Band 6 – Page 22

CORE*, FUR1 2011 VCAA 7 MC

Anthony invested $15 000 in an account. It earned `rtext(%)` interest per annum, compounding monthly.

The amount of interest that is earned in the third year of the investment is given by

A. `15\ 000 (1 + r / 1200)^3 - 15\ 000 (1 + r / 1200)^2`

B. `15\ 000 (1 + r / 1200)^36 - 15\ 000 (1 + r / 1200)^24`

C. `15\ 000 (1 + r / 100)^3 - 15\ 000 (1 + r / 100)^2`

D. `15\ 000 (1 + r / 100)^36 - 15\ 000 (1 + r / 100)^24`

E. `15\ 000 (1 + r / 1200)^4 - 15\ 000 (1 + r / 1200)^3`

Show Answers Only

`B`

Show Worked Solution

`text(Interest in 3rd year)`

♦♦ Mean mark 33%.

`=\ text(Value after 3 yrs) – text(Value after 2 years)`

`text(Using)\ \ A = PR^n,\ \ text(where)\ \ A = 15\ 000, and`

`R`	`= (1 + r / {12 xx 100})`
	`= (1 + r / 1200)`

`text(In first 3 years:)`

`A_3`	`= 15\ 000 (1 + r / 1200)^(12 xx 3)`
	`= 15\ 000 (1 + r / 1200)^36`

`text(In first 2 years:)`

`A_2`	`= 15\ 000 (1 + r / 1200)^(12 xx 2)`
	`= 15\ 000 (1 + r / 1200)^24`

`:.\ text(Interest in 3rd year)`

`= 15\ 000 (1 + r/1200)^36 – 15\ 000 (1 + r/1200)^24`

`=> B`

CORE*, FUR1 2012 VCAA 9 MC

Peter took out a reducing balance loan where interest was calculated monthly. He planned to repay this loan fully, with eight equal monthly payments of $260.

Peter missed the fourth payment, but made a double payment of $520 in the fifth month. He then continued to make payments of $260 for the remaining three months.

Which graph could show the balance of the loan each month over the eight-month period?

Show Answers Only

`A`

Show Worked Solution

`text(By elimination,)`

`text(As Peter missed the 4th payment, the balance of the)`

♦♦ Mean mark 20%.

`text(loan will increase after 4 months due to interest.)`

`:.\ text(Eliminate B, C, D and E.)`

`=> A`

CORE*, FUR1 2013 VCAA 8 MC

A car is purchased for $25 000. The value of the car is to be depreciated each year by 20% using the reducing balance method.

In the fourth year, the car will depreciate in value by

A. $2048

B. $2560

C. $5000

D. $10 240

E. $14 760

Show Answers Only

`B`

Show Worked Solution

`text(Depreciated value after 3 years)`

♦♦ Mean mark 29%.
MARKERS’ COMMENT: A common mistake was to not read the question carefully and give the depreciated value of the car after 4 years.

`= 25\ 000 xx (1 – 0.2)^3`

`= 12\ 800`

`:.\ text(Depreciation in 4th year)`

`= 12\ 800 xx 0.2`

`= $2560`

`=> B`

CORE*, FUR1 2014 VCAA 9 MC

Leslie borrowed $35 000 from a bank.

Interest is charged at the rate of 4.75% on the reducing monthly balance.

The loan is to be repaid with 47 monthly payments of $802.00 and a final payment that is to be adjusted so that the loan will be fully repaid after exactly 48 monthly payments.

Correct to the nearest cent, the amount of the final payment will be

A. $0.39

B. $3.57

C. $802.00

D. $802.39

E. $805.57

Show Answers Only

`E`

Show Worked Solution

`text(Calculate the FV of the loan after 47 payments)`

♦♦♦ Mean mark 18%.
MARKERS’ COMMENT: A majority of students found the amount still owed after the second last payment but failed to add interest during the last month.

`text(By TVM Solver:)`

`N`	`= 47`
`I(%)`	`= 4.75text(%)`
`PV`	`= -35\ 000`
`PMT`	`= 802`
`FV`	`= text(?)`
`text(P/Y)`	`= text(C/Y) = 12`

`:.\ text(Balance owing after 47 payments)\ =$802.39…`

`text(Final payment)`

`=\ text(Balance after 47 payments + month’s interest)`

`=802.39… + (802.39… xx 4.75/12 xx 100text{%})`

`=$805.57`

`=> E`

GRAPHS, FUR1 2007 VCAA 7 MC

The graph above shows a relationship between `y` and `x^3`.

The graph that shows the same relationship between `y` and `x` is

Show Answers Only

`C`

Show Worked Solution

`text(The gradient of the linear graph of)`

♦♦♦ Mean mark 26%.

`y\ text(versus)\ x^3 = 1/3`

`:.\ y = 1/3x^3`

`text(Test if the coordinates in each graph satisfy)`

`text(the equation)\ \ y=1/3x^3.`

`text(Consider)\ C,`

`text(Substitute)\ (1, 1/3)\ text(into equation,)`

`1/3 = 1/3 xx 1^3\ \ \ text{(true)}`

`text(All other options can be shown to not satisfy equation.)`

`=> C`

GRAPHS, FUR1 2007 VCAA 6 MC

Russell is a wine producer. He makes both red and white wine.

Let `x` represent the number of bottles of red wine he makes and `y` represent the number of bottles of white wine he makes.

This year he plans to make at least twice as many bottles of red wine as white wine.

An inequality representing this situation is

A. `y <= x + 2`

B. `y <= 2x`

C. `y >= 2x`

D. `x <= 2y`

E. `x >= 2y`

Show Answers Only

`E`

Show Worked Solution

`x = text(no. red wine bottles)`

♦♦ Mean mark 34%.

`y = text(no. white wine bottles)`

`text(The constraint states that the number of)`

`text(of bottles of red wine will be at least twice)`

`text(the number of bottles of white wine.)`

`x >= 2y`

`=> E`

GRAPHS, FUR1 2008 VCAA 4 MC

When shopping, Betty can use either Easypark or Safepark to park her car.

At Easypark, cars can be parked for up to 8 hours per day.

The fee structure is as follows.

`text(Fee)={(text($5.00,), 0 < text(hours) <= 2),(text($8.00,), 2 < text(hours) <= 5),(text($11.00,), 5 < text(hours) <= 8) :}`

Safepark charges fees according to the formula

`text(Fee) = $2.50 xx text(hours)`

Betty wants to park her car for 5 hours on Monday and 3 hours on Tuesday.

The minimum total fee that she can pay for parking for the two days is

A. `$7.50`

B. `$11.00`

C. `$15.50`

D. `$16.00`

E. `$20.00`

Show Answers Only

`C`

Show Worked Solution

`text(Fee for 5 hours)`

♦♦ Mean mark 29%.
MARKER’S COMMENT: Almost half of students failed to take into account that the same car park did not have to be used on each day, and incorrectly answered D.

`text(Easypark = $8)`

`text(Safepark = $2.50) xx 5 = $12.50`

`text(Fee for 3 hours)`

`text(Easypark = $8)`

`text(Safepark = $2.50) xx 3 = $7.50`

`:.\ text(The minimum total fee is)`

`$8 + $7.50 = $15.50`

`=> C`

GEOMETRY, FUR1 2006 VCAA 9 MC

Points `M` and `P` are the same distance from a third point `O`.

The bearing of `M` from `O` is 038° and the bearing of `P` from `O` is 152°.

The bearing of `P` from `M` is

A. between 000° and 090°

B. between 090° and 180°

C. exactly 180°

D. between 180° and 270°

E. between 270° and 360°

Show Answers Only

`D`

Show Worked Solution

♦♦ Mean mark 31%.

`/_ MOP`	`= 152 − 38`
	`= 114^@`

`text(Let)\ R\ text(be directly south of)\ O`

`/_POR`	`= 180 − (38 + 114)\ \ text{(straight line)}`
	`= 28^@`

`text(Given)\ OM = OP`

`text(Diagram shows that)\ M\ text(is further east than)\ P.`

`:.\ text(Bearing of)\ P\ text(from)\ M\ text(just over 180°)`

`=> D`

Measurement, 2UG 2007 HSC 28c

A piece of plaster has a uniform cross-section, which has been shaded, and has dimensions as shown.

Use two applications of Simpson’s rule to approximate the area of the cross-section. (3 marks)
The total surface area of the piece of plaster is `7480.8\ text(cm²)`.
Calculate the area of the curved surface as shown on the diagram. (2 marks)

Show Answers Only

`50.4\ text(cm²)`
`3500\ text(cm²)`

Show Worked Solution

(i)

`A`	`= h/3 [y_0 +4y_1 + y_2]\ text(… applied twice)`
	`= 3.6/3 (5 + 4 xx 4.6 + 3.7) + 3.6/3 (3.7 + 4 xx 2.8 + 0)`
	`= 32.52 + 17.88`
	`= 50.4\ text(cm²)`

(ii) `text(Total Area) = 7480.8\ text(cm²)`

`text(Area of Base)`	`= 14.4 xx 200`
	`= 2880\ text(cm²)`

`text(Area of End)`	`= 5 xx 200`
	`= 1000\ text(cm²)`

`text(Area of sides)`	`= 2 xx 50.4\ \ \ text{(from (i))}`
	`= 100.8\ text(cm²)`

`:.\ text(Area of curved surface)`

`= 7480.8 – (2880 + 1000 + 100.8)`

`= 3500\ text(cm²)`

Algebra, STD2 A1 2007 HSC 28b

This shape is made up of a right-angled triangle and a regular hexagon.

The area of a regular hexagon can be estimated using the formula `A = 2.598H^2` where `H` is the side-length.

Calculate the total area of the shape using this formula. (3 marks)

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

Show Answers Only

`22.784\ text(cm²)`

Show Worked Solution

`text(Area) = 2.598H^2`

`text(Using Pythagoras)`

`H^2`	`= 2^2 + 2^2`
	`= 8`
`H`	`= sqrt 8`

`:.\ text(Area of hexagon)`	`= 2.598 xx (sqrt 8)^2`
	`= 20.784\ text(cm²)`

`text(Area of triangle)`	`= 1/2 bh`
	`= 1/2 xx 2 xx 2`
	`= 2\ text(cm²)`

`:.\ text(Total Area)`	`= 20.784 + 2`
	`= 22.784\ text(cm²)`

Probability, 2UG 2007 HSC 28a

Two unbiased dice, `A` and `B`, with faces numbered `1`, `2`, `3`, `4`, `5` and `6` are rolled.

The numbers on the uppermost faces are noted. This table shows all the possible outcomes.

A game is played where the difference between the highest number showing and the lowest number showing on the uppermost faces is calculated.

What is the probability that the difference between the numbers showing on the uppermost faces of the two dice is one? (1 mark)
In the game, the following applies.
What is the financial expectation of the game? (3 marks)
If Jack pays `$1` to play the game, does he expect a gain or a loss, and how much will it be? (1 mark)

Show Answers Only

`5/18`
`$0.75`
`text(If Jack pays $1 to play, he should expect)`
`text(a loss of $0.25.)`

Show Worked Solution

(i) `text(# Outcomes with a difference of 1)`

`= 10`

`:.\ P text{(diff of 1)} = 10/36 = 5/18`

(ii) `P text{(no difference)} = 6/36 = 1/6`

`P text{(2, 3, 4 or 5)}`	`= 1 – [P(0) + P(1)]`
	`= 1 – [5/18 + 1/6]`
	`= 1 – 8/18`
	`= 5/9`

`:.\ text(Financial Expectation)`

`= (1/6 xx 3.50) – (5/18 xx 5) + (5/9 xx 2.80)`

`= $0.75`

(iii) `text(If Jack pays $1 to play, he should expect)`

`text(a loss of $0.25.)`

GEOMETRY, FUR1 2007 VCAA 5 MC

A block of land has an area of 4000 m².

When represented on a map, this block of land has an area of 10 cm².

On the map 1 cm would represent an actual distance of

A. `10\ text(m)`

B. `20\ text(m)`

C. `40\ text(m)`

D. `400\ text(m)`

E. `4000\ text(m)`

Show Answers Only

`B`

Show Worked Solution

`text(Area scale factor) = k^2`

♦♦♦ Mean mark 18%.
MARKER’S COMMENT: A majority of students obtained the the “area” scale factor (k² = 400) but failed to convert this to the corresponding linear scale factor.

