Find the equation of the tangent to the curve `y = x^2` at the point where `x = 3`. (2 marks)
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Find the equation of the tangent to the curve `y = x^2` at the point where `x = 3`. (2 marks)
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`6x – y – 9 = 0 `
`y` | `=x^2` |
`dy/dx` | `=2x` |
`text{Need to find equation with m = 6, through (3,9) }`
`text(When) \ x = 3, y = 9, dy/dx = 6`
`text(Using)\ \ \ y-y_1` | `= m (x-x_1)` |
`y – 9` | `= 6(x – 3)` |
`y – 9` | `= 6x – 18` |
`6x – y – 9` | `=0` |
`:.\ text( Equation of the tangent is 6x – y – 9 = 0)`
Which of the following is equal to `1/(2sqrt5\ - sqrt3)`?
(A) `(2sqrt5\ - sqrt3)/7`
(B) `(2sqrt5 + sqrt3)/7`
(C) `(2sqrt5\ - sqrt3)/17`
(D) `(2sqrt5 + sqrt3)/17`
`D`
`1/(2sqrt5\ – sqrt3) xx (2sqrt5 + sqrt3)/(2sqrt5 + sqrt3)`
`= (2sqrt5 + sqrt3)/( (2sqrt5\ – sqrt3)(2sqrt5 + sqrt3) )`
`= (2sqrt5 + sqrt3)/{(2sqrt5)^2 – (sqrt3)^2)`
`= (2sqrt5 + sqrt3)/17`
`=> D`
What is `4.097 84` correct to three significant figures?
(A) `4.09`
(B) `4.10`
(C) `4.097`
(D) `4.098`
`B`
`4.10`
`=> B`
The points `A(–2, –1)`, `B(–2, 24)`, `C(22, 42)` and `D(22, 17)` form a parallelogram as shown. The point `E(18, 39)` lies on `BC`. The point `F` is the midpoint of `AD`.
(i) | `text(Need to find the equation of)\ AD` |
`m_(AD)` | `= (y_2\ – y_1)/(x_2\ – x_1)` |
`= (17+1)/(22+2)` | |
`= 18/24` | |
`= 3/4` |
`text(Equation of)\ AD\ text(has)\ m=3/4 text(, through)\ A text{(–2,–1)}`
`text(Using)\ y\ – y_1` | `= m (x\ – x_1)` |
`y+1` | `=3/4 (x +2)` |
`4y+4` | `=3x+6` |
`3x-4y+2` | `=0` |
(ii) | `B(–2,24)` |
`AD\ text(is)\ \ 3x -4y+2 =0` |
`_|_ text(dist)` | `= | (ax_1 + by_1 + c)/sqrt(a^2 + b^2) |` |
`= | (3(–2)\ – 4(24) + 2)/sqrt(3^2 + (–4)^2 )|` | |
`= | (-6\ -96 +2)/sqrt25 |` | |
`= | -100/5 |` | |
`= 20\ text(units)` |
`:.\ _|_ text(dist of)\ B\ text(from)\ AD = 20\ text(units … as required)`
(iii) | `E(18,39)\ \ \ C(22,42)` |
`EC` | `=sqrt ( (x_2\ – x_1)^2 + (y_2\ – y_1)^2 )` |
`= sqrt ( (22\ – 18)^2 + (42\ – 39)^2 )` | |
`= sqrt (4^2 + 3^2)` | |
`= sqrt25` | |
`= 5\ text(units)` |
`:.\ text(The length of)\ EC\ text(is 5 units.)`
(iv) | `text(Area of trapezium)` | `=1/2 h (a+b)` |
`=1/2 h (EC + FD)` |
`EC = 5\ text(units)`
`text(Need to find)\ FD`
`text(S)text(ince)\ FD = 1/2 xx AD\ \ \ ( F\ text(is midpoint) )`
`A (–2,–1)\ \ D(22,17)`
`AD` | `=sqrt( (22 + 2) + (17 + 1)^2 )` |
`= sqrt (24^2 + 18^2)` | |
`= sqrt (900)` | |
`= 30` |
`=> FD = 1/2 xx 30 = 15`
`text(S)text(ince)\ ABCD\ text(is a parallelogram)`
`h` | `= _|_ text(distance in part)\ text{(i)}` |
`= 20` |
`:.\ text(Area of trapezium)` | `=1/2 xx 20 (5 + 15)` |
`=200\ text(u²)` |
The cubic `y = ax^3 + bx^2 + cx + d` has a point of inflection at `x = p`.
Show that `p= - b/(3a)`. (2 marks)
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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Show)\ \ p= – b/(3a)`
`y` | `=ax^3 + bx^2 + cx + d` |
`y prime` | `=3ax^2 + 2bx + c` |
`y″` | `=6ax + 2b` |
`text(Given P.I. occurs when)\ \ x = p`
`=> y″=0\ \ text(when)\ \ x=p`
`:.\ 6ap + 2b` | `=0` |
`6ap` | `=-2b` |
`p` | `= -(2b)/(6a)` |
`=-b/(3a)\ \ \ text(… as required)` |
Differentiate `(sinx -1)^8`. (2 marks)
`8cosx (sinx -1)^7`
`y= (sinx – 1)^8`
`dy/dx` | `=8 (sinx -1)^7 xx d/dx (sinx -1)` |
`=8 (sinx – 1)^7 xx cosx` | |
`=8cosx (sinx – 1)^7` |
Evaluate `ln3` correct to three significant figures. (1 mark)
`1.10\ \ \ text{(to 3 sig. figures)}`
`ln3` | `=1.09861…` |
`=1.10\ \ \ text{(to 3 sig. figures)}` |
What are the solutions of `2x^2-5x-1 = 0`?
`D`
`2x^2-5x – 1 = 0`
`text(Using)\ x = (-b +- sqrt( b^2-4ac) )/(2a)`
`x` | `= (5 +- sqrt{\ \ (-5)^2-4 xx 2 xx(-1) })/ (2 xx 2)` |
`= (5 +- sqrt(25 + 8) )/4` | |
`= (5 +- sqrt(33) )/4` |
`=> D`
Warrick has a net income of $590 per week. He has created a budget to help manage his money.
