Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống
1
/ 40 trang
THÔNG TIN TÀI LIỆU
Thông tin cơ bản
Định dạng
Số trang
40
Dung lượng
477,76 KB
Nội dung
NUMBER SEQUENCES 109 6. The first, tenth and last terms of an arithmetic progression are 9, 40.5, and 425.5 respectively. Find (a) the number of terms, (b) the sum of all the terms and (c) the 70th term. [(a) 120 (b) 26 070 (c) 250.5] 7. On commencing employment a man is paid a salary of £7200 per annum and receives annual increments of £350. Deter- mine his salary in the 9th year and calculate the total he will have received in the first 12 years. [£10 000, £109 500] 8. An oil company bores a hole 80 m deep. Estimate the cost of boring if the cost is £30 for drilling the first metre with an increase in cost of £2 per metre for each succeeding metre. [£8720] 14.4 Geometric progressions When a sequence has a constant ratio between suc- cessive terms it is called a geometric progression (often abbreviated to GP). The constant is called the common ratio, r Examples include (i) 1,2,4,8, where the common ratio is 2 and (ii) a, ar, ar 2 , ar 3 , where the common ratio is r If the first term of a GP is ‘a’ and the common ratio is r,then the n th term is : ar n−1 which can be readily checked from the above exam- ples. For example, the 8th term of the GP 1, 2, 4, 8, is 12 7 D 128,sincea D 1andr D 2 Let a GP be a, ar, ar 2 , ar 3 , ar n1 then the sum of n terms, S n D a Car Car 2 C ar 3 CÐÐÐCar n1 1 Multiplying throughout by r gives: rS n D ar C ar 2 C ar 3 C ar 4 C ar n1 C ar n 2 Subtracting equation (2) from equation (1) gives: S n rS n D a ar n i.e. S n 1 r D a1 r n Thus the sum of n terms, S n = a.1 − r n / .1 − r/ which is valid when r<1 Subtracting equation (1) from equation (2) gives S n = a.r n − 1/ .r − 1/ which is valid when r>1 For example, the sum of the first 8 terms of the GP 1, 2, 4, 8, 16, is given by: S 8 D 12 8 1 2 1 , since a D 1andr D 2 i.e. S 8 D 1256 1 1 D 255 When the common ratio r of a GP is less than unity, the sum of n terms, S n D a1 r n 1 r , which may be written as S n D a 1 r ar n 1 r Since r<1, r n becomes less as n increases, i.e. r n ! 0asn !1 Hence ar n 1 r ! 0asn !1. Thus S n ! a 1 r as n !1 The quantity a 1 r is called the sum to infinity, S 1 , and is the limiting value of the sum of an infinite number of terms, i.e. S ∞ = a .1 − r/ which is valid when 1 <r<1 110 ENGINEERING MATHEMATICS For example, the sum to infinity of the GP 1 C 1 2 C 1 4 C is S 1 D 1 1 1 2 ,sincea D 1andr D 1 2 ,i.e.S 1 D 2 14.5 Worked problems on geometric progressions Problem 9. Determine the tenth term of the series 3, 6, 12, 24, 3, 6, 12, 24, is a geometric progression with a common ratio r of 2. The n 0 th term of a GP is ar n1 ,wherea is the first term. Hence the 10th term is: 32 101 D 32 9 D 3512 D 1536 Problem 10. Find the sum of the first 7 terms of the series, 1 2 ,1 1 2 ,4 1 2 ,13 1 2 , 1 2 , 1 1 2 , 4 1 2 , 13 1 2 , is a GP with a common ratio r D 3 The sum of n terms, S n D ar