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PATTERNS, FUR1 2013 VCAA 4 MC

The vertical distance, in m, that a hot air balloon rises in each successive minute of its flight is given by the geometric sequence

`64.0,\ \ 60.8,\ \  57.76\ …`

The total vertical distance, in m, that the balloon rises in the first 10 minutes of its flight is closest to

A.   `38`   

B.   `40`  

C.  `473` 

D.  `514`  

E. `1280`  

Show Answers Only

`D`

Show Worked Solution

`text(Series 64.0, 60.8, 57.76 …)`

`text(GP where)\ \  a` `=64.0`
`r` `=60.8/64.0=0.95`
`S_n` `=(a(1-r^n))/(1−r)` 
`S_10` `=(64 (1−0.95^10))/(1−0.95)`  
  `=513.616…`

`=> D`

PATTERNS, FUR1 2013 VCAA 2 MC

The graph above shows the first six terms of a sequence.

This sequence could be

A.  an arithmetic sequence that sums to one.

B.  an arithmetic sequence with a common difference of one.

C.  a Fibonacci-related sequence whose first term is one.

D.  a geometric sequence with an infinite sum of one.

E.  a geometric sequence with a common ratio of one. 

Show Answers Only

`E`

Show Worked Solution

`text(Series is 1, 1, 1, …)`

♦ Mean mark 41%.

`text(The only possibility within the choices)`

`text(is a geometric sequence where)\ \  r=1.`

`=> E`

CORE, FUR1 2013 VCAA 12-13 MC

The time series plot below displays the number of guests staying at a holiday resort during summer, autumn, winter and spring for the years 2007 to 2012 inclusive.
 

CORE, FUR1 2013 VCAA 12-13 MC_1
 

 Part 1

Which one of the following best describes the pattern in the time series?

A.  random variation only

B.  decreasing trend with seasonality

C.  seasonality only

D.  increasing trend only

E.  increasing trend with seasonality 

 

Part 2

The table below shows the data from the times series plot for the years 2007 and 2008. 
 

CORE, FUR1 2013 VCAA 12-13 MC_2
 

Using four-mean smoothing with centring, the smoothed number of guests for winter 2007 is closest to

A.  `85`

B.  `107`

C.  `183`

D.  `192`

E.  `200` 

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`text(The pattern in the time series is seasonal only,)`

`text(with peaks appearing in Summer. There is no)`

`text(apparent year-to-year trend.)`

`=> C`

 

`text(Part 2)`

`text{Mean of guests (Season 1-4)}`

`=(390+126+85+130)/4`
`=182.75`

 

`text{Mean of guests (Season 2-5)}`

`=(126+85+85+130+460)/4`
`=200.25`

 

`:.\ text(Four-mean smoothing with centring)`

`=(182.75+200.25)/2`
`=191.5`

`=> D`

CORE, FUR1 2013 VCAA 11 MC

A least squares regression line is fitted to data in a scatterplot, as shown below.
  

CORE, FUR1 2013 VCAA 11 MC

The corresponding residual plot is closest to

CORE, FUR1 2013 VCAA 11 MC_ab

CORE, FUR1 2013 VCAA 11 MC_cd

CORE, FUR1 2013 VCAA 11 MC_e

 

Show Answers Only

`A`

Show Worked Solution

`text(By Elimination)`

`text(The first 2 and last 3 data points are below regression line,)`

`text(and are therefore negative residuals.)`

`:.\ text(Cannot be)\ B, D\ text(or)\ E.`

 

`text(The first two data points will have similar negative)`

`text(residuals.)`

`:.\ text(Cannot be)\ C.`

`=> A`

CORE, FUR1 2013 VCAA 10 MC

The data in the scatterplot below shows the width, in cm, and the surface area, in cm², of leaves sampled from 10 different trees. The scatterplot is non-linear.
 

CORE, FUR1 2013 VCAA 10 MC

To linearise the scatterplot, (width)2 is plotted against area and a least squares regression line is then fitted to the linearised plot.

The equation of this least squares regression line is

(width)2 = 1.8 + 0.8 × area

Using this equation, a leaf with a surface area of 120 cm² is predicted to have a width, in cm, closest to 

A.     9.2

B.     9.9

C.   10.6

D.   84.6

E.   97.8

Show Answers Only

`B`

Show Worked Solution

`text(Substituting)`

`text(Width)^2` `=1.8+0.8 xx area`                           
  `=1.8+(0.8×120)`
  `=97.8`
`:.\ text(Width)` `=\sqrt97.8`
  `=9.889…\ text(cm)`

`=> B`

CORE, FUR1 2013 VCAA 7 MC

For a city, the correlation coefficient between

  • population density and distance from the centre of the city is `r` = – 0.563
  • house size and distance from the centre of the city is `r` = 0.357.

Given this information, which one of the following statements is true?

  1. Around 31.7% of the variation observed in house size in the city can be explained by the variation in distance from the centre of the city.
  2. Population density tends to increase as the distance from the centre of the city increases.
  3. House sizes tend to be larger as the distance from the centre of the city decreases.
  4. The slope of a least squares regression line relating population density to distance from the centre of the city is positive.
  5. Population density is more strongly associated with distance from the centre of the city than is house size.
Show Answers Only

`E`

Show Worked Solution

`text(Population density)`

♦ Mean mark 49%.

`r^2=(-0.563)^2=0.3169…~~31.7%`

`text(House size)`

`r^2=(0.357)^2 = 0.1274…~~12.7%`

 

`text(In A, 12.7% of the variability is explained. Not true.)`

`text(In B,)\ r = –0.563. text(The negative correlation means population)`

`text(density increases as distance decreases. Not true.)`

`text(In C, statement assumes a negative correlation which is not the case)`

`text(given)\ r = 0.357.\ \ text(Not true.)`

`text(In D, negative correlation would have a negatively slope. Not true.)`

`text{In E, true since}\ \ | \ r^2\  |\ \ text{for house size is greater.}`

`=>\ E`

Quadratic, EXT1 2008 HSC 4c

2008 4c

The points  `P(2ap, ap^2)`,  `Q(2aq, aq^2)`  lie on the parabola  `x^2 = 4ay`. The tangents to the parabola at  `P`  and  `Q`  intersect at  `T`. The chord  `QO`  produced meets  `PT`  at  `K`, and  `/_PKQ`  is a right angle.

  1. Find the gradient of  `QO`, and hence show that  `pq = –2`.   (2 marks)
  2. The chord  `PO`  produced meets  `QT`  at  `L`. Show that  `/_PLQ`  is a right angle.   (1 mark)
  3. Let  `M`  be the midpoint of the chord  `PQ`. By considering the quadrilateral  `PQLK`, or otherwise, show that  `MK = ML`.   (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)  `Q(2aq, aq^2),\ \ \ \ O(0,0)`

`m_(QO)` `= (y_2 – y_1)/(x_2 – x_1)`
  `= (aq^2 – 0)/(2aq – 0)`
  `= q/2`

`text(Gradient of tangent at)\ P = p`

 

`text(We know the extension of chord)\ \ QO _|_ PT\ \ \ text{(given)}`

`m_(QO) xx m_(PT)` `= -1`
`q/2 xx p` `= -1`
`:. pq` `= -2\ \ \ text(… as required.)`

 

(ii)  `m_(PO) = (ap^2-0)/(2ap-0) = p/2`

`m\ text(of tangent)\ QT = q`

`m_(PO) xx m_(QT)` `= p/2 xx q`
  `= (pq)/2`
  `= (-2)/2\ \ \ text{(from (i))}`
  `= -1`

 

`:.\ /_PLQ\ text(is a right angle.)`

 

(iii)  `PQ\ text(subtends 2 right-angles at)\ L\ text(and)\ K`

`=>\ PQLK\ text(is a cyclic quad, with)\ \ PQ`

`text(a diameter.)`

 

`=> text(Midpoint of)\ PQ=M\ \ \ text{(centre of the circle)}`

`:.\ MK = ML\ \ text{(radii)}`

Combinatorics, EXT1 A1 2008 HSC 4b

Barbara and John and six other people go through a doorway one at a time.

