Band 6 – Page 19

GEOMETRY, FUR2 2011 VCAA 4

A lighthouse tower, shaded on Figure 5 below, is in the shape of a truncated cone.

It has circular cross-sections that decrease uniformly from a radius of 3.5 metres at ground level to a radius of 2 metres at the walkway.

The height of the lighthouse tower is 18 metres.

The angle marked as `alpha` is the angle that the outer wall of the lighthouse tower makes with the horizontal at ground level.

Determine the size of the angle `alpha`.

Write your answer in degrees correct to one decimal place. (1 mark)

The lighthouse tower is part of a cone. The height of this cone is `h` metres and the base radius is 3.5 metres, as shown in Figure 6.

1. Determine `h`, the height of this cone, in meters. (2 marks)
2. Determine the volume of the lighthouse tower.
  
  Write your answer to the nearest cubic metre. (1 mark)

Show Answers Only

`85.2^@`
1. `42\ text(m)`
2. `438\ text(m³)`

Show Worked Solution

`tan alpha`	`= 18/1.5`
`:. alpha`	`= tan^-1 12`
	`= 85.23…`
	`= 85.2^@\ text{(1 d.p.)}`

♦♦♦ Mean mark for all parts (combined) 17%.
MARKER’S COMMENT: A common mistake had the base of this triangle equal to 3.5m.

b.i.

`tan 85.2^@`	`= h/3.5`
`:. h`	`= 3.5 xx tan 85.2^@`
	`= 41.68…`
	`= 42\ text{m (nearest m)}`

ii.

`text(Volume of large cone)`

`= 1/3 pi r^2 h`

`= 1/3 pi xx (3.5)^2 xx 42`

`= 538.78…\ text(m³)`

`text(Volume of Upper Cone)`

`= 1/3 xx pi xx 2^2 xx h_(uc)`

`= 4/3 pi xx (42 – 18)`

`= 100.53…\ text(m³)`

`:.\ text(Volume of lighthouse tower)`

`= 538.78… – 100.53…`

`= 438.2…`

`= 438\ text{m³ (nearest m³)}`

MATRICES, FUR2 2014 VCAA 3

There are three candidates in the election: Ms Aboud (`A`), Mr Broad (`B`) and Mr Choi (`C`).

Mr Choi may need to withdraw from the election at the end of May.

Matrix `T`, shown below, shows the percentage of voters who change their preferred candidate, from month to month, before Mr Choi would withdraw from the election.

`{:(qquadqquadqquadqquadtext(this month)),(qquadqquadqquad\ {:(A,qquadB,qquadC):}),(T = [(0.75,0.10,0.20),(0.05,0.80,0.40), (0.20,0.10,0.40)]{:(A),(B),(C):}qquadtext(next month)):}`

Matrix `T_1`, shown below, shows the percentage of voters who change their preferred candidate, from May to June, after Mr Choi would withdraw from the election.

`{:(qquadqquadqquadqquadqquadquadtext(May)),(qquadqquadqquadquad{:(A,qquadB,qquadC):}),(T_1 = [(0.75,0.15,0.6),(0.25,0.85,0.4),(0,0,0)]{:(A),(B),(C):}qquadtext(June)):}`

Consider the voters who preferred Mr Broad in May and who were expected to prefer Mr Choi in June.

What percentage of these voters are now expected to prefer Mr Broad in June? (1 mark)

The state matrix that indicates the number of voters who are expected to have a preference for each candidate in January, `S_1`, is given below.

`S_1 = [(6000), (3840), (2160)]{:(A), (B), (C):}`

If Mr Choi withdraws, how many votes is Mr Broad expected to receive in the election in June?

Write your answer, correct to the nearest vote. (2 marks)

Show Answers Only

`text(50%)`
`6451`

Show Worked Solution

a. `text(Before)\ C\ text(withdraws, 10% are expected to)`

♦♦♦ Mean mark for (a)-(b) was 13% (combined).
MARKER’S COMMENT: In (a), identifying the sub-group expected to change their vote proved very difficult for most students.

`text(change from)\ B\ text(to)\ C.`

`text(After)\ C\ text(withdraws,)\ A\ text(and)\ B\ text(have an)`

`text(increase of 5% each.)`

`:. 50text(% of votes leaving)\ B\ text(to)\ C,\ text(will now)`

`text(prefer)\ B.`

MARKER’S COMMENT: Few students correctly modified the transition matrix from `T_1` to `T_2` in this question.

b.	`S_5`	`= S_(text(May))`
		`= T^4S_1`
		`= [(0.75,0.10,0.20), (0.05,0.80,0.40),(0.20,0.10,0.40)]^4[(6000),(3840),(2160)]`
		`= [(4454), (5154), (2392)]{:(A), (B), (C):}`

`S_6`	`= S_(text(June))`
	`= T_1 xx S_5`
	`= [(0.75,0.15,0.6), (0.25,0.85,0.4), (0,0,0)][(4454), (5154), (2392)]`
	`= [(5549), (6451), (0)]{:(A), (B), (C):}`

`:.\ text(Mr. Broad is expected to receive 6451)`

`text(votes in June.)`

CORE, FUR2 2006 VCAA 3

The heights (in cm) and ages (in months) of the 15 boys are shown in the scatterplot below.

Fit a three median line to the scatterplot. Circle the three points you used to determine this three median line. (2 marks)
Determine the equation of the three median line. Write the equation in terms of the variables height and age and give the slope and intercept correct to one decimal place. (2 marks)
Explain why the three median line might model the relationship between height and age better than the least squares regression line. (1 mark)

Show Answers Only

`text(height =)\ 70.7 + 0.7 xx text(age)`
`text(The three median line is not as influenced)`
`text{by extreme values such as (20, 93).}`

Show Worked Solution

♦ Average mean mark for all parts 36%.

b. `text{Using the points (23,86) and (35, 94),}`

`text(Slope)`	`= (94 – 86)/(35 – 23)`
	`= 0.666…`
	`=0.7\ \ text{(1 d.p.)}`

`:.\ text(height =)\ c + 0.7 xx text(age)`

`text{Substituting (35,94) into the equation,}`

`94`	`=c + 0.66… xx 35`
`:.c`	`=94-23.33…`
	`=70.66…`
	`=70.7\ \ text{(1 d.p.)}`

`:.text(height)\ = 70.7 + 0.7 xx text(age)`

c. `text(The three median line is not as influenced)`

`text{by extreme values such as (20, 93).}`

The number of hours spent doing homework each week (homework hours) and the number of hours spent watching television each week (television hours) were recorded for a group of 20 Year 12 students.

The results are displayed in the table below and a scatterplot constructed as shown.

The relationship between homework hours per week and television hours per week is clearly nonlinear.

A reciprocal transformation applied to the variable, television hours, can be used to linearise the scatterplot.

Apply this reciprocal transformation to the data and determine the equation of the least squares regression line that allows homework hours to be predicted from the reciprocal of television hours.

Write the coefficients correct to two decimal places. (2 marks)
If a student spends 12 hours per week watching television, use the least squares regression line to predict the number of hours that the student spends doing homework. Give your answer correct to one decimal place. (1 mark)

Show Answers Only

`text(Homework hours) = 5.12 + 102.90 xx 1/(text{TV hours})`
`13.7`

Show Worked Solution

a. `text(By calculator,)`

♦♦♦ Avg mean mark for both parts 16%.
MARKER’S COMMENT: A majority of students didn’t understand the term “reciprocal”.

`text(Homework hours) = 5.12 + 102.90 xx 1/(text{TV hours})`

b.	`text(Homework hours)`	`= 5.12 + 102.90 xx 1/12`
		`= 13.695`
		`=13.7\ \ text{(to 1 d.p.)}`

CORE, FUR2 2011 VCAA 4

The average age of women at first marriage in years (average age) and average yearly income in dollars per person (income) were recorded for a group of 17 countries.

The results are displayed in Table 2. A scatterplot of the data is also shown.

The relationship between average age and income is nonlinear.

A log transformation can be applied to the variable income and used to linearise the scatterplot.

Apply this log transformation to the data and determine the equation of the least squares regression line that allows average age to be predicted from log (income).

Write the coefficients for this equation, correct to two decimal places, in the spaces provided. (2 marks)
Use this equation to predict the average age of women at first marriage in a country with an average yearly income of $20 000 per person.

Write your answer correct to one decimal place. (1 mark)

Show Answers Only

`text(Average age) = 2.39 + 5.89 xx log (text(income))`
`text(27.7 years)`

Show Worked Solution

a. `text(By calculator after applying the log transformation,)`

`text(Average age) = 2.39 + 5.89 xx log (text(income))`

♦♦ Parts (i) and (ii) had a combined mean mark 0f 34%.
MARKER’S COMMENT: Students struggled with the log transformation, incorrect data entry and choosing the correct dependent variable.

b. `text(The average age at first marriage with an)`

`text(average yearly income of $20 000)`

`=2.39+5.89 xx log(20\ 000)`

`=27.7\ text{years (1 d.p.)}`

CORE, FUR2 2012 VCAA 2

The maximum temperature and the minimum temperature at this weather station on each of the 30 days in November 2011 are displayed in the scatterplot below.

The correlation coefficient for this data set is `r = 0.630`.

The equation of the least squares regression line for this data set is

maximum temperature = `13 + 0.67` × minimum temperature

Draw this least squares regression line on the scatterplot above. (1 mark)
Interpret the vertical intercept of the least squares regression line in terms of maximum temperature and minimum temperature. (1 mark)
Describe the relationship between the maximum temperature and the minimum temperature in terms of strength and direction. (1 mark)
Interpret the slope of the least squares regression line in terms of maximum temperature and minimum temperature. (1 mark)
Determine the percentage of variation in the maximum temperature that may be explained by the variation in the minimum temperature.

Write your answer, correct to the nearest percentage. (1 mark)

On the day that the minimum temperature was 11.1 °C, the actual maximum temperature was 12.2 °C.

Determine the residual value for this day if the least squares regression line is used to predict the maximum temperature.

Write your answer, correct to the nearest degree. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`text(On average, when the minimum temperature is)`
`text(0 °C, the maximum temperature is 13 °C)`
`text(Moderate and positive.)`
`text(On average, it is predicted that the maximum)`
`text(temperature increases by 0.67 °C for every 1 °C)`
`text(increase in the minimum temperature.)`
`40text(%)`
`-8^@text(C)`

Show Worked Solution

a. `text(The two widest points in this data range are,)`

`text{(0, 13) and (20, 26.4).}`

b. `text(On average, when the minimum temperature is)`

`text(0 °C, the maximum temperature is 13 °C.)`

♦ Parts (i) to (vi) have an average mean mark of 41%.

c. `text(Given)\ r = 0.630,`

`text(Strength: moderate)`

`text(Direction: positive)`

d. `text(On average, it is predicted that \the maximum)`

`text(temperature increases by 0.67 °C for every 1 °C)`

`text(increase in the minimum temperature.)`

e.	`r^2`	`= 0.630^2`
		`=0.3969`
		`=40text{% (nearest %)}`

f. `text(When the minimum temperature was)\ 11.1 text(°C),`

MARKER’S COMMENT: Students had particular difficulty with this part, with many using the incorrect calculation of 12.2 – 11.1 = 1.1.

`text(Predicted Value)`	`= 13 + 0.67 xx 11.1`
	`=20.437…`

`:.\ text(Residual)`	`= 12.2 − 20.437…`
	`= – 8.237…`
	`= – 8\ text{°C (nearest degree)}`

CORE, FUR2 2013 VCAA 4

The time series plot below shows the average pay rate, in dollars per hour, for workers in a particular country for the years 1991 to 2005.

A three median line will be used to model the increasing trend in the average pay rate shown in this time series.

The explanatory variable to be used is year.

Three medians will be used to draw the three median line.
1. On the time series plot above, mark the location of each of the three medians with a cross (`X`). (2 marks)
2. Draw the three median line on the time series plot. (1 mark)
Calculate the slope of the three median line.

Write your answer, correct to one decimal place. (1 mark)
Interpret the slope of the three median line in terms of the relationship between the variables average pay rate and year. (1 mark)

Show Answers Only

i. & ii. `text(See Worked Solutions)`
`0.8`
`text(The average pay rate increases, on average,)`

`text(by $0.80 each year between 1991 and 2005.)`

Show Worked Solution

a.i. & ii. `text(The three medians points are)`

♦ Mean mark 45%.
MARKER’S COMMENT: Few students found the correct middle point.

`(1993,13.7), (1998,16.7), (2003,21.8)`

♦♦ Parts (ii) and (iii) produced an overall mean mark of 32%.

b.	`text(Slope)`	`=(y_2-y_1)/(x_2-x_1)`
		`=(21.8 − 13.7)/(2003 − 1993)`
		`= 0.81`
		`=0.8\ \ text{(to 1 d.p.)}`

MARKER’S COMMENT: Answering “an increase of 0.8 each year” received no marks (refer to actul units).

c. `text(The average pay rate increases, on average,)`

`text(by $0.80 each year between 1991 and 2005.)`

CORE, FUR2 2014 VCAA 4

The scatterplot below shows the population density, in people per square kilometre, and the area, in square kilometres, of 38 inner suburbs of the same city.

For this scatterplot, `r^2 = 0.141`

Describe the association between the variables population density and area for these suburbs in terms of strength, direction and form. (1 mark)
The mean and standard deviation of the variables population density and area for these 38 inner suburbs are shown in the table below.
1. One of these suburbs has a population density of 3082 people per square kilometre.
  
  Determine the standard `z`-score of this suburb’s population density.
  
  Write your answer, correct to one decimal place. (1 mark)

Assume the areas of these inner suburbs are approximately normally distributed.

How many of these 38 suburbs are expected to have an area that is two standard deviations or more above the mean?

Write your answer, correct to the nearest whole number. (1 mark)
How many of these 38 inner suburbs actually have an area that is two standard deviations or more above the mean? (1 mark)

Show Answers Only

`text(The association is weak, negative and linear.)`
1. `− 0.8`
2. `1`
3. `2`

Show Worked Solution

♦♦ Mean mark 30%.
MARKER’S COMMENT: The most common error was to describe the association as positive.

a.	`r`	`= −sqrt(0.141)`
		`= −0.3755`

`:.\ text(The association is weak, negative and linear.)`

b.i	`z`	`=(x-bar x)/s`
		`= (3082 − 4370)/1560`
		`= − 0.82…`
		`= – 0.8\ \ text{(to 1 d.p.)}`

b.ii. `2.5text(%) xx 38 =0.95=1\ text{(nearest whole)}`

b.iii. `text{Area size (2 std deviation above mean)}`

♦♦ Very few students answered this part correctly.

`= 3.4 + 2 xx 1.6`

`=6.6\ text(km²)`

`:.\ text{From the graph, 2 inner suburbs have an area}`

`text(2 standard deviations above the mean.)`

GRAPHS, FUR1 2008 VCAA 9 MC

The shaded region in the graph below represents the feasible region for a linear programming problem.

Which objective function, `Z`, has its maximum value at the point `M`?

A. `Z = x + y`

B. `Z = x - y`

C. `Z = 3x + y`

D. `Z = 3x - 2y`

E. `Z = x + 4y`

Show Answers Only

`A`

Show Worked Solution

`text(Find the coordinates of)\ M,`

♦♦♦ Mean mark 17%.
MARKER’S COMMENT: Even though `Z=x+4y` (option E) has a larger value at `M` than option A, its maximum value occurs at the vertex (0, 60).

`y`	`=100-2.5x\ …\ (1)`
`y`	`=60 – 1/2x\ …\ (2)`
`(1) – (2)`
`0`	`=40 – 2x`
`x`	`=20`

`text(Substitute)\ \ x=20\ \ text(into)\ \ (2)`

`y=50`

`:. M(20,50)`

`text{Testing the boundary coordinates of (0, 60),`

`text{(20, 50) and (40, 0) in each option,`

`text(Consider)\ A,`

`text{At (0, 60),}\ \ Z=0+60=60`

`text{At (20, 50),}\ \ Z=20+50=70`

`text{At (40, 0),}\ \ Z=40+0=40`

`:. Z=x+y\ \ text{has its maximum value at}\ M.`

`=> A`

GRAPHS, FUR1 2012 VCAA 9 MC

The cost of posting a parcel that weighs 500 grams or less depends on its weight, as shown on the graph below.

The total cost of posting three separate parcels weighing respectively 100 grams, 150 grams and 210 grams is $11.

The total cost of posting three separate parcels weighing respectively 60 grams, 110 grams and 350 grams is $8.

