(2021年整理)2019学年镇海中学高三上期中
2019届浙江省宁波市镇海中学高三上学期期中考试数学试卷及解析
2019届宁波市镇海中学高三上学期期中考试数学试卷注意事项：1．答题前，先将自己的姓名、准考证号填写在试题卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。
2．选择题的作答：每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，写在试题卷、草稿纸和答题卡上的非答题区域均无效。
3．非选择题的作答：用签字笔直接答在答题卡上对应的答题区域内。
写在试题卷、草稿纸和答题卡上的非答题区域均无效。
4．考试结束后，请将本试题卷和答题卡一并上交。
一、单选题1．设全集，集合，则集合A． B．C． D．2．某几何体的三视图如图所示，图中的四边形都是边长为的正方形，两条虚线互相垂直，则该几何体的体积是A． B． C． D．3．记为等差数列的前项和,若, 则A． B． C． D．4．4．满足线性约束条件23,23,{0,x yx yxy+≤+≤≥≥的目标函数z x y=+的最大值是A．1 B．32C．2 D．35．已知函数，则函数的图象为A． B．C． D．6．若、是两个相交平面，则在下列命题中，真命题的序号为①若直线，则在平面内一定不存在与直线平行的直线．②若直线，则在平面内一定存在无数条直线与直线垂直．③若直线，则在平面内不一定存在与直线垂直的直线．④若直线，则在平面内一定存在与直线垂直的直线．A．①③ B．②③ C．②④ D．①④7．已知，那么A． B． C． D．8．已知正项等比数列满足，若存在两项，使得，则的最小值为A． B． C． D．9．已知双曲线的左右焦点分别为，为双曲线上一点，为双曲线渐近线上一点，均位于第一象限，且，，则双曲线的离心率为A． B． C． D．10．如图，在三棱柱中，底面为边长为的正三角形，在底面的射影为中点且到底面的距离为，已知分别是线段与上的动点，记线段中点的轨迹为，则等于(注：表示的测度，本题中若分别为曲线、平面图形、空间几何体，分别对应为其长度、面积、体积）。
2019学年镇海中学高三上期中
2019学年镇海中学高三上期中一、选择题：每小题4分，共40分1. 已知集合{}2450A x x x =∈--≤Z ，{}0ln 2B x x =<<，则A B 的元素的个数（ ）A ．2B ．3C ．4D ．72. 若a ，b ，c ∈R ，且a b >，则下列不等式中一定成立的是（ ）A ．ac bc >B ．()20a b c ->C ．11a b<D ．22a b -<-3. 已知n S 是等差数列{}n a 的前n 项和，且24S =，418S =，则6S 等于（ ）A ．50B ．42C ．38D ．364. 函数()243xx f x =的图象大致为（ ）5. 如图是一个几何体的三视图，则这个几何体的表面积是（ ）A ．76B ．84C．76+D．84+6. 将函数()y f x =的图象向右平移6π个单位长度后，得到 ()sin 26g x x π⎛⎫=+ ⎪⎝⎭，则()y f x =的函数解析式为（ ）A ．()cos 2f x x =-B ．()sin 26f x x π⎛⎫=- ⎪⎝⎭C ．()cos 2f x x =D ．()cos 26f x x π⎛⎫=- ⎪⎝⎭7. 设命题p ：()lg 210x -≤，命题q ：()10x a x a-+≤-，若q 是p 的必要不充分条件，则实数a 的取值范围是（ ）A ．10,2⎡⎤⎢⎥⎣⎦B ．10,2⎛⎫ ⎪⎝⎭C ．10,2⎡⎫⎪⎢⎣⎭D ．∅BD俯视图侧视图正视图8. 已知22ππαβ-<-<，sin 2cos 1αβ-=，cos 2sin αβ+=sin 6πβ⎛⎫+= ⎪⎝⎭（ ）ABC． D． 9. 已知椭圆和双曲线有相同的焦点1F ，2F ，设点P 是该椭圆和双曲线的一个公共点，且123F PF π∠=，若椭圆和双曲线的离心率分别为1e ，2e ，则2212e e +的最小值为（ ）AB．4+C．2+ D．110. 设a ，b 为正实数，且121322a b a b +++=，则12a b +的最大值与最小值之和为（ ）A ．2B ．92C ．132D ．9二、填空题：单空题每题4分，多空题每题6分11. 抛物线22y x =的焦点坐标为 ，准线方程为 ．12. 已知点()1,0A ，()0,2B ，点(),P a b 在线段AB 上，则直线AB 的斜率为 ， ⋅a b 的最大值为 ．13. 若实数(),x y 满足约束条件20201x y x y y --≤⎧⎪+-≥⎨⎪≤⎩，则2x y -的最小值为 ，的最小值为 ．14. 已知长方体1111ABCD A B C D -中，112AA =，1AB =，AD ，则直线1AA 与平面1A BD 所成的角为 ，若空间的一条直线l 与直线1AA 所成的角为4π，则直线l 与平面1A BD 所成的最大角为 ．15. 已知{}n a 是等比数列且0n a >，243546225a a a a a a ++=，则4a 的最大值为 ． 16. 已知圆O ：221x y +=，设点P 是恒过点()0,4的直线l 上任意的一点，若在该圆上存在点A 使得3OPA π∠≤，则直线l 的斜率k 的取值范围为 ．17. 已知()11,A x y ，()22,B x y 为单位圆上两点，且满足12OA OB ⋅=，则1122x y x y +++的取值范围为 ．三、解答题：5小题，共74分18. 已知()sin cos sin 222x x x f x a ⎛⎫=⋅++ ⎪⎝⎭．（1）求实数a 的值；（2）若44f a f a ππ⎛⎫⎛⎫++-= ⎪ ⎪⎝⎭⎝⎭，求2141tan παα⎛⎫-+ ⎪⎝⎭+的值．19. 在锐角ABC △中，角A ，B ，C 所对的边分别为a ，b ，c ．已知3b =，2239a c c =-+．（1）求A ；（2）求22sin sin B C +的取值范围．20. 如图，在三棱锥P ABC -中，PAB △与ABC △都为等腰直角三角形，PA PB ⊥，AB AC ⊥，M 为AC的中点，且PM AC =．（1）求二面角P AB C --的大小；（2）求直线PM 与平面PBC 所成角的正弦值．M PCBA21. 已知数列{}n a 的前n 项和为{}n S ，且满足：12a =，()*132n n a S n +=-+∈N ．（1）求数列{}n a 的通项公式；（2）数列{}n b 满足12b =-，()()()2131n n n b b n n n a +-+=+，求数列{}n b 的通项公式．22. 在平面直角坐标系中，已知()2,0F ，()2,P t -，若线段FP 的中垂线l 与抛物线()2:20C y px p =>总是相切．（1）求抛物线C 的方程；（2）若过点()2,1Q 的直线l '交抛物线C 于M 、N 两点，过M ，N 分别作抛物线的切线1l ，2l 相交于点A ，1l ，2l 分别与y 轴交于点B ，C ；（i ）证明：当l '变化时，ABC △的外接圆过定点，并求出定点的坐标； （ii ）求ABC △的外接圆面积的最小值．。
【100所名校】2019届浙江省宁波市镇海中学高三上学期期中考试数学试题(解析版)
好教育云平台 名校精编卷 第1页（共4页） 好教育云平台 名校精编卷 第2页（共4页）2019届浙江省宁波市镇海中学 高三上学期期中考试数学试题数学注意事项：1．答题前，先将自己的姓名、准考证号填写在试题卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。
2．选择题的作答：每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑，写在试题卷、草稿纸和答题卡上的非答题区域均无效。
3．非选择题的作答：用签字笔直接答在答题卡上对应的答题区域内。
写在试题卷、草稿纸和答题卡上的非答题区域均无效。
4．考试结束后，请将本试题卷和答题卡一并上交。
一、单选题1．设全集U =R ，集合A ={x|x ≥3},B ={x|x ≤0<5}，则集合(C U A )∩B = A ．{x|0≤x ≤3} B ．{x|0<x <3} C ．{x|0<x ≤3} D ．{x|0≤x <3}2．某几何体的三视图如图所示，图中的四边形都是边长为2的正方形，两条虚线互相垂直，则该几何体的体积是A ．8−π3B ．163C ．8−π6D ．2033．记S n 为等差数列{a n }的前n 项和,若S 9=45,a 3+a 8=12, 则a 7= A ．10 B ．9 C ．8 D ．74．4．满足线性约束条件23,23,{ 0,0x y x y x y +≤+≤≥≥的目标函数z x y =+的最大值是A ．1B ．32C ．2D ．3 5．已知函数f (x )=x 2−In|x|x，则函数f (x )的图象为A ．B ．C ．D ．6．若α、β是两个相交平面，则在下列命题中，真命题的序号为①若直线m ⊥α，则在平面β内一定不存在与直线m 平行的直线．②若直线m ⊥α，则在平面β内一定存在无数条直线与直线m 垂直． ③若直线m ⊂α，则在平面β内不一定存在与直线m 垂直的直线． ④若直线m ⊂α，则在平面β内一定存在与直线m 垂直的直线． A ．①③ B ．②③ C ．②④ D ．①④ 7．已知sin (π6−α)=√23，那么cos2α+√3sin2α=A ．109 B ．−109 C ．−59 D ．598．已知正项等比数列{a n }满足a 7=a 6+2a 5，若存在两项a m ,a n ，使得a m ⋅a n =16a 12，则1m+9n 的最小值为 A ．32 B ．114 C ．83 D ．103 9．已知双曲线C:x 2a 2−y 2b 2=1(a >0,b >0)的左右焦点分别为F 1,F 2，P 为双曲线C 上一点，Q 为双曲线C 渐近线上一点，P,Q 均位于第一象限，且2QP ⃑⃑⃑⃑⃑ =PF 2⃑⃑⃑⃑⃑⃑⃑ ，QF 1⃑⃑⃑⃑⃑⃑⃑ ⋅QF 2⃑⃑⃑⃑⃑⃑⃑ =0，则双曲线C 的离心率为A ．√3−1B ．√3+1C ．√13−2D ．√13+210．如图，在三棱柱ABC −A 1B 1C 1中，底面为边长为2的正三角形，B 1在底面的射影为AC 中点且B 1到底面的距离为√3，已知E,F 分别是线段AB 1与CA 1上的动点，记线段EF 中点M 的轨迹为L ，则|L |等于(注：|L |表示L 的测度，本题中L 若分别为曲线、平面图形、空间几何体，分别对应为其长度、面积、体积）此卷只装订不密封班级 姓名 准考证号 考场号 座位号A．1B．√102C．√32D．√394二、填空题11．中国古代数学著作《九章算术》中有一个这样的问题：“某贾人擅营，月入益功疾（注：从第2月开始，每月比前一月多入相同量的铜钱，3月入25贯，全年（按12个月计）共入510贯“，则该人每月比前一月多入_________________贯，第12月营收贯数为_________________.12．y=sin(2x+π6)的最小正周期为_________________，为了得到函数y=sin(2x+π6)的图象，可以将函数y=cos2x的图象向左最小移动_______个单位13．已知直线l1:ax+(a+2)y+1=0,l2:x+ay+2=0，其中a∈R，若l1⊥l2，则a=______，若l1//l2，则a=__________.14．已知x,y∈R，且4x2+y2+xy=1，则4x2+y2的最小值_________，此时x的值为___________.15．已知两不共线的非零向量a ,b⃑满足|a|=2,|a−b⃑|=1,则向量a与b⃑夹角的最大值是__________.16．已知数列{a n}为等差数列，其前n项和为S n，且2a1+3a3=S6，给出以下结论：①a10= 0②S10最小③S7=S12④S19=0，正确的有_________________.17．设函数f(x)={|12x−4|+1,x≤1x(x−2)2+a,x>1，若存在互不相等的4个实数x1,x2,x3,x4，使得f(x1)x1=f(x2) x2=f(x3)x3=f(x4)x4=7，则a的取值范围为__________.三、解答题18．已知函数f(x)=sin x3cos x3+√3cos2x3（1）求函数f(x)图象对称中心的坐标；（2）如果ΔABC的三边a,b,c满足b2=ac，且边b所对的角为B，求f(B)的取值范围．19．已知数列{a n}的前n项和为S n，且S n=2a n−32n,n∈N∗，b n=a n−12n（1）求证：数列{b n}为等比数列，并求出数列{a n}的通项公式；（2）是否存在实数λ，对任意m,n∈N∗，不等式S m>λb n恒成立？若存在，求出λ的取值范围，若不存在请说明理由．20．如图，四棱锥P−ABCD的底面ABCD为平行四边形，平面PAB⊥平面ABCD，PB=PC,∠ABC=45°，点E是线段PA上靠近点A的三等分点（1）求证：AB⊥PC（2）若ΔPAB是边长为2的等边三角形，求直线DE与平面PBC所成角的正弦值21．如图，O为坐标原点，点F为抛物线C1:x2=2py(p>0)的焦点，且抛物线C1上点P处的切线与圆C2:x2+y2=1相切于点Q（1）当直线PQ的方程为x−y−√2=0时，求抛物线C1的方程；（2）当正数p变化时，记S1,S2分别为ΔFPQ,ΔFOQ的面积，求S1S2的最小值。
【数学】浙江省镇海中学2018-2019学年高三上学期期中考试数学试卷(精校版)
镇海中学2018学年第一学期期中考试高三年级数学试卷一、选择题：（本大题共10个小题,每小题4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．设全集U R =，集合{}{}|,|053A x x B x x ==≤<≥，则集合()U C A B （ ）A ．{}|03x x ≤≤B ．{}|03x x <<C ．{}|03x x <≤D ．{}|03x x ≤<2．某几何体的三视图如图所示，图中的四边形都是边长为2的正方形，两条虚线互相垂直，则该几何体的体积是（ ）A ．83π-B ．163C ．86π-D ．2033．记n S 为等差数列{}n a 的前n 项和，若93845,12S a a =+=，则7a 等于（ ）A ．10B ．9C ．8D ．74．满足线性约束条件23230,0x y x y x y +≤⎧⎪+≤⎨⎪≥≥⎩的目标函数z x y =+的最大值是（ ）A ． 1B ．32C ．2D ．35．已知函数()2||In x f x x x=-，则函数()f x 的图象为（ ）A ．B ．C ．D ．6．若,αβ是两个相交的平面，则下列命题中，真命题的序号为（ ） ①若直线m α⊥，则在平面β内，一定不存在与直线m 平行的直线 ②若直线m α⊥，则在平面β内，一定存在无数条直线与直线m 垂直 ③若直线m α⊆，则在平面β内，不一定存在与直线m 垂直的直线 ④若直线m α⊆，则在平面β内，一定存在与直线m 垂直的直线A ．①③B ．②③C ．②④D ．①④7．已知sin 63πα⎛⎫-=-⎪⎝⎭，那么cos22αα=（ ） A ．109B ．109-C ．59-D ．598．已知正项等比数列{}n a 满足7652a a a =+，若存在两项,m n a a ，使得2116m n a a a ⋅=，则19m n+的最小值为（ ）A ．32B ．114C ．83D ．1039．已知双曲线()2222:10,0x y C a b a b-=>>的左右交点分别为12,F F ，P 为双曲线C 上一点，Q 为双曲线渐近线C 上一点，,P Q 均位于第一象限，且2122,0QP PF QF QF →→→→=⋅=，则双曲线C 的离心率为（ ）A 1B 1C 2D 210．如图，在三棱柱111ABC A B C -中，底面为边长为2的正三角形，1B 在底面的射影为AC 中点且1B 到,E F 分别是线段1AB 与1CA 上的动点，记线段EF 中点M 的轨迹为L ，则L 等于（ ）(注：L 表示L 的测度，本题中L 若分别为曲线、平面图形、空间几何体，分别对应为其长度、A ．1BCD二、填空题：（本大题共7个小题,多空题每题6分，单空题每小题4分,共36分．）11．中国古代数学著作《九章算术》中有一个这样的问题：“某贾人擅营，月入益功疾（注：从第2月开始，每月比前一月多入相同量的铜钱，3月入25贯，全年（按12个月计）共入510贯“，则该人每月比前一月多入_________________贯，第12月营收贯数为_________________. 12．sin 26y x π⎛⎫=+⎪⎝⎭的最小正周期为_________________，为了得到函数sin 26y x π⎛⎫=+⎪⎝⎭的图象，可以将函数cos 2y x =的图象向左最小移动_________________个单位13．已知直线()12:210,:20l ax a y l x ay +++=++=，其中a R ∈，若12l l ⊥，则a =_________________，若12//l l ，则a =_________________.14．已知,x y R ∈，且2241x y xy ++=，则224x y +的最小值_________________，此时x 的值为_________________.15．已知两个不共线的非零向量,a b →→，满足2,1a a b →→→=-=，则向量,a b →→夹角的最大值是_________________.16．已知数列{}n a 为等差数列，其前n 项和为n S ，且13623a a S +=，给出以下结论：①100a =②10S 最小③712S S =④190S =，正确的有_________________.17．设函数()()2|124|1,12,1x x f x x x a x -+≤⎧⎪=⎨-+>⎪⎩，若存在互不相等的4个实数1234,,,x x x x，使得()()()()123412347f x f x f x f x x x x x ====，则a 的取值范围为_________________. 三、解答题：（本大题共个5小题,共74分．解答应写出文字说明、证明过程或演算步骤）18．(本小题满分14分)已知函数()2sin cos 333x x x f x =+ （1）求函数()f x 图象对称中心的坐标；（2）如果ABC ∆的三边,,a b c 满足2b ac =，且边b 所对的角为B ，求()f B 的取值范围．19．(本小题满分15分)已知数列{}n a 的前n 项和为n S ，且32,2n n n S a n N *=-∈，12n n nb a =- （1）求证：数列{}n b 为等比数列，并求出数列{}n a 的通项公式；（2）是否存在实数λ，对任意,m n N *∈，不等式m nS b λ>恒成立？若存在，求出λ的取值范围，若不存在请说明理由．-的底面ABCD为平行四边形，平面PAB⊥平面ABCD，20．(本小题满分15分)如图，四棱锥P ABCDPB PC ABC︒=∠=，点E是线段PA上靠近点A的三等分点,45⊥（1）求证：AB PC∆是边长为2的等边三角形，求直线DE与平面PBC所成角的正弦值（2）若PAB21．(本小题满分15分)如图，O 为坐标原点，点F 为抛物线()21:20C x py p =>的焦点，且抛物线1C 上点P 处的切线与圆222:1C x y +=相切于点Q（1）当直线PQ的方程为0x y -=时，求抛物线1C 的方程； （2）当正数p 变化时，记12,S S 分别为,FPQ FOQ ∆∆的面积，求12S S 的最小值22．(本小题满分15分)已知a R ∈，函数()1x f x e ax -=-在点()1,1a -处与x 轴相切 （1）求a 的值，并求()f x 的单调区间（2）当1x >时，()()1f x m x Inx >-，求实数m 的取值范围参考答案：1——5DDBCD 6——10CABCD11.5;70 12.56ππ； 13.0或3-；2或-1 14.45±； 15.030 16.①③④ 17.(6,18)18.（1）2()sin()33f x x π=++，对称中心3(,0),22k k Z ππ-+∈（2）()f B ∈+⎦19.（1）证明略；1122n n na -=+（2）92λ<20.（1）略 （2）sin θ=21.（1）21:C x =（2）解析如下：2001110111200022111111001200011221121121(,),(,)20100:1:02-1==-=-=-222=-11=-2Q P P Q x P x Q x y px x y Q l x x y y x x P l x y p p l l x p px px px y y x x x x x x x x p px x y p y x x <<⎧>⎨<⎩+=--=⎧⎪⎪⎧⎪⎪∴⇒⎨⎨⎪⎪-=⎩⎪⎪⎩⎧+=⎪⎨⎪⎩∴设根据对称性：仅取和研究过处切线方程过处切线方程与重合221)p=10101111112111122)()1222()()PQ x x x x x x x x y x px x ==-=-=-=-=-=-F 到PQ l距离为：221141p x d +==2211121121221122211112222111221121122421112211224222211112122()(1)2412212()(1)442(1)(1)441=-2141,(1)4421421(1)(1)4(1p S x x px x p S x p x x S px x p x p S p x x x x y p y x p x x x x p S S p x x p x x p =⨯-⨯+=⨯⨯-⨯+∴==-⨯+⨯⎧+=⎪⎨⎪⎩∴+==-∴=-⨯=-⨯-221121(1)1)x x =-⨯-将21221)x p=带入上式：122222222212221(1)221)11)1)1,1(1)21331111,S S p p t p t S t t t t t S t t t t p 令当且仅当：=-⨯-=-⨯==>=-++===-++≥---==22.（1）解析过程略()()11,,1a =+∞-∞增区间：减区间：（2）12m ≤解析如下：11''1''''''12''11(1)ln ()(1)ln ,1(1)0()0,(1)01()1ln ,1(1)0(1)0,(1)0(),11(1)120,212()(1)ln x x x x x x e x m x xh x e x m x x x h h x h x h x e m x m x x h h h m mh x e x x x h m m m h x e x m x x ----->->-=--->=>≥-=--->=>≥=-->∴=-≥∴≤≤=---时，令由于欲使需要欲使需要接下来证明：'1''12'''123,11()1ln ,1(),12(),1x x x x x h x e m x m x x m mh x e x x x m mh x e x x x --->-=--->=-->=++>①当102m <≤时'''123''1''2'1'12()0()(1)1201()1ln (1)0()(1)ln (1)0102x x x x mm h x e x x m mh x e h m x x x h x e m x m h x h x e x m x x h m ----=++>∴=-->=-≥-∴=--->=∴=--->=∴<≤成立②当0m ≤时1()0,(1)ln 0()(1)ln ,112x f x e x m x x f x m x x x m -=->-≤∴>->≤综上：备注：此题考查的是导数的端点效应。
2018-2019学年浙江省宁波市镇海中学高三(上)期中数学试卷
