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SMK PUTRAJAYA PRESINT 14(1) Skema Jawapan
SPM Matematik Tambahan Tingkatan 5 - Pra SPM Add Math Kertas 1 2013 1. (a) a, c (b) {2, 4, 6, 8} 2. (a) Many-to-one relation
(b) f : x x2 3. f(x) = x + 6, g(x) = x2 – 3x + k
(a) Let y = f –1(5) f (y) = 5 y + 6 = 5 y = –1 Thus, f –1(5) = –1 (b) gf–1(5) = 12 g(–1) = 12 (–1)2 – 3(–1) + k = 12 4 + k = 12 k = 8
4.
5. 4 (b)
6. x < 1 or x > 4
7.
8.
9. (a)
x + y = 6 x + y – 6 = 0 (b) RP = RQ
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x2 + (y – 6)2 = (x – 6)2 + y2 x2 + y2 – 12y + 36 = x2 – 12x + 36 + y2 12x – 12y = 0 x – y = 0 y = x The equation of the locus of point R is y = x.
10. (a) m= 3 (b) 3y=-x + 8 11. N = 8, Σx = 32, Σx2 = 146
(a)
(b) Standard deviation
12. (a) 0.525 rad
(b) 8.352 cm 13.
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14. 55h 15. (a)
m= 8 (b) r= 2
16. d = 5 12023
17. Geometric progression: 1, x2, x4, x6, ....
(a) Common ratio = = x2
(b)
3 – 3x2 = 1 3x2 = 2
= 0.8165
18.
19. 12
20. 3235
21. 5 4
22. x = 0°, 70° 32'/70.53º, 180°, 289° 28'/289.47º, 360° 23. (a) Each team will have 5 students.
Number of ways of selecting the first team = 15C5 Number of ways of selecting the second team = 10C5
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Number of ways of selecting the third team = 5C5 Total number of ways = 15C5 × 10C5 × 5C5 = 756 756 (b) Number of ways of arranging 5 students for a group photograph = 5! = 120
24. (a) P(passes both subjects)
(b) P(passes only one of the subjects)
25. (a) 0.2631
(b) 0.2416
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SMK PUTRAJAYA PRESINT 14(1) Skema Jawapan
SPM Matematik Tambahan Tingkatan 5 - Pra SPM Add Math Kertas 2 2013 1. x + 2y – 1 = 0
x = 1 – 2y …… 1 x2 + y2 + 2xy – 25 = 0 …… 2 Substitute 1 into 2 (1 – 2y)2 + y2 + 2y(1 – 2y) – 25 = 0 1 – 4y + 4y2 + y2 + 2y – 4y2 – 25 = 0 y2 – 2y – 24 = 0 (y + 4)(y – 6) = 0 y = –4 or 6 Substitute the values of y into 1 . When y = –4, x = 1 – 2(–4) = 9 When y = 6, x = 1 – 2(6) = –11 The solutions are x = 9, y = –4 and x = –11, y = 6.
2. (a) 65.83 (b) 336
3. (a)
= 3x² – 8x + 4
y = = x3 – 4x2 + 4x + c At point (1, 1), x = 1 and y = 1. 1 = 13 – 4(1)2 + 4(1) + c c = 0 The equation of the curve is y = x3 – 4x2 + 4x.
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(b)
4. (a) Muthu annual salaries form a geometric progression with a = RM18 000, r = 1.05.
Annual salary in the year 2007 = T6 = RM18 000 × 1.055 = RM22 973 (to the nearest ringgit)
(b) Tn > 36 000 18 000 × 1.05n – 1 > 36 000 1.05n – 1 > 2 (n – 1) log10 1.05 > log10 2
n – 1 > n – 1 > 14.21 n > 15.21 The minimum value of n is 16.
(c) Total salary from the years 2002 to 2007
= RM122 434 (to the nearest ringgit)
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5. (a)
(b)
(c)
6. (a), (b)
7. (a)
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(b) Let the coordinates of point D
Thus, the coordinates of point D are (–5, 13). (c) The locus of point P is a circle with AC as its diameter. APC forms a right-angled triangle. Let the coordinates of point P be (x, y). CP is perpendicular to AP. Gradient of CP × Gradient of AP = –1
(y – 3)(y – 1) = (x + 5)(x – 1) y2 – 4y + 3 = (x2 + 4x – 5) x2 + y2 + 4x – 4y – 2 = 0 Thus, the equation of the locus of P is x2 + y2 + 4x – 4y – 2 = 0.
8. (a)
(b)
Perimeter of the shaded region = AC + BC + Arc AB = 3 + 7.937 + 8.672 = 19.609 cm (c) Area of sector AOB
= × 122 × 0.7227 = 52.0344 cm2
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Area of ΔOBC = × 9 × 7.937 = 35.7165 cm2 Area of the shaded region = Area of sector AOB – Area of ΔOBC = 52.0344 – 35.7165 = 16.32 cm2
9. (a)
(b) (i) 3 4
(ii) 3
10. (a)
Given = 2x y = ∫ 2x dx y = x2 + c The curve passes through Q(1, 3). At Q(1, 3), x = 1 and y = 3. 3 = 12 + c c = 2 Hence, the equation of the curve is y = x2 + 2.
