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2007 MJC Prelim Paper 2 Solutions

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

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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) g f ( ,1] [0, ) R D   . Therefore, fg does not exist. (iv) We need g [0,1] R . ln 0 1 x x 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 ( ) y x

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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 2 2 15000 1.005 300 1.005 300 1.005 3 rd 2 15000 1.005 300 1.00 300 1.005 3 3 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 n n 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 n n n 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
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 x y d 4 4 2 d 2 y x x x (i) d When 2, 4, 1, ln32 d y x y x The equation of tangent is ln32 2 y x When 0, 2 ln32 Coordinates of 2 ln32,0 y x P The equation of normal is ln32 2 y x   When 0, 2 ln32 Coordinates of 2 ln32,0 y x Q Length of 2ln32 PQ (ii) 2 2 ln 2 2 d 1 ln 2 1 d 2 xy xy d d d 2 0.1 0.2 dt d d x x t d d d d d d 1 2 ln 1 0.2 2 9 0.0496 (to 3 s.f) xy xy t t 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 y x xy x y t t t When 1 1 2 , , ln 3 6 9 x y d d d 2 0.1 d d d y y x t x t x When 1 , 6 x d 2 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 y x xy x y t t t (to 3 s.f.) Qn Suggested Solutions 4 Vectors (i) Since P lies on l , 4 3 2 5
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