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Trigonometry, 2ADV T3 SM-Bank 8 MC

The diagram below shows one cycle of a circular function.
 

The amplitude and period of this function are respectively

  1. `3\ \ text(and)\ \ 2 `
  2. `3\ \ text(and)\ \ (pi)/(2)`
  3. `4\ \ text(and)\ \ (pi)/(4)`
  4. `3\ \ text(and)\ \ 4`
Show Answers Only

`D`

Show Worked Solution

`text(Graph centres around)\ \ y = 1`

`text(Amplitude) \ = 3`

`text(Period:) = 4`

`=> D`

CORE, FUR1-NHT 2019 VCAA 5-7 MC

The birth weights of a large population of babies are approximately normally distributed with a mean of 3300 g and a standard deviation of 550 g.

Part 1

A baby selected at random from this population has a standardised weight of  `z = – 0.75`

Which one of the following calculations will result in the actual birth weight of this baby?
 

  1. `text(actual birth weight)\ = 550 - 0.75 × 3300`
  2. `text(actual birth weight)\ = 550 + 0.75 × 3300`
  3. `text(actual birth weight)\ = 3300 - 0.75 × 550`
  4. `text(actual birth weight)\ = 3300 + 0.75/550`
  5. `text(actual birth weight)\ = 3300 - 0.75/550`

 

Part 2

Using the 68–95–99.7% rule, the percentage of babies with a birth weight of less than 1650 g is closest to

  1. 0.14%
  2. 0.15%
  3. 0.17%
  4. 0.3%
  5. 2.5%

 

Part 3

A sample of 600 babies was drawn at random from this population.

Using the 68–95–99.7% rule, the number of these babies with a birth weight between 2200 g and 3850 g is closest to

  1. 111
  2. 113
  3. 185
  4. 408
  5. 489
Show Answers Only

`text(Part 1:)\ \ C`

`text(Part 2:)\ \ B`

`text(Part 3:)\ \ E`

Show Worked Solution

`text(Part 1)`

`text(Actual weight)` `= text(mean) + z xx text(std dev)`
  `= 3300 – 0.75 xx 550`

`=> C`

 

`text(Part 2)`

`z-text(score)` `= (x – barx)/5`
  `= (1650 – 3300)/550`
  `= −3`

 

`:. P(x < 1650)` `= P(z < −3)`
  `= 0.3/2`
  `= 0.15\ text(%)`

`=>B`
 

`text(Part 3)`

`ztext(-score)\ (2200) = (2200 – 3300)/550 = −2`

`ztext(-score)\ (3850) = (3850 – 3300)/550 = 1`
 


 

`text(Percentage)` `= (47.5 + 34)text(%) xx 600`
  `= 81.5text(%) xx 600`
  `= 48text(%)`

`=>\ E`

CORE, FUR1-NHT 2019 VCAA 1-2 MC

The histogram and boxplot shown below both display the distribution of the birth weight, in grams, of 200 babies.
 

Part 1

The shape of the distribution of the babies’ birth weight is best described as

  1. positively skewed with no outliers.
  2. negatively skewed with no outliers.
  3. approximately symmetric with no outliers.
  4. positively skewed with outliers.
  5. approximately symmetric with outliers.

 

Part 2

The number of babies with a birth weight between 3000 g and 3500 g is closest to

  1. 30
  2. 32
  3. 37
  4. 74
  5. 80
Show Answers Only
  1. `text(Part 1:)\ E`
  2. `text(Part 2:)\ D`
Show Worked Solution

`text(Part 1)`

`text(Approximately symmetric with outliers.)`

`=>\ E`

 

`text(Part 2)`

`text(Column representing 3000 – 3500g) ~~  37text(%)`

`text(37%) xx 200 = 74\ text(babies)`

`=> D`

Algebra, MET2-NHT 2019 VCAA 3 MC

If  `x + a`  is a factor of  `8x^3 - 14x^2 - a^2 x`, where  `a ∈ R text(\{0})`, then the value of  `a`  is

  1.  7
  2.  4
  3.  1
  4. –2
  5. –1
Show Answers Only

`D`

Show Worked Solution
`f(–a)` `= 8(–a)^3 – 14(–a)^2 – a^2(–a)`
`0` `= -8a^3 – 14a^2 + a^3`
`0` `= -7a^3 – 14a^2`
`0` `= -7a^2 (a + 2)`
`a` `= -2`

Graphs, MET2-NHT 2019 VCAA 2 MC

The diagram below shows one cycle of a circular function.
 

The amplitude, period and range of this function are respectively

  1. `3, 2 \ \ text(and)\ \ [-2, 4]`
  2. `3, (pi)/(2) \ \ text(and)\ \ [-2, 4]`
  3. `4, 4 \ \ text(and) \ \ [0, 4]`
  4. `4, (pi)/(4) \ \ text(and) \ \ [-2, 4]`
  5. `3, 4 \ \ text(and) \ \ [-2, 4]`
Show Answers Only

`E`

Show Worked Solution

`text(Graph centres around)\ \ y = 1`

`text(Amplitude) \ = 3`

`:. \ text(Range) \ = [1 – 3, 1 + 3] = [-2, 4]`

`text(Period:) = 4`

`=> E`

Calculus, 2ADV C3 2019 MET1 4

Given the function  `f(x) = log_e (x-3) + 2`,

  1. State the domain and range of `f(x)`.  (1 mark)

  2. i.  Find the equation of the tangent to the graph of  `f(x)` at  `(4, 2)`.  (2 marks)

     

    ii. On the axes below, sketch the graph of the function  `f(x)`, labelling any asymptote with its equation.

     

        Also draw the tangent to the graph of  `f(x)`  at  `(4, 2)`.  (4 marks)
     

Show Answers Only
  1. `x >3`

     

    `y in R`

  2. i.  `y = x-2`

     

    ii. `text(See Worked Solutions)`

Show Worked Solution
a.   `text(Domain)` `: \ x > 3`
  `text(Range)` `: \ y in R`

 

b.i.   `g(x)` `= log_e (x-3) + 2`
  `g^{\prime}(x)` `= 1/(x-3)`
  `g^{\prime}(4)` `= 1`

 
`text(Equation of tangent),\ m = 1\ \ text(through)\ (4, 2):`

`y-2` `= 1(x-4)`
`y` `= x-2`

 

b.ii.  

Calculus, 2ADV C2 2019 MET1 1a

Let  `y = (2e^(2x) - 1)/e^x`.

Find  `(dy)/(dx)`.  (2 marks)

Show Answers Only

`(dy)/(dx) = 2e^x + e^(-x)`

Show Worked Solution

`text(Method 1)`

`y` `= 2e^x – e^(-x)`
`(dy)/(dx)` `= 2e^x + e^(-x)`

 

`text(Method 2)`

`(dy)/(dx)` `= (4e^(2x) ⋅ e^x – (2e^(2x) – 1) e^x)/(e^x)^2`
  `= (4e^(3x) – 2e^(3x) + e^x)/e^(2x) `
  `= (2e^(2x) + 1)/e^x`

Algebra, MET1-NHT 2019 VCAA 5a

Let  `h:[-3/2, oo) -> R,\ h(x) = sqrt(2x + 3) - 2.`

Find the value(s) of `x` such that  `[h(x)]^2 = 1`.  (2 marks)

Show Answers Only

`3 or\ text(−1)`

Show Worked Solution
  `[h(x)]^2 = 1` `\ => \ h(x) = +- 1`
  `sqrt(2x + 3)` `= 1` `sqrt(2x + 3) – 2` `= -1`
  `sqrt(2x + 3)` `= 3` `sqrt(2x + 3)` `= 1`
  `2x + 3` `= 9` `2x + 3` `= 1`
  `x` `= 3` `x` `= -1`

 
`:. x = 3 or -1\ \ \ text{(both in the domain of}\ h)`

Calculus, MET1-NHT 2019 VCAA 4

A function `g` has rule  `g(x) = log_e (x -3) + 2`.

  1. State the maximal domain of `g` and the range of `g` over its maximal domain.  (2 marks)
  2. i.  Find the equation of the tangent to the graph of `g` at  `(4, 2)`.  (2 marks)

     

    ii. On the axes below, sketch the graph of the function `g`, labelling any asymptote with its equation.

     

        Also draw the tangent to the graph of `g` at  `(4, 2)`.  (4 marks)
     

Show Answers Only
  1. `x in (3, oo)`

     

    `y in R`

  2. i.  `y = x – 2`
    ii. 
     
Show Worked Solution
a.   `text(Domain)` `: \ x in (3, oo)`
  `text(Range)` `: \ y in R`

 

b.i.   `g(x)` `= log_e (x – 3) + 2`
  `g^{prime} (x)` `= 1/(x – 3)`
  `g^{prime} (4)` `= 1`

 
`text(Equation of tangent),\ m = 1\ \ text(through)\ (4, 2):`

`y – 2` `= 1(x – 4)`
`y` `= x – 2`

 

b.ii.  

