Kunci Sbmptn - Prediksi Dan Pembahasan Soal Hots Sbmptn 2019.pdf

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√𝑥 2 + 4𝑥 ≤ 2√3 𝑥 > √10𝑥 − 25

𝐷1 𝐷2 𝐷1

𝐷2

𝐷1 ∪ 𝐷2 = {𝑥 ∈ ℝ ∣ 𝑥 ≥ 0} 𝐷1 ∩ 𝐷2 = ∅ 𝐷1 ⊆ 𝐷2 𝐷2 ⊆ 𝐷1 5 ∈ 𝐷1 ∪ 𝐷2

𝑥 2 + 4𝑥 ≤ 12 𝑥 2 + 4𝑥 − 12 ≤ 0 (𝑥 + 6)(𝑥 − 2) ≤ 0 {𝑥 ∈ ℝ ∣ −6 ≤ 𝑥 ≤ 2} 𝑥 2 + 4𝑥 ≥ 0 𝑥(𝑥 + 4) ≥ 0 {𝑥 ∈ ℝ ∣ 𝑥 ≤ −4} ∪ {𝑥 ∈ ℝ ∣ 𝑥 ≥ 0} 𝐷1 𝐷1 = {𝑥 ∈ ℝ ∣ −6 ≤ 𝑥 ≤ −4} ∪ {𝑥 ∈ ℝ ∣ 0 ≤ 𝑥 ≤ 2} 𝑥 2 > 10𝑥 − 25 𝑥 2 − 10𝑥 + 25 > 0 (𝑥 − 5)2 > 0 {𝑥 ∈ ℝ ∣ 𝑥 ≠ 5} 10𝑥 − 25 ≥ 0 25 𝑥≥ 10 5 {𝑥 ∈ ℝ ∣ 𝑥 ≥ 2}

𝐷2 5

𝐷2 = {𝑥 ∈ ℝ ∣ 𝑥 ≥ 2} ∪ {𝑥 ∈ ℝ ∣ 𝑥 ≠ 5} 𝐷1 ∩ 𝐷2 = ∅

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𝑥∈ℝ

cot(2𝑥) =

𝑥∈ℝ

cot(𝑥) =

𝑥∈ℝ

(cos(𝑥)−sin(𝑥))2 sin(2𝑥)

+1

cos(𝑥) sin(𝑥)

sin(𝑥)

1

tan(𝑥) = cos(𝑥)

cot(𝑥) = tan(𝑥)

cot(𝑥) =

cos(𝑥) sin(𝑥)

cos(2𝑥) cos2 (𝑥) − sin2 (𝑥) (cos(𝑥) + sin(𝑥))(cos(𝑥) − sin(𝑥)) cot(2𝑥) = = = sin(2𝑥) 2 sin(𝑥) cos(𝑥) 2 sin(𝑥) cos(𝑥)

𝑓(𝑥) = 3𝑥 − 1 𝑓(3) + ⋯ + 𝑓(50) = 3.575 3.675 3.775 3.875 3.975

𝑥

𝑓(1) + 𝑓(2) +

𝑓(1) = 3 ⋅ 1 − 1 = 2 𝑓(2) = 3 ⋅ 2 − 1 = 5 𝑓(3) = 3 ⋅ 3 − 1 = 8 𝑓(1), 𝑓(2), 𝑓(3), ⋯ 𝑎 = 𝑓(1) = 2 50 𝑢50 = 𝑓(50) = 3 ⋅ 50 − 1 = 50 50 (𝑎 + 𝑢50 ) = 25 ⋅ (2 + 149) = 3.775 𝑆50 = 2

149

𝑦 = 𝑎𝑥 2 + 𝑏𝑥 𝑎+𝑏 =

(4,4)

4 2 1 −2 −4

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5

𝑥=4 𝑦 ′ = 2𝑎𝑥 + 𝑏 𝑦 ′ (4) = 5 = 2𝑎 ⋅ 4 + 𝑏 8𝑎 + 𝑏 = 5 𝑏 = 5 − 8𝑎 (4,4) 4 = 𝑎 ⋅ 16 + 𝑏 ⋅ 4 4𝑎 + 𝑏 = 1 4𝑎 + (5 − 8𝑎) = 1 −4𝑎 = −4 𝑎=1 𝑏 = 5 − 8 = −3 𝑎 + 𝑏 = 1 − 3 = −2

𝑥∈ℝ 𝑥

𝑥

2 ≤3 2−𝑥 ≤ 3𝑥 2−𝑥 ≤ 3−𝑥 2𝑥 ≤ 3−𝑥

𝑥∈ℝ

𝑥≥0

2𝑥 ≤ 3𝑥 log(2𝑥 ) ≤ log(3𝑥 ) 𝑥 log(2) ≤ 𝑥 log(3) 𝑥(log(2) − log(3)) ≤ 0 log(2) − log(3) < 0

