SULIT 347/1 347/1 MATEMATIK TAMBAHAN KERTAS 1 OGOS 013 Jam NAMA : KELAS : PEPERIKSAAN PRASPM SEKOLAH-SEKOLAH MENENGAH 013 MATEMATIK TAMBAHAN Kertas 1 Dua Jam JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 347/1 INFORMATION FOR CANDIDATES MAKLUMAT UNTUK CALON all semua one satu Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 3 347/1 ALGEBRA CALCULUS / KALKULUS Negeri Sembilan SPM 013
SULIT 4 347/1 STATISTICS / STATISTIK GEOMETRY / GEOMETRI Negeri Sembilan SPM 013
SULIT 5 347/1 TRIGONOMETRY / TRIGONOMETRI θ j θ [Lihat sebelah] Negeri Sembilan SPM 013
SULIT 6 347/1 Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 7 347/1 all semua 1 1 Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 8 347/1 3 3 Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 9 347/1 4 4 5 Negeri Sembilan SPM 013 [Lihat sebelah] 5
SULIT 10 347/1 6 6 Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 11 347/1 7 8 7 Negeri Sembilan SPM 013 [Lihat sebelah] 8
9 SULIT 1 347/1 9 10 10 Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 13 347/1 11 1 11 1 [Lihat sebelah] Negeri Sembilan SPM 013
13 SULIT 14 347/1 13 Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 15 347/1 14 15sin x = k, 90 sin x = k, 90 Negeri Sembilan SPM 013 [Lihat sebelah] 14 15
SULIT 16 347/1 16OA =4i 3jOB = i+j OA = 4i 3jOB = i + j AB AB 16 Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 17 347/1 17 Negeri Sembilan SPM 013 [Lihat sebelah] 17
SULIT 18 347/1 18 18 Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 19 347/1 19 0 Negeri Sembilan SPM 013 [Lihat sebelah] 19 0
SULIT 0 347/1 1 f(x)dx =3 f(x)dx =3 1 f(x)dx (x f(x))dx Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 1 347/1 P Q R S T 7 8 9 Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 347/1 3 3 Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 3 347/1 4 4 Negeri Sembilan SPM 013 [Lihat sebelah]
SULIT 4 347/1 5 5 END OF QUESTION PAPER KERTAS SOALAN TAMAT Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Price (RM) for the year Ingredient Price index based on the year Weightage Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Section A, Section B Section allsection AfourSection Btwo Section C Negeri Sembilan SPM 013
KEBARANGKALIAN HUJUNG ATAS Q BAGI TABURAN NORMAL Negeri Sembilan SPM 013
Rumus-rumus berikut boleh membantu anda menjawab soalan. Simbol-simbol yang diberi adalah yang biasa digunakan. Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
Negeri Sembilan SPM 013
