已知二次函数f(x)=ax^2+bx+1(a>0) 设方程f(x)=x的两个实数根为x1和x2

题目

已知二次函数f(x)=ax^2+bx+1(a>0) 设方程f(x)=x的两个实数根为x1和x2
(1) 如果 a=2 且x1

答案

f(x) = 2x^2 + bx + 1 = x,
2x^2 + (b-1)x + 1 = 0.
(b-1)^2 - 8 > 0,
(b-1)^2 > 2*2^(1/2)
b > 1 + 2^(3/2)
或
b < 1 - 2^(3/2).
设g(x) = f(x) - x = 2x^2 + (b-1)x + 1
曲线g(x)是开口向上的抛物线.
(1)
g(x1) = 0 > g(2) < 0 = g(x2) < g(4)
8 + 2(b-1) + 1< 0 < 32 + 4(b-1) + 1.
b < -7/2,b > -29/4
-29/4 < b < -7/2.b > 1 + 2^(3/2)或b < 1 - 2^(3/2).
综合,有
-29/4 < b < -7/2
（2）
|x2 - x1| < 2,
4 > (x2 - x1)^2 = (x2 + x1)^2 - 4x2x1 = (b/2)^2 - 4(1/2) = b^2/4 - 2,
b^2 < 24
|b| < 2*6^(1/2)
-2*6^(1/2) < b < 2*6^(1/2).
-2 < x2 - x1 < 2,
-2 = 0 - 2 < x1 - 2 < x2 < x1 + 2 < 2 + 2 = 4
-2 < 0 < x1 < 2 < 4.
0 < g(-2) = 8 -2(b-1) + 1 = 11 - 2b,b < 11/2
0 < g(4) = 32 + 4(b-1) + 1 = 29 + 4b,b > -29/4
-29/4 < b < 11/2.
又,b > 1 + 2^(3/2)或b < 1 - 2^(3/2).
综合,有
1 + 2^(3/2) < b < 11/2或-29/4 < b < 1 - 2^(3/2)