan x = -2`
`text{(which occurs in the 1st and 4th quadrants).}`
`:.\ 2\ text(solutions)`
`=> B`
The scatterplot shows the relationship between expenditure per primary school student, as a percentage of a country’s Gross Domestic Product (GDP), and the life expectancy in years for 15 countries.
--- 1 WORK AREA LINES (style=lined) ---
What is the interquartile range? (1 mark)
--- 1 WORK AREA LINES (style=lined) ---
Would this country be an outlier for this set of data? Justify your answer with calculations. (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
--- 1 WORK AREA LINES (style=lined) ---
--- 1 WORK AREA LINES (style=lined) ---
`text(correlation between the two variables.)`
`text(of 127.3. This line of best fit is only accurate)`
`text(in a lower range of GDP expediture.)`
i. | `text(It indicates there is a strong positive)` |
`text(correlation between the two variables)` |
ii. | `text(IQR)` | `= Q_U\ – Q_L` |
`= 22.5\ – 8.4` | ||
`= 14.1` |
iii. `text(An outlier on the upper side must be more than)`
`Q_u\ +1.5xxIQR`
`=22.5+(1.5xx14.1)`
`=\ text(43.65%)`
`:.\ text(A country with an expenditure of 47.6% is an outlier).`
iv. |
v. `text(Life expectancy) ~~ 73.1\ text{years (see dotted line)}`
`text(Alternative Solution)`
`text(When)\ x=18`
`y=1.29(18)+49.9=73.12\ \ text(years)`
vi. | `text(At 60% GDP, the line predicts a life)` |
`text(expectancy of 127.3. This line of best)` | |
`text(fit is only predictive in a lower range)` | |
`text(of GDP expenditure.)` |
The cost of hiring an open space for a music festival is $120 000. The cost will be shared equally by the people attending the festival, so that `C` (in dollars) is the cost per person when `n` people attend the festival.
--- 1 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
i.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
ii. |
iii. `C = (120\ 000)/n`
`n\ text(must be a whole number)`
iv. `text(Limitations can include:)`
`•\ n\ text(must be a whole number)`
`•\ C > 0`
v. `text(If)\ C = 94:`
`94` | `= (120\ 000)/n` |
`94n` | `= 120\ 000` |
`n` | `= (120\ 000)/94` |
`= 1276.595…` |
`:.\ text(C)text(ost cannot be $94 per person,)`
`text(because)\ n\ text(isn’t a whole number.)`
i.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
ii. |
iii. `C = (120\ 000)/n`
iv. `text(Limitations can include:)`
`•\ n\ text(must be a whole number)`
`•\ C > 0`
v. `text(If)\ C = 94`
`=> 94` | `= (120\ 000)/n` |
`94n` | `= 120\ 000` |
`n` | `= (120\ 000)/94` |
`= 1276.595…` |
`:.\ text(C)text(ost cannot be $94 per person,)`
`text(because)\ n\ text(isn’t a whole number.)`
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:
(i) | `($13\ 380)/100 = 133.8` |
`:.\ text(Stamp duty)` | `= 134 xx $3` |
`= $402` |
(ii) | `text(Total loan)` | `= $13\ 380 + 30 + 402` |
`= $13\ 812` |
`text(Total interest)\ (I)` | `= Prn` |
`= 13\ 812 xx 7.5/100 xx 3` | |
`= 3107.70` |
`text(Total to repay)` | `= 3107.70 + 13\ 812` |
`= 16\ 919.70` |
`text(# Repayments) = 3 xx 12 = 36`
`:.\ text(Monthly repayment)` | `= (16\ 919.70)/36` |
`= 469.9916…` | |
`= $470\ text{(nearest dollar)}` |
(iii) | `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` |
(iv) | `text(Comprehensive insurance covers Alex)` |
`text(for damage done to his own car as well.)` |
The times taken for 160 music downloads were recorded, grouped into classes and then displayed using the cumulative frequency histogram shown.
On the diagram, draw the lines that are needed to find the median download time. (2 marks)
--- 0 WORK AREA LINES (style=lined) ---
In Mathsville, there are on average eight rainy days in October.
Which expression could be used to find a value for the probability that it will rain on two consecutive days in October in Mathsville?
`D`
`P text{(rains)} = 8/31\ \ \text{(independent event for each day)}`
`text{Since each day has same probability:}`
`P(R_1 R_2) = 8/31 xx 8/31`
`=> D`
How many kilobytes are there in 2 gigabytes?
(A) `2^20`
(B) `2^21`
(C) `2^30`
(D) `2^31`
`B`
`2\ text(GB)` | `= 2 xx 2^20\ text(kB)` |
`= 2^21\ text(kB)` |
`=> B`
The point `P(2at, at^2)` lies on the parabola `x^2 = 4ay` with focus `S`.
The point `Q` divides the interval `PS` internally in the ratio `t^2 :1`.
