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Plane Geometry, 2UA 2004 HSC 2b

In the diagram, `ABC`  is an isosceles triangle with  `AB = AC`  and  `/_BAC = 38^@`. The line `BC` is produced to `D`. 

Find the size of `/_ACD`. Give reasons for your answer.   (2 marks)

Show Answers Only

`109^@`

Show Worked Solution

Plane Geometry, 2UA 2004 HSC 2b Answer

`/_ABC` `= 1/2 (180-38)\ \ \ text{(base angle of isosceles}\ Delta ABC text{)}`
  `= 71^@`

 

`:.\ /_ACD` `= 71 + 38\ \ \ text{(exterior angle of}\ Delta ABC text{)}`
  `= 109^@`

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

`/_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`

 

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`
  `= 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(m)^2`

 

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

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

`= (9 sqrt 3 + 45 pi)/4\ \ text(m)^2`

Financial Maths, 2ADV M1 2006 HSC 3c

On the first day of the harvest, an orchard produces 560 kg of fruit. On the next day, the orchard produces 543 kg, and the amount produced continues to decrease by the same amount each day.

  1. How much fruit is produced on the fourteenth day of the harvest?  (2 marks)

  2. What is the total amount of fruit that is produced in the first 14 days of the harvest?  (1 mark)

  3. On what day does the daily production first fall below 60 kg?  (2 marks)

Show Answers Only
  1. `text(339 kg of fruit is produced on 14th day.)`
  2. `text(6293 kg of fruit is produced in the 1st 14 days.)`
  3. `text(On the 31st day, production will first drop below 60 kg.)`
Show Worked Solution

i.   `T_1 = a = 560`

`T_2 = a + d = 543`

`=>\ text(AP where)\ a = 560, d = -17`

`vdots`

`T_14` `= a + 13d`
  `= 560 – (13 xx 17)`
  `= 339`

 

`:.\ text(339 kg of fruit is produced on 14th day.)`

 

ii.  `S_14 = text(total fruit produced in 1st 14 days)`

`S_14` `= n/2 [2a + (n – 1)d]`
  `= 14/2 [2 xx 560 – (14 – 1) xx 17]`
  `= 7 [1120 – 221]`
  `= 6293`

 

`:.\ text(6293 kg of fruit is produced in the 1st 14th days.)`

 

iii.  `text(Find)\ n\ text(such that)\ T_n < 60`

`T_n = a + (n – 1)d` `< 60`
`560 – 17(n – 1)` `< 60`
`560 – 17n + 17` `< 60`
`17n` `> 517`
`n` `> 30.41…`

 
`:.\ text(On the 31st day, production will first drop)`

`text(below 60 kg.)`

Calculus, 2ADV C3 2006 HSC 2c

Find the equation of the tangent to the curve  `y = cos 2x`  at the point whose `x`-coordinate is  `pi/6`.  (3 marks)

Show Answers Only

`y = – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`

Show Worked Solution

`y = cos 2x`

`dy/dx = -2 sin 2x`

`text(When)\ \ x = pi/6`

`y` `= cos (2 xx pi/6)`
  `= cos (pi/3)`
  `= 1/2`

 

`dy / dx` `= -2 sin (pi/3)`
  `= -2 xx sqrt 3 / 2`
  `= – sqrt 3`

 
`text(Equation of tangent,)\ \ m = – sqrt 3, text(through)\ \ (pi/6, 1/2):`

`y – y_1` `= m(x – x_1)`
`y – 1/2` `= – sqrt 3 ( x – pi/6)`
`y – 1/2` `= – sqrt 3 x + (sqrt 3 pi)/6`
`y` `= – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`

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)`

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 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`

Trigonometry, 2ADV T1 2006 HSC 1d

2006 1d

 
Find the value of `theta` in the diagram. Give your answer to the nearest degree.   (2 marks)

Show Answers Only

`18°`

Show Worked Solution

`text(Using the sine rule:)`

`sin theta / 5` `= (sin 33°)/9`
`sin theta` `= (5 xx sin 33°)/9= 0.30257…`
`:. theta` `= 17.612…`
  `= text{18°  (nearest degree)}`

Calculus, 2ADV C2 2007 HSC 2ai

Differentiate with respect to `x`:

`(2x)/(e^x + 1)`  (2 marks)

Show Answers Only

`{2(e^x + 1 – xe^x)}/(e^x + 1)^2`

Show Worked Solution

`y = (2x)/(e^x + 1)`

`u` `= 2x` `\ \ \ \ \v` `= e^x + 1`
`u prime` `= 2` `\ \ \ \ \ v prime` `= e^x`
`(dy)/(dx)` `= (u prime v – uv prime)/v^2`
  `= {2(e^x + 1) – 2x(e^x)}/(e^x + 1)^2`
  `= (2e^x + 2 – 2x * e^x)/(e^x + 1)^2`
  `= {2(e^x + 1 – xe^x)}/(e^x + 1)^2`

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.

Probability, STD2 S2 2006 HSC 26c

A new test has been developed for determining whether or not people are carriers of the Gaussian virus.

Two hundred people are tested. A two-way table is being used to record the results.
 

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

  2.  A person selected from the tested group is a carrier of the virus.

     

     What is the probability that the test results would show this?  (2 marks) 

  3.  For how many of the people tested were their test results inaccurate?  (1 mark)

Show Answers Only
  1. `98`
  2. `37/43`
  3. `28`
Show Worked Solution
i.  `A` `= 200-(74 + 12 + 16)`
  `= 98`

 

ii.  `P` `= text(# Positive carriers)/text(Total carriers)`
  `= 74/86`
  `= 37/43`

 

iii.  `text(# People with inaccurate results)`

`= 12 + 16`

`= 28`

Calculus, EXT1* C3 2005 HSC 6c

2005 6c
 

The graphs of the curves  `y = x^2`  and  `y = 12 - 2x^2`  are shown in the diagram.

