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state the reason. a B(20, 0.6) b B(30, 0.95) c B(40, 0.13) d B(50, 0.06) 212 2 Find the smallest possible value of n for which the following binomial distributions can be well-approximated by a normal distribution. a B(, 0.024) n b B(, 0.15) n c B(, 0.52) n d B(, 0.7) n 3 Describe the binomial distribution that can be approximated by the normal distribution N(14, 10.5). 4 By first evaluating np and npq, use a suitable approximation and continuity correction to find. the discrete random variable ~ B(100, 0.7) X P( X < 75) for 5 The discrete random variable ~ B(50, 0.6) Y. Use a suitable approximation and continuity correction to find P( Y > 26). 6 A biased coin is tossed 160 times. The number of heads obtained, H, follows a binomial distribution where E( H ) = 100. Find: a b the value of p and the variance of H the approximate probability of obtaining more than 110 heads. 7 One card is selected at random from each of 40 packs. Each pack contains 52 cards and includes 13 clubs. Let C be the number of clubs selected from the 40 packs. a Show that the variance of C is 7.5. b Obtain an approximation for the value of <P( C 8), and justify the use of this approximation. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution 8 In a large survey, 55% of the people questioned are in full-time employment. In a random sample of 80 of these people, find: a b c the expected number in full-time employment the standard deviation of the number in full-time employment the approximate probability that fewer than half of the sample are in full-time employment. 9 A company manufactures rubber and plastic washers in the ratio 4:1. The washers are randomly packed into boxes of 25. a Find the probability that a randomly selected box contains: i exactly 21 rubber washers ii exactly 10 plastic washers. b A retail |
pack contains 2000 washers. Find the expectation and variance of the number of rubber washers in a retail pack. c Using a suitable approximation, find the probability that a retail pack contains at most 1620 rubber washers. 10 In a certain town, 63% of homes have an internet connection. a In a random sample of 20 homes in this town, find the probability that: i ii exactly 15 have an internet connection exactly nine do not have an internet connection. b Use a suitable approximation to find the probability that more than 65% of a random sample of 600 homes in this town have an internet connection. 213 11 17% of the people interviewed in a survey said they watch more than two hours of TV per day. A random sample of 300 of those who were interviewed is taken. Find an approximate value for the probability that at least one-fifth of those in the sample watch more than two hours of TV per day. 12 An opinion poll was taken before an election. The table shows the percentage of voters who said they would vote for parties A, B and C. Party Votes (%) A 36 B 41 C 23 Find an approximation for the probability that, in a random sample of 120 of these voters: a exactly 50 said they would vote for party B b more than 70 but fewer than 90 said they would vote for party B or party C. 13 Boxes containing 24 floor tiles are loaded into vans for distribution. In a load of 80 boxes there are, on average, three damaged floor tiles. Find, approximately, the probability that: a b there are more than 65 damaged tiles in a load of 1600 boxes in five loads, each containing 1600 boxes, exactly three loads contain more than 65 damaged tiles. 14 It is known that 2% of the cheapest memory sticks on the market are defective. a In a random sample of 400 of these memory sticks, find approximately the probability that at least five but at most 11 are defective. b Ten samples of 400 memory sticks are tested. Find an approximate value for the probability that there are fewer than 12 defective memory sticks in more than seven of the samples. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PS 15 Randomly selected members of the public were asked whether they approved of plans to |
build a new sports centre and 57% said they approved. Find approximately the probability that more than 75 out of 120 people said they approved, given that at least 60 said they approved. PS 16 A fair coin is tossed 400 times. Given that it shows a head on more than 205 occasions, find an approximate value for the probability that it shows a head on fewer than 215 occasions. PS 17 An ordinary fair die is rolled 450 times. Given that a 6 is rolled on fewer than 80 occasions, find approximately the probability that a 6 is rolled on at least 70 occasions. Checklist of learning and understanding ● A continuous random variable can take any value, possibly within a range, and those values occur by chance in a certain random manner. ● The probability distribution of a continuous random variable is represented by a function called a probability density function or PDF. ● A normally distributed random variable X is described by its mean and variance as X ~ N( )2 µ σ., ● The standard normal random variable is ~ N(0, 1) Z. ● When X ~ N( )2 µ σ then, Z = X µ has a standard normal distribution, and the standardised value − σ z = x µ − tells us how many standard deviations x is from the mean. σ ~ B(,, where =µ np and σ =2 )2σµ, ) X enough to ensure that n p can be approximated by N(. > 5 np and > 5 nq ● 214 npq, provided that n is large ● > 5 and np of np and nq result in better approximations. > 5 nq are the necessary conditions for making this approximation, and larger values ● Continuity corrections must be made when a discrete distribution is approximated by a continuous distribution. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution END-OF-CHAPTER REVIEW EXERCISE 8 1 A continuous random variable, X, has a normal distribution with mean 8 and standard deviation σ.. Given that 0.9772, find 9.5) P( P( 5) X X > = < 2 The variable Y is normally distributed. Given that σ 10 µ= 3 and P( Y < 10) |
= 0.75, find P( Y > 6). 3 In Scotland, in November, on average 80% of days are cloudy. Assume that the weather on any one day is independent of the weather on other days. i Use a normal approximation to find the probability of there being fewer than 25 cloudy days in Scotland in November (30 days). ii Give a reason why the use of a normal approximation is justified. [3] [4] [4] [1] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q2 June 2011 4 At a store, it is known that 1 out of every 9 customers uses a gift voucher in part-payment for purchases. A randomly selected sample of 72 customers is taken. Use a suitable approximation and continuity correction to find the probability that at most 6 of these customers use a gift voucher in part-payment for their [5] purchases. 5 A survey shows that 54% of parents believe mathematics to be the most important subject that their children study. Use a suitable approximation to find the probability that at least 30 out of a sample of 50 parents believe mathematics to be the most important subject studied. [5] 6 Two normally distributed continuous random variables are X and Y. It is given that and. On the same diagram, sketch graphs showing the probability density functions of X that ~ N(2.0, 0.5 )2 and of Y. Indicate the line of symmetry of each clearly labelled graph. ~ N(1.5, 0.2 )2 Y X 7 The random variable X is such that ~ N(82, 126) X. i A value of X is chosen at random and rounded to the nearest whole number. Find the probability that this whole number is 84. ii Five independent observations of X are taken. Find the probability that at most one of them is greater than 87. iii Find the value of k such that P(87 < X k < ) = 0.3. [3] 215 [3] [4] [5] Cambridge International AS & A Level Mathematics 9709 Paper 63 Q5 November 2012 8 a A petrol station finds that its daily sales, in litres, are normally distributed with mean 4520 and standard deviation 560. i Find on how many days of the year (365 days) the daily sales can be expected to exceed 3900 litres. [4] The daily sales at another petrol station are X litres, where X is normally distributed with mean m and standard deviation 560. It is given that. 0. |
122 8000) P( X > = ii Find the value of m. iii Find the probability that daily sales at this petrol station exceed 8000 litres on fewer than 2 of 6 randomly chosen days. b The random variable Y is normally distributed with mean µ and standard deviation σ. Given that σ = 2 3 µ, find the probability that a random value of Y is less than µ2. [3] [3] [3] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q7 November 2015 9 V and W are continuous random variables. ~ N(9, 16) 2 P( × given that W P( 8) 8). V V < < = and W ~ N(6, )2σ. Find the value of σ, [4] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 10 The masses, in kilograms, of ‘giant Botswana cabbages’ have a normal distribution with mean µ and standard deviation 0.75. It is given that 35.2% of the cabbages have a mass of less than 3 kg. Find the value of µ and the percentage of cabbages with masses of less than 3.5 kg. 11 The ages of the vehicles owned by a large fleet-hire company are normally distributed with mean 43 months 6 years old is 0.28. and standard deviation σ. The probability that a randomly chosen vehicle is more than 4 1 Find what percentage of the company’s vehicles are less than two years old. 12 The weights, X grams, of bars of soap are normally distributed with mean 125 grams and standard deviation 4.2 grams. i Find the probability that a randomly chosen bar of soap weighs more than 128 grams. ii Find the value of k such that P( k X < < 128) = 0.7465. iii Five bars of soap are chosen at random. Find the probability that more than two of the bars each weigh more than 128 grams. [5] [5] [3] [4] [4] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q7 November 2009 PS 13 Crates of tea should contain 200 kg, but it is known that 1 out of |
45 crates, on average, is underweight. A sample of 630 crates is selected at random. a Find the probability that more than 12 but fewer than 17 crates are underweight. b Given that more than 12 but fewer than 17 crates are underweight, find the probability that more than 14 crates are underweight. 216 PS 14 Once a week, Haziq rows his boat from the island where he lives to the mainland. The journey time, X minutes, is normally distributed with mean µ and variance σ2. a Given that P(20 ø < X 30) = 0.32 and that P( X < 20) = 0.63, find the values of µ and σ2. b The time taken for Haziq to row back home, Y minutes, is normally distributed and P( Y < 20) = 0.6532. Given that the variances of X and Y are equal, calculate: [5] [5] [4] [3] the mean time taken by Haziq to row back home i ii the expected number of days over a period of five years (each of 52 weeks) on which Haziq takes more [3] than 25 minutes to row back home. PS 15 The time taken, T seconds, to open a graphics programme on a computer is normally distributed with mean 20 and standard deviation σ. Given that P( T > 13 | T ø 27) = 0.8, find the value of: a σ b k for which P( T > k ) = 0.75. PS 16 A law firm has found that their assistants make, on average, one error on every 36 pages that they type. A random sample of 90 typed documents, with a mean of 62 pages per document, is selected. Given that there are more than 140 typing errors in these documents, find an estimate of the probability that there are fewer than 175 typing errors. [5] [3] [6] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cross-topic review exercise 3 CROS-TPI EVW X CROSS-TOPIC REVISION EXERCISE 3 1 a The following table shows the probability distribution for the random variable X. x P X x =( ) 0 1 k |
1 3 10 2 3 20 3 1 20 i Show that k 2=. ii Calculate XE( ) and XVar( ). iii Find the probability that two independent observations of X have a sum of less than 6. b The following table shows the probability distribution for the random variable.1 1 0.2 2 0.3 3 0.4 =. If one independent observation of each random variable is made, find the probability that X Y 3 + The random variable X has a geometric distribution such that P( P( X X = = 2) 5) = 3 3 8. Find XP( 3)<. The variable X has a normal distribution with mean µ and standard deviation XP(, find the value of µ and of and that ùXP( 27.45) 32.83) 0.409 0.834 = = <.σ.σ Given that The length of time, in seconds, that it takes to transfer a photograph from a camera to a computer can be modelled by a normal distribution with mean 4.7 and variance 0.7225. Find the probability that a photograph can be transferred in less than 3 seconds. The mass of a berry from a particular type of bush is normally distributed with mean 7.08 grams and standard deviation σ. It is known that 5% of the berries have a mass of exactly 12 grams or more. a Find the value of.σ b Find the proportion of berries that have a mass of between 6 and 8 grams. The time taken, in minutes, to fit a new windscreen to a car is normally distributed with mean µ and standard deviation 16.32. Given that three-quarters of all windscreens are fitted in less than 45 minutes, find: a the value of µ b the proportion of windscreens that are fitted in 35 to 40 minutes. 2 3 4 5 6 [2] [3] [2] [3] [3] [4] [3] [3] [3] [3] [3] 217217 7 The mid-day wind speed, in knots, at a coastal resort is normally distributed with mean 12.8 and standard deviation.σ a Given that 15% of the recorded wind speeds are less than 10 knots, find the value of σ. b Calculate the probability that exactly two out of 10 randomly selected recordings are less than 10 knots. c Using a suitable approximation, calculate an estimate of the probability that at least 13 out of 100 randomly selected recordings are |
over 15.5 knots. [3] [3] [4] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 8 A technical manual contains 10 pages of text, 7 pages of diagrams and 3 pages of colour illustrations. Four different pages are selected at random from the manual. Let X be the number of pages of colour illustrations selected. a Draw up the probability distribution table for X. b Find: i XE( ) ii the probability that fewer than two pages with colour illustrations are selected, given that at least one page with colour illustrations is selected. 9 A fair six-sided die is numbered 1, 1, 2, 3, 5 and 8. The die is rolled twice and the two numbers obtained are added together to give the score, X. a Find XE( ). b Given that the first number rolled is odd, find the probability that X is an even number. 10 The following table shows the probability distribution table for the random variable Q. q P )= 18 3 x x − + 3 4 218218 a Find the value of x. b Evaluate QVar( ). 11 Research shows that 17% of children are absent from school on at least five days during winter because of ill health. A random sample of 55 children is taken. a Find the probability that exactly 10 of the children in the sample are absent from school on at least five days during winter because of ill health. [4] [2] [2] [4] [2] [3] [2] [2] b Use a suitable approximation to find the probability that at most seven children in the sample are absent from school on at least 5 days during winter because of ill health. c Justify the approximation made in part b. [4] [1] 12 The ratio of adult males to adult females living in a certain town is 17 : 18, and gender, do not have a driving license. of these adults, independent of 2 9 a Show that the probability that a randomly selected adult in this town is male and has a driving license is equal to 17 45. [1] b Find the probability that, in a randomly selected sample of 25 adults from this town, from 8 to 10 inclusive are females who |
have a driving license. 13 A fair eight-sided die is numbered 2, 2, 3, 3, 3, 4, 5 and 6. The die is rolled up to and including the roll on which the first 2 is obtained. Let X represent the number of times the die is rolled. a Find XE( ). b Show that P( X > 4) = 27 64. [4] [1] [1] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cross-topic review exercise 3 c The die is rolled up to and including the roll on which the first 2 is obtained on 20 occasions. Find, by use of a suitable approximation, the probability that 4X > on at least half of these 20 occasions. d Fully justify the approximation used in part c. 14 A student wishes to approximate the distribution of ~ B(240, X p by a continuous random variable Y that ) has a normal distribution. a Find the values of p for which: i approximating X by Y can be justified ii YVar( ). 45< b Find the range of values of p for which both the approximation is justified and YVar( ). 45< [4] [1] [3] [3] [2] 219219 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PRACTICE EXAM-STYLE PAPER Time allowed is 1 hour and 15 minutes (50 marks) 1 A mixed hockey team consists of five men and six women. The heights of individual men are denoted by hm metres and the heights of individual women are denoted by hw metres. It is given that =hm Σ 2Σ =hw 16.25 and 9.08. Σ =hw 9.84, a Calculate the mean height of the 11 team members. 2Σ. b Given that the variance of the heights of the 11 team members is 0.0416 m2, evaluate hm 2 A and B are events such that |
′ ∩ = ) A B ′ ) ] = 0.364, as shown in the Venn diagram opposite., A B ∩ ′ = 0.196 P( P( ) P[( A ∪ B 0.286 and 1 A a Find the value of x and state what it represents. 0.196 x 0.286 b Explain how you know that events A and B are not mutually exclusive. c Show that events A and B are independent. B 0.364 3 Meng buys a packet of nine different bracelets. She takes two for herself and then shares the remainder at random between her two best friends. a How many ways are there for Meng to select two bracelets? b If the two friends receive at least one bracelet each, find the probability that one friend receives exactly one bracelet more than the other. 220 4 Every Friday evening Sunil either cooks a meal for Mina or buys her a take-away meal. The probability that he buys a take-away meal is 0.24. If Sunil cooks the meal, the probability that Mina enjoys it is 0.75, and if he buys her a take-away meal, the probability that she does not enjoy it is x. This information is shown in the following tree diagram. The probability that Sunil buys a take-away meal and Mina enjoys it is 0.156. 0.75 Enjoys the meal Cooks a meal a Find the value of x. b Given that Mina does not enjoy her Friday meal, find the probability that Sunil cooked it. 0.24 Buys a meal Does not enjoy the meal Enjoys the meal x Does not enjoy the meal 5 The following histogram summarises the total distance covered on each of 123 taxi journeys provided for customers of Jollicabs during the weekend. 30 20 10 10 8 Distance (km) 11 12 13 14 15 a Find the upper boundary of the range of distances covered on these journeys. b Estimate the number of journeys that covered a total distance from 8 to 13 kilometres. c Calculate an estimate of the mean distance covered on these 123 journeys. Copyright Material - Review Only - Not for Redistribution [2] [3] [2] [1] [2] [1] [4] [2] [3] [1] [2] [3] Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy |
- Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Practice exam-style paper 6 In each of a series of independent trials, a success occurs with a constant probability of 0.9. a The probability that none of the first n trials results in a failure is less than 0.3. Find the least possible value of n. b State the most likely trial in which the first success will occur. c Use a suitable approximation to calculate an estimate of the probability that fewer than 70 successes occur in 80 trials. 7 The following stem-and-leaf diagram shows the number of shots taken by 10 players to complete a round of golf Key: 6 1 represents 61 shots a Given that the median number of shots is 74.5 and that the mean number of shots is 75.4, find the value of x and of y. The numbers of golf shots are summarised in a box-and-whisker diagram, as shown. 16.8 cm b cm b Given that the whisker is 16.8 cm long, find the value of b, if the width of the box is b cm. c Explain why the mode would be the least appropriate measure of central tendency to use as the average value for this set of data. 8 To conduct an experiment, a student must fit three capacitors into a circuit. He has eight to choose from but, unknown to him, two are damaged. He fits three randomly selected capacitors into the circuit. The random variable X is the number of damaged capacitors in the circuit. a Draw up the probability distribution table for X. b Calculate XVar( ). c The student discovers that exactly one of the capacitors in the circuit is damaged but he does not know which one. He removes one capacitor from the circuit and replaces it with one from the box, both selected at random. Find the probability that the circuit now has at least one damaged capacitor in it. [2] [1] [4] [3] [3] [1] [3] [3] [4] 221 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 THE STANDARD NORMAL DISTRIBUTION FUNCTION If Z is normally distributed with mean 0 and variance 1 |