No. Suggested solution and marking scheme Sub Marks Total Marks 1. (a) 3. (b) 4 (c) many to many relation ( hubungan banyak kepada banyak ), m m, mm. ( a) 6 ( b) k 3 B1: k(6) + 3 = 7 or 6k = 4 1 1 1 1 3 3 1 3. fg 1x 13 B: 6 ( x 3 ) 5 4. 1 B1: g x 3 3 p B: SOR 3 and B1: SOR 3 or 5. x, x 8 POR ( 3 ) POR ( 3 ) p (Both) or equivalent. p 3 3 3 3 3 3 B: x - or 8-8 x B1: ( x )( x 8) 0 or ( x )( x 8) 0 6 (a) k = 8 (b) a = (c) x = 1 1 1 1 3 7 3 x or x = 1.5 B: 3 4 x B1: Seen 3 4 4 3 x or 3x or 3 or OR using logarithms method: equivalent. x = 1.5 (accept -1.499 1.501) 4 3 or equivalent. B: (x+)0.4771 +x (0.4771) = 1.908 or 1.9081x =.86 B1: log10 3 x+ + log10 7 x = log10 81 1 ( (accept any base) Negeri Sembilan SPM 013 3 3
8. x 1 B: x 8 9 x or x + 8 = 9x or equivalent B1: x 8 x 8 log 3 or log 3 log 3 3 x x 3 3 9. 17 or 17 terms or n = 17 B1: 13 + (n 1)(5) = 67 or by listing method: (Must list all the terms and correct ) 13, 8, 3,, 7, 1, 17,, 7, 3, 37, 4, 47, 5, 57, 6, 67 10. 3 3 a, r or equivalent (Both correct ) 7 7 B : ar 3 4 and ar 6 (Both ) B1 : 3 6 1 ar 4 or ar 13 3 3 11. h = 41, k = 11 0.7 B: S = or S = 1 0.01 7 8 B1: r = 0.01 or or 99 11 7 3 or S = 99 8 41 3 or S = 11 11 3 3 1. h =100, k = 7 (both ) B3: h =100 or k = 7 1 k 1 B: log 10 h or k (10) 3 or 10 0 1 1 B1: log10 y log10 x log10 h accept Y X 4 4 13. ( a ) x y 1 or equivalent 5 ( 6 ) (b) 10x + 1y + 11 = 0 1 3 4 B: x 10x 5 y x y 1y 36 or equivalent B1: x 10x 5 y or x y 1y 36 14. 41.81 o, 138.19 o or 41 49, 138 11 B3 : sin x = 3, sin x = 4 ( both) 4 4 B : (3 sin x )(sin x + 4) = 0 B1 : 3( 1 sin x ) 10sin Negeri x+ 5 Sembilan = 0 SPM 013
15. (a) k 1 1 3 (b) 1 k 0 0 B1: cos180 cos x sin180 sin x or cos x 16. (a) 5 5 i 4 j,accept 4 3 B1: 4 i 3 j i j or equivalent (b) 41 or 6.403 1 17. (a) AC 6 a 4 b (b) 9 BD a b B: 1 3 6a ( 6 a 4 b ) or 4b ( 6 a 4 b ) 4 4 B1: 1 3 ( 6 a 4 b ) or ( 6 a 4 b ) 4 4 1 3 4 18. 4 1 (a) or 1 accept 1.333 3 3 (b) 30 1 3 1 4 3 B1 : or 9 1 6 4 3 19. 9 ( 6 x 1 ) dy B: 3 u 6 dx 4 dy or 1 (3)(6x 1) (6) dx 4 B1: 3 u 4 dy du or 6 du dx 3 3 0. p 3 B: 6p(-1) 1 (-1) = 8 or equivalent B1 : dy dx dy 6 px 1x or 8 dx 3 3 Negeri Sembilan SPM 013
1. (a) 6 1 4 (b) 1 3 x B : B1 : 1 4 4 1 3 xdx f ( x) dx 1 4. (a) 4030 (b) 430 B1 : 3! or 6! or equivalent 1 3 3 9 (a) 35 3 3 B1 : 5 7 4 18 (b) 35 B1 : 3 3 4 5 7 5 7 4 4 k u 3 B : k = 4 3u 3 3 B1 : 10 ( 5 3 u ) 5 (a) 1.67 4 B1 : 63. 48 1 (b) k = 0.39 B1 : 0.39 Negeri Sembilan SPM 013
