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Algebra, NAP-A3-NC05

Flynn is making a pattern with soccer balls.
 

 
This table shows the number of balls he needs for each shape in the pattern.
 

  Pattern # 1 2 3 4 5
  Number of balls 1 4 9 16 ?

 
How many balls will Flynn need for Pattern #5?

`17` `23` `25` `30`
 
 
 
 
Show Answers Only

`25`

Show Worked Solution

`text(Solution 1)`

`text(By drawing the next shape, pattern #5 is)`

`:. 25\ text(balls needed for Pattern #5.)`

 

`text(Solution 2)`

`text(The table shows a pattern where the balls in each)`

`text(shape are the square of the pattern number:)`

`:.\ text(Balls in pattern #5) = 5^2 = 25`

Statistics, NAP-A3-CA04

This table summarises the time Tutty spent training her parrot over five days.
 

 
What was the average (mean) time for training the parrot each day?

`text(30 minutes)` `text(56 minutes)` `text(66 minutes)` `text(280 minutes)`
 
 
 
 
Show Answers Only

`text(56 minutes)`

Show Worked Solution
`text(Average)` `= (25 + 55 + 60 + 94 + 46)/5`
  `= 280/5`
  `= 56\ text(minutes)`

Algebra, NAP-A4-NC01

Flynn is making a pattern with soccer balls.
 

 
This table shows the number of balls he needs for each shape in the pattern.
 

  Pattern # 1 2 3 4 5
  Number of balls 1 4 9 16 ?

 
How many balls will Flynn need for Pattern #5?

`17` `23` `25` `30`
 
 
 
 
Show Answers Only

`25`

Show Worked Solution

`text(Pattern #5 is)`

`:. 25\ text(balls needed for Pattern #5.)`

Statistics, NAP-A4-CA04

This table summarises the time Tutty spent training her parrot over five days.
 

 
What was the average (mean) time for training the parrot each day?

`text(30 minutes)` `text(56 minutes)` `text(66 minutes)` `text(280 minutes)`
 
 
 
 
Show Answers Only

`text(56 minutes)`

Show Worked Solution
`text(Average)` `= (25 + 55 + 60 + 94 + 46)/5`
  `= 280/5`
  `= 56\ text(minutes)`

Statistics, NAP-A4-CA03

A dealership sells new and used cars.

The graph shows the price of 2 similar cars and their age in years
 

 
Which one of these statements is true?

 
 Car Q is older and less expensive than Car P.
 
 Car P is newer and less expensive than Car Q.
 
 Car P is older and more expensive than Car Q.
 
 Car Q is newer and more expensive than Car P.
Show Answers Only

`text(Car Q is older and less expensive than Car P.)`

Show Worked Solution

`text(Car)\ Q\ text(is further right on)\ xtext(-axis)`

`=> Q\ text(is older)`

`text(Car)\ Q\ text(is lower on)\ ytext(-axis)`

`=> Q\ text(is less expensive)`

 

`:. text(Car)\ Q\ text(is older and less expensive than Car)\ P.`

Number, NAP-A4-CA01

At 7 am the temperature in Merewether was 15.9°C.

At midday it was 12.8°C warmer.

At 7 pm it was 13.9°C cooler than at midday.

What was the temperature at 7 pm?

`14.8^@text(C)` `15^@text(C)` `17^@text(C)` `42.6^@text(C)`
 
 
 
 
Show Answers Only

`14.8^@text(C)`

Show Worked Solution

`text(Temperature at 6 pm)`

`= 15.9 + 12.8 – 13.9`

`= 14.8^@text(C)`

Number, NAP-B4-CA04

Shelly and Carly collect dolls.

The ratio of the number of dolls Shelly owns compared to Carly is  3 : 2.

Shelley owns 12 dolls.

How many dolls does Carly own?

`2` `6` `8` `18`
 
 
 
 
Show Answers Only

`8`

Show Worked Solution

`text(Ratio 3 : 2)`

`text(Let)\ \ x=\ text(Number of Carly’s dolls)`

`x/12` `= 2/3`
`x` `= (2 xx 12)/3`
  `= 8\ text(dolls)`

Quadratic, EXT1 2017 HSC 14b

Let  `P(2p, p^2)`  be a point on the parabola  `x^2 = 4y`.

The tangent to the parabola at `P` meets the parabola  `x^2 = −4ay`, `a > 0`, at `Q` and `R`. Let `M` be the midpoint of `QR`.

  1. Show that the `x` coordinates of `R` and `Q` satisfy
  2. `qquadx^2 + 4apx - 4ap^2 = 0`.  (2 marks)

  3. Show that the coordination of `M` are  `(−2ap, −p^2(2a + 1))`.  (2 marks)
  4. Find the value of `a` so that the point `M` always lies on the parabola  `x^2 = −4y`.  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `1 + sqrt2, a > 0`
Show Worked Solution

(i)   `x^2 = 4y,\ \ =>y = (x^2)/4`

`(dy)/(dx) = x/2`

`text(At)\ P(2p, p^2),`

`(dy)/(dx) = p`

`text(Equation of tangent:)`

`y = px – p^2`

 

`R and Q\ text(at intersection)`

`y` `= px – p^2\ …\ (1)`
`y` `= −(x^2)/(4a)\ …\ (2)`

 

`text(Subtract)\ (1) – (2)`

`px – p^2 + (x^2)/(4a)` `= 0`
`x^2 + 4apx – 4ap^2` `= 0\ …\ text(as required)`

 

(ii)   `x^2 + 4apx – 4ap^2 = 0`

♦♦ Mean mark 30%.
`x=` `\ (−4ap ± sqrt((4ap)^2 – 4 · 1 · (−4ap^2)))/2`
`=`  `\ (−4ap ± sqrt(16ap^2(a + 1)))/2`
`=`  `\ −2ap ± 2psqrt(a(a + 1))`

 

`=> xtext(-coordinate of)\ M\ text(is)\ −2ap.`

 

`M\ text(lies on tangent)\ \ y = px – p^2`

`:.y` `= p(−2ap) – p^2`
  `= −2ap^2 – p^2`
  `= −p^2(2a + 1)`

`:. M\ text(has coordinates)\ \ (−2ap, −p^2(2a + 1))`

 

(iii)   `text(If)\ M\ text(always lies on)\ \ x^2 = −4y`

♦♦ Mean mark 23%.
`(−2ap)^2` `= −4(−p^2(2a + 1))`
`4a^2p^2` `= 4p^2(2a + 1)`
`a^2` `= 2a + 1`
`a^2 – 2a – 1` `= 0`
`:. a` `= (+2 ± sqrt(4 + 4 · 1· 1))/2`
  `= 1 ± sqrt2`
  `= 1 + sqrt2, \ a > 0`

Mechanics, EXT2* M1 2017 HSC 13c

A golfer hits a golf ball with initial speed `V\ text(ms)^(−1)` at an angle `theta` to the horizontal. The golf ball is hit from one side of a lake and must have a horizontal range of 100 m or more to avoid landing in the lake.
 

     

Neglecting the effects of air resistance, the equations describing the motion of the ball are

`x = Vt costheta`

`y = Vt sintheta - 1/2 g t^2`,

where `t` is the time in seconds after the ball is hit and `g` is the acceleration due to gravity in `text(ms)^(−2)`. Do NOT prove these equations.

  1. Show that the horizontal range of the golf ball is
     
         `(V^2sin 2theta)/g` metres.  (2 marks)

  2. Show that if  `V^2 < 100 g`  then the horizontal range of the ball is less than 100 m.  (1 mark)

It is now given that  `V^2 = 200 g`  and that the horizontal range of the ball is 100 m or more.

