East Spring Secondary 4051/01 MS

12025 ESSS 4NA Preliminary Examination Additional Mathematics Paper 1 (4051/01) MARKING SCHEME QnSuggested SolutionsMarksTotal MarksRemarks1a(i)Since5sin13 , then 513 is in first quadrant.By Pythagoras’ theorem, 22213514414412rropptanadj5tan12 B111a(ii)sin22sincos51221313120169 M1A121bThe principal value of 12cos2 is an angle in the 2nd quadrant in the interval 10cosx.Since 2cos42, the principal value of 12cos2 is 344. B112333321411214(8)14(2)14(2)(24)xxxxxxM1A23A1 Either (2)xor 2(24)xx is correct.

23aComparing equation with general form: Centre of the circle Radius of circle M1A1A133b715yxWhen 2,x72151yThe centre of circle lies on line since it is shown that y = 1 when substitute 2x in the line equation 715yx. B11

34 -32++¯Putting both cases together, -3253Answer: M1M1 A13 1 mark for either correct solution

45 Therefore M1M1M1A14 Long division seenFactorisation seenEither correct values of A or B 6a22222222261023533235223925243292243292(shown)22xxxxxxxxx M1A12

56bMinimum value of = Minimum value of x when it occurs is Coordinates of minimum point B116cThe curve does not lie completely above or below the x-axis because the coefficient of x² is positive. So, the curve is U-shaped and the minimum value of y < 0 (minimum point below x-axis). So, the curve cuts thex-axis twice. B117222122222(multiplyby2)22201,22,2(22)224(1)(2)2(1)2216222422222xxxxxxxabcxxxx M1M1M1A14 -quadratic eqn-Use of formula-simplifying 16to 48asincossinComparing8sin7cossin8,7abRRab222287113tan7841.18592517Rabbao8sin7cos113sin(41.2)(1d.p.)M1M1A13

68bMaximum value of Minimum value of M1M1A13 1 mark for at least 2 correct answers

79Find V in terms of h: Subst into VTo find rate of change in height, Height, (Subst. h = 9 cm) M1M1M1M1M1A16

810aM1A1210b4242426dx 45112=dx2451254521(2(4)5)4(4)5(2(2)5)4(2)521131193213.763832513.76(2d.p.)xxxxxx M1M1M1A1411 2221 1oooooooosec4tan6 tan14tan6tan4tan50tan1tan50tan1ortan5tan(5)tan(1)78.69045,180180,36078.6900675,258.6900675135,31578.7,135,258.7,3151d.p.xxxxxxxxxxxxxxx∘ ∘∘ ∘∘M1M1M1M1A15

912When (tangent is parallel to the x-axis).Subst x = 0 into y, y = 0Subst , Coordinates of the points are M1M1M1M1A1, A16

1013 M1M1M1M1M1MA1714(a)tan= 3yx(equilateral triangle, use trigo ratio)3yx Note:tan= 33 = 3yxOr Since triangle ASR is equilateral triangle, AS = SR = PQ = 62x cm. BS = 6(62)2xx.By Pythagoras’ Theorem, 22222222(2)433(,0)xyxxxyxyyxxyM1