The first three terms of a geometric sequence are
`0.125, 0.25, 0.5`
The fourth term in this sequence would be
A. `0.625`
B. `0.75`
C. `0.875`
D. `1`
E. `1.25`
Aussie Maths & Science Teachers: Save your time with SmarterEd
The first three terms of a geometric sequence are
`0.125, 0.25, 0.5`
The fourth term in this sequence would be
A. `0.625`
B. `0.75`
C. `0.875`
D. `1`
E. `1.25`
`D`
`text(GP sequence is 0.125, 0.25, 0.5)`
`a` | `=0.125` |
`r` | `=t_(2)/t_(1)=0.25/0.125=2` |
`T_4` | `=ar^3` |
`=0.125 xx 2^3` | |
`=1` |
`=> D`
The height (in cm) and foot length (in cm) for each of eight Year 12 students were recorded and displayed in the scatterplot below.
A least squares regression line has been fitted to the data as shown.
Part 1
By inspection, the value of the product-moment correlation coefficient `(r)` for this data is closest to
Part 2
The explanatory variable is foot length.
The equation of the least squares regression line is closest to
Part 3
The plot of the residuals against foot length is closest to
`text(Part 1:)\ B`
`text(Part 2:)\ B`
`text(Part 3:)\ B`
`text(Part 1)`
`text(The correlation is positive and strong.)`
`text(Eliminate)\ C, D\ text(and)\ E.`
`r= 0.98\ text(is too strong. Eliminate)\ A.`
`=> B`
`text(Part 2)`
`text(The intercept with the height axis)\ (ytext{-axis)}`
`text{is below 167 because that is the height when}`
`text{foot length = 20 cm.}`
`text(Eliminate)\ C, D\ text(and)\ E.`
`text(The gradient is approximately 1.3, by observing)`
`text(the increase in height values when the foot)`
`text(length increases from 20 to 22 cm.)`
`=> B`
`text(Part 3)`
`text(First residual is positive. Eliminate)\ A, D, E.`
`text(The next 3 residuals are negative. Eliminate)\ C`
`=> B`
The histogram below displays the distribution of the percentage of Internet users in 160 countries in 2007.
Part 1
The shape of the histogram is best described as
A. approximately symmetric.
B. bell shaped.
C. positively skewed.
D. negatively skewed.
E. bi-modal.
Part 2
The number of countries in which less than 10% of people are Internet users is closest to
A. `10`
B. `16`
C. `22`
D. `32`
E. `54`
Part 3
From the histogram, the median percentage of Internet users is closest to
A. `10text(%)`
B. `15text(%)`
C. `20text(%)`
D. `30text(%)`
E. `40text(%)`
`text(Part 1:)\ C`
`text(Part 2:)\ E`
`text(Part 3:)\ C`
`text(Part 1)`
`text(The shape of the histogram has a definite tail)`
`text(on the right side which means it is positively)`
`text(skewed.)`
`=> C`
`text(Part 2)`
`text(The histogram shows that 32% of countries fall)`
`text(between 0–5%, and 22% fall between 5–10%.)`
`:.\ text(Users below 10%)`
`= 32 + 22`
`= 54 text(%)`
`=> E`
`text(Part 3)`
`text(Total countries = 160)`
`text(Adding bars from the left hand side, there are)`
`text{80 countries in the first 4 bars (i.e. half of 160).}`
`:.\ text(Median is closest to 20%)`
`=> C`
From a point `A` due south of a tower, the angle of elevation of the top of the tower `T`, is 23°. From another point `B`, on a bearing of 120° from the tower, the angle of elevation of `T` is 32°. The distance `AB` is 200 metres.
(i)
(ii) `text(Find)\ \ OT = h`
`text(Using the cosine rule in)\ Delta AOB :`
`200^2 = OA^2 + OB^2 – 2 * OA * OB * cos 60\ …\ text{(*)}`
`text(In)\ Delta OAT,\tan 23^@= h/(OA)`
`=> OA= h/(tan 23^@)\ …\ (1)`
`text(In)\ Delta OBT,\ tan 32^@= h/(OB)`
`=> OB= h/(tan 32^@)\ \ \ …\ (2)`
`text(Substitute)\ (1)\ text(and)\ (2)\ text(into)\ text{(*)}`
`200^2` | `= (h^2)/(tan^2 23^@) + (h^2)/(tan^2 32^@) – 2 * h/(tan 23^@) * h/(tan 32^@) * 1/2` |
`= h^2 (1/(tan^2 23^@) + 1/(tan^2 32^@) + 1/(tan23^@ * tan32^@) )` | |
`= h^2 (4.340…)` | |
`h^2` | `= (40\ 000)/(4.340…)` |
`= 9214.55…` | |
`:. h` | `= 95.99…` |
`= 96\ text(m)\ \ \ text{(to nearest m)}` |
The parallel boxplots below summarise the distribution of population density, in people per square kilometre, for 27 inner suburbs and 23 outer suburbs of a large city.
Which one of the following statements is not true?
`C`
`text(The chart of the inner suburbs has both a higher IQR and)`
`text(range than the outer suburbs. This supports the argument)`
`text(that densities are NOT more variable in the outer suburbs,)`
`text(making C an untrue statement.)`
`text(All other statements can be shown to be true using)`
`text(quartile and median comparisons.)`
`=> C`
The following table shows the data collected from a sample of seven drivers who entered a supermarket car park. The variables in the table are:
distance – the distance that each driver travelled to the supermarket from their home
Part 1
The mean, `barx`, and the standard deviation, `s_x`, of the variable, distance, are closest to
A. `barx = 2.5\ \ \ \ \ \ \s_x = 3.3`
B. `barx = 2.8\ \ \ \ \ \ \s_x = 1.7`
C. `barx = 2.8\ \ \ \ \ \ \s_x = 1.8`
D. `barx = 2.9\ \ \ \ \ \ \s_x = 1.7`
E. `barx = 3.3\ \ \ \ \ \ \s_x = 2.5`
Part 2
The number of categorical variables in this data set is
A. `0`
B. `1`
C. `2`
D. `3`
E. `4`
Part 3
The number of female drivers with three children in the car is
A. `0`
B. `1`
C. `2`
D. `3`
E. `4`
`text(Part 1:) \ C`
`text(Part 2:)\ D`
`text(Part 3:)\ B`
`text(Part 1)`
`text(By calculator)`
`text(Distance)\ \ barx` | `=2.8` |
`s_x` | `≈1.822` |
`=>C`
`text(Part 2)`
`text(Categorical variables are sex, type)`
`text(of car, and post code.)`
`=>D`
`text(Part 3)`
`text(1 female driver has 3 children.)`
`=>B`
The time spent by shoppers at a hardware store on a Saturday is approximately normally distributed with a mean of 31 minutes and a standard deviation of 6 minutes.