2018-2019学年浙江省宁波市镇海中学高三（上）期中数学试卷试题数：22.满分：1501.（单选题.5分）设全集U=R.集合A={x|x≥3}.B={x|0≤x ＜5}.则集合（∁U A ）∩B=（ ）A.{x|0＜x ＜3}B.{x|0≤x ＜3}C.{x|0＜x≤3}D.{x|0≤x≤3}2.（单选题.5分）某几何体的三视图如图所示（单位：cm ）.图中的四边形都是边长为2的正方形.两条虚线互相垂直.则该几何体的体积（单位：cm 3）是（ ）A. 163B. 203C. 8−π6D. 8−π3 3.（单选题.5分）记S n 为等差数列{a n }的前n 项和.若S 9=45.a 3+a 8=12.则a 7等于（ ）A.10B.9C.8D.74.（单选题.5分）满足线性约束条件 {2x +y ≤3x +2y ≤3x ≥0y ≥0.的目标函数z=x+y 的最大值是（ ）A.1B. 32C.2D.3.则函数y=f（x）的大致图象为（）5.（单选题.5分）已知函数f(x)=x2−ln|x|xA.B.C.D.6.（单选题.5分）若α.β是两个相交的平面.则下列命题中.真命题的序号为（）① 若直线m⊥α.则在平面β内.一定不存在与直线m平行的直线② 若直线m⊥α.则在平面β内.一定存在无数条直线与直线m垂直③ 若直线m⊆α.则在平面β内.不一定存在与直线m垂直的直线④ 若直线m⊆α.则在平面β内.一定存在与直线m垂直的直线A. ① ③B. ② ③C. ② ④D. ① ④7.（单选题.5分）已知 sin (π6−α)=−√23 .那么 cos2α+√3sin2α =（ ） A. 109B. −109C. −59D. 598.（单选题.5分）已知正项等比数列{a n }满足：a 7=a 6+2a 5.若存在两项a m 、a n .使得a m a n =16a 12.则 1m + 9n 的最小值为（ ）A. 32B. 83C. 114D.不存在9.（单选题.5分）已知双曲线C ： x 2a 2 - y 2b 2 =1（a ＞0.b ＞0）的左右焦点分别为F 1.F 2.P 为双曲线C 上一点.Q 为双曲线C 渐近线上一点.P.Q 均位于第一象限.且2 QP ⃗⃗⃗⃗⃗ = PF 2⃗⃗⃗⃗⃗⃗⃗ . QF 1⃗⃗⃗⃗⃗⃗⃗ •QF 2⃗⃗⃗⃗⃗⃗⃗ =0.则双曲线C 的离心率为（ ）A. √3 -1B. √3+1C. √13 -2D. √13 +210.（单选题.5分）如图.在三棱柱ABC-A 1B 1C 1中.底面为边长为2的正三角形.B 1在底面的射影为AC 中点且B 1到底面的距离为 √3 .已知E.F 分别是线段AB 1与CA 1上的动点.记线段EF 中点M 的轨迹为L.则|L|等于（ ）（注：|L|表示L 的测度.本题中L 若分别为曲线、平面图形、空间几何体.分别对应为其长度、面积、体积）A.1B. √102C. √32D. √39411.（填空题.4分）中国古代数学著作《九章算术》中有一个这样的问题：“某贾人擅营.月入益功疾（注：从第2月开始.每月比前一月多入相同量的铜钱.3月入25贯.全年（按12个月计）共入510贯“.则该人每月比前一月多入___ 贯.第12月营收贯数为___ ．12.（填空题.4分）y=sin（2x+ π6）的最小正周期为___ .为了得到函数y=sin（2x+ π6）的图象.可以将函数y=cos2x的图象向左最小移动___ 个单位．13.（填空题.4分）已知直线l1：ax+（a+2）y+1=0.l2：x+ay+2=0.其中a∈R.若l1⊥l2.则a=___ .若l1 || l2.则a=___ ．14.（填空题.4分）已知x.y∈R.且4x2+y2+xy=1.则4x2+y2的最小值___ .此时x的值为___ ．15.（填空题.4分）已知两非零向量a，b⃗满足|a|=2 . |a−b⃗|=1 .则向量a，b⃗夹角的最大值是___ ．16.（填空题.4分）已知数列{a n}为等差数列.其前n项和为S n.且2a1+3a3=S6.给出以下结论：① a10=0 ② S10最小③ S7=S12④ S19=0.正确的有___ ．17.（填空题.4分）设函数f(x)={|12x−4|+1，x≤1x(x−2)2+a，x＞1.若存在互不相等的4个实数x1.x2.x3.x4.使得f(x1)x1=f(x2)x2=f(x3)x3=f(x4)x4=7 .则a的取值范围为___ ．18.（问答题.12分）已知函数f(x)=sin x3cos x3+√3cos2x3．（1）求函数f（x）图象对称中心的坐标；（2）如果△ABC的三边a.b.c满足b2=ac.且边b所对的角为B.求f（B）的取值范围．19.（问答题.15分）已知数列{a n}的前n项和为S n.且S n=2a n−32n ，n∈N∗ . b n=a n−12n（1）求证：数列{b n}为等比数列.并求出数列{a n}的通项公式；（2）是否存在实数λ.对任意m.n∈N*.不等式S m＞λb n恒成立？若存在.求出λ的取值范围.若不存在请说明理由．20.（问答题.15分）如图.四棱锥P-ABCD的底面ABCD为平行四边形.平面PAB⊥平面ABCD.PB=PC.∠ABC=45°.点E是线段PA上靠近点A的三等分点．（Ⅰ）求证：AB⊥PC；（Ⅱ）若△PAB是边长为2的等边三角形.求直线DE与平面PBC所成角的正弦值．21.（问答题.15分）如图.O为坐标原点.点F为抛物线C1：x2=2py（p＞0）的焦点.且抛物线C1上点P处的切线与圆C2：x2+y2=1相切于点Q．（Ⅰ）当直线PQ的方程为x-y- √2 =0时.求抛物线C1的方程；（Ⅱ）当正数p变化时.记S1.S2分别为△FPQ.△FOQ的面积.求S1S2的最小值．22.（问答题.15分）已知a∈R.函数f（x）=e x-1-ax在点（1.1-a）处与x轴相切．（1）求a的值.并求f（x）的单调区间；（2）当x＞1时.f（x）＞m（x-1）lnx.求实数m的取值范围．2018-2019学年浙江省宁波市镇海中学高三（上）期中数学试卷参考答案与试题解析试题数：22.满分：1501.（单选题.5分）设全集U=R.集合A={x|x≥3}.B={x|0≤x＜5}.则集合（∁U A）∩B=（）A.{x|0＜x＜3}B.{x|0≤x＜3}C.{x|0＜x≤3}D.{x|0≤x≤3}【正确答案】：B【解析】：先根据补集的定义求出集合A的补集∁U A.然后和集合B进行交集运算.可求（∁U A）∩B．【解答】：解：因为A={x|x≥3}.所以∁U A={x|x＜3}.所以（∁U A）∩B={x|0≤x＜3}．故选：B．【点评】：本题的考点是集合的补集和交集运算.比较基础．2.（单选题.5分）某几何体的三视图如图所示（单位：cm）.图中的四边形都是边长为2的正方形.两条虚线互相垂直.则该几何体的体积（单位：cm3）是（）A. 163B. 203C. 8−π6D. 8−π3【正确答案】：B【解析】：由三视图知原几何体是一个棱长为2的正方体挖去一四棱锥得到的.根据所提供的数据可求出正方体、锥体的体积.从而得到答案．【解答】：解：由三视图知原几何体是一个棱长为2的正方体挖去一四棱锥得到的.该四棱锥的底为正方体的上底.高为1.如图所示：所以该几何体的体积为23- 13 ×22×1= 203．故选：B．【点评】：本题考查三视图.考查柱体、锥体的体积计算.解决该类问题的关键是由三视图还原得到原几何体.画三视图的要求为：“长对正.高平齐.宽相等”．3.（单选题.5分）记S n为等差数列{a n}的前n项和.若S9=45.a3+a8=12.则a7等于（）A.10B.9C.8D.7【正确答案】：B【解析】：利用等差数列的通项公式与求和公式即可得出．【解答】：解：设等差数列{a n}的公差为d.∵S9=45.a3+a8=12.∴9a1+ 9×82d=45.2a1+9d=12.联立解得a1=-3.d=2.则a7=-3+6×2=9．故选：B．【点评】：本题考查了等差数列的通项公式与求和公式.考查了推理能力与计算能力.属于中档题．4.（单选题.5分）满足线性约束条件 {2x +y ≤3x +2y ≤3x ≥0y ≥0.的目标函数z=x+y 的最大值是（ ）A.1B. 32C.2D.3【正确答案】：C 【解析】：先根据约束条件画出可行域.再利用几何意义求最值.只需求出直线z=x+y 过点B （1.1）时.z 最大值即可．【解答】：解：先根据约束条件画出可行域.当直线z=x+y 过点B （1.1）时.z 最大值为2．故选：C ．【点评】：本题主要考查了简单的线性规划.以及利用几何意义求最值.属于基础题．5.（单选题.5分）已知函数 f (x )=x 2−ln|x|x .则函数y=f （x ）的大致图象为（ ）A.B.C.D.【正确答案】：A【解析】：可得函数为奇函数.排除选项.利用特殊点的位置判断选项即可．【解答】：解：由题意可得函数的定义域为（-∞.0）∪（0.+∞）.函数f(x)=x2−ln|x|x.可得f（-x）≠±f（x）.故函数为非奇非偶函数.排除：B、C；当x=- 1e 时.y= 1e2+lne−11e= 1e2−e＜0.排除D；综上可得选项A符合题意.故选：A．【点评】：本题考查函数的性质.由函数的性质入手是解决问题的关键.属中档题．6.（单选题.5分）若α.β是两个相交的平面.则下列命题中.真命题的序号为（）① 若直线m⊥α.则在平面β内.一定不存在与直线m平行的直线② 若直线m⊥α.则在平面β内.一定存在无数条直线与直线m垂直③ 若直线m⊆α.则在平面β内.不一定存在与直线m垂直的直线④ 若直线m⊆α.则在平面β内.一定存在与直线m垂直的直线A. ① ③B. ② ③C. ② ④D. ① ④【正确答案】：C【解析】：若直线m⊥α.则在平面β内.一定存在无数条直线与直线m垂直；若直线m⊆α.则在平面β内.一定存在与直线m垂直的直线．【解答】：解：由α.β是两个相交的平面.知：在① 中.若直线m⊥α.则在平面β内.存在与直线m平行的直线.故① 错误；在② 中.若直线m⊥α.则在平面β内.一定存在无数条直线与直线m垂直.故② 正确；在③ 中.若直线m⊆α.则在平面β内.一定存在与直线m垂直的直线.故③ 错误；在④ 中.若直线m⊆α.则在平面β内.一定存在与直线m垂直的直线.故④ 正确．故选：C．【点评】：本题考查命题真假的判断.考查空间中线线、线面、面面间的位置关系等基础知识.考查运算求解能力.是中档题．7.（单选题.5分）已知sin(π6−α)=−√23.那么cos2α+√3sin2α =（）A. 109B. −109C. −59D. 59【正确答案】：A【解析】：由题意利用辅助角公式、诱导公式.二倍角公式.求得cos2α+√3sin2α的值．【解答】：解：∵已知sin(π6−α)=−√23.∴ cos2α+√3sin2α =2sin（2α+ π6）=2cos（π3-2α）=2[1-2 sin2(π6−α) ]=2（1-2× 29）= 109.故选：A．【点评】：本题主要考查辅助角公式、诱导公式.二倍角公式的应用.属于基础题．8.（单选题.5分）已知正项等比数列{a n}满足：a7=a6+2a5.若存在两项a m、a n.使得a m a n=16a12.则1m + 9n的最小值为（）A. 32 B. 83 C. 114 D.不存在 【正确答案】：C【解析】：设{a n }的公比为q （q ＞0）.由等比数列的通项公式化简a 7=a 6+2a 5.求出q.代入a m a n =16a 12化简得m.n 的关系式.由“1”的代换和基本不等式求出式子的范围.验证等号成立的条件.由m 、n 的值求出式子的最小值．【解答】：解：设正项等比数列{a n }的公比为q.且q ＞0. 由a 7=a 6+2a 5得：a 6q=a 6+2a 6q. 化简得.q 2-q-2=0.解得q=2或q=-1（舍去）. 因为a m a n =16a 12.所以 (a 1q m−1) (a 1q n−1) =16a 12. 则q m+n-2=16.解得m+n=6.所以 1m +9n = 16 （m+n ）（ 1m +9n ）= 16 （10+ nm +9m n ）≥ 16(10+2√nm•9m n ) = 83. 当且仅当 nm=9mn时取等号.此时 {nm =9m nm +n =6.解得 {m =32n =92 . 因为m n 取整数.所以均值不等式等号条件取不到.则 1m +9n ＞ 83 . 验证可得.当m=2、n=4时. 1m +9n 取最小值为 114 . 故选：C ．【点评】：本题考查等比数列的通项公式.利用“1”的代换和基本不等式求最值问题.考查化简、计算能力.注意等号的成立的条件.属于易错题． 9.（单选题.5分）已知双曲线C ： x 2a 2 - y 2b 2 =1（a ＞0.b ＞0）的左右焦点分别为F 1.F 2.P 为双曲线C 上一点.Q 为双曲线C 渐近线上一点.P.Q 均位于第一象限.且2 QP ⃗⃗⃗⃗⃗ = PF 2⃗⃗⃗⃗⃗⃗⃗ . QF 1⃗⃗⃗⃗⃗⃗⃗ •QF 2⃗⃗⃗⃗⃗⃗⃗ =0.则双曲线C 的离心率为（ ） A. √3 -1 B. √3+1 C. √13 -2 D. √13 +2【正确答案】：C【解析】：设Q （m. bam ）.（m ＞0）.P （s.t ）.F 1（c.0）.F 2（c.0）.由向量共线的坐标表示.以及点满足双曲线方程.向量垂直的条件：数量积为0.可得a.b.c 的关系.再由离心率公式.计算可得所求值．【解答】：解：设Q （m. ba m ）.（m ＞0）.P （s.t ）.F 1（-c.0）.F 2（c.0）.QP ⃗⃗⃗⃗⃗ = 12 PF 2⃗⃗⃗⃗⃗⃗⃗ .可得 m+12c 1+12=s. bma1+12=t. 由P 在双曲线上.可得 (m+12c)294a 2 -b 2m 294a 2b 2=1.化为c 2+4mc=9a 2. m=9a 2−c 24c. 由 QF 1⃗⃗⃗⃗⃗⃗⃗ •QF 2⃗⃗⃗⃗⃗⃗⃗ =0.可得 bma m−c • bm am+c =-1. 即c 2-m 2=b 2m 2a 2. 即c 2=m 2•a 2+b 2a 2=m 2• c 2a 2 . 可得m=a. 则4ca=9a 2-c 2.由e= ca 可得e 2+4e-9=0. 可得e=-2+ √13 （负的舍去）. 故选：C ．【点评】：本题考查双曲线的方程和性质.主要是渐近线和离心率的求法.考查向量共线和垂直的条件.考查运算化简能力.属于中档题．10.（单选题.5分）如图.在三棱柱ABC-A 1B 1C 1中.底面为边长为2的正三角形.B 1在底面的射影为AC 中点且B 1到底面的距离为 √3 .已知E.F 分别是线段AB 1与CA 1上的动点.记线段EF 中点M 的轨迹为L.则|L|等于（ ）（注：|L|表示L 的测度.本题中L 若分别为曲线、平面图形、空间几何体.分别对应为其长度、面积、体积）A.1B.√102C. √32 D.√394【正确答案】：D【解析】：取特殊点得到M 的轨迹为平行四边形区域.建立空间直角坐标系.再由三角形面积求解．【解答】：解：当E 位于B 1.A.而F 在A 1C 上移动时.M 的轨迹为平行于A 1C 的两条线段. 当F 位于A 1.C.而E 在AB 1上移动时.M 的轨迹为平行于AB 1的两条线段． 其它情况下.M 的轨迹构成图中平行四边形内部区域．以O 为坐标原点.分别以OC.OB 1所在直线为x.z 轴建立空间直角坐标系. 则A （-1.0.0）.B 1（0.0. √3 ）.C （1.0.0）.A 1 （-1. √3 . √3 ）. ∴ AB 1⃗⃗⃗⃗⃗⃗⃗ =(1，0，√3) . CA 1⃗⃗⃗⃗⃗⃗⃗ =(−2，√3，√3) . 则 |AB 1⃗⃗⃗⃗⃗⃗⃗ |=2 . |CA 1⃗⃗⃗⃗⃗⃗⃗ |=√10 .cos ＜AB 1⃗⃗⃗⃗⃗⃗⃗ ，CA 1⃗⃗⃗⃗⃗⃗⃗ ＞ = AB 1⃗⃗⃗⃗⃗⃗⃗⃗ •CA 1⃗⃗⃗⃗⃗⃗⃗⃗|AB 1⃗⃗⃗⃗⃗⃗⃗⃗ |•|CA 1⃗⃗⃗⃗⃗⃗⃗⃗ | = √1020 ．则异面直线AB 1与CA 1所成角的正弦值为sinθ= √1−(√1020)2=√39020． ∴|L|=2× 12 × |12AB 1⃗⃗⃗⃗⃗⃗⃗ | × |12CA 1⃗⃗⃗⃗⃗⃗⃗ | = 1×√102×√39020=√394． 故选：D ．【点评】：本题考查棱柱的结构特征.考查空间想象能力和思维能力.利用特殊点得到M的轨迹是解答该题的关键.是中档题．11.（填空题.4分）中国古代数学著作《九章算术》中有一个这样的问题：“某贾人擅营.月入益功疾（注：从第2月开始.每月比前一月多入相同量的铜钱.3月入25贯.全年（按12个月计）共入510贯“.则该人每月比前一月多入___ 贯.第12月营收贯数为___ ．【正确答案】：[1]5; [2]70【解析】：由题意可知.每个月的收入构成等差数列{a n}.由已知列式求得首项与公差.再由等差数列的通项公式求得a12．【解答】：解：由题意可知.每个月的收入构成等差数列{a n}.设公差为d．则a3=25.S12=510．∴a1+2d=25.12a1+ 12×112d=510.解得a1=15.d=5.∴a12=15+11×5=70．∴该人每月比前一月多入5贯.第12月营收贯数为70．故答案为：5；70．【点评】：本题考查了等差数列的通项公式与求和公式.考查了推理能力与计算能力.是基础题．12.（填空题.4分）y=sin（2x+ π6）的最小正周期为___ .为了得到函数y=sin（2x+ π6）的图象.可以将函数y=cos2x的图象向左最小移动___ 个单位．【正确答案】：[1]π; [2] 56π【解析】：根据正弦图象性质即可求出最小正周期.结合三角函数平移变换规律即可得到结论【解答】：解：函数y=sin(2x+π6)的最小正周期T= 2π2=π；将函数y=cos2x=sin（2x+ π2）的图象向左平移φ个单位（φ＞0）.可得：y=sin（2x+2φ+π2）；∴2φ+ π2 = π6+2kπ.则φ=k π−π6.φ＞0∴k=1时.可得φ= 5π6故答案为：π；5π6．【点评】：本题主要考查三角函数性质和平移变换的规律应用.属于基础题．13.（填空题.4分）已知直线l1：ax+（a+2）y+1=0.l2：x+ay+2=0.其中a∈R.若l1⊥l2.则a=___ .若l1 || l2.则a=___ ．【正确答案】：[1]0或-3; [2]-1或2【解析】：利用直线与直线垂直、直线与直线平行的性质直接求解．【解答】：解：∵直线l1：ax+（a+2）y+1=0.l2：x+ay+2=0.其中a∈R.l1⊥l2.∴a×1+（a+2）a=0.解得a=0或a=-3；当l1 || l2时. a1=a+2a≠12.解得a=-1或a=2．故答案为：0或-3.-1或2．【点评】：本题考查实数值的求法.考查直线与直线垂直、直线与直线平行的性质等基础知识.考查运算求解能力.考查函数与方程思想.是基础题．14.（填空题.4分）已知x.y∈R.且4x2+y2+xy=1.则4x2+y2的最小值___ .此时x的值为___ ．【正确答案】：[1] 45 ; [2] ±√1010【解析】：由重要不等式4x2+y2≥4xy.解不等式可得4x2+y2的最小值和x的值．【解答】：解：由4x2+y2≥4xy.即有xy≤ 4x 2+y2 4.由4x2+y2+xy=1.可得1≤4x2+y2+ 4x2+y24.可得4x2+y2≥ 45.当且仅当2x=y即x=± √1010时.取得等号.则4x2+y2的最小值为45.故答案为：45 .± √1010．【点评】：本题考查基本不等式的运用：求最值.考查不等式的解法和运算能力.属于基础题．15.（填空题.4分）已知两非零向量a，b⃗满足|a|=2 . |a−b⃗|=1 .则向量a，b⃗夹角的最大值是___ ．【正确答案】：[1] π6【解析】：设向量a，b⃗夹角为θ.由余弦定理求得cosθ= 3+x24x.再利用基本不等式求得cosθ取得最小值.即可求得θ的最大值．【解答】：解：∵两非零向量a，b⃗满足|a|=2 . |a−b⃗|=1 .设向量a，b⃗夹角为θ.由于非零向量a，b⃗以及a−b⃗构成一个三角形.设| b⃗ |=x.则由余弦定理可得1=4+x2-4x•cosθ.解得cosθ= 3+x24x =3x+x4≥ √32.当且仅当x= √3时.cosθ取得最小值为√32.角θ取得最大值为π6.故答案为π6．【点评】：本题主要考查两个向量的加减法的法则.以及其几何意义.余弦定理以及基本不等式的应用.属于中档题．16.（填空题.4分）已知数列{a n}为等差数列.其前n项和为S n.且2a1+3a3=S6.给出以下结论：① a10=0 ② S10最小③ S7=S12④ S19=0.正确的有___ ．【正确答案】：[1] ① ③ ④【解析】：先求出a1=-9d.再表示出求和公式.即可判断．【解答】：解：设等差数列{a n}的公差为d.∵2a1+3a3=S6.∴5a1+6d=6a1+15d.化为：a1+9d=0.即a10=0.给出下列结论：① a10=0.正确；② S 10=10a 1+10(10−1)d2=-45d.可能大于0.也可能小于0.因此不正确；③ S 12-S 7=12a 1+ 12×112 d-7a 1- 7×62d=5a 1+45d=5（a 1+9d ）=0.正确． ④ S 19=19(a 1+a 19)2=19a 10=0.正确； 其中正确结论的个数是 ① ③ ④ ． 故答案为： ① ③ ④【点评】：本题考查了等差数列的通项公式与求和公式及其性质.考查了推理能力与计算能力.属于中档题．17.（填空题.4分）设函数 f (x )={|12x −4|+1，x ≤1x (x −2)2+a ，x ＞1.若存在互不相等的4个实数x 1.x 2.x 3.x 4.使得f (x 1)x 1=f (x 2)x 2=f (x 3)x 3=f (x 4)x 4=7 .则a 的取值范围为___ ．【正确答案】：[1]（6.18）【解析】：由题意可得f （x ）=7x 有4个不同实根.讨论x≤1时.x ＞1时.由解方程和运用导数判断单调性和极值、最值.解不等式即可得到所求范围．【解答】：解：由f (x 1)x 1=f (x 2)x 2=f (x 3)x 3=f (x 4)x 4=7 .可得f （x ）=7x 有4个不同实根. 当x≤1时.f （x ）=|12x-4|+1=7x. 解得x= 35 或x= 519 .故当x ＞1时.f （x ）=7x 有2个不同实根. 设g （x ）=f （x ）-7x=x （x-2）2-7x+a （x ＞1）. g′（x ）=（3x+1）（x-3）.当1＜x ＜3时.g′（x ）＜0.g （x ）递减； 当x ＞3时.g′（x ）＞0.g （x ）递增． 则g （x ）min =g （3）=a-18.又g （1）=a-6. 由a-18＜0.且a-6＞0. 解得6＜a ＜18． 即a 的范围是（6.18）． 故答案为：（6.18）．【点评】：本题考查函数和方程的转化思想.考查分类讨论思想方法.以及导数的运用：求单调区间和极值、最值.考查运算能力.属于中档题．18.（问答题.12分）已知函数f(x)=sin x3cos x3+√3cos2x3．（1）求函数f（x）图象对称中心的坐标；（2）如果△ABC的三边a.b.c满足b2=ac.且边b所对的角为B.求f（B）的取值范围．【正确答案】：【解析】：（1）运用二倍角公式和两角和的正弦公式.由正弦函数的对称中心.解方程可得所求；（2）运用三角形的余弦定理和基本不等式.可得12≤co sB＜1.即有0＜B≤ π3.运用正弦函数的图象和性质.即可得到所求范围．【解答】：解：（1）f(x)=sin x3cos x3+√3cos2x3．= 12sin23x+√32(cos23x+1) .=sin（23x+π3）+ √32.令23x+π3=kπ（k∈Z）.解得：x= 3kπ2−π2（k∈Z）.所以函数的图象的对称中心为：（3kπ2−π2，√32）（k∈Z）．（2）由于b2=ac.所以：cosB= a 2+c2−b22ac≥2ac−ac2ac=12.则：0 ＜B≤π3．所以：π3＜2B3+π3≤5π9.则：√32＜sin(2B3+π3)≤1 .所以：√3＜f(B)≤1+√32．则：f（B）的取值范围为：（√3，1+√32]．【点评】：本题考查三角函数的恒等变换的运用.考查正弦函数的图象和性质.同时考查解三角形的余弦定理和基本不等式的运用．19.（问答题.15分）已知数列{a n}的前n项和为S n.且S n=2a n−32n ，n∈N∗ . b n=a n−12n（1）求证：数列{b n}为等比数列.并求出数列{a n}的通项公式；（2）是否存在实数λ.对任意m.n∈N*.不等式S m＞λb n恒成立？若存在.求出λ的取值范围.若不存在请说明理由．【正确答案】：【解析】：（1）直接利用递推关系式求出数列的通项公式.进一步证明数列为等比数列．（2）利用（1）的结论.进一步利用分组法和恒成立问题求出实数λ的取值范围．【解答】：证明：（1）已知数列{a n}的前n项和为S n.且S n=2a n−32n，n∈N∗ . ①当n=1时. a1=32.则：当n≥2时. S n−1=2a n−1−32n−1. ②① - ② 得：a n=2a n-2a n-1- 32n + 62n.整理得：a n=2a n−1−32n.所以：a n−12n =2(a n−1−12n−1) .故：a n−12na n−1−12n−1=2（常数）.故：数列{a n}是以a1−12=1为首项.2为公比的等比数列．故：a n−12n=1•2n−1=2n−1 .所以：a n=2n−1+12n．由于：b n=a n−12n=2n−1 .所以：b nb n−1=2n−12n−2=2（常数）．故：数列{b n}为等比数列．（2）由（1）得：a n=2n−1+12n.所以：S n=(1+21+22+⋯+2n−1) +（121+122+⋯+12n）.= (2n−1)2−1+12(1−12n)1−12.= 2n−12n.假设存在实数λ.对任意m.n∈N*.不等式S m＞λb n恒成立.即：λb n＜(S m)min .由于：S m=2m−12m为增函数 .故当m=1时. S1=32.所以：λ＜2n−1•32.当n=1时. λ＜32．故存在实数λ.且λ＜32．【点评】：本题考查的知识要点：数列的通项公式的求法及应用.分组法在数列求和中的应用.主要考查学生的运算能力和转化能力.属于中档题．20.（问答题.15分）如图.四棱锥P-ABCD的底面ABCD为平行四边形.平面PAB⊥平面ABCD.PB=PC.∠ABC=45°.点E是线段PA上靠近点A的三等分点．（Ⅰ）求证：AB⊥PC；（Ⅱ）若△PAB是边长为2的等边三角形.求直线DE与平面PBC所成角的正弦值．【正确答案】：【解析】：（Ⅰ）作PO⊥AB于O.连接OC.可得PO⊥面ABCD．由△POB≌△POC.∠ABC=45°.得OC⊥AB.即得AB⊥面POC.可证得AB⊥PC．（Ⅱ）以O 为原点建立空间坐标系. P(0，0，√3)，B(1，0，0)，C(0，1，0)，A(−1，0，0) .利用向量求解．【解答】：解：（Ⅰ）作PO⊥AB 于O… ① .连接OC.∵平面PAB⊥平面ABCD.且面PAB∩面ABCD=AB.∴PO⊥面ABCD ．…（2分）∵PB=PC .∴△POB≌△POC .∴OB=OC .又∵∠ABC=45°.∴OC⊥AB… ②又PO∩CO=O .由 ① ② .得AB⊥面POC.又PC⊂面POC.∴AB⊥PC ．…（6分）（Ⅱ）∵△PAB 是边长为2的等边三角形.∴ PO =√3，OA =OB =OC =1 ． 如图建立空间坐标系. P(0，0，√3)，B(1，0，0)，C(0，1，0)，A(−1，0，0)设面PBC 的法向量为 n ⃗ =(x ，y ，z) .PB ⃗⃗⃗⃗⃗ =(1，0，−√3)，BC ⃗⃗⃗⃗⃗ =(−1，1，0) .由 {n ⃗ •PB ⃗⃗⃗⃗⃗ =x −√3z =0n ⃗ •BC ⃗⃗⃗⃗⃗ =−x +y =0.令 x =√3 .得 n ⃗ =(√3，√3，1) ；AP ⃗⃗⃗⃗⃗ =(1，0，√3)，AE ⃗⃗⃗⃗⃗ =13AP ⃗⃗⃗⃗⃗ =(13，0，√33) . CB ⃗⃗⃗⃗⃗ =DA ⃗⃗⃗⃗⃗ =(1，−1，0) ． DE ⃗⃗⃗⃗⃗ =DA ⃗⃗⃗⃗⃗ +AE ⃗⃗⃗⃗⃗ =(43，−1，√33) .设DE 与面PBC 所成角为θ.sinθ=|cos〈n ⃗ ，DE ⃗⃗⃗⃗⃗ 〉|=n⃗ •DE ⃗⃗⃗⃗⃗⃗ |n ⃗ ||DE ⃗⃗⃗⃗⃗⃗ |=4√33−√3+√33√169+1+39×√3+3+1=√37∴直线DE 与平面PBC 所成角的正弦值 √37 ．