  1. Find the points of intersection of the two curves.  (1 mark)

  2. The shaded region between the curves and the `y`-axis is rotated about the `y`-axis. By splitting the shaded region into two parts, or otherwise, find the volume of the solid formed.  (3 marks)

Show Answers Only
  1. `text{(2, 4), (–2, 4)`
  2. `24pi\ \ text(u³)`
Show Worked Solution
i.    `y` `= x^2` `\ …\ (1)`
  `y` `= 12 − 2x^2` `\ …\ (2)`

 

`text(Substitute)\ \ y = x^2\ \ text(into)\ (2)`

`x^2` `= 12 − 2x^2`
`3x^2 − 12` `= 0`
`3(x^2 − 4)` `= 0`
`x` `= ±2`
`text(When)` `\ x = 2,` `\ y = 4`
`text(When)` `\ x = text(−2),` `\ y = 4`

 
`:.\ text{Intersection at (2, 4), (−2, 4)}`
 

ii.  `text{In (1),}\ \ x^2=y`

`text{In (2),}\ \ \ y` `= 12 − 2x^2`
`2x^2` `= 12 − y`
`x^2` `= (12 − y)/2`
  `= 6 − 1/2y`

 
`:.\ text(Volume)`

`= pi int_0^4 y\ dy + pi int_4^12 6 − 1/2y\ dy`
`= pi[y^2/2]_0^4 + pi[6y − y^2/4]_4^12`
`= pi[16/2 − 0] + pi[(6 xx 12 − 12^2/4) − (6 xx 4 − 4^2/4)]`
`= 8pi + pi[36 − 20]`
`= 8pi + 16pi`
`= 24pi\ \ text(u³)`

Calculus, 2ADV C3 2005 HSC 6b

A tank initially holds 3600 litres of water. The water drains from the bottom of the tank. The tank takes 60 minutes to empty.

A mathematical model predicts that the volume, `V`  litres, of water that will remain in the tank after  `t`  minutes is given by
  

`V = 3600(1 − t/60)^2,\ \ text(where)\ \ 0 ≤ t ≤ 60`.
 

  1. What volume does the model predict will remain after ten minutes?  (1 mark)

  2. At what rate does the model predict that the water will drain from the tank after twenty minutes?  (2 marks)

  3. At what time does the model predict that the water will drain from the tank at its fastest rate?  (2 marks)

Show Answers Only
  1. `text(2500 L)`
  2. `80\ text(liters per minute)`
  3. `0`
Show Worked Solution

i.    `V = 3600(1 − t/60)^2`

`text(When)\ t = 10,`

`V` `= 3600(1 − 10/60)^2`
  `= 3600 xx (5/6)^2`
  `= 2500\ text(L)`

 

ii.   `V = 3600(1 -t/60)^2`

`text(Using chain rule:)`

`(dV)/dt` `= 3600 xx 2 xx (1 – t/60) xx d/dt(1 – t/60)`
  `= 7200(1 – t/60) xx -1/60`
  `= −120(1 – t/60)`

 

`text(When)\ \ t =20`

`(dV)/dt` `= −120(1 – 20/60)`
  `= −80`

 

`:.\ text(After 20 minutes, the water will drain)`

`text(at 80 litres per minute.)`

 

iii. `(dV)/dt` `= −120(1 − t/60)`
    `= −120 + 2t`
  `(d^2V)/dt^2` `= 2`

 
`text(S)text(ince)\ (d^2V)/dt^2\ text(is a constant, no S.P.’s)`

 

`text(Checking limits of)\ \ 0 ≤ t ≤ 60`

`text(At)\ t = 0,`

`(dV)/dt = −120(1-0) = −120\ text(L/min)`

`text(At)\ t = 60,`

`(dV)/dt = −120(1 − 60/60) = 0\ text(L/min)`

 

`:.\ text(The model predicts water will drain)`

`text(out the fastest when)\ \ t = 0.`

L&E, 2ADV E1 2005 HSC 5a

Use the change of base formula to evaluate  `log_3 7`, correct to two decimal places.  (1 mark)

Show Answers Only

`1.77\ \ text{(to 2 d.p.)}`

Show Worked Solution
`log_3 7` `= (log_10 7)/(log_10 3)`
  `= 1.771…`
  `= 1.77\ \ text{(to 2 d.p.)}`

Calculus, 2ADV C3 2005 HSC 4b

A function  `f(x)`  is defined by  `f(x) = (x + 3)(x^2- 9)`.

  1. Find all solutions of  `f(x) = 0`  (2 marks)

  2. Find the coordinates of the turning points of the graph of  `y = f(x)`, and determine their nature.  (3 marks)

  3. Hence sketch the graph of  `y = f(x)`, showing the turning points and the points where the curve meets the `x`-axis.  (2 marks)

  4. For what values of `x` is the graph of  `y = f(x)`  concave down?  (1 mark)

Show Answers Only
  1. `−3 or 3`
  2. `text{53.2 cm  (to 1 d.p.)}`
  3. `text(See worked solutions)`
  4. `x < −1`
Show Worked Solutions
i.    `f(x)` `= (x + 3)(x^2 − 9)`
    `= (x + 3)(x +3)(x − 3)`
  `:. f(x)` `= 0\ text(when)\ \ x=–3\ text(or)\ 3`

 

ii.   `f (x)` `= (x +3)(x^2 − 9)`
    `= x^3 − 9x + 3x^2 − 27`
    `= x^3 + 3x^2 − 9x − 27`
  `f′(x)` `= 3x^2 + 6x − 9`
  `f″(x)` `= 6x + 6`

 

`text(S.P.’s  when)\ \ f′(x) = 0`

`3x^2 + 6x − 9` `= 0`
`3(x^2 + 2x − 3)` `= 0`
`3(x − 1)(x + 3)` `= 0`

 

`text(At)\ x =1`

`f(1)` `= (4)(−8)=−32`
 `f″(1)` `= 6 + 6=12>0`
`:.\ text(MIN at)\ (1, −32)` 

 

`text(At)\ x = −3`

`f(-3)` `= 0`
`f″(−3)` `= (6 xx −3) + 6 = −12 <0`
`:.\ text(MAX at)\ (−3, 0)`

 

iii.   Geometry and Calculus, 2UA 2005 HSC 4b Answer

 

iv.  `f(x)\ \ text(is concave down when)`

`f″(x)` `< 0`
`6x + 6` `< 0`
`6x` `< −6`
`x` `< −1`

Trigonometry, 2ADV T1 2005 HSC 4a

 Trig Calculus, 2UA 2005 HSC 4a

A pendulum is 90 cm long and swings through an angle of 0.6 radians. The extreme positions of the pendulum are indicated by the points `A` and `B` in the diagram.