, the table gives the value of ( )zΦ for each value of z, where Φ (z) Φ ( ) z = P( Z < z. ) Use Φ − ( z ) = − Φ for negative values of z.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0. |
8554 0.8577 0.8599 0.8621 222 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 1.5 1.6 1.7 1.8 1.9 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 2.2 2.3 2.4 2.5 2.6 2.7 2.8 |
2.9 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986 Critical values for the normal distribution The table gives the value of z such that P( <Z ) =, where Z ~ N(0, 1). ADD 12 12 12 11 11 10 10 16 16 15 15 14 14 13 12 11 10 20 20 19 19 18 17 16 15 14 13 12 10 24 24 23 22 22 20 19 18 16 15 14 12 11 10 28 28 27 26 25 24 23 21 19 18 16 14 13 11 10 32 32 31 30 29 27 26 24 22 20 19 16 15 13 11 10 36 36 35 34 32 31 29 27 25 23 21 18 17 14 13 11.75 0.90 0.95 0.975 0.99 0.995 0.9975 0.999 0.9995 0.674 1.282 1.645 1.960 2.326 2.576 2.807 3.090 3.291 Copyright Material - Review Only - Not for Red |
15 a 70 2 b Boundaries at 4, 12, 24, 28 grams. Densities ∝ 28, 33, 17.5. c 310 We can be certain only that 0.1 0.4. 0.8 0.4<<a and that b< < 159, b = 636 b 23.5kg 10 11 = a a 4 hd 5 n 12 33cm 13 p = q29, = 94 Exercise 1C 1 a Points plotted at (1.5, 0), (3, 3), (4.5, 8), (6.5, 32), (8.5, 54), (11, 62), (13, 66). b i 23 ii 7.8s 2 a 19.5cm b Width (cm) No. books ) cf( <9.5 <14.5 <19.5 < 29.5 < 39.5 < 44.5 0 3 16 41 65 70 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Points plotted at (9.5, 0), (14.5, 3), (19.5, 16), 11 a Points plotted at (1.0, 0), (1.5, 60), (29.5, 41), (39.5, 65), (44.5, 70). i 34 or 35 ii ≈ 33.25 to 44.5cm c 3 a Points plotted at (0.10, 0), (0.35, 16), (0.60, 84), (0.85, 134), (1.20, 156) for A. Points plotted at (0.10, 0), (0.35, 8), (0.60, 52), (0.85, 120), (1.20, 156) for B. b i ≈ 107 for engine A; ≈ 87 for engine B. ii ≈ 108 c ≈ 42 a 17; cfs 20 and 37 are precise. 4 b c i 12 =k 4.7 to 4.8 ii 28 d It has the highest frequency density. 5 a i 64 ii 76 6 7 8 224 b ≈ 7.4g c (12, 304) |
a = 32, b = 45, c = 15, d = 33 a 65 b 24 a Ratio of under 155cm to over 155cm is 3:1 for boys and 1:3 for girls. b 81 or 82 c There are equal numbers of boys and girls below and above this height. d Polygon or curve through (140, 0), (155, 25), (175, 50). 9 a Points plotted at (18, 0), (20, 27), (22, 78), (25, 89), (29, 94), (36, 98), (45, 100). b 27 years and 4 or 5 months c i 1000 ii All age groups are equally likely to find employment. Either with valid reasoning; e.g. underestimate because older graduates with work experience are more attractive to employers. 10 Points plotted at (4.4, 0), (6.6, 5), (8.8, 12), (12.1, 64), (15.4, 76), (18.7, 80) for new cars. Points plotted at (4, 0), (6, 5), (8, 12), (11, 64), (14, 76), (17, 80) for > 100 000 km. Polygons 17 cars; curves ≈ 16 cars. (2.0, 182), (2.5, 222), (3.0, 242) for diameters. Points plotted at (2.0, 0), (2.5, 8), (3.0, 40), (3.5, 110), (4.0, 216), (4.5, 242) for lengths. b Least =n c 0; greatest =n 28. Diameter and length for individual pegs are not shown. Best estimate is ‘between 171 and 198 inclusive’. The length and diameter of each peg should be recorded together, then the company can decide whether each is acceptable or not. Exercise 1D 1 a Any suitable for qualitative data. b Pie chart, as circle easily recognised, or a sectional percentage bar chart. 3 4 2 Histogram; area of middle three columns > half total column area. 3 a Numbers can be shown in compact form on three rows; bar chart requires 17 bars, all with frequencies 0 or 1. b Sum 100 = be offered for sale. shows that 11 boxes of 100 tiles could 4 a 7 months b a 5 Percentage cf graph; passes below the point (12, 100). Hist |
ogram: Frequency density may be mistaken for frequency. Pie chart: does not show numbers of trees. b Pictogram: short, medium, tall; two, three and four symbols, each for six trees, plus a key. Shows 12, 18, 24 and a total of 54 trees. 6 a 1. Score (%) Frequency 30–39 40–49 50–59 60–69 70–79 80–89 90–99 3 5 6 15 5 4 2 b 2. Grade Frequency C 8 B 26 A 6 Any three valid, non-zero frequencies that sum to 40. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy c Raw: stem-and-leaf diagram is appropriate. Tables 1 and 2 do not show raw marks, so these diagrams are not appropriate. Table 1: Any suitable for grouped discrete data; e.g. histogram. Table 2: Any suitable for qualitative data. E.g. He worked for less than 34 hours in 49 weeks, and for more than 34 hours in 3 weeks. It may appear that Tom worked for more than 34 hours in a significant number of weeks. Histogram: boundaries at 9, 34 and 44; densities ∝ 98 and 15. 7 a b c Pie chart: sector angles ≈ Bar chart: frequencies 49 and 3. 339.2 and ° 20.8. ° Sectional percentage bar chart: ≈ 94.2 and 5.8%. a Some classes overlap (are not continuous). 8 b Refer to focal lengths as, say, A to E in a key. Pie chart: sector angles 77.1, 128.6,77.1, ° ° ° ° 51.4, 25.7. ° Bar chart or vertical line graph: heights 18, 30, 18, 12, 6. Pictogram: symbol for 1, 3 or 6 lenses. 9 Country C SL Ma G Mo % of population 14.4 8.9 3.8 17.7 27.4 Answers End-of-chapter review exercise 1 1 i 50 ii Boundaries at 20, 30, 40, 45, 50, 60, 70 g. Frequency densities ∝ 2, 3, 10, 12, 5, 1. 2 3 16.5, 3 and 18cm a 6 b Quantitative and |
continuous 4 a 6 b Five additional rows for classes 0–4, 20–24, 25–29, 30–34, 35–39. 5 6 a = b9, = 2 a 48 b 0.7 cm 7 a 120, 180 and 90 b 6.75cm c There is a class between them (not continuous). 8 a 30 days for region A, 31 days for region B. 225 b Bindu: unlikely to be true but we cannot tell, as the amount of sunshine on any particular day is not shown. Janet: true (max. for region A is 106h; min. for region B is 138h). 0 in hundred thousands 10 5 People living in poverty as % of country population 15 20 25 30 Chile Sri Lanka Malaysia Georgia Mongolia Mongolia, for example, has the lowest number, but the highest percentage, of people living in poverty. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 9 i Points plotted at: (20.5, 10), (40.5, 42), (50.5, 104), (60.5, 154), (70.5, 182), (90.5, 200) or (20, 10), (40, 42), (50, 104), (60, 154), (70, 182), (90, 200) or (21, 10), (41, 42), (51, 104), (61, 154), (71, 182), (91, 200). ii 174 to 180 iii 58,59 or 60 2 Measures of central tendency Prerequisite knowledge 1 Mean 5, median 4.3, mode 3.9. = = = 2 1.94 Exercise 2A 1 a No mode b 16, 19 and 21 2 4 ‘The’ is the mode. 3 7 for x; −2 for y. 14–20 for x; 3–6 for y. 5 Most popular size(s) can be pre-cut to serve 226 customers quickly, which may result in less wastage of materials. 6 216 7 69 8 73 c 4 13 40 c 88 Exercise 2B a 50 1 b 7.1 7 a = ±p a 23.25 9 or −10 b =q b 1062 |
Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Answers 6 7 8 x = 12.4; b = –3 a = b4, = 5 a (–1.8, 2.8) b (26, –6) c TE(5.2, –1.2) → (19, –2) ET(5.2, –1.2) → (–9, 14) Location is dependent on order of transformations. 9 p = q40; = 12 000; $75000 8 ‘Average’ could refer to the mean, the median or the mode. Median > 150; Mean < 150. 150 is close to lower boundary of modal class. Claim can be neither supported nor refuted. 9 There is no mode. Mean ($1000 000) is distorted by the expensive home. Median ($239000) is the most useful. Appears unfair; the smaller the amount invested, the higher the percentage profit. 281 g/cm2 x 10 10 a b p i r 54 40, 12, q = = = Reflection in a horizontal line through cf value of 30. Exercise 2E a 15 1 b Median; it is greater than the mean (12.4). c For example, being unable to pay a bill because of low earnings. 2 a 11.5 b Negatively skewed; Median 6; mode = = 3 a b 10.9 median <. t = 8 Median is central to the values but occurs less frequently than all others. Mode is the most frequently occurring value but is also the highest value. 4 5 c Two incorrect a ≈ 4.4 min b 2.8 and 6.4 min Points plotted at (0, 0), (0.2, 16), (0.3, 28), (0.5, 120), (0.7, 144), (0.8, 148). Median 0.4kg = a 92 b 32 6 Mode (15) and median (16) unaffected; mean 7 decreases from 16 to 14.75. a Points plotted at (85, 0), (105, 12), (125, 40), (145, 94), (165, 157), (195, 198), (225, 214), (265, 220). Polygon and curve give median ≈ 150 days. b Likely to use whichever is the greatest. Estimate of mean (152.84) appears advantageous. |
(They could consider using the greatest possible mean, 164.41.) ii Median safe current = median unsafe current 11 a First-half median is in 1–2; second-half median is in 4 – 5. b i 3 ii First half data are positively skewed (least possible mean is 100.8 s). 12 a Points plotted at (0, 0), (26, 15), (36, 35), (50, 60), (64, 75), (80, 80). Median 39%= New points at (0, 0), (16, 9.23), (26, 15), (40, 42.1), (54, 64.2), (80, 80). 227 b 20 13 a i f 5 6 7 8 9 10 11 x ii Mode mean median 8. = = = b c No effect on mode or median. Mean increases to 9. Curve positively skewed. – 11; No effect on mode or median. Curve =b negatively skewed. 14 Any symmetrical curve with any number of modes (or uniform). 15 a Symmetrical; mean median mode = = b Chemistry: negatively skewed; Physics: positively skewed. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 End-of-chapter review exercise 2 a Mean medianand mode 1. < b Mean medianand mode >. 2 3 c n = a b =Mean medianand mode. 22; 623g =Mode 13 Median 28 = c 55 4 a 15.15 b 13.3 5 a 6 b i 14 ii 25 25 a Proof 6 7 b $0.18; it is an estimate of the mean amount paid. 8 a Mode indicates the most common response. Median indicates a central response (one of the options or half-way between a pair). 228 b Allows for a mean response, which indicates which option the average is closest to. 9 a 1 b No; it is the smallest value and not at all central. c 11 d Positively skewed; mode median mean. < < 10 i Boundaries at 0.05, 0.55, 1.05, 2.05, 3.05, 4.55h. Frequency densities ∝ 22, 30, 18, 30 |
, 14. ii 2.1h 11 16.4 12 81 13 14 15 a b Mode = 0, mean 1, median 0 = = Mean; others might suggest that none of the items are damaged. a 4006 − 2980 = $1026 b $3664 1.95 3 Measures of variation Prerequisite knowledge 1 16cm 2 a 4.5 b 27.3 Exercise 3A Box plots given by: smallest … … … … Item (units), as appropriate. Q2 3 Q Q 1 largest / b 35 and 20 d 96 and 59 1 a 25 and 17 c 65 and 25 e 8.5 and 5.6 2 a Range = 3.3; IQR 1.75 = b Negative 3 a 41 and 18 b 9... 28... 37... 46... 50 / Marks. c =Q 3 –2 2 Q Q 1 4 a Yes, if the range alone is considered. b Hockey: 11... 13... 17... 20... 24 / Fouls. Football: 10... 18.5... 20... 22.5... 23 / Fouls. with the same scale. Fewer fouls on average in hockey but the numbers varied more than in football. 5 a b Ranges and IQRs are the same (35 and 18) but their marks are quite different. One of median (33/72) or mean (33/72) and one of range or IQR. 6 a Points plotted at (35, 0), (40, 20), (45, 85), (50, 195), (55, 222), (70, 235), (75, 240). 35... 43.1... 46.6... 49.3... 75 / Speed (km/h), parallel to speed axis. b Positive skew 7 a i Males: 0... 0... 3... 14... 39 / Trips abroad. Females: 3... 5... 12... 20... 22 / Trips abroad. Same scale. ii Males: range = 39; IQR 14; median 3 = = Females: range 19; IQR 15; median 12 = = = On average, females made more trips abroad than males. Excluding the male who made 39 trips, variation for males and females is similar. b No, there are no data on the number of different countries visited. 8 9 a Ω 0.130 th percentile ≈ 68≈ c a 52 cm2 b c 15.2 to 16.0 cm2 b d ≈ ≈ 0.345 |
0.095 Ω Ω 4.0... 25.8... 33.2... 38.8... 56.0 / Area (cm2). d Area < 6.3cm2 or area 58.3cm2. > Estimate ≈ 8 (any from 0 to 15) Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 10 a Points plotted at (–1.5, 0), (–1.0, 24), (–0.5, 70), 11 x = 12, y = 18 Answers (0, 131), (0.5, 165), (1.0, 199), (1.5, 219), (2.5, 236). Gudrun is 22 years old. Variance increases from 69.12 to 72.88 years2. None of the original 50 staff have been replaced. b 30 12 a Mean decreases by 11.6cm. b 89.9 and ° °1.3 c 18% a 10 11 12 a Points plotted at (0, 0), (4, 2), (11, 21), b c d (17, 44), (20, 47), (30, 50). i ≈ 0.06g ii ≈ 0.12 g 40 ≈n Variation is quite dramatic (from 0 up to a possible 3% of mass). Mushrooms are notoriously difficult to identify (samples may not all be of the same type). Toxicity varies by season. 13 Should compare averages and variation (and skewness) and assess effectiveness in reducing pollution level for health benefits. Exercise 3B a 1 Mean 37.5, SD 12.4 = = 2 3 4 b a Mean 0.45, SD 9.23 = = Var(B) Var(C) Var(P) = = = 96 b The three values are identical. No; mean marks are not identical (B 33, C 53 and P 63). = = = Mean 1 or1.69; variance 1.64 = = 24 35 a b Mean 2; SD 0.803 = =. Q Q 2, soIQR 0 = = = 3 1 That the middle 50% of the values are identical. 5 a Girls: mean 40, SD 13.0 min Boys |
and (4, 2) iii (2, 2), (4, 4), (6, 6) b X Y, and Z are not mutually exclusive, b = 2, c = 6 3 5 ii 13 25 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 7 a C H 12 7 13 3 10 b i 44% a 10 11 8 9 ii 5 8 b 6 11 8 c 7 22 10 a Students who study Pure Mathematics and Statistics but not Mechanics. b i 89 100 ii 6 25 c Mechanics, Statistics, Pure Mathematics 11 a No; P( X Y ) ∩ ≠ b 0.9 12 a A and C 13 a 2 75 14 15 a 0.6 a A 232 0 or equivalent. c 0.7 b 0.22 b 31 75 b 0.4 B c 29 75 9 3 1 2 7 5 0 C b 19; they had not visited Burundi. c d They had visited Angola or Burundi but not Cameroon; 15. 2 9 1 Exercise 4C 1 2 a 1 36 a 0.012 2 3 4 5 6 0.42 a 0.84 a 0.343 b 0.85 b 0.441 b 1 4 c 1 9 b 0.782 7 8 9 a b a a i 0.544 ii 0.3264 iii 0.4872 The result in any event has no effect on probabilities in other events. E.g. winning one event may increase an athlete’s confidence in others. 3 5 8 24 b Untrue. Any number from 0 to 10 may be delivered; 9 is the average. b 0.125 c 0.49 10 a 9 25 or 0.36 b 111 400 or 0.2775 11 a 0.84 12 a 1 4 13 14 27 512 a 0.1 15 a i −k 5 25 16 49 625 b a =k 8; 1 6 i b i 0 Exercise 4D 0.63 1 2 3 4 5 0.28 a 0.32 0.35 ) P( B 0.7 a i b i a 28 b 0.9744 b 3 8 b 0.15 c 0.3 ii 3 −k2 25 ii ii 2 9 1 |
108 b 0.48 ii 0.4 P( C ) ii 0.5 iii 0.06 D S 13 6 7 2 S ′S Totals 6 7 13 13 2 15 19 9 28 D ′D Totals Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy b No; 6 28 ≠ 19 28 × 13 28 6 Yes; 20 80 = 7 a P( ) A = b No; ≠ 1 2 × 32 80 9 16 9 16 50 80 × 3 4, P( B ) = 3 4, P and B both occur when, for example, 1 and 2 are rolled; P( 0 8 a P( Y, P( and ) X Y = 1 12 Yes; 1 12 = 1 4 × 1 3 b No; X and Y both occur when, for example, 1 ≠∩X Y and 5 are rolled; 27 64 V W ∩ 1 16, P(, P( 1 8 W P( V No; 1 16 ≠ 1 8 × 27 64 10 a B B′ Totals 60 50 110 48 42 90 108 92 200 M M ′ Totals b Ownership is not independent of gender; e.g. for M and B: 60 200 ≠ 108 200 × 110 200. c Females 54.3%, males 55.6%. If ownership were independent of gender, these percentages would be equal. 11 a = 1860, b = 4092, c = 1488 12 Southbound vehicles; 54 207 69 207 18 207 × = 36 207 = 54 207 × 138 207 or Exercise 4E 10 13 2 3 3 4 11 19 3 4 4 7 12 19 12 23 Those who expressed an interest in exactly two (or more than one) career, or any other appropriate description. 5 16 ii b i Answers a a 20 39 1 5 b b 8 39 47 57 or 0.825 a 10% of the staff are part-time females. b a = 0.2, b = 0.4, c = 0.3 4 7 i c 3 5 a Proof ii 3 4 iii 4 9 9 11 21 b P(3) = 0.08, P(2) = 0.16, P(1) = 0.75 5 6 7 8 10 c d |
c 0.14 4 s =P( S s ) 5 a Proof b r 0 4 81 1 28 81 237 or 0.25 = 1 5 3 0.2 2 49 81 0 1 2 3 × 40 100 ) P( S × 75 100 ). P = P( and ) P( S b Yes; e.g. 30 100 P = a 312 or 531 441 b 310 or 59 049 c 37 or 2187 14 15 a 3 628 800 b 7 257 600 to show =P( R r ) 0.226 0.446 0.275 0.0527 c 0.774 6 a Proof b v =P( V v ) c 69 91 0 24 91 1 45 91 2 20 91 3 2 91 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 7 Number of red grapes selected ( Number of green grapes selected ( {0, 1} ); R R ∈ ); G G ∈ {4, 5} 18 a =k 315 1012 b 21 46 R G 5 + = 8 d P( = D d ) 9 x 0 0.1 0 1 0.6 1 2 0.3 2 P( = X x ) 0.4096 0.4608 0.1296 Hair colour and handedness are independent. a Proof 10 2 1 16 3 2 16 4 3 16 5 2 16 6 3 16 7 2 16 8 2 16 10 1 16 1 1 14 2 6 14 3 6 14 4 1 14 b x P( = X x ) P( X > 6) = 5 16 11 a 0 b n P( = N n ) c Symmetrical 12 =k 1 27 13 a =c 1 86 b 61 86 14 a 0.374 238 b =N 0 is more likely than P( each time a book is selected. ) =N 4; ) P( ′ > N N 15 a Proof b x P( = X x ) XP( is prime) = 16 a P(heads) 0.2= 0 1 12 7 12 1 4 12 2 4 12 3 3 12 b The number of tails obtained, but many others are possible, such as 2H and 0.5H. P( T H> |
) = 0.896 17 a s =P( S s ) b 1 3 1 17 36 2 9 36 3 10 36 Exercise 6B E( 1 X ) = =p E( Y a b 0.2 2.1; Var( X ) = 0.93 2 3 4 5 6 7 ) 1.84; SD( = Y ) = 0.946 E( T ) = 5, Var( T ) 11.5 = m = 16; Var( V ) = 31.3956 831 ) =RVar( a 11; Var( = W ) = 79.8 a E(grade) profit. = 3.54; SD(grade) 1.20; A smallish = =SD 1.20; variability of the profit. E(grade) 2.46, SD(grade) 1.20 = = b Both are unchanged. 8 a x P( = X x ) 1 1 36 2 3 36 3 3 36 4 5 36 5 3 36 6 10 12 15 20 30 9 36 2 36 4 36 2 36 2 36 2 36 b E( X ) 8 = X > E( X )] = 1 3 ; P[ 5 12 49 41 48 ) X = c Var( 9 a h or 49.9 0 1 2 3 P( = H h ) 0.343 0.441 0.189 0.027 b 900 times E( G ) = 0.8; E( B ) 1.2 = 10 a b 2 : 3; It is the same as the ratio for the number of girls to boys in the class. 336 725 c =GVar( ) 0.463 or 11 a Proof b c ) 1.125 =RE( =GE( ) 1.5 12 a $340 b If the successful repayment rate is below 70%. 13 14 a Proof a 1, 2, 3, 5. b =n 35 s =P( S s ) 1 4 12 2 1 12 3 4 12 5 3 12 b P( S > 3 2 ) 4 = 7 12 c =SVar( ) 2 17 48 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 15 a Proof b x P( = X x ) c ) Var( X ) E( X = 1 4 0 |
1 256 1 12 256 2 54 256 3 108 256 4 81 256 ; the probability of not obtaining B with each spin. End-of-chapter review exercise 6 1 2 X 9 14 ; E( = k ) 1 45 =XVar( 196 ) = 2 or 2.36 5 14 or 1.23 13 or =q 48 a =q b 34 3 a $6675 4 5 6 7 b 8 9 b 4.27 6 11 0.909 2 5 3 5 a i ii b j =P( J j ) a =S 1 0 0.3 1 2 0.6 0.1 s =P( S s ) 0 1 1 36 13 36 2 2 36 3 3 36 4 5 7 8 9 11 14 15 19 24 2 36 2 36 2 36 1 36 2 36 2 36 2 36 1 36 2 36 1 36 ) =SE( 5 31 36 a 0, 1, 2, 4, 5. or 5.86 b x Pb 1 or =b 13 30 6 Answers 10 i Proof ii Score P(Score) 0 24 70 2 30 70 4 13 70 6 3 70 11 12 13 14 15 and 2.78 or 0.4 iv iii 1 6 7 2 5 = k 5 3 ; P Y( > 4 ) = 2 3 25 28 a =x 11 22 127 i Proof b ii x P( = X x ) iii 1 3 i Proof ii x P 120 60 40 30 24 20 171 7 15 13 1 3 1 45 2 45 3 45 4 45 5 45 6 45 7 45 8 45 9 45 239 iii 13 1 4 9 iv 3 or 13.3 or 0.444 7 The binomial and geometric distributions Prerequisite knowledge 1 105 2 1 64 + 9 64 + 27 64 + 27 64 = 1 Exercise 7A a 0.0016 1 c 0.0256 2 a 0.0280 c 0.710 3 a 0.0904 c 0.163 b 0.4096 d 0.0272 b 0.261 d 0.552 b 0.910 d 0.969 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: |
Probability & Statistics 1 4 a 0.121 b 0.000933 c 0.588 10 a 6.006 d 0.403 a 0.246 0.0146 0.254 a 0.140 0.177 a 0.599 0.349 a 0.291 a 0.330 a 0.15625 or 5 6 7 8 9 10 11 12 13 14 15 9 17 16 e 0.499 b 0.296 b 0.000684 b 0.257 b 0.648 b 0.878 16 6 18 23 5 32 b 0.578 19 a 0.0098 b a = 208, b = 3 c 68 20 a =p 0.5; the probability of more than 5 m of rainfall in any given month of the monsoon season. b The probability of more than 5 m of rainfall in any given month in the monsoon season is unlikely to be constant or Whether one month has more than 5 m of rainfall is unlikely to be independent of whether another has. 21 a 0.6561 b 0.227 22 0.244 240 Exercise 7B 1 a 1, 0.8 and 0.894 b 13.2, 5.94 and 2.44 2 3 4 5 6 c 65.7, 53.874 and 7.34 d 14.1, 4.14 and 2.04 a 2 and 1.5 a 0.752 b 0.311 b 0.519 c 0.367 n = 50, p = n = p3, 42, p = = 0.9 a a n = w 0.4 b 0.109 7 12 b 0.0462 0 1 2 3 P( = )W w 0.001 0.027 0.243 0.729 7 a E.g. X is not a discrete variable or there are more than two possible outcomes. b E.g. Selections are not independent. c E.g. X can only take the value 0 or X is not a variable. 8 9 =n 18; 0.364 0.75, p k = = 5157 b 5.93 and 5.93 c Proof d 0.197 11 a 46 b 3.68 c i 0.566 ii 0.320 Exercise 7C a 0.0524 1 a 0.148 a 0.125 a 0.0465 a 0.24 a a i 0.032 i 0.0315 2 3 4 5 6 7 8 9 10 11 b 0. |
91808 b 0.901 b 0.875 b 0.482 b 0.922 ii 0.0016 ii 0.484 c 0.4096 c 0.0672 c 0.0280 b 0.2016 iii 0.440 b Faults occur independently and at random. a 0.21 a 0.364 0.0433 b 0.21 b 0.547 c 0.21 a Not suitable; trials not identical (p not constant). b Not suitable; success dependent on previous two letters typed or X cannot be equal to 1 or 2 or p is not constant. c It is suitable. d Not suitable; trials not identical (p not constant). 12 0.096 13 0.176 14 0.977 or 335 343 15 16 a 0.0965 or 0.103 125 1296 b 0.543 Exercise 7D 14 81 Mode 1, mean 2 = = 6 and 0.335 a 16 b 0.00366 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Answers − 2 ; n = 10 7 a Thierry 10 or 0.0574 45 784 With replacement, so that selections are independent. 8 b a b i or 0.105 ii or 0.0625 27 256 1 16 ) 500; = 9 E( X b = 1001 21 =k n3 = 2 a 12 11 k 12 b i 0.263 ii 0.866 iii 0.0199 13 i 0.735 ii n = 144; k = 6 10 a Any representation of the following sequence. 14 36 : 30 : 25 1st toss 2nd toss 3rd toss T T T T H 2 + 4 0.5 + 6 0.5 8 0.5 + +... Anouar Zane b c 0.5 2 3 End-of-chapter review exercise 7 − 1 n 1 n − n a 0.147 1 2 b 0.00678 3 i 1 4 ii 0.0791 or iii 0.09375 or 81 1024 3 32 4 a 4 9 b 0.394 5 648 5 a 0.59049 b |
