Page 27 - Buku Siap OSN Matematika SMP 2015(1)
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Aljabar
f (x + 1) = f(x) – 12
f (x + 1) = [2000 – (x – 1)(12)] – 12
f (x + 1) = [2000 – (12x – 12)] – 12
f (x + 1) = 2000 – 12x + 12 – 12
f (x + 1) = 2000 – 12x
Maka:
f(100) = f(99 + 1) x = 99
f(99 + 1) = 2000 – 12x
f(99 + 1) = 2000 – 12 99
f(100) = 2000 – 1188 = 812
Jadi, nilai f(100) = 812.
3. Jika f adalah fungsi sehingga f(xy) = f(x – y) dan f(6) = 1, maka f(–2) – f(4) =..
Jawab:
Faktor positif dari 6 = {1, 2, 3, 6}
f(xy) = f(x – y)
f(6 1) = f(6 – 1)
f(6) = f(5)
1 = f(5) f(5) = 1
f(xy) = f(x – y)
f(5 1) = f(5 – 1)
f(5) = f(4)
1 = f(4) f(4) = 1
f(xy) = f(x – y)
f(2 3) = f(2 – 3)
f(6) = f(–1)
18 Wahyu