  1. Show that  `pi/12 <= theta <= (5pi)/12`.  (2 marks)

  2. Find the greatest height the ball can achieve.  (2 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `text(See Worked Solution)`
  4. `25(2 – sqrt 3)\ text(metres)`
Show Worked Solution

i.   `text(Find)\ \ t\ \ text(when)\ \ y = 0:`

`1/2 g t^2` `= Vtsintheta`
`1/2 g t` `= Vsintheta`
`t` `= (2Vsintheta)/g`

 

`text(Horizontal range)\ (x)\ text(when)\ \ t = (2Vsintheta)/g :`

`x` `= V · (2Vsintheta)/g costheta`
  `= (V^2 2sintheta costheta)/g`
  `= (V^2sin2theta)/g\ … text(as required)`

 

ii.   `text(If)\ \ V^2 < 100 g`

♦ Mean mark 44%.
`x` `< (100 g sin2theta)/g`
`x` `< 100 sin2theta`

 

`text(S)text(ince)\ −1 <= 2theta <= 1,`

`x < 100\ text(metres)`

 

iii.   `V^2 = 200g,\ \ x >= 100`

`(200 g · sin2theta)/g` `>= 100`
`sin2theta` `>= 1/2`

`:. pi/6 <= 2theta <= (5pi)/6`

`:. pi/12 <= theta <= (5pi)/12\ …\ text(as required)`

 

iv.   `text(Max height occurs when)`

♦♦ Mean mark 35%.
`t` `= 1/2 xx text(time of flight)`
  `= (Vsintheta)/g`

 
`text(Find)\ \ y\ \ text(when)\ \ t = (Vsintheta)/g`

`y` `= V · (Vsintheta)/g · sintheta – 1/2 g ((Vsintheta)/g)^2`
  `= (V^2 sin^2theta)/g – 1/2 · (V^2 sin^2 theta)/g`
  `= (V^2 sin^2theta)/(2g)`

 

`text(Max height when)\ theta = (5pi)/12\ (text(steepest angle)), V^2 = 200 g\ (text(given))`

`y_text(max)` `= (200 g · sin^2 ((5pi)/12))/(2g)`
  `= 100 sin^2 ((5pi)/12)`
  `= 50(1 – cos((5pi)/6))`
  `= 50(1 + sqrt3/2)`
  `= 25(2 + sqrt 3)\ text(metres)`

Plane Geometry, 2UA 2017 HSC 16c

In the triangle `ABC`, the point `M` is the mid-point of `BC`. The point `D` lies on `AB` and

`BD = DA + AC`.

The line that passes through the point `C` and is parallel to `MD` meets `BA` produced at `E`.

Copy or trace this diagram into your writing booklet.

  1. Prove that `Delta ACE` is isosceles.  (3 marks)
  2. The point `F` is chosen on `BC` so that `AF` is parallel to `DM`.
    `qquad` 
    Show that `AF` bisects `/_ BAC`.  (2 marks)

 

 

 

 

 

 

 

 

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

`text(Prove)\ Delta ACE\ text(is isosceles)`

♦♦♦ Mean mark 21%.

`text(S) text(ince)\ DM text(||) EC,`

`∠BDM = ∠BEC\ \ text{(corresponding angles)}`

`∠DBM\ \ text(is common)`

`:. Delta BDM\ text(|||)\ Delta BEC\ \ text{(AAA)}`

`text(Using ratios of similar)\ Delta text(s):`

`(BM)/(BD)` `= (BC)/(BE)`
`(BM)/(BC)` `= (BD)/(BE) = 1/2\ \ text{(}M\ text(is midpoint of)\ BC text{)}`

 

`=> D\ text(is the midpoint of)\ BE.`

`=> BD = DE`

`AE` `= DE – DA`
`AC` `= BD – DA\ text{(given)}`
  `= DE – DA`
`:.  AE` `= AC`

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

 

(ii)  

`text(Show)\ AF\ text(bisects)\ /_ BAC`

♦♦ Mean mark 31%.

`/_ BAF = /_ AEC\ \ \ text{(corresponding angles)}`

`text{From part (i)}`

`/_ AEC` `= /_ ECA\ \ \ text{(}Delta AEC\ text{is isosceles)}`
`/_ FAC` `= /_ ECA\ \ \ text{(alternate angles)}`
`:. /_ BAF` `= /_ FAC`

`:. AF\ text(bisects)\  /_ BAC\ \ \ text(… as required)`

Calculus, 2ADV C3 2017 HSC 16a

John’s home is at point `A` and his school is at point `B`. A straight river runs nearby.

The point on the river closest to `A` is point `C`, which is 5 km from `A`.

The point on the river closest to `B` is point `D`, which is 7 km from `B`.

The distance from `C` to `D` is 9 km.

To get some exercise, John cycles from home directly to point `E` on the river, `x` km from `C`, before cycling directly to school at `B`, as shown in the diagram.
 


  

The total distance John cycles from home to school is `L` km.

  1. Show that  `L = sqrt (x^2 + 25) + sqrt (49 + (9 - x)^2)`.   (1 mark)

  2. Show that if  `(dL)/(dx) = 0`, then  `sin alpha = sin beta`.   (3 marks)

  3. Find the value of `x` that makes  `sin alpha = sin beta`.   (2 marks)

  4. Explain why this value of `x` gives a minimum for `L`.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `45/12`
  4. `text(See Worked Solutions)`
Show Worked Solution
i.  

`text(Using Pythagoras:)`

`L` `= AE + EB`
  `= sqrt (5^2 + x^2) + sqrt (7^2 + (9 – x)^2)`
  `= sqrt(25 + x^2) + sqrt (49 + (9 – x)^2)\ text(… as required)`

 

ii.  `text(From diagram):`

♦ Mean mark 41%.

`sin alpha = x/sqrt(25 + x^2) and sin beta = (9 – x)/sqrt(49 + (9 – x)^2)`

`L` `= sqrt(25 + x^2) + sqrt (49 + (9 – x)^2)`
`(dL)/(dx)` `= (2x)/sqrt(25 + x^2) – (2(9 – x))/sqrt(49 + (9 – x)^2)`

 

`text(If)\ \ (dL)/(dx) = 0,`

`=> (2x)/sqrt(25 + x^2)` `= (2(9 – x))/sqrt(49 + (9 – x)^2)`
`x/sqrt(25 + x^2)` `= (9 – x)/sqrt(49 + (9 – x)^2)`
`sin alpha` `= sin beta\ text(… as required)`

 

iii.  `text(If)\ sin alpha = sin beta,\ text(then)\ alpha = beta and`

♦♦ Mean mark 29%.

`Delta ACE\ text(|||)\ Delta BDE`

`text(Using corresponding sides of similar triangles:)`

`x/5` `= (9 – x)/7`
`7x` `= 45 – 5x`
`12x` `= 45`
`:. x` `= 45/12\ text(km)`
♦♦♦ Mean mark 9%.

 

iv.  

`text(If point)\ B\ text(is reflected across the)`

`text(river),\ AEB\ text(will be a straight line.)`

`text(If any other point is chosen,)\ AEB`

`text(would not be straight and the distance)`

`text(would be longer.)`

Measurement, 2UG 2017 HSC 30e

A solid is made up of a sphere sitting partially inside a cone.

The sphere, centre `O`, has a radius of 4 cm and sits 2 cm inside the cone. The solid has a total height of 15 cm. The solid and its cross-section are shown.
 


 

What is the volume of the cone, correct to the nearest cm³?  (3 marks)

Show Answers Only

`113\ text{cm³  (nearest cm³)}`

Show Worked Solution

`V = 1/3 xx text(base of cone × height)`

`text(Consider the base area of the cone,)`

♦♦ Mean mark 34%.