Calculus, MET1-NHT 2019 VCAA 3

  1. Evaluate  `int_2^7 1/(x + sqrt 3)\ dx`  and  `int_2^7 1/(x - sqrt 3)\ dx`.  (2 marks)
  2. Show that  `1/2 (1/(x - sqrt 3) + 1/(x + sqrt 3)) = x/(x^2 - 3)`.  (1 mark)
  3. Use your answers to part a. and part b. to evaluate  `int_2^7 x/(x^2 - 3)\ dx`  in the form  `1/a log_e(b)`, where  `a` and `b` are positive integers.  (1 mark)
Show Answers Only
  1. `log_e((7 – sqrt 3)/(2 – sqrt 3))`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `1/2 log_e 46`
Show Worked Solution
a.   `int_2^7 1/(x + sqrt 3)\ dx` `= [log_e (x + sqrt 3)]_2^7`
    `= log_e(7 + sqrt 3) – log_e (2 + sqrt 3)`
    `= log_e ((7 + sqrt 3)/(2 + sqrt 3))`

 

`int_2^7 1/(x – sqrt 3)\ dx` `= [log_e (x – sqrt 3)]_2^7`
  `= log_e (7 – sqrt 3) – log_e (2 – sqrt 3)`
  `= log_e ((7 – sqrt 3)/(2 – sqrt 3))`

 

b.   `1/2(1/(x – sqrt 3) + 1/(x + sqrt 3))` `= 1/2 ((x + sqrt 3 + x – sqrt 3)/((x – sqrt 3)(x + sqrt 3)))`
    `= 1/2 ((2x)/(x^2 – 3))`
    `= x/(x^2 – 3)\ \ text(… as required)`

 

c.   `int_2^7 x/(x^2 – 3)` `= 1/2 int_2^7 1/(x – sqrt 3) + 1/(x + sqrt 3)\ dx`
    `= 1/2[log_e ((7 – sqrt 3)/(2 – sqrt 3)) + log_e ((7 + sqrt 3)/(2 + sqrt 3))]`
    `= 1/2 log_e (((7 – sqrt 3)(7 + sqrt 3))/((2 – sqrt 3)(2 + sqrt 3)))`
    `= 1/2 log_e ((49 – 3)/(4 – 3))`
    `= 1/2 log_e 46`

Calculus, MET1-NHT 2019 VCAA 1a

Let  `y = (2e^(2x) - 1)/e^x`.

Find  `(dy)/(dx)`.  (2 marks)

Show Answers Only

`(dy)/(dx) = 2e^x + e^(-x)`

Show Worked Solution

`text(Method 1)`

`y` `= 2e^x – e^(-x)`
`(dy)/(dx)` `= 2e^x + e^(-x)`

 

`text(Method 2)`

`(dy)/(dx)` `= (4e^(2x) ⋅ e^x – (2e^(2x) – 1) e^x)/(e^x)^2`
  `= (4e^(3x) – 2e^(3x) + e^x)/e^(2x) `
  `= (2e^(2x) + 1)/e^x`

Statistics, EXT1 S1 EQ-Bank 23

A light manufacturer knows that 6% of the light bulbs it produces are defective.

Light bulbs are supplied in boxes of 20 bulbs. Boxes are supplied in pallets of 120 boxes.

Calculate the probability that

  1. A box of light bulbs contains exactly 3 defective bulbs.  (1 mark)

  2. A box of light bulbs contains at least 1 defective bulb.  (1 mark)

  3. A pallet contains between 90 and 95 (inclusive) boxes with at least 1 defective bulb (use the probability table attached).  (3 marks)

Show Answers Only
  1. `0.086\ \ (text(to 3 d.p.))`
  2. `0.710\ \ (text(3 d.p.))`
  3. `0.1416`
Show Worked Solution

i.   `P(D) = 0.06, \ P(barD) = 0.94`

`P(D = 3)` `= \ ^20C_3(0.06)^3(0.94)^17`
  `= 0.086\ \ (text(to 3 d.p.))`

 

ii.    `P(D >= 1)` `= 1 – P(D = 0)`
    `= 1 – \ ^20C_0(0.06)^0(0.94)^20`
    `= 0.710\ \ (text(to 3 d.p.))`

 

iii.   `text(Let)\ \ X = text(number of boxes where)\ \ D >= 1`

`text(Let)\ \ overset^p = text(proportion of boxes where)\ \ D >= 1`

`E(overset^p) = p = 0.710`

`text(Var)(overset^p) = (0.710(1 – 0.710))/120 = 0.0017158`

`sigma(overset^p) = 0.04142`
 

`overset^p\ ~\ N(0.710, 0.04142)`

`text(If)\ \ X = 90 \ => \ overset^p = 90/120 = 0.75`

`text(If)\ \ X = 95 \ => \ overset^p = 95/120 = 0.79167`
 

`ztext(-score)\ (X = 90) = (0.75 – 0.710)/(0.04142) = 0.9657`

`ztext(-score)\ (X = 95) = (0.79167 – 0.710)/(0.04142) = 1.972`
 

`P(90 <= X <= 95)` `= P(0.97 <= z <= 1.97)`
  `= 0.9756 – 0.8340`
  `= 0.1416`

Statistics, EXT1 S1 EQ-Bank 20

Netball Australia records show that 10% of all registered players are over the age of 25.

  1. A random survey of 100 netball players was carried out to find out how many were over 25 years of age.

     

    Assuming the sample proportion is normally distributed, calculate the expected mean and standard deviation of this group.  (2 marks)

  2. Using the probability table attached, estimate the probability that at least 15 players surveyed will be over 25 years of age.  (2 marks)

Show Answers Only
  1. `0.03`
  2. `0.0475`
Show Worked Solution
i.   `E(hat p)` `= p = 0.1`
  `text(Var)(hat p)` `= (p(1 – p))/n`
    `= (0.1(0.9))/100`
    `= 0.0009`
  `sigma(hat p)` `= sqrt(0.0009)`
    `= 0.03`

 

ii.  `Ptext{(at least 15 players are over 25)} = P(hat p >= 0.15)`

`hat p\ ~\ N(mu,sigma)\ ~\ N(0.1, 0.03)`

`P(hat p >= 0.15)` `= P(z >= (0.15 – 0.10)/0.03)`
  `= P(z >= 1.67)`
  `= 1 – 0.9525`
  `= 0.0475`

Statistics, SPEC2-NHT 2019 VCAA 6

A paint company claims that the mean time taken for its paint to dry when motor vehicles are repaired is 3.55 hours, with a standard deviation of 0.66 hours.

Assume that the drying time for the paint follows a normal distribution and that the claimed standard deviation value is accurate.

  1. Let the random variable  `barX`  represent the mean time taken for the paint to dry for a random sample of 36 motor vehicles.

     

     Write down the mean and standard deviation of  `barX`.   (2 marks)

At a car crash repair centre, it was found that the mean time taken for the paint company's paint to dry on randomly selected vehicles was 3.85 hours. The management of this crash repair centre was not happy and believed that the claim regarding the mean time taken for the paint to dry was too low. To test the paint company's claim, a statistical test was carried out.

  1. Write down suitable null and alternative hypotheses `H_0` and `H_1` respectively to test whether the mean time taken for the paint to dry is longer than claimed.   (1 mark)
  2. Write down an expression for the  `p`  value of the statistical test and evaluate it correct to three decimal places.   (2 marks)
  3. Using a 1% level of significance, state with a reason whether the crash repair centre is justified in believing that the paint company's claim of mean time taken for its paint to dry of 3.55 hours is too low.  (1 mark)
  4. At the 1% level of significance, find the set of sample mean values that would support the conclusion that the mean time taken for the paint to dry exceeded 3.55 hours. Give your answer in hours, correct to three decimal places.  (2 marks)
  5. If the true time taken for the paint to dry is 3.83 hours, find the probability that the paint company's claim is not rejected at the 1% level of significance, assuming the standard deviation for the paint to dry is still 0.66 hours. Give your answer correct to two decimal places.  (1 mark)
Show Answers Only
  1. `0.11`
  2. `H_0 : \ mu = 3.55`
    `H_1 : \ mu > 3.55`
  3. `0.003 \ text{(to 3 decimal places)}`
  4. `text(S) text(ince) \ \ p < 0.01 , H_0 \ text(should be rejected at the 1% level.)`
  5. `barX > 3.806`
  6. `0.41`
Show Worked Solution
a.   `E(barX)` `= 3.55`
`sigma(barX)` `= (sigma)/(sqrtn)`
  `= (0.66)/(sqrt36)`
  `= 0.11`

