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Probability, MET2 2007 VCAA 7 MC

The random variable `X` has a normal distribution with mean 11 and standard deviation 0.25.

If the random variable `Z` has the standard normal distribution, then the probability that `X` is less than 10.5 is equal to

  1. `text(Pr) (Z > 2)`
  2. `text(Pr) (Z < – 1.5)`
  3. `text(Pr) (Z < 1)`
  4. `text(Pr) (Z >= 1.5)`
  5. `text(Pr) (Z < – 4)`
Show Answers Only

`A`

Show Worked Solution
`text(Pr) (X < 10.5)` `= text(Pr) (Z < (10.5 – 11)/0.25)`
  `= text(Pr) (Z < – 2)`
  `= text(Pr) (Z > 2)`

 
`=> A`

Algebra, MET2 2007 VCAA 3 MC

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

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

Algebra, MET2 2007 VCAA 2 MC

Let  `g(x) = x^2 + 2x - 3 and f(x) = e^(2x + 3).`

Then  `f(g(x))`  is given by 

  1. `e^(4x + 6) + 2 e^(2x + 3) - 3`
  2. `2x^2 + 4x - 6`
  3. `e^(2x^2 + 4x + 9)`
  4. `e^(2x^2 + 4x - 3)`
  5. `e^(2x^2 + 4x - 6)`
Show Answers Only

`D`

Show Worked Solution

`text(Solution 1)`

`text(Define)\ \ f(x) and g(x)\ \ text(on CAS)`

`f(g(x)) = e^(2x^2 + 4x – 3)`

`=>   D`
 

`text(Solution 2)`

`f(g(x))` `=e^(2 xx (x^2 + 2x – 3)+3)`
  `= e^(2x^2 + 4x – 3)`

`=>D`

Algebra, MET2 SM-Bank 8 MC

For the polynomial  `P(x) = x^3 − ax + 4,\ \ P( – 3) = – 5.`

The value of  `a`  is

A.   `− 12`

B.    `− 5`

C.    `– 3`

D.       `3`

E.       `6`

Show Answers Only

`E`

Show Worked Solution
`(-3)^3 -a(-3)+4` `= -5`
`-27+3a+4`  `= -5`
`3a`  `=18`
`a`  `= 6`

 
`⇒ E`

Algebra, MET2 SM-Bank 7 MC

If  `x-2`  is a factor of  `2x^3 - 10x^2 + 6x + a`  where  `a in R text{\}{0},`  then the value of `a` is

A.   `-68`

B.   `-20`

C.     `-2`

D.        `2`

E.      `12`

Show Answers Only

`E`

Show Worked Solution

`text(S)text(ince)\ \ x-2\ \ text(is a factor,)`

`=> P(2)=0`

`P(2)` `= 2 · 2^3 – 10 · 2^2 + 6 · 2 + a`
`0`  `= 16-40+12+a`
`a` `=12`

 

`=>  E`

Graphs, MET2 2008 VCAA 8 MC

The graph of the function  `f: D -> R,\ f(x) = (x - 3)/(2 - x),` where `D` is the maximal domain has asymptotes

  1. `x = 3,\ \ \ \ \ \ \ \ \ \ y = 2`
  2. `x = -2,\ \ \ \ \ y = 1`
  3. `x = 1,\ \ \ \ \ \ \ \ \ \ y = -1`
  4. `x = 2,\ \ \ \ \ \ \ \ \ \ y = -1`
  5. `x = 2,\ \ \ \ \ \ \ \ \ \ y = 1`
Show Answers Only

`D`

Show Worked Solution

`text(Use proper fraction tool on CAS:)`

`[text(CAS: propFrac) ((x – 3)/(2 – x))]`

`f(x) = -1 – 1/(x – 2)`

`:.\ text(Asymptotes:)\ \ x = 2, y = – 1`

`=>   D`

Algebra, MET2 2008 VCAA 7 MC

The inverse of the function  `f: R^+ -> R,\ f(x) = 1/sqrt x - 3`  is

  1. `{:f^-1: R^+ -> R, qquad qquad qquad qquad f^-1(x) = (x + 3)^2:}`
  2. `{:f^-1: R^+ -> R, qquad qquad qquad qquad f^-1(x) = 1/x^2 + 3:}`
  3. `{:f^-1: (3, oo) -> R, qquad qquad qquad f^-1 (x) = (-1)/(x - 3)^2:}`
  4. `{:f^-1: text{(−3, ∞)} -> R, qquad qquad f^-1 (x) = 1/(x + 3)^2:}`
  5. `{:f^-1: text{(−3, ∞)} -> R, qquad qquad f^-1 (x) = -1/x^2 - 3:}`
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ y = f(x)`

`text(Inverse:  swap)\ \ x harr y`

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

 

`text(Domain)\ (f^-1(x))` `= text(Range)\ (f)`
  `= (– 3, oo)`

`=>   D`

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.