(i) `text(Show)\ \ Q = ((2at)/(1 + t^2), (2at^2)/(1 + t^2))`
`P (2at, at^2),\ S (0, a)`
`PS\ text(is divided internally in ratio)\ t^2: 1`
`Q` | `= ((nx_1 + mx_2)/(m + n), (ny_1 + my_2)/(m + n))` |
`= ((1(2at) + t^2(0))/(t^2 + 1), (1(at^2) + t^2 (a))/(t^2 + 1))` | |
`= ((2at)/(1 + t^2), (2at^2)/(1 + t^2))\ \ text(… as required.)` |
(ii) | `m_(OQ)` | `= (y_2\ – y_1)/(x_2\ – x_1)` |
`= ((2at^2)/(1 + t^2))/((2at)/(1 + t^2)) xx (1+t^2)/(1+t^2)` | ||
`= (2at^2)/(2at)` | ||
`= t` |
(iii) | `text(Show)\ Q\ text(lies on a fixed circle radius)\ a` |
`text(S)text(ince)\ Q\ text(passes through)\ (0, 0)` |
`=>\ text(If locus of)\ Q\ text(is a circle, it has)`
`text(diameter)\ QT\ text(where)\ T(0, 2a)`
`text(Show)\ \ QT _|_ OQ`
`text{(} text(angles on circum. subtended by)`
`text(a diameter are)\ 90^@ text{)}`
`m_(OQ) = t\ \ \ \ text{(see part (ii))}`
`text(Find)\ m_(QT),\ \ text(where:)`
`Q((2at)/(1 + t^2), (2at^2)/(1 + t^2)),\ \ \ \ \ T (0,2a)`
`m_(QT)` | `= (y_2\ – y_1)/(x_2\ – x_1)` |
`= ((2at^2)/(1 + t^2)\ – 2a)/((2at)/(1 + t^2)\ – 0)` | |
`= (2at^2\ – 2a (1 + t^2))/(2at)` | |
`= – (2a)/(2at)` | |
`= – 1/t` |
`m_(QT) xx m_(OT) = -1/t xx t = -1`
`=> QT _|_ OQ`
`=>O,\ T,\ Q\ text(lie on a circle.)`
`:.\ text(Locus of)\ Q\ text(is a fixed circle,)`
`text(centre)\ (0, a),\ text(radius)\ a`
The binomial theorem states that
`(1 + x)^n = sum_(r = 0)^(n) ((n),(r)) x^r` for all integers `n >= 1`.
(i) `text(Show)\ sum_(r=1)^n ((n),(r)) rx^r = nx (1 + x)^(n\ – 1)`
`text(Using)\ (1 + x)^n = sum_(r=0)^n ((n),(r)) x^r :`
`(1 + x)^n = ((n),(0)) + ((n),(1)) x + ((n),(2)) x^2 + … + ((n),(n)) x^n`
`text(Differentiate both sides)`
`n(1 + x)^(n\ – 1) = ((n),(1)) + ((n),(2)) 2x + … + ((n),(n)) nx^(n\ – 1)`
`text(Multiply both sides by)\ x`
`nx (1 + x)^(n\ – 1)` | `= ((n),(1)) x + ((n),(2)) 2x^2 + … + ((n),(n)) nx^n` |
`= sum_(r=1)^n ((n),(r)) rx^r\ \ text(… as required.)` |
(ii) `text(Show)\ \ sum_(r=1)^n ((n),(r)) r^2 = n (n+1) 2^(n\ – 2)`
`text(Using part)\ text{(i)}`
`nx (1 + x)^(n\ – 1) = ((n),(1)) x + ((n),(2)) 2x^2 + … + ((n),(n)) nx^n`
`text(Differentiate LHS)`
`nx (n\ – 1)(1 + x)^(n\ – 2) + n(1 + x)^(n\ – 1)`
`= n (1 + x)^(n\ – 2) [x (n\ – 1) + (1 + x)]`
`= n (1 + x)^(n\ – 2) (xn\ – x + 1 + x)`
`= n (xn + 1)(1 + x)^(n\ – 2)`
`text(Differentiate RHS)`
`((n),(1)) + ((n),(2)) 2^2 x + ((n),(3)) 3^2 x^2 + … + ((n),(n)) n^2 x^(n\ – 1)\ text{… (*)}`
`text(Substitute)\ \x = 1\ text(into both sides)`
`((n),(1)) + ((n),(2)) 2^2 + … + ((n),(n)) n^2` | `= n (n + 1)(1 + 1)^(n\ – 2)` |
`sum_(r=1)^n ((n),(r)) r^2` | `= n (n + 1) 2^(n\ – 2)` |
(iii) `text(Show for)\ \ n >= 4`
`((n),(2)) 2^2 + ((n),(4)) 4^2 + ((n),(6)) 6^2 + … + ((n),(n)) n^2 = n (n + 1) 2^(n\ – 3)`
`text{From part(ii) … (*) we know}`
`((n),(1)) + ((n),(2)) 2^2 x + … + ((n),(n)) n^2 x^(n\ – 1) = n (xn + 1)(1 + x)^(n\ – 2)`
`text(When)\ \ x = 1`
`((n),(1)) + ((n),(2)) 2^2 + … + ((n),(n)) n^2 = n (n + 1) 2^(n\ – 2)\ \ \ \ \ …\ (1)`
`text(When)\ x = –1`
`((n),(1))\ – ((n),(2)) 2^2 + ((n),(3)) 3^2\ – …\ – ((n),(n)) n^2` |
`= n (-n + 1)* 0^(n\ – 2)=0\ \ \ \ \ …\ (2)` |
`text{(Note that}\ n≠2\ \ text{because}\ 0^0\ text{is undefined)}` |
`(1)\ – (2)`
`2[((n),(2)) 2^2 + ((n),(4)) 4^2 + … + ((n),(n)) n^2]` | `= n (n + 1) 2^(n\ – 2) – 0` |
`((n),(2)) 2^2 + ((n),(4)) 4^2 + … + ((n),(n)) n^2` | `= 1/2 n (n + 1) 2^(n\ – 2)` |
`= n (n + 1) 2^(n\ – 3)` |
`:.\ text(Above statement true for integers)\ n >= 4.`
The equation `e^t = 1/t` has an approximate solution `t_0 = 0.5`
(i) | `e^t` | `= 1/t` |
`e^t\ – 1/t` | `= 0` |
`text(Let)\ \ f(t)` | `= e^t\ – 1/t` |
`f prime (t)` | `= e^t + 1/(t^2)` |
`f(0.5)` | `=e^0.5-1/0.5` |
`=–0.3512…` | |
`f′(0.5)` | `=e^0.5 +1/0.5^2` |
`=5.6487…` |
`text(Applying Newton where)\ \ t_0 = 0.5`
`t_1` | `= 0.5\ – (f(0.5))/(f prime (0.5)` |
