Band 6 – Page 16

Probability, 2ADV S1 2007 MET1 6

Two events, `A` and `B`, from a given event space, are such that `P(A) = 1/5` and `P(B) = 1/3`.

Calculate `P(A′ ∩ B)` when `P(A ∩ B) = 1/8`. (1 mark)

--- 6 WORK AREA LINES (style=lined) ---
Calculate `P(A′ ∩ B)` when `A` and `B` are mutually exclusive events. (1 mark)

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

Show Answers Only

`5/24`
`1/3`

Show Worked Solution

i. `text(Sketch Venn diagram:)`

♦♦ Mean mark 31%.
MARKER’S COMMENTS: Students who drew a Venn diagram or Karnaugh map were the most successful.

`:.P(A′ ∩ B)`	`=P(B) – P(A ∩B)`
	`=1/3 – 1/8`
	`=5/24`

♦♦ Mean mark 23%.

ii.

`text(Mutually exclusive means)\ \ P(A ∩ B)=0,`

`:. P(A′ ∩ B) = 1/3`

Probability, 2ADV S1 2016 MET2 19 MC

Consider the discrete probability distribution with random variable `X` shown in the table below.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ -1\ \ \ & \ \ \ \ 0\ \ \ \ & \ \ \ \ b\ \ \ \ & \ \ \ \ 2b\ \ \ \ &\ \ \ \ 4\ \ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & a & b & b & 2b & 0.2 \\
\hline
\end{array}

The smallest and largest possible values of `text(E)(X)` are respectively

`-0.8 and 1`
`-0.8 and 1.6`
`0 and 2.4`
`0 and 1`

Show Answers Only

`D`

Show Worked Solution

`text(Smallest)\ E(X)\ \ text(occurs when)\ \ a=0.8,`

♦♦♦ Mean mark 15%.

`:.\ text(Smallest)\ E(X)`	`=0.8 xx -1 + 0.2 xx 4`
	`=0`

`text(Consider the value of)\ b,`

`text(Sum of probabilities) = 1`

`:. 0 <= 4b <= 0.8 -> 0 <= b <= 0.2`

`text(Largest)\ E(X)\ \ text(occurs when)\ \ a=0, and b=0.2,`

`:.\ text(Largest)\ E(X)`

`=0.2 xx 0 + 0.2 xx 0.2+(2xx0.2)xx(2xx0.2)+0.2 xx 4`

`=0.04 + 0.16 + 0.8`

`=1`

`=> D`

Measurement, STD2 M6 2011 HSC 24c

A ship sails 6 km from `A` to `B` on a bearing of 121°. It then sails 9 km to `C`. The
size of angle `ABC` is 114°.

What is the bearing of `C` from `B`? (1 mark)

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Find the distance `AC`. Give your answer correct to the nearest kilometre. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
What is the bearing of `A` from `C`? Give your answer correct to the nearest degree. (3 marks)

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

Show Answers Only

`055^@`
`13\ text(km)`
`261^@`

Show Worked Solution

STRATEGY: Important: Draw North-South parallel lines through major points to make the angle calculations easier!

`text(Let point)\ D\ text(be due North of point)\ B`

`/_ABD=180-121\ text{(cointerior with}\ \ /_A text{)}\ =59^@`

`/_DBC=114-59=55^@`

`:. text(Bearing of)\ \ C\ \ text(from)\ \ B\ \ text(is)\ 055^@`

ii. `text(Using cosine rule:)`

`AC^2`	`=AB^2+BC^2-2xxABxxBCxxcos/_ABC`
	`=6^2+9^2-2xx6xx9xxcos114^@`
	`=160.9275…`
`:.AC`	`=12.685…\ \ \ text{(Noting}\ AC>0 text{)}`
	`=13\ text(km)\ text{(nearest km)}`

iii. `text(Need to find)\ /_ACB\ \ \ text{(see diagram)}`

MARKER’S COMMENT: The best responses clearly showed what steps were taken with working on the diagram. Note that all North/South lines are parallel.

`cos/_ACB`	`=(AC^2+BC^2-AB^2)/(2xxACxxBC)`
	`=((12.685…)^2+9^2-6^2)/(2xx(12.685..)xx9)`
	`=0.9018…`
`/_ACB`	`=25.6^@\ text{(to 1 d.p.)}`

`text(From diagram,)`

`/_BCE=55^@\ text{(alternate angle,}\ DB\ text(||)\ CE text{)}`

`:.\ text(Bearing of)\ A\ text(from)\ C`

	`=180+55+25.6`
	`=260.6`
	`=261^@\ text{(nearest degree)}`

Measurement, STD2 M1 2017 HSC 29a*

A new 200-metre long dam is to be built.

The plan for the new dam shows evenly spaced cross-sectional areas.

Using the Trapezoidal rule, show that the volume of the dam is approximately 44 500 m³. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
It is known that the catchment area for this dam is 2 km².

Assuming no wastage, calculate how much rainfall is needed, to the nearest mm, to fill the dam. (2 marks)

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

Show Answers Only

`text(See Worked Solution)`
`22\ text{mm (nearest mm)}`

Show Worked Solution

`V`	`~~ 50/2[360 + 2(300 + 270 + 140) + 0]`
	`~~ 25(1780)`
	`~~ 44\ 500\ text(m³)`

ii. `text(Convert 2 km² → m²:)`

♦♦♦ Mean mark 11%.

`text(2 km²)`	`= 2000\ text(m × 1000 m)`
	`= 2\ 000\ 000\ text(m²)`

`text(Using)\ \ V=Ah\ \ text(where)\ \ h= text(rainfall):`

`44\ 500`	`= 2\ 000\ 000 xx h`
`:.h`	`= (44\ 500)/(2\ 000\ 000)`
	`= 0.02225…\ text(m)`
	`= 22.25…\ text(mm)`
	`= 22\ text{mm (nearest mm)}`

Probability, NAP-J2-34

Mick has a bag of marbles. His marbles are orange, white, blue and green.

Mick picks one marble from his bag.

Which of the following could be the probability that the marble he picks is green.

`13/3`	`1.52`	`3/13`	`6`

Show Answers Only

`3/13`

Show Worked Solution

`text(Any probability must be between 0 and 1.)`

`:.\ text(Only possibility is)\ 3/13`

Measurement, NAP-J2-32

Kim made these solid prisms out of identical cubes.

Which prism has the largest volume?

Show Answers Only

Show Worked Solution

`text{The prism with the largest volume (36 cubes) is:}`

Number and Algebra, NAP-J2-30

Madison uses the number sentence 15 × 12 = 180 to solve a problem.

Which of the following could be the problem?

	Madison buys 15 showbags. How much does each showbag cost?
	Madison spends $15 on 180 showbags. How much does she spend?
	Madison buys 15 showbags that cost $12 each. How many showbags does she buy?
	Madison buys 12 showbags that cost $15 each. How much does she spend?

Show Answers Only

`text(Madison buys 12 showbags that cost $15 each.)`

`text(How much does she spend?)`

Show Worked Solution

`text($15 per showbag × 12 showbags = $180)`

`:.\ text(Madison buys 12 showbags that cost $15 each.)`

`text(How much does she spend?)`

Number and Algebra, NAP-J2-28

Caroline has 56 match sticks.

She uses 6 match sticks to make two small triangles.

What is the largest number of small triangles like this that Caroline can make with her 56 match sticks?

`17`	`18`	`19`	`20`

Show Answers Only

`18`

Show Worked Solution

`text(3 match sticks are used for 1 triangle.)`

`text(Number of triangles)`	`= 56 ÷ 3`
	`= 18\ text(remainder 2)`

`:.\ text(18 triangles.)`

Measurement, NAP-J2-27

A bottle of milk can fill 4 glasses exactly.

Which of the following holds the most milk?

Show Answers Only

Show Worked Solution

`text(Converting each option to glasses)`

`text(Option 1: 3 × 4 = 12 glasses)`

`text(Option 2: 1 × 4 + 8 = 12 glasses)`

`text(Option 3: 2 × 4 + 1 = 9 glasses)`

`text(Option 4: 2 × 4 + 5 = 13 glasses)`

Number and Algebra, NAP-J2-25

Groucho, Harpo and Zeppo are weighing their grapes.

Harpo's grapes weigh more than Zeppo's, but less than Groucho's grapes.

Which of these could be the weight of Harpo's grapes?

0.11 kg	0.32 kg	0.6 kg	0.7 kg

Show Answers Only

`text(0.32 kg)`

Show Worked Solution

`text(0.32 kg is greater than 0.13 kg and less)`

`text(than 0.4 kg)`

Geometry, NAP-J2-23

Conrad followed these instructions to create a drawing:

Draw a triangle with 2 equal sides.
Use one of the sides of this shape to draw another shape with 5 equal sides.
Draw a line of symmetry through your completed drawing.

Which of these did he draw?

Show Answers Only

Show Worked Solution

Geometry, NAP-J2-22

Cleopatra uses balls and sticks to make a model of a triangular pyramid.

Each stick is the same length.

She decides to change her model into a cube.

How many more balls and sticks will Cleopatra need to make the cube?

	more balls
	more sticks

Show Answers Only

`text(4 more balls)`

`text(6 more sticks)`

Show Worked Solution

`text(A cube requires 8 balls and 12 sticks.)`

`:.\ text(Cleopatra needs 4 more balls and 6 more sticks.)`

Number and Algebra, NAP-J2-21

At lunch time, a hamburger shop sold 260 hamburgers.

At dinner time on the same day, the shop sold 487 hamburgers.

Which number sentence shows how many more hamburgers were sold at dinnertime than at lunchtime?

`+\ 487 = 260`

`487 - 260 =`

`- 487 = 260`

`260 + 487 = `

Show Answers Only

`487 – 260 =`

Show Worked Solution

`487 – 260 =`

Polynomials, EXT2 2018 HSC 16c

Let `alpha, beta, gamma` be the zeros of `p(x) = x^3 + px + q`, where `p` and `q` are real and `q != 0`.