n 1 r 1 Hence S 7 D 1 2 3 7 1 3 1 D 1 2 2187 1 2 D 546 1 2 Problem 11. The first term of a geometric progression is 12 and the fifth term is 55. Determine the 8’th term and the 11’th term The 5th term is given by ar 4 D 55, where the first term a D 12 Hence r 4 D 55 a D 55 12 and r D 4 55 12 D 1.4631719 The 8th term is ar 7 D 121.4631719 7 D 172.3 The 11th term is ar 10 D 121.4631719 10 D 539.7 Problem 12. Which term of the series: 2187, 729, 243, is 1 9 ? 2187, 729, 243, is a GP with a common ratio r D 1 3 and first term a D 2187 The n 0 th term of a GP is given by: ar n1 Hence 1 9 D 2187 1 3 n1 from which 1 3 n1 D 1 92187 D 1 3 2 3 7 D 1 3 9 D 1 3 9 Thus n 1 D 9, from which, n D 9 C1 D 10 i.e. 1 9 is the 10th term of the GP Problem 13. Find the sum of the first 9 terms of the series: 72.0, 57.6, 46.08, The common ratio, r D ar a D 57.6 72.0 D 0.8 also ar 2 ar D 46.08 57.6 D 0.8 The sum of 9 terms, S 9 D a1 r n 1 r D 72.01 0.8 9 1 0.8 D 72.01 0.1342 0.2 D 311 .7 Problem 14. Find the sum to infinity of the series 3, 1, 1 3 , 3, 1, 1 3 , is a GP of common ratio, r D 1 3 NUMBER SEQUENCES 111 The sum to infinity, S 1 D a 1 r D 3 1 1 3 D 3 2 3 D 9 2 D 4 1 2 Now try the following exercise Exercise 53 Further problems on geomet- ric progressions 1. Find the 10th term of the series 5, 10, 20, 40, [2560] 2. Determine the sum of the first 7 terms of the series 0.25, 0.75, 2.25, 6.75, [273.25] 3. The first term of a geometric progression is 4 and the 6th term is 128. Determine the 8th and 11th terms. [512, 4096] 4. Which term of the series 3, 9, 27, is 59 049? [10 th ] 5. Find the sum of the first 7 terms of the series 2, 5, 12 1 2 , (correct to 4 signifi- cant figures). [812.5] 6. Determine the sum to infinity of the series 4, 2, 1, [8] 7. Find the sum to infinity of the series 2 1 2 , 1 1 4 , 5 8 , 1 2 3 14.6 Further worked problems on geometric progressions Problem 15. In a geometric progression the sixth term is 8 times the third term and the sum of the seventh and eighth terms is 192. Determine (a) the common ratio, (b) the first term, and (c) the sum of the fifth to eleventh terms, inclusive (a) Let the GP be a, ar, ar 2 , ar 3 , , ar n1 The 3rd term D ar 2 and the sixth term D ar 5 The 6th term is 8 times the 3rd Hence ar 5 D 8 ar 2 from which, r 3 D 8and r D 3 p 8 i.e. the common ratio r = 2 (b) The sum of the 7th and 8th terms is 192. Hence ar 6 C ar 7 D 192. Since r D 2, then 64a C 128a D 192 192a D 192, from which, a, the first term = 1 (c) The sum of the 5th to 11th terms (inclusive) is given by: S 11 S 4 D ar 11 1 r 1 ar 4 1 r 1 D 12 11 1 2 1 12 4 1 2 1 D 2 11 1 2 4 1 D 2 11 2 4 D 2408 16 D 2032 Problem 16. A hire tool firm finds that their net return from hiring tools is decreasing by 10% per annum. If their net gain on a certain tool this year is £400, find the possible total of all future profits from this tool (assuming the tool lasts for ever) The net gain forms a series: £400 C£400 ð0.9 C £400 ð0.9 2 C , which is a GP with a D 400 and r