2007 TPJC Prelim P1 solution

2007 TPJC Prelim P1 solution - Solutions for H2 Mathematics...

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1 Solutions for H2 Mathematics 9740/01 Preliminary Examination 2007 Q1. (a) Case 1: Contains and two pairs 30 ! 2 ! 2 ! 5 Case 2: Contains and one triplets 20 ! 3 ! 5 Case 3: Does not contain 10 ! 3 ! 2 ! 5 Total no. of different codes 60 10 20 30 (b) No. of codes 4 1 ! 2 ! 3 Q2. r qx px y 2 r x q x p y ) 1 ( ) 1 ( 2 = r x q x x p ) 1 ( ) 1 2 ( 2 = ) ( ) 2 ( 2 r q p x q p )] ( ) 2 ( [ 2 2 r q p x q p y = ) 2 2 2 ( ) 2 4 ( 2 2 r q p x q p By comparing coefficients of x, 4 p + 2 q = 3 r – 2 i.e. 4 p + 2 q – 3 r = – 2 --- (1) Comparing x 2 - terms and constant: 2 p = q r 2 p + 2 q + 2 r = 1 i.e. 2 p q + r = 0 --- (2) 2 p + 2 q + 2 r = 1 --- (3) Using G.C.,  1 0 2 2 2 2 1 1 2 3 2 4 r q p 1 0 2 2 2 2 1 1 2 3 2 4 1 r q p 19 10 , 19 3 , 38 7 r q p

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2 Q3. i) u 1 = 99, u 2 = 96, u 3 = 93, u 4 = 90 ii) u 1 = 102 – 3(1) u 2 = 102 – 3(2) u 3 = 102 – 3(3) u 4 = 102 – 3(4) u n = 102 – 3n where f(n) = 3n iii) Let P n be the statement u n = 102 – 3n When n = 1, LHS = 99, RHS = 99 P 0 is true. Assume P k is true i.e. u k = 102 – 3k When n = k + 1, u k+1 = u k – 3 = 102 – 3k – 3 = 102 – 3(k + 1) P k is true P k+1 is true Hence by induction, P n is true. Q4. i)
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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 TPJC Prelim P1 solution - Solutions for H2 Mathematics...

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