2024年成考专升本《高等数学一》每日一练试题11月26日

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11/26
<p class="introTit">单选题</p><p>1、对于微分方程<img src="https://img2.meite.com/questions/202212/0163886b75a6de5.png" />,利用待定系数法求其特解y*时,下列特解设法正确的是()。</p><ul><li>A:y*=(Ax+B)e<sup>x</sup></li><li>B:y*=x(Ax+B)e<sup>x</sup></li><li>C:y*=Ax<sup>3</sup>e<sup>x</sup></li><li>D:y*=x<sup>2</sup>(Ax+B)e<sup>x</sup></li></ul><p>答 案:D</p><p>解 析:特征方程为r<sup>2</sup>-2r+1=0,特征根为r=1(二重根),<img src="https://img2.meite.com/questions/202212/0163886bbd8c2a4.png" />,a=1为特征根,原方程特解为<img src="https://img2.meite.com/questions/202212/0163886bda4453b.png" />。</p><p>2、当x→0时,x+x<sup>2</sup>+x<sup>3</sup>+x<sup>4</sup>是()的等价无穷小量。  </p><ul><li>A:x</li><li>B:x<sup>2</sup></li><li>C:x<sup>3</sup></li><li>D:x<sup>4</sup> </li></ul><p>答 案:A</p><p>解 析:本题考查的知识点为无穷小量阶的比较。 <img src="https://img2.meite.com/questions/202408/1566bda5492aa84.png" />  </p><p>3、曲线<img src="https://img2.meite.com/questions/202303/17641428768e28c.png" />与其过原点的切线及y轴所围面积为()  </p><ul><li>A:<img src='https://img2.meite.com/questions/202303/176414289f110e7.png' /></li><li>B:<img src='https://img2.meite.com/questions/202303/17641428a6b84fa.png' /></li><li>C:<img src='https://img2.meite.com/questions/202303/17641428acae8ee.png' /></li><li>D:<img src='https://img2.meite.com/questions/202303/17641428b1e166c.png' /></li></ul><p>答 案:A</p><p>解 析:设(x0,y0)为切点,则切线方程为<img src="https://img2.meite.com/questions/202303/17641428de828b2.png" />联立<img src="https://img2.meite.com/questions/202303/17641428eb5aa98.png" />得x0=1,y0=e,所以切线方程为y=ex,故所求面积为<img src="https://img2.meite.com/questions/202303/1764142916eb7e4.png" /></p><p class="introTit">主观题</p><p>1、求微分方程<img src="https://img2.meite.com/questions/202212/03638ac38e5fd44.png" />的通解。</p><p>答 案:解:对应的齐次方程为<img src="https://img2.meite.com/questions/202212/03638ac39bb126c.png" />。特征方程<img src="https://img2.meite.com/questions/202212/03638ac3abef614.png" />,特征根<img src="https://img2.meite.com/questions/202212/03638ac3bb18486.png" />齐次方程通解为<img src="https://img2.meite.com/questions/202212/03638ac3c81e1f0.png" />原方程特解为<img src="https://img2.meite.com/questions/202212/03638ac3d7442b3.png" />,代入原方程可得<img src="https://img2.meite.com/questions/202212/03638ac3ea0899f.png" />,因此<img src="https://img2.meite.com/questions/202212/03638ac3fe7cb5d.png" />。<br />方程通解为<img src="https://img2.meite.com/questions/202212/03638ac40c27e0d.png" /></p><p>2、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>3、求<img src="https://img2.meite.com/questions/202211/2963856b9fa7d50.png" /></p><p>答 案:解:<img src="https://img2.meite.com/questions/202211/2963856bb1eca82.png" /><img src="https://img2.meite.com/questions/202211/2963856bc769de2.png" /></p><p class="introTit">填空题</p><p>1、幂级数<img src="https://img2.meite.com/questions/202303/176414046fa95c8.png" />的收敛半径为()</p><p>答 案:3</p><p>解 析:所给幂级数通项为<img src="https://img2.meite.com/questions/202303/176414049a92cf3.png" />则<img src="https://img2.meite.com/questions/202303/17641404aae13b4.png" /><img src="https://img2.meite.com/questions/202303/17641404b621a4a.png" />所以收敛半径R=3</p><p>2、<img src="https://img2.meite.com/questions/202405/166645bc8980e33.png" />()  </p><p>答 案:<img src="https://img2.meite.com/questions/202405/166645bc90e8937.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202405/166645bc95a949d.png" /></p><p>3、设y=5+lnx,则dy=()。</p><p>答 案:<img src="https://img2.meite.com/questions/202211/16637457c2bf6b0.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202211/16637457c7adbbf.png" /></p><p class="introTit">简答题</p><p>1、计算<img src="https://img2.meite.com/questions/202408/1566bda95c7d643.png" />    </p><p>答 案:<img src="https://img2.meite.com/questions/202408/1566bda960bd750.png" /><img src="https://img2.meite.com/questions/202408/1566bda964a1ff8.png" /></p><p>解 析:本题考查的知识点为定积分的计算。</p>
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