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Get Mastering Advanced Pure Mathematics PDF

By Geoff Buckwell (auth.)

ISBN-10: 0333620496

ISBN-13: 9780333620496

ISBN-10: 1349135518

ISBN-13: 9781349135516

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Without evaluating ex and ~' 1 l (iii) - +- (ii) ex+~ ex ~ (vi) ex4 + ~4 (v) Solution Dividing the equation by 4, we get x2 -~x+~=0 "'+A- __ J__J. "" p4-4 Hence ex~=~ We have the answers to (i) ~and (ii) ~ (iii) .!. + .!. = ~ + ex = ~ ~ .!. = 3 ex ~ ex~ 4 · 4 (iv) ex2 + ~ 2 cannot be evaluated immediately. However (ex+ ~) 2 = ex2 + 2ex~ + ~ 2 Rearranging, ex2 + ~2 = (ex+ ~) 2 - 2ex~ = m2-2 X~ =k (v) 3 ex + ~ suggests working out (ex + ~) 3 3 (ex+ ~) = ex 3 + 3ex2 ~ + 3ex~2 + ~ 3 3 = ex3 + ~ 3 + 3ex~(ex + ~) Hence ex3 + ~ 3 =(ex+ ~) 3 - 3ex~(ex + ~) = 3 {i) -3 X~ X~ = -649 (vi) ex4 + ~4 = (ex2)2 + (~2)2 = (ex2 + ~2f _ 2 ex2~2 = (16I)2 - 2 X (1)2 31 4 = -256 It is possible, by a technique of transforming the equation, to find a new equation where the roots are related to the roots of a given equation.

4 Solve the equation 2x 3 + 3x 2 + 3x + 1 = 0. = 2x 3 + 3x 2 + 3x + 1 =1 =2+3+3+1=9 = 16 + 12 + 6 + 1 = 35 Let f(x) f(O) f(1) f(2) Clearly, as x increases, f(x) is getting bigger and not closer to zero. Try negative values of x. f(-1) = -2 + 3-3 + 1 = -1 f(-2) = -16+ 12- 6+ 1 = -9 f(-3) = -54+27- 9+ 1 = -35 Here you can see that f(x) is becoming more negative. However, since and f(O) = 1 f(-1) = -1 there must be a value of x between 0 and -1 for which f(x) = 0. Try x = -!. = -! ) = x +! is a factor.

Solution If you were not told this was a geometric series, you could deduce nothing from being given just the first two terms. e. 5 9 (4 sig. 5 -1) (4 sig. 6 The second term of a geometric series is 8, and the fifth term is 64. Find the first term and the common ratio. Solution Here (i) ar = 8 and 4 = 64 4 64 ar ar (ii) + (i) (ii) (cancel by a) 8 r3 = 8 ar Hence r=2 Substitution into (i) gives a =4 The first term is 4, the common ratio is 2. 7 How many terms are there in the geometric series 9 324 4+8+ 16 + ...

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Mastering Advanced Pure Mathematics by Geoff Buckwell (auth.)


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