2007 MJC Prelim Paper 1 Solutions

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

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MJC/2007 JC2 Prelim Exam Paper 1 Suggested Marking Scheme/H2 Maths (9740)/Maths Dept Page 1 of 13 2007 H2 M ATHS (9740) JC 2 P RELIM E XAM P APER 1 S UGGESTED S OLUTIONS Qn Suggested Solutions 1 Arithmetic Progression         10 22 6 2 5 2 9 2 21 8 3 0 1 37 5 37 2 ST a d a d ad T               Solving (1) and (2), 3, 8   For series H , First term = 13, common difference = 24     50 50 2 13 49 24 2 30050 S    Qn Suggested Solutions 2 Binomial Series 22 2 2 1 f( ) ( 1) ( 2) 1 ( 1) 2 ( 1)( 2) ( 2) ( ( 2) A B C x x x x x x A x x B x C x xx  By ‘cover-up’ rule, 1 9 C . Comparing coefficients in 2 x , 1 0 9 A C A   . Comparing coefficients in x , 1 20 3 A B C B . 1 1 1 1 ( 2) 9( 3( 9( 2) x x x x x x   1 2 1 1 1 1 (1 ) ) ) 9 3 18 2 x 23 1 ...) 9 1 (1 2 3 4 ...) 3 1 ...) ...(*) 18 2 4 8 x x x x x x x x x   1 3 9 23 ... 2 4 8 16 x x x From (*), Coefficient of the term in 2 1 1 1 1 (2 9 3 18 4 n n xn    2 1 4 2 2 3 3 6 n n  .
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MJC/2007 JC2 Prelim Exam Paper 1 Suggested Marking Scheme/H2 Maths (9740)/Maths Dept Page 2 of 13 Qn Suggested Solutions 3 Differential Equations When radius is r , Volume of bubble blown, 3 4 3 Vr Consider d d d d d d V V r t r t  , At the instant where the radius is r cm,   32 d 4 d r rr t   23 d 4 d r t  2 3 43 dd 3 r rt r  3 4 ln ln 3 r t C     3 3 4 3 3 4 e e t t rA When t = 0, V = 0, 0    3 3 4 1e t r    --- (1) When V = 6  , 3 8 r , sub into (1), 33 44 8 1 e e 87 tt 0.178 min t  Alternative Solution   3 d d V r t Since 34 V V r r     3 d 3 1 d 4 4 VV rV t (variable separable) 1 d 1 4 3 d 4 V Vt 11 ln 4 3 V t C 3 ' 4 3 4 3 3 4 ln 4 3 3 ln 4 3 ' 4 4 3 e , where e 3 e , where e t C t t V t C V t C V A A VA BB rB     
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MJC/2007 JC2 Prelim Exam Paper 1 Suggested Marking Scheme/H2 Maths (9740)/Maths Dept Page 3 of 13 Qn Suggested Solutions 4 Integration by parts/Integration by substitution (a) sec d sec tan d x x When 2, 1 sec o 2 cs 2 4 x   When 2, 1 2 sec cos 2 3 x 2 2 2 2 d 1 x x 3 2 4 2 sec 1  3 2 4 2 tan 3 4 2 sec d   3 4 2 ln sec tan        2 ln 2 3 ln 2 1 32 2 ln 21    (b) 2 cos d x x x   1 1 cos2 d 2 x x x   1 cos2 d 2 x x x x 2 cos d x x x cos2 d x x x 11 sin 2 sin 2 d 22 x x x x  sin 2 24 x x x A
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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 1 Solutions - 2007 H2 MATHS (9740) JC...

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