0.40951 c 0.242 6 a 10 7 8 b 0.00772 or c 0.00162 a 2–12 b a 137 × 2–16 i 0.0706 ii 0.0494 iii 0.1176 b The students wear earphones independently and at random. 9 i 0.993 ii =n 22 8 The normal distribution Prerequisite knowledge 1 23.4 and 11.232 2 n = 32, p = 0.35 Exercise 8A a False 1 d False 2 a i σ σ> P Q b True e True c False f False 241 ii Median for P < median for Q. iii IQR for for Q. IQR P > b i Same as range of P. ii iii No; High values of W are more likely than low values or negatively skewed Pμ μQμW Women & men 170 Height (cm) Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics Apple juice Peach juice 340 Volume (ml) b Peach juice curve wider and shorter than apple juice curve; equal areas; both symmetrical; both centred on 340 ml UK USA 3.3 3.4 Mass (kg) b USA curve wider, shorter and centred to the right of UK curve; equal areas; both symmetrical. 242 6 a Proof b σ X = 1.11 > σ Y = 0.663 Exercise 8B a 0.715 1 c 0.937 e 0.207 g 0.0401 i 0.975 2 a 0.0606 c 0.0400 e 0.190 g 0.770 i 0.719 X Y 2.4 2.6 b 0.993 d 0.531 f 0.0224 h 0.495 j 0.005 b 0.380 d 0.0975 f 0.211 h 0.948 j 0.066 k 1.333 0.600 =k =k =k =k 1.71 0.473 =c – 0.674 – 1.473 =c =c =c =c 2.10 – 0.500 3.09 0.497.371 – 0.380 0.111 =k =k 1.884 = |
k =k =k 1.035 0.003 =c =c 1.245 – 2.14 =c =c =c –1.90 1.96 Exercise 8C a 0.726 1 b 0.191 c 0.629 2 a 0.919 and 0.0808 b 0.613 and 0.387 c 0.964 and 0.0359 d 0.0467 and 0.953 f 0.954 h 0.319 j b d b d 0.0994 =b 15.5 = 23.6 d =g =j 42.7 17.5 e 0.285 and 0.715 g 0.423 0.231 i a c 35.0 18.5 =a =c e e a = 86.8 11.4 =f =h 0.0513 9.80 c 0.933 3 4 5 6 7 σ = 2.68 12.6 8 µ= 9 µ = 58.8, σ = 14.7 10 µ = 93.8, σ = 63.8 = 5.00, σ = 6.40; 0.0620 = = 7.08, σ = 1.95; 0.933 5.78, σ = 2.13; 0.372 11 µ 12 µ 13 µ 14 0.831 Exercise 8D 0.662 1 2 3 a 0.191 b 74 a Small = 28.60%; medium 49.95%; = = large 21.45% =k 58.0 or 58.1 b Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 4 5 6 7 8 µ= 7.57 a 0.567 240 9 days a =b →9.09 5000 b 0.874 b 82.0 m c 0.136 c 0.257 c 0.118 b =n 14 b 0.189 b 0.0228 4.8 9 σ = 3.33 10 µ= 91.2; 28.8% 11 σ = 3.88 12 a σ = 1.83 a µ= 25.0 a 0.683 1.64, c = σ a 0.950 a 0.659 a 0.284 |
13 14 15 16 17 b 23 b =n 1000 b 0.0456 µ = 6.39 Exercise 8E a Yes; µ 1 b No; nq c Yes; µ d No; np = 2 12, σ= = <1.5 5 2 5.2, = σ = <3 5 = 4.524 2 3 4 5 6 a 209 =n =n 11 B(56, 0.25) c b d 34 =n =n 17 0.837 0.844 a p = 0.625; Var( H ) = 37.5 b 0.0432 7 a Proof b 0.292; np = 10 5 and > nq = 30 5 > b 4.45 ii 0.0118 8 9 a 44 a b i 0.187 E( X ) 1600; Var( = X ) = 320 c 0.874 10 a i 0.105 ii 0.135 b 0.145 0.0958 a 0.0729 a 0.239 a 0.789 11 12 13 14 15 0.100 16 17 0.748 0.660 b 0.877 b 0.0787 b 0.920 Answers 243 End-of-chapter review exercise 8 1 0.841 2 3 4 5 6 0.824 i 0.590 ii np = 24 5 and > nq = > 6 5 0.287 0.239.5 2.0 7 i 0.035 ii 0.471 iii a =k 103 i 315 or 316 8 ii 7350 iii 0.840 b 0.933 9 σ = 2.35 10 µ = 3.285; 61.3% 11 5.69% 12 i 0.238 ii =k 116 iii 0.0910 13 a 0.408 b 0.483 c 0.156 14 a µ = 17.5, 2 σ = 58.0 b i 17.0 min ii →38.2 38 days 15 a σ = 7.24 b =k 15.1 16 0.936 Cross-topic review exercise 3 1 i Proof a, Var( ) = 3 4 or 0.9975 ii E( X iii 399 400 X ) = 63 80 or 0.7875 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy |
Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Practice exam-style paper 1 a 1.72 m b 16.75 2 a b =x 0.154; the value of A )∩ or A B ) ∩ ≠ 0 or equivalent. A B P( P( P( and ) B c Proof 3 a 36 b 5 9 a 0.35 or 0.693 b 95 137 a 11km 4 5 b 61 c 10.8km 6 a 12 b First trial c 0.176 7 a b c 4 x y7, = 6.6 = =b It is neither central nor representative or 8 of the 10 values are less than 89. 0 10 28 1 15 28 2 3 28 8 a x P( = X x ) or 0.402 b c 45 112 11 15 2 3 4 5 6 7 or 0.325 b 13 40 19 27 µ = 25.8, σ = 7.27 0.0228 a σ = 2.99 b 26.1% or 26.2% a µ= 34.0 b 11.9% a σ = 2.70 b 0.276 c 0.822 8 a x P( = X x ) 0 140 285 1 120 285 2 24 285 3 1 285 b i ii 3 5 24 29 or 0.828 244 9 10 11 b a a 6 2 3 2 3 =x 5 209 324 a 0.135 b or 0.645 b 0.253 c np> and 5 nq > 5 12 a Proof b 0.432 13 a =XE( ) 4 b Proof c 0.315 d np = 8.4375 5 and > nq = 11.5625 5 > 14 a i 1 48 p< < ii 0 < p < 1 4 47 48 or b 1 48 p< < 1 4 or 3 4 3 4 < p < 1 p< < 47 48 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy G ssary The following abbreviation and symbols are used in this book. No. ≈ ≠ ∝ ∴ ≡ Meaning Number of is approximately equal to is not equal to is proportional to therefore is identical to A Arrangements: see |
permutations Average: any of the measures of central tendency, including the mean, median and mode B Binomial distribution: a discrete probability distribution of the possible number of successful outcomes in a finite number of independent trials, where the probability of success in each trial is the same C Categorical data: see qualitative data Class: a set of values between a lower boundary and an upper boundary Class boundaries: the two values (lower and upper) between which all the values in a class of data lie Class interval: the range of values from the lower boundary to the upper boundary of a class Class mid-value (or midpoint): the value exactly half-way between the lower boundary and the upper boundary of a class Class width: the difference between the upper boundary and the lower boundary of a class Coded: adjusted throughout by the same amount and/or by the same factor Combinations: the different selections that can be made from a set of objects Complement: a number or quantity of something required to make a complete set Continuity correction: an adjustment made when a discrete distribution is approximated by a continuous distribution Continuous data: data that can take any value, possibly within a limited range Cumulative frequency: the total frequency of all values less than a particular value Cumulative frequency graph: a graphical representation of the number of readings below a given value made by plotting cumulative frequencies against upper class boundaries for all intervals D Dependent (events): events that cannot occur without being affected by the occurrence of each other Discrete data: data that can take only certain values E Elementary event: an outcome of an experiment Equiprobable: events or outcomes that are equally likely to occur Expectation: the expected number of times an event occurs Extreme value: an observation that lies an abnormal distance from other values in a set of data F Factorial: the product of all positive integers less than or equal to any chosen positive integer Fair: not favouring any particular outcome, object or person Favourable: leading to the occurrence of a required event Frequency: the number of times a particular value occurs Frequency density: frequency per standard interval G Geometric distribution: a discrete probability distribution of the possible number of trials required to obtain the first successful outcome in an infinite number of independent trials, where the probability of success in each trial is the same Grouped frequency table: a frequency table in which values are grouped into classes H Histogram: a diagram consisting of touching columns whose areas are proportional to frequencies I Independent (events): events that can occur without being affected by the occurrence of |
each other 245 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Interquartile range: the range of the middle half of the values in a set of data; the numerical difference between the upper quartile and the lower quartile K Key: a note that explains the meaning of each value in a diagram L Lower and upper boundary: the smallest and largest values that can exist in a class of continuous data M Mathematical model: a description of a system using mathematical concepts and language Mean: the sum of a set of values divided by the number of values Median: the number in the middle of an ordered set of values Modal class: the class of values with the highest frequency density Mode: the value that occurs most frequently Mutually exclusive (events): events that cannot occur at the same time because they have no common favourable outcomes N Normal curve: a symmetrical, bell-shaped curve Normal distribution: a function that represents the probability distribution of particular continuous random variables as a symmetrical bell-shaped graph O Ordered data: data arranged from smallest to largest (ascending) or largest to smallest (descending) Outliers: extreme values; observations that lie an abnormal distance from other values in a set of data P Parameters: the fixed values that define the distribution of a variable PDF: see probability density function Permutations: the different orders in which objects can be selected and placed Probabilities: measurements on a scale of 0 to 1 of the likelihood that an event occurs Probability density function (PDF): a graph illustrating the probabilities for values of a continuous random variable Probability distribution: a display of all the possible values of a variable and their corresponding probabilities Q Qualitative data: data that take non-numerical values Quantitative data: data that take numerical values Quartile: any of three measures that divide a set of data into four equal parts R Random: occurring by chance and without bias Range: the numerical difference between the largest and smallest values in a set of data Raw data: numerical facts and other pieces of information in their original form Relative frequency: the proportion of trials in which a particular event occurs S Selection: an item or number of items that are chosen Skewed: unsymmetrical Standard deviation: a measure of spread |
cumulative frequency graph 43–4 modal class 28–9 mode 27, 44 addition law 94–5 conditional 108–9, 111–15 dependent events 112–15 experiments, events and outcomes 91–3 of the geometric distribution 180 multiplication law for independent independent events 100–2, 111–12 multiplication law for independent events 100–2 application of 105–6 multiplication law of probability 112–15 mutually exclusive events 94–5 normal curve 190–1 normal distribution 188, 193 approximation to the binomial distribution 208–12 modelling with 205–6 properties of 194 standard normal variable (Z) 195–9 standardising 200–3 tables of values 222 248 parameters of a binomial distribution 167 of a geometric distribution 176 of a normal distribution 193 Pascal's triangle 168, 172 percentiles 58–9 permutations 123, 134 of n distinct objects 125–6 of n distinct objects with restrictions 129–31 of n objects with repetitions 127–8 nPn notation 125 nPr notation 132 problem solving 138–40 of r objects from n objects 132–3 events 100–2, 105–6 multiplication law of probability 112–15 mutually exclusive events 94–5 Venn diagrams 95–7 probability density functions (PDFs) 189–90 probability distributions 150–2 binomial distribution 166–74 geometric distribution 175–82 normal distribution 190–206 qualitative (categorical) data 2 data representation 20 quantitative data 2–3 quartiles 56 grouped data 58–9 ungrouped data 56–7 random selection 91–2 range 55–6 repetitions, permutations with 127–8 restrictions, permutations with 129–31 selection, random 91–2 set notation 95 sigma (Σ) notation 30 skewed data skewed distributions 190 standard deviation 65–8 of the binomial distribution 173–4 calculation from totals 72 coded data 75–80 of combined datasets 72–3 comparison with interquartile range 69 of a discrete random variable 158 of a normal distribution 190–1, 194 standard normal variable (Z) 195–9 standardising a normal distribution 200–3 stem-and-leaf diagrams 3–4 interquartile range 57 median 42 tree diagrams for independent events 100–1 for permutations 125–6 trials 92–3 variables notation 150 see also continuous random variables; discrete random variables variance 65–8 of the binomial distribution 173–4 calculation from totals 72 of coded data 75–80 of combined datasets 72–3 of a discrete random variable 157–8 equivalence of two formulae for 81 of a normal |
distribution 193 variation 55 see also measures of variation Venn diagrams 95–7 Wiles, Andrew 156 box-and-whisker diagrams 60 measures of central tendency 45–6 Z (standard normal variable) 195–9 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistribution R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistribution R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistribution R evie w C opy - C a m bridge U niversity Press - |
R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistribution R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistribution R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistributionike surds so that bn is the same for all the surds. 1.4.2 Rationalising Denominators It is useful to work with fractions which have rational denominators instead of surd denominators. It is possible to rewrite any fraction which has a surd in the denominator as a fraction which has a rational denominator. We will now see how this can be achieved. Any expression of the form pa + pb (where a and b are rational) can be pb changed into a rational number by multiplying by pa can be rationalised by multiplying by pa + pb). This is because pb (similarly pa ¡ � |
� (pa + pb)(pa pb) = a b ¡ ¡ (1.60) which is rational (since a and b are rational). If we have a fraction which has a denominator which looks like pa+pb, then pb achieving a rational we can simply multiply both top and bottom by pa denominator. ¡ or similarly c pa + pb c ¡ pb pa = = = = c pa + pb ¡ ¡ pa pa cpa a pb pb £ cpb b ¡ ¡ pa + pb pa + pb £ cpa + cpb c ¡ pb pa a b ¡ (1.61) (1.62) 1.4.3 Estimating a Surd (NOTE: has anyone got a better way to do this?) It is sometimes useful to know the approximate value of a surd without having to use a calculator. This involves knowing some roots which have integer solutions. For a surd npa, flnd an integer smaller than a with an integer as its nth root and then flnd the next highest integer (which should also be larger than a) with an integer nth root. The surd which you are trying to estimate will be between those two integers. (NOTE: this paragraph sucks, rewrite it so that it can be understood.) For example, when given the surd 3p52 you should be able to tell that it lies somewhere between 3 and 4, because 3p27 = 3 and 3p64 = 4 and 52 is between 27 and 64. In fact 3p52 = 3:73 : : : which is indeed between 3 and 4. 18 The easiest arithmetic procedure10 to flnd the square root of any number N is to choose a number x that is close to the square root, flnd N x and then use x0 = x+ N for the next choice of x. x converges rapidly towards the actual value of the square root - the number of signiflcant digits doubles each time. (NOTE: arithmetic? procedure? converges? rapidly? this language is not basic enough!) 2 x We will now use this method to flnd p55. We know that 72 = 49, therefore the square root of 55 must be close to 7. Let x = 7 then, 55 7 7 + 7;8571 : : : 2 55 7:4285 : : |
: = 7;8571 : : : = 7;4285 : : : = 7;4038 : : : ) x0 = (1.63) (1.64) (1.65) (1.66) ) x0 = 7;4285 : : : + 7;4038 : : : 2 = 7;4162 : : : (1.67) Using a calculator we flnd that p55 = 7;416198 : : :, which is very close to our approximation. 1.5 Accuracy The syllabus requires: write irrational (and rational) solutions rounded to a specified degree of accuracy know when to approximate an irrational by a terminating rational, and when not to express large and small numbers in scientific or engineering notation † † † We already mentioned in section 1.2.2 that certain numbers may take an inflnite amount of paper and ink to write out. Not only is that impossible, but writing numbers out to a high accuracy (too many decimal places) is very inconvenient and rarely gives better answers. For this reason we often estimate the number to a certain number of decimal places or to a given number of signiflcant flgures, which is even better. (NOTE: the notes on rounding need to be better. this is not very good.) Approximating a decimal number to a given number of decimal places is the quickest way to approximate a number. Just count along the number of places you have been asked to approximate the number to and then forget all the numbers after that point. You round up the flnal digit if the number you cut ofi was greater or equal to 5 and round down (leave the digit alone) otherwise. For example, approximating 2;6525272 to 3 decimal places is 2;653 because the flnal digit is rounded up. 10This procedure is known as Heron’s Method, which was used by the Babylonians over 4000 years ago. 19 (NOTE: more on the difierence between DP and SF needed) In a number, each non-zero digit is a signiflcant flgure. Zeroes are only counted if they are between two non-zero digits or are at the end of the decimal part. For example, the number 2000 has 1 signiflcant flgure, but 2000;0 |
has 5 signiflcant flgures. Estimating a number works by removing signiflcant flgures from your number (starting from the right) until you have the desired number of signiflcant flgures, rounding as you go. For example 6;827 has 4 signiflcant flgures, but if you wish to write it to 3 signiflcant flgures it would mean removing the 7 and rounding up, so it would be 6;83. It is important to know when to estimate a number and when not to. It is usually good practise to only estimate numbers when it is absolutely necessary, and to instead use symbols to represent certain irrational numbers (such as …); If it is necessary approximating them only at the very end of a calculation. to approximate a number in the middle of a calculation, then it is often good enough to approximate to a few decimal places. 1.5.1 Scientiflc Notation In science one often needs to work with very large or very small numbers. These can be written more easily in scientiflc notation, which has the general form 10m a £ (1.68) where a is a decimal number between 1 and 10. The m is an integer and if it is positive it represents how many zeros should appear to the right of a. If m is negative then it represents how many times the decimal place in a should 3 10¡ be moved to the left. For example 3;2 represents 0;0032. 103 represents 32000 and 3;2 £ £ If a number must be converted into scientiflc notation, we need to work out how many times the number must be multiplied or divided by 10 to make it into a number between 1 and 10 (i.e. we need to work out the value of the exponent m) and what this number is (the value of a). We do this by counting the number of decimal places the decimal point must move. It is usually enough to estimate a to only a few decimal places. 1.5.2 Worked Examples Worked Example 1 : Manipulating Rational Numbers Question: Simplify the following expressions a) 7 8 + 5 2 b) 11 27 £ 20 3 c) 73 69 ¥ 73 69 Answer: a) Step 1 : Rule of addition Write out the |
rule of addition for rational numbers (1.25) a b + c d = ad + bc bd 20 Step 2 : Fill in the values Fill in the values for a;b;c and d. Here you can read ofi that a = 7, b = 8, c = 5 and £ = 54 16 £ Step 3 : Minimise the denominator 54 16 is the correct answer, but it is not the simplest way to write it. We can see that both 54 and 16 can be divided by 2, so we divide both by 2 and get 27 8, which cannot be simplifled any further. b) Step 1 : Rule of multiplication Write out the rule of multiplication for rational numbers (1.26) a b £ c d = ac bd Step 2 : Fill in the values Fill in the values for a;b;c and d. Here you can read ofi that a = 11, b = 27, c = 20 and d = 3 11 27 £ 20 3 = 11 27 20 3 = 220 81 £ £ There is no number which will divide into both 220 and 81, so 220 81 is the simplest form of the answer. c) Step 1 : Use the division rule Calculate the reciprocal of 73 (1.28) 69 = 69 73, and write out the division rule = ad bc Step 2 : Fill in the values We can read ofi that a = c = 73 and b = d = 69 so 73 69 69 73 £ £ = 5037 5037 = 1 This question could also have been answered in one single line by noticing that the two fractions are the same, and any number divided by itself is one. 1.5.3 Exercises TODO 21 Chapter 2 Patterns in Numbers (NOTE: SH notes:at the moment, this whole chapter needs a lot more inline examples. i think it is too complicated for 16 year olds without examples. even the use of indices is also some of the equations might be over their heads. inconsistent... the use of letters like i, n, m should not be interchanged so much as it is only leading to confusion.) (NOTE: also, i think we are aiming too high. it is possible that when the syllabus says \prove", it really means \show explicitly the flrst few terms and assume the rest of the sequence is the same". so perhaps we should drop a few of the proofs.) 2.1 Sequences |