`text(Need to find)\ x:`

`x^2` `= 4^2 – 2^2 = 16 – 4 = 12`
`x` `= sqrt12\ text(cm)`

 

`:. V` `= 1/3 xx pi xx (sqrt12)^2 xx (15 – 6)`
  `= 1/3 xx pi xx 12 xx 9`
  `= 113.097…`
  `= 113\ text{cm³  (nearest cm³)}`

Statistics, STD2 S5 2017 HSC 29d

All the students in a class of 30 did a test.

The marks, out of 10, are shown in the dot plot.
 


 

  1. Find the median test mark.  (1 mark)

  2. The mean test mark is 5.4. The standard deviation of the test marks is 4.22.

     

    Using the dot plot, calculate the percentage of the marks which lie within one standard deviation of the mean.  (2 marks)

  3. A student states that for any data set, 68% of the scores should lie within one standard deviation of the mean. With reference to the dot plot, explain why the student’s statement is NOT relevant in this context.  (1 mark)

Show Answers Only
  1. `6`
  2. `text(43%)`
  3. `text(The statement assumes the data is normally)`
    `text(distributed which is not the case here.)`
Show Worked Solution
♦ Mean mark 50%.
i.    `text(Median)` `= text(15th + 16th score)/2`
    `= (4 + 8)/2`
    `= 6`

 

ii.   `text(Lower limit) = 5.4 – 4.22 = 1.18`

♦♦ Mean mark 34%.

`text(Upper limit) = 5.4 + 4.22 = 9.62`

`:.\ text(Percentage in between)`

`= 13/30 xx 100`

`= 43.33…`

`= 43text{%  (nearest %)}`

 

iii.   `text(The statement assumes the data is normally)`

♦♦♦ Mean mark 13%.

`text(distributed which is not the case here.)`

Measurement, 2UG 2017 HSC 29a

A new 200-metre long dam is to be built.
The plan for the new dam shows evenly spaced cross-sectional areas.
 


 

  1. Using TWO applications of Simpson’s rule, show that the volume of the dam is approximately  44 333 m³.  (2 marks)
  2. It is known that the catchment area for this dam is 2 km².
    Calculate how much rainfall is needed, to the nearest mm, to fill the dam.  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solution)`
  2. `22\ text{mm  (nearest mm)}`
Show Worked Solution
(i)    `text(Volume)` `~~ h/3 (y_0 + 4y_1 + y_2)\ \ …\ text(applied twice)`
    `~~ 50/3(0 + 4 xx 140 + 270) + 50/3(270 + 4 xx 300 + 360)`
    `~~ 50/3(830) + 50/3(1830)`
    `~~ 44\ 333\ text(m³)\ …\ \ text(as required)`

 

(ii)   `text(Convert 2km² to m²:)`

♦♦♦ Mean mark 11%.
`text(2 km²)` `= 2 xx 1000 xx 1000`
  `= 2\ 000\ 000\ text(m²)`
`:.\ text(Rainfall needed)` `= (44\ 333)/(2\ 000\ 000)`
  `= 0.0221…\ text(m)`
  `= 22.1…\ text(mm)`
  `= 22\ text{mm  (nearest mm)}`

Algebra, 2UG 2017 HSC 28a

Temperature can be measured in degrees Celsius (`C`) or degrees Fahrenheit (`F`).

The two temperature scales are related by the equation  `F = (9C)/5 + 32`.

  1. Calculate the temperature in degrees Fahrenheit when it is  −20 degrees Celsius.  (1 mark)
  2. Solve the following equations simultaneously, using either the substitution method or the elimination method.  (2 marks)
    `qquadF = (9C)/5 + 32`

    `qquadF = C`

     

  3. The graphs of  `F = (9C)/5 + 32`  and  `F = C`  are shown below.
  4.  


  5. What does the result from part (ii) mean in the context of the graph?  (1 mark)     

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Show Answers Only
  1. `−4^@F`
  2. `C = −40, F = −40`
  3. `text(It means the two graphs intersect)`
    `text(at)\ (−40,−40).`

 

 

Show Worked Solution
(i)   `F` `= (9(−20))/5 + 32`
    `= −4^@F`

 

(ii)   `F` `= (9C)/5 + 32` `…\ (1)`
  `F` `= C` `…\ (2)`

 

♦♦ Mean mark 31%.
MARKER’S COMMENT: An area that requires attention.

`text(Substitute)\ \ F = C\ \ text{from (2) into (1)}`

`C` `= (9C)/5 + 32`
`(9C)/5 – C` `= −32`
`(4C)/5 – C` `= −32`
`C` `= −32 xx 5/4 = −40`

 

`text{From (2),}`

`F = −40`

`text{(i.e. when}\ C = −40, F = −40)`

♦♦♦ Mean mark 20%.

 

(iii)   `text(It means the two graphs intersect)`

`text{at (−40,−40).}`

Probability, STD2 S2 2017 HSC 24 MC

A deck of 52 playing cards contains 12 picture cards. Two cards from the deck are drawn at random and placed on a table.

What is the probability, correct to four decimal places, that exactly one picture card is on the table?

A.     `0.0498`

B.     `0.1810`

C.     `0.3550`

D.     `0.3620`

Show Answers Only

`text(D)`

Show Worked Solution

`P(text(exactly 1 picture card))`

`= P(text(picture)) xx P(text(no picture)) + P(text(no picture)) xx P(text(picture))`

`= 12/52 xx 40/51quad+quad40/52 xx 12/51`

♦♦♦ Mean mark 21%.

`= 2((12 xx 40)/(52 xx 51))`

`= 0.36199…`

`=>\ text(D)`

FS Comm, 2UG 2017 HSC 23 MC

How many bits are there in 2 terabytes?

A.     `2^40`

B.     `2^41`

C.     `2^43`

D.     `2^44`

Show Answers Only

`text(D)`

Show Worked Solution

`text(Bytes in 2 terabytes)`

♦♦ Mean mark 34%.

`= 2 xx 2^40\ text(bytes)`

`:.\ text(Bits)` `= 8 xx 2 xx 2^40`
  `= 16 xx 2^40`
  `= 2^4 xx 2^40`
  `= 2^44\ text(bits)`

`=>\ text(D)`

Number and Algebra, NAP-J2-19

A company has 3706 computers in a warehouse.

Another 423 computers are delivered to the warehouse.

Which of these could be used to calculate the correct number of computers in the warehouse?

 
 3000 + 4000 + 700 + 20 + 6 + 3
 
 3000 + 700 + 400 + 20 + 60 + 3
 
 3000 + 700 + 400 + 20 + 6 + 3
 
 3000 + 400 + 70 + 20 + 6 + 3
Show Answers Only

`3000 + 700 + 400 + 20 + 6 + 3`

Show Worked Solution

`3706 + 423`

`= 3000 + 700 + 6 + 400 + 20 + 3`

`=3000 + 700 + 400 + 20 + 6 + 3`

Statistics, NAP-J3-CA17

The bottles in Renee's fridge are pictured below.

Renee decides to make a graph where each bar represents one type of bottle in her fridge.
  

Renee makes an error when creating the graph.

What should Renee do to correct the error?

 
 Make each category bar a different colour.
 
 Change the title to 'Number of bottles in the fridge by volume'.
 
 Change the 'Number of bottles' label to 'Volume of bottles'.
 
 Remove the 'Juice' category since orange juice and apple juice are already shown.
Show Answers Only

`text(Remove the ‘Juice’ category since orange juice and)`

`text(apple juice are already shown.)`

Show Worked Solution

`text(Remove the ‘Juice’ category since orange juice and)`

`text(apple juice are already shown.)`

Number, NAP-J3-CA14

Madison uses the number sentence  15 × 12 = 180  to solve a problem.

Which of the following could be the problem?

 
 Madison buys 15 showbags. How much does each showbag cost?
 
 Madison spends $15 on 180 showbags. How much does she spend?
 
 Madison buys 15 showbags that cost $12 each. How many showbags does she buy?
 