If 2850 shoppers are expected to visit the store on a Saturday, the number of shoppers who are expected to spend between 25 and 37 minutes in the store is closest to
A. 16
B. 68
C. 460
D. 1900
E. 2400
`D`
`barx=31` | `s=6` |
`ztext{-score (25)}` | `=(x-barx)/s` |
`=(25-31)/6` | |
`=–1` |
`z text{-score (37)}` | `=(37-31)/6` |
`=1` |
`∴\ text(# Shoppers)` | `= text(68%) xx 2850` |
`=1938` |
`=> D`
The following ordered stem plot shows the areas, in square kilometres, of 27 suburbs of a large city.
The median area of these suburbs, in square kilometres, is
A. `3.0`
B. `3.1`
C. `3.5`
D. `30.0`
E. `30.5`
`B`
`text(# Data points = 27)`
`text(Median is)\ \ \ (27+1)/2 = text(14th)`
`∴ text(Median)=3.1\ text(km²)`
`=> B`
The table below shows the percentage of households with and without a computer at home for the years 2007, 2009 and 2011.
In the year 2009, a total of `5\ 170\ 000` households were surveyed.
The number of households without a computer at home in 2009 was closest to
A. | `801\ 000` |
B. | `1\ 153\ 000` |
C. | `1\ 737\ 000` |
D. | `3\ 433\ 000` |
E. | `4\ 017\ 000` |
`B`
`text (In 2009, the percentage of households without)`
`text(a computer = 22.3%.)`
`:.\ text (# Households without a computer)`
`= 22.3 text (%) xx 5\ 170\ 000`
`= 1\ 152\ 910`
`rArr B`
The lengths of the left feet of a large sample of Year 12 students were measured and recorded. These foot lengths are approximately normally distributed with a mean of 24.2 cm and a standard deviation of 4.2 cm.
Part 1
A Year 12 student has a foot length of 23 cm.
The student’s standardised foot length (standard `z` score) is closest to
A. –1.2
B. –0.9
C. –0.3
D. 0.3
E. 1.2
Part 2
The percentage of students with foot lengths between 20.0 and 24.2 cm is closest to
A. 16%
B. 32%
C. 34%
D. 52%
E. 68%
`text(Part 1:)\ C`
`text(Part 2:)\ C`
`text(Part 1)`
`bar(x) = 24.2,` | `s=4.2` |
`z text{-score (23)}` | `=(x – bar(x))/s` |
`= (23 – 24.2)/4.2` | |
`= -0.285…` |
`=> C`
`text(Part 2)`
`z text{-score (20)}` | `=(20- 24.2)/4.2` |
`= -1` | |
`z text{-score (24.2)}` | `= 0` |
`text(68% have a)\ z text(-score between –1 and 1)`
`:.\ text(34% have a)\ z text(-score between –1 and 0)`
`=> C`
The passengers on a train were asked why they travelled by train. Each reason, along with the percentage of passengers who gave that reason, is displayed in the segmented bar chart below.
The percentage of passengers who gave the reason ‘no car’ is closest to
A. `text(14%)`
B. `text(18%)`
C. `text(26%)`
D. `text(74%)`
E. `text(88%)`
`A`
`text(Percentage that stated “no car”)`
`=\ text(88% – 74%)`
`=\ text(14%)`
`=> A`
An animal study was conducted to investigate the relationship between exposure to danger during sleep (high, medium, low) and chance of attack (above average, average, below average). The results are summarised in the percentage segmented bar chart below.
The percentage of animals whose exposure to danger during sleep is high, and whose chance of attack is below average, is closest to
A. `4text(%)`
B. `12text(%)`
C. `28text(%)`
D. `72text(%)`
E. `86text(%)`
`E`
`text(The correct percentage is the black bar section)`
`text(above the “high” column heading.)`
`=>E`
The back-to-back ordered stem plot below shows the female and male smoking rates, expressed as a percentage, in 18 countries.
Part 1
For these 18 countries, the lowest female smoking rate is
A. `5text(%)`
B. `7text(%)`
C. `9text(%)`
D. `15text(%)`
E. `19text(%)`
Part 2
For these 18 countries, the interquartile range (IQR) of the female smoking rates is
A. `4`
B. `6`
C. `19`
D. `22`
E. `23`
Part 3
For these 18 countries, the smoking rates for females are generally
A. lower and less variable than the smoking rates for males.
B. lower and more variable than the smoking rates for males.
C. higher and less variable than the smoking rates for males.
D. higher and more variable than the smoking rates for males.
E. about the same as the smoking rates for males.
`text(Part 1:) \ D`
`text(Part 2:) \ B`
`text(Part 3:) \ A`
`text(Part 1)`
`text(Lowest female smoking rate is 15%.)`
`=> D`
`text(Part 2)`
`text(18 data points.)`
`text(Split in half and take the middle point of each group.)`
`Q_L` | `=5 text(th value = 19%)` |
`Q_U` | `= 14 text(th value = 25%)` |
`∴ IQR` | `= 25text(% – 19%) =6text(%)` |
`=> B`
`text(Part 3)`
`text{Smoking rates are lower and less variable (range of}`
`text{females rates vs male rates is 13% vs 30%).}`
`=> A`
The first time a student played an online game, he played for 18 minutes.
Each time he played the game after that, he played for 12 minutes longer than the previous time.
After completing his 15th game, the total time he had spent playing these 15 games was
A. `186` minutes
B. `691` minutes
C. `1206` minutes
D. `1395` minutes
E. `1530` minutes
`E`
`text(Series)\quad18, 18+12, 18+(2×12), …`
`text(AP with)\quad a=18 and d=12`
`S_n` | `=n/2[2a+(n-1)d]` |
`S_15` | `=15/2[2(18)+(15-1)12]` |
`=15/2(36+168)` | |
`=15/2(204)` | |
`=1530` |
`=> E`
The time series plot below displays the number of guests staying at a holiday resort during summer, autumn, winter and spring for the years 2007 to 2012 inclusive.
