2007 HCI Prelims Paper 1 Solution

# 2007 HCI Prelims Paper 1 Solution - 2007 HCI Preliminary...

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2007 HCI Preliminary Examination, H2 Mathematics Paper 1 1. Let be the statement “ n P + = Z n n n u n )! 1 ( 1 P : LHS = RHS = 1 1 = u 1 1 ! 0 = = LHS Therefore, is true. 1 P Assume is true, that is, k P + = Z k some for k k u k )! 1 ( To show : 1 + k P 1 ! 1 + = + k k u k LHS = k k u k k u 1 2 1 + = + (given) ) 1 ( ! 1 )! 1 ( )! 1 ( 1 2 + = + = + = k k k k k k k k k 1 P is true, true also true, by mathematical induction, is true . k P 1 + k P n P + Z n 3. 2 54 ( 2 ) 10 xx x −+ = + > for any x 0 (5 4 ) 0 () x xaxb −− Since 0 > , 0 x 0 a b + + 0 x or axb << But x exist only when x 0. Therefore, 0 x = or ax (ans) b 2007 HCI H2 Mathematics Paper 1

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4 . 88 8 8 h rR R hh r R Rr =⇒ = = 2 81 28 1 6 d3 2 16 d 6 A rrr RR A r r π ππ ⎛⎞ =− = ⎜⎟ ⎝⎠ When d 0 d A r = , 32 32 16 0 16 rr −= ⇒= 2 R r r R h 8cm 2 2 2 0 d A < Thus, A is maximum when 2 R r = . 2 2 16 Maximum 16 4 cm 22 AR R πππ = 5. Re Im O 3 3 d P ( z 1 ) 3 4 π 4 π x 1 y 1 1 1 3 w −≤ r ; Largest value of r = d Deduce cos 3 d π = 4 OR 2 1 2 3 d = 3 2 or 2.12 2 d = Let P represent the complex number z 1 . 1 cos 43 x π = and 1 sin y π = 11 3 2 x y ⇒== 1 33 3 zi =+ + (ans) 2007 HCI H2 Mathematics Paper 1
7. 2 2 1 3 5 2 1 ) 2 )( 1 ( 10 7 + + = + + + + = + + + r r r r C r B r A r r r r = = + + + n r r r r r 1 ) 2 )( 1 ( 10 7 + + + + + + + +

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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 HCI Prelims Paper 1 Solution - 2007 HCI Preliminary...

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