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The network below shows the distances, in metres, between camp sites at a camping ground that has electricity.
The vertices `A` to `I` represent the camp sites.
The minimum length of cable required to connect all the camp sites is 53 m.
The value of `x`, in metres, is at least
The activity network below shows the sequence of activities required to complete a project.
The number next to each activity in the network is the time it takes to complete that activity, in days.
What is the critical path and minimum completion time for this project. (2 marks)
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`text(Scanning forward:)`
| `text(Critical path)` | `= BEIJM` |
| `= 7 + 4 + 3 + 3 + 5` | |
| `= 22\ text(days)` |
The activity network below shows the sequence of activities required to complete a project.
The number next to each activity in the network is the time it takes to complete that activity, in days.
The minimum completion time for this project, in days, is
`text(Scanning forward:)`
| `text(Critical path)` | `= BEIJM` |
| `= 7 + 4 + 3 + 3 + 5` | |
| `= 22\ text(days)` |
`=> E`
The network below shows the distances, in metres, between camp sites at a camping ground that has electricity.
The vertices `A` to `I` represent the camp sites.
The minimum length of cable required to connect all the camp sites is 53 m.
The value of `x`, in metres, is at least
Which one of the following is not a spanning tree for the network above?
| A. |  |
B. |  |
| C. |  |
D. |  |
| E. |  |
Five competitors, Andy (A), Brie (B), Cleo (C), Della (D) and Eddie (E), participate in a darts tournament.
Each competitor plays each of the other competitors once only, and each match results in a winner and a loser.
The matrix below shows the results of this darts tournament.
There are still two matches that need to be played.
`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquad loser),(quadqquadqquadqquadqquadqquad \ A qquad\ B qquad \ C qquad\ D qquad E),(wi\n\n\er qquad{:(A),(B),(C),(D),(E):}[(0,…,0,1,0),(…,0,1,0,1),(1,0,0,…,1),(0,1,…,0,0),(1,0,0,1,0)]):}`
A ‘1’ in the matrix shows that the competitor named in that row defeated the competitor named in that column.
For example, the ‘1’ in row 2, column 3 shows that Brie defeated Cleo.
A ‘…’ in the matrix shows that the competitor named in that row has not yet played the competitor named in that column.
The winner of this darts tournament is the competitor with the highest sum of their one-step and two-step dominances.
Which player, by winning their remaining match, will ensure that they are ranked first by the sum of their one-step and two-step dominances?
The table below shows information about three matrices: `A, B` and `C`.
| `qquad qquad text(Matrix) qquad qquad` | `quad qquad qquad text(Order) quad qquad qquad` | |
| `A` | `2 xx 4` | |
| `B` | `2 xx 3` | |
| `C` | `3 xx 4` |
The transpose of matrix `A`, for example, is written as `A^T`.
What is the order of the product `C^T xx (A^T xx B)^T`?
The element in row `i` and column `j` of matrix `M` is `m_(ij)`.
`M` is a 3 × 3 matrix. It is constructed using the rule `m_(ij) = 3i + 2j`.
`M` is
| A. |
`[(5,7,9),(7,9,11),(11,13,15)]`
|
B. |
`[(5,7,9),(8,10,12),(11,13,15)]`
|
| C. |
`[(5,7,10),(8,10,13),(11,13,16)]`
|
D. |
`[(5,8,11),(7,10,13),(9,12,15)]`
|
| E. |
`[(5,8,11),(8,11,14),(11,14,17)]`
|
Matrices `P` and `W` are defined below.
`P = [(0,0,1,0,0),(0,0,0,0,1),(0,1,0,0,0),(0,0,0,1,0),(1,0,0,0,0)] qquad qquad W = [(A),(S),(T),(O),(R)]`
If `P^n xx W = [(A),(S),(T),(O),(R)]`, the value of `n` could be
The value of a van purchased for $45 000 is depreciated by `k`% per annum using the reducing balance method.
After three years of this depreciation, it is then depreciated in the fourth year under the unit cost method at the rate of 15 cents per kilometre.
