Band 5 – Page 49

CORE, FUR1 2011 VCAA 8 MC

Teresa borrowed $120 000 at an interest rate of 7.67% per annum, compounding monthly.

The loan is to be repaid with equal monthly payments.

She decides to repay the loan by making monthly payments of $1430.

Which of the following statements is true?

She will pay out the loan fully in less than ten years.
The amount of interest that she pays on the loan will increase each year.
After four years the amount that she owes on the loan will be less than $80 000.
Every monthly payment that she makes reduces the amount that she owes on the loan by the same amount.
Monthly payments of $1560 (instead of $1430) will enable her to repay this loan in less than nine years.

Show Answers Only

`E`

Show Worked Solution

`text(Consider E,)`

`text(By TVM solver:)`

`N`	`= ?`
`I (%)`	`= 7.67text(%)`
`PV`	`= – 120\ 000`
`PMT`	`= 1560`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> N = 106.2`

`text{Loan will be paid when}\ \ n<108\ text {months (9 years).}`

`:.\ text(E is true.)`

`text(All other options can be shown to be false.)`

`=> E`

CORE, FUR1 2010 VCAA 12 MC

The time series plot below shows the number of calls each month to a call centre over a twelve-month period.

The plot is to be smoothed using five-point median smoothing.

The smoothed number of calls for month number 10 is closest to

A. `358`

B. `364`

C. `371`

D. `375`

E. `377`

Show Answers Only

`D`

Show Worked Solution

`text(The median for the 5 values from month 8 – 12)`

♦ Mean mark 43%.

`text(is 40 calls.)`

`=> D`

CORE, FUR1 2015 VCAA 13 MC

The quarterly seasonal indices for tractor sales for a supplier are displayed in Table 1.

The quarterly tractor sales in 2014 for this supplier are displayed in Table 2.

The sales data in Table 2 is to be deseasonalised before a least squares regression line is fitted.

The equation of this least squares regression line is closest to

A. deseasonalised sales = 0.32 + 910 × quarter number

B. deseasonalised sales = 370 – 2300 × quarter number

C. deseasonalised sales = 910 + 0.32 × quarter number

D. deseasonalised sales = 2300 – 370 × quarter number

E. deseasonalised sales = 2300 – 0.32 × quarter number

Show Answers Only

`D`

Show Worked Solution

`text(Deseasonalised sales,)`

`text(Using quarter number as the independent,)`

♦ Mean mark 47%.

`text(variable and by CAS,)`

`text(deseasonalised sales)\ = 2300 – 370 xx\ text(quarter number)`

`=> D`

CORE, FUR1 2015 VCAA 8 MC

A dot plot for a set of data is shown below.

Which one of the following boxplots would best represent the dot plot above?

Show Answers Only

`C`

Show Worked Solution

`text(S)text(ince dot plots order data numerically,)`

♦ Mean mark 47%.

`text(Min value = 1001, Max value = 1004)`

`text(Median)`	`=1001\ \ text{(lies between the 12th and 13th}`
	`text{data points given 24 values.)`

`Q_1 = 1001\ \ \ text{(see dot plot above)}`

`Q_3 = (1002+1003)/2=1002.5\ \ \ text{(see dot plot above)}`

`text(Only C satisfies the median,)\ Q_1 and Q_3\ text(conditions.)`

`=> C`

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 2006 VCAA 8 MC

Paula started a stamp collection. She decided to buy a number of new stamps every week.

The number of stamps bought in the `n`th week, `t_n`, is defined by the difference equation

`t_n = t_(n-1) + t_(n-2)\ \ \ text(where)\ \ \ t_1 = 1 and t_2 = 2`

The total number of stamps in her collection after five weeks is

A. `8`

B. `12`

C. `15`

D. `19`

E. `24`

Show Answers Only

`D`

Show Worked Solution

`t_1=1,\ \ \ t_2=2\ \ \ text{(given)}`

♦♦ Mean mark 33%.
STRATEGY: Always ask if the question wants a term or a sum of terms. This will minimise a very common mistake in this topic area.

`:. t_3`	`=t_2 + t_1 = 2+1=3`
`t_4`	`=3+2=5`
`t_5`	`=5+3= 8`

`:.\ text(Total stamps after 5 weeks)`

`=1+2+3+5+8=19`

`rArr D`

PATTERNS, FUR1 2006 VCAA 6 MC

A crystal measured 12.0 cm in length at the beginning of a chemistry experiment.

Each day it increased in length by 3%.

The length of the crystal after 14 days growth is closest to

A. `12.4\ text(cm)`

B. `16.7\ text(cm)`

C. `17.0\ text(cm)`

D. `17.6\ text(cm)`

E. `18.2\ text(cm)`

Show Answers Only

`E`

Show Worked Solution

`text(After day 1, length)\ = 12.0 xx 1.03`

♦♦ Mean mark 25%.
MARKERS’ COMMENT: The question asks for length after 14 days’ growth, not its length on the start of the 14th day.

`text(After day 2, length)\ = 12.0 xx 1.03^2`

`vdots`

`:.\ text(After day 14, length)`

`= 12.0 xx 1.03^14 `

`= 18.15\ text(cm)`

`rArr E`

CORE, FUR1 2006 VCAA 9 MC

A student uses the following data to construct the scatterplot shown below.

To linearise the scatterplot, she applies an `x`-squared transformation.

She then fits a least squares regression line to the transformed data with `y` as the dependent variable.

The equation of this least squares regression line is closest to

A. `y = 7.1 + 2.9x^2`

B. `y = –29.5 + 26.8x^2`

C. `y = 26.8 - 29.5x^2`

D. `y = 1.3 + 0.04x^2`

E. `y = –2.2 + 0.3x^2`

Show Answers Only

`A`

Show Worked Solution

`text(By calculator,)`

♦ Mean mark 42%.

