Band 4 – Page 110

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)

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`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.)`

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)

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\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 Resources, 2UG 2014 HSC 28d

An aerial diagram of a swimming pool is shown.

The swimming pool is a standard length of 50 metres but is not in the shape of a rectangle.

(i) Given `AB=8\ text(cm)`, determine the scale of the diagram such that

1 cm = `x` m (1 mark)

(ii) If the length of a carpark next to the pool measured 5 cm (not shown), how long would it be in real life? (1 mark)

(iii) In the diagram of the swimming pool, the five widths are measured to be:

`CD = 21.88\ text(m)`

`EF = 25.63\ text(m)`

`GH = 31.88\ text(m)`

`IJ = 36.25\ text(m)`

`KL = 21.88\ text(m)`

The average depth of the pool is 1.2 m

Calculate the approximate volume of the swimming pool, in cubic metres. In your calculations, use TWO applications of Simpson’s Rule. (3 marks)

Show Answers Only

(i) `x=6.25\ text(m)`

(ii) `31.25\ text(m)`

(iii) `1775\ text(m³)`

Show Worked Solution

(i)	`\ \ \ \ 8\ text(cm)`	`=50\ text(m)`
	`1\ text(cm)`	`=50/8`
		`=6.25\ text(m)`

`:.x=6.25\ text(m)`

(ii) `text{Using scale from (i)}`

`5\ text(cm)`	`=5 xx 6.25`
	`=31.25\ text(m)`

`:.\ text(The carpark would be 31.25 m long)`

(iii)

`h = 50/4 = 12.5\ text(m)`

♦ Mean mark 50%. Be careful not to give away easy marks!

`A`	`~~ h/3 [y_0 + 4(y_1) + y_2]\ \ text(… applied twice)`
	`~~ 12.5/3 [21.88 + 4(25.63) + 31.88]`
	`+ 12.5/3 [31.88 + 4(36.25) + 21.88]`
	`~~ 12.5/3 [156.28] + 12.5/3 [198.76]`
	`~~ 1,479.33\ text(m²)`

`V`	`= Ah`
	`~~ 1479.33 xx 1.2`
	`~~ 1775.2`
	`~~ 1775\ text(m³)`

Probability, STD2 S2 2014 HSC 28c

A fair coin is tossed three times. Using a tree diagram, or otherwise, calculate the probability of obtaining two heads and a tail in any order. (2 marks)

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`3/8`

Show Worked Solution

`P text{(2 heads, 1 tail)}`

`= P(HHT) + P(HTH) + P(THH)`

`= (1/2 xx 1/2 xx 1/2) + (1/2 xx 1/2 xx 1/2) + (1/2 xx 1/2 xx 1/2)`

`= 1/8 + 1/8 + 1/8`

`= 3/8`

Measurement, STD2 M6 2014 HSC 28b

A radial compass survey of a sports centre is shown in the diagram.

Show that the size of angle `AOB` is 114°. (1 mark)

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Calculate the length of the boundary `AB`, to the nearest metre. (2 marks)

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Find the area of triangle `AOB` in hectares, correct to two significant figures. (3 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`420\ text(m)\ text{(nearest m)}`
`2.8\ text(ha)\ text{(2 sig. figures)}`

Show Worked Solution

`text(Let)\ D\ text(be directly north of)\ O`

`/_AOD = 360\ – 320 = 40^@`

`:.\ /_AOB = 40 + 74 = 114^@\ \ \ text(… as required)`

ii. `text(Using cosine rule)`

`AB^2`	`= AO^2 + BO^2\ – 2 xx AO xx BO xx cos /_AOB`
	`= 287^2 + 211^2\ – 2 xx 287 xx 211 xx cos 114^@`
	`= 126\ 890\ – 121\ 114 (-0.4067…)`
	`= 176\ 151.50…`
`AB`	`= 419.704…`
	`= 420\ text(m)\ text{(nearest m)}`

♦ Mean marks of 42% and 41% for parts (ii) and (iii) respectively.

iii. `text(Using)\ A = 1/2 ab sin C,`

`text(Area)\ Delta AOB`	`= 1/2 xx 287 xx 211 xx sin 114^@`
	`= 27\ 660.786…\ text(m²)`
	`= 2.7660…\ text(ha)\ \ \ \ text{(1 ha = 10 000 m²)}`
	`= 2.8\ text(ha)\ text{(2 sig figures)}`

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.)`

Algebra, STD2 A2 2014 HSC 26f

The weight of an object on the moon varies directly with its weight on Earth. An astronaut who weighs 84 kg on Earth weighs only 14 kg on the moon.

A lunar landing craft weighs 2449 kg when on the moon. Calculate the weight of this landing craft when on Earth. (2 marks)

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`14\ 694\ text(kg)`

Show Worked Solution

`W_text(moon) prop W_text(earth)`

`=> W_text(m) = k xx W_text(e)`

`text(Find)\ k\ text{given}\ W_text(e) = 84\ text{when}\ W_text(m) = 14`

`14`	`= k xx 84`
`k`	`= 14/84 = 1/6`

`text(If)\ W_text(m) = 2449\ text(kg),\ text(find)\ W_text(e):`

`2449`	`= 1/6 xx W_text(e)`
`W_text(e)`	`= 14\ 694\ text(kg)`

`:.\ text(Landing craft weighs)\ 14\ 694\ text(kg on earth)`

Algebra, STD2 A1 2014 HSC 26c

Solve the equation `(5x + 1)/3-4 = 5-7x`. (3 marks)

Show Answers Only

`x = 1`

Show Worked Solution

`(5x + 1)/3-4`	`= 5-7x`
`5x + 1-3(4)`	`= 3(5-7x)`
`5x + 1-12`	`= 15-21x`
`26x`	`= 26`
`:. x`	`= 1`

Measurement, STD2 M6 2014 HSC 26b

Calculate the value of `h` correct to two decimal places. (2 marks)

Show Answers Only

`10.65\ text{(2 d.p.)}`

Show Worked Solution

`sin 28^@`	`= 5/h`
`:. h`	`= 5/(sin 28^@)`
	`= 10.6502…`
	`= 10.65\ \ text{(2 d.p.)}`

Algebra, 2UG 2014 HSC 26a

Expand `4x(7x^4 - x^2)`. (1 mark)

Show Answers Only

`28x^5 – 4x^3`

Show Worked Solution

`4x(7x^4 – x^2)`

`= 28x^5 – 4x^3`

Algebra, STD2 A2 2014 HSC 22 MC

Heather’s car uses fuel at the rate of 6.6 L per 100 km for long-distance driving and 8.9 L per 100 km for short-distance driving.

She used the car to make a journey of 560 km, which included 65 km of short-distance driving.

Approximately how much fuel did Heather’s car use on the journey?

37 L
38 L
48 L
50 L

Show Answers Only

`B`

Show Worked Solution

`text(Fuel used in short distance)`

`= 65/100 xx 8.9\ text(L) = 5.785\ text(L)`

`text(Fuel used in long distance)`

`= 495/100 xx 6.6\ text(L) = 32.67\ text(L)`

`:.\ text(Total Fuel)`	`= 5.785 + 32.67`
	`= 38.455\ text(L)`

`=> B`

Probability, STD2 S2 2014 HSC 19 MC

Jaz has 2 bags of apples.

Bag A contains 4 red apples and 3 green apples.

Bag B contains 3 red apples and 1 green apple.

Jaz chooses an apple from one of the bags.

Which tree diagram could be used to determine the probability that Jaz chooses a red apple?

Show Answers Only

`A`

Show Worked Solution

`text(The tree diagram needs to identify 2 separate events.)`

`text(1st event – which bag is chosen)`

`text(2nd event – choosing a red apple from a particular bag)`

`=> A`

Measurement, STD2 M7 2014 HSC 17 MC

A child who weighs 14 kg needs to be given 15 mg of paracetamol for every 2 kg of body weight.

Every 10 mL of a particular medicine contains 120 mg of paracetamol.

What is the correct dosage of this medicine for the child?

