很急 设函数fx=ax²+bx+c，且f（1）=－a/2，3a＞2c＞2b，求证：

来源：百度知道 编辑：UC知道 时间：2024/06/03 21:45:27
（1）a＞0且－3＜b/a＜－3/4
（2）函数fx在区间（0，2）内至少有一个零点

很急！！！谢谢
还有第三问
设x1，x2是函数fx两零点，则
根号2≤|x1-x2|＜根号57 /4

f(1) = a + b + c = -a/2 => 3a + 2b + 2c = 0

if 3a < 0, then 0 > 3a > 2b > 2C => 3a + 2b + 2c < 0, so this condition is false;

if 3a = 0, then 0= 3a > 2c > 2b => 2c + 2b < 0, this condition is also false.

hence 3a > 0 => a > 0;

OR we have 3a + 2b + 2c = 0 < 9a => a > 0;

then because c > b, 0= 3a + 2b + 2c > 3a + 2b + 2b => 3a + 4b < 0;

Similarly, 3a > 2c => 0= 3a + 2b + 2c < 6a + 2b => b > -3a

(2). Because a > 0, then f(1) = -a/2 < 0, then f(2) = 4a + 2b + c = 3a + 2b + 2c + a - c = a - c

if c < 0 => a - c > 0 ,

if c = 0 => a - c = a > 0,

if c > 0 => f(0) = c > 0.

So no matter what value c takes, f(0)f(1) < 0 OR f(2)f(1) < 0, => at least we have one root in (0,2).

The End