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                          JAWABAN UJIAN NASIONAL 2006 / 2007
                               ...
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     Cara 2: Gunakan persamaan y = a (x – xp)2 + yp
              Titik potong sumbu x      ...
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      (–1, 1) :       (x – 2) (–1 – 2) + (y + 1) (1 + 1) = 13
                      –3 (x – ...
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                                          x       y 2                  2       1        7 3
...
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                                  Refleksi                              Dilatasi
           ...
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19.   Jawab: -

      BG = a 2          ;    BDHF  diwakili oleh garis BH                  ...
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25.   Jawab: D




      27 + 9 + 3 – a3 – a2 – a = 25                                     ...
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29.   Jawab: E
      P(A) = 3/8       ;    P(B) = 6/10




30.   Jawab: D
      Fmod = 14   ...
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JAWAB UAN IPA 2006/2007 P12

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JAWAB UAN IPA 2006/2007 P12

  1. 1. www.aidianet.blogspot.com 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 © Aidia Propitious 1
  2. 2. www.aidianet.blogspot.com 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 © Aidia Propitious 2
  3. 3. www.aidianet.blogspot.com (–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 © Aidia Propitious 3
  4. 4. www.aidianet.blogspot.com 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) © Aidia Propitious 4
  5. 5. www.aidianet.blogspot.com 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 © Aidia Propitious 5
  6. 6. www.aidianet.blogspot.com 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 © Aidia Propitious 6
  7. 7. www.aidianet.blogspot.com 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 – - – - – - - - © Aidia Propitious 7
  8. 8. www.aidianet.blogspot.com 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 = Jika ditemukan kesalahan dalam pembahasan, mohon hubungi reborn4papua@yahoo.com atau 08999812979. Terima kasih. © Aidia Propitious 8

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