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Number and Algebra, NAP-J2-12

Trevor has poured milk into some glasses.

He makes them all exactly one quarter full.

Trevor counts by quarters to find out, in total, how many full cups of milk he has.
 

  Which number does Trevor count next?

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

`1 1/2`

Show Worked Solution

`text(Add)\ 1/4\ text(for each additional cup.)`

`:.\ text(Next number)` `= 1 1/4 + 1/4`
  `= 1 1/2`

 

CHEMISTRY, M4 2009 HSC 20

  1. Calculate the mass of ethanol \(\ce{C2H6O}\) that must be burnt to increase the temperature of 210 g of water by 65°C, if exactly half of the heat released by this combustion is lost to the surroundings.
  2. The heat of combustion of ethanol is 1367 kJ mol −1.  (3 marks)

  3. What are TWO ways to limit heat loss from the apparatus when performing a first-hand investigation to determine and compare heat of combustion of different liquid alkanols?  (1 mark)

Show Answers Only

a.    3.85 grams

b.    Answers could include two of the following:

  • Use of an insulated vessel (Styrofoam cup)
  • Place the vessel as close as safely possible to the Bunsen’s flame.
  • Use a lid for the beaker
Show Worked Solution

a.    \(q=mC \Delta T = 210 \times 4.18 \times 65 = 57\ 057\ \text{J} = 57.057\ \text{kJ}\)

\(\ce{n(C2H5OH) = \dfrac{57.057}{1367} = 0.04174\ \text{mol}}\)

\(\ce{m(C2H5OH) = n \times MM = 0.04174 \times 46.068 = 1.923\ \text{g}}\)

Since half of the heat is lost to environmental surroundings.

\(\ce{m(C2H5OH)_{\text{init}} = 2 \times 1.923= 3.85\ \text{g}}\)
 

b.    Answers could include two of the following:

  • Use of an insulated vessel (Styrofoam cup)
  • Place the vessel as close as safely possible to the Bunsen’s flame.
  • Use a lid for the beaker

CHEMISTRY, M4 2013 HSC 27

A 0.259 g sample of ethanol is burnt to raise the temperature of 120 g of an oily liquid, as shown in the graph. There is no loss of heat to the surroundings.
 

Using the information shown on the graph, calculate the specific heat capacity of the oily liquid. The heat of combustion of ethanol is 1367 kJ mol–1.   (4 marks)

Show Answers Only

\(2.13 \times 10^{3}\ \text{J kg}^{-1}\text{K}^{-1}\)

Show Worked Solution

\(\ce{MM(C2H5OH) = 2 \times 12.01 + 6 \times 1.008 + 16 = 46.068}\)

\(\ce{n(C2H5OH) = \dfrac{\text{m}}{\text{MM}} = \dfrac{0.259}{46.068} = 5.622 \times 10^{-3}\ \text{mol}}\)

\(\text{Heat combustion}(q)\ = 1367 \times 5.622 \times 10^{-3} = 7.685\ \text{kJ} = 7685\ \text{J}\)

Mean mark 59%.
\(q\) \(=mc \Delta T\)  
\(c\) \(=\dfrac{q}{m \Delta T}\)  
  \(=\dfrac{7685}{0.120 \times (50-20)} \)  
  \(=2.13 \times 10^{3}\ \text{J kg}^{-1}\text{K}^{-1} \)  

Statistics, NAP-J3-CA01

Four students recorded the number of times they participated in different sports in one week in the table below.
 

In total, which sport was participated in the most times?

Swimming Soccer Tennis Netball
 
 
 
 
Show Answers Only

`text(Swimming)`

Show Worked Solution

`text(Times each sport was played:)`

`text(Swimming – 11)`

`text(Soccer – 10)`

`text(Tennis – 8)`

`text(Netball – 5)`

`:.\ text(Swimming was participated in the most.)`

L&E, 2ADV E1 SM-Bank 9

Solve  `log_2(6-x)-log_2(4-x) = 2`  for `x`, where  `x < 4`.  (2 marks)

Show Answers Only

`10/3`

Show Worked Solution

`text(Simplify using log laws:)`

`log_2((6-x)/(4-x))` `= 2`
`2^2` `= (6-x)/(4-x)`
`16-4x` `= 6-x`
`3x` `= 10`
`:. x` `= 10/3`

L&E, 2ADV E1 SM-Bank 7

Solve the equation  `2 log_3(5)-log_3 (2) + log_3 (x) = 2`  for  `x.`  (2 marks)

Show Answers Only

`18/25`

Show Worked Solution
`log_3 (5)^2-log_3 (2) + log_3 (x)` `= 2`
`log_3 (25x)-log_3 (2)` `=2`
`log_3 ((25 x)/2)` `= 2`
`(25x)/2` `= 3^2`
`:. x` `= 18/25`

L&E, 2ADV E1 SM-Bank 2 MC

If  `y = log_a (7x - b) + 3`, then `x` is equal to

  1. `1/7 a^(y - 3) + b`
  2. `(y - 3)/(log_a(7 - b))`
  3. `1/7 (a^(y - 3) + b)`
  4. `a^(y - 3) - b/7`
Show Answers Only

`C`

Show Worked Solution
`y – 3` `= log_a (7x – b)`
`a^(y – 3)` `= 7x – b`
`a^(y – 3) + b` `= 7x`
`:. x` `= 1/7 (a^(y – 3) + b)`

`=>   C`

GEOMETRY, FUR1 SM-Bank 32 MC

Makoua and Macapá are two towns on the equator.

The longitude of Makoua is 16°E and the longitude of Macapá is 52°W.

How far apart are these two towns if the radius of Earth is approximately 6400 km?

