Pembahasan no 1 — 5
6. Jawab : D
P(x) = (x — a)4 + (x — b)5 + (x — 3)
Pembagi = x2 — (a + b)x + ab = (x — a)(x — b)
Karena habis dibagi (x — a)(x — b) maka P(a) = 0 dan P(b) = 0
P(a) = (a — b)5 + a — 3 = 0
P(b) = (b — a)4 + b — 3 = 0
(a — b)4 + b — 3 = 0
(a — b)4 = 3 — b
P(a) = (a — b)5 + a — 3 = 0
(a — b)(a — b)4 + a — 3 = 0
(a — b)(3 — b) + a — 3 = 0
3a — ab — 3b + b2 + a — 3 = 0
4a — ab = 3b + 3 – b2
a(4 — b) = 3b + 3 – b2
7. Jawab : E
(0,0081)(x2 + 3x + c) < (0,09) (x2 — 2x + 8)
(0,09)(2x2 + 6x + 2c) < (0,09) (x2 — 2x + 8)
2x2 + 6x + 2c > x2 — 2x + 8
x2 + 8x + 2c — 8 > 0
Agar selalu > 0 maka bentuk ini harus definit positif segingga
a > 0
dan
D < 0
82– 4.1.(2c — 8) < 0
64 — 8c + 32 < 0
96 < 8c
8c > 96
c > 12
8. Jawab : D
x1 + x2 = 3.3log 2 =3log 23 = 3log 8
9x — 3x + 1 — 3 x + 2 — 3.3x + 3 + a = 0
9x — 3x . 31– 3x .32– 3.3x. 33 + a = 0
Misal 3x =y sehingga
y2 – 3y — 9y — 81y + a = 0
y2 – 93y + a = 0
y1.y2 = a
3x1.3x2=a
3x1+x2=a
33log 8=a
8 = a
a = 8
9. Jawab : E
10. Jawab : C