Wednesday, February 28, 2024

Kisi Kisi PAT Matematika Kelas XI no 27

27. Diberikan 𝑓(π‘₯) = π‘₯^2 + 2 dan 𝑔(π‘₯) = π‘₯^2 βˆ’ 2. Hasil dari (𝑔 π‘œ 𝑓)(π‘₯) βˆ’ (𝑓 π‘œ 𝑔)(π‘₯)= ...

a. 8π‘₯^2 + 4

b. 8π‘₯^2 βˆ’ 4

c. βˆ’8π‘₯^2 + 4

d. 4

e. βˆ’4


Jawaban :

𝑓(π‘₯) = π‘₯^2 + 2

𝑔(π‘₯) = π‘₯^2 βˆ’ 2


(𝑔 π‘œ 𝑓)(π‘₯) βˆ’ (𝑓 π‘œ 𝑔)(π‘₯) = g (f (x)) - f (g(x))

(𝑔 π‘œ 𝑓)(π‘₯) βˆ’ (𝑓 π‘œ 𝑔)(π‘₯) = g (π‘₯^2 + 2) - f (π‘₯^2 βˆ’ 2)

(𝑔 π‘œ 𝑓)(π‘₯) βˆ’ (𝑓 π‘œ 𝑔)(π‘₯) = (π‘₯^2 + 2)^2 βˆ’ 2 - ((π‘₯^2 βˆ’ 2)^2 + 2)

(𝑔 π‘œ 𝑓)(π‘₯) βˆ’ (𝑓 π‘œ 𝑔)(π‘₯) = (π‘₯^4 + 4x^2 + 4)βˆ’ 2 - ((π‘₯^4 βˆ’ 4x^2 + 4) + 2)

(𝑔 π‘œ 𝑓)(π‘₯) βˆ’ (𝑓 π‘œ 𝑔)(π‘₯) = π‘₯^4 + 4x^2 + 4 βˆ’ 2 - π‘₯^4 + 4x^2 - 4 - 2

(𝑔 π‘œ 𝑓)(π‘₯) βˆ’ (𝑓 π‘œ 𝑔)(π‘₯) = π‘₯^4 + 4x^2 + 2 - π‘₯^4 + 4x^2 - 6

(𝑔 π‘œ 𝑓)(π‘₯) βˆ’ (𝑓 π‘œ 𝑔)(π‘₯) = 8x^2 - 4


Jawaban : B

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