A.  `4000\ text(km)`

B.  `7600\ text(km)`

C.  `1\ 367\ 200\ text(km)`

D.  `1\ 447\ 600\ text(km)`

E.   `2\ 734\ 400\ text(km)`

Show Answers Only

`B`

Show Worked Solution

2UG-2006-19MC Answer

`text(Angular Difference)` `= 52 + 16`
  `= 68°`

 

`text(Arc)\ AB` `=\ text(Distance between two towns)`
  `= 68/360 xx 2 xx pi xx r`
  `= 68/360 xx 2 xx pi xx 6400`
  `= 7595.67…\ text(km)`

 
`=>  B`

GEOMETRY, FUR1 SM-Bank 18 MC

Stockholm is located at  59°N 18°E  and Darwin is located at  13°S 131°E. 

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

  1. 184 minutes
  2. 188 minutes
  3. 288 minutes
  4. 452 minutes
  5. 596 minutes
Show Answers Only

`D`

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)`

 
`⇒ D`

GEOMETRY, FUR1 SM-Bank 15 MC

Which expression will give the shortest distance, in kilometres, between Mount Isa (20°S  140°E)  and Tokyo (35°N  140°E)?

  1. `15/360 xx 2 xx pi xx 6400`
  2. `55/360 xx 2 xx pi xx 6400`
  3. `125/360 xx 2 xx pi xx 6400`
  4. `140/360 xx 2 xx pi xx 6400`
  5. `305/360 xx 2 xx pi xx 6400`
Show Answers Only

`B`

Show Worked Solution

`text(Tokyo is 35° North of the equator, Mt Isa 20° South)`

`text(Arc length) = 55/360 xx 2 xx pi xx 6400`

`=>  B`

GEOMETRY, FUR2 SM-Bank 25

A ship sails due South from Channel-Port-aux-Basques, Canada,  47°N  59°W  to Barbados, 13°N  59°W.

How far did the ship sail, to the nearest kilometre? Assume that the radius of Earth is 6400 km.    (2 marks)

Show Answers Only

 `3798\ text{km (nearest km)}`

Show Worked Solution
`text(Angular difference in latitude)` `=47` `-13`
  `=34^@`

 
`text{(No difference in longitude)}`
 

`:.\ text(Distance)` `=34/360` `xx 2 pi r`
  `=34/360` `xx 2` `xx pi` `xx 6400`
  `= 3797.836…`
  `= 3798` `text{km   (nearest km)}`

GEOMETRY, FUR2 SM-Bank 15

Osaka is at  34°N, 135°E, and Denver is at  40°N, 105°W. 

  1. Show that there is a 16-hour time difference between the two cities.
    (Ignore time zones.)    (1 mark)
  2. John lives in Denver and wants to ring a friend in Osaka. In Denver it is 9 pm Monday.     

     

    What time and day is it in Osaka then?   (1 mark)

  3. John’s friend in Osaka sent him a text message which happened to take 14 hours to reach him. It was sent at 10 am Thursday, Osaka time.

     

    What was the time and day in Denver when John received the text?    (1 mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `1\ text(pm Tuesday)`
  3. `8\ text(am Thursday)`
Show Worked Solution
a.   `text(Longitude difference)` `= 135 + 105`
    `= 240^@`

 

`text(Time difference)` `= 240/15`
  `= 16\ text(hours   … as required)`

 

b.  `text(Denver is behind Osaka time because it is further west.)`

`:.\ text(Time in Osaka)` `= 9\ text(pm Monday plus 16 hours)`
  `= 1\ text(pm Tuesday)`

 

c.  `text(Denver is 16 hours behind Osaka)`

`:.\ text(John will receive the text at 10 am Thursday less 16)`

`text{plus 14 hours (i.e. 8 am Thursday.)}`

GEOMETRY, FUR2 SM-Bank 14

Pontianak has a longitude of 109°E, and Jarvis Island has a longitude of 160°W.

Both places lie on the Equator

  1. Find the shortest great circle distance between these two places, to the nearest kilometre. You may assume that the radius of the Earth is 6400 km.    (2 marks)
  2. The position of Rabaul is 4° to the south and 48° to the west of Jarvis Island. What is the latitude and longitude of Rabaul?     (1 mark)
Show Answers Only
  1. `10\ 165\ text(km)\ \ \ text{(nearest km)}`
  2. `152^@ text(E)`
Show Worked Solution
a.   `text(Longitude difference)` `= 109 + 160`
    `= 269^@`

 

`=> text(Shortest distance)\ text{(by degree)}` `= 360 – 269`
  `= 91^@`

 

`:.\ text(Shortest distance)` `= 91/360 xx 2 pi r`
  `= 91/360 xx 2 xx pi xx 6400`
  `= 10\ 164.79…`
  `=10\ 165\ text(km)\ text{(nearest km)}`

 

b.   `text(Latitude)`
  `4^@\ text(South of Jarvis Island)`
  `text(S)text(ince Jarvis Island is on equator)`
  `=>\ text(Latitude is)\ 4^@ text(S)`
   
  `text(Longitude)`
  `text(Jarvis Island is)\ 160^@ text(W)`
  `text(Rubail is)\ 48^@\ text(West of Jarvis Island, or 208° West)`
  `text(which is)\ 28^@\ text{past meridian (180°)}`

 

`=>\ text(Longitude)` `= (180\ -28)^@ text(E)`
  `= 152^@ text(E)`

 

`:.\ text(Position is)\ (4^@text{S}, 152^@text{E})`

GEOMETRY, FUR2 SM-Bank 13

Melbourne is located at (38°S, 145°E) and Dubai is located at (24°N, 55°E).

  1. Calculate the difference in longitude between Melbourne and Dubai.  (1 mark)
  2. Show that the time difference between Melbourne and Dubai is 6 hours.  (1 mark)
  3. A plane leaves Melbourne on Friday at 11.30 pm. The flight time to Dubai is 15 hours.

