许昌县三高高一期末综合训练 20160617

2015-2016学年河南省许昌高中、襄城高中、长葛一高、禹州三高联考高一（上）期末数学试卷（理科）一.选择题（每小题5分，共12题，共60分）1. 设集合A={x|−1≤x≤2}，B={x|0≤x≤4}，则A∩B=()A.[0, 2]B.[1, 2]C.[0, 4]D.[1, 4]2. 设l，m是两条不同的直线，α是一个平面，则下列命题正确的是()A.若l⊥m，m⊂α，则l⊥αB.若l⊥α，l // m，则m⊥αC.若l // α，m⊂α，则l // mD.若l // α，m // α，则l // m3. 平行线3x+4y−9=0和6x+8y+2=0的距离是（）A.8 5B.2C.115D.754. 设f(x)={2e x−1,x<2，log3(x2−1),x≥2，，则f[f(2)]的值为()A.0B.1C.2D.35. △ABC是边长为1的正三角形，那么△ABC的斜二测平面直观图△A′B′C′的面积为（）A.√34B.√38C.√68D.√6166. 设f(x)为奇函数，且在(−∞, 0)内是减函数，f(−2)＝0，则xf(x)<0的解集为（）A.(−1, 0)∪(2, +∞)B.(−∞, −2)∪(0, 2)C.(−∞, −2)∪(2, +∞)D.(−2, 0)∪(0，27. 过点(1, 2)，且与原点距离最大的直线方程是( )A.x+2y−5=0B.2x+y−4=0C.x+3y−7=0D.x−2y+3=08. 已知三棱锥的三视图如图所示，其中侧视图为直角三角形，俯视图为等腰直角三角形，则此三棱锥的体积等于（）A.√23B.√33C.2√23D.2√339. 已知点A(2, −3)、B(−3, −2)直线l过点P(1, 1)，且与线段AB相交，则直线l的斜率k的取值范围是（）A.k≥34或k≤−4 B.k≥34或k≤−14C.−4≤k≤34D.34≤k≤410. 已知长方体ABCD−A1B1C1D1中，AB=BC=4，CC1=2，则直线BC1和平面DBB1D1所成角的正弦值为（）A.√32B.√52C.√105D.√101011. 如果实数x、y满足x2+(y−3)2=1，那么yx的取值范围是（）A.[2√2, +∞)B.(−∞, −2√2]C.[−2√2, 2√2]D.(−∞, −2√2]∪[2√2, +∞)12. 已知函数f(x)={|x+1|,x≤0,|log2x|，x>0,若方程f(x)=a有四个不同的解，x2，，x4，且x1<x2<x3<x4，则x3(x1+x2)+1x32x4的取值范围是（）A.(−1, +∞)B.(−1, 1]C.(−∞, 1)D.[−1, 1)二.填空题（每小题5分，共4题，共20分）直线x−2y+5=0与圆x2+y2=8相交于A、B两点，则|AB|=________．若一系列函数的解析式相同，值域相同，但其定义域不同，则称这一系列函数为“同族函数”，试问解析式为y=x2，值域为{1, 2}的“同族函数”共有________个．已知圆柱的侧面展开图是边长为4和6的矩形，则该圆柱的表面积为________．一个四面体的所有棱长都是√2，四个顶点在同一个球面上，则此球的表面积为________．三.解答题（本大题共6个小题，共70分，解答题应写出文字说明.证明过程或演算步骤．）已知全集U=R，A={x|12<2x<4}，B={x|log3x≤2}．（1）求A∩B；（2）求∁U(A∪B)．已知点A(−1, 3)，B(5, −7)和直线l:3x+4y−20=0．（1）求过点A与直线l平行的直线l1的方程；（2）求过A，B的中点与l垂直的直线l2的方程．如图，在三棱锥A−BPC中，AP⊥PC，AC⊥BC，M为AB中点，D为PB中点，且△PMB为正三角形，（1）求证：MD // 平面APC；（2）求证：平面ABC⊥平面APC．如图，四棱锥P−ABCD的底面ABCD是边长为2的菱形，∠BAD＝60∘．已知PB＝PD＝2，PA=√6．(Ⅰ)证明：BD⊥面PAC(Ⅱ)若E为PA的中点，求三菱锥P−BCE的体积．已知函数f(x)=log4(4x+1)+kx(k∈R)．（1）若k=0，求不等式f(x)>12的解集；（2）若f(x)为偶函数，求k的值．已知方程x2+y2−2x−4y+m=0．(1)若此方程表示圆，求m的取值范围；(2)若(1)中的圆与直线x+2y−4=0相交于M、N两点，且OM⊥ON（O为坐标原点），求m；(3)在(2)的条件下，求以MN为直径的圆的方程．参考答案与试题解析2015-2016学年河南省许昌高中、襄城高中、长葛一高、禹州三高联考高一（上）期末数学试卷（理科）一.选择题（每小题5分，共12题，共60分）1.【答案】A【考点】交集及其运算【解析】结合数轴直接求解．【解答】解：由数轴可得A∩B=[0, 2]，故选择A．2.【答案】B【考点】两条直线平行的判定空间中直线与平面之间的位置关系【解析】根据题意，依次分析选项：A，根据线面垂直的判定定理判断．C：根据线面平行的判定定理判断．D：由线线的位置关系判断．B：由线面垂直的性质定理判断；综合可得答案．【解答】解：A，根据线面垂直的判定定理，线要垂直平面内两条相交直线才能满足线面垂直，故不正确；B，由线面垂直的性质可知：平行线中的一条垂直于这个平面则另一条也垂直这个平面，故正确；C，l // α，m⊂α，则l // m或两线异面，故不正确；D，平行于同一平面的两直线可能平行，异面，相交，故不正确．故选B.3.【答案】B【考点】两条平行直线间的距离【解析】先将两平行直线的方程的系数统一，再代入平行线间的距离公式计算即可．【解答】解：两平行直线的距离d=12√A2+B2=√36+64=2．故选B4.【答案】C【考点】函数的求值【解析】考查对分段函数的理解程度，f(2)＝log3(22−1)＝1，所以f（f(2)）＝f(1)＝2e1−1＝2．【解答】解：f[f(2)]＝f[log3(22−1)]＝f(1)＝2e1−1＝2.故选C.5.【答案】D【考点】斜二测画法平面图形的直观图【解析】由原图和直观图面积之间的关系S直观图S原图=√24，求出原三角形的面积，再求直观图△A′B′C′的面积即可．【解答】解：正三角形ABC的边长为1，故面积为√34，而原图和直观图面积之间的关系S直观图S原图=√24，故直观图△A′B′C′的面积为√34×√24=√616故选D．6.【答案】C【考点】函数单调性的性质与判断函数奇偶性的性质与判断【解析】根据函数的奇偶性求出f(2)＝0，x f(x)<0分成两类，分别利用函数的单调性进行求解．【解答】∵f(x)为奇函数，且在(−∞, 0)内是减函数，f(−2)＝0，∴f(−2)＝−f(2)＝0，在(0, +∞)内是减函数∴x f(x)<0则{x>0f(x)<0=f(2)或{x<0f(x)>0=f(−2)根据在(−∞, 0)内是减函数，在(0, +∞)内是减函数解得：x∈(−∞, −2)∪(2, +∞)7.【答案】A【考点】直线的点斜式方程两条直线垂直与倾斜角、斜率的关系【解析】先根据垂直关系求出所求直线的斜率，由点斜式求直线方程，并化为一般式．【解答】解：设A(1, 2)，则OA的斜率等于2，故所求直线的斜率等于−12，由点斜式求得所求直线的方程为y−2=−12(x−1)，化简可得x+2y−5=0，故选A．8.【答案】B【考点】由三视图求体积【解析】三棱锥是底面是等腰直角三角形，腰长是1，．一条侧棱与底面垂直，且这条侧棱的长度是√2，根据三棱锥的体积公式写出体积的表示式，得到结果．【解答】解：∵由三视图知，三棱锥是底面是等腰直角三角形，底边上的高是1，一条侧棱与底面垂直，且这条侧棱的长度是√3，∴三棱锥的体积是13×√3×12×1×2=√33，故选B9.【答案】A【考点】直线的斜率【解析】画出图形，由题意得所求直线l的斜率k满足k≥k PB或k≤k PA，用直线的斜率公式求出k PB和k PA的值，解不等式求出直线l的斜率k的取值范围．【解答】解：如图所示：由题意得，所求直线l的斜率k满足k≥k PB或k≤k PA，即k≥34或k≤4故选：A．10.【答案】C【考点】直线与平面所成的角【解析】要求线面角，先寻找斜线在平面上的射影，因此，要寻找平面的垂线，利用已知条件可得．【解答】解：由题意，连接A1C1，交B1D1于点O∵长方体ABCD−A1B1C1D1中，AB=BC=4∴C1O⊥B1D1∴C1O⊥平面DBB1D1在Rt△BOC1中，OC1=2√2，BC1=2√5∴直线BC1和平面DBB1D1所成角的正弦值为√105故选C．11.【答案】C【考点】求线性目标函数的最值【解析】由题意可得yx表示以(0, 3)为圆心1为半径的圆上的点和原点连线的斜率k，由直线和圆的位置关系数形结合可得．【解答】解：∵实数x、y满足x2+(y−3)2=1，∴yx表示以(0, 3)为圆心1为半径的圆上的点和原点连线的斜率k，当直线与圆相切时，联立x2+(y−3)2=1和y=kx消去y并整理可得(1+k2)x2−6kx+8=0，由△=36k2−32(1+k2)=0可解得k=±2√2，故yx的取值范围是[−2√2, 2√2]，故选：C ．12.【答案】 B【考点】函数的零点与方程根的关系 【解析】作函数f(x)={|x +1|,x ≤0|log 2x|，x >0的图象如下，由图象可得x 1+x 2=−2，x 3x 4=1；1<x 4≤2；从而化简x 3(x 1+x 2)+1x 32x 4，利用函数的单调性求取值范围．【解答】解：作函数f(x)={|x +1|,x ≤0|log 2x|，x >0，的图象如下，由图可知，x 1+x 2=−2，x 3x 4=1；1<x 4≤2； 故x 3(x 1+x 2)+1x 32x 4=−2x 4+x 4，其在1<x 4≤2上是增函数， 故−2+1<−2x 4+x 4≤−1+2；即−1<−2x 4+x 4≤1；故选B ．