2007 PJC Prelim Paper 1_solution

2007 PJC Prelim - 2007 J2 Prelim Paper 1 solution 1 iw z 1 i(1 2 z(1 i)w 6 2i(2 From(1 z iw 1 i(3 Sub(3 into(2 2iw 2 2i w iw 6 2i 3iw w 8 2i 8 2i w

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h r a 2007 J2 Prelim Paper 1- solution 1. i 1 i wz     ------- (1) 2 (1 i) 6 2i zw   ----- (2) From (1), i 1 i     ------- (3) Sub (3) into (2), 2i 2 2i i 6 2i w w w     3i 8 2i ww   82 13 i w i  2 2i w    ; 1i z  2. Using sine rule, sin sin 6 ab 2 sin 3 2 3! b     2 21 6 b (shown) Substitute a = 1 cm and b = 3 cm, 3 6 1 0 2.53, 0.167 or 2.36 rad    Since cannot be negative, reject = -2.53. It is assumed that is small, reject = 2.36 Only 1 possible answer for , i.e . =0.167. 3. 2 11 2 2 1 n r rn rr    1 2 2 2 1 2 3 2 2 23 4 2 2 34 n n n 2 2 2 32 1 2 2 2 2 1 nn       4 2 2 1 2 1 2 1 2 2( 1) 2 ( 1)( 6) 2 n n n n n n Sum equal to zero ( 6) 0 2  1 or 6   Since 2 6    4.     22 1 1 1 2 12 x xx x       1 2 2 1 2 2 1 1 . . . ,| | 1 2 2! 1 1 2 1 2 . . . | 2! 2 x x x x x x x x   2 2 1 2 2 8 2 x x x x   2 3 15 1 1 | | 2 8 2 x x x   9 1 3 1 15 1 8 let ; 1 6 8 2 8 8 64 8 3 3 15 1 2 16 512 6 623 2 512 x           Therefore, 623 6 256 (one possible answer) 9 1 3 1 15 1 8 let 1 6 8 2 8 8 64 8 3 3 15 1 16 512 6 x           Therefore, 623 1536 63 512 623   (alternative answer) 5. 2 2 h ra  2 2 h 2 2 2 44 hh V r h a h a h (shown) Differentiating, 2 2 3 4 dV h a dh 0 dV dh 2 2 3 0 4 h a 2 3 ha ( 2 3 a rejected) 6 b a A B C

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2 x y O -1 3 Substituting back into V ,   3 3 2 2 24 3 4 3 3 3 a aa Va        2 2 3 0 2 d V h dh
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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 PJC Prelim - 2007 J2 Prelim Paper 1 solution 1 iw z 1 i(1 2 z(1 i)w 6 2i(2 From(1 z iw 1 i(3 Sub(3 into(2 2iw 2 2i w iw 6 2i 3iw w 8 2i 8 2i w

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