Band 4 – Page 70

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 4 MC

The head circumference (in cm) of a population of infant boys is normally distributed with a mean of 49.5 cm and a standard deviation of 1.5 cm.

Four hundred of these boys are selected at random and each boy’s head circumference is measured.

The number of these boys with a head circumference of less than 48.0 cm is closest to

A. `3`

B. `10`

C. `64`

D. `272`

E. `336`

Show Answers Only

`C`

Show Worked Solution

`bar x=49.5,\ \ \ s=1.5`

`z text{-score (49.5)`	`=(x-bar x)/s`
	`=(48.0-49.5)/1.5`
	`=– 1`

`:.\ text(Number of boys with a head under 48.0 cm)`

`=16text(%) xx 400`

`=64`

`=> C`

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`

CORE*, FUR1 2007 VCAA 8 MC

The first four terms of a sequence are

`12, 18, 30, 54`

A difference equation that generates this sequence is

A. `t_(n+1)`	`= t_n + 6`	`\ \ \ \ t_1 = 12`
B. `t_(n+1)`	`= 1.5t_n`	`\ \ \ \ t_1 = 12`
C. `t_(n+1)`	`= 0.5t_n + 12`	`\ \ \ \ t_1 = 12`
D. `t_(n+1)`	`= 2t_n - 6`	`\ \ \ \ t_1 = 12`
E. `t_(n+2)`	`= t_(n+1) + t_n`	`\ \ \ \ t_1 = 12, t_2 = 18`

Show Answers Only

`D`

Show Worked Solution

`text(Calculating)\ t_3\ \ text(in each given option)`

`text(eliminates)\ A, B\ text(and)\ C.`

`text(Consider)\ E,`

`t_4 != t_3 + t_2\ \ text(which eliminates)\ E.`

`rArr D`

PATTERNS, FUR1 2007 VCAA 7 MC

The first term, `t_1`, of a geometric sequence is positive.

The common ratio of this sequence is negative.

A graph that could represent the first five terms of this sequence is

Show Answers Only

`E`

Show Worked Solution

`text(General terms are)\ \ \ a, ar, ar^2, …`

`text(By elimination,)`

`a >0,\ text(Eliminate D.)`

`r<0,\ \ text(terms will alternate between positive and)`

`text(negative. Eliminate A, B and C.`

`rArr E`

CORE, FUR1 2007 VCAA 11-13 MC

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

The time series plot below shows the revenue from sales (in dollars) each month made by a Queensland souvenir shop over a three-year period.

Part 1

This time series plot indicates that, over the three-year period, revenue from sales each month showed

A. no overall trend.

B. no correlation.

C. positive skew.

D. an increasing trend only.

E. an increasing trend with seasonal variation.

Part 2

A three median trend line is fitted to this data.

Its slope (in dollars per month) is closest to

A. `125`

B. `146`

C. `167`

D. `188`

E. `255`

Part 3

The revenue from sales (in dollars) each month for the first year of the three-year period is shown below.

If this information is used to determine the seasonal index for each month, the seasonal index for September will be closest to

A. `0.80`

B. `0.82`

C. `1.16`

D. `1.22`

E. `1.26`

Show Answers Only

`text (Part 1:)\ E`

`text (Part 2:)\ C`

`text (Part 3:)\ E`

Show Worked Solution

`text (Part 1)`

`text(The time series plot clearly shows an increasing)`

`text(trend and a seasonal spike and drop over the)`

`text(Summer months.)`

`rArr E`

`text (Part 2)`

♦ Mean mark 37%.
MARKERS’ COMMENT: A common error was to incorrectly use the 6 month and 30 month data points as the median.

`text(From the graph, the median of the bottom third)`

`text(of data points is)\ \ (6.5, 3000).`

`text(From the graph, the median of the top third of)`

`text(data points is)\ \ (30.5, 7000).`

`:.\ text(Gradient of the three median line)`

`=(7000 – 3000)/(30.5 – 6.5)`

`=166.66…`

`rArr C`

`text (Part 3)`

`text(Average monthly sales)\ `	`= (43\ 872)/12`
	`=3656`

`:.\ text(Seasonal index for September)`

`=4597/3656`

`=1.257…`

`rArr E`

CORE, FUR1 2007 VCAA 7-8 MC

The lengths and diameters (in mm) of a sample of jellyfish selected were recorded and displayed in the scatterplot below. The least squares regression line for this data is shown.

The equation of the least squares regression line is

length = 3.5 + 0.87 × diameter

The correlation coefficient is `r = 0.9034`

Part 1

Written as a percentage, the coefficient of determination is closest to

`0.816 text(%)`
`0.903text(%)`
`81.6text(%)`
`90.3text(%)`
`95.0text(%)`

Part 2

From the equation of the least squares regression line, it can be concluded that for these jellyfish, on average

there is a 3.5 mm increase in diameter for each 1 mm increase in length.
there is a 3.5 mm increase in length for each 1 mm increase in diameter.
there is a 0.87 mm increase in diameter for each 1 mm increase in length.
there is a 0.87 mm increase in length for each 1 mm increase in diameter.
there is a 4.37 mm increase in diameter for each 1 mm increase in length.