`= 0.5\ – (–0.3512…)/(5.6487…)` | |
`= 0.5\ – (- 0.062)` | |
`= 0.56\ text{(2 d.p.)}\ \ text(… as required.)` |
(ii) | `y_1` | `= e^(rx),` | `\ \ \ \ \ (dy_1)/(dx) = re^(rx)` |
`y_2` | `= log_e x,` | `\ \ \ \ \ (dy_2)/(dx) = 1/x` |
`text(S)text(ince tangent is common),\ \ (dy_1)/(dx)=(dy_2)/(dx)`
`re^(rx)` | `= 1/x` |
`e^(rx)` | `= 1/(rx)` |
`text(Using part)\ text{(i)},\ rx ~~ 0.56`
`text(At intersection of curves)\ \ y_1 = y_2`
`e^(rx)` | `= log_e x` |
`e^0.56` | `= log_e x` |
`x` | `= e^(e^0.56)` |
`= 5.758 …` |
`text(S)text(ince)\ rx` | `~~ 0.56` |
`=> r` | `~~ 0.56/(5.758 … )` |
`~~ 0.0972 …` | |
`~~ 0.097\ text{(to 3 d.p.)}` |
(i) `text(Find co-efficient of)\ \ x^(2n)`
`text(Expanding)\ \ (1+x)^(4n)`
`((4n),(0)) + ((4n),(1))x + ((4n),(2))x^2 + … + ((4n),(2n))x^(2n) + …`
`:.\ text(Co-efficient of)\ \ x^(2n)\ text(is)\ ((4n),(2n))`
(ii) `text(Show)\ (1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)`
`text(Using)\ (1 + x^2 + 2x)^(2n) = [x(x + 2) + 1]^(2n)`
`[x (x + 2) + 1]^(2n)` |
`= ((2n),(0)) (x(x + 2))^(2n) + ((2n),(1)) (x(x + 2))^(2n\ – 1) + … + ((2n),(2n))` |
`= ((2n),(0)) x^(2n)(x + 2)^(2n) + ((2n),(1)) x^(2n\ – 1) (x + 2)^(2n\ – 1) + … + ((2n),(2n))` |
`= sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)\ text(… as required.)` |
(iii) `((4n),(2n))\ text(is the co-eff of)\ x^(2n)\ text(in expansion)\ (1+x)^(4x)`
`text(S)text(ince)\ (1 + x^2 + 2x)^(2n) = ((x+1)^2)^(2n) = (1 + x)^(4n)`
`=> ((4n),(2n))\ text(is co-efficient of)\ x^(2n)\ text(in expansion)\ (1 + x^2 + 2x)^(2n)`
`text(Using part)\ text{(ii)}`
`(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k))\ x^(2n\ – k) (x + 2)^(2n\ – k)`
`text(Using the given identity,)\ x^(2n)\ text(co-efficients are)`
`k = 0,` | `\ ((2n),(0))((2n\ – 0),(0)) 2^(2n\ – 0)` |
`k = 1,` | `\ ((2n),(1))((2n\ – 1),(1)) 2^(2n\ – 1\ – 1)` |
`vdots` | |
`k = n,` | `\ ((2n),(n))((2n\ – n),(n)) 2^(2n\ – n\ – n)` |
`:.\ ((4n),(2n))` |
`= ((2n),(0))((2n),(0))2^(2n) + ((2n),(1))((2n\ – 1),(1))2^(2n\ – 2) + … + ((2n),(n))((n),(n)) 2^0` |
` = sum_(k=0)^(n)\ ((2n),(k))((2n\ – k),(k)) 2^(2n\ – 2k)\ \ \ \ text(… as required)` |
Explain why the number of possible colour combinations is `r + 1`. (1 mark)
--- 4 WORK AREA LINES (style=lined) ---
Explain why the number of different selections is `((n),(r))`. (1 mark)
--- 4 WORK AREA LINES (style=lined) ---
Using the identity, `n2^(n-1)=sum_(k=1)^n k ((n),(k)),` or otherwise, show that the number of different selections is `(n + 2)2^(n- 1)`. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
i. | `text(1 Ball, # Combinations)` | `= 2\ text{(R or B)}` |
`text(2 Balls, # Combinations)` | `= 3\ text{(BB, BR, RR)}` | |
`text(3 Balls, # Combinations)` | `= 4` |
`text{(BBB, RBB, RRB, RRR)}`
`text(i.e. # Red balls could be 0, 1, 2, 3, …)`
`:.\ text(If)\ r\ text(balls chosen, # Combinations) = r + 1`
ii. | `text(# Selections when choosing)\ (n\ – r)\ text(from)\ n` |
`= ((n),(n\ – r))`
`((n),(n\ – r))` | `= (n!)/((n\ – r)!(n\ – (n\ – r))!)` |
`= (n!)/((n\ – r)! r!` | |
`= ((n),(r))` |
iii. `text(If)\ n\ text(balls are chosen,)`
`text(Let)\ r\ text(balls be red and blue and)`
`(n\ – r)\ text(balls be white labelled.)`
`=> text{# Combinations (Red and blue)} = r + 1`
`=> text{# Combinations (White)} = ((n),(r))`
`text(Any selection of)\ \ r\ \ text(red and blue balls would)`
`text(result in)\ \ (n-r)\ \ text(white balls, with) \ r=0,1,2,…`
`:.\ text{# Selections (for any given}\ r\ text{)}`
`= (r + 1)((n),(r))`
`= r ((n),(r)) + ((n),(r))`
`:.\ text{Total Selections}` | `= sum_(r=1)^n r ((n),(r)) + sum_(r=0)^n ((n),(r))` |
`= n2^(n\ – 1) + 2^n` | |
`= n2^(n\ – 1) + 2*2^(n\ – 1)` | |
`= (n + 2)2^(n\ – 1)\ text(… as required)` |
In the diagram, `ST` is tangent to both the circles at `A`.