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`

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`

Calculus, MET1 2007 ADV 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′` `= 2` `v′` `= e^x`
`(dy)/(dx)` `= (u′v – uv′)/(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)`

Calculus, MET1 2016 VCAA 3

Let  `f: R text{\}{1} -> R`  where  `f(x) = 2 + 3/(x - 1)`.

  1. Sketch the graph of  `f`. Label the axis intercepts with their coordinates and label any asymptotes with the appropriate equation.  (3 marks)

     

  2. Find the area enclosed by the graph of  `f`, the lines  `x = 2`  and  `x = 4`, and the `x`-axis.  (2 marks)
Show Answers Only
  1.  
  2. `4 + 3log_e(3)\ text(units)`
Show Worked Solution
a.   

 

b.    `text(Area)` `= int_2^4 2 + 3(x – 1)^(−1)\ dx`
    `= [2x + 3 log_e(x – 1)]_2^4`
    `= (8 + 3log_e(3)) – (4 + 3log_e(1))`
    `= 4 + 3log_e(3)\ \ text(u²)`

Probability, MET2 2009 VCAA 3

The Bouncy Ball Company (BBC) makes tennis balls whose diameters are normally distributed with mean 67 mm and standard deviation 1 mm. The tennis balls are packed and sold in cylindrical tins that each hold four balls. A tennis ball fits into such a tin if the diameter of the ball is less than 68.5 mm.

  1. What is the probability, correct to four decimal places, that a randomly selected tennis ball produced by BBC fits into a tin?  (2 marks)

BBC management would like each ball produced to have diameter between 65.6 and 68.4 mm.

  1. What is the probability, correct to four decimal places, that the diameter of a randomly selected tennis ball made by BBC is in this range?  (2 marks)
    1. What is the probability, correct to four decimal places, that the diameter of a tennis ball which fits into a tin is between 65.6 and 68.4 mm?  (1 mark)
    2. A tin of four balls is selected at random. What is the probability, correct to four decimal places, that at least one of these balls has diameter outside the desired range of 65.6 to 68.4 mm?  (2 marks)

BBC management wants engineers to change the manufacturing process so that 99% of all balls produced have diameter between 65.6 and 68.4 mm. The mean is to stay at 67 mm but the standard deviation is to be changed.

  1. What should the new standard deviation be (correct to two decimal places)?  (3 marks)

Show Answers Only

a.  `0.9332`

b.  `0.8385`

c.i.  `0.8985`

c.ii.  `0.3482`

d.  `0.54\ text(mm)`

Show Worked Solution

a.   `text(Let)\ \ X = text(diameter),\ \ X ∼ text(N) (67, 1^2)`

`text(Pr) (X < 68.5) = 0.9332\ \ text{(to 4 d.p.)}`

`[text(CAS: normCdf) (– oo, 68.5, 67, 1)]`

 

b.   `text(Pr) (65.6 < X < 68.4)`

`= 0.8385\ \ text{(to 4 d.p.)}`

`[text(CAS: normCdf) (65.6, 68.4, 67, 1)]`

 

c.i.   `text(Pr) (65.6 < X < 68.4 \|\ X < 68.5)`

♦ Mean mark 42%.

`= (text{Pr} (65.6 < X < 68.4))/(text{Pr} (X < 68.5))`

`= (0.838487…)/(0.933193…)`

`= 0.8985\ \ text{(to 4 d.p.)}`

 

  ii.  `text(Let)\ \ Y = text(Number of balls with diameter outside range)`

♦♦ Mean mark 30%.

`Y ∼ text(Bi)(4, 1 – 0.8985…) -> Y ∼ text(Bi) (4, 0.101486)`

`text(Pr) (Y >= 1) = 0.3482\ \ text{(to 4 d.p.)}`

 

d.   `X = text(diameter),\ \ X ∼ text(N) (67, sigma^2)`

♦♦ Mean mark 35%.

 vcaa-graphs-fur2-2009-3di

`text(Pr) (Z < a)` `= 0.005`
`a` `= – 2.5758`

 

`text(Relate 65.6 to its corresponding)\ \ z text(-score):`

`– 2.5758…` `= (65.6 – 67)/sigma`
`:. sigma` `= 0.54\ text(mm)\ \ text{(to 2 d.p.)}`

Calculus, MET2 2009 VCAA 1

Let  `f: R^+ uu {0} -> R,\ f(x) = 6 sqrt x - x - 5.`

The graph of  `y = f (x)`  is shown below.