  1. In how many ways can the eight people go through the doorway if John goes through the doorway after Barbara with no-one in between?   (1 mark)

  2. Find the number of ways in which the eight people can go through the doorway if John goes through the doorway after Barbara.   (1 mark)

Show Answers Only

a.    `5040`

b.    `20\ 160`

Show Worked Solution

a.    `text(Barbara and John can be treated as one person)`

`:.\ text(#Combinations)= 7! = 5040`
 

b.    `text(Solution 1)`

`text(Total possible combinations)=8!`

`text(Barbara has an equal chance of being behind as)`

`text(being in front of John.)`
 

`:.\ text{#Combinations (John before Barbara)}`

`=(8!)/2 =20\ 160`
 

`text(Solution 2)`

`text(If)\ B\ text(goes first,)\ J\ text(has 7 options and the other)`

`text(6 people can go in any order.)`

`=> text(#Combinations) = 7 xx 6!`
 

`text(If)\ B\ text(goes second,)\ J\ text(has 6 options)`

`=> text(#Combinations) = 6 xx 6!`
 

`text(If)\ B\ text(goes third,)\ J\ text(has 5 options)`

`=> text(#Combinations) = 5 xx 6!`

`text(And so on…)`
 

`:.\ text(Total Combinations)`

`= 7 xx 6! + 6 xx 6! + … + 1 xx 6!`

`= 6! ( 7 + 6 + 5 + 4 + 3 + 2 + 1)`

`= 6! ( 28)`

`= 20\ 160`

Calculus, EXT1 C1 2008 HSC 3c

2008 3c
A race car is travelling on the `x`-axis from `P` to `Q` at a constant velocity, `v`.

A spectator is at `A` which is directly opposite `O`, and  `OA = l`  metres. When the car is at `C`, its displacement from `O` is `x` metres and `/_OAC = theta`, with  `- pi/2 < theta < pi/2`.

  1. Show that  `(d theta)/(dt) = (vl)/(l^2 + x^2)`.   (2 marks)

  2. Let `m` be the maximum value of `(d theta)/dt`.  

     

    Find the value of `m` in terms of `v` and `l`.   (1 mark)

  3. There are two values of `theta` for which  `(d theta)/(dt) = m/4`.

     

    Find these two values of `theta`.   (2 marks)

Show Answers Only

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

b.    `v/l`

c.    `+- pi/3`

Show Worked Solution

a.  `text(Show)\ \ (d theta)/(dt) = (vl)/(l^2 + x^2)`

`(d theta)/(dt)` `= (d theta)/(dx) * (dx)/(dt)\ \ \ \ \ \ \ …\ (1)`
`v` `= (dx)/(dt)\ \ \ text{(given)}`

 
`text(Find)\ (d theta)/(dx):`

`tan theta` `= x/l`
`theta` `= tan^-1 (x/l)`
`(d theta)/(dx)` `= l/(l^2+x^2)`

 
`text{Substituting into (1):}`

`(d theta)/(dt)= l/(l^2+x^2) xx v= (vl)/(l^2 + x^2)\ \ \ text(… as required)`
 

b.    `(d theta)/(dt)\ \ text(is a MAX when)\ \ x = 0`

`:. m= (vl)/(l^2 + 0^2)= v/l`
 

c.    `text(Find)\ theta\ text(when)\ \ (d theta)/(dt) = m/4`

MARKER’S COMMENT: Many students lost marks by not giving their answer in the specified range. Be careful!
`(d theta)/(dt)` `= v/(4l)`
`(vl)/(l^2 +x^2)` `= v/(4l)`
`l^2 + x^2` `= 4l^2`
`x^2` `= 3l^2`
`x` `= +- sqrt3 l`

 

`tan theta` `= x/l= +- sqrt 3`
`:. theta` `= +- pi/3,\ \ \ \ \ (- pi/2 < theta < pi/2)`

Proof, EXT1 P1 2008 HSC 3b

Use mathematical induction to prove that, for integers  `n >= 1`,

`1 xx 3 + 2 xx 4 + 3 xx 5 + ... + n(n+2) = n/6 (n + 1)(2n + 7)`.   (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Prove)\ \ 1 xx 3 + 2 xx 4 + 3 xx 5 + … + n(n+2)`

`= n/6 (n+1)(2n + 7)\ \ \ text(for)\ n >= 1`

`text(If)\ \ n = 1`

`text(LHS) = 1 xx 3 = 3`

`text(RHS) = 1/6 (2)(9) = 3 = text(LHS)`

`:.\ text(True for)\ \ n = 1`

 
`text(Assume true for)\ \ n = k`

`text(i.e.)\ \ 1 xx 3 + 2 xx 4 + … + k (k + 2)`

`= k/6 (k + 1)(2k + 7)`

 
`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ \ 1 xx 3 + 2 xx 4 + … + k(k+2) + (k + 1)(k + 3)`

`= ((k+1))/6 (k + 2)(2k + 9)`

`text(LHS)` `= k/6 (k + 1)(2k + 7) + (k + 1)(k + 3)`
  `= 1/6 (k + 1)[k(2k + 7) + 6(k + 3)]`
  `= 1/6 (k + 1)[2k^2 + 7k + 6k + 18]`
  `= 1/6 (k + 1)[2k^2 + 13k + 18]`
  `= 1/6 (k +1)(k + 2)(2k + 9)\ \ \ \ text(… as required)`

 
`=>\ text(True for)\ n = k + 1`

`:.\ text(S) text(ince true for)\ n = 1,\ text(by PMI, true for integral)\ n >= 1.`

Functions, EXT1 F1 2008 HSC 3a

  1.  Sketch the graph of  `y = |\ 2x-1\ |`.   (1 mark)

  2.  Hence, or otherwise, solve  `|\ 2x-1\ | <= |\ x-3\ |`.    (3 marks)

Show Answers Only

a.    

Real Functions, EXT1 2008 HSC 3a Answer

b.    `-2 <= x <= 4/3`

Show Worked Solution
a.    Real Functions, EXT1 2008 HSC 3a Answer

 

b.    `text(Solving for)\ \ |\ 2x-1\ | <= |\ x-3\ |`

`text(Graph shows the statement is TRUE between the)`

`text(points of intersection.)`

`=>\ text(Intersection occurs when)`

`(2x-1)` `= (x-3)\ \ \ text(or)\ \ \ ` `-(2x-1)` `= x-3`
`x` `= -2` `-2x + 1` `= x-3`
    `-3x` `= -4`
    `x` `= 4/3`

 

`:.\ text(Solution is)\ \ {x: -2 <=  x <= 4/3}`

Measurement, STD2 M7 EQ-Bank 22

Bronwyn needs to have 3.0 litres of intravenous liquid given to her over a period of 4 hours. 