Show that `(beta - gamma)^2 = alpha^2 + (4q)/alpha`. (2 marks)
By considering the constant term of a cubic equation with roots `(alpha - beta)^2, (beta - gamma)^2` and `(gamma - alpha)^2`, or otherwise, show that

`(alpha - beta)^2(beta - gamma)^2(gamma - alpha)^2 = −(27q^2 + 4p^3)`. (3marks)
Deduce that if `27q^2 + 4p^3 < 0`, then the equation `p(x) = 0` has 3 distinct real roots. (2 marks)

Show Answers Only

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

Show Worked Solution

i. `p(x) = x^3 + px + q`

`alpha + beta + gamma = 0\ \ =>\ \ alpha = −beta – gamma`

♦ Mean mark 34%.
COMMENT: Solving from the RHS makes this proof much simpler.

`alphabetagamma = −q`

`alphabeta + betagamma + gammaalpha = p`

`text(Show)\ \ (beta – gamma)^2 = alpha^2 + (4q)/alpha`

`text(RHS)`	`= (−beta – gamma)^2 + (4(−alphabetagamma))/alpha`
	`= beta^2 + 2betagamma + gamma^2 – 4betagamma`
	`= beta^2 – 2betagamma + gamma^2`
	`= (beta – gamma)^2`
	`=\ text(LHS)`

ii.	`(beta – gamma)^2`	`= alpha^2 – (4q)/alpha`
		`= (alpha^3 – 4q)/alpha`
		`= ((alpha^3 + px + q) + 3q – palpha)/alpha`
		`= (3q – palpha)/alpha\ \ \ \ (alpha^3 + palpha + q = 0)`

`x`	`= (3q – palpha)/alpha`
`alphax – palpha`	`= 3q`
`alpha`	`= (3q)/(x – p)`

`text(Substituting back into)\ p(x)`

♦♦♦ Mean mark 6%
COMMENT: This question produced just just the 4th lowest mean mark in a tough 2018 paper.

`((3q)/((x + p)))^3 + p((3q)/((x + p))) + q = 0`

`=> 27q^3 + 3pq(x + p)^2 + q(x + p)^3 = 0`

`text(Co-efficient of)\ x^3 = q`

`text(Constant term)`	`= 27q^3 + 3p^3q + qp^3`
	`= 27q^3 + 4p^3q`

`:. (alpha – beta)^2(beta – gamma)^2(gamma – alpha)^2`

`= (−(27q^3 + 4p^3q))/q`

`= −(27q^2 + 4p^3)`

iii. `text(If)\ \ 27q^2 + 4p^3 < 0`

♦♦♦ Mean mark 5%.

`=> (alpha – beta)^2(beta – gamma)^2(gamma – alpha)^2 > 0`

`text(If roots are not distinct, then)\ \ alpha = beta`

`=> (alpha – beta)^2(beta – gamma)^2(gamma – alpha)^2 = 0`

`:.\ text(By contradiction, roots are distinct.)`

`p(x)\ text(is a cubic.)`

`text(If roots are not all real, conjugate root theorem)`

`text(states there must be 2 complex roots.)`

`text(Let roots be:) \ \ alpha, baralpha\ \ text{(complex) and}\ \ beta\ \ text{(real)}`

`(alpha – beta)^2(beta – baralpha)^2(baralpha – alpha)^2`

`= [alphabeta – alphabaralpha – beta^2 + baralphabeta]^2(−2text{Im})^2`

`= [−alphabaralpha – beta^2 + beta(alpha + baralpha)]^2(−2text{Im})^2`

`=\ (text{real})^2(text{Im})^2`

`<0`

`:. text(By contradiction,)\ p(x)\ text(cannot have 2 complex roots.)`

`:. p(x) = 0\ \ text(has 3 real distinct roots.)`

Harder Ext1 Topics, EXT2 2018 HSC 16b

In `DeltaABC`, point `D` is chosen on side `AB` and point `E` is chosen on side `AC` so that `DE` is parallel to `BC` and `(BC)/(DE) = sqrt2`.

The process is repeated two more times. Point `F` is chosen on `BC` and point `G` on `BA` so that `FG` is parallel to `CA` and `(CA)/(FG) = sqrt2`.

Point `H` is chosen on side `CB` and `I` on `CA` so that `HI` is parallel to `BA` and `(BA)/(HI) = sqrt2`.

The segments `FG` and `HI` intersect at `X`, `DE` and `HI` intersect at `Y`, and `DE` and `FG` intersect at `Z`.

Prove that `DY = ZE`. (3 marks)
Find the exact value of the ratio `(YZ)/(BC)`. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`(3sqrt2)/2 – 2`

Show Worked Solution

i. `DeltaABC\ text{|||}\ DeltaGBF\ \ \ text{(AAA)}`

♦♦♦ Mean mark 5%.

`DeltaABC\ text{|||}\ DeltaIHC\ \ \ text{(AAA)}`

`(CA)/(FG) = (BA)/(HI) = sqrt2\ \ (text(given))`

`text(S)text(ince ratios to)\ ΔABC\ text(of similar triangles are the same)`

`=> DeltaGBF ≅ DeltaIHC`

`=> DeltaIHC\ text(is the translation to the right of)\ DeltaGBF`

`text(along)\ BC.`

`=>DZ=YE`

`DY+YZ`	`=YZ+ZE`
`:. DY`	`= ZE`

ii.	`(YZ)/(BC)`	`= (DE – DY – ZE)/(BC)`
		`= (DE)/(BC) – 2 xx (DY)/(BC)\ \ (text{from part (i)})`
		`= 1/sqrt2 – 2 xx (DY)/(BC)`

`text(Find)\ \ (DY)/(BC):`

♦♦♦ Mean mark 3%.
COMMENT: As well reported in the papers at the time, this question, in both parts, proved to be a beast.

`text(Using)\ \ DeltaABC\ text{|||}\ DeltaIHC`

`(HC)/(BC)`	`= 1/sqrt2`
`(BC – BH)/(BC)`	`= 1/sqrt2`
`(BC – DY)/(BC)`	`= 1/sqrt2\ \ \ \ (BH = DY.\ text(Opposite sides)` `text(of parallelogram)\ DYHBtext{)}`
`1 – (DY)/(BC)`	`= 1/sqrt2`
`(DY)/(BC)`	`= 1 – 1/sqrt2`

`:. (YZ)/(BC)`	`= 1/sqrt2 – 2 xx (1 – 1/sqrt2)`
	`= 3/sqrt2 – 2`
	`= (3sqrt2)/2 – 2`

Proof, EXT2 P1 2018 HSC 15c

Let `n` be a positive integer and let `x` be a positive real number.

Show that `x^n - 1 - n(x - 1) = (x - 1)(1 + x + x^2 + … + x^(n - 1) - n)`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Hence, show that `x^n >= 1 + n(x - 1)`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Deduce that for positive real numbers `a` and `b`,

`a^nb^(1-n)>=na + (1-n)b` (2 marks)

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

Show Answers Only

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

Show Worked Solution

i.	`text(RHS)`	`= (x – 1)underbrace{(1 + x + x^2 + … + x^(n – 1) – n)}_{text(GP where)\ \ a = 1,\ r = x}`
		`= (x – 1) ((1(x^n – 1))/(x – 1) – n)`
		`= x^n – 1 – n(x – 1)`
		`=\ text(LHS)`

ii. `text(Let)\ P(x) = (x – 1)(1 + x + x^2 + … + x^(n – 1) – n)`

♦♦♦ Mean mark 11%.

`text(If)\ \ x = 1, P(x) = 0`

`text(If)\ \ 0 < x < 1, \ (x – 1) < 0, \ (1 + x + x^2 + … + x^(n – 1) – n) < 0`

`=> P(x) > 0`

`text(If)\ \ x > 1, \ (x – 1) > 0, \ (1 + x + … + x^(n – 1) – n) > 0`

`=> P(x) > 0`

`x^n – 1 – n(x – 1) >= 0`

`:. x^n >= 1 + n(x – 1)`

iii. `x^n >= 1 + n(x – 1)\ \ text(for)\ \ x ∈ R^+`

♦♦ Mean mark 23%.

`text(S)text(ince)\ a,b ∈ R^+`

`(a/b)^n`	`>= 1 + n(a/b – 1)`
`(a^n)/(b^n) xx b`	`>= b + na – nb,\ \ \ \ (b > 0)`
`a^n b^(1 – n)`	`>= na + (1 – n)b`

Probability, NAP-K2-28

Frank rolls a standard dice once.

Which of the following results is most likely?

	Frank rolls a 6.
	Frank rolls a number less than 3.
	Frank rolls an even number.
	Frank rolls a 5 or 6.

Show Answers Only

`text(Frank rolls an even number)`

Show Worked Solution

`text(Consider the chance of each option:)`

`text{Pr (rolling a 6)} = 1/6`

`text{Pr (rolling a number less than 3)} = 2/6`

`text{Pr (rolling an even number)} = 3/6`

`text{Pr (rolling a 5 or 6)} = 2/6`

`:.\ text(Rolling an even number is the most likely.)`

Measurement, NAP-K2-26

Alvin is making red cordial.

It is made by mixing water with red concentrate.

Alvin adds 1 litre (L) of water to 75 millilitres (mL) of red concentrate.

How much cordial did Alvin make?

`175\ text(mL)`	`751\ text(mL)`	`1075\ text(mL)`	`10\ 075\ text(mL)`

Show Answers Only

`1075\ text(mL)`

Show Worked Solution

`1\ text(L) = 1000\ text(mL)`

`:.\ text(Total cordial)`	`= 1000 + 75`
	`= 1075\ text(mL)`

Statistics, NAP-K2-23 SA

The picture graph shows how many tonnes of fertiliser are needed for 4 gardens.

How many more tonnes of fertiliser does Garden 1 need than Garden 4?

tonnes

Show Answers Only

`10\ text(tonnes)`

Show Worked Solution

`text(Garden 1 tonnes) = 4 + 4 + 3 = 11`

`text(Garden 4 tonnes) = 1`

`:.\ text(Garden 4 needs 10 more tonnes.)`

Number and Algebra, NAP-K2-22 SA

Sanjeev collected football cards. He had collected 35 cards.

At school, Sanjeev and his friends put all their football cards together.

There was a total of 305 cards.

How many football cards, in total, had Sanjeev's friends collected?

Show Answers Only

`270`

Show Worked Solution

`text(Cards collected by Sanjeev’s friends)`

`= 305 – 35`

`= 300 – 30`

`= 270`

Number and Algebra, NAP-K2-20

On Monday, Jeremy went to the doctor and was given 24 tablets.

Jeremy was to take 5 tablets each day starting from Monday.

On which day did Jeremy take the last tablet?