D 0.9 The sum to infinity, S 1 D a 1 r D 400 1 0.9 D £4000 = total future profits Problem 17. If £100 is invested at compound interest of 8% per annum, determine (a) the value after 10 years, (b) the time, correct to the nearest year, it takes to reach more than £300 (a) Let the GP be a, ar, ar 2 , ar n The first term a D £100 and The common ratio r D 1.08 112 ENGINEERING MATHEMATICS Hence the second term is ar D 1001.08 D £108, which is the value after 1 year, the third term is ar 2 D 1001. 08 2 D £116.64, which is the value after 2 years, and so on. Thus the value after 10 years D ar 10 D 100 1.08 10 D £215.89 (b) When £300 has been reached, 300 D ar n i.e. 300 D 1001.08 n and 3 D 1.08 n Taking logarithms to base 10 of both sides gives: lg 3 D lg 1. 08 n D n lg1.08, by the laws of logarithms from which, n D lg 3 lg 1.08 D 14.3 Hence it will take 15 years to reach more than £300 Problem 18. A drilling machine is to have 6 speeds ranging from 50 rev/min to 750 rev/min. If the speeds form a geometric progression determine their values, each correct to the nearest whole number Let the GP of n terms be given by a, ar, ar 2 , ar n1 The first term a D 50 rev/min. The 6th term is given by ar 61 , which is 750 rev/min, i.e., ar 5 D 750 from which r 5 D 750 a D 750 50 D 15 Thus the common ratio, r D 5 p 15 D 1.7188 The first term is a D 50 rev/min, the second term is ar D 501.7188 D 85.94, the third term is ar 2 D 501.7188 2 D 147.71, the fourth term is ar 3 D 501.7188 3 D 253.89, the fifth term is ar 4 D 501.7188 4 D 436.39, the sixth term is ar 5 D 501.7188 5 D 750.06 Hence, correct to the nearest whole number, the 6 speeds of the drilling machine are: 50, 86, 148, 254, 436 and 750 rev/min. Now try the following exercise Exercise 54 Further problems on geomet- ric progressions 1. In a geometric progression the 5th term is 9 times the 3rd term and the sum of the 6th and 7th terms is 1944. Determine (a) the common ratio, (b) the first term and (c) the sum of the 4th to 10th terms inclusive. [(a) 3 (b) 2 (c) 59 022] 2. The value of a lathe originally valued at £3000 depreciates 15% per annum. Cal- culate its value after 4 years. The machine is sold when its value is less than £550. After how many years is the lathe sold? [£1566, 11 years] 3. If the population of Great Britain is 55 million and is decreasing at 2.4% per annum, what will be the population in 5 years time? [48.71 M] 4. 100 g of a radioactive substance disin- tegrates at a rate of 3% per annum. How much of the substance is left after 11 years? [71.53 g] 5. If £250 is invested at compound interest of 6% per annum determine (a) the value after 15 years, (b) the time, correct to the nearest year, it takes to reach £750 [(a) £599.14 (b) 19 years] 6. A drilling machine is to have 8 speeds ranging from 100 rev/min to 1000 rev/min. If the speeds form a geometric progres- sion determine their values, each correct to the nearest whole number. 