Program Peningkatan Prestasi Akademik Spm 2013 Matematik Tambahan

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Jul 26, 2016 - Program Kursus Perguruan Lepasan SPM ini hanya sekali. Peningkatan 0.13 peratus berbanding 15,079 calon pada 2011. Tarikh Keputusan SPM 2012 - 21 Mac, 2013. Mungkin ada insentif tambahan bagi guru-guru berkaitan. Senarai Program Pengajian yang ditawarkan Bagi Sesi Akademik.

Political theory by rajeev bhargava pdf converter. • 2 Nama Pelajar: Tingkatan 5:. 3472/2 Additional Mathematics Sept 2013 PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2013 ADDITIONAL MATHEMATICS Paper 2 (SET B). MARKING SCHEME SULIT 3472/2 • 3 MARKING SCHEME ADDITIONAL MATHEMATICS PAPER 2 N0. SOLUTION MARKS 1 3x y  or 3y x  2 2(3 ) (3 ) 6y y y     2(3 ) ( 1) 6x x x    22 7 6 0y y   22 5 3 0x x   (2 9)( 1) 0y y   (2 3)( 1) 0x x   3 2 x  and 1x  (both) 3 2 y  and 2y  (both P1 K1 Eliminate x/y K1 Solve quadratic equation N1 N1 5 2 (a) (b) 2 2 [ ( )] [(2 1) 4] (2 1) 4 3, 12 11 1 f g x f x m x m n m n n             ( ) 5 1 ( ) 5 3, 5 f x p x y x p y p q        K1(find composite function) N1 N1 K1(find inverse function) N1N1 7 3 (a) (b) y = x  draw the straight line y = x  Number of solutions = 4 P1 cos shape correct. P1 Amplitude = 6 [ Maximum = 3 and Minimum = -3 ] P1 2 full cycle in 0  x  2 or P1 reflection the graph N1 For equation K1 Sketch the straight line N1 6 2   3 -3 • 4 4 (a) (i) (ii) (iii) (b) 2 ( ) 3 4 2 OQ OC CR x y     1 ( ) 2 3 2 BQ BO OC x y     5 BR BQ QR x y    . BR hOC cannot find h not parallel  K1 N1 K1 N1 K1 N1 K1 find h N1 8 5 (a) (b) i) 6 10 60 x x     ii) 2 2 2 2 7 6 10 850 x x      new mean 3(6) 5 23   new standard deviation 3(7) 21  P1 K1 N1 K1 N1 K1 N1 7 • 5 6 (a) (b) 2 2 21 1 1,,. 2 8 32 p p p 2 2 2 2 1 1 8 32 1 1 2 8 p p p p  1 4 r  (i) 11 253200( ) 4 128 8 n n    (ii) 3200 1 1 4 2 4266 3 S   K1 K1 N1 K1K1 N1 K1 N1 8 • 6 7 (a) (b) (c) (i) (ii) (iii) x 1 2 3 4 5 6 10log y 0.26 0.41 0.53 0.66 0.80 0.94 10 10 10log log logy x h k h  10log h = *gradient h = 1.37 10logk h = *y-intercept = 0.88 y = 3.98 N1 6 correct values of log y K1 Plot 10log y vs x. English to sinhala unicode converter.

Correct axes & uniform scale N1 6 points plotted correctly N1 Line of best-fit P1 K1 N1 K1 N1 N1 10 10log y 0.12 0 x • 7 N0. SOLUTION MARKS 8 (a) (b) (c) 21 (6) 26.28 2    = 1.46 rad 6(1.46)ADS  = 8.76 cm Perimeter = 8.76 + 5 + 5 + 8 = 26.76 cm Area of triangle = 17.89 cm 2 Area of rectangle = 40 Area of the shaded region = 17.89 + 40 – 26.28 = 31.61 cm 2 K1 N1 K1 Use s r N1 K1 N1 K1 K1 K1 N1 10 • 8 N0. SOLUTION MARKS 9(a) (b) c) 2 2 16 x x dx dy  At turning point, ( p, 4), 0 dx dy 0 2 16 2  p p 16p 3 = 2 8 13 p, 2 1 p y = ∫ 16x – 2x -2 dx c x x  2 2 16 2 = c x x  2 8 2 At point A ( 2 1, 4 ), 4 = 8( ½) 2 + 4 + c c = -2 The equation of the curve is, 2 2 8 2  x xy 32 2 4 16 xdx yd  At point A( 2 1, 4), 048)8(416 2 2  dx yd Thus,       4, 2 1 A is a minimum point P1 K1 N1 K1 use dxyy  )( 12 K1 integrate correctly K1 N1 K1 K1 N1 10 • 9 N0. SOLUTION MARKS 10 (a) (b) (c) (d) 1 68  yx Thus, A( 0,6) and B( 8, 0) Let D (x,y) CD: DE = 1:3 x =     8 4 32 4 10321   y =     7 4 28 4 10321   Thus D  7,8 4 3 8 6    ABm Thus, 3 4 CEm y - 6 =  0 3 4 x y = 6 3 4  x Area of AOB= 6706 0880 2 1 2856 2 1  N1N1 K1 K1 N1 K1 N1 K1 (use area formula) K1N1 10 • 10 N0. SOLUTION MARKS 11 (a) (i) (ii) (b) (i) (ii) X = Students have a laptop p = 10 7 = 0.7, n = 5 q = 3.0 10 3 10 7 1  P(X= 5) =      5055 7.03.07.05 c = 0.1681 P( X265) =         20 250265 ZP =P(Z > 0.75) = 0.2266 P(X  20 250w = 0.3 From table, 20 250w = -0.524 W = 239.52 g K1 Use P ( X=r ) = rnr r n qpC  N1 K1 K1 Use P ( X=r ) = rnr r n qpC  N1 K1 Use Z =  X N1 K1 K1 N1 10 • 11 N0.

SOLUTION MARKS 12 (a) (b) (c) (d) vinitial = 5 m s -1 v= t 2 – 6t +5 dt dv a  =2t -6 a = 0 t = 3 v = (3) 2 -6(3) + 5 = -4 ms -1 v.