 Madison buys 12 showbags that cost $15 each. How much does she spend?
Show Answers Only

`text(Madison buys 12 showbags that cost $15 each.)`

`text(How much does she spend?)`

Show Worked Solution

`text($15 per showbag × 12 showbags = $180)`

`:.\ text(Madison buys 12 showbags that cost $15 each.)`

`text(How much does she spend?)`

Number, NAP-J3-CA09

Dinesh is a teacher and buys pencils for his class.

He purchases the pencils in two different sized packets.

Dinesh buys 10 packet A's and 4 packet B's.

He then divides all the pencils equally among his 9 students.

How many pencils does each student receive?

`6` `8` `12` `54` `72`
 
 
 
 
 
Show Answers Only

`8`

Show Worked Solution
`text(Total pencils)` `=(10 xx 6)+(4 xx3)`
  `=72`

 

`:.\ text(Pencils per student)` `=72 -: 9`
  `=8`

Number, NAP-J3-CA06 SA

There are 48 Year 7 students at a high school.

Each student is asked if they own a bike or not and the results are recorded.

`3/4` of the students said they owned a bike.

How many Year 7 students at the school own a bike?
Show Answers Only

`36`

Show Worked Solution

`text(Students who own a bike)`

`= 3/4 xx 48`

`= 36`

Algebra, NAP-J3-CA04 SA

A school teacher allocates pieces of cardboard to class groups depending on the number of students in each group.

The table below is used.
  

   
Using the pattern in the table, how many pieces of cardboard should a group of 3 students receive?
Show Answers Only

`9`

Show Worked Solution

`text(The pattern shows that each student receives)`

`text(2 pieces of cardboard.)`

`:.\ text(A group of 3 will be given 9 pieces.)`

Algebra, MET2 2007 VCAA 21 MC

`{x: cos^2(x) + 2cos (x) = 0} =`

  1. `{x : cos (x) = 0}`
  2. `{x : cos(x) = -1/2}`
  3. `{x : cos(x) = 1/2}`
  4. `{x: cos (x) = 0} uu { x : cos (x) = -1/2}`
  5. `{x: cos (x) = 1/2} uu { x : cos (x) = -1/2}`
Show Answers Only

`A`

Show Worked Solution

`text(Factorise:)`

♦♦♦ Mean mark 27%.

`cos (x) (cos x + 2) = 0`
 

 `:. cos x = 0,\ \ text(or)`

`cos x = -2 -> text(no solution)`

`=>   A`

Graphs, MET2 2008 VCAA 18 MC

Let  `f: [0, pi/2] -> R,\ f(x) = sin(4x) + 1.` The graph of  `f`  is transformed by a reflection in the `x`-axis followed by a dilation of factor 4 from the `y`-axis.

The resulting graph is defined by

  1. `g: [0, pi/2] -> R\ \ \ \ \ \ g(x) = -1 - 4 sin (4x)`
  2. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = -1 - sin (16x)`
  3. `g: [0, pi/2] -> R\ \ \ \ \ \ g(x) = 1 - sin (x)`
  4. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = 1 - sin (4x)`
  5. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = -1 - sin (x)`
Show Answers Only

`E`

Show Worked Solution

`f(x) = sin(4x)+1`

♦♦ Mean mark 36%.

`text(Reflecting in the)\ x text(-axis,)`

`=> h(x) = – sin (4x) – 1`

 

`text(Dilation by a factor of 4 from the)\ \ y text(-axis,)`

`=> g(x) = -sin(x)-1`

`text(Dilation factor: adjust the domain from)\ \ [0, pi/2]`

`text(to)\ \ [0xx4, pi/2 xx 4]=[0,2pi].`

`=>   E`

Probability, MET2 2008 VCAA 15 MC

The sample space when a fair die is rolled is `{1, 2, 3, 4, 5, 6}`, with each outcome being equally likely.

For which of the following pairs of events are the events independent?

  1. `{1, 2, 3} and {1, 2}`
  2. `{1, 2} and {3, 4}`
  3. `{1, 3, 5} and {1, 4, 6}`
  4. `{1, 2} and {1, 3, 4, 6}`
  5. `{1, 2} and {2, 4, 6}`
Show Answers Only

`E`

Show Worked Solution
`text(Let)\ \ A` `= {1, 2}`
`B` `= {2, 4, 6}`
`A nn B` `= {2}`
♦♦♦ Mean mark 7%!
MARKER’S COMMENT: Almost two-thirds of students chose option B, which contained mutually exclusive events.
`text(Pr)(A)` `= 1/3`
`text(Pr)(B)` `= 1/2`
`text(Pr)(A nn B)` `= 1/6`

 

`text(S) text(ince)\ \ text(Pr) (A) xx text(Pr) (B)` `= text(Pr)(A nn B)`
`1/3 xx 1/2` `= 1/6`

 

`:. A, B\ \ text(are independent.)`

`=>   E`

Calculus, MET1 SM-Bank 6

The diagram shows the function  `f:(2, oo)→R`,  where  `f(x)= log_e(x - 2).`

In the diagram, the shaded region is bounded by  `f(x)`, the `x`-axis and the line  `x = 7`.

Find the exact value of the area of the shaded region.  (4 marks)

Show Answers Only

`5log_e 5 – 4\ \ \ text(u²)`

Show Worked Solution

`text(Shaded Area)\ (A_1)` `= text(Rectangle) – A_2`
`text(Area of Rectangle)` `= 7 xx log_e 5`

 

`text(Finding the Area of)\ A_2`

`y` `= log_e(x – 2)`
`x – 2` `= e^y`
`x` `= e^y + 2`
`:. A_2` `= int_0^(log_e5) x\ dy`
  `= int_0^(log_e5) e^y + 2\ dy`
  `= [e^y + 2y]_0^(log_e5)`
  `= [(e^(log_e 5) + 2log_e5) – (e^0 + 0)]`
  `= (5 + 2log_e 5) – 1`
  `= 4 + 2log_e 5`

 

`:. A_1` `= 7 log_e5 – (4 + 2log_e 5)`
  `= 5log_e 5 – 4\ \ \ text(u²)`

Probability, MET1 2016 VCAA 8

Let `X` be a continuous random variable with probability density function

`f(x) = {(−4xlog_e(x),0<x<=1),(0,text(elsewhere)):}`

Part of the graph of  `f` is shown below. The graph has a turning point at  `x = 1/e`.

  1. Show by differentiation that  `(x^k)/(k^2)(k log_e(x) - 1)`  is an antiderivative of  `x^(k – 1) log_e(x)`, where `k` is a positive real number.  (2 marks)
  2.  i. Calculate `text(Pr)(X > 1/e)`.  (2 marks)
  3. ii. This content is no longer in the Study Design.
Show Answers Only
  1. `text(See Worked Solutions)`
  2.  i. `1 – 3/(e^2)`
  3. ii. This content is no longer in the Study Design.
Show Worked Solution

a.   `text(Using Product Rule:)`

♦♦ Mean mark part (a) 28%.
MARKER’S COMMENT: Students who expanded before differentiating tended to score more highly.

`d/(dx) ((x^k)/(k^2)(klog_e(x) – 1))`

`=d/(dx)((x^k)/k log_e(x) – (x^k)/(k^2))`

`= x^(k – 1) log_e(x) + 1/k x^(k – 1) – 1/k x^(k – 1)`

`= x^(k – 1) log_e(x)`

`:. intx^(k – 1) log_e(x)\ dx = (x^k)/(k^2)(klog_e(x) – 1)`

 

b.i.   `text(Pr)(x > 1/e)`

♦♦♦ Mean mark 16%.