Show Answers Only

`text (Part 1:)\ C`

`text (Part 2:)\ D`

Show Worked Solution

`text (Part 1)`

`r^2`	`=0.9034^2`
	`=0.8161…`

`rArr C`

`text(Part 2)`

`text(Length)\ =3.5 + 0.87 xx text(diameter)`

`text(Gradient)\ = 0.87`

`text(i.e. the length increases 0.87 mm for each 1 mm)`

`text(increase in diameter.)`

`rArr D`

CORE, FUR1 2007 VCAA 5-6 MC

Samples of jellyfish were selected from two different locations, A and B. The diameter (in mm) of each jellyfish was recorded and the resulting data is summarised in the boxplots shown below.

Part 1

The percentage of jellyfish taken from location A with a diameter greater than 14 mm is closest to

`2text(%)`
`5text(%)`
`25text(%)`
`50text(%)`
`75text(%)`

Part 2

From the boxplots, it can be concluded that the diameters of the jellyfish taken from location A are generally

similar to the diameters of the jellyfish taken from location B.
less than the diameters of the jellyfish taken from location B and less variable.
less than the diameters of the jellyfish taken from location B and more variable.
greater than the diameters of the jellyfish taken from location B and less variable.
greater than the diameters of the jellyfish taken from location B and more variable.

Show Answers Only

`text (Part 1:)\ C`

`text (Part 2:)\ C`

Show Worked Solution

`text (Part 1)`

`text(Consider the boxplot of location A,)`

`Q_3 = 14`

`:.\ text(25% will have a diameter greater than 14 mm.)`

`rArr C`

`text (Part 2)`

`text(The boxplot of location A has a lower median and a)`

`text(higher range than the boxplot of location B.)`

`rArr C`

CORE, FUR1 2007 VCAA 4 MC

The length of 3-month-old baby boys is approximately normally distributed with a mean of 61.1 cm and a standard deviation of 1.6 cm.

The percentage of 3-month-old baby boys with length greater than 59.5 cm is closest to

A. `5text(%)`

B. `16text(%)`

C. `68text(%)`

D. `84text(%)`

E. `95text(%)`

Show Answers Only

`D`

Show Worked Solution

`z text{-score (59.5)}`	`=(x-bar x)/s`
	`=(59.5 – 61.1)/1.6`
	`= – 1`

`:.\ text(% of baby boys longer than 59.5 cm)`

`= 34 + 50`

`=84 text(%)`

`rArr D`

CORE*, FUR1 2008 VCAA 4 MC

In 2008, there are 800 bats living in a park.

After 2008, the number of bats living in the park is expected to increase by 15% per year.

Let `Β_n` represent the number of bats living in the park `n` years after 2008.

A difference equation that can be used to determine the number of bats living in the park `n` years after 2008 is

A. `B_n=1.15B_(n-1)-800`	`\ \ \ \ \ B_0=2008`
B. `B_n=B_(n-1)+1.15xx800`	`\ \ \ \ \ B_0=2008`
C. `B_n=B_(n-1)-0.15xx800`	`\ \ \ \ \ B_0=800`
D. `B_n=0.15B_(n-1)`	`\ \ \ \ \ B_0=800`
E. `B_n=1.15B_(n-1)`	`\ \ \ \ \ B_0=800`

Show Answers Only

`E`

Show Worked Solution

`B_0=800`

`B_1= B_0 + 15 text(%) xx B_0=1.15 B_0`

`B_2= 1.15B_1`

`vdots`

`B_n=1.15 B_(n-1)`

`=> E`

Complex Numbers, SPEC1 2011 VCAA 4

Consider `z = (1 - sqrt 3 i)/(-1 + i),\ \ z in C.`

Find the principal argument of `z` in the form `k pi,\ k in R.` (3 marks)

Show Answers Only

`text(Arg)(z) = 11/12 pi`

Show Worked Solution

`z_1`	`= 1 – sqrt3 i`
	`= sqrt(1^2 + 3)\ text(cis)(−tan^(−1)(sqrt3))`
	`= 2text(cis)(−pi/3)`

`z_2`	`= −1 + i`
	`= sqrt(1^2 + 1^2)\ text(cis)(pi – tan^(−1)(1))`
	`= sqrt2 text(cis)(pi – pi/4)`
	`= sqrt2 text(cis)((3pi)/4)`

`text(Arg)((1 – sqrt3 i)/(−1 + i))`	`= text(Arg)(1 – sqrt3 i) – text(Arg)(−1 + i)`
	`= -pi/3 – (3pi)/4`
	`= -(13pi)/12`

`-(13pi)/12 + 2pi`	`= (-13pi+ 24pi)/12`
	`= (11pi)/12`

PATTERNS, FUR1 2013 VCAA 7 MC

The following are either three consecutive terms of an arithmetic sequence or three consecutive terms of a geometric sequence.

Which one of these sequences could not include 2 as a term?