     

    What will be the time and the day in Dubai when the plane is due to land?  (1 marks)

Show Answers Only
  1. `90^@`
  2. `text(See Worked Solutions)`
  3. `text(8:30 am on Saturday)`
Show Worked Solution

a.   `text(Difference in longitude)`

`= 145 – 55`

`= 90^@`

 

b.   `text(S)text(ince)\ 1^@ = 4\ text(min time difference,)`

`text(Time difference)`

`= 90 xx 4`

`= 360\ text(mins)`

`= 6\ text(hours  … as required)`

 

c.    `text(Arrival time)` `= 11:30 + 15\ text(hours)`
    `= 14:30\ text{(Melbourne time)}`

 

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

`= 14:30 – 6\ text(hours)`

`= 8:30\ text(am on Saturday)`

GEOMETRY, FUR2 SM-Bank 12

This diagram represents Earth. `O` is at the centre, and `A` and `B` are points on the surface.
 

2UG-2005-27b
 

Find the shortest great circle distance from `A` to `B`.

Give your answer in to the nearest km.   (2 marks)

Show Answers Only

`text{4803 km  (nearest km)}`

Show Worked Solution

`A : 35^@text(N)\ 20^@text(E)\ \ \ \ B:8^@text(S)\ 20^@text(E)`

HSC 2005 27b

`text(Angular difference)` `= 35^@ + 8^@`
  `= 43^@`

 

`:.\ text(Distance from)\ A\ text(to)\ B`

`= 43/360 xx 2 xx pi xx r`

`= 43/360 xx 2 xx pi xx 6400`

`= 4803.1…`

`= 4803\ text{km  (nearest km)}`

Probability, MET1 2012 VCAA 8b

The probability density function `f` of a random variable `X` is given by
 

`qquad qquad f(x) = {((x + 1)/12, 0 <= x <= 4), (\ \ 0, text{otherwise}):}`
 

Find the value of `b` such that  `text(Pr) (X <= b) = 5/8.`  (3 marks)

Show Answers Only

`3`

Show Worked Solution
`1/12 int_0^b (x + 1)\ dx` `= 5/8`
`[1/2 x^2 + x]_0^b` `= 15/2`
`1/2b^2 + b` `= 15/2`
`b^2 + 2b – 15` `= 0`
`(b + 5) (b – 3)` `= 0`

 

`:. b = 3,\ \ \ b in (0, 4)`

Calculus, MET1 2006 VCAA 3a

Let  `y = x tan(x)`. Evaluate  `(dy)/(dx)`  when `x = pi/6`.   (3 mark)

Show Answers Only

`(3sqrt3 + 2pi)/9`

Show Worked Solution

`text(Using the product rule:)`

MARKER’S COMMENT: A common error was not knowing the exact trig function values.

`d/(dx)(uv)` `= u^{prime}v + uv^{prime}`
`(dy)/(dx)` `= tan(x) + x/(cos^2(x))`

 

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

`(dy)/(dx)` `= tan(pi/6) + (pi/6)/((cos(pi/6))^2)`
  `= 1/sqrt3 + pi/6 xx (2/sqrt3)^2`
  `= sqrt3/3 + (4pi)/18`
  `= (3sqrt3 + 2pi)/9`

Calculus, MET1 2009 VCAA 1b

For  `f(x) = (cos(x))/(2x + 2)`  find  `f prime (pi)`.   (3 marks)

Show Answers Only

`1/(2 (pi + 1)^2)`

Show Worked Solution

`text(Using Quotient Rule:)`

MARKER’S COMMENT: A majority of students did not substitute  `x=pi`  correctly.
`(g/h)^{prime}` `= (g^{prime} h – gh^{prime})/h^2`
`f^{prime}(x)` `= (-sin (x) (2x + 2)-2 cos (x))/(2x + 2)^2`
`:. f^{prime}(pi)` `= (-sin (pi) (2pi + 2)-2 cos (pi))/(2pi + 2)^2`
  `= (0-2 (-1))/[2 (pi + 1)]^2`
  `= 2/(4(pi + 1)^2)`
  `= 1/(2 (pi + 1)^2)`

Calculus, MET1 SM-Bank 5

The function with rule  `f(x)`  has derivative  `f^{prime}(x) =  cos\ 3x`.

If  `f(pi/6) = 1,`  find  `f(x).`  (3 marks)

Show Answers Only

`f(x)= 1/3 sin\ 3x + 2/3`

Show Worked Solution
`int f(x)\ dx` `=int cos\ 3x\ dx`
  `= 1/3 sin\ 3x + c`
`f(pi/6)`  `= 1/3\ [sin\ (3 xx pi/6) ]+ c`
`1`  `= 1/3\ sin\ pi/2+c`
`c`  `= 2/3`

 
`:.f(x)= 1/3 sin\ 3x + 2/3`

Calculus, MET2 2007 VCAA 20 MC

The average value of the function  `y = tan (2x)`  over the interval  `[0, pi/8]`  is

  1. `2/pi log_e (2)`
  2. `pi/4`
  3. `1/2`
  4. `4/pi log_e 2`
  5. `8/pi`
Show Answers Only

`A`

Show Worked Solution
`y_text(average)` `= 1/(pi/8 – 0) int_0^(pi/8) (tan 2x)\ dx`
  `= 8/pi  int_0^(pi/8) (tan 2x)\ dx`
  `= (2 log_e (2))/pi`

`=>   A`

Probability, MET2 2007 VCAA 18 MC

The heights of the children in a queue for an amusement park ride are normally distributed with mean 130 cm and standard deviation 2.7 cm. 35% of the children are not allowed to go on the ride because they are too short.

