考研數學解答題

1、1三、解答題：三、解答題：15～23小題小題共94分.請將解答寫在答題紙指定位置上請將解答寫在答題紙指定位置上.解答應寫出文字說明、解答應寫出文字說明、證明過程或演算步驟證明過程或演算步驟.(15)(本題滿分10分）設函數，.若與在時是等價無()ln(1)sinfxxaxbxx????3()gxkx?()fx()gx0x?窮小，求的值.abk【答案】11132akb??????【解析】方法一：因為，，233ln(1)()23xxxxox

2、?????33sin()3!xxxox???那么，，23333000(1)()()()ln(1)sin231limlimlim()xxxaaaxbxxoxfxxaxbxxgxkxkx??????????????可得：，所以，100213aabak??????????????11213abk???????????????方法二：由題意得300sin)1ln(lim)()(lim1kxxbxxaxxgxfxx???????203cossin

3、11limkxxbxxbxax??????由分母，得分子，求得c；03lim20??kxx)cossin11(lim0xbxxbxax?????0)1(lim0????ax于是)()(lim10xgxfx??203cossin111limkxxbxxbxx??????）（xkxxxbxxxbxx???????13cos)1(sin)1(lim20203cos)1(sin)1(limkxxxbxxxbxx??????3【解析】兩邊對y積分

4、，得xxyeyyxf)1(2)(????，)()21(2)(2xeyyyxfxx?????)()2(2xeyyx????故，xxexxxf)1()()0(?????求得，)1()(??xexx?故，兩邊關于x積分，得)1()2()(2xeeyyyxfxxx??????????dxxeeyyyxfxx)1()2()(2?????xxdexeyy)1()2(2??????dxeexeyyxxx)1()2(2C)1()2(2??????xxx

5、eexeyyC)2(2????xxxeeyy由，求得yyyyyf2C2)0(22?????.0?C所以.xxxeeyyyxf???)2()(2令，求得.???????????????0)22(0)2(2xyxxxxeyfxeeeyyf??????10yx又，xxxxxxeeeyyf??????2)2(2，，xxyeyf)1(2????xyyef2???當時，，10???yx(01)1xxAf?????0)10(B?????xyf2)10