…（12分）【点评】：本题考查了空间线线垂直的判定.向量法求线面角.属于中档题．21.（问答题.15分）如图.O 为坐标原点.点F 为抛物线C 1：x 2=2py （p ＞0）的焦点.且抛物线C 1上点P 处的切线与圆C 2：x 2+y 2=1相切于点Q ．（Ⅰ）当直线PQ 的方程为x-y- √2 =0时.求抛物线C 1的方程；（Ⅱ）当正数p 变化时.记S 1.S 2分别为△FPQ .△FOQ 的面积.求 S1S 2 的最小值．【正确答案】：【解析】：（Ⅰ）设点P （x 0. x 022p ）.代入直线PQ 的方程得一方程.再根据抛物线在P 处切线斜率为1列一方程.解方程组即可求得p 值；（Ⅱ）易表示出点p 处切线方程.据线圆相切得一方程.再与圆联立方程组可表示出Q 坐标.据弦长公式可表示出|PQ|.利用点到直线的距离公式可表示出点F 到切线PQ 的距离d.则S 1可表示.又 S 2=12|OF ||x Q | = p 2|x 0| .所以 S1S 2 可表示为关于x 0的函数.据函数结构特点利用基本不等式即可求得其最小值．【解答】：解：（Ⅰ）设点P （x 0. x 022p ）.由x 2=2py （p ＞0）得.y= x 22p .求导y′= x p . 因为直线PQ 的斜率为1.所以 x 0p =1且x 0- x 022p - √2 =0.解得p=2 √2 .所以抛物线C 1 的方程为 x 2=4√2y ．（Ⅱ）因为点P 处的切线方程为：y- x 022p = x 0p（x-x 0）.即2x 0x-2py- x 02 =0. 根据切线与圆切.得d=r.即 02√4x 02+4p 2 =1.化简得 x 04=4x 02+4p 2 .由方程组 {2x 0x −2py −x 02=0x 2+y 2=1x 04−4x 02−4p 2=0.解得Q （ 2x 0 . 4−x 022p ）. 所以|PQ|= √1+k 2 |x P -x Q |= √1+x 02p 2 |x 0−2x 0| = √p 2+x 02p |x 02−2x 0| . 点F （0. p 2 ）到切线PQ 的距离是d=202√4x 02+4p 2 = 12 √x 02+p 2 . 所以 S 1=12|PQ |•d = 12 ×√p 2+x 02p |x 02−2x 0| × 12 √x 02+p 2 = x 02+p 24p |x 02−2x 0| . S 2=12|OF ||x Q | = p2|x 0| .而由 x 04=4x 02+4p 2 知.4p 2= x 04−4x 02＞0 .得|x 0|＞2.所以 S 1S 2 = x 02+p 24p |x 02−2x 0|×2|x 0|p = (x 02+p 2)(x 02−2)2p 2 = (4x 02+x 04−4x 02)(x 02−2)2(x 04−4x 02) = x 02(x 02−2)2(x 02−4) = x 02−42+4x 02−4 +3≥2 √2 +3.当且仅当 x 02−42=4x 02−4 时取“=”号.即 x 02=4+2√2 .此时.p= √2+2√2 ．所以 S1S 2 的最小值为2 √2 +3．【点评】：本题主要考查抛物线几何性质、直线与抛物线的位置关系.同时考查解析几何的基本思想方法和运算求解能力.综合性强.难度大．22.（问答题.15分）已知a∈R .函数f （x ）=e x-1-ax 在点（1.1-a ）处与x 轴相切．（1）求a 的值.并求f （x ）的单调区间；（2）当x ＞1时.f （x ）＞m （x-1）lnx.求实数m 的取值范围．【正确答案】：【解析】：（1）求出函数的导数.利用已知条件列出方程.求出a.判断导函数的繁符号.然后求解单调区间．（2）令g （x ）=f （x ）-m （x-1）lnx.x ＞0.求出 g′(x )=e x−1−m (lnx +x−1x )−1 .令h （x ）=g'（x ）.求出导数.通过（i ）若 m ≤12 .（ii ）若 m ＞12 .判断函数的单调性求解最值.然后求解m的取值范围．【解答】：解：（1）函数f （x ）的定义域为R.f'（x ）=e x-1-a 因为f （x ）在点（1.1-a ）处与x 轴相切.所以f'（1）=e 1-1-a=1-a=0∴a=1f'（x ）=e x-1-1当f'（x ）=e x-1-1＞0时.x ＞1.当f'（x ）=e x-1-1＜0时.x ＜1.故函数f （x ）的递增区间为（1.+∞）.递减区间为（-∞.1）；（2）令g （x ）=f （x ）-m （x-1）lnx.x ＞0.则 g′(x )=e x−1−m (lnx +x−1x )−1 .令h （x ）=g'（x ）. ℎ′(x )=e x−1−m (1x +1x 2)（i ）若 m ≤12 .因为当x ＞1时.e x-1＞1. m (1x +1x 2)＜1 .所以h'（x ）＞0.所以h（x）即g'（x）在（1.+∞）上单调递增.又因为g'（1）=0.所以当x＞1时.g'（x）＞0.从而g（x）在[1.+∞）上单调递增.而g（1）=0.所以g（x）＞0.即f（x）＞m（x-1）lnx成立．.可以h'（x）在（0.+∞）上单调递增.（ ii）若m＞12因为h'（1）=1-2m＜0.h'（1+ln（2m））＞0.所以存在x1∈（1.1+ln（2m））.使得h'（x1）=0.且当x∈（1.x1）时.h'（x）＜0.所以h（x）即g′（x）在x∈（1.x1）上单调递减.又因为g′（1）=0.所以当x∈（1.x1）.g′（x）＜0.从而g（x）在[1.x1）上单调递减.而g（1）=0.所以当x∈（1.x1）时.g（x）＜0.即f（x）＞m（x-1）lnx不成立．]．综上所述.m的取值范围为(−∞，12【点评】：本题考查函数的导数的应用.函数的单调性、切线方程函数的最值的求法.考查分析问题解决问题的能力．。
浙江省宁波市镇海中学2019-2020学年高三上学期期中数学试题(解析版)
镇海中学2019学年第一学期期中考试高三年级数学试卷第I 卷(选择题共40分)一、选择题（本大题共10小题，每题4分，共40分）1.已知集合{}2|450,{|0ln 2}A x x x B x x =∈--≤=<<Z ，则A B I 的元素的个数为（ ） A. 2 B. 3C. 4D. 7【答案】C 【解析】 【分析】求出集合,A B ，根据集合的交集运算即可求解. 【详解】}{{}24501,0,1,2,3,4,5A x Z x x =∈--≤=-，{}{}20ln 21B x x x x e =<<=<<{}2,3,4,5A B ∴⋂=所以A B ⋂的元素的个数为4. 故选：C【点睛】本题主要考查集合的交集概念与运算，属于基础题. 2.若,,a b c ∈R 且a b >，则下列不等式中一定成立的是（ ） A. ac bc >B. 2()0a b c ->C.11a b＜ D.22a b -<-【答案】D 【解析】 【分析】根据不等式的性质即可判断.【详解】对于A ，若0c ≤，则不等式不成立；对于B ，若0c =，则不等式不成立； 对于C ，若,a b 均为负值，则不等式不成立；对于D ，不等号的两边同乘负值，不等号的方向改变，故正确； 故选：D【点睛】本题主要考查不等式的性质，需熟练掌握性质，属于基础题. 3.已知n S 是等差数列{}n a 的前n 项和，且244,18S S ==，则6S 等于（ ）A. 50B. 42C. 38D. 36【答案】B 【解析】 【分析】由等差数列前n 项和的性质：232,,n n n n n S S S S S --成等差数列即可求解. 【详解】由24264,,S S S S S --成等差数列， 所以()()62184418S -=+- 所以642S =， 故选：B【点睛】本题主要考查等差数列前n 项的性质，需熟记性质的内容，属于基础题. 4.函数()243xx f x =的图象大致为（ ）A. B.C. D.【答案】A 【解析】 【分析】由函数的奇偶性、极限值以及特殊值，利用排除法即可判断. 【详解】()()f x f x -=Q ，可知函数为偶函数，可排除C ；当x →+∞时，由于指数函数的增长速度快，则()0f x →，可排除B ； 当2x =时，()216162239y f ===<，特殊值法可排除D ； 故选：A【点睛】本题主要考查函数奇偶性等性质的应用，利用函数的性质求函数的大致图像可采用排除法，此题属于中档题.5.如图是一个几何体的三视图，则这个几何体的表面积是（ ）A. 76B. 84C.76+ D.84+【答案】C 【解析】【分析】几何体为侧放的五棱柱，底面为正视图中的五边形，棱柱的高为4. 【详解】由三视图可知几何体为五棱柱，底面为正视图中的五边形，高为4. 所以五棱柱的表面积为(14422244224762⎛⎫⨯-⨯⨯⨯+++++⨯=+⎪⎝⎭故选：C【点睛】本题主要考查三视图，解题的关键是还原几何体，属于基础题. 6.将函数()y f x =的图象向右平移6π个单位长度后，得到()26g x sin x π⎛⎫=+ ⎪⎝⎭，则()y f x =的函数解析式为（ ）A. ()cos2f x x =-B. ()sin 26f x x π⎛⎫=-⎪⎝⎭C. ()cos2f x x =D. ()cos 26f x x π⎛⎫=- ⎪⎝⎭【答案】C 【解析】 【分析】根据函数图象的平移法则即可求解. 【详解】把()26g x sin x π⎛⎫=+⎪⎝⎭的图像向左平移6π个单位可得 ()sin 2sin 2cos 2662f x x x x πππ⎡⎤⎛⎫⎛⎫=++=+= ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦，故选：C【点睛】本题主要考查三角函数图像的平移，需掌握平移法则，平移是对变量x 平移且“左加右减” ，属于基础题.7.设命题():lg 210p x -≤，命题()10x a q x a-+≤-：，若q 是p 的必要不充分条件，则实数a的取值范围是（ ） A. 102⎡⎤⎢⎥⎣⎦， B. 102⎛⎫ ⎪⎝⎭，C. 102⎡⎫⎪⎢⎣⎭，D. ∅【答案】A 【解析】 【分析】首先求出命题p 、q 中不等式的解集，再根据命题之间的关系推出集合的包含关系即可求出参数的取值范围.【详解】解不等式()lg 210x -≤得02-11x <≤，所以112x <≤， 故满足命题p 的集合112P xx ⎧⎫=<≤⎨⎬⎩⎭， 解不等式()10x a x a-+≤-得()()10x a x a --+≤⎡⎤⎣⎦且x a ≠，所以1a x a <≤+故命题q 的集合{}1Q x a x a =<≤+， 若q 是p 的必要而不充分条件，则P n Q即1211a a ⎧≤⎪⎨⎪+≥⎩解得102a ≤≤故选：A【点睛】本题主要考查命题中必要不充分条件，解题的关键是根据命题的关系推出集合的包含关系，属于基础题. 8.已知22ππαβ--＜＜，sin 2cos 1αβ-=，2cos sin αβ+=则3sin πβ⎛⎫-= ⎪⎝⎭（ ）A.B.C.D.【答案】B 【解析】 【分析】两式平方相加利用两角和与差的公式可化为()54sin 3αβ--=，再根据22ππαβ-<-<得出6παβ=+，代入2cos sin αβ+=.【详解】将两个等式两边平方可得2222sin 4sin cos 4cos 1cos 4cos sin 4sin 2ααββααββ⎧-⋅+=⎨+⋅+=⎩， 两式相加可得()54sin 3αβ--=，所以()1sin 2αβ-=， 22ππαβ-<-<Q ，6παβ∴-=，即6παβ=+，代入2cos sin αβ+=3sin 22ββ+=，所以sin 63πβ⎛⎫+= ⎪⎝⎭， 故选：B【点睛】本题主要考查三角函数的化简求值，需熟记两角和与差的公式以及常见的三角函数值，属于中档题.9.已知椭圆和双曲线有相同的焦点12,F F ，设点P 是该椭圆和双曲线的一个公共点，且123F PF π∠=，若椭圆和双曲线的离心率分别为12,e e ，则2212e e +的最小值为（ ）A.B. 4+C. 2D. 1【答案】A 【解析】 分析】设出椭圆与双曲线的标准方程，利用定义可得：12,2m n a m n a +=-=，解出,m n ，利用余弦定理解关于12,e e 的等式，再由基本不等式求出2212e e +的最小值即可.【详解】不妨设椭圆与双曲线的标准方程为：()222210x y a b a b+=>>，()2211221110,0x y a b a b -=>>， 设1PF m =，2PF n =，m n >. 则12,2m n a m n a +=-=，11,m a a n a a ∴=+=- ， 12PF F ∆中，123F PF π∠=，由余弦定理可得2221241cos cos 322m n c F PF mn π+-∠===，化为()()()()22211114a a a a c a a a a ++--=+-，所以2221340a a c +-=，2212134e e ∴+=， ()(2222222112122222121231131144444e e e e e e e e e e ⎛⎫⎛⎫∴+=++=++≥+= ⎪ ⎪⎝⎭⎝⎭当且仅当21e =时，取等号，则2212e e +. 故选：A【点睛】本题主要考查椭圆与双曲线的性质以及基本不等式求最值，考查了学生的计算能力，综合性比较强.10.设a ，b 为正实数，且121322a b a b +++=，则12a b +的最大值和最小值之和为（ ） A. 2 B. 92 C. 132D. 9【答案】C 【解析】 【分析】根据题意可得2122113a b a b ⎡⎤⎛⎫+++= ⎪⎢⎥⎝⎭⎣⎦，再由“1”与12a b +相乘利用基本不等式转化为221212913a b a b⎡⎤⎛⎫++≤+⎢⎥ ⎪⎝⎭⎢⎥⎣⎦，解不等式即可求解. 【详解】由121322a b a b +++=，则2122113a b a b ⎡⎤⎛⎫+++= ⎪⎢⎥⎝⎭⎣⎦，所以1221212213a b a b a b a b ⎡⎤⎛⎫⎛⎫+=++++ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦ 2222121413a b b a a b ⎡⎤⎛⎫=+++++⎢⎥ ⎪⎝⎭⎢⎥⎣⎦22212212591313a b a b ⎡⎤⎡⎤⎛⎫⎛⎫≥++=++⎢⎥⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎢⎥⎣⎦⎣⎦， 当且仅当22a b b a=时，即32a b ==或23时，等号成立，即221212913a b a b⎡⎤⎛⎫++≤+⎢⎥ ⎪⎝⎭⎢⎥⎣⎦，解得12922a b ≤+≤ 所以12a b +的最大值为92；最小值为2； 所以最大值和最小值之和为132. 故选：C【点睛】本题主要考查利用基本不等式求最值，运用基本不等式求最值需验证等号成立的条件，属于中档题.第Ⅱ卷(非选择题共110分)二、填空题（本大题共7小题，多空题每题6分，单空题每题4分，共36分）11.抛物线22y x =的焦点坐标是___________，准线方程是___________.【答案】1(,0)2，12x =-.【解析】试题分析：由题意得，焦点坐标是1(,0)2，准线方程是12x =-，故填：1(,0)2，12x =-.考点：抛物线的标准方程及其性质.12.已知点A （1，0），B （0，2），点(),P a b 在线段AB 上，则直线AB 的斜率为______；⋅a b 的最大值为______．【答案】 (1). 2- (2). 12【解析】 【分析】由直线上两点求斜率公式：1212y y k x x -=-，可求斜率，再用二次函数配方求最值即可求解.【详解】由A （1，0），B （0，2），可得20201AB k -==--，所以直线AB 的斜率为2-， 直线AB ：22y x =-+由点(),P a b 在线段AB 上，所以()2201b a a =-+≤≤，所以()21122224a b a a a ⎡⎤⎛⎫⋅=-+=---⎢⎥ ⎪⎝⎭⎢⎥⎣⎦，所以a b ⋅的最大值为12.故答案为：2-；12【点睛】本题主要考查直线的斜率以及直线方程，需熟记斜率公式以及点斜式方程，属于基础题.13.若实数(),x y 满足约束条件20201x y x y y --≤⎧⎪+-≥⎨⎪≤⎩，则2x y -的最小值为_____ ；______．【答案】 (1). 1(2). 2【解析】 【分析】作出约束条件满足的可行域，然后利用目标函数表示的几何意义即可求解.【详解】作出约束条件20201x y x y y --≤⎧⎪+-≥⎨⎪≤⎩满足的可行域，设2z x y =-，则2y x z =-，由图可知在C 处取得最小值，由201x y y +-=⎧⎨=⎩解得11x y =⎧⎨=⎩，所以()1,1C ， 所以min 2111z =⨯-=，即2x y -的最小值为1；(),x y ，()0,1-两点间的距离，设()0,1-到直线20x y +-=的距离为d ， 则2d==2故答案为：1；2【点睛】本题主要考查简单的线性规划问题，解题的关键是作出可行域，理解目标函数表示的几何意义，属于基础题.14.已知长方体1111ABCD A B C D -中，1112AA AB AD ===，，1AA 与平面1A BD 所成的角为______.【答案】60o 【解析】 【分析】根据等体积法求出点A 到平面1A BD 的距离h ，在直角三角中利用“对边比斜边”即可求解.【详解】设A 到平面1A BD 的距离为h ，在长方体1111ABCD A B C D -中，1112AA AB AD ===，， 则1A D ==2BD ==，1A B ==在1A BD ∆中，由余弦定理15134cos BA D +-∠==，所以1sin BA D ∠=所以111sin 1222A BD S BA D =⋅∠= 因为11A ABD A A BD V V --=，即111133ABD A BD S AA S h ∆⋅⋅=⋅⋅，解得4h =设直线1AA 与平面1A BD 所成的角为θ，则1sin 2h AA θ== 所以60θ=o . 故答案为：60o【点睛】本题主要考查立体几何中的线面角，解题的关键是找到线面所成角，放在三角形中求解，此题也可以建立空间直角坐标系，采用向量法.15.已知{}n a 是等比数列，且0n a >，243546225a a a a a a ++=，则35a a +=__________，4a 的最大值为__________．【答案】 (1). 5 (2). 52【解析】243546225a a a a a a ++=22233553535225()25,05n a a a a a a a a a ⇒++=⇒+=>∴+=Q22354354255()242a a a a a a +∴=≤=⇒≤ ，即4a 的最大值为5216.已知圆22:1O x y +=，设点P 是恒过点（0，4）的直线l 上任意一点，若在该圆上任意点A 满足3OPA π∠≤，则直线l 的斜率k 的取值范围为______．【答案】,⎫⎛+∞-∞⎪ ⎪ ⎣⎭⎝⎦U 【解析】 【分析】由题意在该圆上任意点A 满足3OPA π∠≤，则当直线l 与圆相切时，3OPA π∠=，从而结合图像可知直线的倾斜角的取值范围为566ππα≤≤，由tan k α=即可求解. 【详解】由题意在该圆上任意点A 满足3OPA π∠≤，则当直线l 与圆相切时，3OPA π∠=，设直线的倾斜角为α，由图可知566ππα≤≤因为tan k α=，所以3k ≥或3k ≤-即斜率k 的取值范围为,⎫⎛+∞⋃-∞⎪ ⎪ ⎣⎭⎝⎦，故答案为：,33⎫⎛+∞⋃-∞-⎪ ⎪ ⎣⎭⎝⎦. 【点睛】本题主要考查直线与圆的位置关系以及直线斜率与直线倾斜角的关系，属于基础题. 17.已知单位圆上两点,A B 满足120AOB ∠=o ，点C 是单位圆上的动点，且OC xOA yOB =+u u u r u u u r u u u r，则2x y -的取值范围为_____．【答案】[]22-,【解析】 【分析】由点C 是单位圆上的动点，求得[]1(2)1,12OC OB x y ⋅=--∈-u u u r u u u r ，由此能求出2x y -的取值范围.【详解】Q 单位圆上两点,A B 满足120AOB ∠=o ，点C 是单位圆上的动点，OC xOA yOB =+u u u r u u u r u u u r ，,,OA OB OC ∴u u u r u u u r u u u r 均为单位向量，即221OA OB ==u u u r u u u r ，12OA OB ⋅=-u u u r u u u r ，Q 点C 是单位圆上的动点， ∴OC OB ⋅u u u r u u u r的取值范围是[]1,1-，又Q ()OC OB xOA yOB OB ⋅=+⋅u u u r u u u r u u u r u u u r u u u r[]11(2)1,122xOA OB yOB OB x y x y =⋅+⋅=-+=--∈-u u u r u u u r u u u r u u u r2x y ∴-的取值范围为[]22-,.故答案为：[]22-,【点睛】本题主要考查向量的坐标运算以及向量的数量积，考查学生分析问题的能力，属于向量的综合题，三、解答题（本大题共5小题，共74分）18.已知()222x x x f x sincos sin a ⎛⎫=⋅++ ⎪⎝⎭（1）求实数a 的值；（2）若443f f ππαα⎛⎫⎛⎫++-= ⎪ ⎪⎝⎭⎝⎭，求2141tan παα⎛⎫-+ ⎪⎝⎭+的值． 【答案】（1）12-；（2）516.【解析】 【分析】（1）由二倍角的正、余弦公式以及辅助角公式化简即可求解.（2）由（1）中f （x）24sin x π⎛⎫=- ⎪⎝⎭，把443f f ππαα⎛⎫⎛⎫++-=⎪ ⎪⎝⎭⎝⎭ 代入化简可得43sin πα⎛⎫-= ⎪⎝⎭，再根据两角和与差的公式求出sin ,cos αα的值，代入即可求解. 【详解】（1）()()1112222242x x x f x sin cos sin a sinx cosx a x a π⎛⎫⎛⎫=⋅++=-++=-++ ⎪ ⎪⎝⎭⎝⎭，由于函数的最大值为2， 故102a +=，解得a 12=-． （2）由于f （x）24sin x π⎛⎫=- ⎪⎝⎭，所以443f f ππαα⎛⎫⎛⎫++-= ⎪ ⎪⎝⎭⎝⎭，整理得43sin πα⎛⎫-= ⎪⎝⎭．所以4cos πα⎛⎫-= ⎪⎝⎭，所以224466sin sin ππαα⎛⎫=-+= ⎪⎝⎭．44cos cos ππαα⎛⎫=-+=⎪⎝⎭或，所以2626sin cos αα⎧+=⎪⎪⎨-⎪=⎪⎩或2626sin cos αα⎧=⎪⎪⎨⎪=-⎪⎩，故22122122422211sin cos sin sin cos sin sin cos tan cos cos παααααααααααααα⎛⎫-+⋅⋅+ ⎪+⎝⎭===+++，所以当2626sin cos αα⎧+=⎪⎪⎨⎪=⎪⎩时．144523616sin cos αα-==．当2626sin cos αα⎧=⎪⎪⎨⎪=-⎪⎩时，144523616sin cos αα-==，所以原式516=． 【点睛】本题主要考查三角恒等变换的倍角公式、两角和与差的公式，需熟记公式，属于基础题.19.在锐角ABC ∆中，角A ，B ，C 所对边分别为a ，b ，c ，已知223,39b a c c ==-+．（1）求A ；（2）求22sin sin B C +的取值范围． 【答案】（1）3π；（2）53,42⎛⎤ ⎥⎝⎦. 【解析】 【分析】（1）利用余弦定理即可求解. （2）由23B C π+=，以及两角和与差的公式，则sin 2B +sin 2C ＝112+sin （2B 6π-），再由022032B B πππ⎧⎪⎪⎨⎪-⎪⎩＜＜＜＜，求出6π＜B 2π＜即可求解. 【详解】（1）在锐角△ABC 中，∵b ＝3，a 2＝c 2﹣3c +9， ∴可得c 2+b 2﹣a 2＝bc ，∴由余弦定理可得：cos A 2221222b c a bc bc bc +-===，∴由A 为锐角，可得A 3π=．（2）∵sin 2B +sin 2C ＝sin 2B +sin 2（23π-B ）＝sin 2B +B 12+sin B ）2＝112+（B 12-cos2B ）＝112+sin （2B 6π-）， 又∵022032B B πππ⎧⎪⎪⎨⎪-⎪⎩＜＜＜＜，可得6π＜B 2π＜，∴2B 6π-∈（6π，56π）， ∴sin （2B 6π-）∈（12，1]，∴sin 2B +sin 2C ＝112+sin （2B 6π-）∈（54，32]，即sin 2B +sin 2C 的取值范围是（54，32]．【点睛】本题主要考查余弦定理解三角形以及三角恒等变换两角和与差的公式，解题的关键是利用三角形的内角和求出B 的取值范围，此题属于中档题.20.如图，在三棱锥P ﹣ABC 中，PAB ∆和ABC ∆都为等腰直角三角形，PA PB ⊥，AB AC ⊥，M 为AC 的中点，且PM AC =．（1）求二面角P ﹣AB ﹣C 的大小； （2）求直线PM 与平面PBC 所成角的正弦值．【答案】（1）120o ；（2 【解析】 【分析】（1）取线段AB ，BC 的中点O ，N ，连接PO ，ON ，MN ，PN ，证出PON ∠为P ﹣AB ﹣C 二面角，在PON ∆中利用余弦定理即可求解.（2）由（1）以AO 为x 轴，以ON 为y 轴，过O 作平面ABC 的垂线，以垂线为z 轴建立空间直角坐标系，求出平面PBC 的一个法向量，利用空间向量的数量积即可求出线面角. 【详解】（1）分别取线段AB ，BC 的中点O ，N ，连接PO ，ON ，MN ，PN ，设AC ＝2，则有 在等腰直角△P AB 中，O 是中点， 则有AB ⊥PO ﹣﹣﹣①在等腰直角△ABC 中，点O ，N 分别是AB ， BC中点，则有AB ⊥ON ﹣﹣﹣②由①②可知，AB ⊥平面PON ，又∵MN ∥AB ，∴MN ⊥平面PON ，则有MN ⊥PN ． 