Part 1
Which one of the following best describes the pattern in the time series?
A. random variation only
B. decreasing trend with seasonality
C. seasonality only
D. increasing trend only
E. increasing trend with seasonality
Part 2
The table below shows the data from the times series plot for the years 2007 and 2008.
Using four-mean smoothing with centring, the smoothed number of guests for winter 2007 is closest to
A. `85`
B. `107`
C. `183`
D. `192`
E. `200`
`text(Part 1:)\ C`
`text(Part 2:)\ D`
`text(Part 1)`
`text(The pattern in the time series is seasonal only,)`
`text(with peaks appearing in Summer. There is no)`
`text(apparent year-to-year trend.)`
`=> C`
`text(Part 2)`
`text{Mean of guests (Season 1-4)}`
`=(390+126+85+130)/4` |
`=182.75` |
`text{Mean of guests (Season 2-5)}`
`=(126+85+85+130+460)/4` |
`=200.25` |
`:.\ text(Four-mean smoothing with centring)`
`=(182.75+200.25)/2` |
`=191.5` |
`=> D`
The table below shows the hourly rate of pay earned by 10 employees in a company in 1990 and in 2010.
The value of the correlation coefficient, `r`, for this set of data is closest to
A. `0.74`
B. `0.86`
C. `0.92`
D. `0.93`
E. `0.96`
`E`
`text(By calculator)`
`r=0.962…`
`=> E`
The time, in hours, that each student spent sleeping on a school night was recorded for 1550 secondary-school students. The distribution of these times was found to be approximately normal with a mean of 7.4 hours and a standard deviation of 0.7 hours.
Part 1
The time that 95% of these students spent sleeping on a school night could be
A. less than 6.0 hours.
B. between 6.0 and 8.8 hours.
C. between 6.7 and 8.8 hours.
D. less than 6.0 hours or greater than 8.8 hours.
E. less than 6.7 hours or greater than 9.5 hours.
Part 2
The number of these students who spent more than 8.1 hours sleeping on a school night was closest to
A. 16
B. 248
C. 1302
D. 1510
E. 1545
`text(Part 1:)\ B`
`text(Part 2:)\ B`
`text(Part 1)`
`-2 < z text(-score) < 2,\ \ text(contains 95% of students.)`
`barx=7.4, \ \ \ s=0.7`
`z text(-score of +2)` | `= 7.4+(2×0.7)` |
`= 8.8\ text(hours)` | |
`z text(-score of –2)` | `= 7.4−(2×0.7)` |
`= 6.0\ text(hours)` |
`:. 95 text(% students sleep between 6 and 8.8 hours.`
`=>B`
`text(Part 2)`
`text (Find)\ z text(-score of 8.1 hours)`
`ztext(-score)` | `= (x-barx)/s` |
`= (8.1-7.4)/0.7` | |
`= 1` |
`text(68% students have)\ \ –1 < z text(-score) < 1`
`:. 16 text(% have)\ z text(-score) > 1`
`:.\ text(# Students)` | `= 16text(%) xx1550` |
`= 248` |
`=>B`
The heights of 82 mothers and their eldest daughters are classified as 'short', 'medium' or 'tall'. The results are displayed in the frequency table below.
Part 1
The number of mothers whose height is classified as 'medium' is
A. `7`
B. `10`
C. `14`
D. `31`
E. `33`
Part 2
Of the mothers whose height is classified as 'tall', the percentage who have eldest daughters whose height is classified as 'short' is closest to
A. `text(3%)`
B. `text(4%)`
C. `text(14%)`
D. `text(17%)`
E. `text(27%)`
`text(Part 1:)\ D`
`text(Part 2:)\ C`
`text(Part 1)`
`text(# Mothers classified as medium)`
`=10+14+7\ \ \ text{(from Table)}` |
`=31` |
`=>D`
`text(Part 2)`
`text(# Tall Mothers)` | `=3+11+8` |
`=22` |
`text{# Tall Mothers with short eldest = 3 (from Table)}`
`:.\ text(Percentage)` | `=3/22×100` |
`=13.6363…%` |
`=> C`
The following ordered stem plot shows the percentage of homes connected to broadband internet for 24 countries in 2007.
Part 1
The number of these countries with more than 22% of homes connected to broadband internet in 2007 is
A. `4`
B. `5`
C. `19`
D. `20`
E. `22`
Part 2
Which one of the following statements relating to the data in the ordered stem plot is not true?
A. The minimum is 16%.
B. The median is 30%.
C. The first quartile is 23.5%
D. The third quartile is 32%.
E. The maximum is 38%.
`text(Part 1:)\ C`
`text(Part 2:)\ B`
`text(Part 1)`
`text(There are 19 values greater than 22%)`
`=> C`
`text(Part 2)`
`text(24 data points.)`
`text(Median)` | `= text(12th + 13th)/2` |
`=(29+30)/2` | |
`=29.5` |
`:.\ text(B is incorrect and all other statements can)`
`text(be verified as true.)`
`=> B`
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i. |
ii. `text(Solving for)\ \ |\ 2x – 1\ | <= |\ x – 3\ |`
`text(Graph shows the statement is TRUE)`
`text(between the points of intersection.)`
`=>\ text(Intersection occurs when)`
`(2x – 1)` | `= (x – 3)\ \ \ text(or)\ \ \ ` | `-(2x – 1)` | `= x – 3` |
`x` | `= -2` | `-2x + 1` | `= x – 3` |
`-3x` | `= -4` | ||
`x` | `= 4/3` |
`:.\ text(Solution is)\ \ {x: -2 <= x <= 4/3}`
The polynomial `p(x)` is given by `p(x) = ax^3 + 16x^2 + cx - 120`, where `a` and `c` are constants.
The three zeros of `p(x)` are `– 2`, `3` and `beta`.