A. `–1, 0.5, –0.25`

B. `–1, –3, –5`

C. `5, 12.5, 31.25`

D. `6, 8, 10`

E. `8, 16, 32`

Show Answers Only

`B`

Show Worked Solution

`text(By elimination,)`

`text(In A, GP where)\ \ r=0.5/-1=-0.5`

`:.\ text(Term before −1 is 2.)`

`text(In B, AP where)\ \ d=-2`

`:.\ text{Terms before −1 = −1 − (−2) = 1}`

`text{Term before 1 = 1 − (−2) = 3}`

`:.\ text(Does NOT include 2.)`

`text(Other choices can all be shown to contain 2.`

`=> B`

Calculus, MET2 2013 VCAA 14 MC

Consider the graph of `y = 2^x + c`, where `c` is a real number. The area of the shaded rectangles is used to find an approximation to the area of the region that is bounded by the graph, the `x`-axis and the lines `x = 1` and `x = 5.`

If the total area of the shaded rectangles is 44, then the value of `c` is

A. `14`

B. `-4`

C. `14/5`

D. `7/2`

E. `-16/5`

Show Answers Only

`D`

Show Worked Solution

`text(Finding the shaded area:)`

`44`	`=(2^1 + c) 1 + (2^2 + c) 1 + (2^3 + c) 1 + (2^4 + c) 1`
`44`	`=(2+c) + (4+c) + (8+c) + (16+c)`
`14`	`=4c`
`:. c`	`=7/2`

`=> D`

Probability, MET2 2013 VCAA 10 MC

For events `A` and `B,\ text(Pr)(A ∩ B) = p,\ text(Pr)(A′∩ B) = p - 1/8` and `text(Pr)(A ∩ B prime) = (3p)/5.`

If `A` and `B` are independent, then the value of `p` is

`0`
`1/4`
`3/8`
`1/2`
`3/5`

Show Answers Only

`C`

Show Worked Solution

`text{Pr}(A)`	`= text{Pr}(A ∩ B) + text{Pr}(A ∩ B prime)`
	`= p + (3p)/5`
	`= (8p)/5`

`text{Pr}(B)`	`= text{Pr}(B ∩ A) + text{Pr}(B ∩ A prime)`
	`= p + p – 1/8`
	`= 2p – 1/8`

`text(S)text(ince)\ A and B\ text(are independent events,)`

`text{Pr}(A ∩ B)`	` = text{Pr}(A) xx text{Pr}(B)`
`p`	`=(8p)/5 (2p – 1/8)`
`5p`	`=16p^2-p`
`16p^2-6p`	`=0`
`2p(8p-3)`	`=0`
`:.p`	`=3/8,\ \ \ p!=0`

`=> C`

Calculus, SPEC1 2015 VCAA 6

The acceleration `a` ms¯² of a body moving in a straight line in terms of the velocity `v` ms¯¹ is given by `a = 4v^2.`

Given that `v = e` when `x = 1`, where `x` is the displacement of the body in metres, find the velocity of the body when `x = 2.` (4 marks)

Show Answers Only

`e^5`

Show Worked Solution

`v* (dv)/(dx)`	`= 4v^2`
`(dv)/(dx)`	`= 4v`
`(dx)/(dv)`	`= 1/(4v)`
`x`	`=int 1/(4v) \ dv`
	`=1/4 ln (4v)+c`

`text(When)\ \ x=1, v=e`

`1`	`=1/4 ln(4e)+c`
`:.c`	`=1-1/4 ln(4e)`

`text(Find)\ \ v\ \ text(when)\ \ x=2:`

COMMENT: Strong log and exponential calculation ability required here.

`1/4 ln (4v)+1-1/4 ln(4e)`	`= 2`
`1/4 ln(4v)`	`= 1/4 ln (4e)+1`
`ln(4v)`	`= ln(4e)+4`
`e^(ln4v)`	`= e^(ln(4e) +4)`
`4v`	`= e^(ln(4e)) * e^4`
`4v`	`=4e xx e^4`
`:.v`	`=e^5`

Algebra, MET2 2015 VCAA 20 MC

If `f(x - 1) = x^2 - 2x + 3`, then `f(x)` is equal to

`x^2 - 2`
`x^2 + 2`
`x^2 - 2x + 2`
`x^2 - 2x + 4`
`x^2 - 4x + 6`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ g(x) = f(x – 1)`

`text(Shift)\ g(x)\ text(left 1):`

`g(x + 1)`	`= (x + 1)^2 – 2(x + 1) + 3`
	`= x^2 + 2`

`=> B`

Calculus, MET2 2015 VCAA 19 MC

If `f(x) = int_0^x (sqrt(t^2 + 4))\ dt`, then `f prime (– 2)` is equal to

A. `sqrt 2`

B. `- sqrt 2`

C. `2 sqrt 2`

D. `-2 sqrt 2`

E. `4 sqrt 2`

Show Answers Only

`C`

Show Worked Solution

`f(x)`	`= int_0^x (sqrt(t^2 + 4))\ dt`
`f′(x)`	`= sqrt(t^2 + 4)`
`:. f′(− 2)`	`= sqrt((-2)^2+4)`
	`=sqrt8`
	`=2 sqrt2`

`=> C`

Graphs, MET2 2015 VCAA 17 MC

A graph with rule `f(x) = x^3 - 3x^2 + c`, where `c` is a real number, has three distinct `x`-intercepts