100, 139, 193, 268, 373, 518, 720, 1000 rev/min 14.7 Combinations and permutations A combination is the number of selections of r different items from n distinguishable items when order of selection is ignored. A combination is denoted by n C r or n r where n C r = n! r!.n − r/! NUMBER SEQUENCES 113 where, for example, 4! denotes 4 ð3 ð2 ð1andis termed ‘factorial 4’. Thus, 5 C 3 D 5! 3!5 3! D 5 ð4 ð3 ð2 ð1 3 ð2 ð12 ð1 D 120 6 ð2 D 10 For example, the five letters A, B, C, D, E can be arranged in groups of three as follows: ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE, i.e. there are ten groups. The above calculation 5 C 3 produces the answer of 10 combinations without having to list all of them. A permutation is the number of ways of selecting r Ä n objects from n distinguishable objects when order of selection is important. A permutation is denoted by n P r or n P r where n P r D nn 1n 2 nr C 1 or n P r = n! .n − r /! Thus, 4 P 2 D 43 D 12 or 4 P 2 D 4! 4 2! D 4! 2! D 4 ð3 ð2 2 D 12 Problem 19. Evaluate: (a) 7 C 4 (b) 10 C 6 (a) 7 C 4 D 7! 4!7 4! D 7! 4!3! D 7 ð6 ð5 ð4 ð3 ð2 4 ð3 ð23 ð2 D 35 (b) 10 C 6 D 10! 6!10 6! D 10! 6!4! D 210 Problem 20. Evaluate: (a) 6 P 2 (b) 9 P 5 (a) 6 P 2 D 6! 6 2! D 6! 4! D 6 ð5 ð4 ð3 ð2 4 ð3 ð2 D 30 (b) 9 P 5 D 9! 9 5! D 9! 4! D 9 ð8 ð7 ð6 ð5 ð4! 4! D 15 120 Now try the following exercise Exercise 55 Further problems on permu- tations and combinations Evaluate the following: 1. (a) 9 C 6 (b) 3 C 1 [(a) 84 (b) 3] 2. (a) 6 C 2 (b) 8 C 5 [(a) 15 (b) 56] 3. (a) 4 P 2 (b) 7 P 4 [(a) 12 (b) 840] 4. (a) 10 P 3 (b) 8 P 5 [(a) 720 (b) 6720] 15 The binomial series 15.1 Pascal’s triangle A binomial expression is one that contains two terms connected by a plus or minus sign. Thus pCq, aCx 2 , 2x Cy 3 are examples of binomial expressions. Expanding aCx n for integer values of n from 0 to 6 gives the results shown at the bottom of the page. From the results the following patterns emerge: (i) ‘a’ decreases in power moving from left to right. (ii) ‘x’ increases in power moving from left to right. (iii) The coefficients of each term of the expan- sions are symmetrical about the middle coef- ficient when n is even and symmetrical about the two middle coefficients when n is odd. (iv) The coefficients are shown separately in Table 15.1 and this arrangement is known as Pascal’s triangle. A coefficient of a term may be obtained by adding the two adjacent coefficients immediately above in the previous row. This is shown by the triangles in Table 15.1, where, for example, 1 C 3 D 4, 10 C5 D 15, and so on. (v) Pascal’s triangle method is used for expan- sions of the form a C x n for integer values of n less than about 8 Problem 1. Use the Pascal’s triangle method to determine the expansion of a Cx 7 Table 15.1 (a + x) 0 (a + x) 1 (a + x) 2 (a + x) 3 (a + x) 4 (a + x) 5 (a + x) 6 1 6 15 20 15 6 1 1 5 10 10 5 1 1 4 6 4 1 1 3 3 1 1 2 1 1 1 1 From Table 15.1, the row of Pascal’s triangle cor- responding to a C x 6 is as shown in (1) below. Adding adjacent coefficients gives the coefficients of a Cx 7 as shown in (2) below. 