`y=7.147+2.9387 x^2`

`rArr A`

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 10 MC

The relationship between the variables

size of car (1 = small, 2 = medium, 3 = large)

and

salary level (1 = low, 2 = medium, 3 = high)

is best displayed using

A. a scatterplot.

B. a histogram.

C. parallel boxplots.

D. a back-to-back stemplot.

E. a percentaged segmented bar chart.

Show Answers Only

`E`

Show Worked Solution

`text(A segmented bar chart is required to effectively)`

♦ Mean mark 37%.

`text(display this information given the three)`

`text(sub-categories of each variable.)`

`rArr E`

CORE*, FUR1 2008 VCAA 8 MC

When placed in a pond, the length of a fish was 14.2 centimetres.

During its first month in the pond, the fish increased in length by 3.6 centimetres.

During its `n`th month in the pond, the fish increased in length by `G_n` centimetres, where `G_(n+1) = 0.75G_n`

The maximum length this fish can grow to (in cm) is closest to

A. 14.4

B. 16.9

C. 19.0

D. 28.6

E. 17.2

Show Answers Only

`D`

Show Worked Solution

`text(Initial length) = 14.2\ text(cm)`

♦ Mean mark 45%.

`G_1 = 3.6`

`G_2 = (0.75) G_1 = (0.75) 3.6`

`G_3 = (0.75) G_2 = (0.75^2) 3.6`

`text(Growth is a geometric sequence)`

`underbrace(3.6, \ 3.6(0.75), \ 3.6(0.75)^2, …)_{text(GP where)\ \ a=3.6,\ \ r=0.75}`

`text(S)text(ince)\ \ |\ r\ |< 1,`

`S_oo`	`= a/(1-r)`
	`= (3.6)/(1-0.75)`
	`= 14.4`

`:.\ text(Maximum length of fish)`

`= 14.2 +14.4`

`=28.6\ text(cm)`

`=> D`

Graphs, MET2 2015 VCAA 22 MC

The graphs of the functions with rules `f (x)` and `g (x)` are shown below.

Which one of the following best represents the graph of the function with rule `g(-f(x))?`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ f(x) = -a* |\ x\ |\ \ text(where)\ \ a in R^+`

♦♦ Mean mark 35%.

`:. g(-f(x))`	`= g(a *\|\ x\ \|)`
	`= g(a *\|\ – x\ \|) >= 0`

`:.g(-f(x))\ \ text{is even (symmetrical about the}\ ytext{-axis).}`

`=> B`

Algebra, MET2 2015 VCAA 21 MC

The graphs of `y = mx + c` and `y = ax^2` will have no points of intersection for all values of `m, c` and `a` such that

`a > 0 and c > 0`
`a > 0 and c < 0`
`a > 0 and c > -m^2/(4a)`
`a < 0 and c > -m^2/(4a)`
`m > 0 and c > 0`

Show Answers Only

`D`

Show Worked Solution

`text(Intersect when:)`

`mx + c`	`= ax^2`
`ax^2 – mx – c`	`= 0`

`text(S)text(ince no points of intersection:)`

♦ Mean mark 37%.

`Delta`	`< 0`
`m^2 – 4a(−c)`	`< 0`
`m^2 + 4ac`	`< 0`

`text(Solve for)\ c:`

`:.\ c > (−m^2)/(4a),quada < 0`

`text(or)`

`c < (−m^2)/(4a),quada > 0`

`=> D`

Probability, MET2 2015 VCAA 9 MC

The graph of the probability density function of a continuous random variable, `X`, is shown below.

If `a > 2`, then `text(E)(X)` is equal to

Show Answers Only

`B`

Show Worked Solution

`text(Area under rectangle)`	`= 1`
`1/6(a – 2)`	`= 1`
`:. a`	`= 8`

♦ Mean mark 37%.
MARKER’S COMMENT: Once `a=8` is found, the expected value can also be quickly found halfway between 2 and 8, due to the uniform distribution.

`text(E)(X)`	`= int_2^8 x (1/6)\ dx`
	`=[x^2/12]_2^8`
	`=64/12 – 4/12`
	`=5`

`=> B`

Probability, MET2 2014 VCAA 22 MC

John and Rebecca are playing darts. The result of each of their throws is independent of the result of any other throw. The probability that John hits the bullseye with a single throw is `1/4.` The probability that Rebecca hits the bullseye with a single throw is `1/2.` John has four throws and Rebecca has two throws.

The ratio of the probability of Rebecca hitting the bullseye at least once to the probability of John hitting the bullseye at least once is

`1:1`
`32:27`
`64:85`
`2:1`
`192:175`

Show Answers Only

`E`

Show Worked Solution

`X = #\ text(times John hits bullseye.)`

♦ Mean mark 37%.

`Y = #\ text(times Rebecca hits bullseye.)`

`text(Required ratio:)`

`text(Pr)(Y >= 1)`	`\ :\ text(Pr)(X >= 1)`
`1-text(Pr)(Y=0)`	`\ :\ 1-text(Pr)(X=0)`
`1-(1/2)^2`	`\ :\ 1-(3/4)^4`
`3/4`	`\ :\ 175/256`
`192/256`	`\ :\ 175/256`
`:. 192`	`\ :\ 175`

`=> E`

Calculus, MET2 2014 VCAA 20 MC

The graph of a function, `h`, is shown below.

The average value of `h` is

`4`
`5`
`6`
`7`
`10`

Show Answers Only

`D`

Show Worked Solution

`text(From inspection, the average value of)\ h`

♦ Mean mark 44%.

`text(is either 6 or 7.)`

`h_text(avg) = 7\ text(as area bounded by)\ h(x)`

`text(and)\ y = 7\ text(above)\ y = 7\ text(is equivalent)`

`text(to area bounded below)`

`=> D`

Algebra, MET2 2014 VCAA 17 MC

The simultaneous linear equations `ax - 3y = 5` and `3x - ay = 8 - a` have no solution for

`a = 3`
`a = -3`
both `a = 3` and ` a = -3`
`a in R text(\{3})`
`a in R text(\[−3, 3])`

Show Answers Only

`B`

Show Worked Solution

`text(No solution occurs when:)`

♦ Mean mark 50%.

