2024年成考专升本《高等数学一》每日一练试题04月24日

聚题库
04/24
<p class="introTit">单选题</p><p>1、<img border="0" style="width: 91px; height: 37px;" src="https://img2.meite.com/zzpuce/2023-04/643679c5549d885369.jpeg">( )</p><ul><li>A:-2</li><li>B:-1</li><li>C:1</li><li>D:2</li></ul><p>答 案:D</p><p>解 析:<img title="电气工程师公共基础,章节练习,基础复习,高等数学" src="https://img2.meite.com/question/2022-03/622a869805c91.jpg" alt="电气工程师公共基础,章节练习,基础复习,高等数学" />项A、B、C、D值分别代入,当a=-2代入时,R(A*) = 1。</p><p>2、下列极限正确的是()。</p><ul><li>A:<img src='https://img2.meite.com/questions/202211/2863847c11dc301.png' /></li><li>B:<img src='https://img2.meite.com/questions/202211/2863847c1dd9881.png' /></li><li>C:<img src='https://img2.meite.com/questions/202211/2863847c2a6539c.png' /></li><li>D:<img src='https://img2.meite.com/questions/202211/2863847c388a5c3.png' /></li></ul><p>答 案:C</p><p>解 析:A项,<img src="https://img2.meite.com/questions/202211/2863847c4584e8a.png" />;B项,<img src="https://img2.meite.com/questions/202211/2863847c534a6cd.png" />;C项,<img src="https://img2.meite.com/questions/202211/2863847c617294c.png" />;D项,<img src="https://img2.meite.com/questions/202211/2863847c73cc617.png" />。</p><p>3、过点(1,0,0),(0,1,0),(0,0,1)的平面方程为()。</p><ul><li>A:x+y+z=1</li><li>B:2x+y+z=1</li><li>C:x+2y+z=1</li><li>D:z+y+2z=1</li></ul><p>答 案:A</p><p>解 析:方法一:设所求平面方程为Ax+By+Cz+D=0.由于点(1,0,0),(0,1,0),(0,0,1)在平面上,将上述三点坐标分别代入所设方程,可得A+D=0,B+D=0,C+D=0,即A=B=C=-D,再代回方程可得x+y+z=1。方法二:由于点(1,0,0),(0,1,0),(0,0,1)分别位于x轴、y轴、z轴上,可由平面的截距式方程得出x+y+z=1即为所求平面方程。</p><p class="introTit">主观题</p><p>1、设z=xy<sup>2</sup>+e<sup>y</sup>cosx,求<img src="https://img2.meite.com/questions/202211/16637481f262f04.png" />.</p><p>答 案:解:z=xy<sup>2</sup>+e<sup>y</sup>cosx,<img src="https://img2.meite.com/questions/202211/1663748277d6f76.png" />=2xy+e<sup>y</sup>cosx。</p><p>2、设D是由直线y=x与曲线y=x<sup>3</sup>在第一象限所围成的图形.(1)求D的面积S;<br />(2)求D绕x轴旋转一周所得旋转体的体积V。</p><p>答 案:解:由<img src="https://img2.meite.com/questions/202211/176375ac457995e.png" />,知两曲线的交点为(0,0),(1,1)和(-1,-1),则(1)<img src="https://img2.meite.com/questions/202211/176375ac638d915.png" /><img src="https://img2.meite.com/questions/202211/176375ac7126036.png" /><img src="https://img2.meite.com/questions/202211/176375ac7f4fd17.png" />(2)<img src="https://img2.meite.com/questions/202211/176375ac93635b5.png" /><img src="https://img2.meite.com/questions/202211/176375aca0d2fc7.png" /><img src="https://img2.meite.com/questions/202211/176375acb04c7c5.png" /></p><p>3、求极限<img src="https://img2.meite.com/questions/202211/2963856a48dc287.png" /></p><p>答 案:解:当<img src="https://img2.meite.com/questions/202211/2963856b270530f.png" />时,<img src="https://img2.meite.com/questions/202211/2963856b3fee003.png" />,则<img src="https://img2.meite.com/questions/202211/2963856b4f30989.png" />。</p><p class="introTit">填空题</p><p>1、若级数<img src="https://img2.meite.com/questions/202212/0163885fc36d9dd.png" />条件收敛(其中k>0为常数),则k的取值范围是()。</p><p>答 案:0<k≤l</p><p>解 析:k>1时,级数各项取绝对值,得正项级数<img src="https://img2.meite.com/questions/202212/0163885fe0df37f.png" />,是收敛的p级数,从而原级数绝对收敛;当0<k≤l时,由莱布尼茨交错级数收敛性条件可判明原级数条件收敛,因此应有0<k≤1。</p><p>2、微分方程<img src="https://img2.meite.com/questions/202303/0364019f85a2899.png" />的通解为y=()  </p><p>答 案:<img src="https://img2.meite.com/questions/202303/036401ae0fc3051.png" /></p><p>解 析:将微分方程变量分离,可得<img src="https://img2.meite.com/questions/202303/036401ae31c99e2.png" />两边同时积分<img src="https://img2.meite.com/questions/202303/036401ae49ba074.png" />可得In|y|<img src="https://img2.meite.com/questions/202303/036401ae6ce75e9.png" /><img src="https://img2.meite.com/questions/202303/036401ae7812b6b.png" /></p><p>3、<img src="https://img2.meite.com/questions/202211/306387238d474db.png" />=()。</p><p>答 案:<img src="https://img2.meite.com/questions/202211/30638723985fdf5.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202211/30638723ad2d65d.png" /></p><p class="introTit">简答题</p><p>1、若函数<img src="https://img2.meite.com/questions/202303/1764141c98f2b2a.png" />在x=0处连续。求a。</p><p>答 案:由<img src="https://img2.meite.com/questions/202303/1764141cd6dcca7.png" /> 又因f(0)=a,所以当a=-1时,f(x)在x=0连续。  </p>
相关题库