The total cost of posting two separate parcels weighing respectively 170 grams and 500 grams is

A. `$5`

B. `$7`

C. `$8`

D. `$10`

E. `$12`

Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ x\ \ text(be cost when)\ \ \ 0 <\ text(weight)\ <= 120`

`text(Let)\ \ y\ \ text(be cost when)\ \ \ 200 <\ text(weight)\ <= 500`

`x + 4 + y`	`= 11`
`x + y`	`= 7\ \ …\ \ (1)`
`2x + y`	`= 8\ \ …\ \ (2)`

`(2) – (1)`

`:. x = 1`

`text(Substitute)\ \ x = 1\ \ text(into)\ \ (1)`

`1 + y`	`= 7`
`y`	`= 6`

`:. 0 <\ text(weight)\ <= 120`	`= $1`
`200 <\ text(weight)\ <= 500`	`= $6`

`:.\ text(C)text(ost to post)\ \ text(170g and 500g parcel)`

`= $4 + $6 = $10`

`=> D`

GRAPHS, FUR1 2013 VCAA 8 MC

The shaded region in the graph below represents the feasible region for a linear programming problem.

An objective function `Z = ax + by` has its value maximised at both vertex `M` and vertex `N`.

The values of `a` and `b` could be

A. `a = 15\ and\ b = – 15`

B. `a = 15\ and\ b = 15`

C. `a = 15\ and\ b = 25`

D. `a = 25\ and\ b = 50`

E. `a = 50\ and\ b = – 25`

Show Answers Only

`B`

Show Worked Solution

♦♦ Mean mark 35%.

`Z`	`= ax + by`
`by`	`= -ax + Z`
`y`	`= -a / bx + Z/b…(1)`

`:.\ text(Gradient of objective function)\ = -a / b`

`text{Find gradient of line between}\ M and N,`

`m_(MN)=(y_2 – y_1) / (x_2 – x_1)=(70 – 0) / (0 – 70)=-1`

`:. -a / b`	`= -1`
`a / b`	`= 1 `
` a`	`= b`

`=> B`

GEOMETRY, FUR1 2011 VCAA 9 MC

In the diagram below, `/_ ABD = /_ ACB = theta^@`.

`BD = 24\ text(cm)` and `BC = 40\ text(cm.)`

The area of triangle `ABD` is `100\ text(cm².)`

The area of triangle `ABC`, in cm² is closest to

A. `167`

B. `178`

C. `267`

D. `278`

E. `378`

Show Answers Only

`D`

Show Worked Solution

`/_ ABD = /_ BCA = theta^@\ \ \ text{(given)}`

`/_ BAD\ text(is common)`

`:. Delta BAD\ text(|||)\ Delta CAB\ \ text{(equiangular)}`

`text(Linear scale factor)`

`k= (BC)/(DB) = 40/24 = 5/3`

`:.\ text(Area)\ Delta CAB`	`= k^2 xx text(Area)\ Delta BAD`
	`= (5/3)^2 xx 100`
	`= 277.77…\ \ text(cm²)`

`=> D`

GEOMETRY, FUR1 2015 VCAA 8 MC

A composite shape is made up of a parallelogram and a triangle, as shown below.

Which one of the following is always true?

A. `b = 2a`

B. `a = 2b`

C. `a + b = 180`

D. `a + 2b = 180`

E. `2a - b = 180`

Show Answers Only

`E`

Show Worked Solution

`text(Base angles of isosceles triangle)`

♦♦ Mean mark 25%.

`=(180 – b)°`

`text(Summing up the angles of the isosceles triangle,)`

`b + 2(180^@ – a)`	`=180°`
`b-2a+360°`	`=180°`
`:. 2a – b`	`=180°`

`=> E`

CORE*, FUR1 2015 VCAA 8 MC

Cindy took out a reducing balance loan of $8400 to finance an overseas holiday.

Interest was charged at a rate of 9% per annum, compounding quarterly.

Her loan is to be fully repaid in six years, with equal quarterly payments.

After three years, Cindy will have reduced the balance of her loan by approximately

A. 9%

B. 35%

C. 43%

D. 50%

E. 57%

Show Answers Only

`C`

Show Worked Solution

`text(By TVM Solver,)`

♦♦ Mean mark 33%.

`N`	`=6 xx 4 = 24`
`I (%)`	`=9`
`PV`	`= – 8400`
`PMT`	`=?`
`FV`	`=0`
`text(P/Y)`	`= text(C/Y) = 4`

`:.\ text(Quarterly payment)\ = $456.793…`

`text(By TVM Solver, after 3 years)`

`N`	`=3 xx 4 = 12`
`I (%)`	`=9`
`PV`	`= – 8400`
`PMT`	`=456.793…`
`FV`	`=?`
`text(P/Y)`	`= text(C/Y) = 4`

`:.\ FV\ text{((balance of loan)} = 4757.4076…`

`:.\ text(% reduction)`	`=(8400 – 4757.41)/8400`
	`=0.4336…`
	`=43.36…%`

`=> C`

CORE*, FUR1 2015 VCAA 8 MC

`A_n` is the `n`th term in a sequence.

Which one of the following expressions does not define a geometric sequence?

A. `A_(n + 1) = n`	`\ \ \ \ A_0 = 1`
B. `A_(n + 1) = 4`	`\ \ \ \ A_0 = 4`
C. `A_(n + 1) = A_n + A_n`	`\ \ \ \ A_0 = 3`
D. `A_(n + 1) = –A_n`	`\ \ \ \ A_0 = 5`
E. `A_(n + 1) = 4A_n`	`\ \ \ \ A_0 = 2`

Show Answers Only

`A`

Show Worked Solution

`text(Test all options by looking at the first)`

♦♦ Mean mark 22%.
MARKERS’ COMMENT: A routine way to solve this type of question is to write out the first few terms of each sequence.)

`text(3 terms that each produces.)`

`text(Consider)\ A,`

`A_0=1, \ A_1=0, \ A_2=1`

`text(There is no common ratio in this sequence.)`

`text(All other options can be shown to have a common ratio.)`

`=> A`

CORE, FUR1 2010 VCAA 11 MC

A student uses the following data to construct the scatterplot shown below.

A reciprocal transformation is applied to the `x`-axis and is used to linearise the scatterplot.

With `y` as the dependent variable, the slope of the least squares regression line that is fitted to the linearised plot is closest to

A. `–249`

B. `–25`

C. `0.004`

D. `25`

E. `249`

Show Answers Only

`E`

Show Worked Solution

`text(By calculator,)`

♦♦ Mean mark 35%.

`text(Gradient of)\ y\ text(versus)\ 1/x\ text(is closest to 249.)`

`=> E`

CORE*, FUR1 2013 VCAA 9 MC

The following information relates to the repayment of a home loan of $300 000.

• The loan is to be repaid fully with monthly payments of $2500.
• Interest compounds monthly.
• After the first monthly payment has been made, the amount owing on the loan is $299 000.

Which one of the following statements is true?

After two months, $297 995 is still owing on the loan.
$1000 of interest has been paid in the first month.
The loan will be fully repaid in less than 15 years.
Halfway through the term of the loan, the amount still owing will be $150 000.
Payments of $2750 rather than $2500 per month will reduce the time to repay the loan fully by more than three years.

Show Answers Only

`A`

Show Worked Solution

`text(Find the interest rate using a TVM Solver,)`

♦♦ Mean mark 29%.

`N`	`= 1`
`I (%)`	`= ?`
`PV`	`= -300\ 000`
`PMT`	`= 2500`
`FV`	`= 299\ 000`
`text(P/Y)`	`= text(C/Y) = 12`

`:.I =6(%)`

`text(Test each statement by TVM solver:)`

`text(Consider A,)`

`N`	`= 2`
`I (%)`	`= 6`
`PV`	`= -300\ 000`
`PMT`	`= 2500`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`:.\ FV = $297\ 995`

`=> A`

PATTERNS, FUR 1 2006 VCAA 9 MC

A healthy eating and gym program is designed to help football recruits build body weight over an extended period of time.

Roh, a new recruit who initially weighs 73.4 kg, decides to follow the program.

In the first week he gains 400 g in body weight.

In the second week he gains 380 g in body weight.

In the third week he gains 361 g in body weight.

If Roh continues to follow this program indefinitely, and this pattern of weight gain remains the same, his eventual body weight will be closest to

A. `74.5\ text(kg)`

B. `77.1\ text(kg)`

C. `77.3\ text(kg)`

D. `80.0\ text(kg)`

E. `81.4\ text(kg)`

Show Answers Only

`E`

Show Worked Solution

`text(Checking for a common ratio,)`

♦♦ Mean mark 33%.

`t_2/t_1`	`=380/400=0.95`
`t_3/t_2`	`=361/380=0.95`

`:.\ text(GP where)\ \ \ a`	`=400, and`
`r`	`=0.95`

`text(S)text(ince)\ \ |\ r\ |<1,`

`S_oo`	`=a/(1-r)`
	`=400/(1-0.95)`
	`=8000\ text(g)`

`:.\ text(Net weight gain will be 8 kg which will make Roh’s)`

`text{eventual weight 73.4 + 8 = 81.4 kg.}`

`rArr E`

CORE, FUR1 2006 VCAA 11-13 MC

The following information relates to Parts 1, 2 and 3.

The table shows the seasonal indices for the monthly unemployment numbers for workers in a regional town.

Part 1

The seasonal index for October is missing from the table.

The value of the missing seasonal index for October is

A. `0.93`

B. `0.95`

C. `0.96`

D. `0.98`

E. `1.03`

Part 2

The actual number of unemployed in the regional town in September is 330.

The deseasonalised number of unemployed in September is closest to

A. `310`

B. `344`

C. `351`

D. `371`

E. `640`

Part 3

A trend line that can be used to forecast the deseasonalised number of unemployed workers in the regional town for the first nine months of the year is given by

deseasonalised number of unemployed = 373.3 – 3.38 × month number

where month 1 is January, month 2 is February, and so on.

The actual number of unemployed for June is predicted to be closest to

A. `304`

B. `353`

C. `376`

D. `393`

E. `410`

Show Answers Only

`text (Part 1:)\ D`

`text (Part 2:)\ C`

`text (Part 3:)\ A`

Show Worked Solution

`text (Part 1)`

`text(Oct index)`	`=12-text(sum of other 11)`
	`=12-11.02`
	`=0.98`

`rArr D`

`text (Part 2)`

`text(Deseasonalised number)`	`= text(Actual)/text(Index)`
	`=330/0.94`
	`=351.06…`

`rArr C`

`text (Part 3)`

♦♦ Mean mark 29%.
MARKERS’ COMMENT: 59% of students correctly found the deseasonalised number but failed to convert it to the actual.

`text(Deseasonalised number)`

`=373.3 – 3.38 xx 6`

`=353.02`

`text(Actual number)`	`= text(Deseasonalised) xx text(Index)`
	`=353.02 xx 0.86`
	`=303.59…`

`rArr A`

PATTERNS, FUR1 2013 VCAA 9 MC

Eleven speed bumps are placed on a road.

The speed bumps are placed so that the distance between consecutive speed bumps decreases according to an arithmetic sequence.

The distance between the first and last speed bumps is exactly 100 m.

The smallest distance between consecutive speed bumps is 2 m.

The largest distance, in m, between two consecutive speed bumps is

A. `16`

B. `18`

C. `20`

D. `22`

E. `24`

Show Answers Only

`B`

Show Worked Solution

`text(11 speed bumps over 100m distance has 10 intervals,)`

`text(so the series can be written)`

♦♦ Mean mark 33%.
MARKER’S COMMENT: The not often used formula `S_n=n/2(a+l)` provides the most efficient route to this solution, although `S_n=n/2[2a+(n-1)d]` can also be used after finding the value of `d`.

`a, a+d, a+2d, …, a+9d`

`text(Last distance)\ =a+9d=l=2\quad text{(given)}`

`S_10=100\ text(m)\ \ \ text{(given)}`

`text(Using)\ \ \ S_n`	`=n/2(a+l)`
`100`	`=10/2(a+2)`
`100`	`=5a+10`
`5a`	`=90`
`a`	`=18`

`:. 18text{m is largest distance (d is negative)}`

`=> B`

Calculus, MET2 2013 VCAA 15 MC

Let `h` be a function with an average value of 2 over the interval `[0, 6].`

The graph of `h` over this interval could be

Show Answers Only

`C`

Show Worked Solution

`text(Draw the line)\ \ y=2\ \ text(on each graph.)`

♦♦♦ Mean mark 25%.

`=>\ text(An average value of 2 occurs when the same area)`

`text(is enclosed above and below the graph.)`

`text(Consider)\ C,`

`text(Area)\ A =\ text(Area)\ B`

`=> C`

Calculus, MET2 2015 VCAA 16 MC

Let `f(x) = ax^m` and `g(x) = bx^n`, where `a, b, m` and `n` are positive integers. The domain of `f = text(domain of)\ g = R.`

If `f prime (x)` is an antiderivative of `g(x)`, then which one of the following must be true?

A. `m/n` is an integer

B. `n/m` is an integer

C. `a/b` is an integer

D. `b/a` is an integer

E. `n - m = 2`

Show Answers Only

`D`

Show Worked Solution

`text(Given)\ \ f prime(x)`	`=int g(x)\ dx,`
`amx^(m – 1)`	`= b/(n + 1) x^(n + 1)`

♦♦ Mean mark 22%.

`m – 1 = n + 1\ …\ (1)\ \ \ text{(equate powers)}`

`am = b/(n + 1)\ …\ (2)\ \ \ text{(equate coefficients)}`

`text(Solving simultaneous equations:)`

`m(n + 1) = b/a`

`text(S)text(ince)\ \ m, n + 1\ text(are both integers)\ \ text{(i.e. ∈}\ Z^+text{)},`

`=> m(n + 1) ∈ Z^+`

`:. b/a = m(n + 1) ∈ Z^+`

`=> D`

Graphs, MET2 2015 VCAA 11 MC

The transformation that maps the graph of `y = sqrt(8x^3 + 1)` onto the graph of `y = sqrt(x^3 + 1)` is a

dilation by a factor of `2` from the `y`-axis.
dilation by a factor of `2` from the `x`-axis.
dilation by a factor of `1/2` from the `x`-axis.
dilation by a factor of `8` from the `y`-axis.
dilation by a factor of `1/2` from the `y`-axis.

Show Answers Only

`A`

Show Worked Solution

♦♦ Mean mark 24%.

`text(Let)\ f(x)`	`= sqrt(8x^3 + 1)`
`f(1/2 x)`	`= sqrt(8(1/2 x)^3 + 1)`
	`= sqrt(x^3 + 1)`

`f(1/2 x) -> text(dilate by factor of 2 from)\ ytext(-axis)`

`=> A`

Graphs, MET2 2015 VCAA 3 MC

The rule for a function with the graph above could be

`y = -2(x + b)(x - c)^2(x - d)`
`y = 2(x + b)(x - c)^2(x - d)`
`y = -2(x - b)(x - c)^2(x - d)`
`y = 2(x - b)(x - c)(x - d)`
`y = -2(x - b)(x + c)^2(x + d)`

Show Answers Only

`C`

Show Worked Solution

`text(Polynomial Equation:)`

♦♦♦ Mean mark 20% (lowest in 2015 exam).
MARKER’S COMMENT: `b` is negative. If `b= – 2` for example, the factor is `(x-(– 2))“=(x+2)`.

`y = a(x – b)(x – c)^n(x – d),`

`text(where)\ a < 0`

`n = text(even integer)`

`=> C`

Calculus, MET2 2014 VCAA 21 MC

The trapezium `ABCD` is shown below. The sides `AB, BC` and `DA` are of equal length, `p.` The size of the acute angle `BCD` is `x` radians

The area of the trapezium is a maximum when the value of `x` is

`pi/12`
`pi/6`
`pi/4`
`pi/3`
`(5 pi)/12`

Show Answers Only

`D`

Show Worked Solution

`sin x`	`= h/p`
`:. h`	`= p sinx`

`cosx`	`= w/p`
`:.w`	`= pcos x`

♦♦ Mean mark 28%.

`A_text(trap)`	`= 1/2h xx [p + (p + 2w)]`
	`= (p sin x)/2 [2p + 2pcosx]`
	`= p^2sin x (1 + cosx), \ \ x ∈ (0,pi/2)`

`text(Find when)\ \ (dA)/(dp)=0,\ \ x ∈ (0,pi/2)`

`:. x = pi/3`

`=> D`

Mechanics, EXT2* M1 2008 HSC 7

A projectile is fired from `O` with velocity `V` at an angle of inclination `theta` across level ground. The projectile passes through the points `L` and `M`, which are both `h` metres above the ground, at times `t_1` and `t_2` respectively. The projectile returns to the ground at `N`.