`= −4 int_(1/e)^1 (xlog_e(x))\ dx,\ text(where)\ k = 2`

`= −4[(x^2)/4(2log_e(x) – 1)]_(1/e)^1`

`= −4[1/4(0 -1) – 1/(4e^2)(2log_e(e^(−1)) – 1)]`

`= −4[−1/4 + 1/(4e^2) + 1/(2e^2)]`

`= 1 – 3/(e^2)`

 

b.ii.  This content is no longer in the Study Design.

Probability, MET2 2009 VCAA 17 MC

The sample space when a fair twelve-sided die is rolled is `{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}`. Each outcome is equally likely.

For which one of the following pairs of events are the events independent?

  1. `{1, 3, 5, 7, 9, 11} and {1, 4, 7, 10}`
  2. `{1, 3, 5, 7, 9, 11} and {2, 4, 6, 8, 10, 12}`
  3. `{4, 8, 12} and {6, 12}`
  4. `{6, 12} and {1, 12}`
  5. `{2, 4, 6, 8, 10, 12} and {1, 2, 3}`
Show Answers Only

`A`

Show Worked Solution

`text(Consider option A:)`

♦♦♦ Mean mark 31%.
MARKER’S COMMENT: Many chose option B, which contains mutually exclusive events, for which `text(Pr) (A nn B)=0`

`text(Let)\ \A = {1, 3, 5, 7, 9, 11}`

`text(Let)\ \ B = {1, 4, 7, 10}`

`A nn B = {1, 7}`
 

`text(If independent events,)`

`text(Pr) (A nn B) = text(Pr) (A) xx text(Pr) (B)`

`text(Pr) (A) = 6/12=1/2`

`text(Pr) (B) = 4/12=1/3`

`text(Pr) (A nn B) = 2/12=1/6`

 

`:.\ text(Pr) (A nn B) = text(Pr) (A) xx text(Pr) (B)`

`:. A, B\ \ text(are independent)`

`=>   A`

Calculus, MET2 2011 VCAA 4

Deep in the South American jungle, Tasmania Jones has been working to help the Quetzacotl tribe to get drinking water from the very salty water of the Parabolic River. The river follows the curve with equation  `y = x^2 - 1`,  `x >= 0` as shown below. All lengths are measured in kilometres.

Tasmania has his camp site at  `(0, 0)`  and the Quetzacotl tribe’s village is at  `(0, 1)`. Tasmania builds a desalination plant, which is connected to the village by a straight pipeline.
 

met2-2011-vcaa-q4

  1. If the desalination plant is at the point  `(m, n)`  show that the length, `L` kilometres, of the straight pipeline that carries the water from the desalination plant to the village is given by
  2.    `L = sqrt(m^4 - 3m^2 + 4)`.  (3 marks)
  3. If the desalination plant is built at the point on the river that is closest to the village
  4.  i. find  `(dL)/(dm)`  and hence find the coordinates of the desalination plant  (3 marks)
  5. ii. find the length, in kilometres, of the pipeline from the desalination plant to the village.  (2 marks)

The desalination plant is actually built at  `(sqrt7/2, 3/4)`.

If the desalination plant stops working, Tasmania needs to get to the plant in the minimum time.

Tasmania runs in a straight line from his camp to a point `(x,y)` on the river bank where  `x <= sqrt7/2`. He then swims up the river to the desalination plant.

Tasmania runs from his camp to the river at 2 km per hour. The time that he takes to swim to the desalination plant is proportional to the difference between the `y`-coordinates of the desalination plant and the point where he enters the river.

  1. Show that the total time taken to get to the desalination plant is given by
     

     

    `qquadT = 1/2 sqrt(x^4 - x^2 + 1) + 1/4k(7 - 4x^2)` hours where `k` is a positive constant of proportionality.  (3 marks)

The value of `k` varies from day to day depending on the weather conditions.

  1. If  `k = 1/(2sqrt13)`
  2.  i. find  `(dT)/(dx)`  (1 mark)
  3. ii. hence find the coordinates of the point where Tasmania should reach the river if he is to get to the desalination plant in the minimum time.  (2 marks)
  4. On one particular day, the value of `k` is such that Tasmania should run directly from his camp to the point  `(1,0)`  on the river to get to the desalination plant in the minimum time. Find the value of `k` on that particular day.  (2 marks)
  5. Find the values of `k` for which Tasmania should run directly from his camp towards the desalination plant to reach it in the minimum time.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2.  i. `(sqrt6/2, 1/2)`
  3. ii. `sqrt7/2\ text(km)`
  4. `text(See Worked Solutions)`
  5.  i. `(dT)/(dx) = (x(2x^2 – 1))/(2sqrt(x^4 – x^2 + 1)) – (sqrt13 x)/13`
  6. ii. `(sqrt3/2, −1/4)`
  7. `1/4`
  8. `(5sqrt37)/74`
Show Worked Solution

a.   `text(S)text(ince)\ \ (m,n)\ \ text(lies on)\ \ y=x^2-1,`

♦ Mean mark 47%.

`=> n=m^2-1`

`V(0,1), D(m,m^2 – 1)`

`L` `= sqrt((m – 0)^2 + ((m^2 – 1) – 1)^2)`
  `= sqrt(m^2 + m^4 – 4m^2 + 4)`
  `= sqrt(m^4 – 3m^2 + 4)\ \ text(… as required)`

 

b.i.   `(dL)/(dm) = (2m^2 – 3m)/(sqrt(m^4 – 3m^2 + 4))`

 

`text(Solve:)\ \ (dL)/(dm) = 0quadtext(for)quadm >= 0`

`:. m = sqrt6/2`

`text(Substitute into:)\ \ D(m, m^2 – 1),`

`:. text(Desalination plant at)\ \ (sqrt6/2, 1/2)`

 

♦ Mean mark part (b)(ii) 41%.
b.ii.   `L(sqrt6/2)` `= sqrt(m^4 – 3m^2 + 4)`
    `=sqrt(36/16-3xx6/4+4`
    `=sqrt7/2`

 

c.   `text(Let)\ \ P(x,x^2 – 1)\ text(be run point on bank)`

♦♦♦ Mean mark part (c) 16%.

`text(Let)\ \ D(sqrt7/2, 3/4)\ text(be desalination location)`

`T` `=\ text(run time + swim time)`
  `= (sqrt((x – 0)^2 + ((x^2 – 1) – 0)^2))/2 + k(3/4 – (x^2 – 1))`
  `= (sqrt(x^2 + x^4 – 2x^2 + 1))/2 + k/4(3 – 4(x^2 – 1))`
`:. T` `= (sqrt(x^4 – x^2 + 1))/2 + 1/4k(7 – 4x^2)`

 

d.i.   `(dT)/(dx) = (x(2x^2 – 1))/(2sqrt(x^4 – x^2 + 1)) – (sqrt13 x)/13`

 

d.ii.   `text(Solve:)\ \ (dT)/(dx) = 0`

♦♦ Mean mark part (d)(ii) 33%.

`x = sqrt3/2`

`y=x^2-1=-1/4`

`:. T_(text(min)) \ text(when point is)\ \ (sqrt3/2, −1/4)`

 

e.  `(dT)/(dx) = (x(2x^2 – 1))/(2sqrt(x^4 – x^2 + 1)) – 2kx`

♦♦ Mean mark part (e) 39%.

 

`text(When)\ \ x=1,`

`text(Solve:)\ \ (dT)/(dx)` `=0\ \ text(for)\ k,`
`1/2 -2k` `=0`
`:.k` `=1/4`

 

f.   `text(Require)\ T_text(min)\ text(to occur at right-hand endpoint)\ \ x = sqrt7/2.`

`text(This can occur in 2 situations:)`

♦♦♦ Mean mark 13%.

