2019屆高考數(shù)學(xué)專題三含導(dǎo)函數(shù)的抽象函數(shù)的構(gòu)造精準(zhǔn)培優(yōu)專練理201811081143

1、1培優(yōu)點(diǎn)三培優(yōu)點(diǎn)三含導(dǎo)函數(shù)的抽象函數(shù)的構(gòu)造含導(dǎo)函數(shù)的抽象函數(shù)的構(gòu)造1對于????0fxaa??，可構(gòu)造????hxfxax??例1：函數(shù)??fx的定義域?yàn)镽，()12f??，對任意Rx?，()2fx??，則??24fxx??的解集為（）A()11?B()1???，C()1???，D()????，【答案】B【解析】構(gòu)造函數(shù)????24Gxfxx???，所以()()2Gxfx????，由于對任意Rx?，()2fx??，所以()()20Gxf

2、x?????恒成立，所以????24Gxfxx???是R上的增函數(shù)，又由于()()()112140Gf????????，所以????240Gxfxx????，即??24fxx??的解集為()1???，故選B2對于????0xfxfx??，構(gòu)造????hxxfx?；對于????0xfxfx??，構(gòu)造????fxhxx?例2：已知函數(shù)??yfx?的圖象關(guān)于y軸對稱，且當(dāng)??0x???，????0fxxfx???成立，??0.20.222af

3、?，??log3log3bf???，??33log9log9cf?，則a，b，c的大小關(guān)系是（）Aabc??Bacb??Ccba??Dbac??【答案】D【解析】因?yàn)楹瘮?shù)??yfx?關(guān)于y軸對稱，所以函數(shù)??yxfx?為奇函數(shù)因?yàn)??????xfxfxxfx????????，所以當(dāng)??0x???時(shí)，??????0xfxfxxfx?????????，函數(shù)??yxfx?單調(diào)遞減，當(dāng)??0x???時(shí)，函數(shù)??yxfx?單調(diào)遞減因?yàn)?.2122

4、??，0log31???，3log92?，所以0.230log32log9????，所以bac??故選D3對點(diǎn)增分集訓(xùn)對點(diǎn)增分集訓(xùn)一、選擇題1若函數(shù)??yfx?在R上可導(dǎo)且滿足不等式()()0xfxfx???恒成立，對任意正數(shù)a、b，若ab?，則必有（）A()()afbbfa?B()()bfaafb?C()()afabfb?D()()bfbafa?【答案】C【解析】由已知()()0xfxfx???∴構(gòu)造函數(shù)????Fxxfx?，則()(

5、)()0Fxxfxfx?????，從而??Fx在R上為增函數(shù)。∵ab?，∴()()FaFb?，即()()afabfb?，故選C2已知函數(shù)????Rfxx?滿足??11f?，且??12fx??，則??122xfx??的解集為（）A??11xx????B??1xx???C??11xxx????或D??1xx??【答案】D【解析】構(gòu)造新函數(shù)1()()22xFxfx?????????，則11(1)(1)11022Ff????????????，1