This document contains the answers to questions on a mathematics exam in Indonesia from 2006-2007. It provides the answer choices for 19 multiple choice questions on topics like algebra, geometry, trigonometry, and logic. For each question, it shows the answer choice and provides the steps taken to solve the problem.

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JAWABAN UJIAN NASIONAL 2006 / 2007
MATEMATIKA IPA P12 - A
RABU, 18 APRIL 2007
1. Jawab: C
2. Jawab: B
2
log 3 . 3log 5 = 2log 5 = ab
15
log 20 =
3. Jawab: C
x2 – 5x + 6 = 0
x 1 + x2 = - -
- y1 + y2 = (x1 – 3) + (x2 – 3)
= (x1 + x2) – 6 = 5 – 6
c 6 =–1
x1 . x2 = 6
a 1
y1 . y2 = (x1 – 3) (x2 – 3)
= x1.x2 – 3(x1 + x2) + 9
= 6 – 3(5) + 9 = 0
x2 – (y1 + y2)x + (y1 . y2) = 0  x2 – (–1)x + 0 = 0  x2 + x = 0
4. Jawab: E
Titik Puncak (1, 4)
Titik potong dengan sumbu X (–1, 0) dan (3, 0) (1, 4)
Titik potong dengan sumbu Y (0, 3)
Cara 1: Gunakan persamaan y = a (x – x1) (x – x2) (0, 3)
Titik puncak  y = 4, x = 1
Titik potong sumbu X  x1 = – 1, x2 = 3
4 = a (1 + 1) (1 – 3) = a (–4) (3, 0)
(-1, 0)
4 = – 4a
a=–1
y = – 1 (x + 1) (x - 3) = – (x2 – 2x - 3)
y = – x2 + 2x + 3
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Cara 2: Gunakan persamaan y = a (x – xp)2 + yp
Titik potong sumbu x  x = –1, y = 0
Titik puncak  xp = 1, yp = 4
0 = a (– 1 – 1)2 + 4 = a (4) + 4
0 = 4a + 4  4a = – 4
a=–1
y = –1 (x – 1)2 + 4 = – (x2 – 2x + 1) + 4 = – x2 + 2x – 1
y = –x2 + 2x + 3
5. Jawab: A
f(x) = 3x2 – 4x + 6 ; g(x) = 2x – 1 ; (f o g)(x) = 101
2
(f o g)(x) = f(g(x)) = 3(2x – 1) – 4(2x – 1) + 6
= 3(4x2 – 4x + 1) – 8x + 4 + 6
= 12x2 – 12x + 3 – 8x + 10
101 = 12x2 – 20x + 13
12x2 – 20x – 88 = 0  dibagi 4
3x2 – 5x – 22 = 0
(3x – 11) (x + 2) = 0
11 2
x1 3 ; x2 2
3 3
6. Jawab: E
32x+1 – 28.3x + 9 = 0  32x . 31 – 28 . 3x + 9 = 0
Misal: 3x = A
3A2 – 28A + 9 = 0
(3A – 1) (A – 9) = 0  A = 1/3 ; A=9
3x = 9  x1 = 2 ; 3x = 1/3  x2 = –1
3x1 – x2 = 3(2) – (–1) = 7
7. Jawab: D
(x – 2)2 + (y + 1)2 = 13  Pusat (2, –1) ; Jari-jari (r) = 13
x = –1  (–1 – 2)2 + (y + 1)2 = 13
 9 + y2 + 2y + 1 = 13
 y2 + 2y – 3 = 0
 (y + 3) (y – 1) = 0
 y = –3 ; y = 1  ada 2 titik pada lingkaran: (–1, –3) dan (–1, 1)
Gunakan rumus :(x – a) (x1 – a) + (y – b) (y1 – b) = r2
(–1, –3) : (x – 2) (–1 – 2) + (y + 1) (–3 + 1) = 13
