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Data Analysis, GEN1 2019 NHT 16 MC

A small cafe is open every day of the week except Tuesday.

The table below shows the daily seasonal indices for labour costs at this cafe.

Excluding the day that the cafe is closed, the long-term average weekly labour cost is $9786.

The expected daily labour cost on a Wednesday is closest to

  1.  $1127
  2.  $1315
  3.  $1631
  4.  $1733
  5.  $2022
Show Answers Only

`E`

Show Worked Solution

`text(Average daily labour cost)`

`= 9786/6`

`= 1631`

 

`:.\ text(Wednesday’s expected cost)`

`= 1631 xx 1.24`

`= $2022.44`
 

`=>\ E`

Data Analysis, GEN1 2019 NHT 14 MC

The time series plot below shows the daily number of visitors to a historical site over a two-week period.
 

This time series plot is to be smoothed using seven-median smoothing.

The smoothed number of visitors on day 4 is closest to

  1. 120
  2. 140
  3. 145
  4. 150
  5. 160
Show Answers Only

`D`

Show Worked Solution

`text{The seven data points about day 4 (ascending):}`

`120, 140, 147, 150, 160, 170, 183`

`:.\ text{7 median smoothing (Day 4)} = 150`

`=> D`

Data Analysis, GEN1 2019 NHT 10 MC

A student uses the data in the table below to construct the scatterplot shown below.
 


 

A squared transformation is applied to the variable `x` to linearise the data.

A least squares line is fitted to this linearised data with `x^2` as the explanatory variable.

The equation of this least squares line is closest to

  1. `y = 94.1 - 12.3x^2`
  2. `y = 95.8 - 1.57x^2`
  3. `y = 8.76 - 0.0768x^2`
  4. `y = 107 - 111x^2`
  5. `y = 107 - 0.0768x^2`
Show Answers Only

`B`

Show Worked Solution

`text(Input all data points into CAS:)`

`y = 95.8 – 1.57x^2`

`=> B`

Data Analysis, GEN1 2019 NHT 9 MC

A least squares line of the form  `y = a + bx`  is fitted to a set of bivariate data for the variables `x` and `y`.

For this set of bivariate data,  `barx = 5.50`,  `bary = 5.60`,  `s_x = 3.03`,  `s_y =1.78`  and  `a = 3.1`

The slope of the least squares line, `b`, is closest to

  1.  0.44
  2.  0.45
  3.  0.58
  4.  0.59
  5.  0.76
Show Answers Only

`B`

Show Worked Solution
`bar y` `=a+b barx`
`5.60` `= 3.1 – b(5.50)`
`b` `= (5.60 – 3.1)/5.50`
  `= 04545…`

 
`=> B`

Data Analysis, GEN1 2019 NHT 8 MC

The variables recovery time after exercise (in minutes) and fitness level (below average, average, above average) are

  1. both numerical.
  2. both categorical.
  3. an ordinal variable and a nominal variable respectively.
  4. a numerical variable and a nominal variable respectively.
  5. a numerical variable and an ordinal variable respectively.
Show Answers Only

`E`

Show Worked Solution

`text{Recovery time in minutes → numerical variable}`

`text{Fitness level → ordinal (categories that can be ordered)}`

`=> E`

Data Analysis, GEN1 2019 NHT 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`

Algebra, MET2-NHT 2019 VCAA 14 MC

If  `2log_e(x) - log_e(x + 2) = log_e(y)`, then  `x`  is equal to

  1.  `(y + sqrt{y^2 + 8y})/(2)`
  2.  `(y ± sqrt{y^2 + 8y})/(2)`
  3.  `(y ± sqrt{y^2 - 8y})/(2)`
  4.  `(-1 ± sqrt{4y - 7})/(2)`
  5.  `(-1 + sqrt{4y - 7})/(2)`
Show Answers Only

`A`

Show Worked Solution
`2log_e x – log_e(x + 2)` `= log_e y`
`log_e ({x^2}/{x + 2})`  `= log_e y`

 
`y = (x^2)/(x + 2)`
 
`text(S) text(olve for) \ x :`

`x = (y + sqrt(y^2 + 8y))/(2) \ \ text(only,) \ \ ((y – sqrt{y^2 + 8y})/(2) < 0) `
 
`=> \ A`

Calculus, MET2-NHT 2019 VCAA 13 MC

The graph of  `f(x) = x^3 - 6(b - 2)x^2 + 18x + 6`  has exactly two stationary points for

  1.  `1 < b < 2`
  2.  `b = 1`
  3.  `b = (4 ± sqrt6)/(2)`
  4.  `(4 - sqrt6)/(2) ≤  b ≤ (4 + sqrt6)/(2)`
  5.  `b < (4 - sqrt6)/(2) \ or \ b > (4 + sqrt6)/(2)`
Show Answers Only

`E`

Show Worked Solution
`f(x)` `= x^3 – 6(b – 2) x^2 + 18x + 6`
`f′(x)` `= 3x^3 – 12(b – 2) x + 18`
`Δ` `= [-12(b – 2)]^2 – 4 . 3 . 18`
  `= 144(b – 2)^2 – 216`

 
`text(If 2 solutions) \ => \ Δ > 0`

`text(S) text(olve for) \ b \ text{(by CAS):}`
 
`b < (4 – sqrt6)/(2) \ or \ b > (4 + sqrt6)/(2)`
 
`=> E`

Probability, MET2-NHT 2019 VCAA 10 MC

Let  `f`  be the probability density function  `f : [ 0, (2)/(3)] → R, \ f(x) = kx(2x + 1)(3x -2)(3x + 2)`.
The value of  `k`  is

  1.     `(308)/(405)`
  2.  `–(308)/(405)`
  3.  `–(405)/(308)`
  4.     `(405)/(308)`
  5.     `(960)/(133)`
Show Answers Only

`C`

Show Worked Solution

`text(Solve for)\ k :`

`k int_0^((2)/(3)) x (2x + 1)(3x – 2)(3x + 2)\ dx = 1`

`k = -(405)/(308)`
 

`=> \ C`

Probability, MET2-NHT 2019 VCAA 9 MC

At the start of a particular week, Kim has three red apples and two green apples. She eats one apple everyday. On Monday, Tuesday and Wednesday of that week, she randomly selects an apple to eat. In this three-day period, the probability that Kim does not eat an apple of the same colour on any two consecutive days is

  1.  `(1)/(10)`
  2.  `(1)/(5)`
  3.  `(3)/(10)`
  4.  `(2)/(5)`
  5.  `(6)/(25)`
Show Answers Only

