2. pembahasan soal teta shofi (fix)

8/17/2019 2. Pembahasan Soal TETA SHOFI (Fix)

1/13

Pembahasan Soal

Pilihan Ganda

1. x2

+ (a-3)x + 8 = 0 ; α =

1

2 β dan a>0

α + β =−ba =

−(a−3)1 = 3-a

α ∙ β =c

a =8

1 = 8

α ∙ β = 8

(karena α =

1

2 β)

1

2 β. β

= 8

β2 = 16

β = ±4

α + β = 3-a1

2 β + β = 3-a

3

2 β = 3-a

β =6−2a

3

4 =6−2a

3

a = -3 (tidak memenui)

−4 =6−2a

3

a = ! (memenui)

Jawaban: E

2. x2 - ("+m)x - 8 = 0 ; α =−1

2 β dan a>0

α + β =−ba =

−−(5+m)1 = "+m

α ∙ β =

c

a =

−8

1 = -8


2/13

α ∙ β = -8

(karena α =−1

2 β)−1

2 β. β = -8

β2 = 16

β = ±4

α + β = "+m−1

2 β + β = "+m

1

2 β = "+m

β = 10+2m β = 10+2m

4

= 10+2m - 4

= 10+2m m = -3 m = -#

Jawaban: A

3. $(x) = (%-1)x2 + 2%x + % + 3

a= %-1& '=2%& =%+3

*arat dikatakan denit %,iti$ a>0 dan 0

a>0 0

%-1>0 '2-4a0

%>1 4%2-4(%-1)(%+3)0 4%2-4%2-2%+120

%>6

Jawaban: E 4. $(x) = x2 + (k-")x / 4 = 0

a= 1& '=(k-")& =-4

*arat memiiki akar kem'ar adaa =0

'2-4a=0

(k-")2-4∙1∙(-4)=0

k2

-10k-16=0 (k-8) (k-2)=0

k=8 atau k=2

Jawaban: A". rak x2 + 6x + 8 = 0

x2 + 6x + 8 = 0

(x+4) (x+2) = 0

x= -4 atau x= -2

(-4&0) atau (-2&0)


3/13

Jawaban: C6. tan 300 = tan (360-60)

= tan (-60) = −√ 3

Jawaban: A#. in #" = in (4"+30)

= in 4" , 30 + , 4" in 30

=1

2 √ 2 ∙

1

2 √ 3 +

1

2 √ 2 ∙

1

2

=1

4 √ 2 ( √ 3 + 1)

Jawaban: C

8. in #0 , ∙ in "0 , =−1

2 ( ,(#0+"0) / ,(#0-"0))

=−1

2 ( ,(120) / ,(20))

=−1

2 (−1

2 / ,(20))

=

1

4

+1

2 ,(20)

Jawaban: C

!. in x =3

5 untuk 0 , x !0 , dan , * =

3

5 untuk 0 , x !0

,

, x =4

5 in * =4

5

, (x+*) = , x , * / in x in *

=4

5 3

5 -3

5 4

5

= 0

Jawaban: A

10. in x =3

5 untuk 0 , x !0 , dan , * =

3

5 untuk 0 , x

!0 ,

, x = 45 in * = 45


4/13

4

3

5

x

x

, (x+*) = , x , * + in x in *

=4

5 3

5 +3

5 4

5

=24

25

Jawaban: D11.

Jawaban: A

12.

Jawaban: B

13.

Jawaban: D

14.

2 = 2 + 2

= 42 + 32

= 16 + !

= 2"

= √ 25 = "

adi %an7an = "

2 = 2 + 2

= 42 + 42

= 16 + 16

= 32

= √ 32 = 4 √ 2 & 9 = AC

2 =

4 2

9

9= √ BE2− EO2

= √ (5√ 2 )2 – (52 √ 2)

2

= √50 – 25

2 = √75

2 =5

2√ 6

5= √ AB2+ AE2

= √ x2+ x2

= √ 2 x2

= x √ 2

:in = AE EB

= x x√ 2

=12 √ 2 & = 4",


5/13

Jawaban: D

1".

Jawaban: C

16.

Jawaban: B

1#.

Jawaban: A

18.

Jawaban: C

1!.

