Band 3 – Page 33

GRAPHS, FUR1 2011 VCAA 6 MC

A newsagent has four thick magazines and four thin magazines.

The magazines are stacked one by one into a pile.

The thick magazines are all placed in the bottom part of the file, and the thin magazines are all placed in the top part of the pile.

A graph indicates the changing height of the pile of magazines as each magazine is added to the pile could be

Show Answers Only

`D`

Show Worked Solution

`text(The first 4 magazines increase height in a linear)`

`text(way with a relatively steeper gradient, and the)`

`text(next 4 thin magazines also increase height in a)`

`text(linear fashion but at a lesser gradient.)`

`=> D`

GRAPHS, FUR1 2013 VCAA 4 MC

The Blue Caps cricket club has different prices for its junior and senior subscriptions.

The total cost for two junior subscriptions and one senior subscription is $225.

The cost of a senior subscription is three times the cost of a junior subscription.

The cost of a senior subscription is

A. `$45`

B. `$75`

C. `$90`

D. `$135`

E. `$180`

Show Answers Only

`D`

Show Worked Solution

`text (Let cost for juniors) = x`

`text (Let cost for seniors) = y`

`2x + y`	`= 225\ …\ \ (1)`
`y`	`= 3x\ …\ \ (2)`

`text(Substitute)\ (2)\ \ text(into)\ \ (1)`

`2x + 3x`	`= 225`
`5x`	`= 225`
`x`	`= 45`

`text(Substitute)\ \ x = 45\ \ text(into)\ \ (2)`

`y`	`= 3 xx 45`
	`= 135`

`:.\ text(Senior subscription) = $135`

`=> D`

GEOMETRY, FUR1 2015 VCAA 3 MC

The lengths of the sides of a triangle are 3 cm, 6 cm and 5 cm, as shown below.

The angle, `x`, can be found using

A. `cos(x) = (3^2 + 6^2 - 5^2)/(2 xx 3 xx 6)`

B. `cos(x) = (3^2 + 6^2 + 5^2)/(2 xx 3 xx 6)`

C. `cos(x) = (3^2 + 5^2 - 6^2)/(2 xx 3 xx 6)`

D. `cos(x) = 3/5`

E. `cos(x) = 3/6`

Show Answers Only

`A`

Show Worked Solution

`text(Using the cosine rule:)`

`cosx = (3^2 + 6^2 – 5^2)/(2 xx 3 xx 6)`

`=> A`

CORE*, FUR1 2015 VCAA 1 MC

Fong’s gas bill is $368.40. If he pays this bill on time, it will be reduced by 5%.

In this case, the bill would be reduced by

$1.84
$5.00
$18.42
$184.20
$349.98

Show Answers Only

`C`

Show Worked Solution

`5text(%) xx $368.40 = $18.42`

`=> C`

Calculus, SPEC2 2014 VCAA 11 MC

Let `(dy)/(dx) = x^3 - xy` and `y = 2` when `x = 1`.

Using Euler’s method with a step size of 0.1, the approximation to `y` when `x = 1.1` is

A. 0.9

B. 1.0

C. 1.1

D. 1.9

E. 2.1

Show Answers Only

`D`

Show Worked Solution

`text(When)\ \ x=1,\ \ y=2.`

`text(Using Euler when)\ \ x=1.1`

`y(1.1)`	`~~ 2 + 0.1 xx (1^3 – 1 xx 2)`
	`= 1.9`

`=> D`

CORE*, FUR1 2010 VCAA 4 MC

The price of a one-way airfare between two cities varies each day according to demand.

• On Monday the price is $160.
• The price on Tuesday is 25% greater than the price on Monday.
• The price on Wednesday is 10% less than the price on Tuesday.
• The price on Thursday is 25% less than the price on Wednesday.
• The price on Friday is 20% greater than the price on Thursday.
• The Saturday price is the same as the Friday price.

The price on Saturday is

A. $150

B. $162

C. $176

D. $288

E. $330

Show Answers Only

`B`

Show Worked Solution

`text{Price (Tue)}`	`= 160 xx text(125%)`
	`= 200`
`text{Price (Wed)}`	`= 200 xx 90text(%)`
	`= 180`
`text{Price (Thu)}`	`= 180 xx 75text(%)`
	`=135`
`text{Price (Fri)}`	`= 135 xx 120 text(%)`
	`= 162`

`text(Saturday price is same as Friday.)`

`=> B`

CORE*, FUR1 2010 VCAA 3 MC

Peter received a quote from the Artificial Grass Company for his new front lawn.

The quote is for $1880 plus a Goods and Services Tax (GST) of 10%.

The final amount that Peter pays for the new front lawn is

A. $188

B. $1880

C. $1890

D. $1899

E. $2068

Show Answers Only

`E`

Show Worked Solution

`text(GST)`	`= 1880 xx text(10%)`
	`= 188`

`:.\ text(Total amount)`	`= 1880 + 188`
	`= $2068`

`=> E`

CORE*, FUR1 2010 VCAA 2 MC

$6000 is invested in an account that earns simple interest at the rate of 3.5% per annum.

The total interest earned in the first four years is

A. `$70`

B. `$84`

C. `$210`

D. `$840`

E. `$885`

Show Answers Only

`D`

Show Worked Solution

`P = 6000,\ \ r = 3.5,\ \ T = 4`

`I`	`= (PrT) / 100`
`I`	`= {(6000) (3.5) (4)} / 100`
	`= 840`

`=> D`

CORE*, FUR1 2010 VCAA 1 MC

Sally purchased an electronic game machine on hire purchase. She paid $140 deposit and then $25.50 per month for two years.

The total amount that Sally paid is

A. $191

B. $446

C. $612

D. $740

E. $752

Show Answers Only

`E`

Show Worked Solution

`text(Total paid)`	`= 140 + 25.50 xx 2 xx12`
	`= $752`

`=>E`

CORE*, FUR1 2015 VCAA 2 MC

The second term of a Fibonacci-related sequence is 36 and the third term is 72.

The first term of this sequence is

`2`
`6`
`18`
`36`
`108`

Show Answers Only

`D`

Show Worked Solution

`text(A Fibonacci sequence takes the form)`

`t_3`	`=t_2+t_1`
`:. t_1`	`= t_3 – t_2`
	`= 72 – 36`
	`= 36`

`=> D`

CORE, FUR1 2015 VCAA 11 MC

A log transformation is used to linearise the relationship between the weight of a mouse, in grams, and its age, in weeks.

When a least squares regression line is fitted to the transformed data, its equation is

`text(weight) = – 7 + 30 log_10(text(age))`

This equation predicts that a mouse aged five weeks has a weight, in grams, that is closest to

A. `14`

B. `21`

C. `23`

D. `41`

E. `143`

Show Answers Only

`A`

Show Worked Solution

`text(Weight)`	`= – 7 + 30log_10 5`
	`= 13.969…`

`=> A`

CORE, FUR1 2015 VCAA 2 MC

For an ordered set of data containing an odd number of values, the middle value is always

A. the mean.

B. the median.

C. the mode.

D. the mean and the median.

E. the mean, the median and the mode.

Show Answers Only

`B`

Show Worked Solution

`=> B`

CORE*, FUR1 2006 VCAA 3-4 MC

The following information relates to Parts 1 and 2.

