Kamis, 22 Desember 2011

PENYELESAIAN KODE A


Penyelesaian A

1.        √18 - √32 + √50 + √72 = 3√2 – 4√2  + 5√2  + 6√2 = 10√2 (A)
2.        p + 1 = x p = x – 1
4(x – 1)2 – 2(x – 1) – 3 = 0
4x2 – 8x + 4 – 2x + 2 – 3 = 0
4x2 – 10x + 3 = 0 (C)
3.        (– 8)2 – 4(m – 1)(8m) ≥ 0
64 – 32m2 + 32m ≥ 0
m2 – m – 2 ≤ 0
(m – 2)(m + 1) ≤ 0
m = 2 atau m =  – 1
– 1 ≤ m ≤  2 (C) 
4.        y = a(x – 3)2 – 2
f(0) = 16 16 = a(9) – 2
                    a = 2
y = 2(x – 3)2 – 2 = 2x2 – 12x + 16 (D)                    
5.        Semua peserta Ujian Nasional berdoa sebelum mengerjakan soal (E)
6.        p →  ~q ≡ p →  ~q
~r → q  ≡ ~q → r
                    ~ r          r          
                    ~  p     (A)

7.        x+ y + z = 12                               3x + 2y – z = 8
2x – y + 2z = 12                           2x + 2y +2z = 24
3x + 3z = 24                                 x – 3z = – 16
                             3x + 3z = 24
                             x – 3z =  – 16
                                         x = 2 z = 6 y = 4
x.y.z = 2.4.6 = 48  (B)
8.        2 cos (2x + 1500) = Ö3
cos (2x + 1500) = ½V3
cos (2x + 1500) = cos 300
1.         2x + 1500 = 300 + k.3600                                          2. 2x + 1500 = – 300 + k.3600
2x =  – 1200 + k.3600                                                    2x = – 1800 + k.3600
x = – 600 + k.1800                                                           x = – 900 + k.1800
                         HP= {900, 1200}   (B)

 

9.        Panjang vertikal = 10 tan 300 = 10/3 V3
Sudut puncak kelapa = 45o
(10/3 V3) : sin 450  = x : sin 1200
        10.V3.V3 = 3.x.V2
x = 5√2 (D)


10.    Jarak C ke BDG = 1/3 diagonal ruang = 3 cm (D)
11.    TP = CP =  = 2
Sudut ADB dengan ABC adalah sudut CED = α
Cos α =  =
Tan α = 2√2
12.     5.4.7.6.5 = 4200

13.    1 3 3 1
Paling sedikit dua gambar = 4/8 = ½ (E)
14.    300000 =
300 p + 300w = 320p + 285w
15w = 20p
3w = 4p
 p : w = 3 : 4   (D)
15.    Mo = 85,5 +  = 85,5 + 3 = 88,5 (B)
16.    Tan 1050  = Tan (60 + 45)0
=  .
=
= – 2  –  (B)
17.    Cos 3x + Cos x = 2 Cos 2x cos x = 2 (1 – 2 sin2x) Cos x = 2 (1 – 2.16/25).3/5 = - 42/125 (C)
18.    (x + 1)2 + (y – 3)2 = 1
x2 + y2 + 2x – 6y + 9 =0 (D)
19.     3x – y + (x + 3) – 2(y – 1) – 20 = 0
4x – 3y – 15 = 0 (A)




20.    x4 – 3x3 – 5x2 + x – 6        = (x2 + bx + c)(x2 – x  – 2) + (px + q)
                                         = x4 – x3 – 2x2 + bx3 – bx2 – 2bx + cx2 – cx – 2c + px + q
                                         =  x4 – (1 – b)x3 – (2 + b – c)x2 – (2b + c – p)x – 2c + q
1 – b = 3              2 + b – c = 5        p – c – 2b = 1                         q – 2c = – 6 
b = – 2                             c = – 5                   p = – 8                             q = – 16 
jadi sisanya :  – 8 x – 16 (D)