 

b.   `H_0 : \ mu = 3.55`

`H_1 : \ mu > 3.55`

 

c.   `p` `= text(Pr) (barX > 3.85)`
  `= text(Pr) (z > (3.85 – 3.55)/(0.11))`
  `= text(Pr) (z >2.326)`
  `= 0.003 \ text{(to 3 decimal places)}`

 

d.   `text(S) text(ince)\ \ p < 0.01 , H_0 \ text(should be rejected at the 1% level.)`

`text(i.e. repair centre is justified that the mean time 3.55 hours is too low.)`

 

e.   `text(If) \ \ mu = 3.55`

`(barX – mu)/(sigma)` `> 2.3263`
`barX` `> 2.3263 xx 0.11 + 3.55`
`barX` `> 3.806`

 

f.   `text(Pr) (barX< 3.806 | mu = 3.83)` `= text(Pr) (z < (3.806 – 3.83)/(0.11))`
  `= text(Pr) (z < -0.21818)`
  `= 0.41`

Calculus, SPEC2-NHT 2019 VCAA 3


 

The vertical cross-section of a barrel is shown above. The radius of the circular base (along the `x`-axis) is 30 cm and the radius of the circular top is 70 cm. The curved sides of the cross-section shown are parts of the parabola with rule  `y = (x^2)/(80) - (45)/(4)`. The height of the barrel is 50 cm.
 
a.  i.   Show that the volume of the barrel is given by  `pi int_0^50 (900 + 80 y)\ dy`.  (1 marks)
     ii.  Find the volume of the barrel in cubic centimetres.  (1 marks)    

The barrel is initially full of water. Water begins to leak from the bottom of the barrel such that  `(dV)/(dt) = (-8000pi sqrth)/(A)`  cubic centimetres per second, where after  `t`  seconds the depth of the water is  `h`  centimetres, the volume of water remaining in the barrel is  `V`  cubic centimetres and the uppermost surface area of the water is  `A`  square centimetres.
 
b.     Show that  `(dV)/(dt) = (-400 sqrth)/(4h + 45)`?  (2 marks)
c.      Find  `(dh)/(dt)`  in terms of  `h`. Express your answer in the form  `(-a sqrth)/(pi(b + ch)^2)`, where  `a, b`  and  `c`  are positive integers.  (3 marks)
d.     Using a definite integral in terms of  `h`, find the time, in hours, correct to one decimal place, taken for the barrel to empty.  (2 marks)

Show Answers Only

a.  i. `text(Proof(Show Worked Solution))`

      ii.  `145000 pi \ text(cm)^3`

b.    `text(Proof (Show Worked Solution))`

c.    `(-20 sqrth)/(pi(45 + 4h)^2)`

d.    `9.9\ text(hours)`

Show Worked Solution
a.i. `V` `= pi int x^2 dy`
  `y` `= (x^2)/(80) – (45)/(4)`
  `x^2` `= 80y + 900`
     
  `:. \ V` `= pi int_0^50 (900 + 80y)dy`

 

a.ii. `V` `= pi int_0_50 (900 + 80y) dy`
    `= pi [900y + 40y^2]_0^50`
    `= 145000pi \ text(cm)^3`

 

b.    `A = pi x^2 = pi (900 + 80h)`

`(dV)/(dt)` `= (-8000pi sqrth)/(pi(900 + 80h))`
  `= (-8000 sqrth)/(20(4h + 45))`
  `= (-400 sqrth)/(4h + 45)`

 

c.    `(dV)/(dh) = pi(900 + 80h)`

`(dh)/(dt)` `= (dh)/(dV) ⋅ (dV)/(dt)`
  `= (1)/(pi(900 +80h)) xx (-400 sqrth)/(4h +45)`
  `= (-400 sqrth)/(20pi(4h +45)^2)`
  `= (-20 sqrth)/(pi(45 +4h)^2)`

  
 
d.    `(dt)/(dh) = (-pi(45 + 4h)^2)/(20 sqrth)`

`t` `= -pi int_50^0 ((45 + 4h)^2)/(20 sqrth)\ dh`
  `≈ 35\ 598.6 \ text(seconds)`
  `≈ 9.9 \ text{hours  (to 1 d.p.)}`

Calculus, SPEC2-NHT 2019 VCAA 2

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

  1. State the equations of the asymptotes of the graph of `f`.  (2 marks)
  2. State the coordinates of the stationary points and the point of inflection. Give your answers correct to two decimal places.  (2 marks)
  3. Sketch the graph of `f` from  `x = -6`  to  `x = 6`  (endpoint coordinates are not required) on the set of axes below, labeling the turning points and the point of inflection with their coordinates correct to two decimal places. Label the asymptotes with their equations.  (3 marks)

Consider the function `f_k` with rule  `f_k(x) = (x^2 + x + k)/(x^2-1)`  where  `k ∈ R`.

  1. For what values of `k` will `f_k` have no stationary points?  (2 marks)
  2. For what value of `k` will the graph of `f_k` have a point of  inflection located on the `y`-axis?  (1 marks)
Show Answers Only
i.     `x = 1, x = -1, y = 1`

 ii.   `(-3.73, 0.87) \ text(and) \ (-0.27, -0.87)`

`(-5.52, 0.88)`
iii.

iv.  `-2`

v.  `-1`

Show Worked Solution

i.    `f(x) = 1 + (x + 2)/((x + 1)(x-1))`

`:. \ text(Asymptotes at)\ \ x = 1, x = -1, y = 1`
  

ii.   `f′(x) = (-(x^2 + 4x + 1))/(x^2-1)^2`

`text(Solve:) \ \ f^{′}(x) = 0 \ => \ x = -2 ± sqrt3`

`text(SP’s at) \ (-3.73, 0.87) \ \ text(and)\ \ (-0.27, -0.87)`
 
`f^{″}(x) = (-2x^3-12x^2-6x-4)/(x^2-1)^3`

`text(Solve:) \ \ f^{″}(x) = 0 \ => \ x = -5.52`

`text(POI at) \ \ (-5.52, 0.88)`
 

iii.

 

iv.    `f_k^{′}(x) = (-x^2-2(k+1) x-1)/(x^2-1)^2`

`text(If no SP’s,) \ \ Δ < 0`

`[-2( k + 1)]^2-4(-1)(-1)` `< 0`
`4k^2 + 8k + 4-4` `< 0`
`4k(k + 2)` `< 0`

 
`:. \ -2 < k < 0`
 

v.    `f_k^{″}(x) = (2(x^3 + 3(k + 1) x^2 + 3x + k + 1))/((x^2-1)^3)`

`text(Solve:)\ \ f_k^{″}(x) = 0`

`k = -1`

Complex Numbers, SPEC2-NHT 2019 VCAA 6 MC

`P(z)`  is a polynomial of degree  `n`  with real coefficients where  `z ∈ C`. Three of the roots of the equation  `P(z) = 0`  are  `z = 3 - 2i`, `z = 4`  and  `z = −5i`.

The smallest possible value of  `n`  is

  1.  3
  2.  4
  3.  5
  4.  6
  5.  7
Show Answers Only

`C`

Show Worked Solution

`text(Roots:)\ 4, 3 ± 2i, ±5i\ \ \ (text(conjugate root theorem))`

`:.\ text(Minimum roots = 5)`

`=>\ C`

Complex Numbers, SPEC2-NHT 2019 VCAA 1

In the complex plane, `L` is the with equation `|z + 2| = |z - 1 - sqrt3 i|`.

  1.  Verify that the point (0, 0) lies on `L`.  (1 marks)
  2.  Show that the cartesian form of the equation of  `L`  is  `y = - sqrt3 x`.  (2 marks)
  3.  The line  `L`  can also be expressed in the form  `|z - 1| = |z - z_1|`, where  `z_1 ∈ C`.

     

     Find  `z_1` in cartesian form.  (2 marks)

  4.  Find, in cartesian form, the points(s) of intersection of  `L`  and the graph of  `|z| = 4`.  (2 marks)
  5.  Sketch  `L`  and the graph of  `|z| = 4`  on the Argand diagram below.  (2 marks)
     
     
         
     
  6.  Find the area of the sector defined by the part of  `L`  where  `text(Re)(z) ≥ 0`, the graph of  `|z| = 4`  where  `text(Re)(z) ≥ 0`, and imaginary axis where  `text(Im)(z) > 0`.  (1 marks)
Show Answers Only
  1. `text(Proof(See Worked Solution))`
  2. `text(Proof(See Worked Solution))`
  3. `-(1)/(2) – (sqrt3)/(2) i`
  4. `(2,-2 sqrt3) text(and) (-2, 2 sqrt3)`
  5.  

