Copyright ©www.ujian.orgall right reservedSolusi Latihan Soal UN SMA / MA 2011Program IPAMata Ujian : MatematikaJumlah Soa...
Copyright ©www.ujian.orgall right reserved2x11x112x2x2x9x12−−−=−−+−−−=x211x11−+=5. Jawab : A( )4x26x)x(gf−+=οg(x) = 2 – x ...
Copyright ©www.ujian.orgall right reservedP(x) : (x2 – 8x + 15) sisa ax + bP(x) = (x2 – 8x + 15)k(x) + ax + bP(3) = 3a + b...
Copyright ©www.ujian.orgall right reserved= 23 – 3(– 4).2 = 3215. Jawab : E03mx8x2=++−D = 0(–8)2 – 4.1.(m + 3 ) = 064 – 4m...
Copyright ©www.ujian.orgall right reserved21. Jawab :28dx)6x4(a1=+∫a12x6x2 + = 282a2 + 6a – (2 + 6) = 28a2 + 3a – 18 = 0(a...
Copyright ©www.ujian.orgall right reservedPeluang A dan B berdampingan adalah2124444P =++=25. Jawab : EMerah = 4 putih = 5...
Copyright ©www.ujian.orgall right reserved31. Jawab : DMisal A(a, b) .Karena refleksi terhadap x = –2 dan y = 5 maka koord...
Copyright ©www.ujian.orgall right reservedBtan.Atan1BtanAtan)BAtan(−+=+ 4848481512543.1−+=335615482036=−+=37. Jawab : Bsin...
Copyright ©www.ujian.orgall right reservedRusuk = 6R = titik berat segitiga ABC sehingga32BDBSBR 3131===112BFBRFR 22=+=
Upcoming SlideShare
Loading in …5
×