`C`

Show Worked Solution

`text{Pr (alternate colours)}`

`= text(Pr)(RGR) + text(Pr)(GRG)`

`= (3)/(5) ·(2)/(4) ·(2)/(3) + (2)/(5) ·(3)/(4) ·(1)/(3)`

`= (12)/(60) + (6)/(60)`

`= (3)/(10)`
 

`=> \ C`

Algebra, MET2-NHT 2019 VCAA 8 MC

The simultaneous linear equations  `2y + (m - 1) x = 2`  and  `my + 3x = k`  have infinitely many solutions for

  1.  `m = 3`  and  `k = –2`
  2.  `m = 3`  and  `k = 2`
  3.  `m = 3`  and  `k = 4`
  4.  `m = –2`  and  `k = –2`
  5.  `m = –2`  and  `k = 3`
Show Answers Only

`D`

Show Worked Solution
`2y + (m – 1)x` `= 2\ \ =>\ \ y= -((m-1)/(2)) x + 1`
`my + 3x` `= k\ \ =>\ \ y=-(3)/(m) x + (k)/(m)`

 

`text(Infinite solutions) \ => \ text(gradients and y-intercepts equal)`

`(m – 1)/(2)` `= (3)/(m)`
`m^2 – m – 6` `= 0`
`(m – 3)(m + 2)` `= 0`

 
`m = 3 \ \ text(or)\ \ –2`
 

`text(If) \ \ m = 3 ,`

`(k)/(3) = 1 \ => \ k = 3`
 
`text(If) \ \ m = –2,`

`(k)/(–2) = 1 \ => \ k = –2`
 

`=> \ D`

Algebra, MET2-NHT 2019 VCAA 6 MC

Let  `f : [0, ∞) → R, \ f(x) = x^2 + 1`.

The equation  `f(f(x)) = (185)/(16)`  has real solution(s)

  1.  `x = ± (sqrt13)/(4)`
  2.  `x = (sqrt13)/(4)`
  3.  `x = ± (sqrt13)/(2)`
  4.  `x = (3)/(2)`
  5.  `x = ± (3)/(2)`
Show Answers Only

`D`

Show Worked Solution
`f(f(x))` `= (x^2 + 1)^2 + 1`
`(185)/(16)` `= x^4 + 2x^2 + 2`
`0` `= x^4 + 2x^2 – (153)/(16)`

 
`text{Solve (by CAS):}`

`x = (3)/(2) , \ x ∈ [0, ∞)`

Probability, MET2-NHT 2019 VCAA 5 MC

Consider the probability distribution for the discrete random variable `X` shown in the table below.

The value of  `text(E)(X)`  is

  1.  `(76)/(65)`
  2.  `1`
  3.  `0`
  4.  `(2)/(13)`
  5.  `(86)/(65)`
Show Answers Only

`A`

Show Worked Solution
`1` `= b + b + b + (3)/(5) – b + (3b)/(5)`
`(2)/(5)` `= (13b)/(5)`
`b` `= (2)/(13)`

 

`text(E)(X)` `= -(2)/(13) + 0 + (2)/(13) + 2((3)/(5) – (2)/(13)) +3 ((6)/(65))`
  `= (76)/(65)`

Graphs, MET2-NHT 2019 VCAA 4 MC

The graph of the function  `ƒ : D → R, \ f(x) = (2x -3)/(4 + x)`, where `D` is the maximal domain, has asymptotes

  1.  `x = –4, \ y = 2`
  2.  `x = (3)/(2), \ y = –4`
  3.  `x = –4, \ y = (3)/(2)`
  4.  `x = (3)/(2), \ y = 2`
  5.  `x = 2, \ y = 1`
Show Answers Only

`A`

Show Worked Solution
`f(x)` `= (2x + 8 – 11)/(x +4)`
  `= 2 – (11)/(x + 4)`

 
`:. \ text(Asymptotes at) \ \ x = –4, y = 2 `

Combinatorics, EXT1 A1 2019 MET1 8

A fair standard die is rolled 50 times. Let `W` be a random variable with binomial distribution that represents the number of times the face with a six on it appears uppermost.

  1. Write down the expression for  `P(W = k)`, where  `k in {0, 1, 2, …, 50}`.  (1 mark)

  2. Show that  `(P(W = k + 1))/(P(W = k)) = (50 - k)/(5(k + 1))`.  (2 marks)

Show Answers Only
  1. `\ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50 – k)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

a.  `P(W = k) = \ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50 – k)`

 

b.   `(P(W = k + 1))/(P(W = k))` `= (\ ^50C_(k+1) ⋅ (1/6)^(k+1) ⋅ (5/6)^(49 – k))/(\ ^ 50C_k ⋅ (1/6)^k ⋅ (5/6)^(50-k))`
    `= ((50!)/((49 – k)!(k + 1)!) ⋅ (1/6))/((50!)/((50 – k)! k!) ⋅ (5/6))`
    `= ((50 – k)!k!)/(5(49 – k)!(k + 1)!)`
    `= (50 – k)/(5(k + 1))`

Calculus, 2ADV C4 2019 MET1 7

The shaded region in the diagram below is bounded by the vertical axis, the graph of the function with rule  `f(x) = sin(pix)`  and the horizontal line segment that meets the graph at  `x = a`, where  `1 <= a <= 3/2`.
 


 

Let  `A(a)`  be the area of the shaded region.

Show that  `A(a) = 1/pi-1/pi cos(a pi)-a sin (a pi)`.  (3 marks)

Show Answers Only

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

Show Worked Solution

`text{Strategy 1}`

  `text(Consider the areas above:)`

`int_0^a \sin (pi x)` `\ = text(Area 1 – Area 3)`
  `=[-{1}/{pi} cos (pi x)]_0^a`
  `=-{1}/{pi} cos (a pi)-(-{1}/{pi})`
  `=1/pi-1/picos(a pi)`

 
`text(Area 2 + Area 3 (rectangle) )`

`=-sin(a pi) \times a`

`=-a sin (a pi) \ text{(Area must be +ve)}`
  

`\therefore \ text(Shaded area)` `\ =text(Area 1 + Area 2)`
  `=1/pi-1/pi cos(a pi)-a sin(a pi)`

   

`text{Strategy 2}`

`text(Lower border of shaded area:)\ y = f(a) = sin(a pi)`

`text(Area)` `= int_0^a sin(pi x)-sin (a pi)\ dx`
  `= [-1/pi cos (pi x)-x sin(a pi)]_0^a`
  `= [-1/pi cos(pi a)-a sin (a pi)-(-1/pi-0)]`
  `= 1/pi-1/pi cos (a pi)-a sin(a pi)`

Statistics, EXT1 S1 2019 MET1 6

Jacinta tosses a coin five times.

