2024年成考专升本《高等数学一》每日一练试题06月05日
<p class="introTit">单选题</p><p>1、函数<img src="https://img2.meite.com/questions/202211/296385cd769aee8.png" />单调减少的区间为()。</p><ul><li>A:(-∞,1]</li><li>B:[1,2]</li><li>C:[2,+∞)</li><li>D:[1,+∞)</li></ul><p>答 案:B</p><p>解 析:<img src="https://img2.meite.com/questions/202211/296385cdb8cba03.png" />的定义域为(-∞,+∞),求导得<img src="https://img2.meite.com/questions/202211/296385cdd67cf19.png" />令<img src="https://img2.meite.com/questions/202211/296385cdea4c3e2.png" />得驻点<img src="https://img2.meite.com/questions/202211/296385cdfc01d17.png" />当x<1时,<img src="https://img2.meite.com/questions/202211/296385ce12effde.png" />f(x)单调增加;当1<x<2时,<img src="https://img2.meite.com/questions/202211/296385ce254f76d.png" />,f(x)单调减少;当x>2时,<img src="https://img2.meite.com/questions/202211/296385ce361539c.png" />f(x)单调增加.故单调递减区间为[1,2]。</p><p>2、曲线y<img src="https://img2.meite.com/questions/202303/17641419f9e8ab6.png" />的水平渐近线方程是()</p><ul><li>A:y=2</li><li>B:y=-2</li><li>C:y=1</li><li>D:y=-1</li></ul><p>答 案:D</p><p>解 析:<img src="https://img2.meite.com/questions/202303/1764141a3fc6cd8.png" />所以水平渐近线为y=-1 ps:若<img src="https://img2.meite.com/questions/202303/1764141a70b7b74.png" />,则y=A是水平渐近线,若<img src="https://img2.meite.com/questions/202303/1764141a851ab6e.png" />则x=c是铅直渐近线。
</p><p>3、下列函数在[1,e]上满足拉格朗日中值定理条件的是()。</p><ul><li>A:1/(1-x)</li><li>B:lnx</li><li>C:1/(1-lnx)</li><li>D:<img src='https://img2.meite.com/questions/202211/296385d30cd094d.png' /></li></ul><p>答 案:B</p><p>解 析:AC两项,在[1,e]不连续,在端点处存在间断点(无穷间断点);B项,lnx在[1,e]上有定义,所以在[1,e]上连续,且<img src="https://img2.meite.com/questions/202211/296385d3292f64f.png" />在(1,e)内有意义,所以lnx在(1,e)内可导;D项,定义域为[2,+∞],在[1,2)上无意义。</p><p class="introTit">主观题</p><p>1、设e<sup>x</sup>-e<sup>y</sup>=siny,求y'。</p><p>答 案:解:<img src="https://img2.meite.com/questions/202211/3063870129c17a9.png" /></p><p>2、求函数<img src="https://img2.meite.com/questions/202211/30638706e12f5a2.png" />的极值及凹凸区间和拐点。</p><p>答 案:解:<img src="https://img2.meite.com/questions/202211/30638706ed5abd3.png" />(2)令y'=0,得x<sub>1</sub>=0,x<sub>2</sub>=2。令y''=0,得<img src="https://img2.meite.com/questions/202211/3063870717e6086.png" />。<br />(3)列表如下:<br /><img src="https://img2.meite.com/questions/202211/30638707319f4aa.png" /><br />函数<img src="https://img2.meite.com/questions/202211/306387074267a52.png" />的极小值为y(0)=0,极大值为<img src="https://img2.meite.com/questions/202211/3063870757924c9.png" />函数<img src="https://img2.meite.com/questions/202211/3063870769cdfb8.png" />的凹区间为<img src="https://img2.meite.com/questions/202211/306387078489c52.png" />函数<img src="https://img2.meite.com/questions/202211/306387079b6df7b.png" />的凸区间为<img src="https://img2.meite.com/questions/202211/30638707ae8b225.png" />函数<img src="https://img2.meite.com/questions/202211/30638707c0c95ac.png" />的拐点为<img src="https://img2.meite.com/questions/202211/30638707d78bac6.png" />与<img src="https://img2.meite.com/questions/202211/30638707f184046.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、已知函数<img src="https://img2.meite.com/questions/202211/306386fa886082c.png" />在[-1,1]上满足罗尔定理的条件,那么由定理所确定的<img src="https://img2.meite.com/questions/202211/306386fa9aa6862.png" />=()。</p><p>答 案:</p><p>解 析:<img src="https://img2.meite.com/questions/202211/306386faaeb1276.png" />,解得<img src="https://img2.meite.com/questions/202211/306386fabeef49f.png" />。</p><p>2、设y=f(x)可导,点x<sub>0</sub>=2为f(x)的极小值点,且f(2)=3,则曲线y=f(x)在点(2,3)处的切线方程为()。</p><p>答 案:y=3</p><p>解 析:由于y=f(x)可导,且点x<sub>0</sub>=2为f(x)的极小值点,由极值的必要条件可得<img src="https://img2.meite.com/questions/202211/186376eab73520a.png" />又f(2)=3,可知曲线过点(2,3)的切线方程为<img src="https://img2.meite.com/questions/202211/186376eacb7a1f5.png" /></p><p>3、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 class="introTit">简答题</p><p>1、讨论级数<img src="https://img2.meite.com/questions/202303/17641427ddeef02.png" />敛散性。</p><p>答 案:<img src="https://img2.meite.com/questions/202303/17641427efb7dc9.png" />所以级数收敛。 </p>