已知函數(shù)f(x)=loga(ax2-2x+4-2a)(a>0且a≠1).
(1)當(dāng)a=2時(shí),求函數(shù)f(x)的值域;
(2)若函數(shù)f(x)在(1,+∞)上為增函數(shù),求a的取值范圍.
解:(1)當(dāng)a=2時(shí),
f(x)=log
2(2x
2-2x),
設(shè)
,
則
,
解得u>0,
所以y=log
2u∈R,函數(shù)f(x)的值域?yàn)镽.
(2)設(shè)u(x)=ax
2-2x+4-2a,
使函數(shù)f(x)在(1,+∞)上為增函數(shù),
則a>1時(shí)u(x)在(1,+∞)上為增函數(shù)且u(x)>0,
得
,
解得1<a≤2.
所以a的取值范圍為(1,2].
分析:(1)當(dāng)a=2時(shí),f(x)=log
2(2x
2-2x),設(shè)
,由
,能求出當(dāng)a=2時(shí),函數(shù)f(x)的值域.
(2)設(shè)u(x)=ax
2-2x+4-2a,由函數(shù)f(x)在(1,+∞)上為增函數(shù),知當(dāng)a>1時(shí),u(x)在(1,+∞)上為增函數(shù)且u(x)>0,由此能求出a的取值范圍.
點(diǎn)評(píng):本題考查對(duì)數(shù)函數(shù)的值域和最值,解題時(shí)要認(rèn)真審題,注意對(duì)數(shù)函數(shù)的定義域、值域和單調(diào)性的靈活運(yùn)用.