  1. Find the length of the arc `AB`.   (1 mark)

  2. Find the straight-line distance between the extreme positions of the pendulum.   (2 marks)

  3. Find the area of the sector swept out by the pendulum.   (1 mark)

Show Answers Only
  1. `text(54 cm)`
  2. `text{53.2 cm  (to 1 d.p.)}`
  3. `text(2430 cm)^2`
Show Worked Solution
i.    `text(Arc)\ AB` `= theta/(2 pi) xx 2 pi r`
    `= rtheta`
    `= 90 xx 0.6`
    `= 54\ text(cm)`

 

ii. 

 Trig Calculus, 2UA 2005 HSC 4a Answer

`text(Using the cosine rule:)`

`text(Distance)\ AB\ text(in straight line)`

`AB^2` `= 90^2 xx 90^2-2 xx 90 xx 90 xx cos\ 0.6`
  `= 2829.563…`
`:.AB` `= 53.193…= 53.2\ text{cm  (to 1 d.p.)}`

 

iii. `text(Area of Sector)`

`= 0.6/(2pi) xx pir^2`

`= 0.3 xx 90^2`

`= 2430\ text(cm)^2`

Plane Geometry, 2UA 2005 HSC 3c

2005 3c

In the diagram, `A`, `B`  and  `C`  are the points  `(6, 0), (9, 0)`  and  `(12, 6)` respectively. The equation of the line  `OC`  is  `x - 2y = 0`. The point  `D`  on  `OC`  is chosen so that  `AD`  is parallel to  `BC`. The point  `E`  on  `BC`  is chosen so that  `DE`  is parallel to the `x`-axis.

  1.  Show that the equation of the line `AD` is `y = 2x - 12`.  (2 marks)
  2. Find the coordinates of the point `D`.  (2 marks)
  3. Find the coordinates of the point `E`.  (1 marks)
  4. Prove that  `ΔOAD\ text(|||)\ ΔDEC`.  (2 marks)
  5. Hence, or otherwise, find the ratio of the lengths `AD` and `EC`.  (1 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(8, 4)`
  3. `(11, 4)`
  4. `text(See Worked Solutions)`
  5. `2:1`
Show Worked Solution

(i)   `text(S)text(ince)\ DA\ text(||)\ CB`

`m_(DA)` `= (y_2 − y_1)/(x_2 − x_1)`
  `= (6 − 0)/(12 − 9)`
  `= 2`

 

`:.\ text(Equation of)\ AD, m = 2,\ text(through)\ A(6, 0)`

`y − y_1` `= m(x − x_1)`
 `y − 0` `= 2(x − 6)`
 `y` `= 2x − 12\ \ \ \ …\ text(as required)`

 

(ii)  `D\ text(is at the intersection of)`

`x − 2y` `= 0` `\ \ …\ (1)`
 `y` `= 2x − 12\ ` `\ \ …\ (2)` 

`text(Substitute)\ y = 2x − 12\ text{into (1)}`

`x − 2(2x − 12)` `= 0`
`x − 4x + 24` `= 0`
`text(−3)x + 24` `= 0`
`3x` `= 24`
`x` `= 8`

`text(Substitute)\ x = 8\ text{into (2)}`

`y` `= 2 xx 8 − 12 = 4`
`:.D` `= (8, 4)`

 

(iii)  `text(Distance)\ \ AB=3`

`:. E\ \ text(has coordinates)\ \ (8+3,4)-=(11,4)`

 

(iv)  `text(Prove)\ ΔOAD\ text(|||)\ ΔDEC`

`∠ODA = ∠DCE`

`text{(corresponding angles,}\ AD\ text(||)\ BC)`

`∠OAD = ∠ABE`

`text{(corresponding angles,}\ AD\ text(||)\ BC)`

`∠ABE = ∠DEC`

`text{(corresponding angles,}\ AB\ text(||)\ DE)`

`:.∠OAD = ∠DEC`

`:.ΔOAD\ text(|||)\ ΔDEC\ text{(equiangular)}`

 

(v)     `(AD)/(EC)` `= (OA)/(DE)` 
    `= 6/3` 
    `=2` 

 `text{(corresponding sides of similar triangles)}`

 

`:.\ text(Ratio of lengths)\ AD:EC = 2:1` 

Calculus, 2ADV C3 2006 HSC 5a

A function  `f(x)`  is defined by  `f(x) =2x^2(3-x)`.

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

  2. Find the coordinates of the point of inflection.  (1 mark)

  3. Hence sketch the graph of  `y =f(x)`, showing the turning points, the point of inflection and the points where the curve meets the `x`-axis.  (3 marks)

  4. What is the minimum value of  `f(x)`  for  `–1 ≤ x ≤4`?  (1 mark)

Show Answers Only
  1. `text(Min)\ (0, 0),\ text(Max)\ (2, 8)`
  2. `text(P.I. at)\ (1, 4)`
  3.  
  4. `-32`
Show Worked Solution
i.       `f(x)` `= 2x^2 (3-x)`
  `= 6x^2-2x^3`
`f^{prime} (x)` `= 12x-6x^2`
`f^{″}(x)` `= 12-12x`

 

`text(S.P.’s when)\ f^{′}(x) = 0`

`12x-6x^2` `= 0`
`6x(2-x)` `= 0`

`x = 0 or 2`

`text(When)\ x = 0`

`f(0)` `= 0`
`f^{″}(0)` `= 12-0 = 12 > 0`
`:.\ text(MIN at)\ (0, 0)`

 