The minimum acceptable height correct to the nearest centimetre is

  1. 126
  2. 127
  3. 128
  4. 129
  5. 130
Show Answers Only

`D`

Show Worked Solution

`X = text(height)`

`X ∼ N (130, 2.7^2)`

vcaa-2007-meth2-18

`text(Pr)(X < a)` `= 0.35\ \ \ [text(CAS: invNorm)(0.35, 130, 2.7)]`
`a` `= 128.96`

 
`=>   D`

Calculus, MET2 2007 VCAA 13 MC

For the graph of  `y = 4x^3 + 27x^2-30x + 10`  the subset of `R` for which the gradient is negative is given by the interval

  1. `(0.5, 5.0)`
  2. `(-4.99, 0.51)`
  3. `(-oo, 1/2)`
  4. `(-5, 1/2)`
  5. `(2.25, oo)`
Show Answers Only

`D`

Show Worked Solution
`y` `=4x^3 + 27x^2 – 30x + 10`
`y′` `=12x^2 + 54x – 30`

 

`text(Sketch the graph:)`

vcaa-2007-meth2-13

`:.\ text(Gradient is negative for)\ \  x in (– 5, 1/2).`

`=>   D`

Calculus, MET2 2007 VCAA 12 MC

Let  `f: R -> R`  be a differentiable function such that

  • `f prime(3) = 0`
  • `f prime(x) < 0`  when  `x < 3`  and when  `x > 3`

When  `x = 3`, the graph of  `f`  has a

  1. local minimum
  2. local maximum
  3. stationary point of inflection
  4. point of discontinuity
  5. gradient of 3
Show Answers Only

`C`

Show Worked Solution

`text(Use a gradient table:)`

vcaa-2007-meth2-12

`:.\ text(Point of inflection at)\ \ x = 3.`

`=>   C`

L&E, 2ADV E1 SM-Bank 11 MC

Solve the equation  `e^(4x) - 5e^(2x) + 4 = 0`  for `x`

  1. `x= 1, 4`
  2. `x= – 4, – 1`
  3. `x= 0, log_e 2`
  4. `x= – log_e 2, 0, log_e 2 `
Show Answers Only

`C`

Show Worked Solution

`e^(4x) – 5e^(2x) + 4 = 0`

`text(Let)\ \ X=e^(2x)`

`X^2-5X+4` `=0`
`(X-4)(X-1)` `=0`
`X` `=4 or 1`

 

`:.e^(2x)` `=4` `e^(2x)` `=1`
`2x` `=log_e 4` `x` `=0`
`x` `=(2log_e 2)/2`    
  `=log_e 2`    

`=> C`

Graphs, MET2 2007 VCAA 10 MC

The graph of  `y = kx-3`  intersects the graph of  `y = x^2 + 8x`  at two distinct points for

  1. `k = 11`
  2. `k > 8 + 2 sqrt 3 or k < 8-2 sqrt 3`
  3. `5 <= k <= 6`
  4. `8-2 sqrt 3 <= k <= 8 + 2 sqrt 3`
  5. `k = 5`
Show Answers Only

`B`

Show Worked Solution

`text(Intersection occurs when:)`

`kx-3` `= x^2 + 8x`
`x^2 + (8-k)x + 3` `= 0`

 

`text(For 2 points of intersection:)`

`Delta` `> 0`
`(8-k)^2-4 (3)` `> 0`

 

`:. k < 8-2 sqrt 3\  uu\  k > 8 + 2 sqrt 3`

`=>   B`

Calculus, MET2 2007 VCAA 9 MC

Let  `k = int_-2^-1 1/x\ dx`, then `e^k` is equal to

  1. `log_e(2)`
  2. `1`
  3. `2`
  4. `e`
  5. `1/2`
Show Answers Only

`E`

Show Worked Solution

`text(The integral can be seen as the negative)`

`text(equivalent of the area under)\ \ y=1/x`

`text(between)\ \ x=1 and x=2.`

`k` `= – [log_e x]_1^2`
  `=-(log_e(2)-log_e(1))`
  `= – log_e (2)`
  `= log_e (1/2)`
`:. e^k` `= e^(log_e(1/2))`
  `= 1/2`

 
`=>   E`

Calculus, MET2 2007 VCAA 4 MC

The average rate of change of the function with rule  `f(x) = x^3 - sqrt (x + 1)`  between  `x = 0`  and  `x = 3`  is

A.   `0`

B.   `12`

C.   `26/3`

D.   `25/3`

E.   `8`

Show Answers Only

`C`

Show Worked Solution

`text(Define)\ \ f(x) = x^3 – sqrt (x + 1)\ \ text(on CAS)`

`text(Average ROC)` `= (f(3) – f(0))/(3 – 0)`
  `= 26/3`

`=>   C`

Graphs, MET2 2007 VCAA 1 MC

The linear function  `f: D -> R,\ f(x) = 6 - 2x`  has range  `text([−4, 12]).`

The domain `D` is

  1. `[– 3, 5]`
  2. `[– 5, 3]`
  3. `R`
  4. `[– 14, 18]`
  5. `[– 18, 14]`
Show Answers Only

`A`

Show Worked Solution
`12` `= 6 – 2x`
`x` `= – 3`

 

`- 4` `= 6 – 2x`
`x` `= 5`

 

`:. x in [– 3, 5]`

`=>   A`

Algebra, MET1 SM-Bank 24

The polynomial  `p(x) = x^3-ax + b`  has a remainder of 2 when divided by  `(x-1)`  and a remainder of 5 when divided by  `(x + 2)`.  