The syllabus requires: investigate number patterns, be able to conjecture a pattern and prove those conjectures recognise a linear pattern when there is a constant difference between consecutive terms recognise a quadratic pattern when there is a constant 2nd difference identify ‘‘not real’’ numbers and how they occur (NOTE: i think this would be best taught in quadratic equations as that is the only place they occur in this syllabus) (grade 12) arithmetic and geometric sequences † † † † † Can you spot any patterns in the following lists of numbers? 2;4;6;8; : : : 1;2;4;7; : : : 1;4;9;16; : : : 5;10;20;40; : : : 3;1;4;1;5;9;2; : : : 22 (2.1) (2.2) (2.3) (2.4) (2.5) The flrst is a list of the even numbers, the numbers in the second list flrst difier by one, then by two, then by three. The third list contains the squares of all the integers. In the fourth list, every term is equal to the previous term times two and the last list contains the digits of the number …. These lists are all examples of sequences. In this section we will be studying sequences and how they can be described mathematically. (NOTE: a few real world examples here wouldn’t go amiss.) A sequence is a list of objects (in our case numbers) which have been ordered. We could take as an example a sequence of books. If you put all your books in alphabetical order by the author, that would be a sequence because it is a list of things in order. Someone could look at the sequence and work out how you ordered them if they knew the alphabet. You could rearrange the collection so that it was ordered alphabetically by title. That would be a difierent sequence because the order is difierent. Similarly the sequence of numbers 1;2;3 is difierent to 3;2;1. You could even shu†e up all the books so that the order they were in didn’t follow a pattern, but they would still make a sequence. Notice that not all sequences have to continue forever - what character |
ises a sequence is that it is a list which is ordered. In the alphabetised books example, someone who didn’t know the alphabet would not be able to work out how you had ordered the books. How would you be able to flnd your seats at the theatre or at a stadium if the seats were not ordered? Likewise if you are shown a sequence of numbers, you may not be able to work out what pattern relates them. That might be because there is no pattern, or it might just be that you can’t see it straight away. We will be thinking about sequences in this chapter and it is useful to be able to talk in general about them. We will want to talk about the numbers in the sequence, so rather than having to say something longwinded like \the second term in the sequence is related to the flrst term by this rule....", we give each term in the sequence a name. The flrst term of a sequence is named a1, the second term is named a2 and the nth term is named an. Now we can say \a2 is related to a1 by this rule...". The small n or number like 1 or 2 beside the letter is called a subscript or index but we will refer to it as the subscript It helps us keep everything tidy by using the same letter (in this example, a) for all the terms in a sequence. A sequence does not have to follow a pattern, but when it does we can often write down a formula for the nth term, an. In the example above, 2.3 where the sequence was of all square numbers, the formula for the nth term is an = n2. You can check this by looking at a1 = 12 = 1; a2 = 22 = 4; a3 = 32 = 9; : : : 2.1.1 Arithmetic Sequences Deflnition: A linear arithmetic sequence is a sequence in which each successive term difiers by the same amount. Each term is equal to the previous term plus a constant number. Tn = 1 + k1 where k1 is some constant. We will see in the next example what this Tn constant is and how to determine it. ¡ Say you and 3 friends decide to study for maths and you are seated at a square table. A few minutes later, 2 other friends join you and would like to sit at your table and |
help you study. Naturally you move another table and add it 23 Figure 2.1: Tables moved together to the existing one. Now six of you sit at the table. Another two of your friends join your table and you take a third table and add it to the existing tables. Now 8 of you can sit comfortably. Let assume this pattern continues and we tabulate what is happening. (NOTE: Insert pictures here.) No. Tables (n) No of people seated 1 2 3 4... n 4 = 4 4+2 = 6 4+2+2 = 8 4+2+2+2 = 10... 4+2+2+2+... +2 Formula = 4 + 2(0) = 4 + 2(1) = 4 + 2(2) = 4 + 2(3)... = 4 + 2(n - 1) We can see for 3 tables we can seat 8 people, for 4 tables we can seat 10 people and so on. We started out with 4 people and added two the whole time. Thus for each table added, the number of persons increase with two. Thus, 4;6;8;::: is a sequence and each term (table added), difiers by the same amount (two). More formally, the number we start out with is called a1 and the difierence between each successive term is d. Now our equation for the nth term will be: an = a1 + d(n 1) ¡ (2.6) The general linear sequence looks like a1; a1 + d; a1 + 2d; a1 + 3d; : : :, using the general formula?? How many people can sit in this case around 12 tables? By simply using the derived equation we are looking for where n = 12 and thus a12 an = a1 + d(n a12 = 4 + 2(12 = 4 + 2(11) ¡ ¡ 1) 1) OR = 4 + 22 = 26 24 How many tables would you need for 20 people? an = a1 + d(n 20 = 4 + 2(n 1) ¡ 1) ¡ 20 16 ¡ ¥ 4 = 2(n 2 = n ¡ 1 1 a2 = d A simple test for an arithmetic sequence is to check that a2¡ This is quite an important equation and is a deflnitive test for an arithmetic sequence. If this |
condition does not hold, the sequence is not an arithmetic sequence. It is also important to note the difierence between n and an. n can be compared to a place holder while an is the value at the place ’held’ by n. Like our study table above.Table 1 holds 4 people thus at place n=1 the value of a1 = 4. a1 = a3¡ n an 1 4 2 6 3 8 4 10...... 2.1.2 Quadratic Sequences (NOTE: maybe put in a note about the quadratic equations section, and 1st/2nd difierences in terms of difierentiating wrt n.) A quadratic sequence is a sequence in which the difierences between each consecutive term difier by the In the example sequences same amount, called a constant second difierence. in the introduction, equation (2.2) is a quadratic sequence because the difierence between each term difiers by one each time. We can look at the difierence between each term and see that the difierences form a linear sequence: a a2 ¡ 3 ¡ ¡ a4 ¡ a1 = 2 a2 = 4 a3 = Here you can see clearly that the difierence between each difierence is 1. We call this the constant second difierence and in this case is = 1. The general form of this example of a quadratic sequence (a sequence with constant second difierence) is an = 1 2 (n2 1) ¡ ¡ 1 2 (n ¡ 1) + 1 (2.7) For a general quadratic sequence with constant second difierence D the formula for an is an = D 2 (n2 1) + d(n ¡ ¡ 1 is Dn + d. 1) + a1 (2.8) The difierence between an and an Check for yourself that an ¡ an then again for an an ¡ difierence not equal to 1. ¡ an 1 (setting n = n ¡ ¡ 1 = Dn + d. (Use the formula for an and 1 in the formula) and then work out what 1 is.) Make up your |
own quadratic sequences with a constant second ¡ ¡ 25 Figure 2.2: Tree diagram of series 2.1.3 Geometric Sequences Deflnition: A geometric sequence is a sequence in which every number in the sequence is equal to the previous number in the sequence, multiplied by another constant number. This means that the ratio between consecutive numbers in the sequence is a constant. We will explain what we mean by ratio after looking at this example. What is inuenza (u)? Inuenza, commonly called ’the u’, is caused by the inuenza virus, which infects the respiratory tract (nose, throat, lungs). It can cause mild to severe illness, that most of us get during winter time The main way that inuenza viruses are spread is from person to person in respiratory droplets of coughs and sneezes. (This is called ’droplet spread.’) This can happen when droplets from a cough or sneeze of an infected person are propelled (generally up to 3 feet) through the air and deposited on the mouth or nose of people nearby. It is good practise to cover your mouth when you cough or sneeze to not infect others around you when you have the u. Lets assume you have the u virus and you forgot to cover your mouth when two friends came to visit while you were sick in bed. They leave and the next day, they also have the u. Lets assume that they in turn spread the virus to two of their friends by the same droplet spread the following day. Lets assume this pattern continues and each person infected, infects 2 other friends. We can represent these events in the following manner: (NOTE: Insert pictures here.) Again we can tabulate the events and formulate an equation for the general case: # day (n) 1 2 3 4 5... n # Carrier You spread virus 2 4 8 16...... # Recipients/Carrier 2 4 8 16 32...... Formula 21 8 = 2 x 4 = 2 x 2x2 = 2 x 22 16 = 2 x 8 = 2 x 2x2x2 = 2 x 23 32 = 2 x 16 = 2 x 2x2x2x2 = 2 x 24... = 2 x 2x2x2x...x2 = 2 x 2n 1 ¡ You sneeze and the virus is carried over to 2 people who start the chain ( |
a1 = 2). The next day, each one then infects 2 of their friends. Now 4 people are infected. Each of them infects 2 people the third day and 8 people are infected etc. These events can be written as a geometric sequence: 2;4;8;16;32;::: Note the common factor between the events. Recall from the linear arithmetic sequence how the common difierence between terms were established. In the 26 geometric sequence we can determine the common factor, r by Or more general a2 a1 = a3 a2 = r an+1 an = an+2 an+1 = r (2.9) (2.10) a2 a1 is called the ratio and is used to describe the ’factor difierence’ between the elements of the series. i.e. The ratio between a1 and a2 is 2 From the question in the above example we know a1 = 2 and r = 2 and we 1. Thus in have seen from the table that the nth term is given by an = 2 general, 2n £ ¡ an = a1rn ¡ 1 (2.11) So if we want to know how many people has been infected after 10 days, we need to work out a10 1 ¡ 1 an = a1rn ¡ 210 a10 = 2 29 = 2 512 = 2 £ £ = 1024 £ Or, how many days would pass before 16384 people are infected with the u virus? (NOTE: I’m not sure if SURDs and exponents have been done at this stage. check flrst. This chapter should be taught AFTER exponents and SURDs because the techniques are used!!) 16384 1 1 ¡ an = a1rn ¡ 2n 1 16384 = 2 £ 2 = 2n ¡ ¥ 8192 = 2n 213 = 2n 13 = n n = 14 ¡ 1 ¡ ¡ 1 1 2.1.4 Recursive Equations for sequences When discussing linear and quadratic sequences we noticed that the difierence between two consecutive terms in the sequence could be written in a general way. For linear sequences, where the constant difierence between two consecutive 1 = d for any term in terms was d, we can write this information as an ¡ the sequence. We can rearrange this to an = an 1 + d. This is an expression |
¡ for an in terms of an 1, which is called a recursive equation. So the recursive equation for a linear sequence of constant difierence d is an ¡ ¡ an ¡ an 1 = d ¡ (2.12) 27 We can do the same thing for quadratic sequences. There we noticed that 1 = Dn + d. Then the recursive equation for a quadratic sequence with an ¡ constant second derivative D is an ¡ an ¡ an 1 = Dn + d ¡ (2.13) an 2 wouldn’t include d.) (NOTE: Here we haven’t said explicitly what d is or how to work it out. This bothers me. I think to be honest that you need more information. like maybe an ¡ It is not always possible to flnd a recursive equation for a sequence, even when you know the general way to write down any term an. Can you flnd a recursive equation for a geometric sequence? This is not supposed to be easy! It’s just to get you to have a go at working things out. ¡ Recursive equations are extremely powerful: you can work out every term in the series just by knowing the previous one, and as you can see for the example above, working out an from an 1 can be a much simpler computation than working out an from scratch using a general formula. This means that using a recursive formula when programming a computer to work out a sequence would mean the computer would flnish its calculations signiflcantly quicker. (NOTE: Real world example of this?) ¡ 2.1.5 Extra (NOTE: Jacques had sections on arithmetic/geometric means. i want to add that back in, but make it clear it is non-syllabus. No questions are usually asked in the flnal exam as far as I can see, but it is part of the syllabus.) 2.2 Series (Grade 12) The syllabus requires: (grade 12) prove and calculate the following sums † 1 = n n Xi=1 n i2 = Xi=1 n(2n + 1)(n + 1) 6 n Xi=1 a + (i ¡ 1)d = n 2 (2a + (n 1)) ¡ n Xi=1 1 Xi=1 a:ri ¡ 1 = a:ri ¡ 1 = 1) |
a(rn NOTE: equation 3 is not correct in the official syllabus. is a d missing.) there 28 When we sum terms in a sequence, we get what is called a series. If we only sum a flnite amount of terms, we get a flnite series. We use the symbol Sn to mean the sum of the flrst n terms of a sequence. For example, the sequence of numbers 3;1;4;1;5;9;2; : : : has a flnite series S4 which is simply the flrst 4 terms added together. If we sum inflnitely many terms of a sequence, we get an inflnite series. A sum may be written out using the summation symbol (Sigma). This symbol is the capital \S" (for Sum) in the Greek alphabet. It indicates that you must sum the expression to the right of it P n Xi=m ai = am + am+1 + : : : + an 1 + an ¡ (2.14) ai are the terms in a sequence and here we sum from i = m (as indicated below the summation symbol) up until i = n (as indicated above). We usually just sum from n = 1, which is the flrst term in the sequence. In which case we can notation since they mean the same thing use either Sn or P n Sn = ai = a1 + a2 + : : : + an (2.15) Xi=1 For example, in the following sum 5 i Xi=1 (2.16) we have to add together all the terms in the sequence ai = i from i = 1 up until = 15 (2.17) which gives us 15. Xi=1 2.2.1 Finite Arithmetic Series When we sum a flnite number of terms in an arithmetic sequence, we get a flnite arithmetic series. The simplest arithmetic sequence is when a1 = 1 and d = 0 in the general form (??), in other words all the terms in the sequence are one. ai = d(i = 0(i = 1 ¡ ¡ 1) + a1 1) + 1 a = 1;1;1;1;1; : : : (2.18) If we wish to sum this sequence from i = |
1 to any integer n, we would write n Xi= times (2.19) 29 Since all the terms are equal to one, it means that if we sum to an integer n we will be adding n number of ones together, which is equal to n. n Xi=1 1 = n (2.20) Another simple arithmetic sequence is when a1 = 1 and d = 1, which is the sequence of positive integers ai = d(i = (i 1) + a1 ¡ 1) + 1 ¡ = i (2.21) a = 1;2;3;4;5; : : : If we wish to sum this sequence from i = 1 to any integer n, we would write n Xi=2.22) This is an equation with a very important solution as it gives the answer to the sum of positive integers1. We notice that the largest number may be added to the smallest, then the second largest added to second smallest, giving the same number. If we keep doing this we flnd that all the numbers may be paired up together like this until we reach the middle and there are no more numbers left to pair ofi a1 + a2 + : : : + an = (a1 + an) + (a2 + an 1) + : : : ¡ n 2 times (2.23) If there are an odd number of numbers, then we must not forget to add the unpaired number to the answer at the end. For example 1 + 5) + (2 + 4) + 3 (2.24) = 6 + 6 + 3 = (3 + 3) + (3 + 3) + 3 = 15 We can write this down in general as n Xi=1 i = n 2 (n + 1) (2.25) If we wish to sum any arithmetic sequence, there is no need to work it out term for term as we just have for these examples. We will now show what the general form of a flnite arithmetic series is by starting with the general form of an arithmetic sequence and summing it from i = 1 to any integer n. 1A famous mathematician named Carl Friedrich Gauss discovered this proof when he was only 8 years old. His teacher had decided to give his class a problem which would distract them for the entire day by asking them to add all the numbers from 1 to 100. Young Carl realised how to do this almost instantaneously and |
shocked the teacher with the correct answer, 5050. 30 Writing out the sum of a sequence and then substituting in the general form for an arithmetic sequence gives us n n ai = Xi=1 Xi=1 d(i ¡ 1) + a1 (2.26) If there is a sum inside a sum, we can break it into two separate sums and calculate each part separately. n Xi=1 d(i ¡ n 1) + a1 = di + (a1 ¡ d) (2.27) Xi=1 left |{z} right | {z } If a sum is multiplied by a constant, we can take the constant outside of the d, which is a constant, so we may. The term on the right is a sum of a1 ¡ rewrite that term as P n Xi=1 a1 ¡ d = (a1 ¡ = (a1 ¡ n d) 1 Xi=1 d)n (2.28) Here we used equation (2.20) to arrive at the solution. The term on the left of equation (2.27) is also quite simple. Firstly we can take the constant d out of the sum n n and then we can use equation (2.25) to flnd di = d i Xi=1 Xi=1 n d i = Xi=1 dn 2 (n + 1) (2.29) (2.30) Adding together the solutions to the left and right terms (equations (2.28) and (2.30)) we get the general form of a flnite arithmetic series n Xi=1 d(i ¡ 1) + a1 = n 2 (2a1 + d(n 1)) ¡ (2.31) For example, if we wish to know the series S20 for the arithmetic sequence 1) + 3, we could either calculate each term individually and sum them ai = 7(i ¡ 20 Xi=1 7(i ¡ 1) + 3 = 3 + 10 + 17 + 24 + 31 + 38 + 45 + 52 +59 + 66 + 73 + 80 + 87 + 94 + 101 +108 + 115 + 122 + 129 + 136 = 1390 (2.32) or more sensibly, we could use equation (2.31) noting that d = 7, a1 = 3 and n = 20 so that 20 Xi=1 7( |
i ¡ 1) + 3 = 20 2 (2 £ 3 + 7 19) £ = 1390 (2.33) In this example, it is clear that using (2.31) is beneflcial. 31 2.2.2 Finite Squared Series When we sum a flnite number of terms in a quadratic sequence, we get a flnite quadratic series. The general form of a quadratic series is quite complicated, so we will only look at the simple case when D = 1 and d = a0 = 0 in the general form (??). This is the sequence of squares of the integers ai = i2 = 12;22;32;42;52;62; : : : = 1;4;9;16;25;36 : : : (2.34) If we wish to sum this sequence and create a series, then we write Sn = n Xi=1 i2 = 1 + 4 + 9 + : : : + n2 (2.35) which can be written in general as (NOTE: the syllabus requires that we prove this result! any ideas, without confusing the hell out of a 16 year old? i thought even the other ones were a bit too hard for this level, to be honest.) n Xi=1 i2 = n(2n + 1)(n + 1) 6 (2.36) 2.2.3 Finite Geometric Series When we sum a flnite number of terms in a geometric sequence, we get a flnite geometric series. We know from (??) that we can write out each term of a geometric sequence in a general form. By simply adding together the flrst n terms in the general form we are actually writing out the series Sn = a1 + a1r + a1r2 + : : : + a1rn ¡ 1 We may multiply this by r on both sides, giving us rSn = a1r + a1r2 + a1r3 + : : : + a1rn (2.37) (2.38) You may notice that all the terms are the same in (2.37) and (2.38), except the flrst and last. If we subtract (2.37) from (2.38) we are left with just rSn ¡ Sn(r � |
� Sn = a1 + a1rn 1) = a1(1 + rn) (2.39) dividing by (r sequence since Sn = ¡ n i=1 a:ri ¡ 1 1) on both sides, we have the general form of a geometric 1) (2.40) a(rn r ¡ 1 ¡ P n Xi=1 a:ri ¡ 1 = 32 2.2.4 Inflnite Series Thus far we have been working only with flnite sums, meaning that whenever we determined the sum of a series, we only considered the sum of the flrst n terms. It is the subject of this section to consider what happens when we add inflnitely many terms together. You might think that this is a silly question - surely one will get to inflnity when one sums inflnitely many numbers, no matter how small they are? The surprising answer is that in some cases one will reach inflnity (like when you try to add all the integers together), but in some cases one will get a flnite answer. If you don’t believe this, try doing the following sum on your calculator or computer: 1 32 + :::. You might think that if you keep adding more and more terms you will eventually get larger and larger numbers, but in fact you won’t even get past 1 - try it and see for yourself! 16 + There is a special sigma notation for inflnite series: we write 1i=1 i to indicate the inflnite sum 1 + 2 + 3 + 4 + ::::. When we sum the terms of a series, and the answer we get after each summation gets closer and closer to some number, we say that the series converges. If a series does not converge, we say that it diverges. P There is a rule for knowing instantly which geometric series converge and which diverge. When r, the common ratio, is strictly between -1 and 1, i.e. 1 < r < 1, the inflnite series will converge, otherwise it will diverge. There ¡ is also a formula for working out what the series converges to. The sum of an inflnite series, symbolised by S, is given by the formula 1 S |