Find the value of `beta`. (3 marks)
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`- 5`
`p(x) = ax^3 + 16x^2 + cx – 120`
`text(Roots:)\ \ – 2, \ 3, \ beta`
`-2 + 3 + beta` | `= -B/A` |
`beta + 1` | `= -16/a` |
`beta` | `= -16/a – 1\ \ \ \ \ …\ (1)` |
`-2 xx 3 xx beta` | `= -D/A` |
`-6 beta` | `= 120/a` |
`beta` | `= -20/a\ \ \ \ \ …\ (2)` |
`- 16/a – 1` | `= -20/beta` |
`-16 – a` | `= -20` |
`a` | `= 4` |
`text(Substitute)\ \ a = 4\ \ text(into)\ (1)`
`:. beta` | `= – 16/4 – 1` |
`= -5` |
Evaluate `int_-1^1 1/sqrt(4 - x^2)\ dx`. (2 marks)
`pi/3`
`int_-1^1 1/sqrt(4 – x^2)\ dx`
`= [sin^(-1) (x/2)]_(-1)^1`
`= sin^(-1) (1/2) – sin^(-1) (-1/2)`
`= pi/6 – (- pi/6)`
`= pi/3`
Differentiate `cos^(–1) (3x)` with respect to `x`. (2 marks)
`(-3)/sqrt(1 – 9x^2)`
`y` | `= cos^(-1) (3x)` |
`dy/dx` | `= – 1/sqrt(1 – (3x)^2) * d/(dx) (3x)` |
`= (-3)/sqrt(1 – 9x^2)` |
The polynomial `x^3` is divided by `x + 3`. Calculate the remainder. (2 marks)
`-27`
`P(-3)` | `= (-3)^3` |
`= -27` |
`:.\ text(Remainder when)\ x^3 -: (x + 3) = -27`
The graph shows the predicted population age distribution in Australia in 2008.
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`text(and is not restricted by a 5-year limit.)`
i. | `text{# Females (0-4)}` | `= 0.6 xx 1\ 000\ 000` |
`= 600\ 000` |
ii. | `text(Modal age group)\ =` | `text(35 – 39)` |
iii. | `text{# Males (15-19)}` | `= 0.75 xx 1\ 000\ 000` |
`= 750\ 000` |
`text{# Females (15-19)}` | `= 0.7 xx 1\ 000\ 000` |
`= 700\ 000` |
`:.\ text{Total People (15-19)}` | `= 750\ 000 + 700\ 000` |
`= 1\ 450\ 000` |
iv. | `text(The 80+ group includes all people over 80)` |
`text(and is not restricted by a 5-year limit.)` |
Cecil invited 175 movie critics to preview his new movie. After seeing the movie, he conducted a survey. Cecil has almost completed the two-way table.
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What is the probability that the critic was less than 40 years old and did not like the movie? (2 marks)
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Will this movie be considered a box office success? Justify your answer. (1 mark)
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i. `text{Critics liked and}\ >= 40`
`= 102-65`
`= 37`
`:. A = 37+31=68`
ii. `text{Critics did not like and < 40}`
`= 175-65-37-31`
`= 42`
`:.\ P text{(not like and < 40)}`
`= 42/175`
`= 6/25`
iii. `text(Critics liked) = 102`
`text(% Critics liked)` | `= 102/175 xx 100` |
`= 58.28…%` |
`:.\ text{Movie NOT a box office success (< 65% critics liked)}`
Bob is employed as a salesman. He is offered two methods of calculating his income.
\begin{array} {|l|}
\hline
\rule{0pt}{2.5ex}\text{Method 1: Commission only of 13% on all sales}\rule[-1ex]{0pt}{0pt} \\
\hline
\rule{0pt}{2.5ex}\text{Method 2: \$350 per week plus a commission of 4.5% on all sales}\rule[-1ex]{0pt}{0pt} \\
\hline
\end{array}
Bob’s research determines that the average sales total per employee per month is $15 670.
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i. `text(Method 1)`
`text(Yearly sales)` | `= 12 xx 15\ 670` |
`= 188\ 040` |
`:.\ text(Earnings)` | `= text(13%) xx 188\ 040` |
`= $24\ 445.20` |
ii. `text(Method 2)`
`text(In 1 Year, Weekly Wage)` | `= 350 xx 52` |
`= 18\ 200` |
`text(Commission)` | `= text(4.5%) xx 188\ 040` |
`= 8461.80` |
`text(Total earnings)` | `= 18\ 200 + 8461.80` |
`= $26\ 661.80` |
`:.\ text(Bob should choose Method 2.)`
You are organising an outside sporting event at Mathsville and have to decide which month has the best weather for your event. The average temperature must be between 20°C and 30°C, and average rainfall must be less than 80 mm.
The radar chart for Mathsville shows the average temperature for each month, and the table gives the average rainfall for each month.
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i. `text(One of Feb, Mar, Nov, Dec)`
ii. `text(November)`
Danni is flying a kite that is attached to a string of length 80 metres. The string makes an angle of 55° with the horizontal.
How high, to the nearest metre, is the kite above Danni’s hand?
(A) 46 m
(B) 66 m
(C) 98 m
(D) 114 m
`B`
Luke’s normal rate of pay is $15 per hour. Last week he was paid for 12 hours, at time-and-a-half.
How many hours would Luke need to work this week, at double time, to earn the same amount?
`D`
`text(Amount earned last week)`
`= 12 xx 1.5 xx 15`
`= $270`
`text(Double time rate)` | `= 2 xx 15` |
`= $30 text(/hr)` |
`:.\ text(# Hours at double time)`
`= 270/30`
`= 9\ text(hrs)`
`=> D`
Which expression is equivalent to `12k^3 ÷ 4k`?
`A`
`12k^3 -: 4k` | `=(12k^3)/(4k)` | |
`=3k^2` |
`=> A`
Consider the parabola `x^2 = 8(y\ – 3)`.
