山东省聊城市2012-2013学年高二上学期期中考试预测题(一)

2023-2024学年山东省聊城市高二（上）期中数学试卷一、选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合 1．设a ∈R ，则“直线ax +y ﹣1＝0与直线x +ay +1＝0平行”是“a ＝1”的（ ） A ．充分不必要条件 B ．充要条件C ．必要不充分条件D ．既不充分也不必要条件2．经过两条直线l 1：x +y ＝2，l 2：2x ﹣y ＝1的交点，且直线的一个方向向量v →=(−6，4) 的直线方程为（ ） A ．2x ﹣y ﹣1＝0B ．2x +y ﹣3＝0C ．3x ﹣2y ﹣5＝0D ．2x +3y ﹣5＝03．已知SA ⊥平面ABC ，AB ⊥AC ，SA ＝AB ＝1，BC =√5，则空间的一个单位正交基底可以为（ ） A ．{AB →，12AC →，AS →} B ．{AB →，AC →，AS →} C ．{AB →，12AC →，12AS →} D ．{AS →，AB →，√55BC →}4．椭圆x 216+y 24=1和x 236+y 224=1（ ）A ．长轴长相等B ．短轴长相等C ．焦距相等D ．顶点相同5．已知圆M ：x 2+y 2﹣2ay ＝0（a ＞0）截直线x +y ＝0所得线段的长度是2√2，则圆M 与圆N ：（x ﹣1）2+（y ﹣1）2＝1的位置关系是（ ） A ．内切B ．相交C ．外切D ．相离6．布达佩斯的伊帕姆维泽蒂博物馆收藏的达•芬奇方砖，在正六边形上画了具有视觉效果的正方体图案（如图1），把三片这样的达•芬奇方砖形成图2的组合，这个组合表达了图3所示的几何体．如图3中每个正方体的棱长为1，则点A 到平面QGC 的距离是（ ）A ．14B ．12C ．√22D ．√327．已知圆C ：（x ﹣2）2+y 2＝64，F （﹣2，0）为圆内一点，将圆折起使得圆周过点F （如图），然后将纸片展开，得到一条折痕l ，这样继续下去将会得到若干折痕，观察这些折痕围成的轮廓是一条圆锥曲线，则该圆锥曲线的方程为（ ）A ．x 216+y 212=1B ．x 24+y 2=1C ．x 24+y 23=1D ．x 216+y 24=18．如图，在正方体ABCD ﹣A 1B 1C 1D 1中，O 是AC 中点，点P 在线段A 1C 1上，若直线OP 与平面A 1BC 1所成的角为θ，则sin θ的取值范围是（ ）A ．[√23，√33] B ．[13，12]C ．[√34，√33] D ．[14，13]二、选择题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求.全部选对的得5分,部分选对的得2分,有选错的得О分.9．若直线过点A （1，2），且在两坐标轴上截距的绝对值相等，则直线l 方程可能为（ ） A ．x ﹣y +1＝0B ．x +y ﹣3＝0C ．2x ﹣y ＝0D ．x ﹣y ﹣1＝010．已知点P 在圆C ：x 2+y 2﹣4x ＝0上，直线AB ：y ＝x +2，则（ ） A ．直线AB 与圆C 相交 B ．直线AB 与圆C 相离C ．点P 到直线AB 距离最大值为2√2+2D ．点P 到直线AB 距离最小值为2√2−111．正方体ABCD ﹣A 1B 1C 1D 1的棱长为1，已知平面α⊥AC 1，则关于α截此正方体所得截面的判断正确的是（ ）A ．截面形状可能为正三角形B ．截面形状可能为正方形C ．截面形状可能为正六边形D ．截面面积最大值为√312．已知椭圆C ：x 225+y 29=1，F 1，F 2分别为它的左右焦点，A ，B 分别为它的左右顶点，点P 是椭圆上的一个动点，下列结论中正确的有（ ） A ．存在P 使得∠F 1PF 2=π2 B ．cos ∠F 1PF 2的最小值为−18C ．直线P A 与直线PB 斜率乘积为定值925D ．PF 1⊥PF 2，则△F 1PF 2的面积为9三、填空题：本题共4小题，每小题5分，共20分.13．与圆x 2+y 2﹣2x +4y +3＝0同圆心，且过点（1，1）的圆的方程是 ．14．如图，P A ⊥平面ABCD ，底面ABCD 是正方形，E ，F 分别为PD ，PB 的中点，点G 在线段AP 上，AC 与BD 交于点O ，P A ＝AB ＝2，若OG ∥平面EFC ，则AG ＝ ．15．点P （﹣2，﹣1）到直线l ：（2+λ）x +λy ﹣2﹣λ＝0（λ为任意实数）的距离的最大值是 ． 16．2023年第19届亚运会在中国浙江杭州举行，杭州有很多圆拱的悬索拱桥，经测得某圆拱索桥（如图）的跨度|AB |＝100米，拱高|OP |＝10米，在建造圆拱桥时每隔5米需用一根支柱支撑，则与OP 相距30米的支柱MN 的高度是 米．（注意：√10≈3.162)四、解答题：本题共6小题，第17题10分，其它每题共70分.解答应写出文字说明、证明过程或 17．（10分）已知直线l ：mx ﹣y +1﹣m ＝0和圆C ：x 2+（y ﹣1）＝5． （1）求证：对任意实数m ，直线l 和圆C 总有两个不同的交点； （2）设直线l 和圆C 交于A ，B 两点．若|AB|=√17，求l 的倾斜角．18．（12分）如图，在四棱锥P ﹣ABCD 中，P A ⊥平面ABCD ，PB 与底面所成的角为45°，底面ABCD 为直角梯形，∠ABC ＝∠BAD ＝90°，AD ＝2，P A ＝BC ＝1．（1）求直线PC 与平面PBD 所成角的正弦值；（2）求平面P AB 与平面PCD 所成的锐二面角的余弦值．19．（12分）已知圆C ：x 2+y 2﹣4x ﹣6y +9＝0． （1）过点P （3，5）作圆C 的切线l ，求l 的方程；（2）若圆C 2：x 2+y 2+2x ﹣4y ﹣4＝0与圆C 相交于A 、B 两点，求|AB |． 20．（12分）已知椭圆E ：x 2a 2+y 2b 2=1(a ＞b ＞0)的离心率为√22，上顶点为A （0，1）． （1）求E 的方程；（2）过点P(0，√3)斜率为k 的直线l 与椭圆E 交于不同的两M 、N ，且MN =8√27，求k 的值． 21．（12分）如图，四棱台ABCD ﹣A 1B 1C 1D 1中，上、下底面均是正方形，且侧面是全等的等腰梯形，AB ＝2A 1B 1＝4，E 、F 分别为DC 、BC 的中点，上下底面中心的连线O 1O 垂直于上下底面，且O 1O 与侧棱所在直线所成的角为45°． （1）求证：BD 1∥平面C 1EF ；（2）线段BF 上是否存在点M ，使得直线A 1M 与平面C 1EF 所成的角的正弦值为3√2222，若存在，求出线段BM 的长；若不存在，请说明理由．22．（12分）已知椭圆Γ：x 2a 2+y 2b2=1(a ＞b ＞0)的左、右焦点分别为F 1(−√2，0)和F 2(√2，0)，Γ的下顶点为A ，直线l ：x +y −4√2=0，点M 在l 上． （1）若a ＝2，线段AM 的中点在x 轴上，求M 的坐标；（2）椭圆Γ上存在一个点P （a cos θ，b sin θ）（θ∈[0，2π]），P 到l 的距离为d ，使|PF 1|+|PF 2|+d ＝6，当a 变化时，求d 的最小值．2023-2024学年山东省聊城市高二（上）期中数学试卷参考答案与试题解析一、选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合 1．设a ∈R ，则“直线ax +y ﹣1＝0与直线x +ay +1＝0平行”是“a ＝1”的（ ） A ．充分不必要条件 B ．充要条件C ．必要不充分条件D ．既不充分也不必要条件解：若直线ax +y ﹣1＝0与直线x +ay +1＝0平行，则{a 2−1=0a +1≠0⇒a =1； 若a ＝1，则直线x +y ﹣1＝0与直线x +y +1＝0平行，∴直线ax +y ﹣1＝0与直线x +ay +1＝0平行是a ＝1的充分必要条件． 故选：B ．2．经过两条直线l 1：x +y ＝2，l 2：2x ﹣y ＝1的交点，且直线的一个方向向量v →=(−6，4) 的直线方程为（ ） A ．2x ﹣y ﹣1＝0B ．2x +y ﹣3＝0C ．3x ﹣2y ﹣5＝0D ．2x +3y ﹣5＝0解：根据题意，{x +y =22x −y =1，解可得{x =1y =1，即两直线的交点为（1，1），设A （1，1），设直线上任意一点为M ，其坐标为（x ，y ）， 直线的一个方向向量v →=(−6，4)，则MA →∥v →，则有4（x ﹣1）＝﹣6（y ﹣1），即4x +6y ﹣10＝0，变形可得2x +3y ﹣5＝0， 故要求直线的方程为2x +3y ﹣5＝0． 故选：D ．3．已知SA ⊥平面ABC ，AB ⊥AC ，SA ＝AB ＝1，BC =√5，则空间的一个单位正交基底可以为（ ）A ．{AB →，12AC →，AS →}B ．{AB →，AC →，AS →} C ．{AB →，12AC →，12AS →}D ．{AS →，AB →，√55BC →}解：由于SA ⊥平面ABC ， 所以：SA ⊥AB ，SA ⊥AC ， 由于AB ⊥AC ，AB ＝1，BC =√5， 所以AC ＝2．所以空间的一个单位正交基底可以为{AB →，12AC →，AS →}．故选：A ．4．椭圆x 216+y 24=1和x 236+y 224=1（ ）A ．长轴长相等B ．短轴长相等C ．焦距相等D ．顶点相同解：椭圆x 216+y 24=1中a 2＝16，b 2＝4，故c 2＝16﹣4＝12，x 236+y 224=1中a 2＝36，b 2＝24，故c 2＝36﹣24＝12，故两个椭圆的a ，b 都不相等，而c 相等，故焦距相等． 故选：C ．5．已知圆M ：x 2+y 2﹣2ay ＝0（a ＞0）截直线x +y ＝0所得线段的长度是2√2，则圆M 与圆N ：（x ﹣1）2+（y ﹣1）2＝1的位置关系是（ ） A ．内切B ．相交C ．外切D ．相离解：圆的标准方程为M ：x 2+（y ﹣a ）2＝a 2（a ＞0）， 则圆心为（0，a ），半径R ＝a ， 圆心到直线x +y ＝0的距离d =a2， ∵圆M ：x 2+y 2﹣2ay ＝0（a ＞0）截直线x +y ＝0所得线段的长度是2√2， ∴2√R 2−d 2=2√a 2−a 22=2√a22=2√2，即√a 22=√2，即a 2＝4，a ＝2，则圆心为M （0，2），半径R ＝2，圆N ：（x ﹣1）2+（y ﹣1）2＝1的圆心为N （1，1），半径r ＝1，则MN =√12+12=√2， ∵R +r ＝3，R ﹣r ＝1，∴R ﹣r ＜MN ＜R +r ，即两个圆相交． 故选：B ．6．布达佩斯的伊帕姆维泽蒂博物馆收藏的达•芬奇方砖，在正六边形上画了具有视觉效果的正方体图案（如图1），把三片这样的达•芬奇方砖形成图2的组合，这个组合表达了图3所示的几何体．如图3中每个正方体的棱长为1，则点A 到平面QGC 的距离是（ ）A ．14B ．12C ．√22D ．√32解：建立空间直角坐标系如图，则A （1，1，0），C （0，2，0），G （0，0，2），Q （1，0，2）， GQ →=(1，0，0)，GC →=(0，2，−2)，CA →=(1，−1，0)， 设平面QGC 的一个法向量为n →=(x ，y ，z)，由{n →⋅GQ →=x =0n →⋅GC →=2y −2z =0，取z ＝1，得n →=(0，1，1)， ∴点A 到平面QGC 的距离是|n →⋅CA →||n →|=√2=√22． 故选：C ．7．已知圆C ：（x ﹣2）2+y 2＝64，F （﹣2，0）为圆内一点，将圆折起使得圆周过点F （如图），然后将纸片展开，得到一条折痕l ，这样继续下去将会得到若干折痕，观察这些折痕围成的轮廓是一条圆锥曲线，则该圆锥曲线的方程为（ ）A ．x 216+y 212=1B ．x 24+y 2=1C ．x 24+y 23=1D ．x 216+y 24=1解：F （﹣2，0），C （2，0），点F 关于折痕l 的对称点A 在圆周上，折痕l 为线段AF 的垂直平分线，折痕l 与AC 相交于点P ，如图所示：则有|P A |＝|PF |，可知|PF |+|PC |＝|P A |+|PC |＝|AC |＝8＞|FC |＝4，所以点P 的轨迹是以F ，C 为左、右焦点的椭圆，其中长轴2a ＝8，焦距2c ＝4， 所以点P 的轨迹方程为x 216+y 212=1，即折痕围成轮廓的圆锥曲线的方程为x 216+y 212=1．故选：A ．8．如图，在正方体ABCD ﹣A 1B 1C 1D 1中，O 是AC 中点，点P 在线段A 1C 1上，若直线OP 与平面A 1BC 1所成的角为θ，则sin θ的取值范围是（ ）A ．[√23，√33] B ．[13，12]C ．[√34，√33] D ．[14，13]解：设正方体棱长为1，A 1P A 1C 1=λ（0≤λ≤1）．以D 为原点，分别以DA ，DC ，DD 1为坐标轴建立空间直角坐标系， 则O （12，12，0），P （1﹣λ，λ，1），∴OP →=（12−λ，λ−12，1），∵易证DB 1⊥平面A 1BC 1，∴DB 1→=（1，1，1）是平面A 1BC 1的一个法向量． ∴sin θ＝|cos ＜OP →，DB 1→＞|=1√3√2(λ−12)2+1，当λ=12时sin θ取得最大值√33，当λ＝0或1时，sin θ取得最小值√23． 故选：A ．二、选择题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求.全部选对的得5分,部分选对的得2分,有选错的得О分.9．若直线过点A（1，2），且在两坐标轴上截距的绝对值相等，则直线l方程可能为（）A．x﹣y+1＝0B．x+y﹣3＝0C．2x﹣y＝0D．x﹣y﹣1＝0解：当直线经过原点时，斜率为k=2−01−0=2，所求的直线方程为y＝2x，即2x﹣y＝0；当直线不过原点时，设所求的直线方程为x±y＝k，把点A（1，2）代入可得1﹣2＝k，或1+2＝k，求得k＝﹣1，或k＝3，故所求的直线方程为x﹣y+1＝0，或x+y﹣3＝0；综上知，所求的直线方程为2x﹣y＝0、x﹣y+1＝0，或x+y﹣3＝0．故选：ABC．10．已知点P在圆C：x2+y2﹣4x＝0上，直线AB：y＝x+2，则（）A．直线AB与圆C相交B．直线AB与圆C相离C．点P到直线AB距离最大值为2√2+2D．点P到直线AB距离最小值为2√2−1解：圆C：x2+y2﹣4x＝0，即（x﹣2）2+y2＝4，圆心为C（2，0），半径r＝2，则圆心C到直线AB的距离d=|2+2−0|√1+(−1)2=2√2＞r，所以直线AB与圆C相离，又点P在圆C上，所以点P到直线AB距离最大值为2√2+2，点P到直线AB距离最小值为2√2−2，故正确的有B、C．故选：BC．11．正方体ABCD﹣A1B1C1D1的棱长为1，已知平面α⊥AC1，则关于α截此正方体所得截面的判断正确的是（）A．截面形状可能为正三角形B．截面形状可能为正方形C．截面形状可能为正六边形D．截面面积最大值为√3解：如图所示，当截面为B 1CD 1时，截面为正三角形，选项A 正确；当截面过棱A 1B 1，B 1B ，BC ，CD ，DD 1，D 1A 1的中点时，截面为正六边形，选项C 正确； 当截面为正六边形时，面积最大，因为MN =√2，GH =√22，OE =√(12)2+(√24)2=√64， 所以S ＝2×12×（√22+√2）×√64=3√34，选项D 错误； 与AC 1垂直的截面不可能是正方形，选项B 错误． 故选：AC ．12．已知椭圆C ：x 225+y 29=1，F 1，F 2分别为它的左右焦点，A ，B 分别为它的左右顶点，点P 是椭圆上的一个动点，下列结论中正确的有（ ） A ．存在P 使得∠F 1PF 2=π2B ．cos ∠F 1PF 2的最小值为−18C ．直线P A 与直线PB 斜率乘积为定值925D ．PF 1⊥PF 2，则△F 1PF 2的面积为9解：由椭圆的方程可得a ＝5，b ＝3，所以c ＝4，由题意可得A （﹣5，0），B （5，0），F 1（﹣4，0），F 2（4，0），设上顶点为D （0，3），A 中，DF 1→•DF 2→=（﹣4，﹣3）•（4，﹣3）＝﹣16+9＝﹣7＜0，所以∠F 1PF 2的最大角为钝角， 所以存在P 使得∠F 1PF 2为直角，所以A 正确；B 中，设|PF 1|＝m ，|PF 2|＝n ，由椭圆的定义可得m +n ＝2a ＝10，cos ∠F 1PF 2=m 2+n 2−(2c)22mn =(m+n)2−2mn−642mn =36−2mn 2mn =18mn−1， 因为mn ≤（m+n 2）2＝25，当且仅当m ＝n 时取等号，所以cos ∠F 1PF 2≥1825−1=−725，即cos ∠F 1PF 2的最小值为−725，所以B 不正确； C 中，设P （x 0，y 0），则x 0225+y 029=1，所以y 02=9（1−x 0225），可得k P A •k PB =y 0x 0+5•y 0x 0−5=y 02x 02−25=9(1−x 0225)x 02−25=−925，所以C 不正确；D 中，PF 1⊥PF 2，由B 选项及由勾股定理可得：m 2+n 2＝（2c ）2＝64，即（m +n ）2﹣2mn ＝64， 即2mn ＝100﹣64＝36，所以mn ＝18，所以S △F 1PF 2=12mn ＝9，所以D 正确． 故选：AD ．三、填空题：本题共4小题，每小题5分，共20分.13．与圆x 2+y 2﹣2x +4y +3＝0同圆心，且过点（1，1）的圆的方程是： （x ﹣1）2+（y +2）2＝9 ． 解：圆x 2+y 2﹣2x +4y +3＝0的标准方程为（x ﹣1）2+（y +2）2＝2， 则圆心C （1，﹣2）， ∵圆过点A （1，1）， ∴半径R ＝|AC |＝3，则圆的标准方程为（x ﹣1）2+（y +2）2＝9． 故答案为：（x ﹣1）2+（y +2）2＝9．14．如图，P A ⊥平面ABCD ，底面ABCD 是正方形，E ，F 分别为PD ，PB 的中点，点G 在线段AP 上，AC 与BD 交于点O ，P A ＝AB ＝2，若OG ∥平面EFC ，则AG ＝23．解：由题意建立如图所示的空间直角坐标系， A （0，0，0），因为P A ＝AB ＝2，C （2，2，0），B （2，0，0），D （0，2，0），P （0，0，2），O （1，1，0），因为E ，F 分别是PD ，PB 中点，设G （0，0，b ），设平面EFC 的法向量为n →=（x ，y ，z ）， 因为OG ∥平面EFC ，所以OG →•n →=0，OG →=（﹣1，﹣1，b ）， 所以E （0，1，1），F （1，0，1），则EF →=（1，﹣1，0）， CE →=（﹣2，﹣1，1），则{n →⋅EF →=0n →⋅CE →=0，即{x −y =0−2x −y +z =0，令x ＝1，则y ＝1，z ＝3，所以n →=（1，1，3）， 所以OG →•n →=−1﹣1+3b ＝0，解得b =23， 所以AG ＝b =23． 故答案为：23．15．点P （﹣2，﹣1）到直线l ：（2+λ）x +λy ﹣2﹣λ＝0（λ为任意实数）的距离的最大值是 √10 ． 解：直线l ：（2+λ）x +λy ﹣2﹣λ＝0（λ为任意实数）， 整理得：λ（x +y ﹣1）+（2x ﹣2）＝0， 故{x +y −1=02x −2=0，解得{x =1y =0，故直线l 恒过点Q （1，0），故点P （﹣2，﹣1）到直线l 的最大距离d =√(−2−1)2+(−1−0)2=√10． 故答案为：√10．16．2023年第19届亚运会在中国浙江杭州举行，杭州有很多圆拱的悬索拱桥，经测得某圆拱索桥（如图）的跨度|AB |＝100米，拱高|OP |＝10米，在建造圆拱桥时每隔5米需用一根支柱支撑，则与OP 相距30米的支柱MN 的高度是 6.48 米．（注意：√10≈3.162)解：以O 为原点，以AB 所在直线为x 轴，以OP 所在直线为y 轴建立平面直角坐标系， 设圆心坐标（0，a ），P （0，10），A （﹣50，0）， 则圆拱所在圆的方程为x 2+（y ﹣a ）2＝r 2，所以{(10−a)2=r 2(−50)2+a 2=r 2，解得a ＝﹣120，r 2＝16900， 所以圆的方程为x 2+（y +120）2＝16900．将x ＝﹣30代入圆方程，得：900+（y +120）2＝16900， 因为y ＞0，所以y ＝40√10−120≈40×3.162﹣120＝6.48， 所以MN 的高度是6.48米． 故答案为：6.48．四、解答题：本题共6小题，第17题10分，其它每题共70分.解答应写出文字说明、证明过程或 17．（10分）已知直线l ：mx ﹣y +1﹣m ＝0和圆C ：x 2+（y ﹣1）＝5． （1）求证：对任意实数m ，直线l 和圆C 总有两个不同的交点； （2）设直线l 和圆C 交于A ，B 两点．若|AB|=√17，求l 的倾斜角．（1）证明：由直线l ：mx ﹣y +1﹣m ＝0，得m （x ﹣1）﹣y +1＝0，由{x −1=0−y +1=0，得{x =1y =1，∴直线l ：mx ﹣y +1﹣m ＝0过定点p （1，1），代入圆C ：x 2+（y ﹣1）2＝5，得12+（1﹣1）2＝1＜5，∴点p （1，1）在圆C ：x 2+（y ﹣1）2＝5内部， ∴对任意的m ，直线l 与圆C 总有两个不同的交点．（2）解：直线l 的斜率存在，由|AB|=√17，圆的半径为√5，得圆心到直线l ：mx ﹣y +1﹣m ＝0的距离为√5−174=√32． 则√m 2+1=√32，解得：m =±√3．∴直线l 为y =√3x +1−√3或y =−√3x +1−√3．直线l 的倾斜角为60°或120°．18．（12分）如图，在四棱锥P ﹣ABCD 中，P A ⊥平面ABCD ，PB 与底面所成的角为45°，底面ABCD 为直角梯形，∠ABC ＝∠BAD ＝90°，AD ＝2，P A ＝BC ＝1． （1）求直线PC 与平面PBD 所成角的正弦值；（2）求平面P AB 与平面PCD 所成的锐二面角的余弦值．解：（1）∵P A ⊥面ABCD ，∴P A ⊥AB ，P A ⊥AD ，又∠BAD ＝90°， ∴AB ⊥AD ，∵为PB 与底面所成的角为45°， ∴∠PBA ＝45°，故AB ＝P A ＝1，以A 为坐标原点，AB ，AD ，AP 所在直线分别为x ，y ，z 轴，建立如图所示的空间直角坐标系O ﹣xyz ， 则B （1，0，0），D （0，2，0），P （0，0，1），C （1，1，0）， 则PC →=（1，1，﹣1），PB →=（1，0，﹣1），PD →=（0，2，﹣1）， 设平面PBD 的一个法向量为m →=（x ，y ，z ），则{m →⋅PB →=0m →⋅PD →=0，即{x −z =02y −z =0，取z ＝2，则x ＝2，y ＝1，此时m →=（2，1，2）， 设直线PC 与平面PBD 所成的角为θ， 则sin θ＝|cos ＜m →，PC →＞|＝|m →⋅PC→|PC →||m →|||√3×3|√39． 所以直线PC 与平面PBD 所成角的正弦值为√39． （2）平面P AB 的一个法向量j →=（0，1，0） 设平面PCD 的一个法向量为n →=（x ，y ，z ）， 则{n →⋅PC →=0n →⋅PD →=0，即{x +y −z =02y −z =0， 取y ＝l ，则z ＝2，x ＝l ，此时n →=（1，1，2）， cos ＜n →，j →＞=n →⋅j→|n →||j →|=6×1=√66， 所以平面P AB 与平面PCD 所成的锐二面角的余弦值为√66．19．（12分）已知圆C ：x 2+y 2﹣4x ﹣6y +9＝0． （1）过点P （3，5）作圆C 的切线l ，求l 的方程；（2）若圆C 2：x 2+y 2+2x ﹣4y ﹣4＝0与圆C 相交于A 、B 两点，求|AB |．解：（1）圆C 1方程可化为（x ﹣2）2+（y ﹣3）＝4，则圆心C 1（2，3），半径为2， 由 （3﹣2）2+（5﹣3）2＞4，可知点P 在圆外， 设l 的方程为y ﹣5＝k （x ﹣3），即kx ﹣y +5﹣3k ＝0， 则圆心C 1到直线l 的距离为√1+k 2=2，解得k ＝0或k =−43，∴l 的方程为4x +3y ﹣27＝0或y ＝5．（2）把两圆的方程相减可得直线AB 的方程为6x +2y ﹣13＝0， 则圆心C 到直线AB 的距离d =|6×2+2×3−13|√36+4=√104＜2，直线与圆相交，所以|AB |＝2√4−1016=3√62． 20．（12分）已知椭圆E ：x 2a 2+y 2b2=1(a ＞b ＞0)的离心率为√22，上顶点为A （0，1）．（1）求E 的方程；（2）过点P(0，√3)斜率为k 的直线l 与椭圆E 交于不同的两M 、N ，且MN =8√27，求k 的值． 解：（1）由离心率e =c a =√22，则a =√2c ， 又上顶点A （0，1），知b ＝1，又b 2＝a 2﹣c 2＝1，可知c ＝1，a =√2， ∴椭圆E 的方程为x 22+y 2=1；（2）设直线l ：y =kx +√3，设M （x 1，y 1），N （x 2，y 2）， 则{y =kx +√3x 22+y 2=1，整理得：(1+2k 2)x 2+4√3kx +4=0，Δ=(4√3k)2−4×4×(1+2k 2)＞0，即k 2＞1， ∴x 1+x 2=−4√3k 1+2k2，x 1x 2=41+2k2，∴|MN|=√1+k 2|x 1−x 2|=√1+k 2√(x 1+x 2)2−4x 1x 2=4√(1+k 2)(k 2−1)1+2k2=8√27， 即17k 4﹣32k 2﹣57＝0，解得：k 2＝3或−1917（舍去）， ∴k =±√3．21．（12分）如图，四棱台ABCD ﹣A 1B 1C 1D 1中，上、下底面均是正方形，且侧面是全等的等腰梯形，AB ＝2A 1B 1＝4，E 、F 分别为DC 、BC 的中点，上下底面中心的连线O 1O 垂直于上下底面，且O 1O 与侧棱所在直线所成的角为45°． （1）求证：BD 1∥平面C 1EF ；（2）线段BF 上是否存在点M ，使得直线A 1M 与平面C 1EF 所成的角的正弦值为3√2222，若存在，求出线段BM 的长；若不存在，请说明理由．解：（1）证明：因为OO 1⊥平面ABCD ，以点O 为坐标原点，DA ，OF →，OO 1→的方向分别为x 轴，y 轴，z 轴的正方向，建立如图所示的空间直角坐标系．因为侧棱所在直线与上下底面中心的连线OO 1所成的角为45°，则B （2，2，0），D 1(−1，−1，√2)，C 1(−1，1，√2)，F （0，2，0），E （﹣2，0，0），A 1(1，−1，√2)，所以BD 1→=(−3，−3，√2)，CE 1→=(−1，−1，√2)，EF →=(2，2，0)， 设平面C 1EF 的一个法向量为n →=（x ，y ，z ），则{n →⋅EF →=x +y =0n →⋅C 1E →=x +y +√2z =0，令x ＝1，则n →=（1，﹣1，0）， 因为BD 1→=（﹣3，﹣3，√2），所以n →•BD 1→=0，所以n →⊥BD 1→， 又因为BD 1⊂平面C 1EF ，所以BD 1∥平面 C 1EF ；（2）假设边BC 上存在点M （x ，2，0）满足条件，x ∈[﹣2，2]， 则A 1M →=（x ﹣1，3，−√2），设直线A 1M 与平面C 1EFF 所成角为θ，由题意可得sin θ＝|cos ＜A 1M →，n →＞|=|A 1M →⋅n →||A 1M →|⋅|n →|=|x−4|√2⋅√x 2−2x+12=3√2222， 化简得x 2﹣35x +34＝0，则x ＝1或x ＝34（舍去），即存在点M 符合题意，此时BM ＝1． 22．（12分）已知椭圆Γ：x 2a 2+y 2b 2=1(a ＞b ＞0)的左、右焦点分别为F 1(−√2，0)和F 2(√2，0)，Γ的下顶点为A ，直线l ：x +y −4√2=0，点M 在l 上． （1）若a ＝2，线段AM 的中点在x 轴上，求M 的坐标；（2）椭圆Γ上存在一个点P （a cos θ，b sin θ）（θ∈[0，2π]），P 到l 的距离为d ，使|PF 1|+|PF 2|+d ＝6，当a 变化时，求d 的最小值．解：（1）由题意可得a =2，b =c =√2，所以Γ：x 24+y 22=1，A(0，−√2)，因为AM 的中点在x 轴上， 所以点M 的纵坐标为√2， 将y =√2代入x +y −4√2=0中， 解得x ＝3√2， 则M(3√2，√2)； （2）易知d =|acosθ+bsinθ−42|2=6−2a ，因为椭圆在直线的左下方， 所以acosθ+bsinθ−422=6−2a ，即4√2−√a 2+b 2sin(θ+φ)=6√2−2√2a ， 又a 2＝b 2+2，可得√2a 2−2sin(θ+φ)=2√2a −2√2， 此时√a 2−1sin(θ+φ)=2a −2，|sin(θ+φ)|=√a 2−1≤1，整理得（a ﹣1）（3a ﹣5）≤0， 即1≤a ≤53，所以d =6−2a ≥6−2×53=83． 故d 的最小值为83．。

