2024年成考高起点《数学(理)》每日一练试题01月20日
<p class="introTit">单选题</p><p>1、设集合M={x||x-2|<1},N={x|x>2},则M∩N=()</p><ul><li>A:{x|1<x<3}</li><li>B:{x|x>2}</li><li>C:{x|2<x<3}</li><li>D:{x|1<x<2}</li></ul><p>答 案:C</p><p>解 析:M={x||x-2|<1}解得{x|-1<x-2<1}={x|1<x<3},故M∩N={x|2<x<3}</p><p>2、下列函数中,为奇函数的是()</p><ul><li>A:<img src='https://img2.meite.com/questions/202303/156411642c7ba59.png' /></li><li>B:<img src='https://img2.meite.com/questions/202303/156411643238049.png' /></li><li>C:<img src='https://img2.meite.com/questions/202303/1564116437cc5f5.png' /></li><li>D:<img src='https://img2.meite.com/questions/202303/156411643c1ff56.png' /></li></ul><p>答 案:B</p><p>解 析:当f(-x)=-f(x),函数f(x)是奇函数,只有选项B符合.</p><p>3、过点(-2,2)与直线x+3y-5=0平行的直线是()</p><ul><li>A:x+3y-4=0</li><li>B:3x+y+4=0</li><li>C:x+3y+8=0</li><li>D:3x-y+8=0</li></ul><p>答 案:A</p><p>解 析:所求直线与x+3y-5=0平行,可设所求直线为x+3y+c=0,将点(一2,2)带入直线方程,故-2+3×2+c=0,解得c=-4,因此所求直线为线为x+3y-4=0.</p><p>4、<img border="0" style="width: 58px; height: 37px;" src="https://img2.meite.com/zzpuce/2023-04/643679c8eed3a35616.jpeg">( )</p><ul><li>A:-2</li><li>B:<img border="0" style="width: 24px; height: 37px;" src="https://img2.meite.com/zzpuce/2023-04/643679c9008af39045.jpeg"></li><li>C:<img border="0" style="width: 14px; height: 37px;" src="https://img2.meite.com/zzpuce/2023-04/643679c90917e56279.jpeg"></li><li>D:2</li></ul><p>答 案:C</p><p class="introTit">主观题</p><p>1、设函数f(x)=<img src="https://img2.meite.com/questions/202303/28642286431b211.png" />
(Ⅰ)求f(x)的单调区间;
(Ⅱ)求 f(x)的极值</p><p>答 案:(Ⅰ)函数的定义域为<img src="https://img2.meite.com/questions/202303/28642286bee9cc3.png" /> <img src="https://img2.meite.com/questions/202303/28642286c7d68a9.png" />
(Ⅱ)<img src="https://img2.meite.com/questions/202303/28642286d3444c8.png" />
</p><p>2、在正四棱柱ABCD-A'B'C'D'中,<img src="https://img2.meite.com/questions/202303/2864229a3bc3098.png" />
(Ⅰ)写出向量<img src="https://img2.meite.com/questions/202303/2864229a57ba174.png" />和<img src="https://img2.meite.com/questions/202303/2864229a5e46ac8.png" />关于基底{a,b,c}的分解式;
(Ⅱ)求证:<img src="https://img2.meite.com/questions/202303/2864229a76ba56d.png" />
(Ⅲ)求证:<img src="https://img2.meite.com/questions/202303/2864229a7fdd541.png" />
</p><p>答 案:(Ⅰ)由题意知(如图所示) <img src="https://img2.meite.com/questions/202303/2864229af6b1567.png" />
<img src="https://img2.meite.com/questions/202303/2864229afe90f50.png" />
<img src="https://img2.meite.com/questions/202303/2864229b08314c5.png" />
</p><p>3、在正四棱柱ABCD-A'B'C'D'中,<img src="https://img2.meite.com/questions/202303/28642255fa50503.png" />
(Ⅰ)写出向量<img src="https://img2.meite.com/questions/202303/286422561b1d145.png" />关于基底{a,b,c}的分解式
(Ⅱ)求证:<img src="https://img2.meite.com/questions/202303/286422563d58cde.png" />
(Ⅲ)求证:<img src="https://img2.meite.com/questions/202303/28642256478aacd.png" />
</p><p>答 案:(Ⅰ)由题意知(如图所示) <img src="https://img2.meite.com/questions/202303/286422566983935.png" />
<img src="https://img2.meite.com/questions/202303/28642256740213a.png" /><img src="https://img2.meite.com/questions/202303/286422567c06c5d.png" />
(Ⅱ)<img src="https://img2.meite.com/questions/202303/2864225695c5fbd.png" /><img src="https://img2.meite.com/questions/202303/286422569cdc533.png" />
(Ⅲ)<img src="https://img2.meite.com/questions/202303/28642256a537b6d.png" />
由已知,a,c是正四棱柱的棱,a,b,c两两垂直
<img src="https://img2.meite.com/questions/202303/28642256d1c4379.png" />
</p><p>4、建筑一个容积为8000<img src="https://img2.meite.com/questions/202303/2864224b406cbf6.png" />,深为6m的长方体蓄水池,池壁每<img src="https://img2.meite.com/questions/202303/2864224b5cac16d.png" />的造价为15元,池底每<img src="https://img2.meite.com/questions/202303/2864224b60ac28e.png" />的造价为30元。(I)把总造价y(元)表示为长x(m)的函数;(Ⅱ)求函数的定义域
</p><p>答 案:<img src="https://img2.meite.com/questions/202303/2864224be4311f4.png" /><img src="https://img2.meite.com/questions/202303/2864224bee67713.png" /></p><p class="introTit">填空题</p><p>1、若平面向量a=(x,1),b=(1,-2),且a//b,则x=()
</p><p>答 案:<img src="https://img2.meite.com/questions/202303/286422508f0554f.png" /></p><p>解 析:由于a//b,故<img src="https://img2.meite.com/questions/202303/286422509cd5c14.png" /></p><p>2、lg(tan43°tan45°tan47°)=()
</p><p>答 案:0</p><p>解 析:lg(tan43°tan45°tan47°)=lg(tan43°tan45°cot43°)=lgtan45°=lg1=0</p>