SULIT 347/1 347/1 MATEMATIK TAMBAHAN KERTAS 1 OGOS 013 Jam PEPERIKSAAN PRASPM SEKOLAH-SEKOLAH MENENGAH 013 MATEMATIK TAMBAHAN / ADDITIONAL MATHEMATICS KERTAS 1 / PAPER 1 SKEMA PERMARKAHAN / MARKING SCHEME Negeri Sembilan SPM 013
PEPERIKSAAN PERCUBAAN 013 347/ ADDITIONAL MATHEMATICS Kertas Ogos 1 jam. Duajamtigapuluhminit SKEMA PERMARKAHAN MATEMATIK TAMBAHAN KERTAS PEPERIKSAAN PRASPM TINGKATAN 5, 013 No Solution and mark scheme Sub Marks 1 y = x 4 Or y 4 P1 x = Full Marks (x 4) = 4(x+5) y = y 4 5 x 5 x 1 0 y y 8 0 Solve the equation x = ( 5) ( 5) ( 1 ) 4(1)( 1) x = 5.193, 0.193 y= 6.386, 4.386 Solve the equation ( ) ( ) 4(1)( 8) y = ( 1 ) y = 6.385, 4.385 x=5.193, x = 0.193 5 (a) (i) x 7x 1 0 ( x 3)( x 4) 0 x = 3, x = 4, since <, = 3, = 4. (ii) x 7x 1 0 ( x 3)( x 4) 0 6 3 4 3 x 4 Negeri Sembilan SPM 013
No Solution and mark scheme Sub Marks Full Marks (b) 1 = 3 1 = 1 = (4) + 1 = 9. x x ( 9 ) x ( 9 ) 0 11x 18 0 3 (a) T1 = 1 (160)(80) = 6400 T= 1600 T3=400 P1 r = 1 or 0.5 4 P1 6 Tn> 30 6400 ( 0.5 ) n-1 > 30 log 6400 ( 0.5 ) n-1 > log 30 log 6400 + log( 0.5 ) n-1 > log 30 log( 0.5 ) n-1 > log 30 log 6400 log 30 log 6400 n 1 < log 0.5 n < 4.868 n = 4 listing method or 6400,1600,400,100, 5..all correct n = 4 Negeri Sembilan SPM 013
No Solution and mark scheme Sub Marks Full Marks (b) S 6400 1 0.5 = 8533.33 4 (a) LHS sin x sin xkosx tan x (b) 4 0 3 y = - x x 8 Shape as in the above diagram Amplitude is 4 Number of cycle is 1 Modulus Equation of straight line y = x Straight line is drawn Number of solution 4 P1 P1 P1 Negeri Sembilan SPM 013
No Solution and mark scheme Sub Marks Full Marks 5 a) or b) i) ii) P1 8 Comparing,, 3 k 4 6 (a) (b) (i) 90 x 10 9 1050 10 9 16 // 16.67 3 90 x 11 10 x 0 Negeri Sembilan SPM 013 P1
No Solution and mark scheme Sub Marks Full Marks (ii) x 0 10 11 = 4 // 4.90 7 (a) (b) Plot x 3 4 6 8 10 11 log 10 y 0.6 0.74 0.9 1.1 1.46 1.6 log y against x (Correct axes and uniform scales) 10 10 6 points are correctly plotted. Note: 5 or 4 points correctly plotted Award 1 mark N Line of best fit (c) (i) log10 y (log10 t ) x log10 s Use c log10 log 10 s 0.6 s s = 1.8 0. 0 P1 (ii) Use m log10 t log 10 t 0.115 t 1.30 0.0 Negeri Sembilan SPM 013
No Solution and mark scheme Sub Marks 8 dy a) x 4, dx Full Marks y 9 = 4 (x 4) y = 4x 7 b) Area 1 4 4 10 Area x under curve 4 1 dx 0 3 x = 6 x 4 0 = 14.67 // 44 14 // 3 3 9 Area = 4 14. 67 56 = or 18 or 18. 67 3 3 c) Volume y dy 15 (a) kos = 30 1 9 = y y 9 1 = 8118 3 = 96 SOT = 60 SOT = 1.047 rad (b) Lengkok ST = (1.047)(15) = 15.705 ST=TR PT=TQ PT = 30 15 = 5.981 Negeri Sembilan SPM 013
