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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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