VCAA 2009 1a

  1. State the interval for which the graph of  `f` is strictly decreasing.  (2 marks)
  2. No longer in syllabus.
  3. Points `A` and `B` are the points of intersection of  `y = f (x)`  with the `x`-axis. Point `A` has coordinates  `(1, 0)`  and point `B` has coordinates  `(25, 0)`.

     

    Find the length of `AD` such that the area of rectangle `ABCD` is equal to the area of the shaded region.  (2 marks)
     
    VCAA 2009 1c

  4. The points `P (16, 3)` and `B (25, 0)` are labelled on the diagram.

      
     

          VCAA 2009 1d

     

    1. Find `m`, the gradient of the chord `PB`.   (Exact value to be given.)  (1 mark)
    2. Find  `a in [16, 25]`  such that  `f prime (a) = m`.  (Exact value to be given.)  (2 marks)
Show Answers Only

a.  `x in [9,oo)`

b.   `text(Not in syllabus.)`

c.  `8/3`

d.i.   `m=-1/3`

d.ii.  `a=81/4`

Show Worked Solution
a.   vcaa-2009-1ai
`text(Stationary point when)\ \ f prime (x)` `= 0`
`x` `= 9`

`:.\ text(Strictly decreasing for)\ \ x in [9, oo)`

 

b.  `text(No longer in syllabus.)`

 

c.   vcaa-2009-1ci
`AD` `= y_text(average)`
  `= 1/(25 – 1) int_1^25\ f(x)\ dx`
  `= 8/3\ \ text{[by CAS]}`

 

d.i.   `m_(PB)` `= (3 – 0)/(16 – 25)`
  `:. m_(PB)` `= – 1/3`

 

 d.ii.   `text(Solve)\ \ f prime (a)` `= -1/3\ \ text(for)\ \ a in [16, 25]`
  `:. a` `= 81/4`

Calculus, MET2 2009 VCAA 7 MC

For  `y = e^(2x) cos (3x)`  the rate of change of `y` with respect to `x` when  `x = 0`  is

  1. `0`
  2. `2`
  3. `3`
  4. `– 6`
  5. `– 1`
Show Answers Only

`B`

Show Worked Solution
`y` `= e^(2x) cos (3x)`
`dy/dx` `=e^(2x) xx -3sin(3x) + 2e^(2x) xx cos (3x)`
  `=e^(2x)(-3sin(3x) + 2cos(3x))`

 
`text(When)\ \ x = 0,`

`dy/dx= 2`

`=>   B`

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 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.

Calculus, MET2 2016 VCAA 2

Consider the function  `f(x) = -1/3 (x + 2) (x-1)^2.`

  1.  i. Given that  `g^{′}(x) = f (x) and g (0) = 1`, show that  `g(x) = -x^4/12 + x^2/2-(2x)/3 + 1`.  (1 mark)
  2. ii. Find the values of `x` for which the graph of  `y = g(x)`  has a stationary point.  (1 mark)

The diagram below shows part of the graph of  `y = g(x)`, the tangent to the graph at  `x = 2`  and a straight line drawn perpendicular to the tangent to the graph at  `x = 2`. The equation of the tangent at the point `A` with coordinates  `(2, g(2))`  is  `y = 3-(4x)/3`.

The tangent cuts the `y`-axis at `B`. The line perpendicular to the tangent cuts the `y`-axis at `C`.
 


 

  1.   i. Find the coordinates of `B`.  (1 mark)
  2.  ii. Find the equation of the line that passes through `A` and `C` and, hence, find the coordinates of `C`.  (2 marks)
  3. iii. Find the area of triangle `ABC`.  (2 marks)
  4. The tangent at `D` is parallel to the tangent at `A`. It intersects the line passing through `A` and `C` at `E`.

     


     
     i. Find the coordinates of `D`.  (2 marks)

  5. ii. Find the length of `AE`.  (3 marks)
Show Answers Only
  1.  i. `text(Proof)\ \ text{(See worked solutions)}`
  2. ii. `x = −2, 1`
  3.   i. `(0, 3)`
  4. ii. `y = 3/4 x-7/6`
  5.      `(0, −7/6)`
  6. iii. `25/6\ text(u²)`
  7. `(−1, 25/12)`
  8. `27/20\ text(u)`
Show Worked Solution
a.i.    `g(x)` `= int f(x)\ dx`
    `=-1/3 int (x + 2) (x-1)^2\ dx`
    `=-1/3int(x^3-3x+2)\ dx`
  `:.g(x)` `= -x^4/12 + x^2/2-(2x)/3 + c`

 

`text(S)text(ince)\ \ g(0) = 1,`

`1` `= 0 + 0-0 + c`
`:. c` `= 1`

 
`:. g(x) = -x^4/12 + x^2/2-(2x)/3 + 1\ \ …\ text(as required)`

 

a.ii.   `text(Stationary point when:)`

`g^{′}(x) = f(x) = 0`

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

`:. x = −2, 1`

 

b.i.   

