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2007 PJC Prelim Paper 2_solution

# 2007 PJC Prelim Paper 2_solution - 2007 J2 Prelim Paper 2...

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2007 J2 Prelim Paper 2- solution 1. 6 4 4 3 T T T T 5 3 3 2 ar ar ar ar  2 3 2 1 1 ar r ar r Since , 0; a r 3 2 2 1 0 1 1 0 1 5 1 (n.a.), 2 r r r r r r r   For sum to infinity to exist, | | 1 r i.e. 5 1 2 r . 2 4 3 5 5 1 1 2 S 2. Let n P be the statement 1 1 5 2 3 n n U for , 0 n n . When n = 0, LHS = 0 2 U RHS = 1 1 5 2 3 Hence, the statement is true for n = 0 Assume statement is true for n = k where , 0 k k , i.e. 1 1 5 2 3 k k U 1 1 1 1 To prove is true, i.e. 1 5 2 3 k k k P U 1 1 2 k k LHS U U   1 1 1 2 1 5 2 3 1 1 2 5 2 3 1 1 5 2 3 k k k 1 k k P P is true. Since 0 P is true, 1 k k P P is true, therefore, by mathematical induction 1 1 5 2 3 n n U for all , 0 n n . b) 1 1 5 2 3 3 n n n n Lim U Lim   1 2 n n Lim  is undefined Therefore, sequence is not convergent. 3. 5 i 4 6 2 3 2i=4e w   1 5 i( ) 4 2 24 4 n w e for 0,1,2,3 n Sub 0,1,2,3 n into above, Obtain 5 7 19 31 i( ) i( ) i( ) i( ) 24 24 24 24 2 , 2 , 2 , 2 w e e e e Use the substitution 1 w z , 5 7 19 31 i( ) i( ) i( ) i( ) 24 24 24 24 1 2 , 2 , 2 , 2 e e e e z 5 7 19 31 i( ) i( ) i( ) i( ) 24 24 24 24 1 1 1 1 , , , 2 2 2 2 z e e e e 4. For each 50g packet of STA, Coffee Sugar Milk Amt in g 22 19 9 Therefore, 0.6 0.4 0.3 22 0.3 0.3 0.7 19 0.1 0.3 9 x y z x y z x y 0.6 0.4 0.3 22 0.3 0.3 0.7 19 0.1 0.3 0 9 6 4 3 220 3 3 7 190 1 3 0 90 x y z

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2007 PJC Prelim Paper 2_solution - 2007 J2 Prelim Paper 2...

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