Hua Yi 4E5N A Math Prelim P2 2021 MS

Name: Index Number: Class: HUA YI SECONDARY SCHOOL Preliminary Examination ADDITIONAL MATHEMATICS Paper 2 30 August 2021 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. MARK SCHEME 4E 4E 4049/2

2 1a) p < 0 B1 r < 0 B1 b) (i)           A144 M1044 016 042044 M105142 04 052 2 2 2 2 2        m mm m mmm mm acb mxmx (ii) M1424 and,2 m The curve and the line do not intersect. A1 2a) A24ln12 4ln12ln4 22 3 23 xxx x xxxxxdx d   b)   A1160 sidesboth on 4by divide M13 1lnln3 4 gintegratinfor M13 4ln4ln12 M14ln4ln12 5 1 335 1 2 2 5 1 335 1 2 5 1 25 1 35 1 2                xxxdxxx xxxxdxxx dxxxxdxxx c)   A1717.0or M101ln3,0and04 since gfactorisinfor M101ln34 M104ln12 3 1 2 2 22       x ex xxx xx xxx 3a)           A1179226 8 M16 0424 M1 28 2 6123 0324 183 813                     xx r r xxx xr xx rr rr

3 b)         0 M11792410247 1024 M127 8 7 M14424 44 4 7113               xx x xx r r 4a)                 A16 A15 11 M124612911 241f 93 M10632933 03f .629 be fLet 23 23 23          q p qp qp pq qp xqxpx (x) b)        A12,5 1,3 M102153 M102953 062965 2 23     xxx xxx xxx xxx c) A12 1,5,3 1 M11 056296 056296 32 3 23      yyy xy yyy y yyy 5a) 609.3 M15 1ln2 M1015 2      x x e -x

4 b)   A1units39.5 M1390562.23 units390562.2 M12 609.35 M115 units3 M142 3 2 1 of coordinate- findingfor M12 1,0 2 2 2 609.3 2 2 2            xe dxe Axxy x x 6a)           A1 3cos cos62 M1 3cos sin2cos6cos2 M1 3cos sinsin23coscos2 2 2 22 2       x x x xxx x xxxx dx dy b) A1units/sec5 2 M125 4064.0 25 4 M1 33cos 3cos62 2            dt dx dt dx dx dy   7a) A12 3900 90023 M190022 2 2 22 r rh rhr rhrr     

5 b) M16 5450 3 2 2 3 2 900 M13 2 2 3900 3 3 32 3 2 2 rr rr rr rr rrV             c)   A1cm12600 A14.13 416.13reject 416.13 180 M102 5450 M12 5450 3 2 2 2       V r r r r dr dV   d) 211 M152 2   r dr Vd  Volume is maximum at r = 13.4 Therefore, NO, volume cannot increase further. A1 8a         A12293 22 1292320 93 A13 M133 32 23 23 23 2 3 2              xxxy c Mc cxxxy k cxxxk dxxxky b)     A11 01 M1012 432 M112323 12 1 normal ofGradient 2 2 2 2       x x xx xx xx

6 9a (i) 8cos4sin8 M126cos2sin6sin2cos6 M1cos2sin6 M1sin2cos6         P RQ PQ (ii)   A186.26sin80 565.26 M18 4tan 9442.8 80 M148 22          P R (iii)   A14.63 906565.26 A1m9.16Max      P b) (i)   A1sin122sin8 sin12cossin16 cossin4sin12cossin2cossin18 M1cos2sin6sin2sin2cos22 1sin6cos62 1 panelsolar of Area 2 2 2         (ii)     A1m 16 M143439.63sin1243439.632sin8 panelsolar of Area 2 2   10a)            4,1 M11 1212 M1136127 M1546340 4, be circle ofcenter Let M14 2 311 222 2222        p p ppp pp p y y b)         A15041 units 50 M14516 22 22    yx r c)     each A1254,1and254,1 

7 11a)   A16 A1cossin6 M1cossin3cossin3cossin3cossin3 )sin(3sin3     k BA ABBAABBA BABA b)   A12 63 M12 2 2 36 angleeach for M145cos60sin6 )4560sin(34560sin3    