Jawab A Jawab B Jawab B Jawab D Jawab B Jawab C Jawab C

Fx = x cosx = ux  vx F’x = u’ x  vx + u x  v’ x = 1  cosx + x  sinx = cosx  x sinx F’x+ 2 1  = cosx+ 2 1   x+ 2 1  sinx+ 2 1  = sinx  x+ 2 1  cosx = sinx  x cosx  2 1  cosx

17. Jawab A

Subtitusi : x 2 + ax + b = cosx berpotongan di x = 0 dan x =  x = 0  0 2 + a  0 + b = cos 0  b = 1 x =    2 + a   + b = cos   2 + a   + 1 = 1  a =    2 Sumbu simetri =  2 a = 2 1  +  1

18. Jawab B

3 x 6 x 4 2    = 125 6 2x 6 x 2 2    = 3 5 6    2x 6 x 2 = 3 5 3x 2 = 3 5 x 2 = 5 Diperoleh x 4  = x 2 2  = 2  x 2 – 2  5 = 5 2 1

19. Jawab B

Kurva y = x 3  3x 2  9x Batas: 0  x  2 Cek batas: x = 0  y = 0 x = 2  y = 22 Cek puncak: y  = 0 3x 2  6x  9 = 0 X 2  2x  3 = 0 x + 1 x  3 = 0 x = 1 atau x = 3 Tetapi x = 1 dan x = 3 tidak terletak pada interval 0  x  2 Nilai minimum: min y = 22 Nilai maksimum: max y = 0 20. Jawab E 2 log x  3 logy = 0  2 1 2 logx  3 logy = 0  2 logx  2  3 logy = 0 2 logx + 3 logy 2 = 5  2 logx + 2  3 logy = 5 2  2 logx = 5 2 logx = 2 5 x = 2 5 2 = 4 2 21. Jawab D S 1 Persegi sisi 1, maka Luas S 1 = 1 S 2 persegi dengan sisi x Lihat gambar , 3x = 1  x = 3 1 Luas S 2 = x 2 = 9 1 Rasio r = 1 2 S Luas S Luas = 9 1 Jumlah luas semua persegi =  S = 9 1 1 1  = 8 9

22. Jawab D

L = luas OABC = x  y = x  x  2 2 = x  x 2  4x + 4 = x 3  4x 2 + 4x Luas akan maksimum, jika L  = 0 3x 2  8x + 4 = 0 3x  2x  2 = 0 x = 3 2 atau x = 2 Untuk x = 2  luas = 0 Untuk x = 3 2  luas = x  x  2 2 = 27 32 Maksimum

23. Jawab B

ganjil S  = jumlah suku-suku bernomor ganjil = 1 U + 3 U + 5 U + ... genap S  = jumlah suku-suku bernomor genap = 2 U + 4 U + 6 U + ... 9 = 5 + genap S  genap S  = 4 SNMPTN 2009 Halaman 3 dari 6 halaman + 45 o 45 o x x x x C A O Bx,y =+ Rasio r = r = 5 4 9 = 5 4 1 a  a = 5 9

24. Jawab C

p = 13 14 1  = 13 14 1  13 14 13 14   = 13 14  q = 13 14 1  = 13 14 1  13 14 13 14   = 13 14  p 2 + pq + q 2 = p + q 2  p q = 2 14 2  1 = 55

25. Jawab C

Jika A 1 =      1 3 2 dan B 1 =     3 2 1 , maka A B  1 = B  1  A  1 =     3 2 1       1 3 2 =            3 3 4 BAHASA INDONESIA 26. Jawab: B