【题文】二次函数f(x)满足f (x+1)-f (x)=2x且f (0)=1.⑴求f (x)的解析式;⑵在区间[-1,1]上,y=f (x)的图象恒在y=2x+
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【题文】二次函数f(x)满足f (x+1)-f (x)=2x且f (0)=1.
⑴求f (x)的解析式;
⑵在区间[-1,1]上,y=f (x)的图象恒在y=2x+m的图象上方,试确定实数m的范围.
答案
【答案】(1)
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073735-90429.png)
;(2)
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073735-42797.png)
.
解析
【解析】
试题分析:(1)根据二次函数
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073735-40247.png)
满足条件
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-95246.png)
,及
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-16127.png)
,可求
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-15230.png)
,
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-41744.png)
,从而可求函数
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073735-40247.png)
的解析式;(2)在区间
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-92403.png)
上,
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-76252.png)
的图象恒在
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073737-55978.png)
的图象上方,等价于
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073737-64192.png)
在
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-92403.png)
上恒成立,等价于
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073737-96031.png)
在
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-92403.png)
上恒成立,求出左边函数的最小值,即可求得实数
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073737-85929.png)
的取值范围.
试题解析:(1)由
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-16127.png)
,令
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073737-93003.png)
,得
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-15230.png)
;令
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073737-75879.png)
,得
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-41744.png)
.
设
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073737-35841.png)
,故
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073738-97982.png)
解得
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073738-86431.png)
故
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073735-40247.png)
的解析式为
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073735-90429.png)
.
(2)因为
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-76252.png)
的图像恒在
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073737-55978.png)
的图像上方,所以在
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-92403.png)
上,
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073737-64192.png)
恒成立.即:
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073737-96031.png)
在区间
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-92403.png)
恒成立.所以令
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073738-65766.png)
,故
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073738-80302.png)
在
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073736-92403.png)
上的最小值为
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073738-73353.png)
,∴
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327073735-42797.png)
.
考点:二次函数的性质.
举一反三
【题文】二次函数f(x)满足f (x+1)-f (x)=2x且f (0)=1.
⑴求f (x)的解析式;
⑵在区间[-1,1]上,y=f (x)的图象恒在y=2x+m的图象上方,试确定实数m的范围.
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