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How much money has he put aside altogether to pay the bill? (1 mark)
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How could Warrick reallocate non-essential funds in his budget so he has enough money to pay the bill? Justify your answer with suitable reasons and calculations. (3 marks)
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`text(Clothes and Gifts to pay the bill.)`
i. | `X` | `= 590 ` | `- (175 + 45 + 10 + 15 + 90` |
`+ 40 + 30 + 70 + 50 + 40)` | |||
`= 590 – 565` | |||
`= $25` |
ii. | `text(Weekly Amount)` | `= 40 + 40` |
`= 80` |
`text{Total (3 weeks)}` | `= 3 xx 80` |
`= 240` |
`:.\ text(Warwick has put aside $240 to pay the bill.)`
(iii) | `text(Amount required less amount put aside)` |
`= 620\ – 240` | |
`= $380` |
`text(Extra 2 weeks of savings and telephone)`
`= 2 xx (40 + 40)`
`= $160`
`:.\ text(Funds to be reallocated)`
`= 380\ – 160`
`= 220\ text(over 2 weeks)`
`= $110\ text(per week)`
`text(Non essential items are Entertainment)`
`text(and Clothes and Gifts)`
`text(Amount)` | `= 70 + 50` |
`= $120\ text(per week)` |
`:.\ text(Warwick could reallocate funds for Entertainment)`
`text(and Clothes and Gifts to pay the bill.)`
Part of the floor plan of a house is shown. The plan is drawn to scale.
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i. | `900\ text(mm)` |
ii. | `2000\ text(mm) xx 2000\ text(mm)` |
iii. | `text(Length of Rumpus Room) = AB` |
`AB` | `= 3600 + 90 + 2000 + 90 + 3915` |
`=9695\ text(mm)` |
The two spinners shown are used in a game.
Each arrow is spun once. The score is the total of the two numbers shown by the arrows.
A table is drawn up to show all scores that can be obtained in this game.
⇒ Win `$12` for a score of `4`
⇒ Win nothing for a score of less than `4`
⇒ Lose `$3` for a score of more than `4`
It costs `$5` to play this game. Will Elise expect a gain or a loss and how much will it be?
Justify your answer with suitable calculations. (3 marks)
(i) `X=3+2=5`
(ii) `P(text{score}<4)=6/12=1/2`
(iii) `P(3)=2/3`
(iv) `P(4)=4/12=1/3`
`P(text{score}<4)` | `=6/12=1/2` |
`P(text{score}>4)` | `=2/12=1/6` |
`text(Financial Expectation)`
`=(1/3xx12)+(1/2xx0)-(1/6xx3)-5` |
`=4-0.5-5` |
`=-1.50` |
`:.\ text(Elise should expect a loss of $1.50) `
At another school, students who use mobile phones were surveyed. The set of data is shown in the table.
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Ten new male students are surveyed and all ten are on a plan. The set of data is updated to include this information.
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i. `text(# Students surveyed)=319+261=580`
ii. `Ptext{(Female uses prepaid)}=text(# Females on prepaid)/text(Total females)`
`=172/319` | |
`=0.53918…` | |
`=\ text{54% (nearest %)}` |
iii. `text(% Males on plan)` | `=text(# Males on plan + 10)/text(Total males + 10)` |
`=(103+10)/(261+10)` | |
`=113/271` | |
`=0.4169…` | |
`=\ text{42% (nearest %)}` |
The graph below displays data collected at a school on the number of students
in each Year group, who own a mobile phone.
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Which student is more likely to own a mobile phone?
Justify your answer with suitable calculations. (2 marks)
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i. `text(Year 12 (100%))`
ii. | `text(% Ownership in Year 9)` | `=55/70` |
`=\ text{78.6% (1d.p.)}` | ||
`text(% Ownership in Year 10)` | `=50/60` | |
`=\ text{83.3% (1d.p.)}` |
`:.\ text(The Year 10 student is more likely to own a mobile phone.)`
iii. `text(% Ownership increases as students)`
`text(progress from Year 7 to Year 12.)`
The graph shows tax payable against taxable income, in thousands of dollars.
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i. |
`text(Income on)\ $21\ 000=$3000\ \ \ text{(from graph)}`
ii. `text(Using the points)\ (21,3)\ text(and)\ (39,9)`
`text(Gradient at)\ A` | `= (y_2\ – y_1)/(x_2\ – x_1)` |
`= (9000-3000)/(39\ 000 -21\ 000)` | |
`= 6000/(18\ 000)` | |
`= 1/3\ \ \ \ \ text(… as required)` |
iii. `text(The gradient represents the tax applicable to each dollar)`
`text(Tax)` | ` = 1/3\ text(of each dollar earned)` |
` = 33 1/3\ text(cents per dollar earned)` |
iv. `text( Tax payable up to $21 000 = $3000)`
`text(Tax payable on income between $21 000 and $39 000)`
` = 1/3 (I\ – 21\ 000)`
`:.\ text(Tax payable on)\ \ I` | `= 3000 + 1/3 (I\ – 21\ 000)` |
`= 3000 + 1/3 I\ – 7000` | |
`= 1/3 I\ – 4000` |
A new shopping centre has opened near a primary school. A survey is conducted to determine the number of motor vehicles that pass the school each afternoon between 2.30 pm and 4.00 pm.
The results for 60 days have been recorded in the table and are displayed in the cumulative frequency histogram.
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What problem could arise from the change in the median number of motor vehicles passing the school before and after the opening of the new shopping centre?
Briefly recommend a solution to this problem. (2 marks)
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i. | `X` | `= 25\ -10` |
`= 15` |
ii. |
iii. `text(Median)\ ~~155`
iv. | `text(Problems)` |
`text(- increased traffic delays)` | |
`text(- increased danger to students leaving school)` | |
`text(Solutions)` | |
`text(- signpost alternative routes around school)` | |
`text(- decrease the speed limit in the area)` |
Wind turbines, such as those shown, are used to generate power.
In theory, the power that could be generated by a wind turbine is modelled using the equation
`T = 20\ 000w^3`
where | `T` is the theoretical power generated, in watts |
`w` is the speed of the wind, in metres per second. |
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In practice, the actual power generated by a wind turbine is only 40% of the theoretical power.
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The graph shows both the theoretical power generated and the actual power generated by a particular wind turbine.
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A particular farm requires at least 4.4 million watts of actual power in order to be self-sufficient.
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A more accurate formula to calculate the power (`P`) generated by a wind turbine is
`P = 0.61 xx pi xx r^2 × w^3`
where | `r` is the length of each blade, in metres |
`w` is the speed of the wind, in metres per second. |
Each blade of a particular wind turbine has a length of 43 metres.The turbine operates at a wind speed of 8 m/s.