二.填空题（每小题5分，共4题，共20分） 【答案】2√3【考点】直线与圆的位置关系 【解析】可以直接求出A 、B 然后求值；也可以用圆心到直线的距离来求解． 【解答】解：圆心为(0, 0)，半径为2√2，圆心到直线x −2y +5=0的距离为d =22=√5，故(|AB|2)2+(√5)2=(2√2)2，得|AB|=2√3． 故答案为：2√3． 【答案】 9【考点】函数的概念及其构成要素 【解析】1的原象是正负1；2的原象是正负√2．值域为{1, 2}，由此来判断解析式为y =x 2，值域为{1, 2}的“同族函数”的个数． 【解答】解：1的原象是正负1；2的原象是正负√2．值域为{1, 2}，所以y =x 2的同族函数只有9个，定义域分别为{1, √2}，{−√2, −1}，{√2, −1}，{−√2, 1}， {−√2, −1, 1}，{√2, −1, 1}，{−√2, √2, −1}，{−√2, √2, 1}，{−√2, √2, 1, −1}，共9个 故答案为：9． 【答案】 24+18π或24+8π【考点】旋转体（圆柱、圆锥、圆台） 柱体、锥体、台体的面积求解【解析】已知圆柱的侧面展开图是边长为4和6的矩形，分两种情况：①6=2πr ，②4=2πr ，然后再分别求解． 【解答】解：∵ 圆柱的侧面展开图是边长为4和6的矩形， ①若6=2πr ，则r =3π，∴ 圆柱的表面积为：4×6+2×π×(3π)2=24+18π；②若4=2πr ，r =2π，∴ 圆柱的表面积为：4×6+2×π×(2π)2=24+8π．故答案为：24+18π或24+8π．【答案】 3π【考点】球的表面积和体积 【解析】把四面体补成正方体，两者的外接球是同一个，求出正方体的棱长，然后求出正方体的对角线长，就是球的直径，即可求出球的表面积． 【解答】如图，将四面体补成正方体，则正方体的棱长是1，正方体的对角线长为：√3， 则此球的表面积为：4π×(√32)2=3π三.解答题（本大题共6个小题，共70分，解答题应写出文字说明.证明过程或演算步骤．） 【答案】解：（1）A ={x|12<2x<4}={x|−1<x <2}，B ={x|log 3x ≤2}={x|0<x ≤9， 所以A ∩B ={x|0<x <2}； （2）A ∪B ={x|−1<x ≤9}，C U (A ∪B)={x|x ≤−1或x >9．【考点】交、并、补集的混合运算 【解析】（1）求解指数不等式和对数不等式化简集合A ，B ，然后直接利用交集概念求解； （2）直接利用补集运算求解． 【解答】解：（1）A ={x|12<2x <4}={x|−1<x <2}，B ={x|log 3x ≤2}={x|0<x ≤9， 所以A ∩B ={x|0<x <2}； （2）A ∪B ={x|−1<x ≤9}，C U (A ∪B)={x|x ≤−1或x >9． 【答案】解：（1）3x +4y −20=0的斜率为−34，因为l 1 // l ，所以k 1=−34， 代入点斜式，得y −3=−34(x +1)，化简，得3x +4y −9=0．（2）A ，B 的中点坐标为(2, −2)，因为l 2⊥l ，所以k 2=43， 代入点斜式，得y +2=43(x −2)， 化简，得4x −3y −14=0．【考点】直线的一般式方程与直线的平行关系 直线的一般式方程与直线的垂直关系【解析】（1）根据两直线平行，斜率相等，求出直线的斜率，用点斜式求得直线l 1的方程．（2）A ，B 的中点坐标，根据两直线垂直，斜率之积等于−1，求出直线的斜率，用点斜式求得直线l 2的方程． 【解答】解：（1）3x +4y −20=0的斜率为−34，因为l 1 // l ，所以k 1=−34， 代入点斜式，得y −3=−34(x +1)， 化简，得3x +4y −9=0．（2）A ，B 的中点坐标为(2, −2)，因为l 2⊥l ，所以k 2=43， 代入点斜式，得y +2=43(x −2)， 化简，得4x −3y −14=0．【答案】 证明：（1）∵ M 为AB 中点，D 为PB 中点， ∴ MD // AP ， 又MD ⊄平面APC ， ∴ MD // 平面APC ．（2）∵ △PMB 为正三角形，且D 为PB 中点， ∴ MD ⊥PB ．又由（1）知MD // AP ， ∴ AP ⊥PB ．又已知AP ⊥PC ，PB ∩PC =P∴ AP ⊥平面PBC ，而BC 包含于平面PBC ， ∴ AP ⊥BC ，又AC ⊥BC ，而AP ∩AC =A ， ∴ BC ⊥平面APC ， 又BC 包含于平面ABC∴ 平面ABC ⊥平面PAC ． 【考点】直线与平面平行的判定 平面与平面垂直的判定【解析】（1）∵ M 为AB 中点，D 为PB 中点，由中位线定理得MD // AP ，由线面平行的判定证得MD // 平面APC ； （2）先证得AP ⊥BC ，又有AC ⊥BC ，通过线面垂直的判定证出BC ⊥平面APC ，再由面面垂直的判定证出平面ABC ⊥平面PAC ．【解答】 证明：（1）∵ M 为AB 中点，D 为PB 中点， ∴ MD // AP ，又MD⊄平面APC，∴MD // 平面APC．（2）∵△PMB为正三角形，且D为PB中点，∴MD⊥PB．又由（1）知MD // AP，∴AP⊥PB．又已知AP⊥PC，PB∩PC=P∴AP⊥平面PBC，而BC包含于平面PBC，∴AP⊥BC，又AC⊥BC，而AP∩AC=A，∴BC⊥平面APC，又BC包含于平面ABC∴平面ABC⊥平面PAC．【答案】（1）证明：连接BD，AC交于O点，∵PB＝PD，∴PO⊥BD，又ABCD是菱形，∴BD⊥AC，∵PO⊂平面PAC，AC⊂平面PAC，AC∩PO＝O，∴BD⊥平面PAC．（2）则AC＝2√3，∵△ABD和△PBD的三边长均为2，∴△ABD≅△PBD，∴AO＝PO=√3，∴AO2+PO2＝PA2，∴AC⊥PO，S△PAC=12⋅AC⋅PO＝3，V P−BCE＝V B−PEC=12V B−PAC=12⋅13⋅S△PAC⋅BO=12×13×3×1=12．【考点】棱柱、棱锥、棱台的侧面积和表面积直线与平面垂直【解析】(Ⅰ)连接BD，AC交于O点，分别证明出PO⊥BD，BD⊥AC，根据线面垂直的判定定理证明出BD⊥平面PAC．(Ⅱ)先证明出△ABD≅△PBD，求得PO，根据勾股定理证明出AC⊥PO，求得△PAC的面积，最后根据V P−BCE＝V B−PEC=12V B−PAC求得答案．【解答】（1）证明：连接BD，AC交于O点，∵PB＝PD，∴PO⊥BD，又ABCD是菱形，∴BD⊥AC，∵PO⊂平面PAC，AC⊂平面PAC，AC∩PO＝O，∴BD⊥平面PAC．（2）则AC＝2√3，∵△ABD和△PBD的三边长均为2，∴△ABD≅△PBD，∴AO＝PO=√3，∴AO2+PO2＝PA2，∴AC⊥PO，S△PAC=12⋅AC⋅PO＝3，V P−BCE＝V B−PEC=12V B−PAC=12⋅13⋅S△PAC⋅BO=12×13×3×1=12．【答案】解：（1）f(x)=log4(4x+1)，∵log4(4x+1)>12⇔4x+1>2，∴x>0，即不等式的解集为(0, +∞)．…（2）由于f(x)为偶函数，∴f(−x)=f(x)即log4(4−x+1)−kx=log4(4x+1)+kx，∴2kx=log4(4−x+1)−log4(4x+1)=log44−x+14x+1=−x对任意实数x都成立，所以k=−12…【考点】函数奇偶性的性质对数函数图象与性质的综合应用【解析】（1）根据对数的单调性解对数不等式；（2）根据偶函数的性质求常数k．【解答】解：（1）f(x)=log4(4x+1)，∵log4(4x+1)>12⇔4x+1>2，∴x>0，即不等式的解集为(0, +∞)．…（2）由于f(x)为偶函数，∴f(−x)=f(x)即log4(4−x+1)−kx=log4(4x+1)+kx，∴2kx=log4(4−x+1)−log4(4x+1)=log44−x+14x+1=−x对任意实数x都成立，所以k=−12…【答案】解：(1)(x−1)2+(y−2)2=5−m，∴方程表示圆时，m<5；(2)设M(x1, y1)，N(x2, y2)，则x1=4−2y1，x2=4−2y2，得x1x2=16−8(y1+y2)+4y1y2，∵OM⊥ON，∴x1x2+y1y2=0，∴16−8(y1+y2)+5y1y2=0①，由{x=4−2yx2+y2−2x−4y+m=0，得5y2−16y+m+8=0，∴y1+y2=165，y1y2=8+m5．代入①得m=85．(3)以MN为直径的圆的方程为(x−x1)(x−x2)+(y−y1)(y−y2)=0，即x2+y2−(x1+x2)x−(y1+y2)y=0，∴所求圆的方程为x2+y2−85x−165y=0．【考点】直线与圆的位置关系二元二次方程表示圆的条件圆的一般方程【解析】(1)圆的方程化为标准方程，利用半径大于0，可得m的取值范围；(2)直线方程与圆方程联立，利用韦达定理及OM⊥ON，建立方程，可求m的值；(3)写出以MN为直径的圆的方程，代入条件可得结论．【解答】解：(1)(x−1)2+(y−2)2=5−m，∴方程表示圆时，m<5；(2)设M(x1, y1)，N(x2, y2)，则x1=4−2y1，x2=4−2y2，得x1x2=16−8(y1+y2)+4y1y2，∵OM⊥ON，∴x1x2+y1y2=0，∴16−8(y1+y2)+5y1y2=0①，由{x=4−2yx2+y2−2x−4y+m=0，得5y2−16y+m+8=0，∴y1+y2=165，y1y2=8+m5．代入①得m=85．(3)以MN为直径的圆的方程为(x−x1)(x−x2)+(y−y1)(y−y2)=0，即x2+y2−(x1+x2)x−(y1+y2)y=0，∴所求圆的方程为x2+y2−85x−165y=0．。