The points `B` and `C` are on the larger circle, and the line `BC` is tangent to the smaller circle at `D`. The line `AB` intersects the smaller circle at `X`.
Copy or trace the diagram into your writing booklet
(i) |
`/_ AXD = /_ABD + /_XDB` |
`text{(exterior angle of}\ Delta BXD text{)}` |
(ii) | `/_AXD` | `= /_TAD \ \ text{(angle in alternate segment)}` |
`= /_TAC + /_CAD\ \ text(… as required)` |
(iii) `text(Show)\ \ /_XAD = /_CAD`
`/_ABD + /_XDB` | `=/_TAC + /_CAD\ \ text{(} text(from part)\ text{(i)} text(,)\ text{(ii)} text{)}` |
`text(S)text(ince)\ /_TAC= /_ABD\ text{(angle in alternate segment)}`
`=>/_XDB` | `=/_CAD` |
`/_XDB` | `= /_XAD\ text{(angle in alternate segment)}` |
`:. /_XAD` | `= /_CAD` |
`:.\ AD\ \ text(bisects)\ /_BAC.`
The diagram shows two identical circular cones with a common vertical axis. Each cone has height `h` cm and semi-vertical angle 45°.
The lower cone is completely filled with water. The upper cone is lowered vertically into the water as shown in the diagram. The rate at which it is lowered is given by
`(dl)/(dt) = 10`,
where `l` cm is the distance the upper cone has descended into the water after `t` seconds.
As the upper cone is lowered, water spills from the lower cone. The volume of water remaining in the lower cone at time `t` is `V` cm³.
--- 4 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. `text(Show that)\ V = pi/3 (h^3\ – l^3)`
`text(S)text(ince)\ \ tan45° = r/h=1`
`=>r=h`
`=>\ text(Radius of lower cone) = h`
`:.\ V text{(lower cone)}` | `= 1/3 pi r^2 h` |
`= 1/3 pi h^3` |
`text(Similarly,)`
`V text{(submerged upper cone)} = 1/3 pi l^3`
`V text{(water left)}` | `= 1/3 pi h^3\ – 1/3 pi l^3` |
`= pi/3 (h^3\ – l^3)\ \ \ text(… as required)` |
ii. `text(Find)\ (dV)/(dt)\ text(at)\ l = 2`
`(dV)/(dt)= (dV)/(dl) xx (dl)/(dt)\ …\ text{(1)}`
`=>(dl)/(dt)` | `= 10\ text{(given)}` |
`text(Using)\ \ V` | `= pi/3 (h^3\ – l^3)\ \ \ \ text(from part)\ text{(i)}` |
`=>(dV)/(dl)` | `= -3 xx pi/3 l^2` |
`= -pi l^2` |
`text(At)\ \ l = 2,`
`text{Substitute into (1) above}`
`(dV)/(dt)` | `= -pi xx 2^2 xx 10` |
`= -40 pi\ \ \ text(cm³)//text (sec)` |
iii. `text(Find)\ \ (dV)/(dt)\ \ text(when lower cone has lost)\ 1/8 :`
`text(Find)\ \ l\ \ text(when)\ \ V = 7/8 xx 1/3 pi h^3`
`pi/3 (h^3\ – l^3)` | `= 7/8 xx 1/3 pi h^3` |
`h^3 -l^3` | `= 7/8 h^3` |
`l^3` | `= 1/8 h^3` |
`l` | `= h/2` |
`text(When)\ \ l = h/2 ,`
`(dV)/(dt)` | `= -pi (h/2)^2 xx 10\ \ …\ text{(*)}` |
`= (-5pih^2)/2\ text(cm³)// text(sec)` |
`:. V\ text(is decreasing at the rate of)\ \ (5 pi h^2)/2\ text(cm³)//text(sec).`
A game is played by throwing darts at a target. A player can choose to throw two or three darts.