(i) | `text(Vertex)\ = (0,3)` |
(ii) | `text(Using)\ \ \ x^2` | `= 4ay` |
`4a` | `= 8` | |
`a` | `= 2` |
`:.\ text(Focus) = (0,5)`
(iii) |
(iv) | `text(Intersection when)\ y = 5` |
`=> x^2` | `= 8 (5-3)` |
`x^2` | `= 16` |
`x` | `= +- 4` |
`text(Find shaded area)`
`x^2` | `= 8 (y-3)` |
`y – 3` | `= x^2/8` |
`y` | `= x^2/8 +3` |
`text(Area)` | `= int_-4^4 5\ dx – int_-4^4 x^2/8 + 3\ dx` |
`= int_-4^4 5 – (x^2/8 + 3)\ dx` | |
`= int_-4^4 2 – x^2/8\ dx` | |
`= [2x – x^3/24]_-4^4` | |
`= [(8 – 64/24) – (-8 + 64/24)]` | |
`= 5 1/3 – (- 5 1/3)` | |
`= 10 2/3\ text(u²)` |
Differentiate with respect to `x`:
`sinx/(x+4)`. (2 marks)
`(cosx (x+4) – sin x)/((x + 4)^2)`
`y = sinx/(x + 4)`
`u` | `= sinx` | `\ \ \ \ \ u’` | `= cos x` |
`v` | `= x + 4` | `v’` | `= 1` |
`dy/dx` | `= (u’v – uv’)/v^2` |
`= (cos x (x + 4) – sin x)/(x+4)^2` |
Differentiate with respect to `x`:
`x^2 log_e x` (2 marks)
`x + 2x log_e x`
`y` | `= x^2 log_e x` |
`dy/dx` | `= x^2 * 1/x + 2x * log_e x` |
`= x + 2x log_e x` |
Differentiate with respect to `x`:
`(x^2 + 3)^9` (2 marks)
`18x (x^2 + 3)^8`
`y` | `= (x^2 + 3)^9` |
`dy/dx` | `= 9 (x^2 + 3)^8 2x` |
`= 18x (x^2 + 3)^8` |
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i. | `y` | `= log_e (cos x)` |
`dy/dx` | `= (- sin x)/(cos x)` | |
`= – tan x` |
ii. | `int_0^(pi/4) tan x\ dx` |
`= – [log_e (cos x)]_0^(pi/4)` | |
`= – [log_e(cos (pi/4)) – log_e (cos 0)]` | |
`= – [log_e (1/sqrt2) – log_e 1]` | |
`= – [log_e (1/sqrt2) – 0]` | |
`= – log_e (1/sqrt2)` | |
`= 0.346…` | |
`= 0.35\ \ text{(2 d.p.)}` |
In the diagram, `ABCD` is a quadrilateral. The equation of the line `AD` is `2x- y- 1 = 0`.
(i) | `text(Show)\ BC \ text(||)\ AD` |
`B(0,3),\ \ C(1,5)`
`m_(BC)` | `= (y_2 – y_1)/(x_2 – x_1)` |
`= (5 – 3)/(1 – 0)` | |
`= 2` |
`text(Equation)\ \ AD\ \ text(is)\ \ 2x – y – 1 = 0`
`y` | `= 2x – 1` |
`m_(AD)` | `= 2` |
`:. BC\ text(||) \ AD`
`:. ABCD\ text(is a trapezium)`
(ii) | `text(Given)\ CD\ text(||) \ x text(-axis)` |
`text(Equation)\ CD\ text(is)\ y = 5` | |
`D\ text(is intersection of)` |
`y` | `= 5,\ \ and` |
`2x – y – 1` | `= 0` |
`:. 2x – 5 – 1` | `=0` |
`2x` | `=6` |
`x` | `=3` |
`:.\ D` | `= (3,5)` |
(iii) | `B(0,3),\ \ C(1,5)` |
`text(dist)\ BC` | `= sqrt ( (x_2 – x_1)^2 + (y_2 – y_1)^2 )` |
`= sqrt ( (1-0)^2 + (5-3)^2 )` | |
`= sqrt (1 + 4)` | |
`= sqrt 5\ text(units)` |
(iv) | `text(Show)\ _|_\ text(dist of)\ B\ text(to)\ AD\ text(is)\ 4/sqrt5` |
`B (0,3)\ \ \ \ \ 2x – y – 1 = 0`
`_|_\ text(dist)` | `= | (ax_1 + by_1 + c)/sqrt (a^2 + b^2) |` |
`= |( 2(0) – 1(3) -1 )/sqrt (2^2 + (-1)^2) |` | |
`= | -4/sqrt5 |` | |
`= 4/sqrt 5\ \ \ text(… as required.)` |
(v) | `text(Area)` | `= 1/2 h (a + b)` |
`= 1/2 xx 4/sqrt5 (BC + AD)` |
`BC = sqrt5\ \ text{(part (iii))}`
`A(0,–1),\ \ D(3,5)`
`text(dist)\ AD` | `= sqrt ( (3-0)^2 + (5+1)^2 )` |
`= sqrt (9 + 36)` | |
`= sqrt 45` | |
`= 3 sqrt 5` |
`:.\ text(Area)\ ABCD` | `= 1/2 xx 4/sqrt5 (sqrt5 + 3 sqrt 5)` |
`= 2 / sqrt5 (4 sqrt 5)` | |
`= 8\ text(u²)` |
Let `M` be the midpoint of `(-1, 4)` and `(5, 8)`.
Find the equation of the line through `M` with gradient `-1/2`. (2 marks)
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`x + 2y-14 = 0`
`(-1,4)\ \ \ (5,8)`
`M` | `= ( (x_1 + x_2)/2, (y_1 + y_2)/2)` |
`= ( (-1 + 5)/2, (4 + 8)/2)` | |
`= (2, 6)` |
`text(Equation through)\ (2,6)\ text(with)\ m = -1/2`
`y-y_1` | `= m (x-x_1)` |
`y-6` | `= -1/2 (x-2)` |
`2y-12` | `= -x + 2` |
`x + 2y-14` | `= 0` |
Find the sum of the first 21 terms of the arithmetic series 3 + 7 + 11 + ... (2 marks)
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`903`
`S` | `= 3 + 7 + 11 + …` |
`a` | `= 3` |
`d` | `= 7 – 3 = 4` |
`:. S_21` | `= n/2 [2a + (n – 1) d]` |
`= 21/2 [2 xx 3 + (21 – 1)4]` | |
`= 21/2 [6 + 80]` | |
`= 903` |
Expand and simplify `(sqrt3-1)(2 sqrt3 + 5)`. (2 marks)
`1 + 3 sqrt 3`
`(sqrt 3-1)(2 sqrt 3 + 5)`
`= 2 xx 3 + 5 sqrt 3-2 sqrt 3-5`
`= 1 + 3 sqrt 3`
Solve `|\ 4x - 3\ | = 7`. (2 marks)
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`x = 5/2\ \ text(or)\ x = -1`
`|\ 4x – 3\ | = 7`
`4x – 3` | `= 7` | `\ \ \ \ \ -(4x – 3)` | `= 7` |
`4x` | `= 10` | `-4x + 3` | `= 7` |
`x` | `= 5/2` | `-4x` | `= 4` |
`x` | `= -1` |
`:. x=5/2\ \ text(or)\ \ -1`
Evaluate `2 cos (pi/5)` correct to three significant figures. (2 marks)
`1.62\ text{(3 sig)}`
`2 cos (pi/5)` | `= 1.6180…` |
`= 1.62\ text{(3 sig)}` |
The table below shows the present value of an annuity with a contribution of $1.