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i. | `T=20\ 000w^3` |
`text(If)\ \ w = 7.3` |
`T` | `=20\ 000 xx (7.3)^3` |
`= 7\ 780\ 340\ text(watts)` |
ii. | `text(We know)\ A = 40% xx T` |
`=> A` | `=0.4 xx 20\ 000 xx w^3` |
`=8000w^3` |
iii. | `text(Solution 1)` |
`text(At)\ w=9` | |
`A = text(5.8 million watts)\ \ \ text{(from graph)}` | |
`T = text(14.6 million watts)\ \ \ text{(from graph)}` |
`text(Difference)` | `= text(14.6 million)\ – text(5.8 million)` |
`= text(8.8 million watts)` |
`text(Alternative Solution)` |
`text(At)\ w=9` |
`T` | `= 20\ 000 xx 9^3` |
`= 14\ 580\ 000\ text(watts)` |
`A` | `= 8000 xx 9^3` |
`= 5\ 832\ 000\ text(watts)` |
`text(Difference)` | `=14\ 580\ 000\ – 5\ 832\ 000` |
`=8\ 748\ 000\ \ text(watts)` |
iv. | `text(Find)\ w\ text(if)\ A=4.4\ text(million)` |
`8000w^3` | `= 4\ 400\ 000` |
`w^3` | `= (4\ 400\ 000)/8000` |
`= 550` |
`:. w` | `= root(3)(550)` |
`=8.1932…` | |
`=8.2\ text(m/s)\ \ \ text{(1 d.p.)}` |
`:.\ text(The minimum wind speed required is 8.2 m/s)`
v. | `text(Find)\ P\ text(when)\ w=8\ text(and)\ r=43` |
`P` | `= 0.61 xx pi xx r^2 xx w^3` |
`= 0.61 xx pi xx 43^2 xx 8^3` | |
`= 1\ 814\ 205.92\ text(watts)` |
`text(When speed of wind)\ uarr10%` |
`w′ = 8 xx 110text(%) = 8.8\ text(m/s)` |
`text(Find)\ P\ text(when)\ w′ = 8.8`
`P` | `=0.61 xx pi xx 43^2 xx 8.8^3` |
`= 2\ 414\ 708.08\ text(watts)` |
`text(Increase in Power)` | `=2\ 414\ 708.08\ – 1\ 814\ 205.92` |
`= 600\ 502.16` |
`:.\ text(% Power increase)` | `= (600\ 502.16)/(1\ 814\ 205.92)` |
`= 0.331` | |
`= text(33%)\ \ \ text{(nearest %)}` |
Sri has a gross salary of $56 350. She has tax deductions of $350 for union fees, $2000 in work-related expenses and $250 in donations to charities.
The Medicare levy is 1.5% of her taxable income.
Calculate Sri’s Medicare levy. (3 marks)
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`$806.25`
`text(Deductions)` | `=350+2000+250` |
`=$2600` | |
`text(Taxable)` | `=\ text(Gross Income – deduct)` |
`=56\ 350-2600` | |
`=$53\ 750` | |
`text(Medicare Levy)` | `=\ text(1.5%)\ xx 53\ 750` |
`=$806.25` |
`:.\ text(Sri’s Medicare levy is)\ \ $806.25`
An unbiased coin is tossed 10 times.
A tail is obtained on each of the first 9 tosses.
What is the probability that a tail is obtained on the 10th toss?
`B`
`text(Each toss is an independent event and has an even chance)`
`text(of being a head or tail.)`
`=> B`
On Saturday, Jonty recorded the colour of T-shirts worn by the people at his gym. The results are shown in the graph.
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i. `text(# People)` | `=5+15+10+3+1` |
`=34` |
ii. `P (B\ text{or}\ G)` | `=P(B)+P(G)` |
`=5/34+10/34` | |
`=15/34` |
This advertisement appeared in a newspaper
What is the maximum possible salary per annum for this civil engineer, correct to the nearest dollar? (2 marks)
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`$86\ 396`
`text(Base wage)=1586.70\ text(pw)`
`text(Max weekly base)` | `=$1586.70+(text{3.5%}\ xx 1586.70)` |
`=1586.70+55.53` | |
`=1642.23\ text((nearest cent))` |
`:.\ text(Max annual base)` | `=1642.23xx52` |
`=85\ 395.96` | |
`=$85\ 396\ text{(nearest dollar)}` |
Jay bought a computer for $3600. His friend Julie said that all computers are worth nothing (i.e. the value is $0) after 3 years.
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i. | `S` | `= V_0 – Dn` |
`0` | `= 3600 – D xx 3` | |
`3D` | `= 3600` | |
`D` | `= 3600/3` | |
`= 1200` |
`:.\ text(Annual depreciation = $1200`
ii | `text(Using)\ \ S = V_0 (1-r)^n` | |
`text(where)\ r = text(30%)\ \ text(and)\ \ V_0 = 3600` |
`S` | `=3600 (1 – 30/100)^n` | |
`= 3600 (0.7)^n` |
`(0.7)^n > 0\ text(for all)\ n`
`:.\ text(Salvage value is always)\ >0`
Tayvan is an international company that reports its profits in the USA, Belgium and India at the end of each quarter. The profits for 2008 are shown in the area chart.
(i) | `$8\ 000\ 000\ \ \ text{(from graph)}` |
(ii) | `text{Belgium profit (30 March)}` |
`= $5\ 000\ 000\ – $1\ 000\ 000` | |
`= $4\ 000\ 000` |
The diagram below shows a stem-and-leaf plot for 22 scores.
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i. `text(Mode) = 78`
ii. `22\ text(scores)`
`=>\ text(Median is the average of 11th and 12th scores)`
`:.\ text(Median)` | `= (45 + 47)/2` |
`= 46` |
Billy worked for 35 hours at the normal hourly rate of pay and for five hours at double time. He earned $561.60 in total for this work.
What was the normal hourly rate of pay?
`B`
`text(Let hourly rate) = $X\ text(per hour)`
`35X + 5(2X)` | `= 561.60` |
`45X` | `=561.60` |
`X` | `= 561.60/45` |
`= $12.48` |
`=> B`
A house was purchased in 1984 for $35 000. Assume that the value of the house has increased by 3% per annum since then.
Which expression gives the value of the house in 2009?
`A`
`r =\ text(3%)\ = 0.03`
`n = 25\ text(years)`
`text(Using)\ \ FV = PV(1 + r)^n`
` :.\ text(Value in 2009) = 35\ 000(1+0.03)^25`
`=> A`
Jamie wants to know how many songs were downloaded legally from the internet in the last 12 months by people aged 18–25 years. He has decided to conduct a statistical inquiry.
After he collects the data, which of the following shows the best order for the steps he should take with the data to complete his inquiry?