`text(When)\ x = 2`

`f(2)` `= 2 xx 2^2 (3-2)` `= 8`
`f^{″}(2)` `= 12-(12 xx 2)` `= -12 < 0`
`:.\ text(MAX at)\ (2, 8)`

 

ii.  `text(P.I. when)\ f^{″}(x) = 0`

`12-12x` `= 0`
`12x` `= 12`
`x` `= 1`
`f^{″}(0.5)` `=6>0`
`f^{″}(1.5)` `=-6<0`

`text(S)text(ince concavity changes)\ \ =>\  text(P.I. exists)` 

`f(1)` `= 2 xx 1^2(3-1)`
  `= 4`

`:.\ text(P.I. at)\ (1, 4)`

 

iii.  `f(x)\ text(meets)\ x text(-axis when)\ f(x) = 0`

`2x^2 xx (3-x) = 0`

`x = 0 or 3`

2UA HSC 2006 5a

 

(iv)  `text(The graph clearly shows that in the given range)`

`-1<= x<=4,\ text(the minimum will occur when)\ x = 4`

`:.\ text(Minimum` `= 2 xx 4^2 (3-4)`
  `= -32`

Calculus, 2ADV C3 2005 HSC 2d

Find the equation of the tangent to  `y = log_ex`  at the point  `(e, 1)`.  (2 marks)

Show Answers Only

`y = x/e`

Show Worked Solution

`y = log_ex`

`dy/dx = 1/x`

`text(At)\ \ (e, 1),`

`m = 1/e`
 

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

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

Trig Calculus, 2UA 2005 HSC 2cii

Evaluate `int_0^(pi/6) cos\ 3x\ dx`.  (2 marks)

Show Answers Only

`1/3`

Show Worked Solution
`int_0^(pi/6)\ cos\ 3x\ dx` `= 1/3\ [sin\ 3x]_0^(pi/6)`
  `= 1/3\ [sin\ (3 xx pi/6) − sin\ 0]`
  `= 1/3\ [sin\ pi/2 − sin\ 0]`
  `= 1/3\ [1 − 0]`
  `= 1/3`

Calculus, 2ADV C1 2005 HSC 2bii

Differentiate with respect to `x`:

`x^2/(x-1).`   (2 marks)

Show Answers Only

`(x(x-2))/(x-1)^2`

Show Worked Solution

`y = x^2/(x-1)`

`text(Using)\ \ dy/dx = (u^{′}v-uv^{′})/v^2:`

`u` `= x^2` `v` `= x-1`
`u^{′}` `= 2x` `v^{′}` `= 1`

 

`dy/dx` `= (2x(x-1)-x^2(1))/(x-1)^2`
  `= (2x^2-2x-x^2)/(x-1)^2`
  `= (x^2-2x)/(x-1)^2`
  `= (x(x-2))/(x-1)^2`

Trigonometry, 2ADV T2 2005 HSC 2a

Solve  `cos\ theta =1/sqrt2`  for  `0 ≤ theta ≤ 2pi`.   (2 marks)

Show Answers Only

`pi/4, (7pi)/4`

Show Worked Solution

`cos\ theta = 1/sqrt2,\ \ \ 0 ≤ theta ≤ 2pi`

`text(S)text(ince)\ cos\ pi/4 = 1/sqrt2,\ \ text(and cos)`

`text(is positive in 1st/4th quadrants)`

`theta` `= pi/4, 2pi-pi/4`
  `= pi/4, (7pi)/4`

 

Functions, EXT1* F1 2005 HSC 1e

Find the values of `x` for which `|\ x-3\ | ≤ 1`.   (2 marks)

Show Answers Only

`2 ≤ x ≤ 4`

Show Worked Solution

`|\ x-3\ | ≤ 1`

`text(Solution 1)`

`(x-3)^2 ≤ 1`

`x^2-6x + 9 ≤ 1`

`x^2-6x +8 ≤ 0`

`(x-4)(x-2) ≤ 0`
 

Algebra, 2UA 2005 HSC 1e Answer  

 
`:. 2 ≤ x ≤ 4`

 

`text(Alternative Solution)`

`(x-3)` `≤1` `-(x-3)` ` ≤ 1`
`x` `≤4` `-x +3` `≤ 1`
    `-x` `≤-2`
    `x` `≥ 2`

`:. 2 ≤ x ≤ 4`

Functions, 2ADV F1 2005 HSC 1d

Express  `((2x-3))/2-((x-1))/5`  as a single fraction in its simplest form.  (2 marks)

Show Answers Only

`(8x-13)/10`

Show Worked Solution

`((2x-3))/2-((x-1))/5`

`= (5(2x-3)-2(x-1))/10`

`= (10x-15-2x + 2)/10`

`= (8x-13)/10`

Financial Maths, 2ADV M1 2006 HSC 1f

Find the limiting sum of the geometric series  `13/5 + 13/25 + 13/125 + …`  (2 marks)

Show Answers Only

`13/4`

Show Worked Solution

`13/5 + 13/25 + 13/125`

`=>\ text(GP where)\ \ a=13/5,\ text(and)`

`r = T_2/T_1 = 13/25 ÷ 13/5 = 1/5`

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

`S_oo` `= a/(1-r)`
  `= (13/5)/(1 – 1/5)`
  `= 13/5 xx 5/4`
  `= 13/4`

Data, 2UG 2005 HSC 27a

The area graph shows sales figures for Shoey’s shoe store.

