2024年成考专升本《高等数学一》每日一练试题08月16日

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08/16
<p class="introTit">单选题</p><p>1、设函数f(x)满足<img src="https://img2.meite.com/questions/202303/176413e0e837e78.png" />且f(0)=0,则f(x)=()</p><ul><li>A:<img src='https://img2.meite.com/questions/202303/176413e11e492e3.png' /></li><li>B:<img src='https://img2.meite.com/questions/202303/176413e113de62f.png' /></li><li>C:<img src='https://img2.meite.com/questions/202303/176413e1196224a.png' /></li><li>D:<img src='https://img2.meite.com/questions/202303/176413e10cadab6.png' /></li></ul><p>答 案:D</p><p>解 析:由<img src="https://img2.meite.com/questions/202303/176413e0e837e78.png" />知<img src="https://img2.meite.com/questions/202303/176413e16f36bbb.png" />令<img src="https://img2.meite.com/questions/202303/176413e17d276a1.png" />故<img src="https://img2.meite.com/questions/202303/176413e184a50ac.png" />所以f(u)=u-<img src="https://img2.meite.com/questions/202303/176413e1ab07ac9.png" />由f(0)=0,得C=0.所以<img src="https://img2.meite.com/questions/202303/176413e1c444863.png" /></p><p>2、<img src="https://img2.meite.com/questions/202408/1566bda6fb1a149.png" />,则下列命题中正确的有()。  </p><ul><li>A:<img src='https://img2.meite.com/questions/202408/1566bda6ff20bd7.png' /></li><li>B:<img src='https://img2.meite.com/questions/202408/1566bda702cc73f.png' /></li><li>C:<img src='https://img2.meite.com/questions/202408/1566bda7060088f.png' /></li><li>D:<img src='https://img2.meite.com/questions/202408/1566bda7096b30a.png' /></li></ul><p>答 案:B</p><p>解 析:本题考查的知识点为级数收敛性的定义。 <img src="https://img2.meite.com/questions/202408/1566bda710353c1.png" />  </p><p>3、在空间直角坐标系中,方程<img src="https://img2.meite.com/questions/202212/03638ae66dad330.png" />表示()。</p><ul><li>A:两个平面</li><li>B:双曲柱面</li><li>C:椭圆柱面</li><li>D:圆柱面</li></ul><p>答 案:A</p><p>解 析:由<img src="https://img2.meite.com/questions/202212/03638ae67e14388.png" />得<img src="https://img2.meite.com/questions/202212/03638ae68c0330d.png" />,故为两个平面。</p><p class="introTit">主观题</p><p>1、求<img src="https://img2.meite.com/questions/202211/166374817022206.png" /></p><p>答 案:解:利用洛必达法则,得<img src="https://img2.meite.com/questions/202211/166374817771806.png" /></p><p>2、计算二重积分<img src="https://img2.meite.com/questions/202212/01638853157d8ab.png" />,其中D是由<img src="https://img2.meite.com/questions/202212/0163885321d5b98.png" />和x=4所围的平面区域(在第一象限)。</p><p>答 案:解:图形见下图中阴影部分<img src="https://img2.meite.com/questions/202212/016388534457b4d.png" />由y<sup>2</sup>=x得y=<img src="https://img2.meite.com/questions/202212/016388536ae3331.png" />,则<img src="https://img2.meite.com/questions/202212/016388537b8ad04.png" /></p><p>3、设<img src="https://img2.meite.com/questions/202212/03638aff4f737d0.png" />存在且<img src="https://img2.meite.com/questions/202212/03638aff59d37e0.png" />,求<img src="https://img2.meite.com/questions/202212/03638aff6554eac.png" /></p><p>答 案:解:设<img src="https://img2.meite.com/questions/202212/03638aff7078ded.png" />对<img src="https://img2.meite.com/questions/202212/03638aff830ad7a.png" />两边同时求极限,得<img src="https://img2.meite.com/questions/202212/03638affa03127b.png" />,即<img src="https://img2.meite.com/questions/202212/03638affaad78d4.png" />,得<img src="https://img2.meite.com/questions/202212/03638affb8a0ed1.png" />。</p><p class="introTit">填空题</p><p>1、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>2、曲线<img src="https://img2.meite.com/questions/202211/306386d334837c7.png" />的铅直渐近线方程为()。</p><p>答 案:x=2</p><p>解 析:因为<img src="https://img2.meite.com/questions/202211/306386d34bea9e7.png" />,故曲线的铅直线渐近线方程x=2。</p><p>3、设<img src="https://img2.meite.com/questions/202303/0364019ee6a682b.png" />则dy=()  </p><p>答 案:<img src="https://img2.meite.com/questions/202303/036401acd8502f2.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202303/036401ace51fcbf.png" />故有<img src="https://img2.meite.com/questions/202303/036401acfb369a6.png" /></p><p class="introTit">简答题</p><p>1、设D是由曲线<img src="https://img2.meite.com/questions/202405/166645be83cfdf4.png" />,x=0,y=0所围成的平面图形. (1)求D的面积S; (2)求D绕x轴旋转一周所得旋转体的体积V.  </p><p>答 案:<img src="https://img2.meite.com/questions/202405/166645be8b845c8.png" /></p>
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