The equations of motion of the projectile are

`x = Vtcos theta` and `y = Vtsin theta − 1/2 g t^2`. (Do NOT prove this.)

Show that `t_1 + t_2 = (2V)/g sin theta` AND `t_1t_2 = (2h)/g`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Let `∠LON = α` and `∠LNO = β`. It can be shown that

`tan alpha = h/(Vt_1 cos theta)` and `tan beta = h/(Vt_2 cos theta)`. (Do NOT prove this.)

Show that `tan alpha + tan beta = tan theta`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Show that `tan alpha tan beta = (gh)/(2V^2cos^2theta)`. (1 mark)

--- 5 WORK AREA LINES (style=lined) ---

Let `ON = r` and `LM = w`.

Show that `r = h(cot alpha + cot beta)` and `w = h(cot beta - cot alpha)`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Let the gradient of the parabola at `L` be `tan phi`.

Show that `tan phi = tan alpha - tan beta`. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Show that `w/(tan phi) = r/(tan theta)`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`

Show Worked Solution

i. `text(At)\ \ y = h,`

`-1/2g t^2 + Vtsin theta`	`=h`
`g t^2 − 2Vtsin theta + 2h`	`=0`

`text(Roots are)\ \ t_1\ text(and)\ t_2`

`:. t_1 + t_2`	`= -b/a`
	`= (2V)/g sin theta …\ \ text(as required)`

`:. t_1t_2`	`= c/a`
	`= (2h)/g …\ \ text(as required)`

ii. `text(Show)\ \ tan alpha + tan beta = tan theta`

`text(LHS)`	`= h/(Vt_1cos theta) + h/(Vt_2cos theta)`
	`= (h(t_1 + t_2))/(t_1t_2Vcos theta)`
	`= (h((2V)/g sin theta))/(((2h)/g)Vcos theta),\ \ \ \ text{(using part (i))}`
	`= ((2Vh)/g)/((2Vh)/g) xx (sin theta)/(cos theta)`
	`= tan theta …\ \ text(as required.)`

iii. `text(Show)\ \ tan alpha tan beta = (gh)/(2V^2cos^2 theta)`

`text(LHS)`	`= h/(Vt_1cos theta) xx h/(Vt_2cos theta)`
	`= (h^2)/(t_1t_2V^2cos^2 theta)`
	`= (h^2)/((2h)/g · V^2cos^2 theta)`
	`= (gh)/(2V^2cos^2 theta) …\ \ text(as required.)`

iv. `text(From the diagram,)`

`r = ON = OP + PN`

`text(In)\ DeltaOLP,`

`tan alpha`	`= h/(OP)`
`OP`	`= hcot alpha`
`text(In)\ DeltaNLP,`
`tan beta`	`= h/(PN)`
`PN`	`= hcot beta`

`:.r`	`= hcot alpha + hcot beta`
	`= h(cot alpha + cot beta) …\ \ text(as required.)`

`w`	`= r − (OP + QN)`
	`= r − 2 xx OP,\ \ \ text{(symmetry of parabola)}`
	`= h cot alpha + hcot beta − 2hcot alpha`
	`= h (cot beta − cot alpha) …\ \ text(as required.)`

v. `text(Show)\ \ tan phi = tan alpha − tan beta`

`text(At)\ L, \ t = t_1\ \ text(and)`

`tan phi = (dot y)/(dot x)`

`dot y`	`= Vsin theta − g t`
`dot x`	`= Vcos theta`

`:. tan phi`	`= (Vsin theta – g t_1)/(Vcos theta)`
	`= tan theta – (g t_1)/(Vcos theta)`
	`= tan theta – g/(Vcos theta) xx h/(Vcos theta tan alpha)`
	`= tan theta – (gh)/(V^2cos^2 theta tan alpha)`
	`= tan theta – 2 tan beta,\ \ \ text{(from part (iii))}`
	`= tan alpha + tan beta − 2tan beta,\ \ \ text{(from part (ii))}`
	`= tan alpha – tan beta …\ \ text(as required.)`

vi. `text(Show)\ \ w/(tan phi) = r/(tan theta)`

`w/r`	`= (h(cot beta − cot alpha))/(h(cot alpha + cot beta))`
	`= (1/(tan beta) − 1/(tan alpha))/(1/(tan alpha) + 1/(tan beta)) xx (tan alpha tan beta)/(tan alpha tan beta)`
	`= (tan alpha − tan beta)/(tan beta + tan alpha)`
	`= (tan phi)/(tan theta)`

`:. w/(tan phi) = r/(tan theta) …\ \ text(as required.)`

Calculus, MET2 2013 VCAA 16 MC

The graph of `f: [1, 5] -> R,\ f(x) = sqrt (x - 1)` is shown below.

Which one of the following definite integrals could be used to find the area of the shaded region?

A. `int_1^5 (sqrt (x - 1))\ dx`

B. `int_0^2 (sqrt (x - 1))\ dx`

C. `int_0^5 (2 - sqrt (x - 1))\ dx`

D. `int_0^2 (x^2 + 1)\ dx`

E. `int_0^2 (x^2)\ dx`

Show Answers Only

`E`

Show Worked Solution

`f(5)=2,\ \ f(1)=0`

♦♦♦ Mean mark 21%.

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

`text(Find inverse:)`

`f^-1(x)=x^2 +1`

`:.\ text(Shaded Area)`

`= int_0^2(x^2+1)\ dx – 2 xx 1`

`=int_0^2 (x^2)\ dx`

`=> E`

Graphs, MET2 2013 VCAA 20 MC

A transformation `T: R^2 -> R^2, T ([(x), (y)]) = [(1, 0), (0, -1)] [(x), (y)] + [(5), (0)]` maps the graph of a function `f` to the graph of `y = x^2, x in R.`

The rule of `f` is

`f(x) = -(x + 5)^2`
`f(x) = (5 - x)^2`
`f(x) = -(x - 5)^2`
`f(x) = -x^2 + 5`
`f(x) = x^2 - 5`

Show Answers Only

`A`

Show Worked Solution

`text(Transformations are:)`

`x′`	`=x+5`
`x`	`=x′ -5`
`y′`	`=-y`
`y`	`=-y′`

`-y`	`=(x+5)^2`
`f(x)`	`= – (x+5)^2`
`-y′`	`=-((x′-5)+5)^2`
`y′`	`=(x′)^2`

`=> A`

Binomial, EXT1 2008 HSC 6c

Let `p` and `q` be positive integers with `p ≤ q`.

Use the binomial theorem to expand `(1 + x) ^(p+ q)`, and hence write down the term of
`((1 + x)^(p + q))/(x^q)` which is independent of `x`. (2 marks)
Given that
`((1 + x)^(p + q))/(x^q) = (1 + x)^p(1 + 1/x)^q`,

apply the binomial theorem and the result of part(i) to find a simpler expression for
`1 + ((p),(1))((q),(1)) + ((p),(2))((q),(2)) + … + ((p),(p))((q),(p))`. (3 marks)

Show Answers Only

`\ ^(p + q)C_q`
`\ ^(p + q)C_q`

Show Worked Solution

(i) `(1 + x)^(p + q)`

`=\ ^(p + q)C_0 +\ ^(p + q)C_1 x + … +\ ^(p + q)C_q x^q + … +\ ^(p + q)C_(p + q) · x^(p + q)`

`:.\ text(Independent term of)\ ((1 + x)^(p + q))/(x^q)`

`= (\ ^(p + q)C_q·x^q)/(x^q)`

`=\ ^(p + q)C_q`

(ii) `(1 + x)^p(1 + 1/x)^q`

`= (\ ^pC_0 +\ ^pC_1 x + … +\ ^pC_p x^p)`

`xx (\ ^qC_0 +\ ^qC_1 · 1/x + … +\ ^qC_p · 1/(x^p) + … +\ ^qC_q · 1/(x^q))`

`text(The independent term in this expansion)`

`=\ ^pC_0 ·\ ^qC_0 +\ ^pC_1 ·\ ^qC_1 + … +\ ^pC_p ·\ ^qC_p\ text{(since}\ p ≤ q)`

`= 1 +\ ^pC_1 ·\ ^qC_1 + … +\ ^pC_p ·\ ^qC_p`

`text(S)text(ince)\ ((1 + x)^(p + q))/(x^q) = (1 + x)^p(1 + 1/x)^q,\ text(the independent)`

`text(terms are equal.)`

`:.\ ^(p + q)C_q\ text(is a simpler expression for)`

`1 +\ ^pC_1 ·\ ^qC_1 +\ ^pC_2 ·\ ^qC_2 + … +\ ^pC_p ·\ ^qC_p`

Calculus, 2ADV C3 2004 HSC 10b

The diagram shows a triangular piece of land `ABC` with dimensions `AB = c` metres, `AC = b` metres and `BC = a` metres, where `a ≤ b ≤ c`.

The owner of the land wants to build a straight fence to divide the land into two pieces of equal area. Let `S` and `T` be points on `AB` and `AC` respectively so that `ST` divides the land into two pieces of equal area.

Let `AS = x` metres, `AT = y` metres and `ST = z` metres.

Show that `xy = 1/2 bc`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Use the cosine rule in triangle `AST` to show that

`z^2 = x^2 + (b^2c^2)/(4x^2) − bc cos A.` (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Show that the value of `z^2` in the equation in part (ii) is a minimum when

`x = sqrt((bc)/2)`. (4 marks)

--- 8 WORK AREA LINES (style=lined) ---
Show that the minimum length of the fence is `sqrt(((P − 2b)(P − 2c))/2)` metres, where `P = a + b + c`.

(You may assume that the value of `x` given in part (iii) is feasible.) (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

i.	`text(Area)\ \ ΔABC`	`= 1/2\ bc sin A`
	`text(Area)\ \ ΔAST`	`= 1/2\ xy sin A`

`text(Area)\ \ ΔAST`	`= 1/2 xx text(Area)\ \ ΔABC\ \ \ text{(given)}`
`1/2\ xy sin A`	`= 1/2 xx 1/2\ bc sin A`
`xy sin A`	`= 1/2\ bc sin A`
`:. xy`	`= 1/2\ bc\ \ …\ text(as required.)`

ii. `text(Using the cosine rule in)\ \ ΔAST,`

`z^2 = x^2 + y^2 − 2xy cos A`

`text(S)text(ince)\ \ xy`	`= 1/2\ bc`
`:. y`	`= (bc)/(2x)`

`:. z^2`	`= x^2 + ((bc)/(2x))^2 − 2x((bc)/(2x)) cos A`
	`= x^2 + (b^2c^2)/(4x^2) − bc cos A`

iii.	`z^2`	`= x^2 + ((b^2c^2)/4)x^(−2) − bc cos A`
	`(d(z^2))/(dx)`	`= 2x − (2b^2c^2)/4 x^(−3)= 2x − (b^2c^2)/(2x^3)`
	`(d^2(z^2))/(dx^2)`	`= 2 + (3b^2c^2)/2 x^(−4)= 2 + (3b^2c^2)/(2x^4)`

`text(SP’s occur when)\ \ (d(z^2))/(dx)=0`

`2x − (b^2c^2)/(2x^3)`	`= 0`
`4x^4 − b^2c^2`	`= 0`
`4x^4`	`= b^2c^2`
`x^4`	`= (b^2c^2)/4`
`x^2`	`= (bc)/2`
`x`	`= sqrt((bc)/2),\ \ \ (x > 0)`

` (d^2(z^2))/(dx^2)=2 + (3b^2c^2)/(2x^4)>0\ \ \ text{(for all}\ xtext{)}`

`:.\ text(A minimum value of)\ z^2\ text(when)\ x = sqrt((bc)/2).`

iv. `cos A = (b^2 + c^2 − a^2)/(2bc)`

`text(When)\ \ x = sqrt((bc)/2),`

`:. z^2`	`= x^2 + (b^2c^2)/(4x^2) − bc cos A`
	`= (bc)/2 + (b^2c^2)/(4((bc)/2)) − bc((b^2 + c^2 − a^2)/(2bc))`
	`= (bc)/2 + (bc)/2 − ((b^2 + c^2 − a^2)/2)`
	`= (2bc – b^2 − c^2 +a^2)/2`
	`= (a^2 − (b^2 − 2bc + c^2))/2`
	`= 1/2[a^2 − (b − c)^2]`
	`= 1/2[(a − (b − c))(a + (b − c))]`
	`= 1/2[(a − b + c)(a + b − c)]`
	` = 1/2[(a + b + c − 2b)(a + b + c − 2c)]`
	`= ((P − 2b)(P − 2c))/2\ \ \ text{(using}\ \ P = a + b + c text{)}`

`:.z = sqrt(((P − 2b)(P − 2c))/2)\ \ text(metres)\ \ …\ text(as required.)`

Calculus, 2ADV C3 2004 HSC 9c

Consider the function `f(x) = (log_e x)/x`, for `x > 0`.

Show that the graph of `y = f(x)` has a stationary point at `x = e`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
By considering the gradient on either side of `x = e`, or otherwise, show that the stationary point is a maximum. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Use the fact that the maximum value of `f(x)` occurs at `x = e` to deduce that `e^x ≥ x^e` for all `x > 0`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

i. `f(x) = (log_e x)/x, text(for)\ x > 0`

`text(Using the quotient rule,)`

`f′(u/v)`	`=(u′v-uv′)/v^2`
`f′(x)`	`= (x · d/(dx) (log_e x) − log_e x · d/(dx) (x))/(x^2)`
	`= (x(1/x) − log_e x · 1)/(x^2)`
	`= (1 − log_e x)/(x^2)`

`text(SP’s occur when)\ \ f′(x)=0`

`(1 − log_e x)/(x^2)`	`= 0`
`1 − log_e x`	`= 0`
`log_e x`	`= 1`
`:. x`	`= e`

`:. text(There is a stationary point at)\ \ x = e.`

ii.

`text(When)\ \ x < e, log_e x<1\ \ \ text{(from graph)}`

`1 − log_e x`	`> 0`
`(1 − log_e x)/(x^2)`	`> 0`
`:. f′(x)`	`> 0`

`text(When)\ x > e, log_e x>1\ \ \ text{(from graph)}`

`1 − log_e x`	`< 0`
`(1 − log_e x)/(x^2)`	`< 0`
`:. f′(x)`	`< 0`

`:.\ text(The stationary point at)\ \ x = e\ \ text(is a maximum.)`

iii.	`f(e)`	`= (log_e x)/e`
		`= 1/e`

`:. f(x)\ \ text(has a maximum value at)\ \ (e,1/e)`

MARKER’S COMMENT: Very challenging for most students. Successful students recognised the link to part (ii).

`:. (log_e x)/x`	`≤ 1/e`
`log_e x`	`≤ x/e`
`e log_e x`	`≤ x`
`log_e x^e`	`≤ x`
`e^(log_e x^e)`	`≤ e^x`
`x^e`	`≤ e^x\ \ \ text{(using}\ e^(log x) = x)`
`:. e^x`	`≥ x^e \ \ \ text{(for}\ \ x > 0text{)}`

Calculus, EXT1* C1 2004 HSC 9b

A particle moves along the `x`-axis. Initially it is at rest at the origin. The graph shows the acceleration, `a`, of the particle as a function of time `t` for `0 ≤ t ≤ 5`.

Write down the time at which the velocity of the particle is a maximum. (1 marks)

--- 2 WORK AREA LINES (style=lined) ---
At what time during the interval `0 ≤ t ≤ 5` is the particle furthest from the origin? Give brief reasons for your answer. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

Show Worked Solution

i. `text(The velocity of the particle increases until)`

`t = 2\ \ text{(when}\ \ a=0text{)}.`

`:.\ text(Maximum velocity occurs at)\ \ t = 2.`

ii. `text(It is furthest from the origin when its positive velocity)`

`text(slows and becomes zero. This occurs when the “net” area)`

`text(under the curve is zero).`

`text(The area under the curve and above the)\ x text(-axis)`

`text(between)\ \ t=0 and 2,\ text(is equal to the area above)`

`text(the curve and below the)\ x text(-axis between)\ \ t=2 and 4.`

`:.\ text(Furthest from origin when)\ \ t=4.`

Calculus, 2ADV C4 2004 HSC 8b

The diagram shows the graph of the parabola `x^2 = 16y`. The points `A (4, 1)` and `B (−8, 4)` are on the parabola, and `C` is the point where the tangent to the parabola at `A` intersects the directrix.

