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Statistics, EXT1 S1 SM-Bank 4 MC

When a standard 6-sided die is thrown, the probability that it shows a prime number is  `2/3`.

If 10 standard dice are thrown, the number, `N`, of times a prime number is showing has a binomial distribution.

What is the standard deviation of  `N`, correct to 3 decimal places?

  1. 0.222
  2. 0.471
  3. 1.491
  4. 2.222
Show Answers Only

`C`

Show Worked Solution

`N\ ~\ text(Bin)(n, p)\ ~\ text(Bin)(10, 2/3)`

`text(Var)(N)` `= np(1 – p)`
  `= 10 · 2/3(1 – 2/3)`
  `= 20/9`

 

`:. σ_N` `= sqrt(20/9)`
  `= 1.4907…`

 
`=>\ C`

Statistics, EXT1 S1 2017 MET1 4

In a large population of fish, the proportion of angel fish is  `1/4`.

Let  `hat p`  be the random variable that represents the sample proportion of angel fish for samples of size `n` drawn from the population.

Find the smallest integer value of `n` such that the standard deviation of  `hat p`  is less than or equal to  `1/100`.  (2 marks)

Show Answers Only

`1875`

Show Worked Solution

`text(Let)\ \ X =\ text(number of angel fish)`

`X\ ~\ text(Bin) (n, 1/4)`

`σ\ (hat p)` `= sqrt((p(1 – p))/n)`
  `= sqrt((1/4 xx 3/4)/n)`
  `= sqrt(3/(16n))`

 

`text(Solve:)\ \ \ sqrt(3/(16n))` `<= 1/100`
`3/(16 n)` `<= 1/(10\ 000)`
`(30\ 000)/16` `<= n`
`:. n` `>= 1875`

 
`:. n_text(min) = 1875`

Statistics, EXT1 S1 2012 MET2 3

Steve and Jess are two students who have agreed to take part in a psychology experiment. Each has to answer several sets of multiple-choice questions. Each set has the same number of questions, `n`, where `n` is a number greater than 20. For each question there are four possible options A, B, C or D, of which only one is correct.

  1. Steve decides to guess the answer to every question, so that for each question he chooses A, B, C or D at random.

     

    Let the random variable `X` be the number of questions that Steve answers correctly in a particular set.

    1. What is the probability that Steve will answer the first three questions of this set correctly?  (1 mark)

    2. Use the fact that the variance of `X` is `75/16` to show that the value of `n` is 25.  (1 mark)

  1. The probability that Jess will answer any question correctly, independently of her answer to any other question, is  `p\ (p > 0)`. Let the random variable `Y` be the number of questions that Jess answers correctly in any set of 25.

    If   `P(Y > 23) = 6 xx P(Y = 25)`, show that the value of  `p=5/6`.  (2 marks)

Show Answers Only

a.i.  `1/64`

a.ii.  `text(See Worked Solutions)`

b.  `text(See Worked Solutions)`

Show Worked Solution
a.i.    `Ptext{(3 correct in a row)}` `= (1/4)^3`
    `= 1/64`

 

a.ii.    `text(Var)(X)` `= np(1 – p)`
  `75/16` `= n(1/4)(3/4)`
  `75` `= 3n`
  `:. n` `= 25`

 

b.   `Y ∼\ text(Bin)(25,p)`

♦♦♦ Mean mark part (c) 19%.
`P(Y > 23)` `= 6xx P(Y = 25)`
`P(Y = 24) + P(Y = 25)` `= 6xx P(Y = 25)`
`P(Y = 24)` `= 5xx P(Y = 25)`
`((25),(24))p^24(1 – p)^1` `= 5p^25`
`25p^24(1 – p)` `= 5p^25`
`25p^24-25p^25-5p^25` `=0`
`25p^24-30p^25` `=0`
`5p^24(5 – 6p)` `= 0`

 
`:. p = 5/6,\ \ (p>0)\ \ text(… as required)`

Statistics, EXT1 S1 SM-Bank 3

In a chocolate factory the material for making each chocolate is sent to a machines.

The time, `X` seconds, taken to produce a chocolate by machine is a binomial distribution where it can be shown that  `P(X <= 3) = 9/32`.

A random sample of 10 chocolates is chosen. Find the probability, correct to two decimal places, that exactly 4 of these 10 chocolates took 3 or less seconds to produce.  (2 marks)

Show Answers Only

`0.18`

Show Worked Solution

`text(Let)\ \ C = \ text(number of chocolates that take less than 3 seconds)`

COMMENT: Take care as  `P(X <= 3) = 9/32`  provides the equivalent of  `p` here.

`C ∼\ text(Bin)(n, p) ∼\ text(Bin)(10, 9/32)`

`P(C = 4)`  `=((10),(4)) (9/32)^4 (23/32)^6` 
  `=0.181…`
  `=0.18\ \ \ text{(2 d.p.)}`

Statistics, EXT1 S1 SM-Bank 2

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.

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 three decimal places.  (2 marks)

Show Answers Only

`0.902`

Show Worked Solution

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

`X\ ~\ text(Bin) (n,p)\ ~\ text(Bin) (22, 0.1)`

`P(X>=1)` `=1 – P(X=0)`
  `=1-((n),(0)) (0.1^0)(0.9^22)`
  `=1 – 0.9^22`
  `=0.9015…`
  `=0.902\ \ text{(3 d.p.)}`

Statistics, EXT1 S1 2007 MET1 5

It is known that 50% of the customers who enter a restaurant order a cup of coffee. If four customers enter the restaurant, what is the probability that more than two of these customers order coffee? (Assume that what any customer orders is independent of what any other customer orders.)  (2 marks)

Show Answers Only

`5/16`

Show Worked Solution

`X~\ text(Bin)(n,p)\ ~\ text(Bin)(4, 1/2)`

`P(X > 2)` `= P(X = 3) + P(X = 4)`
  `= ((4),(3))(1/2)^3(1/2) + ((4),(4))(1/2)^4(1/2)^0`
  `= 4 xx 1/16 + 1 xx 1/16` 
  `=5/16`

Statistics, EXT1 S1 2011 MET1 7

A biased coin tossed three times. The probability of a head from a toss of this coin is `p.`

  1. Find, in terms of `p`, the probability of obtaining

    1. three heads from the three tosses  (1 mark)

    2. two heads and a tail from the three tosses.  (1 mark)

  2. If the probability of obtaining three heads equals the probability of obtaining two heads and a tail, find `p`.  (2 marks)

Show Answers Only
    1. `p^3`
    2. `3p^2 (1 – p)`
  2. `0 or 3/4`
Show Worked Solution
a.i.    `text(P) (HHH)` `= p xx p xx p`
    `= p^3`

 

  ii.   `text(P) text{(2 Heads and 1 Tail from 3 tosses)}`

♦ Part (a)(ii) mean mark 41%.

`= ((3), (2)) xx p^2 xx (1 – p)^1`

`= 3 p^2 (1 – p)`

 

b.   `text(If probabilities are equal:)`

♦ Mean mark 48%.
MARKER’S COMMENT: Many students incorrectly assumed `p` could not be zero.
`p^3` `= 3p^2 – 3p^3`
`4p^3 – 3p^2` `= 0`
`p^2 (4p – 3)` `= 0`

 
`:. p = 0 or p = 3/4`

Statistics, EXT1 S1 2016 MET1 4

A paddock contains 10 tagged sheep and 20 untagged sheep. Four times each day, one sheep is selected at random from the paddock, placed in an observation area and studied, and then returned to the paddock.