Thursday	Friday	Saturday	Sunday

Show Answers Only

`text(Friday)`

Show Worked Solution

`text(Number of days)`	`= 24 ÷ 5`
	`= 4\ text(remainder 4)`

`=>\ text(Jeremy has his last tablet on the 5th day.)`

`:.\ text(Friday)`

Statistics, NAP-K2-19

On a school camp, each child chose their meal for dinner.

The table shows how many children chose each meal.

Select all the statements that are true.

	More girls than boys chose spaghetti.
	In total 70 children chose pizza.
	Less than half the girls chose pizza.
	In total, there are the same number of girls on the trip as boys.

Show Answers Only

`text(The true statements are:)`

`text(In total 70 children chose pizza)`

`text(In total, there are the same number)`

`text(of girls on the trip as boys)`

Show Worked Solution

`text(The true statements are:)`

`=>\ text(In total 70 children chose pizza.)`

`=>\ text(In total, there are the same number)`

`text(of girls on the trip as boys.)`

Number and Algebra, NAP-K2-18

Greg plays golf.

During one round, he found 11 golf balls and lost 4.

At the end of the round, Greg had a total of 16 golf balls.

How many golf balls did Greg start with?

1	7	9	23

Show Answers Only

`9`

Show Worked Solution

`text(By trial and error:)`

`text(Consider starting with 9)`

`9 + 11 – 4 = 16`

`:.\ text(Greg starts with 9 golf balls.)`

Geometry, NAP-K2-17

Christy uses some blocks to make a rectangular prism and cube.

How many blocks does Christy use altogether?

17	34	43	47

Show Answers Only

`43`

Show Worked Solution

`text(Blocks in rectangular prism)\ = 2 xx 8 = 16`

`text(Blocks in cube)\ = 3 xx 9 = 27`

`:.\ text(Total blocks)`	`= 16 + 27`
	`= 43`

Harder Ext1 Topics, EXT2 2018 HSC 14d

Three people, `A`, `B` and `C`, play a series of n games, where `n ≥ 2`. In each of the games there is one winner and each of the players is equally likely to win.

What is the probability that player `A` wins every game? (1 mark)
Show that the probability that `A` and `B` win at least one game each but `C` never wins, is

`(2/3)^n - 2(1/3)^n`. (1 mark)
Show that the probability that each player wins at least one game is

`(3^(n - 1) - 2^n + 1)/(3^(n - 1))`. (2 marks)

Show Answers Only

`(1/3)^n`
`text(See Worked Solutions)`
`text(See Worked Solutions)`

Show Worked Solution

i. `text{Pr (A wins every game)} = (1/3)^n`

ii. `text{Pr (C never wins)} = (2/3)^n`

♦ Mean mark 36%.

`text(If C never wins, only 2 scenarios occur where A or B)`

`text(don’t win at least 1 game)`

`→\ text(A wins all or B wins all.)`

`:.\ text{Pr (No C, A and B win at least 1 game)}`

`=\ text{Pr (No C) – Pr (A wins all) – Pr (B wins all)}`

`= (2/3)^n – 2(1/3)^n`

iii. `=>\ text{A, B or C cannot win all games (part(i)).}`

♦♦♦ Mean mark 19%.

`=>\ text(A cannot lose all games, with B and C winning)`

`text{at least 1 each (part (ii)). Similarly for each player.}`

`:.\ text{Pr (each player wins at least 1 game)}`

`= 1 – 3(1/3)^n – 3[(2/3)^n – 2(1/3)^n]`

`= (3^n – 3 – 3 · 2^n + 6)/(3^n)`

`= (3^n – 3 · 2^n + 3)/(3^n)`

`= (3^(n – 1) – 2^n + 1)/(3^(n – 1))\ \ …\ text(as required.)`

Number and Algebra, NAP-K2-11

Jarryd buys 3 water bottles for $1.50 each.

He pays for these water bottles with a $10 note.

How much change should Jarryd receive?

`$1.50`	`$4.50`	`$5.00`	`$5.50`

Show Answers Only

`$5.50`

Show Worked Solution

`text(Change)`	`= $10 – (3 xx $1.50)`
	`= $10 – $4.50`
	`= $5.50`

Number and Algebra, NAP-K2-07 SA

Write a number in the box to make this number sentence true.

`6 + 8 = 10 +`

Show Answers Only

`4`

Show Worked Solution

`6 + 8`	`= 14`
	`= 10 + 4`

Financial Maths, 2ADV M1 2018 HSC 16c

Kara deposits an amount of $300 000 into an account which pays compound interest of 4% per annum, added to the account at the end of each year. Immediately after the interest is added, Kara makes a withdrawal for expenses for the coming year. The first withdrawal is `$P`. Each subsequent withdrawal is 5% greater than the previous one.

Let `$A_n` be the amount in the account after the `n`th withdrawal.

Show that `A_2 = 300\ 000(1.04)^2 - P[(1.04) + (1.05)]`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Show that `A_3 = 300\ 000 (1.04)^3 - P[(1.04)^2 + (1.04)(1.05) + (1.05)^2]`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Show that there will be money in the account when

`qquad (105/104)^n < 1 + 3000/P` (3 marks)

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

Show Answers Only

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

Show Worked Solution

♦ Mean mark 47%.

i.	`A_1`	`= 300\ 000 (1.04) – P`
	`A_2`	`= [300\ 000 (1.04) – P](1.04) – P(1.05)`
		`= 300\ 000 (1.04)^2 – P(1.04) – P(1.05)`
		`= 300\ 000 (1.04)^2 – P[1.04 + 1.05]`

♦ Mean mark part (ii) 30%.

ii.	`A_3`	`= [300\ 000 (1.04)^2 – P(1.04 + 1.05)](1.04) – P(1.05)^2`
		`= 300\ 000 (1.04)^3 – P(1.04)^2 – P(1.04)(1.05) – P(1.05)^3`
		`= 300\ 000 (1.04)^3 – P[(1.04)^2 + P(1.04)(1.05) + (1.05)^2]`

iii.	`A_4`	`= 300\ 000 (1.04)^4 – P[(1.04)^3 + (1.04)^2(1.05) + … + (1.05)^3]`
		`vdots`
	`A_n`	`= 300\ 000 (1.04)^n`
		`- P underbrace{[(1.04)^(n-1) + (1.04)^(n-2) (1.05) + … + (1.04)(1.05)^(n-2) + (1.05)^(n-1)]}_{text(GP),\ a = (1.04)^(n-1),\ r = 1.05/1.04}`

`text(Money in account when)\ \ A_n > 0:`

♦♦♦ Mean mark part (iii) 8%.

`(300\ 000(1.04)^n)/P`	`> (1.04)^(n-1)[((1.05/1.04)^n-1)/((1.05/1.04)-1)]`
`(300\ 000)/P`	`> 1/1.04 [((1.05/1.04)^n-1)/((1.05/1.04)-1)]`
`(300\ 000)/P`	`> ((1.05/1.04)^n – 1)/(1.05 – 1.04)`
`3000/P`	`> (1.05/1.04)^n – 1`
`3000/P + 1`	`> (105/104)^n qquad (text{s}text{ince}\ \ (1.05/1.04)^n = (105/104)^n)`

Probability, 2ADV S1 2018 HSC 16b

A game involves rolling two six-sided dice, followed by rolling a third six-sided die. To win the game, the number rolled on the third die must lie between the two numbers rolled previously. For example, if the first two dice show 1 and 4, the game can only be won by rolling a 2 or 3 with the third die.

What is the probability that a player has no chance of winning before rolling the third die? (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
What is the probability that a player wins the game? (2 marks)

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

Show Answers Only

`4/9`
`5/27`

Show Worked Solution

i. `text(Construct a sample space of the number)`

♦ Mean mark 40%.
COMMENT: Constructing the full sample space is a critical step here..

`text(of possible winning rolls:)`

`text{P(no chance)}`	`= text(number of pairs with no gap)/text(total possibilities)`
	`= 16/36`
	`= 4/9`

ii. `text(The sample space in the table shows:)`

♦♦♦ Mean mark 7%.

`text(→ 8 combinations leave a gap for a single winning number,)`

`text(→ 6 combinations leave a gap for two winning numbers,)`

`vdots`

`:.\ text{P(winning)}`	`= 1/36 [8 xx 1/6 + 6 xx 2/6 + 4 xx 3/6 + 2 xx 4/6]`
	`= 1/36 (8/6 + 12/6 + 12/6 + 8/6)`
	`= 1/36 (40/6)`
	`= 5/27`

Quadratic, 2UA 2018 HSC 8 MC

A radio telescope has a parabolic dish. The width of the opening is 24 m and the distance along the axis from the vertex to the opening is 4 m, as shown in the diagram.

What is the focal length of the parabola?

(A) `1/6\ text(m)`

(B) `1/3\ text(m)`

(C) `6\ text(m)`

(D) `9\ text(m)`

Show Answers Only

`D`

Show Worked Solution

`text(Redraw the parabola on a number plane.)`

♦♦♦ Mean mark 16%.

`text(Substitute)\ \ (12, 4)\ \ text(into)\ \ x = 4ay:`

`12^2`	`= 4 xx a xx 4`
`:. a`	`= 144/16`
	`= 9`

`=> D`

Plane Geometry, EXT1 2018 HSC 14c

In triangle `ABC, BC` is perpendicular to `AC`. Side `BC` has length `a`, side `AC` has length `b` and side `AB` has length `c`. A quadrant of a circle of radius `x`, centered at `C`, is constructed. The arc meets side `BC` at `E`. It touches the side `AB` at `D`, and meets side `AC` at `F`. The interval `CD` is perpendicular to `AB`.

Show that `Delta ABC` and `Delta ACD` are similar. (1 mark)
Show that

`qquad x = (ab)/c`. (1 mark)
From `F`, a line perpendicular to `AC` is drawn to meet `AB` at `G`, forming the right-angled triangle `GFA`. A new quadrant is constructed in triangle `GFA` touching side `AB` at `H`. The process is then repeated indefinitely.
Show that the limiting sum of the areas of all the quadrants is

`qquad (pi ab^2)/(4(2c - a)).` (4 marks)
Hence, or otherwise, show that

`qquad pi/2 < (2c - a)/b`. (1 mark)

Show Answers Only

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

Show Worked Solution

(i) `text(In)\ triangle ABC,`

`/_BCA = 90^@\ \ \ (BC _|_ AC)`

`/_ BCA = /_ ADC\ \ text{(right-angles)}`

`/_ BAC = /_ DAC\ \ text{(common)}`

`:. Delta ABC\ text(|||)\ Delta ACD\ \ text{(equiangular)}`

(ii)	`(CD)/(BC)`	`= (AC)/(AB)`	`text{(corresponding sides of}` `text{similar triangles)}`
	`x/a`	`= b/c`
	`:. x`	`= (ab)/c`

(iii) `text(Area of)\ Q_1 = 1/4 pi x^2`

♦♦♦ Mean mark part (iii) 19%.