  1. Assuming that the coin is fair and given that Jacinta observes a head on the first two tosses, find the probability that she observes a total of either four or five heads.  (2 marks)

  2. Albin suspects that a coin is not actually a fair coin and he tosses it 18 times.

     

    Albin observes a total of 12 heads from the 18 tosses.

  3. Let  `X` = probability of obtaining a head.
  4. Find the range of `X` in which 95% of observations are expected to lie within.  (2 marks)

Show Answers Only
  1.  `1/2`
  2.  `(4/9, 8/9)`
Show Worked Solution

a.   `text(After 2 tosses, 2 heads.)`

`Ptext{(4 or 5}\ H)` `= HHT + HTH + THH + HHH`
  `= (1/2)^3 xx 4`
  `= 1/2`

 

b.    `E(hat p) = p=12/18 = 2/3`

`text(Var)(hatp) = (2/3(1-2/3))/18=1/81`

`sigma(hatp)=1/9`

`text(95% interval)\  (z= +-2)`

`=(hatp-z xx sigma(hatp), \ hatp+z xx sigma(hatp))`

`= (2/3 – 2 xx 1/9, 2/3 + 2 xx 1/9)`

`= (4/9, 8/9)`

Functions, EXT1 F1 2019 MET1 5b

Given the function  `h(x) = sqrt(2x + 3)-2`  for  `h>=-3/2`, find the inverse function  `h^(-1)`  and its domain.   (3 marks)

Show Answers Only

`x>=–2  or  [–2, oo)`

Show Worked Solution

 `y = sqrt (2x + 3)-2`

`text(Inverse: swap)\ \ x ↔ y`

`x` `= sqrt(2y + 3)-2`
`sqrt(2y + 3)` `= x + 2`
`2y + 3` `= (x + 2)^2`
`y` `= 1/2(x + 2)^2-3/2`
`:. h^(-1)` `= 1/2(x + 2)^2-3/2`

 

`text(Domain)\ \ h^(-1)(x)` `= text(Range)\ \ h(x)`
  `= x>=–2  or [–2, oo)`

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 C4 2019 MET1 2

Find  `f(x)`  given that  `f(1) = -7/4`  and  `f ^{\prime}(x) = 2x^2 - 1/4x^(-2/3)`.  (2 marks)

Show Answers Only

`f(x) = 2/3x^3 – 3/4x^(1/3) – 5/3`

Show Worked Solution
`f(x)` `= int 2x^2 – 1/4x^(-2/3) dx`
  `= 2/3x^3 – 1/4 ⋅ 1/(1/3) x^(1/3) + c`
  `= 2/3x^3 – 3/4x^(1/3) + c`

 
`text(Given)\ \ f(1) = -7/4:`

`-7/4` `= 2/3 – 3/4 + c`
`c` `= -5/3`
`:. f(x)` `= 2/3x^3 – 3/4x^(1/3) – 5/3`

Calculus, MET1-NHT 2019 VCAA 1b

Let  `f(x) = x^2 cos(3x)`.
 
Find  `f ^{\prime} (pi/3)`.   (2 marks)

Show Answers Only

`-(2pi)/3`

Show Worked Solution
  `f(x)` `= x^2 cos 3x`
  `f^{\prime}(x)` `= x^2 ⋅ 3(-sin 3x) + 2x cos 3x`
  `f^{\prime}(pi/3)` `= (pi/3)^2 ⋅ 3 (-sin pi) + 2 (pi/3) cos pi`
    `= -(2pi)/3`

Probability, MET1-NHT 2019 VCAA 8

A fair standard die is rolled 50 times. Let `W` be a random variable with binomial distribution that represents the number of times the face with a six on it appears uppermost.

  1. Write down the expression for  `text(Pr)(W = k)`, where  `k in {0, 1, 2, …, 50}`.  (1 mark)

  2. Show that  `(text(Pr)(W = k + 1))/(text(Pr)(W = k)) = (50 - k)/(5(k + 1))`.  (2 marks)

  3. Hence, or otherwise, find the value of `k` for which  `text(Pr)(W = k)`  is the greatest.  (2 marks)

Show Answers Only

  1. `\ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50 – k)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `8`

Show Worked Solution

a.  `text(Pr)(W = k) = \ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50 – k)`

 

b.   `(text(Pr)(W = k + 1))/(text(Pr)(W = k))` `= (\ ^50C_(k+1) ⋅ (1/6)^(k+1) ⋅ (5/6)^(49 – k))/(\ ^ 50C_k ⋅ (1/6)^k ⋅ (5/6)^(50-k))`
    `= ((50!)/((49 – k)!(k + 1)!) ⋅ (1/6))/((50!)/((50 – k)! k!) ⋅ (5/6))`
    `= ((50 – k)!k!)/(5(49 – k)!(k + 1)!)`
    `= (50 – k)/(5(k + 1))`

 

c.  `text(Find)\ k\ text(such that)`

`text(Pr)(W = k + 1)` `< text(Pr)(W = k)`
`50 – k` `< 5(k + 1)`
`6k` `> 45`
`k` `> 7 1/2`

 
`=> text(Pr)(W = 8) > text(Pr)(W = 9)`

`=> text(Pr)(W = 9) > text(Pr)(W = 10)\ …`

`:. text(Pr)(W = 8)\ text(is the greatest.)`

Calculus, MET1-NHT 2019 VCAA 7

The shaded region in the diagram below is bounded by the vertical axis, the graph of the function with rule  `f(x) = sin(pix)`  and the horizontal line segment that meets the graph at  `x = a`, where  `1 <= a <= 3/2`.
 


 

Let  `A(a)`  be the area of the shaded region.

  1. Show that  `A(a) = 1/pi-1/pi cos(a pi)-a sin (a pi)`.   (3 marks)

  2. Determine the range of values of `A(a)`.   (2 marks)

    1. Express in terms of  `A(a)`, for a specific value of `a`, the area bounded by the vertical axis, the graph of  `y = 2(sin(pi x) + sqrt 3/2)`  and the horizontal axis.   (2 marks)

    2. Hence, or otherwise, find the area described in part c.i.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `[2/pi, (2 + 3 pi)/(2pi)]`
    1. `2 A(a)`
    2. `(9 + 4 sqrt 3 pi)/(3 pi)`
Show Worked Solution

a.   `text(Lower border of shaded area:)\ y = f(a) = sin(a pi)`

`text(Area)` `= int_0^a sin(pi x)-sin (a pi)\ dx`
  `= [-1/pi cos (pi x)-x sin(a pi)]_0^a`
  `= [-1/pi cos(pi a)-a sin (a pi)-(-1/pi-0)]`
  `= 1/pi-1/pi cos (a pi)-a sin(a pi)`