Write down the equation of the directrix of the parabola `x^2 = 16y`. (1 mark)
Find the equation of the tangent to the parabola at the point `A`. (2 marks)
Show that `C` is the point `(−6, −4)`. (1 mark)
Given that the equation of the line `AB` is `y = 2 − x/4`, find the area bounded by the line `AB` and the parabola. (2 marks)
Hence, or otherwise, find the shaded area bounded by the parabola, the tangent at `A` and the line `BC`. (3 marks)

Show Answers Only

`y = −4`
`y = x/2 − 1`
`text(Proof)\ \ text{(See Worked Solutions)}`
`18\ \ text(u²)`
`27\ \ text(u²)`

Show Worked Solution

(i)	`x^2`	`= 4ay`
		`= 16y`
	`:.4a`	`= 16`
	`a`	`= 4`

`:.\ text(Equation of the directrix is)\ \ y = −4.`

(ii)	`x^2`	`= 16y`
	`y`	`= (x^2)/(16)`
	`:.(dy)/(dx)`	`= (2x)/(16)=x/8`
	`text(At)\ x = 4`
	`dy/dx`	`= 1/2`

`:.\ text(Equation of the tangent passes through)\ \ A(4, 1)`

`\ text(with)\ \ m = 1/2`

`y − 1`	`= 1/2(x − 4)`
`y − 1`	`= x/2 − 2`
`y`	`= x/2 − 1`

(iii) `C\ text(lies on)\ \ y=-4\ \ text{(directrix)}`

`text(S)text(ince)\ \ C\ \ text(also lies on the tangent,)`

`y`	`= x/2 − 1`
`−4`	`= x/2 − 1`
`−3`	`= x/2`
`:. x`	`= −6`

`:.C\ \ text{is the point (–6, – 4) … as required}`

(iv)	`text(Equation of parabola is)\ \ x^2`	`= 16y`
	`:. y`	`= (x^2)/(16)`

`text(Required area)`

`= int_(−8)^4 (2 − x/4)\ dx − int_(−8)^4 (x^2)/16\ dx`

`= int_(−8)^4 (2 − x/4 − (x^2)/16)\ dx`

`= [2x − (x^2)/8 − (x^3)/48]_(−8)^(\ \ \ 4)`

`= (2(4) − (4^2)/8 − (4^3)/48) − (2(−8) − ((−8)^2)/8 − ((−8)^3)/48)`

`= (8 − 16/8 − 64/48) − (−16 − 64/8 − ((−512)/48))`

`= (6 − 4/3) − (−16 − 8 + 32/3)`

`= 18`

`:.\ text(Area between the parabola and)\ \ AB`

`= 18\ \ text(u²).`

(v) `text(Shaded area = Area of)\ ΔABC − 18\ \ \ text{(from part (iv))}`

`text(Area of)\ \ ΔABC = 1/2 xx b × h`

`h=\ text(⊥ distance of)\ \ C text{(−6, − 4)}\ \ text(from)\ \ AB,`

`text(where AB has equation)`	`4y`	`= 8 − x`
	`x + 4y − 8`	`= 0`

`:.h`	`= \|(ax_1 + by_1 + c)/(sqrt(a^2 + b^2))\|`
	`= \|(1(−6) + 4(−4) + (−8))/(sqrt((1)^2 + (4)^2))\|`
	`= \|(−6 − 16 − 8)/(sqrt(1 + 16))\|`
	`= \|(−30)/sqrt(17)\|`
	`= 30/sqrt17`
`AB`	`= sqrt((−8 − 4)^2 + (4 − 1)^2)`
	`= sqrt(144 + 9)`
	`= sqrt(153)`
	`= 3sqrt17`

`:.\ text(Area of)\ ΔABC`

`= 1/2 × 3sqrt17 × 30/sqrt17`

`= 45\ \ text(u²).`

`:.\ text(Shaded area)`	`= 45 − 18`
	`= 27\ \ text(u²).`

Probability, MET2 2012 VCAA 20 MC

A discrete random variable `X` has the probability function `text(Pr)(X = k) = (1 -p)^k p`, where `k` is a non-negative integer.

`text(Pr)(X > 1)` is equal to

`1 - p + p^2`
`1 - p^2`
`p - p^2`
`2p - p^2`
`(1 - p)^2`

Show Answers Only

`E`

Show Worked Solution

♦♦♦ Mean mark 19%.

`text(Pr)(X > 1)`	`= 1 – text(Pr)(X = 0) – text(Pr)(X = 1)`
	`= 1 – [(1 – p)^0p] – [(1 – p)p]`
	`= 1 – p – (p – p^2)`
	`= 1 – 2p + p^2`
	`= (1 – p)^2`

`=> E`

Polynomials, EXT2 2015 HSC 16b

Let `n` be a positive integer.

By considering `(cos alpha + i sin alpha)^(2n)`, show that
`cos(2n alpha) = cos^(2n) alpha - ((2n), (2)) cos^(2n - 2) alpha sin^2 alpha + ((2n), (4)) cos^(2n - 4) alpha sin^4 alpha - …`
1. `+ … + (-1)^(n - 1) ((2n), (2n - 2)) cos^2 alpha sin^(2n - 2) alpha + (-1)^n sin^(2n) alpha.`
Let `T_(2n) (x) = cos(2n cos^-1 x)`, for `-1 <= x <= 1.` (2 marks)
Show that
`T_(2n)(x) = x^(2n) - ((2n), (2)) x^(2n - 2)(1 - x^2) + ((2n), (4)) x^(2n - 4) (1 - x^2)^2 +`
1. `… + (-1)^n (1 - x^2)^n.` (2 marks)
By considering the roots of `T_(2n) (x)`, find the value of
1. `cos(pi/(4n)) cos((3 pi)/(4n)) …\ cos (((4n - 1) pi)/(4n)).` (3 marks)
Prove that
1. `1 - ((2n), (2)) + ((2n), (4)) - ((2n), (6)) + … + (-1)^n ((2n), (2n)) = 2^n cos ((n pi)/2).` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(-1)^n/2^(2n – 1)`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `text(By De Moivre)`

`(cos alpha + i sin alpha)^(2n)=cos (2n alpha) + i sin (2n alpha)`

`text(By the Binomial Theorem)`

`(cos alpha + i sin alpha)^(2n)`

`=cos^(2n) alpha + ((2n), (1)) cos^(2n – 1) alpha (i sin alpha) + ((2n), (2)) cos^(2n – 2) alpha (isin alpha)^2 +`

`… ((2n), (2n – 2)) cos^2 alpha (isin alpha)^(2n-2) + ((2n), (2n – 1)) cos alpha (i sin alpha)^(2n-1) + (i sin alpha)^(2n)`

`text(Equating real parts,)`

`cos(2n alpha) = cos^(2n) alpha – ((2n), (2)) cos^(2n – 2) alpha sin^2 alpha + … `

`+ (-1)^(n – 1) ((2n), (2n – 2)) cos^2 alpha sin^(2n – 2) alpha + (-1)^n sin^(2n) alpha`

(ii) `T_(2n) (x) = cos (2n cos^-1 x)\ \ text(for)\ \ -1 <= x <= 1`

`text(Given)\ \ alpha = cos^-1 x,\ \ =>x = cos alpha`

`text(Substituting)\ \ x = cos alpha\ \ text{into part (i),}`

`cos (2n cos^-1 x)`

`= x^(2n) – ((2n), (2)) x^(2n – 2) (1 – cos^2 alpha) + ((2n), (4)) x^(2n – 4) (1 – cos^2 alpha)^2 +`

`… + (-1)^n (1 – cos^2 alpha)^n`

`:. T_(2n) (x) = x^(2n) – ((2n), (2)) x^(2n – 2) (1 – x^2) + ((2n), (4)) x^(2n – 4)(1 – x^2)^2 +`

`… + (-1)^n (1 – x^2)^n`

♦♦♦ Mean mark 8%!
STRATEGY: Note that finding the correct roots gained 2 full marks in this part.

(iii) `T_(2n) (x) = 0\ \ text(when)`

`cos (2n cos^-1 x)`	`=0`
`2n cos^-1 x`	`=pi/2, (3 pi)/2, (5 pi)/2, … , ((2k + 1) pi)/2,\ \ k = 0, 1, 2 … , (2n – 1)`
`cos^-1 x`	`= ((2k + 1) pi)/(4n),\ \ k = 0, 1, 2 …, (2n – 1)`
`:. x`	`= cos(pi/(4n)), cos((3 pi)/(4n)), … , cos((4n – 1) pi)/(4n)`

`=>T_(2n) (x) = 0\ \ text(has degree)\ \ 2n`

`=>T_(2n) (x) = 0\ \ text(has)\ \ 2n\ \ text(distinct roots)`

`:.\ text(Product of roots of)\ \ T_(2n) (x) = 0\ \ text(is the constant term)`

`text(divided by the coefficient of)\ \ x^(2n).`

`text(Constant term is)\ \ T_(2n) (0)=(-1)^n`

`text(Coefficient of)\ \ x^(2n)\ \ text(is)\ \ (1 + ((2n), (2)) + ((2n), (4)) + … + 1)`

`:.cos(pi/(4n)) cos((3 pi)/(4n)) cos ((5 pi)/(4n)) …\ cos(((4n – 1) pi)/(4n))`

`=((-1)^n)/((1 + ((2n), (2)) + ((2n), (4)) + … + 1))`

`text(*The denominator can be further simplified to)\ \ 2^(2n-1)\ \ text(by)`

`text(using the binomial expansion of)\ \ (1+x)^(2n)`

♦♦ Mean mark 14%.

(iv) `cos(2n cos^-1 x)= cos ((n pi)/2)\ \ text(when)`

`cos^-1 x=pi/4\ \ \ =>x=1/sqrt2`

`text{Using part (ii)}`

`cos((n pi)/2)`	`=(1/sqrt 2)^(2n) – ((2n), (2)) (1/sqrt 2)^(2n – 2) (1/2) +`
	`((2n), (4)) (1/sqrt 2)^(2n – 4) (1/2)^2 – … + (-1)^n (1/2)^n`
	` = 1/2^n – ((2n), (2)) 1/2^(n-2) 1/2^2 + ((2n), (4)) 1/2^(n-4) 1/2^4 – … + (-1)^n 1/2^n`
	`=1/2^n – ((2n), (2)) 1/2^n + ((2n), (4)) 1/2^n – … + (-1)^n 1/2^n`
`:.2^n cos ((n pi)/2)`	`=1 – ((2n), (2)) + ((2n), (4)) – ((2n), (6)) + … + (-1)^n`

Harder Ext 1 Topics, EXT2 2015 HSC 16a

A table has `3` rows and `5` columns, creating `15` cells as shown.
Counters are to be placed randomly on the table so that there is one counter in each cell. There are `5` identical black counters and `10` identical white counters.
Show that the probability that there is exactly one black counter in each column is `81/1001.` (2 marks)
The table is extended to have `n` rows and `q` columns. There are `nq` counters, where `q` are identical black counters and the remainder are identical white counters. The counters are placed randomly on the table with one counter in each cell.
Let `P_n` be the probability that each column contains exactly one black counter.
Show that `P_n = n^q/(((nq), (q))).` (2 marks)
Find `lim_(n -> oo) P_n.` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(q!)/q^q`

Show Worked Solution

(i) `text(Arrangements of 5 black counters in 15 cells)`

`=\ ^15C_5`

`text(In each column, there are 3 cells to place a black counter.)`

`:.\ text(Ways to place a black counter in each column)`

`\ =3^5=243`

`:.P text{(1 black counter in each column)`

`=243/(\ ^15C_5`

`=81/1001`

(ii) `text{Similarly to part (i),}`

♦♦ Mean mark 16%.

`text(Arrangements of)\ \ q\ \ text(black counters in)\ \ nq\ \ text(cells)`

`=\ ^(nq)C_q`

`text(In each column, there are)\ \ n\ \ text(cells to place a black counter.)`

`:.\ text(Ways to place a black counter in each of)\ \ q\ \ text(columns)`

`\ =n^q`

`:.P text{(1 black counter in each column)`

`=n^q/(\ ^(nq)C_q)`

♦♦♦ Mean mark 5%.
COMMENT: Lowest State mean mark on the 2015 exam.

(iii)	`lim_(n -> oo) P_n`	`= lim_(n -> oo) n^q/(\ ^(nq)C_q)`
		`= lim_(n -> oo)[n^q xx (q!)/((nq)(nq – 1)…(nq – q + 1))]`
		`= lim_(n -> oo)[(n^q q!)/(n^q q(q – 1/n)(q – 2/n) … (q – (q – 1)/n))]`
		`= lim_(n -> oo)[(q!)/(q(q – 1/n)(q – 2/n) … (q – (q – 1)/n))]`
		`= (q!)/q^q`

Proof, EXT2 P1 2015 HSC 15b

Suppose that `x >= 0` and `n` is a positive integer.

Show that `1 - x <= 1/(1 + x) <= 1.` (2 marks)

--- 2 WORK AREA LINES (style=lined) ---
Hence, or otherwise, show that

`1 - 1/(2n) <= n ln (1 + 1/n) <= 1.` (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Hence, explain why

`lim_(n -> oo) (1 + 1/n)^n = e.` (1 mark)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

i.	`1-x^2`	`<=1\ \ \ text(for)\ x>=0`
	`(1-x)(1+x)`	`<=1`
	`(1-x)`	`<=1/(1+x)`

`text(S)text(ince)\ \ 1 + x >= 1\ \ text(when)\ \ x >= 0`

`=>1/(1 + x) <= 1`

`:. 1 – x <= 1/(1 + x) <= 1.`

♦♦ Mean mark 21%.
STRATEGY: The conversion of the middle term from a fraction into a logarithm should flag the need for integration of each term.

ii.	`int_0^(1/n) (1 – x)\ dx`	`<= int_0^(1/n) (dx)/(1 + x) <= int_0^(1/n) 1\ dx`
	`[x – x^2/2]_0^(1/n)`	`<= [ln (1 + x)]_0^(1/n) <= [x]_0^(1/n)`
	`1/n-1/(2n^2)`	`<= ln (1 + 1/n) <= 1/n`
	`1 – 1/(2n) `	`<= n ln (1 + 1/n) <= 1,\ \ \ \ \ (n>=1)`

iii.	`lim_(n -> oo) (1 – 1/(2n))`	`<= lim_(n -> oo){n ln (1 + 1/n)} <= lim_(n -> oo) (1)`
	`1`	`<= lim_(n -> oo) {l ln (1 + 1/n)} <= 1`

♦♦ Mean mark 29%.

`:. lim_(n -> oo) (n ln (1 + 1/n))`	`=1`
`lim_(n -> oo) ln (1 + 1/n)^n`	`=1`
`:.lim_(n -> oo) (1 + 1/n)^n`	`=e`

Harder Ext1 Topics, EXT2 2006 HSC 8b

For `x > 0`, let `f(x) = x^n e^-x`, where `n` is an integer and `n >= 2.`

The two points of inflection of `f(x)` occur at `x = a` and `x = b`, where `0 < a < b`. Find `a` and `b` in terms of `n.` (4 marks)
Show that
1. `(f(b))/(f(a)) = ((1 + 1/sqrt n)/(1 - 1/sqrt n))^n e^(-2 sqrt n).` (2 marks)
Using the following
If `0<=x<=1/sqrt2` then `1 <= ((1 + x)/(1 - x)) e^(-2x) <= e^((4x^3)/3),` (DO NOT prove this)
2. show that `1 <= (f(b))/(f(a)) <= e^(4/(3 sqrt n)).` (2 marks)
What can be said about the ratio `(f(b))/(f(a))` as `n -> oo?` (1 mark)

Show Answers Only

`a = n – sqrt n,\ \ \ \ \ b = n + sqrt n`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`lim_(n -> oo) (f(b))/(f(a)) = 1`

Show Worked Solution

(i)	`f(x)`	`= x^n e^-x,\ \ \ \ n >= 2,\ \ \ x > 0`
	`f′(x)`	`= nx^(n – 1) e^-x – x^n e^-x`
	`f″(x)`	`= n[(n – 1) x^(n -2) e^-x – x^(n – 1) e^-x] – (nx^(n – 1) e^-x – x^n e^-x)`
		`= x^(n – 2) e^-x [n(n – 1) – nx – nx + x^2]`
		`= x^(n – 2) e^-x (x^2-2nx +n^2 – n)`

`text(P.I.’s occur when) \ \ f″(x)=0`

`text(i.e. when)\ \ \ x^2-2nx +n^2 – n = 0,\ \ \ \ (e^-x>0 and x>0)`

`:.x=`	`(2n+-sqrt(4n^2- 41(n^2-n)))/2`
`=`	`\ \ n+- sqrtn`

`text(S)text(ince)\ \ a<b`

`:. a = n – sqrt n\ and\ b = n + sqrt n.`

(ii) `(f(b))/(f(a)) =`	`((n + sqrt n)^n e^-(n + sqrt n))/((n – sqrt n)^n e^-(n – sqrt n))`
`=`	`(n^n(1 + 1/sqrt n)^n)/(n^n(1 – 1/sqrt n)^n) xx e^(-n – sqrt n + n – sqrt n)`
`=`	`((1 + 1/sqrt n)/(1 – 1/sqrt n))^n e^(-2 sqrt n)`

(iii) `text(S)text(ince)\ \ n>=2\ \ \ => 1/sqrt n <=1/sqrt2`

`text(Substituting)\ \ x=1/sqrt n\ \ text(into the inequality)`

`1`	`<= ((1 + 1/sqrt n)/(1 – 1/sqrt n)) e^((-2)/sqrt n) <= e^(4/(3(sqrt n)^3))`
`1^n`	`<= ((1 + 1/sqrt n)/(1 – 1/sqrt n))^n e^((-2)/sqrt n xx n) <= e^(4/(3(sqrt n)^3) xx n)`
`:.1`	`<= (f(b))/(f(a)) <= e^(4/(3 sqrt n))`

(iv)	`lim_(n -> oo) e^(4/(3 sqrt n))`	`=1`
	`:.lim_(n -> oo) (f(b))/(f(a))`	`=1`

Calculus, EXT2 C1 2006 HSC 7b

Let `I_n = int_0^x sec^n t\ dt`, where `0 <= x <= pi/2`.