1 7 21 35 35 21 7 1 (2) 1 6 15 20 15 6 1 (1) The first and last terms of the expansion of a C x 7 are a 7 and x 7 respectively. The powers of ‘a’ decrease and the powers of ‘x’ increase moving from left to right. Hence, .a Y x / 7 = a 7 Y 7a 6 x Y 21a 5 x 2 Y 35a 4 x 3 Y 35a 3 x 4 Y 21a 2 x 5 Y 7ax 6 Y x 7 Problem 2. Determine, using Pascal’s triangle method, the expansion of 2p 3q 5 a Cx 0 D 1 a Cx 1 D a C x a Cx 2 D a Cxa C x D a 2 C 2ax C x 2 a Cx 3 D a Cx 2 a Cx D a 3 C 3a 2 x C 3ax 2 C x 3 a Cx 4 D a Cx 3 a Cx D a 4 C 4a 3 x C 6a 2 x 2 C 4ax 3 C x 4 a Cx 5 D a Cx 4 a Cx D a 5 C 5a 4 x C 10a 3 x 2 C 10a 2 x 3 C 5ax 4 C x 5 a Cx 6 D a Cx 5 a Cx D a 6 C 6a 5 x C 15a 4 x 2 C 20a 3 x 3 C 15a 2 x 4 C 6ax 5 C x 6 THE BINOMIAL SERIES 115 Comparing 2p 3q 5 with a C x 5 shows that a D 2p and x D3q Using Pascal’s triangle method: a Cx 5 D a 5 C 5a 4 x C 10a 3 x 2 C 10a 2 x 3 CÐÐÐ Hence 2p 3q 5 D 2p 5 C 52p 4 3q C 102p 3 3q 2 C 102p 2 3q 3 C 52p3q 4 C 3q 5 i.e. .2p − 3q/ 5 = 32p 5 − 240p 4 q Y 720p 3 q 2 − 1080p 2 q 3 Y 810pq 4 − 243q 5 Now try the following exercise Exercise 56 Further problems on Pascal’s triangle 1. Use Pascal’s triangle to expand x y 7 x 7 7x 6 y C21x 5 y 2 35x 4 y 3 C 35x 3 y 4 21x 2 y 5 C 7xy 6 y 7 2. Expand 2aC3b 5 using Pascal’s triangle. 32a 5 C 240a 4 b C720a 3 b 2 C 1080a 2 b 3 C 810ab 4 C 243b 5 15.2 The binomial series The binomial series or binomial theorem is a formula for raising a binomial expression to any power without lengthy multiplication. The general binomial expansion of a C x n is given by: .a Y x / n = a n Y na n−1 x Y n.n − 1/ 2! a n−2 x 2 Y n.n − 1/.n − 2/ 3! a n−3 x 3 Y ···Y x n where, for example, 3! denotes 3 ð 2 ð 1andis termed ‘factorial 3’. With the binomial theorem n may be a fraction, a decimal fraction or a positive or negative integer. In the general expansion of a Cx n it is noted that the 4th term is: nn 1n 2 3! a n3 x 3 . The number 3 is very evident in this expression. For any term in a binomial expansion, say the r’th term, (r 1) is very evident. It may therefore be reasoned that the r’th term of the expansion .a Y x / n is: n.n − 1/.n − 2/ to.r − 1/terms .r − 1/! a n−.r −1/ x r−1 If a D 1 in the binomial expansion of aCx n then: .1 Y x/ n = 1 Y nx Y n.n − 1/ 2! x 2 Y n.n − 1/.n − 2/ 3! x 3 Y ··· which is valid for 1 <x<1 When x is small compared with 1 then: .1 Y x/ n ≈ 1 Y nx 15.3 Worked problems on the binomial series Problem 3. Use the binomial series to determine the expansion of 2 Cx 7 The binomial expansion is given by: a Cx n D a n C na n1 x C nn 1 2! a n2 x 2 C nn 1n 2 3! a n3 x 3 CÐÐÐ When a D 2andn D 7: 2 Cx 7 D 2 7 C 72 6 x C 76 21 2 5 x 2 C 765 321 2 4 x 3 C 7654 4321 2 3 x 4 C 76543 54321 2 2 x 5 116 ENGINEERING MATHEMATICS C 765432 654321 2x 6 C 7654321 7654321 x 7 i.e. .2 Y x/ 7 = 128 Y 448x Y 672x 2 Y 560x 3 Y 280x 4 Y 84x 5 Y 14x 6 Y x 7 Problem 4. Expand c 1 c 5 using the binomial series c 1 c 5 D c 5 C 5c 4 1 c C 54 21 c 3 1 c 2 C 543 321 