`text(Firstly,)\ \ T\ text(has a local min at)\ \ x = sqrt7/2,`

`text(Solve:)\ \ (x(2x^2 – 1))/(2sqrt(x^4 – x^2 + 1)) – 2kx=0| x = sqrt7/2,\ \ text(for)\ k,`

`:.k = (5sqrt37)/74`

 

`text(S)text(econdly,)\ \ T\ text(is decreasing function over)\ x ∈ (0, sqrt7/2),`

`text(Solve:)\ \ (dT)/(dx) <= 0 | x = sqrt7/2,\ text(for)\ k,`

`:. k > (5sqrt37)/74`

`:. k >= (5sqrt37)/74`

Calculus, MET2 2011 VCAA 3

  1. Consider the function  `f: R -> R, f(x) = 4x^3 + 5x-9`.

     

    1. Find  `f^{′}(x)`  (1 mark)
    2. Explain why  `f^{′}(x) >= 5` for all `x`.  (1 mark)
  2. The cubic function `p` is defined by  `p: R -> R, p(x) = ax^3 + bx^2 + cx + k`, where `a`, `b`, `c` and `k` are real numbers.

     

    1. If `p` has `m` stationary points, what possible values can `m` have?  (1 mark)
    2. If `p` has an inverse function, what possible values can `m` have?  (1 mark)
  3. The cubic function `q` is defined by  `q:R -> R, q(x) = 3-2x^3`.

     

    1. Write down a expression for  `q^(−1)(x)`.  (2 marks)
    2. Determine the coordinates of the point(s) of intersection of the graphs of  `y = q(x)`  and  `y = q^(−1)(x)`.  (2 marks)
  4. The cubic function `g` is defined by  `g: R -> R, g(x) = x^3 + 2x^2 + cx + k`, where `c` and `k` are real numbers.

     

    1. If `g` has exactly one stationary point, find the value of `c`.  (3 marks)
    2. If this stationary point occurs at a point of intersection of  `y = g(x)`  and  `g^(−1)(x)`, find the value of `k`.  (3 marks)
Show Answers Only
    1. `f^{′}(x) = 12x^2 + 5`
    2. `text(See Worked Solutions)`
    1. `m = 0, 1, 2`
    2. `m = 0, 1`
    1. `q^(−1)(x) = root(3)((3-x)/2), x ∈ R`
    2. `(1, 1)`
    1. `4/3`
    2. `−10/27`
Show Worked Solution

a.i.   `f^{′}(x) = 12x^2 + 5`

 

a.ii.  `text(S)text(ince)\ \ x^2>=0\ \ text(for all)\ x,`

♦ Mean mark 47%.
` 12x^2` `>= 0`
`12x^2 + 5` `>=  5`
`f^{′}(x)` `>=  5\ \ text(for all)\ x`

 

b.i.   `p(x) = text(is a cubic)`

♦♦♦ Mean mark part (b)(i) 9%, and part (b)(ii) 20%.
MARKER’S COMMENT: Good exam strategy should point students to investigate earlier parts for direction. Here, part (a) clearly sheds light on a solution!

`:. m = 0, 1, 2`

`text{(Note: part a.ii shows that a cubic may have no SP’s.)}`

 

b.ii.   `text(For)\ p^(−1)(x)\ text(to exist)`

`:. m = 0, 1`

 

c.i.   `text(Let)\ y = q(x)`

`text(Inverse: swap)\ x ↔ y`

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

`:. q^(−1)(x) = root(3)((3-x)/2), \ x ∈ R`

 

c.ii.  `text(Any function and its inverse intersect on)`

   `text(the line)\ \ y=x.`

`text(Solve:)\ \ 3-2x^3` `= xqquadtext(for)\ x,`
`x` `= 1`

 

`:.\ text{Intersection at (1, 1)}`

 

♦ Mean mark part (d)(i) 44%.
d.i.    `g^{′}(x)` `= 0`
  `3x^2 + 4x + c` `= 0`
  `Delta` `= 0`
  `16-4(3c)` `= 0`
  `:. c` `= 4/3`

 

d.ii.   `text(Define)\ \ g(x) = x^3 + 2x^2 + 4/3x + k`

♦♦♦ Mean mark part (d)(ii) 14%.

  `text(Stationary point when)\ \ g^{′}(x)=0`

`g^{′}(x) = 3x^2+4x+4/3`

`text(Solve:)\ \ g^{′}(x)=0\ \ text(for)\ x,`

`x = −2/3`

`text(Intersection of)\ g(x)\ text(and)\ g^(−1)(x)\ text(occurs on)\ \ y = x`

`text(Point of intersection is)\  (−2/3, −2/3)`

`text(Find)\ k:`

`g(−2/3)` `= −2/3\ text(for)\ k`
`:. k` ` = −10/27`

Probability, MET2 2011 VCAA 2

In a chocolate factory the material for making each chocolate is sent to one of two machines, machine A or machine B.

The time, `X` seconds, taken to produce a chocolate by machine A, is normally distributed with mean 3 and standard deviation 0.8.

The time, `Y` seconds, taken to produce a chocolate by machine B, has the following probability density function.
 

`f(y) = {{:(0,y < 0),(y/16,0 <= y <= 4),(0.25e^(−0.5(y - 4)),y > 4):}`
 

  1. Find correct to four decimal places

     

    1. `text(Pr)(3 <= X <= 5)`  (1 mark)
    2. `text(Pr)(3 <= Y <= 5)`  (3 marks)
  2. Find the mean of `Y`, correct to three decimal places.  (3 marks)

The content in Part c(i) and c(ii) is no longer in this course.

  1. It can be shown that  `text(Pr)(Y <= 3) = 9/32`. A random sample of 10 chocolates produced by machine B is chosen. Find the probability, correct to four decimal places, that exactly 4 of these 10 chocolate took 3 or less seconds to produce.  (2 marks)

All of the chocolates produced by machine A and machine B are stored in a large bin. There is an equal number of chocolates from each machine in the bin.

It is found that if a chocolate, produced by either machine, takes longer than 3 seconds to produce then it can easily be identified by its darker colour.

  1. A chocolate is selected at random from the bin. It is found to have taken longer than 3 seconds to produce. 
    Find, correct to four decimal places, the probability that it was produced by machine A.  (3 marks)
Show Answers Only

a.i.  `0.4938`

a.ii. `0.4155`

b.    `4.333`

d.    `0.1812`

e.    `0.4103`

Show Worked Solution

a.i.   `X ∼\ N(3,0.8^2)`

`text(Pr)(3 <= X <= 5) = 0.4938\ \ text{(4 d.p.)}`

 

a.ii.    `text(Pr)(3 <= Y <= 5)` `= text(Pr)(3 <= Y <= 4) + (4 < Y <= 5)`
    `= int_3^4 (y/16)dy + int_4^5(1/4 e^(−1/2(y – 4)))dy`
    `= 0.4155\ \ text{(4 d.p.)}`

 

b.    `text(E)(Y)` `= int_0^4 y(y/16)dy + int_4^∞ 0.25ye^(−1/2(y – 4))dy`
    `= 4.333\ \ text{(3 d.p.)}`

 

The content in Part c(i) and c(ii) is no longer in this course.

d.   `text(Solution 1)`

`text(Let)\ \ W = #\ text(chocolates from)\ B\ text(that take less)`

`text(than 3 seconds)`

`W ∼\ text(Bi)(10, 9/32)`

`text(Using CAS: binomPdf)(10, 9/32,4)`

`text(Pr)(W = 4) = 0.1812\ \ text{(4 d.p.)}`
 

`text(Solution 2)`

`text(Pr)(W = 4)`  `=((10),(4)) (9/32)^4 (23/32)^6` 
  `=0.1812`
♦♦♦ Mean mark part (e) 19%.
MARKER’S COMMENT: Students who used tree diagrams were the most successful.