A farmer plans to breed sheep to sell.

In the first year she starts with 50 breeding sheep.

During the first year, the sheep numbers increase by 84%.

At the end of the first year, the farmer sells 40 sheep.

Part 1

How many sheep does she have at the start of the second year?

A. 2

B. 42

C. 52

D. 84

E. 92

Part 2

If `S_n` is the number of sheep at the start of year `n`, a difference equation that can be used to model the growth in sheep numbers over time is

A. `S_(n+1) = 1.84S_n - 40`	`\ \ \ \ \ text(where)\ \ S_1 = 50`
B. `S_(n+1) = 0.84S_n - 50`	`\ \ \ \ \ text(where)\ \ S_1 = 40`
C. `S_(n+1) = 0.84S_n - 40`	`\ \ \ \ \ text(where)\ \ S_1 = 50`
D. `S_(n+1) = 0.16S_n - 50`	`\ \ \ \ \ text(where)\ \ S_1 = 40`
E. `S_(n+1) = 0.16S_n - 40`	`\ \ \ \ \ text(where)\ \ S_1 = 50`

Show Answers Only

`text (Part 1:)\ C`

`text (Part 2:)\ A`

Show Worked Solution

`text (Part 1)`

`text(Number at the start of the 2nd year)`

`=50 + (84text(%) xx 50) – 40`

`=52`

`rArr C`

`text (Part 2)`

`text(If sheep numbers increase by 84% from the start of)`

`text(each year, and 40 sheep are then sold,)`

`S_(n+1)`	`=S_n + 0.84S_n – 40`
	`=1.84 S_n – 40`

`rArr A`

PATTERNS, FUR1 2006 VCAA 2 MC

The first three terms of a geometric sequence are `6, x, 54.`

A possible value of `x` is

A. `9`

B. `15`

C. `18`

D. `24`

E. `30`

Show Answers Only

`C`

Show Worked Solution

`text(S)text(ince the sequence is geometric,)`

`r`	`=x/6=54/x`
`x^2`	`=54 xx 6`
	`=324`
`:.x`	`=18`

`rArr C`

CORE, FUR1 2006 VCAA 5-6 MC

The distribution of test marks obtained by a large group of students is displayed in the percentage frequency histogram below.

Part 1

The pass mark on the test was 30 marks.

The percentage of students who passed the test is

A. `7text(%)`

B. `22text(%)`

C. `50text(%)`

D. `78text(%)`

E. `87text(%)`

Part 2

The median mark lies between

A. `35 and 40`

B. `40 and 45`

C. `45 and 50`

D. `50 and 55`

E. `55 and 60`

Show Answers Only

`text (Part 1:)\ D`

`text (Part 2:)\ B`

Show Worked Solution

`text (Part 1)`

`text(Adding up the percentage bars above 30)`

`=7+11+14+16+18+12`

`=78text(%)`

`rArr D`

`text (Part 2)`

`text(Adding up the percentage bars from the left,)`

`text(the 50th and 51st percentile lie in the 40–45)`

`text(mark interval.)`

`rArr B`

CORE, FUR1 2006 VCAA 1-3 MC

The back-to-back ordered stemplot below shows the distribution of maximum temperatures (in °Celsius) of two towns, Beachside and Flattown, over 21 days in January.

Part 1

The variables

temperature (°Celsius), and

town (Beachside or Flattown), are

A. both categorical variables.

B. both numerical variables.

C. categorical and numerical variables respectively.

D. numerical and categorical variables respectively.

E. neither categorical nor numerical variables.

Part 2

For Beachside, the range of maximum temperatures is

A. `3°text(C)`

B. `23°text(C)`

C. `32°text(C)`

D. `33°text(C)`

E. `38°text(C)`

Part 3

The distribution of maximum temperatures for Flattown is best described as

A. negatively skewed.

B. positively skewed.

C. positively skewed with outliers.

D. approximately symmetric.

E. approximately symmetric with outliers.

Show Answers Only

`text (Part 1:)\ D`

`text (Part 2:)\ B`

`text (Part 3:)\ E`

Show Worked Solution

`text (Part 1)`

`text(Temperature is numerical,)`

`text(Town is categorical.)`

`rArr D`

`text (Part 2)`

`text(Beachside’s maximum temperature range)`

`=38-15`

`=23°text(C)`

`rArr B`

`text (Part 3)`

`IQR\ text{(Beachside)}`	`=Q_3 – Q_1`
	`=40-33`
	`=7`

`Q_1 – 1.5 xx IQR`	`=33 – 1.5 xx7`
	`=22.5°text(C)`

`:.\ text(Flattown’s maximum temperature readings of)`

`text(18° and 19° are outliers.)`

`rArr E`

PATTERNS, FUR1 2007 VCAA 6 MC

In the first three layers of a stack of soup cans there are 20 cans in the first layer, 19 cans in the second layer and 18 cans in the third layer.

This pattern of stacking cans in layers continues.

The maximum number of cans that can be stacked in this way is

A. `190`

B. `210`

C. `220`

D. `380`

E. `590`

Show Answers Only

`B`

Show Worked Solution

`20, 19, 18, …, 2,1`

`text(AP where)\ \ \ a=20, l=1 and n=20`

`S_n`	`=n/2(a+l)`
`:.S_20`	`=20/2 (20+1)`
	`=210`

`rArr B`

CORE*, FUR1 2007 VCAA 3 MC

The difference equation

`t_(n+1) = at_n + 6 quad text (where) quad t_1 = 5`

generates the sequence

`5, 21, 69, 213\ …`

The value of `a` is

A. – 1

B. 3

C. 4

D. 15

E. 16

Show Answers Only

`B`

Show Worked Solution

`t_2`	`=a t_1 +6`
`:. 21`	`=5a + 6`
`a`	`=3`

`rArr B`

PATTERNS, FUR1 2007 VCAA 2 MC

The yearly membership of a club follows an arithmetic sequence.

In the club’s first year it had 15 members.

In its third year it had 29 members.

How many members will the club have in the fourth year?

A. `8`

B. `22`

C. `36`

D. `43`

E. `57`

Show Answers Only

`C`

Show Worked Solution

`d=(29-15)/2=7`

`T_4`	`=a+3d`
	`=15+3 xx 7`
	`=36`

`rArr C`

CORE, FUR1 2007 VCAA 1-2 MC

The dot plot below shows the distribution of the number of bedrooms in each of 21 apartments advertised for sale in a new high-rise apartment block.