The set of all possible values of `c` is

A. `R`

B. `R^+`

C. `{0, 4}`

D. `(0, 4)`

E. `text{(−∞, 4)}`

Show Answers Only

`D`

Show Worked Solution

`text(Consider the graph of)\ \ f(x) = x^3 – 3x^2`

`text(For three)\ xtext(-intercepts, translate up)`

`text{between (0, 4) units}`

`c ∈ (0,4)`

`=> D`

Calculus, MET2 2015 VCAA 15 MC

If `int_0^5 g(x)\ dx = 20` and `int_0^5 (2g(x) + ax)\ dx = 90`, then the value of `a` is

A. `0`

B. `4`

C. `2`

D. `− 3`

E. `1`

Show Answers Only

`B`

Show Worked Solution

`int_0^5 (2g(x) + ax)\ dx`	`=90`
`2int_0^5 g(x)\ dx + int_0^5 ax\ dx`	`= 90`
`2(20) + a/2[x^2]_0^5`	`=90`
`25/2a`	`= 50`
`:. a`	`= 4`

`=> B`

Probability, MET2 2015 VCAA 13 MC

The function `f` is a probability density function with rule

`f(x) = {(ae^x, 0 <= x <= 1), (ae, 1 < x <= 2), (\ 0, text(otherwise)):}`

The value of `a` is

`1`
`e`
`1/e`
`1/(2e)`
`1/(2e - 1)`

Show Answers Only

`E`

Show Worked Solution

`text(Total area) = 1`

`int_0^1 ae^x\ dx + int_1^2 ae\ dx`

`= 1`

`[ae^x]_0^1 + [ae*x]_1^2`	`=1`
`[ae-a] + [2ae-ae]`	`=1`
`a(2e-1)`	`=1`
`:. a`	`= 1/(2e – 1)`

`=> E`

Probability, MET2 2015 VCAA 12 MC

A box contains five red balls and three blue balls. John selects three balls from the box, without replacing them.

The probability that at least one of the balls that John selected is red is

`5/7`
`5/14`
`7/28`
`15/56`
`55/56`

Show Answers Only

`E`

Show Worked Solution

`text{Pr(at least 1 red)}`	`= 1 – text{Pr(no reds)}`
	`= 1 – text(Pr)(B,B,B)`
	`= 1 – (3/8 xx 2/7 xx 1/6)`
	`= 55/56`

`=> E`

Probability, MET2 2015 VCAA 10 MC

The binomial random variable, `X`, has `text(E)(X) = 2` and `text(Var)( X ) = 4/3.`

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

A. `(1/3)^6`

B. `(2/3)^6`

C. `1/3 xx (2/3)^2`

D. `6 xx 1/3 xx (2/3)^5`

E. `6 xx 2/3 xx (1/3)^5`

Show Answers Only

`D`

Show Worked Solution

`np = 2\ …\ (1)`

`np(1 – p) = 4/3\ …\ (2)`

`text(Solve simultaneous equations:)`

`n = 6,quadp = 1/3`

`:. X ∼\ text(Bi)(6, 1/3)`

`:. text(Pr)(X=1)`	`= ((6),(1)) xx (1/3)^1 xx (2/3)^5`
	`= 6 xx 1/3 xx (2/3)^5`

`=> D`

Calculus, MET2 2015 VCAA 8 MC

The graph of a function `f:\ text{[−2, p]} -> R` is shown below.

The average value of `f` over the interval `text{[−2, p]}` is zero.

The area of the shaded region is `25/8.`

If the graph is a straight line, for `0 <= x <= p`, then the value of `p` is

A. `2`

B. `5`

C. `5/4`

D. `5/2`

E. `25/4`

Show Answers Only

`D`

Show Worked Solution

`text(Shaded Area)`	`=\ text(Area below)\ \ f(x)\ text(from)\ \ 0 to p`
`1/2 p^2`	`= 25/8`
`p^2`	`=25/4`
`:. p`	`= 5/2,quadp > 0`

`=> D`

Graphs, MET2 2015 VCAA 7 MC

The range of the function `f:\ text{(−1, 2]} -> R,\ \ f(x) = -x^2 + 2x-3` is

`R`
`text{(−6, −3]}`
`text{(−6, −2]}`
`text{[−6, −3]}`
`text{[−6, −2]}`

Show Answers Only

`C`

Show Worked Solution

`text(A sketch of the equation:)`

`y`	`=-x^2 + 2x-3`
`dy/dx`	`=-2x+2`

`=>\ text(Concave down with turning point when)\ \ x=1,`

`:. text(Range) = (-6,-2]`

`=> C`

Graphs, MET2 2015 VCAA 5 MC

Part of the graph of `y = f(x)` is shown below.

The corresponding part of the graph of the inverse function `y = f^-1(x)` is best represented by

Show Answers Only

`E`

Show Worked Solution

`f^(-1)\ text(is a reflection in)\ \ y=x,`

`=> E`

Calculus, MET2 2014 VCAA 19 MC

Jake and Anita are calculating the area between the graph of `y = sqrt x` and the `y`-axis between `y = 0` and `y = 4.`

Jake uses a partitioning, shown in the diagram below, while Anita uses a definite integral to find the exact area.