 

b.   `text(S) text(ince)\ 1 <= a <= 3/2,`

`A(1) = 1/pi-1/pi cos(pi)-1 sin (pi) = 2/pi`

`A(3/2) = 1/pi-1/pi cos ((3 pi)/2)-3/2 sin ((3pi)/2) = 1/pi + 3/2 = (2 + 3pi)/(2 pi)`
 

`:.\ text(Range:)\ \ [2/pi, (2 + 3pi)/(2 pi)]`

 

c.i.  `A(a) = int_0^a sin(pi x)-sin (a pi)\ dx`

`A_1` `=2int _0^a sin(pi x) + sqrt 3/2\ dx`   
  `=2int _0^a sin(pi x)-sin((4pi)/3)\ dx,\ \ \ (a=4/3)`  
  `=2int _0^(4/3) sin(pi x)-sin((4pi)/3)\ dx`  
  `=2 xx A(a)`  

 
`:.\ text(When)\ \ a = 4/3,\ \ text(Area) = 2 xx A(a)`

 

c.ii.  `text(When)\ \ a = 4/3`

`text(Area)` `= 2 xx (1/pi-1/pi cos ((4 pi)/3)-4/3 sin ((4 pi)/3))`
  `= 2 xx (1/pi + 1/(2 pi) + 4/3 xx sqrt 3/2)`
  `= 2(3/(2pi) + (2 sqrt 3)/3)`
  `= 3/pi + (4 sqrt 3)/3`
  `= (9 + 4 sqrt 3 pi)/(3 pi)`

Probability, MET1-NHT 2019 VCAA 6a

Jacinta tosses a coin five times.

Assuming that the coin is fair and given that Jacinta observes a head on the first two tosses, find the probability that she observes a total of either four or five heads.  (2 marks)

Show Answers Only

`1/2`

Show Worked Solution

`text(After 2 tosses, 2 heads.)`

`text(Pr)text{(4 or 5}\ H)` `= HHT + HTH + THH + HHH`
  `= (1/2)^3 xx 4`
  `= 1/2`

Algebra, MET1-NHT 2019 VCAA 5b

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

Find the domain and the rule of the inverse function  `h^(-1)`.   (3 marks)

Show Answers Only

`[–2, oo)`

Show Worked Solution

 `y = sqrt (2x + 3)-2`

`text(Inverse: swap)\ \ x ↔ y`

`x` `= sqrt(2y + 3)-2`
`sqrt(2y + 3)` `= x + 2`
`2y + 3` `= (x + 2)^2`
`y` `= 1/2(x + 2)^2 -3/2`
`:. h^(-1)` `= 1/2(x + 2)^2-3/2`

 

`text(Domain)\ \ h^(-1)(x)` `= text(Range)\ h(x)`
  `= [–2, oo)`

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.

  3.     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 2

Find  `f(x)`  given that  `f(1) = -7/4`  and  `f ^{\prime}(x) = 2x^2 - 1/4x^(-2/3)`.  (2 marks)

Show Answers Only

`f(x) = 2/3x^3 – 3/4x^(1/3) – 5/3`

Show Worked Solution
`f(x)` `= int 2x^2 – 1/4x^(-2/3) dx`
  `= 2/3x^3 – 1/4 ⋅ 1/(1/3) x^(1/3) + c`
  `= 2/3x^3 – 3/4x^(1/3) + c`

 
`text(Given)\ \ f(1) = -7/4:`

`-7/4` `= 2/3 – 3/4 + c`
`c` `= -5/3`
`:. f(x)` `= 2/3x^3 – 3/4x^(1/3) – 5/3`

Statistics, EXT1 S1 EQ-Bank 24

The proportion of a population that have brown eyes is 0.35.

Using  `overset^p`  to represent the sample proportion, calculate the smallest sample size, `n`, such that the standard deviation of  `overset^p`  is below 0.02.  (2 marks)

Show Answers Only

`569`

Show Worked Solution

`E(overset^p) = p = 0.35`

`text(Var)(overset^p) = (0.35(1 – 0.35))/n = 0.2275/n`

`sigma(overset^p) = sqrt(0.2275/n)`
 

`text(If)\ \ sigma(overset^p) < 0.02,`

`sqrt(0.2275/n)` `< 0.02`
`0.0004n` `> 0.0275`
`n` `> 0.2275/0.0004`
`n` `> 568.75`

 

`:. text(Smallest)\ \ n = 569`

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 22

It is known that 43% of voters in a city voted for Labor in the election.

If 250 voters from the city are selected at random, use a suitable approximation and the attached probability table to find the probability that at least 120 of those chosen voted for Labor.  (3 marks)

Show Answers Only

`0.0571`

Show Worked Solution

`E(overset^p) = p = 0.43`

`text(Var)(overset^p) = (0.43(1 – 0.43))/250 = 0.0009804`

`sigma(overset^p) = 0.0313`
 

`text(Let)\ \ X = text(number who voted Labor)`

`text(If)\ \ X = 120 \ => \ overset^p = 120/250 = 0.48`

`P(X >= 120)` `= P(z >= (0.48 – 0.43)/0.0313)`
  `= P(z >= 1.58)`
  `= 1 – 0.9429`
  `= 0.0571`

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, EXT1 S1 EQ-Bank 21

A biased coin has a 0.6 chance of landing on heads. The coin is tossed 15 times.

  1. Calculate the probability of obtaining 7, 8 or 9 heads using binomial probability distribution.
  2. Give your answer correct to 3 decimal places.  (2 marks)

  3. Calculate the probability of obtaining 7, 8 or 9 heads using normal approximation to the binomial distribution and the probability table attached.
  4. Give your answer correct to 3 decimal places.  (2 marks)

  5. Could this binomial distribution be reasonably approximated with a normal distribution? Support your answer with a brief calculation.  (1 mark)

Show Answers Only
  1. `0.502\ (text(to 3 d.p.))`
  2. `0.509`
  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `P(7, 8\ text(or)\ 9)` `= \ ^15C_7(0.6)^7(0.4)^8  \ ^15C_8(0.6)^8(0.4)^7 + \ ^15C_9(0.6)^9(0.4)^6`
    `= 0.118056 + 0.177084 + 0.206600`
    `= 0.502\ \ (text(to 3 d.p.))`

 

ii.   `E(overset^p) = p = 0.6`

`text(Var)(overset^p) = (p(1 – p))/n = (0.6 xx 0.4)/15 = 0.016`

`sigma (overset^p) = 0.12649`

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

`P(7, 8, 9) = P(6.5 < X < 9.5)`

`text(If)\ \ X = 6.5 \ => \ overset^p = 6.5/15 = 0.4333`

`text(If)\ \ X = 9.5 \ => \ overset^p = 9.5/15 = 0.6333`

`ztext(-score)\ (X = 6.5) = (0.4333 – 0.6)/0.12649 = −1.32`

`ztext(-score)\ (X = 9.5) = (0.6333 – 0.6)/0.12649 = 0.26`

COMMENT: A quick sketch of the normal distribution curve is often helpful when using probability tables in this context.
 