The value of the van after it travels 30 000 km in this fourth year is $26 166.24
The value of `k` is
The nominal interest rate for a loan is 8% per annum.
When rounded to two decimal places, the effective interest rate for this loan is not
Gen invests $10 000 at an interest rate of 5.5% per annum, compounding annually.
After how many years will her investment first be more than double its original value?
The graph below represents the value of an annuity investment, `A_n`, in dollars, after `n` time periods.
A recurrence relation that could match this graphical representation is
An asset is purchased for $2480.
The value of this asset after `n` time periods, `V_n` , can be determined using the rule
`V_n = 2480 + 45n`
A recurrence relation that also models the value of this asset after `n` time periods is
The time series plot below displays the number of airline passengers, in thousands, each month during the period January to December 1960.
Part 1
During 1960, the median number of monthly airline passengers was closest to
Part 2
During the period January to May 1960, the total number of airline passengers was 2 160 000.
The five-mean smoothed number of passengers for March 1960 is
Table 4 below shows the monthly rainfall for 2019, in millimetres, recorded at a weather station, and the associated long-term seasonal indices for each month of the year.
Part 1
The deseasonalised rainfall for May 2019 is closest to
Part 2
The six-mean smoothed monthly rainfall with centring for August 2019 is closest to
Table 3 below shows the long-term mean rainfall, in millimetres, recorded at a weather station, and the associated long-term seasonal indices for each month of the year.
The long-term mean rainfall for December is missing.
Part 1
To correct the rainfall in March for seasonality, the actual rainfall should be, to the nearest per cent
Part 2
The long-term mean rainfall for December is closest to
In a study, the association between the number of tasks completed on a test and the time allowed for the test, in hours, was found to be non-linear.
The data can be linearised using a log10 transformation applied to the variable number of tasks.
The equation of the least squares line for the transformed data is
log10 (number of tasks) = 1.160 + 0.03617 × time
This equation predicts that the number of tasks completed when the time allowed for the test is three hours is closest to
The data in Table 2 was collected in a study of the association between the variables frequency of nightmares (low, high) and snores (no, yes).
Part 1
The variables in this study, frequency of nightmares (low, high) and snores (no, yes), are
Part 2
The percentage of participants in the study who did not snore is closest to
Part 3
Of those people in the study who did snore, the percentage who have a high frequency of nightmares is closest to
The lifetime of a certain brand of light globe, in hours, is approximately normally distributed.
It is known that 16% of the light globes have a lifetime of less than 655 hours and 50% of the light globes have a lifetime that is greater than 670 hours.
The mean and the standard deviation of this normal distribution are closest to
| A. | mean = 655 hours | standard deviation = 10 hours |
| B. | mean = 655 hours | standard deviation = 15 hours |
| C. | mean = 670 hours | standard deviation = 10 hours |
| D. | mean = 670 hours | standard deviation = 15 hours |
| E. | mean = 670 hours | standard deviation = 20 hours |
The wing length of a species of bird is approximately normally distributed with a mean of 61 mm and a standard deviation of 2 mm.
Using the 68–95–99.7% rule, for a random sample of 10 000 of these birds, the number of these birds with a wing length of less than 57 mm is closest to
`z text(-score)\ (57) = (x – bar x)/s = (57 – 61)/2 = -2`
`:.\ text(Number less than 57)`
`= 2.5text(%) xx 10\ 000`
`= 250`
`=> D`
Data relating to the following five variables was collected from insects that were caught overnight in a trap:
The number of these variables that are discrete numerical variables is
A percentaged segmented bar chart would be an appropriate graphical tool to display the association between month of the year (January, February, March, etc.) and the
The histogram below shows the distribution of weight, in grams, for a sample of 20 animal species. The histogram has been plotted on a `log_10` scale.
The percentage of these animal species with a weight of less than 10 000 g is
The times between successive nerve impulses (time), in milliseconds, were recorded.
Table 1 shows the mean and the five-number summary calculated using 800 recorded data values.