山东省聊城市某重点高中2012-2013学年下学期高二3月模块测试化学试题第I 卷（选择题）一、选择题1．欲将蛋白质从水中析出而又不改变它的性质，应加入 （ ） A.424SO )(NH 溶液 B.酒精溶液 C.23)Pb(NO 溶液 D.NaOH 溶液 2．下列生活中的化学小实验不合理的是 （ ） A. 用米汤检验加碘盐中的碘酸钾(KIO 3) B. 用食用醋除去热水瓶中积存的水垢C. 用纯碱(Na 2CO 3)溶液洗涤沾有油污的器具D. 用灼烧并闻气味的方法区分化纤织物与纯毛织物 3．下列4组连线中，完全正确的一组是 （ ）A ．B ．C ．D ．4．下列化学用语的表述对应正确的是A ．碳酸氢钠的电离方程式：NaHCO 3===Na ＋＋H ＋＋CO 32-B ．用铜做阳极电解氯化铜溶液的离子方程式：Cu 2＋+2Cl2↑C ．硫化钠水解的离子方程式：S 2－＋2H 2O H 2S ＋2OH －D ．碳酸钙的溶解平衡：CaCO 3(s) Ca 2＋(aq) ＋ CO 32－(aq)5．反应A+B →C(△H<0)分两步进行：①A+B →X(△H >0)，②X →C(△H <0)。

下列示意图中，能正确表示总反应过程中能量变化的是A． B． C． D．67①CH3OH(g) + H2O(g)＝CO2(g)＋3H2(g) △H＝＋49.0 kJ·mol－1②CH3OH(g)+ 1/2O2(g)＝CO2(g)＋2H2(g) △H＝－192.9 kJ·mol－1根据上述反应，下列说法正确的是A．反应①中的能量变化如右图所示B．CH3OH转变成H2的过程一定要吸收能量C．1 mol CH3OH充分燃烧放出的热量为192.9 kJD．可推知2H2(g)＋O2(g)＝2H2O(g) △H ＝－483.8 kJ·mol－18．下列说法与盐的水解无关的是①纯碱溶液去油污②实验室配制FeCl3溶液时,往往在FeCl3溶液中加入少量的盐酸③用NaHCO3和Al2(SO4)3两种溶液可作泡沫灭火剂④在NH4Cl溶液中加入金属镁会产生氢气⑤草木灰与氨态氮肥不能混合施用A. ①③④B. ②③⑤C. ③④⑤D.全有关9．铁镍蓄电池又称爱迪生电池，放电时的总反应为：Fe+Ni2O3+3H2O＝Fe(OH)2+2Ni(OH)2下列有关该电池的说法不正确．．．的是A. 电池的电解液为碱性溶液，正极为Ni2O3、负极为FeB. 电池放电时，负极反应为Fe+2OH－-2e－＝ Fe(OH)2C. 电池充电过程中，阴极附近溶液的pH降低D. 电池充电时，阳极反应为2 Ni(OH)2+2OH－-2e－＝Ni2O3+3H2O10．某温度下，在一个2 L的密闭容器中，加入4 mol A和2 mol B进行如下反应：3A(g)+2B(g) 4C(s)+2D(g)，反应一段时间后达到平衡，测得生成 1.6 mol C，则下列说法正确的是（）A．该反应的化学平衡常数表达式是KB．此时B的平衡转化率是40%C．增大该体系的压强，平衡向右移动，化学平衡常数增大D．增加B，平衡向右移动，B的平衡转化率增大11．如右图所示，左侧注射器吸入10mLNO2、N2O4的混合气体,右侧注射器吸入10mL空气，且U型管内两侧液面保持水平位置（液体不与气体反应．．．．．．．．），现将两侧注射器活塞同时快速向内推到5mL，下列法不正确．．．的是A．左侧气体颜色先变深后逐渐变浅B．对于2NO2 N2O4平衡体系，向生成N2O4的方向移动C．压缩后U型管内两侧液面仍然保持水平D.若将两侧注射器活塞重新拉到10mL的位置，平衡后U型管两侧液面仍然保持水平12．下图表示的是难溶氢氧化物在不同pH下的溶解度（S/mol·L－1），下列说法中正确的是A．pH＝3时溶液中铁元素的主要存在形式是Fe3＋B．若Ni(NO3)2溶液中含有少量的Co2＋杂质，可通过调节溶液pH的方法来除去C．若分离溶液中的Fe3＋和Cu2＋，可调节溶液的pH在4左右D．若在含有Cu2＋和Ni2＋的溶液中加入烧碱，Ni(OH)2优先沉淀13．关于下列各装置图的叙述不正确．．．的是①②③④A．用图①装置精炼铜，a极为粗铜，电解质溶液为CuSO4溶液B．图②装置盐桥中KCl的Cl—移向乙烧杯C．图③装置中钢闸门应与外接电源的负极相连获得保护D．图④两个装置中通过导线的电子数相同时，消耗负极材料的物质的量不同14．用Cl2生产某些含氯有机物时会产生副产物HC1。

