2022 SGSS AMath Prelims P2 Ans

2022 A Math 4E/5N PRELIM Paper 2 Marking Scheme Solutions 1a 1b m > 0 , m ≠ 1 2a 2b 3 010)(2 2  zz ee 0)2)(52(  zz ee 92.0 2 5ln 2 5      z z ez AND 2z e (rejected as 0e z  ) 4a f(−2) = −81 4b 0 1)1(7)1(11)1(3f(1) 23   factor.a is )1(  x Using Long division 0)183)(1( 2  xxx

Solutions 1120.0or 79.2 6 768 )3(2 )1)(3(4)8()8( 2    xorx x 5a 3cos DH  cos3DH 5sin DK  sin5DK  sin5cos3  DKDHCF 5b 5c 34 cos( 59.0 )CF     Max 34 cmCF When 59.036 0 59.0 (1 d.p.)        6a When x = 0, y = 2 A (0, 2) 6b When x = 4, y = 4, 4 = 4x +2 m = 0.5 6c 4 4 0 0 4 0 43 2 2 0 2 1 2 22 1 2 22 1 4 24 3 113 Area x dx x dx x x d x x x x units                  7a x 10 20 30 40 lg y -1.221 -0.420 0.374 1.167 7b  Abxy  lglg Accept 1.95 2.05A  34 53 Where 22  R        036.59 3 5tan and 1 1)036.59cos(  A xby 10 

Solutions 2 0.42lg 20 Accept 1.16 1.24 b gradient b     7c Draw the graph: lg 4 xy 410 xA xb   lg4 xA x b  lg4 x x b A   From graph, accept 6.0 6.5x  8a m/s 6.4 4.05 , 0When 4.05 0      evt edt dsv t 8b When v = 0, 4.050  te 0.08 2.53 te t    8c tedt dva  5 2/4.0 ,53.2 When smat  8d t = 0, s = 0 m t = 2.53, s = 3.589 m t = 5, s = 2.966 m Total dist = 3.589 + (3.589 – 2.966) ≈ 4.21 m 8e 9a 2 2 2 5 4 10 5 6 0 5 6 0 ( 6)( 1) 0 1 6 x x x x x x x x x or x                 

Solutions 9b 5 0)4(4)6( 04 046 45 2 2 2 2      c c acbSince cxx cxxx 10a 59049 + 98415x + 73811.25x2 +…. 10b x = 0.08 59049 + 98415(0.08) + 73811.25(0.08)2 ≈ 67400 10c 11a 11b 11c (3, 0), (0, 4)A B  3, 22C       0 0 3, ,22 2 2 P Px y               , 3, 4P Px y  4 3,3 4 OP Tm m  34 ( 3)4y x   4 16 3 9y x   

Solutions 11d Since tangent // to y-axis from centre of circle , 11e 4 3 25y x   x c  3 5 12 2x   3 5 42 2x   13( , ) ( 1, 7), 4, 4x y      