2024年成考专升本《高等数学一》每日一练试题04月12日

<p class="introTit">单选题</p><p>1、<img src="https://img2.meite.com/questions/202212/03638ae588c44f4.png" />（）。</p><ul><li>A：<img src='https://img2.meite.com/questions/202212/03638ae5946a99f.png' /></li><li>B：<img src='https://img2.meite.com/questions/202212/03638ae5a0a586c.png' /></li><li>C：<img src='https://img2.meite.com/questions/202212/03638ae5ab2c9de.png' /></li><li>D：<img src='https://img2.meite.com/questions/202212/03638ae5b7db878.png' /></li></ul><p>答 案：A</p><p>解 析：<img src="https://img2.meite.com/questions/202212/03638ae5cad7d82.png" />。</p><p>2、<img src="https://img2.meite.com/questions/202211/1663744f0c6c11c.png" />＝（）。</p><ul><li>A：e</li><li>B：1</li><li>C：e^-1</li><li>D：－e</li></ul><p>答 案：C</p><p>解 析：由于<img src="https://img2.meite.com/questions/202211/1663744f455aaf1.png" />为连续函数，x＝０在函数的定义区间内，因此可直接将x＝0代入函数求极限，得<img src="https://img2.meite.com/questions/202211/1663744f5725477.png" />。</p><p>3、下列等式成立的是（）。</p><ul><li>A：<img src='https://img2.meite.com/questions/202211/28638480573b3e7.png' /></li><li>B：<img src='https://img2.meite.com/questions/202211/28638480611578b.png' /></li><li>C：<img src='https://img2.meite.com/questions/202211/286384806d4d475.png' /></li><li>D：<img src='https://img2.meite.com/questions/202211/286384807ac325b.png' /></li></ul><p>答 案：D</p><p>解 析：A项，由<img src="https://img2.meite.com/questions/202211/28638480d2b2cbb.png" />，可知<img src="https://img2.meite.com/questions/202211/28638480e0e1db3.png" />；B项，<img src="https://img2.meite.com/questions/202211/28638480ee2b9c1.png" />；C项，<img src="https://img2.meite.com/questions/202211/2863848104d297c.png" />；D项，<img src="https://img2.meite.com/questions/202211/286384811706b9d.png" />。</p><p class="introTit">主观题</p><p>1、设函数f（x）由<img src="https://img2.meite.com/questions/202211/176375a9a462105.png" />所确定，求<img src="https://img2.meite.com/questions/202211/176375a9b68f239.png" /></p><p>答 案：解：方法一：方程两边同时对x求导，得<img src="https://img2.meite.com/questions/202211/176375a9cd89294.png" />即<img src="https://img2.meite.com/questions/202211/176375a9de829b7.png" />故<img src="https://img2.meite.com/questions/202211/176375a9ec6e2cb.png" /><br />方法二：设<img src="https://img2.meite.com/questions/202211/176375a9fdad664.png" />，<br />则<img src="https://img2.meite.com/questions/202211/176375aa1004c84.png" /><img src="https://img2.meite.com/questions/202211/176375aa1f17e53.png" /></p><p>2、设函数<img src="https://img2.meite.com/questions/202211/176375db759e97f.png" />，求f（x）的极大值</p><p>答 案：解：<img src="https://img2.meite.com/questions/202211/176375dba527fd6.png" />当x＜-1或x＞3时，f′（x）＞0，f（x）单调增加；当－1＜x＜3时，f′（x）＜0，f（x）单调减少。<br />故x₁＝－1是f（x）的极大值点，<br />极大值为f（－1）＝5。</p><p>3、计算<img src="https://img2.meite.com/questions/202212/0163880f1f1db81.png" /></p><p>答 案：解：令<img src="https://img2.meite.com/questions/202212/0163880f2c78d6e.png" />当x＝4时，t＝2；当x＝9时，t＝3。则有<img src="https://img2.meite.com/questions/202212/0163880f445e340.png" /><img src="https://img2.meite.com/questions/202212/0163880f6263be5.png" /></p><p class="introTit">填空题</p><p>1、<img src="https://img2.meite.com/questions/202211/306387226310eea.png" />（）。</p><p>答 案：<img src="https://img2.meite.com/questions/202211/306387226d94c7a.png" /></p><p>解 析：<img src="https://img2.meite.com/questions/202211/306387227bd7f3f.png" /><img src="https://img2.meite.com/questions/202211/306387228a7d785.png" /></p><p>2、广义积分<img src="https://img2.meite.com/questions/202212/01638805ac70a46.png" />＝（）。</p><p>答 案：<img src="https://img2.meite.com/questions/202212/01638805b690ea3.png" /></p><p>解 析：<img src="https://img2.meite.com/questions/202212/01638805c7b9370.png" />。</p><p>3、过坐标原点且与平面2x－y＋z＋1＝0平行的平行方程为（）。</p><p>答 案：2x－y＋z＝0</p><p>解 析：已知平面的法线向量为(2，-1，1)，所求平面与已知平面平行<img src="https://img2.meite.com/questions/202211/1663745885379d6.png" />，因此平面方程可设为<img src="https://img2.meite.com/questions/202211/166374589042355.png" />，又平面过原点，故D＝0，即所求平面方程为2x－y＋z＝0。</p><p class="introTit">简答题</p><p>1、证明：当x>0时<img src="https://img2.meite.com/questions/202303/036401a13497bbc.png" />>1+x.  </p><p>答 案：<img src="https://img2.meite.com/questions/202303/036401aff003c5f.png" /></p>