What is the required flow rate in mL per minute?   (2 marks) 

Show Answers Only

`12.5\ text(mL/minute)`

Show Worked Solution

`text(Total liquid)= 3.0 xx 1000= 3000\ text(mL)`

`text(Total minutes)= 4 xx 60= 240`

`:.\ text(Flow Rate)= 3000/240= 12.5\ text(mL/minute)`

Measurement, STD2 M7 EQ-Bank 26

A medication is available in both tablet and liquid form.  A tablet contains 50 mg of the active ingredient while the liquid form contains 60 mg per 10 mL.  

Michael likes taking tablets and Georgia prefers liquid medicines.  If they each need 0.2 g of the active ingredient, what dosages do they take?   (3 marks) 

Show Answers Only

`text(Michael takes 4 tablets and Georgia takes 33.3mL.)`

Show Worked Solution

`text(Michael – tablets)`

`0.2\ text(g) = 200\ text(mg)`

`:.\ text(# Tablets)=200/50= 4`

 
`text(Georgia – liquid)`

`60\ text(mg in)\ 10\ text(mL)`

`1\ text(mg) = 10/60 = 0.166…\ text(mL)`

`200\ text(mg)= 200 xx 0.166…= 33.3\ text(mL)\ \ text{(to 1 d.p.)}`
  

`:.\ text(Michael takes 4 tablets and Georgia takes 33.3 mL.)`

Financial Maths, STD2 F5 EQ-Bank 28

Dominique wants to save $15 000 to use as spending money when she travels overseas in 2 years' time.  

If she invests $3500 at the end of every 6 months into an account earning 4% p.a., compounded half-yearly, will she have enough?

Use the table below to justify your answer.   (2 marks)
 

 

Show Answers Only

`text(Dominique will not have saved)`

`text(enough to reach)\ $15\ 000.`

Show Worked Solution

`text(Interest rate)\ text{(6 monthly)} = text(4%)  -: 2 = text(2%)`

`n = 4\ \ \ text{(6 month periods in 2 years)}`

`=>\ text(FVA factor) = 4.122\ \ \ text{(from Table)}`

`text(FVA)= 3500 xx 4.122= $14\ 427`
   

 `:.\ text(Dominique will not have saved enough to)`

`text(reach)\ $15\ 000.`

Financial Maths, STD2 F5 EQ-Bank 30

Camilla buys a car for $21 000 and repays it over 4 years through equal monthly instalments.

She pays a 10% deposit and interest is charged at 9% p.a. on the reducing balance loan.

Using the Table of present value interest factors below, where `r` represents the monthly interest and `N` represents the number of repayments
 

2UG FM5 S-2 

  1. Calculate the monthly repayment,  `$P`, that Camilla must pay to complete the loan after 4 years  (to the nearest $).   (3 marks)

  2. Calculate the total interest paid over the life of the loan.    (1 mark)

Show Answers Only

a.    `text(Camilla must repay $470 per month)`

b.    `$3660`

Show Worked Solution

a.    `text(Deposit) = 10text(%) xx 21\ 000 = $2100`

`text(Loan Value)= 21\ 000-2100= $18\ 900` 

`text(Monthly interest rate) = text(9%)/12 = 0.0075`

`text(# Repayments) = 4 xx 12 = 48`

`=>\ text(PVA Factor) = 40.18478\ \ text{(from Table)}`

`text(Monthly repayment)\ ($P)` `= (18\ 900)/(40.18478)`
  `= 470.32…= $470\ text{(nearest $)}`

   
`:.\ text(Camilla must repay $470 per month.)`

 

b.    `text(Total Repayments)= 48 xx 470= $22\ 560`
   

`:.\ text(Interest paid over loan)= 22\ 560-18\ 900= $3660`

Functions, EXT1 F2 2008 HSC 2c

The polynomial `p(x)` is given by  `p(x) = ax^3 + 16x^2 + cx-120`, where `a` and `c` are constants.

The three zeros of `p(x)` are `-2,  3` and `beta`.

Find the value of `beta`.   (3 marks) 

Show Answers Only

`-5`

Show Worked Solution

`p(x) = ax^3 + 16x^2 + cx-120`

`text(Roots:)\ \ -2, \ 3, \ beta`

`-2 + 3 + beta` `= -B/A`
`beta + 1` `= -16/a`
`beta` `= -16/a-1\ \ \ \ \ …\ (1)`

 

`-2 xx 3 xx beta` `= -D/A`
`-6 beta` `= 120/a`
`beta` `= -20/a\ \ \ \ \ …\ (2)`

 

MARKER’S COMMENT: Many students displayed significant inefficiencies in solving simultaneous equations.
`- 16/a-1` `= -20/beta`
`-16-a` `= -20`
`a` `= 4`

 
`text(Substitute)\ \ a = 4\ \ text(into)\ (1):`

`beta=-16/4-1= -5`

Mechanics, EXT2* M1 2008 HSC 2b

A particle moves on the `x`-axis with velocity `v`. The particle is initially at rest at  `x = 1`. Its acceleration is given by  `ddot x = x + 4`.

Using the fact that  `ddot x = d/dx (1/2 v^2)`, find the speed of the particle at  `x = 2`.   (3 marks)

Show Answers Only

`sqrt11`

Show Worked Solution
`ddot x` `= d/dx (1/2 v^2)=x+4`
`1/2 v^2` `= int ddot x\ dx`
  `= int x + 4\ dx`
  `= 1/2 x^2 + 4x + c`

 
`text(When)\ \ x = 1, v = 0:`

`0= 1/2 + 4 + c\ \ =>\ \ c=4 1/2`

`:.\ 1/2 v^2` `= 1/2 x^2 + 4x-4 1/2`
`v^2` `= x^2 + 8x-9`

 
`text(When)\ \ x = 2:`

`v^2` `= 2^2 + 8 * 2-9= 11`
`v` `= +- sqrt 11`

 
`:.\ text(When)\ x = 2,\ \ text(Speed) = sqrt 11`

Calculus, EXT1 C2 2008 HSC 2a

Use the substitution  `u = log_e x`  to evaluate  `int_e^(e^2) 1/(x (log_e x)^2)\ dx`.   (3 marks)

Show Answers Only

`1/2`

Show Worked Solution
`u` `= log_e x`
`(du)/(dx)` `= 1/x`
`:. du` `= 1/x\ dx`
`text(When)\ \ \ ` `x = e^2,\ \ ` `u = log_e e^2 = 2`
  `x = e,` `u = 1`

 
`:. int_e^(e^2) 1/(x (log_e x)^2)\ dx`

`= int_1^2 1/(u^2)\ du`

WARNING: Most errors were made in the last stage of substitution in this question. Be careful!

`= int_1^2 u^(-2)\ du`

`= [-1/u]_1^2`

`= [(-1/2) – (-1)]`

`= 1/2`

L&E, EXT1 2008 HSC 1f

Let  `f(x) = log_e [(x - 3)(5 - x)]`.