`k^2`	`= 4000/10`
	`= 400`
`:. k`	`= sqrt(400)`
	`= 20`

`:.\ text(Linear scale factor) = 20,`

`text(i.e.)\ \ 1\ text(cm):20\ text(m)`

`=> B`

GEOMETRY, FUR1 2009 VCAA 9 MC

A vertical pole, `TP`, is 4 metres tall and stands on level ground near a vertical wall.

The wall is 6 metres long and 4 metres high.

The base of the pole, `T`, is 5 metres from one end of the wall at `N` and 4 metres from the other end of the wall at `M`.

The pole falls and hits the wall.

The maximum height above ground level at which the pole could hit the wall is closest to

A. `0\ text(m)`

B. `1.5\ text(m)`

C. `2.3\ text(m)`

D. `2.7\ text(m)`

E. `3.3\ text(m)`

Show Answers Only

`C`

Show Worked Solution

♦♦ Mean mark 27%.
MARKER’S COMMENT: The key concept to finding the solution is to realise that max height occurs when the pole falls at right angles to the wall.

`text(Using cosine rule in)\ Delta MTN\ text(to find)\ theta`

`cos theta`	`= (6^2 + 4^2 − 5^2) / (2 × 6 × 4)`
	`= 0.5625`
`theta`	`= 55.77…^@`

`text(In)\ Delta AMT,`
`sin theta`	`= x/4`
`x`	`= 4 xx sin 55.77…^@`
	`= 3.307…`
`text(When the pole falls, max height) = h`

`text(Using Pythagoras,)`

`4^2`	`= h^2 + 3.307^2`
`h^2`	`= 4^2 − 3.307^2`
	`= 5.0625`
`h`	`= 2.25…\ text(m)`
`=> C`

PATTERNS, FUR1 2009 VCAA 9 MC

There are 10 checkpoints in a 4500 metre orienteering course.

Checkpoint 1 is the start and checkpoint 10 is the finish.

The distance between successive checkpoints increases by 50 metres as each checkpoint is passed.

The distance, in metres, between checkpoint 2 and checkpoint 3 is

A. `225`

B. `275`

C. `300`

D. `350`

E. `400`

Show Answers Only

`D`

Show Worked Solution

`text(9 distances exist between 10 checkpoints)`

♦♦ Mean mark 26%.

`text(9 distances form an AP, where)`

`d = 50,\ \ \ S_9 = 4500`

`S_n`	`= n/2[2a + (n − 1)d]`
`4500`	`= 9/2 [2a + (9 − 1) × 50]`
`4500`	`= 9/2[2a + 400]`
`9a`	`= 2700`
`a`	`= 300`

`:.\ text(Distance between checkpoint 1 and 2)`

`= a=300\ text(m)`

`:.\ text(Distance between checkpoint 2 and 3)`

`= a+d=350\ text(m)`

`=> D`

GEOMETRY, FUR1 2014 VCAA 9 MC

The middle section of a cone is shaded, as shown in the diagram below.

The surface area of the unshaded top section of the cone is 180 cm².

The surface area of the middle section of the cone, in square centimetres, is

A. `80`

B. `120`

C. `300`

D. `320`

E. `500`

Show Answers Only

`D`

Show Worked Solution

`text(Ratio of lengths of)`

♦♦ Mean mark 33%.

`text(Upper Cone : Top Section, is)`

`15:9 = 5:3`

`∴\ text(Ratio of Areas)`	`=5^2:3^2`
	`=25:9`

`text(Let Area of Upper Cone)\ = x`

`x/180`	`= 25/9`
`x`	`= 180 xx 25/9`
	`= 500\ \ text(cm²)`

`∴\ text(Area of shaded middle section)`

`= 500 – 180`

`= 320\ \ text(cm²)`

`=> D`

GEOMETRY, FUR1 2012 VCAA 9 MC

The solid `OPQR`, as shown below, is one-eighth of a sphere of radius 15 cm.

The point `O` is the centre of the sphere and the points `P, Q` and `R` are on the surface of the sphere.

`∠POQ = ∠QOR = ∠ROP = 90°`

The total surface area of the solid `OPQR`, in cm², is closest to

A. 619

B. 648

C. 706

D. 884

E. 1767

Show Answers Only

`D`

Show Worked Solution

`text (Surface Area of sphere) = 4pir^2`

♦♦ Mean mark 31%.

`:.\ text (S.A. of)\ \ PQR\ \ text{(on surface)}`

`=1/8 xx 4 pi r^2`

`= 1/8 xx 4 xx pi xx 15^2`

`= 353.43\ text(cm²)`

`text (Surface Area of sector)\ POQ`

`= 1/4pir^2`

`= 1/4pi × 15^2`

`= 176.71\ text(cm²)`

`:. text (Surface Area of solid)\ OPQR`

`= 353.43 + 3 × 176.71`

`= 883.56\ text(cm²)`

`rArr D`

Algebra, STD2 A2 2007 HSC 18 MC

Chris started to make this pattern of shapes using matchsticks.

If the pattern of shapes is continued, which shape would use exactly 486 matchsticks?

Shape 96
Shape 97
Shape 121
Shape 122

Show Answers Only

`C`

Show Worked Solution

`text(Equation rule:)`

`M = 4S + 2`

`text(Find)\ \ x\ \ text(when)\ \ M = 486:`

`486`	`= 4S+2`
`4S`	`= 484`
`S`	`=121`

`:.\ text(The 121st shape uses 486 matchsticks.)`

`=> C`

CORE, FUR1 2014 VCAA 12 MC

The seasonal index for heaters in winter is 1.25.

To correct for seasonality, the actual heater sales in winter should be

A. reduced by 20%.

B. increased by 20%.

C. reduced by 25%.

D. increased by 25%.

E. reduced by 75%.

Show Answers Only

`A`

Show Worked Solution

`text(Deseasonalised Sales)`

♦♦ Mean mark 26%.
MARKER’S COMMENT: A majority of students appeared to answer this question by inspection only which was not possible. Don’t be tricked!

`=\ text(Actual Sales)/text(Seasonal index)`

`=\ text(Actual Sales)/1.25`

`=80 text(%) xx text(Actual Sales)`

`∴\ text(Winter Sales should be reduced by 20%)`

`=> A`

CORE, FUR1 2011 VCAA 12 MC

The seasonal index for headache tablet sales in summer is 0.80.

To correct for seasonality, the headache tablet sales figures for summer should be

A. reduced by 80%

B. reduced by 25%

C. reduced by 20%

D. increased by 20%

E. increased by 25%

Show Answers Only

`E`

Show Worked Solution

`text(Deseasonalised Sales)`

♦♦♦ Mean mark 10%!
MARKERS’ COMMENT: The key to solving this problem is to rearrange the “Seasonal Index” formula in the formula sheet.

`= \ \ text(Actual Sales) / text(Seasonal index)`

`= \ \ text(Actual Sales) / 0.8`

`= \ \ 1.25 xx text(Actual Sales)`

`:.\ text(Deseasonalised Sales need Actual Sales to)`

`text(be increased by 25%)`

`=> E`

PATTERNS, FUR1 2010 VCAA 8 MC

The `n`th term in a geometric sequence is `t_n`.

The common ratio is greater than one.

A graph that could be used to display the terms of this sequence is

Show Answers Only

`A`

Show Worked Solution

`text(GP where)\ \ r>1`

♦♦♦ Mean mark 16%!

`text(General term is)\ \ \ t_n=ar^(n – 1)`

`text(Consider)\ \a>0,`

`text(the graph of)\ \t_n\ \text(will be positive and)`

`text(increase exponentially, but no graph shows this.)`

`text(Consider)\ \a<0,`

`text(the graph will be negative and increase)`

`text(exponentially in a negative direction.)`

`=> A`

CORE, FUR1 2014 VCAA 8 MC

A single back-to-back stem plot would be an appropriate graphical tool to investigate the association between a car’s speed, in kilometres per hour, and the

A. driver’s age, in years.

B. car’s colour (white, red, grey, other).

C. car’s fuel consumption, in kilometres per litre.

D. average distance travelled, in kilometres.

E. driver’s sex (female, male).

Show Answers Only

`E`

Show Worked Solution

`text(In a back-to-back stem plot, the numerical speed)`

♦♦ Mean mark 28%.

`text(data has to be plotted against categorical data)`

`text(with two options.)`

`∴\ text{Driver’s sex (M or F)}`

`=> E`

♦♦ Mean mark 29%.
MARKER’S NOTE: Postcodes here are categorical variables. Ask yourself “Does it make sense to calculate the mean of this variable?” If the answer is “No”, the variable is categorical.

`text(Categorical variables are sex, type)`

`text(of car, and post code.)`

`=>D`

`text(Part 3)`

`text(1 female driver has 3 children.)`

`=>B`

Measurement, 2UG 2008 HSC 28b

A tunnel is excavated with a cross-section as shown.

Find an expression for the area of the cross-section using TWO applications of Simpson’s rule. (2 marks)
The area of the cross-section must be 600 m². The tunnel is 80 m wide.
If the value of `a` increases by 2 metres, by how much will `b` change? (2 marks)

Show Answers Only

`(2h)/3 (4a + b)`
`b\ text(decreases by 8 m.)`

Show Worked Solution

(i)	`A`	`~~ h/3 [y_0 + 4y_1 + y_2] + h/3 [y_0 + 4y_1 + y_2]`
		`~~ h/3 [0 + 4a + b] + h/3 [b + 4a + 0]`
		`~~ (2h)/3 (4a + b)`

(ii)	`text(Given)\ \ A = 600\ text(m²)`
	`text(If 80 m wide) \ => h = 20`

`A`	`= (2h)/3 (4a + b)`
`600`	`= ((2 xx 20))/3 (4a + b)`
`4a + b`	`= (600 xx 3)/40`
`b`	`= 45 – 4a`

`:.\ text(If)\ a\ text(increases by 2 m,)\ b\ text(will)`

`text(decrease by 8 m.)`

Statistics, STD2 S5 2008 HSC 28a

The following graph indicates `z`-scores of ‘height-for-age’ for girls aged 5 – 19 years.

What is the `z`-score for a six year old girl of height 120 cm? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Rachel is 10 ½ years of age.

(1) If 2.5% of girls of the same age are taller than Rachel, how tall is she? (1 mark)

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

(2) Rachel does not grow any taller. At age 15 ½, what percentage of girls of the same age will be taller than Rachel? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
What is the average height of an 18 year old girl? (1 mark)

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

For adults (18 years and older), the Body Mass Index is given by

`B = m/h^2` where `m = text(mass)` in kilograms and `h = text(height)` in metres.

The medically accepted healthy range for `B` is `21 <= B <= 25`.

What is the minimum weight for an 18 year old girl of average height to be considered healthy? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
The average height, `C`, in centimetres, of a girl between the ages of 6 years and 11 years can be represented by a line with equation

`C = 6A + 79` where `A` is the age in years.

(1) For this line, the gradient is 6. What does this indicate about the heights of girls aged 6 to 11? (1 mark)

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

(2) Give ONE reason why this equation is not suitable for predicting heights of girls older than 12. (1 mark)

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

Show Answers Only

`1`
(1) `155\ text(cm)`

(2) `text(84%)`
`163\ text(cm)`
`55.8\ text(kg)`
(1) `text(It indicates that 6-11 year old girls)`

`text(grow, on average, 6cm per year)`

(2) `text(Girls eventually stop growing, and the)`

`text(equation doesn’t factor this in.)`

Show Worked Solution

i.	`z text(-score) = 1`

ii.	(1)	`text(If 2 ½ % are taller than Rachel)`
		`=> z text(-score of +2)`
		`:.\ text(She is 155 cm)`

	(2)	`text(At age)\ 15\ ½,\ 155\ text(cm has a)\ z text(-score of –1)`
		`text(68% between)\ z = 1\ text(and)\ –1`
		`=> text(34% between)\ z = 0\ text(and)\ –1`
		`text(50% have)\ z >= 0`

		`:.\ text(% Above)\ z text(-score of –1)`
		`= 50 + 34`
		`= 8text(4%)`

`:.\ text(84% of girls would be taller than Rachel at age)\ 15 ½.`

iii.	`text(Average height of 18 year old has)\ z text(-score = 0)`
	`:.\ text(Average height) = 163\ text(cm)`

iv.	`B = m/h^2`
	`h = 163\ text(cm) = 1.63\ text(m)`

`text(Given)\ \ 21 <= B <= 25,\ text(minimum healthy)`

`text(weight occurs when)\ B = 21`

`=> 21`	`= m/1.63^2`
`m`	`= 21 xx 1.63^2`
	`= 55.794…`
	`= 55.8\ text(kg)\ text{(1 d.p.)}`

v.	(1)	`text(It indicates that 6-11 year old girls, on average, grow)`
		`text(6 cm per year.)`
	(2)	`text(Girls eventually stop growing, and the equation doesn’t)`
		`text(factor this in.)`

Measurement, STD2 M6 2008 HSC 25c

Pieces of cheese are cut from cylindrical blocks with dimensions as shown.

