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Measurement, STD2 M1 2015 HSC 28a

The diagram shows an annulus.
 

2015 28a

 
Calculate the area of the annulus.  (1 mark)

Show Answers Only

`≈ 50.26…\ text(cm²)`

Show Worked Solution

`text(Area of annulus)`

`= pi(R^2 − r^2)`

`= pi(5^2 − 3^2)`

`= pi(25 − 9)`

`= 16pi\ text(cm²)`

`≈ 50.26…\ text(cm²)`

Statistics, STD2 S1 2015 HSC 27d

In a small business, the seven employees earn the following wages per week:

\(\$300, \ \$490, \ \$520, \ \$590, \ \$660, \ \$680, \ \$970\)

  1.  Is the wage of $970 an outlier for this set of data? Justify your answer with calculations.  (3 marks)

  2.  Each employee receives a $20 pay increase.

     

     What effect will this have on the standard deviation?  (1 mark)

Show Answers Only

i.    \(\text{See Worked Solutions.} \)

ii.    \(\text{The standard deviation will remain the same.}\)

Show Worked Solution

i.    \(300, 490, 520, 590, 660, 680, 970\)

\(\text{Median}\) \(= 590\)
\(Q_1\) \(= 490\)
\(Q_3\) \(= 680\)
\(IQR\) \(= 680-490 = 190\)

 

\(\text{Outlier if \$970 is greater than:} \)

\(Q_3 + 1.5 x\times IQR = 680 + 1.5 \times 190 = \$965 \) 

\(\therefore\ \text{The wage \$970 per week is an outlier.}\)

♦ Mean mark (i) 39%.


ii.    \(\text{All values increase by \$20, but so too does the mean.} \)

\(\text{Therefore the spread about the new mean will not change} \)

\(\text{and therefore the standard deviation will remain the same.} \)

Measurement, STD2 M7 2015 HSC 27a

At a particular time during the day, a tower of height 19.2 metres casts a shadow. At the same time, a person who is 1.65 metres tall casts a shadow 5 metres long.

  

What is the length of the shadow cast by the tower at that time?  (2 marks)

Show Answers Only

`58\ text{m}`

Show Worked Solution

`text(Both triangles have right angles and a common)`

`text(angle to the ground.)`

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

 

`text(Let)\ x =\ text(length of tower shadow)`

`x/5` `= 19.2/1.65\ \ text{(corresponding sides of similar triangles)}`

 
`x` `= (5 xx 19.2)/1.65`  
  `= 58.1818…`  
  `= 58\ text{m  (nearest m)}`  

FS Comm, 2UG 2015 HSC 26g

Pat’s mobile phone plan is shown.

2015 26g

Last month Pat:

• made calls to the value of `$561`
• sent `152` SMS messages
• sent `37` MMS messages
• used `1.7` GB of data.

What was the total of Pat’s phone bill for last month?  (3 marks)

Show Answers Only

`$96.36`

Show Worked Solution

`text(C)text(ost of excess calls)`

`= 561 − 550`

`= $11`

`text(Excess data)`

`1.7\ text(GB)` `= 1.7 xx 2^(10)\ text(MB)`
`text(Excess)` `= 1.7 xx 2^(10) − 500`
  `= 1240.8\ text(MB)`

 

`:.\ text(C)text(ost of excess data)`

`= 1240.8 xx $0.0293`

`= $36.355…`

`= $36.36`

 

`text(SMS and MMS messages within plan)`

`:.\ text(Total phone bill)`

`= $49 + $11 + $36.36`

`= $96.36`

Measurement, STD2 M1 2015 HSC 26f

Approximately 71% of Earth’s surface is covered by water. Assume Earth is a sphere with a radius of 6400 km.

Calculate the number of square kilometres covered by water.  (2 marks)

Show Answers Only

`3.7 xx 10^8\ text(km²)\ \ text{(nearest km²)}`

Show Worked Solution

`text(Surface area of Earth)`

`= 4pir^2`

`= 4pi xx 6400^2`

 

`:.\ text(Surface covered by water)`

`= text(71%) xx 4pi xx 6400^2`

`= 365\ 450\ 163.7…`

`= 3.7 xx 10^8\ text(km²)\ \ text{(nearest km²)}`

Probability, STD2 S2 2015 HSC 26e

The table shows the relative frequency of selecting each of the different coloured jelly beans from packets containing green, yellow, black, red and white jelly beans.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Colour} \rule[-1ex]{0pt}{0pt} & \textit{Relative frequency} \\
\hline
\rule{0pt}{2.5ex} \text{Green} \rule[-1ex]{0pt}{0pt} & 0.32 \\
\hline
\rule{0pt}{2.5ex} \text{Yellow} \rule[-1ex]{0pt}{0pt} & 0.13 \\
\hline
\rule{0pt}{2.5ex} \text{Black} \rule[-1ex]{0pt}{0pt} & 0.14 \\
\hline
\rule{0pt}{2.5ex} \text{Red} \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} \text{White} \rule[-1ex]{0pt}{0pt} & 0.24 \\
\hline
\end{array}

  1. What is the relative frequency of selecting a red jelly bean?  (1 mark)

  2. Based on this table of relative frequencies, what is the probability of NOT selecting a black jelly bean?  (1 mark)

Show Answers Only
  1. \(0.17\)
  2. \(0.86\)
Show Worked Solution

i.  \(\text{Relative frequency of red}\)

\(= 1-(0.32 + 0.13 + 0.14 + 0.24)\)

\(= 1-0.83\)

\(= 0.17\)

 

ii.  \(P\text{(not selecting black)}\)

\(= 1-P\text{(selecting black)}\)

\(= 1-0.14\)

\(= 0.86\)

Financial Maths, STD2 F4 2015 HSC 26d

A family currently pays $320 for some groceries.

Assuming a constant annual inflation rate of 2.9%, calculate how much would be paid for the same groceries in 5 years’ time.  (2 marks)

Show Answers Only

`$369.17\ \ text{(nearest cent)}`

Show Worked Solution
`FV` `= PV(1 + r)^n`
  `= 320(1.029)^5`
  `= $369.1703…`
  `= $369.17\ \ text{(nearest cent)}`

Data, 2UG 2015 HSC 26a

A farmer used the ‘capture‑recapture’ technique to estimate the number of chickens he had on his farm. He captured, tagged and released 18 of the chickens. Later, he caught 26 chickens at random and found that 4 had been tagged.

What is the estimate for the total number of chickens on this farm?  (2 marks)

Show Answers Only

`117`

Show Worked Solution

`text(Let)\ C = text(total chickens on farm)`

`text(Capture)` `= 18/C`
`text(Recapture)` `= 4/26`
`⇒ 18/C` `= 4/26`
`4C` `= 18 xx 26`
`C` `= (18 xx 26)/4`
  `= 117`

 

`:.\ text(There are an estimated 117 chickens)`

`text(on the farm.)`

Financial Maths, STD2 F1 2015 HSC 25 MC

An insurance company offers customers the following discounts on the basic annual premium for car insurance.

2015 25 mc

If a customer is eligible for more than one discount, subsequent discounts are applied to the already discounted premium. The combined compulsory third party (CTP) and comprehensive insurance discount is always applied last.

Jamie has three insurance policies, including combined CTP and comprehensive insurance, with this company. He has used this company for 8 years and he has never made a claim.

The basic annual premium for his car insurance is $870.

How much will Jamie need to pay after the discounts are applied?

  1.    $482.44
  2.    $515.50
  3.    $541.60
  4.    $557.60
Show Answers Only

`C`

Show Worked Solution

`text(Multi-policy discount)`

`text(New premium)` `= 870 − (text(15%) xx 870)`
  `= $739.50`

 
`text(No claim bonus)`

`text(New premium)` `= 739.50 − (text(20%) xx 739.50)`
  `= $591.60`

 
`text(Combined CTP bonus)`

`text(New premium)` `= 591.60 − 50`
  `= $541.60`

`⇒ C`

Algebra, STD2 A1 2015 HSC 24 MC

Consider the equation  `(2x)/3-4 = (5x)/2 + 1`.

Which of the following would be a correct step in solving this equation?

  1. `(2x)/3-3 = (5x)/2`
  2. `(2x)/3 = (5x)/2 + 5`
  3. `2x-4 = (15x)/2 + 3`
  4. `(4x)/6-8 = 5x + 2`
Show Answers Only

`B`

Show Worked Solution
`(2x)/3-4` `= (5x)/2 + 1`
`(2x)/3` `= (5x)/2 + 5`

 
`=>B`

Algebra, STD2 A1 2015 HSC 23 MC

The number of ‘standard drinks’ in various glasses of wine is shown.
 

A woman weighing 62 kg drinks three small glasses of white wine and two large glasses of red wine between 8 pm and 1 am.

Using the formula for calculating blood alcohol below, what would be her blood alcohol content (BAC) estimate at 1 am, correct to three decimal places?
 

`BAC_text(Female) = (10N-7.5H)/(5.5M)`
 

where    `N` is the number of standard drinks consumed

`H` is the number of hours drinking

`M` is the person's mass in kilograms
 

  1. `0.030`
  2. `0.037`
  3. `0.046`
  4. `0.057`
Show Answers Only

`D`

Show Worked Solution
`N` `= 3 xx 0.9 + 2 xx 1.5`
  `= 5.7\ text(standard drinks)`
`H` `=\ text(5 hours)`
`M` `=\ text(62 kg)`
   
`:.BAC_f` `= (10 xx 5.7-7.5 xx 5)/ (5.5 xx 62)`
  `= 0.05718…`

`⇒ D`

Calculus, EXT1* C1 2015 HSC 14a

In a theme park ride, a chair is released from a height of  `110`  metres and falls vertically. Magnetic brakes are applied when the velocity of the chair reaches  `text(−37)`  metres per second.
 

2015 2ua 14a
 

The height of the chair at time `t` seconds is `x` metres. The acceleration of the chair is given by   `ddot x = −10`. At the release point,  `t = 0, x = 110 and dot x = 0`.

