trial kedah 2014 spm add math k2 skema

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Bahan Pecutan Akhir Add Math SPM

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  • 1. 1Nama Pelajar : Tingkatan 5 : .3472/2Additional MathematicsAugust 2014MODUL PENINGKATAN PRESTASI TINGKATAN 5TAHUN 2014ADDITIONAL MATHEMATICSPaper 2( MODUL 2 )MARKING SCHEME

2. 2SULIT 3472/2MARKING SCHEMEADDITIONAL MATHEMATICS PAPER 2 2014N0. SOLUTION MARKS1x 2y 1 or12xy2 2 x x 502 x 25x 5 x 5x 5 and x 5 (both)y 2 and y 3 (both)P1K1 Eliminate x/yK1 Solve quadratic equationN1N152(a)(b)(i)(ii)78 (1)(2)128T 1010(1)(2 1)2 110231023 (3)(7)(5)107415SV 1023 0.8818.4K1N1K1K1N1K1N173(a)(b)y =xdraw the straight line y =xNumber of solutions = 3P1 cos shape correct.P1 Amplitude = 2 [ Maximum = 1and Minimum = -1 ]P1112cycle in 0 x orN1 For equationK1 Sketch the straight lineN16-4 3. 34(a)(b)1010100xx22 224 10101160xx 10 326.5meanor24224 K1N1K1N1K1N1 N175(a)(b)5 1255535 5353log log 1loglog 13log log 3log 3125K VVKK VKVKV i)1( )21 132 8 24 24x kf xmkm mm and k ii)1 1( ) 38 220pp K1K1N1K1K1N1K1N18 4. 46(a)(b)(3 1)(3 1)( ) 3 13 1'( ) 3x xf x xxf x i)2232 (2) 3(2)2dykx xdxkk ii)12112 (2)21112normal mcy x K1N1K1N1P1K1N17 5. 57(a)(b)(c)(i)(ii)(iii)x 1 2 3 4 5 62yx3.5 5.5 7.5 9.5 11.5 13.52yx2yx= kx+pkk = *gradientk = 2.0pk= *y-interceptp = 3.0y = 40N1 6 correctvalues of2yxK1 Plot2yxvs x.Correct axes &uniform scaleN1 6 points plottedcorrectlyN1 Line of best-fitP1K1N1K1N1N1101.50x 6. 6N0. SOLUTION MARKS8(a)i)ii)iii)b)2(9 )3TS x2PTQR TR TP PQQRPS PT TSMS MR RSPS kMSPS PT TS 6x 8y =k =4PS 4MS and S is a common point or equivalent2 21 36( ) 8( )2 4PS = 45K1 (TS or QR )N1K1N1K1N1K1N1K1N18 9 49 4y x yx y 3 4 29 43 42322TRx yx yx yxy 6x 8y3( 2 )2k x y= 6x= 4y 6x 8y 7. 79a)b)c)223 43 4 0( 1)( 4) 04, 16(4,16)x xx xx xx yK 42yx22023032(4)(2)83283163x dxxcm 162416242 21(4) (12)364216 4642 2256 16642 256ydyy K1 for solvingquad.eqn.N1N1K1 use area ofrectangle - ( y) dxK1 integratecorrectlyand Sub.the limitcorrectlyN1K1K1 correct limitK1 integratecorrectlyN110Area BVolume A 8. 8N0. SOLUTION MARKS10(a)(b)(c)60o1.047 rad8(1.047) OB S or 8(2.095) BC S OR 8(3.142) AC S = 8.38 = 16.76 = 25.14Perimeter = 8.38+16.76+8 or Perimeter = 25.14 + 8= 33.14 = 33.14Area of OAB = 2 1(8) (1.047)2= 33.50 cm2Area of triangle OAB = 2 1(8) sin 602= 27.71Area of the shaded region = 33.50 27.71= 5.79 cm2P1N1K1 Use s rN1K1N1K1 Use formula1 22A r K1K1N110 9. 9N0. SOLUTION MARKS11(a)(i)(ii)(b)(i)(ii)X= Students passed Mathematicsp = 0.85 , q = 1- 0.85 = 0.15 , n = 6P(X =6) = 6 6 06 c (0.85) (0.15)=0.3772P (Y2) = 1 P(Y=0) P(Y = 1)Or = P(Y 2) P(Y 3) ......... P(Y 6)= 1 - 6 1 51c (0.15) (0.85) - 6 0 60 c (0.15) (0.85)=0.2235= 52 , =10P( 40 < X < 60 ) = P (40 5210< Z