2007 MJC Prelim Paper 2 Solutions

2007 MJC Prelim Paper 2 Solutions - 1 2007 H2 MATHS (9740)...

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MJC/2007 JC2 Preliminary Examination/9740/Paper 2 Solutions 1 2007 H2 M ATHS (9740) JC 2 PRELIMINARY EXAM (PAPER 2) S UGGESTED S OLUTIONS SECTION A: PURE MATHEMATICS Qn Suggested Solutions 1 Functions (i) R f = [ 2, )  (ii) intersection point   2, 2 or   1.41,1.41 . (iii) gf ( ,1] [0, ) RD   . Therefore, fg does not exist. (iv) We need g [0,1] R . ln 0 1 xx    Therefore, g [1,e] D , K = 1. (v) R fg =   2,0 x y 0 -2 -2 y = x   2, 2 y = f( x ) 1 f ( ) yx
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MJC/2007 JC2 Preliminary Examination/9740/Paper 2 Solutions 2 Qn Suggested Solutions 2 Geometric Progressions (i) Amount owed at the end of second month      15000 300 1.005 300 1.005 $14545.87   (ii) month Amt at the beginning Amt at the end 1 st 15000   15000 300 1.005 2 nd    15000 1.005 300 1.00       22 15000 1.005 300 1.005 300 1.005 3 rd      2 15000 1.005 300 1.0 300 1.005         33 2 15000 1.005 300 1.005 300 1.005 300 1.005 n th         1 15000 1.005 300 1.005 300 1.005 ... 300 1.005 nn n  Amt of the end of n th month               1 2 15000 1.005 300 1.005 300 1.005 ... 300 1.005 15000 1.005 300 1.005 1.005 ... 1.005 1.005 1.005 1 15000 1.005 300 1.005 1 nnn n n n n         15000 1.005 60300 1.005 1 60300 45300 1.005 (shown) n n n (iii)   60300 45300 1.005 0 201 1.005 51 201 ln 51 ln1.005 57.3 n n n n    
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MJC/2007 JC2 Preliminary Examination/9740/Paper 2 Solutions 3 Assuming no interest, no. of months required = 15000 50 300 No. of extra months required 8 months Qn Suggested Solutions 3 Parametric Equations   2 d 1 d 1 2 , 4 d 2 d 2 xy d 4 4 2 d2 y x x x (i) d When 2, 4, 1, ln32 d y x The equation of tangent is ln32 2 yx     When 0, 2 ln32 Coordinates of 2 ln32,0 P    The equation of normal is   ln32 2     When 2 ln32 Coordinates of Q    Length of 2ln32 PQ (ii)       2 2 ln 2 2 d 1 ln 2 1 xy xy d d d 2 0.1 0.2 dt d d x xt        dd d d d d 12 ln 1 0.2 29 0.0496 (to 3 s.f) xy xy tt     or 1 2 1 ln 10 9 5 Alternative Solution:
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MJC/2007 JC2 Preliminary Examination/9740/Paper 2 Solutions 4   d d d d d d yx xy x y t t t  When 1 1 2 , , ln 3 6 9 xy      d d d 2 0.1 d d d y y x t x t x  When 1 , 6 x   d2 0.1 1.2 1 d 6 y t       d d d 1 2 1.2 ln 0.1 0.0496 d d d 6 9 xy x y t t t (to 3 s.f.) Qn Suggested Solutions 4 Vectors (i) Since P lies on l , 43 2 5 OP   for some R .
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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 MJC Prelim Paper 2 Solutions - 1 2007 H2 MATHS (9740)...

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