2024年高职单招《数学(中职)》每日一练试题12月30日

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<p class="introTit">单选题</p><p>1、已知向量a=(1,2),b=(2,3),c=(-2,2),则(a-b)·c=()  </p><ul><li>A:4</li><li>B:-4</li><li>C:0</li><li>D:-2</li></ul><p>答 案:C</p><p>解 析:因为a=(1,2),b=(2,3),所以a-b=(1-2,2-3)=(-1,-1).又c=(-2,2),所以(a-b)·c=-1×(-2)+(-1)×2=0.</p><p>2、已知双曲线<img src="https://img2.meite.com/questions/202409/1866ea3dff44954.png" />的离心率<img src="https://img2.meite.com/questions/202409/1866ea3e0907f5a.png" />,则双曲线C的渐近线方程为()  </p><ul><li>A:y=±2x</li><li>B:<img src='https://img2.meite.com/questions/202409/1866ea3e12f235b.png' /></li><li>C:<img src='https://img2.meite.com/questions/202409/1866ea3e1b1d2a8.png' /></li><li>D:<img src='https://img2.meite.com/questions/202409/1866ea3e2354fe5.png' /></li></ul><p>答 案:A</p><p>解 析:由双曲线<img src="https://img2.meite.com/questions/202409/1866ea3dff44954.png" />可得 α<sup>2</sup>= m,b<sup>2</sup> =1 因为双曲线C的离心率<img src="https://img2.meite.com/questions/202409/1866ea3e0907f5a.png" />,所以<img src="https://img2.meite.com/questions/202409/1866ea49c2df1be.png" /><img src="https://img2.meite.com/questions/202409/1866ea49d132f1a.png" />解得m =4,所以a=2,所以双曲线C的渐近线方程为<img src="https://img2.meite.com/questions/202409/1866ea49e42ccf2.png" /></p><p>3、<img src="https://img2.meite.com/questions/202409/2166ee7abfc44a1.png" /></p><ul><li>A:1</li><li>B:-1</li><li>C:√3</li><li>D:-√3</li></ul><p>答 案:A</p><p>解 析:<img src="https://img2.meite.com/questions/202409/2166ee7ac74e389.png" /></p><p>4、不等式<img src="https://img2.meite.com/questions/202409/1066dff18a9a9df.png" />的解集是()</p><ul><li>A:<img src='https://img2.meite.com/questions/202409/1066dff18551a76.png' /></li><li>B:<img src='https://img2.meite.com/questions/202409/1066dff190bc5d1.png' /></li><li>C:<img src='https://img2.meite.com/questions/202409/1066dff199d61bd.png' /></li><li>D:<img src='https://img2.meite.com/questions/202409/1066dff1a2abbaf.png' /></li></ul><p>答 案:C</p><p>解 析:方程3x<sup>2</sup>-7x+2=0的两根分别为<img src="https://img2.meite.com/questions/202409/1066dff1a9692e0.png" /><img src="https://img2.meite.com/questions/202409/1066dff1b03b512.png" /><img src="https://img2.meite.com/questions/202409/1066dff199d61bd.png" /></p><p class="introTit">填空题</p><p>1、已知sinα=3cosα,那么tan2α的值为()</p><p>答 案:<img src="https://img2.meite.com/questions/202409/1366e3e35490a81.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202409/1366e3e359b39e2.png" /><img src="https://img2.meite.com/questions/202409/1366e3e35e81ae0.png" /></p><p>2、已知球的直径为2,则该球的体积是()  </p><p>答 案:<img src="https://img2.meite.com/questions/202409/1866ea2dee84dd3.png" /></p><p>解 析:易得球的半径为1,故球的体积为<img src="https://img2.meite.com/questions/202409/1866ea2dfab0399.png" /></p><p>3、如图,在正方体ABCD-A<sub>1</sub>B<sub>1</sub>C<sub>1</sub>D<sub>1</sub>中,异面直线D<sub>1</sub>C与BD 所成角的大小为() <img src="https://img2.meite.com/questions/202409/1366e3a1df8ea40.png" />  </p><p>答 案:60°</p><p>解 析:在正方体ABCD-A<sub>1</sub>B<sub>1</sub>C<sub>1</sub>D<sub>1</sub>中,连接B<sub>1</sub>D<sub>1</sub>与B<sub>1</sub>C,如图.易得BD//B<sub>1</sub>D<sub>1</sub>,所以<img src="https://img2.meite.com/questions/202409/1366e3aa3039039.png" />为异面直线D<sub>1</sub>C与BD所成的角.易知<img src="https://img2.meite.com/questions/202409/1366e3aa5ce9736.png" />是正三角形,所以<img src="https://img2.meite.com/questions/202409/1366e3aa651d5cf.png" />=60°,所以异面直线 D<sub>1</sub>C与 BD 所成角的大小为 60°. <img src="https://img2.meite.com/questions/202409/1366e3a9f12eec4.png" />  </p><p>4、在由1,2,3组成的不多于三位的自然数(可以有重复数字)中任意抽取一个,恰好抽中两位自然数的概率是()  </p><p>答 案:<img src="https://img2.meite.com/questions/202409/1266e250d0586f5.png" /></p><p>解 析:由1,2,3组成的一位自然数有3个,两位自然数有3<sup>2</sup>= 9(个),三位自然数有3<sup>3</sup>= 27(个),故恰好抽中两位自然数的概率是<img src="https://img2.meite.com/questions/202409/1266e2510e2a816.png" /></p><p class="introTit">简答题</p><p>1、已知焦点在y轴上的抛物线过 P(2,2).(1)求抛物线的标准方程;<br />(2)已知直线l:y=x+b(b≠0)与抛物线交于点 A,B,若以 AB 为直径的圆过原点O,求直线l的方程.  </p><p>答 案:(1)设抛物线的方程为 x<sup>2</sup>= 2py, 因为抛物线过 P(2,2), 所以4=4p,解得p=1, 所以抛物线的标准方程为x<sup>2</sup>= 2y. (2)设A(x<sub>1</sub>,y<sub>1</sub>),B(x<sub>2</sub>,y<sub>2</sub>). 由<img src="https://img2.meite.com/questions/202409/1966eb83282daf6.png" />得x<sup>2</sup>-2x-2b=0, 所以<img src="https://img2.meite.com/questions/202409/1966eb833dbf880.png" />,x<sub>1</sub>+x<sub>2</sub>= 2,x<sub>1</sub>x<sub>2</sub> =-2b. 因为以 AB 为直径的圆过原点O, 所以<img src="https://img2.meite.com/questions/202409/1966eb83775522b.png" />,则<img src="https://img2.meite.com/questions/202409/1966eb8382270b9.png" /> 所以x<sub>1</sub>x<sub>2</sub>+y<sub>1</sub>y<sub>2</sub> = 0, 所以x<sub>1</sub>x<sub>2</sub>+(x<sub>1</sub>+b)(x<sub>2</sub>+b)=2x<sub>1</sub>xz<sub>2</sub>+b(x<sub>1</sub>+x<sub>2</sub>)+b<sup>2</sup>=0, 所以-2b+b<sup>2</sup>=0,解得b=2(b=0舍去),符合题意. 所以直线l的方程为y=x+2.  </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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