`B\ text(is the)\ y text(-intercept of)\ \ y = 3-4/3 x`

`:. B (0, 3)`

 

b.ii.   `m_text(norm) = 3/4, \ text(passes through)\ \ A(2, 1/3)`

   `text(Equation of normal:)`

`y-1/3` `=3/4(x-2)`
`y` `=3/4 x -7/6`
   

`:. C (0, −7/6)`

 

b.iii.   `text(Area)` `= 1/2 xx text(base) xx text(height)`
    `= 1/2 xx (3 + 7/6) xx 2`
    `= 25/6\ text(u²)`

 

♦ Mean mark part (c)(i) 43%.
MARKER’S COMMENT: Many students gave an incorrect `y`-value here. Be careful!
c.i.    `text(Solve)\ \ \ g^{′}(x)` `= -4/3\ \ text(for)\ \ x < 0`
  `=> x` `= −1`
  `g(-1)` `=-1/12+1/2+2/3+1`
    `=25/12`

 
`:. D (−1, 25/12)`

 

c.ii.   `text(T) text(angent line at)\ \ D:`

`y-25/12` `=- 4/3(x+1)`
`y` `=- 4/3x + 3/4`

 

`DE\ \ text(intersects)\ \ AE\ text(at)\ E:`

♦♦ Mean mark 32%.
`-4/3 x + 3/4` `= 3/4 x-7/6`
`25/12 x` `= 23/12`
`x` `=23/25`

 
`:. E (23/25, -143/300)`

 

`:. AE` `= sqrt((2-23/25)^2 + (1/3-(-143/300))^2)`
  `= 27/20\ text(units)`

Calculus, MET2 2016 VCAA 1

Let  `f: [0, 8 pi] -> R, \ f(x) = 2 cos (x/2) + pi`.

  1. Find the period and range of  `f`.  (2 marks)
  2. State the rule for the derivative function  `f prime`.  (1 mark)
  3. Find the equation of the tangent to the graph of  `f` at  `x = pi`.  (1 mark)
  4. Find the equations of the tangents to the graph of  `f: [0, 8 pi] -> R,\ \ f(x) = 2 cos (x/2) + pi`  that have a gradient of 1.  (2 marks)
  5. The rule of  `f prime` can be obtained from the rule of  `f` under a transformation `T`, such that
      
    `qquad T: R^2 -> R^2,\ T([(x), (y)]) = [(1, 0), (0, a)] [(x), (y)] + [(−pi), (b)]`
     

     

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

  6. Find the values of  `x, \ 0 <= x <= 8 pi`, such that  `f(x) = 2 f prime (x) + pi`.  (2 marks)
Show Answers Only
  1. `text(Period:)\ 4 pi; qquad text(Range:)\ [pi – 2, pi + 2]`
  2. `f prime (x) = −sin (x/2)`
  3. `y = −x + 2 pi`
  4. `y = x – 2 pi and y = x – 6 pi`
  5. `a = 1/2 and b = -pi/2`
  6. `x = (3 pi)/2, (7 pi)/2, (11 pi)/2, (15 pi)/2`
Show Worked Solution

a.   `text(Period)= (2pi)/n = (2 pi)/(1/2) = 4pi`

MARKER’S COMMENT: Including round brackets rather than square ones was a common mistake.

`text(Range:)\ [pi – 2, pi + 2]`
 

b.   `f prime (x) = text(−sin) (x/2)`

 

c.   `[text(CAS: tangentLine)\ (f(x), x, pi)]`

`y = -x + 2 pi`

 

d.   `text(Solve)\ \ f prime (x) = 1\ \ text(for)\ x in [0, 8 pi]`

♦ Mean mark part (d) 50%.

