Band 4 – Page 98

GEOMETRY, FUR1 2015 VCAA 4 MC

Town A is due west of town B.

Town C is due south of town B.

The bearing of town A from town C is

A. between `000^@` and `090^@`

B. between `090^@` and `180^@`

C. exactly `135^@`

D. between `180^@` and `270^@`

E. between `270^@` and `360^@`

Show Answers Only

`E`

Show Worked Solution

`=> E`

Calculus, SPEC2 2014 VCAA 13 MC

Using the substitution `u = sqrt(x + 1)` then `int_0^2(dx)/((x + 2) sqrt (x + 1))` can be expressed as

A. `int_1^sqrt 3 1/(sqrt u (u^2 + 1))\ du`

B. `int_0^2 2/(u^2 + 1)\ du`

C. `int_1^3 1/(sqrt u (u + 1))\ du`

D. `1/4 int_0^2 1/(u^2(u^2 + 1))\ du`

E. `2 int_1^sqrt 3 1/(u^2 + 1)\ du`

Show Answers Only

`E`

Show Worked Solution

`u`	`=sqrt(x + 1)`
`(du)/(dx)`	`= 1/(2 sqrt (x + 1))\ \ =>\ \ dx=2sqrt(x + 1)\ du`

`u^2`	`= x + 1`
`u^2 + 1`	`= x + 2`

`u(0)`	`= sqrt (0 + 1) = 1`
`u(2)`	`= sqrt (2 + 1) = sqrt 3`

`:. int_0^2 (dx)/((x + 2) sqrt(x + 1))`

`=int_1^sqrt3 (2sqrt(x + 1))/((u^2 + 1) xx sqrt(x + 1))\ du`

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

`=> E`

Calculus, SPEC2 2014 VCAA 12 MC

If `(dy)/(dx) = sqrt((2x^6 + 1))` and `y = 5` when `x = 1`, then the value of `y` when `x = 4` is given by

`int_1^4(sqrt((2x^6 + 1)) + 5)\ dx`
`int_1^4sqrt((2x^6 + 1))\ dx`
`int_1^4 sqrt((2x^6 + 1))\ dx + 5`
`int_1^4sqrt((2x^6 + 1))\ dx - 5`
`int_1^4(sqrt((2x^6 + 1)) - 5)\ dx`

Show Answers Only

`C`

Show Worked Solution

`F(b)-F(a) `	`= int_a^b sqrt((2x^6 + 1))\ dx`
`F(b)`	`= int_a^b sqrt((2x^6 + 1))\ dx + F(a)`
`:. F(4)`	`= int_1^4 sqrt((2x^6 + 1))\ dx + F(1)`
	`= int_1^4 sqrt((2x^6 + 1))\ dx + 5`

`=> C`

CORE, FUR1 2010 VCAA 13 MC

A garden supplies outlet sells water tanks. The monthly seasonal indices for the revenue from the sale of water tanks are given below.

The seasonal index for September is missing.

The revenue from the sale of water tanks in September 2009 was $104 500.

The deseasonalised revenue for September 2009 is closest to

A. `$42\ 800`

B. `$74\ 100`

C. `$104\ 500`

D. `$141\ 000`

E. `$147\ 300`

Show Answers Only

`B`

Show Worked Solution

`text(Seasonal index for September)`	`=12 -\ text(sum of others)`
	`=12-10.59`
	`=1.41`

`:.\ text(Deseasonalised Revenue for September)`

`= (text(actual))/(text(index))`

`= (104\ 500)/(1.41)`

`= $74\ 113.48`

`=> B`

CORE, FUR1 2015 VCAA 12 MC

The time series plot below charts the number of calls per year to a computer help centre over a 10-year period.

