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 satisfy the equations 2 2 0 and 2 2 0x y z x y z      . So    2 2 2 2 0 (0) 0x y z c x y z c        . Hence l lies in p3 for all c. biii [1] 0d  8(i) [5] 1 1 tan ( ) At 1, tan ( 1) 4 and 2( 1) 2 44 yx xy y              

5   0 1 1 2 001 21 1 02 012 1 1 2 1Area of region 0 tan ( ) d2 8 4 tan ( ) d64 1 1tan ( ) ln 164 2 1 ln 264 4 2 R x x xx x x x x x x                                     OR     0 4 2 0 4 2 2 1Area of region 1 1 0 tan( ) dy2 8 4 ln sec64 4 0 ln 264 4 1 ln 264 4 2 Ry y                                8(ii) [2] [2] After transformation, 2( ) 2 84 22 2 yx x          1After transformation, tan ( ) 8yx  22 0 4 Volume 1 tan d 2 4 8 4.35 y yy                    9 [4] 2d 2 ( 1)ed xy yxx    --- (1) Let 2e xzy 22dd e 2edd xxzy yxx   Sub. into (1), 2d 2 ( 1)ed xy yxx    22 de 2 e 1d xx y yxx    d 1d z xx   ( 1) dz x x   2 2 xz x c    2 2e 2 x xy x c    

6 9(i) [1] When 0, 1, 1.x y c    2 2The particular solution is e 1 . 2 x xyx     9(ii) [4] 2 2 2 2 2 2 32 2 2 2 32 d 2 ( 1)ed dd 2 e 2( 1)e (2 1)edd dd 2 2e 2(2 1)e 4 edd x x x x x x x y yxx yy xxxx yy xxxx                      23 23 23 23 d d dWhen 0, 1, 1, 1, 2d d d 12Using Maclaurin's expansion, 1 ... 2! 3! 11 1 ... 23 y y yxy x x x y x x x x x x                9iii [2] Replace x by -2x in the standard series expansion of ex and perform an expansion for 2 2e1 2 x x x   up to and including the term in x3. Compare the coefficients of this series with that of the expansion in (ii) to verify the correctness. 10a [7] 2 2 1 360 120 3V r h h r     From diagram, 2 2 2r h l 2 2 2 2 2 2 2 2 2 2 2 2 24 24 2 () 360 129600 S rl S r l r r h rr r r r             Differentiate w.r.t. r , we get 23 3 d 25920024 d SSr rr  For stationary values of S , set d 0d S r  2 3 6 32 259200 648004 rr r    6 3 64800 4.33 (3 s.f.)r     and 2 360 6.11 (3 s.f.)h r 10b i [3] d 3cos d 5sin yt xt Equation of l

7 5sin 3sin ( 5cos )3cos 5 25 tan sin 3sin33 5 16or tan sin33 ty t x t t y x t t t y x t t        10b ii [4] 16At , 0. cos 5 16At , 0. sin 3 A y x t B x y t       Mid-point of AB, M has coordinates 88cos , sin53 tt  22 22 22 88Let cos and sin53 Then 1 88 53 25 9 64 x t y t xy xy                 11a i [2] 11a ii [2]   63hd 2 a o xx                     x y x y 8 5 8 5 8 3 8 3

8 11b i [2]   2 23y a x   11b ii [3] In order for gk to exist, RDkg Domain of g is    ,,55,    2 k( ) 2 3x a x   , 7x Range of k is   ,325a So, 5325 a 08.0or25 2a 11b iii [3] When f ( ) g( )xx , 5 or 5.59 (3 s.f.)xx Replace with , 5 or 5.59 (3 s.f.)x x x x   12( a) [4]       22 4 7 7 4 7 2 7arg arg arg 3 i 6 21 * 3 i 3 i 2 3i 2 7 2 7 z zz               x y (2,3) (0, 4a+3)

9 (b) [5] iarg i 1 63 1arg i 63 32arg 3 2 6 3 w wi wi                        12b i [2] Least value of 1 i 3 z   1 32 3 43 23      12b ii [2] 2 1 1 2arg i or arg i33 33 zz                   ( 3,1)C  2 1 ,1 3   2 3 