# skema add math p2 trial spm zon a 2012

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3472/2 Matematik Tambahan Kertas 2 2 jam Sept 2012 SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2012 MATEMATIKTAMBAHAN Kertas 2 Dua jam tiga puluh minit JANGANBUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU MARKING SCHEME Skema Pemarkahan inimengandungi 13 halamanbercetak 2 ADDITIONAL MATHEMATICS MARKING SCHEME TRIAL SPM exam Zon A Kuching 2012 PAPER 2QUESTION NO. SOLUTIONMARKS 1x = 1+ 2 yP1 22 Eliminatex ory (1+2y)

5 y+ (1+ 2 y) y

6 = 0K1 Solve the quadratic 2K1equation by quadratic (5) (5) 4(1)(5) y =formula @ completing the 2(1)square must be shown y = 0.854, 5.854N1 Note : orOW 1if the working of solving x = 2.708, 10.708N1quadratic equation is not shown. 5 5 2 (a) (b) 2 2K1 f(x) = bax 2apxap a =2N1 2 K1 2ap = 6or bap= 9 N13 b = 27/2

N1

p =2 3 27 | , |y 2 2 . - N1(3/2, 27/2)max and (0, 9) 9 N1The shape of the function. x 5 2 7 3QUESTION NO. SOLUTIONMARKS 3 (a) (b) a = 5, d = 8 K1 77 = 5 + (n 1)(8) K1 n = 10N1 3 a = 5 , d = 8K1 10 S10 =|2(5 ) + (10 1)(8 )| K1 2 = 410N1 4 the total cost= RM90.20N1 4 7 4 (a) (b)(i) sin 2x cos 2xK1 1 1 cos 2x 2sin x cos x =2 K1 1 (1 2sinx) = cot x N1 3 y y = 2 cos 3x 2 x 0

2 2 5 4QUESTION NO. SOLUTIONMARKS (ii) Shape of cosine curveP1 AmplitudeP1 PeriodP1 2 cos3x = 2k K1 k = 1, 1 N1 8 5 (a) fx = 30.5(3) + 50.5(7) + 70.5(6) + 90.5(2) +110.5(2) = 1270 K1 1270K1 x = 20 = 63.5N1 3 (b) 22 2 2 2 2(30.5) (3) + (50.5) (7) + (70.5) (6) + (90.5) (2) + (110.5) (2) ( =3 + 7 + 6 + 2 + 2K1 2 (63.50) 91265K1 =4032.25 20 = 531N1 3 6 6 (a) OP = OK + KPK1 = a + h(a + b) K1 OP = (1 h)a + hbN1 3 5QUESTION NO. SOLUTIONMARKS (b) 3

hK1 = 4h + 81 h (4h 1)(h + 3) = 0 K1 1 h = N1 4 PL : KL= 3 : 4 N1 4 7 =1.500.30 = 1.5 n = 3N1 0.20.30.40.5 Correct both axes (Uniform scale)K1 All points are plotted correctlyN1 Line of best fitN1 log10 yQ7 (a) Each set of values correct(log10 y must be at least 2 decimal places)N1, N1 2.2 log 10 y = n 2 log 10 x+ log 10 (p + 1)K1 where Y = log 10 y, X = log 10 x, 2.0 m = nandc = log 10 (p + 1) (c)(i) X = log 10 6.9 = 0.8388 = 0.84 Y = 1.56 = log 10 y y = 36.31N1 1.8 1.6 1.4 1.2 n (ii)= gradient 2 n 2 0.80 (iii)log 10 (p + 1) = Y-intercept log 10 (p + 1) = 0.30K1 p = 0.9953 N1 .(0.80, 1.50) 1.0 0.8 0.6 0.4 (0, 0.30) 0.2 log10 x 0 0.1 0.60.7 0.8 0.9 1.0 log10 x 0.300.600.700.780.901.00 N1 log10 y 0.751.191.361.461.65 61.80N1 7QUESTION NO. SOLUTIONMARKS 8 (a) (b) (c) 3or3 13or mnormal=K1 11 +c K1 + N1 3 B(4, 0)K1 4