Twelve pieces are packed in a rectangular box. There are three rows with four pieces of cheese in each row. The curved surface is face down with the pieces touching as shown.

What are the dimensions of the rectangular box? (4 marks)

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To save packing space, the curved section is removed.
What is the volume of the remaining triangular prism of cheese? Answer to the nearest cubic centimetre. (2 marks)

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Show Answers Only

`41\ text(cm) xx 21\ text(cm) xx 15\ text(cm)`
`506\ text(cm)³\ text{(nearest whole)}`

Show Worked Solution

i. `text(Box height) = 15\ text(cm)`

♦ Mean mark 45%.

`text{(radius of the arc)}`

`text(Box width)`	`= 3 xx 7`
	`= 21\ text(cm)`
`text(Box length)`	`= 4x`

`text(Using cosine rule)`

`c^2`	`= a^2 + b^2 – 2ab cos C`
`x^2`	`= 15^2 + 15^2 – 2 xx 15 xx 15 xx cos 40^@`
	`= 450 – 344.7199…`
	`= 105.2800…`
`x`	`= 10.2606…`

`text(Box length)`	`= 4 xx 10.2606…`
	`= 41.04…`

`:.\ text(Dimensions are)\ \ 41\ text(cm) xx 21\ text(cm) xx 15\ text(cm)`

ii. `text(Volume) = Ah`

♦♦♦ Mean mark 22%.

`h = 7\ text(cm)`

`text(Find)\ A:`

`A`	`= 1/2 ab sin C`
	`= 1/2 xx 15 xx 15 xx sin 40^@`
	`= 72.3136…`

`:. V`	`= 72.3136… xx 7`
	`= 506.195…`
	`= 506\ text(cm³)\ \ text{(nearest whole)}`

Probability, STD2 S2 2008 HSC 24b

Three-digit numbers are formed from five cards labelled 1, 2, 3, 4 and 5.

How many different three-digit numbers can be formed? (1 mark)

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If one of these numbers is selected at random, what is the probability that it is odd? (1 mark)

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How many of these three-digit numbers are even? (1 mark)

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What is the probability of randomly selecting a three-digit number less than 500 with its digits arranged in descending order? (2 marks)

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Show Answers Only

`60`
`3/5`
`24`
`1/15`

Show Worked Solution

i. `text(# Different numbers)`

♦ Mean mark 45%.

`= 5 xx 4 xx 3`

`= 60`

ii. `text(The last digit must be one of the)`

`text(5 numbers, of which 3 are odd)`

`:.\ text{P(odd)} = 3/5`

iii. `text{P(even)} = 1- text{P(odd)} = 2/5`

♦ Mean mark 48%.

`:.\ text(Number of even numbers)`

`= 2/5 xx 60`

`= 24`

iv. `text(The numbers that satisfy the criteria:)`

♦♦♦ Mean mark 10%.

`432, 431, 421, 321`

`:.\ text{P(selection)} = 4/60 = 1/15`

Algebra, 2UG 2008 HSC 19 MC

The height of a particular termite mound is directly proportional to the square root of the number of termites.

The height of this mound is 35 cm when the number of termites is 2000.

What is the height of this mound, in centimetres, when there are 10 000 termites?

(A) `16`

(B) `78`

(C) `175`

(D) `875`

Show Answers Only

`B`

Show Worked Solution

`text(Height)\ (h)`	`prop sqrt( text(Termites)\ (T)`
`=>h`	`= k sqrt T`

`text(Given)\ \ h = 35\ \ text(when)\ \ T = 2000`

`35`	`= k xx sqrt 2000`
`k`	`= 35/ sqrt 2000`
	`= 0.7826…`

`text(Find)\ \ h\ \ text(when)\ \ T = 10\ 000`

`h`	`= 0.7826… xx sqrt (10\ 000)`
	`= 78.26…\ text(cm)`

`=> B`

Calculus, 2ADV C4 2008 HSC 9c

A beam is supported at `(-b, 0)` and `(b, 0)` as shown in the diagram.

It is known that the shape formed by the beam has equation `y = f(x)`, where `f(x)` satisfies

	`f^{″}(x)`	`= k (b^2-x^2),\ \ \ \ \ `(`k` is a positive constant)
and	`f^{′}(-b)`	`= -f'(b)`.

Show that `f^{′}(x) = k (b^2x-(x^3)/3)`. (2 marks)

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How far is the beam below the `x`-axis at `x = 0`? (2 marks)

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Show Answers Only

`text(Proof)\ text{(See Worked Solutions)}`
`(5kb^4)/12\ text(units)`

Show Worked Solution

i.	`text(Show)\ \ f^{′}(x) = k (b^2x-x^3/3)`

`f^{″}(x)`	`= k (b^2 – x^2)`
`f^{′}(x)`	`= int k (b^2-x^2)\ dx`
	`= k int b^2-x^2\ dx`
	`= k (b^2x-x^3/3) + c`

`text(S)text(ince S.P. exists at)\ \ x = 0`

`=> f^{′}(x)`	`= 0\ \ text(when)\ \ x = 0`
`0`	`= k (b^2 * 0-0) + c`
`c`	`= 0`

`:.\ f^{′}(x) = k (b^2x-x^3/3)\ \ \ text(… as required)`

ii.	`f(x)`	`= int f^{′}(x)\ dx`
		`= k int b^2x-x^3/3\ dx`
		`= k ((b^2x^2)/2-x^4/12) + c`

`text(We know)\ \ f(x) = 0\ \ text(when)\ \ x = b`

`=> 0`	`= k ( (b^2*b^2)/2-b^4/12) + c`
`c`	`= -k ( (6b^4)/12-b^4/12)`
	`= -k ( (5b^4)/12 )`
	`= -(5kb^4)/12`

`:.\ text(When)\ \ x = 0, text(the beam is)\ \ (5kb^4)/12\ \ text(units)`

`text(below the)\ x text(-axis.)`

Combinatorics, EXT1 A1 2014 HSC 14b

Two players `A` and `B` play a game that consists of taking turns until a winner is determined. Each turn consists of spinning the arrow on a spinner once. The spinner has three sectors `P`, `Q` and `R`. The probabilities that the arrow stops in sectors `P`, `Q` and `R` are `p`, `q` and `r` respectively.

The rules of the game are as follows:

• If the arrow stops in sector `P`, then the player having the turn wins.

• If the arrow stops in sector `Q`, then the player having the turn loses and the other player wins.

• If the arrow stops in sector `R`, then the other player takes a turn.

Player `A` takes the first turn.

Show that the probability of player `A` winning on the first or the second turn of the game is `(1 − r) (p + r)`. (2 marks)

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Show that the probability that player `A` eventually wins the game is `(p + r)/(1 + r)`. (3 marks)

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Show Answers Only

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

Show Worked Solution

i. `text(Show)\ P text{(}A\ text(wins)\ T_1\ text(or)\ T_2 text{)} = (1 – r)(p + r)`

`P (A\ text(wins on)\ T_1) = p`

`P (A\ text(wins on)\ T_2)`

`= P text{(} A\ text(lands on)\ R,\ B\ text(lands on)\ Q text{)}`

`= rq`

`:.\ P text{(}A\ text(wins)\ T_1\ text(or)\ T_2 text{)}`

♦♦ Mean mark 25%.
HINT: Expand out the solution `(1-r)(p+r)“=p+r-rp-r^2` to get a better idea of what you need to prove.

`= p + rq`

`text(S)text(ince)\ q = 1 – (p + r),`

`P text{(} A\ text(wins)\ T_1\ text(or)\ T_2 text{)}`

`= p + r [1 – (p + r)]`

`= p + r – rp – r^2`

`= (1 – r)(p + r)\ \ \ text(… as required.)`

ii. `text(Show)\ P text{(}A\ text(wins eventually) text{)} = (p + r)/(1 + r)`

`P text{(} A\ text(wins)\ T_1\ text(or)\ T_2 text{)} = (1 – r)(p + r)`

`P text{(} text(No result)\ T_1\ text(or)\ T_2 text{)} = r * r = r^2`

`:.\ P text{(} A\ text(wins eventually) text{)}`

`= underbrace{(1 – r)(p + r) + r^2 (1 – r)(p + r) + r^2*r^2 (1 – r)(p + r) + …}_{text(GP where)\ \ a = (1 – r)(p + r),\ \ r = r^2}`

`text(S)text(ince)\ \ 0 < r < 1 \ \ =>\ \ 0 < r^2 < 1`

♦♦♦ Mean mark 11%. Lowest in the 2014 HSC exam!
HINT: The use of ‘eventually’ in the question should flag the possibility of solving by using `S_oo` in a GP.

`:.\ S_oo`	`= a/(1 – r)`
	`= ((1 – r)(p + r))/(1 – r^2)`
	`= ((1 – r)(p + r))/((1 – r)(1 + r))`
	`= (p + r)/(1 + r)\ \ \ text(… as required)`

Plane Geometry, EXT1 2009 HSC 7c

Consider the billboard below. There is a unique circle that passes through the top and bottom of the billboard (points `Q` and `R` respectively) and is tangent to the street at `T`.

Let `phi` be the angle subtended by the billboard at `S`, the point where `PQ` intersects the circle.

Copy the diagram into your writing booklet.

Show that `theta < phi` when `P` and `T` are different points, and hence show that `theta` is a maximum when `P` and `T` are the same point. (3 marks)
Using circle properties, find the distance of `T` from the building. (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`sqrt (h (a + h))`

Show Worked Solution

(i)

♦♦ Both parts proved extremely challenging for students.
MARKER’S COMMENT: Very few students recognised that `phi` is the external angle of `Delta PSR`.

`text(Show)\ \ theta < phi\ \ text(when)\ \ P != T`

`text(Let)\ \ /_PRS = alpha`

`phi`

`= theta + alpha\ \ text{(exterior angle of}\ Delta PSR text{)}`

`:.\ theta < phi,\ alpha != 0`

`text(When)\ P\ text(and)\ T\ text(are the same point,)`

`S,\ P, and T\ \ text(are the same point.)`

`text(i.e. when)\ \ /_PRS = alpha = 0`

`:. theta\ text(will be a maximum)\ (= phi)`

`text(when)\ P = T.`

(ii) `text(Square of tangent)\ =\ text(product of)`

`text(intercepts from an external point.)`

`text(i.e.)\ \ OT^2`	`= OR xx QR`
	`= h (a + h)`
`:.\ OT`	`= sqrt (h (a + h))`

Geometry and Calculus, EXT1 2009 HSC 7b

A billboard of height `a` metres is mounted on the side of a building, with its bottom edge `h` metres above street level. The billboard subtends an angle `theta` at the point `P`, `x` metres from the building.

Use the identity `tan (A - B) = (tan A - tan B)/(1 + tanA tanB)` to show that
1. `theta = tan^(-1) [(ax)/(x^2 + h(a + h))]`. (2 marks)
The maximum value of `theta` occurs when `(d theta)/(dx) = 0` and `x` is positive.
Find the value of `x` for which `theta` is a maximum. (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`sqrt(h (a + h)),\ x > 0`

Show Worked Solution

(i) `text(Consider angles)\ \ A and B\ \ text(on the graph:)`

♦♦ Mean mark data not available for parts of questions although Q7 as a whole scored <30%.
MARKER’S COMMENT: Answers that included a diagram and clearly labelled angles were generally successful.