Darcy plays two games. In Game 1, he chooses to throw two darts, and wins if he hits the target at least once. In Game 2, he chooses to throw three darts, and wins if he hits the target at least twice.
The probability that Darcy hits the target on any throw is `p`, where `0 < p < 1`.
--- 4 WORK AREA LINES (style=lined) ---
--- 3 WORK AREA LINES (style=lined) ---
--- 3 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. `text(Show)\ P text{(wins Game 1)} = 2p\ – p^2`
`P text{(Hits)} = P text{(H)} = p`
`P text{(Miss)} = P text{(M)} = 1 – p`
`P text{(Wins G1)}` | `= 1 – P text{(MM)}` |
`= 1 – (1 – p)^2` | |
`= 1 – 1 + 2p – p^2` | |
`= 2p – p^2\ \ \ text(… as required)` |
ii. `text(Show)\ \ P text{(wins Game 2)} = 3p – 2p^3 :`
`text(Darcy wins G2 if he hits 3 times, or twice.)`
`Ptext{(Hits 3 times)}=\ ^3C_3 (p)^3=p^3`
`Ptext{(Hits twice)}=\ ^3C_2 (p)^2 (1-p)=3p^2-3p^3`
`P text{(Wins G2)}` | `= p^3 + 3p^2-3p^3` |
`= 3p^2 – 2p^3\ \ \ text(… as required)` |
iii. `text(Prove more likely to win G1 vs G2)`
`text(i.e. Show)\ \ \ ` | `2p – p^2 > 3p^2 – 2p^3` | |
`2p^3 – 4p^2 + 2p` | `> 0` | |
`2p (p^2 – 2p + 1)` | `> 0` | |
`2p (p – 1)^2` | `> 0` |
`=> text(TRUE since)\ \ (p-1)^2>0\ \ text(and)\ \ 0 < p < 1`
`:.\ text(More likely for Darcy to win Game 1.)`
iv. `text(If twice as likely to win G1 vs G2)`
`2p\ – p^2` | `= 2(3p^2 – 2p^3)` |
`= 6p^2 – 4p^3` | |
`4p^3 – 7p^2 + 2p` | `= 0` |
`p(4p^2 – 7p + 2)` | `= 0` |
`p` | `= (–(–7) +- sqrt((–7)^2\ – 4 xx 4 xx 2))/(2 xx 4)` |
`= (7 +- sqrt(49\ – 32))/8` | |
`= (7 +- sqrt17)/8` |
`text(S)text(ince)\ \ 0<p<1,`
`p = (7\ – sqrt17)/8`
In the diagram, `Q(x_0, y_0)` is a point on the unit circle `x^2 + y^2 = 1` at an angle `theta` from the positive `x`-axis, where `− pi/2 < theta < pi/2`. The line through `N(0, 1)` and `Q` intersects the line `y = –1` at `P`. The points `T(0, y_0)` and `S(0, –1)` are on the `y`-axis.
(i) |
`text(Show)\ SP = (2 cos theta)/(1\ – sin theta)`
`Delta SPN \ text(|||) \ Delta TQN\ \ \ text{(given)}`
`(SP)/(SN) = (TQ)/(TN)\ \ \ text{(corresponding sides of similar triangles)}`
`/_TQO = theta\ \ \ text{(alternate,}\ TQ\ text(||)\ x text{-axis)}`
`sin theta` | `= (OT)/1 => OT = sin theta` | |
`=>` | `TN` | `= 1\ – sin theta` |
`cos theta` | `= (TQ)/1` | |
`=>` | `TQ` | `= cos theta` |
`SN` | `= 2\ \ \ text{(diameter of unit circle)}` |
`:. (SP)/2` | `= cos theta/(1\ – sin theta)` |
`SP` | `= (2 cos theta)/(1\ – sin theta)\ \ \ text(… as required)` |
(ii) `text(Show)\ \ costheta/(1\ – sin theta) = sec theta + tan theta`
`text(RHS)` | `= 1/(costheta) + (sintheta)/(costheta)` |
`=(1 + sin theta)/cos theta xx cos theta/cos theta` | |
`= (costheta(1 + sintheta))/(cos^2theta)` | |
`= (costheta(1 + sin theta))/(1\ – sin^2 theta)` | |
`= (costheta(1 + sin theta))/((1 + sin theta)(1\ – sin theta)` | |
`= costheta/(1\ – sin theta)\ \ \ text(… as required)` |
(iii) `text(Show)\ \ /_SNP = theta/2 + pi/4`
`/_TOQ = 90\ – theta`
`text(S)text(ince)\ ON = OQ = 1\ text{(unit circle)}`
`=> Delta ONQ\ text(is isosceles)`
`:.\ /_SNP` | `= 1/2 (180\ – (90\ – theta))\ \ \ \ text{(angle sum of}\ Delta ONQ text{)}` |
`= 90\ – 45 + theta/2` | |
`= 45 + theta/2` | |
`= pi/4 + theta/2\ \ \ text(… as required)` |
(iv) `text(Show)\ sec theta + tan theta = tan (theta/2 + pi/4)`
`text(S)text(ince)\ /_SNP = pi/4 + theta/2\ \ \ \ text{(} text(part)\ text{(iii)} text{)}`
`=> tan\ /_SNP = tan (pi/4 + theta/2)`
`text(Also,)\ tan\ /_SNP` | `= (SP)/(SN)` |
`= ((2costheta)/(1\ – sin theta))/2` | |
`= (cos theta)/(1\ – sin theta)` | |
`= sec theta + tan theta\ \ \ \ text{(part (ii))}` |
`:.\ sec theta + tan theta = tan (pi/4 + theta/2)\ \ \ text(… as required)`
(v) | `sec theta + tan theta` | `= sqrt3,\ \ \ (-pi/2 < theta < pi/2)` |
`tan (pi/4 + theta/2)` | `= sqrt3\ \ \ \ text{(part (iv))}` | |
`tan (pi/3)` | `= sqrt 3` |
`text(S)text(ince tan is positive in)\ 1^text(st) // 3^text(rd)\ text(quads)`
`theta/2 + pi/4` | `= pi/3,\ (4pi)/3` |
`theta/2` | `= pi/12\ \ \ \ (-pi/2 < theta < pi/2)` |
`:.\ theta` | `= pi/6\ text(radians)` |
In the diagram, the vertices of `Delta ABC` lie on the circle with centre `O`. The point `D` lies on `BC` such that `Delta ABD` is isosceles and `/_ABC = x`.
