2024年成考专升本《高等数学二》每日一练试题06月10日

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06/10
<p class="introTit">判断题</p><p>1、若<img src="https://img2.meite.com/questions/202307/1364af5780970e6.png" />,则<img src="https://img2.meite.com/questions/202307/1364af578624075.png" />。()  </p><p>答 案:错</p><p>解 析:<img src="https://img2.meite.com/questions/202212/06638ef8852e1bd.png" />所以<img src="https://img2.meite.com/questions/202212/06638ef88b66bc1.png" />  </p><p class="introTit">单选题</p><p>1、设函数z=f(x<sup>2</sup>-y<sup>2</sup>),f(u)二阶可导,则<img src="https://img2.meite.com/questions/202212/06638ea817945c8.png" />=()  </p><ul><li>A:<img src='https://img2.meite.com/questions/202212/06638ea840964a8.png' /></li><li>B:<img src='https://img2.meite.com/questions/202212/06638ea848582a3.png' /></li><li>C:<img src='https://img2.meite.com/questions/202212/06638ea84e7fa03.png' /></li><li>D:<img src='https://img2.meite.com/questions/202212/06638ea85450245.png' /></li></ul><p>答 案:C</p><p>解 析:<img src="https://img2.meite.com/questions/202212/06638ea86b8e519.png" /></p><p>2、函数<img src="https://img2.meite.com/questions/202212/06638ef7a76509a.png" />的单调递增区间是().</p><ul><li>A:(-5,5)</li><li>B:(-∞,0)</li><li>C:(0,+∞)</li><li>D:(∞,+∞)</li></ul><p>答 案:C</p><p>解 析:<img src="https://img2.meite.com/questions/202212/06638ef7c7a68e4.png" />,由y'>0得x>0,所以函数<img src="https://img2.meite.com/questions/202212/06638ef7ddca041.png" />在(0,+∞)上单调递增.</p><p class="introTit">主观题</p><p>1、设函数z=z(x,y)由sin(x+y)+e<sup>z</sup>=0确定,求<img src="https://img2.meite.com/questions/202212/05638d8c1cc4720.png" />.</p><p>答 案:解:设F(x,y,z)=sin(x+y)+e<sup>z</sup>.则<img src="https://img2.meite.com/questions/202212/05638d8c350b955.png" />则<img src="https://img2.meite.com/questions/202212/05638d8c433eca4.png" /></p><p>2、求<img src="https://img2.meite.com/questions/202212/07639038ae5e4b0.png" />.</p><p>答 案:解:<img src="https://img2.meite.com/questions/202212/07639039c813814.png" /></p><p class="introTit">填空题</p><p>1、己知y=x<sup>3</sup>-ax在x=1的切线平行于直线5x-y+1=0,则a=()  </p><p>答 案:-2</p><p>解 析:<img src="https://img2.meite.com/questions/202212/06638eea5b9374a.png" />a=-2  </p><p>2、设函数<img src="https://img2.meite.com/questions/202212/05638d97f3c2789.png" />,在点x=0处的极限存在,则a=().</p><p>答 案:1</p><p>解 析:函数在点x=0处的极限存在,故有<img src="https://img2.meite.com/questions/202212/05638d980657b7f.png" />,<img src="https://img2.meite.com/questions/202212/05638d9812f1291.png" />,故a=1.</p><p class="introTit">简答题</p><p>1、计算<img src="https://img2.meite.com/questions/202303/076406da7743065.png" />  </p><p>答 案:<img src="https://img2.meite.com/questions/202303/076406e36bd4a94.png" /></p><p>2、计算<img src="https://img2.meite.com/questions/202212/06638eb43edc03c.png" />  </p><p>答 案:由洛必达法则有<img src="https://img2.meite.com/questions/202212/06638eb44c28870.png" />  </p>
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