`text(Show)\ \ theta = tan^(-1) [(ax)/(x^2 + h(a + h))]`

`tan A`	`= (a + h)/x`
`tan B`	`= h/x`

`tan (A – B)`	`= ((a + h)/x – h/x)/(1 + ((a + h)/x)(h/x)) xx (x^2)/(x^2)`
	`= (x(a + h) – xh)/(x^2 + h(a + h))`
	`= (ax)/(x^2 + h(a + h)`

`text(S)text(ince)\ \ theta`	`= A – B`
`theta`	`= tan^(-1) [(ax)/(x^2 + h(a + h))]\ \ \ text(… as required.)`

(ii) `text(Max when)\ \ (d theta)/(dx) = 0\ \ text(and)\ \ x > 0`

`text(Let)\ \ u`	`= (ax)/(x^2 + h(a + h))`
`theta`	`= tan^(-1) u`
`(du)/(dx)`	`=(a[x^2 + h (a + h)] – ax * 2x)/([x^2 + h (a + h)]^2)`
	`=(-ax^2 + ah (a + h))/([x^2 + h (a + h)]^2)`

`(d theta)/(dx)`	`=(d theta)/(du) * (du)/(dx)`
	`= 1/(1 + u^2) * (du)/(dx)`
	`= 1/(1 + [(ax)/(x^2 + h(a + h))]^2) xx (-ax^2 + ah (a + h))/[x^2 + h (a + h)]^2`
	`=(-ax^2 + ah(a + h))/([x^2 + h (a + h)]^2 + a^2 x^2)`

`text(Note that)\ \ (d theta)/(dx) = 0\ \ text(when)`

`-ax^2 + ah (a + h)`	`= 0`
`ax^2`	`= ah (a + h)`
`x^2`	`= h (a + h)`
`x`	`= sqrt (h (a + h))\ \ \ \ (x > 0)`

Binomial, EXT1 2009 HSC 6b

Sum the geometric series
1. `(1 + x)^r + (1 + x)^(r + 1) + ... + (1 + x)^n`
and hence show that
1. `((r),(r)) + ((r + 1),(r)) + ... + ((n),(r)) = ((n + 1),(r + 1))`. (3 marks)
Consider a square grid with `n` rows and `n` columns of equally spaced points.
The diagram illustrates such a grid. Several intervals of gradient `1`, whose endpoints are a pair of points in the grid, are shown.
(1) Explain why the number of such intervals on the line `y = x` is equal to `((n),(2))`. (1 mark)
(2) Explain why the total number, `S_n`, of such intervals in the grid is given by
1. `S_n = ((2),(2)) + ((3),(2)) + ... + ((n - 1),(2)) + ((n),(2)) + ((n - 1),(2)) +`
  1. `... + ((3),(2)) + ((2),(2))`. (1 mark)
Using the result in part (i), show that
1. `S_n = (n(n - 1)(2n - 1))/6`. (3 marks)

Show Answers Only

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

Show Worked Solution

(i) `(1 + x)^r + (1 + x)^(r + 1) + … + (1 + x)^n`

♦♦ Exact data unavailable although mean marks for all Q6 was < 25%.
MARKER’S COMMENT: A common mistake was to assume the number of terms in the series was `n`.

`=> text(GP where)\ \ a`	`= (1 + x)^r`
`r`	`= (1 + x)`
`#\ text(Terms)`	`= n – r + 1`

`text(Sum)`	`= (a (r^n – 1))/(r – 1)`
	`= ((1 + x)^r [(1 + x)^(n\ – r + 1) – 1])/((1 + x) – 1)`
	`= ((1 + x)^(n\ – r + 1 + r) – (1 + x)^r)/x`
	`= ((1 + x)^(n + 1) – (1 + x)^r)/x`

`text(Show)`

`((r),(r)) + ((r + 1),(r)) + … + ((n),(r)) = ((n + 1),(r + 1))`

`text(Consider series)\ (1 + x)^r + (1 + x)^(r + 1) + … + (1 + x)^n`

`text(Co-efficient of)\ \ x^r = ((r),(r)) + ((r + 1),(r)) + … + ((n),(r))`

`text(Consider sum of series)\ \ \ ((1 + x)^(n + 1)\ – (1 + x)^r)/x`

`text(Co-efficient of)\ \ x^r`	`=\ text(co-efficient of)\ \ x^(r + 1)\ \ text(in numerator)`
	`= ((n +1),(r + 1))`

`:.\ text(S)text(ince co-efficients are equal)`

`((r),(r)) + ((r + 1),(r)) + … + ((n),(r)) = ((n + 1),(r + 1))`

`text(… as required.)`

(ii)(1)	`text(All intervals start and finish at different points.)`
	`text(Any)\ n xx x\ text(grid has)\ n\ text(points on the)\ y = x\ text(diagonal.)`
	`:.\ text(Possible intervals) = ((n),(2))`

(2)	`text(The longest diagonal of)\ n xx n\ text(grid is)`
	`text(the)\ y = x\ text(diagonal with)\ ((n),(2))\ text(intervals.)`
	`text(Each side of this, there is one less point)`
	`text(on the diagonals, with)\ ((n – 1),(2))\ text(intervals.)`
	`text(This continues until there are only 2 points)`
	`text(on the diagonal with)\ ((2),(2))\ text(intervals.)`

`:.\ S_n`	`= ((2),(2)) + ((3),(2)) + … + ((n – 1),(2)) + ((n),(2))`
	`+ ((n – 1),(2)) + … + ((2),(2))`
	`text(… as required.)`

♦♦♦ Exact data unavailable although very few completed part (iii) correctly.
MARKER’S COMMENT: Note that even if part (i) wasn’t solved, the proof as stated in the question can be used to solve part (iii).

(iii)

`text(Show)\ \ S_n = (n(n – 1)(2n – 1))/6`

`S_n`	`= { ((2),(2)) + ((3),(2)) + … + ((n – 1),(2)) + ((n),(2))}`
	`+ {((n – 1),(2)) + … + ((2),(2))}`
	`= ((n + 1),(3)) + ((n),(3))\ \ \ text{(from part (i))}`
	`= ((n + 1)n(n – 1))/(3 * 2 * 1) + (n* (n – 1)(n – 2))/(3 2 * 1)`
	`= 1/6 n (n – 1) (n + 1 + n – 2)`
	`= (n(n – 1)(2n – 1))/6\ \ \ text(… as required.)`

Calculus, EXT1 C1 2009 HSC 5b

The cross-section of a 10 metre long tank is an isosceles triangle, as shown in the diagram. The top of the tank is horizontal.

When the tank is full, the depth of water is 3 m. The depth of water at time `t` days is `h` metres.

Find the volume, `V`, of water in the tank when the depth of water is `h` metres. (1 mark)

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Show that the area, `A`, of the top surface of the water is given by `A = 20 sqrt3 h`. (1 mark)

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The rate of evaporation of the water is given by `(dV)/(dt) = - kA`, where `k` is a positive constant.

Find the rate at which the depth of water is changing at time `t`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
It takes 100 days for the depth to fall from 3 m to 2 m. Find the time taken for the depth to fall from 2 m to 1 m. (1 mark)

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

Show Answers Only

`10 sqrt 3 h^2\ \ text(m³)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`-k\ \ \ text(metres per day)`
`100\ text(days)`

Show Worked Solution

MARKER’S COMMENT: Students who drew a diagram and included their working calculations on it were the most successful.

(i)
	`text(Let)\ A = text(area of front)`

`tan 30^@`	`= h/x`
`x`	`= h/(tan 30^@)`
	`= sqrt3 h`

`:.\ A`	`= 2 xx 1/2 xx sqrt 3 h xx h`
	`= sqrt 3 h^2\ \ text(m²)`

`V`	`= Ah`
	`= sqrt 3 h^2 xx 10`
	`= 10 sqrt3 h^2\ \ text(m³)`

(ii)	`text(Area of surface)`
	`= 10 xx 2 sqrt 3 h`
	`= 20 sqrt 3 h\ \ text(m²)`

(iii)	`(dV)/(dt)`	`= -kA`
		`= -k\ 20 sqrt3 h`

`V`	`= 10 sqrt3 h^2`
`(dV)/(dh)`	`= 20 sqrt 3 h`

`text(Find)\ (dh)/(dt)`

MARKER’S COMMENT: Half marks awarded for stating an appropriate chain rule, even if the following calculations were incorrect. Show your working!

`(dV)/(dt)`	`= (dV)/(dh) * (dh)/(dt)`
`(dh)/(dt)`	`= ((dV)/(dt))/((dV)/(dh))`
	`= (-k * 20 sqrt 3 h)/(20 sqrt 3 h)`
	`= -k`

`:.\ text(The water depth is changing at a rate)`

`text(of)\ -k\ text(metres per day.)`

♦♦♦ Exact data for part (iv) not available.
COMMENT: Interpreting a constant rate of change was very poorly understood!

(iv)	`text(S)text(ince)\ \ (dh)/(dt)\ \ text(is a constant, each metre)`
	`text(takes the same time.)`

`:.\ text(It takes 100 days to fall from 2 m to 1 m.)`

Plane Geometry, 2UA 2014 HSC 15b

In `Delta DEF`, a point `S` is chosen on the side `DE`. The length of `DS` is `x`, and the length of `ES` is `y`. The line through `S` parallel to `DF` meets `EF` at `Q`. The line through `S` parallel to `EF` meets `DF` at `R`.

The area of `Delta DEF` is `A`. The areas of `Delta DSR` and `Delta SEQ` are `A_1` and `A_2` respectively.

Show that `Delta DEF` is similar to `Delta DSR`. (2 marks)
Explain why `(DR)/(DF) = x/(x + y)`. (1 mark)
Show that
1. `sqrt ((A_1)/A) = x/(x + y)`. (2 marks)
Using the result from part (iii) and a similar expression for
1. `sqrt ((A_2)/A)`, deduce that `sqrt A = sqrt (A_1) + sqrt (A_2)`. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Corresponding sides of similar)\ Delta text(s)`
`text(are in the same ratio.)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `text(Need to show)\ Delta DEF\ text(|||) \ Delta DSR`

`/_FDE\ text(is common)`

`/_DSR = /_DEF = theta\ \ text{(corresponding angles,}\ RS\ text(||)\ FE text{)}`

`:.\ Delta DEF \ text(|||) \ Delta DSR\ \ text{(equiangular)}`

(ii) `(DR)/(DF) = (DS)/(DE) = x/(x + y)`

`text{(Corresponding sides of similar triangles)}`

♦♦ Mean mark 27%
COMMENT: The critical step in solving part (iii) is realising that you need the areas of 2 non-right angled triangles and therefore the formula `A=½\ ab sin C` is required.

(iii) `text(Show)\ sqrt((A_1)/A) = x/(x + y)`

`text(Using Area)`	`= 1/2 ab sin C`
`A_1`	`= 1/2 xx DR xx x xx sin alpha`
`A`	`= 1/2 xx DF xx (x + y) xx sin alpha`

`(A_1)/A`	`= (1/2 * DR * x * sin alpha)/(1/2 * DF * (x + y) * sin alpha)`
	`= (DR * x)/(DF * (x + y)`
	`= (x * x)/((x + y)(x + y))\ \ \ \ text{(using part(ii))}`
	`= (x^2)/((x + y)^2)`
`:.\ sqrt ((A_1)/A)`	`= x/((x + y))\ \ \ text(… as required.)`

(iv) `text(Consider)\ Delta DFE\ text(and)\ Delta SQE`

`/_FED`

`= theta\ text(is common)`

`/_FDE`

`= /_QSE = alpha\ \ `

`text{(corresponding angles,}\ DF\ text(||)\ QS text{)}`

`:.\ Delta DFE\ text(|||)\ Delta SQE\ \ text{(equiangular)}`

`(QE)/(FE) = (SE)/(DE) = y/(x +y)`

`text{(corresponding sides of similar triangles)}`

♦♦ Mean mark 26%

`A_2`	`= 1/2 xx QE xx y xx sin theta`
`A`	`= 1/2 xx FE xx (x + y) xx sin theta`
`(A_2)/A`	`= (QE * y)/(FE * (x + y))`
	`= (y^2)/((x + y)^2)`
`sqrt ((A_2)/A)`	`= y/((x + y))`

`text(Need to show)\ sqrt A = sqrt (A_1) + sqrt (A_2)`

`sqrt(A_2)/sqrtA`	`= y/((x + y))`
`sqrt (A_2)`	`= (sqrtA * y)/((x + y))`

`text(Similarly, from part)\ text{(iii)}`

`sqrt (A_1) = (sqrtA * x)/((x + y))`

`sqrt (A_1) + sqrt (A_2)`	`= (sqrt A * x)/((x + y)) + (sqrt A * y)/((x + y))`
	`= (sqrt A (x + y))/((x + y))`
	`= sqrt A\ \ \ text(… as required.)`

Trigonometry, 2ADV T2 2014 HSC 7 MC

How many solutions of the equation `(sin x - 1)(tan x + 2) = 0` lie between `0` and `2 pi`?

Show Answers Only

`B`

Show Worked Solution

♦♦♦ Mean mark 25%, making it the toughest MC question in the 2014 exam.
COMMENT: Note that the “2 solutions” answer relies on the sum of an infinity of zeros not equalling zero. This concept created unintended difficulty in this question.