许昌三校联考高一上学期第四次考试 化学试卷 可能用到的相对原子质量： N 14 H 1 C 12 O 16 Cl 35.5 S 32 Na 23 一选择题：（每个小题只有一个选项符合题意，3×=60分） 1 ①Si ②Al ③Cl ④Fe ⑤Na A．①②⑤ B．①② C．①②③ D．①②③④⑤ 2. 下列关于物质用途的说法中，不正确的是 A．硅酸钠可用于制备硅胶和木材防火剂等B．氧化铝熔点高可用作耐火材料 C．晶体硅可用于制造光导纤维 D．氯水和臭氧可用于自来水消毒 3. 下列关于氧化还原反应的叙述中，正确的是 A．有单质参加的反应一定是氧化还原反应 B．燃烧反应必须有氧气作氧化剂 C．2Fe+3Cl2===2FeCl3 ； Fe+SFeS 由此可以判断氧化性：Cl2>S D．Mg在化学反应中失去2个电子，而Al在化学反应中失去3个电子 ，因此还原性 Al>Mg 4．NA表示阿伏加德罗常数，下列说法正确的是 A．3.55gCl2通入足量水中充分反应，转移电子数为NA B．18 g NH4+含有质子数为10NA C．13.8g NO2与足量水反应时，转移电子的数目为0.2NA D．标准状况下，相同体积的SO3、HCl、NO含有的分子数相同 5．下列实验操作中正确的是（ ） 蒸馏操作时，温度计水银球靠近蒸馏烧瓶的底部，且向蒸馏液体中加入几块沸石，以防止暴沸 蒸发操作时，应使混合物中的水分完全蒸干后，才能停止加热 焰色反应每次实验都要先用稀硫酸洗净铂丝 过滤、结晶、灼烧、萃取、分液和蒸馏等都是常用的分离混合物的方法 A．Na放入水中，产生气体：2Na + 2H2O ＝ 2Na+ + 2OH－ + H2↑ B．在澄清的石灰水中通入过量SO2：SO2＋OH-==HSO3 C．Al片与少量NaOH溶液反应，产生气体：2Al + 2OH－+2H2O＝2AlO2— +3H2↑ D.次氯酸钙溶液中通入少量SO2气体：Ca2++2ClO+SO2+H2O=CaSO3↓+2HClO 下列是 A．B．下列实验方法正确的是 A．要使AlCl3溶液中Al3+全部沉淀出来可使用氨水 ．某气体能使品红溶液褪色，该气体一定为SO2 NaHCO3溶液除去混在Cl2中的HCl气体 D．用Cl2除去Fe2(SO4)3溶液中的少量FeSO4 9．取一定量的饱和氯水与CaCO3粉末混合，观察到氯水的黄绿色褪去，静置后取上层澄清溶液四份，分别进行以下实验： ①一份滴加盐酸，产生大量气泡；②一份滴加NaOH溶液，出现白色沉淀 ③一份用酒精灯加热出现白色沉淀； ④将有色布条浸入第四份溶液，布条立即褪色 根据以上判断下列说法中错误的是（ ） A．整个实验过程中未涉及氧化还原反应B．由①、②可知澄清溶液中的含有Ca（HCO3）2 C．②、③中的白色沉淀为CaCO3D．由④可知澄清溶液中含有HClO C．固体溶解后，将溶液转移到容最瓶中，然后向容最瓶中直接加水稀释到刻度线 D．配制溶液时容量瓶中原来有少量蒸馏水不会影响实验结果 11.将装有60 mLNO2、NO混合气的量筒倒立于水槽中，反应后气体体积缩小为40mL，则 混合气体中NO2和NO的体积比是A. 53B. 23C. 11D. 21 12.实验室中可用如下反应制取氯气：2KMnO4+16HCl（浓）＝ 2KCl+2MnCl2+5Cl2↑+8H2O A．KMnO4＜Cl2B． mL12.8 mol/L的盐酸与足量的KMnO4反应，可生成1.6 molCl2 C．此反应中氧化剂与还原剂的物质的量之比为1 D． 13.水溶液X中只可能溶有K＋、Mg2＋、Al3＋、、、、、中的若干种离子。

许昌市三校联考高一上学期第三次考试物理试题（90分钟 100分）一、选择题（1—8题是单选题，每题4分，9—12题是多选题;全部选对得4分，少选得2分，错选或不选得0分，共48分）1．运动学是我们高中物理的重要组成部分，下列关于运动学中的概念、物理方法的理解正确的是（）A．速度、质量、加速度、路程的运算时都满足平行四边形法则B．紧急刹车时，物体相对车厢向前滑行了x＝2m，测出x的参考系是车厢C．宇航员在航天器中因失重而处于漂浮状态，所以没有惯性D．用弹簧秤竖直悬挂一物体，弹簧秤的示数一定等于物体重力的大小2．在物理学的发展进程中，科学的物理思想与方法对物理学的发展起到了重要作用，下列关于物理学史及物理学思想的说法中正确的是（）A．牛顿第一定律是利用逻辑思维对事实进行分析的产物,能用实验直接验证B．在推导匀变速直线运动位移公式时，把整个运动过程划分成很多小段，每一小段近似看作匀速直线运动，再把各小段位移相加，这里运用了理想模型法C．在探究加速度与力和质量关系的实验中，“将砝码和砝码盘的重力近似等于细线的拉力”利用了等效替代的思想D．伽利略认为自由落体运动是初速度为零的匀加速直线运动3．距地面5m高处的屋檐，每隔0.2s由静止滴落一滴水滴，不考虑空气阻力，重力加速度g取10m/s2.当第一滴水滴落时，空中共有几滴水（包括刚滴出的水滴）（）A．3 B．4 C．5 D．64．A、B、C 三个物体在拉力F的作用下，一起向右匀速运动，运动过程保持相对静止，不计空气阻力，则运动过程中物体B 所受力的个数为（）A．3 B．4 C．5 D．65．汽车在路段做匀加速直线运动，该段路每隔 15m 设置一盏路灯，从经过某一盏路灯开始计时，5s 后经过第二盏路灯，又过 3s 经过第三盏路灯（可将汽车视为质点），汽车运行加速度大小为（）A．0.5m/s2 B．1m/s2 C．1.5m/s2 D．2m/s26. 如图所示弹簧测力计、绳和滑轮的重量均不计，绳与滑轮间的摩擦力不计，物体的重力都是1kg，在图(甲)、(乙)、(丙)三种情况下，弹簧测力计的读数分别是F1、F2、F3， g取10m/s2；以下判断正确的是（）A．F1 <F2=10N B．F1 =F3=10N C．F1 >F3＝10N D．F2 >F3＝10N7．水平公路上有一辆满载苹果的卡车，遇红灯刹车过程的位移随时间速度的图象如图所示，图象是一条抛物线，若车内有一苹果a如图，则（）A．质点做匀减速运动，初速度大小为5m/sB ．汽车运动的加速度大小是2m/s2C ．经过t=6s ，质点的位移大小为24mD ．a 受到周围苹果对它作用力等于a 的重力8．如图所示，质量为m 的小球套在竖起固定的光滑圆环上，在圆环的最高点有一个光滑小孔，一根轻绳的下端系着小球，上端穿 过小孔用力拉住；开始时绳与竖直方向夹角为θ，小球处于静止 状态,现缓慢拉动轻绳，使小球沿光滑圆环上升一小段距离，则下列关系正确的是（ ）A ．绳与竖直方向的夹角为450时,F=2mgB ．小球沿光滑圆环上升过程中,轻绳拉力逐渐增大C ．小球沿光滑圆环上升过程中,小球所受支持力大小不变D ．小球沿光滑圆环上升过程中,小球所受支持力大小增大9. a 、b 两辆汽车做直线运动的速度时间图像如图所示，开始计 时时，b 车在 a 车前面10m 处，关于两物体的运动情况以下 说法正确的是（ ）A ．a 、b 两车均做匀变速直线运动B ．第一秒末a 车运动方向发生改变C ．经过s t34=，a 车与 b 车速度相等D ．2s 内a 、b 两车路程相等10.如图所示，A 球、B 球质量均为 m ，A 、B 两球用弹簧 k 1 连接处于静止状 态，B 球与地面用弹簧 k 2 连接并用绳系在B 与地面之间使弹簧 k 2 处于 压缩状态，此时绳上的拉力大小为 2mg ，则将绳剪断瞬间，A 、B 两球的 加速度为（ ）A. a A =2gB. a A =0C. a B =2gD. a B =g11.如图所示，水平细杆上套一环A ，环A 与球B 间用一轻质绳相连，质量分别 为0.2kg 、0.4kg ，由于B 球受到风力作用，A 与B 球一起向右匀速运动．已知细绳与竖直方向的夹角为θ=370．（已知6.037sin 0=，8.037cos 0=，2/10s m g =）则下列说法中正确的是（ ）A ．水平风力大小为3NB ．B 球受绳子拉力为7NC. 杆对A 球的支持力随着风力的增加而增加木块始终保持静止状态，下列说法中正确的是（D ．若将力F 竖直向上作用在斜面上，使斜面体竖直向上做匀加速运动，则无论加速度多大，小物块一定不会相对斜面滑动二、填空及实验（共14分）13.（4分）如图为“探究求合力的方法”的实验装置． PO 为橡皮条，两个测力计通过绳子拉橡皮条一端，使 橡皮条水平。