Find the values of  `a`  and  `b`.   (3 marks)

Show Answers Only
`a` `= 4`
`b` `= 5`
Show Worked Solution
`p(x)` `= x^3-ax + b`
`P(1)` `= 2`
`1-a + b` `= 2`
`b` `= a+1\ \ \ …\ text{(1)}`
`P (-2)` `= 5`
`-8 + 2a + b` `= 5`
`2a + b` `= 13\ \ \ …\ text{(2)}`

 

`text(Substitute)\ \ b = a+1\ \ text(into)\ \ text{(2)}`

`2a + a+1` `= 13`
`3a` `= 12`
`:. a` `= 4`
`:. b` `= 5`

Algebra, MET1 SM-Bank 23

The graph of  `P(x) = x^2 + ax + b`  cuts the `x`-axis when  `x=2.`  When  `P(x)`  is divided by  `x + 1`, the remainder is 18.

Find the values of  `a`  and  `b`.   (3 marks)

Show Answers Only

`a = -7\ \ text(and)\ \ b = 10`

Show Worked Solution

`P(x) = x^2 + ax + b`

`text(S)text(ince the graph cuts the)\ xtext(-axis at)\ \ x = 2,`

`P(2)` `=0`  
`2^2 + 2a + b` `= 0`  
`2a + b` `= -4`       `…\ (1)`

 
`P(-1) = 18,`

`(-1)^2-a + b` `= 18`  
`-a + b` `= 17`    `…\ (2)`

 
`text(Subtract)\ \ (1) − (2),`

`3a` `= -21`
`a` `= -7`

 
`text(Substitute)\ \ a = -7\ \ text{into (1),}`

`2(-7) + b` `= -4`
`b` `= 10`

 

`:.a = -7\ \ text(and)\ \ b = 10`

Graphs, MET2 2008 VCAA 22 MC

The graph of the function  `f` with domain  `[0, 6]`  is shown below.

VCAA 2008 22mc

Which one of the following is not true?

  1. The function is not continuous at  `x = 2`  and  `x = 4.`
  2. The function exists for all values of `x` between `0` and `6.`
  3. `f(x) = 0`  for  `x = 2`  and  `x = 5.`
  4.  The function is positive for  `x ∈ [0, 5).`
  5. The gradient of the function is not defined at  `x = 4.`
Show Answers Only

`C`

Show Worked Solution

`f(x) > 0\ \ text(for)\ \ x = 2`

`:.\ text(Option)\ \  C\ \ text(is not true)`

`=>   C`

Probability, MET2 2008 VCAA 14 MC

The minimum number of times that a fair coin can be tossed so that the probability of obtaining a head on each trial is less than 0.0005 is

A.     `8`

B.     `9`

C.   `10`

D.   `11`

E.   `12`

Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ X = text(Number of heads,)`

`X ∼ text(Bi) (n, 1/2)`

`text(Pr) (X = n)` `< 0.0005`
`((n), (n)) (1/2)^n (1/2)^0` `< 0.0005`
`n` `> 10.97`

`:. n_min = 11`

`=>   D`

Probability, MET2 2008 VCAA 13 MC

According to a survey, 30% of employed women have never been married.

If 10 employed women are selected at random, the probability (correct to four decimal places) that at least 7 have never been married is

A.   `0.0016`

B.   `0.0090`

C.   `0.0106`

D.   `0.9894`

E.   `0.9984`

Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ X =\ text(Number who have not been married)`

`X∼\ text(Bi) (10, 0.3)`

`text(Pr) (X >= 7)\ \ \ [text(CAS: binomCdf) (10, 0.3, 7, 10)]`

`= 0.0106`

`=>   C`

Probability, MET2 2008 VCAA 5 MC

Let `X` be a discrete random variable with a binomial distribution. The mean of `X` is 1.2 and the variance of `X` is 0.72

The values of `n` (the number of independent trials) and `p` (the probability of success in each trial) are

A.   `n = 4,\ \ \ \ \ p = 0.3`

B.   `n = 3,\ \ \ \ \ p = 0.6`

C.   `n = 2,\ \ \ \ \ p = 0.6`

D.   `n = 2,\ \ \ \ \ p = 0.4`

E.   `n = 3,\ \ \ \ \ p = 0.4`

Show Answers Only

`E`

Show Worked Solution

`X∼\ text(Bi) (n, p)`

`text(E) (X)` `= 1.2`
`np` `= 1.2\ …\ (1)`
`text(Var) (X)` `= 0.72`
`np (1 – p)` `= 0.72\ …\ (2)`

 

`text(Solve simultaneous equations:)`

`:. n = 3,\ \ p = 0.4`

`=>   E`

Calculus, MET2 2008 VCAA 3 MC

The average value of the function with rule  `f(x) = log_e (3x + 1)`  over the interval  `[0, 2]`  is

  1. `(log_e(7))/2`
  2. `log_e(7)`
  3. `(7 log_e (7))/3 - 2`
  4. `(7 log_e (7) - 6)/6`
  5. `(35 log_e(7) - 12)/18`
Show Answers Only

`D`

Show Worked Solution
MARKER’S COMMENT: Almost a quarter of students found the average rate of change!
`y_text(avg)` `= 1/(2 – 0) int_0^2 log_e (3x + 1)\ dx`
  `= (7 log_e(7) – 6)/6`

 
`=>   D`

Calculus, MET2 2008 VCAA 1 MC

VCAA 2008 1mc

The area under the curve  `y = sin (x)`  between  `x = 0`  and  `x  = pi/2`  is approximated by two rectangles as shown.

This approximation to the area is

A.   `1`

B.   `pi/2`

C.   `((sqrt 3 + 1) pi)/12`

D.   `0.5`

E.   `((sqrt 3 + 1) pi)/6`

Show Answers Only

`C`

Show Worked Solution

`text(Rectangle width) = pi/6`

`text(Area)` `~~ pi/6 [sin (pi/6) + sin (pi/3)]`
  `~~pi/6(sqrt3/2 + 1/2)`
  `~~ (pi (sqrt 3 + 1))/12`

`=>   C`

Algebra, 2UG AM3 SM-Bank 06

Solve these simultaneous equations to find the values of  `x`  and  `y`.

