2025年高职单招《数学(中职)》每日一练试题05月13日
<p class="introTit">单选题</p><p>1、已知平面向量a=(1,-1),b=(-1,2),则a+b=()
</p><ul><li>A:(0,1)</li><li>B:(-2,3)</li><li>C:(2,-3)</li><li>D:1</li></ul><p>答 案:A</p><p>解 析:a+b=(1,-1)+(-1,2)=(0,1)。</p><p>2、下图表示程序框图中的( ).
<img src="https://img2.meite.com/questions/202501/06677b9c0f7a177.png" />
</p><ul><li>A:判断框</li><li>B:处理框</li><li>C:起止框</li><li>D:输入输出框</li></ul><p>答 案:D</p><p>3、对于任意的实数k,直线y=kx-k+1恒过点P,则点P的坐标为()
</p><ul><li>A:(-1,-1)</li><li>B:(-1,1)</li><li>C:(1,-1)</li><li>D:(1,1)</li></ul><p>答 案:D</p><p>解 析:由y=kx-k+1可得y-1=k(x-1).令x-1=0得x=1,此时y=1,所以直线y=kx-k+1恒过点 P(1,1).</p><p class="introTit">主观题</p><p>1、已知<img src="https://img2.meite.com/questions/202501/1167822ff243cf5.png" /></p><p>答 案:方法一:矢量图表示法 矢量图表示法如图所示。<img src="https://img2.meite.com/questions/202501/1167822ff89fe55.png" /><img src="https://img2.meite.com/questions/202501/1167822ffb0aba3.png" />
方法二:矢量表示法<img src="https://img2.meite.com/questions/202501/116782300184954.png" /></p><p>解 析:<img src="https://img2.meite.com/questions/202409/1366e3aff07aaa2.png" /><img src="https://img2.meite.com/questions/202409/1366e3affb6a19c.png" /><img src="https://img2.meite.com/questions/202409/1366e3affff2ae2.png" /></p><p class="introTit">填空题</p><p>1、双曲线y<sup>2</sup>-x<sup>2</sup>=2的焦点坐标是()
</p><p>答 案:(0,±2) </p><p>解 析:由y<sup>2</sup>-x<sup>2</sup>=2得<img src="https://img2.meite.com/questions/202409/1866ea983053f6f.png" />,所以双曲线的焦点在y轴上,a<sup>2</sup>=b<sup>2</sup>= 2,所以c<sup>2</sup>=a<sup>2</sup>+b<sup>2</sup>= 4,所以焦点坐标是(0,±2).</p><p>2、如图,在长方体ABCD-A<sub>1</sub>B<sub>1</sub>C<sub>1</sub>D<sub>1</sub>中,AB=1,AD=2,AA<sub>1</sub>=2,则二面角A-DD<sub>1</sub>-B的正切值为()
<img src="https://img2.meite.com/questions/202409/1366e3a1e9e4681.png" />
</p><p>答 案:<img src="https://img2.meite.com/questions/202409/1366e3aabede20b.png" /></p><p>解 析:易得<img src="https://img2.meite.com/questions/202409/1366e3aae9a9b6c.png" />,<img src="https://img2.meite.com/questions/202409/1366e3ab007062c.png" />平面角.在<img src="https://img2.meite.com/questions/202409/1366e3ab1ab51b8.png" />,AB=1AD=2,<img src="https://img2.meite.com/questions/202409/1366e3ab2d83821.png" />,∴二面角A-DD<sub>1</sub>-B的正切值为<img src="https://img2.meite.com/questions/202409/1366e3aabede20b.png" />.</p><p>3、设等差数列{a<sub>n</sub>}的前n项和为S<sub>n</sub>,若S<sub>m-1</sub>,=-2,S<sub>m</sub>=0,S<sub>m</sub><sub>+1</sub>=3,则公差d=()</p><p>答 案:1</p><p>解 析:因为S<sub>m-1</sub>=-2,S<sub>m</sub>=0,S<sub>m+1</sub>=3,所以a<sub>m</sub>=S<sub>m</sub>-S<sub>m-1</sub>=2,a<sub>m+1</sub>=S<sub>m+1</sub>-S<sub>m</sub>=3,所d=a<sub>m+1</sub>-a<sub>m</sub>=1.</p><p class="introTit">简答题</p><p>1、已知椭圆<img src="https://img2.meite.com/questions/202409/1866ea40f2b8734.png" />的右焦点为<img src="https://img2.meite.com/questions/202409/1866ea40fb35447.png" />,长轴长和短轴长之和为12,过点<img src="https://img2.meite.com/questions/202409/1866ea4108ee28a.png" />且倾斜角为<img src="https://img2.meite.com/questions/202409/1866ea41162d299.png" />的直线与椭圆交于 A,B两点.(1)求椭圆的标准方程;<br />(2)求线段 AB 的中点坐标.
</p><p>答 案:(1)由题意知半焦距<img src="https://img2.meite.com/questions/202409/1866ea79fb4d7f2.png" />,2a+2b=12,即a+b=6. 又a<sup>2</sup>-b<sup>2</sup>=c<sup>2</sup>,
所以a=4,b= 2,
所以椭圆的标准方程为<img src="https://img2.meite.com/questions/202409/1866ea7a24d497e.png" />
(2)易得直线AB的方程为<img src="https://img2.meite.com/questions/202409/1866ea7a30db014.png" />,即<img src="https://img2.meite.com/questions/202409/1866ea7a387f559.png" />
由<img src="https://img2.meite.com/questions/202409/1866ea7a4b9bbc1.png" />得13x<sup>2</sup>-24x-4=0.
设A(x<sub>1</sub>,y<sub>1</sub>),B(x<sub>2</sub>,y<sub>2</sub>),线段 AB 中点的坐标为(x<sub>0</sub>,y<sub>0</sub>),
则<img src="https://img2.meite.com/questions/202409/1866ea7a856c16f.png" />
所以<img src="https://img2.meite.com/questions/202409/1866ea7a8b285d0.png" />
所以<img src="https://img2.meite.com/questions/202409/1866ea7a92c300b.png" />
所以线段A8 的中点坐标为<img src="https://img2.meite.com/questions/202409/1866ea7aa1bd986.png" />
</p>