2024年成考专升本《高等数学一》每日一练试题03月29日

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03/29
<p class="introTit">单选题</p><p>1、设f(x,y)为连续函数,则<img src="https://img2.meite.com/questions/202212/016388458273144.png" />()。</p><ul><li>A:<img src='https://img2.meite.com/questions/202212/016388459115f06.png' /></li><li>B:<img src='https://img2.meite.com/questions/202212/016388459bb6225.png' /></li><li>C:<img src='https://img2.meite.com/questions/202212/01638845a7d9710.png' /></li><li>D:<img src='https://img2.meite.com/questions/202212/01638845b2c869d.png' /></li></ul><p>答 案:D</p><p>解 析:积分区域D可以由<img src="https://img2.meite.com/questions/202212/01638845c0baad5.png" />表示,其图形为图中阴影部分<img src="https://img2.meite.com/questions/202212/01638845dc56766.png" />。也可以将D表示为<img src="https://img2.meite.com/questions/202212/01638845ee0d2f7.png" />,故二重积分也可表示为<img src="https://img2.meite.com/questions/202212/01638845fa06d44.png" />。</p><p>2、下列等式成立的是()</p><ul><li>A:<img src='https://img2.meite.com/questions/202303/176413e31518153.png' /></li><li>B:<img src='https://img2.meite.com/questions/202303/176413e325dd838.png' /></li><li>C:<img src='https://img2.meite.com/questions/202303/176413e32d7f25d.png' /></li><li>D:<img src='https://img2.meite.com/questions/202303/176413e333aa927.png' /></li></ul><p>答 案:C</p><p>解 析:由<img src="https://img2.meite.com/questions/202303/176413e365098f6.png" /><img src="https://img2.meite.com/questions/202303/176413e36f2e341.png" /><img src="https://img2.meite.com/questions/202303/176413e377dec04.png" /></p><p>3、在空间直角坐标系中方程y<sup>2</sup>=x表示的是()。</p><ul><li>A:抛物线</li><li>B:柱面</li><li>C:椭球面</li><li>D:平面</li></ul><p>答 案:B</p><p>解 析:空间中,缺少一维坐标的方程均表示柱面,y<sup>2</sup>=x是母线平行于z轴的柱面。</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、求y'+<img src="https://img2.meite.com/questions/202212/03638abf7b42c03.png" />=1的通解.</p><p>答 案:解:<img src="https://img2.meite.com/questions/202212/03638abf8ac5f6c.png" /></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、幂级数<img src="https://img2.meite.com/questions/202303/176414046fa95c8.png" />的收敛半径为()</p><p>答 案:3</p><p>解 析:所给幂级数通项为<img src="https://img2.meite.com/questions/202303/176414049a92cf3.png" />则<img src="https://img2.meite.com/questions/202303/17641404aae13b4.png" /><img src="https://img2.meite.com/questions/202303/17641404b621a4a.png" />所以收敛半径R=3</p><p>2、过原点且垂直于y轴的平面方程为()。</p><p>答 案:y=0</p><p>解 析:过原点且垂直于y轴的平面即x轴所在的平面,方程为y=0。</p><p>3、设y=f(x)可导,点x<sub>0</sub>=2为f(x)的极小值点,且f(2)=3,则曲线y=f(x)在点(2,3)处的切线方程为()。</p><p>答 案:y=3</p><p>解 析:由于y=f(x)可导,且点x<sub>0</sub>=2为f(x)的极小值点,由极值的必要条件可得<img src="https://img2.meite.com/questions/202211/186376eab73520a.png" />又f(2)=3,可知曲线过点(2,3)的切线方程为<img src="https://img2.meite.com/questions/202211/186376eacb7a1f5.png" /></p><p class="introTit">简答题</p><p>1、<img src="https://img2.meite.com/questions/202303/1764140b5eb19d9.png" /></p><p>答 案:<img src="https://img2.meite.com/questions/202303/1764140b998c053.png" /></p>
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