What is the domain of  `f(x)`?   (2 marks)

Show Answers Only

`text(Domain)\ {x:\ 3<x<5}`

Show Worked Solution

`f(x) = log_e [(x – 3)(5 – x)]`

`text(Find the domain)`

`(x – 3)(5 – x)>0`

 

L&E, EXT1 2008 HSC 1f Answer

`text(If)\ \ x = 4,`

`(4 – 3)(5 – 4) = 1 > 0`

`:.\ text(Domain)\ {x:\ 3<x<5}`

Trig Calculus, EXT1 2008 HSC 1e

Evaluate  `int_0^(pi/4) cos theta sin^2 theta\ d theta`.   (2 marks)

Show Answers Only

`sqrt2/12`

Show Worked Solution

`int_0^(pi/4) cos theta sin^2 theta\ d theta`

`= [1/3 sin^3 theta]_0^(pi/4)`

`= 1/3 (sin (pi/4))^3 – 1/3 (sin 0)^3`

`= 1/3 * (1/sqrt2)^3`

`= 1/3 * 1/(2 sqrt2)`

`= 1/(6 sqrt 2) xx sqrt2/sqrt2`

`= sqrt2/12`

Combinatorics, EXT1 A1 2008 HSC 1d

Find an expression for the coefficient of `x^8 y^4` in the expansion of `(2x + 3y)^12`.   (2 marks) 

Show Answers Only

`10\ 264\ 320`

Show Worked Solution

`text(Find co-efficient of)\ x^8 y^4 :`

MARKER’S COMMENT: More errors were made by students who used `T_(k+1)` as the general term rather than `T_k` (both are possible).

`T_k =\ text(General term of)\ (2x + 3y)^12 `

`T_k` `= ((12),(k)) (2x)^(12-k) * (3y)^k`
  `= ((12),(k)) * 2^(12-k) * 3^k * x^(12-k) * y^k`

 
`x^8 y^4\ text(occurs when)\ k = 4`

`T_4= ((12),(4)) * 2^(12-4) * 3^4 * x^8 y^4`

`:.\ text(Co-efficient of)\ x^8y^4= ((12),(4)) * 2^8 * 3^4= 10\ 264\ 320`

Measurement, 2UG 2008 HSC 28b

A tunnel is excavated with a cross-section as shown.
 

 
 

  1. Find an expression for the area of the cross-section using TWO applications of Simpson’s rule.  (2 marks)
  2. The area of the cross-section must be 600 m2. The tunnel is 80 m wide.   
  3. If the value of `a` increases by 2 metres, by how much will `b` change?   (2 marks)

 

Show Answers Only
  1. `(2h)/3 (4a + b)`
  2.  
  3. `b\ text(decreases by 8 m.)`
Show Worked Solution
(i)    `A` `~~ h/3 [y_0 + 4y_1 + y_2] + h/3 [y_0 + 4y_1 + y_2]`
    `~~ h/3 [0 + 4a + b] + h/3 [b + 4a + 0]`
    `~~ (2h)/3 (4a + b)`

 

(ii)    `text(Given)\ \ A = 600\ text(m²)`
  `text(If 80 m wide) \ => h = 20`

 

`A` `= (2h)/3 (4a + b)`
`600` `= ((2 xx 20))/3 (4a + b)`
`4a + b` `= (600 xx 3)/40`
`b` `= 45 – 4a`

 
`:.\ text(If)\ a\ text(increases by 2 m,)\ b\ text(will)`

`text(decrease by 8 m.)`

Statistics, STD2 S5 2008 HSC 28a

The following graph indicates  `z`-scores of ‘height-for-age’ for girls aged  5 – 19 years.
 

  1. What is the `z`-score for a six year old girl of height 120 cm?   (1 mark)

  2. Rachel is 10 ½  years of age. 

     

    (1)  If  2.5% of girls of the same age are taller than Rachel, how tall is she?   (1 mark)

     

    (2)  Rachel does not grow any taller. At age 15 ½, what percentage of girls of the same age will be taller than Rachel?   (2 marks)

  3. What is the average height of an 18 year old girl?  (1 mark)

For adults (18 years and older), the Body Mass Index is given by

`B = m/h^2`  where  `m = text(mass)`  in kilograms and  `h = text(height)`  in metres.

The medically accepted healthy range for  `B`  is  `21 <= B <= 25`.

  1. What is the minimum weight for an 18 year old girl of average height to be considered healthy?   (2 marks)

  2. The average height, `C`, in centimetres, of a girl between the ages of 6 years and 11 years can be represented by a line with equation
     
            `C = 6A + 79`   where `A` is the age in years. 
     
    (1)  For this line, the gradient is 6. What does this indicate about the heights of girls aged 6 to 11?   (1 mark)

     

    (2)  Give ONE reason why this equation is not suitable for predicting heights of girls older than 12.   (1 mark)

Show Answers Only
  1. `1`
  2. (1) `155\ text(cm)`

     

    (2) `text(84%)`

  3. `163\ text(cm)`
  4. `55.8\ text(kg)`
  5. (1) `text(It indicates that 6-11 year old girls)`

     

          `text(grow, on average, 6cm per year)`

     

    (2) `text(Girls eventually stop growing, and the)`

     

          `text(equation doesn’t factor this in.)`

Show Worked Solution
i.    `z text(-score) = 1`

 

ii. (1)   `text(If 2 ½ % are taller than Rachel)`
    `=> z text(-score of +2)`
    `:.\ text(She is 155 cm)`
     
   (2)   `text(At age)\ 15\ ½,\ 155\ text(cm has a)\ z text(-score of –1)`
    `text(68% between)\ z = 1\ text(and)\ –1`
    `=> text(34% between)\ z = 0\ text(and)\ –1`
    `text(50% have)\ z >= 0`
     
    `:.\ text(% Above)\ z text(-score of –1)`
    `= 50 + 34`
    `= 8text(4%)`

 
`:.\ text(84% of girls would be taller than Rachel at age)\ 15 ½.`

 

iii.   `text(Average height of 18 year old has)\ z text(-score = 0)`
  `:.\ text(Average height) = 163\ text(cm)`

 

iv.   `B = m/h^2`
  `h = 163\ text(cm) = 1.63\ text(m)`

 

`text(Given)\ \ 21 <= B <= 25,\ text(minimum healthy)`

`text(weight occurs when)\ B = 21`

`=> 21` `= m/1.63^2`
`m` `= 21 xx 1.63^2`
  `= 55.794…`
  `= 55.8\ text(kg)\ text{(1 d.p.)}`

 

v. (1)   `text(It indicates that 6-11 year old girls, on average, grow)`
    `text(6 cm per year.)`
  (2) `text(Girls eventually stop growing, and the equation doesn’t)`
    `text(factor this in.)`

Financial Maths, STD2 F4 2008 HSC 27c

A plasma TV depreciated in value by 15% per annum. Two years after it was purchased it had depreciated to a value of $2023, using the declining balance method.

What was the purchase price of the plasma TV?   (2 marks)

Show Answers Only

`$2800`

Show Worked Solution

`S = V_0 (1-r)^n`

`2023` `= V_0 (1-0.15)^2`
`2023` `= V_0 (0.85)^2`
`V_0` `= 2023/0.85^2`
  `= 2800`

 

`:.\ text(The purchase price) = $2800`

Measurement, 2UG 2008 HSC 27a

An aircraft travels at an average speed of  913 km/h. It departs from a town in Kenya  (0°, 38°E)  on Tuesday at 10 pm and flies east to a town in Borneo  (0°, 113°E).