Using five-median smoothing, the smoothed number of calls in year 6 was closest to

A. `3500`

B. `3700`

C. `3800`

D. `4000`

E. `4200`

Show Answers Only

`D`

Show Worked Solution

`text(The median of the 5 data points from year 4 – year 8)`

`text(is 4000.)`

`=> D`

CORE, FUR1 2015 VCAA 9 MC

A least squares regression line has been fitted to the scatterplot above to enable distance, in kilometres, to be predicted from time, in minutes.

The equation of this line is closest to

A. distance `= 3.5 + 1.6 ×`time

B. time `= 3.5 + 1.6 ×`distance

C. distance `= 1.6 + 3.5 ×`time

D. time `= 1.8 + 3.5 ×`distance

E. distance `= 3.5 + 1.8 ×`time

Show Answers Only

`A`

Show Worked Solution

`text{Line passes through (0, 3.5) and (50, 82)`

`:.\ text(Gradient) `	`=(y_2-y_1)/(x_2-x_1)`
	`~~(82 – 3.5)/(50 – 0)`
	`~~ 1.57`

`text(Distance is the dependent variable and the)`

`y text(-intercept is approximately 3.5.)`

`=> A`

CORE, FUR1 2015 VCAA 6-7 MC

The following information relates to Parts 1 and 2.

In New Zealand, rivers flow into either the Pacific Ocean (the Pacific rivers) or the Tasman Sea (the Tasman rivers).

The boxplots below can be used to compare the distribution of the lengths of the Pacific rivers and the Tasman rivers.

Part 1

The five-number summary for the lengths of the Tasman rivers is closest to

`32, 48, 64, 76, 108`
`32, 48, 64, 76, 180`
`32, 48, 64, 76, 322`
`48, 64, 97, 169, 180`
`48, 64, 97, 169, 322`

Part 2

Which one of the following statements is not true?

The lengths of two of the Tasman rivers are outliers.
The median length of the Pacific rivers is greater than the length of more than 75% of the Tasman rivers.
The Pacific rivers are more variable in length than the Tasman rivers.
More than half of the Pacific rivers are less than 100 km in length.
More than half of the Tasman rivers are greater than 60 km in length.

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

♦ Mean mark 46%.

`text(Outliers are inputs into a five-number summary,)`

`text(including the maximum and minimum values.)`

`:.\ text(A maximum length of 180 km is part of the Tasman)`

`text(river summary.)`

`=> B`

`text(Part 2)`

`text(Consider)\ D,`

`D\ text(would be true if its median value was less than)`

`text(100 km, which is not the case.)`

`=> D`

CORE, FUR1 2015 VCAA 3 MC

The dot plot below displays the difference between female and male life expectancy, in years, for a sample of 20 countries.

The mean (`barx`) and standard deviation (`s`) for this data are

A. `text(mean)\ = 2.32`	`\ \ \ \ \ text(standard deviation)\ = 5.25`
B. `text(mean)\ = 2.38`	`\ \ \ \ \ text(standard deviation)\ = 5.25`
C. `text(mean)\ = 5.0`	`\ \ \ \ \ text(standard deviation)\ = 2.0`
D. `text(mean)\ = 5.25`	`\ \ \ \ \ text(standard deviation)\ = 2.32`
E. `text(mean)\ = 5.25`	`\ \ \ \ \ text(standard deviation)\ = 2.38`

Show Answers Only

`E`

Show Worked Solution

`text(By calculator.)`

`=> E`

CORE, FUR1 2015 VCAA 1 MC

The stem plot below displays the average number of decayed teeth in 12-year-old children from `31` countries.

Based on this stem plot, the distribution of the average number of decayed teeth for these countries is best described as

negatively skewed with a median of 15 decayed teeth and a range of 45
positively skewed with a median of 15 decayed teeth and a range of 45
approximately symmetric with a median of 1.5 decayed teeth and a range of 4.5
negatively skewed with a median of 1.5 decayed teeth and a range of 4.5
positively skewed with a median of 1.5 decayed teeth and a range of 4.5

Show Answers Only

`E`

Show Worked Solution

`text(Median = 16th value)\ = 1.5`

`text(Range)\ = 4.7-0.2=4.5`

`text(The clear tail to the upper end of values shows that the)`

`text(data is positively skewed.)`

`=> E`

CORE*, FUR1 2006 VCAA 7 MC

The values of the first five terms of a sequence are plotted on the graph shown below.