  1. Using calculus, show that  `x = -5t^2 + 110`.  (2 marks)

  2. How far has the chair fallen when the magnetic brakes are applied?  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `68.45\ text(m)`
Show Worked Solution

i.  `text(Show)\ \ x = -5t^2 + 110`

`ddot x` `= -10`
`dot x` `= int ddot x\ dt`
  `= int -10\ dt`
  `= -10t + c`

 
`text(When)\ \ t = 0,\ dot x = 0`

`:.\ 0` `= -10 (0) + c`
`c` `= 0`
`dot x` `= -10t`
`x` `= int dot x\ dt`
  `= int -10t\ dt`
  `= -5t^2 + c`

 
`text(When)\ \ t = 0,\ x = 110`

`:.\ 110` `= -5 (0^2) + c`
`c` `= 110`

 
`:.\ x = -5t^2 + 110\ \ text(…  as required.)`
 

ii.  `text(Find)\ \t\ \text(when)\ \ dot x = -37`

`-37` `= -10t`
`t` `= 3.7\ \ text(seconds)`

 
`text(When)\ \ t = 3.7`

`x` `= -5 (3.7^2) + 110`
  `= -68.45 + 110`
  `= 41.55`

 
`:.\ text(Distance the chair has fallen)`

`= 110 – 41.55`

`= 68.45\ text(m)`

Financial Maths, STD2 F4 2015 HSC 17 MC

What amount must be invested now at 4% per annum, compounded quarterly, so that in five years it will have grown to  $60 000?

  1. $8919
  2. $11 156
  3. $49 173
  4. $49 316
Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ FV = PV(1 + r)^n`

`r` `= text(4%)/4` `= text(1%) = 0.01\ text(per quarter)`
`n` `= 5 xx 4` `= 20\ text(quarters)`

 

`60\ 000` `= PV(1 + 0.01)^(20)`
`:.PV` `= (60\ 000)/1.01^(20)`
  `= $49\ 172.66…`

`⇒ C`

Probability, STD2 S2 2015 HSC 16 MC

The probability of winning a game is `7/10`.

Which expression represents the probability of winning two consecutive games?

  1. `7/10 xx 6/9`
  2. `7/10 xx 6/10`
  3. `7/10 xx 7/9`
  4. `7/10 xx 7/10`
Show Answers Only

`D`

Show Worked Solution

`text{Since the two events are independent:}`

`P text{(W)}` `= 7/10`
`P text{(WW)}` `= 7/10 xx 7/10`

 
`=>D`

Measurement, 2UG 2015 HSC 14 MC

Stockholm is located at `text(59°N 18°E)` and Darwin is located at `text(13°S 131°E)`. 

What is the time difference between Stockholm and Darwin? (Ignore time zones and daylight saving.)

(A)   `184` minutes

(B)   `288` minutes

(C)   `452` minutes

(D)   `596` minutes

Show Answers Only

`C`

Show Worked Solution

`text(Stockholm is 59°N 18°E,  Darwin is 13°S 131°E)`

`text{Angular difference (longitude)}`

`= 131^@− 18^@`

`= 113^@`
 

`:.\ text(Time difference)` `= 113 xx 4`
  `= 452\ text(minutes)`

`⇒ C`

Financial Maths, STD2 F4 2015 HSC 10 MC

A piece of machinery, initially worth $56 000, depreciates at 8% per annum.

Which graph best shows the salvage value of this piece of machinery over time?
 

2015 10 mc1

2015 10 mc2

Show Answers Only

`A`

Show Worked Solution

`text(By Elimination)`

`text(A depreciation of 8% per annum depreciates the largest)`

`text(amount in year 1 and then gradually depreciates less each)`

`text(subsequent year.)`

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

`text(Consider when)\ t` `= 5`
`text(Salvage Value)` `= V_0(1 − r)^n`
  `= 56\ 000(1 − 0.08)^5`
  `= 36\ 908.5…`

`text(Graph B depreciates too quickly)`

`:.text(Cannot be)\ B`

`⇒ A`

Measurement, STD2 M1 2015 HSC 8 MC

The Louvre Pyramid in Paris has a square base with side length 35 m and a perpendicular height of 22 m.
 

What is the volume of this pyramid, to the nearest m³?

  1. `257\ text(m)^3`
  2. `1027\ text(m)^3`
  3. `8983\ text(m)^3`
  4. `26\ 950\ text(m)^3`
Show Answers Only

`C`

Show Worked Solution
`V` `= 1/3Ah`
`A` `= 35 xx 35`
  `= 1225\ text(m)^2`

 

`:.V` `= 1/3 xx 1225 xx 22`
  `= 8983.33…\ text(m)^3`

 
`=>C`

Statistics, STD2 S1 2015 HSC 6 MC

The times, in minutes, that a large group of students spend on exercise per day are presented in the box‑and‑whisker plot.
 

What percentage of these students spend between 40 minutes and 60 minutes per day on exercise?

  1. 17%
  2. 20%
  3. 25%
  4. 50%
Show Answers Only

`C`

Show Worked Solution

`text{Q}_1 = 40, \ text(Median) = 60`

`:.\ text(% Students between 40 and 60)`

`= 50text{%}-25text{%}`

`=25 text{%}`
 

`=>C`

Statistics, STD2 S1 2015 HSC 4 MC

On a school report, a student’s record of completing homework is graded using the following codes.

C = consistently
U = usually
S = sometimes
R = rarely
N = never

What type of data is this?

  1. Categorical, ordinal
  2. Categorical, nominal
  3. Numerical, continuous
  4. Numerical, discrete
Show Answers Only

`A`

Show Worked Solution

`text(The data has been grouped into categories and)`

`text(because each category can be ranked, it is ordinal.)`

`⇒ A`

Calculus, 2ADV C4 2015 HSC 15c

Water is flowing in and out of a rock pool. The volume of water in the pool at time `t` hours is `V` litres. The rate of change of the volume is given by

`(dV)/(dt) = 80 sin(0.5t)`

At time  `t = 0`, the volume of water in the pool is 1200 litres and is increasing.

  1. After what time does the volume of water first start to decrease?  (2 marks)

  2. Find the volume of water in the pool when  `t = 3`.  (2 marks)

  3. What is the greatest volume of water in the pool?  (1 mark)

Show Answers Only
  1. `2 pi\ text(hours)`
  2. `1349\ text(litres)`
  3. `1520\ text(litres)`
Show Worked Solution

i.   `(dV)/(dt) = 80 sin (0.5t)`

♦ Mean mark (i) 37%.

`text(Volume decreases when)\ \ (dV)/(dt) < 0`

`(dV)/(dt) < 0\ \ text(when)\ \ t > 2 pi\ text(hours)`

`:.\ text(After 2)\pi\ \text(hours, volume starts to decrease.)`

 

ii.   `V` `= int (dV)/(dt)\ dt`
  `= int 80 sin (0.5t)\ dt`
  `= -160 cos (0.5t) + c`

 

`text(When)\ \ t = 0,\ V = 1200`

`1200` `= -160 cos 0 + c`
`c` `= 1360`
`:.\ V` `= -160 cos (0.5t) + 1360`

 

`text(When)\ \ t = 3`

`V` `= -160 cos (0.5 xx 3) + 1360`
  `= 1348.68…\ \ text(litres)`
  `= 1349\ text{litres  (nearest litre)}`

 

iii.  `V = -160 cos (0.5t) + 1360`

♦♦ Mean mark (iii) 27%.

`=>\ text(Greatest volume occurs when)`

`cos (0.5t) = -1`

`:.\ text(Maximum volume)`

`= -160 (-1) + 1360`

`= 1520\ text(litres)`

Plane Geometry, 2UA 2015 HSC 15b

The diagram shows `Delta ABC` which has a right angle at `C`. The point `D` is the midpoint of the side `AC`. The point `E` is chosen on `AB` such that `AE = ED`. The line segment `ED` is produced to meet the line `BC` at `F`.

Copy or trace the diagram into your writing booklet.

  1. Prove that `Delta ACB` is similar to `Delta DCF.`  (2 marks)
  2. Explain why `Delta EFB` is isosceles.  (1 mark)
  3. Show that `EB = 3AE.`  (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)  `text(Prove)\ \ Delta ACB\ \ text(|||)\ \ Delta DCF`

`/_ EAD = /_ ADE = theta\ \ text{(base angles of isosceles}\ \ Delta AED text{)}`

`/_ CDF = /_ ADE = theta\ \ text{(vertically opposite angles)}`

`/_ DCF = /_ ACB = 90°\ \ text{(}/_ FCB\ text{is a straight angle)}`

`:.\ Delta ACB\ \ text(|||)\ \ Delta DCF\ \ text{(equiangular)}`

 

♦ Mean mark 45%.

(ii)  `/_ DFC = 90 – theta\ \ text{(angle sum of}\ \ Delta DCF text{)}`

`/_ ABC = 90 – theta\ \ text{(angle sum of}\ \ Delta ACB text{)}`

`:.\ Delta EFB\ \ text{is isosceles  (base angles are equal)}`

 

(iii)  `text(Show)\ \ EB = 3AE`

♦♦ Mean mark 23%.
`(DC)/(AC)` `= (DF)/(AB)`

`\ \ \ \ \ text{(corresponding sides of}`

`\ \ \ \ \ text{similar triangles)}`
`1/2` `= (DF)/(AB)`  
`2DF` `= AB`  
`2(EF – ED)` `= AE + EB`  
`2(EB – AE)` `= AE + EB` `\ \ \ \ \ text{(given}\ EF = EB, ED = AE text{)}`
`2EB – 2AE` `= AE + EB`  

 

`:. EB = 3AE\ \ text(…  as required)`

Calculus, EXT1* C1 2015 HSC 15a

The amount of caffeine, `C`, in the human body decreases according to the equation

`(dC)/(dt) = -0.14C,` 

where `C` is measured in mg and `t` is the time in hours.

