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Solusi Latihan Soal UN SMA / MA 2011
Program IPS
Mata Ujian : Matematika
Jumlah Soal : 25
1. Jawaban: C
020)1x(12)1x( 222
≤++−+
x4 + 2x2 + 1 – 12x2 – 12 + 20 ≤ 0
x4 – 10x2 + 9 ≤ 0
(x2 – 1)(x2 – 9) ≤ 0
(x + 1)(x – 1)(x + 3)(x – 3) ≤ 0
−1 31
+ − + −
−3
+
−3 ≤ x ≤ −1 atau 1 ≤ x ≤ 3
2. Jawaban: D
0
1xx6
3xx2
2
2
<
−+
−+
0
)1x3)(1x2(
)3x2)(1x(
<
−+
−−
−
2
1
2
3
x −<<− atau 1x3
1
<<
3. Jawaban: C
0)5a(x)1a(x2
=−−−+
x1 + x2 = −a + 1
x1x2 = −a + 5
12xxxx
2
212
2
1 =+
x1.x2(x1 + x2) = 12
(−a + 5)(−a + 1) = 12
a2
– 6a + 5 = 12
a2
– 6a − 7 = 12
(a – 7)(a + 1) = 0
a = 7
4. Jawaban: C
06x5x2
=++
(x + 3)(x + 2) = 0
x1 = −3 ; x2 = −2
α = x1 + 5 = 2
β = x2 + 6 = 4
x2
– (α + β)x + αβ = 0
x2
– 6x + 8 = 0
5. Jawaban: A
x2
– 5x + 7 = 2x – 3
x2 – 7x + 10 = 0
(x – 2)(x – 5) = 0
x = 2 atau x = 5
y = 2x – 3
x = 2 ⎯→ y = 1
x = 5 ⎯→ y = 7
Koordinat titik potong (2, 1) dan (5, 7)
2
3
− 2
1
3
1 1
+ + +− −

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. Jawaban: D
++−
selalu positif maka
)2
– 4a.(a + 4) < 0
< 0
. Jawaban: B
6
a2(ax)x(f 2
−= 4ax)4
a > 0
D < 0
(2a – 4
4a2
– 16a + 16 – 4a2
– 16a
−32a < −16
a > ½
7
510
510
.
510
510 −
2ba
−
−
+
=+
510
550210
−
+−
=
223
5
21015
−=
−
=
a = 3 b = −2
. Jawaban: C
a + b = 1
8
3
2
15 ⎞⎛
3x
1255
=⎟
⎠
⎜
⎝
−
( ) 5
1
5
23x1
=+−
( ) 12x4
55 −−
=
1x28
55 −−
=
8 – 2x = −1
½
9. Jawaban: D
=
log 3. log = ab
9 = 2x → x = 4
a3log2
= ; log3
b5
2 3
5
2
log 5 = ab
5log
5
12
log
5
12
log 2
2
5
=
5log
5log12log
2
22
−
=
5log
5loglog4log
2
222
−+
=
ab
aba2 −+
=
ab
2aab −−
−=
10. Jawa Dban:
81
x3
x x3log
=xlog
243
x
x
xlog
x3log
=
243x xlogx3log
=−
5log
3x x
x3
=
53log
3x =
53log
3logxlog =
log 3. log x = 5 l g 3
0
o
log x = 5
x = 10000

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1. Jawaban: D
CT2
= AC2
– AT2
= 3p2
– p2
= 2p2
CT =
1
60o
A BT
C
p 3
p
2p
CB
CT
60sin =°
CB
2p
2
3
=
6p
3
2
3
2p2
CB ==
12. Jawaban: B
Un – Un − 1 = Un + 1 – Un
Merupakan sifat barisan aritmatika
U1 = 2x + 1
U2 = −x + 21
U3 = 5x + 14
U2 – U1 = U3 – U2
−x + 21 – (2x + 1) = 5x + 14 – (−x + 21)
−3x + 20 = 6x – 7
−9x = −27 → x = 3
a = 2x + 1 = 7
U2 = −x + 21 = 18
b = U2 – a = 11
U3 + U5 + U7 = a + 2b + a + 4b + a + 6b
= 3a + 12b = 21 + 132 = 153
13. Jawaban: C
a, a + b, a + 2b memiliki jumlah 30
a + a + b + a + 2b = 30
3a + 3b = 30
a + b = 10 ⎯→ a = 10 – b
ketiga bilangan menjadi
10 − b, 10 − b + b, 10 − b + 2b
10 − b, 10, 10 + b
Jika bilangan ketiga ditambah 5 maka diperoleh deret geometri
10 − b, 10, 15 + b
Sehingga
10
b15
b10
10 +
=
−
100 = (15 + b)(10 – b)
100 = 150 – 15b + 10b – b2
b2
+ 5b – 50 = 0
(b + 10)(b – 5) = 0
b = −10 atau b = 5
Untuk b = −10 ketiga bilangan adalah 20, 10, 5
Untuk b = 5 ketiga bilangan adalah 5, 10, 20
Jadi, bilangan terkecil adalah 5