`=> m_1 = m_2quadtext(and)quadc_1 != c_2`

`a/3`	`= 3/a`	`text(and)`	`-5/3`	`!= -((8-a)/a)`
`a`	`= ±3`		`5a`	`!= 24-3a`
			`a`	`!=3`

`:. a = −3`

`=> B`

Probability, MET2 2014 VCAA 16 MC

The continuous random variable `X`, with probability density function `p(x)`, has mean 2 and variance 5.

The value of `int_-oo^oo x^2 p(x)\ dx` is

`1`
`7`
`9`
`21`
`29`

Show Answers Only

`C`

Show Worked Solution

♦ Mean mark 46%.

`text(VAR)(X)`	`= text(E)(X^2) – [text(E)(X)]^2`
`5`	`= int_(−∞)^∞ x^2 p(x)\ dx – (2)^2`
`:. 9`	`= int_(−∞)^∞ x^2 p(x)\ dx`

`=> C`

Calculus, MET2 2014 VCAA 15 MC

Zoe has a rectangular piece of cardboard that is 8 cm long and 6 cm wide. Zoe cuts squares of side length `x` centimetres from each of the corners of the cardboard, as shown in the diagram below.

Zoe turns up the sides to form an open box.

The value of `x` for which the volume of the box is a maximum is closest to

`0.8`
`1.1`
`1.6`
`2.0`
`3.6`

Show Answers Only

`B`

Show Worked Solution

`V`	`=\ text(Base × Height)`
	`=x(8 – 2x)(6 – 2x),qquadx ∈ (0,3)`

`text(Solve:)\ \ V′=0,`

♦ Mean mark 44%.

`[text(CAS:)\ ftext(Max) (x(8 – 2x)(6 – 2x),x,0,3)]`

`V_text(max) -> x`

`= 1.13`

`=> B`

Probability, MET2 2014 VCAA 14 MC

If `X` is a random variable such that `text(Pr)(X > 5) = a` and `text(Pr)(X > 8) = b`, then `text(Pr)(X < 5 | X < 8)` is

`a/b`
`(a - b)/(1 - b)`
`(1 - b)/(1 - a)`
`(ab)/(1 - b)`
`(a - 1)/(b - 1)`

Show Answers Only

`E`

Show Worked Solution

`text(Pr)(X < 5) | text(Pr)(X < 8)`

♦ Mean mark 45%.

`=(text(Pr)(X<5 ∩ X<8))/(text(Pr)(X < 8))`

`= (text(Pr)(X < 5))/(text(Pr)(X < 8))`

`= (1 – a)/(1 – b)`

`= (a – 1)/(b – 1)`

`=> E`

Algebra, MET2 2014 VCAA 13 MC

The domain of the function `h`, where `h(x) = cos(log_a (x))` and `a` is a real number greater than `1`, is chosen so that `h` is a one-to-one function.

Which one of the following could be the domain?

`(a^(-pi/2), a^(pi/2))`
`(0, pi)`
`[1, a^(pi/2)]`
`[a^(-pi/2), a^(pi/2))`
`[a^(-pi/2), a^(pi/2)]`

Show Answers Only

`C`

Show Worked Solution

`text(Choose a value of)\ \ a > 1`

♦ Mean mark 42%.

`text(e.g.)\ \ a = 2`

`text(Graph)\ \ h(x) = cos(log_2(x)), text(and test)`

`text(all domain options.)`

`text(Only)\ C\ text(gives a one-to-one function.)`

`=> C`

Graphs, MET2 2008 VCAA 10 MC

The range of the function `f: [pi/8, pi/3) -> R,\ f(x) = 2 sin (2x)` is

`(sqrt 2, sqrt 3]`
`[sqrt 2, 2)`
`[sqrt 2, 2]`
`(sqrt 2, sqrt 3)`
`[sqrt 2, sqrt 3)`

Show Answers Only

`C`

Show Worked Solution

♦ Mean mark 45%.

`:.\ text(Range) = [sqrt 2, 2],`

`=> C`

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.)`

Probability, MET2 2013 VCAA 17 MC

`A` and `B` are events of a sample space.

Given that `text(Pr)(A|B) = p,\ \ text(Pr)(B) = p^2` and `text(Pr)(A) = p^(1/3),\ text(Pr)(B|A)` is equal to

`p`
`p^(4/3)`
`p^(7/3)`
`p^(8/3)`
`p^3`

Show Answers Only

`D`

Show Worked Solution

`text(Pr)(A \| B)`	`= (text(Pr)(A ∩ B))/(text(Pr)(B)`
`p`	`= (text(Pr)(A ∩ B))/(p^2)`
`:. text(Pr)(A ∩ B)`	`= p^3`

♦ Mean mark 49%.

`text(Pr)(B \| A)`	`= (text(Pr)(A ∩ B))/(text(Pr)(A))`
	`= (p^3)/(p^(1/3))`
`:. text(Pr)(B \| A)`	`= p^(8/3)`

`=> D`

Algebra, MET2 2013 VCAA 18 MC

Let `g(x) = log_2(x),\ \ x > 0`

Which one of the following equations is true for all positive real values of `x?`

`2g (8x) = g (x^2) + 8`
`2g (8x) = g (x^2) + 6`
`2g (8x) = (g (x) + 8)^2`
`2g (8x) = g (2x) + 6`
`2g (8x) = g (2x) + 64`

Show Answers Only

`B`

Show Worked Solution

`text(Consider Option)\ B:`

♦♦ Mean mark 35%.

`text(LHS)`	`= 2g(8x)`
	`= 2log_2(8x)`
	`= 2log_2(8) + 2log_2(x)`
	`=2log_2 (2^3)+ 2log_2(x)`
	`= 6 + log_2(x^2)`
	`= g(x^2) + 6`

`=> B`

Calculus, MET2 2013 VCAA 19 MC

Part of the graph of a function `f: [0, oo) -> R, \ f(x) = e^(x sqrt 3) sin (x)` is shown below.