–3 (x – 2) – 2 (y + 1) = 13
–3x + 6 – 2y – 2 = 13
3x + 2y + 9 = 0
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(–1, 1) : (x – 2) (–1 – 2) + (y + 1) (1 + 1) = 13
–3 (x – 2) + 2 (y + 1) = 13
–3x + 6 + 2y + 2 = 13
3x – 2y + 5 = 0
8. Jawab: A
f(x) : (x – 2) sisa 24  f(2) = 24
f(x) : (2x – 3) sisa 20  f(3/2) = 20
– –
- -
; a=2 ; b = 3/2 ; f(a) = 24 ; f(b) = 20
9. Jawab: E
2 2 1 x 67
2x + 2y + 1z = 67.000
3 1 1 y 61
3x + 1y + 1z = 61.000
1x + 3y + 2z = 80.000 1 3 2 z 80
2 2 1 2 2
D= 3 1 1 3 1 (4) + (2) + (9) – (1) – (6) – (12) = –4
1 3 2 1 3
67 2 1 67 2
Dx = 61 1 1 61 1 (134) + (160) + (183) – (80) – (201) – (244) = –48
80 3 2 80 3
2 67 1 2 67
Dy = 3 61 1 3 61 (244) + (67) + (240) – (61) – (160) – (402) = –72
1 80 2 1 80
2 2 67 2 2
Dz = 3 1 61 3 1 (160) + (122) + (603) – (67) – (366) – (480) = –28
1 3 80 1 3
Dx 48 Dy 72 Dz 28
x 12 ; y 18 ; z 7
D 4 D 4 D 4
Harga: x + y + 4z = (12.000) + (18.000) + 4(7.000) = 58.000
10. Jawab: C
2 1 x y 2 7 2 7 3
A ; B ; C Ct
1 4 3 y 3 1 2 1
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x y 2 2 1 7 3
B – A = Ct 
3 y 1 4 2 1
y–4=1  y=5
x + (5) – 2 = 7  x=4
x . y = (4)(5) = 20
11. Jawab: C
1x + 0 y < 0200 4x + 20y = 1760 x + (60) = 200
4x + 20y < 1760 4x + 24y = 0800 – x = 200 – 60
= 140
16y = 960
y= 60
Pendapatan maksimum: 1.000x + 2.000y = 1.000(140) + 2.000(60) = 260.000
12. Jawab: B
0 1 1 2 1 3
RP P R 1 0 1 ; RQ Q R 3 0 3
4 2 2 2 2 0
RP . RQ (1)(3) (1)(-3) (2)(0) 0
cos θ 0
RP RQ 12 12 22 . 32 (-3)2 0 6 2
Θ = 90°
13. Jawab: A
2 0 2 0 0 0
AB B A 2 0 2 ; AC C A 2 0 2
0 0 0 2 0 2
Proyeksi vektor orthogonal AB pada AC :
2 0
2 2
0 0 0
0 2 0 4 0 1
2
2 2 2 j k
2 2 8 2
0 2 2 2 2 2
14. Jawab: D
y = x2 – 3  Persamaan kuadrat  Carilah titik potong dengan sumbu X dan Y
X 0 + 3 – 3
y –3 0 0 A (0, –3) B (+ 3 , 0) C (– 3 , 0)
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Refleksi Dilatasi
sumbu x k 2
(x, y) (x, -y) (x, y) (kx, ky)
(0, -3) (0, 3) (0, 6)
(+ 3 , 0) (+ 3 , 0) (+2 3 , 0)
(– 3 , 0) (– 3 , 0) (–2 3 , 0)
y = a (x – x1) (x – x2)  6 = a (0 - 2 3 ) (0 + 2 3 )
6 = a (-2 3 ) (2 3 )
6 = – 12 a
a=–½
y = –½ (x – 2 3 ) (x + 2 3 )
= –½ (x2 – 12)
= –½x2 + 6
15. Jawab: B
U3 = a + 2b = 36 a + 2b = 36 a + 2(12) = 36
U5 + U7 = (a + 4b) + (a + 6b) a + 5b = 72 - a = 36 – 24