`text(Area of)\ Q_2 => text(find)\ x_1`

`text(Consider)\ Delta ACD and Delta AFH`

`/_ADC = /_AHF\ \ text{(right angles)}`

`/_ CAD = /_FAH\ \ text{(common)}`

`:. Delta ACD\ text(|||)\ Delta AFH\ \ text{(equiangular)}`

`(FH)/(CD)`	`= (AF)/(AC)`	`text{(corresponding sides}` `text{of similar triangles)}`
`x_1/x`	`= (AC – CF)/(AC)`
	`= (b – x)/b`
	`= (b – (ab)/c)/b`	`text{(using part (ii))}`
	`= (cb – ab)/(bc)`
	`= (c – a)/c`
`:. x_1`	`= x((c – a)/c)`

`=> x_2\ text(will be shorter again by the same ratio)`

`x_2`	`= x_1 ((c – a)/c)`
	`= x((c – a)/c)^2`
	`vdots`
`x_n`	`= x((c – a)/c)^n`

`text(Limiting sum of quadrant areas)`

`= Q_1 + Q_2 + … + Q_n`

`= 1/4 pi x^2 + 1/4 {:pi x_1:}^ 2 + … + 1/4 {:pi x_n:}^2`

`= 1/4 pi x^2 + 1/4 pi x^2 ((c – a)/c)^2 + … + 1/4 pi x^2 ((c – a)/c)^(2n)`

`= 1/4 pi x^2 underbrace{[1 + ((c – a)/c)^2 + … + ((c – a)/c)^(2n)]}_{text(GP with)\ a = 1, \ r = ((c-a)/c)^2`

`= 1/4 pi x^2 [1/(1 – ((c-a)/c)^2)]`

`= 1/4 pi ((ab)/c)^2 (c^2/(c^2 – (c – a)^2))`

`= pi/4 xx (a^2 b^2)/(c^2 – c^2 + 2ac – a^2)`

`= pi/4 xx (a^2 b^2)/(a(2c – a))`

`= (pi ab^2)/(4(2c – a))\ \ \ text(… as required)`

(iv) `text(S)text{ince the limiting sum all the quadrants (from part (iii))}`

♦♦ Mean mark part (iv) 21%.

`text(is less than the area of)\ \ Delta ABC:`

`(pi ab^2)/(4(2c – a))`	`< (ab)/2`
`(pi b)/(4(2c – a))`	`< 1/2`
`pi/4`	`< (2c – a)/(2b)`
`pi/2`	`< (2c – a)/b\ \ \ text(… as required)`

Networks, STD2 N3 2012 FUR1 8 MC

Eight activities, `A, B, C, D, E, F, G` and `H`, must be completed for a project.

The network above shows these activities and their usual duration in hours.

The duration of each activity can be reduced by one hour.

To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is

A. `1`

B. `2`

C. `3`

D. `4`

Show Answers Only

`C`

Show Worked Solution

`text(Two critical paths)\ AFH\ text(and)\ BCFH => 18\ text(hours)`

♦♦ Mean mark 30%.

`text(Paths)\ AEG\ text(and)\ BCEG => 17\ text(hours)`

`text(Reducing)\ A\ text(and)\ B\ text(by 1 hour each reduces each)`

`text(path above by 1 hour.)`

`text(Need to reduce)\ AFH\ text(and)\ BCFH\ text(by 1 hour to 16 hours.)`

`=>\ text(Reducing either)\ F\ text(or)\ H\ text(by 1 hour brings the)`

`text(critical path down to 16 hours.)`

`rArr C`

Networks, STD2 N3 2007 FUR2 4

A community centre is to be built on the new housing estate.

Nine activities have been identified for this building project.

The directed network below shows the activities and their completion times in weeks.

Determine the minimum time, in weeks, to complete this project. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Determine the float time, in weeks, for activity `D`. (2 marks)

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

The builders of the community centre are able to speed up the project.

Some of the activities can be reduced in time at an additional cost.

The activities that can be reduced in time are `A`, `C`, `E`, `F` and `G`.

Which of these activities, if reduced in time individually, would not result in an earlier completion of the project? (1 mark)

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

The owner of the estate is prepared to pay the additional cost to achieve early completion.

The cost of reducing the time of each activity is $5000 per week.

The maximum reduction in time for each one of the five activities, `A`, `C`, `E`, `F`, `G`, is `2` weeks.

Determine the minimum time, in weeks, for the project to be completed now that certain activities can be reduced in time. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Determine the minimum additional cost of completing the project in this reduced time. (1 mark)

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

Show Answers Only

`19\ text(weeks)`
`5\ text(weeks)`
`A, E, G`
`text(15 weeks)`
`$25\ 000`

Show Worked Solution

a. `text(Scanning forwards and backwards:)`

`BCFHI\ \ text(is the critical path.)`

♦ Mean mark of all parts (combined) 40%.

`:.\ text(Minimum time)`	`= 4 + 3 + 4 + 2 + 6`
	`= 19\ text(weeks)`

b.	`text(EST of)\ D`	`= 4`
	`text(LST of)\ D`	`= 9`

`:.\ text(Float time of)\ D`	`= 9 – 4`
	`= 5\ text(weeks)`

c. `A, E,\ text(and)\ G\ text(are not currently on the critical path,)`

`text(therefore reducing their time will not result in an)`

`text(earlier completion time.)`

d. `text(Reduce)\ C\ text(and)\ F\ text(by 2 weeks each.)`

`text(However, a new critical path created:)\ BEHI\ \ text{(16 weeks)}`

`:.\ text(Also reduce)\ E\ text(by 1 week.)`

`:.\ text(Minimum completion time = 15 weeks)`

e.	`text(Additional cost)`	`= 5 xx $5000`
		`= $25\ 000`

Networks, STD2 N3 2009 FUR2 4

A walkway is to be built across the lake.

Eleven activities must be completed for this building project.

The directed network below shows the activities and their completion times in weeks.

What is the earliest start time for activity E? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Write down the critical path for this project. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
The project supervisor correctly writes down the float time for each activity that can be delayed and makes a list of these times.

Determine the longest float time, in weeks, on the supervisor’s list. (1 mark)

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

A twelfth activity, L, with duration three weeks, is to be added without altering the critical path.

Activity L has an earliest start time of four weeks and a latest start time of five weeks.

Draw in activity L on the network diagram above. (1 mark)
Activity L starts, but then takes four weeks longer than originally planned.

Determine the total overall time, in weeks, for the completion of this building project. (1 mark)

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

Show Answers Only

`7`
`BDFGIK`
`H\ text(or)\ J\ text(can be delayed for)`
`text(a maximum of 3 weeks.)`
`text(25 weeks)`

Show Worked Solution

a. `7\ text(weeks)`

♦ Mean mark of all parts (combined): 44%.

b. `text(Scanning forwards and backwards)`

`text(Critical Path is)\ BDFGIK`

c. `H\ text(or)\ J\ text(can be delayed for a maximum)`

`text(of 3 weeks.)`

e. `text(The new critical path is)\ BLEGIK.`

`=>\ text(Activity)\ L\ text(now takes 7 weeks.)`

`:.\ text(Time for completion)`

`= 4 + 7 + 1 + 5 + 2 + 6`

`= 25\ text(weeks)`

Networks, STD2 N3 2010 FUR1 8 MC

A project has 12 activities. The network below gives the time (in hours) that it takes to complete each activity.

The critical path for this project is

A. `ADGK`

B. `ADGIL`

C. `BHJL`

D. `CEGIL`

Show Answers Only

`D`

Show Worked Solution

`text(Scanning forward:)`

♦ Mean mark 41%.

`text(Critical path is)\ \ CEGIL`

`=> D`

Networks, STD2 N3 2006 FUR1 9 MC

The network below shows the activities and their completion times (in hours) that are needed to complete a project.

The project is to be optimised by reducing the completion time of one activity only.

This will reduce the completion time of the project by a maximum of

A. 1 hour

B. 3 hours

C. 4 hours

D. 5 hours

Show Answers Only

`C`

Show Worked Solution

`text(Scanning forward:)`

`text(Critical path:)`

♦♦♦ Mean mark 17%.
MARKER’S COMMENT: When choosing an activity to crash, take care that a new critical path is not created.

`=> BDCEHJ\ text{(19 hours)}`

`text(Other routes not through)\ B,`

`ACEHJ\ text{(15 hours),}\ AFJ\ text{(14 hours)}`

`:.\ text(Activity)\ B\ text(could be reduced by 4 hours without)`

`text(a new critical path emerging.)`

`rArr C`

Networks, FUR2 2007 VCE 3

As an attraction for young children, a miniature railway runs throughout a new housing estate.

The trains travel through stations that are represented by nodes on the directed network diagram below.

The number of seats available for children, between each pair of stations, is indicated beside the corresponding edge.

Cut 1, through the network, is shown in the diagram above.

Determine the capacity of Cut 1. (1 mark)
Determine the maximum number of seats available for children for a journey that begins at the West Terminal and ends at the East Terminal. (1 mark)

On one particular train, 10 children set out from the West Terminal.

No new passengers board the train on the journey to the East Terminal.

Determine the maximum number of children who can arrive at the East Terminal on this train. (1 mark)

Show Answers Only

`43`
`22`
`7`

Show Worked Solution

a. `text(The capacity of Cut 1)`

♦♦ Mean mark for all parts (combined) was 33%.
MARKER’S COMMENT: A common error was counting the edge with “10” in the reverse direction (in part a).

`=14 + 8 + 13 + 8`

`= 43`

`text(Maximum seats)`	`=\ text(minimum cut)`
	`= 6 + 7 + 9`
	`= 22`

c. `text{The path (edge weights) of the train setting out with}`

`text(10 children starts with: 11 → 13.)`

`text(At the next station, a maximum of 7 seats are available)`

`text(which remain until the East Terminal.)`

`:.\ text(Maximum number of children arriving is 7.)`

Networks, FUR2 2013 VCE 3

The rangers at the wildlife park restrict access to the walking tracks through areas where the animals breed.