 `4x - 2y = 18`

`3x + 2y = 10`   (3 marks)

Show Answers Only

`x = 4,\ \ y = -1`

Show Worked Solution

`text(Solution 1 – Eliminations)`

`4x – 2y = 18\ …\ text{(i)}`

`3x +2y = 10\ …\ text{(ii)}`

`text{(i)} + text{(ii)}`

`7x` `= 28`
`x` `= 4`

`text(Substitute)\ \ x = 4\ \ text(into)\ text{(i)}`

`4 xx 4 – 2y` `= 18`
`- 2y` `= 2`
`y` `= -1`

`:.\ text(Solution is)\ \ x = 4\ \ text(and)\ y = -1`

 

`text(Alternative Solution – Substitution)`

`4x – 2y = 18\ …\ text{(i)}`

`3x + 2y = 10\ …\ text{(ii)}`

`text{Divide (i) by 2}`

`2x – y` `= 9`
`y` `= 2x – 9`

`text(Substitute)\ \ y = 2x – 9\ \ text(into)\ text{(ii)}`

`3x + 2(2x – 9)` `= 10`
`3x + 4x – 18` `= 10`
`7x` `= 28`
`x` `= 4`

`text(Substitute)\ \ x= 4\ \ text{into (i)}`

`4 xx 4 – 2y` `= 18`
`- 2y` `= 2`
`y` `= -1`

Algebra, MET1 SM-Bank 25

Solve these simultaneous equations to find the values of  `x`  and  `y`.    (3 marks)

`y = 2x + 1`

`x-2y-4 = 0`

Show Answers Only

`x = -2,\ y = -3`

Show Worked Solution

`text(Solution 1 – Substitution)`

`y = 2x + 1\ \ \ \ \ …\ text{(i)}`

`x-2y-4 = 0\ \ \ \ \ …\ text{(ii)}`
 

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

`x-2(2x + 1)-4` `= 0`
`x-4x-2-4` `= 0`
`-3x- 6` `= 0`
`3x` `= -6`
`x` `= -2`

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

`y = 2(–2) + 1 = -3`

`:.\ text(Solution is)\ x = -2,\ y = -3`
 

`text(Alternative Solution – Elimination)`

`y = 2x + 1\ \ \ \ \ …\ text{(i)}`

`x-2y-4 = 0\ \ \ \ \ …\ text{(ii)}`
 

`text(Multiply)\ text{(i)} xx 2`

`2y` `= 4x + 2`
`-4x + 2y-2` `= 0\ \ \ \ \ …\ text{(iii)}`

 
`text{Add  (ii) + (iii)}`

`-3x-6` `= 0`
`x` `= -2`
`y` `= -3\ \ text{(see Solution 1)}`

Calculus, MET1 SM-Bank 28

The function  `f`  has the rule  `f(x) = 1 + 2 cos x`.

  1. Show that the graph of  `y = f(x)`  cuts the `x`-axis at  `x = (2 pi)/3`.   (1 mark)

  2. Sketch the graph  `y = f(x)`  for  `x  in [-pi,pi]`  showing where the graph cuts each of the axes.   (2 marks)

  3. Find the area under the curve  `y = f(x)`  between  `x = -pi/2`  and  `x = (2 pi)/3`.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}` 
  2.  
       
  3. `((7 pi)/6 + sqrt 3 + 1)\ text(u²)`
Show Worked Solution

a.   `f(x) = 1 + 2 cos x`

`f(x)\ text(cuts the)\ x text(-axis when)\ f(x) = 0`

`1 + 2 cos x` `= 0`
`2 cos x` `=-1`
 `cos x` `= -1/2`

 
`:.  x = (2 pi)/3\ …\ text(as required)`

 

b.   2UA HSC 2006 7b

 

c.  `text(Area)` `= int_(-pi/2)^((2 pi)/3) 1 + 2 cos x\ \ dx`
  `= [x + 2 sin x]_(-pi/2)^((2 pi)/3)`
  `= [((2 pi)/3 + 2 sin­ (2 pi)/3)-((-pi)/2 + 2 sin­ (-pi)/2)]`
  `= ((2 pi)/3 + 2 xx sqrt 3/2)-((-pi)/2 +2(- 1))`
  `= (2 pi)/3 + sqrt(3) + pi/2 + 2`
  `= ((7 pi)/6 + sqrt(3) + 2)\ text(u²)`

Graphs, MET1 SM-Bank 27

The graph shown is  `y = A sin bx`.

  1. Write down the value of  `A`.   (1 mark)

  2. Find the value of  `b`.   (1 mark)

  3. Copy or trace the graph into your writing booklet.

     

    On the same set of axes, draw the graph  `y = 3 sin x + 1`  for  `0 <= x <= pi`.   (2 marks)

Show Answers Only
  1. `A = 4`
  2. `b = 2`
  3. `text(See Worked Solutions for sketch)`
Show Worked Solution

a.   `A = 4`

b.  `text(S)text(ince the graph passes through)\ \ (pi/4, 4)`

`text(Substituting into)\ \ y = 4 sin bx`

`4 sin (b xx pi/4)` `=4`
`sin (b xx pi/4)` `= 1`
`b xx pi/4` `= pi/2`
`:. b` `= 2`

  

 MARKER’S COMMENT: Graphs are consistently drawn too small by many students. Aim to make your diagrams 1/3 to 1/2 of a page. 
c.

