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

7  IJC/2014/JC2 9740/01/S2014   Method 2 Let h x be the expression after the final transformation. (a) Before final transformation: 1111 x x   (b) Before 2nd transformation: 11 x x   Before 1st transformation (original expression)  11 22x x  1g 2x x 7(i) 2 15 16u  3 63 64u  4 255 256u  (ii) Considering 2 11 16u 3 11 64u 4 11 256u 2 11 2 n nu    Hence, the conjecture is 2 11 2 n nu   . Let Pn be the statement 2 11 2 n nu   for n  . When n = 1, LHS = u1 = 3 4 RHS 2(1) 131 24   Since LHS = RHS, P1 is true. Assume that kP is true for some k  Z .

Prelims 2 Paper 1 Suggested Solutions  IJC/2014/JC2 9740/01/S2014 i.e. assume 2 11, f o r s o m e 2 k kuk   Z . To prove that 1kP is true, i.e. prove 2( 1) 1 11 2 k ku     . 2 1 22 2 2 22 31 42 13 11 24 2 1311 24 111 24 11 2 k kk kk k k k uu                  Hence, kP is true 1kP is true. Since 1P is true, and kP is true 1kP is true, by Mathematical Induction, nP is true for all n Z .

9  IJC/2014/JC2 9740/01/S2014   (iii)  2 1 22 32 43 54 1 1 12 2( 1) 2(2) 42 2 2 31 42 1111 22 11 22 11 1 44 2 rNN nn rr NN NN N N N N uu uu uu uu uu uu uu                                        (iv) 2 11 1 3 44 2 5 0 N     Using GC Table, N 2 11 1 44 2 N     3 0.05859 4 0.06152 5 0.06226 Therefore, the smallest integer value of N is 4.

Prelims 2 Paper 1 Suggested Solutions  IJC/2014/JC2 9740/01/S2014 8(a)       19 4 250 19 1 42 250 9 11 1 42 0 5 11 122 1 2 1 2 2 2 11 dd 9 14 129d d 2 12 129 s i n 2 2 1122 9 5 s i n 22 142 26 2( ) 4 12 2 24 48 24 p p p xx x x x xx x xx p p p p                                        (b) 1 2 0 d1 x xx    4 2 2 4 2 2 4 2 2 4 2 4 2 cos 2c os sin d 1c o s cos 2cos sin d sin 2cos d cos 2 1 d 1 sin 22 1 sin sin22 4 2 1 124 2 1 24                                                  

11  IJC/2014/JC2 9740/01/S2014   9(i) (ii) (iii) (iv) 2 5OP a  Since OPQB is a parallelogram, 2 5 2 5 OB PQ OQ OP OQ OQ     b ba ab            Area of triangle OAQ 1 2 12 25 12 25 1 2 OA OQ       aa b aa a b ab    Therefore, k is 1 2 . OPB : OAB 2 : 5 Since ab is a unit vector, sin 1s i n 12 s i n 6 0      ab a b ab b  Therefore, 1 3 b