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5𝑥 2 − 𝑥 − 5 = 0 𝑥1 < 𝑥2

1−𝑥

−5 −

1 5

1 20

2𝑥2

1 +𝑥2

𝑥1

𝑥2

2𝑥 ≤ 3𝑥

1 5

5

2𝑥2 𝑥1 + 𝑥2 − 2𝑥2 𝑥1 − 𝑥2 = = 𝑥1 + 𝑥2 𝑥1 + 𝑥2 𝑥1 + 𝑥2 𝑥1 < 𝑥2 𝑥1 − 𝑥2 = −|𝑥1 − 𝑥2 | −1 1 𝑥1 + 𝑥2 = − = 5 5 √(−1)2 − 4 ⋅ 5 ⋅ (− 6) √25 √𝐷 5 𝑥1 − 𝑥2 = −|𝑥1 − 𝑥2 | = − =− =− = −1 5 5 5 1−

𝑥1 − 𝑥2 −1 = = −5 1 𝑥1 + 𝑥2 5

𝑥 log(𝑏) 1 log(𝑎) log(2𝑎 − 2) ( )=( ) log(𝑎) 1 log(𝑏 − 4) 1 35 6

6 37 6 41 6 43 6

log(2𝑎 − 2) = 1 2𝑎 − 2 = 10 𝑎=6 log(𝑏 − 4) = log(6) 𝑏−4=6 𝑏 = 10 𝑥

log(6) = log(10) 21

1

𝑥−𝑥 =

𝑥

1

1

𝑥−𝑥 = 6−6 =

log(6) = 1 𝑥=6

35 6

4

5

5 36 7 36 8 36 9 36 11 36

(1,3), (3,1), (2,2) 4 (1,4), (2,3), (3,2), (4,1)

3 5 𝑃=

𝑛

4

3 4 7 + = 36 36 36

𝑓(𝑥) = 𝑥 𝑛 + 𝑥 𝑛−1 + 𝑥 𝑛−2 + ⋯ + 𝑥 2 + 𝑥 + 1

𝑓 (𝑛) (𝑥) =

𝑛! 0 𝑛𝑛 (𝑛 − 1)𝑛 𝑥 𝑛! 𝑥

𝑓 ′ (𝑥) = 𝑛𝑥 𝑛−1 + (𝑛 − 1)𝑥 𝑛−2 + ⋯ + 3𝑥 2 + 2𝑥 + 1 𝑓 ′′ (𝑥) = 𝑛(𝑛 − 1)𝑥 𝑛−2 + (𝑛 − 1)(𝑛 − 2)𝑥 𝑛−3 + ⋯ + 4 ⋅ 3 ⋅ 𝑥 2 + 3 ⋅ 2 ⋅ 𝑥 + 2 ⋅ 1 𝑓 ′′′ (𝑥) = 𝑛(𝑛 − 1)(𝑛 − 2)𝑥 𝑛−3 + (𝑛 − 1)(𝑛 − 2)(𝑛 − 3)𝑥 𝑛−4 + ⋯ + 5 ⋅ 4 ⋅ 3 ⋅ 𝑥 2 +4⋅3⋅2⋅𝑥+3⋅2⋅1 𝑛 𝑛 𝑓 (𝑛) (𝑥) = 𝑛(𝑛 − 1)(𝑛 − 2) ⋯ 3 ⋅ 2 ⋅ 1 ⋅ 𝑥 0 = 𝑛!

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(2𝑥 + 𝑦 − 4)(𝑥 + 2𝑦 − 5) ≤ 0 { 𝑥≥0 𝑦≥0 (1,2) 𝑓1 (𝑥, 𝑦) = 3𝑥 + 2𝑦 𝑓2 (𝑥, 𝑦) = 𝑥 + 2𝑦 𝑓3 (𝑥, 𝑦) = 5𝑥 + 𝑦 𝑓4 (𝑥, 𝑦) = 𝑥 + 𝑦

(0,4) (0; 2,5) (1,2) (2,0)

(5,0)

(1,2)

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3 t 4 4 t 5 5 t 6 6 t 7 7 t 8

W Q %

V2 t  mcT R t  mcT

R %V 2

t  1 4200  70  10 

R  70   100  100 

t  3600 R

t2  mcT

R %V 2

t2  1 4200  70  10  t2  3150 R 7 t2  t 8

η

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R  80   100  100 

1  2 2  3 3  4 4  5 5  6

𝑄2 ) 𝑄1 1500 = (1 − ) 2500 = 0,4

𝜂1 = (1 −

𝑄2 ) 𝑄1 1750 = (1 − ) 2500 = 0,3 3 3 𝜂2 = 𝜂1 = 𝜂 4 4 𝜂2 = (1 −

1 vmaks 4

1 4 1 4 1 4 1 4 1 4

15 A 15 A 15 A 15 A 15 A

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vmaks  A v   A2  y 2 1 vmaks   A2  y 2 4 1 A    A2  y 2 4 2



1  2 2  A  A  y 4  1 2 A  A2  y 2 16 1 y 2  A2  A2 16 15 y 2  A2 16 15 y A 16 1 y 15 A 4

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

2

1 1 1 1 3  2 1 6      1 Rp 2 3 6 6 6 1 1 Rp R p  1 I

I1 : I 2 : I 3 

V 12   12 A R 1 1 1 1 1 1 1 : :  : : R1 R2 R3 2 3 6

I1 : I 2 : I 3  3: 2 :1

3 12  6 A 6 2 I 2  12   4 A 6 1 I3  12   2 A 6 I1 

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o i BQ 2 aQ  o i BP 2 aP BQ BP