`-> x = 3 pi or 7 pi`

`:. y = x – 2 pi and y = x – 6 pi\ \ [text(CAS)]`

 

e.   `text(Using the transition matrix,)`

♦♦ Mean mark part (e) 27%.
`x_T` `=x-pi`
`x` `=x_T+pi`
`y_T` `=ay+b`
`y` `=(y_T-b)/a`
   

`f(x)= cos (x/2) + pi/2\ \ ->\ \ f′(x) = -sin(x/2)`

`(y_T-b)/a` `=2cos((x_T+pi)/2)+pi`
`y_T` `=2a cos((x_T+pi)/2)+a pi +b`
  `=-2a sin(x_T/2)+a pi + bqquad [text(Complementary Angles)]`
   
`-2a` `=-1`
`:. a` `=1/2`
`1/2 pi +b` `=0`
`:.b` `=-pi/2`

 

f.   `text(Solve)\ \ f(x) = 2 f prime (x) + pi\ \ text(for)\ \ x in [0, 8 pi]`

♦ Mean mark part (f) 50%.
`2 cos (x/2) + pi` `= -2 sin(x/2)+pi`
`tan(x/2)` `=-1`
`x/2` `=(3pi)/4, (7pi)/4, (11pi)/4, (15pi)/4`
`:.x` `= (3 pi)/2, (7 pi)/2, (11 pi)/2, (15 pi)/2`

Probability, MET2 2010 VCAA 15 MC

The discrete random variable `X` has the following probability distribution.
 

VCAA 2010 15mc

 
If the mean of `X` is 1 then

  1. `a = 0.3 and b = 0.1`
  2. `a = 0.2 and b = 0.2`
  3. `a = 0.4 and b = 0.2`
  4. `a = 0.1 and b = 0.5`
  5. `a = 0.1 and b = 0.3`
Show Answers Only

`C`

Show Worked Solution

`text(E)(X) = 1,`

`1 xx b + 2 xx 0.4` `=1`
`b` `=0.2`

 

`text(Sum of probabilities) = 1,`

`a + 0.2 + 0.4` `= 1`
`a` `=0.4`

`=>   C`

Calculus, MET2 2010 VCAA 6 MC

A function `g` with domain `R` has the following properties.

   ●    `g prime (x) = x^2 - 2x`

   ●    the graph of  `g(x)`  passes through the point  `(1, 0)`

`g (x)`  is equal to

A.   `2x - 2`

B.   `x^3/3 - x^2`

C.   `x^3/3 - x^2 + 2/3`

D.   `x^2 - 2x + 2`

E.   `3x^3 - x^2 - 1`

Show Answers Only

`C`

Show Worked Solution
`g(x)` `= int (x^2 – 2x)\ dx`
  `= 1/3 x^3 – x^2 + c`

 

`text(Graph passes through)\ \ (1, 0),`

`0` `= 1/3 (1)^3 – (1)^2 + c`
`c` `= 2/3`
`:. g(x)` `= 1/3 x^3 – x^2 + 2/3`

`=>   C`

Algebra, MET2 2010 VCAA 4 MC

If  `f(x) = 1/2e^(3x)  and  g(x) = log_e(2x) + 3`  then  `g (f(x))` is equal to
 

  1. `2x^3 + 3`
  2. `e^(3x) + 3`
  3. `e^(8x + 9)`
  4. `3(x + 1)`
  5. `log_e (3x) + 3`
Show Answers Only

`D`

Show Worked Solution

`text(Define)\ \ f(x)= 1/2e^(3x), \ g(x)= log_e(2x) + 3`

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

 
`=>   D`

Probability, MET2 2016 VCAA 16 MC

The random variable, `X`, has a normal distribution with mean 12 and standard deviation 0.25

If the random variable, `Z`, has the standard normal distribution, then the probability that `X` is greater than 12.5 is equal to

  1. `text(Pr) (Z < text{− 4)}`
  2. `text(Pr) (Z < text{− 1.5)}`
  3. `text(Pr) (Z < 1)`
  4. `text(Pr) (Z >= 1.5)`
  5. `text(Pr) (Z > 2)`
Show Answers Only

`E`

Show Worked Solution

`text(Pr) (X > 12.5)`

`= text(Pr) (Z > (12.5 – 12)/.25)`

`= text(Pr) (Z > 2)`
 

`=>   E`

Probability, MET2 2016 VCAA 15 MC

A box contains six red marbles and four blue marbles. Two marbles are drawn from the box, without replacement.

The probability that they are the same colour is

  1. `1/2`
  2. `28/45`
  3. `7/15`
  4. `3/5`
  5. `1/3`
Show Answers Only

`C`

Show Worked Solution

`text{Pr(both marbles the same colour)}`

`= text(Pr) (R, R) + text(Pr) (B, B)`

`= 6/10 xx 5/9 + 4/10 xx 3/9`

`= 7/15`
 

`=>   C`

Graphs, MET2 2016 VCAA 8 MC

The UV index, `y`, for a summer day in Melbourne is illustrated in the graph below, where `t` is the number of hours after 6 am.
 

