微專題構(gòu)造函數(shù)法解選填壓軸題

1、1微專題：構(gòu)造函數(shù)法解選填壓軸題高考中要取得高分，關(guān)鍵在于選準(zhǔn)選好的解題方法，才能省時(shí)省力又有效果。近幾年各地高考數(shù)學(xué)試卷中，許多方面尤其涉及函數(shù)題目，采用構(gòu)造函數(shù)法解答是一個(gè)不錯(cuò)的選擇。所謂構(gòu)造函數(shù)法構(gòu)造函數(shù)法是指通過(guò)一定方式，設(shè)計(jì)并構(gòu)造一個(gè)與有待解答問(wèn)題相關(guān)函數(shù)，并對(duì)其進(jìn)行觀察分析，借助函數(shù)本身性質(zhì)如單調(diào)性或利用運(yùn)算結(jié)果，解決原問(wèn)題方法，簡(jiǎn)而言之就是構(gòu)造函數(shù)解答問(wèn)題。怎樣合理的構(gòu)造函數(shù)合理的構(gòu)造函數(shù)就是問(wèn)題的關(guān)鍵，這里我們來(lái)一起探討

2、一下這方面問(wèn)題。幾種導(dǎo)數(shù)的常見(jiàn)構(gòu)造：1對(duì)于，構(gòu)造????xgxf???????xgxfxh??若遇到，則可構(gòu)????0??aaxf????axxfxh??2對(duì)于，構(gòu)造????0??xgxf??????xgxfxh??3對(duì)于，構(gòu)造()()0fxfx??????xfexhx?4對(duì)于[或]，構(gòu)造()()fxfx?()()0fxfx??()()xfxhxe?5對(duì)于，構(gòu)造????0??xfxxf????xxfxh?6對(duì)于，構(gòu)造????0??xfx

3、xf????xxfxh?一、構(gòu)造函數(shù)法比較大小例1已知函數(shù)()yfx?的圖象關(guān)于y軸對(duì)稱且當(dāng)(0)()()0xfxxfx?????成立，，，則的大小關(guān)系是()0.20.22(2)af?Alog3(log3)bf???A33log9(log9)cf?Aabc.Aabc??.Bacb??.Ccba??.Dbac??【解析】因?yàn)楹瘮?shù)()yfx?關(guān)于y軸對(duì)稱所以函數(shù)()yxfx?為奇函數(shù).因?yàn)閇()]()()xfxfxxfx??所以當(dāng)(0)x?

4、??時(shí)[()]()()0xfxfxxfx???函數(shù)()yxfx?單調(diào)遞減當(dāng)(0)x???時(shí)函數(shù)()yxfx?單調(diào)遞減.因?yàn)?.2122??0131og???3192og?所以0.23013219ogog????所以bac??選D.變式:已知定義域?yàn)榈钠婧瘮?shù)的導(dǎo)函數(shù)為，當(dāng)時(shí)，，R()fx()fx0x?()()0fxfxx??若，則下列關(guān)于的大小關(guān)系正確的是（D）111()2(2)ln(ln2)222afbfcf?????abc.Aabc?

5、?.Bacb??.Ccba??.Dbac??例2已知為上的可導(dǎo)函數(shù)，且，均有，則有()fxRxR??()()fxfx??3練習(xí)1已知函數(shù)對(duì)任意的滿足則（）A)(xfy?)22(??，??x()cos()sin0fxxfxx???B.C.D.)4(2)0(?ff?)3(2)0(???ff)4()3(2??ff?)4()3(2?????ff提示：構(gòu)造函數(shù)，選D()()cosfxgxx?二、構(gòu)造函數(shù)法解恒成立問(wèn)題例1若函數(shù)y=在R上可導(dǎo)且滿足

6、不等式恒成立，對(duì)任意正數(shù)、，若)(xf()()0xfxfx???ab，則必有（）ab?ABCD()()afbbfa?()()bfaafb?()()afabfb?()()bfbafa?【解析】由已知∴構(gòu)造函數(shù)，()()0xfxfx???)()(xxfxF?則，從而在R上為增函數(shù)。()Fx??()()0xfxfx???)(xF∴即，故選C。?ab?()()FaFb?()()afabfb?例2已知是定義在（0，∞）上的非負(fù)可導(dǎo)函數(shù)，且滿足≤0

7、，對(duì)任意正數(shù)、)(xf)()(xfxfx??a，若，則必有（）bab?ABCD()()afbbfa?()()bfaafb?()()afabfb?()()bfbafa?【解析】，，故在（0，∞）上是減函數(shù)，xxfxF)()(?0)()()(2????xxfxxfxFxxfxF)()(?由，有，即。故選A。ba?bbfaaf)()(?()()afbbfa?變式1.設(shè)是上的可導(dǎo)函數(shù)，分別為的導(dǎo)函數(shù)，且滿足()()fxgx、R()()fxgx、

8、()()fxgx、，則當(dāng)時(shí)，有（C）()()()()0fxgxfxgx??axb??.()()()()Afxgbfbgx?.()()()()Bfxgafagx?.()()()()Cfxgxfbgb?.()()()()Dfxgxfbga?變式2.設(shè)函數(shù)時(shí)，有（C）bxaxgxfbaxgxf?????則當(dāng)且上均可導(dǎo)在)()(][)()(AB)()(xgxf?)()(xgxf?CD)()()()(afxgagxf???)()()()(bfxg