(A) Display, organise, conclude, analyse
(B) Organise, display, conclude, analyse
(C) Display, organise, analyse, conclude
(D) Organise, display, analyse, conclude
`D`
`text(Process of statistical enquiry requirements)`
`=> D`
The eye colours of a sample of children were recorded.
When analysing this data, which of the following could be found?
`C`
`text(Eye colour is categorical data)`
`:.\ text(Only the mode can be found)`
`=> C`
The step graph shows the charges for a carpark.
Maria enters the carpark at 10:10 am and exits at 1:30 pm.
How much will she pay in charges?
(A) `$6`
(B) `$12`
(C) `$18`
(D) `$24`
`C`
`text(Time in carpark = 3h 20m)`
`text(From graph, charge will be $18)`
`=> C`
A newspaper states: ‘It will most probably rain tomorrow.’
Which of the following best represents the probability of an event that will most probably occur?
(A) `33 1/3 text(%)`
(B) `text(50%)`
(C) `text(80%)`
(D) `text(100%)`
`C`
`text(Probably) =>\ text(likelihood > 50%)`
`text(However 100% = certainty)`
`:.\ text(80% is the answer)`
`=> C`
Two trees on level ground, 12 metres apart, are joined by a cable. It is attached 2 metres above the ground to one tree and 11 metres above the ground to the other.
What is the length of the cable between the two trees, correct to the nearest metre?
`C`
`text(Using Pythagoras)`
`c^2` | `=12^2+9^2` |
`=144+81` | |
`=225` | |
`:.c` | `=15,\ \ c>0` |
`=>C`
The letters A, B and C are used to make a three-letter company name. Each letter is used only once.
How many different company names can be made?
(A) 3
(B) 6
(C) 9
(D) 27
`B`
`text(# Outcomes)` | `=3xx2xx1` |
`=6` |
`=>B`
The dot plot shows the number of push-ups that 13 members of a fitness class can do in one minute.
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Does the addition of this new member to the class change the probability calculated in part (i)? Justify your answer. (1 mark)
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i. `P` | `= text(# Members > 38 push-ups)/text(Total members)` |
`= 7/13` |
ii. `text(Yes.)`
`Ptext{(+ New member)}` | `= text(Members > 38 push-ups)/text(Total members)` |
`= 7/14≠ 7/13` |
Jim buys a photocopier for $22 000.
Its value is depreciated using the declining balance method at the rate of 15% per annum.
What is its value at the end of 3 years? (2 marks)
`$13\ 510.75`
`S` | `= V_0 (1-r)^n` |
`= 22\ 000 (1-0.15)^3` | |
`= 22\ 000 (0.85)^3` | |
`= 13\ 510.75` |
`:.\ text(After 3 years, it is worth)\ $13\ 510.75`
Handmade chocolates are checked for size and shape. Every 30th chocolate is sampled.
Which term best describes this type of sampling?
(A) Census
(B) Random
(C) Stratified
(D) Systematic
`D`
`text(Systematic Sampling)`
`=> D`
In an experiment, a standard six-sided die was rolled 72 times. The results are shown in the table.
Which number on the die was obtained the expected number of times?
(A) 1
(B) 2
(C) 3
(D) 6
`B`
`text(Probability of rolling a specific number)=1/6`
`:.\ text(After 72 rolls, a specific number is expected)`
`1/6xx72=12\ text(times.)`
`=>\ B`
The quadratic equation `x^2-6x+2=0` has roots `alpha` and `beta`.
(i) `alpha+beta=-b/a=6`
(ii) `alpha beta=c/a=2`
(iii) `1/alpha+1/beta=(alpha+beta)/(alpha beta)=3`
The quadratic equation `x^2+3x -1=0` has roots `alpha` and `beta`.
What is the value of `alpha beta+(alpha+beta)` ?
(A) `4`
(B) `2`
(C) `-4`
(D) `-2`
`C`
`alpha beta+(alpha+beta)` | `=c/a+(- b/a)` |
`=-1-3` | |
`=-4` |
An oil rig, `S`, is 3 km offshore. A power station, `P`, is on the shore. A cable is to be laid from `P` to `S`. It costs $1000 per kilometre to lay the cable along the shore and $2600 per kilometre to lay the cable underwater from the shore to `S`.
The point `R` is the point on the shore closest to `S`, and the distance `PR` is 5 km.
The point `Q` is on the shore, at a distance of `x` km from `R`, as shown in the diagram.
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Determine the path for laying the cable in order to minimise the cost in this case. (2 marks)
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i. `text(C)text(ost)` | `=(PRxx1000)+(SRxx2600)` |
`=(5xx1000)+(3xx2600)` | |
`=12\ 800` |
`:. text(C)text(ost is)\ $12\ 800`
ii. `text(C)text(ost)=PSxx2600`
`text(Using Pythagoras:)`
`PS^2` | `=PR^2+SR^2` |
`=5^2+3^2` | |
`=34` | |
`PS` | `=sqrt34` |
`:.\ text(C)text(ost)` | `=sqrt34xx2600` |
`=15\ 160.474…` | |
`=$15\ 160\ \ text{(nearest dollar)}` |
iii. `text(Show)\ \ C=1000(5-x+2.6sqrt(x^2+9))`
`text(C)text(ost)=(PQxx1000)+(QSxx2600)`
`PQ` | `=5-x` |
`QS^2` | `=QR^2+SR^2` |
`=x^2+3^2` | |
`QS` | `=sqrt(x^2+9)` |
`:.C` | `=(5-x)1000+sqrt(x^2+9)\ (2600)` |
`=1000(5-x+2.6sqrt(x^2+9))\ \ text(… as required)` |
iv. `text(Find the MIN cost of laying the cable)`
`C` | `=1000(5-x+2.6sqrt(x^2+9))` |
`(dC)/(dx)` | `=1000(–1+2.6xx1/2xx2x(x^2+9)^(–1/2))` |
`=1000(–1+(2.6x)/(sqrt(x^2+9)))` |
`text(MAX/MIN when)\ (dC)/(dx)=0`
`1000(–1+(2.6x)/(sqrt(x^2+9)))=0`
`(2.6x)/sqrt(x^2+9)` | `=1` |
`2.6x` | `=sqrt(x^2+9)` |
`(2.6)^2x^2` | `=x^2+9` |
`x^2(2.6^2-1)` | `=9` |
`x^2` | `=9/5.76` |
`=1.5625` | |
`x` | `=1.25\ \ \ \ (x>0)` |
`text(If)\ \ x=1,\ \ (dC)/(dx)<0`
`text(If)\ \ x=2,\ \ (dC)/(dx)>0`
`:.\ text(MIN when)\ \ x=1.25`
`C` | `=1000(5-1.25+2.6sqrt(1.25^2+9))` |
`=1000(122)` | |
`=12\ 200` |
`:.\ text(MIN cost is)\ $12\ 200\ text(when)\ x=1.25`
v. `text(Underwater cable now costs $1100 per km)`
`=>\ C` | `=1000(5-x)+1100sqrt(x^2+9)` |
`=1000(5-x+1.1sqrt(x^2+9))` | |
`(dC)/(dx)` | `=1000(–1+1.1xx1/2xx2x(x^2+9)^(-1/2))` |
`=1000(–1+(1.1x)/sqrt(x^2+9))` |
`text(MAX/MIN when)\ (dC)/(dx)=0`
`1000(–1+(1.1x)/sqrt(x^2+9))` | `=0` |
`(1.1x)/sqrt(x^2+9)` | `=1` |
`1.1x` | `=sqrt(x^2+9)` |
`1.1^2x^2` | `=x^2+9` |
`x^2(1.1^2-1)` | `=9` |
`x^2` | `=9/0.21` |
`x` | `~~6.5\ text{km (to 1 d.p.)}` |
`=>\ text(no solution since)\ x<=5` |
`text(If we lay cable)\ PR\ text(then)\ RS`
`=>\ text(C)text(ost)=5xx1100+3xx1000=8500`
`text(If we lay cable directly underwater via)\ PS`
`=>\ text(C)text(ost)=sqrt34xx1100=6414.047…`
`:.\ text{MIN cost is $6414 by cabling directly from}\ P\ text(to)\ S`.