     
 

The graph is most likely to be the graph of

A.   `y = 5 + 5 cos ((pi t)/7)`

B.   `y = 5 - 5 cos ((pi t)/7)`

C.   `y = 5 + 5 cos ((pi t)/14)`

D.   `y = 5 - 5 cos ((pi t)/14)`

E.   `y = 5 + 5 sin ((pi t)/14)`

Show Answers Only

`B`

Show Worked Solution

`text(Median) = (0 + 10)/2 = 5`

`text(Amplitude) = 5`

`text(Period:)\ \ 14` `= (2 pi)/n`
`n` `= pi/7`

 

`:.\ text(Graph:)\ \ y = 5 – 5 cos ((pi t)/7)`

`=>   B`

Probability, MET2 2016 VCAA 7 MC

The number of pets, `X`, owned by each student in a large school is a random variable with the following discrete probability distribution.
 

  

 
If two students are selected at random, the probability that they own the same number of pets is

  1. `0.3`
  2. `0.305`
  3. `0.355`
  4. `0.405`
  5. `0.8`
Show Answers Only

`C`

Show Worked Solution
`text(Pr) (0, 0)` `+ text(Pr) (1, 1)` `+ text(Pr) (2, 2)` `+ text(Pr) (3, 3)`
`= .5^2 ` `+ qquad .25^2` `+ qquad .2^2` `+ qquad .05^2`
`= .355`      

 
`=>   C`

Algebra, MET2 2016 VCAA 5 MC

Which one of the following is the inverse function of  `g: [3, oo) -> R,\ g(x) = sqrt (2x - 6)?`

  1. `g^(-1): [3, oo) -> R,\ g^(-1) (x) = (x^2 + 6)/2`
  2. `g^(-1): [0, oo) -> R,\ g^(-1) (x) = (2x - 6)^2`
  3. `g^(-1): [0, oo) -> R,\ g^(-1) (x) = sqrt (x/2 + 6)`
  4. `g^(-1): [0, oo) -> R,\ g^(-1) (x) = (x^2 + 6)/2`
  5. `g^(-1): R -> R,\ g^(-1) (x) = (x^2 + 6)/2`
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ y = g(x)`

`text(Inverse: swap)\ x ↔ y`

`x` `= sqrt (2y – 6)`
`x^2` `= 2y – 6`
`y` `= (x^2 + 6)/2`

 

`text(Domain)\ (g^(-1)) = text(Range)\ (g) = [0, oo)`

`=>   D`

Calculus, MET2 2016 VCAA 3 MC

Part of the graph  `y = f(x)`  of the polynomial function  `f` is shown below
 


 

`f prime (x) < 0`  for

  1. `x ∈ (−2, 0) uu (1/3, oo)`
  2. `x ∈ (−9, 100/27)`
  3. `x ∈ (−oo, −2) uu (1/3, oo)`
  4. `x ∈ (−2, 1/3)`
  5. `x ∈ (−oo, −2] uu (1, oo)`
Show Answers Only

`C`

Show Worked Solution

`text(Outlining the sections of negative gradient:)`

`f prime (x) < 0`

`=>   C`

Calculus, MET2 2011 VCAA 11 MC

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

  1. `log_e(2)`
  2. `1/3log_e(6)`
  3. `log_e(3125/4) - 3`
  4. `1/3log_e(3125/4) - 3`
  5. `(5log_e(5) - 2log_e(2) - 3)/3` 
Show Answers Only

`=> E`

Show Worked Solution
`y_text(avg)` `= 1/(3 – 0) int_0^3 log_e(x + 2)\ dx`
  `= (5log_e(5) – 2log_e(2) – 3)/3`

 
`=> E`

Calculus, MET2 2011 VCAA 9 MC

The graph of the function  `y = f(x)`  is shown below.
 

met1-2011-vcaa-9-mc
 

Which if the following could be the graph of the derivative function  `y = f′(x)`?

met1-2011-vcaa-9-mc-abc

met1-2011-vcaa-9-mc-de1

Show Answers Only

`=> B`

Show Worked Solution

`text(Stationary points at)\ \ x = – 3, 0`

`:. f′(x)\ \ text(has)\ x text(-intercepts at)\ -3, 0.`

`text(Only)\ B, C or D\ text(possible).`

`text(By inspection of the)\ \ f(x)\ \ text(graph:)`

`f′(x) > 0quadtext(for)quadx < −3`

`text(Only)\ B\ text(satisfies.)`

`=> B`

Algebra, MET2 2011 VCAA 5 MC

The inverse function of  `g: [2,∞) -> R, g(x) = sqrt(2x - 4)`  is

  1. `g^(−1): [2,∞) -> R, g^(−1)(x) = (x^2 + 4)/2`
  2. `g^(−1): [0,∞) -> R, g^(−1)(x) = (2x - 4)^2`
  3. `g^(−1): [0,∞) -> R, g^(−1)(x) = sqrt(x/2 + 4)`
  4. `g^(−1): [0,∞) -> R, g^(−1)(x) = (x^2 + 4)/2`
  5. `g^(−1): R -> R, g^(−1)(x) = (x^2 + 4)/2`
Show Answers Only