  1. Show that  `C = Ae^(-0.14t)`  is a solution to  `(dC)/(dt) = -0.14C,` where ` A` is a constant.

     

    When `t = 0`, there are 130 mg of caffeine in Lee’s body.  (1 mark)

  2. Find the value of `A.`  (1 mark)

  3. What is the amount of caffeine in Lee’s body after 7 hours?   (1 mark)

  4. What is the time taken for the amount of caffeine in Lee’s body to halve?  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `130`
  3. `48.8\ text(mg)`
  4. `4.95\ text(hours)`
Show Worked Solution
i. `C` `= Ae^(-0.14t)`
  `(dC)/(dt)` `= d/(dt) (Ae^(-0.14t))`
    `= -0.14 xx Ae^(-0.14t)`
    `= -0.14\ C`

 
`:.\ C = Ae^(-0.14t)\ \ text(is a solution)`

 

ii.  `text(When)\ \ t = 0,\ C = 130`

`130` `= Ae^(-0.14 xx 0)`
`:.\ A` `= 130`

 

iii.  `text(Find)\ \ C\ \ text(when)\ \ t = 7`

`C` `= 130\ e^(-0.14 xx 7)`
  `= 130\ e^(-0.98)`
  `= 48.79…`
  `= 48.8\ text{mg  (to 1 d.p.)}`

 
`:.\ text(After 7 hours, Lee will have 48.8 mg)`

`text(of caffeine left in her body.)`

 

iv.  `text(Find)\ \ t\ \ text(when caffeine has halved.)`

`text(When)\ \ t = 0,\ \ C = 130`

`:.\ text(Find)\ \ t\ \ text(when)\ \ C = 65`

`65` `= 130 e^(-0.14 xx t)`
`e^(-0.14t)` `= 65/130`
`ln e^(-0.14t)` `= ln\ 65/130`
`-0.14t xx ln e` `= ln\ 65/130`
`t` `= (ln\ 65/130)/-0.14`
  `= 4.951…`
  `= 4.95\ text{hours  (to 2 d.p.)}`

 

`:.\ text(It will take 4.95 hours for Lee’s)`

`text(caffeine to halve.)`

Financial Maths, 2ADV M1 2015 HSC 14c

Sam borrows $100 000 to be repaid at a reducible interest rate of 0.6% per month. Let  `$A_n`  be the amount owing at the end of  `n`  months and  `$M`  be the monthly repayment.

  1. Show that  `A_2 = 100\ 000 (1.006)^2 - M (1 + 1.006).`  (1 mark)

  2. Show that  `A_n = 100\ 000 (1.006)^n - M (((1.006)^n - 1)/0.006).`  (2 marks)

  3. Sam makes monthly repayments of $780. Show that after making 120 monthly repayments the amount owing is $68 500 to the nearest $100.  (1 mark)

Immediately after making the 120th repayment, Sam makes a one-off payment, reducing the amount owing to $48 500. The interest rate and monthly repayment remain unchanged.

  1. After how many more months will the amount owing be completely repaid?  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(Show Worked Solutions)}`
  2. `text(Proof)\ \ text{(Show Worked Solutions)}`
  3. `text(Proof)\ \ text{(Show Worked Solutions)}`
  4. `79\ text(months)`
Show Worked Solution
i.  `A_1` `= 100\ 000 (1.006) – M`
`A_2` `= A_1 (1.006) – M`
  `= [100\ 000 (1.006) – M] (1.006) – M`
  `= 100\ 000 (1.006)^2 – M (1.006) – M`
  `= 100\ 000 (1.006)^2 – M (1 + 1.006)\ \ text(…  as required)`

 

ii.  `A_3 = 100\ 000 (1.006)^3 – M (1 + 1.006 + 1.006^2)`

`vdots`

`A_n = 100\ 000 (1.006)^n – M (1 + 1.006 + … + 1.006^(n-1))`

`=> text(S)text(ince)\ \ (1 + 1.006 + … + 1.006^(n-1))\ text(is a)`

`text(GP with)\ \ a = 1,\ r = 1.006`

`:.\ A_n` `= 100\ 000 (1.006)^n – M ((a (r^n – 1))/(r – 1))`
  `= 100\ 000 (1.006)^n – M ((1 (1.006^n – 1))/(1.006 – 1))`
  `= 100\ 000 (1.006)^n – M (((1.006)^n – 1)/0.006)`

`text(…  as required.)`

 

iii.  `text(If)\ \ M = 780 and n = 120`

`A_120` `= 100\ 000 (1.006)^120 – 780 ((1.006^120 – 1)/0.006)`
  `= 205\ 001.80… – 780 (175.0030…)`
  `= 205\ 001.80… – 136\ 502.34…`
  `= 68\ 499.45…`
  `= $68\ 500\ \ text{(to nearest $100)  …  as required}`

 

iv.  `text(After the one-off payment, amount owing)=$48\ 500`

`:. A_n = 48\ 500 (1.006)^n – 780 ((1.006^n – 1)/0.006)`

`text(where)\ \ n\ \ text(is the number of months after)`

 `text(the one-off payment.)`
 

`text(Find)\ \ n\ \ text(when)\ \ A_n = 0`

`48\ 500 (1.006)^n – 780 ((1.006^n – 1)/0.006) = 0`

`48\ 500 (1.006)^n` `= 780 ((1.006^n – 1)/0.006)`
`48\ 500 (1.006)^n` `= 130\ 000 (1.006^n – 1)`
  `= 130\ 000 (1.006)^n – 130\ 000`
`81\ 500 (1.006)^n` `= 130\ 000`
`1.006^n` `= (130\ 000)/(81\ 500)`
`n xx ln 1.006` `= ln\ 1300/815`
`n` `= (ln\ 1300/815)/(ln\ 1.006)`
  `= 78.055…`

 
`:.\ text(The amount owing will be completely repaid after)`

`text(another 79 months.)`

Probability, 2ADV S1 2015 HSC 14b

Weather records for a town suggest that:

  • if a particular day is wet `(W)`, the probability of the next day being dry is  `5/6`
  • if a particular day is dry `(D)`, the probability of the next day being dry is  `1/2`.

In a specific week Thursday is dry. The tree diagram shows the possible outcomes for the next three days: Friday, Saturday and Sunday.
 

2015 2ua 14b
 

  1. Show that the probability of Saturday being dry is `2/3`.  (1 mark)

  2. What is the probability of both Saturday and Sunday being wet?  (2 marks)

  3. What is the probability of at least one of Saturday and Sunday being dry?  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1/(18)`
  3. `(17)/(18)`
Show Worked Solution

i.  `text{Show}\ \ P text{(Sat dry)} = 2/3`

`P text{(Sat dry)}`

`= P (W,D) + P (D, D)`

`=(1/2 xx 5/6) + (1/2 xx 1/2)`

`= 5/(12) + 1/4`

`= 2/3\ \ text(…  as required)`
 

ii.  `Ptext{(Sat and Sun wet)}`

`= P (WWW) + P (DWW)`

`= (1/2 xx 1/6 xx 1/6) + (1/2 xx 1/2 xx 1/6)`

`= 1/(72) + 1/(24)`

`= 1/(18)`
 

iii.  `Ptext{(At least Sat or Sun dry)}`

`= 1 – Ptext{(Sat and Sun both wet)}`

`= 1 – 1/(18)`

`= (17)/(18)`

Calculus, 2ADV C3 2015 HSC 13c

Consider the curve  `y = x^3 − x^2 − x + 3`.

  1. Find the stationary points and determine their nature.   (4 marks)

  2. Given that the point  `P (1/3, 70/27)`  lies on the curve, prove that there is a point of inflection at  `P`.  (2 marks)

  3. Sketch the curve, labelling the stationary points, point of inflection and `y`-intercept.  (2 marks)

Show Answers Only
  1. `text(MAX at)\ (-1/3, 86/27); \ text(MIN at)\ (1, 2)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3.  
Show Worked Solution
i. `y` `= x^3 – x^2 – x + 3`
  `(dy)/(dx)` `= 3x^2 – 2x – 1`
  `(d^2y)/(dx^2)` `= 6x – 2`

`text(S.P.’s when)\ (dy)/(dx) = 0`

`3x^2 – 2x – 1` `= 0`
`(3x + 1) (x – 1)` `= 0`

`x = -1/3 or 1`

 

`text(When)\ \ x = -1/3`

`f(-1/3)` `= (-1/3)^3 – (-1/3)^2 – (-1/3) + 3`
  `= -1/27 – 1/9 + 1/3 + 3`
  `= 86/27`
`f″(-1/3)` `= (6 xx -1/3) – 2 = -4 < 0`

`:.\ text(MAX at)\ \ (-1/3, 86/27)`

 

`text(When)\ \ x = 1`

`f(1)` `= 1^3 – 1^2 – 1 + 3 =2`
`f″(1)` `= (6 xx 1) – 2 = 4 > 0`

`:.\ text(MIN at)\ \ (1, 2)`

 

ii.  `(d^2y)/(dx^2) = 0\ \ text(when)`

`6x-2` `=0`
`x` `=1/3`

 

`text(Checking change of concavity)`

`text(Concavity changes either side of)\ x = 1/3`

`:.\ (1/3, 70/27)\ \ text(is a P.I.)`

 

iii.  `text(When)\ \ x` `= 0`
`y` `= 3`

Functions, EXT1* F1 2015 HSC 13b

  1. Find the domain and range for the function  `f(x) = sqrt (9 - x^2)`.  (2 marks)

  2. On a number plane, shade the region where the points `(x, y)` satisfy both of the inequalities
     
      `qquad y <= sqrt (9 - x^2)`  and  `y >= x` .  (2 marks)

Show Answers Only
  1. `text(Domain)\ -3 <= x <= 3\ \ \ \ \ \ \ \ text(Range)\ 0<= y <= 3`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.  `f(x) = sqrt(9 – x^2)`

`text(Domain)`

`9 – x^2` `>= 0`
`x^2` `<= 9`

`-3 <= x <= 3`

`text(Range)`

`0 <= y <= 3`

 

♦ Mean mark 34%.
ii.   

Quadratic, 2UA 2015 HSC 12e

The diagram shows the parabola `y = x^2/2` with focus `S (0, 1/2).` A tangent to the parabola is drawn at `P (1, 1/2).`

  1. Find the equation of the tangent at the point `P`.   (2 marks)
  2. What is the equation of the directrix of the parabola?   (1 mark)
  3. The tangent and directrix intersect at `Q`.
    Show that `Q` lies on the `y`-axis.   (1 mark)

  4. Show that `Delta PQS` is isosceles.   (1 mark)
Show Answers Only
  1. `y = x – 1/2`
  2. `y = -1/2`
  3.  
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   

`y = 1/2 x^2`

`(dy)/(dx) = x`
 

`text(When)\ \ x = 1,\ \ (dy)/(dx) = 1`

`text(Equation of tangent,)\ m = 1,\ text(through)\ (1, 1/2):`

`y – y_1` `= m (x – x_1)`
`y – 1/2` `= 1 (x – 1)`
`y – 1/2` `= x – 1`
`y` `= x – 1/2`

 

(ii)  `text(Directrix is)\ \ y = -1/2`
 

(iii)  `Q\ text(is at the intersection of)`

`y = x – 1/2\ \ …\ \ text{(1)}`

`y = -1/2\ \ \ \ …\ \ text{(2)}`

`text{(1) = (2)}`

`x – 1/2` `= -1/2`
`x` `= 0`

 
`:.\ Q\ text(lies on the)\ y text(-axis)\ \ …\ \ text(as required)`

 

(iv)  `text(Show)\ Delta PQS\ text(is isosceles.)`

`text(Distance)\ PS = 1 – 0 = 1`

`Q\ text(has coordinates)\ (0, -1/2)`

`text(Distance)\ SQ = 1/2 + 1/2 = 1`

`:. PS = SQ = 1`

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

Quadratic, 2UA 2015 HSC 12d

For what values of `k` does the quadratic equation `x^2 – 8x + k = 0` have real roots?  (2 marks)

Show Answers Only

`k <= 16`

Show Worked Solution

`x^2 – 8x + k = 0`

`text(Real roots when)\ Delta >= 0`

`b^2 – 4ac` `>= 0`
`(-8)^2 – 4 xx 1 xx k` `>= 0`
`64 – 4k` `>= 0`
`4k` `<= 64`
`k` `<= 16`

`:.\ text(Real roots exists when)\ k <= 16`

Calculus, 2ADV C4 2015 HSC 10 MC

The diagram shows the area under the curve  `y = 2/x`  from  `x = 1`  to  `x = d`.