c 2 1 c 3 C 5432 4321 c 1 c 4 C 54321 54321 1 c 5 i.e. c − 1 c 5 = c 5 − 5c 3 Y 10c − 10 c Y 5 c 3 − 1 c 5 Problem 5. Without fully expanding 3 Cx 7 ,determinethefifth term The r’th term of the expansion a Cx n is given by: nn 1n 2 to r 1 terms r 1! a nr1 x r1 Substituting n D 7, a D 3andr 1 D 5 1 D 4 gives: 7654 4321 3 74 x 4 i.e. the fifth term of 3 Cx 7 D 353 3 x 4 D 945x 4 Problem 6. Find the middle term of 2p 1 2q 10 In the expansion of a C x 10 there are 10 C 1, i.e. 11 terms. Hence the middle term is the sixth. Using the general expression for the r’th term where a D 2p, x D 1 2q , n D 10 and r 1 D 5 gives: 109876 54321 2p 10 –5 1 2q 5 D 25232p 5 1 32q 5 Hence the middle term of 2p 1 2q 10 is: −252 p 5 q 5 Problem 7. Evaluate (1.002) 9 using the binomial theorem correct to (a) 3 decimal places and (b) 7 significant figures 1 Cx n D 1 Cnx C nn 1 2! x 2 C nn 1n 2 3! x 3 CÐÐÐ 1.002 9 D 1 C0.002 9 Substituting x D 0.002 and n D 9 in the general expansion for 1 C x n gives: 1 C0.002 9 D 1 C90.002 C 98 21 0.002 2 C 987 321 0.002 3 CÐÐÐ D 1 C0.018 C0.000144 C0.000000672 CÐÐÐ D 1.018144672 Hence, 1.002 9 D 1.018, correct to 3 decimal places D 1 .018145, correct to 7 significant figures Problem 8. Determine the value of (3.039) 4 , correct to 6 significant figures using the binomial theorem THE BINOMIAL SERIES 117 (3.039) 4 may be written in the form 1 Cx n as: 3.039 4 D 3 C0.039 4 D 3 1 C 0.039 3 4 D 3 4 1 C0.013 4 1 C0.013 4 D 1 C40.013 C 43 21 0.013 2 C 432 321 0.013 3 CÐÐÐ D 1 C0.052 C0.001014 C 0.000008788 CÐÐÐ D 1.0530228 correct to 8 significant figures Hence 3.039 4 D 3 4 1.0530228 D 85.2948, correct to 6 significant figures Now try the following exercise Exercise 57 Further problems on the binomial series 1. Use the binomial theorem to expand a C2x 4 a 4 C 8a 3 x C 24a 2 x 2 C 32ax 3 C 16x 4 2. Use the binomial theorem to expand 2 x 6 64 192x C 240x 2 160x 3 C 60x 4 12x 5 C x 6 3. Expand 2x 3y 4 16x 4 96x 3 y C216x 2 y 2 216xy 3 C 81y 4 4. Determine the expansion of 2x C 2 x 5 32x 5 C 160x 3 C 320x C 320 x C 160 x 3 C 32 x 5 5. Expand p C2q 11 as far as the fifth term p 11 C 22p 10 q C 220p 9 q 2 C 1320p 8 q 3 C 5280p 7 q 4 6. Determine the sixth term of 3p C q 3 13 [34 749 p 8 q 5 ] 7. Determine the middle term of 2a 5b 8 [700 000 a 4 b 4 ] 8. Use the binomial theorem to determine, correct to 4 decimal places: (a) 1.003 8 (b) 0.98 7 [(a) 1.0243 (b) 0.8681] 9. Evaluate (4.044) 6 correct to 3 decimal places. [4373.880] 15.4 Further worked problems on the binomial series Problem 9. (a) Expand 1 1 C2x 3 in ascending powers of x as far as the term in x 3 ,usingthe binomial series. (b) State the limits of x for which the expansion is valid (a) Using the binomial expansion of 1 C x n , where n D3andx is replaced by 2x gives: 1 1 C2x 3 D 1 C2x 3 D 1 C32x C 34 2! 2x 2 C 345 3! 