The difference between the results obtained by Jake and Anita is

A. `0`

B. `22/3`

C. `26/3`

D. `14`

E. `35`

Show Answers Only

`C`

Show Worked Solution

`text(Jake:)`

`A_J`	`~~ 1[1 + 4 + 9 + 16]`
	`~~ 30`

`text(Anita:)`

`A_A = 16 xx 4 – int_0^16 sqrtx\ dx`

`text(or)`

`A_A`	`= int_0^4 y^2\ dy`
	`= 64/3`

`:.\ text(Difference)`	`= 30 – 64/3`
	`= 26/3`

`=> C`

Algebra, MET2 2014 VCAA 18 MC

The graph of `y = kx - 4` intersects the graph of `y = x^2 + 2x` at two distinct points for

`k = 6`
`k > 6 or k < -2`
`-2 <= k <= 6`
`6 - 2 sqrt 3 <= k <= 6 + 2 sqrt 3`
`k = -2`

Show Answers Only

`B`

Show Worked Solution

`text(Intersect when:)\ kx – 4 = x^2 + 2x`

`x^2 + (2 – k)x + 4 = 0`

`text(2 solutions when)\ \ Delta>0,`

`(2 – k)^2 – 4 xx 4`	`> 0`
`k^2-4k-12`	`>0`

`:. k < − 2 \ \ ∪ \ \ k>6`

`=> B`

Graphs, MET2 2014 VCAA 12 MC

The transformation `T: R^2 -> R^2` with rule

`T ([(x), (y)]) = [(-1, 0), (0, 2)] [(x), (y)] + [(1), (-2)]`

maps the line with equation `x - 2y = 3` onto the line with equation

`x + y = 0`
`x + 4y = 0`
`-x - y = 4`
`x + 4y = -6`
`x - 2y = 1`

Show Answers Only

`C`

Show Worked Solution

`xprime`	`= −x + 1`
`1 – xprime`	`= x`

`yprime`	`= 2y – 2`
`(yprime + 2)/2`	`= y`

`text(Substitute)\ \ x,y\ \ text(into)\ \ x – 2y = 3,`

`(1 – xprime) – 2((yprime + 2)/2)`	`= 3`
`−xprime – yprime`	`= 4`

`=> C`

Probability, MET2 2014 VCAA 11 MC

A bag contains five red marbles and four blue marbles. Two marbles are drawn from the bag, without replacement, and the results are recorded.

The probability that the marbles are different colours is

`20/81`
`5/18`
`4/9`
`40/81`
`5/9`

Show Answers Only

`E`

Show Worked Solution

`text(Pr)(RB) + text(Pr)(BR)`	`= 5/9 xx 4/8 + 4/9 xx 5/8`
	`= 5/9`

`=> E`

Algebra, MET2 2014 VCAA 10 MC

Which one of the following functions satisfies the functional equation `f (f(x)) = x` for every real number `x?`

`f(x) = 2x`
`f(x) = x^2`
`f(x) = 2 sqrt x`
`f(x) = x - 2`
`f(x) = 2 - x`

Show Answers Only

`E`

Show Worked Solution

`text(By trial and error,)`

`text(Consider:)\ \ f(x)=2-x`

`f(f(x))`	`=2-(2-x)`
	`=x`

`=> E`

Calculus, MET2 2014 VCAA 8 MC

If `int_1^4 f(x)\ dx = 6`, then `int_1^4 (5 - 2\ f(x))\ dx` is equal to

A. `3`

B. `4`

C. `5`

D. `6`

E. `16`

Show Answers Only

`A`

Show Worked Solution

`int_1^4 5 – 2\ f(x)\ dx`

`= int_1^4 5\ dx – 2 int_1^4 f(x)\ dx`

`= 15 – 2(6)`

`= 3`

`=> A`

Algebra, MET2 2014 VCAA 6 MC

The function `f: D -> R` with rule `f(x) = 2x^3 - 9x^2 - 168x` will have an inverse function for

`D = R`
`D = (7, oo)`
`D = text{(−4, 8)}`
`D = text{(−∞, 0)}`
`D = [text(−)1/2, oo)`

Show Answers Only

`B`

Show Worked Solution

`text(Inverse exists if)\ f(x)\ text(is)`

`:. D = (7, ∞)\ \ text(gives)\ \ f(x)\ text(one-to-one.)`

`=> B`

Probability, MET2 2014 VCAA 5 MC

The random variable `X` has a normal distribution with mean 12 and standard deviation 0.5.

If `Z` has the standard normal distribution, then the probability that `X` is less than 11.5 is equal to

`text(Pr)(Z > – 1)`
`text(Pr)(Z < – 0.5)`
`text(Pr)(Z > 1)`
`text(Pr)(Z >= 0.5)`
`text(Pr)(Z < 1)`

Show Answers Only

`C`

Show Worked Solution

`text(Pr)(X < 11.5)`	`= text(Pr)(Z < (11.5 – 12)/0.5)`
	`= text(Pr)(Z < −1)`
	`= text(Pr)(Z > 1)`

`=> C`

Calculus, MET2 2014 VCAA 4 MC

Let `f` be a function with domain `R` such that `f (5) = 0` and `f prime (x) < 0` when `x != 5.`

At `x = 5`, the graph of `f` has a

local minimum.
local maximum
gradient of 5
gradient of – 5
stationary point of inflection.

