【答案】
分析:(Ⅰ)在函數(shù)y=f(x)的圖象上任意一點(diǎn)Q(x
,y
),設(shè)關(guān)于原點(diǎn)的對(duì)稱(chēng)點(diǎn)為P(x,y),再由中點(diǎn)坐標(biāo)公式,求得Q的坐標(biāo)代入f(x)=x
2+2x即可.
(Ⅱ)將f(x)與g(x)的解析式代入轉(zhuǎn)化為2x
2-|x-1|≤0,再通過(guò)分類(lèi)討論去掉絕對(duì)值,轉(zhuǎn)化為一元二次不等式求解.
(Ⅲ)將f(x)與g(x)的解析式代入可得h(x)=-(1+λ)x
2+2(1-λ)x+1,再用二次函數(shù)法研究其單調(diào)性.
解答:解:(Ⅰ)設(shè)函數(shù)y=f(x)的圖象上任意一點(diǎn)Q(x
,y
)關(guān)于原點(diǎn)的對(duì)稱(chēng)點(diǎn)為P(x,y),
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212627758403160/SYS201310232126277584031021_DA/0.png)
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212627758403160/SYS201310232126277584031021_DA/1.png)
∵點(diǎn)Q(x
,y
)在函數(shù)y=f(x)的圖象上
∴-y=x
2-2x,即y=-x
2+2x,故g(x)=-x
2+2x
(Ⅱ)由g(x)≥f(x)-|x-1|,可得2x
2-|x-1|≤0
當(dāng)x≥1時(shí),2x
2-x+1≤0,此時(shí)不等式無(wú)解.
當(dāng)x<1時(shí),2x
2+x-1≤0,解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212627758403160/SYS201310232126277584031021_DA/2.png)
.
因此,原不等式的解集為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212627758403160/SYS201310232126277584031021_DA/3.png)
.
(Ⅲ)h(x)=-(1+λ)x
2+2(1-λ)x+1
①當(dāng)λ=-1時(shí),h(x)=4x+1在[-1,1]上是增函數(shù),∴λ=-1
②當(dāng)λ≠-1時(shí),對(duì)稱(chēng)軸的方程為x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212627758403160/SYS201310232126277584031021_DA/4.png)
.
�。┊�(dāng)λ<-1時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212627758403160/SYS201310232126277584031021_DA/5.png)
,解得λ<-1.
ⅱ)當(dāng)λ>-1時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212627758403160/SYS201310232126277584031021_DA/6.png)
,解得-1<λ≤0.綜上,λ≤0.
點(diǎn)評(píng):本題主要考查求對(duì)稱(chēng)區(qū)間上的解析式,解不等式及研究函數(shù)的單調(diào)性,屬中檔題.