The diagram shows the trajectory of a ball thrown horizontally, at speed `v` m/s, from the top of a tower `h` metres above the ground level.
The ball strikes the ground at an angle of 45°, `d` metres from the base of the tower, as shown in the diagram. The equations describing the trajectory of the ball are
`x=vt` and `y=h-1/2 g t^2`, (DO NOT prove this)
where `g` is the acceleration due to gravity, and `t` is time in seconds.
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i. `text(Show)\ \ y=0\ \ text(when)\ t=sqrt((2h)/g)\ \ text(seconds:)`
`0` | `=h-1/2 g t^2` |
`1/2 g t^2` | `=h` |
`t^2` | `=(2h)/g` |
`:.t` | `=sqrt((2h)/g)\ \ text(seconds,)\ \ t>=0\ \ text(… as required)` |
ii. `text(Show that)\ d=2h`
`x=vt\ \ \ \ =>\ \ \ dotx=v`
`y=h-1/2 g t^2\ \ \ \ =>\ \ \ doty=-g t`
`text(At)\ t=sqrt((2h)/g)`,
`doty` | `=-gxxsqrt((2h)/g)` |
`=-sqrt(2gh)` |
`text(S)text(ince the ball strikes the ground at)\ 45^@,\ text(we know)`
`tan45^@=` | `|\ doty\ |/dotx` |
`1=` | `sqrt(2gh)/v` |
`v=` | `sqrt(2gh)` |
`text(S)text(ince)\ \ x=d=vt\ \ text(when)\ \ t=sqrt((2h)/g)`
`d` | `=sqrt(2gh)xxsqrt((2h)/g)` |
`=sqrt((2h)^2)` | |
`=2h\ \ text( … as required)` |
The graph shows the velocity of a particle, `v` metres per second, as a function of time, `t` seconds.
(i) `text(Find)\ v \ text(when) t=0`
`v=20\ \ text(m/s)`
(ii) `text(Particle comes to rest at)\ t=10\ text{seconds (from graph)}`
(iii) `text(Acceleration is zero when)\ t=6\ text{seconds (from graph)}`
(iv) |
`text(Area)` | `~~h/3[y_0+y_n+4text{(odds)}+2text{(evens)}]` |
`~~h/3[y_0+y_4+4(y_1+y_3)+2(y_2)]` | |
`~~2/3[20+60+4(50+80)+2(70)]` | |
`~~2/3[740]` | |
`~~493 1/3` |
`:.\ text{Distance travelled is 493 1/3 m (approx.)}`
The acceleration of a particle is given by
`a=8e^(-2t)+3e^(-t)`,
where `x` is the displacement in metres and `t` is the time in seconds.
Initially its velocity is `text(– 6 ms)^(–1)` and its displacement is 5 m.
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i. `text(Show)\ \ x=2e^(-2t)+3e^-t+t`
`a=8e^(-2t)+3e^-t\ \ text{(given)}`
`v=int a\ dt=-4e^(-2t)-3e^-t+c_1`
`text(When)\ t=0, v=-6\ \ text{(given)}`
`-6` | `=-4e^0-3e^0+c_1` |
`-6` | `=-7+c_1` |
`c_1` | `=1` |
`:. v=-4e^(-2t)-3e^-t+1`
`x` | `=int v\ dt` |
`=int(-4e^(-2t)-3e^-t+1)\ dt` | |
`=2e^(-2t)+3e^-t+t+c_2` |
`text(When)\ \ t=0,\ x=5\ \ text{(given)}`
`5` | `=2e^0+3e^0+c_2` |
`c_2` | `=0` |
`:.\ x=2e^(-2t)+3e^-t+t\ \ text(… as required)`
ii. `text(Particle comes to rest when)\ \ v=0`
`text(i.e.)\ \ -4e^(-2t)-3e^-t+1=0`
`text(Let)\ X=e^-t\ \ \ \ =>X^2=e^(-2t)`
`-4X^2-3X+1` | `=0` |
`4X^2+3X-1` | `=0` |
`(4X-1)(X+1)` | `=0` |
`:.\ \ X=1/4\ \ text(or)\ \ X=-1`
`text(When)\ \ X=1/4:`
`e^-t` | `=1/4` |
`lne^-t` | `=ln(1/4)` |
`-t` | `=ln(1/4)` |
`t` | `=-ln(1/4)=ln(1/4)^-1=ln4` |
`text(When)\ \ X=-1:`
`e^-t=-1\ \ text{(no solution)}`
`:.\ text(The particle comes to rest when)\ t=ln4\ text(seconds)`
iii. `text(Find)\ \ x\ \ text(when)\ \ t=ln4 :`
`x=2e^(-2t)+3e^-t+t`
`\ \ =2e^(-2ln4)+3e^-ln4+ln4`
`\ \ =2(e^ln4)^-2+3(e^ln4)^-1+ln4`
`\ \ =2xx4^-2+3xx4^-1+ln4`
`\ \ =2/16+3/4+ln4`
`\ \ =7/8+ln4`
The acceleration of a particle is given by
`ddotx=4cos2t`,
where `x` is the displacement in metres and `t` is the time in seconds.