) 2 ()3 8 AB 25)4 CA 0B12D35 .3 .3 、6 CABD881646 为 A B CD B 8 ( ) 2&23 CABD3 3339 B俯视侧1视正视C.耳5D. 75许昌三校联考高一上学期第四次考试理科数学试卷7.过点(1，2)，且与原点距离最大的直线方程是()已知点A(2,-3)、 B( -3, -2)直线I 过点P(1,1)，且与线段AB 相交，则直线I 的斜率的取值k 范围是 ()一、选择题(每小题 5分，共12题，共60分)1.设集合 A 二{x|_1 < x w 2},B={ x |0 w x w 4},则 A n B=(A . : 0,2 :B • : 1,2 :C . : 0,4 :D . [1,4 :设l , m 是两条不同的直线，.：<.是一个平面，则下列命题正确的是()A .若 I _ m , m 二 X ，贝U I _B .若 I _ ：- , 1〃 m ，则 m_m// :，则 1〃 m平行线3x 4y -9 =0和6x 8y ^0的距离是|2e xA x V2f(x) = 2，则f(f(2))的值为([Iog 3(x -1), xZ2.△ ABC 是边长为1的正三角形，那么△ ( )ABC 的斜二测平面直观图 .A B C 的面积为(-◎0)内是减函数,A . x 2y-5=03A . k 或 k - -443 1k 一 _ 或 k _ __4 4C.若 I// :- , m 二:工，则 I// mD.若 I// :-, 设 fH ，则工代刃的解集2x y _4 = 0C. x 3y -7 = 0D已知三棱锥的三视图如图所示形，俯视图为等腰直角三角形 设 '为奇函数，且在()C-X0)U (2, + co) .x -2y 3 = 0,其中侧视图为直角三角则此三棱锥的体积等于(-呵一 2) u (0H 2) (-2,0) u (02)3D . k 乞4411.如果实数x、y满足等式X2+ （y —3）2= 1,那么三的取值范围是（）B .（―汽―2 2]D .（―汽―2 2] U [2 2 ,+口x< 0'，若方程.迁J二爲有四个不同的解，，，•,叼（叼+ x2）+ ~T—且贝y ■的取值范围是（）A. I -」B. I ■ ■ ■）C. '：■）D. :-叮丨二、填空题（每小题5分，共4题，共20分）13. 直线x —2y+ 5 = 0与圆x2+ y2= 8相交于A B两点，贝U | AE| = _______ .14. 若一系列函数的解析式相同、值域相同，但其定义域不同，则称这些函数为”同族函数”，那么函数解析式为：二：，值域为」的”同族函数”共有 _________________ 个.15. 已知圆柱的侧面展开图是边长为4和6的矩形，则该圆柱的表面积为 _______________ .16. 一个四面体的所有棱长为2，四个顶点在同一球面上，则此球的表面积为 ___________三、解答题（本大题共6个小题，共70分，解答题应写出文字说明、证明过程或演算步骤.）17. （本小题满分10分）10.如图，已知长方体ABCD -A1B1C1D1中，A^B^4,CC^2,则直线BC1和平面DBB1D1所成的正弦值等于（）CC1A. Q B2 • 2C•亘 D .105 • 103C. 一4 M k 三一A. [2 2,+^)C. [ —2 2, 2 2]_ [lx+ 1112.已知函数1' ■'(I)求Ap| B ；(n)求C U(A U B).已知全集U二R, A = x| 1 <2x：4 , B J x|log3x(I)求Ap| B ；(n)求C U(A U B).18. (本小题满分12 分)已知点| 和直线：—.二二⑴求过点•与直线平行的直线的方程；⑵ 求过厂的中点与垂直的直线的方程.19. (本小题满分12分)点,D为如图，在三棱锥A—BPC中,AP丄PC,ACLBC,M 为AB中PB中点,且厶PMB为正三角形•(I )求证：MD//平面APC;(II )求证：平面ABCL平面APC.20. (本小题满分12分)如图，四棱锥P - ABCD的底面ABCD是边长为2的菱形，.BAD = 60：•已知PB二PD=2, PAL6.(I)证明：PC _ BD(I)若E为PA的中点，求三菱锥P- BCE的体积.21. (本小题满分12分)已知函数f(x)=log4 4x1 kx k R .1(1) 若k =0，求不等式f x .-的解集；2(2) 若f x为偶函数，求k的值.22. (本小题满分12分)已知圆x2y2_2x_4y m=0.(1) 此方程表示圆，求m的取值范围；(2) 若(1)中的圆与直线x 2y -4 0相交于M、N两点，且OM _ ON ( O为坐标原点), 求m的值；(3) 在(2)的条件下，求以MN为直径的圆的方程.高一上学期期末考试理科数学试题答案选择题1 A2 B3 B4 C5 D6 C 7A 8 B 9 A 10 C 11 D 12 D填空题13、「14 、9 15三、解答题17、（I）A J x| _1 ：：：24 +半或4 +律16------------------ 2B =1x|0 ：： x 乞9? ------------------------------ 4APlB 二Cx|O :：: x ：： 2? ------------------------------（n）A U B二「x| —1 ：：：x乞9} ------------------- 7 分C U(A U B)J.x| x _ -1 或x 9? --------------------------- 10 分18、（1）设的方程为烏，将•点的坐标代入得•二：：：所以的方程为^ • ------------------ 6 分⑵设的方程为将•的中点代入得池二1 ,所以的方程为，* 二二--------------------------------- 1219、解(I ) TM为AB中点,D为PB中点，••• MD//AP,又MD平面ABC, AP 平面ABC••• MD〃平面APC ---------- 4 分(n ) •••△PMB为正三角形，且D为PB中点，••• MDL PB.又由(I )知MD//AP,二AP L PB.又已知AP L PC,PBA PC=P••• AP L平面PBC而BC包含于平面PBC,••• AP L BC,又AC L BC而AP A AC=A,二BC L平面APC,又BC平面ABCL20、(1)证明：连接BD,AC交于O点T PB^PD PO —BD分21、 (1) f(x)=log 4 4x 1 , ：log 4 4x 1 --.x 0,即不等式的解集为0,；⑵由于 f x 为偶函数，f -x = f (x)即 log 4 4」-1 - kx = log 4 4x 1 i 亠 kx ,_xx4• 2kx=log 4 4 1 - log 4 41 =log^x4 +1所以k —丄222、(1)方程 x 2 y 2 -2x -4y m =0，可化为(x — 1)2 + (y — 2)2= 5- m •••此方程表示圆， ••• 5— m>0, 即卩 m< 5.x 2+ y 2— 2x — 4y + m= 0, x + 2y — 4= 0,消去 x 得(4 — 2y) + y — 2 x (4 — 2y) — 4y + 0，化简得 5y — 16y +8 = 0.16 y 1+ y 2=丁,即 a+ (4 — 2y 1)(4 — 2y 2)= 0, 将①②两式代入上式得 8=0,解之得m=. 5儿 + 力=半 北1 +尤2 = 4 _ 2儿 + 4 _ 2力 =g由⑵可知」贝UJ4 8、◎，8丿又 ABCD 是菱形 .BD _ AC 而 AC - PO = O BD 丄面 PAC ⑵ 由(1) BD 丄面PAC BD 丄 PC_ 3 PAC —2131川叮八'-----12分4x 12 ,•1二一X 对任意实数x 都成立, 12分设 M(x 1, y <), N X 2, y 2)，则yy =罟.②由 OM L ON 得 y 1y 2+ X 1X 2= 0, •-16 — 8( y 1 + y 2) + 5yy 2= 0. 16 m + 816— 8x + 5x -55••• MN 的中点C 的坐标为 S' PEC —R= \0C\=^又I 5(4\ (81 16-12分•••所求圆的方程为x —5 2+ y—5l= ~r.< 5丿< 5丿5。

