一、函數(shù) - 自考試題

1、1一、函一、函數(shù)1、函數(shù)的值域（首先要挖掘隱含的定義域）、函數(shù)的值域（首先要挖掘隱含的定義域）⑴轉(zhuǎn)化為基本函數(shù)，特別是二次函數(shù)；練習(xí)：1、（C97.10）函數(shù)的3cos3sin2????xxy最小值；2、已知：，α、β求范???sin5sin2sin322??R???22coscos??u圍.⑵有理分式型：Ⅰ????????????????????????)2()()1()0(一一一一一一一一一一一一一一一一一一caycaycdxcac

2、dbcadcdcxbaxy練習(xí)：(C95)作函數(shù)的圖象1231???xxyⅡ用△法，注意)0(22??????prqxpxcbxaxy????一一一一一一一一一一一“)“2()1(⑶無理型：???????????????]0[cos(121:)2()413(413332:)1(22???xxxxytxxxy一一一一一一一一一一一一一一一一一一一一一一一一2、函數(shù)的奇偶性（首先定義域必須關(guān)于原點(diǎn)對(duì)稱）、函數(shù)的奇偶性（首先定義域必須關(guān)于原點(diǎn)

3、對(duì)稱）⑴一一一一一一)()()()()()(xfyxfxfxfyxfxf?????????⑵奇函數(shù)0)0()(???fxy一一一一一一一⑶任一個(gè)定義域關(guān)于原點(diǎn)對(duì)稱的函數(shù)一定可以表示成一個(gè)奇函數(shù)和一個(gè)偶函數(shù)之和)(xf即一一2)()()(2)()()(xfxfxfxfxf??????⑷練習(xí)：①(C93)是偶函數(shù)，且（）)()1221()(xfxFx???)(0)(xfxf一一一一A、奇B、偶C、既奇又偶D、非奇非偶②(C94)定義在上的函

4、數(shù)可以表示成奇函數(shù)g(x)與偶函數(shù)h(x)之和，)(????)(xf若，那么（）)110lg()(??xxfA、)21010lg()()(?????xxxhxxg3斜率為1點(diǎn)，點(diǎn))()(0000xyyxxy??????一一)()(0000axayyxaxy?????????一一斜率為1點(diǎn)，)()(0000xyyxxy????????點(diǎn))()(0000axayyxaxy????????????一條曲線若對(duì)滿足，則)(xfy?Rx?)()(

5、xbfxaf???)(xfy?關(guān)于直線對(duì)稱；（由求得）2bax??2)()(xbxax????3對(duì)稱軸對(duì)稱兩條曲線函數(shù)關(guān)于直線對(duì)稱。)()(xbfyxafy????一2abx??（由解得）xbxa???⑴(C90)的圖象，則的表達(dá)式：arctgxy?CCx???????????一一一一一一一一一一一一一2C?⑵(C92)對(duì)任意t均有，則cbxxxf???2)()2()2(tftf???大小關(guān)系為：)4()1()2(fff⑶(C97)的圖

6、象。)1(log221??????????????????xyCyxyx一一一一一一一一一一一一一一一⑷(i)若對(duì)滿足，則的對(duì)稱軸為)(xfy?Rx?)2()2(xfxf???)(xfy?(ii)函數(shù)的對(duì)稱軸為)2()2(xfyxfy????一(iii)為定義在R上的偶函數(shù)，且對(duì)恒成立，則)(xf)3()5(xfxf???Rx?的一個(gè)周期為：)(xfy?⑸(i)若滿足，則的對(duì)稱軸為)(xfy?)3()1(xfxf???)(xfy?(ii