Initially the particle is at the origin with a velocity of `text(1 ms)^(–1)`.
`dotx=2sin2t+1`. (2 marks)
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i. `text(Show)\ dotx` | `=2sin2t+1` |
`dotx` | `=intddotx\ dt` |
`=int4cos2t\ dt` | |
`=2sin2t+c` |
`text(When)\ t=0, \ \ dotx=1\ \ text{(given)}`
`1=2sin0+c`
`c=1`
`:. dotx=2sin2t+1 \ \ \ text(… as required)`
ii. `text(Find)\ t\ text(when)\ dotx=0 :`
`2sin2t+1` | `=0` |
`sin2t` | `=-1/2` |
`=>sin theta=1/2\ text(when)\ theta=pi/6`
`text(S)text(ince)\ \ sin theta\ \ text(is negative in 3rd and 4th quadrants)`
`2t` | `=pi + pi/6` |
`2t` | `=(7pi)/6` |
`t` | `=(7pi)/12` |
`:.\ text(Particle first comes to rest at)\ t=(7pi)/12\ text(seconds)`
iii. `x` | `=intdotx\ dt` |
`=int(2sin2t+1)\ dt` | |
`=t-cos2t+c` |
`text(When)\ t=0,\ x=0\ \ text{(given)}`
`0=0-cos0+c`
`c=1`
`:. x=t-cos2t+1`
The velocity of a particle is given by
`v=1-2cost`,
where `x` is the displacement in metres and `t` is the time in seconds. Initially the particle is 3 m to the right of the origin.
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i. `text(Find)\ \ v\ \ text(when)\ \ t=0`:
`v` | `=1-2cos0` |
`=1-2` | |
`=-1` |
`:.\ text(Initial velocity is)\ -1\ text(m/s.)`
ii. `text(Solution 1)`
`text(Max velocity occurs when)\ \ a=(d v)/(dt)=0`
`a=2sint`
`text(Find)\ \ t\ \ text(when)\ \ a=0 :`
`2sint=0`
`t=0`, `pi`, `2pi`, …
`text(At)\ \ t=0,\ \ v=-1\ text(m/s)`
`text(At)\ \ t=pi,\ \ v=1-2(-1)=3\ text(m/s)`
`:.\ text(Maximum velocity is 3 m/s)`
`text(Solution 2)`
`v=1-2cost`
`text(S)text(ince)\ \ -1` | `<cost<1` |
`-2` | `<2cost<2` |
`-1` | `<1-2cost<3` |
`:.\ text(Maximum velocity is 3 m/s)`
iii. `x` | `=int v\ dt` |
`=int(1-2cost)\ dt` | |
`=t-2sint+c` |
`text(When)\ \ t=0,\ \ x=3\ \ text{(given)}`
`3=0-2sin0+3`
`c=3`
`:. x=t-2sint+3`
iv. `text(Find)\ \ x\ \ text(when)\ \ v=0\ \ text{(first time):}`
`text(When)\ \ v=0 ,`
`0` | `=1-2cost` |
`cost` | `=1/2` |
`t` | `=cos^-1(1/2)` |
`=pi/3\ \ \ text{(first time)}` |
`text(Find)\ \ x\ \ text(when)\ \ t=pi/3 :`
`x` | `=pi/3-2sin(pi/3)+3` |
`=pi/3-2xxsqrt3/2+3` | |
`=pi/3-sqrt3+3\ \ text(units)` |
The mass `M` of a whale is modelled by
`M=36-35.5e^(-kt)`
where `M` is measured in tonnes, `t` is the age of the whale in years and `k` is a positive constant.
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Find the value of `k`, correct to three decimal places. (2 marks)
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i. `M=36-35.5e^(-kt)`
`35.5e^(-kt)=36-M`
`:. (dM)/(dt)` | `=-kxx-35.5e^(-kt)` |
`=kxx35.5e^(-kt)` | |
`=k(36-M)\ \ \ text(… as required)` |
ii. `text(Find)\ \ k`
`text(When)\ \ t=10,\ \ M=20`
`M` | `=36-35.5e^(-kt)` |
`20` | `=36-35.5e^(-10k)` |
`35.5e^(-10k)` | `=16` |
`lne^(-10k)` | `=ln(16/35.5)` |
`-10k` | `=ln(16/35.5)` |
`:. k` | `=-ln(16/35.5)/10` |
`=0.07969…` | |
`=0.080\ \ text{(to 3 d.p.)}` |
iii. `text(As)\ t->oo, e^(-kt)=1/e^(kt)\ ->0,\ \ k>0`
`M->36`
`:.\ text(The whale’s limiting mass is 36 tonnes.)`
A cup of coffee with an initial temperature of 80°C is placed in a room with a constant temperature of 22°C.
The temperature, `T`°C, of the coffee after `t` minutes is given by
`T=A+Be^(-kt)`,
where `A`, `B` and `k` are positive constants. The temperature of the coffee drops to 60°C after 10 minutes.
How long does it take for the temperature of the coffee to drop to 40°C? Give your answer to the nearest minute. (3 marks)
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`text(28 minutes)`
`T=A+Be^(-kt)`
`text(S)text(ince constant room temp is)\ 22°text(C)“=>\ A=22`
`T=22+Be^(-kt)`
`text(At)\ \ t=0,\ T=80`
`80=22+Be^0`
`=>\ B=58`
`:.\ T=22+58e^(-kt)`
`text(At)\ \ t=10,\ T=60`
`60` | `=22+58e^(-10k)` |
`58e^(-10k)` | `=38` |
`e^(-10k)` | `=38/58` |
`lne^(-10k)` | `=ln(38/58)` |
`-10k` | `=ln(38/58)` |
`:.\ k` | `=-1/10ln(38/58)` |
`text(Find)\ \ t\ \ text(when)\ T=40°text(C) :`
`40` | `=22+58e^(-kt)` |
`58e^(-kt)` | `=18` |
`-kt` | `=ln(18/58)` |
`t` | `=-ln(18/58)/k,\ \ text(where) \ k=–1/10ln(38/58)` |
`=10 xx ln(18/58)/ln(38/58)` | |
`=27.6706…` | |
`=28\ text{mins (nearest minute)}` |
`:.\ text(It takes 28 minutes to cool to 40°C.)`
Light intensity is measured in lux. The light intensity at the surface of a lake is 6000 lux. The light intensity, `I` lux, a distance `s` metres below the surface of the lake is given by
`I=Ae^(-ks)`
where `A`, and `k` are constants.