The first three turning points are labelled `T_1, T_2` and `T_3.`

The `x`-coordinate of `T_3` is

`(8 pi)/3`
`(16 pi)/3`
`(13 pi)/6`
`(17 pi)/6`
`(29 pi)/6`

Show Answers Only

`D`

Show Worked Solution

`f(x)`	`=e^(x sqrt 3) sin (x)`
`f′(x)`	`=sqrt3 e^(x sqrt 3) sin (x) + e^(x sqrt 3) cos (x)`

`text(Find)\ \ x\ \ text(when) f′(x) = 0:`

♦ Mean mark 50%.

`sqrt3 e^(x sqrt 3) sin (x)`	`= – e^(x sqrt 3) cos (x)`
`sqrt3 tan(x)`	`=-1`
`tan (x)`	`=- 1/sqrt3`

`x = (5pi)/6,(11pi)/6,(17pi)/6quad \ x∈(0,3pi)`

`=> D`

Calculus, MET2 2013 VCAA 21 MC

The cubic function `f: R -> R, f(x) = ax^3-bx^2 + cx`, where `a, b` and `c` are positive constants, has no stationary points when

`c > b^2/(4a)`
`c < b^2/(4a)`
`c < 4b^2a`
`c > b^2/(3a)`
`c < b^2/(3a)`

Show Answers Only

`D`

Show Worked Solution

`text(If no stationary points,)`

♦♦ Mean mark 29%.

`=>\ text(No solution to)\ \ f{′}(x) = 0`

`f^{′}(x) = 3ax^2 -2bx +c`

`text(No solution when,)`

`Delta`	`< 0`
`(−2b)^2-4(3ac)`	`< 0`
`3ac`	`> b^2`
`:. c`	`> (b^2)/(3a)`

`=> D`

Statistics, MET2 2013 VCAA 22 MC

Butterflies of a particular species die `T` days after hatching, where `T` is a normally distributed random variable with a mean of 120 days and a standard deviation of `sigma` days.

If, from a population of 2000 newly hatched butterflies, 150 are expected to die in the first 90 days, then the value of `sigma` is closest to

7 days
13 days
17 days
21 days
37 days

Show Answers Only

`D`

Show Worked Solution

`T = text(days until death after hatching,)`

`T ∼ N (120,sigma^2)`

`text(Pr)(T <= 90)`

`= 150/2000`

`−1.4395`	`= (90 – 120)/sigma`
`:. sigma`	`= 20.8`

`=> D`

Binomial, EXT1 2008 HSC 6c

Let `p` and `q` be positive integers with `p ≤ q`.

Use the binomial theorem to expand `(1 + x) ^(p+ q)`, and hence write down the term of
`((1 + x)^(p + q))/(x^q)` which is independent of `x`. (2 marks)
Given that
`((1 + x)^(p + q))/(x^q) = (1 + x)^p(1 + 1/x)^q`,

apply the binomial theorem and the result of part(i) to find a simpler expression for
`1 + ((p),(1))((q),(1)) + ((p),(2))((q),(2)) + … + ((p),(p))((q),(p))`. (3 marks)

Show Answers Only

`\ ^(p + q)C_q`
`\ ^(p + q)C_q`

Show Worked Solution

(i) `(1 + x)^(p + q)`

`=\ ^(p + q)C_0 +\ ^(p + q)C_1 x + … +\ ^(p + q)C_q x^q + … +\ ^(p + q)C_(p + q) · x^(p + q)`

`:.\ text(Independent term of)\ ((1 + x)^(p + q))/(x^q)`

`= (\ ^(p + q)C_q·x^q)/(x^q)`

`=\ ^(p + q)C_q`

(ii) `(1 + x)^p(1 + 1/x)^q`

`= (\ ^pC_0 +\ ^pC_1 x + … +\ ^pC_p x^p)`

`xx (\ ^qC_0 +\ ^qC_1 · 1/x + … +\ ^qC_p · 1/(x^p) + … +\ ^qC_q · 1/(x^q))`

`text(The independent term in this expansion)`

`=\ ^pC_0 ·\ ^qC_0 +\ ^pC_1 ·\ ^qC_1 + … +\ ^pC_p ·\ ^qC_p\ text{(since}\ p ≤ q)`

`= 1 +\ ^pC_1 ·\ ^qC_1 + … +\ ^pC_p ·\ ^qC_p`

`text(S)text(ince)\ ((1 + x)^(p + q))/(x^q) = (1 + x)^p(1 + 1/x)^q,\ text(the independent)`

`text(terms are equal.)`

`:.\ ^(p + q)C_q\ text(is a simpler expression for)`

`1 +\ ^pC_1 ·\ ^qC_1 +\ ^pC_2 ·\ ^qC_2 + … +\ ^pC_p ·\ ^qC_p`

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 C4 2004 HSC 10a

Use Simpson’s rule with 3 function values to find an approximation to the area under the curve
`y = 1/x` between `x = a ` and `x = 3a`, where `a` is positive. (2 marks)
Using the result in part (i), show that
1. `ln 3 ≑ 10/9`. (1 mark)

Show Answers Only

`10/9\ text(u²)`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `y = 1/x`

`A`	`~~ h/3(y_0 + y_2 + 4y_1)`
	`~~ a/3(1/a + 1/(3a) + 4(1/(2a)))`
	`~~ a/3(1/a + 1/(3a) + 2/a)`
	`~~ a/3(3/(3a) + 1/(3a) + 6/(3a))`
	`~~ a/3 × 10/(3a)`
	`~~ 10/9`

`:.\ text(Area)\ ~~ 10/9\ text(u²).`

(ii) `text{Area under the curve}\ \ y=1/x`

`= int_a^(3a) 1/x\ dx`

`= [ln x]_(\ a)^(3a)`

`= ln 3a − ln a`

`= ln\ (3a)/a`

`= ln 3`

`text{Simpson’s rule in part (i) found the approximate}`

`text(value of the same area.)`

`:. ln 3 ≑ 10/9.`

Calculus, EXT1* C1 2004 HSC 9b

A particle moves along the `x`-axis. Initially it is at rest at the origin. The graph shows the acceleration, `a`, of the particle as a function of time `t` for `0 ≤ t ≤ 5`.