2UG-2005-27a

  1. Approximately how many school shoes were sold in January?   (1 mark)
  2. For which month does the graph indicate that the same number of school shoes and business shoes was sold?   (1 mark)
  4. Identify ONE trend in this graph, and suggest a valid reason for this trend.   (2 marks)
Show Answers Only
  1. `15\ 000`
  2.  `text(April)`
  3. `text(Possible trends include)`
  4. `text(- Boot sales peak in the Winter months of Jun-Aug)`
  5. `\ \ text(because of the colder weather.)`
    `text(- School shoe sales peak in January as children prepare)`
    `\ \ text(to return to school for the new year.)`
Show Worked Solution

(i)   `text(School shoes sold)`

`≈ 18\ 000 − 3000`

`≈ 15\ 000`

 

(ii)  `text(April)`

 

(iii) `text(Possible trends include)`

`text(- Boot sales peak in the Winter months of Jun-Aug)`

`\ \ text(because of the colder weather.)`

`text(- School shoe sales peak in January as children prepare)`

`\ \ text(to return to school for the new year.)`

Measurement, 2UG 2004 HSC 26b

The location of Sorong is `text(1°S 131°E)` and the location of Darwin is `text(12°S 131°E)`.

  1. What is the difference in the latitudes of Sorong and Darwin?  (1 mark)
  2. The radius of Earth is approximately `text(6400 km.)`

  3. Show that the great circle distance between Sorong and Darwin is approximately `text(1200 km)`.  (2 marks)

 

Show Answers Only
  1. `11°`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)  `text(Difference in latitudes)`

`= 12^@ – 1^@`

`= 11^@`

 

(ii)   `text(S)text(ince longitude is the same,)`

`text(Distance between Sarang and Darwin)`

`= 11/360 xx 2 pi r`

`= 11/360 xx 2 xx pi xx 6400`

`= 1228.7…\ text(km)`

`~~ 1200\ text(km … as required.)`

Data, 2UG 2006 HSC 23d

The graph shows the amounts charged by Company `A` and Company `B` to deliver parcels of various weights.

2UG-2006-23d

  1. How much does Company `A` charge to deliver a `3` kg parcel?  (1 mark)
  2. Give an example of the weight of a parcel for which both Company `A` and Company `B` charge the same amount.  (1 mark)
  4. For what weight(s) is it cheaper to use Company `A`?  (2 marks)
  5. What is the rate per kilogram charged by Company `B` for parcels up to `8` kg?  (1 mark)
Show Answers Only
  1. `$6`
  2. `text(4 kg or 7 kg)`
  3. `x > 7`
  4. `text($1.50 per kg)`
Show Worked Solution

(i)   `$6`

 

(ii)  `text(4 kg or 7 kg)`

 

(iii)  `text(Let)\ x = text(weights where Company)\ A`

`text{is cheaper (kg).}`

`4 < x <= 6`

`x > 7`

 

(iv)   `text(Company)\ B\ text(charges $12 for 8 kg)`

`:.\ text(Rate)` `= 12/8`
  `= $1.50\  text(per kg)`

Data, 2UG 2006 HSC 23b

This radar chart was used to display the average daily temperatures each month for two different towns.

2UG-2006-23b

  1. What is the average daily temperature of Town `B` for April?  (1 mark)
  2. In which month do the average daily temperatures of the two towns have the greatest difference?  (1 mark)
  4. In which months is the average daily temperature in Town `B` higher than in Town `A`?  (1 mark)
Show Answers Only
  1. `18.5^@\ \ text{(approx)}`
  2. `text(February)`
  3. `text(May, June, July, August and September)`
Show Worked Solution

(i)   `18.5^@\ \ \ text{(approx)}`

 

(ii)   `text(February)`

 

(iii)  `text(May, June, July, August and September)`

Measurement, STD2 M6 2005 HSC 25b

2UG-2005-25b

  1. Use Pythagoras’ theorem to show that `ΔABC` is a right-angled triangle.  (1 mark)

  2. Calculate the size of `∠ABC` to the nearest minute.  (2 marks)

Show Answers Only
  1. `text(Proof)`
  2. `67°23^{′}`
Show Worked Solution

i.   `ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2`

`a^2 + b^2` `= 5^2 + 12^2`
  `= 169`
  `= 13^2`
  `= c^2…\ text(as required.)`

MARKER’S COMMENT: Know your calculator process for producing an angle in minutes/seconds. Note >30 “seconds” rounds up to the higher “minute”.

 
ii.  `sin ∠ABC = 12/13`

`:.∠ABC` `= 67.38…°`
  `=67°22^{′}48^{″}`
  `= 67°23^{′}\ \ \ text{(nearest minute)}`

Financial Maths, STD2 F1 2005 HSC 25a

Reece is preparing his annual budget for 2006.

His expected income is:

• $90 every week as a swimming coach
• Interest earned from an investment of $5000 at a rate of 4% per annum.

His planned expenses are:

• $30 every week on transport
• $12 every week on lunches
• $48 every month on entertainment.

Reece will save his remaining income. He uses the spreadsheet below for his budget.
 

2UG-2005-25a
 

  1. Determine the values of `X`, `Y` and `Z`. (Assume there are exactly 52 weeks in a year.)   (3 marks)

At the beginning of 2006, Reece starts saving.

  1. Will Reece have saved enough money during 2006 for a deposit of $2100 on a car if he keeps to his budget? Justify your answer with suitable calculations.   (2 marks)

Show Answers Only

a.    `X =$200, Y = $1560\ text(and) \ Z = $576`

b.    `text(Reece will have saved have saved enough for a $2100 deposit.)`

Show Worked Solution

a.   `text(Interest on Investment) (X)=5000 xx 4%= $200`

`text(Transport)(Y)= 52 × 30= $1560`

`text(Entertainment) (Z)= 48 × 12= $576`

b.    `text(Total Income)= 4680 + 200= $4880`

`text(Total Expenses)= 1560 + 624 + 576= $2760`

`text(Savings)= 4880-2760= $2120`

`:.\ text(Reece will have saved enough for a $2100 deposit.)` 

Probability, STD2 S2 2005 HSC 23c

Moheb owns five red and seven blue ties. He chooses a tie at random for himself and puts it on. He then chooses another tie at random, from the remaining ties, and gives it to his brother.

  1. What is the probability that Moheb chooses a red tie for himself?  (1 mark)

Copy the tree diagram into your writing booklet.
 