     
  6. `(20pi)/(3)`
Show Worked Solution

a.   `text(Substitute)\ \ z = 0 + 0i\ \  text(into both sides:)`

`text(LHS) = |2| = 2`

`text(RHS) = |-1 – sqrt3i| = sqrt{(-1)^2 + (-sqrt3)^2} = 2`

`:. (0,0)\ \ text(lies on both sides.)`

 

b.  `|x + yi + 2|` `= |x + yi – 1 – sqrt3 i|`
  `|(x + 2) + yi|` `= |(x – 1) + (y – sqrt3) i|`
  `sqrt(x^2 + 4x + 4 + y^2` `= sqrt(x^2 – 2x + 1 + y^2 -2 sqrt3 y + 3`
  `x^2 + 4x + 4 + y^2` `= x^2 – 2x + 4 – 2 sqrt3 y + y^2`
  `6x` `= -2 sqrt3 y`
  `y` `= -(3)/(sqrt3) x`
  `y` `= -sqrt3 x`


c.

`m_text(perp) = (sqrt2)/(3) , text(through)\ (1, 0)`

`y = (sqrt3)/(3) (x – 1)\ …\ L_1`

`text(Intersection) \ L\ text(and) \ L_1,`

`text(Solve:) \ (sqrt3)/3 (x-1) = -sqrt3 x\ \ \ text{(by CAS)}`

`=> x = (1)/(4) , y = -(sqrt3)/(4) \ \ \ text{(point}\ Ptext{)}`
 

`P(x_1,y_1) \ text(is midpoint of) \ \ z_1 \ text(and) \ \ (1, 0):`

`(x_1 + 1)/(2) = (1)/(4) \ => \ x_1 = -(1)/(2)`

`(y_1 + 0)/(2) = -sqrt3/(4) \ => \ y_1 = -sqrt3/(2)`

`:. \ z_1 = -(1)/(2) – (sqrt3)/(2) i`

 

d.   `|z| = 4 => \ text(circle, centre) \ (0,0), \ text(radius) = 4`

`x^2 + y^2 = 16\ …\ (1)`

`y = – sqrt3 x\ …\ (2)`

`text(Substitute)\ (2) \ text(into) \ (1)`

`x^2 + 3x^2 = 16`

`x = ±2`

`:. \ text(Intersection at)\ (2, – 2 sqrt3) \ text(and) \ (-2, 2 sqrt3)`

 

e.


 

f.    

`text(Area)` `= (5)/(12) xx  pi xx 4^2`
  `= (20pi)/(3)`

Functions, EXT1 F1 2019 SPEC2-N 2 MC

The curve given by  `x = 3sec(t) + 1`  and  `y = 2tan(t)-1`  can be expressed in cartesian form as

  1.  `((y + 1)^2)/4-((x-1)^2)/9 = 1`
  2.  `((x + 1)^2)/3-((y-1)^2)/2 = 1`
  3.  `((x-1)^2)/3 + ((y + 1)^2)/2 = 1`
  4.  `((x-1)^2)/9-((y + 1)^2)/4 = 1`
Show Answers Only

`D`

Show Worked Solution

`sec^2theta = tan^2theta + 1`

`x = 3sec(t) + 1 \ => \ sec(t) = (x-1)/3`

`y = 2tan(t)-1 \ => \ tan(t) = (y + 1)/2`

`:.((x-1)/3)^2-((y + 1)/2)^2` `= 1`
`((x-1)^2)/9-((y + 1)^2)/4` `= 1`

`=>\ D`

Vectors, EXT1 V1 EQ-Bank 28

A projectile is fired from a canon at ground level with initial velocity `sqrt300` ms−1 at an angle of 30° to the horizontal.

The equations of motion are  `(d^2x)/(dt^2) = 0`  and  `(d^2y)/(dt^2) = −10`.

  1. Show that  `x = 15t`.  (1 mark)

  2. Show that  `y = 5sqrt3t - 5t^2`.  (2 marks)
  3. Hence find the Cartesian equation for the trajectory of the projectile.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `y = sqrt3/3 – (x^2)/45`
Show Worked Solution

i.   

`dotx = sqrt300\ cos30^@ = 10sqrt3 xx sqrt3/2 = 15\ text(ms)^(−1)`

`x = int 15\ dt = 15t + C_1`

`text(When)\ \ t = 0, \ x = 0 \ => \ C_1 = 0`

`:. x = 15t`

 

ii.   `doty = int −10\ dt = −10t + C_1`

`text(When)\ \ t = 0, \ doty = 10sqrt3 sin30^@ = 5sqrt3 \ => \ C_1 = 5sqrt3`

`:. doty = 5sqrt3 – 10t`
 

`y = int doty\ dt = 5sqrt3t – 5t^2 + C_2`

`text(When)\ \ t = 0, \ y = 0 \ => \ C_2 = 0`

`:. y = 5sqrt3t – 5t^2`

 

iii.   `x = 15t \ => \ t = x/15`

`y` `= 5sqrt3 xx x/15 – 5(x/15)^2`  
`:.y` `=sqrt3/3 x – (x^2)/45`  

Combinatorics, EXT1 A1 SM-Bank 9

`(2 - sqrt3)^5 = a + bsqrt3`.

Evaluate `a` and `b` using a binomial expansion.  (2 marks)

Show Answers Only

`a = 362, \ b = -209`

Show Worked Solution
`(2 – sqrt3)^5` `= \ ^5C_0 *2^5 + \ ^5C_1* 2^4(−sqrt3) + \ ^5C_2 *2^3(−sqrt3)^2 + \ ^5C_3 *2^2(−sqrt3)^3`
               `+ \ ^5C_4 *2(−sqrt3)^4 + \ ^5C_5(−sqrt3)^5`
  `= 32 – 80sqrt3 + 240 – 120sqrt3 + 90 – 9sqrt3`
  `= 362 – 209sqrt3`

 
`:. a = 362, \ b = -209`

Vectors, EXT2 V1 2019 SPEC2 4

The base of a pyramid is the parallelogram  `ABCD`  with vertices at points  `A(2,−1,3),  B(4,−2,1),  C(a,b,c)`  and  `D(4,3,−1)`. The apex (top) of the pyramid is located at  `P(4,−4,9)`.

  1. Find the values of  `a, b`  and  `c`.  (2 marks)

  2. Find the cosine of the angle between the vectors  `overset(->)(AB)`  and  `overset(->)(AD)`.  (2 marks)

  3. Find the area of the base of the pyramid.  (2 marks)

Show Answers Only
  1. `a = 6, b = 2, c = −3`
  2. `4/9`
  3. `2sqrt65\ text(u²)`
Show Worked Solution
i.    `overset(->)(AB)` `= (4 – 2)underset~i + (−2 + 1)underset~j + (1 – 3)underset~k`
    `= 2underset~i – underset~j – 2underset~k`

 
`text(S)text(ince)\ ABCD\ text(is a parallelogram)\ => \ overset(->)(AB)= overset(->)(DC)`

`overset(->)(DC) = (a – 4)underset~i + (b – 3)underset~j + (c + 1)underset~k`

`a – 4 = 2 \ => \ a = 6`

`b – 3 = −1 \ => \ b = 2`

`c + 1 = −2 \ => \ c = −3`

 

ii.   `overset(->)(AB) = 2underset~i – underset~j – 2underset~k`

`overset(->)(AD) = 2underset~i + 4underset~j – 4underset~k`

`cos angleBAD` `= (overset(->)(AB) · overset(->)(AD))/(|overset(->)(AB)| · |overset(->)(AD)|)`
  `= (4 – 4 + 8)/(sqrt(4 + 1 + 4) · sqrt(4 + 16 + 16))`
  `= 4/9`

 

iii.    `1/2 xx text(Area)_(ABCD)` `= 1/2 ab sin c`
  `text(Area)_(ABCD)` `= |overset(->)(AB)| · |overset(->)(AD)| *sin(cos^(−1)\ 4/9)`

  

  

`:. text(Area)_(ABCD)` `= 3 · 6 · sqrt65/9`
  `= 2sqrt65\ text(u²)`

Calculus, EXT2 C1 2019 SPEC1-N 4

Evaluate  `int_(e^3) ^(e^4) (1)/(x log_e (x))\ dx`.   (3 marks)

Show Answers Only

`log_e ((4)/(3))`

Show Worked Solution

`text(Let)\ \ u = log_e x`

`(du)/(dx) = (1)/(x) \ => \ du = (1)/(x) dx`

`text(When) \ \ x = e^4 \ => \ u = 4`

`text(When) \ \ x = e^3 \ => \ u = 3`

`int_(e^3) ^(e^4) (1)/(x log_e (x))` `= int_3 ^4 (1)/(u)\ du`
  `= [ log_e u]_3 ^4`
  `= log_e 4 – log_e 3`
  `= log_e ((4)/(3))`

Vectors, EXT2 V1 SM-Bank 24

Show that the points  `A(2, 1, text{−1}), \ B(4, 2, text{−3})`  and  `C(text{−4}, text{−2}, 5)`  are collinear.  (2 marks)

Show Answers Only

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

Show Worked Solution

`vec(AB) = ((4), (2), (text{−3})) – ((2), (1), (text{−1})) = ((2), (1), (text{−2}))`

`vec(AC) = ((text{−4}), (text{−2}), (5)) – ((2), (1), (text{−1})) = ((text{−6}), (text{−3}), (6)) = -3((2), (1), (text{−2}))`

 

`text(S) text(ince)\ vec(AB)\ text(||)\ vec(AC) and A\ text(lies on both lines,)`

`A, B, C\ \ text(are collinear).`

Vectors, EXT2 V1 SM-Bank 23

A cube with side length 3 units is pictured below.
 