2x 3 CÐÐÐ = 1 − 6x Y 24x 2 − 80x 3 Y (b) The expansion is valid provided j2xj < 1, i.e. jx j < 1 2 or − 1 2 < x < 1 2 118 ENGINEERING MATHEMATICS Problem 10. (a) Expand 1 4 x 2 in ascending powers of x as far as the term in x 3 ,usingthe binomial theorem. (b) What are the limits of x for which the expansion in (a) is true? (a) 1 4 x 2 D 1 4 1 x 4 2 D 1 4 2 1 x 4 2 D 1 16 1 x 4 2 Using the expansion of 1 Cx n 1 4 x 2 D 1 16 1 x 4 2 D 1 16 1 C2 x 4 C 23 2! x 4 2 C 234 3! x 4 3 CÐÐÐ D 1 16 1 Y x 2 Y 3x 2 16 Y x 3 16 Y ··· (b) The expansion in (a) is true provided x 4 < 1, i.e. jxj < 4 or − 4 < x < 4 Problem 11. Use the binomial theorem to expand p 4 Cx in ascending powers of x to four terms. Give the limits of x for which the expansion is valid p 4 Cx D 4 1 C x 4 D p 4 1 C x 4 D 2 1 C x 4 1 2 Using the expansion of 1 Cx n , 2 1 C x 4 1 2 D 2 1 C 1 2 x 4 C 1/2 1/2 2! x 4 2 C 1/2 1/2 3/2 3! x 4 3 CÐÐÐ D 2 1 C x 8 x 2 128 C x 3 1024 ÐÐÐ = 2 Y x 4 − x 2 64 Y x 3 512 −··· This is valid when x 4 < 1, i.e. x 4 < 4 or − 4 < x < 4 Problem 12. Expand 1 p 1 2t in ascending powers of t as far as the term in t 3 . State the limits of t for which the expression is valid 1 p 1 2t D 1 2t 1 2 D 1 C 1 2 2t C 1/2 3/2 2! 2t 2 C 1/2 3/2 5/2 3! 2t 3 CÐÐÐ using the expansion for 1 Cx n = 1 Y t Y 3 2 t 2 Y 5 2 t 3 Y ··· The expression is valid when j2tj < 1, i.e. jt j < 1 2 or − 1 2 < t < 1 2 Problem 13. Simplify 3 p 1 3x p 1 Cx 1 C x 2 3 given that powers of x above the first may be neglected [...]... f0 r1 D f0 3 D 3 7 2 46 3.7 19.35 Thus, r3 D 3 D 3C0. 042 D 3. 042 D 3. 04, 46 3.7 correct to 3 significant figure Similarly, r3 D 3. 042 12 D 0, correct to 3 [ 1.386, 1 .49 1] 8 D 19.35 D 13 D 0, correct to 3 decimal [ 2. 742 , 4. 742 ] 9 f r1 D f 3 D 73 f0 x D 3 x C 4 Further problems on Newton’s method 9 ³ 1 14 1 3 2 e3. 84 C 5 cos 3 e1.92 C 5 cos 7.68 x 3 f 3. 042 f0 3. 042 125 126 ENGINEERING MATHEMATICS difference... 3 D 73 3 f 4 D8 1 x 2 2x places 2 3x 3 10x D 14, correct to 4 significant figures [2.313] 9 ³ 1 64 3 x 4 3x 3 C 7x D 12, correct to 3 decimal places [ 1.721, 2. 648 ] 4 3x 4 4x 3 C 7x decimal places 5 3 ln x C 4x D 5, correct to 3 decimal places [1. 147 ] 6 x 3 D 5 cos 2x, correct to 3 significant figures [ 1.693, 0. 846 , 0. 744 ] Â 300e 2Â C D 6, correct to 3 significant 2 figures [2.05] 9 D 59 9 ³ 19 4 C 5 cos... in terms of : (a) 30° 5 5 (b) 75° (c) 225° a b c 6 12 4 ° (b) 84 510 2 Convert to radians: (a) 48 (c) 232° 15’ [(a) 0.838 (b) 1 .48 1 (c) 4. 0 54] 141 4 5 rad (b) rad Convert to degrees: (a) 6 9 7 rad [(a) 150° (b) 80° (c) 105° ] (c) 12 Convert to degrees and minutes: (a) 0.0125 rad (b) 2.69 rad (c) 7. 241 rad [(a) 0° 43 0 (b) 1 54 80 (c) 41 4° 530 ] 18 .4 Worked problems on arc length and sector of a circle... ˛ D 0.0 04 and t D 100, R0 has a value of: (a) 21 .4 (c) 15 33 2 27 3 is equal to: (b) 7 (c) 1 8 9 (b) 29 .4 (d) 0.067 pCx 4 D p4 C4p3 xC6p2 x 2 C4px 3 Cx 4 Using Pascal’s triangle, the third term of p C x 5 is: (b) 5p4 x (a) 10p2 x 3 3 2 (d) 10p3 x 2 (c) 5p x MULTIPLE CHOICE QUESTIONS ON CHAPTERS 1–16 34 The value of (a) 35 log2 (a) 36 7 17 20 1 8 2 5 (b) of 4 1 2 80 1 2 38 (c) 16 1 4 3 (b) 1 4 (c) 3... ENGINEERING MATHEMATICS Comparing this with equation (2) gives: 2e 2e D 2a, i.e a = − 2 2f and 2f D 2b, i.e b = − 2 2 2 2 and c D a C b r , i.e r = a 2 Y b 2 − c 4x 4 2 6 2 , p 2 2C y 3 2 where 8 2 aD D 2, b D rD 22 C 4 C 12 3 2 c 3D0 6 2 D 3 p 3 D 16 D 4 Thus the circle has centre (2, −3) and radius 4, as shown in Fig 18. 