Copy or trace the diagram into your writing booklet.
(i) |
`/_AOC = 2x`
`text{(angles at circumference and}`
`text{centre on arc}\ AC text{)}`
(ii) `text(Prove)\ ACDO\ text(is cyclic)`
`text(S)text(ince)\ Delta ADB\ text(is isosceles)`
`/_DAB = x\ \ \ text{(opposite equal sides in}\ Delta DBA text{)}`
`=> /_ADB` | `= 180\ – 2x\ \ \ text{(angle sum of}\ Delta DAB text{)}` |
`=> /_CDA` | `= 2x\ \ \ text{(}CDB\ text{is a straight angle)}` |
`text(S)text(ince chord)\ AC\ text(subtends)\ /_CDA = 2x`
`text(and)\ /_COA = 2x,`
`:.\ text(Quadrilateral)\ ACDO\ text(must be cyclic.)`
(iii) |
`text(Need to show)\ OM\ text(passes through)\ P,\ text(centre)`
`text(of circle through)\ ACDO.`
`AM` | `= CM\ text{(} M\ text(is midpoint) text{)}` |
`OC` | `= OA\ text{(radii)}` |
`OM\ text(is common)`
`:.\ Delta OAM -= Delta OCM\ \ \ text{(SSS)}`
`:. /_CMO = /_AMO\ \ \ ` | `text{(corresponding angles of}` |
`\ \ text{congruent triangles)}` |
`text(S)text(ince)\ ∠AMC\ text(is straight angle)`
`/_CMO = /_AMO = 90°`
`:.OM\ text(is perpendicular bisector)`
`text(of chord)\ AC.`
`:. OM\ text(passes through)\ P.`
`:.\ P, M,\ text(and)\ O\ text(are collinear.)`
The equation of motion for a particle undergoing simple harmonic motion is
`(d^2x)/(dt^2) = -n^2 x`,
where `x` is the displacement of the particle from the origin at time `t`, and `n` is a positive constant.
--- 5 WORK AREA LINES (style=lined) ---
Find the values of `A` and `B` in the solution `x = A cos nt + B sin nt`. (2 marks)
--- 6 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. | `x` | `= A cos nt + B sin nt` |
`(dx)/(dt)` | `= – An sin nt + Bn cos nt` | |
`(d^2x)/(dt^2)` | `= – An^2 cos nt\ – Bn^2 sin nt` | |
`= -n^2 (A cos nt + B sin nt)` | ||
`= -n^2 x\ \ \ text(… as required)` |
ii. `text(At)\ \ t=0, \ x=0, \ v=2n:`
`x` | `= Acosnt + Bsinnt` |
`0` | `= A cos 0 + B sin 0` |
`:.A` | `= 0` |
`text(Using)\ \ (dx)/(dt) = Bn cos nt`
`2n` | `= Bn cos 0` |
`Bn` | `= 2n` |
`:.B` | `= 2` |
iii. `text(Max distance from origin when)\ (dx)/(dt) = 0`
`(dx)/(dt)` | `= 2n cos nt` |
`0` | `= 2n cos nt` |
`cos nt` | `= 0` |
`nt` | `= pi/2,\ (3pi)/2,\ (5pi)/2` |
`t` | `= pi/(2n),\ (3pi)/(2n), …` |
`:.\ text(Particle is first at greatest distance)`
`text(from)\ O\ text(when)\ t = pi/(2n).`
iv. `text(Solution 1)`
`text(Find the distance travelled from)\ \ t=0\ \→\ \ t=(2pi)/n`
`text{(i.e. 1 full period)}`
`text(S)text(ince)\ \ x=2 sin (nt)`
`=> text(Amplitude)=2`
`:.\ text(Distance travelled)=4 xx2=8\ text(units)`
`text(Solution 2)`
`text(At)\ t = 0,\ x = 0`
`text(At)\ t` | `= pi/(2n)` |
`x` | `= 2 sin (n xx pi/(2n)) = 2` |
`text(At)\ t` | `= (3pi)/(2n)\ \ \ text{(i.e. the next time}\ \ (dx)/(dt) = 0 text{)}` |
`x` | `= 2 sin (n xx (3pi)/(2n)) = -2` |
`text(At)\ t` | `= (2pi)/n` |
`x` | `= 2 sin (n xx (2pi)/n) = 0` |
`:.\ text(Total distance travelled)`
`= 2 + 4 + 2` |
`= 8\ \ text(units)` |
A plane `P` takes off from a point `B`. It flies due north at a constant angle `alpha` to the horizontal. An observer is located at `A`, 1 km from `B`, at a bearing 060° from `B`.