  1. What is the probability that the number of tagged sheep selected on a given day is zero?  (1 mark)

  2. What is the probability that at least one tagged sheep is selected on a given day?  (1 mark)

  3. What is the probability that no tagged sheep are selected on each of six consecutive days?

     

    Express your answer in the form `(a/c)^c`, where `a`, `b` and `c` are positive integers.  (1 mark)

Show Answers Only
  1. `16/81`
  2. `65/81`
  3. `(2/3)^24`
Show Worked Solution

a.   `text(Let)\ \ X =\ text(Number of tagged sheep,)`

`X ~\ text(Bin)(n,p)\ ~\ text(Bin)(4,1/3)`

`P(X = 0)` `= ((4),(0)) xx (1/3)^0 xx (2/3)^4`
  `= 16/81`

 

b.    `P(X >= 1)` `= 1 – P(X = 0)`
    `= 1 – 16/81`
    `= 65/81`

 

c.   `text(Let)\ \ Y =\ text(Days that no tagged sheep selected,)`

`Y ~\ text(Bin)(6,16/81)`

`P(Y = 6)` `= ((6),(6)) xx (16/81)^6 xx (65/81)^0`
  `= (16/81)^6`
  `=(2/3)^24`

Statistics, EXT1 S1 2015 MET2 10

The binomial random variable,  `X`, has  `E(X) = 2`  and  `text(Var)( X ) = 4/3.`

Calculate  `P(X = 1)`.   (3 marks)

Show Answers Only

`64/243`

Show Worked Solution

`np = 2\ …\ (1)`

`np(1 – p) = 4/3\ …\ (2)`

`text(Solve simultaneous equations:)`

`text(Substitute)\ \ np=2\ \ text(into)\ (2)`

`2(1-p)` `=4/3`  
`2-2p` `=4/3`  
`p` `=1/3`  

 
`n = 6,quadp = 1/3`

`:. X ∼\ text(Bin)(6, 1/3)`

 

`:. P(X=1)` `= ((6),(1)) xx (1/3)^1 xx (2/3)^5`
  `= 6 xx 1/3 xx (2/3)^5`
  `=64/243`

Statistics, EXT1 S1 2008 MET2 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 = 3,\ \ \ p = 0.6`

B.   `n = 2,\ \ \ p = 0.6`

C.   `n = 2,\ \ \ p = 0.4`

D.   `n = 3,\ \ \ p = 0.4`

Show Answers Only

`D`

Show Worked Solution

`X∼\ text(Bin) (n, p)`

`mu` `= 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`

`=>   D`

Statistics, EXT1 S1 MET2 2008 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.     `9`

B.   `10`

C.   `11`

D.   `12`

Show Answers Only

`C`

Show Worked Solution

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

`X ∼ text(Bin) (n, p) ∼ text(Bin) (n, 1/2)`

`P(X = n)` `< 0.0005`
`((n), (n)) (1/2)^n (1/2)^0` `< 0.0005`
`1/2^n` `<5/(10\ 000)`
`2^n` `>2000`
`ln 2^n` `>ln 2000`
`n` `>ln2000/ln2`
`n` `> 10.97`

 
`:. n_min = 11`

`=>   C`

Functions, 2ADV F2 SM-Bank 13

 

  1. Show that the function  `y = (1 - e^x)/(1 + e^x)`  is an odd function?  (1 mark) 

  2. Sketch  `y = (1 - e^x)/(1 + e^x)`, labelling all intercepts and asymptotes.  (2 marks)

Show Answers Only
  1. `{text(all real)\ x,  x!=1/4}`

 

 

 

 

 

 

Show Worked Solution

i.   `f(x) = (1 – e^x)/(1 + e^x)`

`f(−x)` `= (1 – e^(−x))/(1 + e^(−x)) xx (e^x)/(e^x)`
  `= (e^x – 1)/(e^x + 1)`
  `= −(1 – e^x)/(1 + e^x)`
  `= −f(x)`

 
`:. f(x)\ text(is ODD.)`

 

ii.   `y = (1 – e^x)/(1 + e^x) xx (e^(−x))/(e^(−x)) = (e^(−x) – 1)/(e^(−x) + 1) = 1 – 2/(e^(−x) + 1)`

`text(As)\ x -> ∞, \ 2/(e^(−x) + 1) -> 2, \ y -> −1`

`text(As)\ x -> – ∞, \ 2/(e^(−x) + 1) -> 0, \ y -> 1`

`text(When)\ x = 0, \ y = 0`
 

Functions, 2ADV F2 EQ-Bank 9

Consider the function  `f(x) = 1/(4x - 1)`.

  1. Find the domain of  `f(x)`.  (1 mark)

  2. Sketch  `f(x)`, showing all asymptotes and intercepts?  (2 marks)

Show Answers Only
  1. `{text(all real)\ x,  x!=1/4}`
  2.   
Show Worked Solution
i.    `4x – 1` `!= 0`
  `x` `!= 1/4`

 
`:.\ text(Domain:)\ {text(all real)\ x,  x!=1/4}`

 

ii.   `text(When)\ \ x = 0, \ y = −1`

`text(As)\ \ x -> ∞, \ y -> 0^+`

`text(As)\ \ x -> −∞, \ y -> 0^−`

Functions, 2ADV F2 SM-Bank 12

Sketch the graph of  `f(x) = (2x+1)/(x-1)`. Label the axis intercepts with their coordinates and label any asymptotes with the appropriate equation.  (4 marks)

  

Show Answers Only

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

COMMENT: Manipulation of the equation makes graphing much easier.

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

`text(As)\ \ x->oo,\ \ y->2(+)`

`text(As)\ \ x->-oo,\ \ y->2(-)`

`text(As)\ \ x->-1 (-),\ \ y->-oo`

`text(As)\ \ x->-1 (+),\ \ y->oo`

 

 

Functions, 2ADV F2 EQ-Bank 11

On the axes below, sketch the graph of   `f(x) = (2x-2)/(x + 1)`.

Label all axis intercepts. Label each asymptote with its equation. (4 marks)

 
met1-2008-vcaa-q2

Show Answers Only

met1-2008-vcaa-q2-answer1

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

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

`text(As)\ \ x->oo,\ \ y->2(-)`

`text(As)\ \ x->-oo,\ \ y->2(+)`

`text(As)\ \ x->-1 (-),\ \ y->oo`

`text(As)\ \ x->-1 (+),\ \ y->-oo`

 

met1-2008-vcaa-q2-answer1

Functions, 2ADV F2 SM-Bank 10 MC

The graph of the function  `f(x) = (x - 3)/(2 - x)`  has asymptotes

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

`D`

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

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

`=>   C`

Functions, 2ADV F2 SM-Bank 10

Consider the function  `f(x) = x/(4 - x^2)`.

  1. Identify the domain of  `f(x)`.   (1 mark)

  2. Sketch the graph  `y = f(x)`, showing all intercepts and asymptotes.   (3 marks)

Show Answers Only
  1. `text(Domain:)\ \ text(all)\ x,\ x != +- 2`
  2.  
     
Show Worked Solution

i.   `4-x^2 !=0`

`=> x!= 2 or -2`

`:.\ text(Domain:)\ \ text(all)\ x,\ x != +- 2`

 

ii. `text(Asymptotes at)\ \ x = +- 2`

`text(Passes through)\ (0,0)`

COMMENT: Note that  `x->2^(–)`  is a short way of writing as `x` approaches 2 from the negative (or left-hand) side. This notation can save time when required.
`text(As)` `\ \ x -> 2^(-),` `\ y -> oo`
  `\ \ x -> 2 ^(+),` `\ y -> – oo`
`text(As)` `\ \ x -> -2^ (-),` `\ y -> oo`
  `\ \ x -> -2^ (+),` `\ y -> -oo`
`text(As)` `\ \ x -> oo,` `\ y -> 0`
  `\ \ x -> – oo,` `\ y -> 0`

 

  EXT1 2013 11d

 

Functions, 2ADV’ F2 2007 HSC 3b

  1. Find the vertical and horizontal asymptotes of the hyperbola

     

    `qquad y = (x − 2)/(x − 4)`  and hence sketch the graph of  `y = (x − 2)/(x − 4)`.  (3 marks)