  1. What is the distance, to the nearest kilometre, between the two towns? (Assume the radius of Earth is 6400 km.) (2 marks)
  2. How long will the flight take? (Answer to the nearest hour.)   (1 mark)
  3. What will be the local time in Borneo when the aircraft arrives? (Ignore time zones.)   (2 marks)

 

Show Answers Only
  1. `8378\ text(km)`
  2. `9\ text(hours)`
  3. `text(12 midday on Wednesday)`
Show Worked Solution
(i)   `text(Angular difference in longitude)`

`= 113 – 38`

`= 75^@`
 

`text(Arc length)` `= 75/360 xx 2 xx pi xx 6400`
  `= 8377.58…`
  `= 8378\ text(km)\ text{(nearest km)}`

 
`:.\ text(The distance between the two towns is 8378 km.)`

 

(ii)    `text(Flight time)` `= text(Distance)/text(Speed)`
    `= 8378/913`
    `= 9.176…`
    `= 9\ text(hours)\ text{(nearest hr)}`

 

(iii)   `text(Time Difference)` `= 75 xx 4`
    `= 300\ text(minutes)`
    `= 5\ text(hours)`

 
`text(Kenya is further East)`

`=>\ text(Kenya is +5 hours)`
 

`:.\ text(Arrival time in Kenya)`

`= text{10 pm (Tues) + 5 hrs + 9 hrs}\ text{(flight)}`

`= 12\ text(midday on Wednesday)`

Statistics, STD2 S1 2008 HSC 26d

The graph shows the predicted population age distribution in Australia in 2008.
 

 

  1. How many females are in the 0–4 age group?   (1 mark)

  2. What is the modal age group?   (1 mark)

  3. How many people are in the 15–19 age group?   (2 marks)

  4. Give ONE reason why there are more people in the 80+ age group than in the 75–79 age group.   (1 mark)

Show Answers Only

a.    `600\ 000`

b.    `35-39`

c.    `1\ 450\ 000`

d.    `text(The 80+ group includes all people over 80)`

`text(and is not restricted by a 5-year limit.)`

Show Worked Solution

a.    `text{# Females} (0-4)= 0.6 xx 1\ 000\ 000= 600\ 000`

b.    `text(Modal age group)\ =35-39`

c.    `text{# Males} (15-19)= 0.75 xx 1\ 000\ 000= 750\ 000`

`text{# Females} (15-19)= 0.7 xx 1\ 000\ 000= 700\ 000`

`:.\ text{Total People} (15-19)` `= 750\ 000 + 700\ 000`
  `= 1\ 450\ 000`

 

d.     `text(The 80+ group includes all people over 80)`
  `text(and is not restricted by a 5-year limit.)`

Probability, STD2 S2 2008 HSC 26b

The retirement ages of two million people are displayed in a table.
 

 

  1. What is the relative frequency of the 51–55 year retirement age?   (1 mark)

  2. Describe the distribution.   (1 mark)

Show Answers Only

a.    `7/400`

b.    `text(Distribution is negatively skewed because as age increases, so does the)`

`text(number of people in each age bracket.)`

Show Worked Solution

a.    `text(Relative frequency)\ (51-55)`

`= text{# People (51-55)}/text(Total People)`

`= (35\ 000)/(2\ 000\ 000)= 7/400`
 

b.    `text(Distribution is negatively skewed because as age increases, so does the)`

`text(number of people in each age bracket.)`

Probability, STD2 S2 2008 HSC 26a

Cecil invited 175 movie critics to preview his new movie. After seeing the movie, he conducted a survey. Cecil has almost completed the two-way table.
 

  1. Determine the value of  `A`.   (1 mark)

  2. A movie critic is selected at random.

     

    What is the probability that the critic was less than 40 years old and did not like the movie?   (2 marks)

  3. Cecil believes that his movie will be a box office success if 65% of the critics who were surveyed liked the movie.

     

    Will this movie be considered a box office success? Justify your answer.   (1 mark)

Show Answers Only

a.    `58`

b.    `6/25`

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

Show Worked Solution

a.    `text{Critics liked and}\ >= 40 = 102-65 = 37`

`A = 37+31=68`

 
b.    `text{Critics did not like and < 40} = 175-65-37-31= 42`

`P text{(not like and  < 40)}= 42/175= 6/25`
 

c.    `text(Critics liked) = 102`

`text(% Critics liked)= 102/175 xx 100= 58.28…%`

`:.\ text{Movie NOT a box office success (< 65% critics liked)}`

Probability, STD2 S2 2008 HSC 25b

In a drawer there are 30 ribbons. Twelve are blue and eighteen are red.

Two ribbons are selected at random.

  1. Copy and complete the probability tree diagram.   (1 mark)
     

     
  2. What is the probability of selecting a pair of ribbons which are the same colour?   (2 marks)

Show Answers Only

a.

b.    `73/145`

Show Worked Solution

a. 

b.    `Ptext{(same colour)}`

`=\ text{P(BB) + P(RR)}`

`= 12/30 xx 11/29 + 18/30 xx 17/29`

`= 132/870 + 306/870= 73/145` 

Probability, STD2 S2 2008 HSC 24b

Three-digit numbers are formed from five cards labelled  1,  2,  3,  4  and  5.

  1. How many different three-digit numbers can be formed?  (1 mark)

  2. If one of these numbers is selected at random, what is the probability that it is odd?   (1 mark)

  3. How many of these three-digit numbers are even?   (1 mark)

  4. What is the probability of randomly selecting a three-digit number less than 500 with its digits arranged in descending order?   (2 marks)

Show Answers Only
  1. `60`
  2. `3/5`
  3. `24`
  4. `1/15`
Show Worked Solution

i.  `text(# Different numbers)`

♦ Mean mark 45%.

`= 5 xx 4 xx 3`

`= 60`

 

ii.  `text(The last digit must be one of the)`

`text(5 numbers, of which 3 are odd)`

`:.\ text{P(odd)} = 3/5`

 

iii. `text{P(even)} = 1- text{P(odd)} = 2/5`

♦ Mean mark 48%.

`:.\ text(Number of even numbers)`

`= 2/5 xx 60`

`= 24`

 

iv.  `text(The numbers that satisfy the criteria:)`

♦♦♦ Mean mark 10%.

`432, 431, 421, 321`

`:.\ text{P(selection)} = 4/60 = 1/15`

Financial Maths, STD2 F1 2008 HSC 24a

Bob is employed as a salesman. He is offered two methods of calculating his income.

\begin{array} {|l|}
\hline
\rule{0pt}{2.5ex}\text{Method 1: Commission only of 13% on all sales}\rule[-1ex]{0pt}{0pt} \\
\hline
\rule{0pt}{2.5ex}\text{Method 2: \$350 per week plus a commission of 4.5% on all sales}\rule[-1ex]{0pt}{0pt} \\
\hline
\end{array}

Bob’s research determines that the average sales total per employee per month is $15 670. 