`text(When)\ (sin x\ – 1)(tan x + 2) = 0`

`(sinx\ – 1) = 0\ \ text(or)\ \ tan x + 2 = 0`

`text(If)\ \ sin x\ – 1`	`= 0`
`sin x`	`= 1`
`x`	`= pi/2,\ \ \ 0 < x < 2 pi`

`text(If)\ \ tan x + 2`	`= 0`
`tan x`	`= -2`

`=>\ text(Note that since)\ \ tan\ pi/2\ \ text(is undefined, there)`

`text(are only 2 solutions when)\ \ tan x = -2`

`text{(which occurs in the 1st and 4th quadrants).}`

`:.\ 2\ text(solutions)`

`=> B`

Statistics, STD2 S4 2014* HSC 30b

The scatterplot shows the relationship between expenditure per primary school student, as a percentage of a country’s Gross Domestic Product (GDP), and the life expectancy in years for 15 countries.

For the given data, the correlation coefficient, `r`, is 0.83. What does this indicate about the relationship between expenditure per primary school student and life expectancy for the 15 countries? (1 mark)

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For the data representing expenditure per primary school student, `Q_L` is 8.4 and `Q_U` is 22.5.

What is the interquartile range? (1 mark)

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Another country has an expenditure per primary school student of 47.6% of its GDP.

Would this country be an outlier for this set of data? Justify your answer with calculations. (2 marks)

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On the scatterplot, draw the least-squares line of best fit `y = 1.29x + 49.9`. (2 marks)

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Using this line, or otherwise, estimate the life expectancy in a country which has an expenditure per primary school student of 18% of its GDP. (1 mark)

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Why is this line NOT useful for predicting life expectancy in a country which has expenditure per primary school student of 60% of its GDP? (1 mark)

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

Show Answers Only

`text(It indicates there is a strong positive)`

`text(correlation between the two variables.)`
`14.1`
`text(Yes, because it’s > 43.65%)`
`73.1\ text(years)`
`text(At 60% GDP, the line predicts a life expectancy)`
`text(of 127.3. This line of best fit is only accurate)`
`text(in a lower range of GDP expediture.)`

Show Worked Solution

i.	`text(It indicates there is a strong positive)`
	`text(correlation between the two variables)`

ii.	`text(IQR)`	`= Q_U\ – Q_L`
		`= 22.5\ – 8.4`
		`= 14.1`

♦ Mean mark 35%

iii. `text(An outlier on the upper side must be more than)`

`Q_u\ +1.5xxIQR`

`=22.5+(1.5xx14.1)`

`=\ text(43.65%)`

`:.\ text(A country with an expenditure of 47.6% is an outlier).`

iv.

v. `text(Life expectancy) ~~ 73.1\ text{years (see dotted line)}`

♦♦ Mean mark 39%

`text(Alternative Solution)`

`text(When)\ x=18`

`y=1.29(18)+49.9=73.12\ \ text(years)`

♦♦♦ Mean mark 0%. The toughest question on the 2014 paper.
COMMENT: Examiners regularly ask students to identify and comment on outliers where linear relationships break down.

vi.	`text(At 60% GDP, the line predicts a life)`
	`text(expectancy of 127.3. This line of best)`
	`text(fit is only predictive in a lower range)`
	`text(of GDP expenditure.)`

Algebra, STD2 A4 2014 HSC 29a

The cost of hiring an open space for a music festival is $120 000. The cost will be shared equally by the people attending the festival, so that `C` (in dollars) is the cost per person when `n` people attend the festival.

Complete the table below by filling in the THREE missing values. (1 mark)
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & & & & 60 & 48\ & 40 \ \\
\hline
\end{array}
Using the values from the table, draw the graph showing the relationship between `n` and `C`. (2 marks)
What equation represents the relationship between `n` and `C`? (1 mark)

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Give ONE limitation of this equation in relation to this context. (1 mark)

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Is it possible for the cost per person to be $94? Support your answer with appropriate calculations. (1 mark)

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

Show Answers Only

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}

ii.

iii. `C = (120\ 000)/n`

`n\ text(must be a whole number)`

iv. `text(Limitations can include:)`

`•\ n\ text(must be a whole number)`

`•\ C > 0`

v. `text(If)\ C = 94:`

`94`	`= (120\ 000)/n`
`94n`	`= 120\ 000`
`n`	`= (120\ 000)/94`
	`= 1276.595…`

`:.\ text(C)text(ost cannot be $94 per person,)`

`text(because)\ n\ text(isn’t a whole number.)`

Show Worked Solution

ii.

♦ Mean mark (iii) 48%

iii. `C = (120\ 000)/n`

♦♦♦ Mean mark (iv) 7%
COMMENT: When asked for limitations of an equation, look carefully at potential restrictions with respect to both the domain and range.

iv. `text(Limitations can include:)`

`•\ n\ text(must be a whole number)`

`•\ C > 0`

v. `text(If)\ C = 94`

`=> 94`	`= (120\ 000)/n`
`94n`	`= 120\ 000`
`n`	`= (120\ 000)/94`
	`= 1276.595…`

♦ Mean mark (v) 38%

`:.\ text(C)text(ost cannot be $94 per person,)`

`text(because)\ n\ text(isn’t a whole number.)`

FS Driving, 2UG 2014 HSC 27a

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

Stamp Duty for this car is calculated at $3 for every $100, or part thereof, of the sale price.
Calculate the Stamp Duty payable. (1 mark)
Alex borrows the total amount to be paid for the car including Stamp Duty and transfer of registration. Interest on the loan is charged at a flat rate of 7.5% per annum. The loan is to be repaid in equal monthly instalments over 3 years.
Calculate Alex’s monthly repayments. (4 marks)
Alex wishes to take out comprehensive insurance for the car for 12 months. The cost of comprehensive insurance is calculated using the following:
Find the total amount that Alex will need to pay for comprehensive insurance. (3 marks)
Alex has decided he will take out the comprehensive car insurance rather than the less expensive non-compulsory third-party car insurance.
What extra cover is provided by the comprehensive car insurance? (1 mark)

Show Answers Only

`$402`
`$470\ text{(nearest dollar)}`
`$985.74`
`text(Comprehensive insurance covers Alex)`
`text(for damage done to his own car as well.)`

Show Worked Solution

♦♦♦ Mean mark 12%
IMPORTANT: “or part thereof ..” in the question requires students to round up to 134 to get the right multiple of $3 for their calculation.

(i)

`($13\ 380)/100 = 133.8`

`:.\ text(Stamp duty)`	`= 134 xx $3`
	`= $402`

(ii)	`text(Total loan)`	`= $13\ 380 + 30 + 402`
		`= $13\ 812`

`text(Total interest)\ (I)`	`= Prn`
	`= 13\ 812 xx 7.5/100 xx 3`
	`= 3107.70`

`text(Total to repay)`	`= 3107.70 + 13\ 812`
	`= 16\ 919.70`

`text(# Repayments) = 3 xx 12 = 36`

`:.\ text(Monthly repayment)`	`= (16\ 919.70)/36`
	`= 469.9916…`
	`= $470\ text{(nearest dollar)}`

(iii)

`text(Base rate) = $845`

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

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

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

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

♦ Mean mark 34%.

(iv)	`text(Comprehensive insurance covers Alex)`
	`text(for damage done to his own car as well.)`

Statistics, STD2 S1 2014 HSC 26e

The times taken for 160 music downloads were recorded, grouped into classes and then displayed using the cumulative frequency histogram shown.

On the diagram, draw the lines that are needed to find the median download time. (2 marks)

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Show Answers Only

Show Worked Solution

♦♦♦ Mean mark 17%.

Probability, STD2 S2 2014 HSC 16 MC

In Mathsville, there are on average eight rainy days in October.

Which expression could be used to find a value for the probability that it will rain on two consecutive days in October in Mathsville?

`8/31 xx 7/30`
`8/31 xx 7/31`
`8/31 xx 8/30`
`8/31 xx 8/31`

Show Answers Only

`D`

Show Worked Solution

`P text{(rains)} = 8/31\ \ \text{(independent event for each day)}`

`text{Since each day has same probability:}`

`P(R_1 R_2) = 8/31 xx 8/31`

`=> D`

♦♦♦ Mean mark 16%.
Lowest mark of any MC question in 2014!

FS Comm, 2UG 2014 HSC 5 MC

How many kilobytes are there in 2 gigabytes?

(A) `2^20`

(B) `2^21`

(C) `2^30`

(D) `2^31`

Show Answers Only

`B`

Show Worked Solution

♦♦♦ Mean mark 24%!

`2\ text(GB)`	`= 2 xx 2^20\ text(kB)`
	`= 2^21\ text(kB)`

`=> B`

Quadratic, EXT1 2014 HSC 13c

The point `P(2at, at^2)` lies on the parabola `x^2 = 4ay` with focus `S`.

The point `Q` divides the interval `PS` internally in the ratio `t^2 :1`.

Show that the coordinates of `Q` are
`x = (2at)/(1 + t^2)` and `y = (2at^2)/(1 + t^2)`. (2 marks)
Express the slope of `OQ` in terms of `t`. (1 mark)
Using the result from part (ii), or otherwise, show that `Q` lies on a fixed circle of radius `a`. (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`m_(OQ) = t`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `text(Show)\ \ Q = ((2at)/(1 + t^2), (2at^2)/(1 + t^2))`

`P (2at, at^2),\ S (0, a)`

`PS\ text(is divided internally in ratio)\ t^2: 1`

`Q`	`= ((nx_1 + mx_2)/(m + n), (ny_1 + my_2)/(m + n))`
	`= ((1(2at) + t^2(0))/(t^2 + 1), (1(at^2) + t^2 (a))/(t^2 + 1))`
	`= ((2at)/(1 + t^2), (2at^2)/(1 + t^2))\ \ text(… as required.)`

(ii)	`m_(OQ)`	`= (y_2\ – y_1)/(x_2\ – x_1)`
		`= ((2at^2)/(1 + t^2))/((2at)/(1 + t^2)) xx (1+t^2)/(1+t^2)`
		`= (2at^2)/(2at)`
		`= t`

(iii)	`text(Show)\ Q\ text(lies on a fixed circle radius)\ a`
	`text(S)text(ince)\ Q\ text(passes through)\ (0, 0)`

`=>\ text(If locus of)\ Q\ text(is a circle, it has)`

`text(diameter)\ QT\ text(where)\ T(0, 2a)`

`text(Show)\ \ QT _|_ OQ`

♦♦ Mean mark 22%

`text{(} text(angles on circum. subtended by)`

`text(a diameter are)\ 90^@ text{)}`

`m_(OQ) = t\ \ \ \ text{(see part (ii))}`

`text(Find)\ m_(QT),\ \ text(where:)`

`Q((2at)/(1 + t^2), (2at^2)/(1 + t^2)),\ \ \ \ \ T (0,2a)`

`m_(QT)`	`= (y_2\ – y_1)/(x_2\ – x_1)`
	`= ((2at^2)/(1 + t^2)\ – 2a)/((2at)/(1 + t^2)\ – 0)`
	`= (2at^2\ – 2a (1 + t^2))/(2at)`
	`= – (2a)/(2at)`
	`= – 1/t`

`m_(QT) xx m_(OT) = -1/t xx t = -1`

`=> QT _|_ OQ`

`=>O,\ T,\ Q\ text(lie on a circle.)`

`:.\ text(Locus of)\ Q\ text(is a fixed circle,)`

`text(centre)\ (0, a),\ text(radius)\ a`

Binomial, EXT1 2011 HSC 7b

The binomial theorem states that

`(1 + x)^n = sum_(r = 0)^(n) ((n),(r)) x^r` for all integers `n >= 1`.

Show that
1. `sum_(r=1)^n ((n),(r)) rx^r = nx (1 + x)^(n\ - 1)`. (2 marks)
By differentiating the result from part (i), or otherwise, show that
1. `sum_(r = 1)^n ((n),(r)) r^2 = n (n + 1) 2^(n\ - 2)`. (2 marks)
Assume now that `n` is even. Show that, for `n >= 4`,
1. `((n),(2)) 2^2 + ((n),(4)) 4^2 + ((n),(6)) 6^2 + ... + ((n),(n)) n^2 = n(n+1)2^(n\ - 3)`.
2. (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

♦ Mean mark 36%.
MARKER’S COMMENT: Better responses provided a narrative through the solution such as “Differentiate both sides” and “Mult. by `x`”, as seen in the solutions.