The edges on the directed network diagram below represent one-way tracks through the breeding areas. The direction of travel on each track is shown by an arrow. The numbers on the edges indicate the maximum number of people who are permitted to walk along each track each day.

Starting at `A`, how many people, in total, are permitted to walk to `D` each day? (1 mark)

One day, all the available walking tracks will be used by students on a school excursion.

The students will start at `A` and walk in four separate groups to `D`.

Students must remain in the same groups throughout the walk.

1. Group 1 will have 17 students. This is the maximum group size that can walk together from `A` to `D`.
  Write down the path that group 1 will take. (1 mark)
2. Groups 2, 3 and 4 will each take different paths from `A` to `D`.
  Complete the six missing entries shaded in the table below. (2 marks)

Show Answers Only

`37`
1. `A B E C D`
2. `text{One possible solution is:}`

Show Worked Solution

a.	`text(Maximum flow)`	`=\ text(Minimum cut through)\ CD and ED`
		`= 24 + 13`
		`= 37`

♦ Mean mark of all parts (combined) was 41%.

`:.\ text(A maximum of 37 people can walk)`

`text(to)\ D\ text(from)\ A.`

b.i. `A B E C D`

b.ii. `text(One solution using the second possible largest)`

`text(group of 11 students and two groups from the)`

`text(remaining 9 students is:)`

Networks, FUR1 2014 VCE 9 MC

A network of tracks connects two car parks in a festival venue to the exit, as shown in the directed graph below.

The arrows show the direction that cars can travel along each of the tracks and the numbers show each track’s capacity in cars per minute.

Four cuts are drawn on the diagram.

The maximum number of cars per minute that will reach the exit is given by the capacity of

A. Cut A

B. Cut B

C. Cut C

D. Cut D

Show Answers Only

`text(Cut D is the minimum cut / max flow)`

`=> D`

Show Worked Solution

`text(Cut A and B don’t separate both)`

♦♦♦ Mean mark 24%.
COMMENT: Note that the “4” is not included in Cut D as it is flowing in the opposite direction.

`text(car parks from the exit.)`

`text{Cut D has the minimum cut / maximum}`

`text{flow (18) of the two cuts remaining.}`

`=> D`

Networks, FUR1 2012 VCE 7 MC

Vehicles from a town can drive onto a freeway along a network of one-way and two-way roads, as shown in the network diagram below.

The numbers indicate the maximum number of vehicles per hour that can travel along each road in this network. The arrows represent the permitted direction of travel.

One of the four dotted lines shown on the diagram is the minimum cut for this network.

The maximum number of vehicles per hour that can travel through this network from the town onto the freeway is

A. `330`

B. `350`

C. `370`

D. `390`

Show Answers Only

`B`

Show Worked Solution

`text(Consider each “minimum cut” line,)`

♦♦♦ Mean mark 23%.

`text(Line 1: doesn’t seperate the town and freeway)`

`text(Line 2: 240 + 110 = 350)`

`text(Line 3: 240 + 60 + 90 = 390)`

`text(Line 4: 280 + 90 = 370)`

`:.\ text(Line 2 gives the maximum flow)`

`rArr B`

Networks, STD2 N2 2009 FUR1 8 MC

An undirected connected graph has five vertices.

Three of these vertices are of even degree and two of these vertices are of odd degree.

One extra edge is added. It joins two of the existing vertices.

In the resulting graph, it is not possible to have five vertices that are

A. all of even degree.

B. all of equal degree.

C. one of even degree and four of odd degree.

D. four of even degree and one of odd degree.

Show Answers Only

`D`

Show Worked Solution

`text(Consider an example of the graph)`

♦♦♦ Mean mark 25%.

`text{described (below):}`

`A\ text(is possible – join)\ V\ text(and)\ Z`

`B\ text(is possible – join)\ V\ text(and)\ Z`

`C\ text(is possible – join)\ W\ text(and)\ Y`

`D\ text(is NOT possible)`

`=> D`

Number, NAP-K3-NC02 SA

Hendrix is driving from Bundaberg to Caloundra, a distance of 306 kilometres.

When Hendrix gets to the Sunshine Coast, he has 36 kilometres left.

What distance has Hendrix travelled when he gets to the Sunshine Coast?

kilometres

Show Answers Only

`270`

Show Worked Solution

`text(Distance travelled)`	`= 306 – 36`
	`= 270\ text(kilometres)`

Number, NAP-K3-NC01

Jazz buys 4 avocados for $1.20 each.

He pays for the avocados with a $5 note.

How much change should Jazz receive?

`$0.20`	`$0.30`	`$3.80`	`$4.80`

Show Answers Only

`$0.20`

Show Worked Solution

`text(Change) = 5-4xx1.20 = $0.20`

Measurement, NAP-K3-CA10

Sisko is making red cordial for his daughter's birthday party.

Red cordial is made by adding red concentrate with water.

Sisko adds 60 millilitres (mL) of red concentrate to 1 litre (L) of water.

How much red cordial has Sisko made?

`text(61 mL)`	`text(160 mL)`	`text(1060 mL)`	`text(10060 mL)`

Show Answers Only

`text(1060 mL)`

Show Worked Solution

`text(Total volume of red cordial)`

`=1\ text(litre) + 60\ text(mL)`

`=1000 + 60`

`=1060\ text(mL)`

Algebra, NAP-K3-CA08

Raphael is 2 years younger than 3 times his sister's age.

If `s` represents his sister's age, which expression represents Raphael's age?

`2 - 3s`	`3s - 2`	`3s + 2`	`2s - 3`

Show Answers Only

`3s – 2`

Show Worked Solution

`s = text(sister’s age)`

`:.\ text(Raphael’s age = )3s – 2`

Geometry, NAP-K3-CA07

Tran uses blocks to make rectangular prisms.

How many blocks does Tran use altogether?

`20`	`23`	`26`	`28`

Show Answers Only

`28`

Show Worked Solution

`text(Total blocks used in both prisms)`

`= 12 + 16`

`= 28`

Statistics, NAP-K3-CA06

Clive and Alvin asked their friends how many books they had read in the past month.

Clive draws a picture graph to show the results for his friends.

Alvin draws a column graph to show the results for his friends.

How many more of Clive's friends read 3-4 books in the last month than Alvin's friends?

`0`	`4`	`6`	`8`

Show Answers Only

`6`

Show Worked Solution

`text(Number of Clive’s friends)`

`= 4 xx 2`

`= 8`

`text(Number of Alvin’s friends = 2)`

`:. 6\ text(more of Clive’s friends.)`

Probability, NAP-K3-CA02

There are 20 raffle tickets, numbered 1 to 20, in a box.

Three prizes are given away by choosing three tickets from the box. One ticket can win one prize only.

The first ticket drawn is number 15 and wins the third prize.

Which of the following is not possible?

	Second prize is won by number 2.
	First prize is won by a prime number.
	Second prize is an even number.
	First prize is won by number 15.

Show Answers Only

`text(First prize is won by number 15.)`

Show Worked Solution

`text(First prize is won by number 15 is impossible because)`

`text(number 15 has already been chosen and won 3rd prize.)`

`text{(Note that there is no replacement of tickets.)}`

Measurement, NAP-K4-CA02

Kim, Bob and Liz each measure the height of the hedge in their front yards.

Kim's hedge is 0.72 metres tall.
Bob's hedge is 815 millimetres tall.
Liz's hedge is 68 centimetres tall

Who has the tallest hedge in their front yard?

Kim	Bob	Liz

Show Answers Only

`text(Bob)`

Show Worked Solution

`text(Convert each to centimetres)`

`text(Kim = 0.72 × 100 = 72 centimetres)`

`text(Bob = 815 ÷ 10 = 81.5 centimetres)`

`text(Liz = 68 centimetres)`

`:.\ text(Bob’s hedge is the tallest)`

Calculus, MET1 SM-Bank 17

The diagram shows a point `T` on the unit circle `x^2+y^2=1` at an angle `theta` from the positive `x`-axis, where `0<theta<pi/2`.

The tangent to the circle at `T` is perpendicular to `OT`, and intersects the `x`-axis at `P`, and the line `y=1` intersects the `y`-axis at `B`

Show that the equation of the line `PT` is `xcostheta+ysin theta=1`. (2 marks)

--- 7 WORK AREA LINES (style=lined) ---
Find the length of `BQ` in terms of `theta`. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Show that the area, `A`, of the trapezium `OPQB` is given by
`A=(2-sintheta)/(2costheta)` (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Find the angle `theta` that gives the minimum area of the trapezium. (3 marks)

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

Show Answers Only

`text{Proof (See Worked Solutions)}`
`(1-sin theta)/cos theta`
`text{Proof (See Worked Solutions)}`
`theta=pi/6\ \ text(radians)`

Show Worked Solution

`text(Find)\ T:`

♦♦ Mean mark 20%

`text(S)text(ince)\ \ cos theta=x/1\ \ \ text(and)\ \ \ sin theta=y/1`

`:. T\ (cos theta, sin theta)`

`text(Gradient of)\ OT=sin theta/cos theta`

`:.\ text(Gradient)\ PT=-cos theta/sin theta\ \ text{(} _|_ text{lines)}`

`text(Equation of)\ PT\ text(where)`

`m=-cos theta/sin theta,\ \ text(and through)\ \ (cos theta, sin theta)`

`text(Using)\ \ y-y_1`	`=m(x-x_1)`
`y-sin theta`	`=-cos theta/sin theta(x-cos theta)`
`y sin theta-sin^2 theta`	`=-x cos theta+cos^2 theta`
`x cos theta+y sin theta`	`=sin^2 theta+cos^2 theta`
`x cos theta+y sin theta`	`=1\ \ \ \ \ text(… as required)`

ii. `text(Find)\ Q:`

`Q\ => text(intersection of)\ xcos theta+y sin theta=1\ \ text(and)\ \ y=1`

`x cos theta+sin theta`	`=1`
`x cos theta`	`=1-sin theta`
`x`	`=(1-sin theta)/cos theta`