5.6 mL
8.75 mL
11.43 mL
17.5 mL

Show Answers Only

`B`

Show Worked Solution

`text(Paracetamol needed)`

`= 14/2 xx 15\ text(mg)`

`= 105\ text(mg)`

`text(S)text(ince 120 mg is contained in 10 mL,)`

`=> 105\ text(mg is contained in)`

`105/120 xx 10\ text(mL)= 8.75\ text(mL)`

`=> B`

Measurement, 2UG 2014 HSC 15 MC

Which expression will give the shortest distance, in kilometres, between Mount Isa `(20^@ text(S)\ 140^@ text(E))` and Tokyo `(35^@ text(N)\ 140^@ text(E))`?

(A) `15/360 xx 2 xx pi xx 6400`

(B) `55/360 xx 2 xx pi xx 6400`

(C) `140/360 xx 2 xx pi xx 6400`

(D) `305/360 xx 2 xx pi xx 6400`

Show Answers Only

`B`

Show Worked Solution

`text(Tokyo is 35° North of the equator, Mt Isa 20° South)`

`text(Arc length) = 55/360 xx 2 xx pi xx 6400`

`=> B`

Measurement, STD2 M1 2014 HSC 12 MC

A path 1.5 metres wide surrounds a circular lawn of radius 3 metres.

What is the approximate area of the path?

7.1 m²
21.2 m²
35.3 m²
56.5 m²

Show Answers Only

`C`

Show Worked Solution

`text(Area of annulus)`

`= pi (R^2-r^2)`

`= pi (4.5^2-3^2)`

`= pi (11.25)`

`=35.3\ text{m² (1 d.p.)}`

`=> C`

Algebra, 2UG 2014 HSC 11 MC

Simplify `6w^4 xx 1/3 w^2`.

`2w^6`
`2w^8`
`18w^6`
`18w^8`

Show Answers Only

`A`

Show Worked Solution

`6w^4 xx 1/3w^2`	`= (6w^4w^2)/3`
	`= 2w^6`

`=> A`

Algebra, STD2 A4 2014 HSC 3 MC

The diagram shows the graph of an equation.

Which of the following equations does the graph best represent?

`y = 3/x + 1`
`y = 3^x + 1`
`y = 3x^2 + 1`
`y = 3x^3 + 1`

Show Answers Only

`C`

Show Worked Solution

`text(Graph is a parabola that passes through)\ (0, 1).`

`=> C`

Proof, EXT2 P2 EQ-Bank 10

Use mathematical induction to prove that

`sum_(r=1)^n r^3 = 1/4 n^2 (n + 1)^2`

`text(for integers)\ n>=1` (3 marks)

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

Show Worked Solution

`text(Need to prove)\ sum_(r=1)^n r^3 = 1/4 n^2 (n + 1)^2\ \ text(integral)\ n>=1 `

`text(i.e.)\ 1^3 + 2^3 + 3^3 + … + n^3 = 1/4 n^2 (n + 1)^2`

`text(When)\ n = 1`

`text(LHS) = 1^3 = 1`

`text(RHS) = 1/4 1^2 (1 + 1)^2 = 1`

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

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

`text(i.e.)\ 1^3 + 2^3 + … + k^3 = 1/4 k^2 (k + 1)^2`

`text(Need to prove true for)\ n = k + 1`

`1^3 + 2^3 + … + k^3 + (k + 1)^3 = 1/4 (k + 1)^2 (k + 2)^2`

`text(LHS)`	`= 1/4 k^2 (k + 1)^2 + (k + 1)^3`
	`= 1/4 (k + 1)^2 [k^2 + 4(k + 1)]`
	`= 1/4 (k + 1)^2 (k^2 + 4k + 4)`
	`= 1/4 (k + 1)^2 (k + 2)^2`
	`=\ text(RHS)`

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

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

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`

Quadratic, EXT1 2009 HSC 2c

The diagram shows points `P(2t, t^2)` and `Q(4t, 4t^2)` which move along the parabola `x^2 = 4y`. The tangents to the parabola at `P` and `Q` meet at `R`.

Show that the equation of the tangent at `P` is `y = tx\ - t^2.` (2 marks)
Write down the equation of the tangent at `Q`, and find the coordinates of the point `R` in terms of `t`. (2 marks)
Find the Cartesian equation of the locus of `R`. (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`y = 2tx\ – 4t^2,\ R (3t, 2t^2)`
`y = 2/9 x^2`

Show Worked Solution

(i)	`x^2`	`= 4y`
	`y`	`= (x^2)/4`
	`dy/dx`	`= x/2`

IMPORTANT: Deriving this equation was specifically asked in 2009 and 2011, as well as being required in 2014. KNOW IT! Completing this in 60 seconds gains 2 full minutes on allocated time.

`text(At)\ \ x = 2t,`

`dy/dx = (2t)/2 = t`

`:.\ text(T)text(angent has)\ \ m = t\ \ text(and passes)`

`text(through)\ \ (2t, t^2).`

`y – t^2`	`= t (x – 2t)`
`y`	`= tx\ – 2t^2 + t^2`
	`= tx\ – t^2\ \ text(… as required)`

(ii)

`text(T)text(angent at)\ \ P (2t, t^2)\ \ text(is)\ \ y = tx\ – t^2`

`=>Q(4t, 4t^2) -= Q(2(2t), (2t)^2)`

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

`y`	`= 2tx\ – (2t)^2`
	`= 2tx\ – 4t^2`

`text(T)text(angents intersect at)\ \ R`

`y`	`= tx\ – t^2\ \ \ \ \ …\ text{(1)}`
`y`	`= 2tx\ – 4t^2\ \ \ \ \ …\ text{(2)}`

`text(Intersection when)\ text{(1)} = text{(2)}`

`tx\ – t^2`	`= 2tx\ – 4t^2`
`tx`	`= 3t^2`
`x`	`= 3t`

`text(Substitute)\ \ x = 3t\ \ text(into)\ text{(1)}`

`y`	`= t (3t)\ – t^2`
	`= 2t^2`

`:.\ R (3t, 2t^2)`

(iii)

`text(Find locus of)\ \ R`

`x`	`= 3t`	`\ \ \ \ \ …\ text{(1)}`
`y`	`= 2t^2\ \ \ \ \ …\ text{(2)}`

`text{From (1),}\ \ \ \ t = x/3`

`text(Substitute)\ \ t = x/3\ text(into)\ text{(2)}`

`y`	`= 2 (x/3)^2`
	`= 2/9 x^2`

`:.\ text(Locus of)\ \ R\ \ text(is)\ \ y = 2/9 x^2`

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)`

Calculus, EXT1 C1 2013 HSC 13a

A spherical raindrop of radius `r` metres loses water through evaporation at a rate that depends on its surface area. The rate of change of the volume `V` of the raindrop is given by

`(dV)/(dt) = -10^(-4) A`,

where `t` is time in seconds and `A` is the surface area of the raindrop. The surface area and the volume of the raindrop are given by `A = 4pir^2` and `V = 4/3 pi r^3` respectively.

Show that `(dr)/(dt)` is constant. (1 mark)

--- 5 WORK AREA LINES (style=lined) ---
How long does it take for a raindrop of volume `10^(–6)` m³ to completely evaporate? (2 marks)

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

Show Answers Only

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

Show Worked Solution

i.	`text(Show)\ \ (dr)/(dt)\ \ text(is a constant)`

`(dV)/(dt) = (dV)/(dr) * (dr)/(dt)\ \ \ …\ text{(1)}`

`V`	`= 4/3 pi r^3`
`:. (dV)/(dr)`	`= 4 pi r^2`
`(dV)/(dt)`	`= -10^(-4) A\ \ text{(given)}`

`text(Substituting into)\ text{(1)}`

`-10^(-4) A`	`= 4 pi r^2 xx (dr)/(dt)`
	`= A xx (dr)/(dt)`
`:.\ (dr)/(dt)`	`= -10^(-4)\ \ text(… as required)`

ii.	`V`	`= 10^(-6)\ text(m³)`
	`4/3 pi r^3`	`= 10^(-6)`
	`r^3`	`= (3 xx 10^(-6))/(4pi)`
	`r`	`= root(3)((3 xx 10^(-6))/(4pi))`

`text(S)text(ince the radius decreases at a constant rate,)`

♦♦ Mean mark 31%

`t=(root(3)((3 xx 10^(-6))/(4pi)))/(10^(-4))`

`\ \ =62.035 …`

`\ \ =62\ text(seconds)\ text{(nearest whole)}`

`:.\ text(It takes 62 seconds for the raindrop)`

`text(to evaporate.)`

Mechanics, EXT2* M1 2013 HSC 12e

A particle moves along a straight line. The displacement of the particle from the origin is `x`, and its velocity is `v`. The particle is moving so that `v^2 + 9x^2 = k`, where `k` is a constant.