Show that `I_n = (sec^(n - 2) x tan x)/(n - 1) + (n - 2)/(n - 1) I_(n - 2).` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Hence find the exact value of

`int_0^(pi/3) sec^4 t\ dt.` (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`2 sqrt 3`

Show Worked Solution

STRATEGY: The choice of `u=sec^(n-2)t` and `v=sec^2 t` was extremely important in being able to answer this question. Take note!

i. `I_n`	`=int_0^x sec^n t\ dt,\ \ \ 0 <= x < pi/2`
	`=int_0^x sec^(n – 2)t sec^2 t\ dt`

`text(Integrating by parts)`

`u`	`=sec^(n-2)t,`	`u′`	`=(n-2) sec^(n-2)t tan\ t`
`v`	`=tan\ t,`	`v′`	`=sec^2 t`

`I_n`	`=[tan t sec^(n – 2)t]_0^x – int_0^x tan t (n – 2) sec^(n – 3) t sec t tan t\ dt`
	`=tan x sec^(n – 2) x – (n – 2) int_0^x sec^(n – 2) t tan^2 t\ dt`
	`=tan x sec^(n – 2) x – (n – 2) int_0^x sec^(n – 2) t (1+sec^2 t)\ dt`
	`=tan x sec^(n – 2) x – (n – 2) int_0^x sec^n t\ dt + (n – 2) int_0^x sec^(n – 2)t\ dt`
	`=tan x sec^(n – 2) x-(n-2)I_n+(n-2)I_(n-2)`

`I_n + (n – 2) I_n`	`= tan x sec^(n – 2) x + (n – 2) I_(n – 2)`
`(n – 1)I_n`	`= tan x sec^(n – 2) x + (n – 2) I_(n – 2)`
`:.I_n `	`= (tan x sec^(n – 2) x)/(n – 1) + ((n – 2)/(n – 1)) I_(n – 2)`

ii. `int_0^(pi/3) sec^4 t\ dt =`	`(tan\ pi/3 sec^2\ pi/3)/3 + 2/3 int_0^(pi/3) sec^2 t\ dt`
`=`	`(sqrt 3 xx 4)/3 + 2/3 [tan t]_0^(pi/3)`
`=`	`(4 sqrt 3)/3 + 2/3 (sqrt 3 – 0)`
`=`	`2 sqrt 3`

Harder Ext1 Topics, EXT2 2006 HSC 4d

In the acute-angled triangle `ABC, \ K` is the midpoint of `AB, \ L` is the midpoint of `BC` and `M` is the midpoint of `CA`. The circle through `K, L` and `M` also cuts `BC` at `P` as shown in the diagram.

Copy or trace the diagram into your writing booklet.

Prove that `KMLB` is a parallelogram. (1 mark)
Prove that `/_ KPB = /_ KML.` (1 mark)
Prove that `AP _|_ BC.` (2 marks)

Show Answers Only

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

Show Worked Solution

(i)

`(AK)/(KB) = (AM)/(MC) = 1/1`

`:. KM\ text(\|\|)\ BC`	`\ \ \ text{(parallel lines cut in the}` `\ \ \ \ text{same proportion)`

`text(Similarly,)\ \ (CL)/(LB) = (CM)/(MA) = 1/1`

`:. ML\ text(||)\ AB`

`:. KMLB\ \ text(is a parallelogram)`

(ii)

`/_ BPK`

`= /_ KML`

`text{(exterior angle of a cyclic}`
`text{quadrilateral}\ \ KMLP text{)}`

(iii)	`/_ KBP`	`= /_ KML\ \ \ text{(opposite angles of a parallelogram)}`
	`:. /_ KBP`	`= /_ KPB\ \ \ text{(both equal}\ \ /_ KML text{)}`

`:. Delta BKP\ \ text(is isosceles)`

`text(S)text(ince)\ \ BK = KP=KA\ \ \ text{(given}\ K\ \ text(is the midpoint of)\ \ ABtext{)}`

`=>K\ \ text(is the centre of a circle, diameter)\ \ AB,`

`text(that passes through)\ \ A, B and P`

`/_ APB = 90^@\ \ \ \ text{(angle in semi-circle)}`

`:. AP _|_ BC.`

Harder Ext1 Topics, EXT2 2007 HSC 8c

The diagram shows a regular `n`-sided polygon with vertices `X_1, X_2, …, X_n`. Each side has unit length. The length `d_k` of the ‘diagonal’ `X_n X_k` where `k = 1, 2, …, n - 1` is given by

`d_k = (sin\ (k pi)/n)/(sin\ pi/n).` (Do NOT prove this.)

Show, using the identity,
`sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n - 1) pi)/n = cot\ pi/(2n),` (Do NOT prove this.)
1. that `d_1 + … + d_(n - 1) = 1/(2 sin^2\ pi/(2n)).` (2 marks)
Let `p` be the perimeter of the polygon and
`q = 1/n (d_1 + … + d_(n - 1))`.
Show that `p/q = 2 (n sin\ pi/(2n))^2.` (2 marks)
Hence calculate the limiting value of `p/q` as `n -> oo.` (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`pi^2/2`

Show Worked Solution

(i) `d_k = (sin\ (k pi)/n)/(sin\ pi/n)`

`d_1 + d_2 + … + d_(n – 1)`

`=(sin\ pi/n)/(sin\ pi/n) + (sin\ (2 pi)/n)/(sin\ pi/n) + … + (sin\ ((n – 1) pi)/n)/(sin\ pi/n)`

`=1/(sin\ pi/n) xx cot\ pi/(2n)`

`=(cos\ pi/(2n))/(2 sin\ pi/(2n) cos\ pi/(2n) xx sin\ pi/(2n))`

`=1/(2 sin^2\ pi/(2n))`

(ii)	`p/q`	`=n/(1/n (d_1 + d_2 + … + d_(n – 1))`
		`=n^2 xx (2 sin^2\ pi/(2n))/1`
		`=2 (n sin\ pi/(2n))^2`

MARKER’S COMMENT: Applying the result `lim_(theta->0)\ sin theta/theta = 1` was poorly done.

(iii)	`lim_(n -> oo)\ p/q`	`= lim_(n -> oo)\ 2(n sin\ pi/(2n))^2`
		`=2n^2 xx pi^2/(4n^2) xx lim_(n -> oo)\ ((sin\ pi/(2n))^2)/(pi^2/(4n^2))`
		`=pi^2/2 xx lim_(n -> oo)\ (sin\ pi/(2n))/(pi/(2n)) xx lim_(n -> oo)\ (sin\ pi/(2n))/(pi/(2n))`
		`=pi^2/2 xx 1 xx 1\ \ \ \ \ \ text{(noting that as}\ \ n->oo, pi/(2n)->0text{)}`
		`=pi^2/2`

Complex Numbers, EXT2 N2 2007 HSC 8b

Let `n` be a positive integer. Show that if `z^2 != 1` then

`1 + z^2 + z^4 + … + z^(2n - 2) = ((z^n - z^-n)/(z - z^-1)) z^(n - 1)`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
By substituting `z = cos theta + i sin theta` where `sin theta != 0`, into part (i), show that

`1 + cos 2 theta + … + cos (2n - 2) theta + i[sin 2 theta + … + sin (2n - 2) theta]`

`= (sin n theta)/(sin theta) [cos (n - 1) theta + i sin (n - 1) theta].` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Suppose `theta = pi/(2n)`. Using part (ii), show that

`sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n - 1) pi)/n = cot\ pi/(2n).` (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

i. `1 + z^2 + z^4 + … + z^(2n – 2),\ z^2 != 1`

`text(GP where)\ a = 1,\ \ r = z^2,\ \ n\ text(terms):`

`S_n`	`=(1((z^2)^n – 1))/(z^2 – 1)`
	`=(z^(2n) – 1)/(z^2 – 1)`
	`=((z^n – z^-n))/(z – z^-1) xx z^n/z`
	`=((z^n – z^-n)/(z – z^-1))z^(n – 1)`

ii.	`z`	`= cos theta + i sin theta`
	`z^n`	`= cos n theta + i sin n theta\ \ …\ text(etc)\ \ \ \ text{(De Moivre)}`
	`z^-n`	`= cos( -n theta) + i sin (-n theta)`
		`= cos n theta – i sin n theta`

`text(LHS)`	`= 1 + (cos 2 theta + i sin 2 theta) + (cos 4 theta + i sin 4 theta) + `
	`… + (cos(2n – 2) theta + i sin (2n – 2) theta)`
	`= 1 + cos 2 theta + cos 4 theta + … + cos (2n – 2) theta + `
	`i (sin 2 theta + sin 4 theta + … + sin (2n – 2) theta)`

`text{Using part (i):}`

`text(LHS)`	`=((cos n theta + i sin n theta – cos n theta + i sin n theta))/(cos theta + i sin theta – cos theta + i sin theta) xx`
	`[cos (n – 1) theta + i sin (n – 1) theta]`
	`=(2 i sin n theta)/(2 i sin theta) [cos (n – 1) theta + i sin (n – 1) theta]`
	`=(sin n theta)/(sin theta) [cos (n – 1) theta + i sin (n – 1) theta]\ \ text(… as required.)`

iii. `text{Equating the imaginary parts in part (ii):}`

`sin 2 theta + sin 4 theta + … + sin 2 (n – 1) theta = (sin n theta sin (n – 1) theta)/(sin theta)`

`text(When)\ \ theta = pi/(2n),`

`sin\ (2 pi)/(2n) + sin\ (4 pi)/(2n) + … + sin\ (2(n – 1) pi)/(2 n) = (sin\ (n pi)/(2n) sin\ ((n – 1) pi)/(2n))/(sin\ pi/(2n))`

`:. sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n – 1) pi)/n`

`=(sin\ pi/2)/(sin\ pi/(2n)) xx sin\ ((n – 1) pi)/(2n)`

`=1/(sin\ pi/(2n)) sin (pi/2 – pi/(2n))`

`=(cos\ pi/(2n))/(sin\ pi/(2n))`

`=cot\ pi/(2n)`

Calculus, EXT2 C1 2007 HSC 8a

Using a suitable substitution, show that

`int_0^a f(x)\ dx = int_0^a f(a - x)\ dx.` (1 mark)

--- 6 WORK AREA LINES (style=lined) ---
A function `f(x)` has the property that `f(x) + f(a - x) = f(a).`

Using part (i), or otherwise, show that

`int_0^a f(x)\ dx = a/2\ f(a).` (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

i. `text(Show)\ \ int_0^a f(x)\ dx = int_0^a f(a – x)\ dx.`

`text(Let)\ \ x = a – u,\ \ dx = -du`

`text(When)\ \ x = 0,\ \ u = a`

`text(When)\ \ x = a,\ \ u = 0`

`:. int_0^a f(x)\ dx`	`=int_a^0 f(a – u) (-du)`
	`=int_0^a f(a – u)\ du`
	`=int_0^a f(a – x)\ dx\ \ text{.. as required}`

MARKER’S COMMENT: Integrating `f(a)` was poorly done in part (ii). Pay careful attention to this in the Worked Solution.

ii. `f(x) = f(a) – f(a – x)`

`int_0^a f(x)\ dx`	`=int_0^a [f(a) – f(a – x)]\ dx`
	`=int_0^a f(a)\ dx – int_0^a f(a – x)\ dx`
	`=int_0^a f(a)\ dx – int_0^a f(x)\ dx`
`2 int_0^a f(x)\ dx`	`=int_0^a f(a)\ dx`
	`=[f(a) xx x]_0^a`
`int_0^a f(a)\ dx`	`=(f(a))/2 (a – 0)`
	`=a/2\ f(a)`

Proof, EXT2 P1 2007 HSC 7a

Show that `sin x < x` for `x > 0.` (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Let `f(x) = sin x - x + x^3/6`.

Show that the graph of `y = f(x)` is concave up for `x > 0.` (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
By considering the first two derivatives of `f(x)`,show that `sin x > x - x^3/6` for `x > 0.` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

i.	`text(Let)\ \ g(x)`	`=sin x-x`
	`g′(x)`	`=cosx-1<=1\ \ \ text(for all)\ x>0`

MARKER’S COMMENT: A geometric proof using arc length and a right angled triangle caused problems as few students could deal with the case `x>pi`.

`=>g(x)\ \ text(is a decreasing function)`

`text(When)\ \ x=0,\ \ g(0)=0`

`text(Considering)\ \ g(x)\ \ text(when)\ \ x>0,`

`g(x)`	`<0`
`sinx -x`	`<0`
`sin x`	`<x\ \ \ text(for all)\ x>0`

ii.	`f(x)`	`=sin x – x + x^3/6`
	`f prime (x)`	`=cos x – 1 + x^2/2`
	`f ″ (x)`	`=x – sin x`
	`:.\ f″ (x)`	`> 0\ \ \ \ text{(using part (i))}`

`:. f(x)\ \ text(is concave up for)\ \ x > 0.`

iii. `f″(x)>0\ \ \ \ text{(part (ii))}`

`=>f′(x)\ \ text(is an increasing function)`

`text(When)\ \ x=0,\ \ f′(0)=0`

`=>f′(x)>0\ \ \ text(for)\ \ x>0`

`:. f(x)\ \ text(has a positive gradient that steepens)`

`text(for)\ \ x>0, and f(0)=0`

`f(x)`	`>0`
`sin x – x + x^3/6`	`>0`

`:.sin x > x – x^3/6\ \ \ text(for)\ \ x>0`

Calculus, EXT2 C1 2012 HSC 10 MC

Without evaluating the integrals, which one of the following integrals is greater than zero?

`int_(−pi/2)^(pi/2) x/(2 + cos x)\ dx`
`int_(−pi)^pi x^3 sin x\ dx`
`int_(−1)^1 (e^(−x^2) − 1)\ dx`
`int_(−2)^2 tan^(−1)(x^3)\ dx`

Show Answers Only

`B`

Show Worked Solution

`text{Consider (A) and (D)}`

`f(x)=-f(-x)\ \ =>\ text(ODD functions where)`

`int_(−a)^a f(x)\ dx = 0`

`text{Consider (C)}`

`e^(−x^2)<1\ \ text(for all)\ x\ \ => e^(−x^2) − 1<0`

`:. text(Its graph is below the)\ x text(-axis and any integral)`

`text(will be negative)`

`text{Consider (B)}`

`text{(B)}\ text(is an even function where,)`

`x^3 sinx>=0\ \ text(for)\ \ \ -pi<=x<=pi`

`=>B`

Calculus, EXT2 C1 2014 HSC 10 MC

Which integral is necessarily equal to `int_(−a)^a f(x)\ dx`?