1 1K1 = 2 N1 =1unit 3 volume of revolutionK1K1 = 1 41 =K1 = 1unit 3 N1 4 10 QUESTION NO. SOLUTIONMARKS 9 (a) r=5K1 PR = 8.660 K1 23.66 cm N1 3 (b)

|K1 s = 5

| 3 . =10.47N1 2 8(c)Area of shaded region of semi circle 1 12 2 =

(10) ( )

(5) ( )

K1 2 2 2 =117.825 / 117.8096 cm Area of shaded region PQR 12 | 1 = (10)

|

(5)(10) sin 60

K1 2 3 . 2 K1 2 =30.716 / 30.7093 cm Total area of shaded region =117.825+30.716K1 2 =148.541 / 148.5192 cm N1 5 10 QUESTION NO. SOLUTIONMARKS 10 (a)(i) (ii) 2(4) + 3(0) 2(0) + 3(8)K1 or 3 + 23 + 2 8 24 | C = , |N1 55 . 824 + x + y 5 5 K1 = 2 oror(2, 0) 22 1224 | N1 A,| 5 5 . 4 (b)1 2412 24812248 24 = 0() + ( )(0) + (4)( ) + ( )(0)0( )()(4)(0)( )( )(0) 25 5555 55 5 K1 N1 = 19.2 2 (c) mAB = 3K1 y = 3x12N1 (d)2 2 ( x

(4))

+ ( y

0)K1 2 2 x+ y+ 8x = 0N1 10 9QUESTION NO. SOLUTIONMARKS 11(a) (i) (ii) (b) (i) (ii) (iii) 8 2 3 3 5 K1 C3( ) ( ) 55 = 0.2787N1 8 2 6 3 2 8 2 7 32 8K1K1 1 C6( ) ( ) C7( ) ( ) ( ) 55555 0.9052N1 5 0.75P1 h

163 = 0.6K1 16 h = 153.4N1 P(1 o z o 0.75) or1P(Z > 1)P(Z > 0.75)orR(1)R(0.75) K1 0.6147 N1 5 10 10QUESTION NO. SOLUTIONMARKS 12 (a) 6t + 16 = 0 K1 2N1 t =2 3 2 K1 s = 3t+ 16t 2 22 K1 s = 3( )+ 16( ) 3 3 1 N1 = 21 m 3 5 (b)t(3t + 16) = 0K1 1 N1 t =5 3 2 (c) | s83

s0 | + | s4

s83 | K1 2 2 K1 = | 21 30 | + |1621 3 | = 26 23 N1 3 11QUESTION NO. SOLUTIONMARKS 13 (a) (b) x = 135 N1 y = 7.20N1 2 K1 414.75n + 829.5 = 420n + 819 K1 n =2N1 3 (c) RM 8.80 K1 . RM 6.37 N1 2 (d) I 11 = 155K1 08 155 K1I11 =100 10

138.25 112.12N1 3 10 12QUESTION NO. SOLUTIONMARKS 14 (a) (b) (c) 9.79 = K1 sinACBsin 60 ACB = 68.97 / 6858' N1 2 ECD =18068.97 =111.03 /11122K1 2 2 2 ED= 4.5+ 9 2(4.5)(9)cos 111.03 K1 ED = 11.42 cm N1 3 BAC = 1806068.97 = 51.03 / 5122 P1 1 Area of

triangle ABC = 9.79sin 51.03K1 2 = 33.94 1K1 Area oftriangle CDE =94.5sin 111.03 2 = 18.90 Area ofquadrilateral ABCD =33.94 + 18.90 K1 52.84N1 5 10 2468101214 13 Answer for question 15 (a) I. x+yo 18 N1 II.4x+3 yc24 N1 y III. yo2x N1 (b)Refer to the graph, 20 18 1 or 2graph(s) correct 3 graphs correct Correct area N1 (c) (i)x = 8 N1 N1 K1 16 (ii) (6, 12) P1 Maximum Profit = RM [1.50(6) + 2(12)] K1 14 12 (6, 12) - =RM 33.00 N1 10 10 8 6 R 4 2 x 01618