Part 1
The difference, in milliseconds, between the mean time and the median time is
Part 2
Of these 800 times, the number of times that are longer than 300 milliseconds is closest to
Part 3
The shape of the distribution of these 800 times is best described as
Consider the curve defined parametrically by `x = arcsin (t)` `y = log_e(1 + t) + 1/4 log_e (1-t)` where `t in [0, 1)`. --- 4 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) ---
Consider the function defined by `f(x) = {({:mx + n,:}, x < 1), ({: frac{4}{(1 + x^2)},:}, x >= 1):}` where `m` and `n` are real numbers. --- 5 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) ---
Let `f(x) = tan^(-1) (3x - 6) + pi`.
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| a. | `f^{\prime}(x)` | `= (d/(dx) (3x – 6))/(1 + (3x – 6)^2)` |
| `= 3/(9x^2 – 36x + 37)` |
b. `f^{\prime\prime}(x) = (3(18x – 36))/(9x^2 – 36x + 37)^2`
`f^{\prime\prime}(x) = 0\ \ text(when)\ \ 18x – 36 = 0 \ => \ x = 2`
`text(If)\ \ x < 2, 18x – 36 < 0 \ => \ f^{\prime\prime}(x) < 0`
`text(If)\ \ x > 2, 18x – 36 > 0 \ => \ f^{\prime\prime}(x) > 0`
`text(S) text(ince)\ \ f^{\prime\prime}(x)\ \ text(changes sign about)\ \ x = 2,`
`text(a POI exists at)\ \ x = 2`
| c. |  |
Let `f(x) = arctan (3x - 6) + pi`. --- 2 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) ---
b. `f^{\prime\prime}(x) = (3(18x – 36))/(9x^2 – 36x + 37)^2` `f^{\prime\prime}(x) = 0\ \ text(when)\ \ 18x – 36 = 0 \ => \ x = 2` `text(If)\ \ x < 2, 18x – 36 < 0 \ => \ f^{\prime\prime}(x) < 0` `text(If)\ \ x > 2, 18x – 36 > 0 \ => \ f^{\prime\prime}(x) > 0` `text(S) text(ince)\ \ f^{\prime\prime}(x)\ \ text(changes sign about)\ \ x = 2,` `text(a POI exists at)\ \ x = 2` Show Worked Solution
a.
`f^{\prime}(x)`
`= (d/(dx) (3x – 6))/(1 + (3x – 6)^2)`
`= 3/(9x^2 – 36x + 37)`
♦ Mean mark part (b) 42%.
c.
Let `underset ~ a = 2 underset ~i - 3 underset ~j + underset ~k` and `underset ~b = underset ~i + m underset ~j - underset ~k`, where `m` is an integer.
The projection of `underset ~a` onto `underset ~b` is `-11/18 (underset ~i + m underset ~j - underset ~k)`.
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Let `underset ~ a = 2 underset ~i-3 underset ~j + underset ~k` and `underset ~b = underset ~i + m underset ~j-underset ~k`, where `m` is an integer. The vector resolute of `underset ~a` in the direction of `underset ~b` is `-11/18 (underset ~i + m underset ~j-underset ~k)`. --- 6 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) ---
Solve the inequality `3 - x > 1/|x - 4|` for `x`, expressing your answer in interval notation. (3 marks)
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`3 – x > 1/|x – 4|`
`|x – 4| (3 – x) > 1`
`text(If)\ \ x – 4 > 0, x > 4`
| `(x – 4) (3 – x)` | `> 1` |
| `3x – x^2 – 12 + 4x` | `> 1` |
| `-x^2 + 7x – 13` | `> 0` |
`Delta = 7^2 – 4 ⋅ 1 ⋅ 13 = -3 < 0`
`=>\ text(No Solutions)`
`text(If)\ \ x – 4 < 0, x < 4`
| `-(x – 4) (3 – x)` | `> 1` |
| `x^2 – 7x + 12` | `> 1` |
| `x^2 – 7x + 11` | `> 0` |
| `x` | `= (7 +- sqrt(7^2 – 4 ⋅ 1 ⋅ 11))/2` |
| `= (7 +- sqrt 5)/2` |
`text(Combining solutions)`
`(x < (7 – sqrt 5)/2 ∪ x > (7 + sqrt 5)/2) nn x < 4`
`x ∈ (– oo, (7 – sqrt 5)/2)`
Find the cube roots of `1/sqrt 2 - 1/sqrt 2 i`. Express your answers in modulus-argument form. (3 marks)
Find the cube roots of `1/sqrt 2 - 1/sqrt 2 i`. Express your answers in polar form using principal values of the argument. (3 marks)
A 2 kg mass is initially at rest on a smooth horizontal surface. The mass is then acted on by two constant forces that cause the mass to move horizontally. One force has magnitude 10 N and acts in a direction 60° upwards from the horizontal, and the other force has magnitude 5 N and acts in a direction 30° upwards from the horizontal, as shown in the diagram below.
