![SULIT 3 3472/1
Question Solution/ Marking Scheme Answer Marks
8 117 27
2
B2: − 4 3
9 9
31 − 4
B1: or 3 or 133− 4 2 or 117
9
9 1 5 7
B3: + 4
3 6 6
log p p log q q 1 1
B2: + or +
log p 2 log q 2 log p 2 log q 2
B1: log 2 pq = log 2 p + log 2 q
10 (a) -2 1
10
(b) B2: S10 = [2(−6) + 9(3)] (b) 75 3
2
B1: a = −6 and d = 3
11 (a) B2: r 3 = 27 (a) 3 3
B1: ar = 6 or ar 4 = 162 (b) 2 1
12 3
B2: S ∞ = 8
1 − (− 1 )
2
1
4 3
1
B1: r = −
2
13 3 x + 11 8 + 3y
B2 : = 2 and = 17 ( −1, 20) 3
4 4
3 x + 11 8 + 3y
B1 : = 2 or = 17
4 4
14 B2 : p = −1.2 or q = 6.3 p = -1.2
B1 : 3
q = 6.3
5.1 − 1.5
p= or 5.1 = q − 1.2 or 1.5 = q − 1.2(4)
1− 4
B2: h=7 or k=8
h=7
15 3
B1: BA = hi − (k − 3) j k=8
3472/1 Additional Mathematics Paper 1
SULIT](https://image.slidesharecdn.com/addmaths1-100920222120-phpapp01/85/Add-Maths-1-3-320.jpg)
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- 1. Nama Pelajar : ………………………………… Tingkatan 5 : ……………………. 3472/1 Additional Mathematics Paper1 September 2010 PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA SEKOLAH MENENGAH NEGERI KEDAH DARUL AMAN PEPERIKSAAN PERCUBAAN SPM 2010 ADDITIONAL MATHEMATICS MARKING SCHEME Paper 1 .
- 2. SULIT 2 3472/1 SPM Trial Examination 2010 Kedah Darul Aman Marking Scheme Additional Mathematics Paper 1 Question Solution/ Marking Scheme Answer Marks 1 (a) 8 1 (b) 0 ≤ f ( x) ≤ 8 1 2 2+n (a) 13 2 (a) B1: =5 3 (b) 3 x − 13 1 3 (a) B1: 4 g ( x) + 2 = 5 x − 2 5x − 4 (a) 2 4 5(4 x + 2) − 4 1 (b) B1: =4 (b) 2 4 2 4 2 1 k B2: 2 − = 2 3 9 9 9 3 k B1 : α + 2α = − or α × 2α = 9 9 5 (a) 3 1 (b) 11 1 (c) -7 1 6 B2: 6 6 −2≤ x ≤ -2 5 3 5 B1: (5 x − 6)( x + 2) 7 B2: x + 3 = −4(2 x + 3) 5 x= − 3 B1 : 2 x + 3 or 2 −4( 2 x + 3) 3 3472/1 Additional Mathematics Paper 1 SULIT
- 3. SULIT 3 3472/1 Question Solution/ Marking Scheme Answer Marks 8 117 27 2 B2: − 4 3 9 9 31 − 4 B1: or 3 or 133− 4 2 or 117 9 9 1 5 7 B3: + 4 3 6 6 log p p log q q 1 1 B2: + or + log p 2 log q 2 log p 2 log q 2 B1: log 2 pq = log 2 p + log 2 q 10 (a) -2 1 10 (b) B2: S10 = [2(−6) + 9(3)] (b) 75 3 2 B1: a = −6 and d = 3 11 (a) B2: r 3 = 27 (a) 3 3 B1: ar = 6 or ar 4 = 162 (b) 2 1 12 3 B2: S ∞ = 8 1 − (− 1 ) 2 1 4 3 1 B1: r = − 2 13 3 x + 11 8 + 3y B2 : = 2 and = 17 ( −1, 20) 3 4 4 3 x + 11 8 + 3y B1 : = 2 or = 17 4 4 14 B2 : p = −1.2 or q = 6.3 p = -1.2 B1 : 3 q = 6.3 5.1 − 1.5 p= or 5.1 = q − 1.2 or 1.5 = q − 1.2(4) 1− 4 B2: h=7 or k=8 h=7 15 3 B1: BA = hi − (k − 3) j k=8 3472/1 Additional Mathematics Paper 1 SULIT
- 4. SULIT 4 3472/1 Question Solution/ Marking Scheme Answer Marks 16 3 5h B2: = 2 6h − 8 3 3 3 5h B1: = k 2 6h − 8 or AB // AC 17 B2 : 75 0 or 195 0 75 0 ,1950 3 B1 : x − 45 0 = 30 0 or 150 0 18 1 2 1 2 B2 : (3) (2) = (4) θ 1.125 rad 3 2 2 1 2 1 2 B1: (3) (2) or (4) θ 2 2 19 1 1 (a) B1: m = − (a) y = − x+8 2 2 2 1 (b) B1: 0 = − x + 8 (b) (16, 0) 2 2 20 B3: − 6(0 + 1)(4 − 0) + 2(4 − 0) 2 8 4 B2: 2(2 x + 1)(4 − 3x)(−3) + 2(4 − 3 x) 2 B1: 2(2 x + 1)(4 − 3x)(−3) or 2(4 − 3x) 2 21 (2) 2 8(2) −1 x2 8 B2 : 11 = − +c y= + +5 3 2 −1 2 x x2 8 x −1 B1: or 2 −1 3472/1 Additional Mathematics Paper 1 SULIT
- 5. SULIT 5 3472/1 Question Solution/ Marking Scheme Answer Mark 22 B2 : δy = −24(2) −3 × p -3p 3 dy B1 : = −24 x − 3 or δx = p dx 23 B1: 10C3 (0.15) 3 (0.85) 7 0.1298 2 24 (a) 35 1 (b) B1: 7 C3 − 5C1 or 2 ×5 C 2 + 5C3 (b) 30 2 25 100 − 92 (a) 1.6 2 (a) B1 : 5 88 − 92 (b)0.7881 2 (b) B1 : 5 END OF MARKING SCHEME 3472/1 Additional Mathematics Paper 1 SULIT