2024年成考专升本《高等数学一》每日一练试题09月23日

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09/23
<p class="introTit">单选题</p><p>1、<img src="https://img2.meite.com/questions/202212/03638af57dc3578.png" />()。</p><ul><li>A:<img src='https://img2.meite.com/questions/202212/03638af58d9bad0.png' /></li><li>B:<img src='https://img2.meite.com/questions/202212/03638af597298c7.png' /></li><li>C:<img src='https://img2.meite.com/questions/202212/03638af5a083e48.png' /></li><li>D:<img src='https://img2.meite.com/questions/202212/03638af5a9e9bc2.png' /></li></ul><p>答 案:B</p><p>解 析:根据<img src="https://img2.meite.com/questions/202212/03638af5bbb5a1c.png" />,可得<img src="https://img2.meite.com/questions/202212/03638af5ceddb5f.png" />。</p><p>2、函数<img src="https://img2.meite.com/questions/202211/2863848501caa9e.png" />的连续区间是()。</p><ul><li>A:(∞,2)<img src='https://img2.meite.com/questions/202211/28638484ebb9d5c.png' />(2,1)<img src='https://img2.meite.com/questions/202211/28638484ed1f185.png' />(1,+∞)</li><li>B:[3,+∞)</li><li>C:(∞,2)<img src='https://img2.meite.com/questions/202211/28638485094fc55.png' />(2,+∞)</li><li>D:(∞,1)<img src='https://img2.meite.com/questions/202211/286384850a28820.png' />(1,+∞)</li></ul><p>答 案:B</p><p>解 析:函数在定义域内是连续的,故<img src="https://img2.meite.com/questions/202211/28638485202a043.png" />,得<img src="https://img2.meite.com/questions/202211/286384852bd4d86.png" />.故函数的连续区间为[3,+∞)。</p><p>3、若级数<img src="https://img2.meite.com/questions/202212/01638854cfa0e9b.png" />收敛,则<img src="https://img2.meite.com/questions/202212/01638854db021ec.png" />()。</p><ul><li>A:发散</li><li>B:条件收敛</li><li>C:绝对收敛</li><li>D:无法判定敛散性</li></ul><p>答 案:C</p><p>解 析:级数绝对收敛的性质可知,<img src="https://img2.meite.com/questions/202212/01638854ef9f908.png" />收敛,则<img src="https://img2.meite.com/questions/202212/01638855016b01e.png" />收敛,且为绝对收敛。</p><p class="introTit">主观题</p><p>1、求微分方程<img src="https://img2.meite.com/questions/202212/03638ac2f04259b.png" />的通解。</p><p>答 案:解:原方程对应的齐次方程为<img src="https://img2.meite.com/questions/202212/03638ac300089f5.png" />,特征方程及特征根为r<sup>2</sup>-4r+4=0,r<sub>1,2</sub>=2,齐次方程的通解为<img src="https://img2.meite.com/questions/202212/03638ac33159fca.png" />。在自由项<img src="https://img2.meite.com/questions/202212/03638ac3410667c.png" />中,a=-2不是特征根,所以设<img src="https://img2.meite.com/questions/202212/03638ac35b7ee07.png" />,代入原方程,有<img src="https://img2.meite.com/questions/202212/03638ac36d49e52.png" />,故原方程通解为<img src="https://img2.meite.com/questions/202212/03638ac37e8351d.png" />。</p><p>2、设函数f(x)=x-lnx,求f(x)的单调增区间.</p><p>答 案:解:函数f(x)的定义域为(0,+∞)。令y=f(x),则<img src="https://img2.meite.com/questions/202211/166374ad22c2d4a.png" />令y'=0,解得x=1。当0<x<1时,y'<0;当x>1时,y'>0。<br />因此函数f(x)的单调增区间为(1,+∞)。</p><p>3、求微分方程y'-<img src="https://img2.meite.com/questions/202212/03638ac17eaa5ba.png" />=lnx满足初始条件<img src="https://img2.meite.com/questions/202212/03638ac1908c65f.png" />=1的特解。</p><p>答 案:解:P(x)=<img src="https://img2.meite.com/questions/202212/03638ac1a3efa02.png" />,Q(x)=lnx,则<img src="https://img2.meite.com/questions/202212/03638ac1b4c4f05.png" />所以<img src="https://img2.meite.com/questions/202212/03638ac1c719a0f.png" />将<img src="https://img2.meite.com/questions/202212/03638ac1d4d883b.png" />=1代入y式,得C=1.故所求特解为<img src="https://img2.meite.com/questions/202212/03638ac2042e0a4.png" />。</p><p class="introTit">填空题</p><p>1、<img src="https://img2.meite.com/questions/202408/1666bec712bc791.png" />  </p><p>答 案:6</p><p>解 析:<img src="https://img2.meite.com/questions/202408/1666bec71abc7a2.png" /></p><p>2、已知<img src="https://img2.meite.com/questions/202211/3063872438d1a7f.png" />,则<img src="https://img2.meite.com/questions/202211/30638724485b269.png" />=()。</p><p>答 案:<img src="https://img2.meite.com/questions/202211/3063872454cd35d.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202211/3063872462b1da2.png" /><img src="https://img2.meite.com/questions/202211/306387246f61d8b.png" /><img src="https://img2.meite.com/questions/202211/306387247f5c8bb.png" /></p><p>3、幂级数<img src="https://img2.meite.com/questions/202212/0163885f2b44696.png" />的收敛区间(不考虑端点)是()。</p><p>答 案:(-2,2)</p><p>解 析:<img src="https://img2.meite.com/questions/202212/0163885f422728f.png" />,因此R=<img src="https://img2.meite.com/questions/202212/0163885f4ecae22.png" />=2,所以<img src="https://img2.meite.com/questions/202212/0163885f5e66413.png" />的收敛区间为(-2,2)。</p><p class="introTit">简答题</p><p>1、<img src="https://img2.meite.com/questions/202408/1666bf0648328cc.png" />  </p><p>答 案:<img src="https://img2.meite.com/questions/202408/1666bf064c8f37a.png" /></p>
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