| a. |  |
`text(Resolving forces vertically:)`
| `2g` | `= 5 sin 30 + 10 sin 60 + R` |
| `2g` | `= 5/2 + 5 sqrt 3 + R` |
| `R` | `= 2g – 5/2 – 5 sqrt 3\ text(N)` |
| b. | `2a` | `= 10 cos 60 – 5 cos 30` |
| `2a` | `= 5 – (5 sqrt 3)/2` | |
| `:.a` | `= 5/2 – (5 sqrt 3)/4\ text(ms)^(-2)` |
| c. | `text(Distance)` | `= ut + 1/2 at^2` |
| `= 0+ 1/2 (5/2 – (5 sqrt 3)/4) × 4^2` | ||
| `= 20 – 10 sqrt 3\ text(m)` |
Part of the graph of `y = f(x)`, where `f:(0, ∞) -> R, \ f(x) = xlog_e(x)`, is shown below.
The graph of `f` has a minimum at the point `Q(a, f(a))`, as shown above.
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i. Find the value of `k` for which `y = 2x` is a tangent to the graph of `g`. (1 mark)
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ii. Find all values of `k` for which the graphs of `g` and `g^(-1)` do not intersect. (2 marks)
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a. `y = xlog_e x`
| `(dy)/(dx)` | `= x · 1/x + log_e x` |
| `= 1 + log_e x` |
`text(Find)\ x\ text(when)\ (dy)/(dx) = 0:`
| `1 + log_e x` | `= 0` |
| `log_e x` | `= -1` |
| `x` | `= 1/e` |
| `y` | `= 1/e log_e (e^(-1))` |
| `= -1/e` |
`:. Q(1/e, -1/e)`
| b. | `int 2x log_e(x) + x\ dx` | `= x^2 log_e (x) + c` |
| `2 int x log_e(x)\ dx` | `= x^2 log_e (x)-intx\ dx + c` | |
| `:. int x log_e(x)\ dx` | `= (x^2 log_e (x))/2-(x^2)/4 \ \ (c = 0)` |
| c. |  |
`text(When)\ \ x log_e x = 0 \ => \ x = 1`
`=> b = 1`
| `:.\ text(Area)` | `= −int_(1/e)^1 x log_e(x)\ dx` |
| `= [(x^2)/4-(x^2 log_e(x))/2]_(1/e)^1` | |
| `= (1/4-0)-(1/(4e^2)-(log_e(e^(-1)))/(2e^2))` | |
| `= 1/4-(1/(4e^2) + 1/(2e^2))` | |
| `= 1/4-3/(4e^2) \ text(u)^2` |
d.i. `text(When)\ \ f^{prime}(x) = m_text(tang) = 2,`
| `1 + log_e(x)` | `= 2` |
| `x` | `= e` |
`text(T)text(angent meets)\ \ g(x)\ \ text(at)\ \ (e, 2e)`
| `g(e)` | `= f(e) + k` |
| `2e` | `= e log_e e + k` |
| `:.k` | `= e` |
d.ii. `text(Find the value of)\ k\ text(when)\ \ y = x\ \ text(is a tangent to)\ g(x):`
`text(When)\ \ f^{prime}(x) = 1,`
| `1 + log_e(x)` | `= 1` |
| `x` | `= 1` |
`text(T)text(angent occurs at)\ (1, 1)`
`g(1) = f(1) + k \ => \ k = 1`
`:.\ text(Graphs don’t intersect for)\ k ∈ (1, ∞)`
Consider the function `f(x) = x^2 + 3x + 5` and the point `P(1, 0)`. Part of the graph `y = f(x)` is shown below.