Let `u` km be the distance from `B` to the plane and let `r` km be the distance from the observer to the plane. The point `G` is on the ground directly below the plane.
--- 10 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. `text(Show)\ r = sqrt(1 + u^2 – u cos alpha)`
`text(In)\ Delta PGB:`
`cos alpha` | `= (BG)/u` |
`BG` | `= u cos alpha` |
`sin alpha` | `= (PG)/(u)` |
`PG` | `= u sin alpha` |
`text(In)\ Delta PGA,\ text(using Pythagoras):`
`AG^2` | `= r^2\ – PG^2` |
`= r^2\ – u^2 sin^2 alpha` |
`text(Using cosine rule in)\ Delta ABG:`
`AG^2` | `= BG^2 + AB^2\ – 2 xx BG xx AB xx cos 60^@` |
`r^2\ – u^2 sin^2 alpha` | `= u^2 cos^2 alpha + 1\ – 2 (u cos alpha) xx 1 xx 1/2` |
`r^2` | `= u^2 cos^2 alpha + u^2 sin^2 alpha + 1\ – u cos alpha` |
`= u^2 (cos^2 alpha + sin^2 alpha) + 1\ – u cos alpha` | |
`= u^2 + 1\ – u cos alpha` | |
`r` | `= sqrt(u^2 + 1\ – u cos alpha)\ \ text(… as required)` |
ii. `text(Find)\ \ (dr)/(dt)\ text(when)\ t =5`
`(dr)/(dt) = (dr)/(du) xx (du)/(dt)`
`r` | `= (u^2 + 1\ – u cos alpha)^(1/2)` |
`(dr)/(du)` | `= 1/2 (u^2 + 1\ – u cos alpha)^(-1/2) xx (2u\ – cos alpha)` |
`= (2u\ – cos alpha)/(2 sqrt(u^2 + 1\ – u cos alpha))` |
`(du)/(dt)= 360\ text{km/hr (plane’s speed)}`
`text(After 5 mins,)`
`u` | `= 5/60 xx 360 = 30\ text(km)` |
`(dr)/(dt)` | `= ((2 xx 30)\ – cos alpha)/( 2 sqrt(30^2 + 1\ – 30 cos alpha)) xx 360` |
`= (180 (60\ – cos alpha))/sqrt(901\ – 30 cos alpha)\ \ text(km/hr)` |
--- 5 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. `text(General term)`
`\ ^12C_k * (2x^3)^(12\ – k) (-1/x)^k`
`=\ ^12C_k * (–1)^k *2^(12\ – k) * x^(36\ – 3k) * x^(-k)`
`=\ ^12C_k * (–1)^k*2^(12\ – k) * x^(36\ – 4k)`
`text(Constant term occurs when)`
`36\ – 4k` | `= 0` |
`k` | `= 9` |
`:.\ text(Constant term)` | `=\ ^12C_9 * (–1)^9*2^3` |
`= – (12!)/(3!9!) xx 8` | |
`= – 1760` |
ii. `text(General term of)\ (2x^3\ – 1/x)^n`
`\ ^nC_k * (2x^3)^(n\ – k) (–1/x)^k`
`=\ ^nC_k * 2^(n\ – k) * x^(3n\ – 3k) * (–1)^k * x^(-k)`
`=\ ^nC_k * (–1)^k*2^(n\ -k) * x^(3n\ – 4k)`
`text(Constant term when)\ \ 3n\ – 4k = 0.`
`text(i.e.)\ \ k=3/4n`
`text(S)text(ince)\ n\ text(and)\ k\ text(must be integers,)\ \ n\ \ text(must)`
`text(be a multiple of 4.)`
`text(Let)\ \ f(x) = x - (x^2)/2 + (x^3)/3`
--- 4 WORK AREA LINES (style=lined) ---
--- 3 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
ii. Let `g(x) = ln (1 + x)`.