Part 1

The mode of this distribution is

A. `1`

B. `2`

C. `3`

D. `7`

E. `8`

Part 2

The median of this distribution is

A. `1`

B. `2`

C. `3`

D. `4`

E. `5`

Show Answers Only

`text (Part 1:)\ A`

`text (Part 2:)\ B`

Show Worked Solution

`text (Part 1)`

`rArr A`

`text (Part 2)`

`text(The median of 21 data points is the 11th value.)`

`:.\ text(Median) = 2`

`rArr B`

CORE*, FUR1 2008 VCAA 1 MC

A sequence is generated by a first-order linear difference equation.

The first four terms of this sequence are 1, 3, 7, 15.

The next term in the sequence is

A. 17

B. 19

C. 22

D. 23

E. 31

Show Answers Only

`E`

Show Worked Solution

`1,3,7,15\ ...`

`text(Sequence has the pattern,)`

`Τ_(n+1)=2T_n+1`

`:. T_5`	`= 2 xx T_4 +1`
	`=2 xx 15 +1`
	`=31`

`=>E`

CORE*, FUR1 2008 VCAA 7 MC

The sequence `12, 15, 27, 42, 69, 111 …` can best be described as

A. fibonacci-related

B. arithmetic with `d > 1`

C. arithmetic with `d < 1`

D. geometric with `r > 1`

E. geometric with `r < 1`

Show Answers Only

`A`

Show Worked Solution

`12, 15, 27, 42, 69, 111\ …`

`T_3 = T_1 + T_2`

`T_4 = T_3 + T_2`

`=> A`

PATTERNS, FUR1 2008 VCAA 3 MC

The values of the first seven terms of a geometric sequence are plotted on the graph above.
Values of `a` and `r` that could apply to this sequence are respectively

(A) `a=90`	`\ \ \ \ r= – 0.9`
(B) `a=100`	`\ \ \ \ r= – 0.9`
(C) `a=100`	`\ \ \ \ r= – 0.8`
(D) `a=100`	`\ \ \ \ r=0.8`
(E) `a=90`	`\ \ \ \ r=0.9`

Show Answers Only

`B`

Show Worked Solution

`text(By elimination,)`

`T_1=a=100`

`text(Eliminate A and E.)`

`T_2=ar`	`=-90`
`:.100r`	`=-90`
`r`	`=-0.90`

`text(Eliminate C and D.)`

`=> B`

Probability, MET2 2015 VCAA 14 MC

Consider the following discrete probability distribution for the random variable `X.`

The mean of this distribution is

`2`
`3`
`7/2`
`11/3`
`4`

Show Answers Only

`D`

Show Worked Solution

`text(Find)\ p:`

`p + 2p + 3p + 4p + 5p`	`= 1`
`:. p`	`= 1/15`

`text(E)(X)`	`= 1 xx p + 2(2p) + 3(3p) + 4(4p) + 5(5p)`
	`= 55p`
	`= 55 xx (1/15)`
	`= 11/3`

`=> D`

Calculus, MET2 2015 VCAA 4 MC

Consider the tangent to the graph of `y = x^2` at the point `(2, 4).`

Which of the following points lies on this tangent?

A. `text{(1, −4)}`

B. `(3, 8)`

C. `text{(−2, 6)}`

D. `(1, 8)`

E. `text{(4, −4)}`

Show Answers Only

`B`

Show Worked Solution

`text(Find equation of tangent line)`

`text(at)\ x = 2:`

`y = 4x – 4qquadtext([CAS: tangentLine) (x^2,x,2)]`

`(3,8)\ \ text(is on)\ y = 4x – 4`

`=> B`

Algebra, MET2 2014 VCAA 9 MC

The inverse of the function `f: R^+ -> R,\ f(x) = 1/sqrt x + 4` is

A.	`f^-1: (4, oo) -> R`	`f^-1(x) = 1/(x - 4)^2`
B.	`f^-1: R^+ -> R`	`f^-1(x) = 1/x^2 + 4`
C.	`f^-1: R^+ -> R`	`f^-1(x) = (x + 4)^2`
D.	`f^-1:\ text{(−4, ∞)} -> R`	`f^-1(x) = 1/(x + 4)^2`
E.	`f^-1:\ text{(−∞, 4)} -> R`	`f^-1(x) = 1/(x - 4)^2`

Show Answers Only

`A`

Show Worked Solution

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

`text(Inverse: swap)\ x ↔ y`

`x`	`= 1/sqrty + 4`
`x – 4`	`= 1/sqrty`
`sqrty`	`= 1/(x – 4)`
`y`	`= 1/((x – 4)^2) = f^(−1)(x)`

`text(Domain)(f^(−1)) = text(Range)\ (f) = (4,∞)`

`=> A`

Graphs, MET2 2014 VCAA 2 MC

The linear function `f: D -> R, f (x) = 4 - x` has range `text{[−2, 6)}`.

The domain `D` of the function is

`text{[−2, 6)}`
`text{(−2, 2]}`
`R`
`text{(−2, 6]}`
`text{[−6, 2]}`

Show Answers Only

`D`

Show Worked Solution

`:. D = (−2,6)`

`=> D`

Graphs, MET2 2014 VCAA 1 MC

The point `P\ text{(4, −3)}` lies on the graph of a function `f`. The graph of `f` is translated four units vertically up and then reflected in the `y`-axis.

The coordinates of the final image of `P` are

`text{(−4, 1)}`
`text{(−4, 3)}`
`text{(0, −3)}`
`text{(4, −6)}`
`text{(−4, −1)}`

Show Answers Only

`A`

Show Worked Solution

`text(Using mapping notation:)`

`P(4,−3)\ {:(y + 4),(vec(quad(1)quad)):}\ (4,1)\ {:(−x),(vec(\ (2)\ )):}\ pprime(−4,1)`

`=> A`

MATRICES, FUR1 2014 VCAA 2 MC

`y - z`	`= 8`
`5x - y`	`= 0`
`x + z`	`= 4`

The system of three simultaneous linear equations above can be written in matrix form as

A.	`[[0,1,-1],[0,5,-1],[1,0,1]][[x],[y],[z]]=[[8],[0],[4]]`	B.	`[[0,1,-1],[5,-1,0],[1,0,1]][[x],[y],[z]]=[[8],[0],[4]]`

C.	`[[1,-1],[5,-1],[1,1]][[x],[y],[z]]=[[8],[0],[4]]`	D.	`[[0,5,1],[1,-1,0],[-1,0,1]][[x],[y],[z]]=[[8],[0],[4]]`

E.	`[[0,5,0],[-1,-1,0],[1,1,0]][[x],[y],[z]]=[[8],[0],[4]]`

Show Answers Only

`B`

Show Worked Solution

`=>B`

Calculus, EXT1 C1 2008 HSC 4a

A turkey is taken from the refrigerator. Its temperature is 5°C when it is placed in an oven preheated to 190°C.

Its temperature, `T`° C, after `t` hours in the oven satisfies the equation

`(dT)/(dt) = -k(T − 190)`.