河南省许昌市第三高级中学2019-2020学年高一期末考试数学试卷一、单选题（共20题；共40分）1.人们为了书写方便，常常引入“连乘”符号，已知数列的通项公式，若对任意的恒成立，则正整数k=（）A. 5B. 6C. 7D. 82.以下程序的功能是()S＝1；for i＝1：1：10S＝(3^i)*S；endSA. 计算3×10的值B. 计算355的值C. 计算310的值D. 计算1×2×3×…×10的值3.依次连接正六边形各边的中点，得到一个小正六边形，再依次连接这个小正六边形各边的中点，得到一个更小的正六边形，往原正六边形内随机撒一粒种子，则种子落在最小的正六边形内的概率为（）A. B. C. D.4.已知函数f（x）满足，当x∈[1，4]时，f（x）=lnx，若在区间内，曲线g（x）=f（x）﹣ax与x轴有三个不同交点，则实数a的取值范围是（）A. B. C. D.5.f（x）是定义在R上图形关于y轴对称，且在[0，+∞）上是减函数，下列不等式一定成立的是（）A. f[ ]＜f（）B. f[﹣cos60°]＜f（tan30°）C. f[﹣（cos60°）2]≥f（）D. f[﹣sin45°]＞f（﹣3a+2）6.已知f（x）= ，则有关x的方程f2（x）+bf（x）+c=0有5个不等实根的充分条件是（）A. b＜﹣2 且c＞0B. b＜﹣2 且c＜0C. b＜﹣2 且c=0D. b≥﹣2 且c=07.从甲、乙两班分别任意抽出10名学生进行英语口语测验，其测验成绩的方差分别为S12= 13.2，S22=26．26，则A. 甲班10名学生的成绩比乙班10名学生的成绩整齐B. 乙班10名学生的成绩比甲班10名学生的成绩整齐C. 甲、乙两班10名学生的成绩一样整齐D. 不能比较甲、乙两班10名学生成绩的整齐程度8.已知y=f（x）是偶函数，而y=f（x+1）是奇函数，且对任意0≤x≤1，都有f（x）≥0，f（x）是增函数，则a=f（2010），b=f（），c=﹣f（）的大小关系是（）A. b＜c＜aB. c＜b＜aC. a＜c＜bD. a＜b＜c9.设数列是公比为2，首项为1的等比数列，S n是它的前n项和，对任意的，点在( )A. 直线y=2x-1上B. 直线y=x+2上C. 直线y=x-2上D. 直线y=2x+1上10.对于无穷数列，给出下列命题：①若数列既是等差数列，又是等比数列，则数列是常数列.②若等差数列满足，则数列是常数列.③若等比数列满足，则数列是常数列.④若各项为正数的等比数列满足，则数列是常数列.其中正确的命题个数是（）A. 1B. 2C. 3D. 411.已知函数，当时，，若定义在上的有三个不同的零点，则实数的取值范围是（）A. B. C. D.12.两千多年前，古希腊毕达哥拉斯学派的数学家曾经在沙滩上研究数学问题．他们在沙滩上画点或用小石子表示数，按照点或小石子能排列的形状对数进行分类．如下图中实心点的个数5，9，14，20，…为梯形数．根据图形的构成，记此数列的第2013项为a2013，则a2013﹣5=（）A. 2019×2013B. 2019×2012C. 1006×2013D. 2019×100613.函数f（x）= ，若实数a满足f（f（a））=1，则实数a的所有取值的和为（）A. 1B. ﹣C. ﹣﹣D. ﹣214.已知函数f（x）= ，若关于x的方程f（x）=kx2有4个不同的实数解，则k的取值范围是（）A. k≥1B. k＞1C. 0＜k＜1D. 0＜k≤115.如图，在空间四边形中，两条对角线互相垂直，且长度分别为4和6，平行于这两条对角线的平面与边分别相交于点，记四边形的面积为y，设，则（）A. 函数的值域为B. 函数的最大值为8C. 函数在上单调递减D. 函数满足16.若函数y=x2﹣3x﹣4的定义域为[0，m]，值域为[，-4]，则m的取值范围是（）A. （0，4]B. [，-4]C. [,3]D. [，+∞]17.如图，直线与单位圆相切于点，射线从出发，绕着点逆时针旋转，在旋转的过程中，记（），所经过的单位圆内区域（阴影部分）的面积为，记，则下列选项判断正确的是（）A. 当时，B. 对任意，且，都有C. 对任意，都有D. 对任意，都有18.已知函数，则的解集为（）A. B. C. D.19.已知函数，若正实数m，n（）满足，且在区间上的最大值为4，则（）A. B. C. D.20.已知定义域为的函数满足，当时，，设在上的最大值为，且的前项和为，若对任意的正整数均成立，则实数的取值范围为（）A. B. C. D.二、填空题（共10题；共10分）21.在数列{a n}中，a1+2a2++22a3+…2n﹣1a n=（n•2n﹣2n+1）t对任意n∈N*成立，其中常数t＞0．若关于n的不等式+ + +…+ ＞的解集为{n|n≥4，n∈N*}，则实数m的取值范围是________．22.如图，在平面直角坐标系xOy中，点A在x轴正半轴上，∠，设∠AOB ＝，则OA＝________（用表示）；若，则＝________.23.设，，若，则的最小值为________.24.对于函数y=f（x），若存在定义域D内某个区间[a，b]，使得y=f（x）在[a，b]上的值域也是[a，b]，则称函数y=f（x）在定义域D上封闭，如果函数f（x）= （k≠0）在R上封闭，那么实数k的取值范围是________．25.已知函数f（x）=（2﹣x）e x﹣ax﹣a，若不等式f（x）＞0恰好存在两个正整数解，则实数a的取值范围是________．26.在锐角△ABC中，角A、B、C的对边分别为a、b、c，若a2=b2+bc，则的取值范围是________．27.定义在R上的函数f（x），若对任意x0=x1﹣x2且x1≠x2，若对任意的x1，x2，都有＜0，则称函数f（x）为“T函数”，给出下列函数：（1）y=e﹣3x﹣x；（2）y=﹣x3+3x﹣3x+1；（3）y= ；（4）y=﹣x﹣sinx．其中“T函数”的个数________．28.已知函数，，若对任意，，当时都有，则实数b的取值范围为________．29.已知分别是定义在R上的奇函数和偶函数，且．若存在，使得等式成立，则实数a的取值范围是________．30.设函数，则使得成立的x的取值范围是________.三、解答题（共5题；共50分）31.一个正四棱台的斜高是12cm，侧棱长是13cm，侧面积是720cm2．求它的上、下底面的边长．32.已知抛物线y＝－x2－2x＋3与x轴交于A、B两点，点M在此抛物线上，点N在y轴上，以A、B、M、N为顶点的四边形为平行四边形，求点M的坐标.33.如图，半径为4m的水轮绕着圆心O逆时针做匀速圆周运动，每分钟转动4圈，水轮圆心O距离水面2m，如果当水轮上点P从离开水面的时刻（P0）开始计算时间．（1）将点P距离水面的高度y（m）与时间t（s）满足的函数关系；（2）求点P第一次到达最高点需要的时间．34.设函数f（x）=log4（4x+1）+ax（a∈R）．（1）若函数f（x）是定义在R上的偶函数，求a的值；（2）若不等式f（x）+f（﹣x）≥mt+m对任意x∈R，t∈[﹣2，1]恒成立，求实数m的取值范围．35.儿童乘坐火车时，若身高不超过1.1 m，则无须购票；若身高超过1.1 m但不超过1.4 m，可买半票；若超过1.4 m，应买全票．试写出一个购票算法程序．答案解析部分一、单选题1.【答案】A2.【答案】B3.【答案】B4.【答案】D5.【答案】C6.【答案】C7.【答案】A8.【答案】A9.【答案】D10.【答案】C11.【答案】D12.【答案】D13.【答案】C14.【答案】B15.【答案】D16.【答案】C17.【答案】C18.【答案】C19.【答案】B20.【答案】B二、填空题21.【答案】[ ）22.【答案】；23.【答案】24.【答案】（﹣∞，﹣1）∪（1，+∞）25.【答案】26.【答案】（，）27.【答案】328.【答案】29.【答案】[ ，]30.【答案】三、解答题31.【答案】解：设上底面边长为xcm，下底面边长为ycm，由题意得；即，解得；所以上底面边长为10cm，下底面边长为20cm．32.【答案】解:根据题意求得抛物线与x轴的交点分别为(－3,0)、(1,0)不妨设A(－3,0)、B(1,0)，由已知，设M(a，b)、N(0，n)，根据平行四边形两条对角线互相平分的性质，可得两条对角线的中点重合．按A、B、M、N两两连接的线段分别作为平行四边形的对角线进行分类，有以下三种情况：①若以AB为对角线，可得a＋0＝－3＋1，解得a＝－2；②若以AN为对角线，可得a＋1＝－3＋0，解得a＝－4；③若以BN为对角线，可得a＋(－3)＝1＋0，解得a＝4.因为点M在抛物线上，将其横坐标的值分别代入抛物线的解析式，可得M(－2,3)或M(－4，－5)或M(4，－21)．33.【答案】（1）解：以O为原点建立如图所示的直角坐标系．由于水轮绕着圆心O做匀速圆周运动，可设点P到水面的距离y（m）与时间t（s）满足函数关系，∵水轮每分钟旋转4圈，∴．∴．∵水轮半径为4 m，∴A=4．∴．当t=0时，y=0．∴．∴．（2）解：由于最高点距离水面的距离为6，∴．∴．∴．∴t=5+15k（k∈Z）．∴当k=0时，即t=5（s）时，点P第一次达到最高点．34.【答案】（1）解：由函数f（x）是定义在R上的偶函数，得f（x）=f（﹣x）恒成立，则，∴，∴（2a+1）x=0恒成立，则2a+1=0，故（2）解：= ．当且仅当x=0时取等号，∴mt+m≤1对任意t∈[﹣2，1]恒成立，令h（t）=mt+m，由，解得，故实数m的取值范围是．35.【答案】解：程序如下：INPUT“身高h(h>0)”；hIF h<＝1.1THENPRINT“免费乘车”ELSEIF h<＝1.4THENPRINT“半票乘车”ELSEPRINT“全票乘车”END IFEND IFEND。

许昌市三校联考高一上学期第三次考试英语试卷（考试时间：120分钟总分：150分）第一部分阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A、B、C、和D）中，选出最佳选项，并在答题卡上将该项涂黑。

