2024年成考高起点《数学(文史)》每日一练试题08月09日
<p class="introTit">单选题</p><p>1、若函数f(x)=1+<img src="https://img2.meite.com/questions/202303/296423a8d0db03d.png" />在(0,+∞)上是减函数,则()</p><ul><li>A:a>1</li><li>B:a>2</li><li>C:1<a<2</li><li>D:0<a<1</li></ul><p>答 案:D</p><p>解 析:由已知条件函数f(x)=1+<img src="https://img2.meite.com/questions/202303/296423a8d0db03d.png" />在(0,+∞)上是减函数,及对数函数<img src="https://img2.meite.com/questions/202303/296423a9706abcb.png" />的性质可得底数0<a<1</p><p>2、函数<img src="https://img2.meite.com/questions/202303/296423e9f4c42ba.png" />与y<img src="https://img2.meite.com/questions/202303/296423e9fce2f9b.png" />的图像之间的关系是
</p><ul><li>A:关于原点对称</li><li>B:关于x轴对称</li><li>C:关于直线 y=1对称</li><li>D:关于y轴对称</li></ul><p>答 案:D</p><p>解 析:<img src="https://img2.meite.com/questions/202303/296423ea4eab0e4.png" /><img src="https://img2.meite.com/questions/202303/296423ea5d59f26.png" />关于y轴对称,</p><p>3、从15名学生中选出两人担任正、副班长,不同的选举结果共有()
</p><ul><li>A:30种</li><li>B:90种</li><li>C:210种</li><li>D:225种</li></ul><p>答 案:C</p><p>解 析:由已知条件可知本题属于排列问题,<img src="https://img2.meite.com/questions/202303/296423b1cca0339.png" /></p><p>4、设<img src="https://img2.meite.com/questions/202404/20662380b45ed12.png" />()。</p><ul><li>A:2a<sup>2</sup>+1</li><li>B:2a<sup>2</sup>-1</li><li>C:2a-1</li><li>D:2a+1</li></ul><p>答 案:D</p><p>解 析:本题主要考查的知识点为对数函数的性质。 <img src="https://img2.meite.com/questions/202404/20662380bca0220.png" /></p><p class="introTit">主观题</p><p>1、如图:已知在△ADC中,∠C=90°,∠D=30°,∠ABC=45°,BD=20,求AC(用小数表示,保留一位小数)
<img src="https://img2.meite.com/questions/202303/296423e7b58f8d3.png" />
</p><p>答 案:如图 <img src="https://img2.meite.com/questions/202303/296423e7cbcb7ab.png" />
<img src="https://img2.meite.com/questions/202303/296423e7dd72c9c.png" />
</p><p>2、在△ABC中,B=120°,C=30°,BC=4,求△ABC的面积.</p><p>答 案:因为A= 180°-B-C=30°,所以AB = BC=4.因此△ABC的面积<img src="https://img2.meite.com/questions/202303/1564111c8c97743.png" /></p><p>3、已知等差数列<img src="https://img2.meite.com/questions/202303/296423eaf9717d6.png" />前n项和<img src="https://img2.meite.com/questions/202303/296423eb032d219.png" />
(Ⅰ)求通项<img src="https://img2.meite.com/questions/202303/296423eb1a4ebf5.png" />的表达式
(Ⅱ)求<img src="https://img2.meite.com/questions/202303/296423eb26c2214.png" />的值
</p><p>答 案:(Ⅰ)当n=1时,由<img src="https://img2.meite.com/questions/202303/296423eb432a645.png" />得<img src="https://img2.meite.com/questions/202303/296423eb5068b03.png" /> <img src="https://img2.meite.com/questions/202303/296423eb59a45cd.png" />
<img src="https://img2.meite.com/questions/202303/296423eb6100c03.png" />
也满足上式,故<img src="https://img2.meite.com/questions/202303/296423eb755b7df.png" />=1-4n(n≥1)
(Ⅱ)由于数列<img src="https://img2.meite.com/questions/202303/296423eb93e2df0.png" />是首项为<img src="https://img2.meite.com/questions/202303/296423eba5a3367.png" />公差为d=-4的等差数列,所以<img src="https://img2.meite.com/questions/202303/296423ebc29c045.png" />是首项为<img src="https://img2.meite.com/questions/202303/296423ebe5ba947.png" />公差为d=-8,项数为13的等差数列,于是由等差数列前n项和公式得:
<img src="https://img2.meite.com/questions/202303/296423ec1ac9811.png" /><img src="https://img2.meite.com/questions/202303/296423ec20a013e.png" />
</p><p>4、设函数f(x)<img src="https://img2.meite.com/questions/202303/296423a66bbdb95.png" />且f'(-1)=-36
(Ⅰ)求m
(Ⅱ)求f(x)的单调区间</p><p>答 案:(Ⅰ)由已知得f'=<img src="https://img2.meite.com/questions/202303/296423a74f41b7f.png" /> 又由f'(-1)=-36得
6-6m-36=-36
故m=1.
(Ⅱ)由(Ⅰ)得f'(x)=<img src="https://img2.meite.com/questions/202303/296423a792d1aff.png" />
令f'(x)=0,解得<img src="https://img2.meite.com/questions/202303/296423a7b14cf3f.png" />
当x<-3时,f'(x)>0;
当-3<x<2时,f'(x)<0;
当x>2时,f'(x)>0;
故f(x)的单调递减区间为(-3,2),f(x)的单调递增区间为(-∞,-3),(2,+∞)
</p><p class="introTit">填空题</p><p>1、点(4,5)关于直线y=x的对称点的坐标为()</p><p>答 案:(5,4)</p><p>解 析:点(4,5)关于直线y=x的对称点为(5,4).</p><p>2、函数f(x)=<img src="https://img2.meite.com/questions/202303/296423b497a799f.png" />在区间[-3,3]上的最大值为()
</p><p>答 案:4</p><p>解 析:这题考的是高次函数的最值问题,可用导数来求函数在区间[-3,3]上的最值。<img src="https://img2.meite.com/questions/202303/296423b5082e922.png" /> 列出表格<img src="https://img2.meite.com/questions/202303/296423b517619c0.png" />
由上表可知函数在[-3,3]上,在x=1点处有最大值为4.
</p>