2014 H2 Maths Prelims 2 P1 Soln

Prelims 2 Paper 1 Suggested Solutions  IJC/2014/JC2 9740/01/S2014 Qn Solution 1 Let x, y, z be the number of apples, oranges and pineapples respectively. 50xyz 0.8 0.6 1.2 40xy z 1.6 0.3 1.2 53xyz From GC, Adam bought 23 apples, 18 oranges and 9 pineapples. 2(i) Since x, y, z are the first three terms of a geometric progression, 2 2 y z x y y xz yx z    Since z, x, y are three consecutive terms of an arithmetic progression, 2 x zy x x yz     Solving the 2 above equations, 2 22 2 2 2 2 2 2 Dividing throughout by : 2 2 0 (shown) y yzz yy z z y zz yy zz yy             (ii) Geometric progression has common ratio zr y .

3  IJC/2014/JC2 9740/01/S2014   Solving 2 20zz yy      gives 1 or 2r  . Since r < 1, so sum to infinity of geometric progression does not exist. 3(i)   22 2 224 211 3xx xx     2 13x Since  2 10x ,  2 13 0x  for all real values of x. Since 2 24 0xx  for all real values of x,     2 24 3 02 xx x x       3 02 x x   2 o r 3xx  (ii) Using (ii),     2 24 3 02 xx x x    2x  or 3,x  Since 0x  , reject 2x  33xx   or 3x 4(i) 4 3iz  3i 2 arg 3 i 6   -2 3 ++ _

Prelims 2 Paper 1 Suggested Solutions  IJC/2014/JC2 9740/01/S2014 i4 62ez   2i4 62e k z   11 2 11 i4642e k z   , 0, 1,2k  11 3 1 1 1 1 1 2 3iiii42 4 4 2 4 4 2 4 4 2 42e ,2e ,2e ,2ez    (ii) The cartesian equation is 22 2xy 5(i) Let     2 22 5 221 1 x AB xx xx     . Hence  2251 2x AxB x    . Let 1x ,  251 21 B  2B  Let 2x ,  2252 12 A  1A      2 22 51 2 221 1 x xx xx     Im 1 i42 42e  Re O 11 1 i42 42e  12 3 i42 42e      11 3 i42 42e 

5  IJC/2014/JC2 9740/01/S2014   (ii)     2 2 2 5 21 12 2 1 x xx x x       12 22 1x x     =      1 21 2321 2 1 222 ! x xx            2 21 12 1 2 322 4 xx xx        251 5 4 9 24 8 xx    (iii) For expansion of  2 1 x   to be valid, 1x  1x  For expansion of 1 1 2 x   to be valid, 12 x  2x  Hence for the expansion of   2 2 5 21 x x x   to be valid, 1x  , 11 x  . 6 (a) (i) y x

Prelims 2 Paper 1 Suggested Solutions  IJC/2014/JC2 9740/01/S2014 (ii) (b) Method 1 After 1st transformation:  1gg 2x x  After 2nd transformation: 11gg 22x x       After final transformation: 11g1 g 22x x      111g 1 2 x x    11 1gg 22 2 2 xx xx               1g 2x x y x yk (, 0 )a fyx

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