     
 

  1. Calculate the magnitude of vector `vec(AG)`.  (1 mark)

  2. Find the acute angle between the diagonals `vec(AG)` and `vec(BH)`.  (3 marks)

Show Answers Only
  1. `3 sqrt 3\ text(units)`
  2. `70^@32′`
Show Worked Solution

i.   `A(3, 0 , 0), \ \ G(0, 3, 3)`

  `vec(AG)` `= ((0), (3), (3)) – ((3), (0), (0)) = ((text{−3}), (3), (3))`
  `|\ vec(AG)\ |` `= sqrt (9 + 9 + 9)`
    `= 3 sqrt 3\ text(units)`

 

ii.    `H (3, 3, 3)`
  `vec(BH) = ((3), (3), (3))`
`vec(AG) ⋅ vec(BH)` `= |\ vec(AG)\ | ⋅ |\ vec(BH)\ |\ cos theta`
`((text{−3}), (3), (3)) ⋅ ((3), (3), (3))` `= sqrt (9 + 9 + 9) ⋅ sqrt (9 + 9 + 9) cos theta`
`-9 + 9 + 9` `= 27 cos theta`
`cos theta` `= 1/3`
`theta` `= 70.52…`
  `= 70^@32′`

Vectors, EXT2 V1 SM-Bank 21

  1. Find the equation of the vector line  `underset~v`  that passes through  `Atext{(5, 2, 3)}`  and  `B(7, 6, 1)`.  (1 mark)

  2. A sphere has centre  `underset~c`  at  `text{(2, 3, 5)}`  and a radius of  `5sqrt2`  units.
    Find the points where the vector line  `underset~v`  meets the sphere.  (3 marks)

Show Answers Only
  1. `underset~v = ((5),(2),(3)) + lambda((1),(2),(−1))`
  2. `((2),(–4),(6)), \ ((7),(6),(1))`
Show Worked Solution

i.   `overset(->)(BA) = ((7 – 5),(6 – 2),(1 – 3)) = ((2),(4),(−2)) = 2((1),(2),(−1))`

`underset~v = ((5),(2),(3)) + lambda((1),(2),(−1))`
 

ii.   `text(General point)\ underset~v:`

`x = 5 + lambda`

`y = 2 + 2lambda`

`z = 3 – lambda`
 

`text(Equation of sphere,)\ underset~c = (2, 3, 5),\ text(radius)\ 5sqrt2:`

`(x – 2)^2 + (y – 3)^2 + (z -5)^2` `= (5sqrt2)^2`
`(lambda + 3)^2 + (2lambda – 1)^2 + (−lambda – 2)^2` `= 50`
`lambda^2 + 6lambda + 9 + 4lambda^2 – 4lambda + 1 + lambda^2 + 4lambda + 4` `= 50`
`6lambda^2 + 6lambda + 14` `= 50`
`6lambda^2 + 6lambda – 36` `= 0`
`6(lambda + 3)(lambda – 2)` `= 0`
`lambda` `= –3\ text(or)\ 2`

 
`text(When)\ \ lambda = –3,`

`text(Intersection) = ((5),(2),(3)) – 3((1),(2),(−1)) = ((2),(–4),(6))`

`text(When)\ \ lambda = 2,`

`text(Intersection) = ((5),(2),(3)) + 2((1),(2),(−1)) = ((7),(6),(1))`

Functions, 2ADV F1 SM-Bank 53

  1. If  `1/(root3(7+pi)) = (7+pi)^x`, find  `x`.  (1 mark)

  2. Calculate the value of  `1/(root3(7+pi))`  to 3 significant figures.   (1 mark)

Show Answers Only
  1. `-1/3`
  2. `0.462`
Show Worked Solution

i.  `1/(root3(7+pi)) = (7+pi)^(-1/3)`

ii.   `1/(root3(7+pi))` `=0.4619…`
    `=0.462\ \ text{(to 3 sig. fig.)}`

Functions, 2ADV F1 SM-Bank 52

Find `a`  and  `b`  such that  `a,b`  are real numbers and
 

`(6sqrt3-sqrt5)/(2sqrt5)= a + b sqrt15`.  (2 marks)

Show Answers Only

`a= -1/2, \ b=3/5`

Show Worked Solution
`(6sqrt3-sqrt5)/(2sqrt5)` `=(6sqrt3-sqrt5)/(2sqrt5) xx (2sqrt5)/(2sqrt5)`  
  `=(2sqrt5(6sqrt3 – sqrt5))/(4 xx5)`  
  `=(12sqrt15-10)/20`  
  `=- 1/2 + 3/5 sqrt15`  

 
`:. a= -1/2, \ b=3/5`

Functions, 2ADV F1 SM-Bank 51

Find `a`  and  `b`  such that  `a,b`  are real numbers and 
 

`(sqrt3-2)/(2sqrt3)= a + b sqrt3`.  (2 marks)

Show Answers Only

 `a = 1/2, \ b = – 1/3`

Show Worked Solution
`(sqrt3-2)/(2sqrt3)` `= (sqrt3-2)/(2sqrt3) xx (2sqrt3)/(2sqrt3)`
  `= (2sqrt3(sqrt3-2))/(4 xx 3)`
  `= (6-4sqrt3)/12`
  `=1/2 – 1/3 sqrt3`

 
`:.\ a = 1/2, \ b = – 1/3`

Vectors, EXT1 V1 EQ-Bank 27


 

`OABC`  are the vertices of a rhombus.

Prove, using vectors, that its diagonals are perpendicular.  (2 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution

`text(S) text(ince)\ \ OABC\ \ text(is a rhombus)`

`| underset~a | = | underset~c |`
 
`text(Diagonals are) \ \ overset(->)(OB) \ \ text(and) \ \ overset(->)(AC) , text(where)`

`overset(->)(OB) = underset~a + underset~c`

`overset(->)(AC) = underset~c – underset~a`
  

`(underset~c – underset~a) · (underset~a + underset~c)` `= underset~c* underset~a + underset~c* underset~c – underset~a* underset~a – underset~a *underset~c`
  `=| underset~c |^2 – | underset~a |^2`
  `= 0`

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

Vectors, EXT2 V1 SM-Bank 18

A sphere is represented by the equation

`x^2 - 4x + y^2 + 8y + z^2 - 3z + 2 = 0`

  1. Determine the centre  `underset~c`  and radius of the sphere.  (2 marks)

  2. Find the vector equation of the sphere.  (1 mark)

Show Answers Only
  1.  `underset~c = ((2),(-4),({3}/{2})) \ , \ text(radius) = (9)/(2)`
  2. `| \ underset~r – ((2),(-4),({3}/{2})) | = (9)/(2)`
Show Worked Solution

i.  `x^2 – 4x + y^2 + 8y + z^2 – 3z + 2 = 0`

`(x-2)^2 + (y+4)^2 + (z-{3}/{2})^2 + 2 – (89)/(4) = 0`

`(x-2)^2 + (y+4)^2 + (z-{3}/{2})^2 = (81)/(4)`
 
`:. \ underset~c = ((2),(-4),({3}/{2})) \ , \ text(radius) = (9)/(2)`
 

ii. `text(Vector equation:)`

`| \ underset~r – ((2),(-4),({3}/{2})) | = (9)/(2)`

Vectors, EXT2 V1 SM-Bank 11

Given  `lambda_1underset~a + lambda_2underset~b = [(50),(−45),(−8)]`, find  `lambda_1` and `lambda_2`  if 
 

`underset~a = [(2),(−3),(4)]`  and  `underset~b = [(3),(−2),(−3)]`.  (2 marks)

Show Answers Only

`lambda_1 = 7, lambda_2 = 12`

Show Worked Solution

`lambda_1underset~a + lambda_2underset~b = [(2lambda_1 + 3lambda_2),(−3lambda_1 – 2lambda_2),(4lambda_1 – 3lambda_2)]`