Write down the time at which the velocity of the particle is a maximum. (1 marks)

--- 2 WORK AREA LINES (style=lined) ---
At what time during the interval `0 ≤ t ≤ 5` is the particle furthest from the origin? Give brief reasons for your answer. (2 marks)

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

Show Answers Only

Show Worked Solution

i. `text(The velocity of the particle increases until)`

`t = 2\ \ text{(when}\ \ a=0text{)}.`

`:.\ text(Maximum velocity occurs at)\ \ t = 2.`

ii. `text(It is furthest from the origin when its positive velocity)`

`text(slows and becomes zero. This occurs when the “net” area)`

`text(under the curve is zero).`

`text(The area under the curve and above the)\ x text(-axis)`

`text(between)\ \ t=0 and 2,\ text(is equal to the area above)`

`text(the curve and below the)\ x text(-axis between)\ \ t=2 and 4.`

`:.\ text(Furthest from origin when)\ \ t=4.`

Trigonometry, 2ADV T2 2004 HSC 9a

Consider the geometric series `1 − tan^2 theta + tan^4 theta − …`

When the limiting sum exists, find its value in simplest form. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
For what values of `theta` in the interval

`−pi/2 < theta < pi/2` does the limiting sum of the series exist? (2 marks)

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

Show Answers Only

`cos^2 theta`
`− pi/4 < theta < pi/4`

Show Worked Solution

i. `1 − tan^2 theta + tan^4 theta − …`

`=>\ text(GP where)\ \ a=1,\ \ r=T_2/T_1= − tan^2 theta`

`:. S_∞`	`= 1/(1 − (−tan^2 theta))`
	`= 1/(1 + tan^2 theta)`
	`= 1/(sec^2 theta)`
	`= cos^2 theta`

ii.	`text(Find)\ \ theta\ \ text(such that)\ \ \ \|\ r\ \|`	`< 1`
	`\|−tan^2 theta\ \|`	`< 1`
	` tan^2 theta`	`< 1`
	`−1 < tan theta`	`< 1`
	`:. − pi/4 < theta`	`< pi/4`

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)

--- 4 WORK AREA LINES (style=lined) ---
By writing a geometric series for the value of all Betty’s investments, calculate the final value of Luis’ trust fund. (3 marks)

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

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.)`

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)

--- 5 WORK AREA LINES (style=lined) ---
What is the probability of scoring `16` or more on the first turn? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
What is the probability that the sum of the scores in the first two turns is less than 45? (2 marks)

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

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.`

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).`

Calculus, MET2 2012 VCAA 22 MC

The graph of a differentiable function `f` has a local maximum at `(a, b)`, where `a < 0` and `b > 0`, and a local minimum at `(c, d)`, where `c > 0` and `d < 0.`

The graph of `y = -|f (x - 2)|` has

A. a local minimum at `(a - 2, − b)` and a local maximum at `(c - 2, d)`

B. local minima at `(a + 2, − b)` and `(c + 2, d )`

C. local maxima at `(a + 2, b)` and `(c + 2, − d)`

D. a local minimum at `(a - 2, − b)` and a local maximum at `(a - 2, − d)`

E. local minima at `(c + 2, − d)` and `(a + 2, − b)`

Show Answers Only

`B`

Show Worked Solution

`text(The progressive graphs below are one way to)`

♦♦ Mean mark 32%.

`text(explain the shape of)\ \ y = -|f (x – 2)|`

`=> B`

Algebra, MET2 2012 VCAA 19 MC

A function `f` has the following two properties for all real values of `theta.`

`f(pi - theta) = -f(theta)` and `f(pi - theta) = -f(- theta)`

A possible rule for `f` is

`f(x) = sin (x)`
`f(x) = cos (x)`
`f(x) = tan (x)`
`f(x) = sin (x/2)`
`f(x) = tan (2x)`

Show Answers Only

`B`

Show Worked Solution

`text(Solution 1:)`

♦ Mean mark 45%.

`text(By elimination,)`

`f(pi – theta) = -f(theta)`

`text(Opposite signs in the 1st and 2nd quadrant)`

`:.\ text(Cannot be)\ sin\ \ text{(Eliminate}\ A\ text(and)\ D)`

`f(pi – theta) = -f(- theta)`

`text(Opposite signs in the 2nd and 4th quadrant)`

`:.\ text(Cannot be)\ tan\ \ text{(Eliminate}\ C\ text(and)\ E)`

`=> B`

`text(Solution 2:)`

`text(The given conditions imply the function is even.)`

`cos(x)\ \ text(is the only even function given.)`

`=> B`

Calculus, MET2 2012 VCAA 16 MC

The graph of a cubic function `f` has a local maximum at `(a, text{−3)}` and a local minimum at `(b, text{−8)}.`

The values of `c`, such that the equation `f(x) + c = 0` has exactly one solution, are

`3 < c < 8`
`c > -3 or c < -8`
`-8 < c < -3`
`c < 3 or c > 8`
`c < -8`

Show Answers Only

`D`

Show Worked Solution

`text(Sketch a possible graph:)`

♦♦ Mean mark 34%.

`text(For one solution:)`

`text(Shift graph up less than 3)`

`c < 3,\ \ text(or)`

`text(Shift graph up more than 8)`

`c > 8`

`=> D`

Calculus, MET2 2012 VCAA 18 MC

The tangent to the graph of `y = log_e(x)` at the point `(a, log_e(a))` crosses the `x`-axis at the point `(b, 0)`, where `b < 0.`

Which of the following is false?