`int_0^a f(x) − f(−x)\ dx`
`int_0^a f(x) − f(a − x)\ dx`
`int_0^a f(x − a) + f(−x)\ dx`
`int_0^a f(x − a) + f(a − x)\ dx`

Show Answers Only

`D`

Show Worked Solution

`int_(−a)^a f(x)\ dx= int_0^a f(x)\ dx + int_(−a)^0 f(x)\ dx`

`text(Consider)\ \ int_0^a f(x)\ dx`

♦♦ Mean mark 29%.

`text(Let)\ u = a − x\ \ => x = a − u\ \ text(and)\ \ dx = −du`

`int_0^a f(x)\ dx`	`= int_0^a f(a − u) − du`
	`=int_a^0 f(a-u)\ du`
	`= int_0^a f(a − x)\ dx`

`text(Consider)\ \ int_(−a)^0 f(x)\ dx`

`text(Let)\ u = x + a\ \ => x = u − a\ \ text(and)\ \ dx = du`

`int_(−a)^0 f(x)\ dx`	`= int_0^a f(u − a)\ du`
	`= int_0^a f(x − a)\ dx`
`:.int_(−a)^a f(x)\ dx`	`= int_0^a f(x − a)+ f(a − x)\ dx`

`=> D`

Harder Ext1 Topics, EXT2 2013 HSC 10 MC

A hostel has four vacant rooms. Each room can accommodate a maximum of four people.

In how many different ways can six people be accommodated in the four rooms?

4020
4068
4080
4096

Show Answers Only

`A`

Show Worked Solution

`text(If NO maximum)`

♦♦♦ Mean mark 15%.
STRATEGY: The group of 5 can be in any of the four rooms, and note that the remaining person can then be in any of the 3 remaining rooms.

`text(# Ways) = 4^6 = 4096`

`text(However, we can’t have 5 or 6 people in a room)`

`text(# Ways to have 6 people in 1 room) = 4`

`text(# Ways to have 5 people in 1 room)`

`=\ ^6C_5 ^4C_1\ ^3C_1`

`=72`

`:.\ text(Total ways with a maximum of 4)`

`=4096-4-72`

`=4020`

`=> A`

Harder Ext1 Topics, EXT2 2009 HSC 8c

A game is being played by `n` people, `A_1, A_2, ..., A_n`, sitting around a table. Each person has a card with their own name on it and all the cards are placed in a box in the middle of the table. Each person in turn, starting with `A_1`, draws a card at random from the box. If the person draws their own card, they win the game and the game ends. Otherwise, the card is returned to the box and the next person draws a card at random. The game continues until someone wins.

Let `W` be the probability that `A_1` wins the game.

Let `p = 1/n and q = 1 - 1/n`.

Show that `W = p + q^n W.` (1 mark)
Let `m` be a fixed positive integer and let `W_m` be the probability that `A_1` wins in no more than `m` attempts.
Use `e^(-n/(n - 1)) < (1 - 1/n)^n < e^-1,`
to show that, if `n` is large, `W_m/W` is approximately equal to `1 - e^-m.` (3 marks)

Show Answers Only

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

Show Worked Solution

(i) `text(Chance of drawing own card is)\ \ p = 1/n`

`text(Chance of not drawing own card is)\ \ q = 1 – p = 1 – 1/n`

`P(A_1\ text{wins in round 1)}`	`= p`
`P(A_1\ text{wins in round 2)}`	`= q^n*p`
`P(A_1\ text{wins in round 3)}`	`= q^nq^np`
`:.P(A_1\ \ text{wins)}`	`= p + pq^n + pq^(2n) + …`

`=> text(G.P. where)\ \ a=p,\ \ r=q^n,\ \ and\ \ 0<q^n<1`

`W`	`=p/(1-q^n)`
`W-W q^n`	`=p`
`:.W`	`=p+W q^n`

(ii) `W_m`	`=p + q^n p + q^(2n) p + … + q^((m – 1)n) p`
	`=p ((1 – q^(nm)))/(1 – q^n)\ \ \ \ \ \ text{(G.P. with}\ m\ text{terms)}`
`W_m/W`	`=p ((1 – q^(nm)))/(1 – q^n) xx (1 – q^n)/p`
	`=1 – q^(nm)`
	`= 1 – (1 – 1/n)^(nm)`

`text(Considering)\ \ \ e^(-n/(n – 1)) < (1 – 1/n)^n < e^-1`

`text{As}\ \ n -> oo, \ \ n/(n-1)->1, \ \ e^(-n/(n – 1))->e^-1, and`

`:.(1-1/n)^n`	`~~e^-1`
`(1-1/n)^(mn)`	`~~e^-m`
`:.\ W_m/W`	`~~ 1 – e^-m`

Proof, EXT2 P2 2009 HSC 8a

Using the substitution `t = tan\ theta/2`, or otherwise, show that

`qquad cot theta + 1/2 tan\ theta/2 = 1/2 cot\ theta/2.` (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Use mathematical induction to prove that, for integers `n >= 1`,

`qquad sum_(r = 1)^n 1/2^(r - 1) tan x/2^r = 1/2^(n - 1) cot\ x/2^n - 2 cot x.` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Show that `lim_(n -> oo) sum_(r = 1)^n 1/2^(r - 1) tan\ x/2^r = 2/x - 2 cot x.` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence find the exact value of
`qquad tan\ pi/4 + 1/2 tan\ pi/8 + 1/4 tan\ pi/16 + ….` (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`4/pi`

Show Worked Solution

i. `t=tan\ theta/2, \ \ \ sin theta=(2t)/(1+t^2),\ \ \ cos theta=(1-t^2)/(1+t^2)`

`text(Prove)\ \ cot theta + 1/2 tan\ theta/2 = 1/2 cot\ theta/2`

`text(LHS)`	`=cot theta + 1/2 tan\ theta/2`
	`=cos theta/sin theta+ 1/2 tan\ theta/2`
	`=(1-t^2)/(2t)+t/2`
	`=(1-t^2+t^2)/(2t)`
	`=1/(2t)`
	`=1/2 cot\ theta/2`
	`=\ text(RHS)`

ii. `text(If)\ \ n = 1`

`text(LHS)`	`=1/2^0 tan\ x/2^1=tan\ x/2`
`text(RHS)`	`=1/2^0 cot\ x/2 – 2 cot x`
	`=cot\ x/2 – 2 cot x`

`text{Using part (i)},\ \ 1/2 tan\ theta/2`	`= 1/2 cot\ theta/2 – cot theta,`
`:.tan\ theta/2`	`= cot\ theta/2 – 2 cot theta`

`text(RHS)`	`=tan\ x/2`
	`=\ text(LHS)`
`:.\ text(True for)\ \ n=1`

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

`text(i.e.)\ \ sum_(r = 1)^k 1/2^(r – 1) tan\ x/2^r = 1/2^(k – 1) cot\ x/2^k – 2 cot x.`

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

`text(i.e.)\ \sum_(r = 1)^(k + 1) 1/2^(r – 1) tan x/2^r = 1/2^k cot x/2^(k + 1) – 2 cot x`

`text(LHS)`	`=sum_(r = 1)^(k) 1/2^(r – 1) tan x/2^r + 1/2^k tan x/2^(k + 1)`
	`=1/2^(k – 1) cot x/2^k – 2 cot x + 1/2^k tan x/2^(k + 1)`
	`=1/2^(k – 1) (cot x/2^k + 1/2 tan x/2^(k +1)) – 2 cot x,\ \ \ \ text{(Let}\ \ theta=x/2^ktext{)}`
	`=1/2^(k – 1)(cot\ theta + 1/2 tan\ theta/2) – 2 cot x`
	`=1/2^(k – 1)(1/2 cot\ theta/2) – 2 cot x,\ \ \ \ text{(from part (i))}`
	`=1/2^k cot\ theta/2 – 2 cot x`
	`=1/2^k cot x/2^(k + 1) – 2 cot x`
	`=\ text(RHS)`

`=>text(True for)\ \ n = k + 1\ \ text(if it is true for)\ \ n = k.`

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

iii. `lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan x/2^r `

`=lim_(n -> oo) (1/2^(n-1) cot x/2^n – 2 cot x)`

`=lim_(n -> oo) (2/x * x/2^n * 1/(tan x/2^n) – 2 cot x)`

`=2/x xx lim_(n -> oo) ((x/2^n)/(tan x/2^n)) – 2 cot x`

`=>text(As)\ \ n -> oo,\ \ x/2^n=theta->0, and`

`=>lim_(theta-> 0) (theta)/(tan theta) =1`

`:.lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan x/2^r =2/x – 2 cot x`

iv. `tan\ pi/4 + 1/2 tan\ pi/8 + 1/4 tan\ pi/16 + …`

`=lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan (pi/2)/2^r`

`=(2/(pi/2)) – 2 cot pi/2`

`=4/pi`

Complex Numbers, EXT2 N2 2009 HSC 7b

Let `z = cos theta + i sin theta.`

Show that `z^n + z^-n = 2 cos n theta`, where `n` is a positive integer. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Let `m` be a positive integer. Show that

`(2 cos theta)^(2m) = 2 [cos 2 m theta + ((2m), (1)) cos (2m - 2) theta + ((2m), (2)) cos (2m - 4) theta`

`+ … + ((2m), (m - 1)) cos 2 theta] + ((2m), (m)).` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Hence, or otherwise, prove that

`int_0^(pi/2) cos^(2m) theta\ d theta = pi/(2^(2m + 1)) ((2m), (m))`

where `m` is a positive integer. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

i.	`z`	`= cos theta + i sin theta`
	`z^n`	`= cos n theta + i sin n theta\ \ \ \ text{(De Moivre)}`
	`z^-n`	`= cos (-n theta) + i sin (-n theta)\ \ \ \ text{(De Moivre)}`
		`= cos n theta – i sin n theta`
	`z^n + z^-n`	`= cos n theta + i sin n theta + cos n theta – i sin n theta`
		`= 2 cos n theta,\ \ \ \ n > 0`

ii. `z + z^-1 = 2 cos theta`

`:.(2 cos theta)^(2m)`

`=(z + z^-1)^(2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 1) z^-1 + ((2m), (2)) z^(2m – 2) z^-2+`

` … + ((2m), (2m – 1)) z^1 z^-(2m – 1) + z^-(2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 2) + ((2m), (2)) z^(2m – 4)+`

` … + ((2m), (2m – 1)) z^-(2m – 2) + z^(-2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 2) + ((2m), (2)) z^(2m – 4) + … + ((2m), (m)) z^(2m-2m) …`

`+ ((2m), (2)) z^-(2m – 4) + ((2m), (1)) z^-(2m – 2) + z^(-2m)`

`=(z^(2m) + z^(-2m)) + ((2m), (1)) (z^(2m – 2) + z^-(2m – 2)) + ((2m), (2))`

`(z^(2m – 4) + z^-(2m – 4)) + … + ((2m), (m – 1)) (z + z^-1) + ((2m), (m))`

`=2 [cos 2 m theta + ((2m), (1)) cos (2m – 2) theta + ((2m), (2)) cos (2m – 4) theta`

`+ … + ((2m), (m – 1)) cos 2 theta] + ((2m), (m))`

iii. `int_0^(pi/2) cos^(2m) d theta`

`=1/(2^(2m)) int_0^(pi/2) (2 cos theta)^(2m)`

`=1/(2^(2m)) int_0^(pi/2)[2(cos 2 m theta + ((2m), (1)) cos (2m – 2) theta + ((2m), (2))`

`cos (2m – 4) theta + … + ((2m), (m – 1)) cos 2 theta) + ((2m), (m))] d theta`

`=1/(2^(2m)) [2((sin 2 m theta)/(2m) + ((2m), (1)) (sin (2m – 2) theta)/(2m – 2)`

`+ … + ((2m), (m – 1)) (sin 2 theta)/2) + ((2m), (m)) theta]_0^(pi/2)`

`=1/(2^(2m)) [2(0 + 0 + … + 0) + ((2m), (m)) pi/2 – (0)]`

`=pi/(2^(2m + 1)) ((2m), (m))`

Harder Ext1 Topics, EXT2 2009 HSC 6c

The diagram shows a circle of radius `r`, centred at the origin, `O`. The line `PQ` is tangent to the circle at `Q`, the line `PR` is horizontal, and `R` lies on the line `x = c`.

Find the length of `PQ` in terms of `x, y and r.` (1 mark)
The point `P` moves such that `PQ = PR`.
Show that the equation of the locus of `P` is
1. `y^2 = r^2 + c^2 - 2cx.` (2 marks)
Find the focus, `S`, of the parabola in part (ii). (2 marks)
Show that the difference between the length `PS` and the length `PQ` is independent of `x.` (2 marks)

Show Answers Only

`PQ = sqrt (x^2 + y^2 – r^2)`
`text(Proof)\ \text{(See Worked Solutions)}`
`(r^2/(2c), 0)`
`text(Proof)\ \text{(See Worked Solutions)}`

Show Worked Solution

(i)	`OP`	`= sqrt (x^2 + y^2)`
	`PQ^2`	`= OP^2 – OQ^2`
	`PQ^2`	`= x^2 + y^2 – r^2`
	`:.\ PQ`	`= sqrt (x^2 + y^2 – r^2)`

(ii) `text(When)\ \ PQ = PR`

`sqrt (x^2 + y^2 – r^2)`	`=c-x`
`x^2 + y^2 – r^2`	`= (c – x)^2`
`x^2 + y^2 – r^2`	`= c^2 – 2cx + x^2`
`y^2 – r^2`	`= c^2 – 2cx`

`:.\ y^2 = r^2 + c^2 – 2cx\ \ text(is the locus of)\ \ P`

(iii) `text(Rearranging the locus of)\ \ P\ \ text(in the form)`

♦♦♦ “Few” students answered part (iii) correctly (exact data unavailable).

`(y-y_0)^2`	`=4a(x-x_0)`
`y^2`	`= r^2 + c^2 – 2cx`
	`= -2c(x – (r^2 + c^2)/(2c))`

`=>\ text(The parabola is lying on its side and opening to the left.)`

`text(Vertex) = ((r^2 + c^2)/(2c),0)`

`text(Focal length) \ \ \ \ 4a`	`=-2c`
`a`	`=- c/2`

`:.S\ text(has coordinates)\ \ ((r^2 + c^2)/(2c) – c/2, 0) -=(r^2/(2c), 0)`

(iv) `text(The directrix of the parabola has the equation)`

♦♦♦ “Few” students answered part (iv) correctly (exact data unavailable).
MARKER’S COMMENT: Very few used the efficient focus-directrix definition here.

`x`	`=(r^2+c^2)/(2c) + c/2`
	`=(r^2+2c^2)/(2c)`

`text(The definition of a parabola requires that)`

`PS`	`=PM`
	`=(r^2+2c^2)/(2c)-x`

`text(S)text(ince)\ \ \ PQ`	`=PR\ \ \ \ \ \ text{(from part (ii))}`
	`=c-x`

`=> PS-PQ`	`=(r^2+2c^2)/(2c)-x-(c-x)`
	`=r^2/(2c)`

`:. PS-PQ\ \ text(is independent of)\ x.`

Integration, EXT2 2010 HSC 8

Let

`A_n = int_0^(pi/2) cos^(2n) x\ dx` and `B_n = int_0^(pi/2) x^2cos^(2n)x\ dx`,

where `n` is an integer, `n ≥ 0`. (Note that `A_n > 0`, `B_n > 0`.)

Show that
`nA_n = (2n − 1)/2 A_(n − 1)` for `n ≥ 1`. (2 marks)
Using integration by parts on `A_n`, or otherwise, show that
1. `A_n = 2n int_0^(pi/2) x sin x cos^(2n − 1) x\ dx` for `n ≥ 1`. (1 mark)
Use integration by parts on the integral in part (ii) to show that
1. `(A_n)/(n^2) = ((2n − 1))/n B_(n − 1) − 2B_n` for `n ≥ 1`. (3 marks)
Use parts (i) and (iii) to show that
1. `1/(n^2) = 2((B_(n − 1))/(A_(n − 1)) − (B_n)/(A_n))` for `n ≥ 1`. (1 mark)
Show that
1. `sum_(k = 1)^n 1/(k^2) = (pi^2)/6 − 2 (B_n)/(A_n)`. (2 marks)
Use the fact that
`sin x ≥ 2/pi x` for `0 ≤ x ≤ pi/2` to show that
1. `B_n ≤ int_0^(pi/2) x^2(1 − (4x^2)/(pi^2))^n dx`. (1 mark)
Show that
`int_0^(pi/2) x^2(1 − (4x^2)/(pi^2))^n dx = (pi^2)/(8(n + 1)) int_0^(pi/2) (1 − (4x^2)/(pi^2))^(n + 1) dx`. (1 mark)
From parts (vi) and (vii) it follows that
1. `B_n ≤ (pi^2)/(8(n + 1)) int_0^(pi/2)(1 − (4x^2)/(pi^2))^(n + 1) dx`.
Use the substitution `x = pi/2 sin t` in this inequality to show that
1. `B_n ≤ (pi^3)/(16(n + 1)) int_0^(pi/2) cos^(2n + 3)t\ dt ≤ (pi^3)/(16(n + 1)) A_n`. (2 marks)
Use part (v) to deduce that
1. `(pi^2)/6 − (pi^3)/(8(n + 1)) ≤ sum_(k = 1)^n 1/(k^2) < (pi^2)/6`. (1 mark)
What is
`lim_(n → ∞) sum_(k = 1)^n 1/(k^2)`? (1 mark)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`(pi^2)/6`

Show Worked Solution

♦♦ Mean mark part (i) 21%.