Show Answers Only

`E`

Show Worked Solution

`:.\ text(S)text(ince concavity doesn’t change,)`

`text(point of inflection at)\ \ x = 5.`

`=> E`

Calculus, MET2 2014 VCAA 3 MC

The area of the region enclosed by the graph of `y = x (x + 2) (x − 4)` and the `x`-axis is

A. `128/3`

B. `20/3`

C. `236/3`

D. `148/3`

E. `36`

Show Answers Only

`D`

Show Worked Solution

`text(Area)`	`= int_-2^0 x(x + 2)(x – 4)\ dx – int_0^4 x(x + 2)(x – 4)\ dx`
	`= 148/3`

`=> D`

Mechanics, EXT2* M1 2008 HSC 7

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

The equations of motion of the projectile are

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

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

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

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

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

iv. `text(From the diagram,)`

`r = ON = OP + PN`

`text(In)\ DeltaOLP,`

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

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

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

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

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

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

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

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

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

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

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

MATRICES, FUR1 2014 VCAA 3 MC

Regular customers at a hairdressing salon can choose to have their hair cut by Shirley, Jen or Narj.

The salon has 600 regular customers who get their hair cut each month.

In June, 200 customers chose Shirley (S) to cut their hair, 200 chose Jen (J) to cut their hair and 200 chose Narj (N) to cut their hair.

The regular customers’ choice of hairdresser is expected to change from month to month as shown in the transition matrix, `T`, below.

`{:(qquad qquad qquad qquad text(this month)), (qquad qquad qquad quad {:(S, quad quad J, qquad N):}), (T = [(0.75, 0.10, 0.10), (0.10, 0.75, 0.15), (0.15, 0.15, 0.75)] {:(S), (J), (N):} qquad text(next month)):}`

In the long term, the number of regular customers who are expected to choose Shirley is closest to

A. `150`

B. `170`

C. `185`

D. `195`

E. `200`

Show Answers Only

`B`

Show Worked Solution

`text(Consider)\ \ n =50,`

`[(0.75, 0.10, 0.10), (0.10, 0.75, 0.15), (0.15, 0.15, 0.75)]^50 [(200), (200), (200)] = [(171.4), (203.6), (225)]`

`=>B`

MATRICES, FUR1 2014 VCAA 4 MC

Two hundred and fifty people buy bread each day from a corner store. They have a choice of two brands of bread: Megaslice (M) and Superloaf (S).

The customers’ choice of brand changes daily according to the transition diagram below.

On a given day, 100 of these people bought Megaslice bread while the remaining 150 people bought Superloaf bread.

The number of people who are expected to buy each brand of bread the next day is found by evaluating the matrix product

Show Answers Only

`C`

Show Worked Solution

`text(The)\ 2 xx 2\ text(matrix columns need)`

`text(to add up to 100%)`

`:.\ text(Eliminate)\ A and B.`

`text(The transition diagram corresponds)`

`text(to the matrix equation in)\ C.`

`=>C`

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, 2ADV C3 2004 HSC 10b

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

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

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

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

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

`z^2 = x^2 + y^2 − 2xy cos A`

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

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

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

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

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

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

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

iv. `cos A = (b^2 + c^2 − a^2)/(2bc)`

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

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

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

Calculus, 2ADV C3 2004 HSC 9c

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

`text(Using the quotient rule,)`

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

`text(SP’s occur when)\ \ f′(x)=0`

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

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

ii.

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

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

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

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

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

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

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

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

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

Calculus, 2ADV C4 2004 HSC 8b

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

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

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

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

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

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

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

`text(Required area)`

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

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

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

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

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

`= (6 − 4/3) − (−16 − 8 + 32/3)`

`= 18`

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

`= 18\ \ text(u²).`

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

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

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

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

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

`:.\ text(Area of)\ ΔABC`

`= 1/2 × 3sqrt17 × 30/sqrt17`

`= 45\ \ text(u²).`

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

Trigonometry, 2ADV T2 2004 HSC 8a

Show that `cos theta tan theta = sin theta`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Hence solve `8 sin theta cos theta tan theta = text(cosec)\ theta` for `0 ≤ theta ≤ 2pi`. (2 marks)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`pi/6, (5pi)/6`

Show Worked Solution

i. `text(Prove)\ \ cos theta tan theta = sin theta`

`text(LHS)`	`= cos theta tan theta`
	`= cos theta ((sin theta)/(cos theta))`
	`= sin theta`
	`=\ text{RHS}`

ii.	`8 sin theta cos theta tan theta`	`= text(cosec)\ theta`
	`:. 8 sin theta(sin theta)`	`= text(cosec)\ theta,\ \ \ \ text{(part (i))}`
	`8 sin^2 theta`	`= 1/(sin theta)`
	`8 sin^3 theta`	`= 1`
	`sin^3 theta`	`= 1/8`
	`sin theta`	`= 1/2`
	`:. theta`	`= pi/6, (5pi)/6.\ \ \ \ text{(for}\ \ 0 ≤ theta ≤ 2pi text{)}`

Financial Maths, 2ADV M1 2004 HSC 7c

Betty decides to set up a trust fund for her grandson, Luis. She invests $80 at the beginning of each month. The money is invested at 6% per annum, compounded monthly.

The trust fund matures at the end of the month of her final investment, 25 years after her first investment. This means that Betty makes 300 monthly investments.

