3. Matematika X.mipa (peminatan)

KUNCI JAWABAN PENILAIAN AKHIR SEMESTER (PAS) TAHUN 2018

Mata Pelajaran Kelas

: MATEMATIKA PEMINATAN : X.MIPA

1. A 2. C 3. B 4. A 5. D 6. E 7. E 8. D 9. C 10. D

31. (

32𝑎−3 𝑏 2 23 𝑎−5 𝑏 −6

11. B 12. A 13. D 14. A 15. D 16. B 17. E 18. A 19. C 20. E

−1

)

=

=

1 32𝑎−3 𝑏2 ( 3 −5 −6 ) 2 𝑎 𝑏

23 𝑎−5 𝑏 −6 25 𝑎−3 𝑏 2

=

21. C 22. B 23. E 24. B 25. A 26. B 27. E 28. A 29. C 30. B

23 𝑎−5 𝑏 −6 32𝑎−3 𝑏 2

1

= 23−5 𝑎−5−(−3) 𝑏 −6−2 = 2−2 𝑎−2 𝑏 −8 = 4𝑎2 𝑏8

32. 4𝑥−2𝑦+1 = 82𝑥−𝑦 22(𝑥−2𝑦+1) = 23(2𝑥−𝑦) 2(𝑥 − 2𝑦 + 1) = 3(2𝑥 − 𝑦) 2𝑥 − 4𝑦 + 2 = 6𝑥 − 3𝑦 2𝑥 − 6𝑥 − 4𝑦 + 3𝑦 = −2 −4𝑥 − 𝑦 = −2 …(1) 3𝑥+𝑦+1 = 92𝑥−𝑦−4 3𝑥+𝑦+1 = 32(2𝑥−𝑦−4) 𝑥 + 𝑦 + 1 = 2(2𝑥 − 𝑦 − 4) 𝑥 + 𝑦 + 1 = 4𝑥 − 2𝑦 − 8 𝑥 − 4𝑥 + 𝑦 + 2𝑦 = −8 − 1 −3𝑥 + 3𝑦 = −9 …(2) 𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑦 𝑑𝑎𝑟𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 (1) 𝑑𝑎𝑛 (2)

−4𝑥 − 𝑦 = −2 | X 3 −3𝑥 + 3𝑦 = −9 | X 1 −12𝑥 − 3𝑦 = −6 −3𝑥 + 3𝑦 = −9 + −15𝑥 = −15 𝑥=1 𝑥 = 1 substitusi ke persamaan (1) −4(1) − 𝑦 = −2 −4 − 𝑦 = −2 −4 + 2 = 𝑦 −2 = 𝑦 Jadi nilai 𝑥 = 1 dan 𝑦 = −2

33.

2

log 2 𝑥 − 2log𝑥 3 = 4

𝑦 = 2log 𝑥 atau 𝑦 = 2log 𝑥 4 = 2log 𝑥 atau −1 = 2log 𝑥

2

log 2 𝑥 − 3 2log𝑥 − 4 = 0 Misal 𝑦 = 2log 𝑥, maka diperoleh: 𝑦 2 − 3𝑦 − 4 = 0 (𝑦 − 4)(𝑦 + 1) = 0 𝑦 = 4 atau 𝑦 = −1

34. √34−6𝑥 > 3 𝑥

4-6x

1

𝑥 = 24 = 16 atau 𝑥 = 2−1 = 2 1

Jadi HP = { 2 ,16 }

2 −𝑥−1

> 2x2-2x-2

2x2 + 4x -6 < 0 X2+2x-3 < 0 (x - 1)(x + 3) < 0 -3< x < 1

35. Tabel titik bantu f(x) = 3x – 1 x f(x) = 3x – 1 (x, y)

-1

0

1

2

2 3

0

2

8

(0, 0)

(1,2)

(2,8)



(-1, 

2 ) 3

y 8

2 X -1

1

2

_ 1 1 1 1