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i. `text(Table factor when)\ \ n = 4,\ r = text(2%) \ => \ 3.8077`
`:.\ PVA\ text{(Fiona)}` | `= 3000 xx 3.8077` |
`= $11\ 423.10` |
ii. `text(Table factor when)\ n = 2, r = text(4%)`
`=> 1.8861`
`:.\ PVA\ text{(John)}` | `= 6000 xx 1.8861` |
`= $11\ 316.60` |
`:.\ text(Fiona will be better off because her)\ PVA`
`text(is higher.)`
The take-off point `O` on a ski jump is located at the top of a downslope. The angle between the downslope and the horizontal is `pi/4`. A skier takes off from `O` with velocity `V` m s−1 at an angle `theta` to the horizontal, where `0 <= theta < pi/2`. The skier lands on the downslope at some point `P`, a distance `D` metres from `O`.
The flight path of the skier is given by
`x = Vtcos theta,\ y = -1/2 g t^2 + Vt sin theta`, (Do NOT prove this.)
where `t` is the time in seconds after take-off.
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i. `text(Show)\ \ y = x tan theta – (gx^2)/(2V^2) sec^2 theta`
`x` | `= Vt cos theta` |
`t` | `= x/(V cos theta)\ \ \ …\ text{(1)}` |
`text(Subst)\ text{(1)}\ text(into)\ y = -1/2 g t^2 + Vt sin theta`
`y` | `= -1/2 g (x/(Vcostheta))^2 + V sin theta (x/(Vcostheta))` |
`= (-gx^2)/(2V^2 cos^2 theta) + x * (sin theta)/(cos theta)` | |
`= x tan theta – (gx^2)/(2V^2) sec^2 theta\ \ \ text(… as required.)` |
ii. `text(Show)\ D = 2 sqrt 2 (V^2)/g\ cos theta (cos theta + sin theta)`
`text(S)text(ince)\ P\ text(lies on line)\ y = -x`
`-x` | `=x tan theta – (gx^2)/(2V^2) sec^2 theta` |
`-1` | `=tan theta – (gx)/(2V^2) sec^2 theta` |
`(gx)/(2V^2) sec^2 theta` | `= tan theta + 1` |
`x (g/(2V^2))` | `=(sin theta)/(cos theta) * cos^2 theta + 1 * cos^2 theta` |
`x` | `=(2V^2)/g\ (sin theta cos theta + cos^2 theta)` |
`=(2V^2)/g\ cos theta (cos theta + sin theta)` |
`text(Given that)\ \ cos(pi/4)` | `= x/D = 1/sqrt2` |
`text(i.e.)\ \ D` | `= sqrt 2 x` |
`:.\ D = 2 sqrt 2 (V^2)/g\ cos theta (cos theta + sin theta)`
`text(… as required.)`
iii. `text(Show)\ (dD)/(d theta) = 2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)`
`D` | `= 2 sqrt 2 (V^2)/g\ (cos^2 theta + cos theta sin theta)` |
`(dD)/(d theta)` | `= 2 sqrt 2 (V^2)/g\ [2cos theta (–sin theta) + cos theta cos theta + (– sin theta) sin theta]` |
`= 2 sqrt 2 (V^2)/(g) [(cos^2 theta – sin^2 theta) – 2 sin theta cos theta]` | |
`= 2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)\ \ \ text(… as required)` |
iv. `text(Max/min when)\ (dD)/(d theta) = 0`
`2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)` | `= 0` |
`cos 2 theta – sin 2 theta` | `= 0` |
`sin 2 theta` | `= cos 2 theta` |
`tan 2 theta` | `= 1` |
`2 theta` | `= pi/4` |
`theta` | `= pi/8` |
`(d^2D)/(d theta^2)` | `= 2 sqrt 2 (V^2)/g\ [-2 sin 2theta – 2 cos 2 theta]` |
`= 4 sqrt 2 (V^2)/g\ (-sin 2 theta – cos 2 theta)` |
`text(When)\ \ theta = pi/8:`
`(d^2 D)/(d theta^2)` | `= 4 sqrt 2 (V^2)/g\ (- sin (pi/4) – cos (pi/4))` |
`= 4 sqrt 2 (V^2)/g\ (- 1/sqrt2 – 1/sqrt2) < 0` | |
` =>\ text(MAX)` |
`:.\ D\ text(has a maximum value when)\ theta = pi/8`
One end of a rope is attached to a truck and the other end to a weight. The rope passes over a small wheel located at a vertical distance of 40 m above the point where the rope is attached to the truck.
The distance from the truck to the small wheel is `L\ text(m)`, and the horizontal distance between them is `x\ text(m)`. The rope makes an angle `theta` with the horizontal at the point where it is attached to the truck.
The truck moves to the right at a constant speed of `text(3 m s)^(-1)`, as shown in the diagram.
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i. |
`text(Show)\ \ (dL)/(dx) = cos theta`
`text(Using Pythagoras,)`
`L^2` | `=40^2 + x^2` |
`L` | `=(40^2 + x^2)^(1/2)` |
`(dL)/(dx)` | `=1/2 * 2x * (40^2 + x^2)^(-1/2)` |
`=x/ sqrt((40^2 + x^2))` | |
`=x/L` | |
`=cos theta\ \ \ text(… as required.)` |
ii. `text(Show)\ \ (dL)/(dt) = 3 cos theta`
`(dL)/(dt)` | `= (dL)/(dx) * (dx)/(dt)` |
`= cos theta * 3` | |
`= 3 cos theta\ \ \ text(… as required)` |
Milk taken out of a refrigerator has a temperature of 2° C. It is placed in a room of constant temperature 23°C. After `t` minutes the temperature, `T`°C, of the milk is given by
`T = A-Be ^(-0.03t)`,
where `A` and `B` are positive constants.