(i)	`u`	`=cos^(2n − 1)x`
	`du`	`=-(2n-1) sinx cos^(2n − 2) x\ dx`
	`v`	`=sin x`
	`dv`	`=cos x\ dx`

`A_n`	`= int_0^(pi/2) cos x cos^(2n − 1)x\ dx`
	`= [sin x cos^(2n − 1)x]_0^(pi/2) − int_0^(pi/2) sin x*-(2n-1) sinx cos^(2n − 2) x\ dx`
	`= 0 + (2n − 1) int_0^(pi/2) sin^2 x cos^(2n − 2) x\ dx`
	`= (2n − 1) int_0^(pi/2) (1 − cos^2x)cos^(2n − 2)x\ dx`
	`= (2n − 1)(int_0^(pi/2) cos^(2n − 2)x\ dx − int_0^(pi/2) cos^(2n)x\ dx)`
	`= (2n − 1) A_(n − 1) − (2n − 1) A_n`

`A_n + (2n − 1) A_n`	`= (2n − 1) A_(n − 1)`
`2nA_n`	`= (2n − 1) A_(n − 1)`
`:.nA_n`	`= (2n − 1)/2 A_(n − 1)\ \ text(for)\ \ n ≥ 1.`

♦♦ Mean mark part (ii) 32%.

(ii)	`u`	`=cos^(2n) x`
	`du`	`=-2n\ sinx cos^(2n − 1) x\ dx`
	`v`	`= x`
	`dv`	`= dx`

`A_n`	`= int_0^(pi/2) 1 xx cos^(2n)x\ dx`
	`= [x cos^(2n) x]_0^(pi/2) − int_0^(pi/2) x * -2n\ sinx cos^(2n − 1) x\ dx`
	`= 0 + 2n int_0^(pi/2) x sin x cos^(2n − 1) x\ dx`
	`= 2n int_0^(pi/2) x sin x cos^(2n − 1)x\ dx\ \ \ text(for)\ \ n ≥ 1`

♦♦♦ Mean mark part (iii) 8%.

(iii)	`u`	`=sin x cos^(2n-1) x`
	`du`	`=cosx cos^(2n-1)x -(2n-1) cos^(2n-2)x *sin^2x\ dx`
		`=cos^(2n)x – (2n-1)cos^(2n-2) x* (1-cos^2 x)\ dx`
		`=cos^(2n)x – (2n-1)cos^(2n-2)x +(2n-1)cos^(2n)\ dx`
		`=2n cos^(2n)x-(2n-1)cos^(2n-2)x`
	`v`	`= x^2/2`
	`dv`	`= x\ dx`

`A_n`	`= 2n int_0^(pi/2) x sin x cos^(2n − 1)x\ dx`
	`= 2n[x^2/2 sin x cos^(2n − 1) x]_0^(pi/2)`
	`− 2n int_0^(pi/2) x^2/2 (2n cos^(2n)x-(2n-1)cos^(2n-2)x)\ dx`
	`= 0 −2n[n int_0^(pi/2) x^2 cos^(2n) x\ dx − (2n − 1)/2 int_0^(pi/2) x^2 cos^(2n − 2) x \ dx]`
	`= -2n(nB_n-(2n-1)/2 B_(n-1))`
	`= -2n^2B_n+n(2n-1)B_(n-1)`
`:.(A_n)/(n^2)`	`= (2n − 1)/n B_(n − 1) − 2B_n\ \ \ text(for)\ \ n ≥ 1`

♦♦♦ Mean mark part (iv) 13%.

(iv)	`(A_n)/(n^2)`	`= ((2n −1)B_(n − 1))/n − 2B_n\ \ \ text(for)\ \ n ≥ 1`
	`:.1/(n^2)`	`= ((2n − 1)B_(n − 1))/(nA_n) − (2B_n)/(A_n)`
		`= ((2n − 1)B_(n − 1))/((2n − 1)/2 A_(n − 1)) − (2B_n)/(A_n)\ \ \ \ \ text{(using part (i))}`
		`= (2B_(n − 1))/(A_(n − 1)) − (2B_n)/(A_n)`
		`= 2((B_(n − 1))/(A_(n − 1)) − (B_n)/(A_n))\ \ \ text(for)\ \ n ≥ 1`

♦♦♦ Mean mark part (v) 9%.

(v)	`sum_(k = 1)^n 1/(k^2)`	`= 1/(1^2) + 1/(2^2) + 1/(3^2) + … + 1/(n^2)`
	`sum_(k = 1)^n 1/(k^2)`	`= 2((B_0)/(A_0) − (B_1)/(A_1)) + 2((B_1)/(A_1) − (B_2)/(A_2)) + … + 2((B_(n − 1))/(A_(n − 1)) − (B_n)/(A_n))`
		`= 2(B_0)/(A_0) − 2(B_n)/(A_n)`

`A_0 = int_0^(pi/2) dx = [x]_0^(pi/2) = pi/2`

`B_0 = int_0^(pi/2)x^2\ dx = [(x^3)/3]_0^(pi/2) = (pi^3)/24`

`sum_(k = 1)^n 1/(k^2)`	`= 2 xx ((pi^3)/24)/(pi/2) − 2(B_n)/(A_n)`
	`= (pi^2)/6 − 2(B_n)/(A_n)`

♦♦♦ Mean mark part (vi) 8%.

(vi)	`B_n`	`= int_0^(pi/2) x^2 cos^(2n)x\ dx`
	`B_n`	`= int_0^(pi/2) x^2 (1 − sin^2 x)^n\ dx`

`text(S)text(ince)\ \ sin x ≥ 2/pi x,\ \ 0 ≤ x ≤ pi/2`

`:.B_n`	`≤ int_0^(pi/2) x^2 (1 − (2/pi x)^2)^n\ dx`
	`≤ int_0^(pi/2) x^2 (1 − (4x^2)/(pi^2))^n\ dx`

(vii)	`u`	`=x,\ \ \ du=dx`
	`dv`	`=x(1-(4x^2)/(pi^2))^n\ dx`
	`v`	`=(- pi^2)/(8(n+1)) (1-(4x^2)/(pi^2))^(n+1)`

♦♦♦ Mean mark part (vii) 2%.

`int_0^(pi/2) x * x(1 − (4x^2)/(pi^2))^n dx`

`= [(-pi^2)/(8(n + 1))(1 − (4x^2)/(pi^2))^(n + 1) * x]_0^(pi/2) – int_0^(pi/2) (-pi^2)/(8(n + 1))(1 − (4x^2)/(pi^2))^(n + 1)\ dx`

`= (-pi^2)/(8(n + 1)) (0 − 0) + (pi^2)/(8(n + 1)) int_0^(pi/2) (1 − (4x^2)/(pi^2))^(n + 1) dx`

`= (pi^2)/(8(n + 1)) int_0^(pi/2) (1 − (4x^2)/(pi^2))^(n + 1) dx`

♦♦♦ Mean mark part (viii) 4%.

(viii)	`text(Using)\ \ x`	`=pi/2 sin t\ \ \ \ dx=pi/2 cos t\ dt`
	`text(When)\ \ x`	`=0,\ \ \ t=0`
	`text(When)\ \ x`	`=pi/2,\ \ \ t=pi/2`

`B_n`	`≤ (pi^2)/(8(n + 1)) int_0^(pi/2) (1 − (4x^2)/(pi^2))^(n + 1)\ dx`
	`≤ (pi^2)/(8(n + 1)) int_0^(pi/2) (1 − sin^2 t)^(n + 1) *pi/2 cos t\ dt`
	`≤ (pi^3)/(16(n + 1)) int_0^(pi/2) (cos^2 t)^(n + 1) cos t\ dt`
	`≤ (pi^3)/(16(n + 1)) int_0^(pi/2) cos^(2n + 3) t\ dt`

`text(Consider the RHS of the inequality)`

`text(RHS)`	`=(pi^3)/(16(n + 1)) int_0^(pi/2) cos^(2n + 3) t\ dt`
	`≤(pi^3)/(16(n + 1)) int_0^(pi/2) cos^(2n) t\ dt,\ \ \ \ text{(as cos t ≤ 1)}`
	`≤(pi^3)/(16(n + 1)) A_n\ \ \ \ text(… as required)`

♦♦♦ Mean mark part (ix) 3%.

(ix) `sum_(k = 1)^n 1/(k^2) = (pi^2)/6 − 2(B_n)/(A_n)\ \ \ \ text{(from part (v))}`

`:.sum_(k = 1)^n 1/(k^2) < (pi^2)/6\ \ \ \ (A_n > 0`, `B_n > 0)`

`text{Using part (viii)}`

`B_n`	`≤(pi^3)/(16(n + 1)) A_n`
`(2B_n)/A_n`	`≤(pi^3)/(8(n + 1)) `

`:.sum_(k = 1)^n 1/(k^2) ≥ (pi^2)/6 − (pi^3)/(8(n + 1))`

`:. (pi^2)/6 − (pi^3)/(8(n + 1)) ≤ sum_(k = 1)^n 1/(k^2) < (pi^2)/6`

♦♦♦ Mean mark part (x) 19%.

(x) `text(S)text(ince)\ (pi^3)/(8(n + 1)) → 0\ text(as)\ n → ∞`

`lim_(n → ∞) sum_(k = 1)^n 1/(k^2) = (pi^2)/6`

Polynomials, EXT2 2010 HSC 7c

Let `P(x) = (n − 1)x^n − nx^(n − 1) + 1`, where `n` is an odd integer, `n ≥ 3`.

Show that `P(x)` has exactly two stationary points. (1 mark)
Show that `P(x)` has a double zero at `x = 1`. (1 mark)
Use the graph `y = P(x)` to explain why `P(x)` has exactly one real zero other than `1`. (2 marks)
Let `α` be the real zero of `P(x)` other than `1`.
Given that `2^x>=3x-1` for `x>=3`, or otherwise, show that `-1 < α ≤ -1/2`. (2 marks)
Deduce that each of the zeros of `4x^5 − 5x^4 + 1` has modulus less than or equal to `1`. (2 marks)

Show Answers Only

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

Show Worked Solution

(i)	`P′(x)`	`= n(n − 1)x^(n − 1) − n(n − 1)x^(n − 2)`
		`= n(n − 1)x^(n − 2)(x − 1)`

`P′(x) = 0\ \ text(when)\ \ x = 0\ \ text(or)\ \ x = 1.`

`:.P(x)\ text(has exactly two stationary points.)`

(ii)	`P(1)`	`= (n − 1) xx 1 − n xx 1 + 1 = 0, and`
	`P′(1) `	`= 0`

`:. P(x)\ text(has a double zero at)\ \ x=1`

(iii) `P(x)\ text{has exactly two stationary points at}\ \ x=0 and 1`

♦♦ Mean mark part (iii) 26%.

`text(S)text(ince)\ \ P(0) = 1 and P(1)=0`

`=>\ text{A local maximum occurs at (0,1)}`

`text(Noting that)\ \ P(x)`	`-> oo\ \ text(as)\ \ x-> oo, and`
`P(x)`	`-> -oo\ \ text(as)\ \ x-> -oo`

`:.\ text(The double zero at)\ \ x=1,\ text{means that}\ \ P(x)`

`text(has exactly one real zero other than)\ \ x=1.`

♦♦♦ Mean mark part (iv) 3%.

(iv)	`P(-1)`	`=(n-1)(-1)^n-n xx (-1)^(n-1) + 1`
		`=(n-1)(-1) – nxx1 +1\ \ \ \ text{(given}\ n\ text{is odd)}`
		`=-2n+2`
		`<0\ \ \ text{(for odd}\ n>=3text{)}`

`P(-1/2)`	`= (n − 1)(-1/2)^n − n(-1/2)^(n − 1)+1`
	`= -(n − 1) xx 1/(2^n) − n/(2^(n − 1)) + 1\ \ \ \ text{(given}\ n\ text{is odd)}`
	`= (-n + 1 − 2n + 2^n)/(2^n)`
	`= (2^n − (3n − 1))/(2^n)`
	`>=0\ \ \ \ \ text{(given}\ \ 2^n>=3n-1\ \ text{for}\ \ n>=3, and 2^n > 0)`

`text(S)text(ince)\ \ P(x)\ \ text(is continuous, when it changes sign, it cuts the)\ x text(-axis)`

`:. -1 < α ≤ -1/2`

(v) `P(x) = 4x^5 − 5x^4 + 1\ \ text(is of the form)`

♦♦♦ Mean mark part (v) 2%.

`P(x) = (n − 1)x^n − nx^(n − 1) + 1`

`:.P(x)\ \ text(has 5 zeros)\ \=> 1, 1, alpha, beta, bar beta\ \ \ \ (text(where)\ beta\ text{is not real})`

`text(The zeros 1 and α have a modulus ≤ 1.)`

`text(Consider the product of the roots)`

`11alphabetabar beta`	`=- 1/4`
`alphabetabar beta`	`=- 1/4`
`\|\ alphabetabar beta\ \|`	`=\|\ – 1/4\ \|`
`\|\ alpha\ \|*\|\ beta\ \|^2`	`= 1/4`

`text(S)text(ince)\ \ |\ α\ | > 1/2,\ \ |\ beta\ |<1`

`:.text(Each of the zeros of)\ x^5 − 5x^4 + 1\ text(has a modulus)\ <=1.`

Harder Ext1 Topics, EXT2 2010 HSC 7a

In the diagram `ABCD` is a cyclic quadrilateral. The point `K` is on `AC` such that `∠ADK = ∠CDB`, and hence `ΔADK` is similar to `ΔBDC`.

Copy or trace the diagram into your writing booklet.

Show that `ΔADB` is similar to `ΔKDC`. (2 marks)
Using the fact that `AC = AK + KC`,
show that `BD xx AC = AD xx BC + AB xx DC`. (2 marks)
A regular pentagon of side length `1` is inscribed in a circle, as shown in the diagram.

Let `x` be the length of a chord in the pentagon.
Use the result in part (ii) to show that `x = (1 + sqrt5)/2`. (2 marks)

Show Answers Only

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

Show Worked Solution

(i)

Harder Ext1 Topics, EXT2 2010 HSC 7a Answer

`text(S)text(ince)\ \ ∠ADB`	`=∠ADK +∠KDB, and`
`∠KDC`	`=∠CDB +∠KDB`
`=>∠ADB`	`= ∠KDC`
`∠ABD`	`= ∠ACD\ \ \ text{(angles in the same segment on arc}\ ADtext{)}`

`∠ADK`	`= ∠CDB\ \ \ `	`text{(corresponding angles of similar}`
		`text{triangles,}\ ΔADK\ text(\|\|\|)\ ΔBDC text{)}`

`:.ΔADB\ text(|||)\ ΔKDC\ \ text{(equiangular)}`

(ii) `text(S)text(ince)\ ΔADB\ text(|||)\ ΔKDC`

`(BD)/(DC) = (AD)/(DK) = (AB)/(KC)\ \ text{(corresponding sides of similar triangles)}`

`:.AB xx DC = KC xx BD\ \ \ \ …\ (1)`

♦♦ Mean mark part (ii) 31%.

`text(Similarly, using)\ \ ΔADK\ text(|||)\ ΔBDC\ \ text{(given)}`

`(AD)/(BD)=(AK)/(BC)`

`:. BC xx AD = AK xx BD\ \ \ \ …\ (2)`

`text{Add (1) + (2)}`

`AB xx DC + AD xx BC`	`=KC xx BD+AK xx BD`
	`=BD(AK+KC)`
`:.BD xx AC`	`=AD xx BC+AB xx DC`

(iii) `text(Each diagonal of the pentagon is)\ x.`

♦♦ Mean mark part (iii) 24%.

