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CORE, FUR2-NHT 2019 VCAA 4

The scatterplot below plots the variable life span, in years, against the variable sleep time, in hours, for a sample of 19 types of mammals.
 


 

On the assumption that the association between sleep time and life span is linear, a least squares line is fitted to this data with sleep time as the explanatory variable.

The equation of this least squares line is

life span = 42.1 – 1.90 × sleep time

The coefficient of determination is 0.416

  1. Draw the graph of the least squares line on the scatterplot above.  (1 mark)
  2. Describe the linear association between life span and sleep time in terms of strength and direction.  (2 marks)
  3. Interpret the slope of the least squares line in terms of life span and sleep time.  (2 marks)
  4. Interpret the coefficient of determination in terms of life span and sleep time.  (1 mark)
  5. The life of the mammal with a sleep time of 12 hours is 39.2 years.

     

    Show that, when the least squares line is used to predict the life span of this mammal, the residual is 19.9 years.  (2 marks)

Show Answers Only
  1.  

  2. `text(Strength is moderate.)`

     

    `text(Direction is negative.)`

  3. `text(The gradient of –1.9 means that life span decreases by)` 

     

    `text(1.9 years for each additional hour of sleep time.)`

  4. `text(41.6% of the variation in life span can be explained by the)`

     

    `text(variation in sleep time.)`

  5. `text(Proof(See Worked Solution))`
Show Worked Solution

a.    `text{Graph endpoints (0, 42.1) and (18, 7.9)}`
 


 

b.   `text(Strength is moderate.)`

`text(Direction is negative.)`
 

c.    `text(The gradient of –1.9 means that life span decreases by)`

`text(1.9 years for each additional hour of sleep time.)`

 

d.    `text(41.6% of the variation in life span can be explained by the )`

`text(variation in sleep time.)`
 

e.    `text(Predicted value)` `= 42.1 – 1.9 xx 12`
  `= 19.3 \ text(years)`

 

`text(Residual)` `= text(actual) – text(predicted)`
  `= 39.2 – 19.3`
  `= 19.9 \ text(years)`

CORE, FUR2-NHT 2019 VCAA 3

The life span, in years, and gestation period, in days, for 19 types of mammals are displayed in the table below.
 

 

  1. A least squares line that enables life span to be predicted from gestation period is fitted to this data.  (1 mark)

     

    Name the explanatory variable in the equation of this least squares line.

  2. Determine the equation of the least squares line in terms of the variables life span and gestation period.

     

    Write your answers in the appropriate boxes provided below.

     

    Round the numbers representing the intercept and slope to three significant figures.  (2 marks)
      
        =    +    ×   
     

  3. Write the value of the correlation rounded to three decimal places.   (1 mark)
     
         `r =`  
Show Answers Only
  1. `text(gestation period)`
  2. `text(life span) = 7.58 + 0.101 xx \ text(gestation period)`
  3. `0.904`
Show Worked Solution

a.    `text(gestation period)`
 

b.    `text(Input data points into CAS:)`

`text(life span) = 7.58 + 0.101 xx \ text(gestation period)`
 

c.    `r = 0.904 \ text{(by CAS)}`

CORE, FUR2-NHT 2019 VCAA 1

The table below displays the average sleep time, in hours, for a sample of 19 types of mammals.
 


 

  1. Which of the two variables, type of mammal or average sleep time, is a nominal variable?  (1 mark)
  2. Determine the mean and standard deviation of the variable average sleep time for this sample of mammals.

     

    Write your answers in the boxes provided below.

     

    Round your answer to one decimal place.  (1 mark)
     

         mean = hours                   standard deviation = hours
     

  3. The average sleep time for a human is eight hours.

     

    What percentage of this sample of mammals has an average sleep time that is less than the average sleep time for a human.

     

    Round your answer to one decimal place.  (1 mark)

  4. The sample is increase in size by adding in the average sleep time of the little brown bat.

     

    Its average sleep time is 19.9 hours.

     

    By how many many hours will the range for average sleep time increase when the average sleep time for the little brown bat is added to the sample?  (1 mark)

Show Answers Only
  1. `text(type of mammal)`
  2. `text(mean)= 9.2 \ text(hours)`

     

    `sigma = 4.2 \ text(hours)`

  3. `31.6text(%)`
  4. `5.4 \ text(hours)`
Show Worked Solution

a.    `text(type of mammal is nominal)`

 
b.    `text(mean)= 9.2 \ text(hours) \ \ text{(by CAS)}`

`sigma = 4.2 \ text(hours) \ \ text{(by CAS)}`
 

c.    `text(Percentage)` `= (6)/(19) xx 100`
  `= 0.3157 …`
  `= 31.6text(%)`

 

d.    `text(Range increase)` `= 19.9 – 14.5`
  `= 5.4 \ text(hours)`

Statistics, EXT1 S1 EQ-Bank 10

Four cards are placed face down on a table. The cards are made up of a Jack, Queen, King and Ace.