  2. Hence, or otherwise, find the values of  `x`  for which  `(x − 2)/(x − 4) ≤ 3`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `x < 4\ text(and)\ x ≥ 5`
Show Worked Solution

i.    `y = (x − 2)/(x − 4)`

`text(Vertical asymptote at)\ \ x = 4`

`lim_(x → ∞) (x − 2)/(x − 4)` `= lim_(x → ∞) (1 − 2/x)/(1 − 4/x)=1`

`ytext(–intercept)\ = 1/2`

`xtext(–intercept)\ = 2`

 

Geometry and Calculus, EXT1 2007 HSC 3b Answer

 

ii.  `text(Find)\ \ x\ \ text(so that)\ \ (x − 2)/(x − 4) ≤ 3`

`(x − 2)/(x − 4)` `= 3`
`x − 2` `= 3x − 12`
`2x` `= 10`
`x` `= 5`

 
`=>(5, 3)\ \ text(is the intersection of)\ \ y = 3\ and\ y = (x − 2)/(x − 4)`
 

`:. (x − 2)/(x − 4) ≤ 3\ \ text(when)\ \ x < 4\ \ text(and)\ \ x ≥ 5.`

Functions, 2ADV’ F2 2012 HSC 13b

  1. Find the horizontal asymptote of the graph  `y=(2x^2)/(x^2 + 9)`.   (1 mark)

  2. Without the use of calculus, sketch the graph  `y=(2x^2)/(x^2 + 9)`, showing the asymptote found in part (i).    (2 marks)

Show Answers Only
  1. `text(Horizontal asymptote at)\ y = 2`
  2.  
    Geometry and Calculus, EXT1 2012 HSC 13b Answer
Show Worked Solution
i.    `y` `= (2x^2)/(x^2 +9)`
    `= 2/(1 + 9/(x^2))`

 

`text(As)\ \ x -> oo,\ y ->2`

`text(As)\ \ x -> – oo,\ y -> 2`

`:.\ text(Horizontal asymptote at)\ y = 2`

 

ii.    `text(At)\ \ x = 0,\ y = 0`

`f(x) = (2x^2)/(x^2 + 9) >= 0\ text(for all)\ x`

`f(–x) = (2(–x)^2)/((–x)^2 + 9) = (2x^2)/(x^2 + 9) = f(x)`

`text(S)text(ince)\ \ f(x) = f(–x) \ \ =>\ text(EVEN function)`

Geometry and Calculus, EXT1 2012 HSC 13b Answer

Functions, 2ADV’ F2 2015 HSC 5 MC

What are the asymptotes of `y = (3x)/((x + 1)(x + 2))`

A.    `y = 0,` `x = −1,` `x = −2`
B.    `y = 0,` `x = 1,` `x = 2`
C.    `y = 3,` `x = −1,` `x = −2`
D.    `y = 3,` `x = 1,` `x = 2`
Show Answers Only

`A`

Show Worked Solution

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

`text(Asymptotes at)\ \ x = −1\ \ text(and)\ \ x = −2`

`text(As)\ \ x → ∞, y → 0^+`

`text(As)\ \ x → −∞, y → 0^−`

`:.\ text(Horizontal asymptote at)\ \ y = 0`

`⇒ A`

Functions, 2ADV’ F2 2017 HSC 5 MC

Which graph best represents the function  `y = (2x^2)/(1 - x^2)`?
 

A. B.
       
C. D.
Show Answers Only

`D`

Show Worked Solution
`y` `= (2x^2)/((1 – x^2))`
  `= −((2 – 2x^2 – 2))/((1 – x^2))`
  `= −(2(1 – x^2))/((1 – x^2)) – 2/((1 – x^2))`
  `= −2 – 2/((1 – x^2))`

 

`text(As)\ \ x -> oo,\ \ y -> −2`

`:. text(Horizontal asymptote at)\ \ y = −2`

`⇒D`

Calculus, EXT1 C3 2017 SPEC1 8

A slope field representing the differential equation  `dy/dx = −x/(1 + y^2)`  is shown below.

  1. Sketch the solution curve of the differential equation corresponding to the condition  `y(−1) = 1`  on the slope field above and, hence, estimate the positive value of `x` when  `y = 0`. Give your answer correct to one decimal place.  (2 marks)
  2. Solve the differential equation  `(dy)/(dx) = (−x)/(1 + y^2)`  with the condition  `y(−1) = 1`. Express your answer in the form  `ay^3 + by + cx^2 + d = 0`, where `a`, `b`, `c` and `d` are integers.  (2 marks)

Show Answers Only
  1.  

  2. `2y^3 + 6y + 3x^2 – 11 = 0`
Show Worked Solution
a.   

♦♦ Mean mark part (a) 32%.
MARKER’S COMMENT: Solution curve should follow slope ticks and not cross them.

 

b.    `(1 + y^2)(dy)/(dx)` `= −x`
  `int 1 + y^2 dy` `= −int x\ dx`
  `y + (y^3)/3` `= −(x^2)/2 + C, C ∈ R`

 
`text(Substituting)\ (-1,1):`

`1 + (1^3)/3` `= −((−1)^2)/2 + C`
`1 + 1/3` `= −1/2 + C`
`:. C` `= 11/6`

 

`y + 1/3y^3` `= −1/2x^2 + 11/6`
`6y + 2y^3` `= −3x^2 + 11`

 
`:. 2y^3 + 6y + 3x^2 – 11 = 0`

Calculus, EXT1 C3 SM-Bank 5

Bacteria are spreading over a Petri dish at a rate modelled by the differential equation

`(dP)/(dt) = P/2 (1 - P),\ 0 < P < 1`

where  `P`  is the proportion of the dish covered after  `t`  hours.

Given  `2/(P(1 - P)) = 2/P + 2/(1 - P),`

  1. Show by integration that  `(t - c)/2= log_e(P/(1 - P))`, where  `c`  is a constant of integration.  (2 marks)

  2. If half of the Petri dish is covered by the bacteria at  `t = 0`, express  `P`  in terms of  `t`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `P = e^(t/2)/(1 + e^(t/2))`
Show Worked Solution

a.  `(dt)/(dP) = 2/(P(1 – P)) = 2/P + 2/(1 – P)`

`t` `= int 2/P + 2/(1 – P)\ dP`
  `= 2 log_e |P| – 2log_e|1 – P| + c`
`(t-c)/2` `=log_e|P| – log_e|1-P|`
  `=log_e |(P)/(1-P)|`
  `= log_e (P/(1 – P))`

  
`text(S)text(ince)\ \ 0 < P < 1 :\ |P| = P\ and\ |1 – P| = 1 – P`


b.  `text(When)\ \ t=0,\ P=0.5`

`(-c)/2` `= log_e (0.5/0.5)`
`c` `= log_e (1)`
  `=0`

 

`t/2` `= ln (P/(1 – P))`
`e^(t/2)` `= P/(1 – P)`
`e^(t/2) (1 – P)` `= P`
`e^(t/2) – Pe^(t/2)` `= P`
`e^(t/2)` `= P(1 + e^(t/2))`
`:. P` `= e^(t/2)/(1 + e^(t/2))`

Calculus, EXT1 C3 2018 SPEC2 10 MC

The differential equation that best represents the direction field above is

A.  `(dy)/(dx) = (2x + y)/(y - 2x)`

B.  `(dy)/(dx) = (x + 2y)/(2x - y)`

C.  `(dy)/(dx) = (2x - y)/(x + 2y)`

D.  `(dy)/(dx) = (x - 2y)/(y - 2x)`

Show Answers Only

`A`

Show Worked Solution

`text(When)\ \ x=0, \ m=1`

`text(When)\ \ y=0, \ m=-1`

`=>  A`

Calculus, EXT1 C3 2014 SPEC2 14 MC

The differential equation that is best represented by the above direction field is