2012 2ua 10 mc

What value of `d` makes the shaded area equal to `2`?

  1. `e`
  2. `e + 1`
  3. `2e`
  4. `e^2`
Show Answers Only

`A`

Show Worked Solution

`int_1^d 2/x\ dx = 2`

`[2 log_e x]_1^d = 2`

`2 log_e d – 2 log_e 1` `= 2`
`2 log_e d` `= 2`
`log_e d` `= 1`
`:. d` `= e`

`=> A`

L&E, 2ADV E1 2015 HSC 8 MC

The diagram shows the graph of  `y = e^x (1 + x).`
 

How many solutions are there to the equation  `e^x (1 + x) = 1 - x^2`?

  1. `0`
  2. `1`
  3. `2`
  4. `3`
Show Answers Only

`C`

Show Worked Solution

`text(The graphs intersect at 2 points)`

`:. e^x(1 + x) = 1 – x^2\ text(has 2 solutions)`

`=> C`

Calculus, 2ADV C4 2015 HSC 7 MC

The diagram shows the parabola  `y = 4x - x^2`  meeting the line  `y = 2x`  at  `(0, 0)`  and  `(2, 4)`.
 

2015 2ua 7 mc
 

Which expression gives the area of the shaded region bounded by the parabola and the line?

  1. `int_0^2 x^2 - 2x\ dx`
  2. `int_0^2 2x - x^2\ dx`
  3. `int_0^4 x^2 - 2x\ dx`
  4. `int_0^4 2x - x^2\ dx`
Show Answers Only

`B`

Show Worked Solution
`text(Shaded Area)` `= int_0^2 4x – x^2\ dx – int_0^2 2x\ dx`
  `= int_0^2 4x – x^2 – 2x\ dx`
  `= int_0^2 2x – x^2\ dx`

 

`=> B`

Trig Calculus, 2UA 2015 HSC 6 MC

What is the value of the derivative of  `y = 2 sin 3x - 3 tan x`  at  `x = 0`?

(A)  `-1`

(B)     `0`

(C)     `3`

(D)  `-9`

Show Answers Only

`C`

Show Worked Solution
`y` `= 2 sin 3x – 3 tan x`
`(dy)/(dx)` `= 6 cos 3x – 3 sec^2 x`

 

`text(At)\ x = 0`

`(dy)/(dx)` `= 6 cos 0 – 3 sec^2 0`
  `= 6 (1) – 3/(cos^2 0)`
  `= 6 – 3`
  `= 3`

 `=> C`

Calculus, 2ADV C4 2015 HSC 5 MC

Using the trapezoidal rule with 4 subintervals, which expression gives the approximate area under the curve  `y = xe^x`  between  `x = 1`  and  `x = 3`?

  1. `1/4(e^1 + 6e^1.5 + 4e^2 + 10e^2.5 + 3e^3)`
  2. `1/4(e^1 + 3e^1.5 + 4e^2 + 5e^2.5 + 3e^3)`
  3. `1/2(e^1 + 6e^1.5 + 4e^2 + 10e^2.5 + 3e^3)`
  4. `1/2(e^1 + 3e^1.5 + 4e^2 + 5e^2.5 + 3e^3)`
Show Answers Only

`B`

Show Worked Solution

`y = xe^x`

2UA HSC 2015 5mc

`A` `~~ h/2[y_0 + 2y_1 + 2y_2 + 2y_3 + y_4]`
  `~~ 1/4[e^1 + 3e^1.5 + 4e^2 + 5e^2.5 + 3e^3]`

 `=> B`

Plane Geometry, 2UA 2006 HSC 10b

A rectangular piece of paper `PQRS` has sides `PQ = 12` cm and `PS = 13` cm. The point `O` is the midpoint of `PQ`. The points `K` and `M` are to be chosen on `OQ` and `PS` respectively, so that when the paper is folded along `KM`, the corner that was at `P` lands on the edge `QR` at `L`. Let `OK = x` cm and `LM = y` cm.

Copy or trace the diagram into your writing booklet.

  1. Show that `QL^2 = 24x`.  (1 mark)
  2. Let `N` be the point on `QR` for which `MN` is perpendicular to `QR`.
  3. By showing that `Delta QKL\ text(|||)\ Delta NLM`, deduce that `y = {sqrt 6 (6 + x)}/sqrt x`.  (3 marks)
  4. Show that the area, `A`, of `Delta KLM` is given by
    1. `A = {sqrt 6 (6 + x)^2}/(2 sqrt x)`  (1 mark)
    2.  
  5. Use the fact that `12 <= y <= 13` to find the possible values of `x`.  (2 marks)
  6. Find the minimum possible area of `Delta KLM`.  (3 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)}`
  4. `2 2/3 <= x <= 6`
  5.  
  6. `53.33…\ text(cm²)`
Show Worked Solution
(i)     

 `text(Show)\ QL^2 = 24x`

`KP = KL = 6 + x`

`KQ = 6 – x`

`text(Using Pythagoras)`

`QL^2` `= KL^2 – KQ^2`
  `= (6 + x)^2 – (6 – x)^2`
  `= 36 + 12x + x^2 – 36 + 12x – x^2`
  `= 24x\ …\ text(as required)`

 

(ii)  `text(Show)\ y = {sqrt 6 (6 + x)}/sqrt x`

`text(In)\ Delta QKL`

`/_LQK=90°\ \ text{(given)}`

`text(Let)\ /_QKL = theta`

`:. /_QLK = 90 – theta\ \ text{(angle sum of}\ Delta QKL text{)}`

`text(In)\ Delta NLM`

`/_LNM=90°\ \ text{(given)}`

`/_NLM` `= 180 – (90 + 90 – theta)\ \ (/_QLN\ text(is a straight angle))`
  `= theta`

`:. Delta QKL\ text(|||)\ Delta NLM\ \ \ text{(equiangular)}`

`:. y/(MN)` `= (KL)/(QL)\ \ text{(corresponding sides of}`
  `text{similar triangles)}` 
`y/12` `= (6 + x)/sqrt(24x)`
`y` `= (12(6 + x))/(2 sqrt (6x))`
  `= (6 (6 + x))/(sqrt x xx sqrt 6) xx sqrt 6/sqrt 6`
  `= (sqrt 6 (6 + x))/sqrt x\ …\ text(as required)`

 

(iii)  `text(Area)\ DeltaKLM` `= 1/2 xx y xx KL`
  `= 1/2 xx (sqrt 6 (6 + x))/sqrt x xx (6 + x)`
  `= (sqrt 6 (6 + x)^2)/(2 sqrt x)\ …\ text(as required.)`

 

(iv)  `text(Given that)\ \ \ \ \ \ \ \ 12 <= y <= 13`

`12 <= (sqrt 6 (6 + x))/sqrt x <= 13`

`text(Consider)\ (sqrt 6 ( 6 + x))/sqrt x` `>= 12`
`(6 (6 + x)^2)/x` `>= 144`
`(6 + x)^2` `>= 24x`
`36 + 12x + x^2` `>= 24x`
`x^2 – 12x + 36` `>= 0`
`(x – 6)^2` `>= 0`
`:. x` `>= 6`

`text(However, we know)\ OP=6,\ text(and)\ x <= 6`.

`:. x = 6\ text(satisfies both conditions)`

`text(Consider)\ (sqrt 6 (6 + x))/sqrt x` `<= 13`
`(6 (6 + x)^2)/x` `<= 169`
`6 (6 + x)^2` `<= 169x`
`6 (36 + 12x + x^2)` `<= 169x`
`216 + 72x + 6x^2` `<= 169x`
`6x^2 – 97x + 216` `<= 0`

 

`text(Using the quadratic formula)`

`x` `= (-b +- sqrt (b^2 -4ac))/(2a)`
  `= (97 +- sqrt ((-97)^2 -4 xx 6 xx 216))/(2 xx 6)`
  `= (97 +- sqrt 4225)/12`
  `= (97 +- 65)/12`
  `= 13 1/2 or 2 2/3`

`:. 2 2/3 <= x <= 13 1/2`

`text(However, we know)\ x <= 6,\ text(so)`

`2 2/3 <= x <= 6\ text(satisfies both conditions)`

`:.\ text(All possible values of)\ x\ text(are)`

`2 2/3 <= x <= 6`.

