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Solusi-latihan-un-sma-2011-matematika-IPA

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                                 Solusi Latihan Soal UN SMA / MA 2011
                                              Program IPA
                                        Mata Ujian : Matematika
                                    Jumlah Soal : 25


1.       Jawab D
          2 x 2 + x − 15 ≥ 0
          (2 x − 5)( x + 3) ≥ 0

             +               −        +

                  −3             21
                                  2


         x ≤ −3 atau x ≥ 2 1
                           2


2.       Jawab A
          2 − 5x ≥ 3
           x −2
          2 − 5x − 3 ≥ 0
           x −2
          2 − 5 x − 3x + 6 ≥ 0
                x −2
          − 8x + 8 ≥ 0
            x−2
             −           +            −

                   1              2

         1≤x<2



3.       Jawab D
         Eliminasi (1) dan (2) diperoleh
         x − 2y + z = 0
         x+y+z=6
         −3y = −6 ⇒ y = 2
         Eliminasi (2) dan (3) diperoleh
         x+y+z=6
         x − y + 2z = 5
         2y − z = 1 ⇒ 4 − z = 1 ⇒ z =3


4.       Jawab C
         Misalkan x = harga mangga
                  y = harga jeruk
                  z = harga anggur
         Maka: 2x + 2y + z = 70.000             … (1)
                 x + 2y + 2z = 90.000           … (2)
                 2x + 2y + 3z = 130.000         … (3)




     Eliminasi (3) dan (1)
                   2x + 2y + 3z = 130.000

                   2x + 2y + z = 70.000         −
                                  2z = 60.000
                                  Z = 30.000
     Eliminasi (1) dan (2)
                   2x + 2y + z = 70.000
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                           x + 2y + 2z = 90.000                      −
                     x − z = −20.000
                     x = −20.000 + z = 10.000
      Subtitusikan x dan z ke (1)
                                                                          20.000 + 2y + 30.000 = 70.000
                                                                                   y = 10.000


5.        Jawab C
          (1 + 3 2) − (4 −                     50 )
          = (1 + 3 2) − (4 − 5 2 )
          = 1+3 2 −4+5 2
          = 8 2 −3


6.        Jawab C
                    −1        −1          3
               (a    3   ⋅b    2   ⋅ c)
                                                           = 1 ⋅ 1 ⋅6
                                                                      3
                  −1       −1            3                   3 23
           =    (a 3     ⋅b 2       ⋅ c) 2
                                                           = 1 ⋅ (6)
                                                                     3
                −1      −3          3
                                                             3 2
          =    a 2    ⋅b 4       ⋅ c2
                                                           = 1 ⋅3
                                                                  3
                 −1           −3          3                  3
          = 9     2    ⋅ 16    4   ⋅ 36 2                       2
                     1             3                 3
                                                           = 3
                 2 −           4 −                2        =9
          =    (3 ) 2      ⋅ (2 ) 4           ⋅ (6 ) 2
                 −1        −3        3
          = 3         ⋅2        ⋅6


7.        Jawab A
                 2 + 5 x −3
          32 x                  = 27 2 x + 3
               2 x 2 +5x −3
           3                       = (3 3 ) 2 x + 3
          2x2 + 5x − 3 = 6x + 9
          2x2 − x − 12 = 0
          Akar-akar α dan β.
          αβ = c = −6
                a


8.        Jawab E
          AB + AC + BC = 8
          x+x+x                 2 =8
          (2 +           2 )x=8

          x =         8
                    2+ 2

                      8                  2−     2
                =
                    2+ 2                 2−     2
                      8 (2 −         2)
                =
                           2
                = (8 − 4 2 ) cm

9.        Jawab E
          Diketahui log 2 = a dan log 3 = b,
          maka log 72 = log (23 ⋅ 32)
                      = log 23 + log32
                      = 3 log 2 + 2 log3
                      = 3 a + 2b

10.       Jawab B
                                3 log15
          5
              log15 =            3 log5

                                3 log5        + 3 log3
                          =              3 log5

                                3 log2 ⋅ 2 log5          + 3 log3
                          =              3 log2 ⋅ 2 log5
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                      x ⋅y +1
                  =     x⋅y

11.   Jawab B
      Panjang : lebar = 5 : 4
      ⇒ Panjang = 5x
      dan lebar = 4x
      luas 180 m2 ⇒ 5x ⋅ 4x = 180
      ⇒ x2 = 9
      ⇒x=3
      Diagonal =        (5 x )2 + ( 4 x )2
                  =x     41
                  = 3 41

12.   Jawab A
      x2 + (m −2)x + 9 = 0 akar-akar nyata
      D≥0
      (m − 2)2 − 4 ⋅ 1 ⋅ 9 ≥ 0
      m2 − 4m + 4 − 36 ≥ 0
      m2 − 4m − 32 ≥ 0
      (m + 4)(m − 8) ≥ 0
                               +    −      +

      m ≤ −4 atau m ≥ 8          −4   8


13.   Jawab D
      Hasil kali a k a r - a k a r 6x2 − 2x + 3 = 0
      x1 x2 = c = 3 = 1
              a   6      2

14.   Jawab D
      (k + 2) x2 − (2k − 1)x + k − 1 = 0
      akar-akar nyata dan sama
      D=0
      (−(2k − 1))2 − 4 ⋅ (k + 2) ⋅ (k − 1) = 0
      4k2 − 4k + 1 − 4 (k2 + k − 2) = 0
      − 8k + 9 = 0
      k= 9
              8
      Jumlah akar-akar =            x1 + x2
           − (2k − 1)
      =−
              k + 2
        18 − 1
                    8
      = 8         ⋅
         9 + 2      8
         8
      = 18 − 8
        9 + 16
      = 10
        25
      = 2
          5