(i) `text(Show)\ sum_(r=1)^n ((n),(r)) rx^r = nx (1 + x)^(n\ – 1)`

`text(Using)\ (1 + x)^n = sum_(r=0)^n ((n),(r)) x^r :`

`(1 + x)^n = ((n),(0)) + ((n),(1)) x + ((n),(2)) x^2 + … + ((n),(n)) x^n`

`text(Differentiate both sides)`

`n(1 + x)^(n\ – 1) = ((n),(1)) + ((n),(2)) 2x + … + ((n),(n)) nx^(n\ – 1)`

`text(Multiply both sides by)\ x`

`nx (1 + x)^(n\ – 1)`	`= ((n),(1)) x + ((n),(2)) 2x^2 + … + ((n),(n)) nx^n`
	`= sum_(r=1)^n ((n),(r)) rx^r\ \ text(… as required.)`

(ii) `text(Show)\ \ sum_(r=1)^n ((n),(r)) r^2 = n (n+1) 2^(n\ – 2)`

♦♦ Mean mark 36%.

`text(Using part)\ text{(i)}`

`nx (1 + x)^(n\ – 1) = ((n),(1)) x + ((n),(2)) 2x^2 + … + ((n),(n)) nx^n`

`text(Differentiate LHS)`

`nx (n\ – 1)(1 + x)^(n\ – 2) + n(1 + x)^(n\ – 1)`

`= n (1 + x)^(n\ – 2) [x (n\ – 1) + (1 + x)]`

`= n (1 + x)^(n\ – 2) (xn\ – x + 1 + x)`

`= n (xn + 1)(1 + x)^(n\ – 2)`

`text(Differentiate RHS)`

`((n),(1)) + ((n),(2)) 2^2 x + ((n),(3)) 3^2 x^2 + … + ((n),(n)) n^2 x^(n\ – 1)\ text{… (*)}`

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

`((n),(1)) + ((n),(2)) 2^2 + … + ((n),(n)) n^2`	`= n (n + 1)(1 + 1)^(n\ – 2)`
`sum_(r=1)^n ((n),(r)) r^2`	`= n (n + 1) 2^(n\ – 2)`

(iii) `text(Show for)\ \ n >= 4`

`((n),(2)) 2^2 + ((n),(4)) 4^2 + ((n),(6)) 6^2 + … + ((n),(n)) n^2 = n (n + 1) 2^(n\ – 3)`

♦♦♦ Toughest question in the 2011 exam. Mean mark of 2%. The lowest mean of any Ext1 question since that data became available post-2009!

`text{From part(ii) … (*) we know}`

`((n),(1)) + ((n),(2)) 2^2 x + … + ((n),(n)) n^2 x^(n\ – 1) = n (xn + 1)(1 + x)^(n\ – 2)`

`text(When)\ \ x = 1`

`((n),(1)) + ((n),(2)) 2^2 + … + ((n),(n)) n^2 = n (n + 1) 2^(n\ – 2)\ \ \ \ \ …\ (1)`

`text(When)\ x = –1`

`((n),(1))\ – ((n),(2)) 2^2 + ((n),(3)) 3^2\ – …\ – ((n),(n)) n^2`

`= n (-n + 1)* 0^(n\ – 2)=0\ \ \ \ \ …\ (2)`

`text{(Note that}\ n≠2\ \ text{because}\ 0^0\ text{is undefined)}`

`(1)\ – (2)`

`2[((n),(2)) 2^2 + ((n),(4)) 4^2 + … + ((n),(n)) n^2]`	`= n (n + 1) 2^(n\ – 2) – 0`
`((n),(2)) 2^2 + ((n),(4)) 4^2 + … + ((n),(n)) n^2`	`= 1/2 n (n + 1) 2^(n\ – 2)`
	`= n (n + 1) 2^(n\ – 3)`

`:.\ text(Above statement true for integers)\ n >= 4.`

Polynomials, EXT1 2013 HSC 14c

The equation `e^t = 1/t` has an approximate solution `t_0 = 0.5`

Use one application of Newton’s method to show that `t_1 = 0.56` is another approximate solution of `e^t = 1/t`. (2 marks)
Hence, or otherwise, find an approximation to the value of `r` for which the graphs `y = e^(rx)` and `y = log_e x` have a common tangent at their point of intersection. (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`0.097\ text{(to 3 d.p.)}`

Show Worked Solution

MARKER’S COMMENT: Many students had difficulty expressing a valid function while too many others carelessly used `t_0=0.56` rather than 0.5.

(i)	`e^t`	`= 1/t`
	`e^t\ – 1/t`	`= 0`

`text(Let)\ \ f(t)`	`= e^t\ – 1/t`
`f prime (t)`	`= e^t + 1/(t^2)`
`f(0.5)`	`=e^0.5-1/0.5`
	`=–0.3512…`
`f′(0.5)`	`=e^0.5 +1/0.5^2`
	`=5.6487…`

`text(Applying Newton where)\ \ t_0 = 0.5`

`t_1`	`= 0.5\ – (f(0.5))/(f prime (0.5)`
	`= 0.5\ – (–0.3512…)/(5.6487…)`
	`= 0.5\ – (- 0.062)`
	`= 0.56\ text{(2 d.p.)}\ \ text(… as required.)`

(ii)	`y_1`	`= e^(rx),`	`\ \ \ \ \ (dy_1)/(dx) = re^(rx)`
	`y_2`	`= log_e x,`	`\ \ \ \ \ (dy_2)/(dx) = 1/x`

♦♦♦ 2nd toughest question on 2013 paper with mean mark 15%.
MARKER’S COMMENT: Few students used the common tangent information to equate the derivative functions.

`text(S)text(ince tangent is common),\ \ (dy_1)/(dx)=(dy_2)/(dx)`

`re^(rx)`	`= 1/x`
`e^(rx)`	`= 1/(rx)`

`text(Using part)\ text{(i)},\ rx ~~ 0.56`

`text(At intersection of curves)\ \ y_1 = y_2`

`e^(rx)`	`= log_e x`
`e^0.56`	`= log_e x`
`x`	`= e^(e^0.56)`
	`= 5.758 …`

`text(S)text(ince)\ rx`	`~~ 0.56`
`=> r`	`~~ 0.56/(5.758 … )`
	`~~ 0.0972 …`
	`~~ 0.097\ text{(to 3 d.p.)}`

Binomial, EXT1 2013 HSC 14b

Write down the coefficient of `x^(2n)` in the binomial expansion of `(1 + x)^(4n)`. (1 mark)
Show that
1. `(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n\ - k)(x + 2)^(2n\ - k)`. (2 marks)
It is known that
`x^(2n\ - k) (x + 2)^(2n\ - k) = ((2n\ - k),(0)) 2^(2n\ - k) x^(2n\ - k) + ((2n\ - k),(1)) 2^(2n\ - k\ - 1) x^(2n\ - k + 1)`
2. 1. `+ ... + ((2n\ - k),(2n\ - k)) 2^0 x^(4n\ - 2k)`. (Do NOT prove this.)
Show that
1. `((4n),(2n)) = sum_(k = 0)^(n) 2^(2n\ - 2k) ((2n),(k))((2n\ - k),(k))`. (3 marks)

Show Answers Only

`((4n),(2n))`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `text(Find co-efficient of)\ \ x^(2n)`

`text(Expanding)\ \ (1+x)^(4n)`

`((4n),(0)) + ((4n),(1))x + ((4n),(2))x^2 + … + ((4n),(2n))x^(2n) + …`

`:.\ text(Co-efficient of)\ \ x^(2n)\ text(is)\ ((4n),(2n))`

(ii) `text(Show)\ (1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)`

♦♦ Mean mark 30%.

`text(Using)\ (1 + x^2 + 2x)^(2n) = [x(x + 2) + 1]^(2n)`

`[x (x + 2) + 1]^(2n)`

`= ((2n),(0)) (x(x + 2))^(2n) + ((2n),(1)) (x(x + 2))^(2n\ – 1) + … + ((2n),(2n))`

`= ((2n),(0)) x^(2n)(x + 2)^(2n) + ((2n),(1)) x^(2n\ – 1) (x + 2)^(2n\ – 1) + … + ((2n),(2n))`

`= sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)\ text(… as required.)`

(iii) `((4n),(2n))\ text(is the co-eff of)\ x^(2n)\ text(in expansion)\ (1+x)^(4x)`

`text(S)text(ince)\ (1 + x^2 + 2x)^(2n) = ((x+1)^2)^(2n) = (1 + x)^(4n)`

`=> ((4n),(2n))\ text(is co-efficient of)\ x^(2n)\ text(in expansion)\ (1 + x^2 + 2x)^(2n)`

`text(Using part)\ text{(ii)}`

`(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k))\ x^(2n\ – k) (x + 2)^(2n\ – k)`

`text(Using the given identity,)\ x^(2n)\ text(co-efficients are)`

♦♦♦ Toughest question in the 2013 exam with Mean mark 12%!
MARKER’S COMMENT: Simply stating `(1+x^2+2x)^(2n)“=(1+x)^(4n)` received 1 full mark, showing that even initial working can gain marks.

`k = 0,`	`\ ((2n),(0))((2n\ – 0),(0)) 2^(2n\ – 0)`
`k = 1,`	`\ ((2n),(1))((2n\ – 1),(1)) 2^(2n\ – 1\ – 1)`
`vdots`
`k = n,`	`\ ((2n),(n))((2n\ – n),(n)) 2^(2n\ – n\ – n)`

`:.\ ((4n),(2n))`

`= ((2n),(0))((2n),(0))2^(2n) + ((2n),(1))((2n\ – 1),(1))2^(2n\ – 2) + … + ((2n),(n))((n),(n)) 2^0`

` = sum_(k=0)^(n)\ ((2n),(k))((2n\ – k),(k)) 2^(2n\ – 2k)\ \ \ \ text(… as required)`

Combinatorics, EXT1 A1 2010 HSC 7c

A box contains `n` identical red balls and `n` identical blue balls. A selection of `r` balls is made from the box, where `0 <= r <= n`.

Explain why the number of possible colour combinations is `r + 1`. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Another box contains `n` white balls labelled consecutively from `1` to `n`. A selection of `n − r` balls is made from the box, where `0 <= r <= n`.

Explain why the number of different selections is `((n),(r))`. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
The `n` red balls, the `n` blue balls and the `n` white labelled balls are all placed into one box, and a selection of `n` balls is made.

Using the identity, `n2^(n-1)=sum_(k=1)^n k ((n),(k)),` or otherwise, show that the number of different selections is `(n + 2)2^(n- 1)`. (3 marks)

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Show Answers Only

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

Show Worked Solution

♦♦♦ Mean mark 4%.
COMMENT: Solving this part required high level logical reasoning which proved extremely challenging for the vast majority of candidates.

i.	`text(1 Ball, # Combinations)`	`= 2\ text{(R or B)}`
	`text(2 Balls, # Combinations)`	`= 3\ text{(BB, BR, RR)}`
	`text(3 Balls, # Combinations)`	`= 4`

`text{(BBB, RBB, RRB, RRR)}`

`text(i.e. # Red balls could be 0, 1, 2, 3, …)`

`:.\ text(If)\ r\ text(balls chosen, # Combinations) = r + 1`

ii.

`text(# Selections when choosing)\ (n\ – r)\ text(from)\ n`

`= ((n),(n\ – r))`

♦♦♦ Mean mark 18%.

`((n),(n\ – r))`	`= (n!)/((n\ – r)!(n\ – (n\ – r))!)`
	`= (n!)/((n\ – r)! r!`
	`= ((n),(r))`

♦♦♦ Mean mark part (iii) 2%. Beast alert – equal lowest mean mark of any part of any question since this data has been available post-2009.

iii. `text(If)\ n\ text(balls are chosen,)`

`text(Let)\ r\ text(balls be red and blue and)`

`(n\ – r)\ text(balls be white labelled.)`

`=> text{# Combinations (Red and blue)} = r + 1`

`=> text{# Combinations (White)} = ((n),(r))`

`text(Any selection of)\ \ r\ \ text(red and blue balls would)`

`text(result in)\ \ (n-r)\ \ text(white balls, with) \ r=0,1,2,…`

`:.\ text{# Selections (for any given}\ r\ text{)}`

`= (r + 1)((n),(r))`

`= r ((n),(r)) + ((n),(r))`

`:.\ text{Total Selections}`	`= sum_(r=1)^n r ((n),(r)) + sum_(r=0)^n ((n),(r))`
	`= n2^(n\ – 1) + 2^n`
	`= n2^(n\ – 1) + 2*2^(n\ – 1)`
	`= (n + 2)2^(n\ – 1)\ text(… as required)`

Plane Geometry, EXT1 2010 HSC 5c

In the diagram, `ST` is tangent to both the circles at `A`.

