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Kuncisoal mtk-un-smk-prwsta
- 1. Copyright ©www.ujian.org all rightreserved 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 + + +− −
- 2. Copyright ©www.ujian.org all rightreserved . 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
- 3. Copyright ©www.ujian.org all rightreserved Copyright ©www.ujian.org all right reserved 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
- 4. Copyright ©www.ujian.org all rightreserved 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
- 5. Copyright ©www.ujian.org all rightreserved 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