21.       2a + b =8
– 3a + b = – 7
5a = 15
a = 3
b = 2
jadi didanya : 3x + 2 (E)
22.    6(5p + 4) – 3 = 81
30p + 24 = 84
p =   2  (D)
23.    5yx + 3y = 2x – 1
2x – 5yx = 3y + 1
x =   = , jadi a = –3 dan b = 5 maka 2a + b = – 6 + 5 = – 1 ( A)

24.         =  
                                         =                  
                                         =  6          (B)

25.      =  =            ( C )
26.    f(x) = cos5 (4x – 2)
f’(x) = 5cos4(4x – 2).sin (4x – 2)(– 4) = – 10 cos3 (4x – 2) sin (8x – 4).  (E)                         
27.    L = a2 + 4at = 108
V= a2t = a2(108 – a2)/4a = 27a – a3/4
V’ = 27 – ¾ a2 = 0
         27 = ¾ a2
         a2 = 4.9
         a  = 6    (A) 



28.    y = x3 – 3x2 + x + C
– 3 = 8 – 12 + 2 + C
C =  – 1
Jadi persamaan kurva itu :y = x3 – 3x2 + x – 1  ( B )

29.    ò sin5 x. cos x dx = ò sin5 x d(sin x) = 1/6  sin6 x + C    (A)
30.    V = π  =  π{( + 16 + 18)} = 40  π      (A)

31.    102 + 105 + 108 + ... + 399
399 = 102 + (n – 1)3
297 = (n – 1)3
99 = n – 1
n = 100
Sn = 50.501 = 25.050 (C)
32.    ar5 : ar2 = 10 2/3 : 36
r3 = 8/27
r = 2/3
a = 9.9
S~ = 9.9/(1 – 2/3) = 9.9.3 = 243 (B)
33.      
Titik
A
B
C
Koordinat
(30,0)
(0,63)
(8,55)
500x + 300y
15.000.000
18.900.000
4.000.000+16.500.000 =20.500.000



34.    3 = 2 + 2p p = ½
5 = q + 4 q = 1
p + q` = 3/2  (A)
35.    6 – 4 – 2x = 0 x = 1 (E)
36.    Proyeksi vektor ortogonal  pada   = (4  + 8 – 24)/24 {2 – 2+ 4}
= – + - 2                                                            


                                         
37.    rotasi dengan pusat (0,0) sejauh +90o  y = – x’ dan x  =  y’
                                         x – 2y + 4 = 0 y’ + 2x’ + 4 = 0
pencerminan terhadap garis y = x y’ = x’’ dan x’  = y’’
                             x’’ + 2y’’ + 4 = 0
Jadi bayangannya adalah x + 2y + 4 = 0

38.    Lbayangan = ½ .6.4 = 12  
39.     5log(x2 – 7x – 1)  –  5log (4 – 3x) = 0
x2 – 7x – 1 = 4 – 3x
x2 – 4x – 5 = 0
(x – 5)(x + 1) = 0
x = 5 (tm) atau x = – 1   (D )
40.    32x+1 –10.3x + 3 = 0, diketahui akar- akarnya adalah x1 dan x2 dimana           x1 <  x2. Nilai dari 2x1 – 3x2
3a2 – 10a + 3 = 0
(3a – 1)(a – 3) = 0
a = 1/3 atau a = 3
x = – 1 atau x = 1
      2x1 – 3x2 = – 2 – 3 =  – 5

Senin, 19 Desember 2011

PENYELESAIAN KODE B

Penyelesaian KODE B:
1.   √18 + √32 - √50 + √72 = 3√2 + 4√2 – 5√2 + 6√2 = 8√2(D)
2.        p + 2 = x p = x – 2
3(x – 2)2 – 2(x – 2) – 5 = 0
3x2 – 12x + 12 – 2x + 4 – 5 = 0
3x2 – 14x + 11 = 0 (C)
3.        (– 8)2 – 4(k – 1)(8k) < 0
64 – 32k2 + 32k < 0
k2 – k – 2 > 0
(k – 2)(k + 1) > 0
k = 2 atau k =  – 1
k< – 1 atau k > 2 (D) 
 4.        y = a(x – 3)2 – 2
f(0) = 25 25 = a(9) – 2
                    a = 3
y = 3(x – 3)2 – 2 = 3x2 – 18x + 25 (D)
                    
