湖北省武汉市武昌区2015-2016学年度高二年级第二学期期末调研考试理科数学及参考答案和评分细则pdf版

PN2218911||||122?????PNPM.所以||||MNPN的最小值为22.………………………………………………………… 12分22.(本小题满分12分)已知函数axxxf???)1ln()(,R?a.(Ⅰ)求函数)(xf的单调区间;(Ⅱ)若不等式xxfe1)(??对),0[???x恒成立,求实数a的取值范围.解析:(Ⅰ)1111)(?????????xaaxaxxf,1??x.(ⅰ)若0?a,则0)(??xf,所以)(xf在),1(???上单调递增;(ⅱ)若0?a,则当ax11???时,0)(??xf;当ax11???时,0)(??xf.所以,)(xf在)11,1(a???上单调递增;在),11(????a上单调递减.………………4分(Ⅱ)记xxaxxxfxge1)1ln(e1)()(????????)0(?x.则“不等式xxfe1)(??)0(?x恒成立”等价于“0)(min?xg)0(?x恒成立”.因为xaxxge11)(?????,而1e??xx,所以axaxxg????????2111)(.(ⅰ)若2?a,则0)(??xg,从而)(xg在),0[??上单调递增.所以0)0()(min??gxg,符合题意.(ⅱ)若2?a,因为0)1(1)1(e)1(1e)(222??????????xxxxgxx,所以xaxxge11)(?????在),0[??上单调递增.所以02)0()(min?????agxg.所以,存在0x,使得00xx??时,0)(??xg;当0xx?时,0)(??xg.此时,)(xg在),0(0x上单调递减,在),(0??x上单调递增.所以)()(0minxgxg?.因为)(xg在),0(0x上单调递减,所以0)0()()(0min???gxgxg,不合题意.综上,所求实数a的取值范围为]2,(??.……………………………………………12分