East Spring Secondary 4051/02 MS

2025 ESSS 4N Prelim A Math Paper 2 1 MARK SCHEMEQnSolutionMarks12222221(1)70(2)sub (1) into (2)(1)(1)703360201 or 22 or 12,1 or 1,2xyxyxyyyyyyyyyyyxxxyxyM1- substitution M1-simplifying eqnM1 for substiuting to find the other variableA1, A1 for each set of x and y values 2(a)422 13360180 2cab B1B1B1(b) Min point = – 5 Max point =4 B1 – sine graphB1- correct min and max valueB1- 2 cycles (to show π and 2π on the x axis minimally)3(i)222225404(1)(25)08200210yxpxpbacppppp M1-discriminantM1- correct sub of a,b, c

2025 ESSS 4N Prelim A Math Paper 2 2 M1- simplifying A1(ii)By comparing constant2513ppSince 3pis not within the range of 210p, his claim is not correctB1- identifying p = – 3 and using (i) inequality to explain 4(i)Since curve has stationary points at 2 and 2xx,So when d0dyx,2d(2)(2)d (4)ykxxxkx B1- identifying of stationary pointsB1- forming of equation(ii)222333 d(4)d 44 43sub 1 and 2(1)24(1)31243112(1)3sub 2 and 3(2)34(2)38383163(2)3(1)(2)11162333611916 ykxxkxkykxkdxkxykxcxykkckkckcxykkckkckckkkk   3sub 3 into (1)13kk M1-integrating with c includedM1- for eqn 1M1 for eqn 2M1 for solving k

2025 ESSS 4N Prelim A Math Paper 2 3 M1 for solving cA1 for eqn(iii)a = 29B15(i)1(51)(352)sin621525352sin1212sin13512135sin135135156125 =1643935 41XTUXTUXTUXTU M1- area of triangle formulaM1- expansion of surdsM1-simplifying of surdsM1-conjugate surdsM1A1(ii)52.0 and 128.0B16(i)3232(3)0(3)(3)(3)120931535(1)(2)10(2)(2)(2)12104230215(2)(1)(2)55155102sub 2 into (1)11fpqpqpqfpqpqpqppppq M1-for (3)0fM1M1- for (2)10fM1A1

2025 ESSS 4N Prelim A Math Paper 2 4 A1(ii)Let 2322()(3)()by comparing coefficent:1constant: 312 4:32 3(1)2 5()(3)(54)()(3)(4)(1)fxxaxbxcxaccxbabbfxxxxfxxxxM1 for comparing coefficient or long division to find quadratic factorM1 for 254xxA17(i)221LHS=sin11sinsin 1sin1sin 1sin cos xxxxxxxRHS M1 for 1sinxM1 for 21sinxA1(ii)Let x = 2x and 024x2cos21cos21 or cos2120,2,4 or 2,33,,22xxxxxx M1M1M1A2 – for all 3 answers[A1 for 2 correct answers]8(i)2222(20)(0)2455245497 or 7()pppppprejM1M1A1(ii)12OAm B1(iii)112sub m2,14,772(14)235ABmxyyxyx M1A1(iv)B= (0, 35)140735,22(7,21)NN M1

2025 ESSS 4N Prelim A Math Paper 2 5 Let C= (x, – 2x) M1M1-equating the gradientA1(v) 20140701 073514021(490245)2367.5 units M1A19(i)322sub 0340(34)0(4)(1)0(4,0) and (1,0) yxxxxxxxxxEFM1- sub y=0M1A1, A1(ii)2364dyxxdxB1(iii)23644(36)00() or 360 2xxxxxrejxxM1 - 0dydxM1A1(iv)232424324sub 24(2)4121Area34 3122341843238xyxxxdxxxx M1 for 232434 xxxdxM1 for area of triangleM1 for integratingA1

2025 ESSS 4N Prelim A Math Paper 2 6