[ { "page": 1, "content": "Modul Pintas Tingkatan 5 Peperiksaan Percubaan SPM 2018 Skema Jawapan Matematik Kertas 2 1449/2" }, { "page": 2, "content": "No Marking Scheme / Skema markah Marks Submarks Total Marks 1 y  3 – x x < 3 y < 3 1 mark 1 mark 1 mark 3 marks 2 ( a )  AFB or  BFA or  DEC or  CED 1 mark 3 marks ( b )  AFB = 2 200 150 1  sin 1 mark = 97 · 18 ̊ or 97 ̊11' 1 mark 3 60 × 40 × 100 8000 240000 1 mark 1 mark 4 marks 30 × RM2 · 50 or RM75 RM25 1 mark 1 mark 4 x – 20 = y + 20 or equivalent x + 22 = 2( y – 22) or equivalent x = y + 40 or y = x – 40 or x = 2 y – 66 or y = x – 40 or x xy 66  Amelia, x = RM146 or Bala, y = RM106 RM252 1 mark 1 mark 1 mark 1 mark 1 mark 5 marks 5. ( a ) False / Palsu 1 mark 5 marks ( b ) Premise 1: If the length of a side of an equilateral triangle is 8 cm, then the perimeter of the triangle is 24 cm Premis 1: Jika panjang sisi suatu segi tiga sama ialah 8 cm, maka perimeter bagi segi tiga itu ialah 24 cm . 2 marks ( c ) n 2 + 2 n , n = 1, 2, 3, 4, ... Accept n 2 + 2 n for 1 mark 2 marks 6. ( a ) 2 692   x xx 1 mark 4 marks x 2 − 9 x – 10 = 0 1 mark ( b ) ( x – 10)( x + 1) = 0 96 1 mark 1 mark 7. ( a ) e + b = 96 0 · 55 e + 2 · 21 b = 117 · 54 1 mark 1 mark 5 marks ( b )                         54117 96 21.255.0 11 b e                           54117 96 155.0 1212 )550)(1()212)(1( 1 b e e = 57 b= 39 1 mark 1 mark 1 mark" }, { "page": 3, "content": "No Marking Scheme / Skema markah Marks Submarks Total Marks 8. ( a ) (  4, 5) ( b ) 4 5 04 05   OQ m 1 ma rk 1 mark 5 marks cxy  4 5 1 mark c  )4(4 55 1 mark 104 5  xy or 4054  xy 1 mark 9 ( a ) {(1, A), (2, A), (3, A), (4, A), (5, A), (6, A), (1, B), (2, B), (3, B), (4, B), (5, B), (6 , B), (1, C), (2, C), (3, C), (4, C), (5, C), (6, C)} Allow 2 mistakes for 1 mark 2 mark s 6 marks ( b ) (i) {(4, A), (4, B), (4, C)} 6 1 18 3  1 mark 1 mark (ii) {(2, B ), (2, C ), (3, B ), (3, C ), (5, B ), (5, C )} 1 mark 3 1 18 6  1 mark 10 ( a ) 77 222360 240      or 57 222360 240      or 37 222360 240      or 57 222360 120      or 37 222360 120      or 57 222  or 37 222  1 mark 6 marks 77 222360 240      + 57 222360 240      + 37 222360 240      + 57 222360 120      + 37 222360 120      + ( 4 × 2 ) + ( 2×3 ) or equivalent OR 77 222360 240      + ( 57 222  ) + ( 37 222  ) + (4 × 2) + ( 7) + (7) 1 mark 21 1393 or 21 1966 or 93.62 1 mark ( b ) 2 57 22  or 2 2360 270      or (7  7) or ( 4  4 ) or (2  2) 2 57 22  − ( 7×7 ) + ( 4×4 ) + 2 2360 270      + ( 2  2 ) 7 140 or 7 281 or 40 ∙14 1 mark 1 mark 1 mark or equivalent ." }, { "page": 4, "content": "No Marking Scheme / Skema markah Marks Submarks Total Marks 11 ( a ) 2 mark s 6 marks ( b ) 63 30  u 12 ( c ) ( t – 3 )(30) + 2 1 ( 12 – t )(30)= 180 t = 6 1 mark 1 mark 1 mark 1 mark 12. ( a ) ( b ) Graph Axes are drawn in the correct direction, uniform scale for − 4  x  4 and − 36  y  36. 