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Solusi Latihan Soal UN SMA / MA 2011
Program IPA
Mata Ujian : Matematika
Jumlah Soal : 40
1. Jawab: E
~p ∨ q salah maka
~ p = S jadi p = B
q = S
(A) p ⇒ q = B ⇒ S = S
(B) p ⇔ q = B ⇔ S = S
(C) p ∧ q = B ∧ S = S
(D) ~q ⇒ ~p = B ⇒ S = S
(E) p ∨ q = B ∨ S = B
2. Jawab: E
1033 x11x
=+ −+
10
3
3
3.3 x
1
1x
=+
10
p
3
p3 =+
3p2 – 10p + 3 = 0
(p – 3)(3p – 1) = 0
p = 3 p = 1/3
3x = 31 3x = 3−1
x = 1 atau x = −1
3. Jawab : C
2log2)3x2log()1xlog(2 ++≤−
4log)3x2log()1xlog( 2
++≤−
)12x8log()1x2xlog( 2
+≤+−
x2 – 2x + 1 ≤ 8x + 12
x2 – 10 – 11 ≤ 0
(x – 11)(x + 1) ≤ 0
–1 11
– ++
syarat :
x – 1 > 0 2x + 3 > 0
x > 1 2
3
x −>
–1 11
2
3
− 1
1 < x ≤ 11
4. Jawab : D
f(x) = 3x + 1
4x
3x2
)x(g
+
−
=
2x
3x4
)x(g 1
−
−−
=→ −
( ) )x(fg)x(h 1−
= ο
1
2x
3x4
3)x)(gf()x(h 11
+⎟
⎠
⎞
⎜
⎝
⎛
−
−−
== −−
ο

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2x
11x11
2x
2x
2x
9x12
−
−−
=
−
−
+
−
−−
=
x2
11x11
−
+
=
5. Jawab : A
( )
4x2
6x
)x(gf
−
+
=ο
g(x) = 2 – x → g–1(x) = 2 – x
)x)(ggf()x(f 1−
= οο
x2
8x
x2
8x
4)x2(2
6x2 −
=
−
+−
=
−−
+−
=
6. Jawab : B
Puncak (2,18) → y – 18 = a(x – 2)2
Melalui (5,0) → 0 – 18 = a(5 – 2)2
–18 = 9a → a = –2
y – 18 = –2(x – 2)2
= –2(x2 – 4x + 4)
y = –2x2 + 8x + 10
7. Jawab : B
5x)4p(x)1p()x(f 2
+−+−=
a2
b
x −=
)1p(2
)4p(
1p
−
−−
=−
2(p2 – 2p + 1) = –p + 4
2p2 – 3p – 2 = 0
(p – 2)(2p + 1) = 0
p = 2 p = –1/2
8. Jawab : A
x + 3y = 10 → m1 = 3
1
− ; m2 = 3
40)2y()5x( 22
=++−
2
m1R)5x(m2y +±−=+
9140)5x(32y +±−=+
2015x32y ±−=+
3x – y – 17 ± 20
3x – y + 3 = 0 atau 3x – y – 37 = 0
9. Jawab : E
01y6x8yx 22
=++−+
pusat ( 2
1
2
1
,A −− )= (4,–3)
a = 4 b = –3
5a + 2b = 20 – 6 = 14
10. Jawab : D
P(x) : (x2 –9) sisa 5x + 4
P(x) = (x2 – 9) h(x) + 5x + 4 ⎯→P(3) = 19
P(x) : (x – 5) sisa = 7 ⎯→ P(5) = 7

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P(x) : (x2 – 8x + 15) sisa ax + b
P(x) = (x2 – 8x + 15)k(x) + ax + b
P(3) = 3a + b = 19
P(5) = 5a + b = 7
2a = –6 ⎯→ a = –3 ; b = 22
sisa = –3x +22
11. Jawab : D
010x6x7kxx 234
=−+−+
x1 = 1 maka
1 + k – 7 + 6 – 10 = 0 ⎯→ k = 10
x1 + x2 + x3 + x4 = –b/a
1 + x2 + x3 + x4 = –k
x2 + x3 + x4 = –k – 1 = – 11
12. Jawab : B
x + 2y = 16
x + y = 12 _
y = 4 dan x = 8
f(x, y) = 4x + 10y
f(0, 8) = 80
f(12, 0) = 48
f(8, 4) = 32 + 40 = 72
fmaks = 80
13. Jawab : C
2x + 3y + 4z = 20
x – y + 2z = 5
4x + 5y + z = 17
7x + 7y + 7z = 42
x + y + z = 6
14. Jawab : E
04x2x2
=−−
x1 + x2 = 2 x1.x2 = –4
( ) )xx(x.x3xxxx 2121
3
21
3
2
3
1 +−+=+
12 16
12
x
y
(8, 4)
8

