【题文】已知二次函数f(x)=ax
2+bx,f(x+1)为偶函数,函数f(x)的图象与直线y=x相切.
(I)求f(x)的解析式;
(II)已知k的取值范围为[
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093652-10347.gif)
,+∞),则是否存在区间[m,n](m<n),使得f(x)在区间[m,n]上的值域恰好为[km,kn]?若存在,请求出区间[m,n];若不存在,请说明理由.
【答案】解:(1)∵f(x+1)为偶函数,∴f(-x+1)=f(x+1),
即a(-x+1)
2+b(-x+1)=a(x+1)
2+b(x+1)恒成立,
即(2a+b)x=0恒成立,∴2a+b=0,∴b=-2a,∴f(x)=ax
2-2ax,
∵函数f(x)的图象与直线y=x相切,
∴二次方程ax
2-(2a+1)x=0有两相等实数根,∴Δ=(2a+1)
2-4a×0=0,
∴a=
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093653-12864.gif)
,f(x)=-
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093653-41245.gif)
x
2+x. ......5分
(2)∵f(x)=-
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093653-41245.gif)
(x-1)
2+
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093654-89901.gif)
≤
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093654-89901.gif)
,
∴[km,kn]?(-∞,
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093653-41245.gif)
],∴kn≤
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093653-41245.gif)
,又k≥
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093652-10347.gif)
,∴n≤
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093654-66829.gif)
≤
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093654-38178.gif)
,
又[m,n]? (-∞,1],f(x)在[m,n]上是单调增函数,
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093655-77937.gif)
即-
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093655-61136.gif)
即m,n为方程-
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093653-41245.gif)
x
2+x=kx的两根,解得x
1=0,x
2=2-2k.∵m<n且k≥
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093652-10347.gif)
.
故当
![](http://img.shitiku.com.cn/uploads/allimg/20200327/20200327093652-10347.gif)
≤k<1时,[m,n]="[0,2-2k];" 当k>1时,[m,n]=[2-2k,0]; 当k=1时,[m,n]不存在.