 

(v)  `A = (sqrt 6 (6 + x)^2)/(2 sqrt x)`

`text(Using the quotient rule)`

`(dA)/(dx)` `= (u prime v – uv prime)/v^2`
  `= {2 sqrt 6 (6 + x) xx 2 sqrt x – sqrt 6 (6 + x)^2 xx 1/2 xx 2 xx x^(-1/2)}/(2 sqrt x)^2`
  `= {4 sqrt x sqrt 6 (6 + x) – sqrt 6 (6 + x)^2 * 1/sqrt x}/(4x)`
  `= {4 sqrt 6 x (6 + x) – sqrt 6 (6 + x)^2}/(4x sqrt x)`
  `= (sqrt 6 (6 + x) (4x – 6 – x))/(4x sqrt x)`
  `= (sqrt 6 (6 + x) (3x – 6))/(4x sqrt x)`

`text(Max or min when)\ (dA)/(dx) = 0`

`sqrt 6 (6 + x) (3x – 6) = 0`

`:. 3x = 6\ \ ,\ \ x ≠ -6`

`x = 2`

`text(However,)\ x = 2\ text(lies outside the range)`

`text(of possible values)\ \ 2 2/3 <= x <= 6`

`:.\ text(Check limits)`

`text(At)\ \ x = 2 2/3`

`A` `= (sqrt 6 (6 + 2 2/3)^2)/(2 sqrt (2 2/3))`
  `= 56.33…\ text(cm²)`

 

`text(At)\ \ x = 6`

`A` `= (sqrt 6 (6 + 6)^2)/(2 sqrt 6)`
  `= 72.0\ text(cm²)`

 

`:.\ text(Minimum area of)\ Delta KLM\ text(is 56.33… cm²)`

Integration, 2UA 2006 HSC 10a

Use Simpson’s rule with three function values to find an approximation to the value of
 
          `int_0.5^1.5 (log_e x )^3\ dx`.
 

Give your answer correct to three decimal places.  (2 marks)

Show Answers Only

`text(–0.044)\ \ \ text{(to 3 d.p.)}`

Show Worked Solution

`int_0.5^1.5 (log_e x )^3 dx`

`A` `~~ h/3 [y_0 + y_2 + 4(y_1)]`
  `~~0.5/3 [-0.3330… + 0.0666… + 0]`
  `~~ -0.04439…`
  `~~-0.044\ \ \ text{(to 3 d.p.)}`

Calculus, 2ADV C4 2006 HSC 9b

During a storm, water flows into a 7000-litre tank at a rate of `(dV)/(dt)` litres per minute, where `(dV)/(dt) = 120 + 26t-t^2` and `t` is the time in minutes since the storm began.

  1. At what times is the tank filling at twice the initial rate?  (2 marks)

  2. Find the volume of water that has flowed into the tank since the start of the storm as a function of `t`.  (1 mark)

  3. Initially, the tank contains 1500 litres of water. When the storm finishes, 30 minutes after it began, the tank is overflowing.

     

    How many litres of water have been lost?  (2 marks)

Show Answers Only
  1. `t = 6 or 20\ text(minutes)`
  2. `V = 120t + 13t^2 – 1/3t^3`
  3. `800\ text(litres)`
Show Worked Solution

i.  `(dV)/(dt) = 120 + 26t-t^2`

`text(When)\ t = 0, (dV)/(dt) = 120`

`text(Find)\ t\ text(when)\ (dV)/(dt) = 240`

`240 = 120 + 26t-t^2`

`t^2-26t + 120 = 0`

`(t-6)(t-20) = 0`

`t = 6 or 20`

`:.\ text(The tank is filling at twice the initial rate)`

`text(when)\ t = 6 and t = 20\ text(minutes)`

 

ii.  `V` `= int (dV)/(dt)\ dt`
  `= int 120 + 26t-t^2\ dt`
  `= 120t + 13t^2-1/3t^3 + c`

 
`text(When)\ t = 0, V = 0`

`=>  c = 0`

`:. V= 120t + 13t^2-1/3t^3`

 

iii.  `text(Storm water volume into the tank when)\ t = 30`

`= 120(30) + 13(30^2)-1/3 xx 30^3`

`= 3600 + 11\ 700-9000`

`= 6300\ text(litres)`
 

`text(Total volume)` `= 6300 + 1500`
  `= 7800\ text(litres)`

 

`:.\ text(Overflow)` `= 7800-7000`
  `= 800\ text(litres)`

Quadratic, 2UA 2006 HSC 9a

Find the coordinates of the focus of the parabola  `12y = x^2 - 6x - 3`.  (2 marks)

Show Answers Only

`(3, 2)`

Show Worked Solution
`12y` `= x^2 – 6x – 3`
  `= x^2 – 6x + 9 – 12`
`12y` `= (x – 3)^2 – 12`
`(x – 3)^2` `= 12y + 12`
  `= 12(y + 1)`

 

`:.\ text{Vertex is (3, –1)}`

`4a` `= 12`
`a` `= 3`

 

`:.\ text(Coordinates of focus are)\ (3, 2).`

Quadratic, 2UA 2006 HSC 7c

  1. Write down the discriminant of  `2x^2 + (k - 2)x + 8`  where  `k`  is a constant.  (1 mark)
  2. Hence, or otherwise, find the values of  `k`  for which the parabola  `y = 2x^2 + kx + 9` does not intersect the line  `y = 2x + 1`.  (2 marks)

 

Show Answers Only
  1. `k^2 – 4k – 60`
  2. `-6 < k < 10`
Show Worked Solution

(i)  `2x^2 + (k – 2)x + 8`

`Delta` `= b^2 – 4ac`
  `= (k – 2)^2 – 4 xx 2 xx 8`
  `= k^2 – 4k + 4 – 64`
  `= k^2 – 4k – 60`

 

(ii)  `y` `= 2x^2 + kx + 9` `\ \ text{…  (1)}`
`y` `= 2x + 1` `\ \ text{…  (2)}`

`text(Substitute)\ y = 2x + 1\ text{into (1)}`

`2x + 1 = 2x^2 + kx + 9`

`2x^2 + kx – 2x + 8 = 0`

`2x^2 + (k – 2)x + 8 = 0\ …\ text{(∗)}`

 

`text{The graphs will not intercept if (∗) has}`

`text(no roots, i.e.)\ \ Delta <0`

`k^2 – 4k – 60` `< 0`
`(k – 10) (k + 6)` `< 0`

HSC quadratic

`text(From the graph, no intersection when)`

`-6 < k < 10`

Quadratic, 2UA 2006 HSC 7a

Let  `alpha`  and  `beta`  be the solutions of  `x^2 - 3x + 1 = 0`.

  1. Find  `alpha beta`.  (1 mark)
  2. Hence find  `alpha + 1/alpha`.  (1 mark)  

 

Show Answers Only
  1. `1`
  2. `3`
Show Worked Solution

(i)  `x^2 – 3x + 1 = 0`

`alpha beta` `= c/a`
  `= 1`

 

(ii)  `alpha + 1/alpha` `= alpha + beta\ \ \ text{(using}\ beta = 1/alpha\ text{from part (i))}`
  `= -b/a`
  `= (-(-3))/1`
  `= 3`

Calculus, EXT1* C1 2006 HSC 6b

A rare species of bird lives only on a remote island. A mathematical model predicts that the bird population, `P`, is given by

`P = 150 + 300 e^(-0.05t)`

where `t` is the number of years after observations began.

  1. According to the model, how many birds were there when observations began?  (1 mark)

  2. According to the model, what will be the rate of change in the bird population ten years after observations began?  (2 marks)

  3. What does the model predict will be the limiting value of the bird population?  (1 mark)

  4. The species will become eligible for inclusion in the endangered species list when the population falls below `200`. When does the model predict that this will occur?  (2 marks)

Show Answers Only
  1. `450\ text(birds)`
  2. `-9.1\ text{(to 1 d.p.)}`
  3. `150\ text(birds)`
  4. `text(After 35.83… years)`
Show Worked Solution

i.  `P = 150 + 300 e^(-0.05t)`

`text(When)\ \ t = 0`

`P` `= 150 + 300 e^0`
  `= 450`

 
`:.\ text(There were 450 birds when observations began.)`
 

ii.  `(dP)/dt` `= 300 xx (-0.05) xx e^(-0.05t)`
  `= -15 e^(-0.05t)`

 
`text(When)\ \ t = 10`

`(dP)/dt` `= -15 e^(-0.05 xx 10)`
  `= -15 e^(-0.5)`
  `= -9.097…`
  `= -9.1\ text{(to 1 d.p.)}`

 

`:.\ text(After 10 years, the bird population will be)`

`text(decreasing at a rate of 9.1 birds per year.)`

 

iii.  `text(As)\ t rarr oo`

`300 e^(-0.05t) rarr 0`

`:. P = 150 + 300 e^(-0.05t) rarr 150`

`:.\ text(The model predicts a limiting population)`

`text(of 150 birds.)`

 

iv.  `text(Find)\ t\ text(when)\ P < 200`

`150 + 300 e^(-0.05t)` `< 200`
`300 e^(-0.05t)` `< 50`
`e^(-0.05t)` `< 50/300`
`ln e^(-0.05t)` `< ln­ 1/6`
`-0.05t` `< ln­ 1/6`
`t` `> (ln­ 1/6)/(-0.05t)`
`t` `> 35.83…`

 

`:.\ text(The model predicts the population)`

`text(will fall below 200 after 35.83… years.)`

Plane Geometry, 2UA 2006 HSC 6a

In the diagram, `AD` is parallel to `BC`, `AC` bisects `/_BAD` and `BD` bisects `/_ABC`. The lines `AC` and `BD` intersect at `P`.

Copy or trace the diagram into your writing booklet.

  1. Prove that `/_BAC = /_BCA`.  (1 mark)
  2. Prove that `Delta ABP ≡ Delta CBP`.  (2 marks)
  3. Prove that `ABCD` is a rhombus.  (3 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)  

`text(Prove)\ /_BAC = /_BCA`

`/_BCA` `= /_CAD\ \ \ text{(alternate angles,}\ BC\ text(||)\ AD text{)}`
`/_CAD` `= /_BAC\ \ \ text{(}AC\ text(bisects)\ /_BAD text{)}`
`:. /_BAC` `= /_BCA\ …\ text(as required)`

 

(ii)  `text(Prove)\ Delta ABP ≡ Delta CBP`

`/_BAC` `= /_BCA\ \ \ text{(from part (i))}`
`/_ABP` `= /_CBP\ text{(}BD\ text(bisects)\ /_ABC text{)}`
`BP\ text(is common)`

 

`:. Delta ABP ≡ Delta CBP\ \ text{(AAS)}`

 

(iii)  `text(Using)\ Delta ABP ≡ Delta CBP`

`AP = PC\ \ text{(corresponding sides of congruent triangles)}`

`/_BPA = /_BPC\ \ text{(corresponding angles of congruent triangles)}`

`text(Also,)\ /_BPA = /_BPC = 90^@`

`text{(}/_APC\ text{is a straight angle)}`

 

`text(Considering)\ Delta APD and Delta APB`

`/_DAP` `= /_BAP\ text{(}AC\ text(bisects)\ /_BAD text{)}`
`/_DPA` `= 90^@\ text{(vertically opposite angles)}`
`:. /_DPA` `= /_BPA = 90^@`

`PA\ text(is common)`

`:. Delta APD ≡ Delta APB\ \ text{(AAS)}`

`BP = PD\ \ text{(corresponding sides of congruent triangles)}`

 

 `:. ABCD\ text(is a rhombus as its diagonals are)`

`text(perpendicular bisectors.)`

Calculus, 2ADV C4 2006 HSC 5b

  1. Show that `d/dx log_e (cos x) = -tan x.`  (1 mark)


  2.   
     