A. `1 < a < e`

B. The gradient of the tangent is positive

C. `a > e`

D. The gradient of the tangent is `1/a`

E. `a > 0`

Show Answers Only

`A`

Show Worked Solution

`text(Consider Option)\ A:`

♦♦ Mean mark 30%.

`text(T)text{angent at}\ x=1\ text{(from graph),}\ \ b>0.`

`text(T)text{angent at}\ \ x=e\ \ text{is}\ \ y=1/e x,\ \ b=0.`

`text(Graphing the tangents:)`

`text(For negative)\ xtext{-intercept (i.e.}\ \ b<0text{)}`

`a > e`

`:. A\ text(is false.)`

`=> A`

Algebra, MET2 2012 VCAA 17 MC

A system of simultaneous linear equations is represented by the matrix equation

`[(m, 3), (1, m + 2)] [(x), (y)] = [(1), (m)].`

The system of equations will have no solution when

A. `m = 1`

B. `m = -3`

C. `m in text({1, −3})`

D. `m in Rtext(\{1})`

E. `m in {1, 3}`

Show Answers Only

`B`

Show Worked Solution

`text(Change the matrix equation to two)`

♦ Mean mark 43%.

`text(simultaneous linear equations,)`

`mx+3y`	`=m+3m`
`y`	`=- m/3 x+ 4/3 m\ …\ (1)`
`x+(m+2)y`	`=1+m(m+2)`
`y`	`=- 1/(m+2) x + (m^2+2m+1)/(m+2)\ …\ (2)`

`text(No solution if lines are parallel:)`

`- 1/(m+2)`	`= – m/3`
`m^2 + 2m -3`	`=0`
`(m+3)(m-1)`	`=0`

`m=-3 or 1`

`text(Note that if)\ \ m=1,\ \ text(the two lines are identical.)`

`:.\ text(No solution when)\ \ m = -3`

`=> B`

Calculus, MET2 2012 VCAA 8 MC

The function `f: R -> R,\ f(x) = ax^3 + bx^2 + cx`, where `a` is a negative real number and `b` and `c` are real numbers.

For the real numbers `p < m < 0 < n < q`, we have `f(p) = f(q) = 0` and `f prime (m) = f prime (n) = 0.`

The gradient of the graph of `y = f(x)` is negative for

`(text(−∞), m) uu (n, oo)`
`(m, n)`
`(p, 0) uu (q, oo)`
`(p, m) uu (0, q)`
`(p, q)`

Show Answers Only

`A`

Show Worked Solution

`text(The graph crosses the)\ x text(-axis at)`

♦ Mean mark 49%.

`x=p, x=0 and x=q.`

`text(Turning points at)\ \ x=m and x=n.`

`text(Sketching the function:)`

`:. f′(x) < 0quadtext(for)quadx < mquad∪quadx > n`

`=> A`

Calculus, MET2 2012 VCAA 4 MC

Given that `g` is a differentiable function and `k` is a real number, the derivative of the composite function `g (e^(kx))` is

A. `kg prime (e^(kx)) e^(kx)`

B. `kg (e^(kx))`

C. `ke^(kx)\ g(e^(kx))`

D. `ke^(kx)\ g prime (e^(x))`

E. `1/k e^(kx) g prime (e^(kx))`

Show Answers Only

`A`

Show Worked Solution

`text(Using chain rule:)`

♦ Mean mark 45%.

`y`	`= g(e^(kx))`
`(dy)/(dx)`	`= g′(e^(kx)) xx (ke^(kx))`
	`=kg′(e^(kx)) e^(kx)`

`=> A`

Calculus, MET2 2013 VCAA 12 MC

CHAIN RULE

Let `y = 4 cos (x)` and `x` be a function of `t` such that `(dx)/(dt) = 3e^(2t)` and `x = 3/2` when `t = 0.`

The value of `(dy)/(dt)` when `x = pi/2` is

A. `0`

B. `3 pi log_e (pi/2)`

C. `-4 pi`

D. `-2 pi`

E. `-12e`

Show Answers Only

`C`

Show Worked Solution

`(dy)/(dt)`	`= (dy)/(dx) xx (dx)/(dt)`
	`= -4 sin x xx 3e^(2t)`

`x`	`= int 3e^(2t)\ dt`
	`= 3/2 e^(2t) + c`

♦♦ Mean mark 35%.

`text(Given that)\ \ x = 3/2\ \ text(when)\ \ t = 0,`

`3/2`	`=3/2 e^0+c`
`:. c`	`=0`
`:.x`	`= 3/2 e^(2t)`

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

`pi/2`	`=3/2 e^(2t)`
`e^(2t)`	`=pi/3`
`t`	`=1/2 ln(pi/3)`

`:. (dy)/(dt)`	`=-4 sin\ pi/2 xx 3e^ln(pi/3)`
	`=-4 xx 1 xx 3(pi/3)`
	`=-4pi`

`=> C`

Calculus, MET2 2013 VCAA 11 MC

If the tangent to the graph of ` y = e^(ax), a != 0,\ \ text(at)\ \ x = c` passes through the origin, then `c` is equal to

A. `0`

B. `1/a`

C. `1`

D. `a`

E. `-1/a`

Show Answers Only

`B`

Show Worked Solution

`y = e^(ax)\ \ \ => dy/dx = ae^(ax)`

♦ Mean mark 47%.

`text(At)\ \ x = c,`

`y = e^(ac)\ \ \ => (dy)/(dx) = ae^(ac)`

`text(T)text(angent passes through)\ \ (0,0) and (c,e^(ac))`

`:.\ text(Gradient of tangent) = (e^(ac)-0)/(c-0) = e^(ac)/c`

`text(Equating the gradients,)`

`ae^(ac)`	`=e^(ac)/c`
`ac`	`=1`
`c`	`=1/a`

`=> B`

Algebra, MET2 2013 VCAA 7 MC

The function `g: text{[−a, a]} -> R, \ g(x) = sin (2(x - pi/6))` has an inverse function.

