已知函數(shù)f(x)=logax,g(x)=loga(2x+m-2),其中x∈[1,2],a>0且a≠1,m∈R.
(I)當m=4時,若函數(shù)F(x)=f(x)+g(x)有最小值2,求a的值;
(Ⅱ)當0<a<l時,f(x)≥2g(x)恒成立,求實數(shù)m的取值范圍.
【答案】
分析:(I)將m=4代入F(x),求出其定義域,先判斷其為增函數(shù),根據(jù)題意函數(shù)F(x)=f(x)+g(x)有最小值2,列出等式,求a的值;
(Ⅱ)0<a<l,求出其定義域,可以令h(x)=4x
2+(4m-9)x+(m-2)
2,對其進行配方,分類討論,求出h(x)的最小值,讓其大于0即可求實數(shù)m的取值范圍;
解答:解:(I)由題意,m=4時,F(xiàn)(x)=f(x)+g(x)=log
ax+log
a(2x+2)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/0.png)
,
又x∈[1,2],則2x
2+2x∈[4,12].
而函數(shù)F(x)=f(x)+g(x)有最小值2,
∴a>1,解得a=2;
(Ⅱ)由題意,0<a<1時,∵f(x)≥2g(x),
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/1.png)
⇒
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/2.png)
,
⇒
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/3.png)
,
令h(x)=4x
2+(4m-9)x+(m-2)
2=4[x-(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/4.png)
-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/5.png)
)]
2+(m-2)
2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/6.png)
,
(1)當0<m<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/7.png)
時,1<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/8.png)
-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/9.png)
<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/10.png)
,
函數(shù)h(x)min=(m-2)
2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/11.png)
≥0,
解得m無解;
(2)當m≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100200180822422/SYS201311031002001808224019_DA/12.png)
時,函數(shù)h(x)在x∈[1,2]上的單調(diào)遞減,
則h(x)
min=h(1)=m
2-1≥0⇒m≥1.
綜上,實數(shù)m的取值范圍為[1,+∞).
點評:此題主要考查對數(shù)函數(shù)的性質(zhì)及其應用,解題的過程中利用到了轉(zhuǎn)化的思想,考查的知識點比較大,是一道難題;