AIf Confucius（孔子）were still alive today and could celebrate his September 28 birthday with a big cake, there would be a lot of candles. He’d need a fan or a strong wind to help him put them out. While many people in China will remember Confucius on his special day, few people in the United States will give him a passing(短暂的) thought. It’s nothing personal. Most Americ ans don’t even remember the birthdays of their own national heroes.But this doesn’t mean that Americans don’t care about Confucius. In many ways he has become a bridge that foreigners must cross if they want to reach a deeper understanding of China.In the past two decades, the Chinese studies programs have gained huge popularity in Western universities. More recently, the Chinese government has set up Confucius Institutes in more than 80 countries. These schools teach both Chinese language and culture. The main courses of Chinese culture usually included Chinese art, history and philosophy（哲学）. Some social scientists suggest that Westerners should take advantages of the ancient Chinese wisdom to make up for the drawbacks of Westerners philosophy. Students in the United States, at the same time, are racing to learn Chinese. So they will be ready for life in a world where China is an equal power with the United States. Businessmen who hope to make money in China are reading books about Confucius to understand their Chinese customers.So the old thinker’s ideas are still alive and well.Today China attracts the West more than ever, and it will need more teachers to introduce Confucius and Chinese culture to the West.As for the old thinker, he will not soon be forgotten by people in the West, even if his birthday is forgotten.1. The opening paragraph is mainly intended to ________A. provide some key facts about ConfuciusB. attract the readers’ interest in the subjectC. show great respect for the ancient thinkerD. prove the popularity of modern birthday celebrations2. We can learn from Paragraph 4 that American students______A. have a great interest in studying ChineseB. take an active part in Chinese competitionsC. try to get high scores in Chinese examsD. fight for a chance to learn Chinese3. What is the best title for the passage?A. Forgotten Wisdom in AmericaB. Huge Fans of the Chinese LanguageC. Chinese Culture for WesternersD. Old Thinker with a Big Future4. The passage is most likely to appear in ___________.A. a biography lessonB. a history paperC. a newspaperD. a philosophy textbookBThe class of eighth grade at a Los Angeles middle school tap their rulers and nod their heads to the rhythm of the rap video shown on a screen. Their algebra（代数）teacher, LaMar Queen, uses rhyme to help them memorize seemingly difficult items to improve their grades.Queen, 26, is now known at Los Angeles Academy Middle School as the rap teacher, but his fame has spread beyond the 2，200student school in this neighborhood. He's won a national award and shown teachers and parents how to use rap to teach children. “Math is a bad word for many students, but if we put it in a form that kids enjoy, they'll learn，” Queen said. “Rap is what t he kids respond to. They don't have a problem memorizing the songs at all.”Queen's math rap came about by chance. Two months after starting at LA Academy, he was frustrated（泄气的）when kids said his class was boring. They thought that he looked like the singer Kanye West, so they challenged him to rap.Queen has been rapping since the seventh grade. With his students' challenge on his mind, he pushed aside work on his lesson plans and wrote a rap song Slope Intercept.Word of his rapping soon reached the school's main office. Principal Maria Borges went to look into it, and came out smiling.” It interests the kids，” she said.” Kids seem to know all the rap songs, and it's helpful for them to remember math rules.” “Some kids who aren't even in Mr Queen's class go around singing his songs，” said Kejon Closure, 13, who went from a C average to an A.In the raps, Queen defines（定义）a math concept and works through sample problems step by step. He keeps a relaxed manner when singing with his students. Queen also tries to inspire them. His lyrics（歌词）tell students “to get As on your papers”，and “to be respectful and listen to your parents”．Sometimes the students appear in the videos as a reward for good grades and behavior. 5．According to the passage, LaMar Queen has been ________ since he began to teach students at Los Angeles Academy.A．a popular teacher B．a music teacherC．a mathematics teacher D．a famous singer6. LaMar Queen has won a national award because ________．A．he is good at composing rap songsB．his new musical production was shown on TVC．he created a unique teaching method in mathD．he gave students a lot of confidence7. This passage may be taken from China Daily in the column of ________．A．society B．educationC．entertainment D．biographyCThere was a famous temple on a high hill in Assam. The priest (神父）of this temple was widely respected. When he grew old, he started searching for a younger priest to replace him after his death. However， much to his disappointment, he could not find any suitable person. As the priest lay on his deathbed, he called his assistant and told him，“After my death, make sure that only a 'human being ，replaces me as the priest of this temple. ” As soon as he said these words, he died.Information soon travelled far and wide. A day was set for the selection of the successor (继任者）.That day, many people started coming to the temple. The route to the temple was full of thorns (荆棘）and stones. By the time most people reached the temple, they had had cuts on their feet and hands.By afternoon, as the selection process was about to end, one young man walked slowly into the temple. The assistant noticed him and said，"Young man, you are very late. ” The man replied, “I know I am late, so I will not participate in the competition. ” But the assistant was curious to know about this man. He asked again, “But how did you hurt yourself so badly? Didn't you follow the same route as the others?”“I did,“replied the man, “But I thought I must remove the thorns and sharp stones from the path so that when people come to pray in this temple they won't get hurt. ”Hearing this, the assistant smiled and said, "Congratulations, you have been chosen.，’ This greatly annoyed the other participants(参加者). “ How can you choose him as he has not gone through any of the tests?” they demanded.The assistant replied, “Our old priest used to say that only a human being, knows how to watch for other people's interests. All of you climbed the same path. But only this man thought about the others and cleared the path so that no one would get hurt. By this definition (定义），only he qualifies as a human being ，and therefore only he should be the successor to the great old priest. "8. What happened immediately after the old priest died?A. The assistant took the old priest's position.B. Many people climbed to the temple to say goodbye to him.C. The assistant shut the temple and started mourning for the old priest.D. Word spread that there was an urgent need for the old priest's successor.9. Why was the young man late for the competition?A. He had forgotten his way and thus had taken a longer route to the temple.B. He removed all the sharp stones and thorns from the path leading to the temple.C. He offered water to the other participants on their way to the temple.D. He had forgotten all about the competition in the morning.10. How did the other participants react to the assistant's decision?A. They took it easy and learnt a lesson from the incident.B They praised the assistant for being so wise and fair.C. They got angry and opposed the assistant's decision.D. They felt ashamed of themselves and left immediately.11. What was the old priest's definition of a "human being”？A. Someone who cares about other people's interests.B. Someone who succeeds in life and earns a lot of money.C. Someone who removes sharp stones that block the path.D. Someone who knows how to make other people enjoy life.DRyan Jones is only eight months old, but he is already “talking” with his parents. When lie is hungry, he opens and closes his hand. This means milk. He also knows the signs for his favorite toy and the word more.Ryan is not deaf, and his parents are not deaf, but his mother and father are teaching him to sign. They say a word and make a sign at the same time. They repeat this again and again. When Ryan learns a new sign, his family is very excited. Ryan’s parents think that he will be a happier baby because he can communicate with them.Ryan’s pa rents are teaching Ryan to sign because of a man named Joseph Garcia. Although Garcia was not from a deaf family, he decided to learn American Sign Language (ASL). First, he took courses in ASL. Then he got a job helping deaf people communicate with hearing people. In his work, he saw many deaf parents sign to their infants. He noticed that these babies were able to communicate much earlier than hearing children.Garcia decided to try something new. He taught ASL to parents who were not deaf. The families started to teach signs to their infants when they were six or seven months old. These babies started using signs about two months later.Some people worry about signing to babies. They are afraid that these babies won’t feel a need to talk. Maybe they will develop spoken language later than other babies. However, research does not show this. In fact, one study found just the opposite. Signing babies actually learned to speak earlier than other children. As they grow older, these children are moreinterested in books. They also score higher on intelligence tests.There is still a big question for parents: Which are the best signs to teach their babies? Some parents make their own signs. Other parents want to teach ASL. It can be useful because manypeople u nderstand it. There’s no clear answer, but we do know this: All signing babies and their families are talking quite a lot!12. Why are Ryan’s parents teaching him to sign?A. They want to understand him.B. They are doing an experiment.C. He prefers signing to speaking.D. He was born with hearing problems.13. What caused Garcia to start his courses?A. His dream to make ASL popular.B. His wish to help as many deaf people as possible.C. His discovery that parents fail to communicate earlier.D. His concern that parents fail to communicate with their babies.14. Some people say no to signing to babies because they think_____.A. it is bad for their intelligence.B. it is a waste of time and money.C. it will slow their language developmentD. it will affect their future school performance.15. What would be the best title for the text?A. American Sign LanguageB. Children developmentC. Global languageD. Baby talk第二节（共5小题，每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项，选项中有两项为多余选项。