Calculus, MET1 SM-Bank 21

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)`

Calculus, MET1 SM-Bank 20

If   `f(x)= 2 sin 3x - 3 tan x`, find  `f^{prime}(0)`.   (2 marks)

Show Answers Only

`3`

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

  
`text(When)\ \ x = 0,`

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

Algebra, MET1 SM-Bank 24

The rule for  `f`  is  `f(x) = e^x-e^(-x)`.

Show that the inverse function is given by

    `f^(-1)(x) = log_e((x + sqrt(x^2 + 4))/2)`  (3 marks)

Show Answers Only

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

Show Worked Solution

`y = e^x-e^(-x)`

`text(Inverse: swap)\  x harr y`

`x` `= e^y-1/(e^y)`
`xe^y` `= e^(2y)-1`
`e^(2y)-xe^y-1` `= 0`

 

`text(Let)\ \ A = e^y`

`:.A^2-xA-1 = 0`
 

`text(Using the quadratic formula)`

`A` `=(x ± sqrt((-x)^2-4 · 1 · (-1)))/(2 · 1)`
  `=(x ± sqrt(x^2 + 4))/2`

 

`text(S)text(ince)\ \ (x-sqrt(x^2 + 4))/2<0\ \ text(and)\ \ e^y>0,`

`:.e^y` `= (x + sqrt(x^2 + 4))/2`
`log_e e^y` ` = log_e((x + sqrt(x^2 + 4))/2)`
`y` `= log_e((x + sqrt(x^2 + 4))/2)`
`:.f^(-1)(x)` `= log_e((x + sqrt(x^2 + 4))/2)\ \  …\ text(as required)`

Algebra, MET1 SM-Bank 23

The function  `f: [0,oo) → R`  with rule  `f(x) = 1/(1 + x^2)`  is drawn below.

Inverse Functions, EXT1 2004 HSC 5b

  1. Copy or trace this diagram into your writing booklet.
  2. On the same set of axes, sketch  `y=f^(-1)(x)`  where  `f^(-1)` is the inverse function of  `f(x)`.   (1 mark)

  3. Find the domain of the inverse  `f^(-1)`.   (1 mark)

  4. Find an expression for  `y = f^(-1)(x)`  in terms of  `x`.   (2 marks)

  5. The graphs of  `y = f(x)`  and  `y = f^(-1)(x)`  meet at exactly one point  `P`.Let  `α`  be the `x`-coordinate of  `P`. Explain why  `α`  is a root of the equation
  6.      `x^3 + x-1 = 0`.   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `0 < x ≤ 1`
  3. `y = sqrt((1-x)/x), y > 0`
  4. `text(See Worked Solutions)`
Show Worked Solution
a.  

Inverse Functions, EXT1 2004 HSC 5b Answer

b.   `text(Range of)\ \ f(x):\ (0,1]`

`:.\ text(Domain of)\ \ f^(-1)(x):  (0,1]`

 

c.  `f(x) = 1/(1 + x^2)`

`text(Inverse: swap)\  x harr y`

`x` `= 1/(1 + y^2)`
`x(1 + y^2)` `= 1`
`1 + y^2` `= 1/x`
`y^2` `= 1/x-1`
  `= (1-x)/x`
`y` `= ± sqrt((1-x)/x)`

 

`:.y = sqrt((1-x)/x), \ \ y >= 0`

 

d.   `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`

`text(i.e. where)`

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

 

`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`

`text(it is a root of)\ \ \ x^3 + x-1 = 0`

Calculus, MET1 SM-Bank 21

The rule for function  `f` is  `f(x) = e^(-x^2)`.  The diagram shows the graph  `y = f(x)`.

 Inverse Functions, EXT1 2010 HSC 3b

The graph has two points of inflection. 

  1. Find the `x` coordinates of these points.   (2 marks)

  2. Explain why the domain of `f(x)` must be restricted if `f(x)` is to have an inverse function.    (1 mark)

  3. Find the rule for the inverse function `f^(-1)` if the domain of `f(x)` is restricted to  `x ≥ 0.`   (2 marks)

  4. Find the domain for `f^(-1)`.    (1 mark)

  5. Sketch the curve  `y = f^(-1) (x)`.   (1 mark)

Show Answers Only
  1. `x = +- 1/sqrt2`
  2. `text(There can only be 1 value of)\ y\ text(for each value of)\ x.`
  3. `f^(-1)x = sqrt(ln(1/x))`
  4. `0 <= x <= 1`
  5. Inverse Functions, EXT1 2010 HSC 3b Answer

Show Worked Solution
a.    `y` `= e^(-x^2)`
  `dy/dx` `= -2x * e^(-x^2)`
  `(d^2y)/(dx^2)` `= -2x (-2x * e^(-x^2)) + e ^(-x^2) (-2)`
    `= 4x^2 e^(-x^2)-2e^(-x^2)`
    `= 2e^(-x^2) (2x^2-1)`

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

`2e^(-x^2) (2x^2-1)` `= 0` 
 `2x^2-1` `= 0` 
 `x^2` `= 1/2`
 `x` `= +- 1/sqrt2` 
COMMENT: It is also valid to show that `f(x)` is an even function and if a P.I. exists at `x=a`, there must be another P.I. at `x=–a`.
`text(When)\ \ ` `x < 1/sqrt2,` `\ (d^2y)/(dx^2) < 0`
  `x > 1/sqrt2,` `\ (d^2y)/(dx^2) > 0`

 
`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = 1/sqrt2`
 

`text(When)\ \ ` `x <-1/sqrt2,` `\ (d^2y)/(dx^2) > 0`
  `x >-1/sqrt2,` `\ (d^2y)/(dx^2) < 0`