No Solution and mark scheme Sub Marks Full Marks Perimeter kawasan berlorek =ST + TR + RQ + TQ +SP + PT = 15.705 + 15.705+15+15+ 5.981 + 5.981 = 113.37 (c) Luassektor OST = 1 (15) (1.047) Luas SOT = luas OTR = 117.7875 1 Luassegitiga OPQ = (5.981) 15 = 389.715 Luasakawasanberlorek = luas OPQ (luas OST) = 389.715 ( 117.7875 ) = 154.14 10 (a) or = 8 unit (b) = (-1, 3) (c) P1 10 or Negeri Sembilan SPM 013
No Solution and mark scheme Sub Marks (d) Use PA = 5 P1 (x - 3) (y 5) = 5 Full Marks x y 6x 10y 9 0 11 (a) (i0 p = 10 1 // 0.1, q = 10 9 // 0.9 P 0. 1 3 0. 9 7 10 ( X 3 ) C3 = 0.0574 (ii) P ( X ) 1 [ P ( X 0) P ( X 1) P ( X )] 10 0 10 10 1 9 10 1 C 0. 1 0. 9 C 0. 1 0. 9 C 0. 1 0. 9 0 or1 0.3487 0.3874 0. 1937 = 0.070 1 8 P1 10 (b) (i) 3. 6 3. 0 z 0. 5 P or P z 1. 0.88493 // 0.8849 (ii) Seen -0.385 m 3. 0 0.385 0. 5 m =.8075 P1 Negeri Sembilan SPM 013
No Solution and mark scheme Sub Marks Full Marks 1 (a) a = 9 4 t = 0 9 t= 4 9 9 V mak 9 4 4 1 81 V mak 10 or or 8 8 10.13 (b) 9t s t 3 3 c t =, s 1 // 3 1 ort = 3, s // d s 3 s 5 9 // 9.83 m 6 38 // 1.67 3 45 //.5 9t t (c) s 0 3 7 4 t = 0 7 t // 6.75 s 4 (d) v 9 t t 0 3 9 t // 4.5 s Negeri Sembilan SPM 013
No Solution and mark scheme Sub Marks Full Marks 13 (a) Q Guna I = 1 100 Q 0 p= 1.60, q = 1.10, r = 160 P1P1P1 (b) I = 11(400) 15(00) 110(150) 160(50) 400 00 150 50 = 16.3 (c) Q13 100 16.3 5 Q13 = 31.575 (d) 16.3 100 85 100 = 107.36 100 Negeri Sembilan SPM 013
No Solution and mark scheme Sub Marks 14 (a) Full Marks // (b) Draw a line correctly Draw all the lines correctly Correct region R (c) (i) 350 (ii) Profit P = 1x + 15y Maximum point ( 0, 450 ) Maximum profit = 1 ( 0 )+15( 450 ) = RM 6750 15 (i) PR 8 18.4 (8)(18.4) kos = 17.37 70 0 6. 5 17.37 (ii) 0 sin PRQ sin107 0 PRQ 0.97 (b) Q P 10 R QPR = 5.03 0 0 0 180 107 0.97 0 ' PQP 0 0 180 (5.03) Negeri Sembilan SPM 013
No Solution and mark scheme Sub Marks = 75.94 0 Full Marks ' 0 0 QP R 180 53.03 17.97 0 QR sin17.97 0 QR = 14.3 6. 5 sin 0.97 0 Luas P ' QR = 1 = 4.01 6.5 14.3 sin 31.06 0 Negeri Sembilan SPM 013
Graf Q7 log10 y Plot lg y against x (Correct axes and uniform scales) P1 1.8 6 points are correctly plotted N or 5 or 4 points correctly plotted 1.6 Line of best fit 1.4 1. 1.0 0.8 0.6 0.4 0. 0 4 6 8 10 1 x Negeri Sembilan SPM 013
Q14 y 500 450 400 350 300 50 00 150 R 100 50 50 100 150 00 50 300 350 400 450 x Negeri Sembilan SPM 013