14 y 4 2 4, b D D 2 rD 2 4x C 6y 4 2 4 p 8D 2 2 D1 −2 0 −2 −3 4 2y C 8 D 0... progression is 18 and the twelfth term is 46 (c) 4. 27 s 43 44 (c) 2 3 1 Â l2 l1 (c) l2 3 2 The eighteenth term is: (a) 72 (b) 74 (c) 68 42 (a) (b) 0.312 (d) 3.209 8x 2 C 13x 6 D x C p qx 3 The values of p and q are: (a) p D 2, q D 4 (b) p D 3, q D 2 (c) p D 2, q D 8 (d) p D 1, q D 8 If log2 x D 3 then: (a) 1.33 ð 10 41 of expansion ˛ is given by: is: (d) 16 (c) x D 9 40 1 4 (d) 88 (a) x D 8 39 5 16 is equal... x 6 f 2 D 22 2 6D f 4 D 42 4 6 D C6 4 It can be seen from these results that the value of f x changes from 4 at f 2 to C6 at f 4 , indicating that a root lies between 2 and 4 This is shown more clearly in Fig 16.1 16.2 The Newton–Raphson method The Newton–Raphson formula, often just referred to as Newton’s method, may be stated as follows: f (x ) 8 2 f (x ) = x −x−6 4 −2 0 2 4 x 4 −6 Figure 16.1 if... 5 (c) 1 16 25 (d) 43 m2 The engineering expression to: (a) 4 (b) 2 4 (c) 16 ð 4 2 is equal 8ð2 4 1 22 (d) 1 30 In a system of pulleys, the effort P required to raise a load W is given by P D aW C b, where a and b are constants If W D 40 when P D 12 and W D 90 when P D 22, the values of a and b are: (a) a D 5, b D 1 (b) a D 1, b D 28 4 1 (d) a D 1 , b D 4 (c) a D 3 , b D 8 5 31 16 4 7 (a) 18 1 32 (d)... 180 0. 749 D 0. 749 0.915° D 0.915 ð 60 nearest minute, hence 0 ° D 42 .915° D 550 , correct to the 0. 749 radians = 42 ° 55 (b) Since 1 rad D 3 3 rad D 4 4 D ° 180 180 3 180 4 ° then ° D 135° Problem 7 Express in radians, in terms of , (a) 150° (b) 270° (c) 37.5° Since 180° D rad then 1° D 180/ , hence 5p rad 180 6 3p rad rad D (b) 270° D 270 180 2 5p 75 (c) 37.5° D 37.5 rad D rad rad D 180 360 24 (a) 150°... 0.96r or (1 0. 04) r and 1.02 h or (1 C 0.02)h (a) 0. 04 r]2 [ 1 C 0.02 h] D r2h 1 5 ³1C x 2 x Now 1 1 2x ³ 1 C 10x (b) 1 3x 4 p 19 1 C 5x ³1C x (c) p 3 6 1 2x 0. 04 2 D1 0. 04 2 1 C 0.02 2 0. 04 C 0. 04 2 D 1 0.08 , neglecting powers of small terms Hence new volume ³ r2h 1 8 If x is very small such that x 2 and higher powers may be neglected, determine the power series for p p 31 xC4 3 8 x 4 x 5 15 1Cx 3 . 1 or n P r = n! .n − r /! Thus, 4 P 2 D 4 3 D 12 or 4 P 2 D 4! 4 2! D 4! 2! D 4 ð3 ð2 2 D 12 Problem 19. Evaluate: (a) 7 C 4 (b) 10 C 6 (a) 7 C 4 D 7! 4! 7 4 ! D 7! 4! 3! D 7 ð6 ð5 4 ð3 ð2 4 ð3 ð23 ð2 D. C 76 21 2 5 x 2 C 765 321 2 4 x 3 C 765 4 4 321 2 3 x 4 C 765 4 3 5 4 321 2 2 x 5 116 ENGINEERING MATHEMATICS C 765 4 32 65 4 321 2x 6 C 765 4 321 765 4 321 x 7 i.e 117 (3.039) 4 may be written in the form 1 Cx n as: 3.039 4 D 3 C0.039 4 D 3 1 C 0.039 3 4 D 3 4 1 C0.013 4 1 C0.013 4 D 1 C40.013 C 4 3 21 0.013 2 C 4 32 321 0.013 3 CÐÐÐ D