又AB ＝2，则 MN ＝1，又PM ＝AC ＝2，则有PN ==，又OP ＝ON ＝1，由三角形余弦定理可知，12cos PON ∠=-， ∴∠PON ＝120o ，即二面角P ﹣AB ﹣C 的大小为120o ． （2）建立如图所示的空间直角坐标系，过点P 作PD ⊥ON 交NO 延长线于点D ，设AB ＝AC ＝2，的则有A （﹣1，0，0），C （﹣1，2，0），B （1，0，0），M （﹣1，1，0）， 由（1）可知，∠POD ＝180°﹣∠PON ＝60°，又∵OP ＝1，∴122OD PD ==，． ∴1002D ⎛⎫- ⎪⎝⎭，，，102P ⎛- ⎝⎭，．∴312PM ⎛=- ⎝⎭u u u u r ，， 设平面PBC 的一个法向量为()n x y z =r ，，，则有0n BP n BC ⎧⋅=⎪⎨⋅=⎪⎩u u u r r u u u r r ，又∵112BP ⎛=-- ⎝⎭u u u r ，，()220BC =-u u u r ，，，∴102220x y z x y ⎧--=⎪⎨⎪-+=⎩，∴(n =r． 设直线PM 与平面PBC 所成角为θ，则有：10sin θ=故直线PM 与平面PBC【点睛】本题主要考查立体几何的二面角以及运用空间直角坐标系求线面角，求二面角步骤“作、证、求”，关键作出二面角；同时此题考查了学生的计算能力，属于基础题. 21.已知数列{}n a 的前n 项和为n S ，且满足：()*11232n n a a S n N+==-+∈，．（1）求数列{}n a 的通项公式；（2）数列{}n b 满足12b =-，()()()2132n n n b b n n n a +-+=+，求数列{}n b 通项公式．【答案】（1）(2)nn a =--；（2）(2)nn b n-=. 【解析】 【分析】（1）由n S 与n a 的关系可求得数列{a n }是等比数列，再由等比数列的通项公式即可求解. （2）由（1）把n a 代入可得()()12322nn nn bb n n++--=-+，裂项化简即可求解.【详解】（1）数列{a n }的前n 项和为S n ，且满足：()*11232n n a a S n N +==-+∈，．当n ≥2时，a n ＝﹣3S n ﹣1+2，两式相减得：a n +1＝﹣2a n ， 所以数列{a n }是以2为首项﹣2为公比的等比数列．所以(2)nn a =--．（2）由于()()()2132n n n b b n n n a +-+=+，所以()()12322nn nn bb n n++--=-+，由于()()()((122[2)3223212(2)(2)(2)[22)111nn n n n n nn n n nn n n n n n n n +⎤--+-⎤⎡+----⎛⎫⎦⎤-=⋅--=+--=+=- ⎪⎥⎢⎦+++++⎝⎭⎦⎣， 所以()()11221n nn nbb n n++---=-+，所以(2)nn b n-=．【点睛】本题主要考查等比数列的通项公式，需掌握等比数列的定义以及通项公式，属于中档题.22.在平面直角坐标系中，已知()()2,0,2,F P t -，若线段FP 的中垂线l 与抛物线C ：()220y px p =>总是相切．（1）求抛物线C 的方程；（2）若过点Q （2，1）的直线l ′交抛物线C 于M ，N 两点，过M ，N 分别作抛物线的切线12,l l 相交于点A ．12,l l 分别与y 轴交于点B ，C ．（ i ）证明：当'l 变化时，ABC ∆的外接圆过定点，并求出定点的坐标 ； （ ii ）求ABC ∆的外接圆面积的最小值．【答案】（1）28y x =；（2）（i ）证明见解析；（ii ）14417π. 【解析】【分析】 （1）根据F （2，0），P （﹣2，t ）得FP 的中点为（0，2t ），，讨论t 的值，当t ≠0时，求出线段FP 的中垂线l ，代入抛物线方程y 2＝2px ，0∆=即可求解.（2）设过点Q （2，1）的直线l ′的方程为x ﹣2＝m （y ﹣1），代入抛物线的方程y 2＝8x ， 求出y 1+y 2＝8m ，y 1y 2＝8m ﹣16，对y 2＝8x 两边求导得2y •y ′＝8，即y ′4y=，求出,M N 处的切线方程，再求出,B C ，设出外接圆的方程即可求出定点；由上一问可求出半径，配方求半径的最小值即可求解.【详解】（1）F （2，0），P （﹣2，t ），可得FP 的中点为（0，2t ）， 当t ＝0时，FP 的中点为原点， 当t ≠0时，直线FP 的斜率为4t -，线段FP 的中垂线l 的斜率为4t， 可得中垂线l 的方程为y 4t =x 2t +，代入抛物线方程y 2＝2px ， 可得216t x 2+（4﹣2p ）x 24t +=0， 由直线和抛物线相切可得△＝（4﹣2p ）2﹣16＝0，解得p ＝4，则抛物线的方程为y 2＝8x ；（2）（i ）证明：可设过点Q （2，1）的直线l ′的方程为x ﹣2＝m （y ﹣1），即x ＝my +2﹣m ， 代入抛物线的方程y 2＝8x ，可得y 2﹣8my ﹣16+8m ＝0，设M （218y ，y 1），N （228y ，y 2），则y 1+y 2＝8m ，y 1y 2＝8m ﹣16， 由y 2＝8x ，两边对x 求导可得2y •y ′＝8，即y ′4y=， 可得M 处的切线方程为y ﹣y 114y =（x 218y -），化为y 1y ＝4x 212y +，①同理可得N 处的切线方程为y 2y ＝4x 222y +，② 由①②可得y 122y y +==4m ，x 128y y ==m ﹣2，即A （m ﹣2，4m ）， 又l 1，l 2分别与y 轴交于点B （0，12y ），C （0，22y ）， 设过A ，B ，C 的外接圆的方程为x 2+y 2+Dx +Ey +F ＝0，（D 2+E 2﹣4F ＞0）， 即有()()21122222042042216240y y E F y y E F m m D m mE F ⎧++=⎪⎪⎪++=⎨⎪⎪-++-++=⎪⎩结合y 1+y 2＝8m ，y 1y 2＝8m ﹣16，可得D ＝﹣m ﹣2，E ＝﹣4m ，F ＝4m ﹣8，可得△ABC 的外接圆方程为x 2+y 2﹣（m +2）x ﹣4my +4m ﹣8＝0，可得m （4﹣x ﹣4y ）+（x 2+y 2﹣2x ﹣8）＝0，由2244280x y x y x +=⎧⎨+--=⎩可得40x y =⎧⎨=⎩或28172417x y ⎧=-⎪⎪⎨⎪=⎪⎩， 则当l ′变化时，△ABC 的外接圆过定点（4，0）和（2817-，2417）；（ii ）△ABC 的外接圆的半径r2===可得当m 617=时，r=， 则△ABC 外接圆面积的最小值为14417π． 【点睛】本题主要考查直线与抛物线的位置关系，考查学生的分析计算能力，综合性比较强. 的的。
精品解析：【全国百强校】浙江省宁波市镇海中学2019届高三上学期期中考试数学试题(解析版)
浙江省镇海中学2018-2019学年高三上学期期中考试数学试卷一、选择题：（本大题共10个小题,每小题4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.设全集U R =，集合{}{}|3,|05A x x B x x =??，则集合()U C A B?（ ）A. {}|03x x＃ B. {}|03x x <<C. {}|03x x <?D. {}|03x x ?【答案】D 【解析】 【分析】先根据补集的定义求出集合A 的补集U C A ，然后和集合B 进行交集运算，可求()U C A B Ç 【详解】因为A={x|x≥3}， 所以U C A ={x|x ＜3}，所以（U C A ）∩B═{x|0≤x＜3}． 故选：D ．【点睛】本题的考点是集合的补集和交集运算，比较基础．2.某几何体的三视图如图所示，图中的四边形都是边长为2的正方形，两条虚线互相垂直，则该几何体的体积是（ ）A. 83p -B. 163C. 86p -D. 203【答案】D 【解析】 【分析】由三视图知原几何体是一个棱长为2的正方体挖去一四棱锥得到的，根据所提供的数据可求出正方体、锥体的体积，从而得到答案．【详解】由三视图知原几何体是一个棱长为2的正方体挖去一四棱锥得到的，该四棱锥的底为正方体的上底，高为1， 如图所示：所以该几何体的体积为23﹣13×22×1=203． 故选：D ．【点睛】本题考查三视图，考查柱体、锥体的体积计算，解决该类问题的关键是由三视图还原得到原几何体，画三视图的要求为：“长对正，高平齐，宽相等”． 3.记n S 为等差数列{}n a 的前n 项和，若945S =，3812a a +=，则7a 等于（ ）A. 10B. 9C. 8D. 7 【答案】B 【解析】由题意可得：955945,5S a a ==\=，由等差数列的性质可得：385666512,7a a a a a a +=+=+=\=， 该数列的公差：652d a a =-=，故76729a a d =+=+=. 本题选择B 选项.4.满足线性约束条件23,23,0,0x y x y x y ì+?ïï+?ïí³ïï³ïî的目标函数z x y =+的最大值是 （ ） A. 1 B. 32C. 2D. 3 【答案】C 【解析】画出可行域如图阴影部分所示，易得1,1A （）z x y ＝＋在1,1A （）处取得最大值2max z ＝ 故选C点睛：本题主要考查线性规划中利用可行域求目标函数的最值，属简单题.求目标函数最值的一般步骤是“一画、二移、三求”：（1）作出可行域（一定要注意是实线还是虚线）；（2）找到目标函数对应的最优解对应点（在可行域内平移变形后的目标函数，最先通过或最后通过的顶点就是最优解）；（3）将最优解坐标代入目标函数求出最值.视频5.已知函数()2In x f x x x=-，则函数()f x 的图象为（ ）A.B.C.D.【答案】D 【解析】 【分析】写出分段函数，分段求导后利用导函数的符号或导函数的零点判断函数f （x ）的图象的形状．【详解】()2ln x f x x x =-=()2200lnxx x x ln x x x x ì-ïïí-ï-ïî，＞，＜， 当x ＜0时，()()212ln x f x x x--¢=-=()3221x ln x x-+-．令g （x ）=2x 3﹣1+ln （﹣x ）， 由()3216160x g x x x x =¢+=+=，得x =- 当x∈（﹣∞，-x ）＞0，当x∈（-0）时，g′（x ）＜0． 所以g （x）有极大值为32(1g -=?-+=416033ln --＜．又x 2＞0，所以f′（x ）的极大值小于0． 所以函数f （x ）在（﹣∞，0）上为减函数．当x ＞0时，()212lnx f x x x -¢=-=3221x lnx x -+．令h（x）=2x3﹣1+lnx，()2160h x xx¢=+＞．所以h（x）在（0，+∞）上为增函数，而h（1）=1＞0，h（12）=﹣3204ln-＜．又x2＞0，所以函数f′（x）在（0，+∞）上有一个零点，则原函数有一个极值点．综上函数f（x）的图象为D中的形状．故选：D．【点睛】函数图象的辨识可从以下方面入手：（1）从函数的定义域，判断图象的左右位置；从函数的值域，判断图象的上下位置；（2）从函数的单调性，判断图象的变化趋势；（3）从函数的奇偶性，判断图象的对称性；（4）从函数的特征点，排除不合要求的图象.6.若a、b是两个相交平面，则在下列命题中，真命题的序号为（）①若直线m a^，则在平面b内一定不存在与直线m平行的直线．②若直线m a^，则在平面b内一定存在无数条直线与直线m垂直．③若直线m aÌ，则在平面b内不一定存在与直线m垂直的直线．④若直线m aÌ，则在平面b内一定存在与直线m垂直的直线．A. ①③B. ②③C. ②④D. ①④【答案】C【解析】试题分析：对于①，若直线m a^，如果a，b互相垂直，则在平面b内，存在与直线m平行的直线，所以①是错误的；对于②，若直线m a^，则直线m垂直于平面a内的所有直线，则在平面b内，一定存在无数条直线与直线m垂直，所以②正确；对于③，若直线m aÌ，则在平面b内，一定存在与直线m垂直的直线，所以③是错误的；对于④，若直线m aÌ，则在平面b内，一定存在与直线m垂直的直线，所以④是正确的．故应选C．考点：1、直线与平面之间的位置关系．7.已知sin 6pa 骣琪-琪桫cos22a a =（ ） A.109 B. 109- C. 59- D. 59【答案】A 【解析】分析：先把cos 22a a 变形为2sin 26p a 骣琪+琪桫，而22626p ppa a 骣琪+=--琪桫，故可以利用诱导公式和二倍角公式求解.详解：因为cos 222sin 26pa a a 骣琪=+琪桫，故cos 222sin 22cos 2266pppa a a a 轾轾骣骣犏犏琪琪=--=-琪琪犏犏桫桫臌臌21024sin 69p a 骣琪=--=琪桫，故选A.点睛：本题考查诱导公式和两角和差的余弦、正弦公式的逆用，属于基础题.解题中注意根据正弦、余弦前面的系数选择合适的辅助角变形，另外在求值过程中注意寻找已知的角和未知的角之间的联系. 8.已知正项等比数列{}n a 满足7652a a a =+，若存在两项,m n a a ，使得2116m n a a a ?，则19m n+的最小值为（ ） A.32 B. 114 C. 83 D. 103【答案】B 【解析】 【分析】设{a n }的公比为q （q ＞0），由等比数列的通项公式化简a 7=a 6+2a 5，求出q ，代入a m a n =16a 12化简得m ，n 的关系式，由“1”的代换和基本不等式求出式子的范围，验证等号成立的条件，由m 、n 的值求出式子的最小值．【详解】设正项等比数列{a n }的公比为q ，且q ＞0， 由7652a a a =+得：6a q=6a +62a q， 化简得，q 2﹣q ﹣2=0，解得q=2或q=﹣1（舍去）， 因为a m a n =16a 12，所以()()1111m n a q a q --=16a 12，则q m+n ﹣2=16，解得m+n=6，所以19m n +=16（m+n ）（19m n +）=16（10+9n m m n +）≥1106骣琪+琪桫=83，当且仅当9n m m n =时取等号，此时96n m m n m n ì=ïíï+=î，解得3292m n ì=ïïíï=ïî， 因为m n 取整数，所以均值不等式等号条件取不到，则19m n +＞83， 验证可得，当m=2、n=4时，19m n +取最小值为114， 故选：B ．【点睛】本题考查等比数列的通项公式，利用“1”的代换和基本不等式求最值问题，考查化简、计算能力，注意等号的成立的条件，属于易错题．9.已知双曲线()2222:10,0x y C a b a b-=>>的左右焦点分别为12,F F ，P 为双曲线C 上一点，Q 为双曲线C 渐近线上一点，,P Q 均位于第一象限，且22QP PF =，120QF QF ?，则双曲线C 的离心率为（ ）1122 【答案】C 【解析】分析：设()(),0Q at bt t >，根据12FQF D 为直角三角形可以得到1t =，再根据22QP PF =得2323c am b m ì+=ïïíï=ïî，代入双曲线方程可得到离心率. 详解：设()(),0Q at btt >，(),P m n ，注意到1290FQF ??，从而OQ c =，故22222b t a t c +=即1t =，故(),QP m a n b =--，()2,PF c m n =--.又2222m a c m n b n ì-=-ïí-=-ïî，解得2323c am b m ì+=ïïíï=ïî，代入双曲线方程，则有()222224199c a b a b +-=，2ca=，故选C. 点睛：离心率的计算，关键在合理构建关于,,a b c 的等量关系，本题中Q 的坐标与,,a b c 有关联，这种关联可以通过向量关系式转化到P ，最后根据P 在双曲线上可以得到离心率的大小.10.如图，在三棱柱111ABC A B C -中，底面为边长为2的正三角形，1B 在底面的射影为AC 中点且1B 到底,E F 分别是线段1AB 与1CA 上的动点，记线段EF 中点M 的轨迹为L ，则L 等于（ ）(注：L 表示L 的测度，本题中L 若分别为曲线、平面图形、空间几何体，分别对应为其长度、面积、体积）A. 1B.C.D.【答案】D 【解析】由题意画出图形，取特殊点得到M 的轨迹为平行四边形区域，再建立空间坐标系求出面积即可． 【详解】当E 位于B 1（或A ），而F 在A 1C 上移动时，M 的轨迹为平行于A 1C 的一条线段， 当F 位于A 1（或C ），而E 在AB 1上移动时，M 的轨迹为平行与AB 1的一条线段． 其它情况下，M 的轨迹构成图中平行四边形内部区域． 设异面直线AB 1与CA 1所成角为θ， ∴|L|=2×12|12AB 1|•|12CA 1|•sin θ=14|AB 1|•|CA 1|•sin θ．以O 为原点，OB 、OC 、O 1B 为x 轴，y 轴，z 轴建立空间坐标系， 则()()((11A 01,0C 0,1,0,?B A ---，，， ∴()(110,1332AB AC ==-，，，∴11210AB AC ==，111110cos θAB AC AB AC ==,sinθ=∴|L|=124创故选：D【点睛】本题考查棱柱的结构特征，考查空间想象能力和思维能力，利用特殊点得到M 的轨迹是解答该题的关键，是压轴题．二、填空题：（本大题共7个小题,多空题每题6分，单空题每小题4分,共36分．）11.中国古代数学著作《九章算术》中有一个这样的问题：“某贾人擅营，月入益功疾（注：从第2月开始，每月比前一月多入相同量的铜钱，3月入25贯，全年（按12个月计）共入510贯“，则该人每月比前一月多入_________________贯，第12月营收贯数为_________________. 【答案】 (1). 5 (2). 70 【解析】 【分析】设每个月的收入为等差数列{a n }．公差为d ．可得a 3=25，S 12=510．利用等差数列的通项公式与求和公式即【详解】设每个月的收入为等差数列{a n }．公差为d ． 则a 3=25，S 12=510． ∴a 1+2d=25，12a 1+12112´d=510，解得a 1=15，d=5， ∴12a = a 1+11d=15+55=70 故答案为：5,70【点睛】本题考查了等差数列的通项公式与求和公式，考查了推理能力与计算能力，属于中档题． 12.sin 26y x p骣琪=+琪桫的最小正周期为_________________，为了得到函数sin 26y x p 骣琪=+琪桫的图象，可以将函数cos 2y x =的图象向左最小移动_______个单位 【答案】 (1). p (2).56p【解析】 【分析】利用正弦型周期公式得到最小周期性，先根据诱导公式进行化简y=cos2x 为正弦函数的类型，再由左加右减上加下减的原则可确定平移的方案． 【详解】sin 26y x p骣琪=+琪桫的最小正周期为22p p =，由题意y=cos2x=sin （2x +2p ）， 函数y=sin （2x +2p ）的图象经过向左平移56p ，得到函数y=sin [2（x+56p ）+2p ]=1326sin x p 骣琪+琪桫=sin 26x p 骣琪+琪桫的图象， 故答案为：π，56p ．【点睛】本题主要考查三角函数的平移．三角函数的平移原则为左加右减上加下减，注意x 的系数的应用，以及诱导公式的应用．13.已知直线()12:210,:20l ax a y l x ay +++=++=，其中a R Î，若12l l ^，则a =______，若12//l l ，则a =__________.【答案】 (1). 0或3- (2). 2或-1 【解析】 【分析】根据直线垂直的等价条件进行求解即可，根据直线的平行关系求出a 的值. 【详解】∵l 1⊥l 2，∴a+a（a+2）=0， 即a （a+3）=0，解得a=0或a=﹣3， ∵l 1∥l 2，∴a 2﹣a ﹣2=0，解得：a=2或a=﹣1， 经检验均适合题意， 故答案为：0或3- ，2或-1【点睛】两直线位置关系的判断： 1111:0l A x B y C ++=和2222:0l A x B y C ++=的平行和垂直的条件属于常考题型，如果只从斜率角度考虑很容易出错，属于易错题题型，应熟记结论： 垂直： 12120A AB B +=； 平行： 1221A B A B =，同时还需要保证两条直线不能重合，需要检验. 14.已知,x y R Î，且2241x y xy ++=，则224x y +的最小值_________，此时x 的值为___________.【答案】 (1). 45 (2). ±【解析】 【分析】利用均值不等式可得224224x y x yxy +?，即22222241444x y x y xy xy +=++?+从而得到224x y +的最小值及相应的x 值.【详解】∵224224x y x y xy +?，∴2244x y xy +£，当且仅当2x=y 时，等号成立， 又2214x y xy =++，∴22222241444x y x y xy x y +=++?+， ∴22445x y +?，即224x y +的最小值45由22241x y x y xy ì=ïí++=ïî，解得：x =?故答案为：45，±【点睛】本题考查基本不等式以及一元二次不等式的解法，属中档题．15.已知两不共线的非零向量,a b 满足2a =,1a b -=,则向量a 与b 夹角的最大值是__________. 【答案】6p 【解析】 【分析】设向量,a b 夹角为q ，由余弦定理求得23cos 4x xq +=，再利用基本不等式求得cos q 取得最小值，即可求得q 的最大值，得到结果.【详解】因为两非零向量,a b 满足2a =,1a b -=,设向量,a b 夹角为q ， 由于非零向量,a b 以及a b -构成一个三角形，设b x =， 则由余弦定理可得2144cos x x q =+-，解得233cos 44x x x x q ++==?，当且仅当x cos q所以q 的最大值是6p ，故答案是6p.【点睛】该题考查的是有关向量夹角的大小问题，在解题的过程中，涉及到的知识点有余弦定理，基本不等式，注意当什么情况下取得最值，再者就是需要明确角取最大值的时候其余弦值最小.16.已知数列{}na 为等差数列，其前n 项和为n S ，且13623a a S +=，给出以下结论：①100a =②10S 最小③712S S =④190S =，正确的有_________________. 【答案】①③④ 【解析】 【分析】先求出a 1=﹣9d ，再表示出求和公式，即可判断．【详解】设等差数列{a n }的公差为d ，∵2a 1+3a 3=S 6，∴5a 1+6d=6a 1+15d ， 化为：a 1+9d=0，即a 10=0， 给出下列结论：①a 10=0，正确； ②S 10=10a 1+()101012d-=﹣45d ，可能大于0，也可能小于0，因此不正确；③S 12﹣S 7=12a 1+12112´d ﹣7a 1﹣762´d=5a 1+45d=5（a 1+9d ）=0，正确． ④S 19=()119192a a +=19a 10=0，正确；其中正确结论的个数是①③④． 故答案为：①③④【点睛】本题考查了等差数列的通项公式与求和公式及其性质，考查了推理能力与计算能力，属于中档题．17.设函数()()21241,12,1x x f x x x a x ì-+?ï=íï-+>î，若存在互不相等的4个实数1234,,,x x x x ，使得()()()()123412347f x f x f x f x x x x x ====，则a 的取值范围为__________.【答案】(6,18) 【解析】 【分析】由题意可得f （x ）=7x 有4个不同实根，讨论x ≤1时，x ＞1时，由解方程和运用导数判断单调性和极值、最值，解不等式即可得到所求范围．【详解】由()11f x x =()22f x x =()33f x x =()44f x x =7，可得f （x ）=7x 有4个不同实根，当x ≤1时，f （x ）=|12x ﹣4|+1=7x ，解得x=35或x=519， 故当x ＞1时，f （x ）=7x 有2个不同实根， 设g （x ）=f （x ）﹣7x=x （x ﹣2）2﹣7x+a （x ＞1）， g′（x ）=（3x+1）（x ﹣3），当1＜x ＜3时，g′（x ）＜0，g （x ）递减； 当x ＞3时，g′（x ）＞0，g （x ）递增． 则g （x ）min =g （3）=a ﹣18，又g （1）=a ﹣6， 由a ﹣18＜0，且a ﹣6＞0， 解得6＜a ＜18． 故答案为：()6,18．【点睛】本题考查函数和方程的转化思想，考查分类讨论思想方法，以及导数的运用：求单调区间和极值、最值，考查运算能力，属于中档题．三、解答题：（本大题共个5小题,共74分．解答应写出文字说明、证明过程或演算步骤）18.已知函数()2sincos 333x x xf x = （1）求函数()f x 图象对称中心的坐标；（2）如果ABC D 的三边,,a b c 满足2b ac =，且边b 所对的角为B ，求()f B 的取值范围．【答案】（1）对称中心3,22k k Z p p 骣琪-+?琪桫（2）()f B ? 【解析】 【分析】（1）运用二倍角公式和两角和的正弦公式，由正弦函数的对称中心，解方程可得所求；（2）运用三角形的余弦定理和基本不等式，可得12≤cosB＜1，即有0＜B≤3p，运用正弦函数的图象和性质，即可得到所求范围．【详解】（1）()2333x x xf x sin cos =．=12212323sin x cos x +桫，=sin （233x p +）令233x k p p +=（k∈Z）， 解得：x=322k p p -（k∈Z），所以函数的图象的对称中心为：3,22k k Z p p 骣琪-+?琪桫．（2）由于b 2=ac ，所以：cosB=22221222a cb ac acac ac+--?