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1、汉阳高数汉阳高数 B 卷答案卷答案一:填空1,1;2 ,; 3,; 4,1(lnln )lnfxxx1( 1)(1)!(1)nnnx4sin14x二:选择:1,A; 2,B; 3,C; 4,D三:求解:1:原式11122002 222lim(1)lim(1)x exxex exx e xxxxxxx ex ee 2:解:3323( )(1)( )(1)( )3( )(1)( ),(1)3 (1)3fxxg xxg xx g xxg xfg3: 解:2222 1111222 11122222221(ln )(ln ) |(ln )42(ln )42ln42 ln|(ln )22eeeeeeexd
2、xxxxdxexxdxxexdxexxxdxe4:解: 由于 f(x)在任意点可导,故,且 ,42(1)0,(1)56,( )53,(1)53fffxxaxb fab(1)156fab 由以上两式有:30,85ab 5:解:202233023 02 30( )(cos )(sin )5(sin )|(cos )|sin3cos2236f x dxxx dxxx dxxxxx 6:解:单调下降,2( ),( )00,0( )0, ( )xI xxeI xxxI xI x时,单调上升,故为极小值点,极小值为。0,( )0, ( )xI xI x时0x ( )I x(0)0I四:解:设,易知,又在2
3、3131 0( ).231nnaaaf xa xxxxn(1)0,(0)0ff( )f x连续,在可导,由罗尔中值定理:至少使,即0,1(0,1)(0,1) ( )0f,即原方程在至少有一个实根。2 012.0n naaaa(0,1)五:解:21111(1 0)lim( )lim1;(1 0)lim( )lim xxxxff xxff xaxbab 由题意:,故,(1 0)(1 0)ff1ab,2 11( )(1)1(1)limlim211xxf xfxfxx 111( )(1)1(1)(1)limlimlim111xxxf xfaxba xfaxxx 由题意由有:,由上有:(1)(1)2ff
4、a2,1ab 六:解:由洛比达法则:,由于,故sin 200ln(1)2 0coslimlimln(1)axxxxtdtxax x20limln(1)0 xx ,继续对上述极限运用洛比达法则有,上述极限 0lim(cos )01 xaxa 2 1 cosln(1)00022sin1sin1limlimlim.222 1xxxxxxxx xx x七:解:。2222200(1 cos )3asydxatdta八:解:222224 03003324|3xvy dxx dx九:解:只需证,设。则(1)ln(1)xxarctgx( )(1)ln(1)f xxxarctgx。故在单调上2 2210,( )ln(1) 1ln(1)011xxfxxxxx ( )f x(0,)升。