  1. Based on his research, how much could Bob expect to earn in a year if he were to choose Method 1?   (2 marks)

  2. If Bob were to choose a method of payment based on the average sales figures, state which method he should choose in order to earn the greater income. Justify your answer with appropriate calculations.   (3 marks)

Show Answers Only
  1. `$24\ 445.20`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

a.    `text(Method 1)`

`text(Yearly sales)= 12 xx 15\ 670= 188\ 040`

`:.\ text(Earnings)= text(13%) xx 188\ 040= $24\ 445.20`

b.    `text(Method 2)`

`text(In 1 Year, Weekly Wage)= 350 xx 52= 18\ 200`

`text(Commission)= text(4.5%) xx 188\ 040= 8461.80`

`text(Total earnings)= 18\ 200 + 8461.80= $26\ 661.80`

`:.\ text(Bob should choose Method 2.)`

Statistics, STD2 S1 2008 HSC 23e

In a survey, 450 people were asked about their favourite takeaway food. The results are displayed in the bar graph.

2008 23e
 
How many people chose pizza as their favourite takeaway food?   (2 marks)

Show Answers Only

`175`

Show Worked Solution

`text(Number of people who chose pizza)`

COMMENT: This question required measurement of the actual image on the exam. The same methodology works here.

`= text{Length of pizza section}/text{Total length of bar} xx 450`

`~~ 7/18 xx 450~~ 175`
 

`:.\ 175\ text(people chose pizza.)`

Measurement, STD2 M7 2008 HSC 23c

An alcoholic drink has 5.5% alcohol by volume. The label on a 375 mL bottle says it contains 1.6 standard drinks.

  1. How many millilitres of alcohol are in a 375 mL bottle?   (1 mark)

  2. It is recommended that a fully-licensed male driver should have a maximum of one standard drink every hour.

     

    Express this as a rate in millilitres per minute, correct to one decimal place.   (2 marks)

Show Answers Only

a.    `20.625\ text(mL)`

b.    `3.9\ text(mL/min)`

Show Worked Solution

a.    `text(Alcohol)= 5.5/100 xx 375= 20.625\ text(mL)`

b.    `text(S)text(ince 1.6 standard drinks = 375 mL)`

`rightarrow\ text(1 standard drink)= 375/1.6= 234.375\ text(mL)`

`:.\ text(Rate)= 234.375/60= 3.90625~~3.9\ text(mL/min)`
 

Measurement, STD2 M1 2008 HSC 23b

The capacity of a bottle is measured as 1.25 litres correct to the nearest 10 millilitres.

What is the percentage error for this measurement?   (1 mark)

Show Answers Only

`text(0.4%)`

Show Worked Solution

`text(A) text(bsolute error) = 5\ text(mL)`

`:.\ text(% error)` `= 5/1250 xx 100`
  `=\ text(0.4%)`

Measurement, STD2 M7 2008 HSC 20 MC

A point `P` lies between a tree, 2 metres high, and a tower, 8 metres high. `P` is 3 metres away from the base of the tree.

From `P`, the angles of elevation to the top of the tree and to the top of the tower are equal.
 

What is the distance, `x`, from `P` to the top of the tower?

  1. 9 m
  2. 9.61 m
  3. 12.04 m
  4. 14.42 m
Show Answers Only

`D`

Show Worked Solution

`text(Triangles are similar)\ \ text{(equiangular)}`

`text(In smaller triangle:)`

`h^2` `= 2^2 + 3^2`
  `= 13`
`h` `= sqrt 13`
   
`x/sqrt13` `= 8/2\ \ \ text{(sides of similar Δs in same ratio)}`
`x` `= (8 sqrt 13)/2`
  `= 14.422…`

 
`=>  D`

Probability, STD2 S2 2008 HSC 16 MC

A bag contains some marbles. The probability of selecting a blue marble at random from this bag is  `3/8`.

Which of the following could describe the marbles that are in the bag?

  1. `3`  blue,  `8`  red
  2. `6`  blue,  `11`  red
  3. `3`  blue,  `4`  red,  `4`  green
  4. `6`  blue,  `5`  red,  `5`  green 
Show Answers Only

`D`

Show Worked Solution

`P(B) = 3/8`

`text(In)\ A,\ \ ` `P(B) = 3/11`
`text(In)\ B,\ \ ` `P(B) = 6/17 `
`text(In)\ C,\ \ ` `P(B) = 3/11`
`text(In)\ D,\ \ ` `P(B) = 6/16 = 3/8`

`=>  D`

Financial Maths, STD2 F4 2008 HSC 15 MC

Ali is buying a speedboat at Betty’s Boats.
 

VCAA 2008 15 mc
 

What is the amount of interest Ali will have to pay if he chooses to buy the boat on terms?

  1. $3200 
  2. $5600
  3. $19 200
  4. $21 600
Show Answers Only

`B`

Show Worked Solution

`text(Deposit)= text(15%) xx 16\ 000= $2400`

`text(Payments)= 320 xx 5 xx 12= $19\ 200`

`text(Total paid)= 2400 + 19\ 200= $21\ 600`

`:.\ text(Interest)= 21\ 600-16\ 000=$5600`
  

`=>  B`

Statistics, STD2 S1 2008 HSC 13 MC

The height of each student in a class was measured and it was found that the mean height was 160 cm.

Two students were absent. When their heights were included in the data for the class, the mean height did not change.

Which of the following heights are possible for the two absent students?

  1. 155 cm and 162 cm
  2. 152 cm and 167 cm
  3. 149 cm and 171 cm
  4. 143 cm and 178 cm
Show Answers Only

`C`

Show Worked Solution

`text(S) text(ince the mean doesn’t change)`

`=>\ text(2 absent students must have a)`

 `text(mean height of 160 cm.)`

`text(Considering each option given,)`

`(149 + 171) -: 2 = 160`

`=>  C`

Statistics, STD2 S4 2008 HSC 12 MC

A scatterplot is shown.
 

Which of the following best describes the correlation between  \(R\)  and  \(T\)?

  1. Positive
  2. Negative 
  3. Positively skewed
  4. Negatively skewed
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Correlation is positive.}\)

\(\text{NB. The skew does not directly relate to correlation.}\)

\(\Rightarrow  A\)

Measurement, STD2 M1 2008 HSC 11 MC

The diagram shows the floor of a shower. The drain in the floor is a circle with a diameter of 10 cm.

What is the area of the shower floor, excluding the drain?
 

 
 

  1. 9686 cm²
  2. 9921 cm²
  3. 9969 cm²
  4. 10 000 cm²
Show Answers Only

`B`

Show Worked Solution
COMMENT: Students should see that answers are all in cm², and therefore use cm as the base unit for their calculations. 
`text(Area)` `=text(Square)-text(Circle)`
  `= (100 xx 100)-(pi xx 5^2)= 10\ 000-78.5398…`
  `= 9921.46…\ text(cm)^2`

 
`=>  B`

Statistics, STD2 S1 2008 HSC 10 MC

The marks for a Science test and a Mathematics test are presented in box-and-whisker plots.
 

 Which measure must be the same for both tests?

  1. Mean
  2. Range
  3. Median
  4. Interquartile range
Show Answers Only

`D`

Show Worked Solution

`text(IQR)=text(Upper Quartile)-text(Lower Quartile)`

`text{In both box plots, IQR = 3 intervals (against bottom scale)}`

`=>  D`

Algebra, STD2 A1 2008 HSC 9 MC

What is the value of  `sqrt((x+2y)/(8y))`   if  `x = 5.6`  and  `y = 3.1`, correct to 2 decimal places? 

