2024年高职单招《数学(中职)》每日一练试题10月31日

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10/31
<p class="introTit">单选题</p><p>1、命题<img src="https://img2.meite.com/questions/202409/0966deaccadc992.png" />的否定是()</p><ul><li>A:<img src='https://img2.meite.com/questions/202409/0966deacda16607.png' /></li><li>B:<img src='https://img2.meite.com/questions/202409/0966deace0ef30a.png' /></li><li>C:<img src='https://img2.meite.com/questions/202409/0966deace77c667.png' /></li><li>D:<img src='https://img2.meite.com/questions/202409/0966deacee58ab5.png' /></li></ul><p>答 案:C</p><p>解 析:因为命题<img src="https://img2.meite.com/questions/202409/0966deaccadc992.png" /> 是特称命题,所以其否定是全称命题,即<img src="https://img2.meite.com/questions/202409/0966dead097f1a7.png" /></p><p>2、把曲线<img src="https://img2.meite.com/questions/202409/2166ee68159071b.png" />向右平移<img src="https://img2.meite.com/questions/202409/2166ee681e2c171.png" />个单位长度,所得曲线的函数解析式为()</p><ul><li>A:y=sin3x</li><li>B:<img src='https://img2.meite.com/questions/202409/2166ee6823df346.png' /></li><li>C:<img src='https://img2.meite.com/questions/202409/2166ee682a35bbc.png' /></li><li>D:<img src='https://img2.meite.com/questions/202409/2166ee6830b7b77.png' /></li></ul><p>答 案:A</p><p>解 析:将曲线<img src="https://img2.meite.com/questions/202409/2166ee683919ce0.png" />向右平移<img src="https://img2.meite.com/questions/202409/2166ee684326971.png" />个单位长度,得到<img src="https://img2.meite.com/questions/202409/2166ee684a7d558.png" />的图象,所以所得曲线的函数解析式为y=sin3x.</p><p>3、函数<img src="https://img2.meite.com/questions/202409/1266e2968d940d4.png" />的图像过定点()</p><ul><li>A:(0,-1)</li><li>B:(1,-1)</li><li>C:(1,1)</li><li>D:(1,0)</li></ul><p>答 案:B</p><p>解 析:令x=1,则y=log<sub>a</sub>1-1=-1,所以函数<img src="https://img2.meite.com/questions/202409/1266e296976f4d5.png" /> 的图像过定点(1,-1).  </p><p>4、已知函数f(x)=x<sup>2</sup>+ax+2在区间(-∞,1]内单调递减,则实数a的取值范围是()</p><ul><li>A:<img src='https://img2.meite.com/questions/202409/1166e14d8f1cce2.png' /></li><li>B:<img src='https://img2.meite.com/questions/202409/1166e14d94bd1f9.png' /></li><li>C:<img src='https://img2.meite.com/questions/202409/1166e14d99ee491.png' /></li><li>D:<img src='https://img2.meite.com/questions/202409/1166e14d9f0a8a1.png' /></li></ul><p>答 案:A</p><p>解 析:易得函数f(x)=x<sup>2</sup>+ax+2的图像开口向上,对称轴为直线<img src="https://img2.meite.com/questions/202409/1166e14da791fac.png" /><img src="https://img2.meite.com/questions/202409/1166e14dacd854f.png" /><img src="https://img2.meite.com/questions/202409/1166e14db1e24e2.png" /><img src="https://img2.meite.com/questions/202409/1166e14db641e63.png" /></p><p class="introTit">填空题</p><p>1、若2<sup>a</sup>=3,b=log<sub>3</sub>4,则ab=()</p><p>答 案:2</p><p>解 析:<img src="https://img2.meite.com/questions/202409/1266e288869065b.png" /></p><p>2、如图,画一个边长为4的正方形,再将这个正方形各边的中点相连得到第2个正方形,依此类推,这样一共画了10个正方形,则第10个正方形的面积为() <img src="https://img2.meite.com/questions/202409/1866ea837912d6a.png" /></p><p>答 案:<img src="https://img2.meite.com/questions/202409/1866ea8383ce7b7.png" /></p><p>解 析:由题意知,第一个正方形的边长a1=4,面积T1=16;第二个正方形的边长<img src="https://img2.meite.com/questions/202409/1866ea83932e02b.png" />面积T2=8;第三个正方形的边长a3=2,面积T3=4;第四个正方形的边长<img src="https://img2.meite.com/questions/202409/1866ea839c73a9d.png" />面积T4=2,……,由此可发现规律,第n个正方形的面积<img src="https://img2.meite.com/questions/202409/1866ea83a3d2669.png" /></p><p>3、<img src="https://img2.meite.com/questions/202409/1266e2a07fec4f2.png" />的展开式中各项系数之和为()  </p><p>答 案:1</p><p>解 析:令x=1,得<img src="https://img2.meite.com/questions/202409/1266e2a7d117d01.png" />,即<img src="https://img2.meite.com/questions/202409/1266e2a07fec4f2.png" />的展开式中各项系数之和为1.</p><p>4、若双曲线<img src="https://img2.meite.com/questions/202409/1866ea3fc68f76b.png" />的虚轴长为8,渐近线方程为<img src="https://img2.meite.com/questions/202409/1866ea3fd35492c.png" />,则双曲线C的方程为()  </p><p>答 案:<img src="https://img2.meite.com/questions/202409/1866ea725e51abb.png" /></p><p>解 析:由题可得<img src="https://img2.meite.com/questions/202409/1866ea728576feb.png" />解得<img src="https://img2.meite.com/questions/202409/1866ea728dc498f.png" />所以双曲线的方程为<img src="https://img2.meite.com/questions/202409/1866ea725e51abb.png" /></p><p class="introTit">简答题</p><p>1、已知p:x<sup>2</sup>-8x-20≤0,q:1-m≤x≤1+m(m>0).若p是q的必要不充分条件,求实数m的范围.</p><p>答 案:由x<sup>2</sup>-8x-20≤0得-2≤x≤10,即p:-2≤x≤10. ∵p是q的必要不充分条件,q:1-m≤x≤1+m(m>0),<img src="https://img2.meite.com/questions/202409/0966dead3908be8.png" /></p><p>2、若一元二次不等式<img src="https://img2.meite.com/questions/202409/1066dff30435ede.png" />恒成立,求实数a的取值范围.  </p><p>答 案:<img src="https://img2.meite.com/questions/202409/1066dff30e2d19d.png" /><img src="https://img2.meite.com/questions/202409/1066dff31381f7a.png" /></p>
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