Show that the particle moves in simple harmonic motion with period `(2pi)/3`. (2 marks)

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

Show Answers Only

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

Show Worked Solution

`v^2 + 9x^2`	`= k`
`v^2`	`= k\ – 9x^2`
`1/2 v^2`	`= 1/2k\ – 9/2 x^2`

`text(For SHM,)\ \ ddot x = -n^2x`

`ddot x`	`= d/(dx) (1/2v^2)`
	`= -9x`
	`= -3^2 x \ \ \ text(… as required)`

`text(Period)\ (T)\ text(of SHM) = (2pi)/n`

`text(Here,)\ \ n=3`

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

Geometry and Calculus, EXT1 2013 HSC 12d

The point `P(t, t^2 + 3)` lies on the curve `y = x^2 + 3`. The line `l` has equation `y = 2x\ – 1`. The perpendicular distance from `P` to the line `l` is `D(t)`.

Show that
1. `D(t) = (t^2\ - 2t + 4)/sqrt5`. (2 marks)
Find the value of `t` when `P` is closest to `l`. (1 mark)
Show that, when `P` is closest to `l`, the tangent to the curve at `P` is parallel to `l`. (1 mark)

Show Answers Only

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

Show Worked Solution

(i)

`text(Show)\ \ D(t) = (t^2\ – 2t + 4)/sqrt5`

`D(t)`	`= _\|_\ text(dist of)\ (t, t^2 + 3)\ text(from)\ 2x\ – y\ – 1 = 0`
`D`	`= \|(ax_1 + by_1 + c)/sqrt(a^2 + b^2)\|`
	`= \|(2(t)\ – 1 (t^2 + 3)\ – 1)/sqrt(2^2 + (-1)^2)\|`
	`= \|(2t\ – t^2\ – 3\ – 1)/sqrt5\|`
	`= \|(-(t^2\ – 2t + 4))/sqrt5\|`

Mean mark 40%
MARKER’S COMMENT: The biggest difficulty in part (i) was to explain the removal of the absolute value sign.

`text(S)text(ince)\ \ t^2\ – 2t + 4\ \ text(is positive definite)`

`text{(i.e.} \ Delta < 0\ text(and co-efficient of)\ t^2 > 0 text{)},`

`t^2\ – 2t + 4\ \ text(is positive for all)\ t.`

`:.\ D(t) = (t^2\ – 2t + 4)/sqrt5\ \ text(… as required)`

(ii)

`text(MAX or MIN when)\ \ (dD)/(dt) = 0`

`(dD)/(dt) = 1/sqrt5 (2t\ – 2)`	`= 0`
`2t\ – 2`	`= 0`
`t`	`= 1`

`(d^2D)/(dt^2) = 2/sqrt5 > 0\ =>\ text(MIN)`

`:.\ P\ text(is closest to)\ l\ text(when)\ t = 1`

(iii)	`y`	`= x^2 + 3`
	`dy/dx`	`= 2x`
`text(When)\ x = t`

`dy/dx = 2t`

`text(At)\ t = 1`

`dy/dx = 2,`

`=> m_text(tangent) = 2\ \ and\ \ m_l = 2`

`:.\ text(T)text(angent is parallel to)\ \ l\ \ \ \ text(… as required)`

Calculus, EXT1 C3 2013 HSC 12b

The region bounded by the graph `y = 3 sin\ x/2` and the `x`-axis between `x = 0` and `x = (3pi)/2` is rotated about the `x`-axis to form a solid.

Find the exact volume of the solid. (3 marks)

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

Show Answers Only

`(9pi)/2 ((3pi)/2 + 1) text(u³)`

Show Worked Solution

`y`	`= 3 sin\ x/2`
`y^2`	`= 9 sin^2\ x/2`

`text(Using:)\ \ sin^2x= 1/2 (1 – cos 2x)`

`:. V`	`= pi int_0^((3pi)/2) 9 sin^2\ x/2\ dx`
	`= (9pi)/2 int_0^((3pi)/2) (1\ – cosx)\ dx`
	`= (9pi)/2 [x\ – sinx]_0^((3pi)/2)`
	`= (9pi)/2 [((3pi)/2\ – sin\ (3pi)/2)\ – 0]`
	`= (9pi)/2 ((3pi)/2 + 1)\ text(u³)`

Trigonometry, EXT1 T3 2013 HSC 12a

Write `sqrt3cos x - sin x` in the form `2 cos (x + alpha)`, where `0 < alpha < pi/2`. (1 mark)

--- 6 WORK AREA LINES (style=lined) ---
Hence, or otherwise, solve `sqrt3 cos x = 1 + sin x`, where `0 < x < 2pi`. (2 marks)

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

Show Answers Only

`2cos (x + pi/6)`
`pi/6,\ (3pi)/2`

Show Worked Solution

i. `text(Write)\ sqrt3 cosx\ – sinx\ text(in form)`

`2cos (x + alpha),\ \ \ 0 < alpha < pi/2`

`2 (cosx cos alpha\ – sinx sin alpha)`	`= sqrt 3 cosx\ – sinx`
`cosx cos alpha\ – sinx sin alpha`	`= sqrt3/2 cos x\ – 1/2 sinx`

`=> cos alpha`	`= sqrt3/2`
`=> sin alpha`	`= 1/2`
`alpha`	`= pi/6`

`:.\ 2cos (x + pi/6) = sqrt3 cosx\ – sinx`

ii. `text(Solve)\ sqrt3 cosx = 1 + sinx,\ \ \ 0 < x < 2pi`

`sqrt3 cosx\ – sinx`	`= 1`
`2 cos (x + pi/6)`	`= 1\ \ \ text{(from part (i))}`
`cos (x + pi/6)`	`= 1/2`
`cos(pi/3)`	`=1/2`

`text(S)text(ince cos is positive in)\ 1^text(st) // 4^text(th)\ text(quadrants)`

`x + pi/6`	`= pi/3,\ 2pi\ – pi/3\ \ \ \ \ (0 < x < 2pi)`
`:.\ x`	`= pi/6,\ (3pi)/2`

Calculus, EXT1 C2 2013 HSC 11f

Use the substitution `u = e^(3x)` to evaluate `int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`. (3 marks)

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

Show Answers Only

`1/3 (tan^(-1)e\ – pi/4)`

Show Worked Solution

MARKER’S COMMENT: Many students did not calculate in radians and incorrectly got an answer of 8.2. BE CAREFUL!
Note that converting your answer to 0.14 is also correct but not required.

`text(Let)\ \ u = e^(3x)`

`(du)/(dx)`	`= 3e^(3x)`
`:.dx`	`= (du)/(3e^(3x))`

`text(When)`	`\ x = 1/3,`	`\ u = e^(3 xx 1/3) = e`
	`\ x = 0,`	`\ u = e^0 = 1`

`:.int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`

`=int_1^e (e^(3x))/(u^2 + 1) xx (du)/(3e^(3x))`

`= 1/3 int_1^e 1/(u^2 + 1)\ du`

`= 1/3 [tan^(-1)u]_1^e`

`= 1/3 [tan^(-1) e\ – tan^(-1) 1]`

`= 1/3 (tan^(-1)e\ – pi/4)`

Trig Calculus, EXT1 2013 HSC 11e

Find `lim_(x -> 0) (sin\ x/2)/(3x)`. (1 mark)

Show Answers Only

`1/6`

Show Worked Solution

`lim_(x -> 0) (sin\ x/2)/(3x)`

`= lim_(x -> 0) (sin\ x/2)/(6 xx x/2)`

`= 1/6 xx lim_(x -> 0) (sin\ x/2)/(x/2)`

`= 1/6`

Geometry and Calculus, EXT1 2013 HSC 11d

Consider the function `f(x) = x/(4\ - x^2)`.