1 = 1 Xi=1 a1:ri ¡ 1 = a1 ¡ 1 r 1 < r < 1 ¡ (2.41) where a1 is the flrst term of the series, and r is the common ratio. (NOTE: the syllabus requires us to PROVE this series! how can we do that without a notion of a limit? again the syllabus talks nonsense.) We can see how this comes about by looking at (2.40), as. We can ignore the rn term since a small number raised to the power of inflnity is inflnitely small. Try this yourself by typing in a number between -1 and 1 into your calculator and square it, continuing to square the answers thereafter; your calculator will eventually decide that the answer is zero. 1 < r < 1 and n = 1 ¡ 2.3 Worked Examples (NOTE: I think maybe the worked examples should follow the relevant section. Check if the general layout is set.) 1. Classify the following as arithmetic sequence or geometric sequence: 15;19;23; : : : For arithmetic sequence, We have to check for a common difierence or a common ratio. a2 ¡ a2 ¡ a3 ¡ a1 = a3 ¡ a1 = 19 ¡ a2 = 23 ¡ a2 = d 15 = 4 19 = 4 33 Thus, a2 ¡ sequence and d = 4 a1 = a3 ¡ a2 = 4 and we can say that 15;19;23; : : : is an arithmetic 2. Classify the following as arithmetic sequence or geometric sequence: 5;10;20; : : : For arithmetic sequence, We have to check for a common difierence or a common ratio. a2 ¡ a2 ¡ a3 ¡ Thus, a2 ¡ a1 = a3 ¡ a1 = 10 ¡ a2 = 20 ¡ = a3 ¡ a1 6 a2 = d 5 = 5 10 = 10 sequence. a2 and we can say that 5;10;20; : : : is not an arithmetic Test for geometric sequence: a2 a1 a2 a1 a3 a2 = a3 = r a2 = 10 5 = 2 = 20 10 = 2 Thus, a2 a1 sequence. = a3 a2 and r = 2 and we can |
say that 5;10;20; : : : is a geometric 3. Determine d and a9 for the following arithmetic sequence: 17;14;11; : : : It is given that 17;14;11; : : : is an arithmetic sequence, thus a2 ¡ 14 ¡ d = a2 = d 14 = 3 ¡ a1 = a3 ¡ 17 = 11 ¡ 3 ¡ To determine a9 we use an = a1 + d(n Thus: ¡ 1) with n = 9 1) ¡ 3)(9 an = a1 + d(n a9 = 17 + ( = 17 ¡ = 17 ¡ 7 = ¡ 4. Determine r and a7 for the following geometric sequence: 81; ¡ 3(8) 24 1) ¡ 27; 9; : : : ¡ a2 a1 = a3 a2 = ¡ = 27 81 9 27 ¡ = = = r 1 3 1 3 ¡ ¡ a2 a1 a3 a2 34 To determine a7 we use an = a1rn Thus: 1 with n = 7 ¡ an = a1rn ¡ 1 a7 = (81)( 1 )7 ¡ 1(6) 1 3 ¡ (34)(3¡ (34:3¡ 6) 6) (34 ¡ 2) (3. The third term of a geometric sequence is equal to 1 and the 5th term is 16. Find r and the seventh term Given: a3 = 1 and a5 = 16 We also know an = a1rn Thus: ¡ 1 a3 = a1r3 ¡ 1 = a1r2 1 a5 = a1r5 ¡ 16 = a1r4 1 Also: Dividing (2) by (1): a1r4 a1r2 = 16 1 16 = r2 r = 4 To flnd a7 we use a7 = a1r7 1 but flrst we need a1. From (1) we know: ¡ 35 1 = a1r2 1 = a1:(4)2 a1 = 1 16 1 a7 = a1r7 ¡ :46 = 1 16 4096 16 = 256 = 6. The fourth term of an arithmetic sequence is 1 1 2 and the the 8th term is 1 2. Find the second term. Given |
: a4 = 3 Thus we can use an = a1 + d(n 2, a8 = 1 2 and this is an arithmetic sequence. 1) ¡ a4 = 3 a8 = 1 2 = a1 + d(4 2 = a1 + d(8 ¡ ¡ 1) = a1 + 3d and 1) = a1 + 7d Subtract the one equation from the other to get rid of a1 and solve for da1 + 3d) 4d 1 4 (a1 + 7d) ¡ ¡ ¡ 1 4 )(7) 1 2 = a1 + ( ¡ 7 1 4 ) 2 = a1 ¡ a1 = 9 4 a2 = 9 a2 = 8 ( 1 4 )(2 4 ¡ 4 = 2 1) ¡ 2.4 Exercises 1. Classify the following as arithmetic sequence or geometric sequence: 1 3 ¡ ; : : : i) ii) iii) iv) v) ; ; ;0;1; : : : 1; 3 1 2 6 22;2;1. Find a7 for each of the series above. 3. Determine which term in the series 14;8;2; : : : is equal to 34? 4. Which ¡ 36 term in 2;6;18; : : : is equal to 486? 5. In a geometric series, a4 = 2 6. In an arithmetic series, a3 = 7. The third term of a Geometric series is equal to minus three eights and the seventh term is equal to 3 8. Insert 4 numbers between 4 and 972 to form a geometric series. 3 and a6 = 3 ¡ 2 and a8 = 23. Determine a1 and d. 128. Find a5. 2. Find a2. 37 Chapter 3 Functions 3.1 Functions and Graphs The syllabus requires: † † † † † (grade 12) formal definition of the function concept able to switch between words, tables, graphs and formula to represent the relation between variables generate graphs using point to point plotting to test conjectures on relations between x and y for the situations (NOTE: this list we do the trig has changed in the latest syllabus... functions in the trig section. it makes more sense this way.) be warned. y = ax + b y = ax + b a > 0; = ax+b + |
c (NOTE: page 25 of the syllabus lists some more situations, but they are corrupted. the syllabus and add them to this list) we need to get an uncorrupted version of identify the domain and range, axes intercepts, turning points (max/min), asymptotes, shape and symmetry, periodicity and amplitude, rates of change, increasing/decreasing ranges and continuity. and can sketch graphs using these characteristics (grade 12) can generate graphs of function inverses. in particular a > 0; a = 1 y = ax + b y = ax y = ax2 y = sin(x) 38 6 6 † (grade 12) decide which inverses are functions and if necessary the restriction to make it a function (NOTE: functions are neither conceptually simple nor very interesting - so this intro needs to be very sexy (and i know mine probably isn’t, so rewriting is good). try to reword so as to not use 1st and 3rd person.) Most people don’t know it but they’ve come across functions all their lives. In fact, our very existence is tied to certain, very special functions called the laws of nature. Even ignoring those, though, it would be di–cult to go through a day without coming into contact with all sorts of functions. We can say that the idea of a function is one of the most basic and powerful ideas in the mathematics. Functions everywhere? But who’s ever heard of such a thing? Where does the word even come from? Well, they are everywhere, and, once you begin to see them, functions will be the easiest concept in mathematics. Where are these functions? Well, the menu in a restaurant is a function. So are the prices in a supermarket. Today’s temperature. Your height, your age, your weight. These are all functions. A function is just a way of attaching or relating one thing to another. A menu attaches prices to the food in a restaurant, and a supermarket attaches prices to the things it sells. We need to notice one very important fact, which is that these functions can give only one price to each item. We would certainly get angry if a restaurant charged two difierent prices for the same dish. However, it’s perfectly natural for a restaurant to charge the same price for difierent dishes. Similarly, one person cannot have two difierent heights, but two |
people can have the same height. 3.1.1 Variables, Constants and Relations A variable is a label which we allow to change and become any element of some set of numbers. For example, on a menu in a restaurant \price" is a variable on the set of real numbers, since for any menu item the manager can choose any price he or she feels like (with the aim of staying in business). Most often, a variable will be a letter which can take on any value in some set of numbers. In this textbook we will only use real variables, which may take on the value of any real number. Though a variable is free to vary, if we wish we can specify that the variable takes on a speciflc value, in which case we say that we assign a value to the variable. In fact, we do this all the time when working with variables. When we say \what if we set the price to R50", we are just assigning the value \R50" to the variable \price". You have probably already done this quite frequently in algebra, when you say \let x be 1". A constant is a variable which is flxed. We may not know the value of this constant, but this is a number which does not change throughout any problem. The \speed of light" is a variable which is always 300 000km per second, i.e. it is a constant. Such constant variables occur most frequently in the laws of physics. Variables on their own are very abstract, so don’t worry if it is slightly confusing. They become much more understandable when we start to relate them to each other. \Price" on a menu may not be a constant, but it must be tied to the items on that menu. For each item, we have a specifled price. We can 39 think of \item" as a variable in its own right, and then the menu does nothing but tell us the relationship between the two variables \item" and \price". In general, a relation is an equation which relates two variables. For example, y = 5x and y2 + x2 = 5 are relations. In both examples x and y are variables and 5 is a constant, but for a given value of x the value of y will be very difierent in each relation. Our example of a restaurant menu shows that relations between variables take on varied representations. Besides writing them as form |
ulae, we most often come across relations in words, tables and graphs. Instead of writing y = 5x, we could also say \y is always flve times as big as x". We could also give the following table: (NOTE: Working on a Latex-less machine, so table will come later) (I put in a table but not sure if it’s ok - Jothi) x 2 6 8 13 15 y = 5x 10 30 40 65 75 Some of you may object that this table isn’t very satisfactory, as the same table could represent almost any relation between x and y. However, when using tables we normally cheat and just assume that the obvious relationship in the table is the relationship. Finally, we look at graphs (NOTE: surely thisneeds to wait until later? sorry - structuring major headache here) 3.1.2 Deflnition of a Function (grade 12) A function is a relation for which there is only one value of y corresponding to any value of x. We sometimes write y = f (x), which is notation meaning ’y is a function of x’. This deflnition makes complete sense when compared to our real world examples | each person has only one height, so height is a function of people; on each day, in a speciflc town, there is only one average temperature. However, some very common mathematical constructions are not functions. For example, consider the relation x2 + y2 = 4. This relation describes a circle of radius 2 centred at the origin, as in flgure 3.1. If we let x = 0, we see that y2 = 4 and thus either y = 2 or y = 2. Since there are two y values which are ¡ possible for the same x value, the relation x2 + y2 = 4 is not a function. There is a simple test to check if a relation is a function, by looking at its graph. This test is called the vertical line test. If it is possible to draw any vertical line (a line of constant x) which crosses the relation more than once, then the relation is not a function. If more than one intersection point exists, then the intersections correspond to multiple values of y for a single value of x. We can see this with our previous example of the circle by looking at its graph again in flgure |
3.1. We see that we can draw a vertical line, for example the dotted line in the drawing, which cuts the circle more than once. Therefore this is not a function. 40 Figure 3.1: Graph of y2 + x2 = 4 In a function y = f (x), y is called the dependent variable, because the value of y depends on what you choose as x. We say x is the independent variable, since we can choose x to be any number. 3.1.3 Domain and Range of a Relation The domain of a relation is the set of all the x values for which there exists at least one y value according to that relation. The range is the set of all the y values, which can be obtained using at least one x value. If the relation is of height to people, then the domain is all living people, while the range would be about 0:1 to 3 metres | no living person can have a height of 0m, and while strictly it’s not impossible to be taller than 3 metres, no one alive is. An important aspect of this range is that it does not contain all the numbers between 0.1 and 3, but only six billion of them (as many as there are people). As another example, suppose x and y are real valued variables, and we have the relation y = 2x. Then for any value of x, there is a value of y, so the domain of this relation is the whole set of real numbers. However, we know that no matter what value of x we choose, 2x can never be less than or equal to 0. Hence the range of this function is all the real numbers strictly greater than zero. These are two ways of writing the domain and range of a function, set notation and interval notation. (NOTE: the syllabus does not say which notation method to use. we should flnd out, and only use the one if possible. there is no need to add further confusion. then move the unused notation to Extra.) Set Notation First we introduce the symbols > ; < ;. > means ’is greater than’ and • means ’is greater than or equal to’. So if we write x > 5, we say that x is ‚ greater than 5 and if we write x y, we mean that x can be greater than or means ’is less than or equal equal to y. Similarly, < means � |
�is less than’ and to’. Instead of saying that x is between 6 and 10, we often write 6 < x < 10. This directly means ’six is less than x which in turn is less than ten’. ‚ ‚ • ; 41 A set of certain x values has the following form: x : conditions, more conditions f g (3.1) We read this notation as \the set of all x values where all the conditions are satisfled". For example, the set of all positive real numbers can be written as which reads as \the set of all x values where x is a real number x : x f and is greater than zero". (NOTE: have we even explained what > ; < ; ‚ mean yet? remember... this book must not assume that anyone has seen this stufi before.) R; x > 0 g • 2 ; We use the same notation (with the letter y instead of x) for the range of the function. Interval Notation (NOTE: rewrite this subsubsection. flrst describe what the brackets mean, and then introduce the concept of a union. all these concepts are new... so we must describe everything in detail.) Here we write an interval in the form ’lower bracket, lower number, comma, upper number, upper bracket’. We can use two types of brackets, square ones [; ] or round ones (; ). A square bracket means including the number at the end of the interval whereas a round bracket means excluding the number at the end of the interval. It is important to note that this notation can only be used for all real numbers in an interval. It cannot be used to describe integers in an interval or rational numbers in an interval. So if x is a real number greater than 2 and less than or equal to 8, then x is any number in the interval (2;8] (3.2) It is obvious that 2 is the lower number and 8 the upper number. The round bracket means ’excluding 2’, since x is greater than 2, and the square bracket means ’including 8’ as x is less than or equal to 8. Now we come to the idea of a union, which is used to combine things. The. Here we use it to combine two or more intervals. For x < 10, then the set symbol for union is example, if x is a real number such that |
1 < x of all the possible x values is 3 or 6 • • [ (1;3] [ [6;10) (3.3) sign means the union(or combination) of the two intervals. We use where the the set and interval notation and the symbols described because it is easier than having to write everything out in words. [ 3.1.4 Example Functions In this section we will look at several examples of functions. Here we will let go of our real-world examples, and look exclusively at real valued functions, because only in such cases do we see the full use and power of functional mathematics. While it is instructive to see a menu or people’s height as a function, it is not very interesting. On the other hand, all of advanced physics and statistics depend on real valued functions. Very little is more important than gaining an 42 intuitive grasp of real functions, and we will spend the remainder of this chapter doing just that. When considering real valued functions, our major tool is drawing graphs. In the flrst place, if we have two real variables, x and y, then we can assign values to them simultaneously. That is, we can say \let x be 5 and y be 3". Just as we write \let x = 5" for \let x be 5", we have the shorthand notation \let (x; y) = (5; 3)" for \let x be 5 and y be 3". We usually think of the real numbers as an inflnitely long line, and picking a number as putting a dot on that line. If we want to pick two numbers at the same time, we can do something similar, but now we must use two dimensions. What we do is use two lines, one for x and one for y, and rotate the one for y, as in diagram (NOTE: insert diagram). We call this the Cartesian plane. (NOTE: This whole y and f (x) thing needs to be cleared up | I would do it here, but then it’s also discussed above in the deflnition of a function. I think the problem comes with the varying uses of y, and I think a physicist would be better than a mathematician to clear this up. Personally, rigorously, I don’t really know what’s going on with this notation.) The great beauty of doing this is that it allows us to \draw" functions, in |
a very abstract way. Let’s say that we were investigating the function f (x) = 2x. We could then consider all the points (x; y) such that y = f (x), i.e. y = 2x. For example, (1; 2); (2:5; 5); and (3; 6) would all be such points, whereas (3; 5) 3. If we put a dot at each of those points, and then at would not since 5 every similar one for all possible values of x, we would obtain the graph shown in (NOTE: put in). = 2 £ The form of this graph is very pleasing | it is a simple straight line through the middle of the plane. Now some of you may have guessed this graph long before we plotted it, but the point is that the technique of \plotting", which we have followed here, is the key element in understanding functions. To show you why, we will now consider whole classes of functions, and we will relate them by the simple fact that their graphs are nearly identical. Straight Line Functions These functions have the general form f (x) = ax + b (3.4) where a and b are constants. The value of a is called the gradient or slope and tells us how steep the line is (the larger the number, the steeper the line). If a is greater than zero it means the line increases from left to right (slopes upwards), if it is smaller than zero the line increases from right to left (slopes downwards). b is called the y-intercept and tells us where the line goes through the y-axis. For example the function f (x) = 2x + 3 has a gradient of 2 and a y-intercept of 3. This means that the line cuts through the y-axis at a value of 3 and slopes upwards. We can calculate the values of y for certain values of x and then plot them in a graph (see flgure 3.2). x : y = 2x + 3 : -5 -7 -4 -5 -3 -3 -2 -1 11 5 13 43 6 12 Figure 3.2: Graph of f (x) = 2x + 3 However we only need two points to plot a straight line graph. The easiest points to use are the x-intercept (where the line cuts the x-axis) and the ya intercept. The x-intercept |
occurs when y = 0, so it is always equal to b. So if asked to plot a straight line, there is no need to calculate lots of y values, you just need to flnd the x and y intercepts and draw a line through them. ¡ Parabolic Functions A parabola looks like a hill, either upside down (for a \positive" parabola) or right way up (for a \negative" one), which is the same on both sides, as in the diagrams (NOTE: put in): You may have noted that when we say the parabola is \the same on both sides", we are just stating that these functions are horizontally symmetric. This means that if you ip them from left to right along a speciflc line, which is called the line of symmetry, they look the same. This line of symmetry is sometimes called the axis of symmetry. Parabolic functions are functions of the form f (x) = ax2 + bx + c (3.5) where a, b and c are constants. The a involves the shape of the parabola and says how steep the curves are. If a is positive, then the hill is upside-down. If a is negative, then the hill is the right way up. c is the y-intercept, which is where the parabola cuts the y axis. b has to do with the shift in the parabola to the left or the right. Two important features of the parabola are its turning point and line of symmetry(described above). The turning point says how high the hill is. If the hill is the right way up, then the turning point is the maximum value of the parabola, and if it is upside-down, then the turning point is the minimum value. Now the above form of the parabola is the standard form. It can also be 44 written in the form f (x) = a(x p)2 + q ¡ (3.6) where the two new constants p and q give the turning point (p;q) of the parabola. This form of the parabola can be obtained from the standard form by completing the square (See algebra). The value p of the turning point is actually the line of symmetry. So if p = 3, then x = 3 is the line of symmetry(which is always vertical for the parabola). q is the maximum or |