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i. `I=Ae^(-ks)`
`text(Find)\ A,\ text(given)\ I=6000\ text(at)\ s=0`
`6000` | `=Ae^0` |
`:.\ A` | `=6000` |
ii. `text(Find)\ k\ text(given)\ I=1000\ text(at) s=6`
`1000` | `=6000e^(-6xxk)` |
`e^(-6k)` | `=1/6` |
`lne^(-6k)` | `=ln(1/6)` |
`-6k` | `=ln(1/6)` |
`k` | `=- 1/6 ln(1/6)` |
`=0.2986…` | |
`=0.30\ \ \ text{(2 d.p.)}` |
iii. `text(Find)\ \ (dI)/(ds)\ \ text(at)\ \ s=6`
`I` | `=6000e^(-ks)` |
`:.(dI)/(ds)` | `=-6000ke^(-ks)` |
`text(At)\ s=6,`
`(dI)/(ds)` | `=-6000ke^(-6k),\ \ \ text(where)\ k=- 1/6 ln(1/6)` |
`=-298.623…` | |
`=-299\ \ text{(nearest whole number)}` |
`:. text(At)\ s=6,\ text(the light intensity is decreasing)`
`text(at 299 lux per metre.)`
Radium decays at a rate proportional to the amount of radium present. That is, if `Q(t)` is the amount of radium present at time `t`, then `Q=Ae^(-kt)`, where `k` is a positive constant and `A` is the amount present at `t=0`. It takes 1600 years for an amount of radium to reduce by half.
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How many years will it be before the amount of radium reaches the safe level. (2 marks)
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i. `Q=Ae^(-kt)`
`text(When)\ \ t=0,\ \ Q=A`
`text(When)\ \ t=1600,\ \ Q=1/2 A`
`:.1/2 A` | `=A e^(-1600xxk)` |
`e^(-1600xxk)` | `=1/2` |
`lne^(-1600xxk)` | `=ln(1/2)` |
`-1600k` | `=ln(1/2)` |
`k` | `=(-ln(1/2))/1600` |
`=0.0004332\ \ text{(to 4 sig. figures)}` |
ii. `text(Find)\ \ t\ \ text(when)\ \ Q=1/3 A :`
`1/3 A` | `=A e^(-kt)` |
`e^(-kt)` | `=1/3` |
`lne^(-kt)` | `=ln(1/3)` |
`-kt` | `=ln(1/3)` |
`:.t` | `=(-ln(1/3))/k,\ \ \ text(where)\ \ \ k=(-ln(1/2))/1600` |
`=(ln(1/3) xx1600)/ln(1/2)` | |
`=2535.940…` |
`:.\ text(It will take 2536 years.)`
Professor Smith has a colony of bacteria. Initially there are 1000 bacteria. The number of bacteria, `N(t)`, after `t` minutes is given by
`N(t)=1000e^(kt)`.
Show that `k=0.0347` correct to four decimal places. (1 mark)
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i. `N=1000e^(kt)\ text(and given)\ N=2000\ text(when)\ t=20`
`2000` | `=1000e^(20xxk)` |
`e^(20k)` | `=2` |
`lne^(20k)` | `=ln2` |
`20k` | `=ln2` |
`:.k` | `=ln2/20` |
`=0.0347\ \ text{(to 4 d.p.) … as required}` |
ii. `text(Find)\ \ N\ \ text(when)\ t=120`
`N` | `=1000e^(120xx0.0347)` |
`=64\ 328.321..` | |
`=64\ 328\ \ text{(nearest whole number)}` |
`:.\ text(There are)\ 64\ 328\ text(bacteria when t = 120.)`
iii. `text(Find)\ (dN)/(dt)\ text(when)\ t=120`
`(dN)/(dt)` | `=0.0347xx1000e^(0.0347t)` |
`=34.7e^(0.0347t)` |
`text(When)\ \ t=120`
`(dN)/(dt)` | `=34.7e^(0.0347xx120)` |
`=2232.1927…` | |
`=2232\ \ text{(nearest whole)}` |
`:.\ (dN)/(dt)=2232\ text(bacteria per minute at)\ t=120`
iv. `text(Find)\ \ t\ \ text(such that)\ N=100,000`
`=>100\ 000` | `=1000e^(0.0347t)` |
`e^(0.0347t)` | `=100` |
`lne^(0.0347t)` | `=ln100` |
`0.0347t` | `=ln100` |
`t` | `=ln100/0.0347` |
`=132.7138…` | |
`=133\ \ text{(nearest minute)}` |
`:.\ N=100\ 000\ text(when)\ t=133\ text(minutes.)`
Trout and carp are types of fish. A lake contains a number of trout. At a certain time, 10 carp are introduced into the lake and start eating the trout. As a consequence, the number of trout, `N`, decreases according to
`N=375-e^(0.04t)`,
where `t` is the time in months after the carp are introduced.
The population of carp, `P`, increases according to `(dP)/(dt)=0.02P`.
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i. `text(Carp introduced at)\ \ t=0`
`N=375-e^0=374`
`:.\ text(There was 374 trout when carp were introduced.)`
ii. `text(Trout population will be zero when)`
`N` | `=375-e^(0.04t)=0` |
`e^(0.04t)` | `=375` |
`0.04t` | `=ln375` |
`t` | `=ln375/0.04` |
`=148.173 …` | |
`=148\ text{months (nearest month)}` |
`:.\ text(After 148 months, the trout population will be zero.)`
iii. |
iv. `text(We need) \ |(dN)/(dt)|=(dP)/(dt)`
`text(Given)\ N=375-e^(0.04t)`
`(dN)/(dt)=-0.04e^(0.04t)`
`text(Find)\ P\ text(in terms of)\ t`
`text(Given)\ (dP)/(dt)=0.02P`
`=> P=Ae^(0.02t)`
`text(Find)\ A\ \ =>text(when)\ t=0,\ P=10`
`10` | `=Ae^0` |
`:.A` | `=10` |
`=>P` | `=10xx0.02e^(0.02t)` |
`=0.2e^(0.02t)` |
`text(Given that)\ \ (dP)/(dt)=|(dN)/(dt)|`
`0.2e^(0.02t)` | `=0.04e^(0.04t)` |
`5e^(0.02t)` | `=e^(0.04t)` |
`e^(0.04t)/e^(0.02t)` | `=5` |
`e^(0.04t-0.02t)` | `=5` |
`lne^(0.02t)` | `=ln5` |
`0.02t` | `=ln5` |
`t` | `=ln5/0.02` |
`=80.4719…` | |
`=80\ text{months (nearest month)}` |
v. `text(Find)\ t\ text(when)\ N=P`
`text(i.e.)\ \ 375-e^(0.04t)` | `=10e^(0.02t)` |
`e^(0.04t)+10e^(0.02t)-375` | `=0` |
`text(Let)\ X=e^(0.02t),\ text(noting)\ \ X^2=(e^(0.02t))^2=e^(0.04t)`
`:.\ X^2+10X-375` | `=0` |
`(X-15)(X+25)` | `=0` |
`X=15\ \ text(or)\ \ –25`
`text(S)text(ince)\ X=e^(0.02t)`
`e^(0.02t)` | `=15\ \ \ \ (e^(0.02t)>0)` |
`lne^(0.02t)` | `=ln15` |
`0.02t` | `=ln15` |
`t` | `=ln15/0.02` |
`=135.4025…` | |
`=135\ text(months)` |
The velocity of a particle moving along the `x`-axis is given by
`v=8-8e^(-2t)`,
where `t` is the time in seconds and `x` is the displacement in metres.