高二级上学期期中考试题数学本试卷共8页，22小题，满分150分，考试时间120分钟。

第一部分选择题（共60分）一、单选题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1．已知直线l 1：2x ＋my ＝2，l 2：m 2x ＋2y ＝1，且l 1⊥l 2，则m 的值为( )A ．0B ．－1C ．0或1D ．0或－12．若一个圆锥的轴截面是面积为1的等腰直角三角形，则该圆锥的侧面积为( )A.2π B ．22π C ．2πD ．4π3．把正方形ABCD 沿对角线AC 折起，当以A ，B ，C ，D 四点为顶点的三棱锥体积最大时，直线BD 和平面ABC 所成角的大小为( )A ．90°B ．60°C ．45°D ．30°4．若过点（2，1）的圆与两坐标轴都相切，则圆心到直线230x y --=的距离为（ ）A B C D 5．下列命题中，正确的是( )A ．任意三点确定一个平面B ．三条平行直线最多确定一个平面C ．不同的两条直线均垂直于同一个平面，则这两条直线平行D ．一个平面中的两条直线与另一个平面都平行，则这两个平面平行6．已知M (3，23)，N (－1，23)，F (1，0)，则点M 到直线NF 的距离为( )A. 5 B ．23 C ． 22D ．3 37．已知各顶点都在一个球面上的正四棱柱(其底面是正方形，且侧棱垂直于底面)高为4，体积为16，则这个球的表面积是( )A ．20πB ．16πC ．32πD ．24π8．直线:20l x y ++=分别与x 轴、y 轴交于A 、B 两点，点P 在圆22(2)2x y -+=上， 则ABP △面积的取值范围是（ ） A ．[]26，B ．[]48，C ．D ．⎡⎣二、多选题：本题共4小题，每小题5分，共20分.9．若220x x --<是2x a -<<的充分不必要条件，则实数a 的值可以是( ) A ．1B ．2C ．3D ．410．已知,αβ是两个不重合的平面，,m n 是两条不重合的直线，则下列命题正确的是（ ） A ．若//m n m α⊥，，则n α⊥ B ．若//,m n ααβ⋂=，则//m n C ．若m α⊥，m β⊥，则//αβ D ．若,//,m m n n αβ⊥⊥，则//αβ 11．若直线过点(1,2)A ，且在两坐标轴上截距的绝对值相等，则直线l 方程可能为( ) A ．10x y -+=B ．30x y +-=C ．20x y -=D ．10x y --=12．已知四棱锥P ABCD -，底面ABCD 为矩形，侧面PCD ⊥平面ABCD ，BC =CD PC PD ===.若点M 为PC 的中点，则下列说法正确的为（ ）A ．BM ⊥平面PCDB ．//PA 面MBDC ．四棱锥M ABCD -外接球的表面积为36π D ．四棱锥M ABCD -的体积为6第二部分非选择题（90分）三、填空题:本题共4小题,每小题5分，共20分.13．命题“20210x x x ∃<-->，”的否定是______________．14．已知直线l 1的方程为23y x =-+，l 2的方程为42y x =-，直线l 与l 1平行且与l 2在y 轴上的截距相同，则直线l 的斜截式方程为________________．15．若直线:l y kx =与曲线:1M y =+有两个不同交点，则k 的取值范围是________________．16．已知三棱锥S -ABC 的所有顶点都在球O 的球面上，SC 是球O 的直径．若平面SCA ⊥平面SCB ，SA ＝AC ，SB ＝BC ，三棱锥S -ABC 的体积为9，则球O 的体积为____________．四、解答题:共70分.解答应写出文字说明、证明过程或演算步骤.17．(本小题满分10分)已知直线l 1的方程为x ＋2y －4＝0，若l 2在x 轴上的截距为32，且l 1⊥l 2.(1)求直线l 1与l 2的交点坐标；(2)已知直线l 3经过l 1与l 2的交点，且在y 轴上的截距是在x 轴上的截距的2倍，求l 3的方程．18．(本小题满分12分)四棱锥P-ABCD 的底面ABCD 为直角梯形，AB ∥CD ，AB ⊥AD ，AB ＝12CD ＝1，P A ⊥平面ABCD ，P A ＝AD ＝ 3.(1)求证：PD ⊥AB ；(2)求四棱锥P-ABCD 的体积．19．(本小题满分12分)已知圆C 的圆心坐标为(a ，0)，且圆C 与y 轴相切． (1)已知a ＝1，M (4，4)，点N 是圆C 上的任意一点，求|MN |的最小值；(2)已知a ＜0，直线l 的斜率为43，且与y 轴交于点20,3⎛⎫- ⎪⎝⎭.若直线l 与圆C 相离，求a 的取值范围．20．(本小题满分12分)在直三棱柱ABC-A 1B 1C 1中，AB ＝5，AC ＝3，BC ＝4，点D 是线段AB 上的动点．(1)当点D 是AB 的中点时，求证：AC 1∥平面B 1CD ；(2)线段AB 上是否存在点D ，使得平面ABB 1A 1⊥平面CDB 1？若存在，试求出AD 的长度；若不存在，请说明理由．21. (本小题满分12分) 如图，多面体ABCDEF 中，四边形ABCD 是菱形，060ABC ∠=，FA ⊥平面ABCD ，//,2 2.FA ED AB FA ED ===求二面角F BC A --的大小的正切值；求点E 到平面AFC 的距离；求直线FC 与平面ABF 所成的角的正弦值.22. (本小题满分12分)已知圆22+=9：O x y ，过点()0,2P -任作圆O 的两条相互垂直的弦AB 、CD ，设M 、N 分别是AB 、CD 的中点，（1）直线MN 是否过定点? 若过，求出该定点坐标，若不过，请说明理由； （2）求四边形ACBD 面积的最大值，并求出对应直线AB 、CD 的方程.高二级上学期期中考试题 数学答案及说明一、选择题：1.D ，2.A ，3.C ，4.B ，5.C ，6.B ，7.D ，8.A ，9.BCD ，10.ACD ，11.ABC ，12.BC.二、填空题：13.0x ∀<，2210x x --≤；14.y ＝－2x －2；15.13,24⎡⎫⎪⎢⎣⎭；16.36π.题目及详细解答过程：一、单选题(本题共8小题，每小题5分，共40分)1．已知直线l 1：2x ＋my ＝2，l 2：m 2x ＋2y ＝1，且l 1⊥l 2，则m 的值为( ) A ．0 B ．－1 C ．0或1 D ．0或－1 解析：因为l 1⊥l 2，所以2m 2＋2m ＝0，解得m ＝0或m ＝－1. 答案：D2．若一个圆锥的轴截面是面积为1的等腰直角三角形，则该圆锥的侧面积为( ) A.2π B ．22π C ．2π D ．4π 解析：设底面圆的半径为r ，高为h ，母线长为l ，由题可知，r ＝h ＝22l ，则12(2r )2＝1，r ＝1，l ＝2.所以圆锥的侧面积为πrl ＝2π. 答案：A3．把正方形ABCD 沿对角线AC 折起，当以A ，B ，C ，D 四点为顶点的三棱锥体积最大时，直线BD 和平面ABC 所成角的大小为( )A ．90°B ．60°C ．45°D ．30°解析：当三棱锥D ­ABC 体积最大时，平面DAC ⊥平面ABC .取AC 的中点O ，则∠DBO 即为直线BD 和平面ABC 所成的角．易知△DOB 是等腰直角三角形，故∠DBO ＝45°.答案：C4．若过点（2，1）的圆与两坐标轴都相切，则圆心到直线230x y --=的距离为（ ）A B C D 【答案】B【解析】由于圆上的点()2,1在第一象限，若圆心不在第一象限， 则圆与至少与一条坐标轴相交，不合乎题意，所以圆心必在第一象限， 设圆心的坐标为(),a a ，则圆的半径为a ，圆的标准方程为()()222x a y a a -+-=.由题意可得()()22221a a a -+-=，可得2650a a -+=，解得1a =或5a =，所以圆心的坐标为()1,1或()5,5，圆心到直线的距离均为121132555d ⨯--==； 圆心到直线的距离均为22553255d ⨯--== 圆心到直线230x y --=的距离均为22555d -==； 所以，圆心到直线230x y --=25. 故选：B ．5．下列命题中，正确的是( ) A ．任意三点确定一个平面 B ．三条平行直线最多确定一个平面C ．不同的两条直线均垂直于同一个平面，则这两条直线平行D ．一个平面中的两条直线与另一个平面都平行，则这两个平面平行 解析：由线面垂直的性质，易知C 正确． 答案：C6．已知M (3，23)，N (－1，23)，F (1，0)，则点M 到直线NF 的距离为( ) A. 5 B ．23 C ． 22D ．3 3解析：易知NF 的斜率k ＝－3，故NF 的方程为y ＝－3(x －1)，即3x ＋y －3＝0. 所以M 到NF 的距离为|33＋23－3|（3）2＋12＝2 3. 答案：B7．已知各顶点都在一个球面上的正四棱柱(其底面是正方形，且侧棱垂直于底面)高为4，体积为16，则这个球的表面积是( )A ．20πB ．16πC ．32πD ．24π解析：由题意知正四棱柱的底面积为4，所以正四棱柱的底面边长为2，正四棱柱的底面对角线长为22，正四棱柱的对角线为2 6.而球的直径等于正四棱柱的对角线，即2R ＝2 6.所以R ＝ 6.所以S 球＝4πR 2＝24π. 答案：D8．直线:20l x y ++=分别与x 轴、y 轴交于A 、B 两点，点P 在圆22(2)2x y -+=上，则ABP △面积的取值范围是（ ） A ．[]26，B ．[]48，C ．232⎡⎤⎣⎦，D ．2232⎡⎤⎣⎦，【答案】A 【解析】直线20x y ++=分别与x 轴，y 轴交于A ，B 两点，()()2,0,0,2A B ∴--,则22AB =.点P 在圆22(2)2x y -+=上，∴圆心为（2，0），则圆心到直线的距离1202222d ++==.故点P 到直线20x y ++=的距离2d 的范围为2,32⎡⎤⎣⎦，则[]22122,62ABP S AB d d ==∈△.故答案为A.二、多选题(每题5分，共20分)9．若220x x --<是2x a -<<的充分不必要条件，则实数a 的值可以是( ) A ．1B ．2C ．3D ．4【答案】BCD【解析】：由220x x --<，解得12x -<<．又220x x --<是2x a -<<的充分不必要条件，(1∴-，2)(2-，)a ，则2a ．∴实数a 的值可以是2，3，4．故选：BCD ．10．已知,αβ是两个不重合的平面，,m n 是两条不重合的直线，则下列命题正确的是（ ） A ．若//m n m α⊥，，则n α⊥ B ．若//,m n ααβ⋂=，则//m n C ．若m α⊥，m β⊥，则//αβ D ．若,//,m m n n αβ⊥⊥，则//αβ 【答案】ACD 【解析】若m α⊥，则,a b α∃⊂且a b P =使得m a ⊥，m b ⊥，又//m n ，则n a ⊥，n b ⊥，由线面垂直的判定定理得n α⊥，故A 对； 若//m α，n αβ=，如图，设m AB =，平面1111D C B A 为平面α，//m α，设平面11ADD A 为平面β，11A D n αβ⋂==，则m n ⊥，故B 错；垂直于同一条直线的两个平面平行，故C 对；若,//m m n α⊥，则n α⊥，又n β⊥，则//αβ，故D 对； 故选：ACD ．11．若直线过点(1,2)A ，且在两坐标轴上截距的绝对值相等，则直线l 方程可能为( ) A ．10x y -+= B ．30x y +-= C ．20x y -= D ．10x y --=【答案】ABC【解析】：当直线经过原点时，斜率为20210k -==-，所求的直线方程为2y x =，即20x y -=； 当直线不过原点时，设所求的直线方程为x y k ±=，把点(1,2)A 代入可得12k -=，或12k +=，求得1k =-，或3k =，故所求的直线方程为10x y -+=，或30x y +-=； 综上知，所求的直线方程为20x y -=、10x y -+=，或30x y +-=． 故选：ABC ．12．已知四棱锥P ABCD -，底面ABCD 为矩形，侧面PCD ⊥平面ABCD ，23BC =，26CD PC PD ===.若点M 为PC 的中点，则下列说法正确的为（ ）A ．BM ⊥平面PCDB ．//PA 面MBDC ．四棱锥M ABCD -外接球的表面积为36π D ．四棱锥M ABCD -的体积为6 【答案】BC【解析】作图在四棱锥P ABCD -中：为矩形，由题：侧面PCD ⊥平面ABCD ，交线为CD ，底面ABCDBC CD ⊥，则BC ⊥平面PCD ，过点B 只能作一条直线与已知平面垂直，所以选项A错误；连接AC 交BD 于O ，连接MO ，PAC ∆中，OM ∥PA ，MO ⊆面MBD ，PA ⊄面MBD ，所以//PA 面MBD ，所以选项B 正确；四棱锥M ABCD -的体积是四棱锥P ABCD -的体积的一半，取CD 中点N ，连接PN ，PN CD ⊥，则PN平面ABCD ，32PN =，四棱锥M ABCD -的体积112326321223M ABCD V -=⨯⨯⨯⨯=所以选项D 错误.矩形ABCD 中，易得6,3,3AC OC ON ===，PCD 中求得：16,2NM PC ==在Rt MNO 中223MO ON MN =+=即： OM OA OB OC OD ====，所以O 为四棱锥M ABCD -外接球的球心，半径为3， 所以其体积为36π，所以选项C 正确， 故选：BC三、填空题(每题5分，共20分)13．命题“20210x x x ∃<-->，”的否定是______． 【答案】0x ∀<，2210x x --≤【解析】因为特称命题的否定是全称命题，所以，命题20210x x x ∃<-->，， 则该命题的否定是：0x ∀<，2210x x --≤ 故答案为：0x ∀<，2210x x --≤．14．已知直线l 1的方程为23y x =-+，l 2的方程为42y x =-，直线l 与l 1平行且与l 2在y 轴上的截距相同，则直线l 的斜截式方程为________________．解析：由斜截式方程知直线l 1的斜率k 1＝－2，又l ∥l 1，所以l 的斜率k ＝k 1＝－2.由题意知l 2在y 轴上的截距为－2，所以l 在y 轴上的截距b ＝－2.由斜截式方程可得直线l 的方程为y ＝－2x －2.答案：y ＝－2x －215．若直线:l y kx =与曲线()2:113M y x =+--有两个不同交点，则k 的取值范围是________________．解析：曲线M ：y ＝1＋1－（x －3）2是以(3，1)为圆心，1为半径的，且在直线y ＝1上方的半圆．要使直线l 与曲线M 有两个不同交点，则直线l 在如图所示的两条直线之间转动，即当直线l 与曲线M 相切时，k 取得最大值34；当直线l 过点(2，1)时，k 取最小值12.故k 的取值范围是13,24⎡⎫⎪⎢⎣⎭. 答案：13,24⎡⎫⎪⎢⎣⎭16．已知三棱锥S -ABC 的所有顶点都在球O 的球面上，SC 是球O 的直径．若平面SCA ⊥平面SCB ，SA ＝AC ，SB ＝BC ，三棱锥S -ABC 的体积为9，则球O 的体积为____________．解析：如图，连接OA ，OB .由SA ＝AC ，SB ＝BC ，SC 为球O 的直径，知OA ⊥SC ，OB ⊥SC .又由平面SCA ⊥平面SCB ，平面SCA ∩平面SCB ＝SC ，知OA ⊥平面SCB . 设球O 的半径为r ，则OA ＝OB ＝r ，SC ＝2r ，所以三棱锥S ­ABC 的体积为311323r V SC OB OA ⎛⎫=⨯⋅⋅= ⎪⎝⎭，即r 33＝9.所以r ＝3.所以3344336.33=O V r πππ=⨯=球答案：36π四、解答题(每题5分，共70分)17．(本小题满分10分)已知直线l 1的方程为x ＋2y －4＝0，若l 2在x 轴上的截距为32，且l 1⊥l 2.(1)求直线l 1与l 2的交点坐标；(2)已知直线l 3经过l 1与l 2的交点，且在y 轴上的截距是在x 轴上的截距的2倍，求l 3的方程． 解：(1)设l 2的方程为2x －y ＋m ＝0，..........1分因为l 2在x 轴上的截距为32，所以3－0＋m ＝0，m ＝－3，即l 2：2x －y －3＝0.....3分联立⎩⎪⎨⎪⎧x ＋2y －4＝0，2x －y －3＝0，得⎩⎪⎨⎪⎧x ＝2，y ＝1.所以直线l 1与l 2的交点坐标为(2，1)．..........5分 (2)当l 3过原点时，l 3的方程为y ＝12x ..........6分当l 3不过原点时，设l 3的方程为12x y a a +=...........7分 又直线l 3经过l 1与l 2的交点，所以2112a a+=， 得52a =，l 3的方程为2x ＋y －5＝0...........8分 综上，l 3的方程为y ＝12x 或2x ＋y －5＝0...........10分18．(本小题满分12分)四棱锥P-ABCD 的底面ABCD 为直角梯形，AB ∥CD ，AB ⊥AD ，AB ＝12CD ＝1，PA ⊥平面ABCD ，PA ＝AD ＝ 3.(1)求证：PD ⊥AB ；(2)求四棱锥P-ABCD 的体积．18.解：(1)证明：因为PA ⊥平面ABCD ，AB ⊂平面ABCD ，所以PA ⊥AB ，..........1分又因为AB ⊥AD ，AD ∩PA ＝A ，..........3分 所以AB ⊥平面PAD ，..........4分又PD ⊂平面PAD ，..........5分所以AB ⊥PD ...........6分 (2)解：S 梯形ABCD ＝12(AB ＋CD )·AD ＝332，.......8分又PA ⊥平面ABCD ，..........9分所以V 四棱锥P-ABCD ＝13×S 梯形ABCD ·PA ＝13×332×3＝32...........12分19．(本小题满分12分)已知圆C 的圆心坐标为(a ，0)，且圆C 与y 轴相切． (1)已知a ＝1，M (4，4)，点N 是圆C 上的任意一点，求|MN |的最小值； (2)已知a ＜0，直线l 的斜率为43，且与y 轴交于点20,3⎛⎫- ⎪⎝⎭.若直线l与圆C 相离，求a 的取值范围．19.解：(1)由题意可知，圆C 的方程为(x －1)2＋y 2＝1...........2分又|MC |＝（4－1）2＋（4－0）2＝5，..........4分 所以|MN |的最小值为5－1＝4...........5分(2)因为直线l 的斜率为43，且与y 轴相交于点20,3⎛⎫- ⎪⎝⎭，所以直线l 的方程为y ＝43x －23.即4x －3y －2＝0..........7分因为直线l 与圆C 相离，所以圆心C (a ，0)到直线l 的距离d ＞r . 则224243a a ->+.........9分又0a <，所以245a a ->-，解得2a >-..........11分 所以a 的取值范围是(－2，0)．.........12分20．(本小题满分12分)在直三棱柱ABC-A 1B 1C 1中，AB ＝5，AC ＝3，BC ＝4，点D 是线段AB 上的动点． (1)当点D 是AB 的中点时，求证：AC 1∥平面B 1CD ；(2)线段AB 上是否存在点D ，使得平面ABB 1A 1⊥平面CDB 1？若存在，试求出AD 的长度；若不存在，请说明理由．20.解：(1)证明：如图，连接BC 1，交B 1C 于点E ，连接DE ，则点E 是BC 1的中点，又点D 是AB 的中点，由中位线定理得DE ∥AC 1，.........1分 因为DE ⊂平面B 1CD ，.........2分AC 1⊄平面B 1CD ，.........3分所以AC 1∥平面B 1CD ..........4分(2)解：当CD ⊥AB 时，平面ABB 1A 1⊥平面CDB 1........5分 证明：因为AA 1⊥平面ABC ，CD ⊂平面ABC ， 所以AA 1⊥CD ..........6分又CD ⊥AB ，AA 1∩AB ＝A ，.........7分所以CD ⊥平面ABB 1A 1，因为CD ⊂平面CDB 1，.........8分 所以平面ABB 1A 1⊥平面CDB 1，.........9分故点D 满足CD ⊥AB 时，平面ABB 1A 1⊥平面CDB 1......10分 因为AB ＝5，AC ＝3，BC ＝4，所以AC 2＋BC 2＝AB 2， 故△ABC 是以角C 为直角的三角形， 又CD ⊥AB ，所以AD ＝95..........12分22. (本小题满分12分) 如图，多面体ABCDEF 中，四边形ABCD 是菱形，060ABC ∠=，FA ⊥平面ABCD ，//,2 2.FA ED AB FA ED ===求二面角F BC A --的大小的正切值；求点E 到平面AFC 的距离；求直线FC 与平面ABF 所成的角的正弦值.21.解： 作于点G ，连接FG ， 四边形ABCD 是菱形,,，为等边三角形,,-----1分平面ABCD ，平面ABCD ，，又，，平面AFG ，BC FG ∴⊥-----2分 G∴为二面角的平面角，------3分----------------------------4分连接AE ，设点E 到平面AFC 的距离为h ， 则， ----------------------5分即，也就是，--------------------6分解得：； ------------------------------------------------7分(3)作CH AB ⊥于点H ，连接FH ，ABC ∆为等边三角形，H ∴为AB 的中点，221,3,5,AH CH FH FA AH ===+= FA ⊥平面ABCD ，CH ⊂平面ABCD ，FA CH ∴⊥，----8分 又,CH AB AB AF A ⊥⋂=，CH ∴⊥平面ABF ，-----9分CFH ∴∠为直线FC 与平面ABF 所成的角，-------10分36sin 422CH CFH CF ∴∠===．-----------------12分 22.(本小题满分12分)已知圆22+=9：O x y ，过点()0,2P -任作圆O 的两条相互垂直的弦AB 、CD ，设M 、N 分别是AB 、CD 的中点，（1）直线MN 是否过定点?若过，求出该定点坐标，若不过，请说明理由； （2）求四边形ACBD 面积的最大值，并求出对应直线AB 、CD 的方程.22.解：（1）当直线AB CD 、的斜率存在且不为0，设直线AB 的方程为:()()()112220,,,,y kx k A x y B x y =-≠------------1分由2229+=y kx x y =-⎧⎨⎩得：()221450k x kx +--=--------------------2分 点()0,2P -在圆内,故0∆>. 又 1212222422,21211M M Mx x k k x x x y kx k k k +∴+=∴===-=-+++ 即 2222,11kM k k ⎛⎫- ⎪++⎝⎭--------------------3分AB CD ⊥以1k -代换k 得22222,11k k N k k ⎛⎫-- ⎪++⎝⎭22222222111.22211MNk k k k k k k k k k -+-++∴==+++---------------4分∴直线MN 的方程为：222212121k k y x k k k -⎛⎫+=- ⎪++⎝⎭化简得2112k y x k-=-，故直线MN 恒过定点()01-，--------------------5分 当直线AB CD 、的斜率不存在或为0时，显然直线MN 恒过定点()01-， 综上，直线MN 恒过定点()01-，--------------------.6分 (2) 解法一:圆心O 到直线AB的距离1d =AB ==分 （或由第（1）问得：21AB x =-==以1k -代换k 得CD =）AB CD ⊥∴以1k -代换k 得:CD =分12ACBD S AB CD ∴=⋅==分14=≤= 当且仅当221,1k k k==±时,取等号,故四边形ACBD 面积的最大值为14，--------------------11分对应直线AB 、CD 分别为2,2y x y x =-=--或2,2y x y x =--=-----------12分 解法二：设圆心O 到直线AB 、CD 的距离分别为12,d d 、则22222211229,9AB r d d CD r d d =-=-=-=---------------------7分AB CD ⊥222124d d OP ∴+==--------------------8分()()()2222121221991821818414ACBD S AB CD d d d d OP ∴=⋅=≤-+-=-+=-=-=--------------------10分当且仅当12d d =，即1k =±时,取等号,故四边形ACBD 面积的最大值为14，--------------------11分对应直线AB 、CD 分别为2,2y x y x =-=--或2,2y x y x =--=---------12分。

山东省聊城市2023-2024学年高二上学期期中语文试题及参考答案一、现代文阅读(35分)(一)现代文阅读Ⅰ(本题共5小题，17分)阅读下面的文字，完成1-5题。

材料一：要说当今热度最高的网络语言，“内卷”当仁不让，各大主流媒体关于“内卷”的讨论层出不穷，简直“万物皆可卷”。

豆瓣话题“你所在专业或从事行业有哪些‘内卷’现象”下，已有千万级的浏览量，数百条回答涉及求职、晋升、育儿、升学、婚恋各个领域。

那么，到底什么是“内卷”呢？据考证，“内卷”原本还是一个社会学概念，最早由人类学家吉尔茨在《农业内卷化——印度尼西亚的生态变化过程》一文中提出，意思是“边际效用递减、没有发展的增长”。

后来，中国的历史社会学家黄宗智用“内卷”来研究明清时期长江三角洲的小农经济，据此解释为什么当时的社会运行没有出现大突破。

如今作为网络热词的“内卷”，显然没有那么深奥的学术含义，而是带上了调侃的意味，其最早流行可能源于几张清华学霸的照片，有同学边骑自行车边看电脑，因其过分努力的行为给旁观学生带来焦虑和压力，被称为“清华卷王”。

