Question

1.已知二次函数f（x）=ax2+bx，ab为常数且a≠0，满足条件f（1-x)=f（x+1）且方程f（x）=2x有等根

当x为何值f（x）不大于-3

Answer 1

f（1-x)=f（x+1）说明函数的对称轴为x=1,也就是-b/2a=1
b = - 2a
f（x）=2x可化为：
4ax²+2bx-2x=0 ,因为b=-2a
所以
4ax²-4ax-2x=0
2ax²-2ax-x=0 因为方程有重根，
所以a=-1/2,b=1
f(x)= - (1/2)x²+x
f(x)≤ - 3 可化为：
- (1/2)x²+x≤ - 3
x²-2x-6≥0
x≥1+√7,或x≤1-√7

Answer 2

f(1-x) = a(1-x)² + b(1 -x) = ax² -(2a+b)x + a+b
f(x+1) = a(x+1)² + b(x+1) = ax² + (2a+b)x + a+b
f(1-x) = f(x+1)
(2a+b)x = - (2a+b)x
2a+b = 0, b = -2a
f(x) = ax² -2ax
f(x) = 2x, ax² -2ax = 2x
ax² -2(a+1)x = 0
x[x - 2(a+1)]= 0
x = 0, x = 2(a+1) = 0, a = -1
b = 2
f(x) = -x² +2x