`P(7, 8, 9)` `= P(−1.32 < z < 0.26)`
  `= P(z < 1.32) – P(z < −0.26)`
  `= 0.9066 – 0.3974`
  `= 0.509`

 

iii.   `np = 15 xx 0.6 = 9`

COMMENT: `np>10, nq>10` is also common in assessing if normal approximation is reasonable.

`nq=n(1 – p) = 15 xx 0.4 = 6`

`text(S)text(ince)\ \ np > 5 and nq > 5,`

`=>\ text(Normal approximation is reasonable.)`

Combinatorics, EXT1 A1 EQ-Bank 14

A delivery company has 1095 packages to deliver on a given day.

It has 17 delivery vans that will deliver all packages. If one van delivers more packages than all other vans, the company pays the driver a $100 bonus.

What is the minimum number of packages a van could deliver and still win the $100 bonus.  (2 marks)

Show Answers Only

`66`

Show Worked Solution

`text(Pigeonholes)\ (k) = 17`

COMMENT: Note that “By PHP” refers to by pigeonhole principle.

`text(Pigeons)\ (n) = 1095`
 
`(n)/(k) = (1095)/(17) = 64\ text(remainder 7)`

`text(S)text(ince 7 vans could deliver 65 packages and the rest 64 packages)`

`=> \ text(By PHP, the minimum packages to win the $100 bonus = 66)`

Combinatorics, EXT1 A1 EQ-Bank 13

A sock drawer contains blue, white and green socks.

If individual socks are randomly chosen from the drawer, what is the minimum number that must be selected to ensure there are at least three pairs?   (2 marks)

Show Answers Only

`8`

Show Worked Solution

`text(Consider the worst-case scenario:)`

COMMENT: Note that “By PHP” refers to by pigeonhole principle.

`text{(i.e. the most socks chosen without 3 pairs)}`

`text(It is possible to choose 7 socks and only have 2 pairs)`

`=> \ text(5B, 1W, 1G  (2 pairs))`

`=> \ text(3B, 3W, 1G  (2 pairs))`
 
`text(Choosing the 8th sock produces 3 pairs in any scenario.)`

`:. \ text(By PHP, a minimum of 8 selections ensures 3 pairs.)`

Combinatorics, EXT1 A1 EQ-Bank 12

Eleven numbers are randomly chosen from the set of integers, `S`, where

`S = {1, 2, 3, 4, ..., 20}`

Prove that the sum of two of the eleven numbers randomly selected must equal 21.   (2 marks)

Show Answers Only

`text(Proof (Show Worked Solution))`

Show Worked Solution

`text(Rearranging)\ S\ text(into 10 pairs that sum to 21:)`

COMMENT: Note that “By PHP” refers to by pigeonhole principle.

`{1, 20}, {2, 19}, {3, 18}, …. , {10, 11}`
 
`text(Select one number from each pair)`

`=>\ text(10 numbers where two do not sum to 21`
 
`text(The 11th number chosen must complete a pair that sums to 21)`

`:. \ text(By PHP, two of the eleven numbers must sum to 21.)`

Combinatorics, EXT1 A1 SM-Bank 11

A multiple choice quiz asks students 4 questions. Each question has three possible answers, a, b or c, and students must attempt each question.

How many students must do the quiz to ensure that at least two sets of answers are identical?    (2 marks)

Show Answers Only

`82`

Show Worked Solution

`text(Total possible answer combinations)`

COMMENT: Note that “By PHP” refers to by pigeonhole principle.

`= 3 xx 3 xx 3 xx 3`

`= 81`
 

`:. \ text(By PHP, 82 students must do the quiz. `

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`

Mechanics, SPEC2-NHT 2019 VCAA 5

A pallet of bricks weighing 500 kg sits on a rough plane inclined at an angle of  `α°` to the horizontal, where  `tan(α°) = (7)/(24)`. The pallet is connected by a light inextensible cable that passes over a smooth pulley to a hanging container of mass `m` kilograms in which there is 10 L of water. The pallet of bricks is held in equilibrium by the tension `T` newtons in the cable and a frictional resistance force of 50 `g` newtons acting up and parallel to the plane. Take the weight force exerted by 1 L of water to be `g` newtons.
  


 

  1. Label all forces acting on both the pallet of bricks and the hanging container on the diagram above, when the pallet of bricks is in equilibrium as described.   (1 mark)
  2. Show that the value of `m` is 80.  (3 marks)

Suddenly the water is completely emptied from the container and the pallet of bricks begins to slide down the plane. The frictional resistance force of 50 `g` newtons acting up the plane continues to act on the pallet.

  1. Find the distance, in metres, travelled by the pallet after 10 seconds.  (3 marks)
  2. When the pallet reaches a velocity of  `3\ text(ms)^-1`, water is poured back into the container at a constant rate of 2 L per second, which in turn retards the motion of the pallet moving down the plane. Let  `t`  be the time, in seconds, after the container begins to fill. 
  3.   i. Write down, in terms of  `t`, an expression for the total mass of the hanging container and the water it contains after `t` seconds. Give your answer in kilograms.  (1 mark)
  4.  ii. Show that the acceleration of the pallet down the plane is given by  `(text(g)(5 - t))/(t + 290)\ text(ms)^-2`  for  `t ∈[0, 5)`.  (2 marks)
  5. iii. Find  the velocity of the pallet when  `t = 4`. Give your answer in metres per second, correct to one decimal place.  (2 marks)
Show Answers Only
  1.  
    `qquad`
  2. `text(Proof(Show Worked Solution))`
  3. `(25 text(g))/(29)`
  4.   i. `80 + 2t`
     ii. `text(Proof (Show Worked Solution))`
    iii. `3.4\ text(ms)^-1`
Show Worked Solution

a.

 
b.   `text(Resolving vertical forces on container:)`

`T – (m + 10)g = 0 \ …\ (1)`

`text(Resolving forces on plane:)`
 


 

`tan α = (7)/(24) \ => \ sin α = (7)/(25)`
 

`text(Solve for m:)`

`(m + 10)g` `= 500 text(g) · (7)/(25) – 50 text(g)`
`m + 10` `= 140 – 50`
`:. \ m` `= 80`

 

c.   `text(Resolving vertical forces on container:)`

`T – 80 g = 80 a \ …\ (1)`

`text(Resolving forces on plane:)`

`500 g sin α – (T + 50 g) = 500 a`

`90 g – T = 500 a \ …\ (2)`

`text(Add) \ (1) + (2)`

`10 g` `= 580 a`
`a` `= (g)/(58)`
`s` `= ut + (1)/(2) at^2`
  `= 0 + (1)/(2) · (g)/(58) + 10^2`
  `= (25g)/(29)`