由此生发的网络用法层出不穷。

可以作为名词，如“人生是张饼，内卷是宿命”；可以作为动词，如“今天你又内卷了吗”；也可以作为形容词使用，如“这个人也太内卷了”。

从学术意义上的“内卷”到网络语言中的“内卷”，它经历了一个语义指称对象泛化，同时语义内涵不断丰富的过程。

在网络语言中，“内卷”主要表达以下几种隐含意义：竞争的白热化导致竞争的无序化。

人们在日常生活的方方面面拼尽全力，以使自已在社会上获取竞争优势，但最终却会导致竞争的无序化，让整个群体在这种“不良竞争”中受损。

比如外卖小哥原本有10分钟送餐时间，结果有人通过提速、占道等不良竞争行为，提前几分钟抵达，公司通过大数据计算，觉得他们还能再快，就会进一步缩短送餐时间。

在这种情况下，外卖小哥尽管已经非常努力，甚至因闯红灯等行为有生命危险，但规定的送餐时间却越来越短，工作环境越来越恶化。

2023-2024学年山东省普高联考高二（上）期中数学试卷一、单项选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1．已知点A （3，2，3），B （1，1，4），则A 、B 的中点的坐标为（ ） A ．(1，12，−12)B ．(2，32，72)C ．（4，3，7）D ．(−1，−12，12)2．已知直线l 1：2x +2y ﹣5＝0，l 2：4x +ny +1＝0，若l 1∥l 2，则n 的值为（ ） A ．﹣6B ．6C ．4D ．﹣43．过点A （1，1）的直线l 与圆M ：x 2+y 2﹣6x ＝0相交的所有弦中，弦长最短为（ ） A ．5B ．2C ．√5D ．44．已知空间四边形OABC ，其对角线是OB ，AC ，M ，N 分别是对边OA ，BC 的中点，点G 在线段MN 上，且MG ＝3GN ，用基底向量OA →，OB →，OC →表示向量OG →应是（ ） A ．OG →=18OA →+38OB →+38OC →B ．OG →=18OA →−38OB →+38OC →C ．OG →=16OA →+13OB →+13OC →D ．OG →=16OA →−13OB →+13OC →5．已知实数x ，y 满足方程x 2+y 2﹣2x ＝0，则y+1x+1的最大值是（ ）A ．34B ．43C ．0D ．126．战国时期成书《经说》记载：“景：日之光，反蚀人，则景在日与人之间”．这是中国古代人民首次对平面镜反射的研究，体现了传统文化中的数学智慧．在平面直角坐标系xOy 中，一条光线从点（2，3）射出，经y 轴反射后与圆x 2﹣6x +y 2+4y +12＝0相切，则反射光线所在直线的斜率为（ ） A ．−43或−34B ．17C ．57D ．567．已知中心在原点，半焦距为4的椭圆x 2m 2+y 2n 2=1(m ＞0，n ＞0，m ≠n)被直线方程2x ﹣y +9＝0截得的弦的中点横坐标为﹣4，则椭圆的标准方程为（ ） A ．x 28+y 24=1 B ．x 232+y 216=1C ．x 28+y 24=1或y 28+x 24=1D ．x 232+y 216=1或y 232+x 216=18．苏州有很多圆拱的悬索拱桥（如寒山桥），经测得某圆拱索桥（如图）的跨度AB ＝100米，拱高OP ＝10米，在建造圆拱桥时每隔5米需用一根支柱支撑，则与OP 相距30米的支柱MN 的高度是（ ）米．（注意：√10取3.162）A ．6.48B ．4.48C ．2.48D ．以上都不对二、多项选择题：本题共4小题，每小题5分，共20分.在每小题给出的四个选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9．空间直角坐标系中，已知O （0，0，0），OA →=(−1，2，1)，OB →=(−1，2，−1)，OC →=(2，3，−1)，则（ ） A ．|AB →|=2B ．△ABC 是直角三角形C ．与OA →平行的单位向量的坐标为(√66，−√63，−√66)D ．{OA →，OB →，OC →}可以作为空间的一组基底10．在如图所示的三棱锥O ﹣ABC 中，OA ＝OC ＝OB ＝1，OA ⊥面OBC ，∠BOC =π3，下列结论正确的为（ ）A ．直线AB 与平面OBC 所成的角为45° B ．二面角O ﹣BC ﹣A 的正切值为√33C ．O 到面ABC 的距离为√217D ．异面直线OC ⊥AB11．已知直线l ：kx ﹣y +2k ＝0（k ∈R ）和圆O ：x 2+y 2＝8，则（ ） A ．直线l 恒过定点（2，0）B ．存在k 使得直线l 与直线l 0：x ﹣2y +2＝0垂直C ．直线l 与圆O 相交D ．若k ＝1，则圆O 上到直线l 的距离为√2的点有四个12．已知抛物线y 2＝4x ，焦点F ，过点P （1，1）作斜率互为相反数的两条直线分别交抛物线于A ，B 及C ，D 两点．则下列说法正确的是（ ） A ．抛物线的准线方程为x ＝﹣1 B ．若|AF |＝5，则直线AP 的斜率为1 C ．若PA →=3BP →，则直线AB 的方程为y ＝xD ．∠CAP ＝∠BDP三、填空题：本题共4小题，每小题5分，共20分.13．过P （﹣1，a ）、Q （a +1，4）两点的直线的倾斜角为45°，那么实数a ＝ ．14．a →=(1，−1，2)，b →=(−2，1，0)，c →=(−3，1，k)，若a →，b →，c →共面，则实数k ＝ ． 15．古希腊数学家阿波罗尼斯在《圆锥曲线论》中记载了用平面截圆锥得到圆锥曲线的方法．如图，将两个完全相同的圆锥对顶放置（两圆锥的顶点和轴都重合），已知两个圆锥的底面直径均为4，侧面积均为2√5π．记过两个圆锥轴的截面为平面α，平面α与两个圆锥侧面的交线为AC ，BD ．已知平面β平行于平面α，平面β与两个圆锥侧面的交线为双曲线C 的一部分，且C 的两条渐近线分别平行于AC ，BD ，则该双曲线C 的离心率为 ．16．如图，已知菱形ABCD 中，AB ＝2，∠BAD ＝120°，E 为边BC 的中点，将△ABE 沿AE 翻折成△AB 1E （点B 1位于平面ABCD 上方），连接B 1C 和B 1D ，F 为B 1D 的中点，则在翻折过程中，AE 与B 1C 的夹角为 ，点F 的轨迹的长度为 ．四、解答题：本大题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤. 17．（10分）已知点A （1，2，﹣1），B （2，k ，﹣3），C （0，5，1），向量a →=(−3，4，5)． （1）若AB →⊥a →，求实数k 的值；（2）求向量AC →在向量a →方向上的投影向量．18．（12分）已知△ABC 的顶点A （5，1），B （1，3），C （4，4）． （1）求AB 边上的高所在直线的方程； （2）求△ABC 的外接圆的方程．19．（12分）如图，在长方体ABCD ﹣A 1B 1C 1D 1中，M 为BB 1上一点，已知BM ＝2，CD ＝3，AD ＝4，AA 1＝5．（1）求直线A 1C 和平面ABCD 的夹角； （2）求点A 到平面A 1MC 的距离．20．（12分）已知定点A （1，﹣2），点B 为圆（x +1）2+（y +4）2＝4上的动点． （1）求AB 的中点C 的轨迹方程；（2）若过定点P(12，−2)的直线l 与C 的轨迹交于M ，N 两点，且|MN|=√3，求直线l 的方程．21．（12分）如图，该几何体是由等高的半个圆柱和14个圆柱拼接而成．C ，E ，D ，G 在同一平面内，且CG＝DG ．（1）证明：平面BFD ⊥平面BCG ；（2）若直线GC 与平面ABG 所成角的正弦值为√105，求平面BFD 与平面ABG 所成角的余弦值．22．（12分）“工艺折纸”是一种把纸张折成各种不同形状物品的艺术活动，在我国源远流长，某些折纸活动蕴含丰富的数学知识，例如：用一张圆形纸片，按如下步骤折纸（如图）：步骤1：设圆心是E，在圆内异于圆心处取一定点，记为F；步骤2：把纸片折叠，使圆周正好通过点F（即折叠后图中的点A与点F重合）；步骤3：把纸片展开，并留下一道折痕，记折痕与AE的交点为P；步骤4：不停重复步骤2和3，就能得到越来越多的折痕．现取半径为4的圆形纸片，设点F到圆心E的距离为2√3，按上述方法折纸．以线段EF的中点为原点，线段EF所在直线为x轴建立平面直角坐标系xOy，记动点P的轨迹为曲线C．（1）求C的方程；（2）设轨迹C与x轴从左到右的交点为点A，B，点P为轨迹C上异于A，B，的动点，设PB交直线x＝4于点T，连结AT交轨迹C于点Q．直线AP、AQ的斜率分别为k AP、k AQ．（ⅰ）求证：k AP•k AQ为定值；（ⅱ）证明直线PQ经过x轴上的定点，并求出该定点的坐标．2023-2024学年山东省普高联考高二（上）期中数学试卷参考答案与试题解析一、单项选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1．已知点A （3，2，3），B （1，1，4），则A 、B 的中点的坐标为（ ） A ．(1，12，−12)B ．(2，32，72)C ．（4，3，7）D ．(−1，−12，12)解：因为A （3，2，3），B （1，1，4），所以中点M(3+12，2+12，3+42)=(2，32，72)． 故选：B ．2．已知直线l 1：2x +2y ﹣5＝0，l 2：4x +ny +1＝0，若l 1∥l 2，则n 的值为（ ） A ．﹣6B ．6C ．4D ．﹣4解：因为l 1∥l 2，所以42=n 2≠1−5⇒n =4．故选：C ．3．过点A （1，1）的直线l 与圆M ：x 2+y 2﹣6x ＝0相交的所有弦中，弦长最短为（ ） A ．5B ．2C ．√5D ．4解：将A （1，1）代入x 2+y 2﹣6x ，得到12+12﹣6×1＜0，所以点A 在圆内， 再根据x 2+y 2﹣6x ＝0可得圆心坐标M （3，0），可知当l 与AM 垂直时，弦长最小， 因为AM =√5，即最短弦长为的一半为√32−(√5)2=2，所以最短弦长为2×2＝4． 故选：D ．4．已知空间四边形OABC ，其对角线是OB ，AC ，M ，N 分别是对边OA ，BC 的中点，点G 在线段MN 上，且MG ＝3GN ，用基底向量OA →，OB →，OC →表示向量OG →应是（ ）A ．OG →=18OA →+38OB →+38OC →B ．OG →=18OA →−38OB →+38OC →C ．OG →=16OA →+13OB →+13OC →D ．OG →=16OA →−13OB →+13OC →解：∵OG →=OM →+MG →=OM →+34MN →=OM →+34(MO →+OC →+CN →)=OM →+34MO →+34OC →+34×12CB →=14OM →+34OC →+38(OB →−OC →)=18OA →+38OB →+38OC → 故选：A ．5．已知实数x ，y 满足方程x 2+y 2﹣2x ＝0，则y+1x+1的最大值是（ ）A ．34B ．43C ．0D ．12解：C 的方程x 2+y 2﹣2x ＝0可化为（x ﹣1）2+y 2＝1， 它表示圆心（1，0），半径为1的圆，y+1x+1表示圆上的点与点P （﹣1，﹣1）的连线的斜率k ， 设过圆上点与点P （﹣1，﹣1）的直线方程为y +1＝k （x +1）， 则圆心（1，0）到直线y +1＝k （x +1）的距离d =|2k−1|√k +1≤1，可得0≤k ≤43，即最大值为43，故选：B ．6．战国时期成书《经说》记载：“景：日之光，反蚀人，则景在日与人之间”．这是中国古代人民首次对平面镜反射的研究，体现了传统文化中的数学智慧．在平面直角坐标系xOy 中，一条光线从点（2，3）射出，经y 轴反射后与圆x 2﹣6x +y 2+4y +12＝0相切，则反射光线所在直线的斜率为（ ） A ．−43或−34B ．17C ．57D ．56解：根据题意，设B 与点（2，3）关于y 轴的对称，则B 的坐标为（﹣2，3）， 则反射光线经过点B ，且与圆x 2﹣6x +y 2+4y +12＝0相切，设反射光线所在直线的方程为：y﹣3＝k（x+2），即kx﹣y+2k+3＝0，圆x2﹣6x+y2+4y+12＝0的标准方程为（x﹣3）2+（y+2）2＝1，则圆心为（3，﹣2），半径r＝1，由圆心（3，﹣2）到反射光线的距离等于半径可得：√1+k2=1，即12k2+25k+12＝0，解得k=−43或k=−34．故选：A．7．已知中心在原点，半焦距为4的椭圆x2m2+y2n2=1(m＞0，n＞0，m≠n)被直线方程2x﹣y+9＝0截得的弦的中点横坐标为﹣4，则椭圆的标准方程为（）A．x28+y24=1B．x232+y216=1C．x28+y24=1或y28+x24=1D．x232+y216=1或y232+x216=1解：设直线2x﹣y+9＝0与椭圆相交于A（x1，y1），B（x2，y2）两点，由{x12m2+y12n2=1x22 m2+y22n2=1，得(x1+x2)(x1−x2)m2+(y1+y2)(y1−y2)n2=0，得k=y1−y2x1−x2=−n2m2×x1+x2y1+y2=2，弦的中点坐标是M（﹣4，1），直线AB的斜率k＝2，所以n2m2=12，m2=2n2，又m2﹣n2＝16，所以m2＝32，n2＝16，椭圆的标准方程为x232+y216=1．故选：B．8．苏州有很多圆拱的悬索拱桥（如寒山桥），经测得某圆拱索桥（如图）的跨度AB＝100米，拱高OP＝10米，在建造圆拱桥时每隔5米需用一根支柱支撑，则与OP相距30米的支柱MN的高度是（）米．（注意：√10取3.162）A．6.48B．4.48C．2.48D．以上都不对解：以O为原点，以AB所在直线为x轴，以OP所在直线为y轴建立平面直角坐标系，设圆心坐标（0，a），P（0，10），A（﹣50，0），则圆拱所在圆的方程为x 2+（y ﹣a ）2＝r 2， ∴{(10−a)2=r 2(−50)2+a 2=r 2，解得a ＝﹣120，r 2＝16900， ∴圆的方程为x 2+（y +120）2＝16900．将x ＝﹣30代入圆方程，得：900+（y +120）2＝16900， ∵y ＞0，∴y =40√10−120≈40×3.162﹣120＝6.48． 故选：A ．二、多项选择题：本题共4小题，每小题5分，共20分.在每小题给出的四个选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9．空间直角坐标系中，已知O （0，0，0），OA →=(−1，2，1)，OB →=(−1，2，−1)，OC →=(2，3，−1)，则（ ） A ．|AB →|=2B ．△ABC 是直角三角形C ．与OA →平行的单位向量的坐标为(√66，−√63，−√66)D ．{OA →，OB →，OC →}可以作为空间的一组基底 解：因为OA →=(−1，2，1)，OB →=(−1，2，−1)，所以AB →=OB →−OA →=(0，0，−2)，所以|AB →|=2，选项A 正确； 又因为OC →=(2，3，−1)，所以BC →=OC →−OB →=(3，1，0)， 所以AB →⋅BC →=0，所以△ABC 是直角三角形，选项B 正确； 因为|OA →|=√1+4+1=√6， 所以与OA →平行的单位向量的坐标为：±OA →|OA →|=±(√66，−√63，−√66)，选项C 错误； 假设OA →，OB →，OC →共面，则存在唯一的有序数对（x ，y ）使OA →=xOB →+yOC →，即（﹣1，2，1）＝x （﹣1，2，﹣1）+y （2，3，﹣1）＝（﹣x +2y ，2x +3y ，﹣x ﹣y ）， 所以{−1=−x +2y 2=2x +3y 1=−x −y ，此方程组无解，故OA →，OB →，OC →不共面，故可作为空间一组基底，选项D 正确． 故选：ABD ．10．在如图所示的三棱锥O ﹣ABC 中，OA ＝OC ＝OB ＝1，OA ⊥面OBC ，∠BOC =π3，下列结论正确的为（ ）A ．直线AB 与平面OBC 所成的角为45° B ．二面角O ﹣BC ﹣A 的正切值为√33C ．O 到面ABC 的距离为√217D ．异面直线OC ⊥AB解：选项A ，因为OA ⊥面OBC ，故∠ABO 为直线AB 与平面OBC 所成的角， 又OA ＝OC ＝OB ＝1，所以tan ∠ABO ＝1，故直线AB 与平面OBC 所成的角是45°，故A 正确； 选项B ，取BC 中点为D ，连接OD ，AD ，因为OA ＝OB ＝OC ＝1，OA ⊥平面OBC ，∠BOC =π3，所以AB =AC =√2，BC =1，OD ⊥BC ，AD ⊥BC ， 因为OD ∩AD ＝D ，所以BC ⊥平面AOD ，故∠ODA 为二面角O ﹣BC ﹣A 的平面角，则tan ∠ODA =OA OD =2√33， 故二面角O ﹣BC ﹣A 的正切值为2√33，故B 错误；选项C ，因为AB =AC =√2，BC =1，所以AD =√72，设O 到面ABC 的距离为h ，则由V A ﹣OBC ＝V O ﹣ABC ，可得：13×√34×1=13×12×√72×ℎ，解得ℎ=√217，故C 正确；选项D ，若OC ⊥AB ，又OC ⊥OA ，且AB ∩OA ＝A ，则OC ⊥面OAB ， 则有OC ⊥OB ，与∠BOC =π3矛盾，故D 错误．故选：AC ．11．已知直线l ：kx ﹣y +2k ＝0（k ∈R ）和圆O ：x 2+y 2＝8，则（ ） A ．直线l 恒过定点（2，0）B ．存在k 使得直线l 与直线l 0：x ﹣2y +2＝0垂直C ．直线l 与圆O 相交D ．若k ＝1，则圆O 上到直线l 的距离为√2的点有四个解：由直线l ：kx ﹣y +2k ＝0，整理成k （x +2）﹣y ＝0，则直线恒过定点（﹣2，0），故A 错误； 若直线l ：kx ﹣y +2k ＝0与直线l 0：x ﹣2y +2＝0垂直， 则k +2＝0，解得k ＝﹣2，故B 正确；因为（﹣2）2+0＝4＜8，所以定点（﹣2，0）在圆O ：x 2+y 2＝8内部， 所以直线l 与圆O 相交，故C 正确； 当k ＝1时，直线l 化为x ﹣y +2＝0， 圆心O 到直线的距离d =|2|√2=√2，圆O 半径2√2， 因为d ＜r 且d =12r ，所以圆O 到直线l 距离为√2的点有三个，故D 错误．故选：BC ．12．已知抛物线y 2＝4x ，焦点F ，过点P （1，1）作斜率互为相反数的两条直线分别交抛物线于A ，B 及C ，D 两点．则下列说法正确的是（ ） A ．抛物线的准线方程为x ＝﹣1 B ．若|AF |＝5，则直线AP 的斜率为1 C ．若PA →=3BP →，则直线AB 的方程为y ＝xD ．∠CAP ＝∠BDP解：对于选项A ：因为抛物线方程为y 2＝4x ，可得该抛物线的准线方程为x ＝﹣1，故选项A 正确； 对于选项B ：不妨设A （x 0，y 0），因为|AF |＝5，所以x 0+p2=x 0+1=5，x 0=4，解得y 0＝±4， 又P （1，1），则直线AP 的斜率为4−14−1=1或−4−14−1=−53，故选项B 错误； 对于选项C ：不妨设A （x 1，y 1），B （x 2，y 2），因为P （1，1），所以BP →=(1−x 2，1−y 2)，PA →=(x 1−1，y 1−1)， 因为PA →=3BP →，所以{3(1−x 2)=x 1−13(1−y 2)=y 1−1，得{x 1=4−3x 2y 1=4−3y 2．因为y 12=4x 1，所以(4−3y 2)2=4(4−3x 2)，即3y 22−8y 2=−4x 2， 因为y 22=4x 2，所以4y 22−8y 2=0，y 2=0或y 2＝2，当y 2＝0时，x 2＝0，解得x 1＝4，y 1＝4； 当y 2＝2时，x 2＝1，解得x 1＝1，y 1＝﹣2，此时直线AB 的斜率不存在，直线CD 的斜率为0，不符合题意；则A （4，4），B （0，0），此时直线AB 的方程为y ＝x ，故选项C 正确． 对于选项D ：易知直线AB ，CD 的斜率存在，不妨设直线AB ：y ＝k （x ﹣1）+1， 则直线CD ：y ＝﹣k （x ﹣1）+1，A （x 1，y 1），B （x 2，y 2），C （x 3，y 3），D （x 4，y 4）， 联立{y =k(x −1)+1y 2=4x ，即{x =1k (y −1)+1y 2=4x，消去x 并整理得y 2−4k y +4k −4=0，因为P （1，1）在抛物线内部，所以Δ＞0， 由韦达定理得y 1+y 2=4k ，y 1y 2=4k−4，因为|AP|=√1+1k 2|y 1−1|，|BP|=√1+1k2|y 2−1|， 所以|AP|⋅|BP|=(1+1k 2)|(y 1−1)(y 2−1)|=(1+1k2)|y 1y 2−(y 1+y 2)+1| =(1+1k 2)|4k −4−4k +1|=3(1+1k2)， 同理得|CP|⋅|DP|=3[1+1(−k)2]=3(1+1k 2)，所以|AP |•|BP |＝|CP |•|DP |，即|AP||DP|=|CP||BP|，又∠CP A ＝∠BPD ，所以△APC ∽△BPD ，则∠CAP ＝∠BDP ，故选项D 正确． 故选：ACD ．三、填空题：本题共4小题，每小题5分，共20分.13．过P （﹣1，a ）、Q （a +1，4）两点的直线的倾斜角为45°，那么实数a ＝ 1 ． 解：过P （﹣1，a ）、Q （a +1，4）两点的直线的倾斜角为45°， 则k PQ ＝tan45°＝1，又k PQ =4−aa+2=1⇒a =1． 故答案为：1．14．a →=(1，−1，2)，b →=(−2，1，0)，c →=(−3，1，k)，若a →，b →，c →共面，则实数k ＝ 2 ． 解：因为a →，b →，c →共面，所以存在x ，y ∈R ，使得c →=xa →+yb →， 又因为a →=(1，−1，2)，b →=(−2，1，0)，c →=(−3，1，k)， 所以（﹣3，1，k ）＝x （1，﹣1，2）+y （﹣2，1，0）， 所以{−3=x −2y1=−x +y k =2x ，解得x ＝1，y ＝2，k ＝2．故答案为：2．15．古希腊数学家阿波罗尼斯在《圆锥曲线论》中记载了用平面截圆锥得到圆锥曲线的方法．如图，将两个完全相同的圆锥对顶放置（两圆锥的顶点和轴都重合），已知两个圆锥的底面直径均为4，侧面积均为2√5π．记过两个圆锥轴的截面为平面α，平面α与两个圆锥侧面的交线为AC ，BD ．已知平面β平行于平面α，平面β与两个圆锥侧面的交线为双曲线C 的一部分，且C 的两条渐近线分别平行于AC ，BD ，则该双曲线C 的离心率为 √5 ．解：以AC ，BD 的交点在平面β内的射影为坐标原点，两圆锥的轴在平面β内的射影为y 轴，在平面β内与x轴垂直的直线为x轴，建立平面直角坐标系．根据题意可设双曲线C的方程为x2a2−y2b2=1（a＞0，b＞0）．∵两个圆锥的底面直径均为4，则底面半径为2，又侧面积均为2√5π，∴一个圆锥的母线长为√5．则双曲线C的渐近线方程为y＝±2x，即ba=2．∴双曲线的离心率为e=ca=√c2a2=√a2+b2a2=√1+(ba)2=√5．故答案为：√5．16．如图，已知菱形ABCD中，AB＝2，∠BAD＝120°，E为边BC的中点，将△ABE沿AE翻折成△AB1E （点B1位于平面ABCD上方），连接B1C和B1D，F为B1D的中点，则在翻折过程中，AE与B1C的夹角为90°，点F的轨迹的长度为π2．解：在菱形ABCD中，∠BAD＝120°，E为边BC的中点，所以AE⊥BC，在翻折过程中，有AE⊥B1E，AE⊥CE，因为B1E∩CE＝E，B1E、CE⊂平面B1CE，所以AE⊥平面B1CE，又B1C⊂平面B1CE，所以AE⊥B1C，即AE与B1C的夹角为90°；分别取AB ，AB 1的中点M 和N ，连接EM ，EN ，FN ，因为N ，F 分别为AB 1和B 1D 的中点， 所以FN =12AD ，FN ∥AD ，又E 为BC 的中点，所以CE =12BC =12AD ，CE ∥AD ，所以FN ＝CE ，FN ∥CE ，所以点F 的轨迹与点N 的轨迹相同，即从点M 到点N 的轨迹，因为AE ⊥平面B 1CE ，所以点B 1的轨迹是以E 为圆心，BE 为半径的圆， 所以点N 的轨迹是以AE 的中点为圆心，BE 2为半径的圆， 所以点N 的轨迹长度为12×2π×BE2=π×12=π2，即点F 的轨迹长度为π2．故答案为：90°，π2．四、解答题：本大题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤. 17．（10分）已知点A （1，2，﹣1），B （2，k ，﹣3），C （0，5，1），向量a →=(−3，4，5)． （1）若AB →⊥a →，求实数k 的值；（2）求向量AC →在向量a →方向上的投影向量．解：（1）由题意，AB →=(1，k −2，−2)，a →=(−3，4，5)， 因为AB →⊥a →，所以AB →⋅a →=0，即﹣3+4k ﹣8﹣10＝0，得k =214． （2）由题意，AC →=(−1，3，2)，a →=(−3，4，5)，所以向量AC →在向量上a →上的投影向量为：(AC →⋅a →|a →|)a →|a →|=3+12+10√9+16+253√210，2√25，√22)=(−32，2，52)．18．（12分）已知△ABC 的顶点A （5，1），B （1，3），C （4，4）． （1）求AB 边上的高所在直线的方程；（2）求△ABC 的外接圆的方程． 解：（1）∵A （5，1），B （1，3）， ∴直线AB 的斜率k AB =1−35−1=−12， ∴AB 边上的高所在直线的斜率为2， ∵AB 边上的高所在直线过点C （4，4），∴AB 边上的高所在直线的方程为y ﹣4＝2（x ﹣4），即2x ﹣y ﹣4＝0． （2）∵CA →=(1，−3)，CB →=(−3，−1)， ∴CA →⋅CB →=0，即△ABC 为以角C 为直角的直角三角形， 故△ABC 的外接圆以AB 中点（3，2）为圆心，|AB|2=12√(1−5)2+(3−1)2=√5为半径，∴△ABC 的外接圆的方程为（x ﹣3）2+（y ﹣2）2＝5．19．（12分）如图，在长方体ABCD ﹣A 1B 1C 1D 1中，M 为BB 1上一点，已知BM ＝2，CD ＝3，AD ＝4，AA 1＝5．（1）求直线A 1C 和平面ABCD 的夹角； （2）求点A 到平面A 1MC 的距离．解：（1）依题意：AA 1⊥平面ABCD ，连接AC ，则A 1C 与平面ABCD 所成夹角为∠A 1CA ，∵AA 1＝5，AC =√32+42=5， ∴△A 1CA 为等腰三角形， ∴∠A 1CA =π4，∴直线A 1C 和平面ABCD 的夹角为π4，（2）（空间向量），如图建立坐标系，则A （0，0，0），C （3，4，0），A 1（0，0，5），M （3，0，2）， ∴AC →=（3，4，0），A 1C →=（3，4，﹣5），MC →=（0，4．﹣2）， 设平面A 1MC 的法向量n →=（x ，y ，z ），由{n →⋅A 1C →=3x +4y −5z =0n →⋅MC →=4y −2z =0，可得n →=（2，1，2）， ∴点A 到平面A 1MC 的距离d =|AC →⋅n →||n →|=3×2+4×1√2+1+2=103．20．（12分）已知定点A （1，﹣2），点B 为圆（x +1）2+（y +4）2＝4上的动点． （1）求AB 的中点C 的轨迹方程；（2）若过定点P(12，−2)的直线l 与C 的轨迹交于M ，N 两点，且|MN|=√3，求直线l 的方程．解：定点A （1，﹣2），点B 为圆（x +1）2+（y +4）2＝4上的动点． （1）设点C 的坐标为（x ，y ），则点B 的坐标为（2x ﹣1，2y +2）， ∵点B 为圆（x +1）2+（y +2）2＝4上的动点，∴（2x ﹣1+1）2+（2y +2+4）2＝4，即x 2+（y +3）2＝1， ∴AB 的中点C 的轨迹方程为x 2+（y +3）2＝1；（2）当直线l的斜率存在时，设直线l的方程为y+2=k(x−12 )，∵圆的半径r＝1且|MN|=√3，∴圆心到直线的距离d=1 2，∴d=|1−k2|√1+k=12，解得k=34，∴直线l的方程为y+2=34(x−12)，即6x﹣8y﹣19＝0；当直线l的斜率不存在时，直线l的方程为x=1 2，此时|MN|=√3，满足条件；综上，直线l的方程为x=12或6x﹣8y﹣19＝0．21．（12分）如图，该几何体是由等高的半个圆柱和14个圆柱拼接而成．C，E，D，G在同一平面内，且CG＝DG．（1）证明：平面BFD⊥平面BCG；（2）若直线GC与平面ABG所成角的正弦值为√105，求平面BFD与平面ABG所成角的余弦值．解：（1）证明：如图，连接CE，DG，因为该几何体是由等高的半个圆柱和14个圆柱拼接而成，CG＝DG，所以∠ECD＝∠DCG＝45°，所以∠ECG＝90°，所以CE⊥CG，因为BC∥EF，BC＝EF，所以四边形BCEF 为平行四边形， 所以BF ∥CE ， 所以BF ⊥CG ，因为BC ⊥平面ABF ，BF ⊂平面ABF ， 所以BC ⊥BF ，因为BC ，CG ⊂平面BCG ，BC ∩CG ＝C ， 所以BF ⊥平面BCG ， 因为BF ⊂平面BFD ， 所以平面BFD ⊥平面BCG ．（2）如图，以A 为坐标原点建立空间直角坐标系，设AF ＝2，AD ＝t ，则A （0，0，0），B （0，2，0），F （2，0，0），D （0，0，t ），G （﹣1，1，t ），C （0，2，t ），则AB →=(0，2，0)，AG →=(−1，1，t)，GC →=(1，1，0)， 设平面ABG 的一个法向量为m →=(x ，y ，z)， 则{m →⋅AB →=0，m →⋅AG →=0，所以{m →⋅AB →=(x ，y ，z)⋅(0，2，0)=2y =0m →⋅AG →=(x ，y ，z)⋅(−1，1，t)=−x +y +tz =0，令z ＝1，y ＝0，x ＝t ，所以m →=(t ，0，1)，记直线GC 与平面ABG 所成的角为θ，则sinθ=|cos〈GC →，m →〉|=|GC →⋅m →||GC →||m →|=|t|√2×√t +1=√105，解得t ＝2（负值舍去），即AD ＝2，设平面BFD 的一个法向量为n →=(x′，y′，z′)，FB →=(−2，2，0)，FD →=(−2，0，2)，则{n →⋅FB →=0n →⋅FD →=0即{−2x ′+2y ′=0−2x′+2z′=0，令x ′＝1，则n →=(1，1，1)， 所以cos ＜m →，n →＞=m →⋅n →|m →||n →|=√2+1⋅√1+1+1=35×3=√155，所以平面BFD 与平面ABG 所成角的余弦值为√155． 22．（12分）“工艺折纸”是一种把纸张折成各种不同形状物品的艺术活动，在我国源远流长，某些折纸活动蕴含丰富的数学知识，例如：用一张圆形纸片，按如下步骤折纸（如图）： 步骤1：设圆心是E ，在圆内异于圆心处取一定点，记为F ；步骤2：把纸片折叠，使圆周正好通过点F （即折叠后图中的点A 与点F 重合）； 步骤3：把纸片展开，并留下一道折痕，记折痕与AE 的交点为P ； 步骤4：不停重复步骤2和3，就能得到越来越多的折痕．现取半径为4的圆形纸片，设点F 到圆心E 的距离为2√3，按上述方法折纸．以线段EF 的中点为原点，线段EF 所在直线为x 轴建立平面直角坐标系xOy ，记动点P 的轨迹为曲线C ． （1）求C 的方程；（2）设轨迹C 与x 轴从左到右的交点为点A ，B ，点P 为轨迹C 上异于A ，B ，的动点，设PB 交直线x ＝4于点T ，连结AT 交轨迹C 于点Q ．直线AP 、AQ 的斜率分别为k AP 、k AQ ． （ⅰ）求证：k AP •k AQ 为定值；（ⅱ）证明直线PQ 经过x 轴上的定点，并求出该定点的坐标．解：（1）因为|PE|+|PF|=|PA|+|PE|=4＞|EF|=2√3， 所以点P 的轨迹是以E ，F 为焦点，且长轴长2a ＝4的椭圆， 焦距2c =|EF|=2√3， 此时b 2＝a 2﹣c 2＝1， 则轨迹C 方程为x 24+y 2=1；（2）证明：（i ）不妨设P （x 1，y 1），Q （x 2，y 2），T （4，m ）， 由题可知A （﹣2，0），B （2，0），第21页（共21页） 则k AP =y 1x 1+2，k AQ =k AT =m−04−(−2)=m 6， 因为k BP =k BT =y 1x 1−2=m 2， 所以m =2y 1x 1−2， 所以k AP ⋅k AQ =y 1x 1+2⋅m 6=y 1x 1+2⋅y 13(x 1−2)=y 123(x 12−4)，① 因为点P 在椭圆上，所以x 124+y 12=1，② 联立①②，解得k AP •k AQ =−112， 故k AP •k AQ 为定值；（ii ）证明：不妨设直线PQ 的方程为x ＝ty +n ，P （x 1，y 1），Q （x 2，y 2），联立{x =ty +nx 24+y 2=1，消去x 并整理得（t 2+4）y 2+2tny +n 2﹣4＝0， 由韦达定理得{y 1+y 2=−2tn t 2+4y 1y 2=n 2−4t 2+4， 由（i ）知k AP ⋅k AQ =−112， 即y 1x 1+2⋅y 2x 2+2=y 1y 2(ty 1+n+2)(ty 2+n+2)=−112， 整理得n 2−44n 2+16n+16=−112， 解得n ＝1或n ＝﹣2（舍去），所以直线PQ 的方程为x ＝ty +1，故直线PQ 经过定点（1，0）．。