Show that `f prime (x) > 0` for all `x` in the domain of `f(x)`. (2 marks)
Sketch the graph `y = f(x)`, showing all asymptotes. (2 marks)

Show Answers Only

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

Show Worked Solution

(i) `f(x) = x/(4\ – x^2)`

`u`	`= x`	`\ \ \ \ \ v`	`= 4\ – x^2`
`u prime`	`= 1`	`\ \ \ \ \ \ \ \ \ \ v prime`	`= -2x`

`f prime (x)`	`= (u prime v\ – u v prime)/(v^2)`
	`= (1* (4\ – x^2)\ – x(-2x))/((4\ – x^2)^2)`
	`= (4 + x^2)/((4\ – x^2)^2)`

`text(Domain is all)\ x,\ x != +- 2`

`text(Consider)\ f prime (x)`

`4 + x^2`	`> 0\ \ \ text{(in domain)}`
`(4\ – x^2)^2`	`> 0\ \ \ text{(in domain)}`

`:.\ f prime (x) > 0\ text(for all)\ x,\ x != +- 2`

(ii)

`text(Asymptotes at)\ x = +- 2`

`text(Passes through)\ (0,0)`

COMMENT: Note that `x->2(–)` is a short way of writing as `x` approaches 2 from the negative (or left-hand) side. This notation can save time when required.

`text(As)`	`\ x -> 2 (-),`	`\ y -> oo`
	`\ x -> 2 (+),`	`\ y -> – oo`

`text(As)`	`\ x -> -2 (-),`	`\ y -> oo`
	`\ x -> -2 (+),`	`\ y -> -oo`

`text(As)`	`\ x -> oo,`	`\ y -> 0`
	`\ x -> – oo,`	`\ y -> 0`

Calculus, EXT1 C2 2013 HSC 11b

Find `int 1/sqrt (49 - 4x^2)\ dx`. (2 marks)

Show Answers Only

`1/2 sin^(-1) ((2x)/7) + c`

Show Worked Solution

`int 1/sqrt(49 – 4x^2)\ dx`

`= int 1/(2 sqrt(49/4 – x^2))\ dx`

`= 1/2 int 1/sqrt((7/2)^2 – x^2)\ dx`

`= 1/2 sin^(-1) ((2x)/7) + c`

Trigonometry, EXT1 T1 2013 HSC 9 MC

The diagram shows the graph of a function.

Which function does the graph represent?

`y = cos^(-1) x`
`y = pi/2 + sin^(-1) x`
`y = - cos^(-1) x`
`y = - pi/2\ - sin^(-1) x`

Show Answers Only

`B`

Show Worked Solution

`text(By elimination,)`

`text(The graph passes through)\ \ (1, pi)`

`text(The only equation to satisfy this point is)`

`y = pi/2 + sin^(-1) x`

`=> B`

Combinatorics, EXT1 A1 2013 HSC 7 MC

A family of eight is seated randomly around a circular table.

What is the probability that the two youngest members of the family sit together?

`(6!\ 2!)/(7!)`
`(6!)/(7!\ 2!)`
`(6!\ 2!)/(8!)`
`(6!)/(8!\ 2!)`

Show Answers Only

`A`

Show Worked Solution

`text(Fix youngest person in 1 seat,)`

`text(Total combinations around table) = 7!`

`text(Combinations with youngest side by side) =2!6!`

`:.\ text{P(sit together)} = (6!\ 2!)/(7!)`

`=> A`

Calculus, EXT1 C2 2013 HSC 5 MC

Which integral is obtained when the substitution `u = 1 + 2x` is applied to `int x sqrt(1 + 2x)\ dx`?

`1/4 int (u - 1) sqrt u\ du`
`1/2 int (u - 1) sqrt u\ du`
`int (u - 1) sqrt u\ du`
`2 int (u - 1) sqrt u\ du`

Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ u`	`= 1 + 2x`
`:.x`	`= 1/2 (u – 1)`
`(du)/(dx)`	`= 2`
`:.dx`	`= 1/2\ du`

`int x sqrt(1 + 2x)\ dx`

`=int 1/2 (u – 1) xx u^(1/2) xx 1/2\ du`

`= 1/4 int (u – 1) sqrt u\ du`

`=> A`

Functions, EXT1 F2 2013 HSC 4 MC

Which diagram best represents the graph `y = x (1- x)^3 (3- x)^2`?

Show Answers Only

`D`

Show Worked Solution

`y = x(1 – x)^3 (3 – x)^2`

`text(By elimination)`

`text(Consider when)\ \ x < 0,`

`y = text{(–ve)} xx text{(+ve)} xx text{(+ve)} < 0`

`:.\ text(Cannot be)\ A\ text(or)\ C`

`text(Consider the cubic factor)\ \ (1 – x)^3,`

`text(The graph must have a stationary point at)\ x = 1`

`:.\ text(Cannot be)\ B`

`=> D`

Binomial, EXT1 2010 HSC 7b

The binomial theorem states that

`(1 + x)^n = ((n),(0)) + ((n),(1))x + ((n),(2))x^2 + ((n),(3))x^3 + ... + ((n),(n))x^n.`

Show that
1. `2^n = sum_(k = 0)^n ((n),(k))`. (1 mark)
Hence, or otherwise, find the value of
1. `((100),(0)) + ((100),(1)) + ((100),(2)) + ... + ((100),(100))`. (1 mark)
Show that
1. `n2^(n\ - 1) = sum_(k = 1)^n k ((n),(k))`. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`2^100`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

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

`text(Let)\ \ x = 1`

`(1 + 1)^n`	`= ((n),(0)) + ((n),(1))1 + ((n),(2))1^2 + … + ((n),(n))1^n`
`2^n`	`= ((n),(0)) + ((n),(1)) + … + ((n),(n))`
	`= sum_(k=0)^n ((n),(k))\ text(… as required)`

(ii) `((100),(0)) + ((100),(1)) + ((100),(2)) + … + ((100),(100)) = 2^100`

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

♦ Mean mark 37%

(iii) `(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) = 1((n),(1)) + 2x ((n),(2)) + 3x^2 ((n),(3)) + … + nx^(n\ – 1) ((n),(n))`

`text(Let)\ \ x = 1`

`n (1 + 1)^(n\ – 1)`	`= 1 ((n),(1)) + 2 ((n),(2)) + 3 ((n),(3)) + … + n ((n),(n))`
`n 2^(n\ – 1)`	`= sum_(k = 1)^n k ((n),(k))\ text(… as required)`

Trigonometry, EXT1 T3 2010 HSC 6a

Show that `cos(A - B) = cos A cos B(1 + tan A tan B)`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Suppose that `0 < B < pi/2` and `B < A < pi`.
Deduce that if `tan Atan B = − 1`, then `A\ - B = pi/2`. (1 mark)

--- 4 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)\ \ cos(A\ – B) = cosA cosB (1 + tanA tanB)`

`text(RHS)`	`= cosA cosB (1 + (sinA sinB)/(cosA cosB))`
	`= cosA cos B + sinA sin B`
	`= cos(A – B)\ text(… as required)`

ii. `text(Given)\ \ tanA tanB = -1`

`cos (A – B)`	`= cosA cosB (1\ – 1)`
`cos (A – B)`	`= 0`
`A – B`	`= cos^(-1) 0`
	`= pi/2, (3pi)/2, …`

`text(S)text(ince)\ \ \ 0 < B < pi/2\ \ text(and)\ \ \ B < A < pi,`

`=> A\ – B = pi/2`

Trig Ratios, EXT1 2010 HSC 5a

A boat is sailing due north from a point `A` towards a point `P` on the shore line.

The shore line runs from west to east.

In the diagram, `T` represents a tree on a cliff vertically above `P`, and `L` represents a landmark on the shore. The distance `PL` is 1 km.

From `A` the point `L` is on a bearing of 020°, and the angle of elevation to `T` is 3°.

After sailing for some time the boat reaches a point `B`, from which the angle of elevation to `T` is 30°.