 

e.   
`text(Pr)(A | L)` `= (text(Pr)(AL))/(text(Pr)(L))`
  `= (0.5 xx 0.5)/(0.5 xx 0.5 + 0.5 xx 23/32)`
  `= 0.4103\ \ text{(4 d.p.)}`

Calculus, MET2 2016 VCAA 4

  1. Express  `(2x + 1)/(x + 2)`  in the form  `a + b/(x + 2)`,  where `a` and `b` are non-zero integers.  (2 marks)
  2. Let  `f: R text(\){−2} -> R,\ f(x) = (2x + 1)/(x + 2)`.
  3.  i. Find the rule and domain of  `f^(-1)`, the inverse function of  `f`.  (2 marks)
  4. ii. Part of the graphs of  `f` and  `y = x`  are shown in the diagram below.
     
       
     
  5.    Find the area of the shaded region.  (1 mark)
  6. iii. Part of the graphs of  `f`  and  `f^(-1)`  are shown in the diagram below.

 


 

   Find the area of the shaded region.  (1 mark)

  1. Part of the graph of  `f`  is shown in the diagram below.
     
       
     

     

    The point  `P(c, d)`  is on the graph of  `f`.

     

    Find the exact values of `c` and `d` such that the distance of this point to the origin is a minimum, and find this minimum distance.   (3 marks)

Let  `g: (−k, oo) -> R, g(x) = (kx + 1)/(x + k)`, where `k > 1`.

  1. Show that  `x_1 < x_2`  implies that  `g(x_1) < g(x_2),` where  `x_1 in (−k, oo) and x_2 in (−k, oo)`.  (2 marks)
  2. Let `X` be the point of intersection of the graphs of  `y = g (x) and y = −x`.
  3.   i. Find the coordinates of `X` in terms of `k`.  (2 marks)
  4.  ii. Find the value of `k` for which the coordinates of `X` are  `(-1/2, 1/2)`.  (2 marks)
  5. iii. Let  `Ztext{(− 1, − 1)}, Y(1, 1)`  and `X` be the vertices of the triangle `XYZ`. Let  `s(k)`  be the square of the area of triangle `XYZ`.
     

     

       
     

     

    Find the values of `k` such that  `s(k) >= 1`.  (2 marks)

  6. The graph of `g` and the line  `y = x`  enclose a region of the plane. The region is shown shaded in the diagram below.
     

     

     

    Let  `A(k)`  be the rule of the function `A` that gives the area of this enclosed region. The domain of `A` is  `(1, oo)`.

  7.  i. Give the rule for  `A(k)`.  (2 marks)
  8.  ii. Show that  `0 < A(k) < 2`  for all  `k > 1`.  (2 marks)
Show Answers Only
  1. `2 + {(−3)}/(x + 2)`
  2.   i. `f^(-1) (x) = (-3)/(x – 2) – 2,\ \ x in R text(\){2}`
  3.  ii. `4 – 3 ln(3)\ text(units²)`
  4. iii. `8 – 6 log_e (3)\ text(units²)`
  5. `c = sqrt 3 – 2, \ d = 2 – sqrt 3`
  6. `text(min distance) = (2 sqrt 2 – sqrt 6)`
  7. `text(Proof)\ \ text{(See Worked Solutions)}`
  8.   i. `X (−k + sqrt (k^2 – 1), k – sqrt (k^2 – 1))`
  9.  ii. `k = 5/4`
  10. iii. `k in (1, 5/4]`
  11.  i. `A(k) = (k^2 – 1) log_e ((k – 1)/(k + 1)) + 2k`
  12. ii. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

a.   `text(Solution 1)`

`(2x + 1)/(x + 2)` `= 2 – 3/(x + 2)`
  `= 2 + {(-3)}/(x + 2) qquad [text(CAS: prop Frac) ((2x + 1)/(x + 2))]`

 

`text(Solution 2)`

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

 

b.i.  `text(Let)\ \ y = f(x)`

`text(For Inverse: swap)\ \ x ↔ y`

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

 

`text(Range of)\ f(x):\ \ y in R text(\){2}`

`:. f^(-1) (x) = (-3)/(x – 2) – 2, \ x in R text(\){2}`

 

b.ii.   `text(Find intersection points:)`

`f(x)` `= x`
`(2x+1)/(x+2)` `=x`
`2x+1` `=x^2+2x`
`:. x` `= +- 1`
`:.\ text(Area)` `= int_(-1)^1 (f(x) – x)\ dx`
  `=int_(-1)^1 (2 – 3/(x + 2) – x)\ dx`
  `= 4 – 3 ln(3)\ text(u²)`

 

b.iii.    `text(Area)` `= int_(-1)^1 (f(x) – f^(-1) (x)) dx`
    `=2 xx (4 – 3 ln(3))\ \ \ text{(twice the area in (b)(ii))}`
    `= 8 – 6 log_e (3)\ text(u²)`
♦♦♦ Mean mark part (c) 25%.

 

c.   
  `text(Let)\ \ z` `= OP, qquad P(c, -3/(c + 2) + 2)`
  `z` `= sqrt (c^2 + (2 – 3/(c + 2))^2), \ c > -2`

`text(Stationary point when:)`

`(dz)/(dc) = 0, c > -2`

`:.\ c = sqrt 3 – 2 overset and (->) d = 2 – sqrt 3`

`:. text(Minimum distance) = (2 sqrt 2 – sqrt 6)`

 

d.   `text(Given:)\ \ -k < x_1 < x_2`

♦♦♦ Mean mark part (d) 8%.

`text(Must prove:)\ \ g(x_2) – g(x_1) > 0`

`text(LHS:)`

`g(x_2) – g(x_1)`

`= (kx_2 +1)/(x_2+k) – (kx_1 +1)/(x_1+k)`

`=((kx_2 +1)(x_1+k)-(kx_1 +1)(x_2+k))/((x_2+k)(x_1+k))`

`=(k^2(x_2-x_1) – (x_2-x_1))/((x_2+k)(x_1+k))`

`=((k^2-1)(x_2-x_1))/((x_2+k)(x_1+k))`

 

`x_2 – x_1 >0,\ and \ k^2-1>0`

`text(S)text(ince)\ \ x_2>x_1> -k,`

`=> -x_2<-x_1<k`

`=>k+x_1 >0, \ and \ k+x_2>0`

 

`:.g(x_2) – g(x_1) >0`

`:. g(x_2) > g(x_1)`

 

♦♦ Mean mark part (e)(i) 32%.
e.i.    `g(x)` `= -x`
  `(kx +1)/(x+k)` `=-x`
  `kx+1` `=-x^2-xk`
  `x^2+2k+1` `=0`
  `:. x` `=(-2k +- sqrt(4k^2-4))/2`
    `= sqrt (k^2 – 1) – k\ \ text(for)\ \ x > -k`

 

`:. X (-k + sqrt(k^2 – 1),\ \ k – sqrt(k^2 – 1))`

 

e.ii.   `text(Equate)\ \ x text(-coordinates:)`

♦ Mean mark part (e)(ii) 44%.
`-k + sqrt(k^2 – 1)` `= -1/2`
`sqrt(k^2 – 1)` `=k-1/2`
`k^2-1` `=k^2-k+1/4`
`:. k` `= 5/4`

 

e.iii.   `s(k)` `= (1/2 xx YZ xx XO)^2`
    `= 1/4 xx (YZ)^2 xx (XO)^2`

 

`ZO = sqrt(1^2+1^2) = sqrt2`

♦♦♦ Mean mark (e)(iii) 6%.