2UG-2005-23c
 

  1. Complete your tree diagram by writing the correct probability on each branch.  (2 marks)
  2. Calculate the probability that both of the ties are the same colour.  (2 marks)

Show Answers Only
  1. `5/12`
  2.  
  3. `31/66`
Show Worked Solution
i. `P(R)` `= (#\ text(red ties))/(#\ text(total ties))`
    `= 5/12`

 

ii.  

 

iii. `Ptext((same colour))`

`= P(text(RR)) + P(text(BB))`

`= 5/12 × 4/11\ \ +\ \ 7/12 × 6/11`

`= 20/132 + 42/132`

`= 31/66`

Algebra, STD2 A4 2004 HSC 26a

  1. The number of bacteria in a culture grows from 100 to 114 in one hour.

     

    What is the percentage increase in the number of bacteria?   (1 mark)

  2. The bacteria continue to grow according to the formula  `n = 100(1.14)^t`, where `n` is the number of bacteria after `t` hours.

     

    What is the number of bacteria after 15 hours?   (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Time in hours $(t)$} \rule[-1ex]{0pt}{0pt} & \;\; 0 \;\;  &  \;\; 5 \;\;  & \;\; 10 \;\;  & \;\; 15 \;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of bacteria ( $n$ )} \rule[-1ex]{0pt}{0pt} & \;\; 100 \;\;  &  \;\; 193 \;\;  & \;\; 371 \;\;  & \;\; ? \;\; \\
\hline
\end{array}

  1. Use the values of `n` from  `t = 0`  to  `t = 15`  to draw a graph of  `n = 100(1.14)^t`.

     

    Use about half a page for your graph and mark a scale on each axis.   (4 marks)

  2. Using your graph or otherwise, estimate the time in hours for the number of bacteria to reach 300.   (1 mark)

Show Answers Only
  1. `text(14%)`
  2. `714`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(8.4 hours)`
Show Worked Solution

i.   `text(Percentage increase)`

COMMENT: Common ADV/STD2 content in new syllabus.

`= (114 -100)/100 xx 100`

`= 14text(%)`

 

ii.  `n = 100(1.14)^t`

`text(When)\ \ t = 15,`

`n` `= 100(1.14)^15`
  `= 713.793\ …`
  `= 714\ \ \ text{(nearest whole)}`

 

iii. 

 

iv.  `text(Using the graph)`

`text(The number of bacteria reaches 300 after)`

`text(approximately 8.4 hours.)`

Data, 2UG 2004 HSC 24a

The following graphs have been constructed from data taken from the Bureau of Meteorology website. The information relates to a town in New South Wales.

The graphs show the mean 3 pm wind speed (in kilometres per hour) for each month of the year and the mean number of days of rain for each month (raindays).

2004 24a

  1. What is the mean 3 pm wind speed for September?  (1 mark)
  2. Which month has the lowest mean 3 pm wind speed?  (1 mark)
  3. In which three-month period does the town have the highest number of raindays?  (1 mark)
  4. Briefly describe the pattern relating wind speed with the number of raindays for this town. Refer to specific months.  (2 marks)

 

Show Answers Only
  1. `text(15 km/h)`
  2. `text(February)`
  3. `text(Jan – Mar)`
  4. `text(The mean number of rain days tends to be higher)`
    `text(when the wind speed is lower and vice versa.)`
    `text(For example, the highest number of mean rain)`
    `text(days is in Feb, which is also the month of the lowest)`
    `text(mean wind speed.)`

 

 

Show Worked Solution

(i)    `text(15 km/h)`

(ii)   `text(February)`

(iii)  `text(Jan – Mar)`

(iv)  `text(The mean number of rain days tends to be higher)`

`text(when the wind speed is lower and vice versa.)`

`text(For example, the highest number of mean rain)`

`text(days is in Feb, which is also the month of the lowest)`

`text(mean wind speed.)`

Measurement, STD2 M1 2005 HSC 23b

A clay brick is made in the shape of a rectangular prism with dimensions as shown.
 

  1. Calculate the volume of the clay brick.   (1 mark)

Three identical cylindrical holes are made through the brick as shown. Each hole has a radius of 1.4 cm.  
 

  1. What is the volume of clay remaining in the brick after the holes have been made? (Give your answer to the nearest cubic centimetre.)   (3 marks)

  2. What percentage of clay is removed by making the holes through the brick? (Give your answer correct to one decimal place.)   (1 mark)

Show Answers Only

a.   `text(1512 cm)^3`

b.   `text{1364 cm}^3`

c.   `text{9.8%}`

Show Worked Solution

a.    `V= l × b × h= 21 × 8 × 9= 1512\ text(cm)^3`

b.    `text(Volume of each hole)`

`= pir^2h= pi × 1.4^2 × 8= 49.260…\ text(cm)^3`

 `:.\ text(Volume of clay still in brick)`

`= 1512 − (3 × 49.260…)`

`= 1364.219…= 1364\ text{cm}^3\ text{(nearest whole)}`

c.     `text(Percentage of clay removed)`

`= ((3 × 49.260…))/1512 × 100`

`= 9.773…= 9.8 text{%   (1 d.p.)}`

Measurement, 2UG 2005 HSC 12 MC

The shaded region represents a block of land bounded on one side by a road.

2UG-2005-12MC

What is the approximate area of the block of land, using Simpson’s rule?

(A)   `680\ text(m²)`

(B)   `760\ text(m²)`

(C)   `840\ text(m²)`

(D)   `1360\ text(m²)`

Show Answers Only

`A`

Show Worked Solution

2UG-2005-12MC Answer

`A` `≈ h/3[y_0 + 4y_1 + y_2]`
  `≈ 20/3 [19 + (4 × 15) + 23]`
  `≈ 20/3 [102]`
  `≈ 680\ text(m²)`

`=>  A`

Probability, STD2 S2 2005 HSC 11 MC

The diagram shows a spinner.
 


 

The arrow is spun and will stop in one of the six sections.

What is the probability that the arrow will stop in a section containing a number greater
than 4?