Show that

`qquad BP = (sqrt3 tan 3°)/(tan20°)`. (3 marks)
Find the distance `AB`. Give your answer to 1 decimal place. (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`2.5\ text(km)\ \ text{(to 1 d.p.)}`

Show Worked Solution

(i) `text(Show)\ \ BP = (sqrt3 tan 3°)/(tan 20°)`

`text(In)\ Delta ATP`

`tan 3°`	`= (TP)/(AP)`
`=> AP`	`= (TP)/(tan 3)`

`text(In)\ Delta APL`

`tan 20°`	`= 1/(AP)`
`=> AP`	`= 1/tan 20`

`:. (TP)/(tan3)`	`= 1/(tan20)`
`TP`	`= (tan3°)/(tan20°)\ \ \ text(…)\ text{(} text(1)text{)}`

`text(In)\ \ Delta BTP`

`tan 30°`	`= (TP)/(BP)`
`1/sqrt3`	`= (TP)/(BP)`
`BP`	`= sqrt3 xx TP\ \ \ \ \ text{(using (1) above)}`
	`= (sqrt3 tan3°)/(tan20°)\ \ \ text(… as required)`

(ii)	`AB`	`= AP\ – BP`
	`AP`	`= 1/(tan20°)\ \ \ text{(} text(from part)\ text{(i)} text{)}`

`:.\ AB`	`= 1/(tan 20°)\ – (sqrt3 tan3°)/(tan20°)`
	`= (1\ – sqrt3 tan 3)/(tan20°)`
	`= 2.4980…`
	`= 2.5\ text(km)\ text{(to 1 d.p.)`

Quadratic, EXT1 2010 HSC 4c

The diagram shows the parabola `x^2 = 4ay`. The point `P(2ap, ap^2)`, where `p != 0`, is on the parabola.

The tangent to the parabola at `P`, `y = px − ap^2`, meets the `y`-axis at `L`.

The point `M` is on the directrix, such that `PM` is perpendicular to the directrix.

Show that `SLMP` is a rhombus. (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Solution 1)`

`text(Show)\ \ SLMP\ \ text(is a rhombus)`

`L\ \ text(occurs when tangent meets)\ x text(-axis)`

`y = 0 – ap^2`

`:.L(0,–ap^2),\ \ M(2ap, –a)`

`m_(PS)`	`= (y_2\ – y_1)/(x_2\ – x_1)`
	`= (ap^2\ – a)/(2ap\ – 0)`
	`= (p^2\ – 1)/(2p)`
`m_(ML)`	`= (-a + ap^2)/(2ap\ – 0`
	`= (p^2\ – 1)/(2p)`

`text(S)text(ince)\ \ `	`PS \ text(\|\|)\ ML\ \ \ text{(same gradient)}`
`text(and)\ \ `	`PM \ text(\|\|)\ SL\ \ \ text{(vertical)}`
`=>`	`\ \ SLMP\ text(is a parallelogram)`

`text(S)text(ince)\ \ `	`PS = PM\ \ \ text{(definition of a parabola)}`
`=>`	`\ \ SLMP\ text(is a rhombus … as required.)`

`text(Alternate Solution)`

`L(0,–ap^2),\ \ M(2ap, –a)`

`m_(SM)`	`= (y_2\ – y_1)/(x_2\ – x_1) = (-a\ -a)/(2ap\ – 0) = – 1/p`
`m_(LP)`	`= p`

`text(S)text(ince)\ \ m_(SM) xx m_(LP) = p xx – 1/p = -1`

`=>\ text(Diagonals are perpendicular)`

`text(Midpoint)\ LP`	`= ( (0 + 2ap)/2,\ (-ap^2 + ap^2)/2)`
	`= (ap, 0)`
`text(Midpoint)\ MS`	`= ((0 + 2ap)/2,\ (a\ – a)/2)`
	`= (ap,\ 0)`

`=>\ text(Diagonals bisect)`

`:.\ text(S)text(ince diagonals are)\ _|_\ text(bisectors,)`

`SLMP\ text(is a rhombus … as required.)`

Trigonometry, EXT1 T3 2010 HSC 4b

Express `2 cos theta + 2 cos (theta + pi/3)` in the form `R cos (theta + alpha)`,

where `R > 0` and `0 < alpha < pi/2`. (3 marks)

--- 12 WORK AREA LINES (style=lined) ---
Hence, or otherwise, solve `2 cos theta + 2 cos (theta + pi/3) = 3`,
for `0 < theta < 2pi`. (2 marks)

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

Show Answers Only

`2 sqrt 3 cos (theta + pi/6)`
`(5pi)/3`

Show Worked Solution

i.	`2 cos theta + 2 cos (theta + pi/3)`
	`= 2 cos theta + 2 (cos theta cos (pi/3)\ – sin theta sin (pi/3))`
	`= 2 cos theta + 2 cos theta xx 1/2\ – 2 sin theta xx sqrt3/2`
	`= 2 cos theta + cos theta\ – sqrt3 sin theta`
	`= 3 cos theta\ – sqrt3 sin theta`

`R cos (theta + alpha) = R cos theta cos alpha – R sin theta sin alpha`

`R cos alpha`	`= 3\ \ \ \ \ `	`R sin alpha`	`= sqrt3`
`cos alpha`	`= 3/R\ \ \ \ \ `	`sin alpha`	`= sqrt3/R`

`tan alpha`	`= sin alpha/cos alpha = sqrt3/3 = 1/sqrt3`
`tan\ pi/6`	`=1/sqrt3`
`:. alpha`	`=pi/6\ \ \ \ \ (0 < alpha < pi/2)`

`R^2`	`= 3^2 + (sqrt3)^2`
	`= 9 + 3`
`R`	`= sqrt 12 = 2 sqrt3`

`:.\ 2 cos theta + 2 cos (theta + pi/3) = 2 sqrt 3 cos (theta + pi/6)`

♦ Mean mark part (ii) 49%
MARKER’S COMMENT: Many students did not check their answers against the stated domain for `theta`.

ii.	`2 cos theta + 2 cos(theta + pi/3)`	`= 3`
	`2 sqrt 3 cos (theta + pi/6)`	`= 3`
	`cos (theta + pi/6)`	`= 3/(2sqrt3) = sqrt3/2`
	`cos^(-1) (sqrt3/2)`	`= pi/6`

`text(S)text(ince cos is positive in 1st and 4th quadrants,)`

`theta + pi/6`	`= pi/6,\ 2pi\ – pi/6`
`:. theta`	`= (5pi)/3\ \ \ \ \ (0 < theta < 2pi)`

Mechanics, EXT2* M1 2010 HSC 4a

A particle is moving in simple harmonic motion along the `x`-axis.

Its velocity `v`, at `x`, is given by `v^2 = 24 − 8x − 2x^2`.

Find all values of `x` for which the particle is at rest. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Find an expression for the acceleration of the particle, in terms of `x`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Find the maximum speed of the particle. (2 marks)

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

Show Answers Only

`x = –6\ \ text(or)\ \ 2`
`-4\ – 2x`
`4 sqrt 2`

Show Worked Solution

(i)

`v^2 = 24\ – 8x\ – 2x^2`

`text(Find)\ \ x\ \ text(when)\ \ v=0`

`24 – 8x – 2x^2`	`= 0`
`x^2 + 4x – 12`	`= 0`
`(x + 6)(x – 2)`	`= 0`
`x= -6\ \ text(or)\ \ 2`

`:.\ text(Particle at rest when)\ \ x = –6\ \ text(or)\ \ 2`

(ii)	`ddot x`	`= d/(dx) (1/2 v^2)`
		`= d/(dx) (12 – 4x – x^2)`
		`= -4 – 2x`

(iii) `text(Solution 1)`

`text(Max speed when)\ \ ddot x = 0`

`-4 – 2x`	`= 0`
`2x`	`= -4`
`x`	`= -2`

`text(At)\ \ x = -2`

MARKER’S COMMENT: While most students found `x=-2` correctly, too many made errors substituting this back in or failed to finish the answer by taking the square root. BE CAREFUL! .

`v^2`	`= 24 – 8(–2) – 2(–2)^2`
	`= 24 + 16 – 8`
	`= 32`
`v`	`= +- sqrt32`
	`= +- 4 sqrt2`

`:.\ text(Maximum speed is)\ \ 4 sqrt2`

`text(Alternate Answer)`

`text(Maximum speed occurs when)`

`x = (-6+2)/2 = –2`

`text(At)\ \ x = –2`

`v^2`	`= 32\ \ \ text{(see working above)}`
`v`	`= +- 4 sqrt 2`

`:.\ text(Maximum speed is)\ 4sqrt2`

Inverse Functions, EXT1 2010 HSC 3b

Let `f(x) = e^(-x^2)`. The diagram shows the graph `y = f(x)`.