`YZ=2 xx ZO = 2sqrt2`

`(YZ)^2 = 8`

`(XO)^2` `=(-k + sqrt(k^2 – 1))^2 – (k – sqrt (k^2 – 1))^2`
  `=2(-k + sqrt(k^2 – 1))^2`
   

 `text(Solve)\ \ s(k) >= 1\ \ text(for)\ \ k >= 1,`

`1/4 xx 8 xx 2(-k + sqrt(k^2 – 1))^2` `>=1`
`(-k + sqrt(k^2 – 1))^2` `>=1/4`
`k-sqrt(k^2-1)` `>=1/2`
`k-1/2` `>= sqrt(k^2-1)`
`k^2-k+1/4` `>=k^2-1`
`:.k` `<= 5/4`

`:.  1<k<= 5/4`

 

♦♦ Mean mark (f)(i) 28%.
f.i.    `A(k)` `= int_(-1)^1 (g(x) – x)\ dx,\ \ k > 1`
    `= int_(-1)^1 ((kx+1)/(x+k) -x)\ dx`
    `= int_(-1)^1 (k+ (1-k^2)/(x+k) -x)`
    `=[kx + (1-k^2) log_e(x+k)-x^2/2]_(-1)^1`
    `=(k+(1-k^2)log_e(1+k)-1/2)-(-k+(1-k^2)log_e(k-1)-1/2)`
    `=2k+(1-k^2)log_e ((1+k)/(k-1))`
    `= (k^2 – 1) log_e ((k – 1)/(k + 1)) + 2k`

 

♦♦♦ Mean mark part (f)(ii) 4%.
f.ii.  
  `0` `< A(k) < text(Area of)\ Delta ABC`
  `0` `< A(k) < 1/2 xx AC xx BO`
  `0` `< A(k) < 1/2 sqrt(2^2 + 2^2) xx (sqrt(1^2 + 1^2))`
  `0` `< A(k) < 1/2 xx 2 sqrt 2 xx sqrt 2`
  `:. 0` `< A(k) < 2`

Probability, MET2 2016 VCAA 3

A school has a class set of 22 new laptops kept in a recharging trolley. Provided each laptop is correctly plugged into the trolley after use, its battery recharges.

On a particular day, a class of 22 students uses the laptops. All laptop batteries are fully charged at the start of the lesson. Each student uses and returns exactly one laptop. The probability that a student does not correctly plug their laptop into the trolley at the end of the lesson is 10%. The correctness of any student’s plugging-in is independent of any other student’s correctness.

  1. Determine the probability that at least one of the laptops is not correctly plugged into the trolley at the end of the lesson. Give your answer correct to four decimal places.  (2 marks)
  2. A teacher observes that at least one of the returned laptops is not correctly plugged into the trolley.

     

    Given this, find the probability that fewer than five laptops are not correctly plugged in. Give your answer correct to four decimal places.  (2 marks)

The time for which a laptop will work without recharging (the battery life) is normally distributed, with a mean of three hours and 10 minutes and standard deviation of six minutes. Suppose that the laptops remain out of the recharging trolley for three hours.

  1. For any one laptop, find the probability that it will stop working by the end of these three hours. Give your answer correct to four decimal places.  (2 marks)

A supplier of laptops decides to take a sample of 100 new laptops from a number of different schools. For samples of size 100 from the population of laptops with a mean battery life of three hours and 10 minutes and standard deviation of six minutes, `hat P` is the random variable of the distribution of sample proportions of laptops with a battery life of less than three hours.

  1. Find the probability that `text(Pr) (hat P >= 0.06 | hat P >= 0.05)`. Give your answer correct to three decimal places. Do not use a normal approximation.  (3 marks)

It is known that when laptops have been used regularly in a school for six months, their battery life is still normally distributed but the mean battery life drops to three hours. It is also known that only 12% of such laptops work for more than three hours and 10 minutes.

  1. Find the standard deviation for the normal distribution that applies to the battery life of laptops that have been used regularly in a school for six months, correct to four decimal places.  (2 marks)

The laptop supplier collects a sample of 100 laptops that have been used for six months from a number of different schools and tests their battery life. The laptop supplier wishes to estimate the proportion of such laptops with a battery life of less than three hours.

  1. Suppose the supplier tests the battery life of the laptops one at a time.

     

    Find the probability that the first laptop found to have a battery life of less than three hours is the third one.  (1 mark)

The laptop supplier finds that, in a particular sample of 100 laptops, six of them have a battery life of less than three hours.

  1. Determine the 95% confidence interval for the supplier’s estimate of the proportion of interest. Give values correct to two decimal places.  (1 mark)
  2. The supplier also provides laptops to businesses. The probability density function for battery life, `x` (in minutes), of a laptop after six months of use in a business is
     

     

    `qquad qquad f(x) = {(((210 - x)e^((x - 210)/20))/400, 0 <= x <= 210), (0, text{elsewhere}):}`
     

    1. Find the mean battery life, in minutes, of a laptop with six months of business use, correct to two decimal places.  (1 mark)
    2. This content is no longer in the Study Design.
Show Answers Only
  1. `0.9015`
  2. `0.9311`
  3. `0.0478`
  4. `0.658`
  5. `8.5107`
  6. `1/8`
  7. `p in (0.01, 0.11)`
    1. `170.01\ text(min)`
    2. This content is no longer in the Study Design.
Show Worked Solution

a.   `text(Solution 1)`

`text(Let)\ \ X = text(number not correctly plugged),`

`X ~ text(Bi) (22, .1)`

`text(Pr) (X >= 1) = 0.9015\ \ [text(CAS: binomCdf)\ (22, .1, 1, 22)]`

 

`text(Solution 2)`

`text(Pr) (X>=1)` `=1 – text(Pr) (X=0)`
  `=1 – 0.9^22`
  `=0.9015\ \ text{(4 d.p.)}`

 

 b.   `text(Pr) (X < 5 | X >= 1)`

MARKER’S COMMENT: Early rounding was a common mistake, producing 0.9312.

`= (text{Pr} (1 <= X <= 4))/(text{Pr} (X >= 1))`

`= (0.83938…)/(0.9015…)\ \ [text(CAS: binomCdf)\ (22, .1, 1,4)]`

`= 0.9311\ \ text{(4 d.p.)}`

 

c.   `text(Let)\ \ Y = text(battery life in minutes)`

MARKER’S COMMENT: Some working must be shown for full marks in questions worth more than 1 mark.

`Y ~ N (190, 6^2)`

`text(Pr) (Y <= 180)= 0.0478\ \ text{(4 d.p.)}`

`[text(CAS: normCdf)\ (−oo, 180, 190,6)]`

 

d.   `text(Let)\ \ W = text(number with battery life less than 3 hours)`

♦ Mean mark part (d) 33%.

`W ~ Bi (100, .04779…)`

`text(Pr) (hat P >= .06 | hat P >= .05)` `= text(Pr) (X_2 >= 6 | X_2 >= 5)`
  `= (text{Pr} (X_2 >= 6))/(text{Pr} (X_2 >= 5))`
  `= (0.3443…)/(0.5234…)`
  `= 0.658\ \ text{(3 d.p.)}`

 

e.   `text(Let)\ \ B = text(battery life), B ~ N (180, sigma^2)`

`text(Pr) (B > 190)` `= .12`
`text(Pr) (Z < a)` `= 0.88`
`a` `dot = 1.17499…\ \ [text(CAS: invNorm)\ (0.88, 0, 1)]`
`-> 1.17499` `= (190 – 180)/sigma\ \ [text(Using)\ Z = (X – u)/sigma]`
`:. sigma` `dot = 8.5107`

 

f.    `text(Pr) (MML)` `= 1/2 xx 1/2 xx 1/2`
    `= 1/8`

 

g.   `text(95% confidence int:) qquad quad [(text(CAS:) qquad qquad 1 – text(Prop)\ \ z\ \ text(Interval)), (x = 6), (n = 100)]`

`p in (0.01, 0.11)`

 

h.i.    `mu` `= int_0^210 (x* f(x)) dx`
  `:. mu` `dot = 170.01\ text(min)`

 

h.ii.    This content is no longer in the Study Design.