绝密★启用前山东省聊城市2012-2013学年高一上学期“七校联考”期末检测历史试题考试时间：100分钟；第I卷（选择题）一、选择题”(韦庆远《中材料所述的政治模式是．郡县制 C．分封制 D．行省制（周朝）800年的统治中，影响之深远，常使历史学家难于区分，C．①②④ D．②③④《晋书》以为统一后又有②郡县制代替分封制是历史的进步③元代④废分封、行郡县有利于中央对地方的控制．③④ C．①③ D．②④下的封地分给其他子弟。

这项措施的影响是A ．宗法制得以恢复B ．中央集权得到了加强C ．分封制被彻底取消D ．地方取得较大自主权5.两汉孝廉家世可考知者128人统计分类表由上表可见，汉代察举制A ．削弱了身份制、世袭制B ．兼顾了各阶层利益C ．体现了公平公正原则D ．沿袭了世卿世禄制6.下列选项中，通过限制地方来加强中央的措施有 ①西汉“推恩令”、刺史制度 ②唐朝三 省六部制 ③宋朝设立文臣知州、通判 ④明朝废丞相、设殿阁（ ） A.①② B.③④ C.②③ D.①③7.唐朝与宋朝加强君主权力的措施中最为相似的是A.设立六部B.分化相权C.分散地方官的权力D.解除统兵大将的兵权8.五代以来，君主七朝八姓。

越匡胤“黄袍加身”建立北宋后的百余年间，未发生类似的 现象，一般认为，北宋统治体制的变革是重要原因。

下列各项中最能反映其体制变革的 一项是A ．采用文官取代武将任地方长官B ．从中央到地方实行财政军分权C ．降低将帅之地位，疏远将兵关系D ．削弱相权，另设枢密院管理军事9.纵观明代历史，经常发生皇帝不理政务，甚至二十年不上朝的现象。

但即使皇帝不上朝理政，国家机器也能依靠一班大臣和一整套政务流程维持正常运转。

这是因为 A ．“中朝”的决策 B ． 内阁的作用 C ．六部掌握实权 D ．军机处的设置10.下面四幅图反映了中国历史上皇权势力不断变化的总趋势，正确是A B C D 11.中国近代以鸦片战争为开端，主要是因为A.中国第一次被西方国家打败B.长期闭关锁国状态被打破皇权年代 皇 权年代 皇 权年代 皇 权年代C.社会性质开始发生变化D.民族矛盾取代了阶级矛盾12.“争天下，打天下，穷爷们天不怕来地不怕；杀到天津卫，朝廷快让位；杀到杨柳青，天子吓得发了懵。

绝密★启用前山东省聊城市2012-2013学年高一上学期“七校联考”期末检测英语试题考试时间：100分钟；C．confusing；confused D．confused；confusing11．He looks forwards every afternoon to _____ the flower-lined garden.A. visitB. paying a visitC. walk inD. walking in12．She is _______ at her boyfriend.A. disappointedB. disappointingC. disappointD. to be disappointed 13．---Don't put the waste on the ground.---Oh, I'm very sorry. I ________ the dustbin（垃圾箱） there.A. don't seeB. isn't seeingC. didn't seeD. haven't seen14．I don’t like the pattern of the trousers.____ , the color doesn’t suit me.A. HoweverB. InsteadC. BesidesD. But15．—Mary，new copies of 《Harry Potter》 are on sale now．—Great! Let’s go and buy this afternoon．A．it B．that C．any D．one二、完型填空16．It is interesting how NASA(美国航空航天管理局)chose their astronauts for landing them on the moon. They chose men 16 the age of twenty and thirty-five. There were about fifty of them, Many were 17 air pilots. 18 were scientists with two or three degrees. NASA telephoned each man they were going to choose; told him the plans and the 19 they might get in. They then asked him if he was willing to be trained as an astronaut. “How could any man 20 such an exciting job?” One of them said, “Dangerous? Of course. It’s dangerous 21 most exciting”The health and physical condition of 22 was, of course, very necessary. 23 those in very good health and physical condition were chosen.While being trained to be astronauts. They went through many 24 . They studied the star and the moon, and they also studied geology, the science of rocks. This was necessary 25 astronauts would have to look for rocks on the moon. They would try to find rocks which might help to tell the 26 of the moon. They were all 27 to fly in helicopters （直升飞机）。