(b)
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Area of the shaded region
(c) Volume of revolution
11. (a)
(ii) Mean = np = 10 × 0.75 = 7.5
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(b)
12. (a)
(b) Area of triangle ABC
= (7.8)(4.574) sin 43° = 12.17 cm2 (c) (i)
(ii) Area of new triangle ABC′
= (7.8)(4.574) sin 67° = 16.42 cm2
13. (a) 118.5 (b) (i) 147.1 (ii) RM10.71 (c) n = 128 (d) RM52
14. (a) 20 cm s–1
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(b) 7 cm s–2
(c) (d) 72 cm
15. (a) I : x + y ≤ 100 II : x ≤ 3y III : y – x ≤ 30 (b)
(c) (i) 75 tiles (ii) RM528
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Rumusan Kertas SEKOLAH SMK PUTRAJAYA PRESINT 14(1)
CIKGU Ng Seng Chew
SUBJEK SPM Matematik Tambahan Tingkatan 4,5
NAMA KERTAS PENTAKSIRAN Untitled (Kertas 1)
Tingkatan 4 Tingkatan 5
56% 44%
14 Soalan 11 Soalan
Mudah Sederhana Sukar Tahun Lepas Sendiri
4% 80% 4% 12% -
1 Soalan 20 Soalan 1 Soalan 3 Soalan -
Indeks Kesukaran
2.12
SPM Matematik Tambahan Tingkatan 4,5 1 → (25 Soalan Diperlukan)
Jumlah Soalan: 25
Topik Mudah Sederhana Sukar Tahun Lepas Sendiri
Form 4 - 4.01 Functions 1 2 - - -
Form 4 - 4.02 Quadratic Equations - 2 - - -
Form 4 - 4.03 Quadratic Functions - - 1 - -
Form 4 - 4.05 Indices and Logarithms - 1 - 1 -
Form 4 - 4.06 Coordinate Geometry - 2 - - -
Form 4 - 4.07 Statistics - 1 - - -
Form 4 - 4.08 Circular Measures - 1 - - -
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Form 4 - 4.09 Differentiation - 2 - - -
Form 5 - 5.01 Progressions - 1 - 2 -
Form 5 - 5.02 Linear Law - 1 - - -
Form 5 - 5.03 Integration - 2 - - -
Form 5 - 5.04 Vectors - 1 - - -
Form 5 - 5.05 Trigonometric Functions - 1 - - -
Form 5 - 5.06 Permutations and Combinations - 1 - - -
Form 5 - 5.07 Probability - 1 - - -
Form 5 - 5.08 Probability Distributions - 1 - - -
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Rumusan Kertas SEKOLAH SMK PUTRAJAYA PRESINT 14(1)
CIKGU Ng Seng Chew
SUBJEK SPM Matematik Tambahan Tingkatan 4,5
NAMA KERTAS PENTAKSIRAN Pra SPM Add Math Kertas 2 2013 (Kertas 2)
Tingkatan 4 Tingkatan 5
40% 60%
6 Soalan 9 Soalan
Mudah Sederhana Sukar Tahun Lepas Sendiri
- 26.67% 53.33% 20% -
- 4 Soalan 8 Soalan 3 Soalan -
Indeks Kesukaran
2.73
SPM Matematik Tambahan Tingkatan 4,5 2 → (15 Soalan Diperlukan)
Jumlah Soalan: 15
Topik Mudah Sederhana Sukar Tahun Lepas Sendiri
Section A → 6 Soalan Diperlukan
Form 4 - 4.04 Simultaneous Equations - 1 - - -
Form 4 - 4.07 Statistics - 1 - - -
Form 5 - 5.01 Progressions - - - 1 -
Form 5 - 5.03 Integration - - 1 - -
Form 5 - 5.04 Vectors - - - 1 -
Form 5 - 5.05 Trigonometric Functions - - 1 - -
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Section B → 5 Soalan Diperlukan
Form 4 - 4.06 Coordinate Geometry - - 1 - -
Form 4 - 4.08 Circular Measures - - 1 - -
Form 5 - 5.02 Linear Law - - 1 - -
Form 5 - 5.03 Integration - - - 1 -
Form 5 - 5.08 Probability Distributions - - 1 - -
Section C → 4 Soalan Diperlukan
Form 4 - 4.10 Solution of Triangles - - 1 - -
Form 4 - 4.11 Index Numbers - 1 - - -
Form 5 - 5.09 Motion Along a Straight Line - 1 - - -
Form 5 - 5.10 Linear Programming - - 1 - -
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