 

d.i.   `m = 80 + 2t`
 

d.ii.   `text(Resolving vertical forces on container:)`

`T – (80+2t)g = (80+2t)a \ …\ (1)`

`text(Resolving forces on plane:)`

`90g – T = 500a \ …\ (2)`

`text(Add)\ (1) + (2)`

`(90 – 80 – 2t)g` `= (500 + 80 + 2t)a`
`(10 – 2t)g` `=(580 + 2t)a`
`a` `= (g(5 – t))/(t + 290) ms^-2`

 

d.iii.   `(dv)/(dt) = (g(5 – t))/(t + 290)`

`v = int (dv)/(dt)\ dt = 295 log_e ((t + 290)/(290)) – g t + c`

 
`text(When)\ t = 0, v = 3\ \ text{(given)} \ => \ c = 3`

`:. \ v = 295 log_e ((t+290)/(290)) – g t + 3`

`:. \ v(4) = 3.4\ text(ms)^-1`

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`

Statistics, EXT1 S1 EQ-Bank 13

On average, batsmen playing cricket in a T20 competition play a scoring shot two out of every three balls.

In a regular season, a total of 1200 overs that each contain six balls, are bowled.

Estimate how many overs would have at least five scoring shots.  (3 marks)

Show Answers Only

`421`

Show Worked Solution

`text(Let)\ \ X = text(number of scoring shots in a six ball over)`

`X\ ~\ text(Bin) (6, 2/3)`

`P(X >= 5)` `= P(X = 5) + P(X = 6)`
  `=\ ^6 C_5 ⋅ (2/3)^5 (1/3) + \ ^6 C_6 ⋅ (2/3)^6`
  `= 64/243 + 64/729`
  `= 256/729`

 

`text(Let)\ \ Y=\ text(number of overs with at least 5 scoring shots)`

`Y\ ~\ text(Bin)(1200, 256/729)`

`E(Y)` `=np`  
  `=1200 xx 256/729`  
  `=421.39…`  
  `=421\ \ text{(nearest over)}`  

Statistics, EXT1 S1 EQ-Bank 12

A manufacturer of pool cue tips knows that 4% of pool cue tips produced in its factory need to be scrapped.

A random sample of 10 cue tips produced in the factory is examined.

  1. What is the probability that exactly 3 cue tips need to be scrapped?

     

    Give your answer correct to 3 decimal places.  (1 mark)

  2. What is the probability that, at most, 2 cue tips need to be scrapped?

     

    Give your answer correct to 3 decimal places.  (2 marks)

Show Answers Only
  1. `0.006`
  2. `0.993`
Show Worked Solution

i.     `text(Let)\ \ X = text(number of defective cue tips)`

`X\ ~\ text(Bi) (10, 0.04)`

`P(X = 3)` `= \ ^10 C_3 ⋅ (0.04)^3 (0.96)^7`
  `= 0.00577…`
  `=0.006\ \ text{(to 3 d.p.)}`

 

ii.   `P(X <= 2)` `= P(X = 0) + P(X = 1) + P(X = 2)`
    `=\ ^10 C_0(0.04)^0 (0.96)^10 + \ ^10 C_1(0.04) (0.96)^9 + \ ^10 C_2 (0.04)^2 (0.96)^8`
    `= 0.66483 + 0.27701 + 0.05194`
    `= 0.994\ \ text{(to 3 d.p.)}`

Vectors, EXT1 V1 2019 SPEC2-N 4

A snowboarder at the Winter Olympics leaves a ski jump at an angle of `theta` degrees to the horizontal, rises up in the air, performs various tricks and then lands at a distance down a straight slope that makes an angle of 45° to the horizontal, as shown below.

Let the origin `O` of a cartesian coordinate system be at the point where the snowboarder leaves the jump, with a unit vector in the positive `x` direction being represented by  `underset~i`  and a unit vector in the positive `y` direction being represented by  `underset~j`. Distances are measured in metres and time is measured in seconds.

The position vector of the snowboarder  `t`  seconds after leaving the jump is given by

`underset~r (t) = (6t-0.01t^3) underset~i + (6 sqrt3 t-4.9t^2 + 0.01t^3) underset~j , \ t ≥ 0`
 


 

  1. Find the angle  `theta °`.    (2 marks)

  2. Find the speed, in metres per second, of the snowboarder when she leaves the jump at `O`.    (1 mark)

  3. Find the maximum height above `O` reached by the snow boarder. Give your answer in metres, correct to one decimal place.    (2 marks)

  4. Show that the time spent in the air by the snowboarder is  `(60(sqrt3 + 1))/(49)`  seconds.    (3 marks)

Show Answers Only
  1. `60°`
  2. `12 \ text(ms)^-1`
  3. `5.5 \ text(m)`
  4. `text(See Worked Solutions)`
Show Worked Solution

a.   `v(t) = (6-0.03t^2)underset~i + (6 sqrt3-9.8t + 0.03t^2) underset~j`

`text(When) \ t =0,`

`v(t) = 6underset~i + 6 sqrt3 underset~j`

`tan theta = (6 sqrt3)/(6) = sqrt3`

`:. \ theta` `= tan^-1 sqrt3= 60°`

 

b.    `text(Speed)` `= |v(0)|`
    `= sqrt(6^2 + (6 sqrt3)^2)`
    `= 12 \ text(ms)^-1`

 
c.   `text(Max height when) \ underset~j \ text(component of) \ v(t) = 0`

`text(Solve): \ \ 6 sqrt3-9.8t + 0.03t^2 = 0`

`=> t =  1.064 \ text(seconds)`

`text(Max height)` `= 6 sqrt3 (1.064)-4.9(1.064)^2 + 0.01(1.064)^3`  
  `~~5.5\ text(m)`  

 
d.   `text(Time of Flight  ⇒  Solve for)\ \ t\ \ text(when)\ \ y=-x:`

`6 sqrt 3 t-4.9t^2 + 0.01 t^3` `= -(6t-0.01t^3)`
`(6 + 6 sqrt3)t-4.9 t^2` `= 0`
`t(6 + 6 sqrt3-4.9 t)` `= 0`
`4.9 t` `= 6 + 6 sqrt3`
`t` `= (6 + 6 sqrt3)/(4.9)`
  `= (60(sqrt3 + 1))/(49)\ text(seconds)`

Vectors, SPEC2-NHT 2019 VCAA 4

A snowboarder at the Winter Olympics leaves a ski jump at an angle of `theta` degrees to the horizontal, rises up in the air, performs various tricks and then lands at a distance down a straight slope that makes an angle of 45° to the horizontal, as shown below.