The maximum possible value of `a` is

`pi/12`
`1`
`pi/6`
`pi/4`
`pi/2`

Show Answers Only

`A`

Show Worked Solution

`text(If)\ \ y=sin x,\ \ text(the inverse function exists in)`

♦ Mean mark 37%.

`text(the domain)\ \ \ -pi/2<= x <= pi/2`

`text(Find)\ x\ text(such that:)`

`2(x – pi/6) = +- pi/2`

`:. g^-1\ \ text(exists in the domain)\ \ [-pi/12, (5pi)/12]`

`text(S)text(ince the form of the domain is)\ \ [a,-a],`

`a = pi/12`

`=> A`

Proof, EXT2 P1 2015 HSC 15c

For positive real numbers `x` and `y`, `sqrt (xy) <= (x + y)/2`. (Do NOT prove this.)

Prove `sqrt (xy) <= sqrt ((x^2 + y^2)/2)`, for positive real numbers `x` and `y.` (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Prove `root4(abcd) <= sqrt ((a^2 + b^2 + c^2 + d^2)/4)`, for positive real numbers `a, b, c` and `d.` (2 marks)

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

Show Answers Only

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

Show Worked Solution

i. `text(S)text(ince)\ \ (x – y)^2 = x^2 + y^2 – 2xy`

`and \ (x – y)^2 >= 0`

`0`	`≤x^2 + y^2 – 2xy`
`2xy`	`≤x^2 + y^2`
`:.sqrt (xy)`	`≤sqrt ((x^2 + y^2)/2)`

ii. `sqrt(ab) <= sqrt((a^2 + b^2)/2),\ \ \ \ text{(part (i))}`

♦♦ Mean mark 29%.

`sqrt(cd) <= sqrt((c^2 + d^2)/2),\ \ \ \ text{(part (i))}`

`sqrt(ab) sqrt(cd)`	`<=sqrt((a^2 + b^2)/2)*sqrt((c^2 + d^2)/2)`
	`<=sqrt(((a^2 + b^2)/2) * ((c^2 + d^2)/2))`

`sqrt (xy) <= (x + y)/2\ \ \ \ text{(given):}`
`sqrt(ab) sqrt(cd)`	`<=1/2((a^2 + b^2+c^2+d^2)/2)`
`sqrt(abcd)`	`<=(a^2 + b^2+c^2+d^2)/4`
`:.root4(abcd)`	`<=sqrt((a^2 + b^2+c^2+d^2)/4)`

Proof, EXT2 P1 2015 HSC 15b

Suppose that `x >= 0` and `n` is a positive integer.

Show that `1 - x <= 1/(1 + x) <= 1.` (2 marks)

--- 2 WORK AREA LINES (style=lined) ---
Hence, or otherwise, show that

`1 - 1/(2n) <= n ln (1 + 1/n) <= 1.` (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Hence, explain why

`lim_(n -> oo) (1 + 1/n)^n = e.` (1 mark)

--- 4 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.	`1-x^2`	`<=1\ \ \ text(for)\ x>=0`
	`(1-x)(1+x)`	`<=1`
	`(1-x)`	`<=1/(1+x)`

`text(S)text(ince)\ \ 1 + x >= 1\ \ text(when)\ \ x >= 0`

`=>1/(1 + x) <= 1`

`:. 1 – x <= 1/(1 + x) <= 1.`

♦♦ Mean mark 21%.
STRATEGY: The conversion of the middle term from a fraction into a logarithm should flag the need for integration of each term.

ii.	`int_0^(1/n) (1 – x)\ dx`	`<= int_0^(1/n) (dx)/(1 + x) <= int_0^(1/n) 1\ dx`
	`[x – x^2/2]_0^(1/n)`	`<= [ln (1 + x)]_0^(1/n) <= [x]_0^(1/n)`
	`1/n-1/(2n^2)`	`<= ln (1 + 1/n) <= 1/n`
	`1 – 1/(2n) `	`<= n ln (1 + 1/n) <= 1,\ \ \ \ \ (n>=1)`

iii.	`lim_(n -> oo) (1 – 1/(2n))`	`<= lim_(n -> oo){n ln (1 + 1/n)} <= lim_(n -> oo) (1)`
	`1`	`<= lim_(n -> oo) {l ln (1 + 1/n)} <= 1`

♦♦ Mean mark 29%.

`:. lim_(n -> oo) (n ln (1 + 1/n))`	`=1`
`lim_(n -> oo) ln (1 + 1/n)^n`	`=1`
`:.lim_(n -> oo) (1 + 1/n)^n`	`=e`

Mechanics, EXT2 2015 HSC 14c

A car of mass `m` is driven at speed `v` around a circular track of radius `r`. The track is banked at a constant angle `theta` to the horizontal, where `0 < theta < pi/2`. At the speed `v` there is a tendency for the car to slide up the track. This is opposed by a frictional force `mu N`, where `N` is the normal reaction between the car and the track, and `mu > 0`. The acceleration due to gravity is `g`.

Show that
`v^2 = rg((tan theta + mu)/(1 - mu tan theta)).` (3 mark)
At the particular speed `V`, where `V^2 = rg`, there is still a tendency for the car to slide up the track.
Using the result from part (i), or otherwise, show that `mu < 1.` (2 marks)

Show Answers Only

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

Show Worked Solution

(i)

`text(Resolving the forces vertically)`

`N cos theta`	`= mg + mu N sin theta`
`N (cos theta – mu sin theta)`	`= mg\ \ \ …\ (1)`

`text(Resolving the forces horizontally)`

`N sin theta + mu N cos theta`	`=(m v^2)/r`
`N (sin theta + mu cos theta)`	`=(m v^2)/r\ \ \ …\ (2)`

`text(Divide)\ \ (2)÷(1)`

`((m v^2)/r)/(mg)`	`=(sin theta + mu cos theta)/(cos theta – mu sin theta)`
`v^2/(rg)`	`=(sin theta + mu cos theta)/(cos theta – mu sin theta)`
`v^2`	`=rg ((sin theta + mu cos theta)/(cos theta – mu sin theta))`
	`= rg ((tan theta + mu)/(1 – mu tan theta))`

(ii) `text(Given that)\ \ V^2=rg`

♦ Mean mark 46%.