`text{Hence in applying the result in (i) you have}`

`BD = AC = x, AD = DC = CB = 1, BA = x`

`x xx x`	`= 1 xx 1 + x xx 1`
`x^2 − x − 1`	`= 0`
`x`	`= (1 ± sqrt(1 + 4))/2`
	`= (1 ± sqrt(5))/2`
`:.x`	`= (1 + sqrt(5))/2,\ \ \ (x>0)`

Harder Ext1 Topics, EXT2 2010 HSC 5c

A TV channel has estimated that if it spends `$x` on advertising a particular program it will attract a proportion `y(x)` of the potential audience for the program, where

`(dy)/(dx) = ay(1 − y)`

and `a > 0` is a given constant.

Explain why
`(dy)/(dx)` has its maximum value when `y = 1/2`. (1 mark)
Using
`int (dy)/(y(1 − y)) = ln (y/(1 − y)) + c`, or otherwise, deduce that
`y(x) = 1/(ke^(-ax) + 1)` for some constant `k > 0`. (3 marks)
The TV channel knows that if it spends no money on advertising the program then the audience will be one-tenth of the potential audience.
Find the value of the constant `k` referred to in part (c)(ii). (1 mark)
What feature of the graph
`y = 1/(ke^(-ax) + 1)` is determined by the result in part (c)(i)? (1 mark)
Sketch the graph
`y(x) = 1/(ke^(-ax) + 1)` (1 mark)

Show Answers Only

`text(See Worked Solutions.)`
`text{Proof (See Worked Solutions)}`
`9`
`y = 1/2`
`text(See Worked Solutions.)`

Show Worked Solution

(i) `(dy)/(dx) = ay(1 − y)`

♦♦ Mean mark part (i) 32%.

`=>text(Given)\ \ a>0,\ \ ay(1 − y)\ text(is an inverted parabola)`

`text(with zeros at)\ \ y = 0\ \ text(and)\ \ y = 1`

`=>\ text(Parabola symmetry means that a maximum occurs when)`

`y = (0+1)/2=1/2`

`:.(dy)/(dx)\ \ text(has its maximum value when)\ y = 1/2.`

♦ Mean mark part (ii) 41%.

(ii) `dy/dx=ay(1 − y)`

`int (dy)/(y(1 − y))`	`=int a\ dx`
`ax`	`= ln (y/(1 − y))+c`
`e^(ax − c)`	`= y/(1 − y)`
`e^(ax − c) − ye^(ax − c)`	`= y`
`y(1 + e^(ax − c))`	`= e^(ax − c)`
`y`	`= (e^(ax − c))/(1 + e^(ax − c))`
`y`	`= 1/(e^c e^(− ax) + 1)`

`text(Let)\ k = e^c`

`:.y(x) = 1/(ke^(-ax) + 1)\ text(for some constant)\ \ k > 0.`

(iii) `text(When)\ \ x = 0, y = 0.1`

`0.1`	`= 1/(ke^0 + 1)`
`k + 1`	`= 10`
`k`	`= 9`

♦♦ Mean mark part (iv) 25%.

(iv) `text(The gradient the curve is greatest at)\ y = 1/2\ \ \ text{from part (i)}`

`:. text(A point of inflection occurs at)\ y =1/2.`

(v) `text(As)\ \ x->oo,\ \ y->1/(0+1)=1`

♦♦ Mean mark part (v) 12%.

Harder Ext1 Topics, EXT2 2010 HSC 4c

Let `k` be a real number, `k ≥ 4`.

Show that, for every positive real number `b`, there is a positive real number `a` such that `1/a + 1/b = k/(a + b)`. (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`1/a + 1/b`	` = k/(a+b)`
`text(Multiply b.s. by)\ \ ab(a+b)`
`b(a+b)+a(a+b)`	`=kab`
`(a + b)^2`	`=kab`
`a^2 + 2ab + b^2`	`=kab`
`a^2+(2-k)ba+b^2`	`=0`

`text(Given)\ \ b>0,\ \ a\ \ text(is real when)\ \ Delta>=0`

♦♦♦ Mean mark 13%.

`(2-k)^2 b^2-4b^2`	`>=0`
`(4-4k+k^2)b^2-4b^2`	`>=0`
`b^2k(k-4)`	`>=0`
`k`	`>=4`

`:. a \ \ text(is real)`

`text(Using the quadratic formula to solve for)\ \ a`

`a`	`= (-(2-k)b +- sqrt Delta)/2`
	`=1/2 (2-k)b +- sqrt Delta`

`text(S)text(ince)\ \ b>0,\ \ (k-2)>0,\ \ and sqrt Delta>0`

`=>a=1/2 (2-k)b + sqrt Delta>=0`

`text(i.e. there exists a positive real number)\ \ a\ \ text(that satisfies.)`

Conics, EXT2 2010 HSC 3d

The diagram shows the rectangular hyperbola `xy = c^2`, with `c > 0`.

The points `A(c, c)`, `R(ct, c/t)` and `Q(-ct, -c/t)` are points on the hyperbola,
with `t ≠ ±1`.

The line `l_1` is the line through `R` perpendicular to `QA`.
Show that the equation of `l_1` is
1. `y = -tx + c(t^2 + 1/t)`. (2 marks)
The line `l_2` is the line through `Q` perpendicular to `RA`.
Write down the equation of `l_2`. (1 mark)
Let `P` be the point of intersection of the lines `l_1` and `l_2`.
Show that `P` is the point `(c/(t^2), ct^2)`. (2 marks)
Give a geometric description of the locus of `P`. (1 mark)

Show Answers Only

`text{Proof (Show Worked Solutions)}`
`y = tx + c(t^2 − 1/t)`
`text{Proof (Show Worked Solutions)}`
`text{Proof (Show Worked Solutions)}`

Show Worked Solution

(i)

`m_(QA)`	`= (c + c/t)/(c + ct)`
	`= (1 + 1/t)/(1 + t)`
	`= (t + 1)/(t(1 + t))`
	`= 1/t`

`:.\ text(Gradient of perpendicular to)\ QA\ \ (l_1) = -t`

`:.text(Equation of)\ \ l_1`

`y − c/t`	`= -t(x − ct)`
`y`	`= -tx + ct^2 + c/t`
	`= -tx + c(t^2 + 1/t)\ \ \ …\ (1)`

(ii)	`m_(RA)`	`= (c − c/t)/(c − ct)`
		`= (1 − 1/t)/(1 − t)`
		`= (t − 1)/(t(1 − t))`
		`= -1/t`

`:.m\ text(of)\ l_2 = t`

`:.\ text(Equation of)\ l_2`

`y + c/t`	`= t(x + ct)`
`y`	`= tx + ct^2 – c/t`
	`= tx + c(t^2 – 1/t)\ \ \ …\ (2)`

(iii) `text{Solving (1) and (2) simultaneously}`

`-tx + c(t^2 + 1/t)`	`= tx + c(t^2 − 1/t)`
`ct^2+c/t-ct^2+c/t`	`=2tx`
`(2c)/t`	`= 2tx`
`x`	`= c/(t^2)`

`text{Substitute into (2)}`

`y`	`= (ct)/(t^2) + ct^2 − c/t`
	` = ct^2`

`:.P\ text(has coordinates)\ (c/(t^2), ct^2)`

(iv) `text(Using the coordinates of)\ P`

♦♦ Mean mark part (iv) 1%.
MARKER’S COMMENT: Most students found the correct locus, although very few indicated that it was only the branch in the 1st quadrant.

`=>t^2 = c/x,\ \ t^2 = y/c`

`y/c`	`= c/x`
`:.xy`	`=c^2`

`text(S)text(ince the parameter of)\ P\ text(is)\ t^2`

`=>\ text(the locus is in the first quadrant.)`

`:.\ text(The locus of)\ P\ text(is the branch of the rectangular hyperbola)`

`xy = c^2\ text{in the first quadrant (excluding}\ A, t≠±1 text{)}`

Harder Ext1 Topics, EXT2 2010 HSC 3c

Two identical biased coins are each more likely to land showing heads than showing tails.

The two coins are tossed together, and the outcome is recorded. After a large number of trials it is observed that the probability that the two coins land showing a head and a tail is `0.48`.

What is the probability that both coins land showing heads? (2 marks)

Show Answers Only

`0.36`

Show Worked Solution

`text(Let)\ \ p=P(H),\ \ =>(1-p)=P(T)`

♦♦♦ Mean mark 19%.

`text(S)text(ince)\ \ P(H,T)+P(T,H)=0.48`

`p(1-p)+(1-p)p`	`=0.48`
`2p(1-p)`	`=0.48`
`p^2-p+0.24`	`=0`
`(p-0.6)(p-0.4)`	`=0`

`:. p=0.6\ \ \ \ \ text{(given}\ P(H)>P(T)text{)}`

`:.P(H,H)=0.6 xx 0.6 = 0.36`

Polynomials, EXT2 2011 HSC 8c

Let `beta` be a root of the complex monic polynomial

`P(z) = z^n + a_(n - 1) z^(n - 1) + … + a_1z + a_0.`

Let `M` be the maximum value of `|\ a_(n - 1)\ |,\ |\ a_(n - 2)\ |,\ …\ ,\ |\ a_0\ |.`

Show that
`|\ beta\ |^n <= M(|\ beta\ |^(n - 1) + |\ beta\ |^(n - 2) + … + |\ beta\ | + 1).` (2 marks)
Hence, show that for any root `beta` of `P(z)`
1. `|\ beta\ | < 1 + M.` (3 marks)
Let `S(x) = sum_(k = 0)^n c_k(x + 1/x)^k`, where the real numbers `c_k` satisfy `|\ c_k\ | <= |\ c_n\ |` for all `k < n`, and `c_n != 0.`
Using parts (i) and (ii), or otherwise, show that `S(x) = 0` has no real solutions. (3 marks)

Show Answers Only

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

Show Worked Solution

(i) `P(z) = z^n + a_(n – 1) z^(n – 1) + … + a_1z + a_0.`

♦♦♦ Mean mark part (i) 7%.

`M = text(Max)\ (\ |\ a_(n – 1)\ |, |\ a_(n – 2)\ |, … , |\ a_0|\ )`

`P(beta) = 0`

`0`	`=beta^n + a_(n – 1) beta^(n – 1) + … + a_1 beta + a_0`
`beta^n`	`= -(a_(n – 1) beta^(n – 1) + … + a_1 beta + a_0)`
`\|\ beta^n\ \|`	`=\|\ a_(n – 1) beta^(n – 1) + … + a_1 beta + a_0\ \|`
	`<= \|\ a_(n – 1) beta^(n – 1)\ \| + … + \|\ a_1 beta\ \| + \|\ a_0\ \|`
`\|\ beta\ \|^n`	`<= \|\ a_(n – 1)\ \|\|\ beta\ \|^(n – 1) + … + \|\ a_1\ \|\|\ beta\ \| + \|\ a_0\ \|`
	`<= M (\|\ beta\ \|^(n – 1) + … + \|\ beta\ \| + 1)`

(ii) `(1 + |\ beta\ | + … + |\ beta\ |^(n – 1))`

♦♦♦ Mean mark part (ii) 3%.

`=>text(GP where)\ \ a = 1,\ \ r = |\ beta\ |, and n\ text(terms)`

`text(Consider)\ |\ beta\ | > 1`

`1 + |\ beta\ | + … + |\ beta\ |^(n – 1) = (1(|\ beta\ |^n – 1))/(|\ beta\ | – 1)`

`text(Using)\ \ |\ beta\ |^n <= M(|\ beta\ |^(n – 1) + … + |\ beta\ | + 1)\ \ \ text{(part (i))}`

`\|\ beta\ \|^n`	`<= M((\|\ beta\ \|^n – 1))/(\|\ beta\ \| – 1)`
`\|\ beta\ \|^n (\|\ beta\ \| – 1)`	`<= M(\|\ beta\ \|^n – 1)`
`\|\ beta\ \| – 1`	`<= M ((\|\ beta\ \|^n – 1))/\|\ beta\ \|^n`
`\|\ beta\ \|`	`<= 1 + M (1 – 1/\|\ beta\ \|^n)`
`\|\ beta\ \|`	`< 1 + M`

`text(Consider)\ |\ beta\ | <= 1`

`|\ beta\ | < 1 + M,\ \ text(as)\ \ M > 0.`

`:.|\ beta\ | < 1 + M\ \ text(for any root)\ \ beta`

(iii) `S(x)`	`=sum_(k=0)^n c_k(x + 1/x)^k`
	`=c_n(x + 1/x)^n + c_(n – 1)(x + 1/x)^(n – 1) + … + c_1(x + 1/x) + c_0`
`(S(x))/c_n`	`=(x + 1/x)^n + (c_(n-1))/c_n (x + 1/x)^(n – 1) + … + c_1/c_n(x + 1/x) + c_0/c_n`

`text(This is a monic polynomial of the form in part (i))`

♦♦♦ Mean mark part (iii) 2%.

`text(Let)\ \ z = x + 1/x`

`P(z) = z^n + a_(n – 1) z^(n – 1) + … + a_1 z + a_0`

`text(Consider the coefficients)\ \ a_k = c_k/c_n`

`text(S)text(ince)\ \ |\ c_k\ |<=|\ c_n\ |,\ \ \ text{(given)}`

`=>\|\ a_k\ \| <=`	`(\|\ c_k\ \|)/\|\ c_n\ \| <= 1\ \ text(from above)`
`=>M <=`	`1\ \ \ \ text{(since}\ M\ text(is the max of)\ \|\ a_k\ \| text{)}`

`text(If)\ \ beta\ \ text(is a real root of)\ \ P(z) and |\ beta\ | < 1 + M,\ \ text(then)`

`=>|\ beta\ | < 2`

`text(However,)\ \ beta`	`= x + 1/x`
`:. \|\ beta\ \|`	`= \|\ x + 1/x\ \|`
	`= \|\ x\ \| + 1/\|\ x\ \|\ \ \ \ text{(for}\ x\ text{real)}`
	`>= 2`

`:.text(We have)\ \ |\ beta\ | < 2 and |\ beta\ | = |\ x + 1/x\ | >= 2`

`text(This is a contradiction and hence our assumption that)`

`beta\ \ text(is a real root of)\ \ S(x)\ \ text(is incorrect.)`

`text(Alternative strategy to show a contradiction)`

`text(Graphing)\ \ y = x + 1/x`

`text(S)text(ince the turning points occur at)\ \ y=+- 2`

`=>|\ x + 1/x\ | >= 2.`

`:.S(x)\ \ text(has no real solutions.)`

Harder Ext1 Topics, EXT2 2011 HSC 8b

A bag contains seven balls numbered from `1` to `7`. A ball is chosen at random and its number is noted. The ball is then returned to the bag. This is done a total of seven times.

What is the probability that each ball is selected exactly once? (1 mark)
What is the probability that at least one ball is not selected? (1 mark)
What is the probability that exactly one of the balls is not selected? (2 marks)

Show Answers Only

`(6!)/7^6=720/(117\ 649)`
`1 – (6!)/7^6=(116\ 929)/(117\ 649)`
`(3 xx 6!)/7^5=2160/(16\ 807)`

Show Worked Solution

(i) `P text{(each ball is selected once)}`

♦♦ Mean mark part (i) 32%.

`=7/7 xx 6/7 xx 5/7 xx 4/7 xx 3/7 xx 2/7 xx 1/7`

`=(6!)/7^6`

`=720/(117\ 649)`

(ii) `P text{(at least one ball is not selected)}`

`=1 – P text{(each ball once)}`

`=1 – (6!)/7^6`

`=(116\ 929)/(117\ 649)`

(iii) `text{Consider when 1 of the balls is not selected (say #1).}`

♦♦♦ Mean mark part (iii) 2%.

`=>\ text(One of the other 6 balls is selected twice.)`

`text(e.g. 2, 2, 3, 4, 5, 6, 7).`

`text(This can be done in)\ \ (7!)/(2!)\ \ text(ways.)`

`text(With 6 differently numbered pairs possible,)`

`text(This can be done in)\ \ 6 xx (7!)/(2!)\ \ text(ways)`

`text(S)text(ince there are 7 numbers that can be left out,)`

`text(Total number of ways to leave 1 number out)`

`=7 xx 6 xx (7!)/(2!)`

`=21 xx 7!`

`:.P text{(exactly one ball not selected)}`	`=(21 xx 7!)/7^7`
	`=(3 xx 6!)/7^5`
	`=2160/(16\ 807)`