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= 23 – 3(– 4).2 = 32
15. Jawab : E
03mx8x2
=++−
D = 0
(–8)2 – 4.1.(m + 3 ) = 0
64 – 4m – 12 = 0
52 = 4m ⎯→ m = 13
16. Jawab : B
=
+−−+
∞→ 5
7x8x43x2
x
lim 2
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−−++
∞→
= 7x8x49x12x4
x
lim
5
1 22
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ +
=
42
812
5
1
=1
17. Jawab :
)x2cos1(x
x8cosx2sinx2sin
0x
lim
−
−
→ )x2cos1(x
)x8cos1(x2sin
0x
lim
−
−
→
= 2
2
1
2
2
1
)x(.x
)x8(.x2
0x
lim
→
= =32
18. Jawab : D
y = x4 + 6 y = 22
x4 + 6 = 22
x4 = 16 maka x = ±2
x = 2 ⎯→ m = 4x3 = 32
y – 22 = 32(x – 2)
y = 32x – 42
x = –2 ⎯→ m = –4x3 = –32
y – 22 = –32(x + 2)
y = –32x – 42
19. Jawab : C
t
t
t
x
x
t
x2
xt
xt
xt
xt
L = x2 + 4xt = 300
4xt = 300 – x2
t = 75/x – x/4
V = x2.t = 75x – ¼ x3
V’ = 0
75 – ¾ x2 = 0 maka x2 = 100 ⎯→x = 10
20. Jawab :
Misal y = x2 + 3x + 5
3x2
dx
dy
+= maka
3x2
dy
dx
+
=
∫ +++ dx)5x3xsin()9x6( 2
∫ +
+=
3x2
dy
ysin)3x2(3
∫= dyysin3 = – 3cos y + c
= –3 cos (x2 + 3x + 5) + c

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21. Jawab :
28dx)6x4(
a
1
=+∫
a
1
2
x6x2 + = 28
2a2 + 6a – (2 + 6) = 28
a2 + 3a – 18 = 0
(a + 6)(a – 3) = 0
a = – 6 atau a = 3
22. Jawab : D
x2 – 8x + 16 = 9x
x2 – 17x + 16 = 0
(x – 1)(x – 16) = 0
x = 1 atau x = 16
y = 9x y = x2
– 8x + 10
x
y
0 41
L1 = = 4, 5dxx9
1
0
∫
L2 = = 9dx)16x8x(
4
1
2
∫ +−
L = L1 + L2 = 13,5
23. Jawab : C
Berat
Badan
Frekuensi
50 – 52
53 – 55
56 – 58
59 – 61
62 – 64
4
5
3 fk =
14
2
6 (kelas
Q3)
n = 20
i
f
fn
tQ k4
3
b3 ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+= 623
6
1415
5,61 =⎟
⎠
⎞
⎜
⎝
⎛ −
+=
24. Jawab : D
Banyaknya susunan A, B, C, dan D berdampingan adalah 4! = 24
Jika A dan B diharuskan berdampinngan maka banyaknya susunan adalah
• A dan B di kiri
AB CD = 2! 2! = 4
• A dan B di tengah
C AB D = 2! 2! = 4
• A dan B di kanan
CD AB = 2! 2! = 4