绝密★启用前2015-2016学年河南省许昌市三校高一上学期第三次考试数学试卷（带解析）试卷副标题考试范围：xxx ；考试时间：169分钟；命题人：xxx学校:___________姓名：___________班级：___________考号：___________注意事项．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（题型注释）1、偶函数在上单调递增，则与的大小关系是（ ） A ． B ． C ． D ．2、如图，长方体ABCD －A 1B 1C 1D 1中，AA 1＝AB ＝2，AD ＝1，E ，F ，G 分别是DD 1，AB ，CC 1的中点，则异面直线A 1E 与GF 所成角为（ ）A ．B ．C ．D ．3、给出下列命题： （1）直线a 与平面不平行，则a 与平面内的所有直线都不平行； （2）直线a 与平面不垂直，则a 与平面内的所有直线都不垂直；（3）异面直线a 、b 不垂直，则过a 的任何平面与b 都不垂直； （4）若直线a 和b 共面，直线b 和c 共面，则a 和c 共面 其中错误命题的个数为（ ） A ．0B ．1C ．2D ．34、在一个锥体中，作平行于底面的截面，若这个截面面积与底面面积之比为1∶3， 则锥体被截面所分成的两部分的体积之比为（ ） A ．1∶B ．1∶9C ．1∶D ．1∶5、如图所示是一个几何体的三视图，则该几何体的体积为（ ）A ．B ．C ．D ．6、长方体的一个顶点上三条棱长分别是，且它的个顶点都在同一球面上则这个球的表面积是（ ）A ．B ．C ．D ．都不对7、直线平面，，那么过点且平行于直线a 的直线 （ ）A ．只有一条，不在平面内B ．有无数条，不一定在内C ．只有一条，且在平面内D ．有无数条，一定在内8、如图是正方体的平面展开图，则在这个正方体中：①与平行． ②与是异面直线． ③与垂直． ④与是异面直线．以上四个命题中正确的个数是（ ） A ．1B ．2C ．3D ．49、是空间三条不同的直线，则下列命题正确的是（ ）A ．B ．C ．D ．10、如图是一个平面图形的直观图，斜边，则该平面图形的面积是（ ）A ．B ．1C ．D ．211、若直线不平行于平面，则下列结论成立的是（ ）A ．内所有的直线都与异面B ．内不存在与平行的直线C ．内所有的直线都与相交D ．直线与平面有公共点12、已知集合，，则=（ ）A ．B ．C ．D ．第II卷（非选择题）二、填空题（题型注释）13、将边长为2，锐角为60°的菱形ABCD沿较短对角线BD折成四面体ABCD，点E、F分别为AC、BD的中点，则下列命题中正确的是．（将正确的命题序号全填上）①EF∥AB②EF与异面直线AC与BD都垂直③当四面体ABCD的体积最大时，AC=④AC垂直于截面BDE14、已知函数在[-1,）上是减函数, 则实数a的取值范围是_______________15、若不等式对任意恒成立，则的取值范围是16、在长方体中，分别为的中点，则直线与平面的位置关系是_____________．三、解答题（题型注释）17、已知（1）求的值；（2）当（其中，且为常数）时，是否存在最小值，如果存在求出最小值；如果不存在，请说明理由； （3）当时，求满足不等式的的范围．18、二次函数f （x ）满足且f （0）=1．（1）求f （x ）的解析式； （2）若在区间上, 不等式f （x ）2x+m 恒成立，求实数m 的范围．19、如图，在四棱锥中，底面，，，是的中点．（Ⅰ）求和平面所成的角的大小； （Ⅱ）证明平面； （Ⅲ）求二面角的正弦值20、如图，多面体AED-BFC 的直观图及三视图如图所示，M 、N 分别为AF 、BC 的中点．（Ⅰ）求证：MN ∥平面CDEF ； （Ⅱ）求多面体A-CDEF 的体积； （Ⅲ）求证：．21、如图，ABCD 与ADEF 为平行四边形，M ，N ，G 分别是AB ，AD ，EF 的中点．求证：（1）BE ∥平面DMF ； （2）平面BDE ∥平面MNG ．22、如图是古希腊数学家阿基米德的墓碑文，墓碑上刻着一个圆柱，圆柱内有一个内切球，这个球的直径恰好与圆柱的高相等．相传这个图形表达了阿基米德最引以自豪的发现．我们来重温这个伟大发现：（1）求圆柱的体积与球的体积之比； （2）求圆柱的表面积与球的表面积之比．23、（1）计算：（2）已知简单组合体如图，试画出它的三视图（尺寸不作严格要求）参考答案1、D2、D3、D4、D5、A6、B7、C8、B9、B10、D11、D12、A13、②，③，④14、15、16、平行17、（1）0；（2）时，无最小值．（3）18、（1）f（x）=x2-x+1．（2）19、见解析20、（Ⅰ）见解析；（Ⅱ）；（Ⅲ）见解析21、见解析22、见解析23、见解析【解析】1、试题分析：由题意，偶函数在上单调递增，则，．由，得考点：含绝对值对数函数的函数性质，可借助图像数形结合．2、试题分析：如图，连接B1G,B1F ．则异面直线A1E与GF所成角，为∠B1GF．△B1GF中，得∠B1GF=考点：异面直线所成角的算法．3、试题分析：错误命题为（1），（2），（4）（可通过摆放模型制造反例）考点：线与线，线与面的位置关系的判断．会举反例．4、试题分析：由题意可知：面积比为1∶3，则高的比为1∶，所截得的锥体与原锥体的体积比为：1∶．所以锥体被截面所分成的两部分的体积之比为1∶考点：几何体的面积比与相似比和体积比．5、试题分析：由三视图可知为两个半圆柱和一个长方体组成．则体积为考点：平面的性质6、试题分析：由题球的半径为．则考点：长方体与外接球的表面积．7、试题分析：由线与面平行的性质可得C正确．考点：线与面平行的性质8、试题分析：由展开图③，④为正确．（可将图复原为正方体观察可得）．考点：展开与折叠问题，考查空间想象能力．9、试题分析：A含义为垂直与同一直线的两直线平行．（空间中易得反例）排除法得B正确．考点：空间中线与线的位置关系．10、试题分析：，则由斜二测画法，．所以：考点：斜二测画法．11、试题分析：线与面不平行，则相交或在平面内．排除法D正确．考点：线与平面的位置关系．12、试题分析：由题可先解出集合（注意集合M中的限制条件取整数），则．考点：二次不等式的解法和交集运算．13、试题分析：将边长为2，锐角为60°的菱形ABCD沿较短对角线BD折成四面体ABCD，如图，对于①，由图可知，EF与AB异面，故①错误；对于②，∵点E、F分别为AC、BD的中点，AB=BC，DC=DA，∴BE⊥AC，DE⊥AC，BE∩DE=E，∴AC⊥平面BDE，EF?平面BDE，∴AC⊥EF；又BE=DE，F为BD的中点，∴EF⊥BD；∴EF是异面直线AC与BD的公垂线，故②正确；对于③，由②知，AC⊥平面BDE，故④正确；③当四面体ABCD的体积最大时，面ADB垂直面CDB,则AC=综上所述，正确命题的序号是②③④．考点：展开与折叠问题，考查空间想象能力．14、试题分析：令：，因为是减函数，则为增函数．又，．综上考点：复合函数的单调性．15、试题分析：可转化为，由，所以当时考点：恒成立问题可化为二次函数的最值问题．16、试题分析：如图：取A′C′中点O，连结NO，BO，∵M，N分别为AB，A′D′的中点，BONM是平行四边形，∴MN∥BO，∵MN不包含于平面A′BC′，BO⊂平面A′BC′，∴直线MN∥平面A′BC′．考点：线面平行的判定定理．17、试题分析：（1）由问题观察，联系函数的奇偶性可得．（2）求给定区间的最值问题，可化为考查函数在区间上的单调性来解决．由于，中a值不确定，需分类讨论．（3）题目明确了，可借助函数的单调性和奇偶性来解不等式．试题解析：理科[解：（1）由所以f（x）的定义域为：（－1，1），∴f（x）为奇函数，则：．∴（2）设，则∵，∴，∴，当时，在上是减函数，又∴时，有最小值，且最小值为当时，在上是增函数，又∴时，无最小值．（3）由（1）得所以所以X的取值范围是考点：（1）函数的奇偶性．（2）复合函数的单调性及分类思想：（3）函数奇偶性和单调性的综合运用．18、试题分析：（1）可利用待定系数法．已知为二次函数可先设函数解析式，先设，再结合题中的两个条件建立方程求出a,b．（2）恒成立问题．本题由（1）知二次函数对称轴确定，而区间含有参数t，通过以对称轴为界讨论t的取值，从而化为最值问题，找出m的取值范围．试题解析：解：（1）设f（x）=ax2+bx+c，由f（0）=1得c=1，故f（x）=ax2+bx+1∵f（x+1）-f（x）=2x,∴a（x+1）2+b（x+1）+1-（ax2+bx+1）=2x．即2ax+a+b=2x,所以,∴f（x）=x2-x+1．（2）由题意得x2-x+1>2x+m在[t,t+2]上恒成立．即x2-3x+1-m>0在[t,t+2]上恒成立．设g（x）= x2-3x+1-m,其图象的对称轴为直线x=，当t>1．5时，g（x）最小值为g（t）= t2-3t+1-m>0此时，m< t2-3t+1当时，g（x）最小值为g（1．5）= >0此时，当时，g（x）最小值为g（t+2）= t2+t-1-m>0此时m< t2+t-1综上：考点：（1）求函数解析式及方程思想．（2）恒成立问题：即转化为二次函数轴定区间动的最值问题．19、试题分析：（1）线面角的问题，结合图形需找出PB上的点与面PAD的垂线，回到线面角的定义可求出．（2）证线面垂直，基本思路是找线AE与面PCD中的两条相交直线垂直，可证．（3）二面角的问题，一个方法是回到定义．或是利用三垂线定理找出．本题可结合条件利用定义EM PD，AM PD．从而找到并证出．试题解析：（Ⅰ）解：在四棱锥中，因底面，平面，故．又，，从而平面．故在平面内的射影为，从而为和平面所成的角．在中，，故．所以和平面所成的角的大小为．（Ⅱ）证明：在四棱锥中，因底面，平面，故．CD CA,所以CD平面PAC, 所以CD AE,AE PC,所以AE平面PCD,（Ⅲ）过E作EM PD,连结AM,则AM PD,所以∠AME即二面角的平面角，设PA=,AE=，在ABCD中所以AD=在中，，考点：（Ⅰ）线面角的算法．（Ⅱ）线面垂直的证明．（Ⅲ）二面角的算法．20、试题分析：（1））欲证线与面平行可结合题中的中点条件，找线MN与面中的线EC平行得证．（2）本题只给出了三视图，条件没有直接给出，需要通过三视图读出，求四棱锥的体积，需知底面积和高．（3）欲证，看图需转化为证AF与CE所在的平面CEB垂直可得．试题解析：（Ⅰ）证明：由多面体AED-BFC的三视图知，三棱柱AED-BFC中，底面DAE是等腰直角三角形，DA=AE=2,DA平面ABEF，侧面ABEF,ABCD都是边长为2的正方形，连结EB，则M是EB的中点，在中，MN∥EC,且EC平面CDEF, MN平面CDEF, 所以MN∥平面CDEF （Ⅱ）（Ⅲ）,DA∥BC, ,,因为面ABEF是正方形，,,考点：（Ⅰ）线面平行的证法．（Ⅱ）锥体体积的算法．（Ⅲ）线与线垂直的证法．21、试题分析：（1）欲证线面平行常转化为找线与面中的一条直线平行．本题中可结合题中的中点条件，找线BE与面中的线MO平行得证．（2）证面面平行，需运用面与面平行的判定找线与面平行，利用中点条件找出两条相交直线DE和BD与面BDE平行得证．试题解析：（1）如图，连接AE，则AE必过DF与GN的交点O，连接MO，则MO为△ABE的中位线，所以BE∥MO，又BE⊄平面DMF，MO⊂平面DMF，所以BE∥平面DMF．（2）因为N，G分别为平行四边形ADEF的边AD，EF的中点，所以DE∥GN，又DE⊄平面MNG，GN⊂平面MNG，所以DE∥平面MNG．又M为AB中点，所以MN为△ABD的中位线，所以BD∥MN，又BD⊄平面MNG，MN⊂平面MNG，所以BD∥平面MNG，又DE与BD为平面BDE 内的两条相交直线，所以平面BDE∥平面MNG．考点：线面平行与面面平行的判定方法．22、试题分析：（1）本题关键是找出圆柱和球体中高和半径的数量关系：即：h=2R,r=R可求出比．（2）结合（1）中的数量关系，通过圆柱和球的表面积公式可得．试题解析：解：（1）（2）考点：球与圆柱的内切及体积和表面积的算法23、试题分析：（1）由题是对运算能力的考察：主要有对数，指数和根式运算．可通过对数的运算性质和指数运算最后化为基本运算．（2）由直观图为三个底面不同的圆柱组成，通过三视图的定义易得．试题解析：（1）（2）考点：（1）指数和对数的运算．（2）三视图的画法．。

