City UK AS 1051As1051paper2(summer)solutions. ALL ANSWERS CORRECT. VERIFIED BY EXPERTS

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AS1051 CITY UNIVERSITY London BSc Honours Degree in Actuarial Science Part 1 Mathematics for Actuarial Science: Paper 2 (solutions) 2014 Time allowed: 2 hours Full marks may be obtained for c... orrect answers to ALL of the SIX questions in Section A and TWO of the THREE questions in Section B. If more than TWO questions from Section B are answered, the best TWO marks will be credited. All necessary working must be shown. 1 Turn over : : : Section A 1. (a) [4 marks] (seen) A relation on a set A is said to be reflexive if aRa 8a 2 A. A relation is said to be transitive if 8a; b; c 2 A, aRb and bRc ) aRc (b) [4 marks] (unseen) 8P 2 M P − P = 0A and 0 2 R Suppose P − Q = lA for some l 2 R then P − Q = −(P − Q) = −lA = A(−l) and −l 2 R and so QRP. Suppose that P − Q = lA and Q − R = mA for some l; m 2 R. Then P − R = P − Q + Q − R = lA + mA. So P − R = A(l + m) with l + m 2 R. So if PRQ and QRR then PRR. 2. (a) [4 marks] Solve first An+1 = −4An yields Anh = C(−4)n. Now try a solution of the form un = an + b. This leads to an + a + b + 4an + 4b = 2n so a = 2=5 and b = −2=25. Thus A n = C(−4)n + 2 5 n − 2 25 : (b) [4 marks] Using the auxiliary equation λ2 − λ − 1 = 0 we determine that the solution for the homogeneous equation is A n = C 1 + p5 2 !n + D 1 −2p5!n : Now apply A0 = 1 and A1 = 1 to get: C = 1 + p5 2p5 D = p5−1 2p5 and so A n = 1 + p5 2p5 1 + p5 2 !n + p25p−5 1 1 −2p5!n : 2 Turn over : : : This study source was downloaded by 100000824368823 from CourseHero.com on 05-01-2021 05:39:44 GMT -05:00 [Show More]

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