Show that `T = 190 - 185e^(-kt)` satisfies both this equation and the initial condition. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
The turkey is placed into the oven at 9 am. At 10 am the turkey reaches a temperature of 29°C. The turkey will be cooked when it reaches a temperature of 80°C.

At what time (to the nearest minute) will it be cooked? (3 marks)

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

Show Answers Only

`text(See Worked Solutions)`
`12:44\ text(pm)`

Show Worked Solution

i.	`T`	`= 190 − 185e^(-kt)`
	`(dT)/(dt)`	`= -k xx -185e^(-kt)`
		`= -k(190 − 185e^(-kt) − 190)`
		`= -k(T − 190)`

`:.T = 190 − 185e^(-kt)\ text(satisfies equation.)`

`text(When)\ \ t = 0,`

`T`	`= 190 − 185e^0`
	`= 5°`

`:.\ text(Initial conditions are satisfied.)`

ii. `text(When)\ \ t = 1,\ T = 29`

`29`	`= 190 − 185e^(-k)`
`185e^(-k)`	`= 161`
`e^(-k)`	`= 161/185`
`-k`	`= ln\ 161/185`
`:.k`	`= -ln\ 161/185`
	`= 0.1389…`

`text(Find)\ \ t\ \ text(when)\ \ T = 80 :`

`80`	`= 190 − 185e^(-kt)`
`185e^(-kt)`	`= 110`
`e^(-kt)`	`= 110/185`
`-kt`	`= ln\ 110/185`
`:.t=`	`= ln\ 110/185 -: -0.1389…`
	`= 3.741…`
	`= 3\ text(hours)\ 44\ text{mins (nearest minute)}`

`:.\ text(The turkey will be cooked at 12:44 pm.)`

Calculus, EXT1* C1 2004 HSC 7b

At the beginning of 1991 Australia’s population was 17 million. At the beginning of 2004 the population was 20 million.

Assume that the population `P` is increasing exponentially and satisfies an equation of the form `P = Ae^(kt)`, where `A` and `k` are constants, and `t` is measured in years from the beginning of 1991.

Show that `P = Ae^(kt)` satisfies `(dP)/(dt) =kP`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
What is the value of `A`? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Find the value of `k`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Predict the year during which Australia’s population will reach 30 million. (2 marks)

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

Show Answers Only

`kP`
`1.7 × 10^7`
`0.013\ \ text(to 3 decimal places.)`
`2036`

Show Worked Solution

i.	`P`	`= Ae^(kt)`
	`(dP)/(dt)`	`= kAe^(kt)`
		`= kP`

ii. `P = Ae^(kt)`

`text(When)\ \ t = 0, \ P = 1.7 × 10^7`

`1.7 × 10^7`	`= Ae^0`
`1.7 × 10^7`	`= A xx 1`
`:. A`	`= 1.7 × 10^7`

iii. `P = 1.7 × 10^7e^(kt)`

`text(When)\ t = 13, \ P = 2 × 10^7`

`2 × 10^7`	`= 1.7 × 10^7e^(13k)`
`(2 × 10^7)/(1.7 × 10^7)`	`= e^(13k)`
`ln (2/(1.7))`	`= ln e^(13k)`
	`= 13k`
`:.k`	`= 1/13 ln (2/(1.7))`
	`= 0.0125…`
	`= 0.013\ \ \ text{(to 3 d.p.)}`

iv. `P`

`= 1.7 × 10^7e^(kt)`

`text(Find)\ \ t\ \ text(when)\ \ P = 3 × 10^7,`

MARKER’S COMMENT: Many students had the correct calculations but didn’t answer the question by identifying the exact year and lost a valuable mark.

`3 × 10^7`	`= 1.7 × 10^7e^(kt)`
`(3 × 10^7)/(1.7 × 10^7)`	`= e^(kt)`
`ln (3/(1.7))`	`= ln e^(kt)`
`ln (3/(1.7))`	`= kt ln e`
	`= kt`
`:.t`	`= (ln (3/(1.7)))/k`
	`= (ln(3/(1.7)))/(0.0125…)`
	`= 45.433…`
	`= 45.4\ \ text{years (to 1 d.p.)}`

`:.\ text(The population will reach 30 million in 2036.)`

Plane Geometry, 2UA 2004 HSC 6b

The diagram shows a right-angled triangle `ABC` with `∠ABC = 90^@`. The point `M` is the midpoint of `AC`, and `Y` is the point where the perpendicular to `AC` at `M` meets `BC`.

Show that `ΔAYM ≡ ΔCYM`. (2 marks)
Suppose that it is also given that `AY` bisects `∠BAC`. Find the size of `∠YCM` and hence find the exact ratio `MY : AC`. (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`1 : 2sqrt3`

Show Worked Solution

(i)

`text(In)\ ΔAYM\ text(and)\ ΔCYM`

MARKER’S COMMENT: Markers strongly recommend that students copy the diagram in geometry questions and label it along with the workings of their solution.

`∠AMY`	`= ∠CMY = 90^@\ \ \ (MY ⊥ AC)`
`AM`	`=CM\ \ \ text{(given)}`
`YM\ text(is common)`

`:.ΔAYM ≡ ΔCYM\ \ text{(SAS)}`

(ii) `text(Let)\ \ /_BAC = 2 theta`

`∠YAB = ∠YAM = theta\ \ \ \ (AY\ text(bisects)\ ∠BAC)`

`∠YAM`	`= ∠YCM=theta`	`\ \ \ text{(corresponding angles of}`
		`\ \ \ text{congruent triangles)}`
`:.∠YAB= ∠ YAM = ∠YCM=theta`

♦♦ Very few students answered this part correctly.
MARKER’S COMMENT: Students who began by making `∠BAC=2theta` were the most successful.

`text(In)\ \ ΔABC,`

`∠ABC + ∠BAC + ∠YCM`	`= 180^@`
`:.90^@ +2theta+theta`	`= 180^@`
`3theta`	`= 90^@`
`:.theta`	`= 30^@`

`text(In)\ \ Delta MYC,`

`:.tan 30^@`	`= (MY)/(MC)`
`(MY)/(MC)`	`= 1/sqrt3`
`(MY)/(2MC)`	`=1/(2 sqrt3)`
`(MY)/(AC)`	`= 1/(2sqrt3)\ \ \ text{(given}\ \ AC=2MC text{)}`
`:.MY : AC`	`= 1 : 2sqrt3.`

Calculus, 2ADV C3 2004 HSC 5b

A particle moves along a straight line so that its displacement, `x` metres, from a fixed point `O` is given by `x = 1 + 3 cos 2t`, where `t` is measured in seconds.