，则：03B p £＜． 所以：253339B p p p +?＜，2133B sin p 骣琪+?琪桫，()1f B ?．则：f （B ）的取值范围为：2]． 【点睛】本题考查三角函数的恒等变换的运用，考查正弦函数的图象和性质，同时考查解三角形的余弦定理和基本不等式的运用．19.已知数列{}na 的前n 项和为n S ，且32,2n n n S a n N *=-?，12n n nb a =-（1）求证：数列{}n b 为等比数列，并求出数列{}n a 的通项公式； （2）是否存在实数l ，对任意,m n N *Î，不等式m nS b l>恒成立？若存在，求出l 的取值范围，若不存在请说明理由．【答案】（1）证明略；1122n n n a -=+ （2）32l ＜ 【解析】 【分析】（1）直接利用递推关系式求出数列的通项公式，进一步证明数列为等比数列； （2）利用（1）的结论，进一步利用分组法和恒成立问题求出实数λ的取值范围． 【详解】证明：（1）已知数列{a n }的前n 项和为S n ，且*322n n n S a n N =-?，，① 当n=1时，132a =， 则：当n ≥2时，111322n n n S a ---=-，② ①﹣②得：a n =2a n ﹣2a n ﹣1﹣32n +62n ，整理得：1322n n n a a -=-，所以：11112()22n n n n a a ---=-，故：1112212n nn n a a ---=-（常数）， 故：数列{a n }是以1112a -=为首项，2为公比的等比数列．故：1111222n n n n a ---=?，所以：1122n n n a -=+．由于：1122n n n n b a -=-=，所以：121222n n n n b b ---==（常数）．故：数列{b n }为等比数列． （2）由（1）得：1122n n na -=+， 所以：()1211222n n S -=+++++（12111222n +++）， =()111212212112n n 骣??琪-桫+--， =122n n-，假设存在实数λ，对任意m ，n ∈N *，不等式m nS b l＞恒成立， 即：()m min nS b l ＜， 由于：122m m m S =-为增函数， 故当m=1时，132S =，所以：1322n l -×＜，当n=1时，32l ＜．故存在实数λ，且32l ＜．【点睛】本题考查的知识要点：数列的通项公式的求法及应用，分组法在数列求和中的应用，主要考查学生的运算能力和转化能力，属于中档题．20.如图，四棱锥P ABCD -的底面ABCD 为平行四边形，平面PAB ^平面ABCD ，,45PB PC ABC °=?，点E 是线段PA 上靠近点A 的三等分点（1）求证：AB PC ^（2）若PAB D 是边长为2的等边三角形，求直线DE 与平面PBC 所成角的正弦值 【答案】（Ⅰ）见解析;【解析】试题分析：（Ⅰ）由平面PAB ^面ABCD ÞPO ^ 面ABCD 再证POB POC D @DOB OC OCAB AB ?轣轣面POC AB PC 轣 ；（Ⅱ）建立空间坐标系， 求得面PBC 的法向量为()4333,3,1,,1,cos ,37n DE n DE n DE n DE骣×琪==-掎?=琪桫. 试题解析：（Ⅰ）作PO AB ^于O ……①,连接OC ， ∵平面PAB ^平面ABCD ，且PAB ABCD AB ?面面 ，∴PO ^面ABCD.∵PB PC =，∴POB POC D @D ,∴OB OC =， 又∵45ABC °?，∴OC AB ^……②又PO COO ?,由①②，得AB ^面POC ，又PC Ì面POC ，∴AB PC ^.（Ⅱ）∵PAB D 是边长为2的等边三角形，∴1PO OA OB OC ===如图建立空间坐标系，(()()(),1,0,0,0,1,0,1,0,0P B C A - 设面PBC 的法向量为(),,n xy z =，()()1,0,3,1,1,0PB BC =-=- 0{n PBx n BC x y ?-=?-+=，令x =()3,3,1n =()111,0,3,33AP AE AP 骣琪===琪桫，()1,1,0CB DA ==-4,3DE DA AE 骣琪=+=-琪桫，设DE 与面PBC 所成角为qsin cos ,16n DE n DE nDEq -×=狁===∴直线DE 与平面PBC21.如图，O 为坐标原点，点F 为抛物线()21:20C x py p =>的焦点，且抛物线1C 上点P 处的切线与圆222:1C x y +=相切于点Q（1）当直线PQ 的方程为0x y --=时，求抛物线1C 的方程；（2）当正数p 变化时，记12,S S 分别为,FPQ FOQ D D 的面积，求12S S 的最小值。
浙江省镇海中学2019-2020学年第一学期期中考试高三数学试题
镇海中学学年第一学期期中考试高三数学试题卷一、选择题（本大题共小题，每题分，共分）1. 已知集合，则的元素的个数为A. B. C. D.2. 若且，则下列不等式中一定成立的是A. B. C. D.3. 已知是等差数列的前项和，且，则等于A. B. C. D.4. 函数的图像大致为5. 如图是一个几何体的三视图，则这个几何体的表面积是A.B.C.D.6. 将函数的图象向右平移个单位长度后，得到，则的函数解析式为A. B. )C. D. )7. 设命题，命题，若是的必要不充分条件，则实数的取值范围是A. B. C. D.8. 已知，，，则A. B. C. D.9. 已知椭圆和双曲线有相同的焦点，设点是该椭圆和双曲线的一个公共点，且，若椭圆和双曲线的离心率分别为，则的最小值为A. B. C. D.10. 设为正实数，且，则的最大值和最小值之和为A. B. C. D.二、填空题（本大题共小题，多空题每题分，单空题每题分，共分）11. 抛物线方程的焦点坐标为；准线方程为12. 已知点，点在线段上，则直线的斜率为；的最大值为13. 若实数满足约束条件，则的最小值为；的最小值为14. 已知长方体中，，则直线与平面所成的角为；若空间的一条直线与直线所成的角为，则直线与平面所成的最大角为15. 已知数列是等比数列且，，则的最大值为16. 已知圆，设点是恒过点的直线上任意一点，若在该圆上任意点满足，则直线的斜率的取值范围为17. 已知点为单位圆上两点，且满足，则的取值范围为三、解答题（本大题共小题，共分）18. 已知的最大值为（I）求实数的值；（II）若，求的值.19. 在锐角中，角所对边分别为，已知，（I）求；（II）求的取值范围.20. 如图，在三棱锥中，和都为等腰直角三角形，，，为的中点，且（I）求二面角的大小；（II）求直线与平面所成角的正弦值.21. 已知数列 的前 项和为 ，且满足： （I ）求数列 的通项公式；（II ）数列 满足 ， ，求数列 通项公式.22. 在平面直角坐标系中，已知 ， ，若线段 的中垂线 与抛物线B C总是相切.（I）求抛物线的方程；（II）若过点的直线交抛物线于两点，过分别作抛物线的切线相交于点. 分别与轴交于点.（i）证明：当变化时，的外接圆过定点，并求出定点的坐标；（ii）求的外接圆面积的最小值.。
【20套试卷合集】浙江省镇海中学2019-2020学年生物高三上期中模拟试卷含答案
2019-2020学年高三上生物期中模拟试卷含答案（考试时间：90分钟满分：100 ）一、选择题（本大题包括40小题共60分，1～20题，每小题1分；21～40题，每小题2分，每小题只有一个选项符合题意。
）1．下列属于相对性状的是（）A．豌豆的高茎和黄粒B．猫的白毛和狗的黑毛C．人的白化病和虎的白化病D．人的单眼皮和双眼皮2．下列说法不正确的是（）A．个体表现隐性性状可作为推断其基因型是纯合体的依据B．后代全部出现显性性状则可推断双亲是显性纯合体C．生物后代出现隐性性状则其显性亲本一定是杂合体D．生物表现出显性性状则基因型可能是显性纯合体，也可能是显性杂合体3．孟德尔探索遗传规律时，运用了“假说一演绎”法，该方法的基本内涵是：在观察与分析的基础上提出问题后，通过推理和想象提出解决问题的假说，根据假说进行演绎推理，再通过实验证明假说。
下列相关叙述中不正确的是（）A．“为什么F1只有显性性状、F2又出现隐性性状?”属于孟德尔提出的问题之一B．“豌豆在自然状态下一般是纯种”不属于孟德尔假说的内容C．“测交实验”是对推理过程及结果进行的检验D．“生物性状是由遗传因子决定的、体细胞中遗传因子成对存在”不属于假说内容4．下列有关人体细胞分裂的叙述，正确是（）A．人体细胞有丝分裂过程中偶尔也会发生交叉互换现象B．减数第二次分裂前期细胞中的DNA数目与有丝分裂中期的相同C．正常成年人体的所有细胞中，至少含有23条染色体，最多含有92条染色体D．减数第二次分裂的细胞中无同源染色体，染色体数为23条5．下列细胞不含同源染色体的是（）①体细胞②肝细胞③精原细胞④卵原细胞⑤次级精母细胞⑥精子⑦次级卵母细胞⑧精子细胞⑨卵细胞⑩受精卵细胞A．⑤⑥⑦⑧⑨B．①②③④⑤C．⑥⑦⑧⑨⑩D．①②③④⑩6.为研究高光强对移栽幼苗光合色素的影响,某同学用乙醇提取叶绿体色素,用石油醚进行纸层析,右图为滤纸层析的结果(Ⅰ、Ⅱ、Ⅲ、Ⅳ为色素条带)。
2019年浙江省镇海中学期中考试高三数学试题含答案
镇海中学2019学年第一学期期中考试高三年级数学试卷第I卷(选择题共40分)一、选择题（本大题共10小题，每题4分，共40分）1．已知集合A＝{x∈Z|x2﹣4x﹣5≤0}，B＝{x|0＜lnx＜2}，则A∩B的元素的个数为（）A．2 B．3 C．4 D．72．若a，b，c∈R且a＞b，则下列不等式中一定成立的是（）A．ac＞bc B．（a﹣b）c2＞0 C．＜D．﹣2a＜﹣2b 3．已知S n是等差数列{a n}的前n项和，且S2＝4，S4＝18，则S6等于（）A．50 B．42 C．38 D．364．函数的图象大致为（）A5．如图是一个几何体的三视图，则这个几何体的表面积是（）A．76 B.84 C．D．6．将函数y＝f（x）的图象向右平移个单位长度后，得到，则y＝f（x）的函数解析式为（）A．f（x）＝﹣cos2x B．f（x）＝sin（2x）C．f（x）＝cos2x D．f（x）＝cos（2x）7．设命题p：lg（2x﹣1）≤0，命题：，若q是p的必要不充分条件，则实数a的取值范围是（）A．，B．，C．，D．∅8．已知＜＜，sinα﹣2cosβ＝1，，则（）A．B．C．D．9．已知椭圆和双曲线有相同的焦点F1，F2，设点P是该椭圆和双曲线的一个公共点，且∠F1PF2，若椭圆和双曲线的离心率分别为e1，e2，则e12+e22的最小值为（）A．B．C．D．10．设a，b为正实数，且，则的最大值和最小值之和为（）A．2 B．C．D．9第Ⅱ卷(非选择题共110分)二、填空题（本大题共7小题，多空题每题6分，单空题每题4分，共36分）11．抛物线y2＝2x的焦点坐标是▲ ，准线方程是▲ ．12．已知点A（1，0），B（0，2），点P（a，b）在线段AB上，则直线AB的斜率为▲ a•b的最大值为▲ ．13．若实数（x，y）满足约束条件，则2x﹣y的最小值为▲的最小值为▲ ．14．已知长方体ABCD﹣A1B1C1D1中，，，，则直线AA1与平面A1BD所成的角为▲ 若空间的一条直线l与直线AA1所成的角为，则直线l与平面A1BD所成的最大角为▲ ．15．已知{a n}是等比数列，且a n＞0，a2a4+2a3a5+a4a6＝25，则a3+a5＝▲ ，a4的最大值为▲16．已知圆O：x2+y2＝1，设点P是恒过点（0，4）的直线l上任意一点，若在该圆上任意点A满足，则直线l的斜率k的取值范围为▲ ．17．已知点A（x1，y1），B（x2，y2）为单位圆上两点，且满足，则|x1+y1|+|x2+y2|的取值范围为▲ ．三、解答题（本大题共5小题，共74分）18．已知的最大值为．（Ⅰ）求实数a的值（Ⅱ）若，求的值．19．在锐角△ABC中，角A，B，C所对边分别为a，b，c，已知b＝3，a2＝c2﹣3c+9．（Ⅰ）求A（Ⅱ）求sin2B+sin2C的取值范围．20．如图，在三棱锥P﹣ABC中，△P AB和△ABC都为等腰直角三角形，P A⊥PB，AB⊥AC，M为AC的中点，且PM＝AC．（Ⅰ）求二面角P﹣AB﹣C的大小（Ⅱ）求直线PM与平面PBC所成角的正弦值．21．已知数列{a n}的前n项和为S n，且满足：，．（Ⅰ）求数列{a n}的通项公式（Ⅱ）数列{b n}满足b1＝﹣2，，求数列{b n}通项公式．22．在平面直角坐标系中，已知F（2，0），P（﹣2，t），若线段FP的中垂线l与抛物线C：y2＝2px（p＞0）总是相切．（Ⅰ）求抛物线C的方程（Ⅱ）若过点Q（2，1）的直线l′交抛物线C于M，N两点，过M，N分别作抛物线的切线l1，l2相交于点A．l1，l2分别与y轴交于点B，C．（i）证明：当l′变化时，△ABC的外接圆过定点，并求出定点的坐标（ii）求△ABC的外接圆面积的最小值．一、1．C 2．D 3．B 4．A 5．B 6．C 7．A 8．B 9．A 10．C二、11．（，0）x12．﹣2 13．1 14．60015．5 16．[，+∞）∪（﹣∞，] 17．[，]三、18、（Ⅰ），由于函数的最大值为，故，解得a．（Ⅱ）由于f（x），所以，整理得．所以，所以或．或，所以或，故，所以当时．．当时，，所以原式．19．（Ⅰ）在锐角△ABC中，∵b＝3，a2＝c2﹣3c+9，∴可得c2+b2﹣a2＝bc，∴由余弦定理可得：cos A，∴由A为锐角，可得A．（Ⅱ）∵sin2B+sin2C＝sin2B+sin2（B）＝sin2B+（cos B sin B）2＝1（sin2B cos2B）＝1sin（2B），又∵＜＜＜＜，可得＜B＜，∴2B（，），∴sin（2B）（，1]，∴sin2B+sin2C＝1sin（2B）（，]，即sin2B+sin2C的取值范围是（，]．20．（1）分别取线段AB，BC的中点O，N，连接PO，ON，MN，PN，设AC＝2，则有在等腰直角△P AB中，O是中点，则有AB⊥PO﹣﹣﹣①在等腰直角△ABC中，点O，N分别是AB，BC的中点，则有AB⊥ON﹣﹣﹣②由①②可知，AB⊥平面PON，又∵MN∥AB，∴MN⊥平面PON，则有MN⊥PN．又AB＝2，则MN＝1，又PM＝AC＝2，则有PN，又OP＝ON＝1，由三角形余弦定理可知，∠，∴∠PON＝120°，即二面角P﹣AB﹣C的大小为1200．（2）建立如图所示的空间直角坐标系，过点P作PD⊥ON交NO延长线于点D，设AB ＝AC＝2，则有A（﹣1，0，0），C（﹣1，2，0），B（1，0，0），M（﹣1，1，0），由（1）可知，∠POD＝180°﹣∠PON＝60°，又∵OP＝1，∴，．∴，，，，，．∴，，，设平面PBC的一个法向量为，，，则有，又∵，，，，，，∴，∴，，．设直线PM与平面PBC所成角为θ，则有：．故直线PM与平面PBC所成角的正弦值为．21．（Ⅰ）数列{a n}的前n项和为S n，且满足：，．当n≥2时，a n＝﹣3S n﹣1+2，两式相减得：a n+1＝﹣2a n，所以数列{a n}是以2为首项﹣2为公比的等比数列．所以．（Ⅱ）由于，所以，由于，所以，所以．22．（Ⅰ）F（2，0），P（﹣2，t），可得FP的中点为（0，），当t＝0时，FP的中点为原点，当t≠0时，直线FP的斜率为，线段FP的中垂线l的斜率为，可得中垂线l的方程为y x，代入抛物线方程y2＝2px，可得x2+（4﹣2p）x0，由直线和抛物线相切可得△＝（4﹣2p）2﹣16＝0，解得p＝4，则抛物线的方程为y2＝8x；（Ⅱ）（i）证明：可设过点Q（2，1）的直线l′的方程为x﹣2＝m（y﹣1），即x＝my+2﹣m，代入抛物线的方程y2＝8x，可得y2﹣8my﹣16+8m＝0，设M（，y1），N（，y2），则y1+y2＝8m，y1y2＝8m﹣16，由y2＝8x，两边对x求导可得2y•y′＝8，即y′，可得M处的切线方程为y﹣y1（x），化为y1y＝4x，①同理可得N处的切线方程为y2y＝4x，②由①②可得y4m，x m﹣2，即A（m﹣2，4m），又l1，l2分别与y轴交于点B（0，），C（0，），设过A，B，C的外接圆的方程为x2+y2+Dx+Ey+F＝0，（D2+E2﹣4F＞0），即有，结合y1+y2＝8m，y1y2＝8m﹣16，可得D＝﹣m﹣2，E＝﹣4m，F＝4m﹣8，可得△ABC的外接圆方程为x2+y2﹣（m+2）x﹣4my+4m﹣8＝0，可得m（4﹣x﹣4y）+（x2+y2﹣2x﹣8）＝0，由可得或，则当l′变化时，△ABC的外接圆过定点（4，0）和（，）；（ii）△ABC的外接圆的半径r，可得当m时，r的最小值为，则△ABC的外接圆面积的最小值为π．。
【100所名校】2019届浙江省宁波市镇海中学高三上学期期中考试数学试题(含答案)
2019届浙江省宁波市镇海中学 高三上学期期中考试数学试题数学注意事项：1．答题前，先将自己的姓名、准考证号填写在试题卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。
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一、单选题1．设全集U =R ，集合A ={x|x ≥3},B ={x|x ≤0<5}，则集合(C U A )∩B = A ．{x|0≤x ≤3} B ．{x|0<x <3} C ．{x|0<x ≤3} D ．{x|0≤x <3}2．某几何体的三视图如图所示，图中的四边形都是边长为2的正方形，两条虚线互相垂直，则该几何体的体积是A ．8−π3B ．163C ．8−π6D ．2033．记S n 为等差数列{a n }的前n 项和,若S 9=45,a 3+a 8=12, 则a 7= A ．10 B ．9 C ．8 D ．74．4．满足线性约束条件23,23,{ 0,0x y x y x y +≤+≤≥≥的目标函数z x y =+的最大值是A ．1B ．32C ．2D ．3 5．已知函数f (x )=x 2−In|x|x，则函数f (x )的图象为A ．B ．C ．D ．6．若α、β是两个相交平面，则在下列命题中，真命题的序号为①若直线m ⊥α，则在平面β内一定不存在与直线m 平行的直线．②若直线m ⊥α，则在平面β内一定存在无数条直线与直线m 垂直． ③若直线m ⊂α，则在平面β内不一定存在与直线m 垂直的直线． ④若直线m ⊂α，则在平面β内一定存在与直线m 垂直的直线． A ．①③ B ．②③ C ．②④ D ．①④ 7．已知sin (π6−α)=√23，那么cos2α+√3sin2α=A ．109 B ．−109 C ．−59 D ．598．已知正项等比数列{a n }满足a 7=a 6+2a 5，若存在两项a m ,a n ，使得a m ⋅a n =16a 12，则1m+9n 的最小值为 A ．32 B ．114 C ．83 D ．103 9．已知双曲线C:x 2a 2−y 2b 2=1(a >0,b >0)的左右焦点分别为F 1,F 2，P 为双曲线C 上一点，Q 为双曲线C 渐近线上一点，P,Q 均位于第一象限，且2QP ⃑⃑⃑⃑⃑ =PF 2⃑⃑⃑⃑⃑⃑⃑ ，QF 1⃑⃑⃑⃑⃑⃑⃑ ⋅QF 2⃑⃑⃑⃑⃑⃑⃑ =0，则双曲线C 的离心率为A ．√3−1B ．√3+1C ．√13−2D ．√13+210．如图，在三棱柱ABC −A 1B 1C 1中，底面为边长为2的正三角形，B 1在底面的射影为AC 中点且B 1到底面的距离为√3，已知E,F 分别是线段AB 1与CA 1上的动点，记线段EF 中点M 的轨迹为L ，则|L |等于(注：|L |表示L 的测度，本题中L 若分别为曲线、平面图形、空间几何体，分别对应为其长度、面积、体积）此卷只装订不密封班级 姓名 准考证号 考场号 座位号A．1B．√102C．√32D．√394二、填空题11．中国古代数学著作《九章算术》中有一个这样的问题：“某贾人擅营，月入益功疾（注：从第2月开始，每月比前一月多入相同量的铜钱，3月入25贯，全年（按12个月计）共入510贯“，则该人每月比前一月多入_________________贯，第12月营收贯数为_________________.12．y=sin(2x+π6)的最小正周期为_________________，为了得到函数y=sin(2x+π6)的图象，可以将函数y=cos2x的图象向左最小移动_______个单位13．已知直线l1:ax+(a+2)y+1=0,l2:x+ay+2=0，其中a∈R，若l1⊥l2，则a=______，若l1//l2，则a=__________.14．已知x,y∈R，且4x2+y2+xy=1，则4x2+y2的最小值_________，此时x的值为___________.15．已知两不共线的非零向量a ,b⃑满足|a|=2,|a−b⃑|=1,则向量a与b⃑夹角的最大值是__________.16．已知数列{a n}为等差数列，其前n项和为S n，且2a1+3a3=S6，给出以下结论：①a10= 0②S10最小③S7=S12④S19=0，正确的有_________________.17．设函数f(x)={|12x−4|+1,x≤1x(x−2)2+a,x>1，若存在互不相等的4个实数x1,x2,x3,x4，使得f(x1)x1=f(x2) x2=f(x3)x3=f(x4)x4=7，则a的取值范围为__________.三、解答题18．已知函数f(x)=sin x3cos x3+√3cos2x3（1）求函数f(x)图象对称中心的坐标；（2）如果ΔABC的三边a,b,c满足b2=ac，且边b所对的角为B，求f(B)的取值范围．19．已知数列{a n}的前n项和为S n，且S n=2a n−32n,n∈N∗，b n=a n−12n（1）求证：数列{b n}为等比数列，并求出数列{a n}的通项公式；（2）是否存在实数λ，对任意m,n∈N∗，不等式S m>λb n恒成立？若存在，求出λ的取值范围，若不存在请说明理由．20．如图，四棱锥P−ABCD的底面ABCD为平行四边形，平面PAB⊥平面ABCD，PB=PC,∠ABC=45°，点E是线段PA上靠近点A的三等分点（1）求证：AB⊥PC（2）若ΔPAB是边长为2的等边三角形，求直线DE与平面PBC所成角的正弦值21．如图，O为坐标原点，点F为抛物线C1:x2=2py(p>0)的焦点，且抛物线C1上点P处的切线与圆C2:x2+y2=1相切于点Q（1）当直线PQ的方程为x−y−√2=0时，求抛物线C1的方程；（2）当正数p变化时，记S1,S2分别为ΔFPQ,ΔFOQ的面积，求S1S2的最小值。
2020-2021学年浙江省宁波市镇海中学高三(上)期中数学试卷
