Kuncisoal mtk-un-smk-prwsta

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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 3 1+ − + −−3

+

−3 ≤ x ≤ −1 atau 1 ≤ x ≤ 3

2. Jawaban: D

01xx63xx2

2

2

<−+−+

0)1x3)(1x2()3x2)(1x(<

−+−−

− 21

23 x −<<− atau 1x3

1 <<

3. Jawaban: C 0)5a(x)1a(x 2 =−−−+

x1 + x2 = −a + 1 x1x2 = −a + 5

12xxxx 2212

21 =+

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

06x5x 2 =++ (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)

23 − 2

1 31 1

+ + + − −


. 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 – 44a2 – 16a + 16 – 4a2 – 16a−32a < −16 a > ½

7

510510.

510510 −2ba

−

−

+=+

510

550210−

+−=

2235

21015−=

−=

a = 3 b = −2

. Jawaban: C

a + b = 1

8

3

2 15 ⎞⎛3x 1255

=⎟⎠

⎜⎝ −

( )515 23x1 =+−

( ) 12x4 55 −− = 1x28 55 −− =

8 – 2x = −1½

9. Jawaban: D =

log 3. log = ab

9 = 2x → x = 4

a3log2 = ; log3 b52 3 5 2log 5 = ab

5log5

12log

512log

2

2

5 =

5log

5log12log2

22 −=

5log5loglog4log

2

222 −+=

ababa2 −+

=

ab2aab −−

−=

10. Jawa D ban:

81x3

x x3log

= xlog

243xx

xlog

x3log

=

243x xlogx3log =− 5log 3x x

x3

= 53log 3x =

53log 3logxlog = log 3. log x = 5 l g 3

0

olog x = 5 x = 10000



1. Jawaban: D

CT2 = AC2 – AT2

= 3p2 – p2 = 2p2

CT =

1

60o

A BT

C

p 3

p

2p

CBCT60sin =°

CB2p

23=

6p32

32p2CB ==

12. Jawaban: B Un – Un − 1 = Un + 1 – UnMerupakan 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

10b15

b1010 +

=−

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


4 = 8 a = 256

14. Jawaban: C a

9

2r328

256aar

4

95 =→===

188

r8a3

=== 6

ar3 = 8 →

a1 + a7 = a + ar = 1 + 64 = 65

15. Jawaban: B

31642 ...coscos =+θ+θ−θ cos

31

r1=

− a

31

cos1cos2

2 =θ+

θ

3 cos2θ = 1 + cos2θ 2 cos2θ = 1 cos2θ = 1 cos2θ = ½

221cos −=θ

π=°=θ 135 34

16. Ja aban: B w

=+−

−→ 95

164

lim 2x2

x x

92 2x

204 −→x2lim

=x

x

+

1092lim

−=4→x

2 −=+x

17. Jawaban: E

xxx 50 −→xx

2sin7tan8sinlim + 5

315

2578

==−+

=xx

xxxx

18. Jawaban: B 21

21

xx−⎞⎛ += xxy += ⎟⎠

⎜⎝

⎟⎠⎝⎠⎝ 22⎞−

−21

1

x ⎜⎛ +⎟

⎞⎜⎛ +=

221 11.xx1'y

⎟⎟⎠

⎜⎝+

+=

x21.

2

⎞1⎜⎛11

xx

⎟⎟⎠

⎞⎜⎜⎝

⎛ +

+=

x21x2.

xx

121

⎟⎠

⎜=x2

⎟⎞

⎜⎛ +1x2 ⎝ + xx4

19. Jawaban: D

401500p4B−+=

pp

= 5 B = 100 – 200 + 1500 = 1400

B = 4p2 – 40p + 1500 B’ = 0 ⎯→ 8p – 40 = 0 ⎯→ p


c = 0 ⎯→

20. Jawaban: D

bam −= ax + by +

⎯→ 35m1 = 01y3x5 =+−

21m 2 −= ⎯→3y2x =−+ 0

66.

).(1 21

35

215 +

21

21

m.m1mm

tan+

−13

56310=

−+

= 3

−+==α

21. Jawaban: ⎞⎛ + a1a⎟⎟⎠

⎜⎜⎝

=a0

A

−

10b1

AA 1

a = 1 1 + = −b b

22.

B

⎟⎟⎠

⎞⎜⎜⎝

⎛=−

10b1

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 =

120C103 =

21C72 =

3C31 =

4021

12021.3

= 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 kax + y +

a +

Ni21 – 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

75,4616205,45

ffd

x s =+=+=∑∑ x

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