高中數學多項式

1. For constants a and b.
Set (ax^2+4x+b)/(x^2+2)= k (for some real number x), then
(k-a)x^2-4x+(2k-b)=0 has real solutions,
thus
16-4(k-a)(2k-b)>=0
2k^2-(2a+b)k+(ab-4) <=0 ---(A)
The solution of (A) is 1<=k<=4, so
(k-1)(k-4)<=0 ---(B)
Comparing (A) and (B), then
2a+b=10, ab-4=8
so that, (a, b)=(2, 6) or (3, 4)

2. f(99)≡f(-2) (mod 101)
f(-2)=-32+100-94+14=-12
so, f(99) mod 101= -12+101=89

2010-01-10 13:27:20 補充：
a mod 101表示整數a除以101的餘數
如: 123 mod 101= 22
38 mod 101= 38
-50 mod 101= 61

f(x)=(x+2)(4x^2+17x+13)-12
f(99)=101*a-12=101(a-1)+89