`=> (tan theta + mu)/(1 – mu tan theta)=1`

`(tan theta + mu)`	`=(1 – mu tan theta)`
`mu(1+ tan theta)`	`=1-tan theta`
`mu`	`=(1-tan theta)/(1+ tan theta)`
	`=1- (2tan theta)/(1+ tan theta)`

`text(S)text(ince)\ \ tan theta>0\ \ text(for)\ \ 0<theta<pi/2`

`=> (2tan theta)/(1+ tan theta) >0`

`:. mu<1`

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.`

Harder Ext1 Topics, EXT2 2006 HSC 8b

For `x > 0`, let `f(x) = x^n e^-x`, where `n` is an integer and `n >= 2.`

The two points of inflection of `f(x)` occur at `x = a` and `x = b`, where `0 < a < b`. Find `a` and `b` in terms of `n.` (4 marks)
Show that
1. `(f(b))/(f(a)) = ((1 + 1/sqrt n)/(1 - 1/sqrt n))^n e^(-2 sqrt n).` (2 marks)
Using the following
If `0<=x<=1/sqrt2` then `1 <= ((1 + x)/(1 - x)) e^(-2x) <= e^((4x^3)/3),` (DO NOT prove this)
2. show that `1 <= (f(b))/(f(a)) <= e^(4/(3 sqrt n)).` (2 marks)
What can be said about the ratio `(f(b))/(f(a))` as `n -> oo?` (1 mark)

Show Answers Only

`a = n – sqrt n,\ \ \ \ \ b = n + sqrt n`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`lim_(n -> oo) (f(b))/(f(a)) = 1`

Show Worked Solution

(i)	`f(x)`	`= x^n e^-x,\ \ \ \ n >= 2,\ \ \ x > 0`
	`f′(x)`	`= nx^(n – 1) e^-x – x^n e^-x`
	`f″(x)`	`= n[(n – 1) x^(n -2) e^-x – x^(n – 1) e^-x] – (nx^(n – 1) e^-x – x^n e^-x)`
		`= x^(n – 2) e^-x [n(n – 1) – nx – nx + x^2]`
		`= x^(n – 2) e^-x (x^2-2nx +n^2 – n)`

`text(P.I.’s occur when) \ \ f″(x)=0`

`text(i.e. when)\ \ \ x^2-2nx +n^2 – n = 0,\ \ \ \ (e^-x>0 and x>0)`

`:.x=`	`(2n+-sqrt(4n^2- 41(n^2-n)))/2`
`=`	`\ \ n+- sqrtn`

`text(S)text(ince)\ \ a<b`

`:. a = n – sqrt n\ and\ b = n + sqrt n.`

(ii) `(f(b))/(f(a)) =`	`((n + sqrt n)^n e^-(n + sqrt n))/((n – sqrt n)^n e^-(n – sqrt n))`
`=`	`(n^n(1 + 1/sqrt n)^n)/(n^n(1 – 1/sqrt n)^n) xx e^(-n – sqrt n + n – sqrt n)`
`=`	`((1 + 1/sqrt n)/(1 – 1/sqrt n))^n e^(-2 sqrt n)`

(iii) `text(S)text(ince)\ \ n>=2\ \ \ => 1/sqrt n <=1/sqrt2`

`text(Substituting)\ \ x=1/sqrt n\ \ text(into the inequality)`

`1`	`<= ((1 + 1/sqrt n)/(1 – 1/sqrt n)) e^((-2)/sqrt n) <= e^(4/(3(sqrt n)^3))`
`1^n`	`<= ((1 + 1/sqrt n)/(1 – 1/sqrt n))^n e^((-2)/sqrt n xx n) <= e^(4/(3(sqrt n)^3) xx n)`
`:.1`	`<= (f(b))/(f(a)) <= e^(4/(3 sqrt n))`

(iv)	`lim_(n -> oo) e^(4/(3 sqrt n))`	`=1`
	`:.lim_(n -> oo) (f(b))/(f(a))`	`=1`

`text(Using)\ \ \ I`	`= (PrT) / 100`
	`= {(1450) (3.6) (1/12)} / 100`
	`= $4.35`

`:.\ text(Savings)`	`=(36 xx 1446.75) – (18 xx 2697.39)`
	`=52\ 083 – 48\ 553.02`
	`=$3529.98`

`text(GP where)\ \ \ a`	`=15, and`
`r`	`=t_2/t_1=13.5/15=0.9`

`S_n`	`=(a(1-r^n))/(1-r)`
`:.S_14`	`=(15(1-0.9^14))/(1-0.9)`
	`=115.68…`

`text(LHS)`	`= \|\ (x + y)^2 – (x – y)^2\ \|`
	`= \|\ 4xy\ \|`
	`= 4*\|\ xy\ \|`

`:.\ text(Gradient)`	`=(log 869 – log 98)/(9-5)`
	`= 0.2369…`

`f(x)`	`= x^2\ text(satisfies)\ \|\ f(x + y) – f(x – y)\ \|`
	`= 4 sqrt (f(x) f(y))`

`text(RHS)`	`= 4 sqrt(x^2 y^2)`
	`= 4 sqrt((xy)^2)`
	`= 4*\|\ xy\ \|`
	`=\ text(LHS)`

A. `f^-1:\ R^+ -> R`	`f^-1(x) = 1/x^2 - 2`
B. `f^-1: R text(\{0}) -> R`	`f^-1(x) = 1/x^2 - 2`
C. `f^-1: R^+ -> R`	`f^-1(x) = 1/x^2 + 2`
D. `f^-1:\ text{(−2, ∞)} -> R`	`f^-1(x) = x^2 + 2`
E. `f^-1:\ (2, oo) -> R`	`f^-1(x) = 1/(x^2 - 2)`