`2lambda_1 + 3lambda_2 = 50\ \ …\ (1)`  
`4lambda_1 – 3lambda_2 = −8\ \ …\ (2)`  

 
`(1) + (2)`

`6lambda_1` `= 42`
`lambda_1` `= 7`

 
`text(Substitute)\ \ lambda_1=7\ \ text{into (1):}`

`14+3lambda_2` `= 50`
`lambda_2` `= 12`

Vectors, EXT2 V1 SM-Bank 15

Consider the two vector line equations

`underset~(v_1) = ((1),(4),(−2)) + lambda_1((3),(0),(−1)), qquad underset~(v_2) = ((3),(2),(2)) + lambda_2((4),(2),(−6))`

  1. Show that  `underset~(v_1)`  and  `underset~(v_2)`  intersect and determine the point of intersection . (2 marks)

  2. What is the acute angle between the vector lines, to the nearest minute.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `40°29’\ \ (text(nearest minute))`
Show Worked Solution

i.   `text(Solve simultaneously:)`

`1 + 3lambda_1` `= 3 + 4lambda_2` `\ \ …\ (1)`
`4 + 0lambda_1` `= 2 + 2lambda_2` `\ \ …\ (2)`
`−2 – lambda_1` `= 2 – 6lambda_2` `\ \ …\ (3)`

 
`=> lambda_2 = 1\ \ \ text{(from (2))}`

`=>lambda_1 = 2\ \ \ text{(from (1) and (3))}`

`:.\ text(vector lines intersect)`
  

`text(P.O.I.) = ((1),(4),(−2)) + 2((3),(0), (−1)) = ((7),(4),(−4))`

 

ii.   `underset~(v_1) = underset~(a_1) + lambda_1*underset~(b_1)`

`underset~(v_2) = underset~(a_2) + lambda_2*underset~(b_2)`

`costheta` `= (underset~(b_1) · underset~(b_2))/(|underset~b_1||underset~b_2|)`
  `= (12 + 0 + 6)/(sqrt10 sqrt56)`
  `= 0.7606…`

 

`theta` `= 40.479…`
  `= 40°29’\ \ (text(nearest minute))`

Vectors, EXT2 V1 SM-Bank 6

  1. What vector line equation, `underset~r`, corresponds to the Cartesian equation
  2. `qquad (x + 2)/5 = (y - 5)/4`  (1 mark)

  3. Express   `underset~v`  in Cartesian form where,
  4. `qquad underset~v = ((1),(−4)) + lambda((3),(1))`  (1 mark)

Show Answers Only
  1.  `underset~r = ((−2),(5)) + lambda((5),(4))`
  2.  `y = (x – 13)/3`
Show Worked Solution

i.   `underset~r = ((−2),(5)) + lambda((5),(4))`

COMMENT: Ensure you know the format for conversion from Cartesian to vector form (part i)!

 

ii.   `((x),(y)) = ((1),(−4)) + lambda((3),(1))`

`x = 1 + 3lambda \ \ => \ lambda = (x – 1)/3`
`y = −4 + lambda\ \ => \ lambda = y + 4`

 
`y + 4 = (x – 1)/3`

`:. y = (x – 13)/3`

Vectors, EXT2 V1 SM-Bank 4

Determine the equation of the line vector `underset~r`, given it passes through the point  `(7, 1, 0)`  and is parallel to the line joining  `P(2, −1, 2)`  and  `Q(3, 4, 1)`.   (2 marks)

Show Answers Only

`underset~r = ((7),(1),(0)) + lambda((1),(5),(−1))`

Show Worked Solution

`overset(->)(PQ) = ((3),(4),(1)) – ((2),(−1),(2)) = ((1),(5),(−1))`
 

`:. underset~r = ((7),(1),(0)) + lambda((1),(5),(−1))`

Vectors, EXT2 V1 SM-Bank 2

  1. Find values of  `a`, `b`, `c`  and  `d`  such that  `underset~v = ((a),(b)) + 2((c),(d))`  is a vector equation of a line that passes through  `((3),(1))`  and  `((−3),(−3))`.  (2 marks)

  2. Determine whether  `underset~u = ((4),(6)) + lambda((−2),(3))`  is perpendicular to  `underset~v`.  (1 mark)

  3. Express  `underset~u`  in Cartessian form.  (1 mark)

Show Answers Only
  1. `a = 3, b = 1, c = −3, d = −2`, or

     

    `a = −3, b = −3, c = 3, d = 2`

  2. `text(See worked solutions)`
  3. `y = −3/2x + 12`
Show Worked Solution

i.   `text(Method 1)`

`overset(->)(OA) = underset~a = ((3),(1)),\ \ overset(->)(OB) = underset~b = ((−3),(−3))`

`overset(->)(AB)` `= overset(->)(OB) – overset(->)(OA)`
  `= ((−3),(−3)) – ((3),(1))`
  `= ((−6),(−4))`

 

`underset~v` `= underset~a + lambdaunderset~b`
  `= ((3),(1)) + lambda((−6),(−4))`
  `= ((3),(1)) + 2((−3),(−2))`

 
`:. a = 3, b = 1, c = −3, d = −2`

 

`text(Method 2)`

`overset(->)(BA)` `= overset(->)(OA) – overset(->)(OB)`
  `= ((3),(1)) – ((−3),(−3))`
  `= ((6),(4))`

 
`underset~v = ((−3),(−3)) + 2((3),(2))`
  

`:. a = −3, b = −3, c = 3, d = 2`
 

ii.   `underset~u = ((4),(6)) + lambda((−2),(3))`

`underset~v = ((3),(1)) + 2((−3),(−2))`

`((−2),(3)) · ((−3),(−2)) = 6 – 6 = 0`

`:. underset~u ⊥ underset~v`
 

iii.   `((x),(y))= ((4),(6)) + lambda((−2),(3))`

`x = 4 – 2lambda\ \ \ …\ (1)`

`y = 6 + 3lambda\ \ \ …\ (2)`

`text(Substitute)\ \ lambda = (4 – x)/2\ \ text{from (1) into (2):}`

`y` `= 6 + 3((4 – x)/2)`
`y` `= 6 + 6 – (3x)/2`
`y` `= −3/2x + 12`

Vectors, EXT2 V1 SM-Bank 1

Consider the vectors  `underset~u = a underset~i - b underset~j + c underset~k`  and  `underset~v = underset~i - 8underset~j + 4underset~k`.

Find all possible values of  `a, b`  and  `c`  if  `underset~u`  is parallel to  `underset~v`  and  has a magnitude of 3.  (3 marks)

Show Answers Only

`1/3 , 8/3 , 4/3`

`text(or)`

` -1/3 , – 8/3 , – 4/3`

Show Worked Solution
`|underset~v|` `= sqrt(1 + 64 + 16) = 9`
`underset~overset^v` `= underset~v /|underset~v| =  (1)/(9) underset~i – (8)/(9) underset~j + (4)/(9) underset~k \ \ text{(magnitude of 1)}`

 
`text(S) text(ince) \ underset~u  \ text(has a magnitude of 3:)`

`underset~u` `= ± 3 ((1)/(9) underset~i – (8)/(9) underset~j + (4)/(9) underset~k)`
  `= ± (1/3 underset~i – 8/3 underset~j + 4/3 underset~k)`

 

`:. \ a, b, c` `= 1/3 , – 8/3 , 4/3\ \ text{or}\ \ -1/3 , 8/3 , – 4/3`

Vectors, EXT2 V1 SM-Bank 10

  1. Determine the point of intersection of  `underset ~a`  and  `underset ~b`  given.

`qquad underset ~a = ((3), (5), (1)) + lambda ((1), (3), (text{−2})),`  and
 

`qquad underset ~b = ((text{−2}), (2), (text{−1})) + mu ((1), (text{−1}), (2))`  (2 marks)

 
  1. Determine if the point  `(2, text{−2}, 5)`  lies on  `underset ~b`.  (1 mark)

Show Answers Only
  1. `((1), (text{−1}), (5))`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.     `text(At point of intersection:)`

`3 + lambda` `= -2 + mu\ \ text{… (1)}`
`5 + 3 lambda` `= 2 – mu\ \ text{… (2)}`
`1 – 2 lambda` `= -1 + 2 mu\ \ text{… (3)}`

 
`(1) + (2)`

`8 + 4 lambda` `= 0`
`lambda` `= -2,\ \ mu = 3`

 
`text{Intersection (using}\ lambda = –2 text{)}:`
 

`((x), (y), (z)) = ((3 – 2 xx 1), (5 – 2 xx 3), (1 – 2 xx text{−2})) = ((1), (text{−1}), (5))`

