2024年成考专升本《高等数学一》每日一练试题06月09日
<p class="introTit">单选题</p><p>1、设z=x<sup>2</sup>-3y,则dz=()。</p><ul><li>A:2xdx-3ydy</li><li>B:x<sup>2</sup>dx-3dy</li><li>C:2xdx-3dy</li><li>D:x<sup>2</sup>dx-3ydy</li></ul><p>答 案:C</p><p>解 析:z=x<sup>2</sup>-3y,则<img src="https://img2.meite.com/questions/202211/1763759f22820a2.png" />。</p><p>2、已知<img src="https://img2.meite.com/questions/202303/1764141d53ef756.png" />则k=()</p><ul><li>A:0或1</li><li>B:0或-1</li><li>C:0或2</li><li>D:1或-1</li></ul><p>答 案:A</p><p>解 析:<img src="https://img2.meite.com/questions/202303/1764141d985a1a5.png" /><img src="https://img2.meite.com/questions/202303/1764141d9f79aef.png" />所以k=0或k=1.</p><p>3、<img src="https://img2.meite.com/question/import/b5d171cd7ac82036afa9e91d92d023dd.png" /></p><ul><li>A:e<sup>x</sup></li><li>B:e<sup>x -1</sup></li><li>C:e<sup>x</sup>-1</li><li>D:e<sup>x+1</sup></li></ul><p>答 案:A</p><p class="introTit">主观题</p><p>1、求微分方程<img src="https://img2.meite.com/questions/202212/03638b01839ca7b.png" />的通解.</p><p>答 案:解:对应齐次微分方程的特征方程为<img src="https://img2.meite.com/questions/202212/03638b019490467.png" />,解得r<sub>1</sub>=3,r<sub>2</sub>=-2.所以齐次通解为<img src="https://img2.meite.com/questions/202212/03638b01b0cf47a.png" />。设方程的特解设为y*=(Ax+B)e<sup>x</sup>,代入原微分方程可解得,A=<img src="https://img2.meite.com/questions/202212/03638b01cb92f6c.png" />,B=<img src="https://img2.meite.com/questions/202212/03638b01d9339ef.png" />.即非齐次微分方程特解为<img src="https://img2.meite.com/questions/202212/03638b01ea50130.png" />。所以微分方程<img src="https://img2.meite.com/questions/202212/03638b01f95ca36.png" />的通解为<img src="https://img2.meite.com/questions/202212/03638b0209e7294.png" />。</p><p>2、将函数<img src="https://img2.meite.com/questions/202212/016388669caf0e9.png" />展开为x的幂级数,并指出收敛区间(不讨论端点)。</p><p>答 案:解:<img src="https://img2.meite.com/questions/202212/01638866b8810f5.png" />,有<img src="https://img2.meite.com/questions/202212/01638866d5b9017.png" />,即收敛区间为(-4,4)。</p><p>3、求函数y=xe<sup>x</sup>的极小值点与极小值</p><p>答 案:解:方法一:<img src="https://img2.meite.com/questions/202211/176375aa3db2b45.png" />令y'=0,得x=-1。<br />当x<-1时,y'<0;当x>-1时,y'>0。<br />故极小值点为x=-1,极小值为<img src="https://img2.meite.com/questions/202211/176375aa84bb60d.png" />。<br />方法二:,<br />令y'=0,得x=-1,又<img src="https://img2.meite.com/questions/202211/176375aaa1a9d82.png" />,<img src="https://img2.meite.com/questions/202211/176375aab80c42f.png" />。<br />故极小值点为x=-1,极小值为<img src="https://img2.meite.com/questions/202211/176375aacd1921a.png" />。</p><p class="introTit">填空题</p><p>1、<img src="https://img2.meite.com/questions/202211/296385663fd47de.png" />()。</p><p>答 案:<img src="https://img2.meite.com/questions/202211/296385664b18199.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202211/296385665d40f4a.png" /></p><p>2、<img src="https://img2.meite.com/questions/202211/30638722026e1bf.png" />()。</p><p>答 案:<img src="https://img2.meite.com/questions/202211/306387220ce46d3.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202211/306387221aad9d8.png" /></p><p>3、<img src="https://img2.meite.com/questions/202211/296385675962fc3.png" />=()。</p><p>答 案:1</p><p>解 析:<img src="https://img2.meite.com/questions/202211/296385676b1b812.png" />。</p><p class="introTit">简答题</p><p>1、设函数<img src="https://img2.meite.com/questions/202303/036401a023d2795.png" />
</p><p>答 案:<img src="https://img2.meite.com/questions/202303/036401af4921316.png" /></p>