After 25 years, what will be the value of the first $80 invested? (2 marks)

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By writing a geometric series for the value of all Betty’s investments, calculate the final value of Luis’ trust fund. (3 marks)

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Show Answers Only

`$357.20\ \ text(to the nearest cent)`
`$55\ 716.71\ \ text(to the nearest cent)`

Show Worked Solution

i. `A = P(1 + r/(100))^n`

`P`	`= $80`
`n`	`= 300\ \ text(months)`
`r`	`=\ text(6% per annum)`
	`=\ text(0.5% per month)`

`:.A`	`= 80(1 + (0.5)/(100))^300`
	`= 80(1.005)^300`
	`= 357.1975…`
	`= $357.20\ \ text{(nearest cent)}`

ii.	`A_1`	`= 80(1.005)^300`
	`A_2`	`= 80(1.005)^299`
	`A_3`	`= 80(1.005)^298`
		` vdots`
	`A_300`	`= 80(1.005)^1`

`:.\ text(Final value of fund)`

`= 80(1.005)^300 + 80(1.005)^299 + … + 80(1.005)^1`

`= 80[1.005^300 + 1.005^299 + … + 1.005^1]`

`= 80[1.005^1 + 1.005^2 + … + 1.005^300]`

`=>\ text(GP where)\ \ \ a = 1.005, \ r = 1.005, \ n = 300`

`= 80[(a(r^n − 1))/(r − 1)]`

`= 80[(1.005(1.005^300 − 1))/(1.005 − 1)]`

`= 55\ 716.714…`

`= $55\ 716.71\ \ text(to the nearest cent.)`

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)

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What is the value of `A`? (1 mark)

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Find the value of `k`. (2 marks)

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Predict the year during which Australia’s population will reach 30 million. (2 marks)

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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.)`

Financial Maths, 2ADV M1 2004 HSC 7a

Evaluate `sum_(n = 2)^4 n^2`. (1 mark)

Show Answers Only

`29`

Show Worked Solution

`sum_(n = 2)^4 n^2`	`= 2^2 + 3^2 + 4^2`
	`= 4 + 9 + 16`
	`= 29`

Probability, 2ADV S1 2004 HSC 6c

In a game, a turn involves rolling two dice, each with faces marked 0, 1, 2, 3, 4 and 5. The score for each turn is calculated by multiplying the two numbers uppermost on the dice.

What is the probability of scoring zero on the first turn? (2 marks)

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What is the probability of scoring `16` or more on the first turn? (1 mark)

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What is the probability that the sum of the scores in the first two turns is less than 45? (2 marks)

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Show Answers Only

`(11)/(36)`
`1/9`
`1291/1296`

Show Worked Solution

MARKER’S COMMENT: Students who drew up the table for the sample space were “overwhelmingly” more successful in all parts of this question.

`:. P(0) = (11)/(36)`

ii. `P(≥ 16)= 4/36=1/9`

iii. `Ptext{(Sum} < 45) = 1 − Ptext{(Sum} ≥ 45)`

`Ptext{(Sum} ≥ 45)`	`=P(20,25)+P(25,20)+P(25,25)`
	`=(2/36 xx 1/36) + (2/36 xx 1/36)+(1/36 xx 1/36)`
	`=2/1296 + 2/1296+ 1/1296`
	`=5/1296`

`:.Ptext{(Sum} < 45)`	`= 1 − 5/1296`
	`= 1291/1296`

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.`

L&E, 2ADV E1 2004 HSC 6a

Solve the following equation for `x`:

`e^(2x) + 3e^x − 10 = 0`. (2 marks)

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Show Answers Only

`x = ln 2`

Show Worked Solution

`e^(2x) + 3e^x − 10`	`= 0`
`:. (e^x)^2 + 3e^x − 10`	`= 0`

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

`:. X^2 + 3X – 10`	`= 0`
`(X + 5)(X − 2)`	`= 0`
`:. X =2 or -5`

`text(If)\ \ X`	`=2`
`e^x`	`=2`
`ln e^x`	`=ln 2`
`x`	`=ln 2`

`text(If)\ \ X`	`=-5`
`e^x`	`=-5\ \ \ text{(no solution)}`

`:. x=ln 2`

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)

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Sketch the graph of `x` as a function of `t` for `0 ≤ t ≤ pi`. (2 marks)

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Hence, or otherwise, find when AND where the particle first comes to rest after `t = 0`. (2 marks)

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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)

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How many hours of lessons will Clare have completed after her twenty-first lesson? (2 marks)

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During which lesson will Clare have completed a total of 50 hours of driving lessons? (2 marks)

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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, EXT1* C3 2004 HSC 4c

In the diagram, the shaded region is bounded by the curve `y = 2 sec x`, the coordinate axes and the line `x = pi/3`. The shaded region is rotated about the `x`-axis.

Calculate the exact volume of the solid of revolution formed. (3 marks)

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Show Answers Only

`4sqrt3\ pi\ \ text(u³)`

Show Worked Solution

`y`	`= 2 sec x`
`:. y^2`	`= 4 sec^2 x`

`V`	`= pi int_0^(pi/3) y^2\ dx`
	`= pi int_0^(pi/3) 4 sec^2 x\ dx`
	`= 4pi int_0^(pi/3) sec^2 x\ dx`
	`= 4pi[tan x]_0^(pi/3)`
	`= 4pi(tan\ pi/3 − tan 0)`
	`= 4pi(sqrt3 − 0)`
	`= 4sqrt3\ pi\ \ text(u³)`

Calculus, MET2 2012 VCAA 1

A solid block in the shape of a rectangular prism has a base of width `x` cm. The length of the base is two-and-a-half times the width of the base.