2020-2021学年浙江省宁波市镇海中学高三（上）期中数学试卷试题数：22.满分：01.（单选题.4分）已知集合A={x|log2x＜1}.集合B={x|-1≤x≤1}.则A∩B=（）A.[-1.1]B.[-1.2）C.（0.1]D.（-∞.2）2.（单选题.4分）设a=30.7.b=（13）-0.8.c=log0.70.8.则a.b.c的大小关系为（）A.a＜b＜cB.b＜a＜cC.b＜c＜aD.c＜a＜b3.（单选题.4分）已知平面α、β.直线l⊂α.直线m不在平面α上.下列说法正确的是（）A.若α || β.m || β.则l || mB.若α || β.m⊥β.则l⊥mC.若1 || m.α || β.则m || βD.若l⊥m.m || β.则α⊥β4.（单选题.4分）已知x.y满足约束条件{2x−y−1≤0x+y+1≥0y≤1.则Z=|x-3y-2|的取值范围是（）A.[0.7]B.（1.7）C.[0.4]D.[1.4]5.（单选题.4分）已知等比数列{a n}的前n项和为S n.若1a1+1a2+1a3=2 .a2=2.则S3=（）A.8B.7C.6D.46.（单选题.4分）函数f（x）= x(e x−e−x)x2−1的部分图象大致是（）A.B.C.D.7.（单选题.4分）已知函数f（x）=|x|（e x-e-x）.对于实数a.b.“a+b＞0”是“f（a）+f（b）＞0”的（）A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件) .将f（x）的图象上所有点向右平移θ（θ＞0）8.（单选题.4分）已知函数f(x)=2sin(2x+π6对称.则θ的最小值为（）个单位长度.得到的图象关于直线x=π6A. π6B. π3C. π2D.π9.（单选题.4分）已知线段AB 是圆C ：x 2+y 2=4的一条动弦.且 |AB |=2√3 .若点P 为直线x+y-4=0上的任意一点.则 |PA ⃗⃗⃗⃗⃗ +PB ⃗⃗⃗⃗⃗ | 的最小值为（ ） A. 2√2−1 B. 2√2+1 C. 4√2−2 D. 4√2+210.（单选题.4分）已知数列{a n }满足a 0=0.|a i+1|=|a i +1|（i∈N ）.则| ∑a k 20k=1 |的值不可能是（ ） A.2 B.4 C.10 D.1411.（填空题.6分）复数i (1+2i )1+i的虚部为___ .模为___ ． 12.（填空题.6分）已知某空间几何体的三视图如图所示（单位：cm ）.则该几何体的体积（单位：cm 3）是___ ；此几何体各个面中.面积的最大值（单位：cm 2）为___ ．13.（填空题.6分）若（1+x ）（1-2x ）7=a 0+a 1x+a 2x 2+…+a 8x 8.则a 1+a 2+…+a 7的值是___ ；在上述展开式右边的九项中.随机任取不同的三项.假设这三项均不相邻.则有___ 种不同的取法． 14.（填空题.6分）已知数列{a n }.{b n }满足：a 1=1.a n +a n+1=n.b n =a 2n-1.则数列b n =___ ；记S n 为数列{a n }的前n 项和.S 31-S 24=___ ．15.（填空题.4分）函数f （x ）=sin （ωx+φ）的部分图象如图所示.则f （x ）的单调递增区间为___ ．16.（填空题.4分）已知x＞0.y＞0.则2xyx2+4y2+xyx2+y2的最大值为___ ．17.（填空题.4分）四面体ABCD中.AB⊥BC.CD⊥BC.BC=2.且异面直线AB和CD所成的角为60°.若四面体ABCD的外接球半径为√5 .则四面体ABCD的体积的最大值为___ ．18.（问答题.0分）在△ABC中.内角A.B.C的对边分别为a.b.c.若sin2A+sin2C-sin2B= 23 sinAsinC.c=2．（1）求sinB的值；（2）设D在BC边上.且BD=AD=2DC.求△ABC的面积．19.（问答题.0分）如图.在四棱锥S-ABCD中.侧面SCD为钝角三角形且垂直于底面ABCD.底面为直角梯形且∠ABC=90°.AB=AD= 12BC.CD=SD.点M是SA的中点．（1）求证：BD⊥平面SCD；（2）若直线SD与底面ABCD所成的角为60°.求SD与平面MBD所成角的正弦值．20.（问答题.0分）已知数列{a n}满足a1= 12，(a n+1+1)(a n+1)=2a n+1，n∈N∗．（1）求数列{a n}的通项公式；（2）证明：对∀n∈N*.a1a2a3+a2a3a4+…+a n a n+1a n+2＜112．21.（问答题.0分）已知椭圆C：x2a2 + y2b2=1（a＞b＞0）的离心率为√32.且经过点P（- √3 . 12）．（1）求椭圆C的标准方程；（2）设点M是椭圆C上位于第一象限内的动点.A.B分别为椭圆C的左顶点和下顶点.直线MB 与x轴交于点C.直线MA与y轴交于点D.O为椭圆的中心.求三角形OCD的面积的取值范围．22.（问答题.0分）已知函数f（x）=e x+cosx-2.f'（x）为f（x）的导数．（1）当x≥0时.求f'（x）的最小值；（2）当x≥−π2时.xe x+xcosx-ax2-2x≥0恒成立.求a的取值范围．2020-2021学年浙江省宁波市镇海中学高三（上）期中数学试卷参考答案与试题解析试题数：22.满分：01.（单选题.4分）已知集合A={x|log2x＜1}.集合B={x|-1≤x≤1}.则A∩B=（）A.[-1.1]B.[-1.2）C.（0.1]D.（-∞.2）【正确答案】：C【解析】：求出集合A.集合B.由此能求出A∩B．【解答】：解：∵集合A={x|log2x＜1}={x|0＜x＜2}.集合B={x|-1≤x≤1}.∴A∩B={x|0＜x≤1}=（0.1]．故选：C．【点评】：本题考查交集的求法.考查交集定义等基础知识.考查运算求解能力.是基础题．2.（单选题.4分）设a=30.7.b=（1）-0.8.c=log0.70.8.则a.b.c的大小关系为（）3A.a＜b＜cB.b＜a＜cC.b＜c＜aD.c＜a＜b【正确答案】：D【解析】：根据指数函数和对数函数的性质即可求出．）-0.8=30.8.【解答】：解：a=30.7.b=（13则b＞a＞1.log0.70.8＜log0.70.7=1.∴c＜a＜b.【点评】：本题考查了指数函数和对数函数的性质.属于基础题．3.（单选题.4分）已知平面α、β.直线l⊂α.直线m不在平面α上.下列说法正确的是（）A.若α || β.m || β.则l || mB.若α || β.m⊥β.则l⊥mC.若1 || m.α || β.则m || βD.若l⊥m.m || β.则α⊥β【正确答案】：B【解析】：由空间中直线与直线、直线与平面、平面与平面位置关系逐一分析四个选项得答案．【解答】：解：对于A.若α || β.m || β.则l || m或l与m异面.故A错误；对于B.若α || β.m⊥β.则m⊥α.又l⊂α.则l⊥m.故B正确；对于C.若1 || m.α || β.则m || β或m⊂β.故C错误；对于D.若l⊥m.m || β.则α || β或α与β相交.故D错误．故选：B．【点评】：本题考查空间中直线与直线、直线与平面、平面与平面位置关系的判定及应用.考查空间想象能力与思维能力.是中档题．4.（单选题.4分）已知x.y满足约束条件{2x−y−1≤0x+y+1≥0y≤1.则Z=|x-3y-2|的取值范围是（）A.[0.7]B.（1.7）C.[0.4]D.[1.4]【正确答案】：A【解析】：根据二元一次不等式组画出可行域.目标函数几何意义z′=x-3y-2的纵截距相反数.平移目标函数观察Z取值范围．【解答】：解：如图可行域：令z′=x-3y-2.平移直线x-3y-2=0可知当直线过C（0.-1）时.z′取得最大值1.经过B（-2.1）时.z′有最小值-7.Z=|x-3y-2|.所以Z的取值范围：[0.7]【点评】：本题考查线性规划问题.属常规题较简单.解题的关键是画好可行域.弄清z所对应的几何意义．5.（单选题.4分）已知等比数列{a n}的前n项和为S n.若1a1+1a2+1a3=2 .a2=2.则S3=（）A.8B.7C.6D.4【正确答案】：A【解析】：利用已知条件化简.转化求解即可．【解答】：解：1a1+1a2+1a3=a1+a3a1a3+1a2=a1+a2+a3a22=S34=2 .则S3=8．故选：A．【点评】：本题考查等比数列的性质.考查化归与转化的思想．6.（单选题.4分）函数f（x）= x(e x−e−x)x2−1的部分图象大致是（）A.B.C.D.【正确答案】：D【解析】：由函数为偶函数.可排除选项A.由f（2）＞0.可排除BC.即可得到正确答案．【解答】：解：函数的定义域为{x|x≠±1}. f(−x)=−x(e −x−e x)x2−1=f(x) .故函数f（x）为偶函数.其图象关于y轴对称.故排除A；又f(2)=2(e 2−e−2)3＞0 .故排除BC；故选：D．【点评】：本题考查利用函数性质确定函数图象.考查数形结合思想.属于基础题．7.（单选题.4分）已知函数f（x）=|x|（e x-e-x）.对于实数a.b.“a+b＞0”是“f（a）+f（b）＞0”的（）A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件【正确答案】：C【解析】：根据条件判断函数f（x）是奇函数.同时也是增函数.结合函数奇偶性和单调性的性质.利用充分条件和必要条件的定义进行判断即可．【解答】：解：∵f（x）=|x|（e x-e-x）.∴f（-x）=|-x|（e-x-e x）=-|x|（e x-e-x）=-f（x）.即函数f（x）是奇函数.当x≥0.f（x）为增函数.则由a+b＞0得a＞-b.此时f（a）＞f（-b）=-f（b）.即f（a）+f（b）＞0成立.即充分性成立. 若f（a）+f（b）＞0得f（a）＞-f（b）=f（-b）.则a＞-b.即a+b＞0成立.即必要性成立.则“a+b＞0”是“f（a）+f（b）＞0”成立的充要条件.故选：C．【点评】：本题主要考查充分条件和必要条件的判断.结合条件判断函数的奇偶性和单调性是解决本题的关键．8.（单选题.4分）已知函数f(x)=2sin(2x+π6) .将f（x）的图象上所有点向右平移θ（θ＞0）个单位长度.得到的图象关于直线x=π6对称.则θ的最小值为（）A. π6B. π3C. π2D.π【正确答案】：C【解析】：根据三角函数图象平移法则写出平移后的函数解析式.再根据函数图象关于直线x=π6对称求出θ的最小值．【解答】：解：函数f(x)=2sin(2x+π6) .将f（x）的图象上所有点向右平移θ（θ＞0）个单位长度.得y=f（x-θ）=2sin[2（x-θ）+ π6 ]=2sin（2x-2θ+ π6）；又函数y的图象关于直线x=π6对称.即2× π6 -2θ+ π6=kπ+ π2.k∈Z；解得θ=- 12kπ.k∈Z；又θ＞0.所以θ的最小值为π2．故选：C ．【点评】：本题考查了三角函数的图象与性质的应用问题.也考查了图象平移问题.是基础题． 9.（单选题.4分）已知线段AB 是圆C ：x 2+y 2=4的一条动弦.且 |AB |=2√3 .若点P 为直线x+y-4=0上的任意一点.则 |PA ⃗⃗⃗⃗⃗ +PB ⃗⃗⃗⃗⃗ | 的最小值为（ ） A. 2√2−1 B. 2√2+1 C. 4√2−2 D. 4√2+2 【正确答案】：C【解析】：过O 作OD⊥AB .由已知求得|OD|.再求出原点到直线x+y-4=0的距离.求得| PD ⃗⃗⃗⃗⃗ |的最小值.再由 |PA ⃗⃗⃗⃗⃗ +PB ⃗⃗⃗⃗⃗ | =2| PD ⃗⃗⃗⃗⃗ |求解．【解答】：解：如图.P 为直线x+y-4=0上的任意一点.过原点O 作OD⊥AB .由|AB|=2 √3 .可得|OD|=1. ∴| PD ⃗⃗⃗⃗⃗ |=|OP|-1.则 |PD ⃗⃗⃗⃗⃗ |min =√2−1=2√2−1 ． 则| PA ⃗⃗⃗⃗⃗ +PB ⃗⃗⃗⃗⃗ |=| PO ⃗⃗⃗⃗⃗ +OA ⃗⃗⃗⃗⃗ +PO ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ |=2| PO ⃗⃗⃗⃗⃗ +OD ⃗⃗⃗⃗⃗⃗ |=2| PD ⃗⃗⃗⃗⃗ |. ∴ |PA ⃗⃗⃗⃗⃗ +PB ⃗⃗⃗⃗⃗ | 的最小值为 4√2−2 ． 故选：C ．【点评】：本题主要考查了直线与圆相交的性质.考查向量模的最值的求法.理解题意是关键.是中档题．10.（单选题.4分）已知数列{a n }满足a 0=0.|a i+1|=|a i +1|（i∈N ）.则| ∑a k 20k=1 |的值不可能是（ ）A.2B.4C.10D.14【正确答案】：B【解析】：可将数列的递推式两边平方.运用累加法和排除法.可得结论．【解答】：解：|a i+1|=|a i +1|.两边平方可得a i+12=a i 2+2a i +1. 由a 12=a 02+2a 0+1=0+0+1.a 22=a 12+2a 1+1.a 32=a 22+2a 2+1.….a 212=a 202+2a 20+1. 上面的式子累加可得a 212=2（a 1+a 2+…+a 20）+21.则| ∑a k 20k=1 |=| a 212−212|.若| a 212−212 |=2.可得a 21=±5.故A 可能； 若|a 212−212|=4.可得a 21不为整数.故B 不可能； 若| a 212−212 |=10.可得a 21=±1.故C 可能； 若|a 212−212|=14.可得a 21=±7.故D 可能．故选：B ．【点评】：本题考查数列递推式的运用和数列的求和.考查分类讨论思想和判断能力.属于中档题．11.（填空题.6分）复数i (1+2i )1+i 的虚部为___ .模为___ ． 【正确答案】：[1] 32; [2]√102【解析】：利用复数代数形式的乘除运算化简.可得复数的虚部.再由复数模的计算公式求复数的模．【解答】：解：∵ i (1+2i )1+i = −2+i1+i=(−2+i )(1−i )(1+i )(1−i )=−2+2i+i−i 22=−12+32i .∴复数 i (1+2i )1+i 的虚部为 32. |i (1+2i )1+i |=| −12+32i |= √(−12)2+(32)2=√102． 故答案为： 32 ； √102．【点评】：本题考查复数代数形式的乘除运算.考查复数的基本概念.考查复数模的求法.是基础题．12.（填空题.6分）已知某空间几何体的三视图如图所示（单位：cm）.则该几何体的体积（单位：cm3）是___ ；此几何体各个面中.面积的最大值（单位：cm2）为___ ．【正确答案】：[1] 163; [2]4 √2【解析】：首先把三视图转换为几何体的直观图.进一步利用体积公式和三角形的面积公式的应用求出结果．【解答】：解：根据几何体的三视图转换为直观图为：该几何体为四棱锥体E-ABCD．如图所示：所以V E−ABCD=13×2×4×2=163. S△ADE=12×4×√22+22=4√2．故答案为：163；4√2．【点评】：本题考查的知识要点：三视图和几何体的直观图之间的转换.几何体的体积公式的应用.主要考查学生的运算能力和转换能力及思维能力.属于基础题．13.（填空题.6分）若（1+x）（1-2x）7=a0+a1x+a2x2+…+a8x8.则a1+a2+…+a7的值是___ ；在上述展开式右边的九项中.随机任取不同的三项.假设这三项均不相邻.则有___ 种不同的取法．【正确答案】：[1]125; [2]35【解析】：利用二项式定理可知.对已知关系式中的x赋值0与1即可求得a1+a2+…+a8的值．先求出任取不同的三项的所有取法.再求出三项均相邻和只有两项相邻的不同取法.利用间接法即可求解．【解答】：解：∵（1+x）（1-2x）7=a0+a1x+a2x2+…+a8x8.∴a8= C77•（-2）7=-128．令x=0.得（1+0）（1-0）7=a0.即a0=1；令x=1.得（1+1）（1-2）7=a0+a1+a2+…+a7+a8=-2.∴a1+a2+…+a7=-2-a0-a8=-2-1+128=125．在上述展开式右边的九项中.随机任取不同的三项.有C93 =84种不同取法.三项均相邻.有7种不同的取法.两项相邻.有2×6+6×5=42种不同的取法.故三项均不相邻有84-7-42=35种不同的取法．故答案为：125；35．【点评】：本题主要考查二项式定理的应用.组合数公式的应用.属于中档题．14.（填空题.6分）已知数列{a n}.{b n}满足：a1=1.a n+a n+1=n.b n=a2n-1.则数列b n=___ ；记S n为数列{a n}的前n项和.S31-S24=___ ．【正确答案】：[1]n; [2]97【解析】：由a n+a n+1=n.可将n换为n-1.相减可得{a n}的奇数项和偶数项均为公差为1的等差数列.再由数列的分组求和.可得所求和．【解答】：解：a1=1.a n+a n+1=n.可得a2=1-a1=0.将n换为n-1可得a n-1+a n=n-1.n≥2.又a n+a n+1=n.相减可得a n+1-a n-1=1.可得{a n}的奇数项和偶数项均为公差为1的等差数列.可得b n=a2n-1=1+（n-1）=n；a2n=0+n-1=n-1.则S31-S24=a25+a26+a27+a28+a29+a30+a31=（a25+a27+a29+a31）+（a26+a28+a30）=（13+14+15+16）+（12+13+14）=58+39=97．故答案为：n.97．【点评】：本题考查数列的递推式的运用.以及数列的求和方法：分组求和.考查转化思想和运算能力、推理能力.属于中档题．15.（填空题.4分）函数f（x）=sin（ωx+φ）的部分图象如图所示.则f（x）的单调递增区间为___ ．【正确答案】：[1][2k- 54 .2k- 14].k∈Z【解析】：由周期求出ω.由五点法作图求出φ的值.可得函数的解析式.再利用正弦函数的单调性.得出结论．【解答】：解：根据函数f（x）=sin（ωx+φ）的部分图象.可得12• 2πω= 54- 14.∴ω=π．再根据五点法作图.可得π× 14+φ=π.∴φ= 3π4.f（x）=sin（πx+ 3π4）．令2kπ- π2≤πx+ 3π4≤2kπ+ π2.k∈Z.解得 2k- 54≤x≤2k- 14.k∈Z.故函数的增区间为[2k- 54 .2k- 14].k∈Z．故答案为：[2k- 54 .2k- 14].k∈Z．【点评】：本题主要考查由函数y=Asin（ωx+φ）的部分图象求解析式.由周期求出ω.由五点法作图求出φ的值.正弦函数的单调性.属于中档题．16.（填空题.4分）已知x＞0.y＞0.则2xyx2+4y2+xyx2+y2的最大值为___ ．【正确答案】：[1] 2√23【解析】：换元t= xy +2yx.然后结合基本不等式可求t的范围.然后结合函数y=t+ 1t的单调性可求范围.然后2xyx2+4y2+xyx2+y2= 3x3y+6xy3x4+5x2y2+4y4=3xy+6yxx2y2+4y2x2+5=3(xy+2yx)(xy+2yx)2+1= 3t1+t2= 3t+1t.从而可求【解答】：解：因为x＞0.y＞0.令t= xy +2yx.则t ≥2√2 .所以y=t+ 1t 在[2 √2 .+∞）上单调递增.y ≥9√24.则2xyx2+4y2+xyx2+y2= 3x3y+6xy3x4+5x2y2+4y4=3xy+6yxx2y2+4y2x2+5=3(xy+2yx)(xy+2yx)2+1= 3t1+t2= 3t+1t≤9√24= 2√23.当且仅当t= 1t即t=1时取等号.故答案为：2√23【点评】：本题主要考查了利用基本不等式求解最值.解题的关键是应用条件的配凑.属于中档试题．17.（填空题.4分）四面体ABCD中.AB⊥BC.CD⊥BC.BC=2.且异面直线AB和CD所成的角为60°.若四面体ABCD的外接球半径为√5 .则四面体ABCD的体积的最大值为___ ．【正确答案】：[1]2 √3【解析】：构建直三棱柱ABE-CDF.设G.H分别为△ABE.△CDF的外心.连结GH.取其中点O.则O 为直三棱柱ABE-CDF的外接球的球心.也是四面体ABCD的外接球的球心.推导出∠ABE=60°.设三棱柱底面△ABE的外接圆半径为r.则r= √5−1 =2.AE=2rsin60°=2 √3 .由余弦定理推导出12=AB2+BE2-AB•BE≥2AB•BE-AB•BE=AB•BE.由此能求出四面体ABCD的体积的最大值．【解答】：解：四面体ABCD中.AB⊥BC.CD⊥BC.BC=2.且异面直线AB和CD所成的角为60°. 构建直三棱柱ABE-CDF.设G.H分别为△ABE.△CDF的外心.连结GH.取其中点O.则O为直三棱柱ABE-CDF的外接球的球心.也是四面体ABCD的外接球的球心.∵异面直线AB与CD所成角为60°.∴∠ABE=60°.设三棱柱底面△ABE的外接圆半径为r.则r= √5−1 =2.AE=2rsin60°=2 √3 .由余弦定理得：AE2=AB2+BE2-2AB•BE•cos60°.∴AB2+BE2-AB•BE=12.∴12=AB2+BE2-AB•BE≥2AB•BE-AB•BE=AB•BE.∴四面体ABCD的体积：V A−BCD=13V ABE−CDF = 13×12×AB×BE×sin60°×BC = √36×AB×BE≤ 2√3．∴四面体ABCD的体积的最大值为2 √3．故答案为：2√3．【点评】：本题考查四面体的体积的求法.考查空间中线线、线面、面面间的位置关系等基础知识.考查运算求解能力.是中档题．18.（问答题.0分）在△ABC中.内角A.B.C的对边分别为a.b.c.若sin2A+sin2C-sin2B= 23 sinAsinC.c=2．（1）求sinB的值；（2）设D在BC边上.且BD=AD=2DC.求△ABC的面积．【正确答案】：【解析】：（1）利用正弦定理化角为边.再由余弦定理求出cosB.从而求出sinB的值；（2）根据题意画出图形.利用余弦定理求出BD的值.再求△ABC的面积．【解答】：解：（1）△ABC中.sin2A+sin2C-sin2B= 23sinAsinC.由正弦定理得.a2+c2-b2= 23ac.所以cosB= a 2+c2−b22ac=23ac2ac= 13；又B∈（0.π）.所以sinB= √1−sin2B = √1−(13)2= 2√23；（2）如图所示.设BD=AD=2DC=x.由c=AB=2.利用余弦定理得.AD2=AB2+BD2-2AB•BD•cosB. 即x2=22+x2-2×2×x× 13.解得x=3.CD= 12 x= 32.所以△ABC的面积为S△ABC= 12AB•BC•sinB= 12×2×（3+ 32）× 2√23=3 √2．【点评】：本题考查了正弦、余弦定理的应用问题.也考查了运算求解能力.是中档题．19.（问答题.0分）如图.在四棱锥S-ABCD中.侧面SCD为钝角三角形且垂直于底面ABCD.底面为直角梯形且∠ABC=90°.AB=AD= 12BC.CD=SD.点M是SA的中点．（1）求证：BD⊥平面SCD；（2）若直线SD与底面ABCD所成的角为60°.求SD与平面MBD所成角的正弦值．【正确答案】：【解析】：（1）取BC的中点E.连接DE.分别计算BD.CD.利用勾股定理的逆定理证明BD⊥CD.再根据面面垂直的性质得出BD⊥平面SCD；（2）建立空间坐标系.计算平面MBD的法向量n⃗ .计算n⃗和DS⃗⃗⃗⃗⃗ 的夹角得出SD与平面MBD所成角的大小．【解答】：（1）证明：取BC 的中点E.连接DE. 设AB=a.则AD=a.BC=2a.BE= 12 BC=a. ∵∠ABC=90°.AD || BE.AD=BE. ∴四边形ABED 是正方形. ∴BD= √2 a.DE⊥BC .DE=CE=a. ∴C D= √2 a.∴BD 2+CD 2=BC 2.故BD⊥CD .∵平面SCD⊥平面ABCD.平面SCD∩平面ABCD=CD.BD⊂平面ABCD.BD⊥CD . ∴BD⊥平面SCD ．（2）解：过S 作SN⊥CD .交CD 延长线于N.∵平面SCD⊥平面ABCD.平面SCD∩平面ABCD=CD.SN⊂平面SCD.SN⊥CD . ∴SN⊥平面ABCD.∴∠SDN 为直线SD 与底面ABCD 所成的角.故∠SDN=60°. ∵SD=CD= √2 a.∴DN=√2a 2 .SN= √6a2. 以D 为原点.以DB.DC.及平面ABCD 的过点D 的垂线为坐标轴建立空间直角坐标系D-xyz.如图所示.则B （ √2 a.0.0）.D （0.0.0）.A （ √22a.- √22a.0）.S （0.- √22a. √62a ）. ∵M 是SA 的中点.∴M （ √24 a.- √22 a. √64 a ）.∴ DS ⃗⃗⃗⃗⃗ =（0.- √22 a. √62 a ）. DB ⃗⃗⃗⃗⃗⃗ =（ √2 a.0.0）. DM ⃗⃗⃗⃗⃗⃗ =（ √24 a.- √22 a. √64a ）. 设平面MBD 的法向量为 n ⃗ =（x.y.z ）.则 {n ⃗ •DB ⃗⃗⃗⃗⃗⃗ =0n ⃗ •DM ⃗⃗⃗⃗⃗⃗ =0 .即 {√2ax =0√24ax −√22ay +√64az =0 . 令z=2可得 n ⃗ =（0. √3 .2）.∴cos ＜ n ⃗ . DS ⃗⃗⃗⃗⃗ ＞= n ⃗ •DS ⃗⃗⃗⃗⃗ |n ⃗ ||DS ⃗⃗⃗⃗⃗ | = √62a √7×√2a= √2114 . ∴SD 与平面MBD 所成角的正弦值为|cos ＜ n ⃗ . DS ⃗⃗⃗⃗⃗ ＞|= √2114 ．【点评】：本题考查面面垂直的性质.考查空间向量与线面角的计算.属于中档题． 20.（问答题.0分）已知数列{a n }满足a 1= 12，(a n+1+1)(a n +1)=2a n +1，n ∈N ∗ ． （1）求数列{a n }的通项公式；（2）证明：对∀n∈N *.a 1a 2a 3+a 2a 3a 4+…+a n a n+1a n+2＜ 112 ．【正确答案】：【解析】：（1）求得a n+1= an1+a n.判断a n ＞0.两边取倒数.结合等差数列的定义和通项公式.可得所求通项公式；（2）求得a k a k+1a k+2= 12 [ 1(k+1)(k+2) - 1(k+2)(k+3)].再由数列的裂项相消求和和不等式的性质.即可得证．【解答】：解：（1）由a 1= 12，(a n+1+1)(a n +1)=2a n +1，n ∈N ∗ .可得a n+1= an 1+a n.由a 1＞0.可得a n ＞0. 则 1a n+1=1+ 1a n.即1a n+1- 1a n=1.所以{ 1a n}是首项为2.公差为1的等差数列. 则 1a n=2+n-1=n+1.即a n = 1n+1 ；（2）证明：a n = 1n+1 .对k=1.2.3.….a k a k+1a k+2= 1(k+1)(k+2)(k+3)= 12 [ 1(k+1)(k+2)- 1(k+2)(k+3)].所以a1a2a3+a2a3a4+…+a n a n+1a n+2= 12 [ 12×3- 13×4+ 13×4- 14×5+…+ 1(n+1)(n+2)- 1(n+2)(n+3)]= 12 [ 12×3- 1(n+2)(n+3)]= 112- 12(n+2)(n+3)＜112．【点评】：本题考查数列的递推式的运用.以及数列的裂项相消求和.考查转化思想和运算能力、推理能力.属于中档题．21.（问答题.0分）已知椭圆C：x2a2 + y2b2=1（a＞b＞0）的离心率为√32.且经过点P（- √3 . 12）．（1）求椭圆C的标准方程；（2）设点M是椭圆C上位于第一象限内的动点.A.B分别为椭圆C的左顶点和下顶点.直线MB 与x轴交于点C.直线MA与y轴交于点D.O为椭圆的中心.求三角形OCD的面积的取值范围．【正确答案】：【解析】：（1）通过椭圆E的离心率.及椭圆过点P（- √3 . 12）.求得a.b即可得到椭圆方程．（2）设M（x0.y0）.x0＞0.y0＞0.写出直线MA、MB的方程即可得到得D（0. 2 y0x0+2）.C（x0y0+1.0）.所以三角形OCD的面积S= 12 |OC||OD|= x0y0(x0+2)(y0+1)= x0y0x0y0+x0+2y0+2令x0+2y0=t.利用椭圆参数方程无得t的范围即可求解．【解答】：解：（1）由题意{ca =√323 a2+14b2=1.结合a2=b2+c2.解得a2=4.b2=1.c2=3.故.椭圆C的标准方程为：x 24+y2=1．；（2）设M（x0.y0）.x0＞0.y0＞0.a（-2.0）.B（0.-1）.直线MA的方程为：y=y0x0+2(x+2) .令x=0.得D（0. 2 y0x0+2）.直线MB的方程为：y=y0+1x0x−1 .令x=0.得C（x0y0+1.0）.所以三角形OCD的面积S= 12 |OC||OD|= x0y0(x0+2)(y0+1)= x0y0x0y0+x0+2y0+2.令x0+2y0=t.则t2=(x0+2y0)2=x02+4y02+4x0y0 =4+4x0y0. ∴ x0y0=t2−44.∴S=t2−44t2−44+t+2=1+ −4t+2．令x0=2cosθ，y0=sinθ，θ∈(0，π2) .则t=2cosθ+2sinθ=2 √2 sin（θ+π4）.∵ θ∈(0，π2) .∴ θ+π4∈(π4，3π4) .sin（θ+π4）∈（√22.1]．∵函数S=1+ −4t+2在（2.2 √2 ]单调递增.∴ S∈(0，3−2√2 ]．三角形OCD的面积的取值范围为（0.3-2 √2 ]．【点评】：本题考查椭圆方程的求法.直线与椭圆的位置关系的综合应用.三角形面积的求法.考查转化思想以及计算能力.是难题．22.（问答题.0分）已知函数f（x）=e x+cosx-2.f'（x）为f（x）的导数．（1）当x≥0时.求f'（x）的最小值；（2）当x≥−π2时.xe x+xcosx-ax2-2x≥0恒成立.求a的取值范围．【正确答案】：【解析】：（1）求导.判断函数的单调性.进而得到函数的最值；（2）令h（x）=e x+cosx-2-ax.依题意当x≥−π2时.x•h（x）≥0恒成立.然后分a≤1及a＞1讨论.即可得出结论．【解答】：解：（1）f'（x）=e x-sinx.令g（x）=e x-sinx.x≥0.则g'（x）=e x-cosx．当x∈[0.π）时.g'（x）为增函数.g'（x）≥g'（0）=0；当x∈[π.+∞）时.g'（x）≥eπ-1＞0．故x≥0时.g'（x）≥0.g（x）为增函数.故g（x）min=g（0）=1.即f'（x）的最小值为1．时.x•h（x）≥0恒成立．（2）令h（x）=e x+cosx-2-ax.h'（x）=e x-sinx-a.则x≥−π2当a≤1时.若x≥0.则由（1）可知.h'（x）≥1-a≥0.所以h（x）为增函数.故h（x）≥h（0）=0恒成立.即x•h（x）≥0恒成立；，0] .则h''（x）=e x-cosx.若x∈[−π2h'''（x）=e x+sinx在[−π，0]上为增函数.2)=e−π2−1＜0 .又h'''（0）=1. ℎ‴(−π2故存在唯一x0∈(−π，0) .使得h'''（x0）=0．2，x0)时.h'''（x）＜0.h''（x）为减函数；当x∈(−π2x∈（x0.0）时.h'''（x）≥0.h''（x）为增函数．)=e−π2＞0 .h''（0）=0.又ℎ″(−π2，0)使得h''（x1）=0．故存在唯一x1∈(−π2故x∈(−π，x1)时.h''（x1）＞0.h'（x）为增函数；2x∈（x1.0）时.h''（x1）＜0.h'（x）为减函数．)=eπ2+1−a＞0 .h'（0）=1-a≥0.又ℎ′(−π2，0]时.h'（x）＞0.h（x）为增函数.所以x∈[−π2故h（x）≤h（0）=0.即x•h（x）≥0恒成立；当a＞1时.由（1）可知h'（x）=e x-sinx-a在[0.+∞）上为增函数.且h'（0）=1-a＜0.h'（1+a）≥e1+a-1-a＞0.故存在唯一x2∈（0.+∞）.使得h'（x2）=0．则当x∈（0.x2）时.h'（x）＜0.h（x）为减函数.所以h（x）＜h（0）=0.此时x•h（x）＜0.与x•h（x）≥0恒成立矛盾．综上所述.a≤1．【点评】：本题考查利用导数研究函数的单调性.极值及最值.考查不等式的恒成立问题.考查转化思想.分类讨论思想.考查逻辑推理能力.属于中档题．。