7 points and 2 points* plotted accurately Smooth and continuous curv e without straight line(s) and passes through all the 9 correc t points − 4  x  4 and − 36  y  36. Notes : (1) 7 or 8 points plotted correctly, award 1 mark (2) Other scale being used, subtract 1 mark ( c ) 12  y  14 1 · 3  x  1 · 5 Note: Do not accept the values of x and y obtained b y calculation ( d ) Identify the equation y = − 3 x + 2 correctly drawn − 3 · 05  x  − 3 · 25 3 · 65  x  3 · 85 x − 3 2 y 12 − 18 1 mark 1 mark 1 mark 2 marks 1 mark 1 mark 1 mark 2 marks 1 mark 1 mark 12 marks 12 Time / Masa ( Second/ Saat ) 30 u t 3 Speed (ms - 1 )" }, { "page": 5, "content": "No Marking Scheme / Skema markah Marks Submarks Total Marks 13 ( a ) (i) ( 4 , − 9 ) (ii) ( 3, − 1 ) 2 marks 2 marks 12 marks ( b ) (i) ( a ) V is R otation 90 ̊clockwise about the centre ( 2, −2 ) V ialah p utaran 90 ̊ ikut arah jam pada pusat ( 2, −2 ) ( b ) W ia an enlargement with a scale factor 2 and centre L ( 2 , 1 ). W ialah pembesaran , fak tor skala 2 pada pusat L (2, 1) 2 marks 3 marks ( ii ) Area of image = 2 2  area of object Area of shaded region = area of imag e  area of object = 8 4 = 2 2  area of object  area of object area of object = 3 84 = 2 8 2 mark s 1 mark 14 ( a ) Column I 1 mark Column I I 1 mark Column II 1 mark Column II 1 mark 12 marks Column I Column II Column III Column IV Mass ( kg) Jisim (kg) Upper boundry Sempadan atas Frequency Kekerapan Cummulative Frequency Kekerapan longgokan 0 – 0·9 0·95 0 0 1·0 – 1·9 1·95 4 4 2·0 – 2·9 2·95 10 14 3·0 – 3·9 3·95 26 40 4·0 – 4·9 4·95 24 64 5·0 – 5·9 5·95 17 81 6·0 – 6·9 6·95 12 93 7·0 – 7·9 7·95 7 100 ( b ) 100 )7(457)12(456)17(455)24(454)26(453)10(452)4(451  4 · 49 or equivalent 2 marks 1 mark ( c ) Ax es are drawn in the correct direction, uniform scale for 0 ≤ vertical axis ≤ 100 and 0 · 95 ≤ horizontal axis ≤ 7 · 95 All points plotted accurately. Smooth and continuous curve. 1 mark 2 marks 1 mark ( d ) 100 – 66 = 34 1 mark" }, { "page": 6, "content": "No Marking Scheme / Skema markah Marks Submarks Total Marks 15 ( a ) 12 marks Correct shape rectangles QKLM , KJGL and JIHG. 1 mark K – L is join ed by a dashed line . 1 mark QI  IH  QK = KJ  JI 1 mark Measurements accurate up to  0 · 2 cm (one way) and all right angles = 90 o  1 o 2 marks 15 ( b )(i) Correct shape pentagon ABHGN. 1 mark BH  AB  AN  GH 1 mark Measurements accurate up to  0 · 2 cm (one way) and all right angles = 1 mark H 5 cm A B 8 cm 4 cm N 1 cm G 4 cm M G Q J 8 cm 4 cm I 1 cm F H E" }, { "page": 7, "content": "No Marking Scheme / Skema markah Marks Submarks Total Marks 90 o  1 o 15 ( b )(ii) Correct shape rectangles NIPN and BCPN. 1 mark N – P is joined by a dashed line . 1 mark BH = BC  BN = NH 1 mark Measurements accurate up to  0 · 2 cm (one way) and all right angles = 9 0 o  1 o 1 mark 16 (a) 120  W/ B Note : Award 1 mark for 120  or W / B is seen (b) (45  S, 60  W / B ) Award 2 marks for (   S, 60  W / B ) Award 1 mark for (   N, 60  W / B ) (c) (180  – 45  – 45  )  60 or 2 (90  – 45  )  60 or equivalent 5400 nautical miles (d) 120   60  cos 45  (45  + 45  )  60 18 60) × (90 + 45) cos 60 (120  582 · 84 2 marks 3 marks 1 mark 1 mark 1 mark 1 mark 2 marks 1 mark B C H I 4 cm 5 cm P N 8 cm" }, { "page": 8, "content": "- 4 - 3 - 2 - 1 0 1 2 3 4 y - 40 - 30 - 20 - 10 10 20 30 40 x X X Graph for Question 12 Graf untuk Soalan 12" }, { "page": 9, "content": "Graph for Question 14 Graf untuk Soalan 14 8 0 7 0 6 0 5 0 4 0 X X X X 3 0 2 0 X 1 0 10 0 9 0 X O X 0·95 1·9 5 2·95 3·9 5 4·95 5·9 5 6·95 7·95 X" } ]