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© Aidia Propitious 1
JAWABAN UJIAN NASIONAL 2006 / 2007
MATEMATIKA IPA P12 - A
RABU, 18 APRIL 2007
1. Jawab: C
2. Jawab: B
2log 3 . 3log 5 = 2log 5 = ab
15log 20 =
3. Jawab: C
x2 – 5x + 6 = 0
x1 + x2 = - --
x1 . x2 = 61
6
a
c
y1 + y2 = (x1 – 3) + (x2 – 3)
= (x1 + x2) – 6 = 5 – 6
= – 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)
Titik potong dengan sumbu Y (0, 3) Cara 1: Gunakan persamaan y = a (x – x1) (x – x2)
Titik puncak y = 4, x = 1
Titik potong sumbu X x1 = – 1, x2 = 3
4 = a (1 + 1) (1 – 3) = a (–4)
4 = – 4a
a = – 1
y = – 1 (x + 1) (x - 3) = – (x2 – 2x - 3)
y = – x2 + 2x + 3
(1, 4)
(0, 3)
(-1, 0) (3, 0)
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© Aidia Propitious 2
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
(f o g)(x) = f(g(x)) = 3(2x – 1)2 – 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
2 x ; 3
23
3
11x 21
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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© Aidia Propitious 3
(–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
2x + 2y + 1z = 67.000 3x + 1y + 1z = 61.000 1x + 3y + 2z = 80.000
D =
3
1
2
1
3
2
231
113
122
(4) + (2) + (9) – (1) – (6) – (12) = –4
Dx =
3
1
2
80
61
67
2380
1161
1267
(134) + (160) + (183) – (80) – (201) – (244) = –48
Dy =
80
61
67
1
3
2
2801
1613
1672
(244) + (67) + (240) – (61) – (160) – (402) = –72
Dz =
3
1
2
1
3
2
8031
6113
6722
(160) + (122) + (603) – (67) – (366) – (480) = –28
74
28
D
D z ; 18
4
72
D
Dy ; 12
4
48
D
Dx zyx
Harga: x + y + 4z = (12.000) + (18.000) + 4(7.000) = 58.000
10. Jawab: C
12
37 C
13
27 C ;
y3
2y x B ;
41
12A t
80
61
67
z
y
x
231
113
122
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© Aidia Propitious 4
B – A = Ct 12
37
41
12
y3
2y x
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
Pendapatan maksimum: 1.000x + 2.000y = 1.000(140) + 2.000(60) = 260.000
12. Jawab: B
0
3
3
2
0
1
2
3
2
RQRQ ;
2
1
1
2
0
1
4
1
0
RPRP
026
0
0 (-3)3 . 211
(2)(0)(1)(-3) (1)(3)
RQ RP
RQ . RPθcos
22222
Θ = 90°
13. Jawab: A
2
2
0
0
0
0
2
2
0
ACAC ;
0
2
2
0
0
0
0
2
2
ABAB
Proyeksi vektor orthogonal AB pada AC :
k j
2
2
0
2
1
2
2
0
8
040
2
2
0
220
2
2
0
0
2
2
222
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)
4x + 20y = 1760 4x + 24y = 0800 –
16y = 960 y = 60
x + (60) = 200 x = 200 – 60 = 140
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© Aidia Propitious 5
Refleksi
(x, y) xsumbu (x, -y)
Dilatasi
(x, y) 2k (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 U5 + U7 = (a + 4b) + (a + 6b) 144 = 2a + 10b 72 = a + 5b
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 q = ani memakai topi r = memakai payung ~p = hari tidak panas
18. Jawab: B
ACH titik tengah P DP = 1/3 DF EGB titik tengah Q FQ = 1/3 DF DF Diagonal ruang
Jarak PQ = 1 – 1/3 – 1/3 = 1/3 . 6 3 = 2c
a + 2b = 36 a + 5b = 72 -
–3b = –36 b = 12
a + 2(12) = 36 a = 36 – 24 a = 12
p q p q p r ~q v r q r ~r (modus tolens) ~r p r ~p
D
P
Q
F
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© Aidia Propitious 6
19. Jawab: -
BG = a 2 ; BDHF diwakili oleh garis BH ; BH = a 3
HG2 = BH2 + BG2 – 2(BH) (BG) cos B
a2 = (a 3 )2 + (a 2 )2 – 2(a 3 ) (a 2 ) cos B
a2 = 3a2 + 2a2 - 2 6 a2 . cos B
cos B = 63
1
a62
a4
a62
aa52
2
2
22
B = 35,3
20. Jawab: A
C = 45 ; a = p ; b = 2 2 p
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 = 5 p
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 - 1x5 B’ = - ½ . 5 (5x + 1)-½ = - 5/2 (5x + 1) -½
85
1625
1352
5
132 .
)(
)(
23. Jawab: E
4
x2
1 tan . x
x sin . xsin . 2 Limit
x2
1tan . x
)xsin21(1 Limit
0x
2
0x
24. Jawab: C
6 2x cos
6 2x sin 4
6 2x cos 2
6 2x sin 2)x('f
332
1 .
2
1 . 4
6 2(0) cos
6 2(0) sin 4)0('f
B
H G a
a 3 a 2
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© Aidia Propitious 7
25. Jawab: D
27 + 9 + 3 – a3 – a2 – a = 25 a3 + a2 + a – 14 = 0
Gunakan horner:
2 1 1 1 -14
+ + +
2 6 14
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
–-
– - – - - -
(a – 2) (a2 + 3a + 7) = 0
a = 2 ½ a = 1
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© Aidia Propitious 8
29. Jawab: E
P(A) = 3/8 ; P(B) = 6/10
30. Jawab: D
Fmod = 14 ; L = 48,5 ; c = 6 ; fk = 4 + 6 + 9 = 19 ; n = 50
Mod = – –
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