The graph has two points of inflection.
Find the `x` coordinates of these points. (3 marks)
Explain why the domain of `f(x)` must be restricted if `f(x)` is to have an inverse function. (1 mark)
Find a formula for `f^(-1) (x)` if the domain of `f(x)` is restricted to `x ≥ 0`. (2 marks)
State the domain of `f^(-1) (x)`. (1 mark)
Sketch the curve `y = f^(-1) (x)`. (1 mark)
(1) Show that there is a solution to the equation `x = e^(-x^2)` between `x = 0.6` and `x = 0.7`. (1 mark)
(2) By halving the interval, find the solution correct to one decimal place. (1 mark)

Show Answers Only

`x = +- 1/sqrt2`
`text(There can only be 1 value of)\ y`
`text(for each value of)\ x.`
`f^(-1)x = sqrt(ln(1/x))`
`0 <= x <= 1`
(1) `text(Proof)\ \ text{(See Worked Solutions)}`
(2) `0.7\ text{(1 d.p.)}`

Show Worked Solution

(i)	`y`	`= e^(-x^2)`
	`dy/dx`	`= -2x * e^(-x^2)`
	`(d^2y)/(dx^2)`	`= -2x (-2x * e^(-x^2)) + e ^(-x^2) (-2)`
		`= 4x^2 e^(-x^2)\ – 2e^(-x^2)`
		`= 2e^(-x^2) (2x^2\ – 1)`

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

`2e^(-x^2) (2x^2\ – 1)`	`= 0`
`2x^2\ – 1`	`= 0`
`x^2`	`= 1/2`
`x`	`= +- 1/sqrt2`

COMMENT: It is also valid to show that `f(x)` is an even function and if a P.I. exists at `x=a`, there must be another P.I. at `x=–a`.

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

`=>\ text(Change of concavity)`

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

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

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = – 1/sqrt2`

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

(iii)

`y = e^(-x^2)`

`text(Inverse function can be written)`

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

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

`:. f^(-1) (x)=sqrt(ln (1/x))`

♦ Parts (iv) and (v) were poorly answered with mean marks of 39% and 49% respectively.

(iv)

`f(0) = e^0 = 1`

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

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

(v)

(vi)(1) `x = e^(-x^2)`

`text(Let)\ g(x) = x\ – e^(-x^2)`

`g(0.6)`	`=0.6\ – e^(-0.6^2)`
	`=0.6\ – 0.6977 < 0`
`g(0.7)`	`=0.7\ – e^(-0.7^2)`
	`=0.7\ – 0.6126 > 0`
`=>g(x)\ text(changes sign)`

`:.\ g(x)\ \ text(has a root between 0.6 and 0.7)`

`:.\ x = e^(-x^2)\ \ text(has a solution between 0.6 and 0.7)`

♦ Mean mark 37%.
MARKER’S COMMENT: Better responses showed the change in sign between `g(0.65)` and `g(0.7)` as shown in the solution.

(vi)(2)	`g(0.65)`	`=0.65\ – e^(-0.65^2)`
		`=0.65\ – 0.655 < 0`

`:.\ text(A solution lies between 0.65 and 0.7)`

`:.\ x = 0.7\ \ text{(1 d.p.)}`

Combinatorics, EXT1 A1 2010 HSC 3a

At the front of a building there are five garage doors. Two of the doors are to be painted red, one is to be painted green, one blue and one orange.

How many possible arrangements are there for the colours on the doors? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
How many possible arrangements are there for the colours on the doors if the two red doors are next to each other? (1 mark)

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

Show Answers Only

`60`
`24`

Show Worked Solution

i.	`text(# Arrangements)`	`= (5!)/(2!)`
		`= 60`

♦ Mean mark 50%
MARKER’S COMMENT: Drawing a diagram was a successful strategy for many students in this part.

ii.

`text(When 2 red doors are side-by-side,)`

`text(# Arrangements)`	`= 4!`
	`= 24`

Calculus, EXT1 C1 2010 HSC 2d

A radio transmitter `M` is situated 6 km from a straight road. The closest point on the road to the transmitter is `S`.

A car is travelling away from `S` along the road at a speed of `text(100 km h)`⁻¹. The distance from the car to `S` is `x\ text(km)` and from the car to `M` is `r\ text(km)`.

Find an expression in terms of `x` for `(dr)/(dt)`, where `t` is time in hours. (3 marks)

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

Show Answers Only

`(100x)/sqrt(x^2 + 36)\ \ text(km/hr)`

Show Worked Solution

`text(Using Pythagoras,)`

`r^2`	`= x^2 + 6^2`
`r`	`= sqrt(x^2 + 36),\ r > 0`
`(dr)/(dt)`	`= (dx)/(dt) * (dr)/(dx)\ \ \ \ …\ (1)`
`(dx)/(dt)`	`= 100\ \ \ text{(given)}`
`(dr)/(dx)`	`= 1/2 (x^2 + 36)^(-1/2) xx d/(dx) (x^2 + 36)`
	`= x/sqrt(x^2 + 36)`

`text{Substituting into (1)}`
`:.\ (dr)/(dt)`	`= (100x)/sqrt(x^2 + 36)\ \ text(km/hr)`

Functions, EXT1 F2 2010 HSC 2c

Let `P(x) = (x + 1)(x-3) Q(x) + ax + b`,

where `Q(x)` is a polynomial and `a` and `b` are real numbers.

The polynomial `P(x)` has a factor of `x-3`.

When `P(x)` is divided by `x + 1` the remainder is `8`.

Find the values of `a` and `b`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Find the remainder when `P(x)` is divided by `(x + 1)(x-3)`. (1 mark)

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

Show Answers Only

`a = -2,\ b = 6`
` -2x + 6`

Show Worked Solution

i. `P(x) = (x+1)(x-3)Q(x) + ax + b`

`(x-3)\ \ text{is a factor (given)}`

`:. P (3)`	`= 0`
`3a + b`	`= 0\ \ \ …\ text{(1)}`

`P(x) ÷ (x+1)=8\ \ \ text{(given)}`

`:.P(-1)`	`= 8`
`-a + b`	`= 8\ \ \ …\ text{(2)}`

`text{Subtract (1) – (2)}`

`4a`	`= -8`
`a`	`= -2`

`text(Substitute)\ \ a = -2\ \ text{into (1)}`

`-6 + b`	`= 0`
`b`	`= 6`

`:. a= – 2, \ b=6`

ii. `P(x) -: (x + 1)(x-3)`

`= ((x+1)(x-3)Q(x)-2x + 6)/((x+1)(x-3))`

`= Q(x) + (-2x + 6)/((x+1)(x-3))`

`:.\ text(Remainder is)\ \ -2x + 6`

COMMENT: This question requires a fundamental understanding of the remainder theorem.

Calculus, EXT1 C2 2010 HSC 2a

The derivative of a function `f(x)` is given by

`f^{′}(x) = sin^2 x`.

Find `f(x)`, given that `f(0) = 2`. (2 marks)

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

Show Answers Only

`1/2x-1/4 sin 2x + 2`

Show Worked Solution

`f^{′}(x)`	`= sin^2 x`
`f(x)`	`= int sin^2 x\ dx`
	`= int 1/2(1-cos 2x)\ dx`
	`= int 1/2-1/2 cos 2x\ dx`
	`= 1/2 x-1/4 sin 2 x + c`

`text(Given)\ \ f(0) = 2,`

`2`	`= 1/2 xx 0-1/4 sin 0 + c`
`:. c= 2`

`:.\ f(x) = 1/2 x-1/4 sin 2x + 2`

Statistics, EXT1 S1 2010 HSC 1f

Five ordinary six-sided dice are thrown.

What is the probability that exactly two of the dice land showing a four?

Leave your answer in unsimplified form. (1 mark)

Show Answers Only

`\ ^5C_2 (1/6)^2 (5/6)^3`

Show Worked Solution

COMMENT: Surprisingly, half of students sitting the exam got this wrong. Easily the most poorly answered part of Q1 in 2010.