  1.    `2/5`
  2.    `2/3`
  3.    `1/3`
  4.    `1/2`
Show Answers Only

`D`

Show Worked Solution

`P\ text((number greater than 4))`

`= P(7) + P (9)`

`= 2/6 + 1/6`

`= 1/2`

`=>  D`

Financial Maths, STD2 F4 2004 HSC 25a

Tai uses the declining balance method of depreciation to calculate tax deductions for her business. Tai’s computer is valued at $6500 at the start of the 2003 financial year. The rate of depreciation is 40% per annum.

  1. Calculate the value of her tax deduction for the 2003 financial year.  (1 mark)

  2. What is the value of her computer at the start of the 2006 financial year?  (2 marks)

Show Answers Only
  1. `$2600`
  2. `$1404`
Show Worked Solution
i.  `text(Tax deduction)` `= 40 text(%) xx $6500`
  `= $2600`

 

ii. `text(Using)\  S = V_0(1 – r)^n,`

`text(Value at the start of 2006 FY)`

`= 6500(1 – 0.4)^3`

`= $1404`

Algebra, 2UG 2004 HSC 23b

Kirbee is shopping for computer software. Novirus costs `$115` more than
Funmaths. Let `x` dollars be the cost of Funmaths.

  1. Write an expression involving `x` for the cost of Novirus.  (1 mark)
  2. Novirus and Funmaths together cost `$415`. Write an equation involving
  3. `x` and solve it to find the cost of Funmaths.  (2 marks)
Show Answers Only
  1. `N = x + 115`
  2. `text(Funmaths costs $150)`
Show Worked Solution

(i)   `text(Let)\ \ N = text(cost of Novirus)`

`N = x + 115`

 

(ii)  `N + x = 415\ …\ (1)`

`text(Substitute)\ \ N = x + 115\ text{into (1) above}`

`x + 115 + x` `= 415`
`2x` `= 300`
`x` `= 150`

`:.\ text(Funmaths costs $150)`

Linear Functions, 2UA 2004 HSC 2a

The diagram shows the points  `A(text(−1) , 3)`  and  `B(2, 0)`.

The line  `l`  is drawn perpendicular to the  `x`-axis through the point  `B`.
 

Linear Functions, 2UA 2004 HSC 2a 
 

  1. Calculate the length of the interval  `AB`.   (1 mark)
  2. Find the gradient of the line  `AB`.   (1 mark)
  3. What is the size of the acute angle between the line  `AB`  and the line  `l`?   (1 mark)
  4. Show that the equation of the line  `AB`  is  `x + y − 2 = 0`.    (1 mark)
  5. Copy the diagram into your writing booklet and shade the region defined by  `x + y − 2 <= 0`.   (1 mark)
  6. Write down the equation of the line  `l`.   (1 mark)
  7. The point  `C`  is on the line  `l`  such that  `AC`  is perpendicular to  `AB`. Find the coordinates of  `C`.   (2 marks)

 

Show Answers Only
  1. `3 sqrt2\ text(units)`
  2. `-1`
  3. `45^@`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. Linear Functions, 2UA 2004 HSC 2a Answer
  6. `x = 2`
  7. `C\ (2, 6)`
Show Worked Solution
(i)    `A(-1,3)\ \ \ \ \ B(2,0)`
`AB` `= sqrt( (x_2 – x_1)^2 + (y_2 – y_1)^2 )`
  `= sqrt( (2+1)^2 + (0-3)^2 )`
  `= sqrt(9+9)`
  `= sqrt 18`
  `= 3 sqrt 2\ text(units)`

 

(ii)   `text(Gradient of)\ AB` `= (y_2 – y_1)/(x_2 – x_1)`
    `= (0 – 3)/(2 + 1)`
    `= – 1`

 

(iii)   `text(S) text(ince Gradient)\ AB = – 1`

`/_ABO = 45^@`

`:.\ text(Angle between)\ AB\ text(and)\ l`

`= 90 – 45`

`= 45^@`

 

(iv)   `text(Equation of)\ AB\ text(has)\ m = -1,\ text(through)\ \ (2,0)`
`y – y_1` `= m(x – x_1)`
`y – 0` `= -1 (x – 2)`
`y` `= -x + 2`

 

`:.\ x + y – 2 = 0\ \ \ …\ text(as required)`

 

(v)

Linear Functions, 2UA 2004 HSC 2a Answer

`text{Origin (0,0) satisfies the inequality}`

`:.\ text(Shaded area is below)\ x + y – 2 = 0`

 

(vi)   `text(Equation of)\ l`
  `x = 2`

 

(vii)   `m_(AC) xx m_(AB)` `= -1\ \ \ (AC⊥AB)`
  `m_(AC) xx -1` `= -1`
  `m_(AC)` `= + 1`

 

`text(Equation of)\ AC\ text(has)\ \ m = 1,\ text(through)\ (-1, 3),`

`y – y_1` `= m(x – x_1)`
`y – 3` `= 1 (x + 1)`
`y` `= x + 4`

 

`=>C\ text(lies on)\ y = x + 4`

`text(When)\ \ x = 2,\ \ y = 6`

`:. C\ (2, 6)`

Functions, EXT1* F1 2004 HSC 1f

Find the values of `x` for which `|\ x + 1\ |<= 5`.   (2 marks)

Show Answers Only

`-6 <= x <= 4`

Show Worked Solution

`text(Solution 1)`

`|\ x + 1\ |<= 5`

`-5` `≤x+1≤5`
`:.-6` `≤x≤4` 

 

`text(Solution 2)`

`|\ x + 1\ |<= 5`

`(x+1)^2` `<= 5^2`
` x^2 + 2x + 1` `<= 25`
 `x^2 + 2x-24` `<= 0`
`(x + 6)(x-4)` `<= 0`

 

Algebra, 2UA 2004 HSC 1f Answer

`:.\ -6 <= x <= 4`

Measurement, STD2 M6 2006 HSC 9 MC

What is the area of this triangle, to the nearest square metre?
 