`=> D`

Show Worked Solution

`text(Let)\ \ y = g(x)`

`text(Inverse: swap)\ x ↔ y`

`x` `= sqrt(2y – 4)`
`x^2` `= 2y-4`
`2y` `=x^2+4`
`:. y` `= (x^2 +4)/2`

 

`g^(−1)(x) = (x^2 + 4)/2`

 

`text(Domain)\ (g^(−1)) = text(Range)\ g(x) = [0,∞)`

`=> D`

Graphs, MET2 2011 VCAA 1 MC

The midpoint of the line segment joining  `text{(0, − 5)}`  to  `(d,0)`  is

  1. `(d/2,−5/2)`
  2. `(0,0)`
  3. `((d - 5)/2,0)`
  4. `(0,(5 - d)/2)`
  5. `((5 + d)/2,0)`
Show Answers Only

`=> A`

Show Worked Solution
`text(Midpoint)` `= ((0 + d)/2,(−5 + 0)/2)`
  `= (d/2,−5/2)`

 
`=> A`

CORE, FUR2 SM-Bank VCE 2

Spiro is saving for a car. He has an account with $3500 in it at the start of the year.

At the end of each month, Spiro adds another $180 to the account.

The account pays 3.6% interest per annum, compounded monthly.

  1. i. What is the interest rate per month?  (1 mark)
  2. ii. Write a recurrence relation that models Spiro's investment, with `V_n` representing the balance of his account after `n` months.  (1 mark)
  3. What will be the balance of Spiro's account after 3 months?

     

    Write your answer correct to the nearest cent.  (1 mark)

Show Answers Only
    1. `0.3text(% per month)`
    2. `V_0 = 3500,qquadV_(n + 1) = V_n xx 1.003 + 180`
  2. `$4073\ \ (text(nearest $))`
Show Worked Solution
a.i.    `text(Interest rate)` `= 3.6/12`
    `= 0.3text(% per month)`

 

a.ii.    `V_0` `= 3500`
  `V_1`  `= 3500 xx 1.003 + 180`
  `V_2`  `= V_1 xx 1.003 + 180`
  `vdots`   
  `V_(n + 1)`  `= V_n xx 1.003 + 180` 

 

`:.\ text(Recurrence relationship:)`

`V_0 = 3500,qquadV_(n + 1) = V_n xx 1.003 + 180`

 

b.    `V_1` `= 3500 xx 1.003 + 180 = $3690.50`
  `V_2` `= 3690.50 xx 1.003 + 180 = $3881.5715`
`V_3` `= 3881.5715 xx 1.003 + 180` `= $4073.216…`
    `= $4073.22\ \ text{(nearest cent)}`

CORE, FUR2 SM-Bank VCE 1

Joe buys a tractor under a buy-back scheme. This scheme gives Joe the right to sell the tractor back to the dealer through either a flat rate depreciation or unit cost depreciation.

  1. The recurrence relation below can be used to calculate the price Joe sells the tractor back to the dealer `(P_n)`, based on the flat rate depreciation, after `n` years
     

     

    `qquadP_0 = 56\ 000,qquadP_n = P_(n - 1) - 7000`
     

    1. Write the general rule to find the value of `P_n` in terms of `n`.  (1 mark)
    2. Hence, or otherwise, find the time it will take Joe's tractor to lose half of its value.  (1 mark)
  2. If Joe uses the unit cost method to depreciate his tractor, he depreciates $2.75 per kilometre travelled.

     

    1. How many kilometres does Joe's tractor need to travel for half its value to be depreciated? Round your answer to the nearest kilometre.  (1 mark)
    2. Joe's tractor travels, on average, 2500 kilometres per year. Which method, flat rate depreciation or unit cost depreciation, will result in the greater annual depreciation? Write down the greater depreciation amount correct to the nearest dollar.  (1 mark)
Show Answers Only
    1. `P_n = 56\ 000 – 7000n`
    2. `4\ text(years)`
    1. `10\ 182\ text{km  (nearest km)}`
    2. `text(The flat rate depreciation results in an extra)`
      `text($125 depreciation each year.)`
Show Worked Solution
a.i.    `P_1` `= P_0 – 7000`
  `P_2` `= P_0 – 7000 – 7000`
    `= 56\ 000 – 7000 xx 2`
  `vdots`  
  `P_n`  `= 56\ 000 – 7000n` 