What is the initial displacement of the particle? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Sketch the graph of `x` as a function of `t` for `0 ≤ t ≤ pi`. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Hence, or otherwise, find when AND where the particle first comes to rest after `t = 0`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Find a time when the particle reaches its greatest magnitude of velocity. What is this velocity? (2 marks)

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

Show Answers Only

`text(4 m to the right of)\ O.`
`text(See Worked Solutions)`
`t = pi/2\ text(seconds, 2 m to the left of)\ O.`
`text(6 m s)^(−1)`

Show Worked Solution

i. `x = 1 + 3 cos 2t`

`text(When)\ \ t = 0,`

`x`	`= 1 + 3 cos 0`
	`= 1 + 3`
	`= 4`

`:.\ text(Initial displacement is 4 m to the right of)\ O.`

ii. `text(Period)\ = (2pi)/n = (2pi)/2 = pi`

`text(Considering the range)`

`-1`	`<=cos 2t<=1`
`-3`	`<=3cos 2t<=3`
`-2`	`<=1 + 3 cos 2t<=4`

iii.	`x`	`= 1 + 3 cos 2t`
	`:.v`	`= −6 sin 2t`

`text(The particle comes to rest when)\ \ v=0`

`-6 sin 2t`	`= 0`
`sin 2t`	`= 0`
`2t`	`= 0, pi, 2pi…`
`t`	`= 0, pi/2, pi…`

`:.\ text(After)\ \ t=0, text(particle first comes to rest when)`

`t = pi/2\ text(seconds.)`

`text(When)\ t = pi/2,`

`x`	`= 1 + 3 cos 2(pi/2)`
	`= 1 + 3 cos pi`
	`= 1 + 3(−1)`
	`= −2`

`:.\ text(Particle first comes to rest at 2 m to the left of)\ O.`

iv. `x = 1 + 3 cos 2t`

`v`	`= -6 sin 2t`
`a`	`= -12 cos 2t`

`text(MAX occurs when)\ \ a=0`

`−12 cos 2t`	`= 0`
`cos 2t`	`= 0`
`2t`	`= pi/2, (3pi)/2, …`
`t`	`= pi/4, (3pi)/4, …`

`:.\ text(Maximum at)\ \ t=pi/4,\ \ (3pi)/4, …\ text(seconds,)`

`text(When)\ \ t = pi/4\ text(seconds,)`

`v`	`= -6 sin 2(pi/4)`
	`= -6 sin(pi/2)`
	`= −6`

`:.\ text(Maximum is 6 m s)^(−1).`

Financial Maths, 2ADV M1 2004 HSC 5a

Clare is learning to drive. Her first lesson is 30 minutes long. Her second lesson is 35 minutes long. Each subsequent lesson is 5 minutes longer than the lesson before.

How long will Clare’s twenty-first lesson be? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
How many hours of lessons will Clare have completed after her twenty-first lesson? (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
During which lesson will Clare have completed a total of 50 hours of driving lessons? (2 marks)

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

Show Answers Only

`text(130 minutes)`
`28\ text(hours)`
`text(30th lesson)`

Show Worked Solution

i. `30, 35, 40, …`

`=>\ text(AP where)\ \ a = 30,\ \ d = 5`

`T_n`	`= a + (n − 1) d`
`T_(21)`	`= 30 + 20(5)`
	`= 30 + 100`
	`= 130`

`:.\ text(Clare’s twenty-first lesson will be 130)`

`text(minutes long.)`

ii. `text(Find)\ \ S_21\ \ text(given)\ \ a = 30, \ d = 5`

`S_n`	`= n/2[2a + (n − 1)d]`
`:.S_21`	`= 21/2[2(30) + 20 × 5]`
	`= 21/2[60 + 100]`
	`= 21/2 × 160`
	`= 1680\ text(minutes)`
	`= 28\ text(hours.)`

(iii) `text(50 hours)`	`= 50 × 60`
	`= 3000\ text(minutes)`

`text(Find)\ \ n,\ \ text(given)\ \ a = 30, \ d = 5, \ S_n = 3000`

`S_n`	`= n/2[2a + (n − 1)d]`
`:. 3000`	`= n/2[2(30) + (n − 1)5]`
`6000`	`= n[60 + 5n − 5]`
`6000`	`= n[55 + 5n]`
`6000`	`= 55n + 5n^2`

`5n^2 + 55n − 6000`	`= 0`
`n^2 + 11n − 1200`	`= 0`

`:.n`	`= (−11 ± sqrt((11)^2 − 4(1)(−1200)))/(2(1))`
	`= (−11 ± sqrt(121 + 4800))/(2)`
	`= (−11 ± sqrt4921)/2`
	`= 29.5749…\ \ \ (n> 0)`

`:.\ text(Clare completes 50 hours of lessons during)`

`text(her 30th lesson.)`

Calculus, MET2 2012 VCAA 15 MC

If `f prime (x) = 3x^2 - 4`, which one of the following graphs could represent the graph of `y = f (x)`?

Show Answers Only

`B`

Show Worked Solution

`f prime (x) = 3x^2 – 4`

`:. f(x)\ text(is a cubic with two stationary points.)`

`=> B`

Probability, MET2 2012 VCAA 12 MC

Demelza is a badminton player. If she wins a game, the probability that she will win the next game is 0.7. If she loses a game, the probability that she will lose the next game is 0.6. Demelza has just won a game.

The probability that she will win exactly one of her next two games is

A. `0.33`

B. `0.35`

C. `0.42`

D. `0.49`

E. `0.82`

Show Answers Only

`A`

Show Worked Solution

`text(After a win in game 1,)`

`text{Pr(exactly one of next two)}`

`= text(Pr)(WWL) + text(Pr)(WLW)`

`= 1 xx 0.7 xx 0.3 + 1 xx 0.3 xx 0.4`

`= 0.33`

`=> A`

CORE*, FUR2 2014 VCAA 1

The adult membership fee for a cricket club is $150.

Junior members are offered a discount of $30 off the adult membership fee.

Write down the discount for junior members as a percentage of the adult membership fee. (1 mark)

Adult members of the cricket club pay $15 per match in addition to the membership fee of $150.

If an adult member played 12 matches, what is the total this member would pay to the cricket club? (1 mark)

If a member does not pay the membership fee by the due date, the club will charge simple interest at the rate of 5% per month until the fee is paid.

Michael paid the $150 membership fee exactly two months after the due date.

Calculate, in dollars, the interest that Michael will be charged. (1 mark)

The cricket club received a statement of the transactions in its savings account for the month of January 2014.

The statement is shown below.

1. Calculate the amount of the withdrawal on 17 January 2014. (1 mark)
2. Interest for this account is calculated on the minimum balance for the month and added to the account on the last day of the month.
  
  What is the annual rate of interest for this account?
  
  Write your answer, correct to one decimal place. (1 mark)

Show Answers Only

`text(20%)`
`$330`
`$15`
1. `$17\ 000`
2. `3.5`

Show Worked Solution

a. `text(Discount for junior members)`

`= 30/150 xx 100text(%)`

`= 20text(%)`

b.	`text(Match Payments)`	`= 12 xx 15=$180`

`:.\ text(Total paid to the club)`

`= 150 + 180`

`= $330`

c.	`I`	`= (PrT)/100`
		`= (150 xx 5 xx 2)/100`
		`= $15`

d.i. `text(Withdrawal on 17-Jan)`

`= 59\ 700 – 42\ 700`

`= $17\ 000`

d.ii. `text(Minimum Jan balance) = $42\ 700`

`47\ 200 xx r xx 1/12`	`= 125.12`
`:. r`	`= (125.12 xx 12)/(42\ 700)`
	`= 0.0351…`

`:.\ text(Annual interest rate) = 3.5text(%)`

Graphs, MET2 2012 VCAA 6 MC

A section of the graph of `f` is shown below.