A gambler bets that she will choose the Queen in a random pick of one of the cards.

If this process is repeated 7 times, express the gambler's success as a Bernoulli random variable and calculate

  1. the mean.  (1 mark)

  2. the variance.  (1 mark)

Show Answers Only
  1. `7/4`
  2. `21/16`
Show Worked Solution

i.     `text(Let)\ \ X = text(number of Queens chosen)`

`X\ ~\ text(Bin) (7,1/4)`

`E(X)` `=np`
  `= 7 xx 1/4`
  `=7/4`

 

ii.   `text(Var)(X)` `= np(1-p)`
    `=7/4(1-1/4)`
    `= 21/16`

Statistics, EXT1 S1 EQ-Bank 1 MC

If `X` equals the number of successes in `n` independent Bernoulli trials, how many distinct values can `X` take?

  1. `\ n-1`
  2. `\ n(n-1)`
  3. `\ n`
  4. `\ n+1`
Show Answers Only

`D`

Show Worked Solution

`text(Distinct values of)\ X\ text(in:)`

`text{1 trial = 2}\ \ (X=0 or 1)`

`text{2 trials = 3}\ \ (X=0, 1 or 2)`

 `vdots`

`n\ text{trials =}\ n+1\ \ (X=0, 1, …, n)`

`=>  D`

GRAPHS, FUR1-NHT 2019 VCAA 2 MC

Two straight lines have the equations  `3x - 2y = 3`  and  `-2x + 5y = 9`.

These lines have one point of intersection.

Another line that also passes through this point of intersection has the equation

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

`B`

Show Worked Solution
`3x – 2y` `= 3\ text{… (1)}`
`-2x + 5y` `= 9\ text{… (2)}`

 
`text{Mult}\ \ (1) xx 2`

`6x – 4y = 6\ text{… (3)}`

`text{Mult}\ \ (2) xx 3`

`-6x + 15y = 27\ text{… (4)}`

`(3) + (4)`

`11y = 33 \ => \ y = 3`

`text(Substitute)\ \ y = 3\ \ text{into (1)}`

`3x – 6 = 3 \ => \ x = 3`

`=>  B`

MATRICES, FUR1-NHT 2019 VCAA 3 MC

In matrix `A`, the element  `a_21 = 7`.

The order of matrix `A` is  `3 xx 2`.

Matrix `B` is the transpose of matrix `A`.

Matrix `B` could be

A. `\ [(3,9,4),(2,7,1)]` B. `\ [(4,7,2),(1,0,5)]` C. `\ [(4,1,3),(7,8,0)]`
D. `\ [(5,6),(7,1),(4,2)]` E. `\ [(9,7),(3,4),(0,2)]`    
Show Answers Only

`B`

Show Worked Solution

`A = (3 xx 2) \ => \ a_21 = 7`

`text(S)text(ince)\ \ B = A^T:`

`B = (2 xx 3) \ => \ b_12=7`

`=>\ B`

MATRICES, FUR1-NHT 2019 VCAA 1 MC

The number of individual points scored by Rhianna (`R`), Suzy (`S`), Tina (`T`), Ursula (`U`) and Vicki (`V`) in five basketball matches `(F, G, H, I, J)` is shown in matrix `P` below.
 

`{:(),(),(P=):}{:(qquadqquadqquad\ text(match)),((quadF,G,H,I,J)),([(2,\ 0,\ 3,\ 1,\ 8),(4,7,2,5,3),(6,4,0,0,5),(1,6,1,4,5),(0,5,3,2,0)]):}{:(),(),({:(R),(S),(T),(U),(V):}):}{:(),(),(text(player)):}`
 

Who scored the highest number of points and in which match?

  1. Suzy in match  `I`
  2. Tina in match  `H`
  3. Vicki in match  `F`
  4. Ursula in match  `G`
  5. Rhianna in match  `J`
Show Answers Only

`E`

Show Worked Solution

`text(Highest points = 8 =)\ e_15`

`e_15 \ => \ text(Rhianna in match)\ J`

`=>\ E`

NETWORKS, FUR1-NHT 2019 VCAA 2 MC

Consider the graph below.
 

 
Euler’s formula will be verified for this graph.

What values of  `e, v`  and  `f`  will be used in this verification?