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i. Find the slope of the line connecting points `P` and `Q` in terms of `a`. (1 mark)
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ii. Find the slope of the tangent to the graph of `f` at point `Q` in terms of `a`. (1 mark)
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iii. Let the tangent to the graph of `f` at `x = a` pass through point `P`.
Find the values of `a`. (2 marks)
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iv. Give the equation of one of the lines passing through point `P` that is tangent to the graph of `f`. (1 mark)
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Find the value of `int_0^3 (8x)/(1 + x^2)\ dx` in the form `a log_e(b)`. (2 marks)
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If `f^{′}(x)= (x^2)/(x^3 + 1)` and `f(1)= log_e 2,` find `f(x)`. (3 marks)
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Evaluate `int_1^4(sqrt x + 1)\ dx`. (3 marks)
Let `y=xsinx.` Evaluate `dy/dx` for `x=pi`. (3 marks)
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`f(x) = 1/sqrt2 sqrtx`, where `x in [0,2]`
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The graph of `y = f(x)`, where `x ∈ [0, 2]`, is shown on the axes below.
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`f^(-1)(x) = 2x^2`
a. `text(Domain)\ \ f^(-1)(x)= text(Range)\ \ f(x)=[0,1]`
`y = 1/sqrt2 x`
`text(Inverse: swap)\ \ x ↔ y`
| `x` | `= 1/sqrt2 sqrty` | |
| `sqrty` | `= sqrt2 x` | |
| `y` | `= 2x^2` |
`:. f^(-1)(x) = 2x^2`
| b. | |
Let `f:[0,2] -> R`, where `f(x) = 1/sqrt2 sqrtx`.
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The graph of `y = f(x)`, where `x ∈ [0, 2]`, is shown on the axes below.
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`f^(-1)(x) = 2x^2`
| a. | `text(Domain)\ \ f^(-1)(x)` | `= text(Range)\ \ f(x)=[0,1]` |
`y = 1/sqrt2 x`
`text(Inverse: swap)\ \ x ↔ y`
| `x` | `= 1/sqrt2 sqrty` |
| `sqrty` | `= sqrt2 x` |
| `y` | `= 2x^2` |
`:. f^(-1)(x) = 2x^2`
| b. | |
| c. | |
| `A` | `= int_0^(1/2) 1/sqrt2 sqrtx-2x^2 dx + int_(1/2)^1 2x^2-1/sqrt2 sqrtx\ dx` |
| `= [sqrt2/3 x^(3/2)-2/3 x^3]_0^(1/2) + [2/3 x^3-sqrt2/3 x^(3/2)]_(1/2)^1` | |
| `= [sqrt2/3 (1/sqrt2)^3-2/3(1/2)^3] + [(2/3-sqrt2/3)-(2/24-sqrt2/3 · 1/(2sqrt2))]` | |
| `= (1/6-1/12) + 2/3-sqrt2/3-(1/12-1/6)` | |
| `= 1/12 + 2/3-sqrt2/3 + 1/12` | |
| `= (1 + 8-4sqrt2 + 1)/12` | |
| `= (5-2sqrt2)/6\ \ text(u²)` |
For a certain population the probability of a person being born with the specific gene SPGE1 is `3/5`.
The probability of a person having this gene is independent of any other person in the population having this gene.