The first order difference equation that could describe the sequence is

A. `t_(n+1) = t_n + 5,`	`\ \ \ \ \ t_1 = 4`
B. `t_(n+1) = 2t_n + 1,`	`\ \ \ \ \ t_1 = 4`
C. `t_(n+1) = t_n - 3,`	`\ \ \ \ \ t_1 = 4`
D. `t_(n+1) = t_n + 3,`	`\ \ \ \ \ t_1 = 4`
E. `t_(n+1) = 3t_n,`	`\ \ \ \ \ t_1 = 4`

Show Answers Only

`B`

Show Worked Solution

`text(By elimination,)`

`text(There is no common difference between terms,)`

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

`text(The equation in B has)\ \ t_2=9,\ \ text(while the equation)`

`text(in C has)\ \ t_2=12.`

`rArr B`

CORE*, FUR1 2006 VCAA 5 MC

A difference equation is defined by

`f_(n+1) - f_n = 5\ \ \ \ \ text (where)\ \ f_1 =– 1`

The sequence `f_1, \ f_2, \ f_3, ...` is

A. `5, 4, 3\ …`

B. `4, 9, 14\ …`

C. `– 1, – 6, – 11\ …`

D. `– 1, 4, 9\ …`

E. `– 1, 6, 11\ …`

Show Answers Only

`D`

Show Worked Solution

`text(Using)\ \ \ f_(n+1) = f_n +5`

`text(Consider)\ \ f_1 = – 1,`

`f_2= – 1 + 5=4`

`f_3 = 4+5=9`

`rArr D`

CORE, FUR1 2006 VCAA 10 MC

The table below displays the total monthly rainfall (in mm) in a reservoir catchment area over a one-year period.

Using three mean moving average smoothing, the smoothed value for the total rainfall in April is closest to

A. `65`

B. `66`

C. `70`

D. `75`

E. `88`

Show Answers Only

`C`

Show Worked Solution

`text(Three mean average for April)`

`=(35+99+75)/3`

`=69.66…`

`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 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)

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

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)

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`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`

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`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`

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`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)}`

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`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`

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`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)`

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`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`

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`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`

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`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`

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`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]}`

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`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`

`text(Shaded Area)`	`= 3 xx (1/2 ab sinC )`
	`= 3 xx (1/2 xx 2 xx 2 xx sin60^@)`
	`=5.196…\ text(cm²)`

`text(Perimeter)`	`= (1/2 xx pi xx 12) + (4 + 12 + 4)`
	`=18.849… + 20`
	`= 38.849…\ text(m²)`

`text{Increase (%)}`	`=(850\ 000 – 600\ 000)/(600\ 000)`
	`=0.4166…`
	`=41.66…text(%)`

`K_31`	`= 0.99 xx 7550 + 250`
	`= 7724.5`
`K_32`	`= 0.99 xx 7724.5 + 250`
	`= 7897.255`

`N`	`=4 xx 20=80`
`I (%)`	`= ?`
`PV`	`= – 300\ 000`
`PMT`	`= 6727.80`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 4`

`:.\ text(Depreciation Rate)`	`= (35\ 000 – 25\ 000)/(40\ 000 – 0)`
	`= $0.25\ text(per km)`

`:.\ text(Interest charged)`	`= 24\ 000 – 15\ 000`
	`= $9000`

`text(Total after 2 years)`	`= 1200(1 + (0.27916…)/100)^24`
	`=$1283.03…`

`:.\ text(Interest earned)`	`= 1283.03 – 1200`
	`= $83.03`