The points `B` and `C` are on the larger circle, and the line `BC` is tangent to the smaller circle at `D`. The line `AB` intersects the smaller circle at `X`.

Copy or trace the diagram into your writing booklet

Explain why `/_AXD = /_ABD + /_XDB.` (1 mark)
Explain why `/_AXD = /_TAC + /_CAD.` (1 mark)
Hence show that `AD` bisects `/_BAC`. (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)

`/_ AXD = /_ABD + /_XDB`

`text{(exterior angle of}\ Delta BXD text{)}`

♦ Mean mark 47%

(ii)	`/_AXD`	`= /_TAD \ \ text{(angle in alternate segment)}`
		`= /_TAC + /_CAD\ \ text(… as required)`

(iii) `text(Show)\ \ /_XAD = /_CAD`

`/_ABD + /_XDB`

`=/_TAC + /_CAD\ \ text{(} text(from part)\ text{(i)} text(,)\ text{(ii)} text{)}`

`text(S)text(ince)\ /_TAC= /_ABD\ text{(angle in alternate segment)}`

♦♦ Mean mark 22%
IMPORTANT: Recognising that angles in the alternate segment are equal is an examiner favourite. LOOK FOR IT!

`=>/_XDB`	`=/_CAD`
`/_XDB`	`= /_XAD\ text{(angle in alternate segment)}`
`:. /_XAD`	`= /_CAD`

`:.\ AD\ \ text(bisects)\ /_BAC.`

Calculus, EXT1 C1 2011 HSC 7a

The diagram shows two identical circular cones with a common vertical axis. Each cone has height `h` cm and semi-vertical angle 45°.

The lower cone is completely filled with water. The upper cone is lowered vertically into the water as shown in the diagram. The rate at which it is lowered is given by

`(dl)/(dt) = 10`,

where `l` cm is the distance the upper cone has descended into the water after `t` seconds.

As the upper cone is lowered, water spills from the lower cone. The volume of water remaining in the lower cone at time `t` is `V` cm³.

Show that `V = pi/3(h^3\ - l^3)`. (1 mark)

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Find the rate at which `V` is changing with respect to time when `l = 2`. (2 marks)

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Find the rate at which `V` is changing with respect to time when the lower cone has lost `1/8` of its water. Give your answer in terms of `h`. (2 marks)

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Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`-40 pi\ \ text(cm³)// text(sec)`
`(-5pih^2)/2\ text(cm³)// text(sec)`

Show Worked Solution

i. `text(Show that)\ V = pi/3 (h^3\ – l^3)`

♦ Mean mark 42%

`text(S)text(ince)\ \ tan45° = r/h=1`

`=>r=h`

`=>\ text(Radius of lower cone) = h`

`:.\ V text{(lower cone)}`	`= 1/3 pi r^2 h`
	`= 1/3 pi h^3`

`text(Similarly,)`

`V text{(submerged upper cone)} = 1/3 pi l^3`

`V text{(water left)}`	`= 1/3 pi h^3\ – 1/3 pi l^3`
	`= pi/3 (h^3\ – l^3)\ \ \ text(… as required)`

ii. `text(Find)\ (dV)/(dt)\ text(at)\ l = 2`

`(dV)/(dt)= (dV)/(dl) xx (dl)/(dt)\ …\ text{(1)}`

`=>(dl)/(dt)`

`= 10\ text{(given)}`

`text(Using)\ \ V`	`= pi/3 (h^3\ – l^3)\ \ \ \ text(from part)\ text{(i)}`
`=>(dV)/(dl)`	`= -3 xx pi/3 l^2`
	`= -pi l^2`

`text(At)\ \ l = 2,`

`text{Substitute into (1) above}`

`(dV)/(dt)`	`= -pi xx 2^2 xx 10`
	`= -40 pi\ \ \ text(cm³)//text (sec)`

iii. `text(Find)\ \ (dV)/(dt)\ \ text(when lower cone has lost)\ 1/8 :`

♦♦♦ Mean mark 12%
MARKER’S COMMENT: Many unsuccessful answers attempted to find an alternate version of `(dV)/(dt)`. Part (ii) directed students directly toward the correct strategy.

`text(Find)\ \ l\ \ text(when)\ \ V = 7/8 xx 1/3 pi h^3`

`pi/3 (h^3\ – l^3)`	`= 7/8 xx 1/3 pi h^3`
`h^3 -l^3`	`= 7/8 h^3`
`l^3`	`= 1/8 h^3`
`l`	`= h/2`

`text(When)\ \ l = h/2 ,`

`(dV)/(dt)`	`= -pi (h/2)^2 xx 10\ \ …\ text{(*)}`
	`= (-5pih^2)/2\ text(cm³)// text(sec)`

`:. V\ text(is decreasing at the rate of)\ \ (5 pi h^2)/2\ text(cm³)//text(sec).`

Statistics, EXT1 S1 2011 HSC 6c

A game is played by throwing darts at a target. A player can choose to throw two or three darts.

Darcy plays two games. In Game 1, he chooses to throw two darts, and wins if he hits the target at least once. In Game 2, he chooses to throw three darts, and wins if he hits the target at least twice.

The probability that Darcy hits the target on any throw is `p`, where `0 < p < 1`.

Show that the probability that Darcy wins Game 1 is `2p- p^2`. (1 mark)

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Show that the probability that Darcy wins Game 2 is `3p^2- 2p^3`. (1 mark)

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Prove that Darcy is more likely to win Game 1 than Game 2. (2 marks)

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Find the value of `p` for which Darcy is twice as likely to win Game 1 as he is to win Game 2. (2 marks)

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Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`p = (7 – sqrt17)/8`

Show Worked Solution

i. `text(Show)\ P text{(wins Game 1)} = 2p\ – p^2`

♦ Mean mark 47%.

`P text{(Hits)} = P text{(H)} = p`

`P text{(Miss)} = P text{(M)} = 1 – p`

`P text{(Wins G1)}`	`= 1 – P text{(MM)}`
	`= 1 – (1 – p)^2`
	`= 1 – 1 + 2p – p^2`
	`= 2p – p^2\ \ \ text(… as required)`

ii. `text(Show)\ \ P text{(wins Game 2)} = 3p – 2p^3 :`

♦ Mean mark 40%.

`text(Darcy wins G2 if he hits 3 times, or twice.)`

`Ptext{(Hits 3 times)}=\ ^3C_3 (p)^3=p^3`

`Ptext{(Hits twice)}=\ ^3C_2 (p)^2 (1-p)=3p^2-3p^3`

`P text{(Wins G2)}`	`= p^3 + 3p^2-3p^3`
	`= 3p^2 – 2p^3\ \ \ text(… as required)`

♦♦♦ Mean mark 17%.
COMMENT: It is critical for students to utilise the limits of `0<p<1` to answer part (iii).

iii. `text(Prove more likely to win G1 vs G2)`

`text(i.e. Show)\ \ \ `	`2p – p^2 > 3p^2 – 2p^3`
	`2p^3 – 4p^2 + 2p`	`> 0`
	`2p (p^2 – 2p + 1)`	`> 0`
	`2p (p – 1)^2`	`> 0`

`=> text(TRUE since)\ \ (p-1)^2>0\ \ text(and)\ \ 0 < p < 1`

`:.\ text(More likely for Darcy to win Game 1.)`

iv. `text(If twice as likely to win G1 vs G2)`

♦♦ Mean mark 21%.
MARKER’S COMMENT: BE CAREFUL in formulating your equation here. Many students multiplied the wrong side by 2.

`2p\ – p^2`	`= 2(3p^2 – 2p^3)`
	`= 6p^2 – 4p^3`
`4p^3 – 7p^2 + 2p`	`= 0`
`p(4p^2 – 7p + 2)`	`= 0`

`p`	`= (–(–7) +- sqrt((–7)^2\ – 4 xx 4 xx 2))/(2 xx 4)`
	`= (7 +- sqrt(49\ – 32))/8`
	`= (7 +- sqrt17)/8`

`text(S)text(ince)\ \ 0<p<1,`

`p = (7\ – sqrt17)/8`

Trig Ratios, EXT1 2011 HSC 5a

In the diagram, `Q(x_0, y_0)` is a point on the unit circle `x^2 + y^2 = 1` at an angle `theta` from the positive `x`-axis, where `− pi/2 < theta < pi/2`. The line through `N(0, 1)` and `Q` intersects the line `y = –1` at `P`. The points `T(0, y_0)` and `S(0, –1)` are on the `y`-axis.

Use the fact that `Delta TQN` and `Delta SPN` are similar to show that
`SP = (2costheta)/(1\ - sin theta)`. (2 marks)
Show that `(costheta)/(1\ - sin theta) = sec theta + tan theta`. (1 mark)
Show that `/_ SNP = theta/2 + pi/4`. (1 mark)
Hence, or otherwise, show that `sectheta + tantheta = tan(theta/2 + pi/4)`. (1 mark)
Hence, or otherwise, solve `sec theta + tan theta = sqrt3`, where `-pi/2 < theta < pi/2`. (2 marks)

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)}`
`pi/6\ text(radians)`

Show Worked Solution

(i)

♦ Mean mark 41%
MARKER’S COMMENT: When questions direct you to use a certain fact for a proof, use that fact!

`text(Show)\ SP = (2 cos theta)/(1\ – sin theta)`

`Delta SPN \ text(|||) \ Delta TQN\ \ \ text{(given)}`

`(SP)/(SN) = (TQ)/(TN)\ \ \ text{(corresponding sides of similar triangles)}`

`/_TQO = theta\ \ \ text{(alternate,}\ TQ\ text(||)\ x text{-axis)}`

	`sin theta`	`= (OT)/1 => OT = sin theta`
`=>`	`TN`	`= 1\ – sin theta`

	`cos theta`	`= (TQ)/1`
`=>`	`TQ`	`= cos theta`

`SN`	`= 2\ \ \ text{(diameter of unit circle)}`
`:. (SP)/2`	`= cos theta/(1\ – sin theta)`
`SP`	`= (2 cos theta)/(1\ – sin theta)\ \ \ text(… as required)`

(ii) `text(Show)\ \ costheta/(1\ – sin theta) = sec theta + tan theta`

♦ Mean mark 35%

`text(RHS)`	`= 1/(costheta) + (sintheta)/(costheta)`
	`=(1 + sin theta)/cos theta xx cos theta/cos theta`
	`= (costheta(1 + sintheta))/(cos^2theta)`
	`= (costheta(1 + sin theta))/(1\ – sin^2 theta)`
	`= (costheta(1 + sin theta))/((1 + sin theta)(1\ – sin theta)`
	`= costheta/(1\ – sin theta)\ \ \ text(… as required)`

(iii) `text(Show)\ \ /_SNP = theta/2 + pi/4`

♦♦♦ Mean mark 11%

`/_TOQ = 90\ – theta`

`text(S)text(ince)\ ON = OQ = 1\ text{(unit circle)}`

`=> Delta ONQ\ text(is isosceles)`

`:.\ /_SNP`	`= 1/2 (180\ – (90\ – theta))\ \ \ \ text{(angle sum of}\ Delta ONQ text{)}`
	`= 90\ – 45 + theta/2`
	`= 45 + theta/2`
	`= pi/4 + theta/2\ \ \ text(… as required)`

(iv) `text(Show)\ sec theta + tan theta = tan (theta/2 + pi/4)`

♦♦ Mean mark 28%

`text(S)text(ince)\ /_SNP = pi/4 + theta/2\ \ \ \ text{(} text(part)\ text{(iii)} text{)}`

`=> tan\ /_SNP = tan (pi/4 + theta/2)`

`text(Also,)\ tan\ /_SNP`	`= (SP)/(SN)`
	`= ((2costheta)/(1\ – sin theta))/2`
	`= (cos theta)/(1\ – sin theta)`
	`= sec theta + tan theta\ \ \ \ text{(part (ii))}`

`:.\ sec theta + tan theta = tan (pi/4 + theta/2)\ \ \ text(… as required)`

♦ Mean mark 45%

(v)	`sec theta + tan theta`	`= sqrt3,\ \ \ (-pi/2 < theta < pi/2)`
	`tan (pi/4 + theta/2)`	`= sqrt3\ \ \ \ text{(part (iv))}`
	`tan (pi/3)`	`= sqrt 3`

`text(S)text(ince tan is positive in)\ 1^text(st) // 3^text(rd)\ text(quads)`

`theta/2 + pi/4`	`= pi/3,\ (4pi)/3`
`theta/2`	`= pi/12\ \ \ \ (-pi/2 < theta < pi/2)`
`:.\ theta`	`= pi/6\ text(radians)`

Plane Geometry, EXT1 2011 HSC 4b

In the diagram, the vertices of `Delta ABC` lie on the circle with centre `O`. The point `D` lies on `BC` such that `Delta ABD` is isosceles and `/_ABC = x`.