Let the origin `O` of a cartesian coordinate system be at the point where the snowboarder leaves the jump, with a unit vector in the positive `x` direction being represented by  `underset~i`  and a unit vector in the positive `y` direction being represented by  `underset~j`. Distances are measured in metres and time is measured in seconds.

The position vector of the snowboarder  `t`  seconds after leaving the jump is given by

`underset~r (t) = (6t-0.01t^3) underset~i + (6 sqrt3 t-4.9t^2 + 0.01t^3) underset~j , \ t ≥ 0`
 


 

  1. Find the angle  `theta °`.    (2 marks)

  2. Find the speed, in metres per second, of the snowboarder when she leaves the jump at `O`.   (1 mark)

  3. Show that the time spent in the air by the snowboarder is  `(60(sqrt3 + 1))/(49)`  seconds.   (3 marks)

  4. Find the total distance the snowboarder travels while airborne. Give your answer in metres, correct to two decimal places.   (2 marks)

Show Answers Only
  1. `60°`
  2. `12 \ text(ms)^-1`
  3. `5.5 \ text(m)`
  4. `text(See Worked Solutions)`
  5. `38.51 \ text{m  (to 2 decimal places)}`
Show Worked Solution

a.   `v(t) = (6-0.03t^2)underset~i + (6 sqrt3-9.8t + 0.03t^2) underset~j`

`text(When) \ t =0,`

`v(t) = 6underset~i + 6 sqrt3 underset~j`

`tan theta = (6 sqrt3)/(6) = sqrt3`

`:. \ theta` `= tan^-1 sqrt3`
  `= 60°`

 

b.    `text(Speed)` `= |v(0)|`
    `= sqrt(6^2 + (6 sqrt3)^2)`
    `= 12 \ text(ms)^-1`

 
c.   `text(Max height when) \ underset~j \ text(component of) \ v(t) = 0`

`text(Solve): \ \ 6 sqrt3-9.8t + 0.03t^2 = 0`

`=> t =  1.064 \ text(seconds)`

`text(Max height)` `= 6 sqrt3 (1.064)-4.9(1.064)^2 + 0.01(1.064)^3`  
  `~~5.5\ text(m)`  

 
d.   `text(Time of Flight  ⇒  Solve for)\ \ t\ \ text(when)\ \ y=-x:`

`6 sqrt 3 t-4.9t^2 + 0.01 t^3` `= -(6t-0.001t^3)`
`(6 + 6 sqrt3)t-4.9 t^2` `= 0`
`t(6 + 6 sqrt3-4.9 t)` `= 0`
`4.9 t` `= 6 + 6 sqrt3`
`t` `= (6 + 6 sqrt3)/(4.9)`
  `= (60(sqrt3 + 1))/(49)\ text(seconds)`


 
e.   `text(Total distance) \ = text(Area under) \ v(t) \ text(graph from)\ \ t = 0 \ \ text(to)\ \ t_1 = (60(sqrt3 + 1))/(49)`

`|v(t)| = sqrt{(6-0.03 t^2)^2 + (6 sqrt3- 9.8t + 0.03 t^2)^2}`

`text(Distance)` `= int_0^(t_1) |v(t)| \ dt`
  `= 38.51 \ text{m (to 2 d.p.)}`

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.  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)

  3.  Find, in cartesian form, the points(s) of intersection of  `L`  and the graph of  `|z| = 4`.   (2 marks)

  4.  Sketch  `L`  and the graph of  `|z| = 4`  on the Argand diagram below.   (2 marks)

     
     
         
     

  5.  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)`

Statistics, SPEC2-NHT 2019 VCAA 18 MC

Consider a random variable `X` with probability density function
 

`f(x) = {(2x,, 0<= x <= 1),(0,, x < 0\ \ text(or)\ \ x > 1):}`
 

If a large number of samples, each of size 100, is taken from his distribution, then the distribution of the sample means, `barX`, will be approximately normal with mean  `E(barX) = 2/3`  and standard deviation  `text(sd)(barX)`  equal to

  1.  `sqrt2/60`
  2.  `sqrt2/6`
  3.  `1/180`
  4.  `1/18`
  5.  `sqrt2/30`
Show Answers Only

`A`

Show Worked Solution

`E(X) = E(barX) = 2/3`

`E(X^2) = int_0^1 x^2 · 2x\ dx = [(x^4)/2]_0^1 = 1/2`

`text(Var)(X)` `= E(X^2)-[E(X)]^2`
  `= 1/2-4/9`
  `= 1/18`

 
`sigma(barX) = (sqrt(1/18))/sqrt100 = 1/(10 xx 3sqrt2) = sqrt2/60`

`=>\ A`

Vectors, SPEC2-NHT 2019 VCAA 14 MC

The position vector  `underset~r(t)`  of a mass of 3 kg after `t` seconds, where  `t >= 0`, is given by  `underset~r(t) = 10tunderset~i + (16t^2 - 4/3t^3)underset~j`.

The force, in newtons, acting on the mass when  `t = 2`  seconds is

  1.  `16underset~j`
  2.  `32underset~j`
  3.  `48underset~j`
  4.  `30underset~i + 144underset~j`
  5.  `16`
Show Answers Only

`C`

Show Worked Solution

`underset~r(t) = 10tunderset~i + (16t^2 – 4/3t^3)underset~j`

`underset~v(t) = 10underset~i + (32t – 4t^2)underset~j`

`underset~a(t) = (32 – 8t)underset~j`

`text(When)\ t = 2,`

`underset~F` `= m underset~a`
  `= 3(32 – 16)underset~j`
  `= 48underset~j`

`=>\ C`

Vectors, SPEC2-NHT 2019 NHT 13 MC

For the vectors  `underset~a = underset~i + 3underset~j - underset~k`,  `underset~b = −underset~i - 4underset~j + 2underset~k`  and  `underset~c = −underset~i - 6underset~j + lambdaunderset~k`  to be linearly dependent, the value of  `lambda`  must be

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show Answers Only

`E`

Show Worked Solution

`text(Linearly dependent if)\ ∃ lambda, m, n,\ text(such that)`

`((−1),(−6),(lambda)) = m((1),(3),(−1)) + n((−1),(−4),(2))`

`text{Solve (by CAS):}`

`lambda = 4, m = 2, n = 3`

`=>\ E`

Vectors, EXT2 V1 SM-Bank 25

The acute angle  `theta`  is the angle between the vectors  `underset~a = −2underset~i + 2underset~j - underset~k`  and  `underset~b = −4underset~i + 4underset~j + 7underset~k`.