    2006 5b
     
    The shaded region in the diagram is bounded by the curve  `y =tan x`  and the lines  `y =x`  and  `x = pi/4.`

     

    Using the result of part (i), or otherwise, find the area of the shaded region.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `(1/2 log_e 2 – pi^2/32)\ text(u²)`
Show Worked Solution
i.   `d/dx log_e (cos x)` `= (-sin x)/cos x`
  `= – tan x\ …\ text(as required)`

 

ii.  `text(Shaded Area)`

`= int_0^(pi/4) tan x\ dx – int_0^(pi/4) x\ dx`

`= int_0^(pi/4) tan x – x\ dx`

`= [-log_e (cos x) – 1/2 x^2]_0^(pi/4)`

`=[(-log_e (cos­ pi/4) – 1/2 xx (pi^2)/16) – (-log_e(cos 0) – 0)]`

`= -log_e­ 1/sqrt 2 – pi^2/32 + log_e1`

`= -log_e 2^(-1/2) – pi^2/32`

`= (1/2 log_e 2 – pi^2/32)\ text(u²)`

Calculus, 2ADV C3 2004 HSC 4b

Consider the function  `f(x) = x^3 − 3x^2`.

  1. Find the coordinates of the stationary points of the curve  `y = f(x)`  and determine their nature.   (3 marks)

  2. Sketch the curve showing where it meets the axes.   (2 marks)

  3. Find the values of  `x`  for which the curve  `y = f(x)`  is concave up.   (2 marks)

Show Answers Only
  1. `text(MAX at)\ (0,0),\ \ text(MIN at)\ (2,-4)`
  2.  
    1. Geometry and Calculus, 2UA 2004 HSC 4b Answer
  3. `f(x)\ text(is concave up when)\ x>1`
Show Worked Solution
(i)    `f(x)` `= x^3 – 3x^2`
  `f'(x)` `= 3x^2 – 6x`
  `f″(x)` `= 6x – 6`

 

`text(S.P.’s  when)\ \ f'(x) = 0`

`3x^2 – 6x` `= 0`
`3x (x – 2)` `= 0`
`x` `= 0\ \ text(or)\ \ 2`

 

`text(When)\ x = 0`

`f(0)` `= 0`
`f″(0)` `= 0 – 6 = -6 < 0`
`:.\ text(MAX at)\ (0,0)`

 

`text(When)\ x = 2`

`f(2)` `= 2^3 – (3 xx 4) = -4`
`f″(2)` `= (6 xx 2) – 6 = 6 > 0`
`:.\ text(MIN at)\ (2, -4)`

 

(ii)   `f(x) = x^3 – 3x^2\ text(meets the)\ x text(-axis when)\ f(x) = 0`
`x^3 – 3x^2` `= 0`
`x^2 (x-3)` `= 0`
`x` `= 0\ \ text(or)\ \ 3`

 Geometry and Calculus, 2UA 2004 HSC 4b Answer

(iii)   `f(x)\ text(is concave up when)`
`f″(x)` `>0`
`6x – 6` `>0`
`6x` `>6`
`x` `>1`

 

`:. f(x)\ text(is concave up when)\ \ x>1`

Trigonometry, 2ADV T1 2004 HSC 4a

2004 4a
 

`AOB`  is a sector of a circle, centre  `O`  and radius 6 cm.

The length of the arc  `AB`  is  `5pi` cm.

Calculate the exact area of the sector  `AOB`.   (2 marks)

Show Answers Only

`15 pi\ text(cm²)`

Show Worked Solution
`text(Arc length)\ ` `= theta/(2 pi) xx 2 pi r = r theta`
`5 pi` `= 6 theta`
`:.\ theta` `= (5pi)/6\ text(radians)`

 

`text(Area of sector)\ AOB`

`= theta/(2pi) xx pi r^2`

`= 1/2 r^2 theta`

`= 1/2 xx 6^2 xx (5pi)/6`

`= 15pi\ text(cm²)`

Trigonometry, 2ADV T1 2004 HSC 3c

Trig Ratios, 2UA 2004 HSC 3c
 

The diagram shows a point  `P`  which is  30 km due west of the point  `Q`.

The point  `R`  is 12 km from  `P`  and has a bearing from  `P`  of  070°. 

  1. Find the distance of  `R`  from  `Q`.   (2 marks)

  2. Find the bearing of  `R`  from  `Q`.   (2 marks)

Show Answers Only
  1. `19.2\ text(km)\ \ \ text{(1 d.p.)}` 
  2. `282^@`
Show Worked Solution

i.   `text(Join)\ \ RQ\ \ text(to form)\ \ Delta RPQ`

Trig Ratios, 2UA 2004 HSC 3c Answer

`/_RPQ = 90 – 70 = 20^@`

`text(Using the cosine rule:)`

`RQ^2` `= PR^2 + PQ^2 – 2 xx PR xx PQ xx cos /_RPQ`
  `= 12^2 + 30^2 – 2 xx 12 xx 30 xx cos 20^@`
  `= 367.421…`
`:.\ RQ` `= 19.168…`
  `= 19.2\ text(km)\ \ text{(1 d.p.)}`

 

ii.   `text(Using sine rule:)`

`(sin /_RQP)/12` `= (sin 20^@)/(19.168…)`
`sin/_RQP` `= (12 xx sin 20^@)/(19.168…)`
  `= 0.214…`
`/_RQP` `= 12.36…^@`
  `= 12^@\ \ \ text{(nearest degree)}`

 

`:.\ text(Bearing of)\ R\ text(from)\ Q`

`=270+12`

`=282^@`

Probability, 2ADV S1 2006 HSC 4c

A chessboard has 32 black squares and 32 white squares. Tanya chooses three different squares at random.

  1. What is the probability that Tanya chooses three white squares?  (2 marks)

  2. What is the probability that the three squares Tanya chooses are the same colour?.  (1 mark)

  3. What is the probability that the three squares Tanya chooses are not the same colour?  (1 mark)

Show Answers Only
  1. `5/42`
  2. `5/21`
  3. `16/21`
Show Worked Solution
i.  `text(P)(WWW)` `= 32/64 xx 31/63 xx 30/62`
  `= 5/42`

 

ii.  `text{P(same colour)}`

`= P(WWW) + P(BBB)`

`= 5/42 + 32/64 xx 31/63 xx 30/62`

`= 5/42 + 5/42`

`= 5/21`

 

iii.  `text{P(not all the same colour)}`

`= 1 – text{P(same colour)}`

`= 1 – 5/21`

`= 16/21`

Calculus, EXT1* C3 2006 HSC 4b

2006 4b

In the diagram, the shaded region is bounded by the parabola  `y = x^2 + 1`, the `y`-axis and the line  `y = 5`.

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

Show Answers Only

`8 pi\ \ text(u³)`

Show Worked Solution

`y = x^2 + 1`

`x^2 = y – 1`

`V` `= pi int_1^5 x^2 \ dy`
  `= pi int_1^5 y-1 \ dy`
  `= pi [y^2/2 – y]_1^5`
  `= pi[(25/2 – 5) – (1/2 – 1)]`
  `= pi[15/2 – (-1/2)]`
  `= 8 pi\ \ text(u³)`

Trigonometry, 2ADV T1 2006 HSC 4a

In the diagram, `ABCD` represents a garden. The sector  `BCD`  has centre `B` and  `/_DBC = (5 pi)/6`

The points `A, B` and `C` lie on a straight line and  `AB = AD = 3` metres.

Copy or trace the diagram into your writing booklet.

  1. Show that  `/_DAB = (2 pi)/3.`  (1 mark)

  2. Find the length of  `BD`.  (2 marks)

  3. Find the area of the garden  `ABCD`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `3 sqrt 3\ \ text(m)`
  3. `(9 sqrt 3 + 45 pi) / 4\ \ text(m²)`
Show Worked Solution
i.   

`text(Show)\ /_DAB = (2 pi)/3`

`/_DBA` `= pi – (5 pi)/6\ \ \ text{(π radians in straight angle}\ ABC text{)}`
  `= pi/6\ text(radians)`

 
`:. /_BDA = pi/6\ text(radians)\ \ \ text{(base angles of isosceles}\ Delta ADB text{)}`

`:. /_DAB` `= pi – (pi/6 + pi/6)\ \ \ text{(angle sum of}\ Delta ADB text{)}`
  `= (2 pi)/3\  text(radians … as required)`

 

ii.  `text(Using the cosine rule:)`

`BD^2` `= AD^2 + AB^2 – 2 xx AD xx AB xx cos {:(2 pi)/3`
  `= 9 + 9 – (2 xx 3 xx 3 xx -0.5)`
  `= 27`
`:. BD` `= sqrt 27`
  `= 3 sqrt 3\ \ text(m)`

 

iii.  `text(Area of)\ Delta ADB` `= 1/2 ab sin C`
  `= 1/2 xx 3 xx 3 xx sin{:(2 pi)/3`
  `= 9/2 xx sqrt3/2`
  `= (9 sqrt 3)/4\ \ text(m²)`

 
`text(Area of sector)\ BCD`

`= {(5 pi)/6}/(2 pi) xx pi r^2`

`= (5 pi)/12 xx (3 sqrt 3)^2`

`= (45 pi)/4\ \ text(m²)`

 

`:.\ text(Area of garden)\ ABCD`

`= (9 sqrt 3)/4 + (45 pi)/4`

`= (9 sqrt 3 + 45 pi)/4\ \ text(m²)`

Linear Functions, 2UA 2006 HSC 3a

In the diagram, `A, B and C` are the points  `(1, 4), (5, –4) and (–3, –1)`  respectively. The line  `AB`  meets the y-axis at `D`.