The rule of `f` could be

`f (x) = tan (x)`
`f (x) = tan (x - pi/4)`
`f (x) = tan(2 (x - pi/4))`
`f (x) = tan(2 (x - pi/2))`
`f (x) = tan(1/2 (x - pi/2))`

Show Answers Only

`C`

Show Worked Solution

`text(Period) = pi/2`

`=>\ text(must be)\ C\ text(or)\ D`

`text(Shift)\ \ y = tan(x)\ \ text(right)\ \ pi/4.`

`=> C`

Algebra, STD2 A1 SM-Bank 2

If `A = P(1 + r)^n`, find `A` given `P = $300`, `r = 0.12` and `n = 3` (give your answer to the nearest cent). (2 marks)

Show Answers Only

`$421.48\ \ text{(nearest cent)}`

Show Worked Solution

`A`	`= P(1 + r)^n`
	`= 300(1 + 0.12)^3`
	`= 300(1.12)^3`
	`= 421.478…`
	`= $421.48\ \ text{(nearest cent)}`

Algebra, STD2 A1 SM-Bank 1

What is the value of `5a^2 - b`, if `a = −4` and `b = 3`. (2 marks)

Show Answers Only

`77`

Show Worked Solution

`5a^2 − b`	`= 5(−4)^2 − 3`
	`= 5 xx 16 − 3`
	`= 77`

Algebra, MET2 2013 VCAA 5 MC

If `f: text{(−∞, 1)} -> R,\ \ f(x) = 2 log_e (1 - x)\ \ text(and)\ \ g: text{[−1, ∞)} -> R, g(x) = 3 sqrt (x + 1),` then the maximal domain of the function `f + g` is

`text{[−1, 1)}`
`(1, oo)`
`text{(−1, 1]}`
`text{(−∞, −1]}`
`R`

Show Answers Only

`A`

Show Worked Solution

`text(Consider)\ \ f(x) = 2 log_e (1 – x):`

`(1-x)`	`>0`
`:. x`	`<1`

`text(Consider)\ \ g(x) = 3 sqrt (x + 1):`

`(x+1)`	`>=0`
`:. x`	`>= -1`

`:.\ text(The maximal domain of)\ \ f + g\ \ text{is [−1, 1)}.`

`=> A`

Graphs, MET2 2013 VCAA 4 MC

Part of the graph of `y = f(x)`, where `f: R -> R,\ \ f(x) = 3-e^x,` is shown below.

Which one of the following could be the graph of `y = f^{-1}(x),` where `f^{-1}` is the inverse of `f?`

Show Answers Only

`E`

Show Worked Solution

`text(From the graph of)\ \ f(x),`

`f^-1\ \ text(has a vertical asymptote at)\ \ x = 3`

`text(and an)\ \ x text(-intercept at)\ \ (2, 0).`

`=> E`

Algebra, MET2 2013 VCAA 1 MC

The function with rule `f(x) = -3 tan(2 pi x)` has period

`2/pi`
`2`
`1/2`
`1/4`
`2 pi`

Show Answers Only

`C`

Show Worked Solution

`text(Period)`	`= pi/n`
	`=pi/(2 pi)`
	`= 1/2`

`=> C`

Functions, EXT1′ F2 2015 HSC 14b

The cubic equation `x^3 – px + q = 0` has roots `alpha, beta` and `gamma`.

It is given that `alpha^2 + beta^2 + gamma^2 = 16` and `a^3 + beta^3 + gamma^3 = -9`.

Show that `p = 8.` (1 mark)
Find the value of `q.` (2 marks)
Find the value of `alpha^4 + beta^4 + gamma^4.` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`3`
`128`

Show Worked Solution

(i) `alpha + beta + gamma = – b/a=0`

`alpha beta + beta gamma + gamma alpha = c/a= -p`

`(alpha + beta + gamma)^2`	`= alpha^2 + beta^2 + gamma^2 + 2(alpha beta + beta gamma + gamma alpha)`
`0`	`= 16 – 2p`
`:. p`	`= 8`

(ii)	`alpha^3 – 8 alpha + q`	`=0\ \ \ …\ (1)`
	`beta^3 – 8 beta + q`	`=0\ \ \ …\ (2)`
	`gamma^3 – 8 gamma + q`	`=0\ \ \ …\ (3)`

`(1)+(2)+(3)`

`(alpha^3 + beta^3 + gamma^3) – 8(alpha + beta + gamma) + 3q = 0`

`-9 – 0 + 3q`	`=0`
`q`	`=3`

COMMENT: Part (iii) proved challenging for many students with the State mean mark just over 50%.

(iii)	`alpha(alpha^3 – 8 alpha + 3)`	`=0\ \ \ …\ (1)`
	`beta(beta^3 – 8 beta + 3)`	`=0\ \ \ …\ (2)`
	`gamma(gamma^3 – 8 gamma + 3)`	`=0\ \ \ …\ (3)`

`(1)+(2)+(3)`

`alpha^4 + beta^4 + gamma^4 – 8(alpha^2 + beta^2 + gamma^2) + 3(alpha + beta + gamma) = 0`

`alpha^4 + beta^4 + gamma^4`	`= 8 xx 16 – 0`
	`= 128`

Calculus, EXT2 C1 2015 HSC 14a

Differentiate `sin^(n - 1) theta cos theta`, expressing the result in terms of `sin theta` only. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, or otherwise, deduce that

`int_0^(pi/2) sin^n theta\ d theta = ((n-1))/n int_0^(pi/2) sin^(n - 2) theta\ d theta`, for `n>1.` (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Find `int_0^(pi/2) sin^4 theta\ d theta.` (1 mark)

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

Show Answers Only

`text(See Worked Solutions)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(3 pi)/16`

Show Worked Solution

i. `d/(d theta) (sin^(n – 1) theta cos theta)`

`=(n – 1) sin^(n – 2) theta cos theta cos theta + sin^(n – 1) theta xx (-sin theta)`

`=(n – 1) sin^(n – 2) theta cos^2 theta – sin^n theta`

`=(n – 1) sin^(n – 2) theta (1 – sin^2 theta) – sin^n theta`

`=(n – 1) sin^(n – 2) theta – (n – 1) sin^n theta – sin^n theta`

`=(n – 1) sin^(n – 2) theta – n sin^n theta`

ii. `text{From part (i)}`

`n sin^n theta = (n – 1) sin^(n – 2) theta – d/(d theta) (sin^(n – 1) theta cos theta)`