2021年宁波市镇海中学高三英语上学期期中试卷及答案
2021年宁波市镇海中学高三英语上学期期中试卷及答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项ADesign a science holiday to win a science holidayIs there a scientific holiday,tour,vacation or experience in your list that you dream of making a reality? We might just be able to help with that! We are the crowd sourcing thoughts and suggestions onnew tours,and the winner will receive two free places on their tour when it's launched.Starting entries 1st May 2020Deadline 31st August 2020ENTER NOW---Write a brief description of your ideal science holiday,tour,vacation or experience.---No matter how big or small,we are looking for brilliant, unique thoughts.It can be a complex multi-location overseas holiday tour,a weekend away deep-diving,or a simple walking tour(and everything in between).---Our current tours are mainly individual-focused but we would love to hear ideas for families and schools.---Creations that have environmentally friendly themes and a low carbon impact are also welcome.---Your entry can be as simple as a great title but writing more specifies will improve the possibility of walking off with the prize.---The competition doesn't close by September,2020,so plenty of time to brainstorm your entry.---We will launch the best idea in late 2020 or at the beginning of 2021. The winner will receive the award once the result is published.If you want to submit more than 3,999 words,include photos/maps with your entry, or submit multiple entries then send it to tours @ . The competition guidelines and other related information are available at .1. When will the winner probably be announced?A. In August,2020.B. In September,2020.C. In January,2021.D. In May,2021.2. What kind of entries might be more likely to win?A. Those producing a low carbon impact.B. Those with a long and unique title.C. Those providing detailed information.D. Those for individual-focused tours.3. What is the purpose of this text?A. To give guidance on a competition.B. To advertise a science holiday.C. To inspire creativity in teenagers.D. To collect the ideas for tours.B"Long time no see." is a very interesting sentence. When I first read this sentence from an American friend's e-mail, I laughed. I thought it was a perfect example of Chinglish.Obviously, it is a word-by-word literal translation of the Chinese greeting with wrong English grammar and structure! Later on, my friend told me that it is a standard American greeting. I was too surprised to believe her. Her words were unbelievable at all. So I did research onGoogle. com. To my surprise, there are over 60 thousand web pages containing "Long time no see." Though it is sort of informal, it is part of the language that Americans use daily. Interestingly, if you type this phrase in Microsoft Word, the software will tell you that the grammar needs to be corrected.Nobody knows the origin of this Chinglish sentence. Some people believe that it came from Charlie Chan's movies. In the 1930s, Hollywood moviemakers successfully created a worldwide famous Chinese detective named "Charlie Chan" on wide screens. Detective Chan liked to teach Americans some Chinese wisdom by quoting Confucius. "Long time no see." was his symbol. Soon after Charlie Chan, "Long time no see." became a popular expression in America thanks to the popularity of these movies.Some scholars compare America to a hugemelting pot. All kinds of cultures are mixed in the pot together, and they change the colour and taste of each other. Language is usually the first thing to be influenced in the mixed pot.You can have some examples from other countries such as pizza from Italian, sushi from Japanese, and déjà vu from French etc. There is a long list! Since Americans admire Chinese culture more and more nowadays, I believe more Chinese words will become American English in the future. In this way, the American's melting pot keeps adding richness and flavour.4. What did the writer himself feel surprised at?A. The Chinglish expression "Long time no see."B. So many literal translations of the expressions used in America.C. "Long time no see." is used as a standard American English greeting.D. Finding out Americans use the expression every day.5. What do the underlined words "melting pot" in Paragraph 4 probably mean?A. Confucius's words.B. Culture mixture.C. A kind of cooked dish.D. American changing cultures.6. According to the passage, what can be inferred?A. Detectives translated the phrase "Long time no see."B. Cultures cannot be changed in the huge melting pot.C. The huge melting pot greatly affects all kinds of languages.D. Hollywood made "Long time no see." popular.7. What is the main idea of the passage?A. Some Chinese expressions are introduced into English.B. You'll not be surprised at a tofu and peanut butter hamburger in a restaurant in America.C. Some American expressions can be used in China.D. American English keeps being enriched by different cultures.CWhy do you check social media? Is it to keep up with everything that your friends and family are doing? Is it to find new trendy spots to eat?Regardless of the reason, you may find yourself with different degrees of envy or discomfort after a quick look at your phone. Then you might be suffering from a phenomenon known as “Fear of Missing Out (FOMO). ”While the phenomenon of FOMO can be traced back for centuries, it had never been the issue as it is today, causing widespread discussion and research. This rise in checking social media is naturally connected to the increasing leading position that social media holds over our lives. Every time someone opens their WeChat Moments, Facebook, Twitter... etc. , they are bombarded with the highlight reels of other peoples' lives. A sunny beach, delicious-looking food, a super cute kitten-they are all uplifting photos, yet they're very likely to bring about more unhappiness than joy. You see, the first thought to come out of your mind may be “Wow, that's so cool/delicious/cute”, but then it takes a hard U-turn. You're thinking: “I wish I were there” or “my life is so boring compared to his/hers.” The more you see, the more likely you are to have these negative feelings.What's worse is the habit many people have of turning to social media in search of happiness when they are feeling down, not realizing that they are just going to end up in a negative cycle of endless disappointment. Montesquieu once said: “If one only wished to be happy, this could be easily accomplished; but we wish to be happier than other people, and this is always difficult, for we believe others to be happier than they are.”FOMO will go hand in hand with dissatisfaction and envy. Appreciate what you already have, because someone else out there in the world would gladly give everything to be you.8. What is the purpose of the questions in paragraph 1?A. To introduce the topic of the passage.B. To explain the function of FOMO.C. To describe the features of FOMO.D. To give the reasons for checking social media.9. How might people feel seeing other's perfect life through social media?A. Joyful.B. Admiring.C. Comfortable.D. Envious.10. What can we conclude from Montesquieu's words in paragraph 3?A. We could turn to social media for happiness.B. We couldn't realise our dream without hard work.C. We couldn't harvest happiness through comparison.D. We could live better than others by showing ourselves online.11. What is the author's attitude towards FOMO?A. Indifferent.B. Objective.C. Doubtful.D. Hopeful.DPaper is an important part of modern life. People use it in school, at work, to make artwork and books, to wrap presents and much more. Trees are the most common material for paper these days.So how do people make paper out of trees today? People first cut trees, load them onto trucks and bring them to a factory. Machines cut open the outer coverings of the trees, and cut the trees into pieces. Those pieces are boiled into a soup. After that, it is hit flat, dried and cut up into sheets of paper.The entire process, from planting a small tree to buying your school notebook, takes a very long time. Just growing the trees takes 10 to 20 years.Making tons of paper from trees can harm the planet. Humans cut down 80, 000 to 160,000 trees around the world every day, and use many of them to make paper. Some of those trees come from tree farms. But people also cut down forests for paper, which means that animals and birds lose their homes.Cutting forests down also contributes to climate change, and paper factories pollute the air. After you throwpaper, it often takes the paper six to nine years to break down. That's why recycling is important. It saves a lot of trees, slows climate change and helps protect endangered animals, birds and all creatures that rely on forests for their homes and food.So if paper isn't good for the environment, why don't people write on something else?The answer: They do. With computers, tablets and cellphones, people use much less paper than in the past. Maybe a day will come when we won't use paper at all — or will save it for very special books and artworks.12. What can we know about making paper out of trees?A. It costs much money.B. It takes a lot of time.C. It is very easy and fast.D. It is dangerous and difficult.13. What is the impact of paper production?A. It promotes the recycling.B. It does harm to the environment.C. It slows down the climate change.D. It protects the animals from losing homes.14. How will we use paper someday in the future according to the text?A. Use it for books only.B. Use the recycled paper.C. Treasure it occasionally.D. Use it for artworks.15. What idea does the author want to express from the text?A. The influence of making paper on environment.B. The wonderful experience of making paper.C. The necessary process of making paper.D. The good reasons for making paper.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2019-2020学年宁波市镇海中学高三英语期中试题及参考答案