How long does it take for the milk to reach a temperature of 10°C? (3 marks)
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`16\ text(mins)`
`T = A – Be^(-0.03t)`
`text(Room temperature constant @)\ 23°`
`=>A = 23`
`:.T = 23 – Be^(-0.03 t)`
`text(At)\ t = 0,\ T = 2`
`2 = 23 – Be^0`
`=>B=21`
`:. T = 23 – 21e^(-0.03t)`
`text(Find)\ \ t\ \ text(when)\ \ T = 10`
`10` | `= 23 – 21e^(-0.03 t)` |
`21e^(-0.03 t)` | `=13` |
`e^(-0.03t)` | `=13/21` |
`ln e^(-0.03t)` | `=ln (13/21)` |
`-0.03 t` | `=ln (13/21)` |
`:. t` | `=(ln (13/21))/(-0.03)` |
`=15.9857…` | |
`=16\ text(mins)\ \ \ text{(nearest minute)}` |
Solve `(x^2 + 5)/x > 6`. (3 marks)
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`0<x<1\ \ text(or)\ \ x>5`
Solve `(x + 2/x)^2 - 6 (x + 2/x) + 9 = 0`. (3 marks)
`x = 1\ text(or)\ 2`
`(x + 2/x)^2 – 6(x + 2/x) + 9 = 0`
`text(Let)\ (x + 2/x) = X`
`:.\ X^2 – 6X + 9` | `= 0` |
`(X – 3)^2` | `= 0` |
`X` | `= 3` |
`:.\ x + 2/x` | `= 3` |
`x^2 + 2` | `= 3x` |
`x^2 – 3x + 2` | `= 0` |
`(x – 1)(x – 2)` | `= 0` |
`:.\ x = 1\ text(or)\ 2`
What is the derivative of `3 sin^(-1)\ x/2`?
(A) `6/sqrt(4 - x^2)`
(B) `3/sqrt(4 - x^2)`
(C) `3/(2sqrt(4 - x^2))`
(D) `3/(4sqrt(4 - x^2))`
`B`
`y = 3 sin^(-1)\ x/2`
`dy/dx = 3 xx 1/sqrt(4 – x^2)`
`=> B`
What is the constant term in the binomial expansion of `(2x - 5/(x^3))^12`?
(A) `((12),(3)) 2^9 5^3`
(B) `((12),(9)) 2^3 5^9`
(C) `-((12),(3)) 2^9 5^3`
(D) `-((12),(9)) 2^3 5^9`
`C`
`text(General term)`
`T_k` | `= ((12),(k)) (2x)^(12-k) * (-1)^k *(5x^(-3))^k` |
`= ((12),(k)) 2^(12-k) * x^(12-k) * (-1)^k * 5^k * x^(-3k)` | |
`= ((12),(k)) (-1)^k * 2^(12-k) * 5^k * x^(12-4k)` |
`text(Constant term when)`
`12 – 4k` | `= 0` |
`k` | `= 3` |
`:.\ text(Constant term)`
`=((12),(3)) (-1)^3 * 2^9 * 5^3`
`= – ((12),(3)) * 2^9 * 5^3`
`=> C`
Which expression is equal to `cos x - sin x`?
(A) `sqrt 2 cos (x + pi/4)`
(B) `sqrt 2 cos (x - pi/4)`
(C) `2 cos (x + pi/4)`
(D) `2 cos (x - pi/4)`
`A`
`R cos (x + alpha) = R cos x cos alpha – R sinx sin alpha`
`:.\ R cosx cos alpha – R sinx sin alpha = cos x – sin x`
`R cos alpha = 1,\ \ \ \ R sin alpha = 1`
`R^2` | `= 1^2 + 1^2` |
`R` | `= sqrt 2` |
`:.\ sqrt 2 cos alpha` | `= 1` |
`cos alpha` | `= 1/sqrt2` |
`alpha` | `= pi/4` |
`:.\ sqrt 2 cos (x + pi/4) = cosx – sinx`
`=> A`
Consider the function `f(x) = (x^4 + 3x^2)/(x^4 + 3)`.
(i) | `f(x)` | `= (x^4 + 3x^2)/(x^4 + 3)` |
`f(–x)` | `= ((–x)^4 + 3(-x)^2)/((–x)^4 + 3)` | |
`= (x^4 + 3x^2)/(x^4 + 3)` | ||
`= f(x)` |
`:.\ text(Even function.)`
(ii) | `y` | `= (x^4 + 3x^2)/(x^4 + 3)` |
`= (1 + 3/(x^2))/(1 + 3/(x^4))` |
`text(As)\ \ ` | `x` | `-> oo` |
`y` | `-> 1` |
`:.\ text(Horizontal asymptote at)\ \ y = 1`
(iii) | `f(x) = (x^4 + 3x^2)/(x^4 + 3)` |
`u` | `= x^4 + 3x^2\ \ \ \ \ ` | `v` | `= x^4 + 3` |
`u prime` | `= 4x^3 + 6x\ \ \ \ \ ` | `v prime` | `= 4x^3` |
`f prime (x)` | `= (u prime v\ – u v prime)/(v^2)` |
`= ((4x^3 + 6x)(x^4 + 3)\ – (x^4 + 3x^2)4x^3)/((x^4 + 3)^2)` | |
`= (4x^7 + 12x^3 + 6x^5 + 18x\ – 4x^7\ – 12x^5)/((x^4 + 3)^2)` | |
`= (-6x^5 + 12x^3 + 18x)/((x^4 + 3)^2)` | |
`= (-6x(x^4\ – 2x^2\ – 3))/((x^4 + 3)^2)` |
`text(S.P. when)\ x = 0\ \ \ text(or) \ \ x^4\ – 2x^2\ – 3 = 0`
`text(Let)\ X = x^2`
`X^2\ – 2X\ – 3` | `= 0` |
`(X\ – 3)(X + 1)` | `= 0` |
`X = 3\ \ text(or)\ \ -1`
`:. x^2` | `= 3` | `text(or)\ \ \ \ \ ` | `x^2 = -1` |
`x` | `= +- sqrt3\ \ \ \ ` | `text{(no solution)}` |
`:.\ text(SPs occur when)\ \ x = 0, – sqrt3, sqrt3`
(iv) | `text(When)\ x = 0,\ ` | `y = 0` |
`text(When)\ x = sqrt3,\ \ \ ` | `y = ((sqrt3)^4 + 3(sqrt3)^2)/((sqrt3)^4 + 3) = 3/2` |
(i) |
(ii) `text(3 solutions)`
(iii) `2 cos 2x = x + 1`
`f(x)` | `= 2 cos 2x\ – x\ – 1` |
`f prime (x)` | `= -4 sin 2x\ – 1` |
`=>f(0.4)` | `= 2 cos 0.8\ – 0.4\ – 1` |
`=-0.0065865 …` | |
`=> f prime(0.4)` | `= -4 sin 0.8\ – 1` |
`=-3.869424 …` |
`text(Find)\ x_1\ text(where)`
`x_1` | `= 0.4\ – (f(0.4))/(f prime(0.4))` |
`= 0.4\ – ((-0.0065865 …)/(-3.869424 …))` | |
`= 0.39829…` | |
`= 0.398\ \ text{(3 d.p.)}` |
The polynomial `p(x) = x^3-ax + b` has a remainder of `2` when divided by `(x-1)` and a remainder of `5` when divided by `(x + 2)`.