A.   `(dy)/(dx) = 1/(x - y)`

B.   `(dy)/(dx) = y - x`

C.   `(dy)/(dx) = 1/(y - x)`

D.   `(dy)/(dx) = x - y`

Show Answers Only

`C`

Show Worked Solution

`text(Consider quadrant 2,)`

`x < 0, \ \ y > 0, \ \  m > 0\ \ => text(Eliminate A and D)`
 

`text(Consider vertical gradients where)\ \ m=oo\ \ => text(Eliminate B)`

 
`=> C`

Calculus, EXT1 C3 2013 SPEC2 12 MC

SPEC2 2013 VCAA 12 MC

The differential equation that best represents the above direction field is

A.   `(dy)/(dx) = x^2 - y^2`

B.   `(dy)/(dx) = y^2 - x^2`

C.   `(dy)/(dx) = −x/y`

D.   `(dy)/(dx) = x/y`

Show Answers Only

`D`

Show Worked Solution

`text(By inspection:)`

`text(When)\ \ x=0\ \ =>\ \ (dy)/(dx) = 0`

`text(When)\ \ y=0\ \ => (dy)/(dx) -> oo`

`:.\ text(Eliminate A, and B)`
 

`text(Along)\ \ y = x\ \ =>\ \ (dy)/(dx) > 0`

`:.\ text(Eliminate C)`

`=> D`

Calculus, EXT1 C3 2012 SPEC2 10 MC

The diagram that best represents the direction field of the differential equation  `(dy)/(dx) = xy`  is

A. B. 
C. D.
Show Answers Only

`A`

Show Worked Solution

`(dy)/(dx) = xy`

`text(When)\ \ x=0 \ or\  y=0\ \ =>\ text(gradient = 0)`

`text(In 1st and 3rd quartile)\ \ =>\ \ text(gradients positive)`

`text(In 2nd and 4th quartile)\ \ =>\ \ text(gradients negative)`

`=> A`

Calculus, EXT1 C3 2011 SPEC2 17 MC

SPEC2 2011 VCAA 17 MC

The differential equation which best represents the above direction field is

A.   `(dy)/(dx) = (y - 2x)/(2y + x)`

B.   `(dy)/(dx) = (2x - y)/(y - 2x)`

C.   `(dy)/(dx) = (2y - x)/(y + 2x)`

D.   `(dy)/(dx) = (y - 2x)/(2y - x)`

E.   `(dy)/(dx) = (x - 2y)/(2y + x)`

Show Answers Only

`A`

Show Worked Solution

`text(When)\ \ x=0\ \ => \ \ text(gradients are all positive)`

Almost half of all students answered incorrectly – mean mark 52%.

`text(Eliminate B and E.)`

`text(When)\ \ y=0\ \ => \ \ text(gradients are all negative)`

`text(Eliminate D.)`

`text(Option A will have zero gradient along)\ \ y=2x\ \ text{(correct)}`

`text(Option C will have zero gradient along)\ \ y=1/2 x\ \ text{(incorrect)}`

`=> A`

Calculus, EXT1 C3 2017 SPEC1-N 7

Let  `(dy)/(dx) = (4 - y)^2`.

Express  `y`  in terms of  `x`, where  `y(0) = 3`.  (3 marks)

Show Answers Only

`y = 4-1/(x + 1)`

Show Worked Solution
`(dy)/(dx)` `=(4-y)^2`
`(dx)/(dy)` `= 1/(4-y)^2`
`x` `= int 1/(4-y)^2\ dy`
  `= int (4-y)^(-2) dy`
  `= (-1)(-1)(4-y)^(-1)+ c`
  `= 1/(4-y) + c`

 
`text(When)\ \ x=0,\ \ y=3:`

`0` `= 1/(4-3) + c`
`:.c` `= -1`

 

`x` `= 1/(4-y) – 1`
`x + 1` `= 1/(4-y)`
`1/(x + 1)` `= 4-y`
`:. y` `= 4-1/(x + 1)`

Calculus, EXT1 C3 2018 VCE 8

A tank initially holds 16 L of water in which 0.5 kg of salt has been dissolved. Pure water then flows into the tank at a rate of 5 L per minute. The mixture is stirred continuously and flows out of the tank at a rate of 3 L per minute.

  1.  Show that the differential equation for `Q`, the number of kilograms of salt in the tank after `t` minutes, is given by
    `qquad (dQ)/(dt) = -(3Q)/(16 + 2t)`  (1 mark)
      
  2.  Solve the differential equation given in part a. to find `Q` as a function of `t`.
      
    Express your answer in the form  `Q = a/(16 + 2t)^(b/c)`, where `a, b` and `c` are positive integers.  (3 marks)
Show Answers Only
  1.  `text(Proof)\ \ text{(See Worked Solutions)}`
  2.  `Q = 32/(16 + 2t)^(3/2)`
Show Worked Solution

a. `Q_0 = 0.5, \ V_0 = 16`

♦ Net mean mark of both parts 44%.

`V(t)` `= 16 + (5 – 3) t`
  `= 16 + 2t`

 
`text(Concentration)\ (C)= Q/V= Q/(16 + 2t)\ text(kg/L)`

`(dQ)/(dt)` `= 0 xx 5 – 3C`
  `= -(3Q)/(16 + 2t)`

MARKER’S COMMENT: Many students took the common factor of 2 from  `16+2t`. This wasn’t necessary and complicated the arithmetic in part b.

 

b.   `-1/(3Q) * (dQ)/(dt) = 1/(16 + 2t)`

`int -1/(3Q)\ dQ` `= int 1/(16 + 2t) dt`
`-1/3 int 1/Q\ dQ` `= 1/2 int 2/(16 + 2t)\ dt`
`-1/3 ln Q` ` = [1/2 ln(16 + 2t)] + c`

 
`text(When)\ \ t=0,\ \ Q=0.5,`

COMMENT: A very challenging test of using exponential and log laws!

`-1/3 ln (1/2)` `= 1/2 ln (16) +c`
`c` `= -1/2 ln (16) -1/3 ln (1/2)`

 

`-1/3 ln Q` `= 1/2 ln(16 + 2t) -1/2 ln(16) – 1/3 ln (1/2)`
`-1/3 ln Q` `= 1/2 ln ((16 + 2t)/16) – 1/3 ln (1/2)`
`-1/3 ln Q` `= ln (((16 + 2t)^(1/2))/4) – ln (2^(-1/3))`
`ln (Q^(-1/3))` `= ln (((16 + 2t)^(1/2))/(2^2 ⋅ 2^(-1/3)))`
`Q^(-1/3)` `= ((16 + 2t)^(1/2))/(2^(5/3))`
`Q` `= (((16 + 2t)^(1/2))/(2^(5/3)))^-3`
  `= (16 + 2t)^(- 3/2)/(2^(-5))`
`:. Q` `= 32/((16 + 2t)^(3/2))`

Calculus, EXT1 C1 2013 VCE 5

A container of water is heated to boiling point (100°C) and then placed in a room that has a constant temperature of 20°C. After five minutes the temperature of the water is 80°C.

  1. Use Newton’s law of cooling  `(dT)/(dt) = -k (T - 20)`, where `T text(°C)` is the temperature of the water at the time `t` minutes after the water is placed in the room, to show that  `e^(-5k) = 3/4.`  (2 marks)

  2. Find the temperature of the water 10 minutes after it is placed in the room.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `65`
Show Worked Solution
a.    `(dT)/(dt)` `= -k (T – 20)`
  `(dt)/(dT)` `= -1/k * 1/(T – 20)`
  `t` `= -1/k int 1/(T – 20)\ dT`
  `-kt` `= log_e (T – 20) + c`
     

`text(When)\ \ t = 0,\ \ T = 100,`

COMMENT: Recognise that exponential G+D models are simply differential equations with known solutions (syllabus p.58)!

