2024年成考专升本《高等数学二》每日一练试题08月25日

<p class="introTit">判断题</p><p>1、若<img src="https://img2.meite.com/questions/202307/1364af5780970e6.png" />，则<img src="https://img2.meite.com/questions/202307/1364af578624075.png" />。（）  </p><p>答 案：错</p><p>解 析：<img src="https://img2.meite.com/questions/202212/06638ef8852e1bd.png" />所以<img src="https://img2.meite.com/questions/202212/06638ef88b66bc1.png" />  </p><p class="introTit">单选题</p><p>1、设函数z＝xe^y，则<img src="https://img2.meite.com/questions/202212/03638b0aea61193.png" />（）.</p><ul><li>A：e^x</li><li>B：e^y</li><li>C：xe^y</li><li>D：ye^x</li></ul><p>答 案：B</p><p>解 析：<img src="https://img2.meite.com/questions/202212/03638b0b0f6d2ec.png" />，<img src="https://img2.meite.com/questions/202212/03638b0b1c8a1b5.png" />.</p><p>2、设<img src="https://img2.meite.com/questions/202212/06638eeb0a112ec.png" />则f（x）在点x＝0处（）.</p><ul><li>A：可导且<img src='https://img2.meite.com/questions/202212/06638eeb199d2a7.png' />＝0</li><li>B：可导且<img src='https://img2.meite.com/questions/202212/06638eeb1b04b30.png' />＝1</li><li>C：不连续</li><li>D：连续但不可导</li></ul><p>答 案：A</p><p>解 析：因为<img src="https://img2.meite.com/questions/202212/06638eeb2e653bc.png" />，所以，f（x）在x＝0处连续；又<img src="https://img2.meite.com/questions/202212/06638eeb4483efe.png" />所以f（x）在点x＝0处可导且<img src="https://img2.meite.com/questions/202212/06638eeb586decf.png" />＝0.</p><p class="introTit">主观题</p><p>1、计算<img src="https://img2.meite.com/questions/202212/06638edfd88067f.png" />．</p><p>答 案：解：<img src="https://img2.meite.com/questions/202212/06638edfe86e57a.png" />.</p><p>2、计算<img src="https://img2.meite.com/questions/202212/05638d582780bce.png" /></p><p>答 案：解：这是<img src="https://img2.meite.com/questions/202212/05638d5859384fb.png" />型极限，可以使用洛必达法则<img src="https://img2.meite.com/questions/202212/05638d586b29a99.png" /></p><p class="introTit">填空题</p><p>1、设函数f(x)在x＝2处连续，且<img src="https://img2.meite.com/questions/202212/06638edd3c64d7a.png" />存在，则f(2)＝（）．</p><p>答 案：1</p><p>解 析：因为<img src="https://img2.meite.com/questions/202212/06638edd5716086.png" />存在，所以<img src="https://img2.meite.com/questions/202212/06638edd6438a74.png" />，即<img src="https://img2.meite.com/questions/202212/06638edd8a49417.png" />．因为f(x)在x＝2处连续，所以f(2)＝1.</p><p>2、<img src="https://img2.meite.com/questions/202408/1966c2f00be454e.png" />  </p><p>答 案：2</p><p>解 析：<img src="https://img2.meite.com/questions/202408/1966c2f0145e6e9.png" /></p><p class="introTit">简答题</p><p>1、已知函数f(x)=ax³-bx²+cx在区间<img src="https://img2.meite.com/questions/202212/07639002a0c63f6.png" />内是奇函数，且当x＝1时，f(x)有极小值<img src="https://img2.meite.com/questions/202212/07639002ba72fe4.png" />，求另一个极值及此曲线的拐点．  </p><p>答 案：f(x)=ax³-bx²+cx，<img src="https://img2.meite.com/questions/202212/07639002da708ed.png" /> 由于f(x)是奇函数，则必有x²的系数为0，即b=0. <img src="https://img2.meite.com/questions/202212/07639003148fbf6.png" />即a+c=<img src="https://img2.meite.com/questions/202212/076390032089cb2.png" />，<img src="https://img2.meite.com/questions/202212/0763900324b36b3.png" />得3a+c=0．解得a=<img src="https://img2.meite.com/questions/202212/076390033fe0499.png" />c=<img src="https://img2.meite.com/questions/202212/0763900347cfddd.png" /> 此时<img src="https://img2.meite.com/questions/202212/07639003591bd3b.png" /> 令<img src="https://img2.meite.com/questions/202212/0763900361d0796.png" />得<img src="https://img2.meite.com/questions/202212/07639003674342f.png" /><img src="https://img2.meite.com/questions/202212/076390036dd8f49.png" />所以<img src="https://img2.meite.com/questions/202212/076390037794ea3.png" />为极大值，<img src="https://img2.meite.com/questions/202212/07639003826209a.png" />得x=0，x<0时，<img src="https://img2.meite.com/questions/202212/0763900396d70f8.png" /> 所以（0，0）为曲线的拐点．</p><p>2、设函数y=sin²x,求<img src="https://img2.meite.com/questions/202212/076390018c80ad7.png" />  </p><p>答 案：<img src="https://img2.meite.com/questions/202212/076390019cc99c9.png" /> 所以<img src="https://img2.meite.com/questions/202212/07639001a8307c3.png" /></p>