15.   Jawab B
      x 2 − 2 x + 3 = 0 adalah x1 dan x 2
                  3             3             2   2
      Nilai x1 x 2 + x1x 2 = x1x 2 ( x1 + x 2 )
      = x1x 2 (( x1 + x 2 )2 − 2 x1x 2 )
      = 3 ⋅ (2 2 − 2 ⋅ 3)
      = −6
16.   Jawab C
      Jumlah kuadrat akar-akar persamaan 2 x 2 − (p + 1) x + 1 = 0 adalah 3
          2       2
      x1 + x 2 = 3
      ( x1 + x 2 )2 − 2 x1x 2 = 3
          −(p + 1) 2
      (           ) − 2⋅ 1 = 3
             2           2
      p2 + 2p + 1
                  −1 = 3
           4
      p2 + 2p + 1
                   =4
            4
      p2 + 2p − 15 = 0
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      (p + 5) (p − 3) = 0
      p = −5 atau p = 3
      Nilai p positif adalah p = 3



17.   Jawab C
       x 2 − x + 2 = 0 akar x1 dan x 2
      ( 2x1 − 2 ) + ( 2x 2 − 2 ) = 2(x1 + x 2 ) − 4
                                 = 2 ⋅ 1 − 4
                                 = −2
      ( 2x1 − 2 ) ⋅ ( 2x 2 − 2 )
      = 4 x1 ⋅ x 2 − 4(x1 + x 2 ) + 4
      =4⋅2−4⋅1+4
      =8
      persamaan kuadrat baru yang akar-akarnya 2x1 − 2 dan 2x 2 − 2 adalah
      x2 + 2x + 8 = 0


18.   Jawab B
      akar-akar 2x2 + 6x − 5 = 0 adalah x1 dan x2
            2     2     2(x1 + x 2 )
       Z1 = x + x =        x1 x 2    = −6 = 12
             1     2                   −5      5
                                               2

      Z 2 = x1 x2 = − 5
                      2
      Persamaan kuadrat akar-akar Z1 dan Z 2 adalah
      x2 − ( Z1 + Z 2 ) x + Z1 Z 2 = 0
      x2 − ( 12 − 5 )x + ( 12 ⋅ (− 5 )) = 0
              5 2           5      2
      x2 + 1 x − 6 = 0
             10
      10x2 + x − 60 = 0


19.   Jawab B
      x2 − (2a + 3)x + 3a = 0
      kedua akar saling berkebalikan
      x2 = 1
            x1
      x1 x2 = 1
      3a = 1
      a= 1
           3


20.   Jawab D
      3x2 − (2m − 8)x − 2 = 0 akar-akar real berlawanan tanda
      x1 = − x2
      x1 + x2 = 0
      −b =0
        a
      b=0
      −(2m − 8) = 0
      m=4
21.   Jawab E
      x2 − 2px + 3p = 0
      Kedua akar mempunyai perbandingan 1 : 3
          x
      ⇒ 2 = 1 ⇒ x2 = 3 x1
          x1     3
                                       c
      •> x1 + x2 = − b = •> x1 ⋅ x2 = a = 3p
                         2p
                     a
      x1 + 3x1 = 2p             x1 ⋅ 3x1 = 3p
      4x1 = 2p                     x12 = p
                                   1 p2 = p
      x1 = 1 p                     4
            2
                                    p=4
                                   2p = 8
22    Jawab C
      f ( x) = 2x 2 − 4 x + 1
                b          −4
      xp = −        = =−        =1
               2a          2⋅ 2
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               y p = 2 ⋅ 12 − 4 ⋅ 1 + 1 = −1
           Koordinat titik balik (1, −1)


23.        Jawab E
           Y = a(x − x1) (x − x2)
           Y = a(x − (−1)) (x − 3)
           Y = a(x + 1) (x − 3)
           Melalui (1,4), maka
           4 = a(1 + 1) (1 − 3)
           a = −1
           Subtitusikan a kepersamaan grafik
           Y = − (x + 1) (x − 3)
           Y = − (x2 − 2x − 3)
           Y = − x2 + 2x + 3


24.        Jawab B
                   sin 150o + sin 120 o
               cos 210o − cos 300o
                    sin(180o − 30o ) + sin(180o − 60o )
           =
                    cos(180o + 30o ) − cos(360o − 60o )
                     sin 30o + sin 60 o
           =
                    − cos 30 o − cos 60o
                     1 + 1         3
           =         2   2
                    −1 3
                       2
                                   −   1
                                       2
           = −1

25.        Jawab D
                                   Aturan sin:        AC = AB
                                                    sin 45   sin 60
                     C
                                                     AC =   AB sin 45
                          o                                  sin 60
                     60
                                                           12 6 ⋅ 1 2
                                                         =         2
                                                               1 3
                                                               2

               o                                o
          75                               45
      A                                             B
                              12   6

				
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