`=>c = -log_e 80,`

`:. -kt` `= log_e (T – 20) – log_e 80`
  `= log_e ((T – 20)/80)`

 
`text(When)\ \ t = 5,\ \ T = 80,`

`-5k = log_e (3/4)`

`:. e^(-5k) = 3/4`

 

(b)   `text(Find)\ \ T\ \ text(when)\ \ t = 10:`

`-10k` `= log_e ((T – 20)/80)`
`(T – 20)/80` `= e^(-10k)`
`(T – 20)/80` `= (e^(-5k))^2`
`(T – 20)/80` `= (3/4)^2`
`:. T` `= (3/4)^2 xx 80 + 20`
  `=(9 xx 80)/16 +20`
  `= 65^@ text(C)`

Calculus, EXT1 C3 2017 SPEC2 9 MC

The gradient of the tangent to a curve at any point  `P(x, y)`  is half the gradient of the line segment joining `P` and the point  `Q(-1, 1)`.

The coordinates of points on the curve satisfy the differential equation

A.   `(dy)/(dx) = (y + 1)/(2(x - 1))`

B.   `(dy)/(dx) = (2(y - 1))/(x + 1)`

C.   `(dy)/(dx) = (x - 1)/(2(y + 1))`

D.   `(dy)/(dx) = (y - 1)/(2(x + 1))`

Show Answers Only

`D`

Show Worked Solution
`m_text(tang)` `= 1/2 m_(PQ)`
`m_(PQ)` `= (y – 1)/(x – (-1))`
  `= (y – 1)/(x + 2)`

 
`:. m_text(tang) = (dy)/(dx) = (y – 1)/(2(x + 1))`

`=>   D`

Calculus, EXT1 C3 2017 SPEC2-N 10 MC

A solution to the differential equation  `(dy)/(dx) = (cos(x + y) - cos(x - y))/(e^(x + y))`  can be obtained from

  1. `int e^y/(sin(y))\ dy = -int (2 sin(x))/e^x\ dx`
  2. `int e^y/(cos(y))\ dy = int 2/e^x\ dx`
  3. `int e^y/(cos(y))\ dy = -int (2 cos(x))/e^x\ dx`
  4. `int e^y/(cos(y))\ dy = int (2 sin(x))/e^x\ dx`
Show Answers Only

`A`

Show Worked Solution
`dy/dx` `=(cos(x + y) – cos(x – y))/(e^(x + y))`
`(dy)/(dx)` `= (cos(x) cos(y) – sin(x) sin(y) – cos(x) cos(y) – sin(x) sin(y))/(e^x ⋅ e^y)`
`e^y *(dy)/(dx)` `= (-2 sin(x) sin(y))/(e^x)`
`e^y/(sin(y)) *(dy)/(dx)` `= (-2 sin(x))/(e^x)`
`:. int e^y/(sin(y))\ dy` `= -int (2 sin(x))/e^x\ dx`

 
`=>   A`

Calculus, EXT1 C3 2018 SPEC2 9 MC

A solution to the differential equation  `(dy)/(dx) = 2/{sin(x + y) - sin(x - y)}`  can be obtained from

  1. `int 1\ dx = int 2 sin(y)\ dy`
  2. `int cos(y)\ dy = int text{cosec}(x)\ dx`
  3. `int cos(x)\ dx = int text{cosec}(y)\ dy`
  4. `int sec(x)\ dx = int sin(y)\ dy`
Show Answers Only

`D`

Show Worked Solution
`(dy)/(dx)` `= 2/{sin(x) cos(y) + sin(y) cos(x) – (sin(x) cos(y) – sin(y) cos(x))}`
  `= 2/{2 sin(y) cos(x)}`
  `= 1/{sin(y) cos(x)}`

 
`sin(y) *(dy)/(dx)= sec(x)`

`int sin (y)\ dy= int sec(x)\ dx`

`=>  D`

Calculus, EXT1 C3 2014 VCE 10 MC

A large tank initially holds 1500 L of water in which 100 kg of salt is dissolved. A solution containing 2 kg of salt per litre flows into the tank at a rate of 8 L per minute. The mixture is stirred continuously and flows out of the tank through a hole at a rate of 10 L per minute.

The differential equation for `Q`, the number of kilograms of salt in the tank after `t` minutes, is given by

A.   `(dQ)/(dt) = 16 - (5Q)/(750 - t)`

B.   `(dQ)/(dt) = 16 - (5Q)/(750 + t)`

C.   `(dQ)/(dt) = 16 + (5Q)/(750 - t)`

D.   `(dQ)/(dt) = (100Q)/(750 - t)`

Show Answers Only

`A`

Show Worked Solution

`(dQ_text(in))/(dV)= 2\ text(kg/L),\ (dV_text(in))/(dt) = 8\ text(L/min)`

`V_0 = 1500,\ Q_0 = 100`

`(dV_text(out))/(dt) = 10\ text(L/min)`
  

`V(t)` `= 1500 + (8 – 10)t`
  `= 1500 – 2t`
  `= 2(750 – t)`

 

`(dQ_text(in))/(dt)` `= 2 xx 8 = 16 text(kg/min)`
`(dQ_text(out))/(dt)` `= Q/(v(t)) xx 10`
  `= (10Q)/(2(750 – t))`
  `= (5Q)/(750 – t)`

 
`:.(dQ)/(dt)= 16 – (5Q)/(150 – t)`

`=> A`

Calculus, EXT1 C3 2013 VCE 13 MC

Water containing 2 grams of salt per litre flows at the rate of 10 litres per minute into a tank that initially contained 50 litres of pure water. The concentration of salt in the tank is kept uniform by stirring and the mixture flows out of the tank at the rate of 6 litres per minute.

If `Q` grams is the amount of salt in the tank `t` minutes after the water begins to flow, the differential equation relating `Q` to `t` is

A.   `(dQ)/(dt) = 20 - (3Q)/(25 + 2t)`

B.   `(dQ)/(dt) = 10 - (3Q)/(25 + 2t)`

C.   `(dQ)/(dt) = 20 - (3Q)/(25 - 2t)`

D.   `(dQ)/(dt) = 10 - (3Q)/(25 - 2t)`

Show Answers Only

`A`

Show Worked Solution
`text(Volume)` `= 50 + (10 – 6)t`
  `= 50 + 4t`

 
`text(Salt in tank at time)\ \ t=Q\ text(grams)`

`:.\ text(Concentration)\ = Q/(50 + 4t)\ text(grams per litre)`
 

`(dQ)/(dt)text(in) = 2 xx 10 = 20\ \ text(g/min)`

`(dQ)/(dt)text(out)` `= 6 xx Q/(50 + 4t)`
  `= (3Q)/(25 + 2t)`

 
`:. (dQ)/(dt) = 20 – (3Q)/(25 + 2t)`

`=> A`

Vectors, EXT1 V1 EQ-Bank 4 MC

The diagram shows a grid of equally spaced lines. The vector  `overset(->)(OA) = underset~a`  and the vector  `overset(->)(OH) = underset~h`. The point `Q` is halfway between `F` and `H`.

Which expression represents the vector `overset(->)(EQ)`?

  1. `−1/4 underset~a + 1/2 underset~h`
  2. `1/2 underset~a - 1/4 underset~h`
  3. `1/4 underset~a + 1/2 underset~h`
  4. `1/4 underset~a + underset~h`
Show Answers Only

`A`

Show Worked Solution

`overset(->)(EQ) = 1/2underset~h – 1/4underset~a`

`=> A`

Vectors, EXT2 V1 2017 SPEC1 10

Consider the vectors  `underset ~a = - underset ~i - 2 underset ~j + 3 underset ~k`  and  `underset ~b = 2 underset ~i + c underset ~j + underset ~k`.