`:.\ text(Length of)\ BQ\ text(is)\ \ (1-sin theta)/cos theta\ text(units)`

iii. `text(Show Area)\ \ OPQB=(2-sin theta)/(2cos theta)`

`A=1/2h(a+b)\ \ text(where)\ \ h=OB=1\ \ a=OP\ \ text(and)`

`b=BQ=(1-sin theta)/cos theta`

`text(Find length)\ OP:`

`P => xcos theta+ysin theta=1 \ text(cuts)\ \ x text(-axis)`

♦♦ Mean mark (iii) 24%

`xcos theta`	`=1`
`x`	`=1/cos theta`
`=>\ text(Length)\ OP`	`=1/cos theta`

`text(Area)\ OPQB`	`=1/2xx1(1/cos theta+(1-sin theta)/cos theta)`
	`=1/2((2-sin theta)/cos theta)`
	`=(2-sin theta)/(2cos theta)\ \ text(u²)\ \ \ text(… as required)`

iv. `text(Find)\ theta\ text(such that Area)\ OPQB\ text(is a MIN)`

`A`	`=(2-sin theta)/(2cos theta)`
`(dA)/(d theta)`	`=(2cos theta(-cos theta)-(2- sin theta)(-2 sin theta))/(4cos^2 theta)`
	`=(4 sin theta-2sin^2 theta-2 cos^2 theta)/(4 cos^2 theta)`
	`=(4sin theta-2(sin^2 theta+cos^2 theta))/(4 cos^2 theta)`
	`=(4sin theta-2)/(4cos^2 theta)`
	`=(2sin theta-1)/(2 cos^2 theta)`

Mean mark (iv) 19%
IMPORTANT: Look for any opportunity to use the identity `sin^2 theta“+cos^2 theta=1` as it is an examiner’s favourite and can often be the key to simplifying difficult trig equations.

`text(MAX or MIN when)\ (dA)/(d theta)=0`

`=>2sin theta-1`	`=0`
`sin theta`	`=1/2`
`theta`	`=pi/6\ \ \ \ \0<theta<pi/2`

`text(Test for MAX/MIN:)`

IMPORTANT: Is the 1st or 2nd derivative test easier here? Students must evaluate and choose. Examiners often make one significantly easier than the other.

`text(If)\ theta=pi/12\ \ (dA)/(d theta)<0`

`text(If)\ theta=pi/3\ \ (dA)/(d theta)>0\ \ =>text(MIN)`

`:.\text(Area)\ OPQB\ text(is a MIN when)\ theta=pi/6`.

Calculus, MET1 SM-Bank 27

A cone is inscribed in a sphere of radius `a`, centred at `O`. The height of the cone is `x` and the radius of the base is `r`, as shown in the diagram.

Show that the volume, `V`, of the cone is given by
`V = 1/3 pi(2ax^2-x^3)`. (2 marks)

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Find the value of `x` for which the volume of the cone is a maximum. You must give reasons why your value of `x` gives the maximum volume. (3 marks)

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Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`x = 4/3 a`

Show Worked Solution

i. `text(Show)\ V = 1/3 pi (2ax^2-x^3)`

`V = 1/3 pi r^2 h`

`text(Using Pythagoras:)`

`(x-a)^2 + r^2`	`= a^2`
`r^2`	`= a^2-(x-a)^2`
	`= a^2-x^2 + 2ax-a^2`
	`= 2ax-x^2`

`:. V`	`= 1/3 xx pi xx (2ax-x^2) xx x`
	`= 1/3 pi (2ax^2-x^3)\ …\ text(as required)`

ii. `(dV)/(dx) = 1/3 pi (4ax-3x^2)`

`(d^2V)/(dx^2) = 1/3 pi (4a-6x)`

`text(Max or min when)\ (dV)/(dx) = 0`

`1/3 pi (4ax-3x^2)`	`= 0`
`4ax-3x^2`	`= 0`
`x(4a-3x)`	`= 0`

`3x`	`= 4a,`	` \ \ \ \ x ≠ 0`
`x`	` =4/3 a`

`text(When)\ \ x = 4/3 a`

`(d^2V)/(dx^2)`	`= 1/3 pi (4a-6 xx 4/3 a)`
	`= 1/3 pi (-4a) < 0`
`=>\ text(MAX)`

`:.\ text(Cone volume is a maximum when)\ \ x = 4/3 a.`

Calculus, MET1 SM-Bank 35

The diagram shows two parallel brick walls `KJ` and `MN` joined by a fence from `J` to `M`. The wall `KJ` is `s` metres long and `/_KJM=alpha`. The fence `JM` is `l` metres long.

A new fence is to be built from `K` to a point `P` somewhere on `MN`. The new fence `KP` will cross the original fence `JM` at `O`.

Let `OJ=x` metres, where `0<x<l`.

Show that the total area, `A` square metres, enclosed by `DeltaOKJ` and `DeltaOMP` is given by
`A=s(x-l+l^2/(2x))sin alpha`. (3 marks)

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Find the value of `x` that makes `A` as small as possible. Justify the fact that this value of `x` gives the minimum value for `A`. (3 marks)

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Hence, find the length of `MP` when `A` is as small as possible. (1 mark)

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Show Answers Only

`text{Proof (See Worked Solutions)}`
`l/sqrt2`
`(sqrt2-1)s\ \ text(metres)`

Show Worked Solution

`A=text(Area)\ Delta OJK+text(Area)\ Delta OMP`

♦♦♦ Students found this question extremely challenging (exact results not available).

`text(Using sine rule)`

`text(Area)\ Delta OJK=1/2\ x s sin alpha`

`text(Area)\ DeltaOMP =>text(Need to find)\ \ MP`

`/_OKJ`	`=/_MPO\ \ text{(alternate angles,}\ MP\ text(\|\|)\ KJtext{)}`
`/_PMO`	`=/_OJK=alpha\ \ text{(alternate angles,}\ MP\ text(\|\|)\ KJtext{)}`

`:.\ DeltaOJK\ text(|||)\ Delta OMP\ \ text{(equiangular)}`

`=>x/s`	`=(l-x)/(MP)\ `	` text{(corresponding sides of similar triangles)}`
`MP`	`=(l-x)/x *s`

`text(Area)\ Delta\ OMP`	`=1/2 (l-x)* MP * sin alpha`
	`=1/2 (l-x)((l-x))/x s sin alpha`

`:. A`	`=1/2 xs sin alpha+1/2 (l-x)((l-x))/x* s sin alpha`
	`=s sin alpha(1/2 x+1/2 (l-x)*((l-x))/x)`
	`=s sin alpha(1/2 x+(l-x)^2/(2x))`
	`=s sin alpha(1/2 x+l^2/(2x)-l+1/2 x)`
	`=s(x-l+l^2/(2x))sin alpha\ \ \ \ text(… as required)`

ii. `text(Find)\ x\ text(such that)\ A\ text(is a minimum)`

MARKER’S COMMENT: Students who could not complete part (i) are reminded that they can still proceed to part (ii) and attempt to differentiate the result given.
Note that `l` and `alpha` are constants when differentiating.

`A`	`=s(x-l+l^2/(2x))sin alpha`
`(dA)/(dx)`	`=s(1-l^2/(2x^2))sin alpha`

`text(MAX/MIN when)\ (dA)/(dx)=0`

`s(1-l^2/(2x^2))sin alpha`	`=0`
`l^2/(2x^2)`	`=1`
`2x^2`	`=l^2`
`x^2`	`=l^2/2`
`x`	`=l/sqrt2,\ \ \ x>0`

`(d^2A)/(dx^2)=s((l^2)/(2x^3))sin alpha`

`text(S)text(ince)\ \ 0<alpha<90°\ \ =>\ sin alpha>0,\ \ l>0\ \ text(and)\ \ x>0`

`(d^2A)/(dx^2)>0\ \ \ =>text(MIN at)\ \ x=l/sqrt2`

iii. `text(S)text(ince)\ \ MP=((l-x))/x s\ \ text(and MIN when)\ \ x=l/sqrt2`

`MP`	`=((l-l/sqrt2)/(l/sqrt2))s xx sqrt2/sqrt2`
	`=((sqrt2 l-l))/l s`
	`=(sqrt2-1)s\ \ text(metres)`

`:.\ MP=(sqrt2-1)s\ \ text(metres when)\ A\ text(is a MIN.)`

Calculus, MET1 SM-Bank 30

A function is given by `f(x) = 3x^4 + 4x^3-12x^2`.

Find the coordinates of the stationary points of `f(x)` and determine their nature. (3 marks)

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Hence, sketch the graph `y = f(x)` showing the stationary points. (2 marks)

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For what values of `x` is the function increasing? (1 mark)

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For what values of `k` will `f(x) = 3x^4 + 4x^3-12x^2 + k = 0` have no solution? (1 mark)

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Show Answers Only

`text(MAX at)\ (0,0)`
`text(MIN at)\ text{(1,–5)}`
`text(MIN at)\ text{(–2,–32)}`
`f(x)\ text(is increasing for)\ -2 < x < 0\ text(and)\ x > 1`
`text(No solution when)\ k > 32`

Show Worked Solution

i.	`f(x)`	`= 3x^4 + 4x^3 -12x^2`
	`f^{′}(x)`	`= 12x^3 + 12x^2-24x`
	`f^{″}(x)`	`= 36x^2 + 24x-24`

`text(Stationary points when)\ f prime (x) = 0`

`=> 12x^3 + 12x^2-24x`	`=0`
`12x(x^2 + x-2)`	`=0`
`12x (x+2) (x -1)`	`=0`

`:.\ text(Stationary points at)\ x=0,\ 1\ text(or)\ –2`

`text(When)\ x=0,\ \ \ \ f(0)=0`
`f^{″}(0)`	`= -24 < 0`
`:.\ text{MAX at (0,0)}`

`text(When)\ x=1`

`f(1)`	`= 3+4\-12 = -5`
`f^{″}(1)`	`= 36 + 24\-24 = 36 > 0`
`:.\ text{MIN at (1,–5)}`

`text(When)\ x=–2`

`f(–2)`	`=3(–2)^4 + 4(–2)^3-12(–2)^2`
	`= 48 -32\-48`
	`= -32`
`f^{″}(–2)`	`= 36(–2)^2 + 24(–2) -24`
	`=144-48-24 = 72 > 0`
`:.\ text{MIN at (–2,–32)}`

ii.

