2007 CJC Prelims Paper 1 Solution

# 2007 CJC Prelims Paper 1 Solution - H2 maths Marking Scheme...

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H2 maths Marking Scheme for Prelim Paper 1 1. c bx ax dx dy d cx bx ax y 2 3 2 2 3 [M1] 0 ) 1 )( 2 ( ) 1 )( 3 ( 1 ) 1 ( ) 1 ( ) 1 ( 7 ) 1 ( ) 1 ( ) 1 ( 4 ) 2 ( ) 4 ( ) 8 ( c b a d c b a d c b a d c b a [A1]  2 5 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 7 4 0 1 2 3 1 1 1 1 1 1 1 1 1 2 4 8 rref Thus a = 1, b = 1, c = -5, d = 2 Equation of curve: y = x 3 + x 2 - 5x + 2 [A1] 2 (i) r=1 1 1 r(r + 1) = 1 2 r=1 2 1 r(r + 1) = 1 2 + 1 6 = 2 3 r=1 3 1 r(r + 1) = 1 2 + 1 6 + 1 12 = 3 4 r=1 4 1 r(r + 1) = 1 2 + 1 6 + 1 12 + 1 20 = 4 5 [B2] (ii) We conjecture that r=1 n 1 r(r + 1) = n n + 1 [B1] (iii) Let P n be the statement: r=1 n 1 r(r + 1) = n n + 1 When n = 1: LHS = u 1 = 1 2 RHS = 1 2 = LHS P 1 is true. [B1] Assume that P k is true for a k Z + i.e. r=1 k 1 r(r + 1) = k k + 1 Prove that P k+1 is also true i.e. r=1 k+1 1 r(r + 1) = k + 1 k + 2 LHS = r=1 k 1 r(r + 1) + 1 (k + 1)(k + 2) [B1] = k k + 1 + 1 (k + 1)(k + 2) = k(k + 2) + 1 (k + 1)(k + 2) = k 2 + 2k + 1 (k + 1)(k + 2) = (k + 1) 2 (k + 1)(k + 2) [B1] = k + 1 k + 2 = RHS P k true P k+1 true also. Since P 1 is true, and P k true P k+1 true, by Math Induction, P n is true for all n Z + [If no proper statements, deduct 1 mark from whole question] 3 (i) 2 2 2 2 2 2 2 2 1 ) 1 ( 1 2 ) 1 ( ) 1 ( ) 1 ( 1 1 n n n n n n n n n U U n n [B1] (ii) 22 1 21 ( 1) N n n nn = 1 3 2 2 1 1 1 ... N N N n N n n U U U U U U U U U U [M1] = 1 1 N U U = 2 ) 1 ( 1 1 N [A2] [M1]

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(iii) 2 2 2 2 2 2 9 15 21 ... 1 2 2 3 3 4  2 2 ) 1 ( 1 2 3 N N N 2 2 2 2 2 2 2 2 ) 1 ( 1 2 ...
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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 CJC Prelims Paper 1 Solution - H2 maths Marking Scheme...

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