 

ii.   `text(If)\ \ (2, text{−2}, text{−10})\ \ text(lies on)\ underset ~b, ∃ mu\ \ text(that satisfies:)`

`-2 + mu` `= 2\ \ text{… (1)}\ => \ mu = 4`
`2 – mu` `= ­text{−2}\ \ text{… (2)}\ => \ mu = 4`
`-1 + 2 mu` `= 5\ \ text{… (3)}\ => \ mu = 3`

 
`=>\  text(No solution)`

`:. (2, text{−2}, 5)\ \ text(does not lie on)\ underset ~b.`

Vectors, EXT2 V1 SM-Bank 9

  1. Find the equation of line vector  `underset ~r`, given it passes through  `(1, 3, –2)`  and  `(2, –1, 2)`.   (2 marks)

  2. Determine if  `underset ~r`  passes through  `(4, –9, 10)`.   (1 mark)

Show Answers Only
  1. `underset ~r = ((1), (3), (-2)) + lambda ((1), (-4), (4)) or`

       
      

    `underset ~r = ((2), (-1), (2)) + lambda ((-1), (4), (-4))`

  2. `text(See Worked Solutions)`
Show Worked Solution

i.     `text(Method 1)`

`text(Let)\ \ A(1, 3, –2) and B(2, –1, 2)`

`vec (AB)` `= ((2), (-1), (2)) – ((1), (3), (-2)) = ((1), (-4), (4))`
`underset ~r` `= ((1), (3), (-2)) + lambda ((1), (-4), (4))`

 

`text (Method 2)`

`vec (BA)` `= ((1), (3), (-2)) – ((2), (-1), (2)) = ((-1), (4), (-4))`
`underset ~r` `= ((2), (-1), (2)) + lambda ((-1), (4), (-4))`

 

ii.   `text(If)\ \ (4, –9, 10)\ \ text(lies on the vector line,)`

`∃ lambda\ \ text(that satisfies:)`

`1 + lambda` `= 4\ \ text{… (1)}`
`3 – 4 lambda` `= -9\ \ text{… (2)}`
`-2 + 4 lambda` `= 10\ \ text{… (3)}`

 

`lambda = 3\ \ text(satisfies all equations)`

`:. (4, –9, 10)\ \ text(lies on the line.)`

Functions, EXT1′ F1 2008 HSC 3a

The following diagram shows the graph of  `y = g(x)`.
 

 
Draw separate one-third page sketches of the graphs of the following:

  1.  `y = |g(x)|`  (1 mark)

  2.  `y = 1/(g(x))`  (2 marks)

Show Answers Only
  1.  
  2.  
Show Worked Solution
i.   

 

ii.   

GRAPHS, FUR2 2019 VCAA 2

Each branch within the association pays an annual fee based on the number of members it has.

To encourage each branch to find new members, two new annual fee systems have been proposed.

Proposal 1 is shown in the graph below, where the proposed annual fee per member, in dollars, is displayed for branches with up to 25 members.
 


 

  1. What is the smallest number of members that a branch may have?  (1 mark)
  2. The incomplete inequality below shows the number of members required for an annual fee per member of $10.

      

    Complete the inequality by writing the appropriate symbol and number in the box provided.   (1 mark)
     

3 ≤ number of members  
 

 

Proposal 2 is modelled by the following equation.

annual fee per member = – 0.25 × number of members + 12.25

  1. Sketch this equation on the graph for Proposal 1, shown below.  (1 mark)

 

 

  1. Proposal 1 and Proposal 2 have the same annual fee per member for some values of the number of members.

      

    Write down all values of the number of members for which this is the case.  (1 mark)

Show Answers Only
  1. `3`
  2. `3 <= text(number of members) <= 10`
  3. `text(See Worked Solutions)`
  4. `9, 17, 25`
Show Worked Solution

a.  `3`
 

b.  `3 <= text(number of members) <= 10`
 

c.  

 

d.    `text(Same annual fee occurs when graphs intersect.)`

`text(Member numbers): 9, 17, 25`

GRAPHS, FUR2 2019 VCAA 1

The graph below shows the membership numbers of the Wombatong Rural Women’s Association each year for the years 2008–2018.
 


 

  1. How many members were there in 2009?  (1 mark)
    1. Show that the average rate of change of membership numbers from 2013 to 2018 was − 6 members per year.  (1 mark)
    2. If the change in membership numbers continues at this rate, how many members will there be in 2021?  (1 mark)
Show Answers Only
  1. `60`
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `14`
Show Worked Solution

a.  `60`

 

b.i.   `text(Members in 2013)` `= 62`
  `text(Members in 2018)` `= 32`
  `text(Average ROC)` `= (32 – 62)/5`
    `= -6\ text(members per year.)`

 

b.ii.   `text(Members in 2021)` `= 32 – 3 xx 6`
    `= 14`

GEOMETRY, FUR2 2019 VCAA 3

The following diagram shows a crane that is used to transfer shipping containers between the port and the cargo ship.
 


 

The length of the boom, `BC`, is 25 m. The length of the hoist, `AB`, is 15 m.

    1. Write a calculation to show that the distance `AC` is 20 m.  (1 mark)
    2. Find the angle `ACB`.

        

      Round your answer to the nearest degree.  (1 mark)

  1. The diagram below shows a cargo ship next to a port. The base of a crane is shown at point `Q`.
     

      

      

    `qquad`
     

      

    The base of the crane (`Q`) is 20 m from a shipping container at point `R`. The shipping container will be moved to point `P`, 38 m from `Q`. The crane rotates 120° as it moves the shipping container anticlockwise from `R` to `P`.

      

    What is the distance `RP`, in metres?

      

    Round your answer to the nearest metre.  (1 mark)

  2. A shipping container is a rectangular prism.

      

    Four chains connect the shipping container to a hoist at point `M`, as shown in the diagram below.
     

      

    `qquad` 
     

      

    The shipping container has a height of 2.6 m, a width of 2.4 m and a length of 6 m.

     

    Each chain on the hoist is 4.4 m in length.

      

    What is the vertical distance, in metres, between point `M` and the top of the shipping container?

      

    Round your answer to the nearest metre.  (2 marks)

Show Answers Only
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `37^@`
  1. `51\ text(m)`
  2. `3\ text(m)`
Show Worked Solution
a.i.   `AC` `= sqrt(BC^2 – AB^2)`
    `= sqrt(25^2 – 15^2)`
    `= sqrt 400`
    `= 20`

 

a.ii.   `tan\  /_ ACB` `= 15/20`
  `/_ ACB` `= tan^(-1) (3/4)`
    `= 36.86…`
    `~~ 37^@`

 

b.  `text(Using cosine rule:)`

`RP^2` `= 20^2 + 38^2 – 2 xx 20 xx 38 xx cos 120^@`
  `= 2604`
`:. RP` `= 51.029…`
  `~~ 51\ text(metres)`

 

c.  

`text(Find)\ h => text(need to find)\ x`

`text(Consider the top of the container)`
 


 

`x` `= sqrt(3^2 + 1.2^2)`
  `~~ 3.2311`

 

`:. h` `= sqrt(4.4^2 – 3.2311^2)`
  `= 2.98…`
  `~~ 3\ text(metres)`

GEOMETRY, FUR2 2019 VCAA 1

The following diagram shows a cargo ship viewed from above.
 

 
The shaded region illustrates the part of the deck on which shipping containers are stored.

  1. What is the area, in square metres, of the shaded region?  (1 mark)

Each shipping container is in the shape of a rectangular prism.

Each shipping container has a height of 2.6 m, a width of 2.4 m and a length of 6 m, as shown in the diagram below.
 

  1. What is the volume, in cubic metres, of one shipping container?  (1 mark)
  2. What is the total surface area, in square metres, of the outside of one shipping container?  (1 mark)
  3. One shipping container is used to carry barrels. Each barrel is in the shape of a cylinder.

      

    Each barrel is 1.25 m high and has a diameter of 0.73 m, as shown in the diagram below.

      

    Each barrel must remain upright in the shipping container
     
     

      

    `qquad qquad`
     
    What is the maximum number of barrels that can fit in one shipping container?  (1 mark)

Show Answers Only
  1. `6700\ text(m²)`
  2. `37.44\ text(m³)`
  3. `72.48\ text(m²)`
  4. `48`
Show Worked Solution
a.   `text(Area)` `= 160 xx 40 + 12 xx 25`
    `= 6700\ text(m²)`

 

b.   `text(Volume)` `= 6 xx 2.4 xx 2.6`
    `= 37.44\ text(m³)`

 

c.   `text(S.A.)` `= 2(6 xx 2.6) + 2 (6 xx 2.4) + 2(2.4 xx 2.6)`
    `= 72.48\ text(m²)`

 

d.   `6 ÷ 0.73 = 8.21` `=> \ text(8 barrels along length)`
  `2.4 ÷ 0.73 = 3.28` `=> \ text(3 barrels along width)`
  `2.6 ÷ 1.25 = 2.08` `=> \ text(2 vertical layers)`

 

`:.\ text(Maximum barrels)` `= 2 xx 8 xx 3`
  `= 48`

NETWORKS, FUR2 2019 VCAA 2

Fencedale High School offers students a choice of four sports, football, tennis, athletics and basketball.