Pembahasan ujian nasional matematika ipa sma 2013

13,763 views

Published on

Published in: Education
  • Be the first to comment

Pembahasan ujian nasional matematika ipa sma 2013

  1. 1. Copyright ©www.ujian.orgall right reservedSolusi Latihan Soal UN SMA / MA 2011Program IPAMata Ujian : MatematikaJumlah Soal : 401. Jawab: E~p ∨ q salah maka~ p = S jadi p = Bq = 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 = B2. Jawab: E1033 x11x=+ −+10333.3 x11x=+10p3p3 =+3p2 – 10p + 3 = 0(p – 3)(3p – 1) = 0p = 3 p = 1/33x = 31 3x = 3−1x = 1 atau x = −13. Jawab : C2log2)3x2log()1xlog(2 ++≤−4log)3x2log()1xlog( 2++≤−)12x8log()1x2xlog( 2+≤+−x2 – 2x + 1 ≤ 8x + 12x2 – 10 – 11 ≤ 0(x – 11)(x + 1) ≤ 0–1 11– ++syarat :x – 1 > 0 2x + 3 > 0x > 1 23x −>–1 1123− 11 < x ≤ 114. Jawab : Df(x) = 3x + 14x3x2)x(g+−=2x3x4)x(g 1−−−=→ −( ) )x(fg)x(h 1−= ο12x3x43)x)(gf()x(h 11+⎟⎠⎞⎜⎝⎛−−−== −−ο
  2. 2. Copyright ©www.ujian.orgall right reserved2x11x112x2x2x9x12−−−=−−+−−−=x211x11−+=5. Jawab : A( )4x26x)x(gf−+=οg(x) = 2 – x → g–1(x) = 2 – x)x)(ggf()x(f 1−= οοx28xx28x4)x2(26x2 −=−+−=−−+−=6. Jawab : BPuncak (2,18) → y – 18 = a(x – 2)2Melalui (5,0) → 0 – 18 = a(5 – 2)2–18 = 9a → a = –2y – 18 = –2(x – 2)2= –2(x2 – 4x + 4)y = –2x2 + 8x + 107. Jawab : B5x)4p(x)1p()x(f 2+−+−=a2bx −=)1p(2)4p(1p−−−=−2(p2 – 2p + 1) = –p + 42p2 – 3p – 2 = 0(p – 2)(2p + 1) = 0p = 2 p = –1/28. Jawab : Ax + 3y = 10 → m1 = 31− ; m2 = 340)2y()5x( 22=++−2m1R)5x(m2y +±−=+9140)5x(32y +±−=+2015x32y ±−=+3x – y – 17 ± 203x – y + 3 = 0 atau 3x – y – 37 = 09. Jawab : E01y6x8yx 22=++−+pusat ( 2121,A −− )= (4,–3)a = 4 b = –35a + 2b = 20 – 6 = 1410. Jawab : DP(x) : (x2 –9) sisa 5x + 4P(x) = (x2 – 9) h(x) + 5x + 4 ⎯→P(3) = 19P(x) : (x – 5) sisa = 7 ⎯→ P(5) = 7
  3. 3. Copyright ©www.ujian.orgall right reservedP(x) : (x2 – 8x + 15) sisa ax + bP(x) = (x2 – 8x + 15)k(x) + ax + bP(3) = 3a + b = 19P(5) = 5a + b = 72a = –6 ⎯→ a = –3 ; b = 22sisa = –3x +2211. Jawab : D010x6x7kxx 234=−+−+x1 = 1 maka1 + k – 7 + 6 – 10 = 0 ⎯→ k = 10x1 + x2 + x3 + x4 = –b/a1 + x2 + x3 + x4 = –kx2 + x3 + x4 = –k – 1 = – 1112. Jawab : Bx + 2y = 16x + y = 12 _y = 4 dan x = 8f(x, y) = 4x + 10yf(0, 8) = 80f(12, 0) = 48f(8, 4) = 32 + 40 = 72fmaks = 8013. Jawab : C2x + 3y + 4z = 20x – y + 2z = 54x + 5y + z = 177x + 7y + 7z = 42x + y + z = 614. Jawab : E04x2x2=−−x1 + x2 = 2 x1.x2 = –4( ) )xx(x.x3xxxx 21213213231 +−+=+12 1612xy(8, 4)8
  4. 4. Copyright ©www.ujian.orgall right reserved= 23 – 3(– 4).2 = 3215. Jawab : E03mx8x2=++−D = 0(–8)2 – 4.1.(m + 3 ) = 064 – 4m – 12 = 052 = 4m ⎯→ m = 1316. Jawab : B=+−−+∞→ 57x8x43x2xlim 2⎟⎟⎠⎞⎜⎜⎝⎛+−−++∞→= 7x8x49x12x4xlim51 22⎟⎟⎠⎞⎜⎜⎝⎛ +=4281251=117. Jawab :)x2cos1(xx8cosx2sinx2sin0xlim−−→ )x2cos1(x)x8cos1(x2sin0xlim−−→= 221221)x(.x)x8(.x20xlim→= =3218. Jawab : Dy = x4 + 6 y = 22x4 + 6 = 22x4 = 16 maka x = ±2x = 2 ⎯→ m = 4x3 = 32y – 22 = 32(x – 2)y = 32x – 42x = –2 ⎯→ m = –4x3 = –32y – 22 = –32(x + 2)y = –32x – 4219. Jawab : Ctttxxtx2xtxtxtxtL = x2 + 4xt = 3004xt = 300 – x2t = 75/x – x/4V = x2.t = 75x – ¼ x3V’ = 075 – ¾ x2 = 0 maka x2 = 100 ⎯→x = 1020. Jawab :Misal y = x2 + 3x + 53x2dxdy+= maka3x2dydx+=∫ +++ dx)5x3xsin()9x6( 2∫ ++=3x2dyysin)3x2(3∫= dyysin3 = – 3cos y + c= –3 cos (x2 + 3x + 5) + c