Use the result in part c.i. to show that `f prime (x) >= g prime (x)` for all `x >= 0`. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
a. | `f(x) = x\ – (x^2)/2 + (x^3)/3` |
`text(Turning points when)\ f prime (x) = 0`
`f prime (x) = 1\ – x + x^2`
`x^2\ – x + 1 = 0`
`text(S)text(ince)\ \ Delta` | `= b^2\ – 4ac` |
`= (–1)^2\ – 4 xx 1 xx 1` | |
`= -3 < 0 => text(No solution)` |
`:.\ f(x)\ text(has no turning points)`
b. | `text(P.I. when)\ f″(x) = 0` |
`f″(x)` | `= -1 + 2x = 0` |
`2x` | `= 1` |
`x` | `= 1/2` |
`text(Check for change in concavity)`
`f″(1/4)` | `= -1/2 < 0` |
`f″(3/4)` | `= 1/2 > 0` |
`=>\ text(Change in concavity)`
`:.\ text(P.I. at)\ \ x = 1/2`
`f(1/2)` | `= 1/2\ – ((1/2)^2)/2 + ((1/2)^3)/3` |
`= 1/2\ – 1/8 + 1/24` | |
`= 5/12` |
`:.\ text(Point of Inflection at)\ (1/2, 5/12)`
c.i. | `text(Show)\ 1\ – x + x^2\ – 1/(1 + x) = (x^3)/(1 + x),\ \ \ x != -1` |
`text(LHS)` | `= (1+x)/(1+x)\ – (x(1+x))/(1+x) + (x^2(1+x))/((1+x))\ – 1/(1+x)` |
`= (1 + x\ – x\ – x^2 + x^2 + x^3\ – 1)/(1+x)` | |
`= (x^3)/(1+x)\ \ \ text(… as required)` |
c.ii. | `text(Let)\ g(x) = ln(1+x)` |
`g prime (x) = 1/(1 + x)` |
`f prime (x)\ – g prime (x)` | `= 1\ – x + x^2\ – 1/(1+x)` |
`= (x^3)/(1 + x)\ \ text{(using part (i))}` |
`text(S)text(ince)\ (x^3)/(1 + x) >= 0\ text(for)\ x >= 0`
`f prime (x)\ – g prime (x) >= 0` |
`f prime (x) >= g prime (x)\ text(for)\ x >= 0` |
d. |
e. | `text(Show)\ d/(dx) [(1 + x) ln (1 + x)\ – (1 + x)] = ln (1 + x)` |
`text(Using)\ d/(dx) uv=uv′+vu′` |
`text(LHS)` | `= (1+x) xx 1/(1 + x) + ln(1+x)xx1 + – 1` |
`= 1+ ln(1+x)\ – 1` | |
`= ln(1+x)` | |
`=\ text(RHS … as required)` |
f. | `text(Area)` | `= int_0^1 f(x)\ – g(x)\ dx` |
`= int_0^1 (x\ – (x^2)/2 + (x^3)/3\ – ln(x+1))\ dx` | ||
`= [x^2/2\ – x^3/6 + (x^4)/12\ – (1 + x) ln (1+x) + (1+x)]_0^1` | ||
`text{(using part (e) above)}` | ||
`= [(1/2 – 1/6 + 1/12 – (2)ln2 + 2) – (ln1 + 1)]` | ||
`= 5/12\ – 2ln2 + 2\ – 1` | ||
`= 1 5/12\ – 2 ln 2\ \ text(u²)` |
Each week Van and Marie take part in a raffle at their respective workplaces.
The probability that Van wins a prize in his raffle is `1/9`. The probability that Marie wins a prize in her raffle is `1/16`.
What is the probability that, during the next three weeks, at least one of them wins a prize? (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
`91/216`
`P text{(Van loses)}` | `= 1 – 1/9 = 8/9` |
`P text{(Marie loses)}` | `= 1 – 1/16 = 15/16` |
`P text{(both lose)}` | `= 8/9 xx 15/16 = 5/6` |
`text{P(At least 1 wins)}`
`= 1\ – P text{(both lose for 3 weeks)}`
`= 1\ – (5/6)(5/6)(5/6)`
`= 1\ – 125/216`
`= 91/216`
The diagram shows the graph of a function `y = f(x)`.
--- 1 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
--- 7 WORK AREA LINES (style=lined) ---
Between 5 am and 5 pm on 3 March 2009, the height, `h`, of the tide in a harbour was given by
`h = 1 + 0.7 sin(pi/6 t)\ \ \ text(for)\ \ 0 <= t <= 12`
where `h` is in metres and `t` is in hours, with `t = 0` at 5 am.
--- 2 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
Find all times between 5 am and 5 pm on 3 March 2009 during which the ship was able to enter the harbour. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
i. `h = 1 + 0.7 sin (pi/6 t)\ \ text(for)\ 0 <= t <= 12`
`T` | `= (2pi)/n\ \ text(where)\ n = pi/6` |
`= 2 pi xx 6/pi` | |
`= 12\ text(hours)` |
`:.\ text(The period of)\ h\ text(is 12 hours.)`
ii. `text(Find)\ h\ text(at low tide)`
`=> h\ text(will be a minimum when)`
`sin(pi/6 t) = -1`
`:.\ h_text(min)` | `= 1 + 0.7(-1)` |
`= 0.3\ text(metres)` |
`text(S)text(ince)\ \ sinx = -1\ \ text(when)\ \ x = (3pi)/2`
`pi/6 t` | `= (3pi)/2` |
`t` | `= (3pi)/2 xx 6/pi` |
`= 9\ text(hours)` |
`:.\ text{Low tide occurs at 2pm (5 am + 9 hours)}`
iii. `text(Find)\ \ t\ \ text(when)\ \ h >= 1.35`
`1 + 0.7 sin (pi/6 t)` | `>= 1.35` |
`0.7 sin (pi/6 t)` | `>= 0.35` |
`sin (pi/6 t)` | `>= 1/2` |
`sin (pi/6 t)` | `= 1/2\ text(when)` |
`pi/6 t` | `= pi/6,\ (5pi)/6,\ (13pi)/6,\ text(etc …)` |
`t` | `= 1,\ 5\ \ \ \ \ \ (0 <= t <= 12)` |
`text(From the graph,)`
`sin(pi/6 t) >= 1/2\ \ \ text(when)\ \ 1 <= t <= 5`
`:.\ text(Ship can enter the harbour between 6 am and 10 am.)`