Sembawang Secondary 4051/02 MS 2025

Prepared By: Ms Adeline LimPage 1 of 2 SEMBAWANG SECONDARY SCHOOLSEC 4NA PRELIMINARY EXAMINATION 2025 ADDITIONAL MATHEMATICS PAPER 2 MARKING SCHEMEQnSolution133222281252522552541025ababaabbabaabb222223517 3680 4(6)4(3)(8) 600xxxxxbacSince the discriminant is less than zero, the line and the curve does not meet3cos302sin60yycoscos30sinsin302sincos60cossin60yyyy3113cossin2sincos2222yyyy31cossinsin3cos22yyyy331cossin22yy331sin22cosyy33tanytan33y4(a)0222kkxx ]2,2,1[kckbaReal roots: 042acb021422kk08442kk4022kk(Shown)

Prepared By: Ms Adeline LimPage 2 of 2 0 1 2 3 4 5 6 7 –1 –2 –3 –4 –5 –6 –7 4(b) 022kk012kk[Leading to critical points: 2,1]kk–12–++xFrom graph:2or 1kk

Prepared By: Ms Adeline LimPage 3 of 2 5(a)Area of triangle XYZ43141634343312YZXY1434312431cmZYX43cm11634123211538221534 cm23ab5(b)By Pythagoras’ Theorem,2XZ=22YZXY=2243431=222 2424334324311=31683316381=319834881=3319481882XZ=68XZ=68417417遬XZ= 217 cm

Prepared By: Ms Adeline LimPage 4 of 2 6(a)cos601 2 222210023210010032QAxQAxxQARBxxPSxPSxPS° 6(b)2sin60323 231003 223 =10034PAxPAxPAxxAxxx°6(c)231003 43(1006)43Let (1006) 041006061005021633AxxdAxdxdAxdxxxx223302Therefore is a maximum value.dAdxA

Prepared By: Ms Adeline LimPage 5 of 2 7(a)22d323 ................. B1ddsub 3 and 0d3(3)2(3)30 .............M127630 630 5 ........................... A1yxhxxyxxhhhh7(b)323253Sub (3,0)(3)5(3)3(3)0 274590 9 ........................... B1yxxxkkkk7(c)2dLet 0d3103 0 ............................. M1(3)(31)013(given) or ................ A13yxxxxxxx7(d)2222222d 3103 dd610 .......................... B1ddWhen 3,80 . It is a maximum point. ......... B1d1dWhen ,80. It is a minimum point . ..........B13dAccept the first derivyxxxyxxyxxyxxative test using change in the gradients.8(a)2222222243100(2)2(1.5)1.5100 ....... M1 ( completing of square)(2)(1.5)16.25centre = ( 2,1.5) ......................... A165radius = or 16.25 units or 4.03 units.........2xyxyxyxy....... A1

Prepared By: Ms Adeline LimPage 6 of 2 8(b)22Sub (6,1)614(6)3(1)100yes, (6,1) lies on the circle. .................. B18(c)(2,1.5) (4,2)1.527gradient............... M12444Sub into 742(4) ....................... M17307430equation of tangent is ............... A177yxcccyx8(d)centre(2,1.5) (4,2)Let (,)422 and 1.5 ....................... M1( concept of midpoint)2244 230 5(0,5) ........................... AABababababB19(a)226sin10cossin() = sincoscossincos6 and sin10610 ...................M1136 or 234 10tan = ........................ M16 1.0303 radians6sittRtRtRtRRRRn10cos136sin(1.03) ............. A1ttt9(b)(i)Since maximum value of sin(1.03)1t, maximum value of x =13611.661, so it is not possible for the mass to reach a distance of 18 mm. ……………………………. B1√

Prepared By: Ms Adeline LimPage 7 of 2 9(b)(ii)Let sin(1.03)1 ................ M11.031.57070.541 s ( to 3sf) .............. A1ttt9(c)Let 136sin(1.03)8 .......................................... M1 sin(1.03)0.68599 1.030.75596, 2.38563 ...........................M1 (angles in 1st and 2nd tttquadrants) 0.274(rejected) , 1.36 s ( to 3sf) ...... A1 t10(a)2261175401401 or 46 or 31,6 4,3xxxxxxxxxyyAB 10(b)4214321232 611d6113264411143113239Area of trapezium1633213.5Shaded area: 13.594.5 unitsxxxxxx