The block has a total surface area of 6480 sq cm.

Show that if the height of the block is `h` cm, `h = (6480-5x^2)/(7x).` (2 marks)
The volume, `V` cm³, of the block is given by `V(x) = (5x(6480-5x^2))/14.`
Given that `V(x) > 0` and `x > 0`, find the possible values of `x`. (2 marks)
Find `(dV)/(dx)`, expressing your answer in the form `(dV)/(dx) = ax^2 + b`, where `a` and `b` are real numbers. (3 marks)
Find the exact values of `x` and `h` if the block is to have maximum volume. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`x ∈ (0,36)`
`(dV)/(dx) = −75/14 x^2 + (16\ 200)/7`
`(120sqrt3)/7`

Show Worked Solution

a. `text(Total surface area) = 6480`

`6480`	`= 2 xx (5/2 x xx h) + 2xx(5/2x^2) + 2 xx (xh)`
`6480`	`= 5xh + 5x^2 + 2xh`
`7xh`	`= 6480-5x^2`
`:. h`	`= (6480-5x^2)/(7x)\ …\ text(as required)`

b. `V (x) > 0quadtext(for)quadx > 0,`

`=>(6480-5x^2)`	`>0`
`5x^2`	`<6480`
`x`	`<36`

`:. x ∈ (0,36)`

c.	`V(x)`	`=(5x(6480-5x^2))/14`
		`=(-25x^3)/14 + (16\ 200x)/7`
	`(dV)/(dx)`	`=- 75/14 x^2 + (16\ 200)/7`

d. `text(S.P. when)\ (dV)/(dx) = 0quadtext(for)quad x > 0`

`text(Solve:)\ \-75/14 x^2 + (16\ 200)/7=0\ \ text(for)\ x,`

`=> x = 12sqrt3`

`text(Substitute)\ \ x = 12sqrt3quadtext(for)quadh,`

`:. h`	`= (6480-5(12sqrt3)^2)/(7(12sqrt3))`
	`= (120sqrt3)/7`

Calculus, MET2 2012 VCAA 21 MC

The volume, `V` cm³, of water in a container is given by `V = 1/3 pi h^3` where `h` cm is the depth of water in the container at time `t` minutes. Water is draining from the container at a constant rate of 300 cm³/min. The rate of decrease of `h`, in cm/min, when `h = 5` is

A. `12/pi`

B. `4/pi`

C. `25 pi`

D. `60/pi`

E. `30 pi`

Show Answers Only

`A`

Show Worked Solution

`V = 1/3 pi h^3`

`(dV)/(dh) = pi h^2`

`(dV)/(dt) = (dV)/(dh) xx (dh)/(dt),`

`(dV)/(dt) = pi h^2 xx (dh)/(dt)`

`text(When)\ \ h = 5,`

`-300`	`= 25 pi xx (dh)/(dt)`
`(dh)/(dt)`	`= -12/pi`

`:.\ text(Rate of decrease is)\ \12/pi\ \ text(cm/min.)`

`=> A`

Calculus, MET2 2012 VCAA 14 MC

The graph of `f: R^+ uu {0} -> R,\ f(x) = sqrt x` is shown below.

In order to find an approximation to the area of the region bounded by the graph of `f`, the `y`-axis and the line `y = 4`, Zoe draws four rectangles, as shown, and calculates their total area.

Zoe's approximation to the area of the region is

A. `14`

B. `21`

C. `29`

D. `30`

E. `64/3`

Show Answers Only

`D`

Show Worked Solution

`text(Width of each rectangle)\ = 1`

`text(Height of each rectangle:)`

`text(When)\ \ y=1 => 1=sqrt x, \ x=1`

`text(When)\ \ y=2\ => 2=sqrt x, \ x=4`

`text(When)\ \ y=3 => x=9`

`text(When)\ \ y=2 => x=16`

`:. A`	`~~ 1[1 + 4 + 9 + 16]`
	`~~ 30\ \ text(u²)`

`=> D`

`b`	`=r xx s_y/s_x`
	`=-0.5675 xx 6.98/2.61`
	`=-1.517…`
`a`	`=bar y – b barx`
	`=23.93 – (- 1.517… xx 4.56)`
	`=30.85…`

`:. W_(n+1)/W_n`	`=0.96`
`W_(n+1)`	`=0.96 W_n`

`z text(-score)`	`=(x-bar x)/s`
	`=(56-67)/10.2`
	`= – 1.0784…`

`text(AP where)\ \ \ a`	`= 15, and`
`d`	`= 19-15=4`

`S_n`	`= n/2 [2a + (n – 1) d]`
`:. S_12`	`= 12/2[2 xx 15+(12-1)4]`
	`= 6[30 + 44]`
	`= 444`

`S_(n+1)`	`=S_n`
`S_(max)`	`=0.2×S_(max)+15`
`0.8 xx S_(max)`	`=15`
`S_(max)`	`=15/0.8`
	`=18.75`