144 = 2a + 10b a = 12
72 = a + 5b –3b = –36
b = 12
16. Jawab: E
a = 80.000.000  sederhanakan menjadi a = 80
r=¾
U3 = a . r2 = (80) (¾)2 = (80) (9/16) = 45
17. Jawab: B
p = hari panas p q p q p r
q = ani memakai topi ~q v r q r ~r (modus tolens)
r = memakai payung ~r p r ~p
~p = hari tidak panas
18. Jawab: B F
1
ACH  titik tengah P  DP = /3 DF
Q
EGB  titik tengah Q  FQ = 1/3 DF
P
DF  Diagonal ruang D
Jarak PQ = 1 – 1/3 – 1/3 = 1/3 . 6 3 = 2c
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19. Jawab: -
BG = a 2 ; BDHF  diwakili oleh garis BH ; BH = a 3
B
HG2 = BH2 + BG2 – 2(BH) (BG) cos B
a2 = (a 3 )2 + (a 2 )2 – 2(a 3 ) (a 2 ) cos B
a 2 a 3
a2 = 3a2 + 2a2 - 2 6 a2 . cos B
5a2 a2 4a2 1
G H cos B = 6  B = 35,3
a 2 6a 2
2 6a 2 3
20. Jawab: A
C = 45 ; a=p ; b= 2 2p
c2 = a2 + b2 – 2ab cos C = (p)2 + (2 2 p)2 – 2 (p) (2 2 p) cos 45
= p2 + 8p2 – 4 2 p2 (½ 2 )
= 9p2 – 4p2
= 5p2
c= 5p
21. Jawab: C
cos 40 + (cos 80 + cos 160) = cos 40 + [2 cos ½ (80 + 160) . cos ½ (80 – 160)
= cos 40 + [2 cos 120 . cos 40 ]
= cos 40 (1 + 2 cos 120)
= cos 40 (1 + 2 (-½))
= 0
22. Jawab: A
A = x2 – x – 6  A’ = 2x – 1
B=4- 5x 1  B’ = - ½ . 5 (5x + 1)-½ = - 5/2 (5x + 1) -½
2(3) 1 5 . 2 16
8
5 5
2 5(3) 1
23. Jawab: E
1 (1 2 sin2 x) 2 . sin x . sin x
Limit Limit 4
x 0 1 x 0 1
x . tan x x . tan x
2 2
24. Jawab: C
f' (x) 2 sin 2x 2 cos 2x 4 sin 2x cos 2x
6 6 6 6
1 1
f' (0) 4 sin 2(0) cos 2(0) 4. . 3 3
6 6 2 2
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25. Jawab: D
27 + 9 + 3 – a3 – a2 – a = 25  a3 + a2 + a – 14 = 0
Gunakan horner:
2 1 1 1 -14 (a – 2) (a2 + 3a + 7) = 0
+ + + a=2
2 6 14 ½a=1
1 3 7 0
26. Jawab: B
Short-cut agar luas persegi panjang maksimum:
- panjang = ½ x = ½ (4) = 2
- lebar = ½ y = ½ (5) = 2½
- luas maksimum = ¼ xy = ¼ (4)(5) = 5
27. Jawab: C
y = x2 ; x+y=6  y=6–x
ykurva = ygaris  x2 = 6 – x  x2 + x – 6 = 0  (x + 3)(x – 2) = 0  x = -3, x = 2
- - -
- - - - - - - -
28. Jawab: D
y = -x2 + 4 ; y = -2x + 1
ykurva = ygaris  -x2 + 4 = -2x + 1  x2 – 2x – 3 = 0
 (x – 3) (x + 1) = 0
 x = 3 , x = -1
– -
– - – - - -
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29. Jawab: E
P(A) = 3/8 ; P(B) = 6/10
30. Jawab: D
Fmod = 14 ; L = 48,5 ; c=6 ; f k = 4 + 6 + 9 = 19 ; n = 50
– –
Mod =
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