♦ Mean mark (HSC) 42%
MARKER’S COMMENT: Be careful to use the correct inequality signs, and not carelessly include `>=` or `<=` by mistake.

iii.	`f(x)\ text(is increasing for)`
	`-2 < x < 0\ text(and)\ x > 1`

iv. `text(Find)\ k\ text(such that)`

♦♦♦ Mean mark (HSC) 12%.

`3x^4 + 4x^3-12x^2 + k = 0\ \ text(has no solution)`

`k\ text(is the vertical shift of)\ \ y = 3x^4 + 4x^3-12x^2`

`=>\ text(No solution if it does not cross the)\ x text(-axis.)`

`:.\ text(No solution when)\ \ k > 32`

Calculus, 2ADV C3 SM-Bank 13

The figure shown represents a wire frame where `ABCE` is a convex quadrilateral. The point `D` is on line segment `EC` with `AB = ED = 2\ text(cm)` and `BC = a\ text(cm)`, where `a` is a positive constant.

`/_ BAE = /_ CEA = pi/2`

Let `/_ CBD = theta` where `0 < theta < pi/2.`

Find `BD` and `CD` in terms of `a` and `theta`. (2 marks)

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Find the length, `L` cm, of the wire in the frame, including length `BD`, in terms of `a` and `theta`. (1 mark)

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Find `(dL)/(d theta)`, and hence show that `(dL)/(d theta) = 0` when `BD = 2CD`. (2 marks)

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Find the maximum value of `L` if `a = 3 sqrt 5`. (1 mark)

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Show Answers Only

`BD = a cos theta,\ \ \ CD = a sin theta`
`L = 4 + a + 2 a cos theta + a sin theta`
`text(Proof)\ text{(See Worked Solutions)}`
`L_max = 19 + 3 sqrt 5`

Show Worked Solution

i. `text(In)\ \ Delta BCD,`

`cos theta`	`= (BD)/a`
`:. BD`	`= a cos theta`
`sin theta`	`= (CD)/a`
`:. CD`	`= a sin theta`

ii.	`L`	`= 4 + 2 BD + CD + a`
		`= 4 + 2a cos theta + a sin theta + a`
		`= 4 + a + 2a cos theta + a sin theta`

iii. `text(Noting that)\ a\ text(is a constant:)`

♦ Mean mark (Vic) 35%.

`(dL)/(d theta)= – 2 a sin theta + a cos theta`

`text(When)\ \ (dL)/(d theta) = 0`,

`- 2 a sin theta+ a cos theta`	`= 0`
`a cos theta`	`= 2 a sin theta`
`:. BD`	`= 2CD\ \ text{(using part (a))}`

iv. `text(SP’s when)\ \ (dL)/(d theta)=0,`

♦♦♦ Mean mark (Vic) 5%.

`- 2 a sin theta+ a cos theta`	`= 0`
`sin theta`	`=1/2 cos theta`
`tan theta`	`=1/2`

`text(If)\ \ tan theta=1/2,\ \ cos theta = 2/sqrt5,\ \ sin theta = 1/sqrt5`

`L_(max)`	`= 4 + a + 2a cos theta + a sin theta`
	`= 4 + (3 sqrt 5) + 2 (3 sqrt 5) (2/sqrt 5) + (3 sqrt 5) (1/sqrt 5)`
	`= 4 + 3 sqrt 5 + 12 + 3`
	`= 19 + 3 sqrt 5\ text(cm)`

NETWORKS, FUR2 2017 VCAA 4

The rides at the theme park are set up at the beginning of each holiday season.

This project involves activities A to O.

The directed network below shows these activities and their completion times in days.

Write down the two immediate predecessors of activity I. (1 mark)

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The minimum completion time for the project is 19 days.

i. There are two critical paths. One of the critical paths is A–E–J–L–N.
Write down the other critical path. (1 mark)

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ii. Determine the float time, in days, for activity F. (1 mark)

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The project could finish earlier if some activities were crashed.

Six activities, B, D, G, I, J and L, can all be reduced by one day.

The cost of this crashing is $1000 per activity.

i. What is the minimum number of days in which the project could now be completed? (1 mark)

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ii. What is the minimum cost of completing the project in this time? (1 mark)

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Show Answers Only

a. `D\ text(and)\ E`

b.i. `A E I L N`

b.ii. `6\ text(days)`

c.i. `17`

c.ii. `$4000`

Show Worked Solution

a. `D\ text(and)\ E\ (text(note the dummy is not an activity.))`

b.i. `A E I L N`

♦♦ Mean mark part (b)(i) 44% and part (b)(ii) 28%.

b.ii.	`text(Float time)`	`= 19-(2 + 3 + 3 + 3 + 2)`
		`= 6\ text(days)`

c.i. `text(Reduce activities:)\ I, J, L\ \ (text(on critical path))`

♦♦ Mean mark part (c)(i) 35%.

`text(New critical path)\ \ A C G N\ \ text(takes 18 days.)`

`:. text(Reduce activity)\ G\ text(also.)`

`text(⇒ this critical path reduces to 17 days.)`

`text(⇒ Minimum Days = 17)`

c.ii. `text(Minimum time requires crashing)\ \ I, J, L\ text(and)\ G`

♦♦♦ Mean mark part (c)(ii) 15%.

`:.\ text(Minimum Cost)`	`= 4 xx 1000`
	`= $4000`

MATRICES, FUR2 2017 VCAA 3

Senior students at a school choose one elective activity in each of the four terms in 2018.

Their choices are communication (`C`), investigation (`I`), problem-solving (`P`) and service (`S`).

The transition matrix `T` shows the way in which senior students are expected to change their choice of elective activity from term to term.

`{:(qquadqquadqquadqquadquadtext(this term)),(qquadqquadqquad\ CqquadquadIqquadquadPqquad\ S),(T = [(0.4,0.2,0.3,0.1),(0.2,0.4,0.1,0.3),(0.2,0.3,0.3,0.4),(0.2,0.1,0.3,0.2)]{:(C),(I),(P),(S):}qquadtext(next term)):}`

Let `S_n` be the state matrix for the number of senior students expected to choose each elective activity in Term `n`.

For the given matrix `S_1`, a matrix rule that can be used to predict the number of senior students in each elective activity in Terms 2, 3 and 4 is

`S_1 = [(300),(200),(200),(300)],qquadS_(n + 1) = TS_n`

How many senior students will not change their elective activity from Term 1 to Term 2? (1 mark)

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Complete `S_2`, the state matrix for Term 2, below. (1 mark)

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Of the senior students expected to choose investigation (`I`) in Term 3, what percentage chose service (`S`) in Term 2? (2 marks)

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What is the maximum number of senior students expected in investigation (`I`) at any time during 2018? (1 mark)

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Show Answers Only

`320`
`S_2 = [(250),(250),(300),(200)]`
`text(25%)`
`250`

Show Worked Solution

a. `text(Students who do not change)`

♦ Mean mark 47%.

`= 0.4 xx 300 + 0.4 xx 200 + 0.3 xx 200 + 0.2 xx 300`

`= 120 + 80 + 60 + 60`

`= 320`

b.	`S_2 = TS_1`	`= [(0.4,0.2,0.3,0.1),(0.2,0.4,0.1,0.3),(0.2,0.3,0.3,0.4),(0.2,0.1,0.3,0.2)][(300),(200),(200),(300)]`
		`= [(250),(250),(300),(200)]`

c.	`S_3`	`= TS_2`
		`= [(260),(240),(295),(205)]`

♦♦♦ Mean mark 13%.
MARKER’S COMMENT: A poorly understood and answered question worthy of careful attention.

`text(Number)\ (I)\ text(in Term 3 = 240)`

`text(Number)\ (S)\ text(in Term 2 = 200)`

`text(S)text(ince 30% move from)\ S\ text(to)\ I\ text(each term:)`

`text(Percentage)`	`= (0.3 xx 200)/240`
	`= 60/240`
	`= 25text(%)`

d.	`S_4`	`= TS_3`
		`= [(261),(239),(294.5),(205.5)]`

♦ Mean mark 43%.

`:. text(Max number of)\ (I)\ text(students is 250.)`

`(text(During term 2))`

CORE, FUR2 2017 VCAA 6

Alex sends a bill to his customers after repairs are completed.

If a customer does not pay the bill by the due date, interest is charged.

Alex charges interest after the due date at the rate of 1.5% per month on the amount of an unpaid bill.

The interest on this amount will compound monthly.

Alex sent Marcus a bill of $200 for repairs to his car.

Marcus paid the full amount one month after the due date.

How much did Marcus pay? (1 mark)

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Alex sent Lily a bill of $428 for repairs to her car.

Lily did not pay the bill by the due date.

Let `A_n` be the amount of this bill `n` months after the due date.

Write down a recurrence relation, in terms of `A_0`, `A_(n + 1)` and `A_n`, that models the amount of the bill. (2 marks)

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Lily paid the full amount of her bill four months after the due date.

How much interest was Lily charged?

Round your answer to the nearest cent. (1 mark)

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Show Answers Only

`$203`
`A_o = 428,qquadA_(n + 1) = 1.015A_n`
`$26.26\ \ (text(nearest cent))`

Show Worked Solution

a.	`text(Amount paid)`	`= 200 + 200 xx 1.5text(%)`
		`= 1.015 xx 200`
		`= $203`

♦ Mean mark part (b) 47%.
MARKER’S COMMENT: A recurrence relation has the initial value written first. Know why `A_n=428 xx 1.015^n` is incorrect.

b. `A_o = 428,qquadA_(n + 1) = 1.015A_n`

c.	`text(Total paid)\ (A_4)`	`= 1.015^4 xx 428`
		`= $454.26`

♦♦ Mean mark part (c) 29%.

`:.\ text(Total Interest)`	`= 454.26-428`
	`= $26.26\ \ (text(nearest cent))`

CORE, FUR2 2017 VCAA 4

The eggs laid by the female moths hatch and become caterpillars.

The following time series plot shows the total area, in hectares, of forest eaten by the caterpillars in a rural area during the period 1900 to 1980.

The data used to generate this plot is also given.

The association between area of forest eaten by the caterpillars and year is non-linear.

A log₁₀ transformation can be applied to the variable area to linearise the data.

