Kuncisoal mtk un smk tekno
1.
Jawab: E
~p ∨ q salah maka
~ p = S jadi p = B
q=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 = B
2.
Jawab: E
3 x +1 + 31− x = 10
31
3 x .31 + x = 10
3
3
3p + = 10
p
2
3p – 10p + 3 = 0
(p – 3)(3p – 1) = 0
p=3
p = 1/3
3x = 31 3x = 3−1
x = 1 atau x = −1
3.
Jawab : C
2 log(x − 1) ≤ log(2x + 3) + 2 log 2
log( x − 1) 2 ≤ log(2 x + 3) + log 4
log( x 2 − 2 x + 1) ≤ log(8x + 12)
x2 – 2x + 1 ≤ 8x + 12
x2 – 10 – 11 ≤ 0
(x – 11)(x + 1) ≤ 0
+
–
–1
syarat :
x–1>0
x>1
+
11
2x + 3 > 0
x > − 32
− 32 –1 1 11
1 < x ≤ 11
4.
Jawab : D
f(x) = 3x + 1
−4x − 3
2x − 3
→ g −1 ( x ) =
g(x ) =
x+4
x−2
−1
h(x ) = g ο f (x)
(
)
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Hak Cipta dilindungi undang-undang
⎛ − 4x − 3 ⎞
h −1 ( x ) = (f ο g −1 )( x ) = 3⎜
⎟ +1
⎝ x−2 ⎠
−12 x − 9 x − 2 −11x − 11
=
+
=
x−2
x−2
x−2
11x + 11
=
2−x
5.
Jawab : A
x+6
2x − 4
g(x) = 2 – x → g–1(x) = 2 – x
f ( x ) = (f οg οg −1 )( x )
(f οg )( x ) =
=
2−x+6
−x + 8 x − 8
=
=
2(2 − x ) − 4
2x
− 2x
6.
Jawab : B
Puncak (2,18) → y – 18 = a(x – 2)2
Melalui (5,0) → 0 – 18 = a(5 – 2)2
–18 = 9a → a = –2
y – 18 = –2(x – 2)2
= –2(x2 – 4x + 4)
y = –2x2 + 8x + 10
7.
Jawab : B
f ( x ) = (p − 1) x 2 + (p − 4) x + 5
b
x=−
2a
−(p − 4)
p −1=
2(p − 1)
2
2(p – 2p + 1) = –p + 4
2p2 – 3p – 2 = 0
(p – 2)(2p + 1) = 0
p = 2 p = –1/2
8.
Jawab : A
x + 3y = 10 → m1 = − 13 ; m2 = 3
( x − 5) 2 + ( y + 2) 2 = 40
y + 2 = m( x − 5) ± R 1 + m 2
y + 2 = 3( x − 5) ± 40 1 + 9
y + 2 = 3x − 15 ± 20
3x – y – 17 ± 20
3x – y + 3 = 0 atau 3x – y – 37 = 0
9.
Jawab : E
x 2 + y 2 − 8x + 6 y + 1 = 0
pusat (− 12 A,− 12 ) = (4,–3)
a = 4 b = –3
5a + 2b = 20 – 6 = 14
10.
Jawab : D
P(x) : (x2 –9) sisa 5x + 4
P(x) = (x2 – 9) h(x) + 5x + 4 ⎯→P(3) = 19
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P(x) : (x – 5) sisa = 7 ⎯→ P(5) = 7
P(x) : (x2 – 8x + 15) sisa ax + b
P(x) = (x2 – 8x + 15)k(x) + ax + b
P(3) = 3a + b = 19
P(5) = 5a + b = 7
2a
= –6 ⎯→ a = –3 ; b = 22
sisa = –3x +22
11.
Jawab : D
x 4 + kx 3 − 7 x 2 + 6x − 10 = 0
x1 = 1 maka
1 + k – 7 + 6 – 10 = 0 ⎯→ k = 10
x1 + x2 + x3 + x4 = –b/a
1 + x2 + x3 + x4 = –k
x2 + x3 + x4 = –k – 1 = – 11
12.
Jawab : B
x + 2y = 16
x + y = 12 _
y = 4 dan x = 8
y
12
8
(8, 4)
x
12
f(x, y) = 4x + 10y
f(0, 8) = 80
f(12, 0) = 48
f(8, 4) = 32 + 40 = 72
fmaks = 80
13.
Jawab : C
2x + 3y + 4z = 20
x – y + 2z = 5
4x + 5y + z = 17
7x + 7y + 7z = 42
x+y+z=6
14.
