導(dǎo)數(shù)--壓軸題訓(xùn)練

1、1導(dǎo)數(shù)導(dǎo)數(shù)壓軸題訓(xùn)練壓軸題訓(xùn)練1（2014湖南）.22（2014湖南）..已知常數(shù)0a?函數(shù)????2ln12xfxaxx????.(1)討論??fx在區(qū)間??0??上的單調(diào)性(2)若??fx存在兩個極值點12xx且????120fxfx??求a的取值范圍.【答案】(1)詳見解析【解析】解:(1)對函數(shù)??fx求導(dǎo)可得????2412afxaxx????????????2224112axaxaxx????????????224112ax

2、aaxx?????因為????2120axx???所以當(dāng)10a??時即1a?時??0fx?恒成立則函數(shù)??fx在??0??單調(diào)遞增當(dāng)1a?時????210aafxxa?????則函數(shù)??fx在區(qū)間??210aaa?????????單調(diào)遞減在??21aaa???????????單調(diào)遞增的.(2)解:(1)對函數(shù)??fx求導(dǎo)可得????2412afxaxx????????????2224112axaxaxx????????????22411

3、2axaaxx?????因為????2120axx???所以當(dāng)10a??時即1a?時??0fx?恒成立則函數(shù)??fx在??0??單調(diào)遞增當(dāng)1a?時????210aafxxa?????則函數(shù)??fx在區(qū)間??210aaa?????????單調(diào)遞減在??21aaa???????????單調(diào)遞增的.3解：（1）因為2()1xfxeaxbx????所以()()2xgxfxeaxb?????又()2xgxea???因為[01]x?，1xee??所

4、以：①若12a?，則21a?，()20xgxea????，所以函數(shù)()gx在區(qū)間[01]上單增，min()(0)1gxgb???②若122ea??，則12ae??，于是當(dāng)0ln(2)xa??時()20xgxea????，當(dāng)ln(2)1ax??時()20xgxea????，所以函數(shù)()gx在區(qū)間[0ln(2)]a上單減，在區(qū)間[ln(2)1]a上單增，min()[ln(2)]22ln(2)gxgaaaab????③若2ea?，則2ae?，

5、()20xgxea????所以函數(shù)()gx在區(qū)間[01]上單減，min()(1)2gxgeab????綜上：()gx在區(qū)間[01]上的最小值為min1121()22ln(2)2222baegxaaabaeeaba???????????????????（2）由(1)0f??10eab?????1bea???，又(0)0f?若函數(shù)()fx在區(qū)間(01)內(nèi)有零點，則函數(shù)()fx在區(qū)間(01)內(nèi)至少有三個單調(diào)區(qū)間由（1）知當(dāng)12a?或2ea?時

6、，函數(shù)()gx即()fx?在區(qū)間[01]上單調(diào)，不可能滿足“函數(shù)()fx在區(qū)間(01)內(nèi)至少有三個單調(diào)區(qū)間”這一要求。若122ea??，則min()22ln(2)32ln(2)1gxaaabaaae???????令3()ln12hxxxxe????（1xe??）則1()ln2hxx???。由1()ln02hxxxe??????所以()hx在區(qū)間(1)e上單增，在區(qū)間()ee上單減max3()()ln1102hxheeeeeee?????