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Pembahasan Latihan Soal Sbmpt n 2013
Tes Kem am puan Dasar Um um (TKDU) – M at em at ika Dasar
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( sebagian dar i 5 Paket Latihan Sbmptn)
1 Jaw ab: A
~ p
q = ~(~ p)
q = p
q2 Jaw ab: E
Dat anya: 15, x, 50, y, 90
15 x 50 y 90 X
5
min
15 15 50 50 90
X 44
5
max
15 50 50 90 90
X 59
5
44 X 59
3 Jaw ab: C
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2 2 2 2 2 2 2 2 2 2 2
x 3x 1 2 x 4 x 3x 28 x 3x 1 2
0 x 4 x 3x 28
x 3x 1 2( x 7) 0 ( x 4)( x 7) x 3x 28
x 3x 1 2x 14 0 x 3x 28 x 3x 28
x x 15 0 x 3x 28
( )
0 ( x 7)( x 4)
D 1 60 59 definit (+ )
Bilangan bulat pada int erval 10 x 10, yang m em enuhi adalah ...
10 ( 9) ( 8) 5 6 7 8 9 8
4 Jaw ab: D
Diket ahui f 1( x 1) 2x 7 3x 7
M aka f 1( x) o ( x 1) 2x 7 3x 7
Diperoleh
1 2x 7
f ( x) o invers( x 1) 3x 7
2x 7
o ( x 1) 3x 7
2( x 1) 7 3( x 1) 7 2x 9 3x 4
7 4
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Diket ahui f (3x4) 1, m aka1 2 9
3x 4 f ( 1) 11 3 4
3x 15
x 5
5 Jaw ab: B
2
x 3x m 0 adalah
dan
2 2 18
( ) 18
m 3 18
m 6
3 3 3
3
( ) 3 ( )
3 3 ( 6) 3 81
6 Jaw ab: D
f(x) = ax2 + 6x + a, nilai m aksim um 8
Nilai m aksim um 3
a 4 D = 8
36 4a24a
= 8
4a2
32a
36 = 0
a2
8a
9 = 0
(a
9) (a + 1) = 0 Karena grafik m em buka kebaw ah
a < 0
a =
1Sehingga sum bu simet rinya adalah x = b
2a
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7 Jaw ab: Bx 7 x 3 x 7 2x 6 x 7 2x 6
3x 1
2 4 250
2 2 250
2 250 2 250 3x 3x 3x 3 x
2 2 250
2 125 2 5 2 5
Dengan dem kian
2x x 2 1
4 2 5
25
8 Jaw ab: C
9 2 ... 222 , 0 A
log
9 3 ... 333 , 0 B
log
3
3 3
3
A
log( ) log A log B B
1 3 log A logB
3 1 3 log A logB
3 6 1 9 9 5 9 0, 555...
9 Jaw ab: D
2 2 2 x 2 2 x 2 x 2 x 2
x 2x x 4
Lim 2
x 2 x 4 x 4
x( x 2) ( x 2) ( x 2)
Lim 2
x 2 ( x 2)
x 2 Lim x 2
x 2 Lim ( x 2)
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10 Jaw ab: ABarisan arit m at ika
12, q, r , s, t , u, 60, ..., x, y, 552
7
U 60
a 6b 60 12 6b 606b 72 b 12
u60 b 601248
x 5522b55224528
x u 52848480
11 Jaw ab: E
3log x9log x81log x ... 10
3
a log x
3 1 9
2 2
3 3
1
log x
U log x 1
r
U log x log x 2
3 1 2
S 10 a
10 1 r
log x 10 1
3 5
log x 5
x 3 243
12 Jaw ab: B
3 2
2 2
f (2x 5) 4x 5x 6x 11 turunkan
2 f (2x 5) 12x 10x 6
f (2x 5) 6x 5x 3
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13 Jaw ab: A4 tan a
3
dan
< a < 23
a kuadran III
sin = (
)cos = (
) Diperoleh …4 sin a
5
3 cos a
5
3 sin( a) cos a
2 5
4 4 cos( a) sin a ( )
2 5 5
JAA = Jum lah akar-akar = 3 4
5 5
= 1
5
KAA = Kali akar-akar = 3
5
4 5 =
12 25
2
x JAA xKAA0
2 1 12
x x 0
5 25
2
25x 5x120
3 a 4
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14 Jaw ab: EYang diarsir kuadran I berart i x
0, y
0Jadi daerah diarsir (daerah B dan D) m em enuhi
x
0, y
0(5x +3 y
15) (x + 2y
4)
015 Jaw ab: E
10 2
10 ! 10 9 8! C
8! 3!
8!
45 2 1
Daerah 5x + 3y
15
x + 2y 4 ( 2x + y 4) ( x + 2y 4)
A 0 0 0
B 0 0 0
x + 2y 4 = 0 5
2
3 4
y
x A
B C
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