  1. Show that the equation of the line  `AB`  is  `2x + y - 6 = 0`.  (2 marks)
  2. Find the coordinates of the point `D`.  (1 mark)
  3. Find the perpendicular distance of the point `C` from the line  `AB`.  (1 mark)
  4. Hence, or otherwise, find the area of the triangle  `ADC`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `D(0, 6)`
  3. `13/sqrt 5\ text(units)`
  4.  
  5. `text(6.5 u²)`
Show Worked Solution

(i)  `text(Show)\ \ AB\ \ text(is)\ \ 2x + y – 6 = 0`

`A (1, 4)\ \ \ B (5, text(–4))`

`m_(AB)` `= (y_2 – y_1) / (x_2 – x_1)`
  `= (-4 – 4) / (5 – 1)`
  `= (-8)/4`
  `= – 2`

 
`:.\ text(Equation of)\ AB, m = -2,\ text(through)\ \ (1,4)`

`y-y_1` `=m(x-x_1)`
`y – 4` `= -2 (x – 1)`
`y – 4` `= -2x + 2`
`2x + y – 6` `= 0\ …\ text(as required)`

 

(ii)  `AB\ text(intersects y-axis at)\ D`

`0 + y – 6` `= 0`
`y` `= 6`

`:. D\ text(has coordinates)\ (0, 6)`

 

(iii)  `C\ text{(–3, –1)}`

`AB\ text(is)\ 2x + y – 6 = 0`

`_|_ text(dist)` `= |\ (ax_1 + by_1 + c)/sqrt (a^2 + b^2)\ |`
  `= |\ (2(−3) + 1(-1) – 6)/sqrt(2^2 + 1^2)\ |`
  `= |\ (-13)/sqrt 5\ |`
  `= 13/sqrt 5\ text(units)`

 

(iv)

 
`text(dist)\ AD` `= sqrt((x_2 – x_1)^2 + (y_2 – y_1)^2)`
  `= sqrt((0 – 1)^2 + (6 – 4)^2`
  `= sqrt (1 + 4)`
  `= sqrt 5`
`text(Area of)\ Delta ADC` `= 1/2 xx b xx h`
  `= 1/2 xx sqrt 5 xx 13/sqrt 5`
  `= 6.5\ text(u²)`

Trigonometry, 2ADV T1 2005 HSC 9b

Trig Ratios, 2UA 2005 HSC 9b
 

The triangle  `ABC`  has a right angle at `B, \ ∠BAC = theta` and  `AB = 6`. The line `BD` is drawn perpendicular to `AC`. The line `DE` is then drawn perpendicular to `BC`. This process continues indefinitely as shown in the diagram.

  1. Find the length of the interval `BD`, and hence show that the length of the interval  `EF`  is  `6 sin^3\ theta`.  (2 marks)

  2. Show that the limiting sum 
     
    `qquad BD + EF + GH + ···`
     
    is given by  `6 sec\ theta tan\ theta`.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution
i.   Trig Ratios, 2UA 2005 HSC 9b Answer

`text(Show)\ EF = 6\ sin^3\ theta`

`text(In)\ ΔADB`

`sin\ theta` `= (DB)/6`
`DB` `= 6\ sin\ theta`
`∠ABD` `= 90 − theta\ \ \ text{(angle sum of}\ ΔADB)`
`:.∠DBE` `= theta\ \ \ (∠ABE\ text{is a right angle)}`

 

`text(In)\ ΔBDE:`

`sin\ theta` `= (DE)/(DB)`
  `= (DE)/(6\ sin\ theta)`
`DE` `= 6\ sin^2\ theta`
`∠BDE` `= 90 − theta\ \ \ text{(angle sum of}\ ΔDBE)`
`∠EDF` `= theta\ \ \ (∠FDB\ text{is a right angle)}`

 

`text(In)\ ΔDEF:`

`sin\ theta` `= (EF)/(DE)`
  `= (EF)/(6\ sin^2\ theta)`
`:.EF` `= 6\ sin^3\ theta\ \ …text(as required)`

 

ii. `text(Show)\ \ BD + EF + GH\ …`

`text(has limiting sum)\ =6 sec theta tan theta`

`underbrace{6\ sin\ theta + 6\ sin^3\ theta +\ …}_{text(GP where)\ \ a = 6 sin theta, \ \ r = sin^2 theta}`
 

`text(S)text(ince)\ \ 0 < theta < 90^@`

`−1` `< sin\ theta` `< 1`
`0` `< sin^2\ theta` `< 1`

 
`:. |\ r\ | < 1`
 

`:.S_∞` `= a/(1 − r)`
  `= (6\ sin\ theta)/(1 − sin^2\ theta)`
  `= (6\ sin\ theta)/(cos^2\ theta)`
  `= 6 xx 1/(cos\ theta) xx (sin\ theta)/(cos\ theta)`
  `= 6 sec\ theta\ tan\ theta\ \ …text(as required.)`

Calculus, EXT1* C1 2005 HSC 9a

A particle is initially at rest at the origin. Its acceleration as a function of time, `t`, is given by

`ddot x = 4sin2t`

  1. Show that the velocity of the particle is given by  `dot x = 2 − 2\ cos\ 2t`.  (2 marks)

  2. Sketch the graph of the velocity for  `0 ≤ t ≤ 2π`  AND determine the time at which the particle first comes to rest after  `t = 0`.  (3 marks)

  3. Find the distance travelled by the particle between  `t = 0`  and the time at which the particle first comes to rest after  `t = 0`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `2pi\ \ text(units)`
Show Worked Solution

i.   `text(Show)\ \ dot x = 2 − 2\ cos\ 2t`

`ddot x` `= 4\ sin\ 2t`
`dot x` `= int ddot x\ dt`
  `= int4\ sin\ 2t\ dt`
  `= −2\ cos\ 2t + c`

 
`text(When)\ t = 0, \ x = 0`

`0 = −2\ cos\ 0 + c`

`c = 2`

`:.dot x = 2 − 2\ cos\ 2t\ \ …text(as required)`

 

ii.  `text(Considering the range)`

`−1` `≤ \ \ \ cos\ 2t` `≤ 1`
`−2` `≤ \ \ \ 2\ cos\ 2t`  `≤ 2` 
`0` `≤ 2 − 2\ cos\ 2t` `≤ 4` 

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

Calculus in the Physical World, 2UA 2005 HSC 9a Answer

`text(After)\ t = 0,\ text(the particle next comes)`

`text(to rest at)\ t = pi.`

 

iii.  `text(Distance travelled)`

`= int_0^pi dot x\ dt`

`= int_0^pi 2 − 2\ cos\ 2t\ dt`

`= [2t − sin\ 2t]_0^pi`

`= [(2pi − sin\ 2pi) − (0 − sin\ 0)]`

`= 2pi\ \ text(units)`

Financial Maths, 2ADV M1 2005 HSC 8c

Weelabarrabak Shire Council borrowed $3 000 000 at the beginning of 2005. The annual interest rate is 12%. Each year, interest is calculated on the balance at the beginning of the year and added to the balance owing. The debt is to be repaid by equal annual repayments of $480 000, with the first repayment being made at the end of 2005.

Let  `A_n`  be the balance owing after the `n`-th repayment.

  1. Show that  `A_2 = (3 × 10^6)(1.12)^2 - (4.8 × 10^5)(1 + 1.12)`.  (1 mark)

  2. Show that  `A_n = 10^6[4 − (1.12)^n]`.  (2 marks)

  3. In which year will Weelabarrabak Shire Council make the final repayment?  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `2017`
Show Worked Solution
i.    `A_1` `= (3 xx 10^6)(1.12) − (4.8 xx 10^5)`
  `A_2` `= A_1(1.12) − (4.8 xx 10^5)`
    `= [(3 xx 10^6)(1.12) − (4.8 xx 10^5)](1.12) − (4.8 xx 10^5)`
    `= (3 xx 10^6)(1.12)^2 − (4.8 xx 10^5)(1.12) − (4.8 xx 10^5)`
    `= (3 xx 10^6)(1.12)^2 − (4.8 xx 10^5)(1 + 1.12)\ \ …text(as required)`

 

ii.  `text(Show)\ A_n = 10^6[4 − (1.12)^n]`

`A_3` `= (3 xx 10^6)(1.12)^3 − (4.8 xx 10^5)(1 + 1.12 + 1.12^2)`
`vdots`  
`A_n` `= (3 xx 10^6)(1.12)^n − (4.8 xx 10^5)(1 + 1.12 + … + 1.12^(n − 1))`
  `=> text(Noting that)\ \ (1+1.12+ … + 1.12^(n-1))\ text(is a)`
  `text(GP where)\ \ a = 1, \ \ r = 1.12`
 `A_n` `= (3 xx 10^6)(1.12)^n − (4.8 xx 10^5)((1(1.12^n − 1))/(1.12 − 1))`
  `= (3 xx 10^6)(1.12)^n − (4.8 xx 10^5)((1(1.12^n − 1))/0.12)`
  `= (3 xx 10^6)(1.12)^n − (4 xx 10^6)(1.12^n − 1)`
  `= 10^6[3(1.12)^n − 4(1.12^n − 1)]`
  `= 10^6[3(1.12)^n − 4(1.12^n) + 4]`
  `= 10^6[4 − (1.12)^n]\ \ …text(as required)`

 

iii.  `text(Find)\ n\ text(such that)\ A_n = 0`

`10^6[4 − (1.12)^n]` `= 0`
`4 − (1.12)^n` `= 0`
`1.12^n` `= 4`
`n xx ln 1.12` `= ln4`
`n` `= (ln4)/ln1.12`
  `= 12.23…\ \ text(years.)`

 

`:.\ text(The final repayment will be made)`

`text(in 2017, the thirteenth year.)`

Calculus, 2ADV C4 2005 HSC 8b

2005 8b

The shaded region in the diagram is bounded by the circle of radius 2 centred at the origin, the parabola  `y = x^2– 3x + 2`, and the `x`-axis.

By considering the difference of two areas, find the area of the shaded region.  (3 marks)

Show Answers Only

`(pi − 5/6)\ \ text(u²)`

Show Worked Solution

`text(Shaded Area = Area in the quarter circle less)`

`text(the area below the parabola between)\ x= 0 and 1.`

`text(Area of)\ 1/4\ text(circle)` `= 1/4 xx pir^2`
  `= 1/4 xx pi xx 2^2`
  `= pi\ \ \ text(u²)`

 

`text(Area below the parabola between)\ x= 0 and 1`

`=int_0^1y\ dx`

`= int_0^1x^2 − 3x + 2\ dx`

`= [x^3/3 − 3/2x^2 + 2x]_0^1`

`= [(1/3 − 3/2 + 2) − 0]`

`= 5/6`

 

`:.\ text(Shaded Area) = (pi − 5/6)\ \ \ text(u²)`

Calculus, 2ADV C3 2005 HSC 8a

2005 8a

A cylinder of radius  `x`  and height  `2h`  is to be inscribed in a sphere of radius  `R`  centred at  `O`  as shown.