 

  1. `text(152 m²)`
  2. `text(283 m²)`
  3. `text(328 m²)`
  4. `text(351 m²)`
Show Answers Only

`C`

Show Worked Solution

`text(Using the Sine rule)`

`A` `= 1/2 ab\ sin C`
  `= 1/2 xx 39 xx 47 xx sin 21^@`
  `=\ text(328.44… m²)`

 
`=>  C`

Probability, 2ADV S1 2004 HSC 1e

A packet contains 12 red,  8 green, 7 yellow and 3 black jellybeans.

One jellybean is selected from the packet at random.

What is the probability that the selected jellybean is red or yellow?   (2 marks)

Show Answers Only

`19/30`

Show Worked Solution

`text(12 R,  8 G,  7 Y,  3 B)`

`text(Total jellybeans) = 30`

`P text{(R or Y)}=\ text{(# Red + Yellow)}/text(Total jellybeans)= (12 + 7)/30= 19/30`

Functions, 2ADV F1 2004 HSC 1c

Solve   `(x-5)/3-(x+1)/4 = 5`.   (2 marks)

Show Answers Only

`83`

Show Worked Solution
`(x-5)/3-(x+1)/4` `= 5`
`12((x-5)/3)-12((x+1)/4)` `= 12 xx 5`
`4x-20-3x-3` `= 60`
`x-23` `= 60`
`:. x` `= 83`

Financial Maths, STD2 F1 2006 HSC 5 MC

A salesman earns $200 per week plus $40 commission for each item he sells.

How many items does he need to sell to earn a total of $2640 in two weeks?

  1. 33
  2. 56
  3. 61
  4. 66
Show Answers Only

`B`

Show Worked Solution

`text(Let items sold) = n`

`text{Wages over 2 weeks}\ (w)`

`= (2 xx 200) + 40n=400 + 40n` 

`text(Find)\ n\ text(when)\ w = 2640:`

`2640` `= 400 + 40n`
`40n` `= 2240`
`n` `= 56`

 
`=>  B`

Measurement, STD2 M6 2006 HSC 3 MC

The angle of depression of the base of the tree from the top of the building is 65°. The height of the building is 30 m.

How far away is the base of the tree from the building, correct to one decimal place?
 


 

  1. 12.7 m
  2. 14.0 m
  3. 33.1 m
  4. 64.3 m
Show Answers Only

`B`

Show Worked Solution
 

`text(Let)\ d =\ text(distance from base to tree)`

`tan25^@` `=d/30`  
`:.d` `=30 xx tan25^@`  
  `=13.98…\ text{m}`  

 
`=>  B`

Probability, STD2 S2 2006 HSC 1 MC

The probability of an event occurring is `9/10.`

Which statement best describes the probability of this event occurring?

  1.    The event is likely to occur.
  2.    The event is certain to occur.
  3.    The event is unlikely to occur.
  4.    The event has an even chance of occurring.
Show Answers Only

`A`

Show Worked Solution

`text(The event is highly likely to occur)`

`text(but not certain.)`

`=>  A`

Probability, STD2 S2 2005 HSC 23a

There are 100 tickets sold in a raffle. Justine sold all 100 tickets to five of her friends. The number of tickets she sold to each friend is shown in the table.
 

  1. Justine claims that each of her friends is equally likely to win first prize.

     

    Give a reason why Justine’s statement is NOT correct.   (1 mark)

  2. What is the probability that first prize is NOT won by Khalid or Herman?   (2 marks)

Show Answers Only
  1. `text(The claim is incorrect because each of her friends)`
    `text(bought a different number of tickets and therefore)`
    `text(their chances of winning are different.)`
  2. `69/100`
Show Worked Solution

i.    `text(The claim is incorrect because each of her friends bought)`

`text(a different number of tickets and therefore their chances of)`

`text(winning are different.)`

 

ii.  `text(Number of tickets not sold to K or H)`

`= 45 + 10 + 14`

`= 69`
 

`:.\ text(Probability 1st prize NOT won by K or H)`

`= 69/100`

Measurement, STD2 M1 2004 HSC 23a

The diagram shows the shape of Carmel’s garden bed. All measurements are in
metres.

  1. Show that the area of the garden bed is 57 square metres.   (2 marks)

  2. Carmel decides to add a 5 cm layer of straw to the garden bed.

     

    Calculate the volume of straw required. Give your answer in cubic metres.   (2 marks)

  3. Each bag holds 0.25 cubic metres of straw.

     

    How many bags does she need to buy?   (2 marks)

  4. A straight fence is to be constructed joining point A to point B.

     

    Find the length of this fence to the nearest metre.   (2 marks)

Show Answers Only

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

b.   `text(2.85 m³)`

c.   `text(She needs to buy 12 bags)`

d.   `8\ text{m  (nearest metre)}`

Show Worked Solution

a.    `text(Area of)\ Delta ABC= 1/2 xx b xx h= 1/2 xx 10 xx 5.1=25.5\ text(m²)`

`text(Area of)\ Delta ACD= 1/2 xx 10 xx 6.3= 31.5\ text(m²)`

`:.\ text(Total Area)= 25.5 + 31.5= 57\ text(m² … as required)`

b.    `V= Ah= 57 xx 0.05= 2.85\ text(m³)`

c.    `text(Bags to buy)= 2.85/0.25= 11.4`

`:.\ text(She needs to buy 12 bags.)`

d.   `text(Using Pythagoras,)`

`AB^2= 6.0^2 + 5.1^2= 36 + 26.01= 62.01`

`AB= 7.874…=8\ text{m  (nearest metre)}`

Measurement, STD2 M7 2005 HSC 4 MC

The diagram is a scale drawing of a butterfly.
 

2UG-2005-4MC 
 

What is the actual wingspan of the butterfly?

  1.    2.5 cm
  2.    3 cm
  3.    15 cm
  4.    8.75 cm
Show Answers Only

`B`

Show Worked Solution

`text(Wingspan is 3 times the scale distance)`

`text(that equals 1 cm.)`

`:.\ text(Wingspan)` `= 3 × 1`
  `= 3\ text(cm)`

`=> B`