minimum value of the function(that is the maximum or minimum value of y). At flrst it might seem di–cult to sketch the graph of a parabola but once a simple procedure is followed, then it becomes easier. When sketching the graph, we need to use some information about it. The only information we have are its shape, x and y-intercepts and its turning point. We start ofi by seeing whether the parabola is an hill that is the right way up or upside-down. Recall that we can flnd this out from the sign of a. Next we calculate the x-intercepts by setting y = 0 and solving the equation 0 = ax2 + bx + c (3.7) which doesn’t always have a solution, meaning that not all parabolas cut the x-axis. These would be hills which never quite make it to the x-axis, or upsidedown hills which are never low enough to touch the x-axis. However, if there is one solution, and it is not zero, then because of the symmetry there must be two solutions which can be both positive, or negative or a plus and a minus one. (NOTE: max/min turning points.) The y-intercept is just c. Last we flnd the turning point of the parabola. One way is to write the equation of the parabola in the form 3.6 and we have b 2a, the line of symmetry and found p and q. Another way is to calculate x = ¡ also the value of p. q is found by putting p in the equation of the parabola. So now we are able to plot some points and join them up to form a parabola. We can create a table of x and y values for the parabola f (x) = x2 9 and then plot them (see flgure 3.3). Note that you could spot the symmetry of the graph by examining the table alone, where we see that x = 1 and x = 1 give the same value for y. Note also that for this parabola b = 0, so the line of the symmetry is the y-axis since the parabola looks the same on both sides of the y-axis. ¡ ¡ x : y = x2 ¡ 9 : -4 7 -3 0 -2 -5 |
-1 -8 0 -9 1 -8 2 -5 3 0 4 7 Hyperbolic Functions Hyperbolas look like 2 parabolas on their side which are mirror reections of each other around the diagonal (NOTE: sketch). Hyperbolic functions look like f (x) = a x + b (3.8) where a and b are constants. Just like for parabolas, a tells us how steep the curves are and b tells us how high the curves are. 45 Figure 3.3: Graph of the parabola f (x) = x2 9 ¡ Since we cannot divide by zero1, it is not possible to have x = 0, so there is no y-intercept. When you go far enough away from the y-axes, the curves start to look like straight lines, and we call them asymptotes. For example we can create a table of x and y values for the hyperbolic function f (x) = 4 x and plot them (see flgure 3.4) x : f (x) = 4 x : -8 - 1 2 -4 -1 -2 -2 -1 -4 - 1 2 - Exponential Functions y = abx + c b > 0 (3.9) 3.2 Exponentials and Logarithms The syllabus requires: (grade 12) switch between log and exp form of an equation (grade 12) derive and use the laws of logs † † (NOTE: need an intro. this should have lots of stufi about how people used exp/logs to multiply numbers by adding them. with a few examples... to show that you don’t need a calculator.) 1(NOTE: i’m sure there are interesting facts about dividing by zero (not that i know of | luke)) 46 Figure 3.4: Graph of the hyperbola f (x) = 4 x 3.2.1 Exponential Functions (NOTE: need an intro. we already covered exponentials in \numbers", but maybe we should move it here instead.) 3.2.2 Logarithmic Functions (NOTE: these Laws need introduced properly with more detailed derivations and examples of their use, highlight each ones importance. rewrite the intro to not include so many new terms... and to read better for a 16 year old.) Logarithms, commonly referred to as Logs, are |
the algebraic inverse of exponents. When we say \inverse function" we mean that the answer becomes the question and the question becomes the answer. For example, in the expression ab = x the \question" is \what is a raised to the b power." The answer is \x." The inverse function would be logax = b or \by what power must we raise a to obtain x." The answer is \b." Many students flnd logarithms di–cult. For now you can be successful if you learn the terminology and come to understand the relationships of the terms. (NOTE: this next graph needs more explanation) Law 1 Since a0 = 1, loga1 = 0 47 Figure 3.5: The Exponential Function f (x) = ex Law 2 Since a1 = a, logaa = 1 Law 3 This one is a bit trickier to see. The law is that logaax = x. If we re-write it as loga(ax) = x we can see that it is ax = (ax), which is, of course, true. We can also then say that logaax = x logaa = x(1) = x. The upshot being that any exponent of the (operand?) can simply be moved to simple multiplication by the log. ¢ Law 4 The laws of exponents am logarithms loga(m ¢ n) = logam+logan and loga( m an = am+n and a m an = am ¡ n ) = logam n translate to the laws of logan respectively. ¡ ¢ Base In the previous examples a is the base. We generally use the \common" base, 10, or the natural base, e. The number e is an irrational number between 2:71 and 2:72. It comes up surprisingly often in Mathematics, but for now su–ce it to say that it is one of the two common bases. While the notation log10(x) and loge(x) may be used, log10(x) is often styled log(x) in Science and loge(x) is normally written as ln(x) in both Science and Mathematics. 48 Figure 3.6: The Logarithmic Function f (x) = ln(x) It is often necessary or convenient to convert a log from one base to another. An Engineer might need an approximate solution to a |
log in a base for which he does not have a table or calculator function, or it may be algebraically convenient to have two logs in the same base. To afiect a change of base, apply the change of base formula: logax = logbx logba (3.10) where b is any base you flnd convenient. Normally a and b are known, therefore logba is normally a known, if irrational, number. 3.3 Extra (NOTE: this is non-syllabus content on absolute value functions, but perhaps the absolute operator should be worked into the main text and this section deleted, as it is quite important.) 3.3.1 Absolute Value Functions (NOTE: i’m pretty sure this is not on the syllabus) The absolute value of x has the following deflnition = x j j ‰ x x ¡ 0 if x ‚ if x < 0 (3.11) 49 Figure 3.7: The Functions f (x) = ln(x) and f (x) = ex are symmetrical about the origin. In other words, the absolute value sign makes the term inside this sign positive. If it is already positive, then there is no change, and otherwise the sign of this term changes. Now an absolute value function has the following general form where a, b and c are constants. x f (x3.12) Let us again consider an absolute value function with the general form y = + c. We must consider two cases separately: b: x a j b j ‚ ¡ x Now, since x is positive and therefore ‚ b and thus, the term inside the absolute value sign b. Thus y = a(x ¡ b) + c = ax + (c ab) ¡ In other words, this is a straight line with slope a and y-intercept c x < b: In this case, the term in the absolute value sign is negative and thus (x ¡ ¡ b) = x + b. Therefore ¡ y = a( x + b) + c = ¡ ax + (c + ab) ¡ which is a straight line with slope a and y-intercept c + ab. ¡ 50 (3.13) ab. ¡ x b j ¡ = (3.14) j 0 Now at x = b the function value is y = a j j consists of half |
of two straight lines with slopes turning point (b;c). + c = c. Therefore function a and a, which meet at the ¡ The function has the axis of symmetry x = b. In other words, the part of the function on one side of the vertical line x = b is the same as the reection about this line of the part of the function on the other side. We can see this as follows: Consider the function values at the points x = b + z and x = b z, where z > 0 (these are two point the same distance from the line x = b). Now the function values at these two points are ¡ and (b + z) f (b + z = az + c j + c b j ¡ f (b ¡ z) = a (b ¡ j z = a j j ¡ = az + c + c b j z) ¡ + c (3.15) (3.16) (3.17) (3.18) (3.19) (3.20) These function values are the same. Therefore, whether we move to the left or the right of the line x = b, the function values remain the same. Therefore x = b is an axis of symmetry. x = b z z az + c (b;c) b + z z b ¡ Figure 3.8: Graph of f (b + z) and f (b of symmetry x = b x z) where f (x) = a j j ¡ + c; with the line Now let us consider two cases: a < 0 and a > 0. If a is positive, then the line on the left of the turning point (with slope a) will have a negative slope and the line on the right (with slope a) will have a positive slope. Thus the graph will be shaped like a V. ¡ 51 Otherwise, if a is negative, then the line on the left has the positive slope a and the line on the right has the negative slope a. Therefore the graph is ¡ an upsidedown V. (b;c) a < 0 (b;c) a > 0 Figure 3.9: Graph of f (b + z) and f (b for a > 0 the other for a < 0. x z) where f (x) = a j j ¡ + c. One case is Notice also that an absolute value function does |
not necessarily have xIt these do exist, then they will be the x-intercepts of the two intercepts. straight lines making up the absolute value function. 52 Chapter 4 Numerics 4.1 Optimisation The syllabus requires: Linear Programming (Grade 11) † 1. Solve linear programming problems by optimising a function in two variables, subject to one or more linear constraints, by numerical search along the boundary of the feasible region. 2. Solve a system of linear equations to find the co-ordinates of the vertices of the feasible region. In everyday life people are interested in knowing the most e–cient way of carrying out a task or achieving a goal. For example, a farmer might want to know how many crops to plant during a season in order to maximise yield (produce) or a stock broker might want to know how much to invest in stocks in order to maximise proflt. These are examples of optimisation problems, where by optimising we mean flnding the maxima or minima of a function. This function we wish to optimise (i.e. maximise or minimise) is called the objective function (we will only be looking at objective functions which are functions of two variables). In the case of the farmer, the objective function is the yield and it is dependent on the amount of crops planted. If the farmer has two crops then we can express the yield as f (x;y) where the variable x represents the amount of the flrst crop planted and y the amount of the second crop planted. For the stock broker, assuming that there are two stocks to invest in, f (x;y) is the amount of proflt earned by investing x rand in the flrst stock and y rand in the second. In practice it is often that constraints, or restrictions, are placed on x and y. The most common of these constraints is the non-negativity constraint. That is, we might require that x 0. For the farmer, it would make little ‚ sense if we were to speak of planting a negative amount of crops and so when 0 must be considered. Other optimising f (x;y) the constraints x constraints might be that the farmer cannot plant more of the second crop than the flrst crop and that no more than 20 units of the flrst crop can be planted; 20. Constraints |
these constraints translate into the inequalities x y and x 0 and y 0 and y ‚ ‚ ‚ ‚ • 53 mean that we can’t just take any x and y when looking for the x and y that optimise our objective function. If we think of the variables x and y as a point (x;y) in the xy plane then we call the set of all points in the xy plane that satisfy our constraints the feasible region. Any point in the feasible region is called a feasible point. y 20 15 10 5 5 10 15 20 x Figure 4.1: The feasible region corresponding to the constraints x x y and x 20. 0, y 0, ‚ ‚ ‚ • For example, the non-negativity constraints x 0 mean that every (x;y) we can consider must lie in the flrst quadrant of the xy plane. The y means that every (x;y) must lie on or below the line y = x constraint x 20 means that x must lie on or to the left of the line x = 20. For these and x constraints the feasibility region is illustrated as the shaded region in Figure 4.1. 0 and y ‚ ‚ ‚ • Constraints that have the form ax + by c or ax + by = c are called linear p2; constraints. Examples of linear constraints are x + y a constraint being linear just means that it requires that any feasible point (x;y) lies on one side of or on a line. Interpreting constraints as graphs in the xy plane is very important since it allows us to construct the feasible region such as in Figure 4.1. We have the following rule for any linear constraint: 2x = 7 and y • • • ¡ 0, ax + by = c ax + by c • If b = 0, feasible points must lie on the line y = a b If b = 0, feasible points must lie on the line x = c=a ¡ x + c b : If b = 0, feasible points must lie on or below the line y = a b ¡ x + c b : If b = 0, feasible points must lie on or to the left of the line x = c=a Once we have determined the feasible region the solution of our problem will be the feasible point where the objective function is a maximum/ minimum. Sometimes there will be more than one feasible point where the objective function |
is a maximum/minimum | in this case we have more than one solution. 54 6 6 4.1.1 Linear Programming The objective function is called linear if it looks like f (x;y) = ax + by where the coe–cients a and b are real numbers. For example, f (x;y) = 10x y is a linear objective function. If the objective function and all of the constraints are linear then we call the problem of optimising the objective function subject to these constraints a linear program. All optimisation problems we will look at will be linear programs. ¡ The major consequence of the constraints being linear is that the feasible region is always a polygon. This is evident since the constraints that deflne the feasible region all contribute a line segment to its boundary (see Figure 4.1). It is also always true that the feasible region is a convex polygon. The objective function being linear means that the feasible point(s) that gives the solution of a linear program always lies on one of the vertices of the feasible region. This is very important since, as we will soon see, it gives us a way of solving linear programs. (NOTE: Should I mention that a linear objective function deflnes a plane? This is crucial to the fact that optimal solutions are obtained at the vertices, though. Do Grade 11s know the equation of a plane? I would like to use the idea that the level sets of planes are lines and in so doing justify the \ruler" method.) We will now see why the solutions of a linear program always lie on the vertices of the feasible region. Firstly, note that if we think of f (x;y) as lying on the z axis, then the function f (x;y) = ax+by (where a and b are real numbers) is the deflnition of a plane. If we solve for y in the equation deflning the objective function then f (x;y) = ax + by x;y) b (4.1) If we consider Equation 4.1 corresponding to f (x;y) = What this means is that if we flnd all the points where f (x;y) = c for any real number c (i.e. f (x;y) is constant with a value of c), then we have the equation of a line. This line we call a |
level line of the objective function (NOTE: Should I use this terminology?). Consider again the feasible region described in Figure 2y with this 4.1. Lets say that we have the objective function f (x;y) = x feasible region. 20 ¡ then we we get the level line y = 1 2 x + 10 which has been drawn in Figure 4.2. 2 + 5), f (x;y) = 0 (y = x Level lines corresponding to f (x;y) = 2 ), f (x;y) = 10 (y = x 10) have also been drawn in. It is very important to realise that these aren’t the only level lines; in fact, there are inflnitely many of them and they are all parallel to each other. Remember that if we look at any one level line f (x;y) has the same value for every point (x;y) that lies on that line. Also, f (x;y) will always have difierent values on difierent level lines. 5) and f (x;y) = 20 (y = x 10 (y = x 2 ¡ 2 ¡ ¡ ¡ If a ruler is placed on the level line corresponding to f (x;y) = 20 in Figure 4.2 and moved down the page parallel to this line then it is clear that the ruler will be moving over level lines which correspond to larger values of f (x;y). So if we wanted to maximise f (x;y) then we simply move the ruler down the page until we reach the \lowest" point in the feasible region|this point will then be the feasible point that maximises f (x;y). Similarly, if we wanted to minimise f (x;y) then the \highest" feasible point will give the minimum value of f (x;y). ¡ 55 y 20 15 10 5 f (x;y) = 20 ¡ 10 f (x;y) = ¡ f (x;y) = 0 f (x;y) = 10 f (x;y) = 20 x 5 10 15 20 Figure 4.2: The feasible region corresponding to the constraints x x represent various level lines of f (x;y). 20 with objective function f (x;y) = x y and x 0, ‚ 2y. The dashed lines 0, y � |
� ¡ • ‚ Since our feasible region is a polygon, these points will always lie on vertices in the feasible region. (NOTE: We could have inflnitely many solutions if the gradient of a constraint = the gradient of the level lines... should I mention this?). The fact that the value of our objective function along the line of the ruler increases as we move it down and decreases as we move it up depends on this particular example. Some other examples might have that the function increases as we move the ruler up and decreases as we move it down. It is a general property, though, of linear objective functions that they will consistently increase or decrease as we move the ruler up or down. Knowing which direction to move the ruler in order to maximise/minimise f (x;y) = ax + by is as simple as looking at the sign of b (i.e. \is b negative, positive or zero?"). If b is positive, then f (x;y) increases as we move the ruler up and f (x;y) decreases as we move the ruler down. The opposite happens for the case when b is negative: f (x;y) decreases as we move the ruler up and f (x;y) increases as we move the ruler down. If b = 0 then we need to look at the sign of a. If a is positive then f (x;y) increases as we move the ruler to the right and decreases if we move the ruler to the left. Once again, the opposite happens for a negative. If we look again at the objective function mentioned earlier, f (x;y) = x 2), then we should flnd that f (x;y) increases as we move the ruler down the page since b = 2 < 0. This is exactly what we found happening in Figure 4.2. 2y (a = 1 and b = ¡ ¡ The main points about linear programming we have encountered so far are ¡ † † † † The feasible region is always a polygon. Solutions occur at vertices of the feasible region. Moving a ruler parallel to the level lines of the objective function up/down to the top/bottom of the feasible region shows us which of the vertices is the solution. The direction in which to move the ruler is determined by the sign of b and also possibly by the sign of a. 56 (NOTE: I would like to |
mention f to determine ‘the direction in which to move the ruler’. Even if I neglect the fact that students certainly know nothing about partial difierentiation, I’m still not sure whether I can mention and work with vectors in the plane...) r These points are su–cient to determine a method for solving any linear program. If we wish to maximise the objective function f (x;y) then: 1. Find the gradient of the level lines of f (x;y) (this is always going to be a b as we saw in Equation 4.1) ¡ 2. Place your ruler on the xy plane, making a line with gradient units on the x-axis and a units on the y-axis) ¡ a b (i.e. b ¡ 3. The solution of the linear program is given by appropriately moving the ruler. Firstly we need to check whether b is negative, positive or zero. (a) If b > 0, move the ruler up the page, keeping the ruler parallel to the level lines all the time, until it touches the \highest" point in the feasible region. This point is then the solution. (b) If b < 0, move the ruler in the opposite direction to get the solution at the \lowest" point in the feasible region. (c) If b = 0, check the sign of a i. If a < 0 move the ruler to the \leftmost" feasible point. This point is then the solution. ii. If a > 0 move the ruler to the \rightmost" feasible point. This point is then the solution. (NOTE: Point 3 is essentially trying to work with ing what it is or that it exists!) f without actually know- r 4.2 Gradient The syllabus requires: † Investigate numerically the average gradient between two points on a curve and develop an intuitive understanding of the concept of the gradient of a curve at a point (NOTE: this is undefined if this should be numerical, or an intro to differentiation. it is best to have it spread over both) perhaps 4.3 Old Content (please delete when flnished) 4.3.1 Problems We often have to solve problems in which there are several variables, which we can change to suit us. We can now develop a method of dealing with such problems. 57 Worked Example 1: Q: A farmer grows wheat and maize. He has |
20 flelds of available land on which he can plant crops. He must grow at least 5 flelds of maize. Also he cannot grow more than twice as much maize as wheat. Draw a graph to show the feasible region showing the possible number of flelds of wheat and maize the farmer can plant. What is the maximum number of flelds of wheat the farmer can plant? A: Step 1: Analyse the problem and assign the variables x and y. Let x be the number of flelds of wheat the farmer plants. Let y be the number of flelds of maize the farmer plants. Step 2: Write down the inequalities which are the restrictions on x and y. x + y 20 (the farmer only has 20 flelds) (4.2) (at least 5 flelds of maize must be planted) (4.3) (it is not possible to have a negative number of flelds of wheat)(4.4) (4.5) (the farmer cannot plant more than twice as much maize as wheat) • y x 5 0 ‚ ‚ 2x y • Remember that every piece of information you are given is important, so check that you have not left out an inequality. Also note that often variables cannot be negative, which give further inequalities as in the case of x 0. Step 3: Solve for y in terms of x where possible. ‚ y y x y x + 20 • ¡ 5 0 ‚ ‚ 2x • (4.6) (4.7) (4.8) (4.9) Step 4: Plot a graph and flnd the feasible region. 24 21 18 15 12 12 15 18 21 24 Figure 4.3: Graph of TODO Step 5: Answer the original question. 58 We need to flnd the maximum number of flelds of wheat which can be planted. This is the maximum x value which is in the feasible region. This occurs at the point (15,5). Thus the maximum x value is 15. Remembering to give the answer in terms of the original question: The farmer can plant a maximum of 15 flelds of wheat. 4.3.2 Maximising or Minimising the Objective Function The objective function is a function of x and y |