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Peluang A dan B berdampingan adalah
2
1
24
444
P =
++
=
25. Jawab : E
Merah = 4 putih = 5
Peluang terambilnya kedua bola merah
6
1
8
3
9
4
=×=
26. Jawab : C
CB.At
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
515
10
1y
1x
50
23
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −+
515
10
5y5
1y2x3
5y = –15 maka y = –3
3x + 2y = 0
3x – 6 = 0 maka x = 2
Jadi 2x + y = 4 – 3 = 1
27. Jawab : D
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
=
23
14
A
1T
AdetAdetk −
=
A
1
Ak =
25
1
)38(
1
A
1
k 22
=
+−
==
28. Jawab : C
kˆ4jˆiˆba +−=+
ρρ
maka 18ba =+
ρρ
14ba =−
ρρ
b.a2baba
222 ρρρρρρ
++=+
b.a2baba
222 ρρρρρρ
−+=−
b.a4baba
22 ρρρρρρ
=−−+
18 – 14 = b.a4
ρρ
b.a
ρρ
= 1
29. Jawab : D
kˆ4jˆ3iˆ2m +−=
ρ
kˆjˆ2iˆn −+−=
ρ
n
n
n.m
p 2
ρ
ρ
ρρ
= n
141
462 ρ
++
−−−
= n2
ρ
−= kˆ2jˆ4iˆ2 +−=
30. Jawab : C
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
y
x
2
3
6
5
'y
'x
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−
+
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
y4
x2
'y
'x
x’ = 2 + x ⎯→ x = x’ – 2
y’ = –4 + y ⎯→ y = y’ + 4
Garis semula 7y4x5 =+
Bayangannya :
5(x’ – 2) + 4(y’ + 4) = 7
5x + 4y = 1

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31. Jawab : D
Misal A(a, b) .
Karena refleksi terhadap x = –2 dan y = 5 maka koordinat A menjadi
A’(2x–a, 2y – b) = ( –4–a, 10 – b)
Sstelah itu koordinat itu diputar 270o terhadap O sehingga menjadi (8, –1)
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
− b10
a4
01
10
1
8
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
−
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
− a4
b10
1
8
8 = 10 – b maka b = 2
–1 = 4 +a maka a = –5
Koordinat A(–5, 2)
32. Jawab : E
189S6 = dan r = 2
189
1r
)1r(a 6
=
−
−
189
1
63.a
= ⎯→ a = 3
U3.U4 = ar2.ar3 =a2r5 = 9.32 = 288
33. Jawab : C
S10 = 1.400.000 S15 = 2.850.000
5(2a+9b) = 1.400.000 2
15
(2a + 14 b) = 2.850.000
2a + 9b = 280.000 2a + 14b = 380.000
Dari kedua persamaan diperoleh b = 20.000
U1 = a = 50.000
34. Jawab : A
60S =∞ 20s genap =∞
402060S ganjil =−=∞
2
1
S
S
r
ganjil
genap
==
∞
∞
60
r1
a
=
−
maka a = 60(1 – r) = 60. ½ = 30
U2 = ar = 15
35. Jawab : B
75o
60o
45o
A
C
B
oo
60sin
AB
45sin
BC
=
6
3
2
60sin
45sin
AB
BC
3
1
o
o
===
36. Jawab : D
4
3
Atan = dan 12
5
Btan =

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Btan.Atan1
BtanAtan
)BAtan(
−
+
=+ 48
48
48
15
12
5
4
3
.
1−
+
=
33
56
1548
2036
=
−
+
=
37. Jawab : B
sin A + sin B = 2 sin ½ (A + B) cos ½ (A – B)
sin 40o + sin 20o = 2 sin 30o cos 10o
= 2.½.cos 10o = cos 10o
38. Jawab : E
2
1
x2cos =
cos 2x = cos 60o
→ 2x = 60o + n.360o
x = 30o + n.180o = 30o
→ 2x = –60o + n.360o
x = –30o + n.180o = 150o
Hp { 30o, 150o}
39. Jawab : D
T
D C
B
A
E
F
α
10
2 2
AD = AB = 22 maka AE = 2
22
AETATE −= 22210 =−=
EF = AB = 22
TF = TE = 22
Karena TE = EF = TF maka ΔTEF sama sisi
Sudut antara TAD dengan ABCD = α = 60o
40. Jawab : E
A B
CD
E F
G
Q
P
H
R
S

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Rusuk = 6
R = titik berat segitiga ABC sehingga
32BDBSBR 3
1
3
1
===
112BFBRFR 22
=+=