2007 IJC Prelims Paper 2 Solution

2007 IJC Prelims Paper 2 Solution - 2007 prelim2 maths...

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2007 prelim2 maths solution H2 paper 2 pg 1 of 6 Let z be a 4 th root of 3i + . Then 1 i 4 6 3 i 2e z π = += 1 2i 6 2e , 0,1 ,2,3 k k ππ  +   == 11 1 i 2 42 4 2 e , 0,1, 2, 3 k zk + 1 4 z 1 i 2 4 2 e , 0,1, 2, 3 k ±+ 2i 2ii 2iii -4 -1 3 (1 .5 , -0.25 ) x y -0.5 1 3 (3 ,) - -1 3 (1 .5 , -2 ) x y 2 -4 (1 .5 , 2 ) 2 - -1 2 6 (4 .5 , -8 ) x y 6
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2007 prelim2 maths solution H2 paper 2 pg 2 of 6 3ai The negative terms of the series is U 2 , U 4 , U 6 , … 35 7 77 2007 , 2007 , 2007 ,... 9 99  - --   New GP with first term 7 2007 1561 9 - =- Common ratio 2 7 49 9 81 -= So sum to infinity exists and 1561 3 951.28 49 1 1 81 a S r - = = - - 3aii 1 2007 n U < 1 71 2007 9 2007 n - -< 1 2007 9 2007 n - < 1 2 9 2007 n - < ( 29 2 1 ln ln 9 2007 n ( 29 2ln 2007 1 7 ln 9 n - - 61.52 n Least n = 62 Alternatively Key into GC: Min(n)=1 Un=2007(-7/9)^(n-1 Umin= 2007 Use Table (trial and error) n=61, |Un| =… >1/2007 = 5.0x10 -4 n=62, |Un| = … <5.0x10 -4 Least n=62 3b 22 24 29 2 24 16 16 9 20 4 7 dd d ++ = +=+ = 15 1 54 2(2) 14 27 90 S  =+   =
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2007 prelim2 maths solution H2 paper 2 pg 3 of 6 4i Let 1 tan yx - = . Then tan = . Differentiate wrt x to get 2 d sec 1 d y y x = . So 2 22 d 111 d sec 1 tan 1 y x y = == ++ .
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This note was uploaded on 08/29/2009 for the course MA 9740 taught by Professor Moe during the Summer '07 term at Singapore Management.

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2007 IJC Prelims Paper 2 Solution - 2007 prelim2 maths...

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