ch3自測題(中值定理及導(dǎo)數(shù)應(yīng)用)

1、1階段自測題（中值定理及導(dǎo)數(shù)應(yīng)用）一、填空題1____0______________???xxxlnlim02函數(shù)的極小值點(diǎn)為xxy2??x2ln1?3曲線在對應(yīng)于的點(diǎn)處的曲率為????????3233ttytx1?t164設(shè)時，是比高階的無窮小，則，0?x12????bxaxex2x?a21?b1?5極限??????????xxxln111lim112?二、單項(xiàng)選擇題1函數(shù)在點(diǎn)處導(dǎo)數(shù)為零是在點(diǎn)取到極值的（D）)(xfy?0xx?)(x

2、f0xx?（Ａ）充分但非必要條件；（Ｂ）必要但非充分條件；（Ｃ）充分必要條件；（Ｄ）非充分、非必要條件。2設(shè)，則點(diǎn)（C）1)()()(lim3?????axafxfaxax?（Ａ）是極大值點(diǎn)；（Ｂ）是極小值點(diǎn)；（Ｃ）不是極值點(diǎn)；（Ｄ）以上結(jié)論都不一定成立3函數(shù)在區(qū)間上（D）xxy?)1[??e（Ａ）不存在最大值，不存在最小值；（Ｂ）最大值是；ee1（Ｃ）最大值是；（Ｄ）最小值是ee1?ee1?4設(shè)有二階連續(xù)導(dǎo)數(shù)，且，則（B）)(xf1)

3、(lim0)0(0??????xxffx（Ａ）是的極大值；)0(f)(xf（Ｂ）是的極小值；)0(f)(xf（Ｃ）是曲線的拐點(diǎn)；))0(0(f)(xfy?3=xxx)arctan2(lim?????2?e（2）)sin()1(cos2lim20xxxxxx????21))(!31()1)(!41211(2lim)sin()1(cos2lim332442020???????????????xoxxxxxxoxxxxxxxzx（3）xexx

4、xxsin)1(211lim220?????212111limsin)1(211lim22220220?????????????xxxxxexxxxx四、設(shè)函數(shù)在上滿足且，證明)(xf)(????)()(xfxf??1)0(?fxexf?)(證明：令則在上連續(xù)，可導(dǎo)，且()()xFxefx??()Fx)(????，所以，又因?yàn)?)()()0xxFxefxefx????????()FxC?0(0)(0)1()1()xFefFxfxe???

5、???五、（1）當(dāng)時，證明)20(??xxxx3sin2tan??證明：令，，，xxxxf3sin2tan)(???0)0(?f3cos2sec)(2????xxxf，，所以，當(dāng)0)0(??f0)1(secsin2sin2tansec2)(32???????xxxxxxf)(xf?單調(diào)遞增，，則當(dāng)單調(diào)遞增，)20(??x0)0()(????fxf)(xf)20(??x，即當(dāng)時，證明0)0(3sin2tan)(?????fxxxxf)20