Jawab : E
x 2 − 2x − 4 = 0
x1 + x2 = 2
x1.x2 = –4
3
3
3
x 1 + x 2 = (x 1 + x 2 ) − 3x 1 .x 2 ( x 1 + x 2 )
= 23 – 3(– 4).2 = 32
16
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15.
16.
Jawab : E
x 2 − 8x + m + 3 = 0
D=0
(–8)2 – 4.1.(m + 3 ) = 0
64 – 4m – 12 = 0
52 = 4m ⎯→ m = 13
Jawab : B
lim
x→∞
17.
1 ⎛ lim
2 x + 3 − 4 x 2 − 8x + 7
= = ⎜⎜
5 ⎝x → ∞
5
Jawab :
lim
lim
sin 2 x − sin 2x cos 8x
=
x→0
x→0
x (1 − cos 2 x )
18.
Jawab : D
y = x4 + 6 y = 22
x4 + 6 = 22
x4 = 16 maka x = ±2
x = 2 ⎯→ m = 4x3 = 32
y – 22 = 32(x – 2)
y = 32x – 42
x = –2 ⎯→ m = –4x3 = –32
y – 22 = –32(x + 2)
y = –32x – 42
19.
Jawab : C
t
xt
t
xt
x2
x
xt
lim
sin 2x (1 − cos 8x )
=
x→0
x (1 − cos 2 x )
t
x
xt
t
L = x2 + 4xt = 300
4xt = 300 – x2
t = 75/x – x/4
V = x2.t = 75x – ¼ x3
V’ = 0
75 – ¾ x2 = 0 maka x2 = 100 ⎯→x = 10
20.
Jawab :
Misal y = x2 + 3x + 5
dy
dy
= 2x + 3 maka dx =
dx
2x + 3
2
(
6
x
+
9
)
sin(
x
+
3
x
+
5
)
dx
∫
dy
2x + 3
= ∫ 3 sin y dy = – 3cos y + c
= ∫ 3(2 x + 3) sin y
= –3 cos (x2 + 3x + 5) + c
⎞ 1 ⎛ 12 + 8 ⎞
⎟⎟ =1
4x 2 + 12 x + 9 − 4x 2 − 8x + 7 ⎟⎟ = ⎜⎜
⎠ 5⎝ 2 4 ⎠
2x. 12 (8x ) 2
x. 12 ( x ) 2
=32
Jawab: E
~p ∨ q salah maka
~ p = S jadi p = B
q=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 = B
2.
Jawab: E
3 x +1 + 31− x = 10
31
3 x .31 + x = 10
3
3
3p + = 10
p
2
3p – 10p + 3 = 0
(p – 3)(3p – 1) = 0
p=3
p = 1/3
3x = 31 3x = 3−1
x = 1 atau x = −1
3.
Jawab : C
2 log(x − 1) ≤ log(2x + 3) + 2 log 2
log( x − 1) 2 ≤ log(2 x + 3) + log 4
log( x 2 − 2 x + 1) ≤ log(8x + 12)
x2 – 2x + 1 ≤ 8x + 12
x2 – 10 – 11 ≤ 0
(x – 11)(x + 1) ≤ 0
+
–
–1
syarat :
x–1>0
x>1
+
11
2x + 3 > 0
x > − 32
− 32 –1 1 11
1 < x ≤ 11
4.
Jawab : D
f(x) = 3x + 1
−4x − 3
2x − 3
→ g −1 ( x ) =
g(x ) =
x+4
x−2
−1
h(x ) = g ο f (x)
(
)
Copyright ©www.ujiannasional.org
Hak Cipta dilindungi undang-undang
⎛ − 4x − 3 ⎞
h −1 ( x ) = (f ο g −1 )( x ) = 3⎜
⎟ +1
⎝ x−2 ⎠
−12 x − 9 x − 2 −11x − 11
=
+
=
x−2
x−2
x−2
11x + 11
=
2−x
5.
Jawab : A
x+6
2x − 4
g(x) = 2 – x → g–1(x) = 2 – x
f ( x ) = (f οg οg −1 )( x )
(f οg )( x ) =
=
2−x+6
−x + 8 x − 8
=
=
2(2 − x ) − 4
2x
− 2x
6.
Jawab : B
Puncak (2,18) → y – 18 = a(x – 2)2
Melalui (5,0) → 0 – 18 = a(5 – 2)2
–18 = 9a → a = –2
y – 18 = –2(x – 2)2
= –2(x2 – 4x + 4)
y = –2x2 + 8x + 10
7.
Jawab : B
f ( x ) = (p − 1) x 2 + (p − 4) x + 5
b
x=−
2a
−(p − 4)
p −1=
2(p − 1)
2
2(p – 2p + 1) = –p + 4
2p2 – 3p – 2 = 0
(p – 2)(2p + 1) = 0
p = 2 p = –1/2
8.