`:. int_0^(pi/2) sin^n theta\ d theta`

`=1/n int_0^(pi/2) ((n – 1) sin^(n – 2) theta – d/(d theta) (sin^(n – 1) theta cos theta)) d theta`

`=1/n int_0^(pi/2) (n – 1) sin^(n – 2) theta\ d theta – 1/n int_0^(pi/2) d/(d theta) (sin^(n – 1) theta cos theta)\ d theta`

`= 1/n int_0^(pi/2) (n – 1) sin^(n – 2) theta\ d theta – 1/n int_0^(pi/2) d/(d theta) (sin^(n – 1) theta cos theta)\ d theta`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta – 1/n [sin^(n – 1) theta cos theta]_0^(pi/2)`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta – 1/n (0 – 0)`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta,\ \ \ \ (n>1)`

iii.	`int_0^(pi/2) sin^4 theta\ d theta`	`= 3/4 int_0^(pi/2) sin^2 theta\ d theta`
		`= 3/4 xx [(2-1)/2 int_0^(pi/2) d theta]`
		`= 3/8 xx [theta]_0^(pi/2)`
		`= (3 pi)/16`

Harder Ext1 Topics, EXT2 2015 HSC 13c

A small spherical balloon is released and rises into the air. At time `t` seconds, it has radius `r` cm, surface area `S = 4 pi r^2` and volume `V = 4/3 pi r^3`.

As the balloon rises it expands, causing its surface area to increase at a rate of `((4 pi)/3)^(1/3)\ \text(cm)^2 text(s)^-1`. As the balloon expands it maintains a spherical shape.

By considering the surface area, show that
1. `(dr)/(dt) = 1/(8 pi r) (4/3 pi)^(1/3).` (2 marks)
Show that
1. `(dV)/(dt) = 1/2 V^(1/3).` (2 marks)
When the balloon is released its volume is `8000\ text(cm³)`. When the volume of the balloon reaches `64000\ text(cm³)` it will burst.
How long after it is released will the balloon burst? (2 marks)

Show Answers Only

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

Show Worked Solution

(i)	`(dS)/(dt)`	`= (dS)/(dr) xx (dr)/(dt)`
	`((4 pi)/3)^(1/3)`	`=8 pi r xx (dr)/(dt)`
	`(dr)/(dt)`	`=((4 pi)/3)^(1/3) xx 1/(8 pi r)`
		`=1/(8 pi r) (4/3 pi)^(1/3)`

(ii) `(dV)/(dt) = (dV)/(dr) xx (dr)/(dt)`

`(dV)/(dt) =`	`4 pi r^2 xx ((4 pi)/3)^(1/3) xx 1/(8 pi r)`
`=`	`r/2 xx ((4 pi)/3)^(1/3)`
`=`	`1/2 (4/3 pi r^3)^(1/3)`
`=`	`1/2 V^(1/3)`

(iii) `text(Find)\ \ t\ \ text(when)\ \ V=64\ 000`

`text(When)\ \ t = 0,\ \ V = 8000`

`(dV)/(dt)`	`=1/2 V^(1/3)`
`(dt)/(dV)`	`=2/V^(1/3)`
`int_0^t\ dt`	`= int_8000^64000 2/V^(1/3)\ dV`
`:.t`	`= [3V^(2/3)]_8000^64000`
	`= 3(1600 – 400)`
	`= 3600\ \ text(seconds)`
	`= 1\ \ text(hour)`

Conics, EXT2 2015 HSC 13a

The hyperbolas `H_1:\ \ x^2/a^2 - y^2/b^2 = 1` and `H_2:\ \ x^2/a^2 - y^2/b^2 = -1` are shown in the diagram.

Let `P(a sec theta, b tan theta)` lie on `H_1` as shown on the diagram.

Let `Q` be the point `(a tan theta, b sec theta)`.

Verify that the coordinates of `Q(a tan theta, b sec theta)` satisfy the equation for `H_2.` (1 mark)
Show that the equation of the line `PQ` is `bx + ay = ab (tan theta + sec theta).` (2 marks)
Prove that the area of `Delta OPQ` is independent of `theta.` (3 marks)

Show Answers Only

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

Show Worked Solution

(i) `text(Substitute)\ \ Q(a tan theta, b sec theta)\ \ text(into)`

`\ \ x^2/a^2 – y^2/b^2 = -1`

`text(LHS)`	`=(a^2 tan^2 theta)/a^2 – (b^2 sec^2 theta)/b^2`
	`=tan^2 theta – sec^2 theta\ \ \ \ \ \ (sec^2 theta = tan^2 theta +1)`
	`=-1`

`:. Q\ \ text(lies on)\ \ H_2`

(ii)	`m_(PQ)`	`=(b tan theta-b sec theta)/(a sec theta- a tan theta)`
		`=(-b(sec theta – tan theta))/(a(sec theta – tan theta))`
		`=-b/a`

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

`(y – b tan theta)`	`=-b/a (x – a sec theta)`
`ay – ab tan theta`	`=-bx + ab sec theta`
`bx + ay`	`=ab (tan theta + sec theta)`

♦ Mean mark 47%.

(iii) `text(Area)\ \ Delta OPQ = 1/2 xx QP xx d,\ \ text(where)`

`d= text(Perpendicular distance from)\ \ O\ \ text(to)\ \ QP`

`d`	`= \|\ (-ab(tan theta + sec theta))/sqrt (a^2 + b^2)\ \|`
	`= (ab (tan theta + sec theta))/sqrt (a^2 + b^2)`

`text(Distance)\ \ QP`

`QP^2`	`=(a sec theta – a tan theta)^2 + (b tan theta – b sec theta)^2`
	`=a^2(sec theta – tan theta)^2+b^2(sec theta – tan theta)^2`
	`=(a^2+b^2)(sec theta – tan theta)^2`
`QP`	`=sqrt(a^2+b^2) *\|\ sec theta – tan theta\ \|`
	`=sqrt(a^2+b^2)(sec theta – tan theta),\ \ \ \ \ (sec theta>=tan theta\ \ text(for)\ \ 0<=theta<=90^@)`

`text(Area)\ \ Delta OPQ`	`=1/2 sqrt(a^2+b^2)(sec theta – tan theta)*(ab (tan theta + sec theta))/sqrt(a^2 + b^2)`
	`= (ab)/2 xx (sec theta – tan theta) (sec theta + tan theta)`
	`= (ab)/2 xx (sec^2 theta – tan^2 theta)`
	`= (ab)/2`

`:.\ text(Area)\ \ Delta OPQ\ \ text(is independent of)\ \ theta.`

Functions, EXT1′ F1 2015 HSC 12c

By writing `((x -2) (x - 5))/(x - 1)` in the form `mx + b + a/(x - 1)`, find the equation of the oblique asymptote of `y = ((x -2) (x - 5))/(x - 1).` (2 marks)
Hence sketch the graph `y = ((x -2) (x - 5))/(x - 1)`, clearly indicating all intercepts and asymptotes. (2 marks)

