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 +270^o → 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)

L_bayangan = 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. 5^2x –26.5^x + 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)