绝密★启用前山东省聊城市“四县六校”2012-2013学年下学期高二期末联考英语试题考试时间：120分钟；1．答题前填写好自己的姓名、班级、考号等信息2．请将答案正确填写在答题卡上第I卷（选择题）一、单项选择1．It is requested that the reporter ____ to blame for the wrong report.A. referring toB. referring to asC. referred to beD. referred to being2．--- What‘s the model plane look like?--- The wings of the plane are _____ of its body.A. more than the length twiceB. twice more than the lengthC. more than twice the lengthD. more twice than the length3．The place _____ the bridge is supposed to be built should be _____ the cross-river traffic is heaviest.A. which; whereB. at which; whichC. at which; whereD. which; in which4．--- What a pity! Leo missed the football match.--- ______. He was busy with his schoolwork.A. So he didB. Neither he didC. Neither did heD. So did he5．--I‘m sorry I made a mistake!---________. Nobody is perfect.A. Take your timeB. You‘re rightC. Whatever you sayD. Take it easy6．The school isn‘t the one I really wanted to go to, but I suppose I‘ll just have to________ it.A. make the best ofB. get away fromC. keep an eye onD. catch up with7．My parents_________ in Hong Kong. They were born there and have lived nowhere else.A. liveB. livedC. were livingD. had lived8．________ children whose parents had died in the earthquake________ sent to live with families in other cities.A. A great number of ; wasB. The great number of ; wasC. A great number of ; wereD. The great number of ; were9．---How about your job hunting?---No luck. Now, I‘ve reached the stage________ I don‘t care what I do.A. whichB. whenC. whereD. that10．Either you or one of your students________ to attend the meeting that is due tomorrow.A. areB. isC. haveD. be11．While________ along the road, Mr. Smith met an old friend of his.A. walkedB. he is walkingC. walkingD. is walking12．When we visited Zhou Zhuang again ten years later, we found it changed so much that we could hardly ________ it.A. requestB. recognizeC. swapD. realize13．The country has________ too many wars.A. packed upB. made senseC. gone byD. gone through14．The report said a bus went out of ____ control in the highway _____ south of the city and rushed into the river.A．the; the B．/; the C．the; / D．/; / 15．Which do you feel like ________ the time on the train, chatting with friends or just reading something?A．killing B．to share C．to kill D．sharing二、完型填空I used to be ashamed of my grandma. I know that's a ____16____ thing to say, but it was true until today, so I have to____17___ it.The____18____ started when my friend Katy found Grandma's false teeth floating in a glass on the bathroom sink. I was so used to seeing them that I____19____ took notice of them. But Katy shouted, laughing and____20____ to talk to them. I had to get down on my knees and____21____ her to shut up so my grandma wouldn't____22____ and get hurt.After that happened, I____23____ there were a million things about Grandma that were embarrassing（令人窘迫）.Once she took Jill and me out to Burger King. ____24____ ordering our hamburgers well-done, she told the person behind the counter, "They'll have two Whoppers (巨无霸) well-to-do. " Jill burst out laughing, but I almost____25____.After a while, I started wishing I could____26____ Grandma in a closet. I even complained to my parents. Both my parents said I had to be careful not to make Grandma feel____27____ in our home. Then last Wednesday, something happened that____28____ everything completely. My teacher told us to help find interesting old people and____29____ them about their____30____for a big Oral History project. I was trying to think of someone when Angie pushed me gently."V olunteer your grandmother," she whispered. "She's____31____ and rich in experience."That was the last thing I ever thought Angie would say about my grandma.This is how I ended up on____32____ today interviewing my own grandmother before the whole school assembly (集合). All my friends and teachers were listening to her____33____ she was a great heroine. I was____34____ of my grandma and hoped she would____35____ know that I had been ashamed of her.16．A. funny B. common C. terrible D. clear17．A. admit B. receive C. refuse D. show18．A. quarrel B. accident C. trouble D. adventure19．A. already B. always C. simply D. hardly20．A. enjoying B. pretending C. imagining D. continuing21．A. warn B. demand C. advise D. beg22．A. mind B. hear C. see D. fall23．A. expected B. declared C. realized D. doubted24．A. Because of B. Except for C. Such as D. Instead of25．A. died B. cheered C. disappeared D. suffered26．A. meet B. avoid C. arrange D. hide27．A. independent B. inconvenient C. unwelcome D. unfamiliar28．A. changed B. finished C. stopped D. Prepared29．A. interview B. report C. tell D. write30．A. news B. lives C. advantages D. achievements31．A. free B. popular C. interesting D. embarrassing32．A. show B. stage C. duty D. time33．A. and then B. even if C. so that D. as if34．A. sure B. proud C. ashamed D. afraid35．A. never B. even C. still D. once三、阅读理解Since the beginning of the year, smog has covered parts of North China. In January, Beijing saw only five days without smog. The rising PM 2.5 readings terrified many people, and some health experts said that whenever the smog gets serious, hospitals receive more patients suffering acute respiratory （呼吸系统）and heart diseases.Later, news of polluted underground water in some provinces scared people who wondered whether the water they drink is safe.So the need to emphasize environmental protection while developing the economy is heard everywhere.Smog especially is a common concern. Like a popular online post said, air may be the only thing that is equal for everyone, despite your income or vocation. People with higher incomes are able to drink only bottled spring water and eat only organic food by paying higher prices, but they breathe the same air as everyone else.At a meeting on Monday, many Representatives have expressed their concerns about the air quality, too. One talked about his experience in Beijing. ―After taking a taxi from the capital airport to my hotel, which took about an hour, I washed my nose and found the inside of my nose was black. We should ask ourselves this question: Why do we want to develop? It's for living a better life. Dirty air is definitely not a better life," he said.China needs to develop its economy and invest(投资) in high-tech. Every Chinese wants a strong country. But without blue sky, clean water and safe food, the achievements in the economy will become meaningless. Space technologies are not to be developed for building a base on Mars so that one day all human beings can migrate to the red planet because they have destroyed Earth.What the public wants is a strong and beautiful China. Former president Hu Jintao spoke at the 18th Party Congress last November saying that great efforts must be made to promote ecological progress and build a beautiful China. The words have shown the central government's resolution to address the environment issue.36．The effect of smog doesn‘t include.A．the rising of PM 2.5 readingsB．more people suffering diseasesC．the increase of people‘s incomeD．patients increased in hospital37．Why smog has become a common concern?A．Because people have to pay higher prices.B．Because nobody can avoid it.C．Because we have to develop industry.D．Because a popular online-post discussed it.38．The underlined word ―they‖ in paragraph 6 refers to.A．human beings B．other plantsC．space technologies D．industrial development39．From the last two paragraphs we can infer that.A．high-tech can completely solve the problem of pollutionB．space technologies should be developed in a large scaleC．we can move to the Mars after the earth has been destroyedD．we must protect the environment while developing economyLoren Gladstone of Toronto is 58, but thinking over how to bequeath (遗赠) his digital property(财产). Doing the paperwork after his parents' death was a challenge. ―When my time comes, I wonder if my children will even know what paper is，‖ he says. As a software developer, his virtual property is both valuable and vital to his business. That reflects a problem. Online lives have increasing economic and emotional value. But testamentary (遗嘱) laws offer confusing and incomplete ways of bequeathing and inheriting (继承) them.Digital property may include software, websites, downloaded content, online gaming identities, social-media accounts and even e-mails. In Britain alone holdings of digital music may be worth over ￡9 billion ($14 billion). A fifth of respondents to a Chinese local-newspaper survey said they had over 5，000 yuan($790) of digital property. And value does not lie only in money.―Anyone with kids under 14 years old probably has two prints of them and the rest are in online galleries，‖says Nathan Lustig of Entrustet, a company that helps people manage digital property.Service providers have different rules—and few state them clearly in their terms and conditions. Many give users a personal right to use an account, but nobody else, even after death. Facebook allows relatives to close an account or turn it into a memorial page. Gmail (run by Google) will provide copies of e-mails to an executor (遗嘱执行人). Music downloaded via iTunes is held under a license which can be abolished on death. Apple declined to comment on the record on this or other policies. All e-mail and data on its iCloud service are deleted on the death of the owner.This has led to cases to court in America. In 2004 the family of Justin Ellsworth, an army man killed in Iraq, took Yahoo! to court in Michigan to get copies of his e-mails. This year, a court in Oregon ruled that another American mother whose son had died could use her dead son's password to enter his Facebook account for a short period. Now five American states have made laws giving executors control over the social-networking accounts of dead users.But this raises the subject of privacy. Passing music on is one thing; not everyone may want their relatives to read their e-mails. Colin Pearson, a London-based lawyer, says access should come only with a clear provision in a will.But laws, wills and password safes may be contrary to the providers' terms of service, especially when the executor is in one country and the data in another. Headaches for the living and lots of lovely work for lawyers.40．Why does Loren begin to think over how to bequeath his digital property at the age of 58?A. Because he is afraid his children don't know what paper is.B. Because there's no complete law dealing with digital property.C. Because his digital property is of great value and importance.D. Because he is worried his children will be taken to court.41．Which of the following can be inferred from the passage?A. Digital property is assessed in terms of nothing except money.B. No laws in America have been made to deal with digital property.C. The relatives may read the e－mail of the dead without permission.D. Lawyers can make money through cases about digital property.42．Facebook, Google and Apple have a similar rule that ________．A. users are offered accounts used by nobody else except users themselvesB. relatives of the dead may close an account or use it at their own willC. the executor may enter the e－mail and read it by themselves at any timeD. the data downloaded by the dead will be copied and then deleted from net43．Which of the following can best serve as the title of the passage?A. Digital InformationB. Testamentary LawsC. Deathless DataD. Vital PropertyHuman cloning technology could be used to reserve heart attacks. Scientists believe that they may be able to treat heart attack by cloning their healthy heart cells and injecting them into the areas of the heart that have been damaged, and other problems may be solved if human cloning and its technology are not forbidden.With cloning, infertile couples could have children. Current treatments for infertility, in terms of percentages, are not very successful. Couples go through physical and emotionally painful procedures for a small chance of having children. Many couples run out of time and money without successfully having children. Human cloning could make it possible for many more infertile couples to have children than ever before.We should be able to clone the bone marrow（骨髓）for children and adults suffering from leukemia （白血病）. This is expected to be one of the first benefits to come from cloning technology. We may learn how to switch cells on and off through cloning and thus be able to cure cancer.Cloning technology can be used to test for and perhaps cure gene-related diseases. The above is just a few examples of what human cloning technology can do for mankind. This new technology promises unprecedented （前所未有的）advancement in medicine if people will release their fears and let the benefit begin.44．What does the underlined word ―infertile‖ (paragraph 2) most probably mean?A. with physical and emotional problemsB. short of time and moneyC. unable to give birth to childrenD. separated from each other for a long time45．According to the text, one of the first expected benefits from cloning technology is in ______.A. the treatment of mental diseasesB. the reserve of heart diseasesC. the cure of gene-related diseasesD. the bearing of babies46．According to the writer, the main problem with the development of human cloning technology is that__________.A. it has brought about benefits so farB. it may be out of human controlC. people still know little about itD. people are afraid of such technologyGetting rid of dirt, in the opinion of most people, is a good thing. However, there is nothing fixed about attitudes to dirt.In the early 16th century, people thought that dirt on the skin was a means to block out disease, as medical opinion had it that washing off dirt with hot water could open up the skin and let ills in. A particular danger was thought to lie in public baths. By 1538, the French king had closed the bath houses in his kingdom. So did the king of England in 1546. Thus began a long time when the rich and the poor in Europe lived with dirt in a friendly way. Henry IV, king of France, was famously dirty. Upon learning that a nobleman had taken a bath, the king ordered that, to avoid the attack of disease, the nobleman should not go out.Though the belief in the merit of dirt was long-lived, dirt has no longer been regarded as a nice neighbor ever since the 18th century. Scientifically speaking, cleaning away dirt is good to health. Clean water supply and hand washing are practical means of preventing disease. Yet, it seems that standards of cleanliness have moved beyond science since World War II. Advertisements repeatedly sell the idea: clothes need to be whiter than white, cloths ever softer, surfaces to shine. Has the hate for dirt, however, gone too far?Attitudes to dirt still differ hugely nowadays. Many first-time parents nervously try to warn their children off touching dirt, which might be responsible for the spread of disease. On the contrary, Mary Ruebush, an American immunologist (免疫学家)，encourages children to play in the dirt to build up a strong immune system. And the latter position is gaining some ground.47．The kings of France and England in the 16th century closed bath houses because________.A. they lived healthily in a dirty environmentB. they thought bath houses were too dirty to stay inC. they believed disease could be spread in public bathsD. they considered bathing as the cause of skin disease48．Which of the following best describes Henry IV‘ s attitude to bathing?A. AfraidB. CuriousC. ApprovingD. Uninterested49．How does the passage mainly develop?A. By providing examples.B. By making comparisons.C. By following the order of time.D. By following the order of importance.50．What is the author‘s purpose in writing the passage?A. To stress the role of dirt.B. To introduce the history of dirt.C. To call attention to the danger of dirt.D. To present the change of views on dirt.America is growing older. Fifty years ago, only 4 out of every 100 people in the United States were 65 or older. Today, 10 out of every 100 Americans are over 65. The aging of the population will affect （影响）American society in many ways — education, medicine, and business. Quietly, the graying of America has made us a very different society— one in which people have a quite different idea of what kind of behavior（行为）is suitable （合适）at various ages. A person's age no longer tells you anything about his/her social position, marriage or health. There's no longer a particular year in which one goes to school or goes to work or gets married or starts a family. The social clock that kept us on time and told us when to go to school ,get a job, or stop working isn't as strong as it used to be. It doesn't surprise us to hear of a 29-year-old university president or a 35-year-old grandmother, or a 70-year-old man who has become a father for the firs time. Public ideas are changing. Many people say, 'I am much younger than my mother - or my father - was at my age.' No one says ‗Act your age‘ anymore. We've stopped looking with surprise at older people who act in youthful ways.51．It can be learnt from the text that the aging of the population in America_________ .A.has made people feel youngerB.has changed people's social positionC. has changed people's understanding of ageD. has slowed down the country's social development52．The underlined word ‗one‘ refers to_________.A. a societyB. AmericaC. a placeD. population53．‗Act your age‘ means people should_________.A.be active when they are oldB.do the right thing at the right ageC. show respect for their parents young or oldD. take more physical exercise suitable to their age54．If a 25-year-old man becomes general manager of a big firm, the writer of the text would most probably consider it ______.A.normalB.wonderfulC.unbelievableD.unreasonableStudents in many countries are learning English. Some of these students are small children. Others are teen-agers. Many are adults. Some learn at school, others by themselves. A few learn English by learning the language over the radio, on TV, or in film. One must work hard to learn another language.Why do all these people want to learn English? It is difficult to answer this question. Many boys and girls learn English at school because it is one of their subjects required for study. They study their own language and maths and English: Some people learn it because it is useful for their work. Many people learn English for their work. Many people learn English for their higher studies, because at college or university some of their books are in English. Other people learn English because they want to read newspapers or magazines in English.55．People learn English _______.A. at schoolB. over the radioC. on TVD. not all in the same way56．Different kinds of people want to learn English _______.A. together with other subjectsB. for different reasonsC. for their workD. for higher studies at colleges57．From this passage we know that _______.A. we can learn English easilyB. English is very difficult to learnC. English is learned by most people in the worldD. English is a useful language but one must work hard to learn58．Which of the following is right?A. We don‘t need to learn any foreign languages.B. We can do well in all our work without English.C. English is the most important subject in schools.D. We should learn English because we need to face the world.第II卷（非选择题）四、单词拼写阅读下面短文，根据以下提示：1）汉语提示，2）首字母提示，3）语境提示，在每个空格内填入一个适当的英语单词，并将该词完整地写在右边相对应的横线上。

2024届山东省聊城市化学高二上期中达标测试试题注意事项：1．答题前，考生先将自己的姓名、准考证号填写清楚，将条形码准确粘贴在考生信息条形码粘贴区。

2．选择题必须使用2B铅笔填涂；非选择题必须使用0．5毫米黑色字迹的签字笔书写，字体工整、笔迹清楚。

3．请按照题号顺序在各题目的答题区域内作答，超出答题区域书写的答案无效；在草稿纸、试题卷上答题无效。

4．保持卡面清洁，不要折叠，不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀。

一、选择题(共包括22个小题。

每小题均只有一个符合题意的选项）1、锂-空气电池是一种新型的二次电池，其放电时的工作原理如图所示，下列说法正确的是A．该电池放电时，正极的反应式为O2＋4e－＋4H＋=2H2OB．该电池充电时，阴极发生了氧化反应：Li＋＋e－=LiC．电池中的有机电解液可以用稀盐酸代替D．充电时，空气极与直流电源正极相连2、下列实验操作规范且能达到目的的是( )目的操作在50 mL酸式滴定管中装入盐酸，调整初始读数为A．取20.00 mL盐酸30.00 mL后，将剩余盐酸放入锥形瓶B．除去锅炉中沉积的CaSO4可用Na2CO3溶液浸泡后，再将不溶物用酸溶解去除C．测定醋酸钠溶液pH 用玻璃棒蘸取溶液，点在湿润的pH试纸上在0.1mol/LNa2S溶液中滴加少量等浓度的ZnSO4溶D．证明K sp(ZnS) > K sp(CuS)液,再加入少量等浓度的CuSO4溶液A．A B．B C．C D．D3、下列各式中，属于正确的电离方程式的是A．HCO3-+H2O H2CO3+OH-B．HCO3-+OH-H2O+CO32-C．NH3+H+NH4+D．NH3·H2O NH4++OH-4、下列有关电解质溶液中微粒的物质的量浓度关系正确的是A．在0.1mol•L-1 Na2SO3溶液溶液中：c（Na+）=2c（SO32―）+ c（HSO3―）+ c（H2SO3）B．在0.1 mol·L-1Na2CO3溶液中：c（OH-）-c（H+）＝c（HCO3-） + 2c（H2CO3）C．向0.2 mol·L-1NaHCO3溶液中加入等体积0.1 mol·L-1NaOH溶液：c（CO32-）＞c（HCO3-）＞c（OH-）＞c（H+）D．常温下，CH3COONa和CaCl2混合溶液中：c（Na+）+ c（Ca2+）=c（CH3COOH）+ c（CH3COO-）+2 c（Cl-）5、要验证甲烷中含有C､H元素，可将其完全燃烧产物依次通过：①浓硫酸②无水硫酸铜③澄清石灰水A．①②③B．②③C．②③①D．③②6、要检验某溴乙烷中的溴元素，正确的实验方法是( )A．加入四氯化碳振荡，观察四氯化碳层是否有棕红色出现B．滴入AgNO3溶液，再加入稀硝酸呈酸性，观察有无浅黄色沉淀生成C．加入NaOH溶液加热，然后加入稀硝酸呈酸性，再滴入AgNO3溶液，观察有无浅黄色沉淀生成D．加入NaOH溶液共热，冷却后滴入AgNO3溶液，观察有无浅黄色沉淀生成7、下列事实中，不能比较氢硫酸与亚硫酸的酸性强弱的是A．氢硫酸不能与碳酸氢钠溶液反应，而亚硫酸可以B．氢硫酸的导电能力低于相同浓度的亚硫酸C．0.10 mol·L－1的氢硫酸和亚硫酸的pH分别为4.5和2.1D．氢硫酸的还原性强于亚硫酸8、某烯烃A，分子式为C8H16，A在一定条件下被氧化只生成一种物质B，则符合上述条件的烃A有（）种（考虑顺反异构）。