Show Answers Only

`y = x − 6`
`text(See Worked Solutions)`

Show Worked Solution

i. `((x – 2) (x – 5))/(x – 1)`	`= (x^2 – 7x + 10)/(x – 1)`
	`= (x(x – 1) – 6 (x – 1) + 4)/(x – 1)`
	`= x – 6 + 4/(x – 1)`

`:.text(Equation of oblique asymptote is)\ \ y = x – 6`

ii. `text(Vertical asymptote at)\ \ x = 1`

`x text(-intercepts)\ \ 2 and 5,\ \ y text(-intercept) = -10`

Polynomials, EXT2 2015 HSC 12b

The polynomial `P(x) = x^4 - 4x^3 + 11x^2 - 14x + 10` has roots `a + ib` and `a + 2ib` where `a` and `b` are real and `b != 0.`

By evaluating `a` and `b`, find all the roots of `P(x).` (3 marks)
Hence, or otherwise, find one quadratic polynomial with real coefficients that is a factor of `P(x).` (1 mark)

Show Answers Only

`1 +- i,\ \ 1 +- 2i`
`x^2 − 2x + 2\ \ or\ \ x^2 − 2x + 5`

Show Worked Solution

(i) `text(S)text(ince coefficients of)\ \ P(x)\ \ text(are real,)`

`=>\ text(Complex roots occur in conjugate pairs)`

`=>\ text(Roots are)\ \ a +- ib\ \ and\ \ a +- 2ib`

`text(Sum of roots) = -b/a=4`

`4`	`=a + ib + a – ib + a + 2ib + a – 2ib`
`4a`	`=4`
`:.a`	`=1`

`text(Products of roots)`

`(a + ib) (a – ib) (a – 2ib) (a – 2ib)`	`= 10`
`(a^2 + b^2) (a^2 + 4b^2)`	`= 10`
`(1 + b^2) (1 + 4b^2)`	`= 10`
`4b^4 + 5b^2 + 1`	`= 10`
`4b^4 + 5b^2 – 9`	`= 0`
`(4b^2 + 9) (b^2 – 1)`	`= 0`

`:.b^2 = 1,\ \ \ \ (b\ \ text{is real})`

`:.b = +- 1`

`:.P(x)\ text(has roots)\ \ \ 1 +- i,\ 1 +- 2i.`

(ii)	`P(x)`	`=(x – 1 – i) (x – 1 + i)(x-1-2i)(x-1+2i)`
		`=(x^2 – 2x + 2)(x^2 – 2x + 5)`

Complex Numbers, EXT2 N1 2015 HSC 12a

The complex number `z` is such that `|\ z\ |=2` and `text(arg)(z) = pi/4.`

Plot each of the following complex numbers on the same half-page Argand diagram.

`z` (1 mark)

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`u = z^2` (1 mark)

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`v = z^2 - bar z` (1 mark)

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

Show Answers Only

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

Show Worked Solution

i. `z =`	`2 text(cis) pi/4`
`=`	`sqrt 2 (1 + i)`

ii. `u =`	`z^2`
`=`	`4 text(cis)\ pi/2`
`=`	`4i`

COMMENT: 12a(iii) had the lowest mean mark (55%) of any part within Q12 and deserves attention.

iii. `v =`	`z^2 – bar z`
`=`	`4i – sqrt 2 (1 – i)`
`=`	`- sqrt 2 + (4 + sqrt 2) i`

Calculus, EXT2 C1 2015 HSC 11f

Show that

`cot theta + text(cosec)\ theta = cot(theta/2).` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, or otherwise, find

`int (cot theta + text(cosec)\ theta)\ d theta.` (1 mark)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`2 ln\ |sin\ theta/2| + c`

Show Worked Solution

i. `cot theta + text(cosec)\ theta =`	`(cos theta)/(sin theta) + 1/(sin theta)`
`=`	`(1 + cos theta)/(sin theta)`
`=`	`(1 + 2 cos^2 (theta/2) – 1)/(2 sin (theta/2) cos (theta/2))`
`=`	`(2 cos^2(theta/2))/(2 sin (theta/2) cos (theta/2))`
`=`	`(cos (theta/2))/(sin (theta/2))`
`=`	`cot (theta/2)`

COMMENT: Part (ii) mean mark 51%.

ii. `int (cot theta + text(cosec)\ theta)\ d theta =`	`int cot (theta/2)\ d theta`
`=`	`int (cos (theta/2))/(sin (theta/2))\ d theta`
`=`	`2 ln\ \|sin\ theta/2\| + c`

Graphs, EXT2 2015 HSC 11e

Find the value of `(dy)/(dx)` at the point `(2, text(−1))` on the curve `x + x^2 y^3 = -2.` (3 marks)

Show Answers Only

`1/4`

Show Worked Solution

`x + x^2 y^3 = -2`

`1 + 2xy^3 + x^2 3y^2* (dy)/(dx)`	`= 0`
`3x^2 y^2* (dy)/(dx)`	`= -(1 + 2xy^3)`
`(dy)/(dx)`	`= (-(1 + 2xy^3))/(3x^2 y^2)`

`text(At)\ \ P(2, text(−1)),`

`(dy)/(dx) =`	`(-(1 – 4))/12`
`=`	`1/4`

Complex Numbers, EXT2 N1 2015 HSC 11b

Consider the complex numbers `z = -sqrt 3 + i` and `w = 3 (cos\ pi/7 + i sin\ pi/7).`

Evaluate `|\ z\ |.` (1 mark)

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Evaluate `text(arg)(z).` (1 mark)

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Find the argument of `z/w.` (1 mark)

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

Show Answers Only

`2`
`(5 pi)/6`
`(29 pi)/42`

Show Worked Solution

i. `\|\ z\ \|`	`= sqrt ((-sqrt3)^2 + 1^2)`
	`= 2`

ii. `text(arg)\ (z) =`	`tan^-1 (1- sqrt 3)`
`=`	`pi – pi/6`
`=`	`(5 pi)/6`

iii. `text(arg) (z/w) =`	`text(arg)\ z – text(arg)\ w`
`=`	`(5 pi)/6 – pi/7`
`=`	`(29 pi)/42`

`text(Area)`	`=1/2h(a+b)`
	`= 1/2xx 60(60 + 80)`
	`= 4200`

`text(GP where)\ \ \ a`	`=2, and`
`r`	`= t_2/t_1=6/2=3`

`t_7`	`= ar^6`
	`= 2 xx 3^6`
	`= 1458`

`text(Commission)`	`= 20/500`
	`= 0.04`
	`= 4text (%)`

`text(Volume difference)`	`= 300 – 200`
	`= 100\ text(L)`

`text(Rate of increase)`	`= (450 – 300)/ (10 – 8)`
	`= 75\ text(L/hr)`

`text(At)\ (6, 4),\ \ P`	`= 2 (6) – 4`
	`= 8`

`text(At)\ (6, 1),\ \ P`	`= 2 (6) – 1`
	`= 11`