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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