5= √ AB2+ AE2

= √ a2+a2

= √ 2a2

= a √ 2

:in = AE

EB=

a

a√ 2=

1

2 √ 2 & = 4",

a

a

x−2

√ 3 x−2−√ 2 x=¿ lim

x→ 2

x−2

√ 3 x−2−√ 2 x x √ 3 x−2+√ 2 x

√ 3 x−2+√ 2 xlim x→ 2

¿

=

lim x→2

( x−2 ) (√ 3 x−2+√ 2 x )

3 x−2−2 x

=

lim x→2

( x−2 ) (√ 3 x−2+√ 2 x )

x−2

( x−1 )3

( x−1 ) ( x+8 )¿

x3−3 x2+3 x−1

x2+7 x−8

=¿ lim x→ 1

¿

lim x→1

¿

4 x3+2 x2−5

8 x

3

− x+2

= lim x→∞

¿

4 x3

x3 +

2 x2

x3 −

5

x3

8 x

3

x3 − x

x3− 2

x3

lim x→∞

¿


6/13

Jawaban: A

20.

Jawaban: E

21. ika 'an*ak 'una eati 12 tankai denan %erentae 10


7/13

1314-

161#-

1!20-

22

46"

1"

18

21

60

108

10"

∑i=1

n

❑ 40 "04

´ x=∑i=1

n

f i ∙ xi

∑i=1

n

f i

=504

40 =12,6

adi& rata-rata %,in *an di%er,e %emain tere'ut 12&6 Jawaban: E

23.

ea ?1

Q1=data ke−25+1

4

?1 teretak %ada kea inter@a 208-211

Q1= L1+

(

1

4 n−f k

Q 1

f Q1

)∙ p

¿207,5+( 1

4 25−6

2 ) ∙4

¿207,5+0,5

¿208

Jawaban: A

Berat

(ram)

Frekuen

si200-203204-20#208-211

212-21"216-21!220-223224-22#

332

14#"

uma 2"


8/13

24.

Pan!an Tubuh"kan(#m)

Frekuensi($ i)

Frekuensi%umulati$

($ k)8-10

11-1314-161#-1!20-2223-2"26-28

"3

126!#8

"8

20263"42"0

uma "0

e = data ke- ( 50+12 ) = data ke-2"&"e teretak di kea inter@a 1#-1!

e = L+( n

2−f k

M e

f M e) ∙ p

= 16,5+(25−20

6 )∙3

= 16,5+2,5

= 1! Jawaban: D

2". :euai *arat-*arat *an ada& *aitu 1. ianan tia diit e'i dari 4002. ianan ena%3. nka ',e diuan& makanka *an da%at menem%ati diit %ertama adaa 4& #& 8& ! = 4nka *an da%at menem%ati diit kedua adaa 1& 2& 4& #& 8& ! =6nka *an da%at menem%ati diit ketia adaa 2& 4& 8 = 3enan menunakan aturan %erkaian& 'an*ak 'ianan *an

da%at di'entuk adaa ai %erkaian dari 4×6×3=72

Jawaban: C


9/13

26. 12" =12!

5 ! (12−5 )! =

12 !

5 !7 !=

12×11×10×9×8×7 !

5 !7 ! =

12×11×10×9×8

5! =

12×11×10×9×8

5×4×3×2×1 =792

Jawaban: B

2#.A AA AAA

:atu eminai uda di%atikan akan men7adi 7uara A& maka teriaem'ian eminai untuk mem%ere'utkan dua %,ii 7uara. an*akkemunkinan uunan 7uara *an ter7adi ada !B2 kemunkinan.

an*ak uunan 7uara = !B2 =9 !

(9−2 ) !=

9 !

7 !=

9×8×7 !

7 ! =9×8=72

Jawaban: B

28. adu

AA

adu A

1 2 3 4 " 6

1 (1&1) (1&2) (1&3) (1&4) (1&") (1&6)2 (2&1) (2&2) (2&3) (2&4) (2&") (2&6)

3 (3&1) (3&2) (3&3) (3&4) (3&") (3&6)4 (4&1) (4&2) (4&3) (4&4) (4&") (4&6)" ("&1) ("&2) ("&3) ("&4) ("&") ("&6)6 (6&1) (6&2) (6&3) (6&4) (6&") (6&6)

ari ta'e diata& teriat 'aCa ada 36 titik am%e dan mata dadu*an munu denan 7uma 10 adaa (4&6)& ("&")& (6&4)& makan() = 3n(:) = 36

eina n ( P )=n( A)

n (S )

= 3

36=

1

12

Jawaban: E

2!. an*ak titik am%e n(:) = 1#3 =

17!

