RI P1 MS

2012 RI H2 Mathematics Preliminary Examination Paper 1 Qn Solution 1 [3] 21x y z   ------ (1) 30xz ------- (2) 0.05 0.8 3(0.9 )y z x 2.7 0.05 0.8 0x y z    ----- (3) Using GC, 2.1, 12.6 and 6.3x y z   2 [2] [3] 22 22 2 2 2 2 3 2 4 6 1 (1 2 )(1 2 ) ( 2)( 3) ( 2)( 3)( 4) 1 ( 2)(2 ) (2 ) (2 ) ... 2! 3! 1 4 12 32 ... xx x x x x x x                 Coefficient of     2 2 3 ... 1 ( 1) ( 1)! is 2 2 ( 1) ( 1)2!! r r r r r r r rxr rr         Expansion is valid for 221x  2 1 2 11 22 x x      Range of validity is 11, 22  . 3 [6]          2 2 2 22 2 22 2 2 22 2 i 5 12i 5 (1) 62 12 (2) Sub in (1) 6 5 5 36 0 9 4 0 2 or 2 2, 3 or 2, 3 3 2i or 3 2i w a b ab ab ab a b b b bb bb b b a b a w                                    2 1 3i 0 1 1 4 1 3i 1 5 12i 22 1 3 2i 2 2 i or 1 i zz z z             

2 (complete the square is acceptable) 4(i) [6] (ii) 1 812 and 2012 812 ( 1)2 601 nu u n n       1 812 and 2002 812 ( 1)14 86 mv v m m        601 86Required sum = 812 2012 (812 2002) 72761022     5(a) [2] 2 1 14 2 2 1 1ln(2e )d ln 2 d 14 12ln 2 d2 1 (2 ) 1ln 2 sin (2 )2 x xx x x x x x c              5(b) [5]   4 2 2 2 22 2 2 1 1 1 1 d d d 1 12 1 2u12 2 2 2 1 2 (1 ) 2ln(1 2 ) 1 2ln 9 ln 4 13 9 1 3 1ln 2ln4 3 2 3 x u u xx uu u u u u uu u                        d 1 1 d2 2 d 2d When 1, 1 1 When 4, 4 2 ux u xu x x uu xu xu            22 2Let (1 ) 1 (1 ) 2 (1 ) Subst 1, 2 Subst 0, 2 u A B u u u u A u B uB u A B            

3 4 2 2 2 2 2 2 2 1 1 1 1 1 d 2 d d d Alternatively, 1 12 1 12 2 (1 ) 112 1 (1 ) 12 ln(1 ) 1 112 ln 3 2 ln 232 312ln 23 x uu u u xx uu u u uu u u                              6 [3] [3] (i) f (1 )yx (ii) O x y ( 1, 2) ) (3, 4) y = 1 1 2 3 1 2(2, ) ) 1y x y y = 1 f ( )x O 1 4( 2, ) 1 2x 4x

4 7(a) [3]       11, 334 31 3344 33 22 / / and is a common point , , are collinear PR OR OP OQRQ a b a b a OA OR OA O O A R         7bi [2] From GC 3 2 xz yz zz    Cartesian equation is 2 3 2 x y z 7bii [2] Any point  ,,x y z on l has x  , 3 2 y  and z  for some real  . So    2 2 2 2 332 2 2 222 0 for all x y z c x y z c c                                   Hence l lies in p3 for all c. OR All points  ,,x y z on l satisf