  1. `0.69`
  2. `2.62`
  3. `2.83` 
  4. `4.77`  
Show Answers Only

`A`

Show Worked Solution
`sqrt((x+2y)/(8y))` `=sqrt((5.6+(2 xx 3.1))/((8 xx 3.1)))`
  `=sqrt(11.8/24.8)=0.6897…`

`=>  A`

Statistics, STD2 S1 2008 HSC 8 MC

What is the median of the following set of scores?
 

 
 

  1. 12
  2. 13
  3. 14
  4. 15
Show Answers Only

`C`

Show Worked Solution

`text(Median)=(n+1)/2=(33+1)/2=\ text (17th score)`

`:.\ text(Median is 14)`

`=>  C`

Measurement, STD2 M6 2008 HSC 5 MC

What is the size of the smallest angle in this triangle?
 

  1. `29^@` 
  2. `47^@`
  3. `58^@`
  4. `76^@`
Show Answers Only

`B`

Show Worked Solution

`text(Smallest angle is opposite smallest side.)`

`cos A` `= (b^2 + c^2-a^2)/(2bc)`
  `= (7^2 + 8^2-6^2)/(2 xx 7 xx 8)`
  `= 0.6875`
`A` `=cos ^(-1)(0.6875)`
`:.\ A` `= 46.567…^@`

 
`=>  B`

Statistics, STD2 S1 2008 HSC 3 MC

The stem-and-leaf plot represents the daily sales of soft drink from a vending machine.

If the range of sales is 43, what is the value of  2008 3 mc  ?

 
 

  1. 4
  2. 5
  3. 24
  4. 25
Show Answers Only

`A`

Show Worked Solution

`text(Range = High) – text(Low) = 43`

`:.\ 67 – text(Low)` `= 43`
`text(Low)` `= 24`

    
`:.\ N = 4`

`=>  A`

Probability, 2ADV S1 2008 HSC 9a

It is estimated that 85% of students in Australia own a mobile phone.

  1. Two students are selected at random. What is the probability that neither of them owns a mobile phone?  (2 marks)

  2. Based on a recent survey, 20% of the students who own a mobile phone have used their mobile phone during class time. A student is selected at random. What is the probability that the student owns a mobile phone and has used it during class time?   (1 mark)

Show Answers Only
  1. `9/400`
  2. `17/100`
Show Worked Solution

i.     `P(M) = 0.85`

COMMENT: `M^c` is syllabus notation for the complement of event `M`.

`P(M^c) = 1-0.85 = 0.15`

`:.\ P(M^c, M^c)` `= 15/100 * 15/100`
  `= 225/(10\ 000)`
  `= 9/400`

 

ii.  `text{P(owns mobile and used it)}`

`= P(M) xx  P\text{(used it)}`

`= 17/20 xx 20/100`

`= 17/100`

Plane Geometry, 2UA 2008 HSC 8b

In the diagram,  `ABCD`  is a parallelogram and  `ABEF`  and  `BCGH`  are both squares.

Copy or trace the diagram into your writing booklet.

  1. Prove that  `CD = BE`.  (1 mark)
  2. Prove that  `BD = EH`.   (3 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)  `text(Prove)\ CD = BE`

`CD` `= AB\ \ text{(opposite sides of parallelogram)}`
`AB` `= BE\ \ text{(sides of square)}`
`:.\ CD` `= BE\ \ \ text(… as required)`

 

(ii)

`text(Prove)\ BD = EH`

`CD` `= BE\ \ text{(from part (i))}`
`BC` `= BH\ \ text{(sides of square)}`

`/_BDC = /_ABD = alpha\ \ text{(} text(alternate,)\ AB\ text(||)\ CD text{)}`

`text(Let)\ /_DBC = beta`

`/_BCD` `= 180 – alpha – beta\ \ \ \ text{(angle sum of}\ Delta BDCtext{)}`
`/_EBH` `= 360 – 90 – 90 – alpha – beta\ \ \ `
   `text{(angles about a point)}`
  `= 180 – alpha – beta`
  `= /_BCD`

`=> Delta DCB ~= Delta EBH\ \ text{(SAS)}`

`:.\ BD = EH\ \ \ ` `text{(corresponding sides of)`
  `\ \ text{congruent triangles)}`

Calculus, 2ADV C3 2008 HSC 8a

Let  `f(x) = x^4 − 8x^2`.

  1. Find the coordinates of the points where the graph of  `y = f(x)`  crosses the axes.  (2 marks)

  2. Show that  `f(x)`  is an even function.   (1 mark)

  3. Find the coordinates of the stationary points of  `f(x)`  and determine their nature.   (4 marks)

  4. Sketch the graph of  `y = f(x)`.   (1 mark)

Show Answers Only

a.    `(-2 sqrt 2,0), (2 sqrt 2, 0)`

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

c.    `text(S.P.s at)\ (0,0), (2,-16), (-2,-16)`

d.    

  1. 2UA HSC 2008 8ai
Show Worked Solution
a.     `f(x) = x^4 – 8x^2`

`y text(-intercept when)\ x = 0`

`:.\ text(Cuts)\ y\ text(axis at)\ (0,0)`

`x\ text(-intercept when)\ f(x) = 0`

`x^4 – 8x^2` `= 0`
`x^2 (x^2 – 8)` `= 0`

`x^2 – 8 = 0\ \ \ \ \ text(or)\ \ \ \ \ x = 0`

`x^2` `= 8`
`x` `= +- sqrt 8 = +- 2 sqrt 2`

 

`:.\ text(Cuts)\ x text(-axis at)\ (-2 sqrt 2,0),\ \ text(and)\ \ (2 sqrt 2, 0)`

`text{(Note that it only touches}\ xtext{-axis at (0,0))}`

 

b.     `f(x)` `= x^4 – 8x^2`
  `f(–x)` `= (–x)^4 – 8(–x)^2`
    `= x^4 – 8x^2`
    `= f(x)`

 

`:.\ f(x)\ text(is an even function.)`

 

c.     `f(x)` `= x^4 – 8x^2`
  `f'(x)` `= 4x^3 – 16x`
  `f″(x)` `=12x^2-16`

 

`text(S.P.  when)\ \ f'(x) = 0`

`4x^3 – 16x` `= 0`
`4x (x^2 – 4)` `= 0`
`x^2 – 4` `=0\ \ \ \ \ x = 0`
`x^2` `= 4`
`x` `= +-2`
`text(At)\ x=2\ \ `  `f(x)`  `=(2)^4 – 8(2)^2 = -16`
  `f″(x)` `=12(2^2)-16>0`

`:.\ text{MIN at (2, –16)}`

`:.\ text{MIN at (–2, –16)},\ \ \ (f(x)\ text(is even))`

 

`text(At)\ x=0\ \ `  `f(x)`  `=(0)^4 – 8(0)^2 = 0`
  `f″(x)` `=12(0^2)-16<0`

`:.\ text{MAX at (0,0)}`

 

d.    

Probability, 2ADV S1 2008 HSC 7c

Xena and Gabrielle compete in a series of games. The series finishes when one player has won two games. In any game, the probability that Xena wins is `2/3` and the probability that Gabrielle wins is `1/3`.