Find the value of  `c, \ c in R`, if the angle between  `underset ~a`  and  `underset ~b`  is  `pi/3`.  (4 marks)

Show Answers Only

`c = -3`

Show Worked Solution
`underset ~a ⋅ underset ~b` `= -1 xx 2 + (-2) xx c + 3 xx 1`
  `= -2 – 2c + 3`
  `= 1 – 2c`

 

`1-2c` `= sqrt((-1)^2 + (-2)^2 + 3^3) *sqrt(2^2 + c^2 + 1^2) xx cos (pi/3)`
`1 – 2c` `= 1/2(sqrt 14 ⋅ sqrt(5 + c^2))`
`2 – 4c` `= sqrt(14(5 + c^2))`
`(2 – 4c)^2` `= 14(5 + c^2)`
`4 – 16c + 16c^2` `= 70 + 14c^2`
`2c^2 – 16c – 66` `= 0`
`c^2 – 8c – 33` `= 0`
`(c – 11)(c + 3)` `= 0`

 
`c = 11 or c = -3`

`text(S)text(ince)\ \ 2 – 4c = sqrt(15(5 + c^2))`

`2 – 4c > 0\ \ =>\ \ c<2`

`:. c = -3`

Vectors, EXT2 V1 2015 VCE 1

Consider the rhombus  `OABC`  shown below, where  `vec (OA) = a underset ~i`  and  `vec (OC) = underset ~i + underset ~j + underset ~k`, and `a` is a positive real constant.
 

VCAA 2015 spec 1a
 

  1. Find  `a.`  (1 mark)

  2. Show that the diagonals of the rhombus  `OABC`  are perpendicular.  (2 marks)

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

i.   `|\ vec(OA)\ | = |\ vec(OC)\ |,`

`:. a` `= sqrt (1^2 + 1^2 + 1^2)`
  `= sqrt 3`

 

ii.   `overset(->)(CB) = overset(->)(OA), \ \ overset(->)(AB) = overset(->)(OC)`

MARKER’S COMMENT: Vector notation was poor in many answers.

`overset(->)(OB) = overset(->)(OC) + overset(->)(CB)`

`= underset~i + underset~j + underset~k + sqrt3 underset~i`

`= (sqrt3 + 1)underset~i + underset~j + underset~k`
 

`overset(->)(AC) = overset(->)(AO) + overset(->)(OC)`

`= −sqrt3underset~i + underset~i + underset~j + underset~k`

`= (1 – sqrt3)underset~i + underset~j + underset~k`
 

`overset(->)(AC) · overset(->)(OB)` `= (1 + sqrt3)(1 – sqrt3) + 1 + 1`
  `= 1 – 3 + 1 + 1`
  `= 0`

 
`:. overset(->)(AC) ⊥ overset(->)(OB)`

Vectors, EXT2 V1 2014 SPEC1 1

Consider the vector  `underset ~a = sqrt 3 underset ~i - underset ~j - sqrt 2 underset ~k`, where  `underset ~i, underset ~j`  and  `underset ~k`  are unit vectors in the positive directions of the `x, y` and `z` axes respectively.

  1.  Find the unit vector in the direction of  `underset ~a`.  (1 mark)

  2.  Find the acute angle that  `underset ~a`  makes with the positive direction of the `x`-axis.  (2 marks)

  3.  The vector  `underset ~b = 2 sqrt 3 underset ~i + m underset ~j - 5 underset ~k`.

     

     Given that  `underset ~b`  is perpendicular to  `underset ~a,` find the value of `m`.  (2 marks)

Show Answers Only
  1. `1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`
  2. `theta = 45^@`
  3. `m = 6 + 5 sqrt 2`
Show Worked Solution
i.    `|underset ~a|` `= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)`
    `= sqrt 6`
`:. hat underset ~a` `= underset ~a/|underset ~a|`
  `= 1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`

 

Mean mark part (b) 51%.

ii.    `underset ~a ⋅ underset ~i` `= sqrt 3 xx 1 = sqrt 3`
  `underset ~a ⋅ underset ~i` `= |underset ~a||underset ~i| cos theta`
    `= sqrt 6 cos theta`
  `sqrt 3` `= sqrt 6 cos theta`
`cos theta` `= 1/sqrt 2`
`:. theta` `= pi/4 = 45^@`

 

iii.   `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`

`6 – m + 5 sqrt 2` `=0`  
`:. m` `=6 + 5 sqrt 2`  

Vectors, EXT2 V1 2013 SPEC1 3

The coordinates of three points are  `A\ ((– 1), (2), (4)), \ B\ ((1), (0), (5)) and C\ ((3), (5), (2)).`

  1. Find  `vec (AB).`  (1 mark)

  2. The points `A, B` and `C` are the vertices of a triangle. 

     

    Prove that the triangle has a right angle at `A.`  (2 marks)

  3. Find the length of the hypotenuse of the triangle.  (1 mark)

Show Answers Only
  1. `2underset~i – 2underset~j + underset~k`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `sqrt 38`
Show Worked Solution
i.   `vec(AB)` `=(1 – −1)underset~i + (0 – 2)underset~j + (5 – 4)underset~k`
    `= 2underset~i – 2underset~j + underset~k`

 

ii.    `overset(->)(AC)` `= (3 – −1)underset~i + (5 – 2)underset~j + (2 – 4)underset~k`
    `= 4underset~i + 3underset~j – 2underset~k`

 

`overset(->)(AB) · overset(->)(AC)` `= 2 xx 4 + (−2) xx 3 + 1 xx (−2)`
  `= 8 – 6 – 2`
  `= 0`

 
`=>  overset(->)(AB) ⊥ overset(->)(AC)`

`:. DeltaABC\ text(has a right angle at)\ A.`
 

iii.    `overset(->)(BC)` `= (3 – 1)underset~i + (5 – 0)underset~j + (2 – 5)underset~k`
    `= 2underset~i + 5underset~j – 3underset~k`

 

`|overset(->)(BC)|` `= sqrt(2^2 + 5^2 + (−3)^2)`
  `= sqrt(4 + 25 + 9)`
  `= sqrt38`

Vectors, EXT2 V1 SM-Bank 8

If  `underset ~a = -2 underset ~i - underset ~j + 3 underset ~k`  and  `underset ~b = -m underset ~i + underset ~j + 2 underset ~k`, where  `m`  is a real constant, find  `m`  such that the vector  `underset ~a - underset ~b`  will be perpendicular to vector  `underset ~b`.   (2 marks)

Show Answers Only

`0 or 2`

Show Worked Solution
`underset ~a – underset ~b` `= (m – 2) underset ~i – 2 underset ~j + underset ~k `
`(underset ~a – underset ~b) ⋅ underset ~b` `= -m(m – 2) – 2(1) + 1(2) = 0`
`0` `= -m^2 + 2m – 2 + 2`
`0` `= 2m – m^2`
`0` `= m(2 – m)`
`:. m` `= 0 \ or \ 2`

Vectors, EXT2 V1 SM-Bank 7

If  `theta`  is the angle between  `underset ~a = sqrt 3 underset ~i + 4 underset ~j - underset ~k`  and  `underset ~b = underset ~i - 4 underset ~j + sqrt 3 underset ~k`, then find  `cos(2 theta)`.   (2 marks)

Show Answers Only

`7/25`

Show Worked Solution
`underset ~a ⋅ underset ~b` `= sqrt 3 xx 1 + 4 xx (-4) + (-1) xx sqrt 3`
  `= -16`

 
`|\ underset~a\ | = sqrt20,\ \ |\ underset~b\ | = sqrt20,`

`cos(theta)` `=(underset ~a ⋅ underset ~b)/(|\ underset~a\ |\ |\ underset~b\ | `
  `= (-16)/(sqrt 20 xx sqrt 20)`
  `= -4/5`

 

`cos (2 theta)` `= 2cos^2theta – 1`
  `= 2(-4/5)^2 – 1`
  `= 7/25`

Vectors, EXT1 V1 SM-Bank 25

Consider the vectors given by  `underset ~a = m underset ~i + underset ~j`  and  `underset ~b = underset ~i + m underset ~j`, where  `m in R`.