Jawab : A
x + 3y = 10 → m1 = − 13 ; m2 = 3
( x − 5) 2 + ( y + 2) 2 = 40
y + 2 = m( x − 5) ± R 1 + m 2
y + 2 = 3( x − 5) ± 40 1 + 9
y + 2 = 3x − 15 ± 20
3x – y – 17 ± 20
3x – y + 3 = 0 atau 3x – y – 37 = 0
9.
Jawab : E
x 2 + y 2 − 8x + 6 y + 1 = 0
pusat (− 12 A,− 12 ) = (4,–3)
a = 4 b = –3
5a + 2b = 20 – 6 = 14
10.
Jawab : D
P(x) : (x2 –9) sisa 5x + 4
P(x) = (x2 – 9) h(x) + 5x + 4 ⎯→P(3) = 19
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Hak Cipta dilindungi undang-undang
P(x) : (x – 5) sisa = 7 ⎯→ P(5) = 7
P(x) : (x2 – 8x + 15) sisa ax + b
P(x) = (x2 – 8x + 15)k(x) + ax + b
P(3) = 3a + b = 19
P(5) = 5a + b = 7
2a
= –6 ⎯→ a = –3 ; b = 22
sisa = –3x +22
11.
Jawab : D
x 4 + kx 3 − 7 x 2 + 6x − 10 = 0
x1 = 1 maka
1 + k – 7 + 6 – 10 = 0 ⎯→ k = 10
x1 + x2 + x3 + x4 = –b/a
1 + x2 + x3 + x4 = –k
x2 + x3 + x4 = –k – 1 = – 11
12.
Jawab : B
x + 2y = 16
x + y = 12 _
y = 4 dan x = 8
y
12
8
(8, 4)
x
12
f(x, y) = 4x + 10y
f(0, 8) = 80
f(12, 0) = 48
f(8, 4) = 32 + 40 = 72
fmaks = 80
13.
Jawab : C
2x + 3y + 4z = 20
x – y + 2z = 5
4x + 5y + z = 17
7x + 7y + 7z = 42
x+y+z=6
14.
Jawab : E
x 2 − 2x − 4 = 0
x1 + x2 = 2
x1.x2 = –4
3
3
3
x 1 + x 2 = (x 1 + x 2 ) − 3x 1 .x 2 ( x 1 + x 2 )
= 23 – 3(– 4).2 = 32
16
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Hak Cipta dilindungi undang-undang
15.
16.
Jawab : E
x 2 − 8x + m + 3 = 0
D=0
(–8)2 – 4.1.(m + 3 ) = 0
64 – 4m – 12 = 0
52 = 4m ⎯→ m = 13
Jawab : B
lim
x→∞
17.
1 ⎛ lim
2 x + 3 − 4 x 2 − 8x + 7
= = ⎜⎜
5 ⎝x → ∞
5
Jawab :
lim
lim
sin 2 x − sin 2x cos 8x
=
x→0
x→0
x (1 − cos 2 x )
18.
Jawab : D
y = x4 + 6 y = 22
x4 + 6 = 22
x4 = 16 maka x = ±2
x = 2 ⎯→ m = 4x3 = 32
y – 22 = 32(x – 2)
y = 32x – 42
x = –2 ⎯→ m = –4x3 = –32
y – 22 = –32(x + 2)
y = –32x – 42
19.
Jawab : C
t
xt
t
xt
x2
x
xt
lim
sin 2x (1 − cos 8x )
=
x→0
x (1 − cos 2 x )
t
x
xt
t
L = x2 + 4xt = 300
4xt = 300 – x2
t = 75/x – x/4
V = x2.t = 75x – ¼ x3
V’ = 0
75 – ¾ x2 = 0 maka x2 = 100 ⎯→x = 10
20.
Jawab :
Misal y = x2 + 3x + 5
dy
dy
= 2x + 3 maka dx =
dx
2x + 3
2
(
6
x
+
9
)
sin(
x
+
3
x
+
5
)
dx
∫
dy
2x + 3
= ∫ 3 sin y dy = – 3cos y + c
= ∫ 3(2 x + 3) sin y
= –3 cos (x2 + 3x + 5) + c
⎞ 1 ⎛ 12 + 8 ⎞
⎟⎟ =1
4x 2 + 12 x + 9 − 4x 2 − 8x + 7 ⎟⎟ = ⎜⎜
⎠ 5⎝ 2 4 ⎠
2x. 12 (8x ) 2
x. 12 ( x ) 2
=32