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Find the value of this constant. (1 mark)
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i. `text(Initial velocity when)\ \ t=0`
`v` | `=8-8e^0` |
`=0\ text(m/s)` | |
`:.\ text(Particle is initially at rest.)` |
ii. `a=d/(dt) (v)=-2xx-8e^(-2t)=16e^(-2t)`
`text(S)text(ince)\ e^(-2t)=1/e^(2t)>0\ text(for all)\ t`.
`=>\ a=16e^(-2t)=16/e^(2t)>0\ text(for all)\ t`.
`:.\ text(Acceleration is positive for all)\ \ t>0`.
iii. `text{S}text{ince the particle is initially at rest, and ALWAYS}`
`text{has a positive acceleration.`
`:.\ text(It moves in a positive direction for all)\ t`.
iv. `text(As)\ t->oo`, `e^(-2t)=1/e^(2t)->0`
`=>8/e^(2t)->0\ text(and)`
`=>v=8-8/e^(2t)->8\ text(m/s)`
`:.\ text(As)\ \ t->oo,\ text(velocity approaches 8 m/s.)`
v. |
The zoom function in a software package multiplies the dimensions of an image by 1.2. In an image, the height of a building is 50 mm. After the zoom function is applied once, the height of the building in the image is 60 mm. After the second application, it is 72 mm.
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i. | `T_1` | `=a=50` |
`T_2` | `=ar^1=50(1.2)=60` | |
`T_3` | `=ar^2=50(1.2)^2=72` |
`=>\ text(GP where)\ \ a=50,\ \ r=1.2`
`\ \ vdots`
`T_9` | `=50(1.2)^8` |
`=214.99` |
`:.\ text{Height will be 215 mm (nearest mm)}`
ii. | `T_n=ar^(n-1)` | `>400` |
`:.\ 50(1.2)^(n-1)` | `>400` | |
`1.2^(n-1)` | `>8` | |
`ln 1.2^(n-1)` | `>ln8` | |
`n-1` | `>ln8/ln1.2` | |
`n` | `>12.405` |
`:.\ text(The height of the building in the 13th image)`
`text(will be higher than 400 mm, which is the 12th)`
`text(time the zoom would be applied.)`
A tree grows from ground level to a height of 1.2 metres in one year. In each subsequent year, it grows `9/10` as much as it did in the previous year.
Find the limiting height of the tree. (2 marks)
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`12\ text(m)`
`a=1.2, \ \ \ r=9/10`
`text(S)text(ince)\ \ |\ r\ |<1,`
`S_oo` | `=a/(1-r)` |
`=1.2/(1-(9/10))` | |
`=12\ text(m)` |
`:.\ text(Limiting height of tree is 12 m.)`
The velocity of a particle moving along the `x`-axis is given by `dotx=10-2t`, where `x` is the displacement from the origin in metres and `t` is the time in seconds. Initially the particle is 5 metres to the right of the origin.
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i. `dotx=10-2t`
`text(Acceleration)=ddotx=d/(dx)dotx=-2`
`:.\ text(The acceleration is constant.)`
ii. `text(Particle comes to rest when)\ dotx=0`
`10-2t=0\ text(when)\ t=5`
`:.\ text(Particle comes to rest after 5 seconds.)`
iii. `text(Show)\ x=26\ text(when)\ t=7`
`x` | `=int dotx\ dt` |
`=int (10-2t)\ dt` | |
`=10t -t^2+c` |
`text(Given)\ t=0\ text(when)\ x=5`
`=> 5` | `=10(0)-0^2+c` |
`c` | `=5` |
`:. x=10t-t^2+5`
`text(At)\ \ t=7`
`x` | `=10(7)-7^2+5` |
`=70-49+5` | |
`=26` |
`:.\ text(The particle is 26 metres to the right of the origin)`
`text(after 7 seconds … as required)`
iv. `text(Find the distance travelled in the first 7 seconds:)`
`text(At)\ t=5`,
`x` | `=10(5)-5^2+5` |
`=50-25+5` | |
`=30\ text(m)` |
`=>\ text{The particle travels 25 m to the right}\ (x=5\ text{to}\ 30)`
`text{then 4 m to the left}\ (x=30\ text{to}\ 26)`
`:.\ text(The total distance travelled in 7 seconds)`
`=25+4`
`=29\ text(m)`
Differentiate with respect to `x`.
`(e^x+1)^2`. (2 marks)
`2e^x(e^x+1)`
`y` | `=(e^x+1)^2` |
`dy/dx` | `=2(e^x+1)^1xxd/(dx) (e^x+1)` |
`=2e^x(e^x+1)` |
Differentiate `ln(5x+2)` with respect to `x`. (2 marks)
`5/(5x+2)`
`y` | `=ln(5x+2)` |
`dy/dx` | `=5/(5x+2)` |
Solve `2^(2x+1)=32`. (2 marks)
`x=2`
`2^(2x+1)` | `=32` |
`2^(2x+1)` | `=2^5` |
`2x+1` | `=5` |
`:. x` | `=2` |
Differentiate `(3+e^(2x))^5`. (2 marks)
`10e^(2x)(3+e^(2x))^4`
`y=(3+e^(2x))^5`
`(dy)/dx` | `=5(3+e^(2x))^4 xx d/(dx)(3+e^(2x))` |
`=5(3+e^(2x))^4 xx 2e^(2x)` | |
`=10e^(2x)(3+e^(2x))^4` |