Find the value(s) of  `m`  if the acute angle between  `underset ~a`  and  `underset ~b`  is 30°.   (2 marks)

Show Answers Only

`sqrt 3, 1/sqrt 3`

Show Worked Solution
`underset ~a *underset ~b` `= m xx 1 + 1 xxm`
  `=2m`
   
`underset ~a *underset ~b` `= |underset ~a||underset ~b| cos 30^@`
  `= sqrt(m^2 + 1) *sqrt(1 + m^2) *cos 30^@`
  `= {(m^2 + 1) sqrt 3}/2`

 

`{(m^2 + 1) sqrt 3}/2` `=2m`  
`m^2 sqrt 3 + sqrt 3` `=4m`  
`m^2 sqrt 3 – 4m + sqrt 3` `=0`  
`(sqrt 3 m)^2 – 4(sqrt 3 m) + 3` `=0`  
`(sqrt 3 m)^2 – 4(sqrt 3 m) + 2^2 – 1` `=0`  
`(sqrt 3 m – 2)^2 – 1` `=0`  
`sqrt 3 m – 2` `= +-1`  
`sqrt 3 m` `= 2 +- 1`  

 
`:. m = (2 +- 1)/sqrt 3 = 3/sqrt 3 or 1/sqrt 3`

`= sqrt 3, 1/sqrt 3`

Vectors, EXT2 V1 2013 VCE 15 MC

Let  `underset~u = 4underset~i - underset~j + underset~k`,  `underset~v = 3underset~j + 3underset~k`  and  `underset~w = −4underset~i + underset~j + underset~k`.

Which one of the following statements is not true?

A.   `|\ underset~u\ | = |\ underset~v\ |`

B.   `|\ underset~u\ | = |\ −underset~w\ |`

C.   `underset~u.underset~v = 0`

D.   `(underset~u + underset~w).underset~v = 12`

Show Answers Only

`D`

Show Worked Solution
`|underset~u|` `= sqrt(16 + 1 + 1)=sqrt18`
`|underset~v|` `= sqrt(9 + 9)=sqrt18`
`|underset~w|` `= sqrt(16+1+1)=sqrt18`

 
`underset~u · underset~v= 4 xx 0 + (−1) xx 3 + 1 xx 3=0`
 

`(underset~u + underset~w) · underset~v` `= ((4 – 4)underset~i + (−1 + 1)underset~j + (1 + 1)underset~k) · underset~v`
  `= 2underset~k · underset~v`
  `= 6`

 
`=> D`

Vectors, EXT1 V1 SM-Bank 2


 

In the diagram above, `LOM` is a diameter of the circle with centre `O`.

`N` is a point on the circumference of the circle.

If  `underset~r = vec(ON)`  and  `underset ~s = vec(MN)`, express  `vec(LN)`  in terms of  `underset~r `  and  `underset~s`.   (2 marks)

Show Answers Only

`vec(LN)= 2 underset~r –  underset~s`

Show Worked Solution

`vec(LN)` `= vec(LM) + vec(MN)`
`vec(LN)` `= vec(LO) + vec(ON)`
  `= 1/2\ vec(LM) + vec(ON)`

 

`:. vec(LM) + underset~s` `= 1/2\ vec(LM) + underset~r`
`1/2\ vec(LM)` `= underset ~r – underset~s`
`vec(LM)` `= 2 underset~r – 2 underset~s`

 

`:. vec(LN)` `= 2 underset~r – 2 underset~s + underset~s`
  `= 2 underset~r –  underset~s`

 

Vectors, EXT1 V1 2011 VCE 10 MC

The diagram below shows a rhombus, spanned by the two vectors  `underset~a`  and  `underset~b`.
 

SPEC2 2011 VCAA 10 MC
 

It follows that

  1. `underset~a*underset~b = 0`
  2. `underset~a = underset~b`
  3. `(underset~a + underset~b)*(underset~a-underset~b) = 0`
  4. `|\ underset~a + underset~b\ | = |\ underset~a-underset~b|`
Show Answers Only

`C`

Show Worked Solution

`text(Consider A:)`

`text(If)\ \ underset~a · underset~b = 0\ \ =>\ \ underset~a ⊥ underset~b\ \ text{(incorrect)}`

`text(Consider B:)`

`underset~a != underset~b\ \ text{(incorrect)}`

`text(Consider C:)`

`underset~a + underset~b` `= overset(->)(OC)`
`underset~a-underset~b` `= overset(->)(BA)`

 
`overset(->)(OC) · overset(->)(BA)=0`

`text{The diagonals of a rhombus are perpendicular  (correct)}`

 
`=> C`

Functions, EXT1 F1 SM-Bank 12

Given  `f(x) = x^3 - x^2 - 2x`, without calculus sketch a separate half page graph of the following functions, showing all asymptotes and intercepts.

  1.   `y = |\ f(x)\ |`  (1 mark)

  2.   `y = f(|x|)`  (2 marks)

  3.    `y = 1/(f(x))`  (2 marks)

Show Answers Only
  1.  
  2.   
  3.   
Show Worked Solution
i.    `f(x)` `= x^3 – x^2 – 2x`
    `= x(x^2 – x – 2)`
    `= x(x – 2)(x + 1)`

 

 

ii.   

 
`y = f(|x|)\ text(is a reflection of)\ y = f(x)\ text(for)\ x > 0`

`text(is the)\ ytext(-axis.)`

 

iii.

Functions, EXT1 F1 SM-Bank 9

  1. Sketch the graph of the function described by the parametric equations

     

          `x = 4t - 7`

     

          `y = 2t^2 + t`  (2 marks)

  2. State the domain and range of the function.  (1 mark)

Show Answers Only
  1.  
  2. `text(Domain:  all)\ x`
    `text(Range)\ {y: −1/8 <= y < ∞}`
Show Worked Solution

i.   `x = 4t – 7 \ \ => \ \ t = (x + 7)/4`

`y` `= 2t^2 + t`
`y` `= 2((x + 7)/4)^2 + ((x + 7)/4)`
`16y` `= 2(x + 7)^2 + 4(x + 7)`
`16y` `= 2x^2 + 28x + 98 + 4x + 28`
`16y` `= 2x^2 + 32x + 126`
`8y` `= x^2 + 16x + 63`
`y` `= 1/8(x + 7)(x + 9)`

 
`=>\ text(Equation is a concave up quadratic with)`

`text(zeros at)\ \ x = −9\ text(and)\ \ x = −7.`
  

 

ii.   `text(Axis at)\ \ x = −8`

`:.\ y_text(min)` `= 1/8(−1)(1)`
  `= −1/8`

 
`text(Domain: all)\ x`

`text(Range:)\ −1/8 <= y < ∞`

Functions, 2ADV F2 SM-Bank 6 MC

The graph of a function  `f(x)`  is obtained from the graph of the function  `g(x) = sqrt (2x - 5)`  by a reflection in the `x`-axis followed by a dilation from the `y`-axis by a factor of  `1/2`.

Which one of the following is the function  `f(x)`?

A.   `f(x) = sqrt (5 - 4x)`

B.   `f(x) = - sqrt (x - 5)`

C.   `f(x) = sqrt (x + 5)`

D.   `f(x) = −sqrt (4x - 5)`

Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ y=sqrt(2x-5)`

`text(1st transformation:)`

`y = – sqrt(2x-5)`

COMMENT: Using “swap” terminology for dilations from the y-axis is simpler and more intelligible for students in our view.

 

`text(2nd transformation:)`

`text(Dilate from)\ y text(-axis by a factor of)\ 1/2`

`=>\ text(Swap)\ \ x → 2x`

`y` `=-sqrt(2(2x)-5)`
  `=- sqrt(4x-5)`
`:. f(x)` `= −sqrt(4x – 5)`

 
`=>   D`

Functions, 2ADV F2 SM-Bank 4 MC

The graph of the function  `f(x) = 3x^(5/2)`  is reflected in the `x`-axis and then translated 3 units to the right and 4 units down.

