Diketahui :

f(x) = 2x+1

g(x) = x²-3x+4

jika a negatif memenuhi persamaan

(fog)(a) = 1 + (gof)(a)

Ditanya :

nilai a² = ?

pembahasan :

g(x) = x²-3x+4

g(a) = a² -3a +4

f(x) = 2x+1

(fog)(a) = f(g(a))

= 2(a²-3a+4)+1

= 2a²-6a+9

f(x) = 2x+1

f(a) = 2a+1

(gof)(a) =

g(f(a) =

(2a+1)²-3(2a+1)+4 =

4a²+4a+1-6a-3+4 =

4a²-2a+2

(fog)(a) = 1 + (gof)(a)

2a²-6a+9 = 1 + 4a²-2a+2

-2a²-4a+6=0

a²+2a-3 =0

(a+3)(a-1) =0

a = -3 atau a = 1

diambil a negatif yaitu:

a = -3

nilai a² = (-3)² = 9

jawaban :

nilai a² adalah 9

opsi C

pelajari soal komposisi:

yomemimo.com/tugas/46930666

yomemimo.com/tugas/37873122

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DETAIL JAWABAN

Kelas : 10

Mapel: Matematika

Bab : Fungsi

kode soal : 2

Kode : 10.2.3

Kata Kunci : fungsi, komposisi