3 ! (17−3 ) !=

17 !

3 !14 !=

17×16×15×14 !

3!14 ! =

17×16×15

3 ! =

17×16×15

3×2 =

4080

6 =680

emunkinan teram'i 2 n,@e

B() = "2 =5 !

2 ! (5−2)!

= 5!

2!3!

=5×4×3 !

2!3!

=5×4

2!

=20

2 =10

emunkinan teram'i 1 'uku 7eni ain


10/13

B() = (8+4)1 = 121 =12 !

1 ! (12−1 ) !=

12!

1!11!=12×11!

1 !11! =

12

1!=12

B( ∩ ) = B() × B() = 10 ×12 = 120

adi %euan teram'i 2 n,@e dan 1 'uku 7eni ain

n(B) =nP( A∩B)

n(S) =

120

680

Jawaban: D

30. n(:) = 4+3+3=10

kemunkinan teram'in*a ',a meran() = 4

B() =n( A)

n(S)

= 4

10

kemunkinan teram'in*a ',a itamn() = 3

B() =n(B)n (S )

= 3

10

7adi %euan teram'in*a ',a mera atau itam adaa

B( ∪B ) = B() + B() =4

10+

3

10=

7

10

Jawaban : A

3.

4.

B= √ BA2− AP2

= √ (12√ 3 )2+122

= √ 576 = 24

, t = AB

BP=

12√ 3

24 =

1√ 3

2 & t = 300&

:in t =1

2 & :in t = A

AB

A

AB =1

2lim x→ 2

x4−1

x2+1

dan lim x→−2

x4−1

x2+1

*=$(x)= x

4−1 x

2+1

Diai %endekatan $(x)= x

4−1

x2+1 %ada aat x mendekati 2 adaa

E 1&8 1&! 1&!!F 1&!!! 1&!!!!* 2&24 2&61 2&!6 2&!! 3&00

= x

4−12 -


11/13

". da dua kemunkinan %aanan keeren *an diam'i.a. Benam'ian A keeren %uti

Benam'ian AA keeren 'iru1 = ke7adian teram'i keeren %uti %ada %enam'ian An(1) = "n(:1) = 6 + " = 11

B(1) =n( A1)n(S1)

= 5

11

2 = ke7adian teram'i keeren 'iru etea %enam'iankeeren %utin(2) = 6keeren %uti %ada %enam'ian A tidak dikem'aikan& makan(:2) = 4 + 6 = 10

B(2) =n( A2)n(S2)

= 6

10

1∩ 2 = ke7adian teram'i keeren %uti %ada %enam'ian A

dan keeren 'iru %ada %enam'ian AA

B(1∩ 2) = B(1)

× B(2) =5

11 ×

6

10= 3

11

'. Benam'ian A keeren 'iruBenam'ian AA keeren %uti3 = ke7adian teram'i keeren 'iru %ada %enam'ian An(3) = 6n(:3) = 6 + " = 11

B(3) = n( A3)n(S3)= 611

4 = ke7adian teram'i keeren %uti etea %enam'iankeeren 'irun(4) = "keeren 'iru %ada %enam'ian A tidak dikem'aikan& makan(:4) = " + " = 10

B(4) =n( A4)n(S4)

= 5

10=1

2


12/13

3∩ 4 = ke7adian teram'i keeren 'iru %ada %enam'ian A

dan keeren %uti %ada %enam'ian AA

B(3∩ 4) = B(3)

× B(4) =6

11 × 1

2= 3

11

adi&

%euan teram'in*a 1 keeren %uti = B(1∩ 2) + B(3

∩ 4)

=3

11+

3

11=

6

11

ESSA&

1. in2x ( ,2x + 2 in2x ) = in2x

in2x ( ,2x + 2 in2

x ) = in2x ,2x + 2 in2

x in2x = 2 inx ,x ,2x + 2 in2x in2x

= 2 inx ( ,x ,2x + inx in2x)

= 2 inx ( , (2x-x))

= 2 inx ,x

= in2x

aka ter'ukti 'aCa in2x ( ,2x + 2 in2x ) = in2x


13/13

2.sin70cos20

cos70sin20 =sin 90+sin 50sin 90−sin 50

=1+sin501−sin 50

2. pembahasan soal teta shofi (fix)

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