The equation of the new graph is

A.   `y = 3(x - 3)^(5/2) + 4`

B.   `y = -3 (x - 3)^(5/2) - 4`

C.   `y = -3 (x + 3)^(5/2) - 1`

D.   `y = -3 (x - 4)^(5/2) + 3`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ y= 3x^(5/2)`

`text(Reflect in the)\ x text(-axis:)`

`y= – 3x^(5/2)`
 

`text(Translate 3 units to the right:)`

`y=- 3(x-3)^(5/2)`
 

`text(Translate 4 units down:)`

`y=- 3(x-3)^(5/2) – 4`
 

`=>   B`

Functions, 2ADV F2 SM-Bank 3

 

The diagram below shows part of the graph of the function with rule

`f (x) = k log_e (x + a) + c`, where `k`, `a` and `c` are real constants.
 

    • The graph has a vertical asymptote with equation  `x = –1`.
    • The graph has a y-axis intercept at 1.
    • The point `P` on the graph has coordinates  `(p, 10)`, where `p` is another real constant.
       

      VCAA 2010 1b

  1. State the value of `a`.  (1 mark)

  2. Find the value of `c`.  (1 mark)

  3. Show that  `k = 9/(log_e (p + 1)`.  (2 marks)

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

i.   `text(Vertical Asymptote:)`

`x = – 1`

`:. a = 1`
 

  ii.   `text(Solve)\ \ f(0) = 1\ \ text(for)\ \ c,`

`c = 1`
 

iii.  `f(x)= k log_e (x + 1) + 1`

`text(S)text(ince)\ \ f(p)=10,`

`k log_e (p + 1) + 1` `= 10`
`k log_e (p + 1)` `= 9`
`:. k` `= 9/(log_e (p + 1))\ text(… as required)`

Algebra, STD2 A4 SM-Bank 2

Moses finds that for a Froghead eel, its mass is directly proportional to the square of its length.

An eel of this species has a length of 72 cm and a mass of 8250 grams.

What is the expected length of a Froghead eel with a mass of 10.2 kg? Give your answer to one decimal place.  (3 marks)

Show Answers Only

`80.1\ text{cm}`

Show Worked Solution

`text(Mass) prop text(length)^2`

`m = kl^2`
 

`text(Find)\ k:`

`8250` `= k xx 72^2`
`k` `= 8250/72^2`
  `= 1.591…`

 
`text(When)\ \ l\ \ text(when)\ \ m = 10\ 200:`

`10\ 200` `= 1.591… xx l^2`
`l^2` `= (10\ 200)/(1.591…)`
`:. l` `= 80.069…`
  `= 80.1\ text{cm  (to 1 d.p.)}`

Probability, 2ADV S1 SM-Bank 3

In a workplace of 25 employees, each employee speaks either French or German, or both.

If 36% of the employees speak German, and 20% speak both French and German.

  1. Calculate the probability one person chosen could speak German if they could speak French. Give your answer to the nearest percent.  (1 mark)

  2. Calculate the probability one person chosen could not speak French if they could speak German. Give your answer to the nearest percent.  (1 mark)

Show Answers Only
  1. `24text(%)`
  2. `44text(%)`
Show Worked Solution

i.   `text(Expressing in a Venn diagram:)`
 


 

`P(G|F)` `= (P(G ∩ F))/(P(F))`
  `= 0.20/0.84`
  `= 0.238…`
  `= 24text(%)`

 

ii.    `P(text(not)\ F|G)` `= (P(text(not)\ F ∩ G))/(P(G))`
    `= 0.16/0.36`
    `= 0.444…`
    `= 44text(%)`

Financial Maths, 2ADV M1 SM-Bank 8

When placed in a pond, the length of a fish was 14.2 centimetres.

During its first month in the pond, the fish increased in length by 3.6 centimetres.

During its `n`th month in the pond, the fish increased in length by `G_n` centimetres, where  `G_(n+1) = 0.75G_n`

Calculate the maximum length this fish can grow to.  (3 marks)

Show Answers Only

`28.6\ text(cm)`

Show Worked Solution

`text(Initial length) = 14.2\ text(cm)`

`G_1 = 3.6`

`G_2 = (0.75) G_1 = (0.75) 3.6`

`G_3 = (0.75) G_2 = (0.75^2) 3.6`

`text(Growth is a geometric sequence)`

`underbrace(3.6, \ 3.6(0.75), \ 3.6(0.75)^2, …)_{text(GP where)\ \ a=3.6,\ \ r=0.75}`

 
`text(S)text(ince)\ \ |\ r\ |< 1,`

`S_oo` `= a/(1-r)`
  `= (3.6)/(1-0.75)`
  `= 14.4`

 
`:.\ text(Maximum length of fish)`

`= 14.2 +14.4`

`=28.6\ text(cm)`

Financial Maths, 2ADV M1 SM-Bank 6

Julie deposits some money into a savings account that will pay compound interest every month.

The balance of Julie’s account, in dollars, after `n` months, `V_n` , can be modelled by the recurrence relation shown below.
 

`V_0 = 12\ 000, qquad V_(n + 1) = 1.0062\ V_n`
 

  1.  Recursion can be used to calculate the balance of the account after one month.

     

    1. Write down a calculation to show that the balance in the account after one month, `V_1`, is  $12 074.40. (1 mark)

    2. After how many months will the balance of Julie’s account first exceed $12 300  (1 mark)

  2.  A rule of the form  `V_n = a xx b^n`  can be used to determine the balance of Julie's account after `n` months.

    1. Complete this rule for Julie’s investment after `n` months by writing the appropriate numbers in the boxes provided below. (1 mark)
       
balance = 
 
  × 
 
  n

 

    1. What would be the value of `n` if Julie wanted to determine the value of her investment after three years?  (1 mark)

Show Answers Only

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

a.ii.  `4\ text(months)`

b.i  `text(balance) = 12\ 000 xx 1.0062^n`

b.ii.  `36`

Show Worked Solution
a.i.   `V_1` `= 1.0062 xx V_0`
    `= 1.0062 xx 12000`
    `= $12\ 074.40\ text(… as required.)`

 

a.ii.   `V_2` `= 1.0062 xx 12\ 074.40 = 12\ 149.26`
  `V_3` `= 1.0062 xx 12\ 149.26 = 12\ 224.59`
  `V_4` `= 1.0062 xx 12\ 224.59 = 12\ 300.38`

 
`:.\ text(After 4 months)`

 
b.i.  `text(balance) = 12\ 000 xx 1.0062^n`

 
b.ii.  `n = 12 xx 3 = 36`

Financial Maths, 2ADV M1 SM-Bank 5 MC

Shirley would like to purchase a new home. She will establish a loan for $225 000 with interest charged at the rate of 3.6% per annum, compounding monthly.

Each month, Shirley will pay only the interest charged for that month.

Let `V_n` be the value of Shirley’s loan, in dollars, after `n` months.

A recurrence relation that models the value of `V_n` is

  1. `V_0 = 225\ 000,qquadV_(n + 1) = 1.003 V_n`
  2. `V_0 = 225\ 000,qquadV_(n + 1) = 1.036 V_n`
  3. `V_0 = 225\ 000,qquadV_(n + 1) = 1.003 V_n - 8100`
  4. `V_0 = 225\ 000,qquadV_(n + 1) = 1.003 V_n - 675`
Show Answers Only

`D`

Show Worked Solution

`text(Monthly interest rate)`

`= text(3.6%)/12 = 0.3text(%) = 0.003`
 

`text(Monthly payment)`

`= 225\ 000 xx 0.3text(%)`

`= $675`
 

`:.\ text(Recurrence Relation is)`

`V_(n + 1) = 1.003V_n – 675`
 

`=> D`