When the equation of the least squares line that can be used to predict log₁₀ (area) from year is determined, the slope of this line is approximately 0.0085385
Round this value to three significant figures. (1 mark)
Perform the log₁₀ transformation to the variable area and determine the equation of the least squares line that can be used to predict log₁₀ (area) from year.
Write the values of the intercept and slope of this least squares line in the appropriate boxes provided below.
Round your answers to three significant figures. (2 marks)

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The least squares line predicts that the log₁₀ (area) of forest eaten by the caterpillars by the year 2020 will be approximately 2.85

Using this value of 2.85, calculate the expected area of forest that will be eaten by the caterpillars by the year 2020.
i. Round your answer to the nearest hectare. (1 mark)

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ii. Give a reason why this prediction may have limited reliability. (1 mark)

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Show Answers Only

a. `0.00854\ (text(3 sig fig))`

b. `log_10(text(area)) = −14.4 + 0.000854 xx text(year)`

c.i. `708\ text(hectares)`

c.ii. `text(This prediction extrapolates significantly from the given)`
`text(data range and as a result, its reliability decreases.)`

Show Worked Solution

a. `0.0085385 = 0.00854\ (text(3 sig fig))`

♦ Mean marks of part (a) and (b) 44%.

`log_10(text(area))`

`= −14.4 + 0.000854 xx text(year)`

♦♦ Mean mark part (c)(i) 29%.
COMMENT: When the question specifies using the value 2.85, use it!

c.i.	`log_10(text(Area))`	`= 2.85`
	`:.\ text(Area)`	`= 10^2.85`
		`= 707.94…`
		`= 708\ text(hectares)`

c.ii. `text(This prediction extrapolates significantly from the given)`

`text(data range and as a result, its reliability decreases.)`

Calculus, 2ADV C3 SM-Bank 7

The graph of `f(x) = sqrt x (1 - x)` for `0<=x<=1` is shown below.

Calculate the area between the graph of `f(x)` and the `x`-axis. (2 marks)

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For `x` in the interval `(0, 1)`, show that the gradient of the tangent to the graph of `f(x)` is `(1 - 3x)/(2 sqrt x)`. (1 mark)

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The edges of the right-angled triangle `ABC` are the line segments `AC` and `BC`, which are tangent to the graph of `f(x)`, and the line segment `AB`, which is part of the horizontal axis, as shown below.

Let `theta` be the angle that `AC` makes with the positive direction of the horizontal axis.

Find the equation of the line through `B` and `C` in the form `y = mx + c`, for `theta = 45^@`. (3 marks)

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Show Answers Only

`4/15\ text(units)^2`
`text(Proof)\ \ text{(See Workes Solutions)}`
`y = -x + 1`

Show Worked Solution

i.	`text(Area)`	`= int_0^1 (sqrt x – x sqrt x)\ dx`
		`= int_0^1 (x^(1/2) – x^(3/2))\ dx`
		`= [2/3 x^(3/2) – 2/5 x^(5/2)]_0^1`
		`= (2/3 – 2/5) – (0 – 0)`
		`= 10/15 – 6/15`
		`= 4/15\ text(units)^2`

ii.	`f (x)`	`= x^(1/2) – x^(3/2)`
	`f prime (x)`	`= 1/2 x^(-1/2) – 3/2 x^(1/2)`
		`= 1/(2 sqrt x) – (3 sqrt x)/2`
		`= (1 – 3x)/(2 sqrt x)\ \ text(.. as required.)`

iii. `m_(AC) = tan 45^@=1`

♦♦♦ Mean mark (Vic) part (iii) 20%.
MARKER’S COMMENT: Most successful answers introduced a pronumeral such as `a=sqrtx` to solve.

`=> m_(BC) = -1\ \ (m_text(BC) _|_ m_(AC))`

`text(At point of tangency of)\ BC,\ f prime(x) = -1`

`(1 – 3x)/(2 sqrt x)`	`=-1`
`1-3x`	`=-2sqrtx`
`3x-2sqrt x-1`	`=0`

`text(Let)\ \ a=sqrtx,`

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

`:. sqrt x`	`=1`	`or`	`sqrt x=- 1/3\ \ text{(no solution)}`
`x`	`=1`

`f(1)=sqrt1(1-1)=0\ \ =>B(1,0)`

`text(Equation of)\ \ BC, \ m=-1, text{through (1,0):}`

`y-0`	`=-1(x-1)`
`y`	`=-x+1`

Algebra, MET2 2017 VCAA 4

Let `f : R → R :\ f (x) = 2^(x + 1)-2`. Part of the graph of `f` is shown below.

The transformation `T: R^2 -> R^2, \ T([(x),(y)]) = [(x),(y)] + [(c),(d)]` maps the graph of `y = 2^x` onto the graph of `f`.

State the values of `c` and `d`. (2 marks)

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Find the rule and domain for `f^(-1)`, the inverse function of `f`. (2 marks)

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Find the area bounded by the graphs of `f` and `f^(-1)`. (3 marks)

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Part of the graphs of `f` and `f^(-1)` are shown below.

Find the gradient of `f` and the gradient of `f^(-1)` at `x = 0`. (2 marks)

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The functions of `g_k`, where `k ∈ R^+`, are defined with domain `R` such that `g_k(x) = 2e^(kx)-2`.

Find the value of `k` such that `g_k(x) = f(x)`. (1 mark)

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Find the rule for the inverse functions `g_k^(-1)` of `g_k`, where `k ∈ R^+`. (1 mark)

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i. Describe the transformation that maps the graph of `g_1` onto the graph of `g_k`. (1 mark)

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ii. Describe the transformation that maps the graph of `g_1^(-1)` onto the graph of `g_k^(-1)`. (1 mark)

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The lines `L_1` and `L_2` are the tangents at the origin to the graphs of `g_k` and `g_k^(-1)` respectively.
Find the value(s) of `k` for which the angle between `L_1` and `L_2` is 30°. (2 marks)

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Let `p` be the value of `k` for which `g_k(x) = g_k^(−1)(x)` has only one solution.
i. Find `p`. (2 marks)

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ii. Let `A(k)` be the area bounded by the graphs of `g_k` and `g_k^(-1)` for all `k > p`.
State the smallest value of `b` such that `A(k) < b`. (1 mark)

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Show Answers Only

a. `c =-1, \ d =-2`

b. `f^(-1)(x) = log_2 (x + 2)-1, \ x ∈ (-2,∞)`

c. `3-2/(log_e(2))\ text(units)²`

d. `f^{prime}(0)= 2log_e(2) and f^(-1)^{prime}(0)= 1/(2log_e(2))`

e. `k = log_e(2)`

f. `g_k^(-1)(x)= 1/k log_e((x + 2)/2)`

g.i. `text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

g.ii. `text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

h. `k=sqrt3/6\ or\ sqrt3/2`

i.i. `p = 1/2`

i.ii. `b=4`

Show Worked Solution

a. `text(Using the matrix transformation:)`

`x^{prime}`	`=x+c\ \ => x=x^{prime}-c`
`y^{prime}`	`=y+d\ \ =>y= y^{prime}-d`

`y^{prime}-d`	`=2^((x)^{prime}-c)`
`y^{prime}`	`= 2^((x)^{prime}-c)+d`

`:. c = -1, \ d = -2`

b. `text(Let)\ \ y = f(x)`

`text(Inverse : swap)\ x\ ↔ \ y`

`x`	`= 2^(y + 1)-2`
`x + 2`	`= 2^(y + 1)`
`y + 1`	`= log_2(x + 2)`
`y`	`= log_2(x + 2)-1`

`text(dom)(f^(-1)) = text(ran)(f)`

`:. f^(-1)(x) = log_2 (x + 2)-1, \ x ∈ (-2,∞)`

c. `text(Intersection points occur when)\ \ f(x)=f^(-1)(x)`

MARKER’S COMMENT: Specifically recommends using `f^(-1)(x)-f(x)` in this type of integral to avoid errors in stating the integral.

`x`

`= -1, 0`

`text(Area)`	`= int_(-1)^0 (f^(-1)(x)-f(x))\ dx`
	`= 3-2/(log_e(2))\ text(units)²`

d.	`f^{prime}(0)`	`= 2log_e(2)`
	`f^{(-1)^prime}(0)`	`= 1/(2log_e(2))`

e. `g_k(x) = 2e^(kx)-2`

`text(Solve:)\ \ g_k(x) = f(x)quad text(for)\ k ∈ R^+`

`:. k = log_e(2)`

f. `text(Let)\ \ y = g_k(x)`

`text(Inverse : swap)\ x\ text(and)\ y`

`x`	`= 2e^(ky)-2`
`e^(ky)`	`=(x+2)/2`
`ky`	`=log_e((x+2)/2)`
`:. g_k^(-1)(x)`	`= 1/k log_e((x + 2)/2)`

♦♦ Mean mark part (g)(i) 31%.

g.i.	`g_1(x)`	`= 2e^x-2`
	`g_k(x)`	`= 2e^(kx)-2`

`:. text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

♦♦ Mean mark part (g)(ii) 30%.

g.ii.	`g_1^(-1)(x)`	`= log_e((x + 2)/2)`
	`g_k^(-1)(x)`	`= 1/klog_e((x + 2)/2)`

`:. text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

h. `text(When)\ \ x=0,`

♦♦♦ Mean mark part (h) 13%.

`g_k^{prime}(0)=2k\ \ => m_(L_1)=2k`

`g_k^{(-1)^prime}(0)=1/(2k)\ \ => m_(L_2)=1/(2k)`

`text(Using)\ \ tan 30^@=|(m_1-m_2)/(1+m_1m_2)|,`

`text(Solve:)\ \ 1/sqrt3`

`=+- ((2k-1/(2k)))/2\ \ text(for)\ k`

`:. k=sqrt3/6\ or\ sqrt3/2`

i.i `text(By inspection, graphs will touch once if at)\ \ x=0,`

♦♦♦ Mean mark part (i)(i) 4%.

`m_(L_1)`	`=m_(L_2)`
`2k`	`=1/(2k)`
`k`	`=1/2,\ \ (k>0)`

`:. p = 1/2`

i.ii `text(As)\ k→oo, text(the graph of)\ g_k\ text(approaches)\ \ x=0`

♦♦♦ Mean mark part (i)(ii) 2%.

`text{(vertically) and}\ \ y=-2\ \ text{(horizontally).}`

`text(Similarly,)\ \ g_k^(-1)\ \ text(approaches)\ \ x=-2`

`text{(vertically) and}\ \ y=0\ \ text{(horizontally).}`

`:. lim_(k→oo) A(k) = 4`

`:.b=4`