  5. 5. Copyright ©www.ujian.orgall right reserved21. Jawab :28dx)6x4(a1=+∫a12x6x2 + = 282a2 + 6a – (2 + 6) = 28a2 + 3a – 18 = 0(a + 6)(a – 3) = 0a = – 6 atau a = 322. Jawab : Dx2 – 8x + 16 = 9xx2 – 17x + 16 = 0(x – 1)(x – 16) = 0x = 1 atau x = 16y = 9x y = x2– 8x + 10xy0 41L1 = = 4, 5dxx910∫L2 = = 9dx)16x8x(412∫ +−L = L1 + L2 = 13,523. Jawab : CBeratBadanFrekuensi50 – 5253 – 5556 – 5859 – 6162 – 64453 fk =1426 (kelasQ3)n = 20iffntQ k43b3 ⎟⎟⎠⎞⎜⎜⎝⎛ −+= 623614155,61 =⎟⎠⎞⎜⎝⎛ −+=24. Jawab : DBanyaknya susunan A, B, C, dan D berdampingan adalah 4! = 24Jika A dan B diharuskan berdampinngan maka banyaknya susunan adalah• A dan B di kiriAB CD = 2! 2! = 4• A dan B di tengahC AB D = 2! 2! = 4• A dan B di kananCD AB = 2! 2! = 4
  6. 6. Copyright ©www.ujian.orgall right reservedPeluang A dan B berdampingan adalah2124444P =++=25. Jawab : EMerah = 4 putih = 5Peluang terambilnya kedua bola merah618394=×=26. Jawab : CCB.At=⎟⎟⎠⎞⎜⎜⎝⎛−−=⎟⎟⎠⎞⎜⎜⎝⎛ −⎟⎟⎠⎞⎜⎜⎝⎛515101y1x5023⎟⎟⎠⎞⎜⎜⎝⎛−−=⎟⎟⎠⎞⎜⎜⎝⎛ −+515105y51y2x35y = –15 maka y = –33x + 2y = 03x – 6 = 0 maka x = 2Jadi 2x + y = 4 – 3 = 127. Jawab : D⎟⎟⎠⎞⎜⎜⎝⎛−−=2314A1TAdetAdetk −=A1Ak =251)38(1A1k 22=+−==28. Jawab : Ckˆ4jˆiˆba +−=+ρρmaka 18ba =+ρρ14ba =−ρρb.a2baba222 ρρρρρρ++=+b.a2baba222 ρρρρρρ−+=−b.a4baba22 ρρρρρρ=−−+18 – 14 = b.a4ρρb.aρρ= 129. Jawab : Dkˆ4jˆ3iˆ2m +−=ρkˆjˆ2iˆn −+−=ρnnn.mp 2ρρρρ= n141462 ρ++−−−= n2ρ−= kˆ2jˆ4iˆ2 +−=30. Jawab : C⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛−+⎟⎟⎠⎞⎜⎜⎝⎛−=⎟⎟⎠⎞⎜⎜⎝⎛yx2365yx⎟⎟⎠⎞⎜⎜⎝⎛+−+=⎟⎟⎠⎞⎜⎜⎝⎛y4x2yxx’ = 2 + x ⎯→ x = x’ – 2y’ = –4 + y ⎯→ y = y’ + 4Garis semula 7y4x5 =+Bayangannya :5(x’ – 2) + 4(y’ + 4) = 75x + 4y = 1
  7. 7. Copyright ©www.ujian.orgall right reserved31. Jawab : DMisal A(a, b) .Karena refleksi terhadap x = –2 dan y = 5 maka koordinat A menjadiA’(2x–a, 2y – b) = ( –4–a, 10 – b)Sstelah itu koordinat itu diputar 270o terhadap O sehingga menjadi (8, –1)⎟⎟⎠⎞⎜⎜⎝⎛−−−⎟⎟⎠⎞⎜⎜⎝⎛−=⎟⎟⎠⎞⎜⎜⎝⎛− b10a4011018⎟⎟⎠⎞⎜⎜⎝⎛+−=⎟⎟⎠⎞⎜⎜⎝⎛− a4b10188 = 10 – b maka b = 2–1 = 4 +a maka a = –5Koordinat A(–5, 2)32. Jawab : E189S6 = dan r = 21891r)1r(a 6=−−189163.a= ⎯→ a = 3U3.U4 = ar2.ar3 =a2r5 = 9.32 = 28833. Jawab : CS10 = 1.400.000 S15 = 2.850.0005(2a+9b) = 1.400.000 215(2a + 14 b) = 2.850.0002a + 9b = 280.000 2a + 14b = 380.000Dari kedua persamaan diperoleh b = 20.000U1 = a = 50.00034. Jawab : A60S =∞ 20s genap =∞402060S ganjil =−=∞21SSrganjilgenap==∞∞60r1a=−maka a = 60(1 – r) = 60. ½ = 30U2 = ar = 1535. Jawab : B75o60o45oACBoo60sinAB45sinBC=63260sin45sinABBC31oo===36. Jawab : D43Atan = dan 125Btan =
  8. 8. Copyright ©www.ujian.orgall right reservedBtan.Atan1BtanAtan)BAtan(−+=+ 4848481512543.1−+=335615482036=−+=37. Jawab : Bsin 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 10o38. Jawab : E21x2cos =cos 2x = cos 60o→ 2x = 60o + n.360ox = 30o + n.180o = 30o→ 2x = –60o + n.360ox = –30o + n.180o = 150oHp { 30o, 150o}39. Jawab : DTD CBAEFα102 2AD = AB = 22 maka AE = 222AETATE −= 22210 =−=EF = AB = 22TF = TE = 22Karena TE = EF = TF maka ΔTEF sama sisiSudut antara TAD dengan ABCD = α = 60o40. Jawab : EA BCDE FGQPHRS
  9. 9. Copyright ©www.ujian.orgall right reservedRusuk = 6R = titik berat segitiga ABC sehingga32BDBSBR 3131===112BFBRFR 22=+=

×