绝密★启用前2015-2016学年河南许昌市三校高一上学期期末联考语文试卷（带解析）试卷副标题考试范围：xxx ；考试时间：106分钟；命题人：xxx学校:___________姓名：___________班级：___________考号：___________注意事项．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（题型注释）1、下列各句中，没有语病的一句是（ ）A ．高速公路上发生交通事故的主要原因是司机违反交通规则或操作不当造成的，交通部门应加强安全宣传，增强司机的安全意识。

B ．心理学家认为，给孩子讲故事有助于培养孩子的创造性思维和语言表达水平。

C ．巴金先生出生于成都一个封建官僚家庭，学生时代接受了五四运动的洗礼，参加过反封建的社会活动。

D ．由于延迟退休或实行弹性退休制、改革事业单位养老保险制度等重大变革，对广大群众来说并不是一下子就能愉快接受的，需要政府部门做好宣传解释工作。

2、依次填入下列各句横线处的成语，最恰当的一组是（ ）①我也是爽快的脾气，我们不说废话了，你先提出你的“合作”条件来，要是可以商谈的话，我一定 回答你。

②谷敬文为了补救自己的失言，便换了一副 的样子说：“南屏山之事，我敢说，我试卷第2页，共11页比旅长还要焦急得多。

”③他们不但是合伙人，更有着 、荣辱与共的亲密关系。

改革开放后，他们共同办企业，更加志同道合，许多事情都能心照不宣，配合默契。

A ．开诚布公推心置腹肝胆相照 B ．开诚布公肝胆相照推心置腹 C ．推心置腹开诚布公肝胆相照 D ．推心置腹肝胆相照开诚布公3、依次填入下面一段文字横线处的语句，衔接最恰当的一组是（ ）没有谁可以活在社会舆论的真空里，“贫二代”如此，“拼二代”亦如此，“富二代”更不例外。

， 。

当然，这其中不乏误解与偏见。

， ，只是，“富二代”的代言人在为这个群体正名的时候， 。

许昌三校联考高一上学期第四次考试英语试卷第一部分阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项。

AAre there any foreigners in your city? How do you address(称呼) them in Englis h when you want to talk to these foreigners? The following may be some simple rules the beginners should follow.1.When talking to strangers there is often no special form of address in English. In most cases,if you want to catch the attention of a stranger it is necessary to use such phrases as “Excuse me”.2.In British English “Sir” and “Madam” are considered to be too formal for most situations.They are used mostly to customers in shops or restaurants .While in American English “Sir”and “Madam” are not so formal and are commonly used between strangers, especially with older peop le whose names you don’t know.3.When you talk to some people you know, you can use their names directly. If you are friends,use their first names; if your relationship is more formal, use “Mr.”, “Mrs.”, “Ms.”, etc, before their family names.4.There are many other forms of address which can be used between friends and strangers. However,many of these are limited in use. For example, “pal” and “mate” can be used between strangers, but are usually only used by men talking to other men.If you meet foreigners, do remember not to address them in wrong ways.1. According to the passage, “Excuse me” is mainly used to _______.A. address a person you don’t knowB. apologize to othersC. catch the attention of a strangerD. suggest good manners2. If John Smith is your best friend, according to the passage you should often call him______.A. JohnB. SmithC. Mr. JohnD. Mr. Smith3. What is the title of this passage?A. Foreigners in Your CityB. How to Introduce OthersC. Politeness EducationD. How to Address ForeignersBDo you want to get home from work knowing you have made a real difference in someone's life? If yes, don't care about sex or age! Come and join us, then you'll make it!Position: Volunteer Social Care Assistant(No Pay with Free Meals)Place: Manchester Hours: Part TimeWe are now looking for volunteers to support people with learning disabilities to live active lives! Only 4 days left. Don't miss the chance of lending your warm hands to help others!Role:You will provide people with learning disabilities with all aspects of their daily lives. You will help them to develop new skills. You will help them to protect their rights and their safety. But your primary concern is to let them know they are valued.Skills and Experience Required:You will have the right values and great listening skills. You will be honest and patient. You will have the ability to drive a car and to communicate in fluent written a nd spoken English since you'll have to help those people with different learning disabilities. Previous care-related experience will be a great advantage for you.4. The text is meant to ______ .A. leave a noteB. send an invitationC. present a documentD. carry an advertisement5. What does the underlined part in Paragraph 1 mean?A. You'll make others' lives more meaningful with this job.B. You'll arrive home just in time from this job.C. You'll earn a good salary from this job.D. You'll succeed in getting this job.6. The volunteers' primary responsibility is to help people with learning disabilities ______.A. to get some financial supportB. to properly protect themselvesC. to lean some new living skillsD. to realize their own importance7. Which of the following can first be chosen as a volunteer?A. The one who can drive a car.B. The one who has done similar work before.C. The one who has patience to listen to others.D. The one who can use English to communicate.CMost people know that Marie Curie was the first woman to win the Nobel Prize, and the first person to win it twice. However, few people know that she was also the mother of a Nobel Prize winner.Born in September, 1987, Irene Curie was the first of the Curies’ two daughters. Along with nine other children whose parents were also famous scholars, Irene studied in their own school, and her mother was one of the teachers. She finished her high school education at the College of Sévigné in Paris.Irene entered the University of Paris in 1914 to prepare for a degree in mathematics and physics. When World War I began, Irene went to help her mother, who was using X-ray facilities （设备） to help save the lives of wounded soldi ers. Irene continued the work by developing X-ray facilities in military hospitals in France and Belgium. Her services were recognized in the form of a Military’s Medal by the French government.In 1918, Irene became her mother’s assistant at the Curie Institute. In December 1924, Frederic Joliot joined the Institute, and Irene taught him the techniques required for his work. They soon fell in love and were married in 1926. Their daughter Helene was born in 1927 and their son Pierre five years later.Like her mother, Irene combined family and career. Like her mother, Irene was awarded a Nobel Prize, along with her husband, in 1935. Unfortunately, also like her mother, she developed leukemia because of her work with radioactivity（辐射能）. Irene Joli ot-Curie died from leukemia on March 17, 1956.8. Why was Irene Curie awarded a Military Medal?A. Because she received a degree in mathematics.B. Because she contributed to saving the wounded.C. Because she won the Nobel Prize with Frederic.D. Because she worked as a helper to her mother.9. Where did Irene Curie meet her husband Frederic Joliot?A. At the Curie Institute.B. At the University of Paris.C. At a military hospital.D. At the College of Sevigne.10.When was the second child of Irene Curie and Frederic Joliot born?A. In 1932.B. In 1927.C. In 1897.D. In 1926.11.In which of the following aspects was Irene Curie different from her mother ?A. Irene worked with radioactivity.B. Irene combined family and career.C. Irene won the Nobel Prize onceD. Irene died from leukemia.DHoney from the African forest is not only a kind of natural sugar, it is also delicious. Most people, and many animals, like eating it. However, the only way for them to get that honey is to find a wild bees' nest and take the honey from it. Often, these nests are high up in trees, and it is difficult t o find them. In parts of Africa, though, people and animals looking for honey have a strange and unexpected helper----a little bird called a honey guide.The honey guide does not actually like honey, but it does like the wax in the beehives (蜂房). The little bird cannot reach this wax, which is deep inside the bees' nest. So, when it finds a suitable nest, it looks for someone to help it. The honey guide gives a loud cry that attracts the attention of both passing animals and people. Once it has their attention, it flies through the forest, waiting from time to time for the curious animal or person as it leads them to the nest. When they finally arrive at the nest, the follower reaches in to get at the delicious honey as the bird patiently waits and watches. Some of the honey, and the wax, always falls to the ground, and this is when the honey guide takes its share.Scientists do not know why the honey guide likes eating the wax, but it is very determined in its efforts to get it. The birds seem to be able to smell wax from a long distance away. They will quickly arrive whenever a beekeeper is taking honey from his beehives, and will even enter churches when beeswax candles are being lit.12. Why is it difficult to find a wild bees' nest?A. It's small in size.B. It's hidden in trees.C. It's covered with wax.D. It's hard to recognize.13. What do the underlined words "the follower" in Paragraph 2 refer to?A. A bee.B. A bird.C. A honey seeker.D. A beekeeper.14. The honey guide is special in the way .A. it gets its foodB. it goes to churchC. it sings in the forestD. it reaches into bees' nests15. What can be the best title for the text?A. Wild BeesB. Wax and HoneyC. Beekeeping in AfricaD. Honey-Lover's Helper第二节（共5小题；每小题2分，满分10分）Public Speaking TrainingGet a coach__16__，so get help. Since there are about a billion companies out there all ready to offer you public speaking training and courses, here are some things to look for when deciding the training that's right for you.Focus on positivesAny training you do to become more effective at public speaking should always focus on the positive aspects of what you already do well. Nothing can hurt confidence more than being told that you aren't doing well. __17__， so good public speaking training should develop those instead of telling you what you shouldn't do.__18__If you find a public speaking course that looks as though it's going to give you lots of dos and don'ts, walk away! Your brain is so full of what you're going to be talking about. __19__. As far as we're concerned, there are basically no hard and fast rules about public speaking. Your audience can be your friends.You are a special person not a cloneMost importantly, good public speaking training should treat you as a special one, with your own personal habits. __20_. Your training course should help you bring out your personality, not try to turn you into someone you're not.A．You aren't like anybody elseB．You already do lots of things wellC．Turn your back on too many rulesD．Check the rules about dos and don'tsE．Whatever the presentation, public speaking is toughF．The one thing you don't want is for them to fall asleepG．So trying to force a whole set of rules into it will just make things worse第二部分英语知识运用（共两节，满分75分）第一节（一）完形填空（共 20 小题；每小题 1.5 分，共 30 分）第二节阅读下面短文，掌握其大意，从每题所给的 A、B、C、D 四个选项中，选出最佳选项，并在答题卡上将该项涂黑。