山东省聊城市2023-2024学年高二上学期11月期中英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读理解One of the most well-known figures of American history, Thomas Jefferson was the third President of the United States and the author of the Declaration of Independence. His life was an inspiring one and at the same time a little bit contradictory (矛盾的) to his statements andJefferson and his timeAmerican SphinxThe Art of Power1．Which book would you recommend regarding Thomas Jefferson’s whole life?A．Jefferson and His Time B．American SphinxC．The Art of Power D．none of the above2．What is the book “American Sphinx” about??A．It is a biography of Thomas Jelferson’s life.B．It is a book about American politics.C．It is an analysis of Thomas Jefferson’s mind.D．It is a collection of short stories.3．One problem about “The Art of Power” is that readers _________.A．may feel bored while they are reading itB．cannot get a complete description of Thomas Jefferson’s lifeC．learn Thomas Jefferson’s political life mainly from historians’ viewpointD．cannot get an objective point of view about Thomas Jefferson’s giftWhen I first set foot in Western Australia’s Pilbara, a landscape holding3.5-billion-year-old clues to the beginning of life, I was very disappointed. The year was 1994.I drove excitedly out of the west coast town of Port Hedland, but all I saw for the first 150 kilometers were a few dead trees and smoky dust across the burnt, flat plain. And the heat!! I’d never experienced anything this terrible before. Or breathed air so thick with biting flies.But as we continued to head south on the highway to. Marble Bar—the hottest town in Australia—some low, broad hills started to rise from the horizon. As we continued down a dirt track into the hills, the burnt plains gave way to grass-covered hills. This grass is called spinifex, an amazing but cruel creation. It grows as bushes up to one meter in diameter (直径),with round, fine leaves with needle-sharp tips. The tips will go through just about any piece of cloth. My guide wore thick gaiters (护腿) to protect his legs. But he had failed to inform me of the risk. Without any gaiters, my skin was covered with needle tips that remained in my legs for months.The land, ultimately, proved worth the discomfort. Here I was walking over some of Earth’s oldest, best-preserved rocks that contain evidence of life from almost the very beginnings of time on our planet.This area had changed much from when it was first formed 3.5 billion years ago. Back then it would have been a black volcanic land, with no color from vegetation. Over the hills I might have seen a green, iron-rich sea under an orange sky heavy with carbon dioxide and without oxygen. Several billion years after that, the world would turn cold and become covered in a global ice sheet, destroying almost every living thing. When it melted away, oxygen levels rose again. Life really got going. Animals slowly occupied the land, so did new types of plants. The greening of our planet began, and a wide variety of organisms (有机体) appeared including. unfortunately for me, spinifex.4．What was the author’s first reaction to the Pilbara landscape?A．Dissatisfaction.B．Excitement.C．Amusement.D．Confusion. 5．What does the writer mean by “The land, ultimately, proved worth the discomfort”?A．It was no easy task to set foot on the land.B．He had never seen spinifex on any other land.C．The trip was valuable despite all the suffering.D．The tips remaining in his flesh were not that uncomfortable.6．What is the significance of the Pilbara landscape?A．It is where life began.B．It is still a black volcanic land.C．It is home to many unique plants.D．It contains almost no living things. 7．Which of the following might be the best title of the passage?A．How Life Started on Earth B．Pilbara: Rich in OrganismsC．An Abandoned but Lively Land D．Journey to a Land across TimeThroughout our daily lives, we have known plenty of people and will know more. But how can we tell if someone is worth our trust? In a paper published recently in the Journal of Personality and Social Psychology, researcher gave us the answer.The researchers asked 401 adults from the United States to fill out a questionnaire measuring their guilt-proneness (内疚倾向) in different situations as well as several other qualities, and then play a short online game. In this game, Player 1 is given 1, which they can choose to give to Player 2. Any money given to Player 2 is then automatically increased to 2.50. Player 2 can then decide whether to keep all of the money or behave in a trustworthy way by returning a part of the money to Player 1. The researchers found more guilt-prone people were more likely to share the money with Player 1. Actually, in follow-up studies, guilt-proneness predicted trustworthiness better than other personality qualities the researchers measured.Why might guilt lead to trustworthy behavior? The researchers found people who were guilt-prone also reported feeling a must to act in ethical (合乎道德的) and responsible ways while interacting (互动) with their partners in the game. People who are guilt-prone tend to avoid engaging in behavior that might harm or disappoint others. If they do something bad, guilt encourages them to try to make things right again.Then, how can we use this research to ascertain whether someone is trustworthy? “One way to do this might be observe how they respond to experience regret,” lead author Emma Levine, assistant professor at the University of Chicago Levine, explains. Another way is to ask them to describe a difficult dilemma they faced in the past, suggests co-author Taya Cohen, associate professor at Carnegie Mellon University. This is particularly effective, Cohen and her colleagues have found, because it allows us to see if they’re concerted about the effects their actions have on others.8．What did guilt-prone player 2 tend to do in the online game?A．Keep all the money.B．Share the money with Player 1.C．Return the money to the researchers.D．Spend the money on themselves.9．Why might guilt lead to trustworthy behavior?A．It encourages people to harm others.B．It makes people feel responsible to act ethically.C．It makes people want to disappoint others.D．It makes people avoid difficult dilemmas.10．What does the underlined word “ascertain” in the last paragraph mean?A．Ask B．Express C．Describe D．Determine 11．How is the text organized and developed?A．By providing background.B．By making a lot of comparisons.C．By answering the raised questions.D．By analyzing effects of guilt-proneness.Children like playing make-believe and enjoy inventing a variety of characters while immersed (沉浸) in imaginative play. Cooking games are a fantastic way to stimulate imaginative play. Children can act out real-life situations and behaviors through pretend play. There are various online cooking games for children as well. For children, they are more than just a fun pastime activity. They also provide a number of developmental and educational benefits.Free online cooking games encourage kids to try new things and think outside the box by letting them experiment with ingredients, recipes, and preparation methods. Such games can help kids develop a growth mindset, creativity, and the freedom to make their own choices.The act of cooking, whether virtual or in the real world, requires a lot of counting and measuring, which promotes the development of fundamental mathematical abilities such as counting and adding. Additionally, many free cooking games online require kids to perform various kitchen tasks or read the recipes or lists of ingredients and then follow the directions, teaching them to read, improving their vocabulary, and boosting their confidence.Most cooking games offer an in-built social space. For example, playing multiplayer or two-player games allows children to interact with others, help each other, laugh together, solve problems in cooperation, and so on. This is especially convenient for timid or introverted kids who have difficulty reaching out to others in the real world. Online games can help these kids develop self-esteem and practice their communication skills, so they eventually feel more confident making friends in real life.However, young children can have difficulty finding age-appropriate games, using trustworthy and safe platforms, or setting limits on their screen time. Therefore, as adults, we must prioritize the kids’ online safety and well-being by helping them select the appropriate games and get the most out of the free cooking games online.12．What is the benefit of playing multiplayer or two-player cooking games?A．Children can communicate with others and learn to be cooperative.B．Children can learn how to cook together.C．Children can compete against each other and improve their skills.D．Children can learn how to use different kitchen tools.13．How do the adults ensure children’s online safety and well-being?A．Adults should not let children play online games.B．Adults should help their children select the suitable games.C．Adults should help their children play online cooking games.D．Adults should not allow children to play creative games.14．What’s the author’s attitude toward online cooking games?A．Negative.B．Neutral.C．Favorable.D．Indifferent. 15．Which of the following titles fits best?A．Virtual Cooking: A New Way to Avoid the Kitchen?B．The Benefits of Online Cooking Games for ChildrenC．Cooking Games: A Waste of Time or a Tool for Building Life Skills?D．Online cooking: A Way to Improves Children’s Social Skills二、七选五How to cultivate emotional awarenessIn school, we learned the periodic table and were told to recite the alphabet, but nobody forced us to take a class on what emotions are and how they work. 16 Yet, research shows that being aware of your emotions is hugely beneficial and people with high emotional awareness have better social and emotional functioning.Emotional awareness is being able to identify and make sense of not only our own emotions but those of others. It’s absolutely essential in maintaining good mental health. But if you’ve spent a lifetime masking your real emotions, being honest with yourself for the first time may prove tough. 17Turn to mindfulnessMindfulness is the process of bringing our attention to: the present moment and becoming more aware of our thoughts. 18 There are many ways to be mindful, frompractising breathwork to sitting in silence. Where emotional awareness is concerned, meditation which can aid clarity and allow us to view our emotions in a more measured way is strongly recommended.Practice daily self-reflection19 You’d better create a habit of checking with yourself every day. Self-reflection cultivates emotional awareness and is key to understanding why you feel the way you do. Activities such as journaling can be really beneficial in learning about your thought processes. It may also help you actively choose healthy coping strategies to manage more challenging emotions. Free-writing, where you write out everything that’s on your mind without judgement, can be used to dig deeper into your thoughts and feelings.Name what you’re feelingSometimes what we really struggle with is putting our emotions into words. 20 Instead of employing basic descriptions like angry, happy, frustrated or sad, consult The Emotion Wheel which identifies eight basic emotions at the centre of the wheel, with dozens of corresponding emotions of various intensities stemming out from it. It gives you a specific name for what you’re feeling. By naming them, we can not only find relief from those uncomfortable feelings but bring more joy and fulfilment into our lives too.A．When was the last time you stopped and truly observed how you feel?B．Turn into your emotions and honestly reflecting on how you feel is important.C．It can be difficult to feel heard by others if we don’t have the language to describe how we’re feeling.D．But it gets easier with practice.E．It’s not always easy to put a finger on exactly what’s wrong, without digging a little deeper.F．It’s a state of calm, non-judgmental state of mind.G．And for many of us, that means understanding how we feel and why can be incredibly difficult.三、完形填空Mom was late once again!!“Sorry, honey. I left the office only a little bit late, but the 21 caused really badtraffic, and I…”I hung up. All too often, situations like this happen. Being a teenager is not 22 , and my parents often make me feel less important than their work.In third grade, I broke my arm at school. When the school nurse called my mom, she was too 23 to answer. When the phone call 24 my father, he sighed and said, “I’m in the middle of my research.” I waited in the nurse’s office for hours before my parents finally 25 to take me to hospital.I wished my parents could spare me some time. To make my parents take me seriously, I26 . I tried to establish my own authority (权威) by 27 their schedule. My mom did everything possible to settle the problem, but nothing 28 until she met that adviser.One day after I shut my door in her face, my mom didn’t complain or cry. 29 , she knocked and said, “I have something to tell you.” I opened the door. “Today I met a really good adviser. He said that every time your daughter is angry, she is just saying: I 30 you.”It occurred to me that my anger was merely a mask for me to 31 my loneliness and disappointment. But facing one’s 32 need is the first step towards being happier. This concept not only helped things between my parents and me but also enabled me to handle conflicts and emotions. I started to pay attention to other people’s 33 . When an angry customer approached me at my volunteering job, he was saying, “I need more service.” When a homeless person shouted to me in a parking lot, he was simply saying, “I need food and help.”I am now able to see 34 the fog of anger and see the real face and heart of other person.Five minutes after I hung up on her, my mom’s car finally appeared through the fog. She rolled down her window and said, “Let’s go to a hotpot place, shall we?” I 35 . “OK, but promise me you won’t be late again.”21．A．rain B．snow C．wind D．fog 22．A．easy B．unusual C．awesome D．exciting 23．A．upset B．scared C．nervous D．busy 24．A．reached B．bothered C．alarmed D．woke 25．A．passed by B．drove away C．showed up D．gave in26．A．chattered B．screamed C．whispered D．apologized 27．A．ignoring B．checking C．following D．completing 28．A．appeared B．mattered C．happened D．worked 29．A．Moreover B．Therefore C．Instead D．Meanwhile 30．A．hate B．need C．believe D．understand 31．A．avoid B．relieve C．show D．cover 32．A．unmet B．urgent C．increasing D．physical 33．A．behavior B．anger C．reaction D．satisfaction 34．A．around B．across C．into D．through 35．A．yelled B．complained C．smiled D．hesitated四、用单词的适当形式完成短文阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。

2024-2025学年山东省聊城市高二历史上册期中达标自测试卷班级：________________ 学号：________________ 姓名：______________一、单选题（每题3分）1.题目：商朝时期，我国政治制度的特点是（）A. 以血缘关系为纽带建立国家政治制度B. 实现了权力的高度集中C. 确立了“家天下”的局面D. 形成了完备的官僚政治体系答案：A解析：商朝时期，政治制度以血缘关系为纽带，形成宗法制和分封制，A项正确；商朝时期尚未实现权力的高度集中，B项错误；夏朝确立“家天下”的局面，C项错误；秦朝形成了完备的官僚政治体系，D项错误。

2.题目：关于秦朝统一后推行的郡县制，下列说法正确的是（）A. 彻底解决了地方割据问题B. 郡县长官由皇帝直接任免C. 体现了分封制的思想D. 强化了君主专制答案：B解析：秦朝统一后推行郡县制，郡县长官由皇帝直接任免，有利于加强中央集权，B项正确；郡县制有利于加强中央集权，但不能彻底解决地方割据问题，A项错误；郡县制体现了中央集权的思想，C项错误；郡县制强化了中央集权，而非君主专制，D项错误。

3.题目：汉武帝时期，儒学取得独尊地位的主要原因是（）A. 儒学有利于维护封建专制统治B. 百家争鸣局面的结束C. 汉武帝重用信奉儒学的人D. 儒学吸取了佛、道的思想答案：A解析：汉武帝时期，儒学取得独尊地位的主要原因是儒学有利于维护封建专制统治，符合汉武帝加强中央集权的需要，A项正确；百家争鸣局面的结束是汉武帝独尊儒术的结果，B项错误；汉武帝重用信奉儒学的人是儒学取得独尊地位的表现，不是原因，C项错误；儒学吸取佛、道思想是在宋明理学时期，D项错误。

4.题目：魏晋南北朝时期，北方民族交融的表现不包括（）A. 汉族人学习胡人的生活方式B. 汉族人学习胡人的政治制度C. 北方汉族人大量南迁D. 少数民族学习汉族的先进文化答案：C解析：魏晋南北朝时期，北方民族交融的表现包括汉族人学习胡人的生活方式、汉族人学习胡人的政治制度、少数民族学习汉族的先进文化等，C项“北方汉族人大量南迁”是北方人口南迁的表现，属于经济重心南移的内容，不属于民族交融的表现，符合题意。

山东省聊城市高二上学期期中数学试卷姓名:________ 班级:________ 成绩:________一、选择题 (共12题；共24分)1. （2分）下列说法中正确的有：已知求得线性回归方程y=bx+a，相关系数r，①若r＞0，则x增大时，y 也相应增大；②若r＜0，则x增大时，y也相应增大；③若r=1，或r=﹣1，则x与y的关系完全对应（有函数关系），在散点图上各个散点均在一条直线上．（）A . ①②B . ②③C . ①③D . ①②③2. （2分）若样本+2，+2，，+2的平均数为10，方差为3，则样本2+3，2+3，… ，2+3，的平均数、方差、标准差是（）A . 19，12，B . 23，12，C . 23，18，D . 19，18，3. （2分） 2015年11月11日，天猫交易额以912.17亿元的成绩刷新了世界纪录．随之快递的订单量也激增．某机构就双十一期间快递公司A的物流速度进行了随机调查，如图是200名受调查者对快递公司A的评分（百分制）的频率分布直方图，则其得分的众数大致为（）A . 65B . 70C . 75D . 804. （2分） (2016高一下·湖南期中) 如图，矩形长为5，宽为3，在矩形内随机撒100颗黄豆，数得落在椭圆内的黄豆数为80颗，以此实验数据为依据可以估算椭圆的面积约为（）A . 11B . 9C . 12D . 105. （2分）(2017·郴州模拟) 某地市高三理科学生有15000名，在一次调研测试中，数学成绩ξ服从正态分布N（100，σ2），已知p（80＜ξ≤100）=0.35，若按成绩分层抽样的方式取100份试卷进行分析，则应从120分以上的试卷中抽取（）A . 5份B . 10份C . 15份D . 20份6. （2分） (2016高二上·昌吉期中) 一次实验：向如图所示的正方形中随机撒一大把豆子，经查数，落在正方形中的豆子的总数为N粒，其中m（m＜N）粒豆子落在该正方形的内切圆内，以此估计圆周率π为（）A .B .C .D .7. （2分）学校为了调查学生在课外读物方面的支出情况，抽取了一个容量为n的样本，其频率分布直方图如图，其中支出在[50，60）的同学有30人，若想在这n人中抽取50人，则在[50，60）之间应抽取的人数为（）A . 10人B . 15人C . 25人D . 30人8. （2分）某班级有50名学生，现要采取系统抽样的方法在这50名学生中抽出10名学生，将这50名学生随机编号1﹣50号，并分组，第一组1﹣5号，第二组6﹣10号，…，第十组46﹣50号，若在第三组中抽得号码为12，则在第八组中抽得号码为（）A . 37B . 38C . 39D . 409. （2分） (2015高二上·安徽期末) 一个袋子中有5个大小相同的球，其中有3个黑球与2个红球，如果从中任取两个球，则恰好取到两个同色球的概率是（）A .B .C .D .10. （2分）把三进制数1021（3）化为十进制数等于（）A . 102B . 34C . 12D . 4611. （2分）在区间内随机取个实数a，则直线，直线与x轴围成的面积大于的概率是（）A .B .C .D .12. （2分） (2018高一下·葫芦岛期末) 葫芦岛市某工厂党委为了研究手机对年轻职工工作和生活的影响情况做了一项调查：在厂内用简单随机抽样方法抽取了30名25岁至35岁的职工，对其“每十天累计看手机时间”（单位：小时）进行调查，得到茎叶图如下．所抽取的男职工“每十天累计看手机时间”的平均值和所抽取的女生“每十天累计看手机时间”的中位数分别是（）A .B .C .D .二、填空题 (共4题；共4分)13. （1分） (2017高二下·南通期中) 随机变量X的概率分布规律为P（X=k）= ，k=1，2，3，4，其中c是常数，则P（ . ＜X＜）的值为________．14. （1分）已知随机变量ξ服从正态分布N（1，4），若p（ξ＞4）=0.1，则p（﹣2≤ξ≤4）=________．15. （1分）(2017·郴州模拟) 两所学校分别有2名，3名学生获奖，这5名学生要排成一排合影，则存在同校学生排在一起的概率为________.16. （1分） (2016高二上·徐水期中) 样本中共有五个个体，其值分别为a，0，1，2，3．若该样本的平均值为1，则样本方差为________．三、解答题 (共6题；共55分)17. （10分）(2017·蚌埠模拟) 当今信息时代，众多中小学生也配上了手机．某机构为研究经常使用手机是否对学习成绩有影响，在某校高三年级50名理科生第人的10次数学考成绩中随机抽取一次成绩，用茎叶图表示如图：（1）根据茎叶图中的数据完成下面的2×2列联表，并判断是否有95%的把握认为经常使用手机对学习成绩有影响？及格（60及60以上）不及格合计很少使用手机经常使用手机合计（2）从50人中，选取一名很少使用手机的同学（记为甲）和一名经常使用手机的同学（记为乙）解一道函数题，甲、乙独立解决此题的概率分别为P1，P2，P2=0.4，若P1﹣P2≥0.3，则此二人适合为学习上互帮互助的“对子”，记X为两人中解决此题的人数，若E（X）=1.12，问两人是否适合结为“对子”？参考公式及数据：，其中n=a+b+c+dP（K2≥k0）0.100.050.025k0 2.706 3.841 5.02418. （15分） (2017高二下·深圳月考) 某中学为了解高一年级学生身高发育情况，对全校名高一年级学生按性别进行分层抽样检查，测得身高（单位：）频数分布表如表、表 .表：男生身高频数分布表身高/频数表：女生身高频数分布表身高/频数（1）求该校高一女生的人数；（2）估计该校学生身高在的概率；（3）以样本频率为概率，现从高一年级的男生和女生中分别选出人，设表示身高在学生的人数，求的分布列及数学期望.19. （5分） (2016高一上·黑龙江期中) 若点（，2）在幂函数f（x）的图象上，点（2，）在幂函数g（x）的图象上，定义h（x）= 求函数h（x）的最大值及单调区间．20. （10分）(2017·烟台模拟) 在某大学自主招生的面试中，考生要从规定的6道科学题，4道人文题共10道题中，随机抽取3道作答，每道题答对得10分，答错或不答扣5分，已知甲、乙两名考生参加面试，甲只能答对其中的6道科学题，乙答对每道题的概率都是，每个人答题正确与否互不影响．（1）求考生甲得分X的分布列和数学期望EX；（2）求甲，乙两人中至少有一人得分不少于15分的概率．21. （5分）一个盒子里装有标号为1，2，3，…，5的5张标签，现随机地从盒子里无放回地抽取两张标签．记X为两张标签上的数字之和．（1）求X的分布列．（2）求X的期望E（X）和方差D（X）．22. （10分） (2016高二上·临川期中) 设函数（1）若b和c分别是先后抛掷一枚骰子得到的点数，求对任意x∈R，f（x）＞0恒成立的概率．（2）若b是从区间[0，8]（3）任取得一个数，c是从[0，6]任取的一个数，求函数f（x）的图象与x轴有交点的概率．参考答案一、选择题 (共12题；共24分)1-1、2-1、3-1、4-1、5-1、6-1、7-1、8-1、9-1、10-1、11-1、12-1、二、填空题 (共4题；共4分)13-1、14-1、15-1、16-1、三、解答题 (共6题；共55分)17-1、17-2、18-1、18-2、18-3、19-1、20-1、20-2、21-1、22-1、22-2、。

绝密★启用前山东省聊城市2012-2013学年高一上学期“七校联考”期末检测语文试题考试时间：100分钟；1．答题前填写好自己的姓名、班级、考号等信息2．请将答案正确填写在答题卡上第I 卷（选择题）一、选择题 ．．．．的一句是（3分） A ．面对世界性的金融危机，各国政府纷纷使出重拳，采取强力措施拯救危机，但能否奏效，还需要各国政府的不懈努力。

B ．教育部日前发出通知，对严重不符合卫生要求的学校食堂要及时进行整改，坚决杜绝食物中毒事件的发生。

C. 清华大学在江苏招生的负责人说，江苏教育部门虽然对2009年等级要求还没有明确表态，但从目前清华大学招办反馈的信息看，学校认为2008年“AA+”这一等级要求比较合理。

D ．我相信我有足够的勇气面对生活中已经发生的一切，我甚至敢于深入到悲剧的核心，在纯粹的荒谬之中停留，但我的生活并没有因此出现奇迹般的变化。

2．下列各组词语中加点字的注音完全正确的一项是（ ）（3分） A ．给．(jǐ)以 粗犷．(gu ǎng) 疾风劲．(jìng )草 咎．(jiù)由自取 B ．偌．(n uò)大 讣．(fù)告 掎．(jǐ)角之势 蜚．(fēi )声文坛 C ．僭．(jiàn )越 症．(zhēng )结 拾．(shè)级而上 厝．(cuò)火积薪 D ．弄．(lòng )堂 刽．(guì)子手 返璞．(pǔ)归真 力能扛．(gāng )鼎 3．下列各组词语中有两个错别字的一项是（ ）（3分） A ．积腋成裘 令人发指 老态龙钟 独当一面 B ．掂斤播两 合盘托出 动辄得咎 厉兵秣马 C ．理屈词穷 断章取义 振聋发聩 计日成功 D ．当仁不让 艰壁清野 疾言厉色 划地为牢 4．依次填人下面一段文字横线处的语句，衔接最恰当的一组是（ ）（3分） 微博的出现极大地降低了言论的门槛，______。

山东省部分市2023-2024学年高二上学期期中考试英语试题分类汇编完形填空山东省聊城市2023-2024学年高二上学期期中英语试题第一节完形填空（共15小题，每小题1分，满分15分）阅读下面短文，选出最佳选项，并在答题卡上将该项涂黑。

Mom was late once again!!“Sorry, honey. I left the office only a little bit late, but the 21 caused really bad traffic, and I…”I hung up. All too often, situations like this happen. Being a teenager is not 22 , and my parents often make me feel less important than their work.In third grade, I broke my arm at school. When the school nurse called my mom, she was too 23 to answer. When the phone call 24 my father, he sighed and said, “I’m in the middle of my research.” I waited in the nurse’s office for hours before my parents finally 25 to take me to hospital.I wished my parents could spare me some time. To make my parents take me seriously, I26 . I tried to establish my own authority (权威) by 27 their schedule. My mom did everything possible to settle the problem, but nothing 28 until she met that adviser.One day after I shut my door in her face, my mom didn’t complain or cry. 29 , she knocked and said, “I have something to tell you.” I opened the door. “Today I met a really good adviser. He said that every time your daughter is angry, she is just saying: I 30 you.”It occurred to me that my anger was merely a mask for me to 31 my loneliness and disappointment. But facing one’s 32 need is the first step towards being happier. This concept not only helped things between my parents and me but also enabled me to handle conflicts and emotions. I started to pay attention to other people’s 33 . When an angry customer approached me at my volunteering job, he was saying, “I need more service.” When a homeles s person shouted to me in a parking lot, he was simply saying, “I need food and help.”I am now able to see 34 the fog of anger and see the real face and heart of other person.Five minutes after I hung up on her, my mom’s car finally appeared th rough the fog. She rolled down her window and said, “Let’s go to a hotpot place, shall we?” I 35 . “OK, but promise me you won’t be late again.”21．A．rain B．snow C．wind D．fog 22．A．easy B．unusual C．awesome D．exciting 23．A．upset B．scared C．nervous D．busy 24．A．reached B．bothered C．alarmed D．woke 25．A．passed by B．drove away C．showed up D．gave in 26．A．chattered B．screamed C．whispered D．apologized 27．A．ignoring B．checking C．following D．completing 28．A．appeared B．mattered C．happened D．worked 29．A．Moreover B．Therefore C．Instead D．Meanwhile 30．A．hate B．need C．believe D．understand 31．A．avoid B．relieve C．show D．cover 32．A．unmet B．urgent C．increasing D．physical 33．A．behavior B．anger C．reaction D．satisfaction 34．A．around B．across C．into D．through 35．A．yelled B．complained C．smiled D．hesitated山东省菏泽市2023-2024学年高二上学期期中英语试题第一节(共15小题；每小题1分, 满分15分)阅读下面短文, 从每题所给的A、B、C、D四个选项中选出可以填入空内处的最佳选项。