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4 = 8
a = 256
14. Jawaban: C
a
9
2r32
8
256
a
a
r 4
9
5
=→===
1
8
8
r
8
a 3
===
6
ar3
= 8 →
a1 + a7 = a + ar = 1 + 64 = 65
15. Jawaban: B
3
1642
...coscos =+θ+θ−θcos
3
1
r1
=
−
a
3
1
cos1
cos2
2
=
θ+
θ
3 cos2
θ = 1 + cos2θ
2 cos2
θ = 1
cos2
θ = 1
cos2
θ = ½
2
2
1
cos −=θ
π=°=θ 135
3
4
16. Ja aban: Bw
=
+−
−
→ 95
16
4
lim 2
x
2x x
92 2
x
2
0
4
−
→x
2lim
=
x
x
+
1092
lim
−=
4→x
2
−=+x
17. Jawaban: E
xxx 50 −→
xx
2sin
7tan8sinlim +
5
3
15
25
78
==
−
+
=
x
x
xx
xx
18. Jawaban: B
2
1
2
1
xx
−
⎞⎛ +=xxy += ⎟
⎠
⎜
⎝
⎟
⎠⎝⎠⎝ 22
⎞−
−
2
1
1
x⎜
⎛
+⎟
⎞
⎜
⎛ +=
2
2
1
1
1.xx
1
'y
⎟⎟
⎠
⎜
⎝
+
+
=
x2
1.
2
⎞1
⎜
⎛11
xx
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
+
=
x2
1x2
.
xx
1
2
1
⎟
⎠
⎜
=
x2
⎟
⎞
⎜
⎛ +1x2
⎝ + xx4
19. Jawaban: D
40
1500
p4
B
−+=
pp
= 5
B = 100 – 200 + 1500 = 1400
B = 4p2
– 40p + 1500
B’ = 0 ⎯→ 8p – 40 = 0 ⎯→ p

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c = 0 ⎯→
20. Jawaban: D
b
a
m −=ax + by +
⎯→
3
5
m1 =01y3x5 =+−
2
1
m2 −=⎯→3y2x =−+ 0
6
6
.
).(1 2
1
3
5
2
15
+
21
21
m.m1
mm
tan
+
−
13
56
310
=
−
+
=3
−+
==α
21. Jawaban:
⎞⎛ + a1a
⎟⎟
⎠
⎜⎜
⎝
=
a0
A
−
10
b1
AA 1
a = 1 1 + = −b
b
22.
B
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=−
10
b1
A 1
( ) ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −
==
−1
a
1 + 1 = −
b = −2
Jawaban: B
A
I 6 4 18
II 4 8 18
X y
6x + 4y ≤ 18 ⎯→ 3x + 2y ≤ 9
4x + 8y ≤ 18 ⎯→ 2x + 4y ≤ 9
x ≥ 0, y ≥ 0
23. Jawaban: B
10 lampu (3 cacat, 7 baik)
dipilih 3 lampu (1 cacat, 2 baik)
banyaknya cara memilih 3 lampu dari 10 :
Banyaknya cara memilih 2 lampu baik dari 7 lampu baik :
Banyaknya cara memilih 1 lampu cacat dari 3 lampu cacat :
1 lampu cacat =
120C10
3 =
21C7
2 =
3C3
1 =
40
21
120
21.3
=Peluang terpilihnya
24. Jawaban: A
……. (1)
2x + 4y – 3z = 1 ………(2) × 3
3x + 6y – 5z = 1 = 0 ….. (3 2
6x + 12y – 9z = 3
6x + 12y – 10z = 0
x + y + 2z = 9 …
) ×
z = 3
pers (1) dikurangi z ma
z = 9 −z
= 9 − 3 = 6
b + c = 6
25. Jawaban: E
lai f x d Fd
Jika ka
x + y +
a +
Ni
21 – 30 2 25,5 −20 −40
31 – 40 4 35,5 −10 −40
41 – 50 4 45 0 0,5
51 – 60 2 55 10 20,5
61 – 70 4 65,5 20 80
16 20
xs = 45,5
d = x – x3
75,46
16
20
5,45
f
fd
xs =+=+=
∑
∑x