  1. Show that the volume of the cylinder is given by
     
         `V = 2pih(R^2 − h^2).` (1 mark)

  2. Hence, or otherwise, show that the cylinder has a maximum volume when  `h = R/sqrt3.`  (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions).}`
  2. `text{Proof (See Worked Solutions).}`
Show Worked Solution
i.    `V` `= pir^2h\ \ \ \ text{(general form)}`
    `= pix^2 xx 2h`

 

`text(Using Pythagoras)`

`R^2` `= h^2 + x^2`
`x^2` `= R^2 − h^2`
`:.V` `= 2pih(R^2 − h^2)\ \ …text(as required.)` 

 

ii. `V` `= 2pih(R^2 − h^2)`
    `= 2piR^2h − 2pih^3`
  `(dV)/(dh)` `= 2piR^2 − 3 xx 2pih^2`
    `= 2piR^2 − 6pih^2`
  `(d^2V)/(dh^2)` `= −12pih`

`text(Max or min when)\ (dV)/(dh) = 0`

`2piR^2 − 6pih^2` `= 0`
`6pih^2` `= 2piR^2`
`h^2` `= R^2/3`
`h` `= R/sqrt3,\ \ h > 0`

`text(When)\ h = R/sqrt3`

`(d^2V)/(dh^2) = −12pi xx R/sqrt3 < 0`

`:.\ text(Volume is a maximum when)\ h = R/sqrt3\ text(units.)`

Calculus, EXT1* C1 2005 HSC 7b

Calculus in the Physical World, 2UA 2005 HSC 7b
 

The graph shows the velocity, `(dx)/(dt)`, of a particle as a function of time. Initially the particle is at the origin. 

  1. At what time is the displacement, `x`, from the origin a maximum?  (1 mark)

  2. At what time does the particle return to the origin? Justify your answer.  (2 marks)

  3. Draw a sketch of the acceleration,  `(d^2x)/(dt^2)`, as a function of time for  `0 ≤ t ≤ 6`.  (2 marks)

Show Answers Only
  1. `2`
  2. `4`
  3. `text(See Worked Solutions.)`
Show Worked Solution

i.   `text(Maximum displacement in graph when)`

`(dx)/(dt)` `= 0`
`:.t` `= 2`

 

ii.  `text(Particle returns to the origin when)\ t = 4.`

`text(The displacement can be calculated by the)`

`text(net area below the curve and since the)`

`text(area above the curve between)\ t = 0\ text(and)\ t = 2`

`text(is equal to the area below the curve between)`

`t = 2\ text(and)\ t = 4,\ text(the displacement returns to)`

`text{the initial displacement (i.e. the origin).}`

 

iii.   Calculus in the Physical World, 2UA 2005 HSC 7b Answer

Financial Maths, 2ADV M1 2005 HSC 7a

Anne and Kay are employed by an accounting firm.

Anne accepts employment with an initial annual salary of  $50 000. In each of the following years her annual salary is increased by $2500.

Kay accepts employment with an initial annual salary of  $50 000. In each of the following years her annual salary is increased by 4%.

  1. What is Anne’s annual salary in her thirteenth year?  (2 marks)

  2. What is Kay’s annual salary in her thirteenth year?  (2 marks)

  3. By what amount does the total amount paid to Kay in her first twenty years exceed that paid to Anne in her first twenty years?  (3 marks)

Show Answers Only
  1. `$80\ 000`
  2. `text{$80 052 (nearest dollar)}`
  3. `text{$13 904 (nearest $)}`
Show Worked Solution

i.   `text(Let)\ T_n = text(Anne’s salary in year)\ n`

`T_1 = a = $50\ 000`

`T_2 = a + d = $52\ 500`

`⇒\ text(AP where)\ a = $50\ 000,\ \ d = $2500`

`T_n = a + (n − 1)d`

`T_13` `= 50\ 000 + (13 − 1) xx 2500`
  `=80\ 000`

 

`:.\ text(Anne’s salary in her 13th year is $80 000.)`

 

ii.  `text(Let)\ K_1 =text(Kay’s salary in year)\ n`

`K_1` `= a` `= 50\ 000`
`K_2` `= ar` `= 50\ 000 xx 1.04 = 52\ 000`
`⇒\ text(GP where)\ \ a = 50\ 000, \ \ r = 1.04`
`K_n` `= ar^(n − 1)`
 `K_13` `= 50\ 000 xx (1.04)^12`
  `= $80\ 051.61…`
  `= $80\ 052\ \ \ text{(nearest dollar)}`

 

iii.  `text(Anne)`

`S_n` `= n/2[2a + (n − 1)d]`
 `S_20` `= 20/2[2 xx 50\ 000 + (20 − 1)2500]`
  `= 10[100\ 000 + 47\ 500]`
  `= $1\ 475\ 000`

 

`text(Kay)`

`S_n` `= (a(r^n − 1))/(r − 1)`
`S_20`  `= (50\ 000(1.04^20 -1))/(1.04 − 1)`
  `= $1\ 488\ 903.929…`

 

`text(Difference)`

`= 1\ 488\ 903.929… − 1\ 475\ 000`

`= $13\ 903.928…`

`= $13\ 904\ \ \  text{(nearest $)}`

 

`:.\ text(Kay’s total salary exceeds Anne’s by)\ $13\ 904`

Algebra, STD2 A4 2006 HSC 28b

A new tunnel is built. When there is no toll to use the tunnel, 6000 vehicles use it each day. For each dollar increase in the toll, 500 fewer vehicles use the tunnel.

  1. Find the lowest toll for which no vehicles will use the tunnel.  (1 mark)

  2. For a toll of $5.00, how many vehicles use the tunnel each day and what is the total daily income from tolls?  (2 marks)

  3. If `d` (dollars) represents the value of the toll, find an equation for the number of vehicles `(v)` using the tunnel each day in terms of `d`.  (2 marks)

  4. Anne says ‘A higher toll always means a higher total daily income’.

     

    Show that Anne is incorrect and find the maximum daily income from tolls. (Use a table of values, or a graph, or suitable calculations.)  (3 marks)

Show Answers Only
  1. `text($12 toll is the lowest for which)`

     

    `text(no vehicles will use the tunnel.)`

  2. `$17\ 500`
  3. `v = 6000 – 500d`
  4. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.  `text(500 less vehicles per $1 toll)`

`12 xx 500 = 6000`

`:. $12\ text(toll is the lowest for which no)`

`text(vehicles will use the tunnel.)`

 

ii.  `text(If the toll is $5)`

`5 xx 500 = 2500\ text(less vehicles)`

`:.\ text(Vehicles using the tunnel)`

`= 6000 – 2500`

`= 3500`

`:.\ text(Daily toll income)` `= 3500 xx $5`
  `= $17\ 500`

 

iii.    `d` `=\ text(toll)`
`v` `=\ text(Number of vehicles using the tunnel)`
    `:. v` `= 6000 – 500d`

 

iv.  `text(Income from tolls)`

`=\ text(Number of vehicles) xx text(toll)`

`= (6000 – 500d) xx d`

`= 6000d – 500d^2`

`= 500d (12 – d)`
 

 

`text(From the graph, the maximum income from tolls)`

`text(occurs when the toll is $6.)`

`:.\ text(Anne is incorrect.)`

 

`text(Alternate Solution)`

`text{The table of values shows that income (I) increases}`

`text(and peaks when the toll hits $6 before decreasing)`

`text(again as the toll gets more expensive.)`

`:.\ text(Anne is incorrect.)`

Financial Maths, STD2 F4 2006 HSC 27c

Kai purchased a new car for $30 000. It depreciated in value by $2000 per year for the first three years.

After the end of the third year, Kai changed the method of depreciation to the declining balance method at the rate of 25% per annum.

  1. Calculate the value of the car at the end of the third year.  (1 mark)

  2. Calculate the value of the car seven years after it was purchased.  (2 marks)

  3. Without further calculations, sketch a graph to show the value of the car over the seven years.

     

    Use the horizontal axis to represent time and the vertical axis to represent the value of the car.  (3 marks)

Show Answers Only
  1. `$24\ 000`
  2. `$7593.75`
  3. `text{See Worked Solutions}`
Show Worked Solution

i.  `text(Using)\ \ S = V_0 – Dn`

`S` `= 30\ 000 – (2000 xx 3)`
  `= $24\ 000`

 

ii.  `text(Using)\ \ S = V_0(1 – r)^n`

`text(where)\ V_0` `= 24\ 000`
`r` `= 0.25`
`n` `= 4`

 

`S` `= 24\ 000(1 – 0.25)^4`
  `= $7593.75`

 
`:.\ text(The value of the car after 7 years is $7593.75)`

 

iii.

Statistics, STD2 S4 2006 HSC 27b

Each member of a group of males had his height and foot length measured and recorded. The results were graphed and a line of fit drawn.
 

  1. Why does the value of the `y`-intercept have no meaning in this situation?  (1 mark)

  2. George is 10 cm taller than his brother Harry. Use the line of fit to estimate the difference in their foot lengths.  (1 mark)

  3. Sam calculated a correlation coefficient of  −1.2  for the data. Give TWO reasons why Sam must be incorrect.  (2 marks)

Show Answers Only
  1. `text(The y-intercept occurs when)\ x = 0.\ text(It has`
    `text(no meaning to have a height of 0 cm.)`
  2. `text(A 10 cm height difference means George should)`
    `text(have a 3 cm longer foot.)`
  3. `text(A correlation co-efficient must be between –1 and 1.)`
    `text(Foot length is positively correlated to a person’s)`
    `text(height and therefore can’t be a negative value.)`
Show Worked Solution

i.  `text(The y-intercept occurs when)\ x = 0.\ text(It has)`

`text(no meaning to have a height of 0 cm.)`

 

ii.  `text(A 20 cm height difference results in a foot length)`

`text(difference of 6 cm.)`
 

`:.\ text(A 10 cm height difference means George should)`

`text(have a 3 cm longer foot.)`

 

iii.  `text(A correlation co-efficient must be between –1 and 1.)`

`text(Foot length is positively correlated to a person’s)`

`text(height and therefore isn’t a negative value.)`