In a randomly selected group of four people, what is the probability that three or more people have the SPGE1 gene? (2 marks)
For a certain population the probability of a person being born with the specific gene SPGE1 is `3/5`. The probability of a person having this gene is independent of any other person in the population having this gene. --- 5 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) ---
A car manufacturer is reviewing the performance of its car model X. It is known that at any given six-month service, the probability of model X requiring an oil change is `17/20`, the probability of model X requiring an air filter change is `3/20` and the probability of model X requiring both is `1/20`.
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| a. |  |
| `text(Pr)(F ∩ O′)` | `= text(Pr)(F) – text(Pr)(F∩ O)` | |
| `= 3/20 – 1/20` | ||
| `= 1/10` |
| b. |  |
| `text(Pr)(F ∩ O′)` | `= n/(m + n) – 1/(m + n)` |
| `1/20` | `= (n – 1)/(m + n)` |
| `m + n` | `= 20n – 20` |
| `m` | `= 19n – 20` |
A car manufacturer is reviewing the performance of its car model X. It is known that at any given six-month service, the probability of model X requiring an oil change is `17/20`, the probability of model X requiring an air filter change is `3/20` and the probability of model X requiring both is `1/20`. --- 5 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) ---
Show Worked Solution
a.
`text(Pr)(F ∩ O′)`
`= text(Pr)(F) – text(Pr)(F∩ O)`
`= 3/20 – 1/20`
`= 1/10`
b.
`text(Pr)(F ∩ O′)`
`= n/(m + n) – 1/(m + n)`
`1/20`
`= (n – 1)/(m + n)`
`m + n`
`= 20n – 20`
`m`
`= 19n – 20`
Solve the equation `2 log_2(x + 5) - log_2(x + 9) = 1`. (3 marks)
Shown below is part of the graph of a period of the function of the form `y = tan(ax + b)`.
Find the value of `a` and the value of `b`, where `a > 0` and `0 < b < 1`. (3 marks)
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Shown below is part of the graph of a period of the function of the form `y = tan(ax + b)`.
The graph is continuous for `x = ∈ [-1, 1]`.
Find the value of `a` and the value of `b`, where `a > 0` and `0 < b < 1`. (3 marks)
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A study on the mobile phone usage of NSW high school students is to be conducted.
Data is to be gathered using a questionnaire.
The questionnaire begins with the three questions shown.
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Prove that `log_3 7` is irrational. (2 marks)
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A light inextensible string hangs over a frictionless pulley connecting masses of 3 kg and 7 kg, as shown below.
| a. |  |
| b. | `sum F = 7g – 3g = 4g` | ` = (7 + 3)a` |
| `a = (4g)/10= (2g)/5` | ||
| `sum F_text(on 3kg) = T – 3g` | `= 3a` | |
| `T – 3g` | `= (6g)/5` | |
| `T` | `= 3g + (6g)/5` | |
| `= (21g)/5\ \ text(N)` |
Particles of mass 3 kg and `m` kg are attached to the ends of a light inextensible string that passes over a smooth pulley, as shown.
If the acceleration of the 3 kg mass is `4.9\ text(m/s)^2` upwards, then
A. `m` = 4.5
B. `m` = 6.0
C. `m` = 9.0
D. `m` = 13.5
E. `m` = 18.0
| `sum F` | `=\ text(total mass × acceleration)` |
| `mg – 3g` | `= (m + 3) xx 4.9` |
| `9.8m-3xx9.8` | `= 4.9m + 14.7` |
| `:. m` | `= 9` |
`=> C`
A light inextensible string passes over a smooth pulley, as shown below, with particles of mass 1 kg and `m` kg attached to the ends of the string.
If the acceleration of the 1 kg particle is 4.9 `text(ms)^(-2)` upwards, then `m` is equal to
`T-(9.8 xx 1) = 4.9 xx 1`
`T=14.7`
| `m xx 9.8-T` | `=m xx 4.9` |
| `4.9m` | `= 14.7` |
| `:. m` | `= 14.7/4.9` |
| `= 3` |
`=> C`
Particles of mass 3 kg and 5 kg are attached to the ends of a light inextensible string that passes over a fixed smooth pulley, as shown above. The system is released from rest.