5.        Semua peserta Ujian Nasional berdoa sebelum mengerjakan soal (E)
6.        ~p q ≡ ~p   q
r ~q  ≡ q ~r
                   r          r          
                      p     (B)
7.        6x – 4y + 2z = 18                         -x + 3y+ z = 6
2x + y 2z = 23                           3x – 2y + z = 9
8x – 3y = 41                                 4x – 5y = 3
                             8x – 3y = 41
                             8x – 10y = 6
                                         y = 5 x = 7 z = – 2
x.y.z = 7.5.(– 2) =  – 70  (A)
8.        2 sin (2x +1500) = 1 
Sin (2x + 1500) = ½
Sin (2x + 1500) = sin 300
1.         2x + 1500 = 300 + k.3600                                          2. 2x + 1500 = 1500 + k.3600
2x =  – 1200 + k.3600                                                    2x = k.3600
x = – 600 + k.1800                                                           x = k.1800
                         HP= {1200, 1800}   (B)
9.        CA = 30 mil dan CB = 40 mil
AB2 = 302 + 402 – 2.30.40 cos 1200
        = 900 + 1600 + 1200
        = 3700
AB = 10√37 (B)


10.    Jarak E ke AFH = 1/3 diagonal ruang = 5√3 cm (B)
11.    DE = CE = √(36 - 9) = 3√3
Sudut ADB dengan ABC adalah sudut CED = α
Cos α = (27 + 27 - 36)/2.9.3 = 1/3
Tan α = 2√2
12.     7.6.5.5.4 = 4200

13.    1 3 3 1
Paling banyak dua angka = 7/8 (E)
14.    300.000 = (285.000p + 320.000w)/(p + w)...........ruas kanan dan kiri dibagi 1.000
300 p + 300w = 285p + 320w
15p = 20w
3p = 4w
 p : w = 4 : 3   (E)
15.    Me = 85,5 + 5(25-23)/10 = 85,5 + 1 = 86.5 (B)
16.    Tan 150  = tan (60 – 45)0
= (√3 - 1)/(1 + √3) . (√3 - 1)/(√3 - 1)
= (4 - 2√3)/2
= 2 – √3 
17.    sin 3x + sin x = 2 sin 2x cos x = 4 sin x cos2 x = 4. 3/5 . 16/25 = 192/125 (D)
18.    (x + 3)2 + (y – 1)2 = 1
x2 + y2 + 6x – 2y + 9 =0
19.     – x + 3y – 2(x – 1) + (y + 3) – 20 = 0
3x – 4y + 15 = 0 (C)
20.    x4 – 3x3 – 5x2 + x – 6        = (x2 + bx + c)(x2 + x – 2) + (px + q)
                                         = x4 + x3 – 2x2 + bx3 + bx2 – 2bx + cx2 + cx – 2c + px + q
                                         =  x4 + (1 + b)x3 – (2 – b – c)x2 – (2b – c – p)x – 2c + q
b + 1 =  – 3              2 – b – c = 5        p + c – 2b = 1                       q – 2c = – 6 
b = – 4                                 c = 1                      p = – 8                            q = – 4 
jadi sisanya : – 8x – 4 (E)

21.       2a + b = 24
– 5a + b = 10
7a = 14
a = 2
b = 20
jadi didanya : 2x + 20 (D)
22.    6(5p + 4) – 3 = – 39
30p + 24 = – 36
p =  – 2  (A)
23.    yx + 2y = 3x – 5
3x – yx = 2y + 5
x = (2y + 5)/(3 - y) = (-2y - 5)(y - 3), jadi a = –2 dan b = 1 maka 3a + b = – 6 + 1 = – 5 ( A)

