2024年成考高起点《数学(理)》每日一练试题01月24日

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01/24
<p class="introTit">单选题</p><p>1、在△ABC中,若lgsinA-lgsinB-lgcos=lg2,则△ABC是()</p><ul><li>A:以A为直角的三角形</li><li>B:b=c的等腰三角形</li><li>C:等边三角形</li><li>D:钝角三角形</li></ul><p>答 案:B</p><p>解 析:判断三角形的形状,条件是用一个对数等式给出先将对数式利用对数的运算法则整理。 ∵lgsinA-lgsinB-lgcos=lg2,由对数运算法则可得,左<img src="https://img2.meite.com/questions/202303/286422954acc4dc.png" /> 两个对数底数相等则真数相等:<img src="https://img2.meite.com/questions/202303/2864229567bb0e0.png" />即2sinBcosC=sinA 在△ABC中,∵A+B+C=180°,∴A=180°-(B+C), <img src="https://img2.meite.com/questions/202303/28642295c273960.png" /><img src="https://img2.meite.com/questions/202303/28642295c819886.png" /><img src="https://img2.meite.com/questions/202303/28642295d337926.png" /><img src="https://img2.meite.com/questions/202303/28642295d958b6e.png" /><img src="https://img2.meite.com/questions/202303/28642295df5ccdf.png" /> 故为等腰三角形</p><p>2、若甲:x>1,乙:<img src="https://img2.meite.com/questions/202303/286422537c285e5.png" />则  </p><ul><li>A:甲是乙的必要条件,但不是乙的充分条件</li><li>B:甲是乙的充分必要条件</li><li>C:甲不是乙的充分条件,也不是乙的必要条件</li><li>D:甲是乙的充分条件,但不是乙的必要条件</li></ul><p>答 案:D</p><p>解 析:<img src="https://img2.meite.com/questions/202303/286422541d9ea18.png" />而<img src="https://img2.meite.com/questions/202303/2864225425630cc.png" />故甲是乙的充分条件,但不是必要条件</p><p>3、已知集合M =(2,3,5,a),N =(1,3,4,b),若M∩N=(1,2,3),则a,b的值为  </p><ul><li>A:a=2,b=1</li><li>B:a=1,b=1</li><li>C:a=1,b= 2</li><li>D:a=1,b=5</li></ul><p>答 案:C</p><p>解 析:M∩N={2,3,5,a} ∩{1,3,4,6} ={1,2,3} 又因为M中无“1”元素,而有“a”元素,只有a=1 而N中无“2”元素,而有“b元素”,只有b=2  </p><p>4、设函数<img src="https://img2.meite.com/questions/202303/15641163f4b1b54.png" />,则f(x+1)=()</p><ul><li>A:x<sup>2</sup>+2x+1</li><li>B:x<sup>2</sup>+2x</li><li>C:x<sup>2</sup>+1</li><li>D:x<sup>2</sup></li></ul><p>答 案:B</p><p>解 析:<img src="https://img2.meite.com/questions/202303/15641167631a1eb.png" /></p><p class="introTit">主观题</p><p>1、为了测河的宽,在岸边选定两点A和B,望对岸标记物C,测得<img src="https://img2.meite.com/questions/202303/2864228db8c0e49.png" />AB=120m,求河的宽 <img src="https://img2.meite.com/questions/202303/2864228dd64bdcb.png" /></p><p>答 案:如图, <img src="https://img2.meite.com/questions/202303/2864228df3f06d3.png" /> ∵∠C=180°-30°-75°=75° ∴△ABC为等腰三角形,则AC=AB=120m 过C做CD⊥AB,则由Rt△ACD可求得CD=<img src="https://img2.meite.com/questions/202303/2864228e8a387f3.png" />=60m, 即河宽为60m  </p><p>2、已知a,b,c成等差数列,a,b,c+1成等比数列.若b=6,求a和c.</p><p>答 案:由已知得<img src="https://img2.meite.com/questions/202303/1564116c009cc19.png" />解得<img src="https://img2.meite.com/questions/202303/1564116c0e039c1.png" /></p><p>3、设函数f(x)=xlnx+x.(I)求曲线y=f(x)在点((1,f(1))处的切线方程;<br />(II)求f(x)的极值.</p><p>答 案:(I)f(1)=1,f'(x)=2+lnx,故f'(1)=2.所以曲线y=f(x)在点(1,f(1))处的切线方程为y=2x-1.(II)令f'(x)=0,解得<img src="https://img2.meite.com/questions/202303/1564116d2d14a94.png" />当<img src="https://img2.meite.com/questions/202303/1564116d3d33026.png" />时,f'(x)<O;当<img src="https://img2.meite.com/questions/202303/1564116d6f6aec3.png" />时,f'(x)>O.故f(x)在区间<img src="https://img2.meite.com/questions/202303/1564116db9a0764.png" />单调递减,在区间<img src="https://img2.meite.com/questions/202303/1564116dc99fc91.png" />单调递增.因此f(x)在<img src="https://img2.meite.com/questions/202303/1564116ddb842d0.png" />时取得极小值<img src="https://img2.meite.com/questions/202303/1564116de4f1b79.png" /></p><p>4、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 class="introTit">填空题</p><p>1、函数<img src="https://img2.meite.com/questions/202303/2864225857f0d64.png" />的图像与坐标轴的交点共有()  </p><p>答 案:2</p><p>解 析:当x=0时,y=<img src="https://img2.meite.com/questions/202303/286422587f2a68b.png" />-2=-1,故函数与y轴交于(0,-1)点,令y=0,则有<img src="https://img2.meite.com/questions/202303/28642258b372965.png" />故函数与x轴交于(1,0) 点,因此函数 <img src="https://img2.meite.com/questions/202303/28642258c6c51c8.png" />与坐标轴的交点共有 2个.</p><p>2、函数<img src="https://img2.meite.com/questions/202303/28642285dd68e2c.png" />的定义域是()</p><p>答 案:<img src="https://img2.meite.com/questions/202303/28642285f367786.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202303/28642285fcec513.png" /><img src="https://img2.meite.com/questions/202303/2864228601aa2da.png" />所以函数<img src="https://img2.meite.com/questions/202303/286422860d8135c.png" />的定义域是<img src="https://img2.meite.com/questions/202303/286422861644b62.png" /></p>
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