Assuming the system remains connected, the speed of the 5 kg mass after two seconds is
A. 4.0 m/s
B. 4.9 m/s
C. 9.8 m/s
D. 19.6 m/s
Find all solutions for `z`, in exponential form, given `z^4 = -2 sqrt3 - 2 i`. (3 marks)
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`text{Convert} \ \ z^4\ \ text{to Mod/Arg:}`
`| z^4 | = sqrt{(2 sqrt3)^2 + 2^2} = 4`
| `tan \ theta` | `= frac{2 sqrt3}{2} = sqrt3` |
| `theta` | `= frac{pi}{3}` |
`text{arg} (z^4) = – (frac{pi}{2} + frac{pi}{3}) = – frac{5 pi}{6}`
`text{By De Moivre:}`
`| z | = root4 (4) = sqrt2`
`text{arg}(z) = -frac{5 pi}{24} + frac{ k pi}{2} \ , \ k = 0 , ± 1 , ± 2`
`k = 0:\ text{arg} (z)=-frac{5 pi}{24}`
`k = 1:\ text{arg} (z)= -frac{5 pi}{24} + frac{pi}{2} = frac{7 pi}{24}`
`k = – 1:\ text{arg} (z)= -frac{5 pi}{24} – frac{pi}{2} = frac{-17 pi}{24}`
`k = 2:\ text{arg} (z)= -frac{5 pi}{24} + pi = frac{19 pi}{24}`
`therefore \ text{Solutions are:} \ sqrt2 e^(- frac{i 17 pi}{24}) \ , \ sqrt2 e^(frac{ -i 5 pi}{24}) \ , \ sqrt2 e^(frac{i 7 pi}{24}) \ , \ sqrt2 e^(frac{i 19 pi}{24})`
If `(x + iy)^3 = e^( - frac{i pi}{2}),\ \ x, y ∈ R`, find a solution in the form `x + i y, \ x, y ≠ 0`. (2 marks)
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Let `z = sqrt3 - 3 i`
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| i. | `z` | `= sqrt3 – 3 i` |
| `|z|` | `= sqrt((sqrt3)^2 + 3^2) = 2 sqrt3` |
| `tan theta` | `= frac{3}{sqrt3}=sqrt3` |
| `theta` | `= frac{pi}{3}` |
| `text{arg} (z)` | `= – frac{pi}{3}` |
`therefore z = 2 sqrt3 \ text{cis} (frac{-pi}{3})`
ii. `z^n + (overset_z)^n = 0`
`[2 sqrt3 \ cos (frac{-pi}{3}) + i sin (frac{-pi}{3})]^n + [ 2 sqrt3 \ cos (frac{-pi}{3}) – i sin (frac{-pi}{3}) ]^n = 0`
`(2 sqrt3)^n [cos (frac{-n pi}{3}) + i sin (frac{-n pi}{3}) + cos (frac{-n pi}{3}) – i sin (frac{-n pi}{3}) = 0`
| `2 \ cos (frac{-n pi}{3})` | `= 0` |
| `cos (frac{n pi}{3})` | `= 0` |
| `frac{n pi}{3}` | `= frac{pi}{2} + k pi \ , \ k = 0, ± 1, ± 2, …` |
| `frac{n}{3}` | `= frac{(2k + 1)}{2}` |
| `n` | `= frac{3 (2k + 1)}{2}` |
`text{Numerator will always be odd ⇒ no solution exists}`
Sketch the region in the complex plane where the inequalities
`| z + overset_z | ≤ 1` and `| z - i | ≤ 1`
hold simultaneously. (3 marks)
`| z + overset_z | ≤ 1 `
| `| x + i y + x – iy |` | `≤ 1` |
| `| 2x |` | `≤ 1` |
| `| x |` | `≤ frac{1}{2}` |
`| z – i | ≤ 1 \ => \ text{Circle, radius = 1 , centre (0, 1)`