BQ 

oi 2 aP oi 2 aQ aP 2 BP  B  2 B aQ 1

 o o  o 

𝜃1 = 0

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  



35

fp 

v  vp v vs

36

fs

37

38

56 96 14 16

[𝐻+]

10−1

𝑎

2

𝑀1 𝑉1

0,05𝑀 𝑥 1𝑚𝑙

𝑉2

100 𝑚𝑙

39





𝑃𝑥𝑉 𝑅𝑥𝑇

2 𝑎𝑡𝑚 𝑥 0,5 𝑙𝑖𝑡𝑒𝑟 0,082 𝑙

𝑎𝑡𝑚 𝐾𝑥 𝑚𝑜𝑙

303 𝐾

𝑚𝑎𝑠𝑠𝑎

4,16 𝑔𝑟𝑎𝑚

𝑛

0,04 𝑚𝑜𝑙

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0,5 𝑚𝑚𝑜𝑙 100 𝑚𝑙



√𝐾𝑠𝑝

√10−10 10−5

2 𝑚𝑚𝑜𝑙 100 𝑚𝑙 2,5 𝑚𝑚𝑜𝑙 100 𝑚𝑙

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   

  𝑒𝑖𝑡

65 𝑥 0,75𝑥 321𝑥60 2

96.500

96.500

42

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2𝜋 3

2𝜋



𝑃 2𝜋 2𝜋⁄ 3



𝑏−0



𝑎− 0

= −2

 −1 ) −5

 

( −1 ) −5

(

−𝑏 − 5



𝑎− 1

 44

=3

√30 √2 √34 √38 H

G

E

 

√𝐻𝐷2

+

𝐻𝑃 = √𝐻𝐷2 + (𝐷𝐴2 + 𝐴𝑃2 ) 𝐻𝑃 = √42 + (42 + 22 ) = 6

 

D

C

𝑃𝑄 = √𝑃𝐵 2 + 𝐵𝑄 2 = 2√2 A 2

√34

sin(3𝑥 − 12)

𝑥 → 4 4 − √20 − 𝑥

lim

sin(3𝑥 − 12)

𝑥 → 4 4 − √20 − 𝑥

⨯

4 + √20 − 𝑥 4 + √20 − 𝑥

Q

R

𝐻𝑅 = √𝐻𝑃2 − 𝑃𝑅 2 = √62 − (√2)

lim

F

𝐷𝑃2

lim

(sin(3𝑥 − 12))(4 + √20 − 𝑥)

16 − (20 − 𝑥) 𝑥→4 (sin 3(𝑥 − 4))(4 + √20 − 𝑥)

lim

𝑥–4

𝑥→4

3(4 + √20 − 4) = 24

45

P

B

α

3

α

2

√3 √3 √3 √3 √3



α

3

α

2 𝑈2



𝑈1 3

α

α

α α

α

α α

1

α

2 1

α

2

1

√3 2

√3 𝑎(𝑟 𝑛 − 1)



𝑟− 1 8 1 ((√3) 2

− 1)

√3 − 1 1 (81 − 1) 2

3−1

×

√3 + 1 √3 + 1

(√3 + 1)

√3

160 3 260 3 290 3 320 3 640 3

𝑈2

α α



𝑈3

α

2

α

α

=

𝜋 𝜋 𝜋 𝜋 𝜋 46

1

𝑈2

2

𝑈1

 2



𝑉 = 𝜋 ∫0 {(9 − 𝑥 2 )2 − (1 + 𝑥 2 )2 } 𝑑𝑥



𝑉 = 𝜋 ∫0 (80 − 20𝑥 2 ) 𝑑𝑥 = 𝜋 (80𝑥 −

2

Y y = 1 + x2

y = 9 – x2 X 2

   

⨯ ⨯ ⨯ ⨯

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20 3

2 𝑥 3 )| = 𝜋 (80(2) − 0

20 3

(2)3 ) =

320 3

√5

1



√ 𝐴2 + 4 1

√ 𝐴2 + 4 5 4

1 4

4

(2𝐴)2 − 𝐶

(2𝐴)2 − 𝐶

𝐴2 − 𝐶 = 25

𝐴2 =

𝐶 + 25 5⁄ 4 1



√ 𝐴2 + 4 1

√ 𝐴2 + 4 10 4

1 4

(3𝐴)2 − 𝐶

√5

𝐴2 − 𝐶 = 45

𝐴2 = 

1

𝐶 + 45 10⁄ 4 𝐶 + 25 𝐶 + 45 = 10⁄ 5⁄ 4 4



48

1 4

(3𝐴)2 − 𝐶

√5





1

3

3

2

2 3 2 3 1 2 1 6 2 3



  9 6

=1

1

3

5

3 1

2 3

6 4

3

2

6

1 2

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50

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    

52

-

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A. B. C. D. E.

A. B. C. D. E.

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  

62

63

   

64

  

65

A. B. C. D. E.

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



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