24. lim  x2 /{1 - ( x2 + 1)} =  lim  x2 /{1 - ( x2 + 1)}.{1 + ( x2 + 1)}/{1 + ( x2 + 1)}                            
        x →0                                 x →0              
                                     = lim  x2 {1 + ( x2 + 1)}/{1 - x2 - 1)}
                                         x →0 
                                     = (1  + 1)/(-1)     
                                     =  – 2          (A)
25. lim   x sin 3x/(1 - cos 4x)  = lim (x sin 3x) / (2 sin2 2x) = (1.3)/(2.2.2) = 3/8 (C)
      x →0                                  x →0         
26.    f(x) = sin3 (3 – 2x)
f’(x) = 3sin2(3 – 2x).cos (3 – 2x)(– 2) = – 3 sin (6 – 4x) sin (3 – 2x)  (E)
27.    L = a2 + 4at = 432
V= a2t = a2(432 – a2)/4a = 108a – a3/4
V’ = 108 – ¾ a2 = 0
         108 = ¾ a2
         a2 = 4.36
         a  = 12    (D) 
28.    y = x3 – 6x2 + 5x + C
– 4 = 8 – 24 + 10 + C
C = 2
Jadi persamaan kurva itu :y = x3 – 6x2 + 5x + 2 ( C )
29.    sin3 x. cos x dx = sin3 x d(sin x) = ¼ sin4 x + C    (A)
30.    V = π (1/5 x5 + 2x3 + 9x) =  π{2(1/5 + 2 + 9)} = 22 2/5 π      (D)
31.    1005 + 1010 + 1015 + ... + 2000
2000 = 1005 + (n – 1)5
995 = (n – 1)5
199 = n – 1
n = 200
Sn = 100.3005 = 300.500
32.    ar5 : ar2 = 2/3 : 18
r3 = 1/27
r = 1/3
a = 9.18
S~ = 9.18/(1 – 1/3) = 9.18.3/2 = 243 (B)
33.     anulir
34.    3 = 2 + 2p p = ½
7 = q + 4 q = 3
p.q` = 3/2  (A)
35.    8 – 2x – 2 = 0 x = 3 (A)
36.    Proyeksi vektor ortogonal a pada  b   = (– 3  + 2 + 10)/9 {i + 2j - 2k}
= i + 2j - 2k                                                      
37.    rotasi dengan pusat (0,0) sejauh +270o y = x’ dan x  = – y’
                                         2x – y + 4 = 0 – 2y’ – x’ + 4 = 0
pencerminan terhadap garis y = x y’ = x’’ dan x’  = y’’
                             – 2x’’ – y’’ + 4 = 0
Jadi bayangannya adalah 2x + y – 4 = 0
38.  hasil translasinya adalah A'(-2,2), B'(1,2) dan C(1,4)
       hasil dilatasinya A"(-4,4), B"(2,4) dan C"(2,8)
Lbayangan = 1/2.6.4 = 12
39.     5log(x2 – 7x – 1)  –  5log (4 – 3x) = 0
x2 – 7x – 1 = 4 – 3x
x2 – 4x – 5 = 0
(x – 5)(x + 1) = 0
x = 5 (tm) atau x = – 1   (D )
40.    5. 52x –26.5x + 5 = 0
5a2 – 26a + 5 = 0
(5a – 1)(a – 5) = 0
a = 1/5 atau a = 5
x = – 1 atau x = 1
       3x1 – 4x2 = – 3 – 4 =  – 7   (B)

Minggu, 18 Desember 2011

KUNCI LATIHAN UJIAN


KUNCI JAWABAN
LATIHAN UJIAN NASIONAL
MATEMATIKA (B)

1
D
6
B
11
A
16
A
21
D
26
E
31
B
36
D
2
C
7
A
12
D
17
D
22
A
27
D
32
B
37
C
3
D
8
B
13
E
18
D
23
A
28
C
33

38
B
4
D
9
D
14
E
19
C
24
A
29
A
34
A
39
D
5
E
10
B
15
B
20
E
25
C
30
D
35
A
40
B


KUNCI JAWABAN
LATIHAN UJIAN NASIONAL
MATEMATIKA (A)

1
A
6
A
11
A
16
B
21
E
26
E
31
C
36
E
2
C
7
B
12
D
17
C
22
D
27
A
32
B
37
A
3
C
8
B
13
E
18
D
23
A
28
B
33
D
38
D
4
D
9
D
14
D
19
A
24
B
29
A
34
A
39
D
5
E
10
D
15
B
20
D
25
C
30
A
35
E
40
B