Find the exact value of  `sin(2theta)`.   (2 marks)

Show Answers Only

`(4sqrt2)/9`

Show Worked Solution

`underset~a = ((−2),(2),(−1)), \ |underset~a| = sqrt(4 + 4 + 1) = 3`

`underset~b = ((−4),(4),(7)), \ |underset~b| = sqrt(16 + 16 + 49) = 9`

`underset~a · underset~b = ((−2),(2),(−1))((−4),(4),(7)) = 8 + 8 – 7 = 9`

`costheta = (underset~a · underset~b)/(|underset~a||underset~b|) = 9/(3 xx 9) = 1/3`
 

`sintheta` `= sqrt8/3`
`sin2theta` `= 2sinthetacostheta`
  `= 2 · sqrt8/3 · 1/3`
  `= (4sqrt2)/9`

Vectors, SPEC2-NHT 2019 VCAA 12 MC

Given that `theta` is the acute angle between the vectors  `underset~a = −2underset~i + 2underset~j - underset~k`  and  `underset~b = −4underset~i + 4underset~j + 7underset~k`, then  `sin(2theta)` is equal to

  1.  `2sqrt2`
  2.  `(4sqrt2)/9`
  3.  `(2sqrt2)/9`
  4.  `(2sqrt2)/3`
  5.  `(4sqrt2)/3`
Show Answers Only

`B`

Show Worked Solution

`underset~a = ((−2),(2),(−1)), \ |underset~a| = sqrt(4 + 4 + 1) = 3`

`underset~b = ((−4),(4),(7)), \ |underset~b| = sqrt(16 + 16 + 49) = 9`

`underset~a · underset~b = ((−2),(2),(−1))((−4),(4),(7)) = 8 + 8 – 7 = 9`

`costheta = (underset~a · underset~b)/(|underset~a||underset~b|) = 9/(3 xx 9) = 1/3`

`sintheta` `= sqrt8/3`
`sin2theta` `= 2sinthetacostheta`
  `= 2 · sqrt8/3 · 1/3`
  `= (4sqrt2)/9`

`=>\ B`

Vectors, SPEC2-NHT 2019 VCAA 11 MC

The vector resolute of  `underset~a = 2underset~i - underset~j + 3underset~k`  that is perpendicular to  `underset~b = underset~i + underset~j - underset~k`  is

  1. `−2/3(underset~i + underset~j - underset~k)`
  2. `−2/3(2underset~i - underset~j + 3underset~k)`
  3. `1/3(8underset~i - underset~j + 7underset~k)`
  4. `underset~i - 2underset~j + 4underset~k`
  5. `underset~i + underset~j + 2underset~k`
Show Answers Only

`C`

Show Worked Solution

`underset~a = ((2),(−1),(3)),\ \ underset~b = ((1),(1),(−1))`

`|underset~b| = sqrt3`

`underset~a · underset~b = 2 – 1 – 3 = −2`
 

`text(Vector resolute)\ underset~a\ text(onto)\ underset~b`

`= (underset~a · underset~overset^b)underset~overset^b`

`= −2/sqrt3 xx 1/sqrt3((1),(1),(−1))`

`= −2/3((1),(1),(−1))`
 

`text(Vector resolute)\ underset~a ⊥ underset~b`

`= underset~a – (underset~a · underset~overset^b)underset~overset^b`

`= ((2),(−1),(3)) + 2/3((1),(1),(−1))`

`= 1/3((8),(−1),(7))`

`=>\ C`

Calculus, SPEC2-NHT 2019 VCAA 10 MC

Euler's method, with a step size of 0.1, is used to approximate the solution of the differential equation  `1/y(dy)/(dx) = cos(x)`, with  `y = 2`  when  `x = 0`.

When  `x = 0.2`, the value obtained for `y`, correct to four decimal places, is

  1.  2.2000
  2.  2.3089
  3.  2.3098
  4.  2.4189
  5.  2.4199
Show Answers Only

`D`

Show Worked Solution
`1/y · (dy)/(dx)` `= cosx`
`(dy)/(dx)` `= ycosx`

 
`text(Using)\ \ y_0 = 2, x_0 = 0, h = 0.1:`

`y_1` `= y_0 + h(y_0cosx_0)`
  `= 2 + 0.1(2cos0)`
  `= 2.2`
`y_2` `= 2.2 + 0.1(2.2cos(0.1))`
  `= 2.4189`

 
`=>\ D`

Calculus, SPEC2-NHT 2019 VCAA 9 MC

With a suitable substitution, `int_1^2sqrt(5x - 1)\ dx` can be expressed as

  1. `5int_1^2sqrtu\ du`
  2. `1/5int_1^2sqrtu\ du`
  3. `5int_4^9sqrtu\ du`
  4. `1/5int_4^9sqrtu\ du`
  5. `5int_4^9sqrt(5u - 1)\ du`
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ u = 5x – 1`

`(du)/(dx) = 5\ \ =>\ 1/5\ du = dx`
 

`text(When)\ \ x = 2, u = 9`

`text(When)\ \ x = 1, u = 4`

`int_1^2sqrt(5x – 1)\ dx = 1/5 int_4^9 sqrtu\ du`

`=>\ D`

Calculus, SPEC2-NHT 2019 VCAA 8 MC

The total area enclosed between the `x`-axis and the graph of `f(x) = |x^3| - x^2 - |x|` is closest to

  1.  −2.015
  2.  −1.008
  3.     1.008
  4.     2.015
  5.     2.824
Show Answers Only

`D`

Show Worked Solution

`text(Graph)\ \ f(x) = |x^3| – x^2 – |x|\ \ \ (text(by CAS))`

`text(Intersection occurs at:)`

`x = 0, ±((1 + sqrt5)/2)`

`=>\ text(Function is even)`

`:.\ text(Area)` `= −2int_0^(1 + sqrt5) f(x)\ dx\ \ (text(area is below axis))`
  `= 2.015`

 
`=>\ D`

Calculus, SPEC2-NHT 2019 VCAA 7 MC

The gradient of the line that is perpendicular to the graph of a relation at any point  `P(x, y)`  is half the gradient of the line joining `P` and the point  `Q(−1,1)`.

The relation satisfies the differential equation

  1. `(dy)/(dx) = (y - 1)/(2(x + 1))`
  2. `(dy)/(dx) = (2(x + 1))/(y + 1)`
  3. `(dy)/(dx) = (2(x - 1))/(y + 1)`
  4. `(dy)/(dx) = (x + 1)/(2(1 - y))`
  5. `(dy)/(dx) = (2(x + 1))/(1 - y)`
Show Answers Only

`E`

Show Worked Solution

`m_(PQ) = (y – 1)/(x + 1) \ => \ m_text(perp) = 1/2((y – 1)/(x + 1))`

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

 
`=>\ E`