# 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
all right reserved

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
all right reserved
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
all right reserved
(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
all right reserved

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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