. We are usually told to maximise or minimise this function. Worked Example 2: Q: Consider the same situation as in worked example 1. The farmer can make a proflt of R100 on every fleld of wheat and R200 on every fleld of maize that he grows. How many flelds of wheat and maize must the farmer plant to maximise his proflt and what is this maximum proflt? Steps 1 - 4 are as in worked example 1. A: Step 5: Deflne the objective function. The objective function, in this case, is the proflt in terms of the number of flelds of wheat and maize (the variables x and y). This is given by Step 6: Solve for y. P = 100x + 200y y = 1 2 ¡ x + P 200 (4.10) (4.11) Step 7: Maximise/minimise the objective function. In this case we need to maximise the objective function which is the proflt P. The greater P the larger the y intercept of the straight line of y as a function 1 2 (line A is an example of of x. However, the slope of the line will always be such a line). ¡ Now to maximise P we need the y-intercept to be as large as possible, but the line must still pass through the feasible region. Thus take a ruler and move it parallel to line A (keeping the slope the same). Move the ruler outwards until it is at the edge of the feasible region. This is line B, which is the line of maximum P. The point on the feasible region through which this line passes (in this case ( 20 3 and y = 40 3 ; 40 3 ). The proflt can be calculated from the objective function as P = R3333. 3 )) is the point giving this proflt (so x = 20 3 = 13 1 3 = 6 2 Step 8: Give the answer in terms of the question. For a maximum proflt of R3333, the farmer must plant 6 2 3 flelds of wheat and 13 1 3 flelds of maize. (NOTE: Further examples need to be included here, particularly add an example which uses discreet variables.) Worked Example 3: |
Q: A delivery company delivers wood to client A and bricks to client B. The company has a total of 5 trucks. A truck cannot travel more than 8 hours per day and it takes 4 hours make the trip to and back from client A and 2 hours for 59 20 15 10 5 A B 5 10 15 20 25 30 Figure 4.4: Graph of TODO client B. To honour an agreement with client B, at least 2 truck loads of bricks must be delivered per day. Also client A needs no more than 9 truck loads of wood per day. The delivery company makes a proflt of R100 per truck load of wood and R150 per truck load of bricks delivered. How many truck loads of wood and bricks should be delivered per day so as to maxmise the proflt? What is this maximum proflt? Note: Client A is in the opposite direction to client B, so each truck can only deliver a full truck load to A or B (a truck cannot take half a load to A and the other half to B). A: Step 1: Let x be the number of truck loads of wood the company delivers to client A per day. Let y be the number of truck loads of bricks the company delivers to client B per day. Step 2: Firstly, the total number of hours of delivery time available is 5 8 hours = 40 hours, since there are 5 trucks, which cannot be driven more than 8 hour per day. Delivery to client A takes 2 hours and delivery to client B takes 4 hours. Therefore £ The other inequalities are 2x + 4y 40 • (4.12) y x x;y 2 9 0 ‚ • ‚ (the company must delivered at least 2 truck loads of bricks per day) (4.13) (4.14) (client A needs no more than 9 truck loads of wood per day) (4.15) (we cannot have a negative number of truck loads) (4.16) 60 Furthermore, we know that x and y must be integers (in other words we cannot have a fractional truck load). These are therefore called discreet variables. Step 3: y = y ‚ x • x;y + 10 x 2 ¡ 2 9 0 ‚ (4.17) (4.18) (4.19) (4.20) Step 4: We now plot the constraints and the feasible region (see the graph at the end). The region enclosed |
by the constraints is the shaded region, but since the variables x and y can only take on positive integer values, the feasible region actually consists of the collection of dots showing the integer values in the shaded region. Step 5: The objective function is the proflt (in Rands), which is P = 100x + 150y (4.21) Step 6: Solving for y gives y = 2 3 ¡ x + P 150 (4.22) Step 7: We need to maximise the proflt P and therefore we need to maximise the y-intercept of the previously deflned straight line. Line A shows an arbitrary 2 straight line with slope 3, which is drawn, for convenience, with intercepts x = 9 and y = 6. If a ruler is moved outwards parallel to this line (i.e. keeping the slope flxed) to the edge of the feasible region, we obtain line B, which passes through the point (8;6).(NOTE: RULERS??? is this really the best way to do this? can we please have some equations of lines!) ¡ Therefore the maximum proflt occurs when x = 8 and y = 6. This proflt (in Rands) is P = 100x + 150y = 100(8) + 150(6) = 1700 (4.23) (4.24) (4.25) Note: We cannot use the point (9; 5 1 edge of the shaded region, because this is 5 1 a half a truck load). 2 ), which is actually the point at the 2 is not an integer (we cannot have Step 8: The maximum proflt is R1700 per day, which is obtained when the company delivers 8 truck loads of wood to client A and 6 truck loads of bricks to client B per day. 61 B 10 8 6 4 2 A 6 2 4 8 10 12 14 16 18 Figure 4.5: Graph of TODO Essay 1 : Difierentiation in the Financial World Author: Fernando Durrell I lived in Cape Town (South Africa) all my life. I attended Thomas Wildschudtt I then attended St. Owen’s Senior School in Junior and Senior Primary Schools. Retreat up to half way through Grade 11 at which point I left for St. Joseph’s Senior School in Rondebosch |
(since St. Owen’s closed permanently at the end of my Grade 11 year). It was always one of my ambitions to attend the University of Cape Town (UCT) because it is a prestigious university. I applied to study medicine at UCT, but was not accepted, and so I enrolled for a science degree at UCT and have never regretted it. (I can stand only so much visible blood.) I wasn’t sure about what I wanted to do with my life so I enrolled for Mathematics, Applied Mathematics, Chemistry and Physics in my flrst year at university. By the end of my flrst year at UCT, I wanted to continue with the Mathematics stream. I completed by Bachelor of Science (BSc) degree with majors (main subjects) Mathematics and Applied Mathematics and then completed my BSc (Honours) degree in Applied I completed by Master of Science degree in Financial Mathematics Mathematics. and am currently registered for the degree of Doctor of Philosophy in Mathematics. Difierentiation in the Financial World Most of us don’t really think about saving our money to buy something in the future - we have to spend it now! Our parents unfortunately (and one day when we’re older most probably we too) have to save for the future: for that possible time when they don’t have a job; they may save to purchase furniture for the house; or a present for your birthday. The most important thing adults save for is retirement - this is when they decide that they want to stop working. Their kids may not be able to care for them because they too may have families or may not have jobs themselves. So, adults have to save money while they are 62 working to support them when they retire. The amount of money they receive after they retire is called their pension - so adults save so that they can receive a pension when they retire. Suppose James is paying R100 every month toward his pension. When James retires, he wants every month to receive a bit more than the R100 he contributed toward his pension (while he was working). If he doesn’t get a bit more than R100 pension every month (when he retires), then he may as well save his money under his bed until he retires. Now, there are many adults like James who are saving for their pension. To whom do all these adults pay their monthly pension savings? They pay their monthly pensions savings to a pension fund. Suppose |
there are ten million adults paying R100 every month to a pension fund. That means that each month the pension fund receives R100x10 000 000 = R1b (i.e. one billion rand) in total each month. Now, each adult, like James, will want to receive a monthly pension which is greater than R100 when they retire. So, the pension fund must ensure that, when a pension fund contributor retires, he/she receives more than R100 pension every month. There are many pension funds in the world, so, if the pension fund James is saving with is going to give him a R110 monthly pension and another pension fund is going to give him a R120 monthly pension, then he is going to save with the latter pension fund. So, pension funds can’t give pensioners too little pension. In fact, they have to give pensioners as big a pension as possible. Now, the pension fund can’t just put the monthly R1bn in the bank and let it earn interest and divide this amongst all pensioners. The government takes a lot of the interest earned by pension funds as tax. So, the pension funds have to make more money, and so they turn to the stock market. But the thing about the stock market is that one can lose a lot of money very quickly if one is not careful. The advantage about putting one’s money in the bank is that, when you come back the next day, your money will still be there. If you invest in the stock market today and you come back tomorrow, then you could have lost a substantial amount of money, but you could also have made a lot of money. So, for the pension fund, depositing the monthly R1b with the bank is appealing, but so is the stock market. The pension has to flnd the best combination of the two (the bank and the stock market). That involves, flnding out how much money to deposit with the bank and how much to invest in the stock market so that the pension fund makes as much money as possible. To solve this problem, involves difierentiation, which is the topic of the next chapter. This is just one way in which difierentiation is used in the flnancial world. 63 Chapter 5 Difierentiation 5.1 Limit and Derivative Calculus is fundamentally difierent from the mathematics that you have |
studied previously. Calculus is more dynamic and less static. It is concerned with change and motion. It deals with quantities that approach other quantities. For that reason it may be useful to have an overview of the subject before beginning its In this section we give a glimpse of some of the main ideas intensive study. of calculus by showing how limits arise when we attempt to solve a variety of problems. 5.1.1 Gradients and limits A traditional slingshot is essentially a rock on the end of a string, which you rotate around in a circular motion and then release. When you release the string, in which direction will the rock travel? Many people mistakenly believe that the rock will follow a curved path. Newton’s First Law of Motion tells us that the path is straight. In fact, the rock follows a path along the tangent line to the circle, at the point of release. If we wanted to determine the path followed by the rock, we could do so, as tangent lines to circles are relatively easy to flnd. Recall, from elementary geometry that a tangent line to a circle is a line that intersects the circle in exactly one point. In this chapter we will be concerned with tangent lines to a variety of functions, as the tangent line gives us the slope of a function at a point. Now let us consider the problem of trying to flnd the equation of the tangent line t to a curve with equation y = f (x) at a given point P. 64 P f(x) Since we know that the point P lies on the tangent line, we can flnd out the equation of t if we know its slope m. The problem is that we need two points to compute the slope and we only have one, namely P on t. To get around the problem we flrst flnd an approximation to m by taking a nearby point Q on the curve and computing the slope mP Q of the secant line P Q. f(x) P x Q a 65 From the flgure we see that mP Q = f (x) x f (a) a ¡ ¡ (5.1) Now imagine that Q moves along the curve toward P.The secant line approaches the tangent line as its limiting position. This means that the slope mP Q of the secant line becomes closer and closer to the slope m of the tangent line as Q approaches |
P. We write m = lim P! Q mP Q and we say that m is the limit of mP Q as Q approaches P along the curve. Since x approaches a as Q approaches P, we could also use Equation (5.1) to write ¡ ¡ The tangent problem has given rise to the branch of calculus called difierential m = lim a! (5.2) x f (x) x f (a) a calculus. 5.1.2 Difierentiating f (x) = xn The central concept of difierential calculus is the derivative. After learning how to calculate derivatives, we use them to solve problems involving rates of change. Deflnition: The derivative of a function f at a number a, denoted by f 0(a), is if this limit exists. f 0(a) = lim 0! h f (a + h) h ¡ f (a) (5.3) Let us use this deflnition to calculate the derivative of f (x) = x2, where n is a positive integer. f (x) ¡ f (x + h) h (x + h)2 h x2 + 2xh + h2 x2 ¡ x2 ¡ h 2xh + h2 h 2x + h f 0(x) = lim 0! h = lim 0 h! = lim 0 h! = lim 0 h! = lim 0 h! = 2x You should repeat this calculation for f (x) = x3 and (if you haven’t spotted a pattern yet!) for f (x) = x4. Then see if you can generalise what you are seeing to write down a formula for f 0(x) where f (x) = xn. (This isn’t a valid 66 mathematical way of arriving at a formula, but if you want to prove the general case you need to use the binomial theorem, which is outside the scope of your syllabus.) Hopefully you calculated the derivative of f (x) = x3 to be 3x2, and shortly after that spotted the pattern for powers of x: d dx (xn) = nxn ¡ 1 5.1.3 Other notations If we use the traditional notation y = f (x) to indicate that the dependent variable is y and the independent variable is x |
, then some common alternative notations for the derivative are as follows: f 0(x) = y0 = dy dx = df dx = d dx f (x) = Df (x) = Dxf (x) The symbols D and d=dx are called difierential operators because they indicate the operation of difierentiation, which is the process of calculating a derivative. It is very important that you learn to identify these difierent ways of denoting the derivative, and that you are consistent in your usage of them when answering questions. Note Though we choose to use a fractional form of representation, dy IS NOT a fraction, i.e. dy with respect to x. Thus, dp is the \operator", operating on some function of x. dx is a limit and dx. dy dx means y difierentiated dx does not mean dy dx means p difierentiated with respect to x. The ‘ d dx ’ ¥ The syllabus requires: † † (grade 12) understand the limit concept in the context of approximating the rate of change or gradient of a function at a point (grade 12) establish derivatives of f (x) = xn from 1st principles and then generalise to obtain the derivative of f (x) = b f (x) = x3 f (x) = x2 1 x f (x) = 5.2 Rules of Difierentiation In order to avoid difierentiating functions from flrst principles, we can establish certain rules. Rule 1 If f is a constant function, f (x) = c, then f 0(x) = 0 67 Rule 1 may also be written as d dx c = 0 (5.4) This result is geometrically evident if one considers the graph of a constant function. This is an horizontal line, which has slope 0. Rule 2: The Power Rule If f (x) = xn, where n is an integer, then f 0(x) = nxn ¡ 1 The Power Rule may also be written as d dx (xn) = nxn ¡ 1 (5.5) (5.6) This rule applies when n is a negative number. For example, the derivative of f (x) = 1 x is f 0(x) = x¡ 2, remembering that 1 x = x |
¡ 1. ¡ Rule 3: Linearity of Difierentiation If c is a constant and both f and g are difierentiable, then d dx (cf ) = c df dx df dx + d dx (f + g) = 5.2.1 Summary d dx c = 0 d dx (xn) = nxn ¡ 1 d dx (cf ) = c df dx dg dx (5.7) (5.8) d dx (f + g) = df dx + dg dx 5.3 Using Difierentiation with Graphs The syllabus requires: † † (grade 12) find equation of a tangent to a graph (grade 12) sketch graph of a cubic function using diff to determine stationary points and their nature. x-axis intercept use factor theorem to determine 5.3.1 Finding Tangent Lines In section 5.1.1 we saw that flnding the tangent to a function is the same as flnding its slope at a particular point. The slope of a function at a point is just its derivative. If we want to flnd a general formula for a tangent to a function, we difierentiate the function. To flnd the slope of the tangent at a particular point, we substitute that point’s x value into the function’s derivative. This will give us a single value, which is the slope of a straight line. We’ll look at one of these problems in the Worked Examples (section 5.4). 68 5.3.2 Curve Sketching Suppose we are given that f (x) = ax3 + bx2 + cx + d and we are asked to sketch the graph of this function. We will use our newfound knowledge of difierentiation to solve this problem. There are FIVE steps to be followed: 1. If a > 0, then the graph is increasing from left to right, and has a maximum and then a minimum. As x increases, so does f (x). If a < 0, then the graph decreasing is from left to right, and has flrst a minimum and then a maximum. f (x) decreases as x increases. 2. Determine the value of the y-intercept by substituting x = 0 into f (x) 3. Determine the x |
-intercepts by factorising ax3 + bx2 + cx + d = 0 and solving for x. First try to eliminate constant common factors, and to group like terms together so that the expression is expressed as economically as possible. Use the factor theorem if necessary. 4. Find the turning points of the function by working out the derivative df dx and setting it to zero, and solving for x. 5. Determine the y-coordinates of the turning points by substituting the x values obtained in the previous step, into the expression for f (x). 6. Step 6 of 5, Draw a neat sketch. The syllabus requires: (grade 12) use the differentiation rules † Dx[f (x) g(x)] = Dx[f (x)] § § Dx[k:f (x)] = k:Dx[f (x)] Dx[g(x)] 5.4 Worked Examples Worked Example 2 : Finding derivatives from flrst prin- ciples Question: Find the derivative of the function f (x) = x2 a. ¡ 8x + 9 at the number Answer: Step 1 : Write out the deflnition From deflnition (5.3) we have f 0(a) = lim 0! h f (a + h) h ¡ f (a) 69 Step 2 : Fill in the function f (x) and multiply out f 0(a) = lim 0! h = lim 0 h! [(a + h)2 ¡ 8(a + h) + 9] h [a2 ¡ ¡ 8a + 9] a2 + 2ah + h2 8a ¡ 8h + 9 ¡ h a2 + 8a 9 ¡ ¡ Step 3 : Simplify f 0(a) = lim 0! h 2ah + h2 h h(2a + h h 8h 8) ¡ ¡ = lim 0 h! = 2a 8 ¡ And you’re done! Worked Example 3 : Finding and using derivatives from flrst principles Question: If f (x) = 4x + 2x2, flnd f 0(x) from flrst principles and hence calculate f 0(2). Answer: Step 1 : Write out deflnition (5.3) and � |
�ll in f (x) f 0(x) = lim! h 0 • = lim h! 0 • f (x + h) h ¡ f (x) ‚ 4(x + h) + 2(x + h)2 h (4x + 2x2) ¡ ‚ Step 2 : Multiply out and simplify 4h + 4xh + 2h2 h ‚ [4 + 4x + 2h] h 0 • f 0(x) = lim! = lim 0 h! = 4 + 4x Step 3 : Substitute the value of x into f 0(x) Since f 0(x) = 4 + 4x then f 0(2) = 4 + 4(2) = 12 70 Worked Example 4 : Using Notation and Rules of Dif- ferentiation Question: Difierentiate the following using the Rules of Difierentiation listed above: a) y = t4 e) Du(um) c) h(x) = x6+x4 b)y = x1000 d) d dr (5r3) Answer: a) Step 1 : Write out the Power Rule, equation (5.5) d dt (tn) = ntn ¡ 1 Step 2 : In this case n = 4 so... y0 = 4t3 ¡ 1 Step 3 : Simplify dy dt = 4t3 b) Step 1 : Write out the Power Rule, equation (5.5) d dx (xn) = nxn ¡ 1 Step 2 : In this case n = 1000 so... y0 = 1000x999 ¡ 1 Step 3 : Simplify y0 = 1000x999 c) Step 1 : Write out the Second Linearity Rule, equation (5.8) dx (f + g) = df d dx + dg dx Step 2 : Write out the Power Rule, equation (5.5) d dx (xn) = nxn ¡ 1 Step 3 : Identify f and g, and difierentiate them separately using the Power Rule f (x) = x6 so f 0(x) = 6x6 1 = 6x5 ¡ g(x) = x4 so g0(x) = 4x4 ¡ 1 = 4x3 Step 4 : Add the derivatives of f and g h0(x) = 6 |