The bipartite graph below illustrates the sports that each student can play.
  
 


  

Each student will be allocated to only one sport.

  1. Complete the table below by allocating the appropriate sport to each student.   (1 mark)

 

         Student                                 Sport                         
    Blake  
    Charli  
    Huan  
    Marco  

 

  1. The school medley relay team consists of four students, Anita, Imani, Jordan and Lola.

      

    The medley relay race is a combination of four different sprinting distances: 100 m, 200 m, 300 m and 400 m, run in that order.

      

    The following table shows the best time, in seconds, for each student for each sprinting distance.
     

      Best time for each sprinting distance (seconds)
         Student             100 m               200 m               300 m               400 m       
      Anita 13.3 29.6 61.8 87.1
      Imani 14.5 29.6 63.5 88.9
      Jordan 13.3 29.3 63.6 89.1
      Lola 15.2 29.2 61.6 87.9

      
     
    The school will allocate each student to one sprinting distance in order to minimise the total time taken to complete the race.

      

    To which distance should each student be allocated?

      

    Write your answers in the table below.  (2 marks)
     

           Student                                Sprinting distance (m)                         
      Anita  
      Imani  
      Jordan  
      Lola  
Show Answers Only
a.           Student                                 Sport                         
    Blake   Tennis
    Charli   Football
    Huan   Basketball
    Marco   Athletics

 

b.           Student                Sprinting distance (m)         
    Anita 400
    Imani 200
    Jordan 100
    Lola 300
Show Worked Solution

a.    `text(Charli must choose football)`

`=>\ text(Blake must choose tennis)`

`=>\ text(Huan must choose basketball etc…)`
 

         Student                                 Sport                         
    Blake   Tennis
    Charli   Football
    Huan   Basketball
    Marco   Athletics

 

b.    `text(Using the Hungarian Algorithm:)`

`text(After row and column reduction,)`
 

     Student        100 m         200 m         300 m         400 m    
    Anita 0 2.3 2.1 1.1
    Imani 0 1.1 2.6 1.7
    Jordan 0 2 3.9 3.1
    Lola 0 0 0 0

 

`text(Final table values:)`
 

     Student        100 m         200 m         300 m         400 m    
    Anita 0 1.2 1 0
    Imani 0 0 1.5 0.6
    Jordan 0 0.9 2.8 2
    Lola 1.1 0 0 0

 

         Student                Sprinting distance (m)         
    Anita 400
    Imani 200
    Jordan 100
    Lola 300

NETWORKS, FUR2 2019 VCAA 1

Fencedale High School has six buildings. The network below shows these buildings represented by vertices. The edges of the network represent the paths between the buildings.
 


 

  1. Which building in the school can be reached directly from all other buildings?  (1 mark) 
  2. A school tour is to start and finish at the office, visiting each building only once

     

    1. What is the mathematical term for this route?  (1 mark)
    2. Draw in a possible route for this school tour on the diagram below.  (1 mark)

 

Show Answers Only
  1. `text(Office)`
    1. `text(Hamiltonian cycle)`
    2. `text(See Worked Solutions)`
Show Worked Solution

a.  `text(Office)`
 

b.i.   `text(Hamiltonian cycle)`

`text{(note a Hamiltonian path does not start and finish}`

  `text{at the same vertex.)}`
 

b.ii.   `text(One of many solutions:)`
 

MATRICES, FUR2 2019 VCAA 2

The theme park has four locations, Air World `(A)`, Food World `(F)`, Ground World `(G)` and Water World `(W)`.

The number of visitors at each of the four locations is counted every hour.

By 10 am on Saturday the park had reached its capacity of 2000 visitors and could take no more visitors.

The park stayed at capacity until the end of the day

The state matrix, `S_0`, below, shows the number of visitors at each location at 10 am on Saturday.
 

`S_0 = [(600), (600), (400), (400)] {:(A),(F),(G),(W):}`
 

  1. What percentage of the park’s visitors were at Water World `(W)` at 10 am on Saturday?  (1 mark)

Let `S_n` be the state matrix that shows the number of visitors expected at each location `n` hours after 10 am on Saturday.

The number of visitors expected at each location `n` hours after 10 am on Saturday can be determined by the matrix recurrence relation below.
 

`{:(qquad qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad text(  this hour)),(qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad qquad qquad quad A qquad quad F qquad \  G \ quad quad W),({:S_0 = [(600), (600), (400), (400)], qquad S_(n+1) = T xx S_n quad quad qquad text(where):}\ T = [(0.1,0.2,0.1,0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]{:(A),(F),(G),(W):}\ text(next hour)):}`
 
 

  1. Complete the state matrix, `S_1`, below to show the number of visitors expected at each location at 11 am on Saturday.  (1 mark)

 
`S_1 = [(\ text{______}\ ), (\ text{______}\ ), (300),(\ text{______}\ )]{:(A),(F),(G),(W):}`
 

  1. Of the 300 visitors expected at Ground World `(G)` at 11 am, what percentage was at either Air World `(A)` or Food World `(F)` at 10 am?  (1 mark)
  2. The proportion of visitors moving from one location to another each hour on Sunday is different from Saturday.

     

    Matrix `V`, below, shows the proportion of visitors moving from one location to another each hour after 10 am on Sunday.

     

    `qquad qquad {:(qquadqquadqquadqquadqquadtext(this hour)),(qquad qquad qquad \ A qquad quad F qquad \  G \ quad quad W),(V = [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]{:(A),(F),(G),(W):}\ text(next hour)):}`

     

     
    Matrix `V` is similar to matrix `T` but has the first two rows of matrix `T` interchanged.

     

    The matrix product that will generate matrix `V` from matrix `T` is

     


    `qquad qquad V = M xx T`

     


    where matrix `M` is a binary matrix.

     

    Write down matrix `M`.  (1 mark)

 
`qquad qquad qquad M = [( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , )]`

Show Answers Only
  1. `20%`
  2. `S_1 = [(300 ), (780), (300),(620)]{:(A),(F),(G),(W):}`
  3. `60%`
  4. `M = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)]`
Show Worked Solution

a.   `text(Total visitors)\ =2000`

`text{Percentage}\  (W)` `= 400/2000`
  `= 20%`

 

b.   `S_1` `= TS_0`
    `= [(0.1, 0.2, 0.1, 0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)][(600),(600),(400),(400)]=[(300),(780),(300),(620)]`

 

c.   `text(Visitors from)\ A to G` `= 0.1 xx 600 = 60`
  `text(Visitors from)\ F to G` `= 0.2 xx 600 = 120`

 

  `:.\ text(Percentage of)\ G\ text(at 11 am)` `= (60 + 120)/300`
    `= 60%`

 

d.   `M = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)]`
 

 
`text(Check:)`

`M xx T = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)][(0.1, 0.2, 0.1, 0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)] = [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]`

Calculus, EXT2 C1 2003 HSC 1b

Use integration by parts to find   `int x^3 log_e x  dx`   (3 marks)

Show Answers Only

`(x^4 log_e x)/(4)-(x^4)/(16) + C`

Show Worked Solution
`u` `= log_e x,` `\ \ \ \ u^{′}` `= (1)/(x)`
`v^{′}` `= x^3,` `v` `= (x^4)/(4)`

 

`int x^3 log_e x  dx` `= uv-int u^{′}v \ dx`
  `= (x^4)/(4)  log_e x-int (1)/(x) ⋅ (x^4)/(4)  dx`
  `= (x^4 log_e x)/(4)-(1)/(4) int x^3 dx`
  `= (x^4 log_e x)/(4)-(x^4)/(16) + C`

Calculus, EXT2 C1 2004 HSC 1a

Use integration by parts to find  `int x e^(3x)  dx`.   (2 marks)

Show Answers Only

`(x e^(3x))/3-(1)/(9)  e^(3x) + C`

Show Worked Solution
`u` `= x` `\ \ \ \ u^{′}` `= 1`
`v^{′}` `= e^(3x)` `v` `= (1)/(3)  e^(3x)`

 

`int x e^(3x)  dx` `= uv-int u^{′} v \ dx`
  `= (xe^(3x))/(3)-(1)/(3) int e ^(3x)  dx`
  `= (x e^(3x))/3-(1)/(9)  e^(3x) + C`