Find the values of `a` and `b`. (3 marks)
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`a` | `= 4` |
`b` | `= 5` |
`p(x)` | `= x^3-ax + b` |
`P(1)` | `= 2` |
`1-a + b` | `= 2` |
`b` | `= a+1\ \ \ …\ text{(1)}` |
`P (-2)` | `= 5` |
`-8 + 2a + b` | `= 5` |
`2a + b` | `= 13\ \ \ …\ text{(2)}` |
`text(Substitute)\ \ b = a+1\ \ text(into)\ \ text{(2)}`
`2a + a+1` | `= 13` |
`3a` | `= 12` |
`:. a` | `= 4` |
`:. b` | `= 5` |
Differentiate `x cos^2 x`. (2 marks)
`-2x cos x sin x + cos^2x`
`y` | `= x cos^2 x` |
`dy/dx` | `= x * – sin x * 2 cos x + 1 * cos^2 x` |
`= -2x cos x sin x + cos^2 x` |
Let `f(x) = ln (x - 3)`. What is the domain of `f(x)`? (1 mark)
`x > 3`
`f(x) = ln (x – 3)`
`text(S)text(ince)\ \ \ ` | `x – 3` | `> 0` |
`x` | `> 3` |
In `Delta DEF`, a point `S` is chosen on the side `DE`. The length of `DS` is `x`, and the length of `ES` is `y`. The line through `S` parallel to `DF` meets `EF` at `Q`. The line through `S` parallel to `EF` meets `DF` at `R`.
The area of `Delta DEF` is `A`. The areas of `Delta DSR` and `Delta SEQ` are `A_1` and `A_2` respectively.
(i) `text(Need to show)\ Delta DEF\ text(|||) \ Delta DSR`
`/_FDE\ text(is common)`
`/_DSR = /_DEF = theta\ \ text{(corresponding angles,}\ RS\ text(||)\ FE text{)}`
`:.\ Delta DEF \ text(|||) \ Delta DSR\ \ text{(equiangular)}`
(ii) `(DR)/(DF) = (DS)/(DE) = x/(x + y)`
`text{(Corresponding sides of similar triangles)}`
(iii) `text(Show)\ sqrt((A_1)/A) = x/(x + y)`
`text(Using Area)` | `= 1/2 ab sin C` |
`A_1` | `= 1/2 xx DR xx x xx sin alpha` |
`A` | `= 1/2 xx DF xx (x + y) xx sin alpha` |
`(A_1)/A` | `= (1/2 * DR * x * sin alpha)/(1/2 * DF * (x + y) * sin alpha)` |
`= (DR * x)/(DF * (x + y)` | |
`= (x * x)/((x + y)(x + y))\ \ \ \ text{(using part(ii))}` | |
`= (x^2)/((x + y)^2)` | |
`:.\ sqrt ((A_1)/A)` | `= x/((x + y))\ \ \ text(… as required.)` |
(iv) `text(Consider)\ Delta DFE\ text(and)\ Delta SQE`
`/_FED` | `= theta\ text(is common)` |
`/_FDE` | `= /_QSE = alpha\ \ ` | `text{(corresponding angles,}\ DF\ text(||)\ QS text{)}` |
`:.\ Delta DFE\ text(|||)\ Delta SQE\ \ text{(equiangular)}`
`(QE)/(FE) = (SE)/(DE) = y/(x +y)`
`text{(corresponding sides of similar triangles)}`
`A_2` | `= 1/2 xx QE xx y xx sin theta` |
`A` | `= 1/2 xx FE xx (x + y) xx sin theta` |
`(A_2)/A` | `= (QE * y)/(FE * (x + y))` |
`= (y^2)/((x + y)^2)` | |
`sqrt ((A_2)/A)` | `= y/((x + y))` |
`text(Need to show)\ sqrt A = sqrt (A_1) + sqrt (A_2)`
`sqrt(A_2)/sqrtA` | `= y/((x + y))` |
`sqrt (A_2)` | `= (sqrtA * y)/((x + y))` |
`text(Similarly, from part)\ text{(iii)}`
`sqrt (A_1) = (sqrtA * x)/((x + y))`
`sqrt (A_1) + sqrt (A_2)` | `= (sqrt A * x)/((x + y)) + (sqrt A * y)/((x + y))` |
`= (sqrt A (x + y))/((x + y))` | |
`= sqrt A\ \ \ text(… as required.)` |
The parabola `y = −2x^2 + 8x` and the line `y = 2x` intersect at the origin and at the point `A`.
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i. |
`text(Need to find)\ x\ text(-co-ord of)\ A`
`y` | `= 2x\ \ \ \ \ …\ text{(i)}` |
`y` | `= -2x^2 + 8x\ \ \ \ \ …\ text{(ii)}` |
`text(Subst)\ y = 2x\ text(from)\ text{(i)}\ text(into)\ text{(ii)}`
`-2x^2 + 8x` | `= 2x` |
`-2x^2 + 6x` | `= 0` |
`-2x (x\ – 3)` | `= 0` |
`:.\ x = 0\ text(or)\ 3`
`:.\ x\ text(-coordinate of)\ A\ text(is 3)`
ii. | `text(Area)` | `= int_0^3 (-2x^2 + 8x)\ dx\ – int_0^3 2x\ dx` |
`= int_0^3 (-2x^2 + 6x)\ dx` | ||
`= [-2/3x^3 + 3x^2]_0^3` | ||
`= [(-2/3(3^3) + 3(3^2))\ – (0 + 0)]` | ||
`= -18 + 27` | ||
`= 9\ text(u²)` |