x5 + 4x3 d) 71 Step 1 : Write out the First Linearity Rule, equation (5.7) dx (cf ) = c df d dx Step 2 : Write out the Power Rule, equation (5.5) d dr (rn) = nrn ¡ 1 Step 3 : In this case n = 3 and c = 5 so... d 1 dr (5r3) = 5 3r3 ¡ £ Step 4 : Simplify d dr (5r3) = 15r2 e) Step 1 : Write out the Power Rule, equation (5.5) Du(un) = nun ¡ 1 Step 2 : In this case n = m so... Du(um) = mum ¡ 1 which cannot be simplifled further Worked Example 5 : Finding tangent lines Question: Find the slope of the tangent to the graph of y(x) = 3x2 + 4x + 1 at x = 5. Answer: Step 1 : Difierentiate y to get a general equation for the tangent to the graph y0(x) = 6x + 4 Step 2 : Substitute the value x = 5 into the tangent equation just calculated y0(5) = 6 5 + 4 = 34 £ So the slope of the tangent line to y(x) at x = 5 is 34. Worked Example 6 : Drawing graphs Question: Draw the graph of f (x) = x3 + 3x2. Answer: Step 1 : Basic shape of graph a is positive so from left to right, the graph has flrst a maximum and then a minimum 72 Step 2 : y intercept y = x3 + 3x2 therefore y(0) = 0. Step 3 : x intercepts x3 + 3x2 = 0 x2(x + 3) = 0 x = 0 or x = 3 ¡ Step 4 : Turning points dy dx = 3x2 + 6x set this to zero 0 = 3x2 + 6x 0 = 3x(x + 2) x = 0 or x = 2 ¡ Step 5 : y-coordinate of the turning points y(0) = 0 and y( Local max at ( ¡ 2) = ( 2)2 = 4 2)3 + 3( ¡ 2; 4) and local min at (0; 0) ¡ ¡ Step 6 : Draw a neat sketch (- |
2; 4) (-3; 0) (0; 0) 73 5.5 Exercises 1. Draw the graph of y = x2 + x 6 for to this curve at x = 3, x = 1 and x = gradient of the curve at each of these points. ¡ x 6. Draw the tangents 5 ¡ 2, and hence flnd a value for the ¡ • • 2. Draw the graph of y = x2 4x ¡ 4 x for 0 hence flnd a value for the gradient of the curve at each of these points. 6. Draw tangents to the curve at x = 4, x = 3 and x = 2 and • • 3. Difierentiate each of the following from flrst principles to flnd dy dx (a) y = 5x (b) y = 9x + 5 (c) y = 3x2 (d) y = x3 (e) y = x2 + 3x x2 + 7 ¡ (f) y = 5x (g) y = 1 x (h) y = 1 x2 2x2 flnd f 0(x) from flrst principles and hence evaluate f 0(4) ¡ 3 flnd f 0(x) from flrst principles and hence evaluate ¡ 2x flnd f 0(x) from flrst principles and hence evaluate 4. If f (x) = 3x 1) and f 0( ¡ 5. If f (x) = 2x2 + 5x 2) 1) and f 0( f 0( ¡ 6. If f (x) = x3 ¡ ¡ f 0(1),f 0(0) and f 0( 1) ¡ 1. Difierentiate the following functions with respect to x: (a) x5 (b) x3 (c) 12x2 (d) 5x4 (e) 3x2 (f) 7 (g) x5=3 (h) x3=4 (i) x2=5 (j) 8x1=4 (k) px (l) px3 74 (m) 2=x (n) 3=x2 2. Find the gradient function dy dx for each of the following: 4 ¡ ( |
a) y = x2 + 7x 7x2 (b) y = x (c) y = x3 + 7x2 (d) y = 3x2 + 7x ¡ (e) y = (x + 3)(x 1) (f) y = (2x + 3)(x + 2. Find the gradient of the following lines at the points indicated: ¡ (a) y = x2 + 4x at (0;0) x2 at (1;4) (b) y = 5x (c) y = 3x3 (d) y = 5x + x3 at ( (e) y = 3x + 1 (f) y = 2x2 1; ¡ x at (1;4) x + 4 ¡ 2x at (2;20) 6) ¡ x at (2;8) ¡ 4. Find the coordinates of the point(s) on the following lines where the gra- dient is given: (a) y = x2, gradient 8 (b) y = x2, gradient (c) y = x2 8 ¡ 4x + 5, gradient 2 x2, gradient 3 ¡ (d) y = 5x ¡ (e) y = x4 + 2, gradient (f) y = x3 + x2 4 ¡ x + 1, gradient 0 ¡ 5. If f (x) = x3 + 4x flnd (a) f (1) (b) f 0(x) (c) f 0(1) (d) f "(x) (e) f "(1) 75 Chapter 6 Geometry (NOTE: we need motivation and history of geometry here. real world examples (and obscure flgures) and some interesting facts (this is rich... e.g. architecture, like carpentry). What is a degree... why computer graphics, manufacturing... is it out of 360... etc.) 6.1 Polygons The syllabus requires: † † † † develop conjectures related to triangles, quadrilaterals and other polygons. using any logical method attempt to justify, explain or prove these conjectures define various polygons (isosceles, equilateral, right angled triangles, trapezium, isosceles trapezium, kite, parallelogram, rectangle, rhombus, square and |
the regular polygons) can tell when polygons are similar. similar equilateral triangles are the line drawn parallel to one side of a triangle divides the other 2 sides proportionally A polygon is a shape or flgure with many straight sides. A polygon has interior angles. These are the angles that are inside the polygon. The number of sides of a polygon equals the number of interior angles. If a polygon has equal length sides and equal interior angles then the polygon is called a regular polygon. (NOTE: the language used in this notation section sometimes aims too high; words like \denote" and \line segment" may be daunting when simpler words may be used. the audience is only 15 and all of this is new to them. everything needs explained in detail, using simple language... and diagrams help too.) We denote a line segment that extends between a point A and some point B by line AB. The length of this line is just AB. So if we say, AB = CD we mean that the length of the line segment from A to B is equal to the length of the line segment from C to D. 76 ~AB is the line segment with length AB and direction from point A to point B. Similarly, ~BA is the line segment with length AB = BA and direction from point B to point A. Suppose we have two line segments AB and BC that join at a point B. We denote the angle B between the line segments by ^B. A line of symmetry divides a shape in such a way that it appears the same on both sides of the line. For example, if you divide a square along its one diagonal then you divide it into two triangles that are exactly the same i.e. they flt perfectly on each other when the square is folded along the diagonal. If a line AB bisects a line CD then AB divides CD into half. A stop sign is in the shape of an octagon, an eight-sided polygon. Some coins are heptagonal and hexagonal. In the UK there are two heptagonal coins. The honeycomb of a beehive consist of hexagonal cells. (NOTE: these are true examples, but maybe best left till the list of names of polygons. examples here should try to motivate the study of polygons... how can we actually use the study of polygons to enrich our lives.) 6.1.1 Triangles A triangle is a three-sided polygon. |
The sum of the angles of a triangle is 180–. The exterior angle of any corner of a triangle is equal to the sum of the two opposite interior angles (NOTE: a diagram for these 2 rules). We have the following triangles: Equilateral All 3 sides are equal and each angle is 60–. (NOTE: need an example diagram) Isosceles Two equal angles occur opposite two equal sides and vice versa. (NOTE: need an example diagram) Right-angled This triangle has a right angle. The side opposite this angle is called the hypotenuse. Pythagoras’s Theorem is often applied to this type of triangle (NOTE: the students have not been subjected to Pythagoras yet... so place in a reference to the relevant chapter/section.) (NOTE: need an example diagram) Scalene This is any other triangle where the sides have difierent lengths and angles are difierent sizes. (NOTE: this is not on the syllabus but its small, and need an example diagram) 77 6.1.2 Quadrilaterals Quadrilaterals are four-sided polygons. The basic quadrilaterals are the trapezium, parallelogram, rectangle, rhombus, square and kite,. Trapezium This quadrilateral has one pair of parallel opposite sides. It may also be called a trapezoid. If the other pair of opposite sides is also parallel then the trapezium is a parallelogram. Another type of trapezium is the isosceles trapezium, where one pair of opposite sides is parallel, the other pair of sides is equal and the angles at the ends of each parallel side are equal. An isosceles trapezium has one line of symmetry and its diagonals are equal in length. (NOTE: need an example diagram) Parallelogram A parallelogram is a special type of trapezium. It is a quadrilateral with two pairs of opposite sides equal. Squares, rectangles and rhombuses are parallelograms. We have the following properties of parallelograms. Both pairs of opposite sides are parallel. Both pairs of opposite sides are equal in length. (NOTE: what does equal mean? is it in length, or direction, or both? we must be more precise in our wording) Both pairs of opposite angles are equal. Both diagonals bisect each other (i.e. they cut each |
other in half). There are not always lines of symmetry. (NOTE: what is a line of symmetry? we haven’t mentioned them before here. is this really a property?) (NOTE: need an example diagram) Rectangle This is a parallelogram with 90– angles. † Both pairs of opposite sides are parallel. Both pairs of opposite sides are equal. All angles are equal to 90–. Both diagonals bisect each other. Diagonals are equal in length. There are two lines of symmetry. (NOTE: need an example diagram) Rhombus This is a parallelogram with adjacent sides equal. † Both pairs of opposite sides are parallel. All sides are equal in length. Both pairs of opposite angles equal. Both diagonals bisect each other at 90–. Diagonals of a rhombus bisect both pairs of opposite angles. There are two lines of symmetry. (NOTE: need an example diagram) Square † This is a rhombus with all four angles equal to 90– or a rectangle with adjacent sides equal. 78 In a square both pairs of opposite sides are parallel. All sides are equal in length. All angles are equal to 90–. Both diagonals bisect each other at rightangles. Diagonals are equal in length and bisect both pairs of opposite angles. There are four lines of symmetry. (NOTE: need an example diagram) Kite A kite is a parallelogram with two pairs of adjacent sides equal. † Other properties of a kite are that the two pairs of adjacent sides are equal. One pair of opposite angles are equal where the angles must be between unequal sides. One diagonal bisects the other diagonal and one diagonal bisects one pair of opposite angles. Diagonals intersect at right-angles. There is one line of symmetry. (NOTE: need an example diagram) 6.1.3 Other polygons There are many other polygons, some of which are given in the table below. (NOTE: need an example diagram) 6.1.4 Similarity of Polygons If two polygons are similar, one is an enlargement of the other. This means that the two polygons will have the same angles and their sides will be in the same proportion. (NOTE: expand this with quick examples.) We can use the symbol to mean is similar to. Two polygons are similar if and only if either or both of the following are » true: |
Area of trapezium: 1 2 £ £ Area of parallelogram and rhombus: base Area of rectangle: length Area of square: length of side (NOTE: everything from here on in Extra is probably acceptable syllabus material, but it is here for now so i can see what needs to be brought back in to the main text. the theorems are not on the syllabus, but we should maybe include them since they use basic geometry techniques... but do not call them theorems, rather use them as in-line examples or worked examples.) perpendicular height length of side breadth £ £ £ Parallelograms 4 4 ABC and alternate angles =; AD alternate angles =; AB ADC: 1. \DAC = \ACB To show that a quadrilateral is a parallelogram, show any one of the flrst four properties or that one pair of opposite sides are equal and parallel. Theorem 1: Given a parallelogram ABCD (with both pairs of opposite sides parallel), prove that the opposite sides and angles are equal. (flgure 4 here) Proof: Join AC. In k BC 2. \BAC = \ACD CD 3. AC = AC (AAS) So AB = CD and DA = BC ABC common sides So · 4 corresponding sides in congruent triangles ^B = ^D; ^A = ^C corresponding angles in congruent triangles Hence opposite sides equal and opposite angles equal. Theorem 2: Given parallelogram ABCD with AC and BD joined and denote their intersection by O. Prove that AC and BD bisect each other. (flgure 5 here) Proof: In alternate CD 3. angles =; AB AB = CD (AAS) So AO = OC; BO = OD corresp sides in congruent triangles Hence AC and BD bisect each other. CD 2. \ABD = \BDC k opposite sides of parm equal So COD: 1. \BAC = \ACD alternate angles =; AB AOB and CDO CDA ABO · 4 4 4 4 4 k k 80 Rectangles To prove that a quadrilateral is a rectangle you can flrst prove that it is a parallelogram and then prove that it has a right-angle. Or you can directly prove that it has four right-angles. Theorem : Given rectangle ABCD prove ACD and that diagonals are |
equal in length. common sides (RHS) So 4 corresp sides in congruent triangles Hence diagonals are equal in (flgure 6 here) Proof: opposite sides equal 2. DC = DC all angles equal So BCD: 1. AD = BC 4 are equal 3. \D = \C AC = BD length. BCD ADC · 4 In 4 Rhombuses AOB and AOD: 1. AB = AD To prove that a quadrilateral is a rhombus you can flrst prove that it is a parallelogram. Then you can prove: all four sides equal, diagonals intersect at right-angles or diagonals bisect corner angles. Theorem : Given rhombus ABCD with diagonals intersecting at point O. Prove that the diagonals intersect at right-angles and that they bisect the corner angles. (flgure 7 here) Proof: In all sides of rhombus equal 2. AO = AO 4 4 diags bisect each other So common sides are equal 3. OB = OD · (SSS) So \AOB = \AOD corresp angles in congruent triangles 4 But BD is a straight line So \AOB = \AOD = 90– sum of angles is 180– So AC and BD intersect therefore diagonals intersect Now \BAO = \DAO \ABO = corresp angles in congruent triangles Similarly, \CBO \BCO = \CDO = \ADO AOD ) \DCO Hence diagonals also bisect the corner angles. COB DOC AOB BOC · 4 · 4 COD AOD AOB ) ) 4 4 · 4 4 4 Squares To prove that a quadrilateral is a square you can prove that it is a rhombus and then prove that it has four right-angles or equal diagonals. You can also prove that it is a rectangle and then prove all four sides equal, diagonals intersect at right-angles or diagonals bisect corner angles. 6.2 Solids The syllabus requires: † † † † † † analyse, describe and represent the properties and relationships of geometric solids by calculating surface area, volume and the effect on these by scaling one or more dimension by k. estimate volume of everyday objects solids to consider: sphere, hemisphere, combinations with cylinders ( |
good Similarly, the function y = f (x + a), is the result of shifting the function by a to the left (the function value at x + a is moved to x). f (x) f (x a) ¡ a f (x + a) f (x) a Figure 6.1: Graph of a function with the translations x x Note that this is just a simple shift either left or right of the entire graph. x + a and x!! a. ¡ Now let us look at the function y b = f (x), where b is some positive constant, which is the result of replacing y by y b in the function y = f (x). This gives us that y = f (x) + b. The function is thus shifted upwards by the constant b. ¡ ¡ We can also replace y by y + b, which again just results in movement in the opposite direction. We can see this because y + b = f (x) implies that y = f (x) b, which shows that f (x) has been shifted downwards by b. ¡ f (x) + b b f (x) f (x) b f (x) b ¡ Figure 6.2: Graph of sin(x) with the translations y y! that this is just a simple shift either up or down of the entire graph. y + b and b! b. Note ¡ Now, what about relations in general? If we take a relation, which depends b (where a and b are positive a and y by y on x and y, and replaced x by x ¡ ¡ 83 ¡ a;y constants), then what would happen? Well, the relation value at the point (x b) would be moved to the point (x;y). Therefore the graph of the relation would be shifted right by a and upwards by b. Similarly, if we replaced x by x + a and y by y + b, then the function would be moved left by a and downwards by b. ¡ Reections x), where f (x) is a known function. This takes Now consider deflning y = f ( ¡ the function value at the point x to the point x. In other words, the function values on one side of the y-axis are moved to the other side of the y-axis. Thus the function is reected about the y-axis. ¡ Alternatively |
we can look at the function y = f (x). This is the same as ¡ saying y = f (x), which reects the function about the x-axis (every positive function value is changed to the corresponding negative function value and vice versa). ¡ f (x) f ( x) ¡ f (x) f (x) ¡ Figure 6.3: Graph of a function with the reections x! f (x). Note that these are just reections in the vertical and horizontal axes,! ¡ x and f (x) ¡ respectively. As before, we can generalise these idea to deal with relations. In all cases, x, then the relation will be reected about the y-axis and if ¡ y then there will be a reection about the x-axis. if we change x to we replace y by ¡ Note: We say that f (x) is symmetric about the y axis if f (x) = f ( x) (in other words, the function and its reection about the y-axis are the same). f (x), then we say that f (x) is symmetric about the x-axis. Similarly, if f (x) = ¡ ¡ 6.5 Stretching and Shrinking Graphs (NOTE: The flgures are correct, but i think the flgures are negating the truth.) We shall now look at what happens to f (x) if we consider the function y = f (ax), where a is a positive constant and a > 1. The point ax on the x-axis is further 84 ). Now the function value at ax is moved x from the y-axis than x (since j j to x, so the function is moved towards the x-axis by a factor of a. Thus the efiect is to shrink f (x) horizontally by a factor of a. ax j j > The function y = f ( x closer to the y-axis than x (as x so the function is stretched horizontally by a factor of a. a ) has the opposite efiect. The point x ). The function value at on the x-axis is a is moved to f (x) f (ax) f ( x a ) f (x) ax x x x a Figure 6.4: Graph of a function with the rescaling x that these are |
just a shrinking and stretching in the horizontal axis. ax and x!! x a. Note (NOTE: this is correct. vertical and horizontal rescalings are difierent... i think the author got confused and thought the same thing happened in each. this needs flxed.) Replacing y by by, where b is a positive constant and b > 1, gives by = f (x) and thus y = 1 b f (x). The function value at any point x is reduced by a factor of b. Therefore the graph shrinks vertically by a factor of b. b to give y b = f (x), we obtain the function y = bf (x). At each x value the function value is increased by a factor of b so the function is stretched vertically. Similarly, if we replace y with y Again these result can be used to deal with relations as well. If x and y are 85 1 b f (x) f (x) f (x) bf (x) Figure 6.5: Graph of a function with the rescaling f (x)! bf (x). Note that these are just a shrinking and stretching in the vertical axis. 1 b f (x) and f (x)! replaced by ax and by in any relation, the efiect is to shrink the graph of this relation by a factor of a horizontally and by a factor of b vertically. Similarly, a and y to y changing x to x b causes the relation to be stretched horizontally and vertically by factors of a and b respectively. 6.6 Mixed Problems If we perform many of these transformations on a given function, then we must combine the difierent efiects. However, it is very important that we efiect these changes in the right order. Here are some examples of mixed problems. 6.7 Equation of a Line The syllabus requires: 86 derive formula for the equation of a line when given 2 points derive formula for the line parallel to a given line and passing through a point derive formula for the inclination of a line † † † 6.8 Circles The syllabus requires: † † † † (grade 12) tangent is perpendicular to the radius (grade 12) the line from the centre of a circle perpendicular to a chord bisects the chord (grade 12) angle subtended by an arc at the centre of a circle is double |
the size of the angle subtended by the same arc at the circle (grade 12) the opposite angle of a cyclic quadrilateral are supplementary the tangent chord theorem (NOTE: really... word for word) thats what it says, 6.8.1 Circles & Semi-circles Circles: A circle centered at the origin with radius r is described by the relation x2 + y2 = r2 (6.2) r) We can see that a circle is not a function, since both (0;r) and (0; satisfy the relation (in other words, the line x = 0 will always cut the circle at two points). ¡ Now, since x2 = r2 y2 and y2 is never negative, it follows that x2 and thus y2 = r2 ¡ r ¡ • x2 and therefore • x ¡ r ¡ • y • r. Therefore the domain of the relation is [ r. Thus the range of the relation is [ r2 r;r]. Similarly, r;r]. • ¡ ¡ Semi-Circles: The equation for a circle x2 + y2 = r2 can also be written as y = §p r2 x2 ¡ (6.3) Now let us we consider the positive and negative square roots separately. These describe semi-circles on either side of the x-axis. Thus the equations for two types of semi-circles are as follows: y = r2 x2 and y = r2 x2 (6.4) p The domain of each of these semi-circles is [ the flrst semi-circle) and [ r;0] (for the second semi-circle). ¡ ¡p r;r] and the range is [0;r] (for ¡ ¡ ¡ 87 Note: These semi-circles are functions, since there is only one y value corresponding to each x value. 6.9 Locus The syllabus requires: † (grade 12) derive the equation of the locus of all points; equidistant from a given point, equidistant from 2 given points, equidistant from a given point and a line parallel to the x or y axis 6.10 Other Geometries The syllabus requires: basic knowledge of spherical geometry, taxicab geometry and fractals † 6.11 Unsorted 6 |
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