Part of the tree diagram for this series of games is shown.
 

 
 

  1. Complete the tree diagram showing the possible outcomes.   (1 mark)
  2. What is the probability that Gabrielle wins the series?   (2 marks)

  3. What is the probability that three games are played in the series?   (2 marks)

Show Answers Only
  1.    
  2. `7/27`
  3. `4/9`
Show Worked Solution
i. 2UA HSC 2008 7ci
MARKER’S COMMENT: A tree diagram with 8 outcomes is incorrect (i.e. no third game is played if 1 player wins the first 2 games).

 
ii.  `P text{(} G\ text(wins) text{)}`

`= P(XGG) + P (GXG) + P (GG)`

`= 2/3 xx 1/3 xx 1/3 + 1/3 xx 2/3 xx 1/3 + 1/3 xx 1/3`

`= 2/27 + 2/27 + 1/9`

`= 7/27`
 

iii.  `text(Strategy 1:)`

`P text{(3 games played)}`

`= P (XG) + P(GX)`

`= 2/3 xx 1/3 + 1/3 xx 2/3`

`= 4/9`
 

`text(Strategy 2:)`

`P text{(3 games)}`

`= 1-[P(XX) + P(GG)]`

`= 1-[2/3 xx 2/3 + 1/3 xx 1/3]`

`= 1-5/9`

`= 4/9`

Trigonometry, 2ADV T1 2008 HSC 7b

2008 7b

The diagram shows a sector with radius `r` and angle `theta` where `0 < theta <= 2pi`.

The arc length is `(10pi)/3`. 

  1.  Show that  `r >= 5/3`.   (2 marks)

  2.  Calculate the area of the sector when  `r = 4`.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `(20pi)/3\ text(u²)`
Show Worked Solution
i.    `text(Show)\ r >= 5/3`
`text(Arc length)\ ` `= r theta\ \ text(where)\ \ 0 < theta <= 2pi`
`r theta` `= (10pi)/3`
`:.theta` `= (10pi)/(3r)`

 

`text(Using)\ 0 <= theta <= 2 pi:`

`0 <= (10pi)/(3r)` `<= 2pi`
`(10pi)/3` `<= 2 pi r`
`5/3` `<= r`

 

`:.\ r >= 5/3\ \ \ text(… as required.)`

 

ii.   `text(Area)` `= 1/2 r^2 theta`
    `= 1/2 xx 4^2 xx (10pi)/(3 xx 4)`
    `= (20pi)/3\ text(u²)`

Calculus, EXT1* C3 2008 HSC 6c

The graph of  `y = 5/(x - 2)`  is shown below.
 

2008 6c
 

The shaded region in the diagram is bounded by the curve  `y = 5/(x - 2)`, the  `x`-axis and the lines  `x = 3`  and  `x = 6`.

Find the volume of the solid of revolution formed when the shaded region is rotated about the  `x`-axis.  (3 marks)

Show Answers Only

`(75pi)/4\ text(u³)`

Show Worked Solution
`y` `= 5/(x – 2)`
`V` `= pi int_3^6 y^2\ dx`
  `= pi int_3^6 (5/(x – 2))^2\ dx`
  `= 25 pi int_3^6 1/((x – 2)^2)\ dx`
  `= 25 pi [(-1)/(x – 2)]_3^6`
  `= 25 pi [-1/4 – (-1)]`
  `=25 pi [3/4]`
  `= (75pi)/4\ text(u³)`

Calculus, 2ADV C4 2008 HSC 5a

The gradient of a curve is given by  `dy/dx = 1-6 sin 3x`. The curve passes through the point  `(0, 7)`.

What is the equation of the curve?   (3 marks)

Show Answers Only

`y = x + 2 cos 3x + 5`

Show Worked Solution
`dy/dx` `= 1-6 sin 3x`
`y` `= int 1-6 sin 3x\ dx`
  `= x + 2 cos 3x + c`

  
`text(Passes through)\ (0,7):`

`=> 0 + 2 cos 0 + c` `= 7`
`2 + c` `= 7`
`c` `= 5`

  
`:.\ text(Equation is)\ \ \ y = x + 2 cos 3x + 5`

Quadratic, 2UA 2008 HSC 4c

Consider the parabola  `x^2 = 8(y\ – 3)`.

  1. Write down the coordinates of the vertex.  (1 mark)
  2. Find the coordinates of the focus.   (1 mark)
  3. Sketch the parabola.   (1 mark)
  4. Calculate the area bounded by the parabola and the line  `y = 5`.   (3 marks)
Show Answers Only

(i)   `(0,3)`

(ii)   `(0,5)`

(iii)  

(iv)  `10 2/3\ text(u²)`

Show Worked Solution
(i)    `text(Vertex)\ = (0,3)`

 

(ii)   `text(Using)\ \ \ x^2` `= 4ay`
  `4a` `= 8`
  `a` `= 2`

`:.\ text(Focus) = (0,5)`

 

(iii) 2UA HSC 2008 4c 

 

(iv)   `text(Intersection when)\ y = 5`
`=> x^2` `= 8 (5-3)`
`x^2` `= 16`
`x` `= +- 4`

`text(Find shaded area)`

`x^2` `= 8 (y-3)`
`y – 3` `= x^2/8`
`y` `= x^2/8 +3`

 

`text(Area)` `= int_-4^4 5\ dx – int_-4^4 x^2/8 + 3\ dx`
  `= int_-4^4 5 – (x^2/8 + 3)\ dx`
  `= int_-4^4 2 – x^2/8\ dx`
  `= [2x – x^3/24]_-4^4`
  `= [(8 – 64/24) – (-8 + 64/24)]`
  `= 5 1/3 – (- 5 1/3)`
  `= 10 2/3\ text(u²)`

Plane Geometry, 2UA 2008 HSC 4a

In the diagram, `XR` bisects `/_PRQ` and `XY\ text(||)\ QR`.

Prove that `Delta XYR` is an isosceles triangle.   (2 marks)

Show Answers Only

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

Show Worked Solution

`text{Since}\ XR\ text{bisects}\ /_PRQ`

`/_XRQ` `= /_YRX = theta`
`/_RXY` `= theta\ \ \ text{(} text(alternate angles,)\ XY\ text(||)\ QR text{)}`

 
`:.\ Delta XYR\ \ text(is isosceles)`

Calculus, 2ADV C4 2008 HSC 3b

  1. Differentiate  `log_e (cos x)`  with respect to  `x`.   (2 marks)

  2. Hence, or otherwise, evaluate  `int_0^(pi/4) tan x\ dx`.   (2 marks)

Show Answers Only

a.    `-tan x`

b.    `-log_e (1/sqrt2)\ \ text(or)\ \ 0.35\ \ text{(2 d.p.)}`

Show Worked Solution
a.     `y` `= log_e (cos x)`
  `dy/dx` `= (-sin x)/(cos x)`
    `=-tan x`

 

b.     `int_0^(pi/4) tan x\ dx`
  `=-[log_e (cos x)]_0^(pi/4)`
  `=-[log_e(cos (pi/4))-log_e (cos 0)]`
  `=-[log_e (1/sqrt2)-log_e 1]`
  `=-[log_e (1/sqrt2)-0]`
  `=-log_e (1/sqrt2)`
  `= 0.346…\ = 0.35\ \ text{(2 d.p.)}`