`P(4) = 1/6`

`P(bar4) = 5/6`

`text(# Combinations of two 4’s) =\ ^5C_2`

`:.\ P text{(exactly two 4s)} =\ ^5C_2 (1/6)^2 (5/6)^3`

L&E, EXT1 2010 HSC 1c

Solve `ln(x + 6) = 2 ln x`. (3 marks)

Show Answers Only

`x = 3`

Show Worked Solution

`ln(x + 6)`	`= 2 lnx`
`ln(x + 6)`	`= ln x^2`
`x^2`	`= x + 6`
`x^2\ – x\ – 6`	`= 0`
`(x\ – 3)(x + 2)`	`= 0`
`:. x = 3\ \ text(or)\ \ -2`

`text(S)text(ince RHS)\ = 2lnx,\ \ x>0`

`:. x=3`

Trigonometry, EXT1 T1 2010 HSC 1b

Let `f(x) = cos^(-1) (x/2)`. What is the domain of `f(x)`? (1 mark)

Show Answers Only

`-2 <= x <= 2`

Show Worked Solution

`f(x) = cos^(-1) (x/2)`

`–1`	`<= x/2`	`<= 1`
`–2`	`<= x`	`<= 2`

`:.\ text(Domain of)\ \ f(x)\ \ text(is)\ \ \ –2 <= x <= 2`

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)

--- 4 WORK AREA LINES (style=lined) ---
Find the rate at which `V` is changing with respect to time when `l = 2`. (2 marks)

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

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

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).`

Geometry and Calculus, EXT1 2011 HSC 4a

Consider the function `f(x) = e^(-x)\ - 2e^(-2x)`.

Find `f prime (x)`. (1 mark)
The graph `y = f(x)` has one maximum turning point.
Find the coordinates of the maximum turning point. (2 marks)
Evaluate `f(ln2)`. (1 mark)
Describe the behaviour of `f(x)` as `x -> oo`. (1 mark)
Find the `y`-intercept of the graph `y = f(x)`. (1 mark)
Sketch the graph `y = f(x)` showing the features from parts (ii) - (v).
You are not required to find any points of inflection. (2 marks)

Show Answers Only

`-e^(-x) + 4e^(-2x)`
`text(Max T.P. at)\ (ln4, 1/8)`
`0`
`text(As)\ x -> oo, f(x) -> 0`
`-1`

Show Worked Solution

(i)	`f(x)`	`= e^(-x)\ – 2e^(-2x)`
	`f prime (x)`	`= -e^(-x) + 4e^(-2x)`

(ii)

`text(T.P. when)\ f prime (x) = 0`

`-e^(-x) + 4e^(-2x) = 0`

`text(Let)\ e^(-x) = X`

`- X + 4X^2`	`= 0`
`X (4X\ – 1)`	`= 0`

`X = 0\ \ text(or)\ \ 1/4`

`text(If)\ \ e^(-x) = 0,\ \ \ text(No solution)`

`text(If)\ \ e^(-x) = 1/4`

`ln e^(-x)`	`= ln (1/4)`
`-x`	`= ln 4^(-1)`
	`= -ln 4`
`x`	`= ln4`

ALGEBRA TIP: Note that `e^ln x=x` can often help calculations. This is easily proven by simply taking the `ln` of both sides of the equation.

`f″(x)`	`= e^(-x)\ – 8e^(-2x)`
`f″(ln4)`	`< 0 => text(MAX)`
`f (ln4)`	`= e^(-ln4)\ – 2e^(-2ln4)`
	`= e^(ln(1/4))\ – 2 e^(ln(1/16))`
	`= 1/4\ – 2(1/16)`
	`= 1/8`

`:.\ text(Maximum T.P. at)\ \ (ln4, 1/8)`

(iii)	`f(ln2)`	`= e^(-ln2)\ – 2e^(-2ln2)`
		`= e^(ln(1/2))\ – 2e^(ln(1/4))`
		`= 1/2\ – 2 xx 1/4`
		`= 0`

(iv)	`f(x)`	`= e^(-x)\ – 2e^(-2x)`
		`= e^(-x) (1\ – 2e^(-x))`
	`text(As)\ x -> oo, \ e^(-x) ->0,\ f(x) -> 0`

(v)

`y text(-intercept at)\ \ f(0)`

`f(0)`	`= e^0\ – 2e^0`
	`= -1`

(vi)

Quadratic, EXT1 2011 HSC 3b

The diagram shows two distinct points `P(t, t^2)` and `Q(1\ - t, (1\ - t)^2)` on the parabola `y = x^2`. The point `R` is the intersection of the tangents to the parabola at `P` and `Q`.

Show that the equation of the tangent to the parabola at `P` is `y = 2tx\ – t^2`. (2 marks)
Using part `text{(i)}`, write down the equation of the tangent to the parabola at `Q`. (1 mark)
Show that the tangents at `P` and `Q` intersect at
`R (1/2, t\ - t^2)`. (2 marks)
Describe the locus of `R` as `t` varies, stating any restriction on the `y`-coordinate. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`y = 2 (1 -t)x\ – (1\ – t)^2`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Locus of)\ R\ text(is vertical line)`
`x = 1/2,\ y<1/4`

Show Worked Solution

(i)

`text(Show tangent at)\ P\ text(is)\ y = 2tx\ – t^2`

`y`	`= x^2`
`dy/dx`	`= 2x`

`x=t\ \ \ \ text(at)\ P`

`dy/dx = 2t`

`text(Find equation with)\ \ m = 2t\ \ text(through)\ \ P(t, t^2)`

`y\ – y_1`	`= m(x\ – x_1)`
`y\ – t^2`	`= 2t ( x\ – t)`
`y`	`= 2tx\ – 2t^2 + t^2`
	`= 2tx\ – t^2\ text(… as required)`

(ii) `text(T)text(angent at)\ Q\ text(has equation)`

MARKER’S COMMENT: Many students derived this equation rather than substituting the new parameter, costing them valuable time. This is a benefit of using the parametric approach.

`y = 2(1\ – t)x\ – (1\ – t)^2`

(iii) `text(Need to show)\ R(1/2,\ t\ – t^2)`

`R\ text(is at intersection of tangents)`

`2tx\ – t^2`	`= 2(1\ – t)x\ – (1\ – t)^2`
`2tx\ – t^2`	`= 2x\ – 2tx\ – 1 + 2t\ – t^2`
`4tx\ – 2x`	`= -1 + 2t\ – t^2 + t^2`
`2x(2t\ – 1)`	`= 2t\ – 1`
`2x`	`= 1`
`x`	`= 1/2`

`text(Using)\ \ y = 2tx – t^2\ \ text(when)\ \ x = 1/2`

`y`	`= 2t(1/2)\ – t^2`
	`= t\ – t^2`

`:.\ R(1/2, t\ – t^2)\ text(… as required)`

(iv) `text(Locus of)\ R`

♦♦ Mean mark of 22%.
MARKER’S COMMENT: Many students stated the locus as `y=t-t^2` rather than realising it had to be a straight line since `x=½`, and that `y=t-t^2` simply restricted the values of `y`.

`text(S)text(ince)\ x = 1/2\ text(is a constant)`

`R\ text(is a vertical line)`

`text(Now,)\ y = t\ – t^2 = t(1\ – t)`

`text(Graphically,)\ \ y\ \ text(has a maximum at)\ \ t = 1/2`

`text(Max)\ \ y = 1/2\ – (1/2)^2 = 1/4`

`=> y < 1/4\ \ text{(tangents can’t meet on parabola)}`

`:.\ text(Locus of)\ R\ text(is vertical line)\ x = 1/2,\ \ y<1/4`

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)`

Combinatorics, EXT1 A1 2011 HSC 2e

Alex’s playlist consists of 40 different songs that can be arranged in any order.