Copy or trace the diagram into your writing booklet.

Explain why `/_AOC = 2x`. (1 mark)
Prove that `ACDO` is a cyclic quadrilateral. (2 marks)
Let `M` be the midpoint of `AC` and `P` the centre of the circle through `A, C, D` and `O`.
Show that `P, M` and `O` are collinear. (1 mark)

Show Answers Only

`text(Angle at the centre of a circle is twice)`
`text(angle of circumference on same arc)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)

`/_AOC = 2x`

`text{(angles at circumference and}`

`text{centre on arc}\ AC text{)}`

♦ Mean mark 38%
STRATEGY: Proving part (ii) by showing opposite angles are supplementary also worked but was more time consuming.

(ii) `text(Prove)\ ACDO\ text(is cyclic)`

`text(S)text(ince)\ Delta ADB\ text(is isosceles)`

`/_DAB = x\ \ \ text{(opposite equal sides in}\ Delta DBA text{)}`

`=> /_ADB`	`= 180\ – 2x\ \ \ text{(angle sum of}\ Delta DAB text{)}`
`=> /_CDA`	`= 2x\ \ \ text{(}CDB\ text{is a straight angle)}`

`text(S)text(ince chord)\ AC\ text(subtends)\ /_CDA = 2x`

`text(and)\ /_COA = 2x,`

`:.\ text(Quadrilateral)\ ACDO\ text(must be cyclic.)`

(iii)

`text(Need to show)\ OM\ text(passes through)\ P,\ text(centre)`

♦♦♦ Mean mark 7%. 2nd hardest question in the 2011 paper!

`text(of circle through)\ ACDO.`

`AM`	`= CM\ text{(} M\ text(is midpoint) text{)}`
`OC`	`= OA\ text{(radii)}`

`OM\ text(is common)`

`:.\ Delta OAM -= Delta OCM\ \ \ text{(SSS)}`

`:. /_CMO = /_AMO\ \ \ `	`text{(corresponding angles of}`
	`\ \ text{congruent triangles)}`

`text(S)text(ince)\ ∠AMC\ text(is straight angle)`

`/_CMO = /_AMO = 90°`

`:.OM\ text(is perpendicular bisector)`

`text(of chord)\ AC.`

`:. OM\ text(passes through)\ P.`

`:.\ P, M,\ text(and)\ O\ text(are collinear.)`

Mechanics, EXT2* M1 2011 HSC 3a

The equation of motion for a particle undergoing simple harmonic motion is

`(d^2x)/(dt^2) = -n^2 x`,

where `x` is the displacement of the particle from the origin at time `t`, and `n` is a positive constant.

Verify that `x = A cos nt + B sin nt`, where `A` and `B` are constants, is a solution of the equation of motion. (1 mark)

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The particle is initially at the origin and moving with velocity `2n`.

Find the values of `A` and `B` in the solution `x = A cos nt + B sin nt`. (2 marks)

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When is the particle first at its greatest distance from the origin? (1 mark)

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What is the total distance the particle travels between `t = 0` and `t = (2pi)/n`? (1 mark)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`A = 0,\ B = 2`
`t = pi/(2n)`
`text(8 units)`

Show Worked Solution

i.	`x`	`= A cos nt + B sin nt`
	`(dx)/(dt)`	`= – An sin nt + Bn cos nt`
	`(d^2x)/(dt^2)`	`= – An^2 cos nt\ – Bn^2 sin nt`
		`= -n^2 (A cos nt + B sin nt)`
		`= -n^2 x\ \ \ text(… as required)`

ii. `text(At)\ \ t=0, \ x=0, \ v=2n:`

`x`	`= Acosnt + Bsinnt`
`0`	`= A cos 0 + B sin 0`
`:.A`	`= 0`

`text(Using)\ \ (dx)/(dt) = Bn cos nt`

`2n`	`= Bn cos 0`
`Bn`	`= 2n`
`:.B`	`= 2`

♦♦ Mean mark part (iii) 47%

iii. `text(Max distance from origin when)\ (dx)/(dt) = 0`

`(dx)/(dt)`	`= 2n cos nt`
`0`	`= 2n cos nt`
`cos nt`	`= 0`
`nt`	`= pi/2,\ (3pi)/2,\ (5pi)/2`
`t`	`= pi/(2n),\ (3pi)/(2n), …`

`:.\ text(Particle is first at greatest distance)`

`text(from)\ O\ text(when)\ t = pi/(2n).`

iv. `text(Solution 1)`

`text(Find the distance travelled from)\ \ t=0\ \→\ \ t=(2pi)/n`

`text{(i.e. 1 full period)}`

♦♦ Mean mark 22%
MARKER’S COMMENT: Many students found the displacement at `t` rather than the distance travelled while another common error found distance as 2 x amplitude.

`text(S)text(ince)\ \ x=2 sin (nt)`

`=> text(Amplitude)=2`

`:.\ text(Distance travelled)=4 xx2=8\ text(units)`

`text(Solution 2)`

`text(At)\ t = 0,\ x = 0`

`text(At)\ t`	`= pi/(2n)`
`x`	`= 2 sin (n xx pi/(2n)) = 2`

`text(At)\ t`	`= (3pi)/(2n)\ \ \ text{(i.e. the next time}\ \ (dx)/(dt) = 0 text{)}`
`x`	`= 2 sin (n xx (3pi)/(2n)) = -2`

`text(At)\ t`	`= (2pi)/n`
`x`	`= 2 sin (n xx (2pi)/n) = 0`

`:.\ text(Total distance travelled)`

`= 2 + 4 + 2`

`= 8\ \ text(units)`

Calculus, EXT1 C1 2012 HSC 14c

A plane `P` takes off from a point `B`. It flies due north at a constant angle `alpha` to the horizontal. An observer is located at `A`, 1 km from `B`, at a bearing 060° from `B`.

Let `u` km be the distance from `B` to the plane and let `r` km be the distance from the observer to the plane. The point `G` is on the ground directly below the plane.

Show that `r = sqrt(1 + u^2 - u cos alpha)`. (3 marks)

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The plane is travelling at a constant speed of 360 km/h.
At what rate, in terms of `alpha`, is the distance of the plane from the observer changing 5 minutes after take-off? (2 marks)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`(180 (60\ – cos alpha))/sqrt(901\ – 30 cos alpha)\ \ text(km/hr)`

Show Worked Solution

i. `text(Show)\ r = sqrt(1 + u^2 – u cos alpha)`

♦ Mean mark 42%
IMPORTANT: Students should always be looking for opportunities to use the identity `sin^2 alpha“+cos^2 alpha=1` to clean up calculations with trig functions.

`text(In)\ Delta PGB:`

`cos alpha`	`= (BG)/u`
`BG`	`= u cos alpha`
`sin alpha`	`= (PG)/(u)`
`PG`	`= u sin alpha`

`text(In)\ Delta PGA,\ text(using Pythagoras):`

`AG^2`	`= r^2\ – PG^2`
	`= r^2\ – u^2 sin^2 alpha`

`text(Using cosine rule in)\ Delta ABG:`

`AG^2`	`= BG^2 + AB^2\ – 2 xx BG xx AB xx cos 60^@`
`r^2\ – u^2 sin^2 alpha`	`= u^2 cos^2 alpha + 1\ – 2 (u cos alpha) xx 1 xx 1/2`
`r^2`	`= u^2 cos^2 alpha + u^2 sin^2 alpha + 1\ – u cos alpha`
	`= u^2 (cos^2 alpha + sin^2 alpha) + 1\ – u cos alpha`
	`= u^2 + 1\ – u cos alpha`
`r`	`= sqrt(u^2 + 1\ – u cos alpha)\ \ text(… as required)`

ii. `text(Find)\ \ (dr)/(dt)\ text(when)\ t =5`

♦♦♦ Mean mark 14%

`(dr)/(dt) = (dr)/(du) xx (du)/(dt)`

`r`	`= (u^2 + 1\ – u cos alpha)^(1/2)`
`(dr)/(du)`	`= 1/2 (u^2 + 1\ – u cos alpha)^(-1/2) xx (2u\ – cos alpha)`
	`= (2u\ – cos alpha)/(2 sqrt(u^2 + 1\ – u cos alpha))`

`(du)/(dt)= 360\ text{km/hr (plane’s speed)}`

`text(After 5 mins,)`

`u`	`= 5/60 xx 360 = 30\ text(km)`
`(dr)/(dt)`	`= ((2 xx 30)\ – cos alpha)/( 2 sqrt(30^2 + 1\ – 30 cos alpha)) xx 360`
	`= (180 (60\ – cos alpha))/sqrt(901\ – 30 cos alpha)\ \ text(km/hr)`

Combinatorics, EXT1 A1 2012 HSC 11f

Use the binomial theorem to find an expression for the constant term in the expansion of

`(2x^3 - 1/x)^12`. (2 marks)

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For what values of `n` does `(2x^3 - 1/x)^n` have a non-zero constant term? (1 mark)

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Show Answers Only

`-1760`
`n\ text(must be a multiple of 4)`

Show Worked Solution

i. `text(General term)`

`\ ^12C_k * (2x^3)^(12\ – k) (-1/x)^k`

`=\ ^12C_k * (–1)^k *2^(12\ – k) * x^(36\ – 3k) * x^(-k)`

`=\ ^12C_k * (–1)^k*2^(12\ – k) * x^(36\ – 4k)`

`text(Constant term occurs when)`

`36\ – 4k`	`= 0`
`k`	`= 9`

`:.\ text(Constant term)`	`=\ ^12C_9 * (–1)^9*2^3`
	`= – (12!)/(3!9!) xx 8`
	`= – 1760`

ii. `text(General term of)\ (2x^3\ – 1/x)^n`

♦♦♦ Mean mark 16%.

`\ ^nC_k * (2x^3)^(n\ – k) (–1/x)^k`

`=\ ^nC_k * 2^(n\ – k) * x^(3n\ – 3k) * (–1)^k * x^(-k)`

`=\ ^nC_k * (–1)^k*2^(n\ -k) * x^(3n\ – 4k)`

`text(Constant term when)\ \ 3n\ – 4k = 0.`

`text(i.e.)\ \ k=3/4n`

`text(S)text(ince)\ n\ text(and)\ k\ text(must be integers,)\ \ n\ \ text(must)`

`text(be a multiple of 4.)`

`I`	`= (PrT)/100`
`584`	`=(P xx 6.2 xx 1/12)/100`
`:. P`	`=(584 xx 100)/(6.2 xx 1/12)`
	`=$113\ 032.25…`

`x xx 1.03 xx 1.03`	`= 46\ 500`
`x (1.03)^2`	`= 46\ 500`
`:. x`	`= (46\ 500) / 1.03^2`
	`= 43\ 830.70…`

`R-C`	`=175`
`150S-350`	`= 175`
`150S`	`=525`
`:. S`	`= $3.50`

`x^2 + 40^2`	`= 50^2`
`x^2`	`= 2500 − 1600`
	`= 900`
`x`	`= 30\ text(cm)`

`d + x`	`= 50\ text{(radius)}`
`:.\ d`	`= 20\ text(cm)`

`text{S.A. (base)}`	`= pir^2`
	`= pi xx 20^2`
	`= 1256.63…`

`text{S.A. (side)}`	`= 2 pi r h`
	`= 2 xx pi xx 20 xx 30`
	`= 3769.91…`

`:.\ text(Total S.A.)`	`= 1256.63… + 3769.91… + 2513.27…`
	`= 7539.81…\ text(cm²)`

`/_ PQR`	` = 180 – 74\ \ \ text{(co-interior,}\ QR text(\|\|) PS text{)}`
	` = 106^@`

`PR^2`	` = 18^2 + 25^2 – 2 xx 18 xx 25 xx cos 106^@`
	` = 324 + 625 + 248.07…`
	` = 1197.07…`
`:. PR`	` = 34.59…\ \ text(cm)`