 
`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x =-1/sqrt2`

 

b.   `text(In)\ f(x), text(there are 2 values of)\ y\ text(for)`
  `text(each value of)\ x.`
  `:.\ text(The domain of)\ f(x)\ text(must be restricted)`
  `text(for)\ \ f^(-1) (x)\ text(to exist).`

 

c.  `y = e^(-x^2)`

`text(Inverse: swap)\  x harr y` 

`x` `= e^(-y^2),\ \ \ x >= 0`
`lnx` `= ln e^(-y^2)`
`-y^2` `= lnx`
`y^2` `= -lnx`
  `=ln(1/x)`
`y` `= +- sqrt(ln (1/x))`

 

`text(Restricting)\ \ x>=0,\ \ =>y>=0`

`:.  f^(-1) (x)=sqrt(ln (1/x))`

 

d.   `f(0) = e^0 = 1`

`:.\ text(Range of)\ \ f(x)\ \ text(is)\ \ 0 < y <= 1`

`:.\ text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ 0 < x <= 1`

 

e. 

Inverse Functions, EXT1 2010 HSC 3b Answer

Graphs, MET1 SM-Bank 20

The rule for  `f` is  `f(x) = x-1/2 x^2`  for  `x <= 1`.  This function has an inverse,  `f^(-1) (x)`.

  1. Sketch the graphs of  `y = f(x)`  and  `y = f^(-1) (x)`  on the same set of axes. (Use the same scale on both axes.)   (2 marks)

  2. Find the rule for the inverse function  `f^(-1) (x)`.    (2 marks)

  3. Evaluate  `f^(-1) (3/8)`.    (1 mark)

Show Answers Only
  1.  
    Inverse Functions, EXT1 2008 HSC 5a Answer
  2. `y = 1-sqrt(1-2x)`
  3. `1/2`
Show Worked Solution
a. 

Inverse Functions, EXT1 2008 HSC 5a Answer
b.   `y = x-1/2 x^2,\ \ \ x <= 1`

 

`text(For the inverse function, swap)\ \ x↔y,`

`x` `= y-1/2 y^2,\ \ \ y <= 1`
`2x` `= 2y-y^2`
`y^2-2y + 2x` `= 0`

 

`text(Using quadratic formula,)`

`y` `= (2 +- sqrt( (-2)^2-4 * 1 * 2x) )/2`
  `= (2 +- sqrt(4-8x))/2`
  `= (2 +- 2 sqrt(1-2x))/2`
  `= 1 +- sqrt (1-2x)`

 

`:. y = 1-sqrt(1-2x), \ \ (y <= 1)`

 

c.    `f^(-1) (3/8)` `= 1-sqrt(1-2(3/8))`
    `= 1-sqrt(1-6/8)`
    `= 1-sqrt(1/4)`
    `= 1-1/2`
    `= 1/2`

Algebra, MET1 SM-Bank 11

Let  `f: R→R`  where  `f(x)= x^3-2`.

Evaluate  `f^(-1)(25),` where  `f^(-1)`  is the inverse function of  `f`.   (2 marks)

Show Answers Only

`3`

Show Worked Solution

`text(Let)\ \ y = x^3-2`

`text(For inverse),\ \ x harr y`

`x` `= y^3-2`
`y^3` `= x + 2`
`y` `= (x + 2)^(1/3)`

 

`:. f^(-1)(25)` `=(25+2)^(1/3)`
  `=3`

Calculus, MET1 2006 ADV 2bi

If  `f^{prime}(x)= (x^2)/(x^3 + 1)`  and  `f(1)= log_e 2,` find  `f(x)`.   (3 marks)

Show Answers Only

`f(x) = 1/3 log_e(x^3 + 1) + 2/3 log_e 2`

Show Worked Solution
`f(x)` `=int f^{prime}(x)\ dx`
  `=int (x^2)/(x^3 + 1)\ dx`
  `= 1/3 int (3x^2)/(x^3 + 1)\ dx`
  `=1/3 log_e |(x^3 + 1)|+c`

 

`text(S)text(ince)\ \ f(1)= log_e 2,`

`1/3 log_e 2+c` `= log_e 2`
`c` `=2/3 log_e 2`
   

`:.\ f(x) = 1/3 log_e(x^3 + 1) + 2/3 log_e 2`

Calculus, MET1 2008 ADV 3b

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

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

Show Answers Only
  1. `-tan x`
  2. `-log_e(1/sqrt2)\  text{or  0.35  (2 d.p.)}`
Show Worked Solution
a.   `y` `= log_e(cos x)`
  `(dy)/(dx)` `= (-sin x)/(cos x)`
    `=-tan x`

 

b.   `int_0^(pi/4) tan x\ dx`

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

`= -[log_e(cos(pi/4))-log_e(cos 0)]`

`= -[log_e(1/sqrt2)-log_e 1]`

`= -[log_e(1/sqrt2)-0]`

`= -log_e(1/sqrt2)`

`= 0.346…`

`= 0.35\ \ (text(2 d.p.))`

Calculus, MET1 2016 ADV 12d

  1. Differentiate  `y = xe^(3x)`.   (1 mark)

  2. Hence find the exact value of  `int_0^2 e^(3x) (3 + 9x)\ dx`.   (2 marks)

Show Answers Only
  1. `e^(3x) (1 + 3x)`
  2. `6e^6`
Show Worked Solution

a.  `y = xe^(3x)`

`text(Using product rule,)`

`(dy)/(dx)` `= x · 3e^(3x) + 1 · e^(3x)`
  `= e^(3x) (1 + 3x)`

 

b.  `int_0^2 e^(3x) (3 + 9x)\ dx`

`= 3 int_0^2 e^(3x) (1 + 3x)\ dx`

`= 3 [x e^(3x)]_0^2`

`= 3 (2e^6-0)`

`= 6e^6`