  1. `e = 5, v = 5, f = 2`
  2. `e = 5, v = 5, f = 3`
  3. `e = 6, v = 5, f = 2`
  4. `e = 6, v = 5, f = 3`
  5. `e = 6, v = 6, f = 3`
Show Answers Only

`D`

Show Worked Solution

`text(Euler):\ \ v + f = e + 2`

`v` `= 5`
`e` `= 6`
`:. f` `= 3`

 
`=>  D`

Vectors, EXT1 V1 EQ-Bank 5 MC

What is the angle between the vectors  `((2),(1))`  and  `((-4),(2))`?

A.   `cos^(-1)(0.06)`

B.   `cos^(-1)(–0.06)`

C.   `cos^(-1)(0.6)`

D.   `cos^(-1)(–0.6)`

Show Answers Only

`D`

Show Worked Solution
`cos theta` `=(underset~a * underset~b)/(|underset~a||underset~b|)`  
  `=(-8+2)/(sqrt(2^2+1^2) xx sqrt((-4)^2+2^2)`  
  `=(-6)/(sqrt5 sqrt20)`  
  `=-0.6`  
`:. theta` `= cos^(-1) (-0.6)`  

  
`=> D`

Graphs, MET2-NHT 2019 VCAA 1

Parts of the graphs of  `f(x) = (x - 1)^3(x + 2)^3`  and  `g(x) = (x - 1)^2(x + 2)^3`  are shown on the axes below.
 


 

The two graphs intersect at three points,  (–2, 0),  (1, 0)  and  (`c`, `d`). The point  (`c`, `d`)  is not shown in the diagram above.

  1. Find the values of `c` and `d`.   (2 marks)
  2. Find the values of `x` such that  `f(x) > g(x)`.   (1 mark)
  3. State the values of `x` for which

     

    i.   `f' (x) > 0`   (1 mark)

     

    ii.  `g' (x) > 0`   (1 mark)

  4. Show that  `f(1 + m) = f(–2 - m)`  for all  `m`.   (1 mark)
  5. Find the values of `h` such that  `g(x + h) = 0`  has exactly one negative solution.   (2 marks)
  6. Find the values of `k` such that  `f(x) + k = 0`  has no solutions.   (1 mark)
Show Answers Only
  1. `c = 2 \ , \ d = 64`
  2. `(–∞, –2) \ ∪ \ (2, ∞)`
  3. i.  `(–(1)/(2), 1) \ ∪ \ (1, ∞)`
    ii. `(–∞, –2) \ ∪ \ (–2, –(1)/(5)) \ ∪ \ (1, ∞)`
  4. `text{Proof (Show Worked Solution)}`
  5.  `-2< h <=1`
  6. `(729)/(64)`
Show Worked Solution

a.    `text(Solve:) \ \ f(x) = g(x)`

`x = 1 , \ –2 \ text(and) \ 2`
 
`f(2) = 1^3 xx 4^3 = 64`
 
`text(Intersection at) \ (2, 64)`

`:. \ c = 2 \ , \ d = 64`

 

b.    `text(Using the graph and intersection at) \ (2, 64):`

`f(x) > g(x) \ \ text(for) \ \ (–∞, –2) \ ∪ \ (2, ∞)`

 

c.i.  `f'(x) > 0 \ \ text(for) \ \ (–(1)/(2), 1) \ ∪ \ (1, ∞)`

c.ii.  `g'(x) > 0 \ \ text(for) \ \ (–∞, –2) \ ∪ \ (–2, –(1)/(5)) \ ∪ \ (1, ∞)`

 

d.     `f(1 + m)` `= (1 + m – 1)^3 (1 + m + 2)^3`
  `= m^3 (m + 3)^3`

 

`f(–2 – m)` `= (–2 – m -1)^3 (-2 – m + 2)^3`
  `= (-m – 3)^3 (-m)^3`
  `= (–1)^3 (m + 3)^3 (–1)^3 m^3`
  `= m^3 (m + 3)^3`

 

e.     `g(x + h)` `= (x + h – 1)^2(x + h + 2)^3`
  `= underbrace{(x – (1 -h))^2}_{text(+ve solution)} * underbrace{(x – (h – 2))^3}_{text(–ve solution)}`

 
`1 – h ≥ 0 \ \ => \ \ h ≤ 1`

`-h – 2 < 0 \ \ =>\ \ h > -2`

`:. -2< h <=1`

 

f.    `f(x) \ \ text(minimum S.P. when) \ \ f ′(x) = 0 \ =>  \ x =-(1)/(2)`

`text(S.P. at) \ \ (-(1)/(2) \ , \ -(729)/(64))`

`:. \ text(No solution if) \ \ k > (729)/(64)`

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 E1 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.)`