2024年成考专升本《高等数学一》每日一练试题04月02日

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04/02
<p class="introTit">单选题</p><p>1、下列各点在球面(x-1)<sup>2</sup>+y<sup>2</sup>+(z-1)<sup>2</sup>=1上的是()。</p><ul><li>A:(1,0,1)</li><li>B:(2,0,2)</li><li>C:(1,1,1)</li><li>D:(1,1,2)</li></ul><p>答 案:C</p><p>解 析:将各个点代入球面公式可知(1,1,1)在球面上。</p><p>2、<img src="https://img2.meite.com/question/import/591d0b5c9f65cc151933621f3f7d06ae.png" /></p><ul><li>A:2x<sup>2+C</sup></li><li>B:x<sup>2+C</sup></li><li>C:1/2x<sup>2+C</sup></li><li>D:x+C</li></ul><p>答 案:C</p><p>3、设<img src="https://img2.meite.com/questions/202211/176375fffdc6fce.png" />与<img src="https://img2.meite.com/questions/202211/176376000a56c70.png" />都为正项级数,且<img src="https://img2.meite.com/questions/202211/1763760017d685f.png" />则下列结论正确的是()。</p><ul><li>A:若<img src='https://img2.meite.com/questions/202211/176376003a44e7a.png' />收敛,则<img src='https://img2.meite.com/questions/202211/1763760048b1d0a.png' />收敛</li><li>B:若<img src='https://img2.meite.com/questions/202211/176376005457117.png' />发散,则<img src='https://img2.meite.com/questions/202211/176376006229a54.png' />发散</li><li>C:若<img src='https://img2.meite.com/questions/202211/1763760080cc99e.png' />收敛,则<img src='https://img2.meite.com/questions/202211/17637600916ac24.png' />收敛</li><li>D:若<img src='https://img2.meite.com/questions/202211/176376009e49fa5.png' />收敛,则<img src='https://img2.meite.com/questions/202211/17637600a929d79.png' />发散</li></ul><p>答 案:C</p><p>解 析:由正项级数的比较判别法可知,若<img src="https://img2.meite.com/questions/202211/17637600ba270c9.png" />都为正项级数,且<img src="https://img2.meite.com/questions/202211/17637600c938b5b.png" />则当<img src="https://img2.meite.com/questions/202211/17637600d887cb8.png" />收敛时,可得知<img src="https://img2.meite.com/questions/202211/17637600e8480be.png" />必定收敛.</p><p class="introTit">主观题</p><p>1、计算<img src="https://img2.meite.com/questions/202109/1661429dcbbed55.png" width="107" /></p><p>答 案:<img src="https://img2.meite.com/questions/202109/1661429de15b12a.png" width="259" /></p><p>2、设<img src="https://img2.meite.com/questions/202211/166374ab053a051.png" />,求y'.</p><p>答 案:解:<img src="https://img2.meite.com/questions/202211/166374acf55bc71.png" /></p><p>3、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 class="introTit">填空题</p><p>1、<img src="https://img2.meite.com/questions/202303/1764141bea38c4f.png" />()</p><p>答 案:<img src="https://img2.meite.com/questions/202303/1764141c03d308e.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202303/1764141c185c3e5.png" /><img src="https://img2.meite.com/questions/202303/1764141c1fe101b.png" /></p><p>2、设a≠0,则<img src="https://img2.meite.com/questions/202212/03638aebb972006.png" />=()。</p><p>答 案:<img src="https://img2.meite.com/questions/202212/03638aebc536366.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202212/03638aebd4a8f1a.png" />。</p><p>3、<img src="https://img2.meite.com/questions/202211/176375d624c404f.png" />()。</p><p>答 案:<img src="https://img2.meite.com/questions/202211/176375d638cc9a8.png" /></p><p>解 析:由不定积分性质,可得<img src="https://img2.meite.com/questions/202211/176375d64c73c87.png" />。</p><p class="introTit">简答题</p><p>1、设f(x)<img src="https://img2.meite.com/questions/202303/1764141f831cbe2.png" />求f(x)的间断点。</p><p>答 案:由题意知,使f(x)不成立的x值,均为f(x)的间断点,故sin(x-3)=0或x-3=0时f(x)无意义,所以方程点为: x-3=<img src="https://img2.meite.com/questions/202303/17641420052b85c.png" /><img src="https://img2.meite.com/questions/202303/176414200b45e3f.png" />  </p>
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