How many arrangements are there for the 40 songs? (1 mark)

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Alex decides that she wants to play her three favourite songs first, in any order.
How many arrangements of the 40 songs are now possible? (1 mark)

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

Show Answers Only

`40!`
`6 xx 37!`

Show Worked Solution

i.	`#\ text(Arrangements) = 40!`

ii.	`#\ text(Arrangements)`	`= 3! xx 37!`
		`= 6 xx 37!`

Trigonometry, EXT1 T1 2011 HSC 2d

Sketch the graph of the function `f(x) = 2arccos x`. Clearly indicate the domain and range of the function. (2 marks)

Show Answers Only

Show Worked Solution

`y`	`= 2 cos^(-1) x`
`y/2`	`= cos^(-1)x`

`text(Domain)\ -1 <= x <= 1`

`text(S)text(ince)\ \ 0 <= y/2 <= pi`

`text(Range)\ 0 <= y <= 2pi`

Combinatorics, EXT1 A1 2011 HSC 2c

Find an expression for the coefficient of `x^2` in the expansion of `(3x - 4/x)^8`. (2 marks)

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

Show Answers Only

`-870\ 912`

Show Worked Solution

`(3x – 4/x)^8 = sum_(k=0)^8\ ^8C_k *(3x)^(8\ – k) *(–4/x)^k`

`text(General term)`	`=\ ^8C_k * 3^(8\ – k) * x^(8\ – k) * (–1)^k4^k x^(-k)`
	`=\ ^8C_k(–1)^k 3^(8\ – k) * 4^k * x^(8\ – 2k)`

`text(Co-efficient of)\ \ x^2=2\ \ text(occurs when,)`

` 8 – 2k`	`= 2`
`2k`	`= 6`
`k`	`= 3`

`:.\ text(Co-efficient of)\ \ x^2`

`=\ ^8C_3 * (–1)^3*3^5 * 4^3`

`= -56 xx 243 xx 64`

`= -870\ 912`

Polynomials, EXT1 2011 HSC 2b

The function `f(x) = cos2x\ - x` has a zero near `x = 1/2`.

Use one application of Newton’s method to obtain another approximation to this zero. Give your answer correct to two decimal places. (3 marks)

Show Answers Only

`0.52\ \ \ text{(to 2 d.p.)}`

Show Worked Solution

`f(x)`	`= cos2x\ – x`
`f prime (x)`	`= -2 sin 2x\ – 1`
`f(0.5)`	`=cos (2xx0.5) – 0.5`
	`=0.0403…`
`f′(0.5)`	`=-2 sin (2 xx 0.5) – 1`
	`=-2.6829…`

`x_1`	`= x_0\ – (f(0.5))/(f prime (0.5))`
	`= 0.5\ – (0.0403…)/(-2.6829…)`
	`= 0.5\ – (-0.01502…)`
	`= 0.51502…`
	`= 0.52\ \ \ text{(to 2 d.p.)}`

Mechanics, EXT2* M1 2012 HSC 14b

A firework is fired from `O`, on level ground, with velocity `70` metres per second at an angle of inclination `theta`. The equations of motion of the firework are

`x = 70t cos theta\ \ \ \ `and`\ \ \ y = 70t sin theta\ – 4.9t^2`. (Do NOT prove this.)

The firework explodes when it reaches its maximum height.

Show that the firework explodes at a height of `250 sin^2 theta` metres. (2 marks)

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Show that the firework explodes at a horizontal distance of `250 sin 2 theta` metres from `O`. (1 mark)

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For best viewing, the firework must explode at a horizontal distance between 125 m and 180 m from `O`, and at least 150 m above the ground.

For what values of `theta` will this occur? (3 mark)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`66.97^@… <= theta <= 75^@\ text(for best viewing)`

Show Worked Solution

i. `text(Show max height is)\ 250 sin^2 theta\ text(metres)`

`text(Firework explodes at max height)`

`text(Find)\ \ t\ \ text(when)\ dot y = 0`

`y = 70t sin theta\ – 4.9t^2`

`dot y = 70 sin theta\ – 9.8t`

`text(When)\ \ dot y = 0`

`70 sin theta\ – 9.8t`	`= 0`
`9.8t`	`=70 sin theta`
`t`	`= 70/9.8 sin theta`

`:.\ y_max`	`= 70 * 70/9.8 sin theta * sin theta\ – 4.9 * (70^2)/(9.8^2) sin^2 theta`
	`= 500 sin^2 theta\ – 250 sin^2 theta`
	`= 250 sin^2 theta\ text(metres)\ \ \ \ \ text(… as required)`

ii. `x = 70t cos theta`

`text(At)\ \ t = 70/9.8 sin theta`

`x`	`= 70 * 70/9.8 sin theta cos theta`
	`= 250 * 2 sin theta cos theta`
	`= 250 sin 2 theta\ text(metres)\ \ \ \ \ text(… as required)`

iii. `text(Best viewing when)`

♦♦ Mean mark 28%
MARKER’S COMMENT: Many students ignored the findings in earlier parts and tried unsuccessfully to solve inequalities that still contained `t`.

`125 <= x <= 180\ \ text(and)\ \ y >= 150`

`text(S)text(ince)\ x = 250 sin 2 theta`

`125`	`<= 250 sin 2 theta <= 180`
`1/2`	`<= sin 2 theta <= 18/25`

`text(In the 1st quadrant)`

`30^@ <= 2 theta <= 46.054…^@`

`15^@ <= theta <= 23.0^@`

`text(In the 2nd quadrant)`

`133.94…^@`	`<= 2 theta <= 150^@`
`67.0^@`	`<= theta <= 75^@`

`text{Using part (i)}`

`text(When)\ theta = 23.0^@`

`y_(max)`	`= 250 xx sin^2 23.0^@`
	`~~ 38 < 150text(m)`

`=> text(“Highest” max height for)\ \ 15°<=theta<=23.0\ \ text(does not satisfy.)`

`text(When)\ theta = 67.0…^@`

`y_text(max)`	`= 250 xx sin^2 67.0^@`
	`~~ 212 > 150 text(m)`

`=> text(“Lowest” max height for)\ \ 67°<=theta<=75°\ \ text(satisfies.)`

`:.\ 67.0^@ <= theta <= 75^@\ text( for best viewing)`.

Plane Geometry, EXT1 2012 HSC 14a

The diagram shows a large semicircle with diameter `AB` and two smaller semicircles with diameters `AC` and `BC`, respectively, where `C` is a point on the diameter `AB`. The point `M` is the centre of the semicircle with diameter `AC`.

The line perpendicular to `AB` through `C` meets the largest semicircle at the point `D`. The points `S` and `T` are the intersections of the lines `AD` and `BD` with the smaller semicircles. The point `X` is the intersection of the lines `CD` and `ST`.

Copy or trace the diagram into your writing booklet.

Explain why `CTDS` is a rectangle. (1 mark)
Show that `Delta MXS` and `Delta MXC` are congruent. (2 marks)
Show that the line `ST` is a tangent to the semicircle with diameter `AC`. (1 mark)

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)

`/_SDT = 90°\ text{(angle in semi-circle)}`

♦ Mean mark 49%

`/_ ASC = 90°\ \ text{(angle in semi-circle)}`

`=> /_ CSD = 90°\ \ text{(} /_ ASD\ text{is a straight angle)}`

`text(Similarly)`

`/_CTB = /_CTD=90°`

`/_SCT = 90°\ \ text{(angle sum of quadrilateral}\ CTDS text{)}`

`text(S)text(ince all angles are right angles,)`

`CTDS\ text(is a rectangle)`

(ii) `ST\ text(and)\ DC\ text(are diagonals of rectangle)`

`text(S)text(ince they bisect and are equal)`

`=> XS = XC`

`SM = CM\ text{(radii)}`

`MX\ text(is common)`

`:.\ Delta MXS -= Delta MXC\ text{(SSS)}`

♦ Mean mark part (iii) 41%
STRATEGY: The congruency proof in part (ii) provided the critical information to answer this efficiently. Keep previous parts of questions front and centre of your mind when working on a solution.

(iii) `/_ XSM = /_ XCM = 90°`

`text{(corresponding angles of congruent triangles)}`

`text(S)text(ince)\ MS _|_ ST\ text(at circumference)`

`text(and)\ MS\ text(is a radius,)`

`=> ST\ text(is a tangent)`

` (10N – 7.5(4))/(6.8(84))`	`< 0.05`
`10N – 30`	`< 0.05 (571.2)`
`10N`	`< 28.56 + 30`
	`< 58.56`
`N`	`< 5.856`

`D(t)`	`= _\|_\ text(dist of)\ (t, t^2 + 3)\ text(from)\ 2x\ – y\ – 1 = 0`
`D`	`= \|(ax_1 + by_1 + c)/sqrt(a^2 + b^2)\|`
	`= \|(2(t)\ – 1 (t^2 + 3)\ – 1)/sqrt(2^2 + (-1)^2)\|`
	`= \|(2t\ – t^2\ – 3\ – 1)/sqrt5\|`
	`= \|(-(t^2\ – 2t + 4))/sqrt5\|`