數(shù)學(xué)《指數(shù)函數(shù) 冪函數(shù) 對(duì)數(shù)函數(shù)》同步練習(xí)

1、指數(shù)函數(shù)、對(duì)數(shù)函數(shù)與冪函數(shù)一、選擇題一、選擇題1.若函數(shù)1???bayx（0?a且1?a）的圖象經(jīng)過(guò)二、三、四象限，則一定有（）.A.10??a且0?bB.1?a且0?bC.10??a且0?bD.1?a且0?b2.已知函數(shù)?????????4)1(4)21()(xxfxxfx，則)3log2(2?f的值為（）.A.31B.61C.121D.2413.若10??a，且函數(shù)xxfalog)(?，則下列各式中成立的是（）.A.)41()31(

2、)2(fff??B.)31()2()41(fff??C.)41()2()31(fff??D.)2()31()41(fff??4.已知0?a，集合2|axxA???，1|??xaxB，若???BA，則實(shí)數(shù)a的取值范圍是（）.A.)2(??B.)10(C.)2()10(???D.)1()10(???5.已知函數(shù))3(log)(2???axxxfa（a0?且1?a）滿足：對(duì)任意實(shí)數(shù)21xx、，當(dāng)221axx??時(shí)，總有0)()(21??xfx

3、f，那么實(shí)數(shù)a的取值范圍為（）.A.（0，3）B.（1，3）C.（0，32）D.（1，32）6.設(shè)函數(shù)xxfalog)(?（10??aa）滿足2)9(?f，則)2(log91?f等于（）.A.2B.2C.2?D.2?7.已知nm??1，令2)(logman?，2logmbn?，)(loglogmcnn?，則（）.A.cba??B.bca??C.cab??D.bac??8.若方程1)21()41(??xx0??a有正數(shù)解，則實(shí)數(shù)a的取值范

4、圍是（）.A.)1(??B.)2(???C.)23(??D.)03(?二、填空題二、填空題∴241)21()21()21()3log3()3log2(3log33log32222????????ff.）3.D（解析：由題意知????????1log10log)(xxxxxfaa，故)41()31(ff?又21log2log)2(aaf???，故)31()2(ff?.）4.C（解析：∵22|axaxA???????當(dāng)1?a時(shí)，0|??xx

5、B又???BA，故02???a，即a2?當(dāng)1?a時(shí)，??B當(dāng)10??a時(shí)，0|??xxB又02???a，故???BA綜上，可知)2()10(????a.）5.D（解析：由題意知函數(shù))(xf在]2(a??上恒成立且單調(diào)遞減，即????????10122aa解得321??a.）6.A（解析：由題意知：9log2a?∴3?a，xxf3log)(?的反函數(shù)為??)(1xfx3，所以23)2(log2log919???f，故選A.）7.D（解析：

6、∵nm??1∴1log0??mn∴0)(loglog?mnn.）8.D（解析：令1]1)21[()21(2)21()21()41()(221????????xxxxxxf當(dāng)x0?時(shí)，)(xf的值域?yàn)椋?，3）方程0)21()41(1????axx有正數(shù)解即axf??)(在（0，??）上能成立，故)30(??a，即)03(??a.）二、填空題二、填空題9.65a10.1(解析：由)2(log)12(log222xxx???得：??????