 

a.ii.    `text(Depreciated value)` `= 56\ 000 ÷ 2=$28\ 000`

`text(Find)\ n,`

`28\ 000` `= 56\ 000 – 7000n`
`7000n` `= 28\ 000`
`:. n` `= 4\ text(years)`

 

b.i.    `text(Distance travelled)` `= ((56\ 000 – 28\ 000))/2.75`
    `= 10\ 181.81…`
    `= 10\ 182\ text{km  (nearest km)}`

 

b.ii.   `text(Annual depreciation of unit cost)`

`= 2500 xx $2.75`

`= $6875`

`text(Annual flat rate depreciation = $7000)`

`text(Difference)\ = 7000 – 6875 = $125`
 

`:.\ text(The flat rate depreciation results in an extra)`

 `text($125 depreciation each year.)`

NETWORKS, FUR2 2016 VCAA 2

The suburb of Alooma has a skateboard park with seven ramps.

The ramps are shown as vertices `T`, `U`, `V`, `W`, `X`, `Y` and `Z` on the graph below.
 

 

The tracks between ramps `U` and `V` and between ramps `W` and `X` are rough, as shown on the graph above.

  1. Nathan begins skating at ramp `W` and follows an Eulerian trail.

     

    At which ramp does Nathan finish?  (1 mark)

  2. Zoe begins skating at ramp `X` and follows a Hamiltonian path.

     

    The path she chooses does not include the two rough tracks.

     

    Write down a path that Zoe could take from start to finish.  (1 mark)

  3. Birra can skate over any of the tracks, including the rough tracks.

     

    He begins skating at ramp `X` and will complete a Hamiltonian cycle.

     

    In how many ways could he do this?  (1 mark) 

Show Answers Only
  1. `U`
  2. `XYTUZVW`
    `XYTUZWV`
  3. `4\ text(ways)`
Show Worked Solution

a.   `text{The Eulerian trail (visits each edge exactly once):}`

“XYZWVZUYTU`

`:. text(Finishes at ramp)\ U.`

 

b.   `text{Hamiltonian Paths (touch each vertex exactly once):}`

`XYTUZVW`

`XYTUZWV`

 

c.   `text(Hamiltonian cycles:)`

♦♦ Mean mark 31%.

`XYTUZVWX`

`XYTUVZWX`

`text(These two cycles can be reversed)`

`text(to add two more possibilities.)`

`XWVZUTYX`

`XWZVUTYX`

`:. 4\ text(ways.)`

GRAPHS, FUR2 2016 VCAA 2

The bonus money is provided by a company that manufactures and sells hockey balls.

The cost, in dollars, of manufacturing a certain number of balls can be found using the equation

cost = 1200 + 1.5 × number of balls

  1. How many balls would be manufactured if the cost is $1650?  (1 mark)
  2. On the grid below, sketch the graph of the relationship between the manufacturing cost and the number of balls manufactured.  (1 mark)

     

  3. The company will break even on the sale of hockey balls when it manufactures and sells 200 hockey balls.

     

    Find the selling price of one hockey ball.  (1 mark) 

Show Answers Only
  1. `300`

  3. `$7.50\ text(per ball)`
Show Worked Solution
a.    `1200 + 1.5 xx n` `= 1650`
  `:. n` `= ((1650 – 1200))/1.5`
    `= 300`

 

b.   

 

c.   `text(When)\ n = 200,`

`C` `= 1200 + 1.5 xx 200`
  `= 1500`

 

`:.\ text(Selling price if break even)`

`= 1500/200`

`= $7.50\ text(per ball)`

GRAPHS, FUR2 2016 VCAA 1

Maria is a hockey player. She is paid a bonus that depends on the number of goals that she scores in a season.

The graph below shows the value of Maria’s bonus against the number of goals that she scores in a season.

  1. What is the value of Maria’s bonus if she scores seven goals in a season?  (1 mark) 
  2. What is the least number of goals that Maria must score in a season to receive a bonus of $2500?  (1 mark)

Another player, Bianca, is paid a bonus of $125 for every goal that she scores in a season.

  1. What is the value of Bianca’s bonus if she scores eight goals in a season?  (1 mark)
  2. At the end of the season, both players have scored the same number of goals and receive the same bonus amount.

     

    How many goals did Maria and Bianca each score in the season?  (1 mark)

Show Answers Only
  1. `$1500`
  2. `15`
  3. `$1000`
  4. `text(28 goals)`
Show Worked Solution

a.   `$1500`

 

b.   `15`

 

c.    `text(Bonus)` `= 8 xx 125`
    `= $1000`

 

d.   `text(Draw the graph of Bianca’s payments on)`

`text(the same graph.)`

`text(The intersection is where both players)`

`text(receive the same bonus amount.)`

`:.\ text(Maria and Bianca each score 28 goals.)`