聊城市某重点高中2012-2013学年下学期高二3月模块测试生物试题第I卷（选择题）一、选择题1．下图是一温带地区在原生（初级）演替过程中两种量值变化曲线，这两条曲线分别表示什么？A、A表示草本植物的种类数量，B表示整个植被的生物量B、A表示整个种群的生物量，B表示草本植物的种类数量C、A表示土壤中有机物总量，B表示整个植被中的种类数量D、A表示食肉类的种类数量，B表示食肉类的生物总量2．关于演替的说法错误的是 ( )A．演替是群落根本性质发生变化的现象B．只有群落的结构受到干扰或破坏，才会出现群落的演替C．演替过程只要不遭到人类的破坏和各种自然力的干扰，其总的趋势是会导致物种多样性的增加D．不论是成型的群落或是正在发展形成过程中的群落，演替现象一直存在着，它贯穿着整个群落发展的始终3．植被随纬度(从南到北)和海拔(从平原到高山)的增加而发生的演替变化是( )A．常绿阔叶林落叶阔叶林针叶阔混交林针叶林苔原荒漠B．苔原荒漠针叶林针叶阔混交林落叶阔叶林常绿阔叶林C．常绿阔叶林针叶林针叶阔混交林落叶阔叶林草原荒漠D．苔原荒漠常绿阔叶林针叶林针叶阔混交林落叶阔叶林4．有关图示的各项叙述中，正确的是（）A．图中有两条捕食链B．狐与鸟之间既存在捕食关系又存在竞争关系C．细菌占有第二、三、四、五营养级D．该图表示生物群落，狐为第三、四营养级5．如下图表示某草地上草、虫、鸟三类生物数量的变化曲线，下列叙述正确的是（）A．甲、乙、丙依次是鸟、虫、草B．生态系统崩溃的原因最可能是鸟类的锐减C．b点时甲的数量下降主要是由天敌的减少引起的D．a点时甲的数量上升主要是由食物增加引起的6．下图表示四类生物细胞的部分结构。

以下相关叙述中不正确的是（）A．Ⅰ一定不是第一营养级B．Ⅱ、Ⅲ是生态系统中最重要的成分C．Ⅳ能将有机物转化为无机物D．Ⅰ、Ⅱ、Ⅲ、Ⅳ构成一个生态系统7．下图甲为某清洁水体在遭受一次性有机物轻度污染后，污染物含量的变化。

聊城市某重点高中2012-2013学年下学期高二3月模块测试历史试题第I卷（选择题）一．选择题1.从促进近代化的角度看，戊戌变法最深远的历史影响是A．确立了维新思想的主导地位B．促进了民主共和思想的形成与发展C．开创了新的时代风气、社会舆论和思想观念D．有助于西方政治学说与儒家思想的结合2.史记.商君列传载:：“商君相秦十年，宗室内贵戚多怨望”。

这主要是因为商鞅变法Ａ、允许工商者入仕做官Ｂ、准许土地自由买卖Ｃ、承认土地归私人所有Ｄ、规定按军功授爵赐田3.关于穆罕默德•阿里改革的表述正确的是（）A、未采取保护民族工业发展的措施B、改革中创办的工业以民族工业最为重要C、穆罕默德•阿里政权取得的一个最重要和最令人惊异的成果是消灭了马木路克D、是一次有利于埃及近代化发展的封建改革4.关于“庆历新政”表述正确的是（）A.以整顿吏治为改革中心B.以增加财政收入为主要目标C.推行时间较长D.改革教育制度，培养变法人才5.梁启超在《变法通议》中写道：同治初年，德相俾斯麦对人说：“三十年后，日本其兴，中国其弱乎?日人之游欧洲者，讨论学业，讲究官制，归而行之。

中人之游欧洲者，询某厂船炮之利，某厂价值之廉，购而用之。

强弱之原，其在此乎?”梁启超引这段话的根本意图是A．比较出中日近代化的异同 B．寻找甲午战争日胜中败的历史原因C．批评洋务派的活动 D．证明中国进行社会政治改革的必要性6.以下经济改革，未触及土地所制的是A.商鞅变法、新中国建立初土地改革B. 英国圈地运动、美国宅地法C. 人民公社运动、家庭联产承包责任制D. 俄国1861年改革、日本明治维新7.下列事件集中全面反映了俄国农奴制危机的是（）A．农奴起义风起云涌 B．工业发展步履维艰C．克里米亚战争的失败 D．十二月党人起义8.《魏书·食货志》记载：“自昔以来，诸州户口，籍贯不实，包藏隐漏，废公罔私。

富强者并兼有余，贫弱者糊口不足。

”北魏孝文帝时为解决上述问题采取的措施有（）①推行均田制②实行三长制③实行俸禄制④废除宗主督护制A．①②③ B．①②④ C．①④D．②④9.穆罕默德·阿里认识到，应向西方寻找发展的金钥匙。

聊城市某重点高中2012-2013学年下学期高二3月模块测试英语试题第I卷（选择题）一、单项选择1．Professor Li has earned a world ____ with his pioneer research in biochemistry.A. nameB. reputationC. influenceD. feature2．Don’t take ________ for granted that your parents should support you all your life.A. thisB. thatC. themD. it3．Many people were afraid to go to swim when they remembered the scenes________ people were eaten by the big white shark.A. in whichB. by whichC. whichD. that4．You can borrow my car ________ you promise not to drive too fast.A. unlessB. even ifC. in caseD. as long as5．Of the two brothers, Bill is ______ younger one, and he is a quiet boy, ______ that most adults like very much.A. a; oneB. a ; the oneC. the; oneD. the ; the one6．The job is not very profitable ________ cash, but I can get valuable experience from it.A. in case ofB. in terms ofC. in return forD. in addition to7．We were told not to touch the equipment in the laboratory unless ________.A. allowed to doB. allowing to doC. allowed toD. allowing to8．According to the rules, students ______ get grades not lower than 85% in any subject in order to get the scholarship.A. willB. shallC. mayD. would9．It’s ____ you behave in difficulties that shows what you are really like.A. thatB. whatC. howD. where10．------You ran into Mr. Li! How did that happen?------Well, _______ of us was looking where we were going. We came around the corner at the same time.A. neitherB. eitherC. bothD. None11．------What's the matter with Tim?------Oh, Tim's cell phone was left in a taxi accidentally, never ________ again.A. to findB. to be foundC. findingD. being found12．Will you ____ it that my birds are looked after well while I'm away?A. see toB. turn toC. turn offD. see off13．The manager of the travel agency was trying hard to figure out how he would attract more customers _____ an extraordinary idea _____ his mind and he immediately knew what to do.A. when; flashed acrossB. as; crossedC. when; hitD. before; struck14．I am quite ______ that the new cosmetics, whether it appears in the form of tablet, capsule, powder or injection, will _______ through the country among young ladies.A. optimistic about; sweepB. optimistic ; sweepC. pessimistic; spreadD. pessimistic about; spread15．--- Is that 120? My brother is seriously ill!---Ok. Help is _________ .A. in the wayB. in many waysC. on the wayD. by the way二、完型填空16．A young man was getting ready to graduate from college. For many months he had 16 a beautiful sports car in a dealer’s showroom, and 17 his father could well 18 it, he told him that was all he wanted.On the morning of his graduation day his father called him into his own study and told him how 19 he was to have such a fine son. He handed his son a beautiful gift box. 20 but slightly disappointed, the young man 21 the box and found a lovely book. 22 , he raised his voice at his father and said, “ 23 all your money you give me a book?” and rushed out of the house 24 the book in the study.He did not contact（联系）his father for a whole year 25 one day he saw in the street an old man who looked like his father. He 26 he had to go back home and see his father.When he arrived at his father’s house, he was told that his father had been in hospital for a week. The moment he was about to 27 the hospital, he saw on the desk the 28 new book ,just as he had left it one 29 ago. He opened it and began to 30 the pages. Suddenly, a car key 31 from an envelope taped behind the book .It had a tag（标签）with dealer’s name, the 32 dealer who had the sports car he had 33 .On the tag was the 34 of his graduation, and the 35 PAID IN FULL.【小题1】A. expected B. enjoyed C. admired D. owned【小题2】A. finding B. proving C. deciding D. knowing【小题3】A. afford B. offer C. keep D. like【小题4】A. encouraged B. comfortable C. proud D. moved【小题5】 A. Nervous B. Serious C. Careful D. Curious【小题6】A. packed B. opened C. picked up D. put aside【小题7】A. Angrily B. Eagerly C. Calmly D. Anxiously【小题8】A. At B. From C. With D. To【小题9】A. tearing B. putting C. forgetting D. leaving【小题10】 A. until B. as C. before D. unless【小题11】 A. learned B. realized C. recognized D. admitted【小题12】A. get to B. search for C. turn to D. leave for【小题13】A. much B. still C. hardly D. quite【小题14】 A. year B. month C. week D. day【小题15】A. clean B. read C. turn D. count【小题16】A. lost B. came C. appeared D. dropped【小题17】 A. old B. same C. special D. new【小题18】A. remembered B. desired C. found D. met【小题19】 A. picture B. place C. date D. sign【小题20】A. word B. information C. messages D. card三、阅读理解17．DiscoverNews magazine of science devoted to the wonders and stories of modern science, written for the 41-45 ADBCACover Price: $59.88Price: $19.95 ($1.66/issue)You Save : $39.93 (67%)Issues : 12 issues/12 monthsSelfPublished by Conde Nast Publications Inc., Self is a handbook devoted to women’s overall physical and mental health. Every issue contains usable articles such as “Style Lab”, in which wearable clothes are mixed and matched on non-models and the “Eat-Right Road Map”, with tips on how to eat properly.Cover Price: $ 35.86Price: $15.00 ($2.5/issue)You Save: $ 20.86(58%)Issues: 6 issues/12 monthsIn StyleIn Style is a guide to the lives and lifestyles of the world’s famous people. The magazine covers the choices people make about their homes, their clothes and their free time activities. With photos and articles, it opens a door to these people’s homes, families, parties and weddings, offering ideas about beauty, fitness and in general, lifestyles. Publisher: The Time Inc. Magazine Company.Cover Price: $47.88Price: $23.88($2.38/issue)You Save: $24.00(50%)Issues: 10 issues/12 monthsWiredThis magazine is designed for leaders in the field of information engineering including top managers and professionals in the computer, business, design and education industries. Published by Conde Nast Publications Inc., Wired often carries articles on how technology changes people’s lives.Cover Price : $ 59.40Price : $ 10.00 ($1.00/issue)You Save: $49.40(84%)Issues: 10 issues/12 months【小题1】Which of the following magazines is published monthly?A．Discover B．Self C．In Style D．Wired【小题2】Which two following magazines are published by the same publisher? A．Wired and In Style B．Discover and In StyleC．Self and Discover D．Self and Wired43．Which magazine offers the biggest price cut?A．In Style B．Wired C．Discover D．Self【小题3】Those who are interested in management and the use of high technology would probably choose ________ .A．In Style B．Self C．Wired D．Discover【小题4】The “Style Lab” in Self provides readers with articles which________.A. offer advice to ordinary women on clothesB. show how a woman can become famousC. introduce places with the best foodD. discuss ways of training models18．Dogs wag(摇摆) their tails in different directions depending on whether they are excited and wanting to move forward or threatened and thinking of moving back, a study has found.Researchers in Italy examined the tail wagging behaviour of 30 dogs, catching their responses to a range of stimuli(刺激物) with video cameras. To conduct the study they chose 15 male dogs and 15 female ones aged between one and six years. The dogs were all family pets whose owners had allowed them to take part in the experiment at Bari University. The dogs were placed in a large wooden box with an opening at the front to allow for them to view various stimuli. They were tested one at a time.The researchers led by Professor Giorgio Vallortigara of the University of Trieste found that when the dogs were shown their owners—a positive experience—their tails wagged energetically to the right side. When they were shown an unfamiliar human they wagged to the right, but with somewhat less enthusiasm. The appearance of a cat again caused a right-hand side wag, although with less intensity again. The appearance of a large unfamiliar dog, similar to a German shepherd, changed the direction of tail wagging to the left. Researchers supposed the dog was thinking of moving back. When the dogs were not shown any stimuli they tended to wag their tails to the left, suggesting they preferred company. While the changes in the tail wagging were not easily noticed without the aid of video, it was thought that the findings could help people judge the mood (心情) of dogs. Computer and video systems, for example, could be used by professional dog trainers to determine the mood of dogs that they were required to approach.【小题1】The video cameras were used to catch the dogs ’responses because .A. it was easie r to catch the dogs’ response changes in the tail waggingB. the dogs were put in the wooden boxes and tested one at a time.C. they enabled the dogs’ owners to know about their dogs’ habitD. the dogs wagged their tails in different directions when they were in different moods【小题2】The underlined word“intensity” in the passage means .A. surpriseB. worryC. excitementD. interest【小题3】When there are no stimuli, a dog will .A. wag to the leftB. wag to the rightC. not wag at allD. wag to the left and then to the right【小题4】The underlined word “they”in the last paragraph refers to .A. the dogsB. the trainersC. the systemsD. the researchers【小题5】The purpose of doing the experiment is .A. to train dogs for their ownersB. to help people judge the mood of dogsC. to help dogs find companyD. to help people choose their pet dogs19．Sports account for a growing amount of income made on the sales of commercial time by television companies. Many television companies have used sports to attract viewers from particular sections of the general public, and then they have sold audiences to advertisers.An attraction of sport programs for the major U.S. media companies is that events are often held on Saturday and Sunday afternoons—the slowest time periods of the week for general television viewing. Sport events are the most popular weekend programs, especially among male viewers who may not watch much television at other times during the week. This means the television networks are able to sell advertising time at relatively high prices during what normally would be dead time for programming. Media corporations also use sports to attract commercial sponsors that might take their advertising dollars elsewhere if television stations did not report certain sports. The people in the advertising departments of major corporations realize that sports attract male viewers. They also realize that most business travelers are men and that many men make family decisions on the purchases of computers, cars and life insurance.Golf and tennis are special cases for television programming. These sports attract few viewers, and the ratings are unusually low. However, the audience for these sports is attractive to certain advertisers. It is made up of people from the highest income groups in the United States, including many lawyers and business managers. This is why television reporting of golf and tennis is sponsored by companies selling high-priced cars, business and personal computer, and holiday trips .This is also why the networks continue to carry these programs regardless of low ratings. Advertisers are willing to pay high fees to reach high-income consumers and those managers who make decisions to buy thousands of “company cars” and comp uter. With such viewers, these programs don’t need high ratings to stay on the air.【小题1】Television sport programs on weekend afternoons .A. result in more sport eventsB. get more viewers to play sportsC. bring more money to the television networksD. make more people interested in television【小题2】Why would weekend afternoons become dead time without sport programs?A. Because there would be few viewersB. Because the advertisers would be off workC. Because television programs would go slowlyD. Because viewers would pay less for watching television【小题3】In many families, men make decisions on .A. holiday tripsB. sports viewingC. television shoppingD. expensive purchases【小题4】The ratings are not important for golf and tennis programs because .A. their advertisers are carmakersB. their viewers are attracted by sportsC. their advertisers target at rich peopleD. their viewers can afford expensive cars【小题5】.What is the passage mainly about?A. Television ratings are determined by male viewers.B. Sports are gaining importance in advertising on television.C. Rich viewers contribute most to television companies.D. Commercial advertisers are the major sponsors of sport events.Apple SeedsCirculation(发行量) : 1 Year, 9 IssuesCover Price: $44.55Price For You: $33.95Product Description: Apple Seeds is an award winning magazine filled with stories for kids aged from 7 to 9. The cover is very soft, providing durability（耐用性）that allows each issue to be enjoyed for many years to come. Besides, there is a big surprise for you --- it’s being sold at a m ore favorable discount than usual. Better LifeCirculation: 1 Year, 12 IssuesCover Price: $44.55Price For You: $15.00Product Description: Designed for those who have a strong interest in personal lifestyle, Better Life is America’s complete home and fam ily service magazine. It offers help with food, recipes, decorating, building, gardening, family health, money management, and education.Humor TimesCirculation: 1 Year, 12 IssuesCover Price: $36.00Price For You: $11.95Product Description: Humor Times Magazine is for those who love to laugh! Full of cartoons and humor columns, it shows up in your mailbox once a month and keeps you smiling all year round! In today’s world, you need a reason to laugh. So let’s find it in Humor Times.News ChinaCirculation: 1 Year, 12 IssuesCover Price: $47.88Price For You: $19.99Product Description: News China Magazine is the English edition of China Newsweek. The magazine covers the latest Chinese domestic news in politics, business, society, environment, culture, sports and travels, etc. It is the first comprehensive news magazine for readers interested in China.20．What do we know about Apple Seeds?A. The soft cover enables it to be read and kept long.B. It can be purchased as an award for your children.C. It offers the biggest discount among all the magazines.D. The magazine is going to surprise you for many years.21．Tom wants to beautify his house, so he may choose _________ .A. Apple SeedsB. Better LifeC. Humor TimesD. News China22．What kind of people may buy News China ?A. People who have an interest in Personal lifestyle of the Chinese.B. People who have a strong sense of humor and love to laugh.C. People who want to enlarge the knowledge of their kids.D. People who are interested in Chin a’s politics, business and culture. 23．Which magazine provides the biggest discount if you buy it for the whole year?A. Apple SeedsB. Better LifeC. Humor TimesD. News ChinaOne evening I went out and left my 17-year-old son in charge of his 8-year-old brother and 4-year-old sister．On this occasion, the work was made less troublesome by the presence of his girlfriend．I left with complete confidence that the older children would do a wonderful job of babysitting the younger children．Later, I discovered that complete confidence was the last thing I should have left home with．I had decided to return home earlier than planned so that my son and his girlfriend could go out．I called home with this happy news．But instead of hearing his cheerful, grateful voice on the other end of the line, all I heard was the sound of a telephone ringing．It was, I should point out, after 10 p．m．,when the two younger children should have been in bed, and when the two older children should have been answering the phone．“I’ll give him a lesson,” I said． I decided they must be outside．Why they might be outside at 10:30 on a winter night I had no idea, but it was the only explanation I could come up with．Finally, in desperation, I called his girlfriend’s house． After what seemed like countless rings, his girlfriend answered．“Yes,” she said brightly, “He’s right here．”He came on the phone．I was not my usual calm, rational（理智的）self． After all, one of the rules of survival for modern parents is that you can’t t rust modern teenagers．“Where are the children?” I said．He said they were with him．They had done nothing wrong．My son had taken the younger children over to his girlfriend’s house just for ice cream and cake．This was too good to be believed． Well, it turns out that I shouldn’t have believed it．It was only part of the truth．The following Saturday evening we were at my parents’ home, celebrating my birthday．My oldest son gave me the children’s gifts．Mounted and framed were a series of lovely color photographs of my children, dressed in their best clothes, and wearing their most wonderful expressions．They are pictures to treasure a lifetime, all taken by the father of my son’s girlfriend．24．The author went out and left her eldest son in charge of the younger childrenbecause ．A．she knew that her eldest son was a good baby-sitterB．she thought it no hard work to take care of the younger onesC．she believed he could do well with his girlfriend’s helpD．she could not find a baby-sitter on that winter night25．When the author called home that evening, she found that ．A．two younger children had already been in bedB．the children were preparing a birthday gift for herC．her son was quarrelling with his girlfriendD．there was no one answering the telephone26．What might the children do that evening?A．They had a birthday party． B．They framed some photographs．C．They had their pictures taken． D．They made some beautiful clothes．27．What does the author intend to tell us by the story?A．Modern teenagers are not worth trusting．B．It is no easy job to look after young children．C．It’s no good to have a girlfriend at an early age．D．Her children have a caring and tender heart．第II卷（非选择题）四、短文改错28．此题要求改正所给短文中的错误。