2019-2020学年宁波市镇海中学高三英语期中试题及参考答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项ADuring ancient times, children didn’t have smartphones, iPad or computers to entertain themselves. Instead, they came up with interesting games to play.★Stone ballsDuring the Qing Dynasty, kicking a stone ball around was a popular sport in the northern part of China, and it was often played in the winter to protect kids from the cold. Stones were carved into small balls and kicked along with feet. In 1999, the sport was included in the 6th National Ethnic Group Traditional Sports Meeting held in Beijing.★Flying kitesKites have quite a long history. The earliest kites were made of wood, instead of paper. Nowadays, the four most famous kites are the Beijing kite, Tianjin kite , Weifang kite and Nantong kite, of which each has distinctive features. The kite which resembles a swallow is a well-known Beijing style.★Hide-and-seekHide-and-seek is a traditional game for children, popular around the nation. There are two ways to play: covering a child’s eyes while other kids run around to tease(戏弄) him or, more commonly, participants hide and one child must try to find them.★Playing diabolosA diabolo is always made of wood or bamboo and has empty space in the center. By juggling(边抛边接) the diabolo on the rope, the high-speed spinning diabolos will make a sound like a whistle. Playing diabolos is an interesting folk game, especially popular in North China. Playing diabolos was also included in the first group of national intangible cultural heritage(非物质文化遗产).1.Why did ancient children often play stone balls in the winter?A.To practice their feet.B.To warm themselves.C.To train their skills.D.To relax themselves.2.Which kites are swallow-shaped?A.Weifang kites.B.Tianjin kites.C.Beijing kites.D.Nantong kites.3.Why does playing diabolos make a sound?A.Because the diabolo’s center is empty.B.Because the high-speed spinning diabolo is light.C.Because the diabolo is equipped with a whistle.D.Because ropes’ surface moves against the diabolo’s.BJanet Fein, aged 84, received her bachelor's degree from the University last week, having waited a long time for her chance to reach that goal.Growing up in the Bronx area ofNew York City, Fein worked at a dress manufacturer after graduating early at the age of 16. After getting married, she spent 18 years staying home with her children. She held several jobs through her life, including 20 years as a secretary at a hospital until her retirement at age 77.Fein has had a full life. But even then, she was not ready to take it easy and rest during a well-earned retirement. She decided to major in sociology because she felt it was “substantial.”Fein took part in a state program that lets people who are 65 and older take free classes at public universities in Texasand kept going to class even as her health conditions worsened. During her studies, it became necessary for her to use a walker to get around and she required oxygen. She also developed knee problems. So, Fein took online classes to finish the last part of her degree requirements.Sheila Rollerson, Fein' scollege advisor, and Carol, the professor told the media that Fein never showed signs of giving up even with all of her difficulties. She would speak up a lot in class and it just made for a more interesting class.Fein has also inspired Renee Brown, one of Fein's caregivers. At 53, Renee plans to begin nursing school to further her career. “Renee, you can do it. If I can do it you can do it, and you will feel so good about it,” Brown remembers what Fein told her.4. What is the second paragraph mainly about?A. Janet Fein's growthB. Janet Fein's aimC. Janet Fein's job experiencesD. Janet Fein's regrets5. What does the underlined phrase “substantial” in paragraph 3 refer to?A. InterestingB. ValuableC. EasyD. Affordable6. What can be learned about Fein according to the passage?A. Fein lived on little money after her retirement.B. Fein had a problem with her arms while at universities.C. The Internet played a role in Fein's university education.D. The good health helped Fein get her bachelor's degree.7. What is Janet Fein like according to the passage?A. Hardworking and humorous.B. Determined and generous.C. Positive and patient.D. Inspiring and perseverant.CThe American poet Louise Gluck, author of 12 collections of poetry, has been awarded the 2020 Nobel Prize in Literature. Born inNew Yorkin 1943, Gluck published her first volume of poetry, “Firstborn”, in 1968, quickly gaining her reputation as a poet. In the decades since, she has become one of the country's most celebrated literary figures. Her work uses the power of myth to deal with some of our darkest human concerns. Her straightforward language always gets at the heart of deep-seated anxieties: loneliness, rejection, death ...Stephanie Burt,an English professor atHarvardUniversity, said, “She's someone who's been able to make emotion states vivid on the page... Few poets have tried as hard as she has not to repeat herself. And her strongest books are really different from one another”. “She offers poetry lovers a lot of inspiration, but she's also on a lot of bookshelves," said Chiasson, a poet, who added she is a generous reader of her fellow writers’work.At Yale, where Gluck is a professor of English, she served for years as judge of the Yale Series of Younger Poets Prize and worked closely with poets she chose for the prize and those she did not, helping them shape their work. “In that very practical way she's had an enormous influence on a great many figures,” said Langdon Hammer, a professor of English at Yale. “She's someone who has been a 'guiding spirit’ for generations of students, writers, and readers.”Gluck described teaching and writing as symbiotic. “I teach not out of selflessness or generosity: I do it because it feeds me,” she said. “It feeds them, too, so it's a happy relationship. I'm sure not all my students feel that way, but some do. I never feel that it takes me from my work: I think it gives me my work.”8. Which of the following topics might Gluck tend to explore in her work?A. Victory.B. Divorce.C. Romance.D. Achievement.9. What quality does Gluck have according to the passage?A. Humorous and intelligent.B. Ambitious and helpful.C. Considerate and optimistic.D. Inspiring and creative.10. What does teaching mean to Gluck?A. A source of wealth.B. A barrier to writing.C. A fountain of creation.D. A stepping stone to fame.11. What is the best title for the text?A. A Guiding SpiritB. A Successful PioneerC. An Adventurous CreatorD. A Hardworking WriterD"Long time no see." is a very interesting sentence. When I first read this sentence from an American friend's e-mail, I laughed. I thought it was a perfect example of Chinglish.Obviously, it is a word-by-word literal translation of the Chinese greeting with wrong English grammar and structure! Later on, my friend told me that it is a standard American greeting. I was too surprised to believe her. Her words were unbelievable at all. So I did research onGoogle. com. To my surprise, there are over 60 thousand web pages containing "Long time no see." Though it is sort of informal, it is part of the language that Americans use daily. Interestingly, if you type this phrase in Microsoft Word, the software will tell you that the grammar needs to be corrected.Nobody knows the origin of this Chinglish sentence. Some people believe that it came from Charlie Chan's movies. In the 1930s, Hollywood moviemakers successfully created a worldwide famous Chinese detective named "Charlie Chan" on wide screens. Detective Chan liked to teach Americans some Chinese wisdom by quoting Confucius. "Long time no see." was his symbol. Soon after Charlie Chan, "Long time no see." became a popular expression in America thanks to the popularity of these movies.Some scholars compare America to a hugemelting pot. All kinds of cultures are mixed in the pot together, and they change the colour and taste of each other. Language is usually the first thing to be influenced in the mixed pot.You can have some examples from other countries such as pizza from Italian, sushi from Japanese, and déjà vu from French etc. There is a long list! Since Americans admire Chinese culture more and more nowadays, I believe more Chinese words will become American English in the future. In this way, the American's melting pot keeps adding richness and flavour.12. What did the writer himself feel surprised at?A. The Chinglish expression "Long time no see."B. So many literal translations of the expressions used in America.C. "Long time no see." is used as a standard American English greeting.D. Finding out Americans use the expression every day.13. What do the underlined words "melting pot" in Paragraph 4 probably mean?A. Confucius's words.B. Culture mixture.C. A kind of cooked dish.D. American changing cultures.14. According to the passage, what can be inferred?A. Detectives translated the phrase "Long time no see."B. Cultures cannot be changed in the huge melting pot.C. The huge melting pot greatly affects all kinds of languages.D. Hollywood made "Long time no see." popular.15. What is the main idea of the passage?A. Some Chinese expressions are introduced into English.B. You'll not be surprised at a tofu and peanut butter hamburger in a restaurant in America.C. Some American expressions can be used in China.D. American English keeps being enriched by different cultures.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2019-2020学年宁波市镇海中学高三英语期中试题及参考答案
2019-2020学年宁波市镇海中学高三英语期中试题及参考答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项AOur Teen Summer Spanish Program is two weeks of fun, educational excitement that helps students learn Spanish fast. Our Spanish summer program allows our students to learn from highly trained, certified teachers and be absorbed in the language and theculture of Costa Rica.Features include:* Intensive(强化的) daily Spanish classes* Extracurricular classes in dance, cooking, music, and handiwork* Outdoor activities including hiking, camping, rafting, and ziplining(高空滑索)* Homestay with a local Costa Rican family* Volunteer work in needy neighborhoodsOur Teaching Methods:We are proud to use TPRS---Total Physical Response Storytelling---in our curriculum. This innovative method uses strange and amusing stories to teach new vocabulary, increase fluency, and get students involved by giving them the opportunity to alter the details themselves. Because of the silliness, creativity, and repetition involved, TPRS allows students to learn easily and remember information effortlesslyMemorizing vocabulary and listening to lectures on grammar are slow, inefficient ways to learn a new language. The best way to truly learn and commit new material to memory is through conversation. In our Spanish classes, students can expect to speak up to 80% of each class. By speaking in the new language freely and consistently, students can see progress faster because they are using the new grammar and vocabulary that they have learned at the same time. This helps the brain remember the new words and grammar structures for future use, making it much easier to progress.1.What does the program do?A.It offers weekly Spanish classesB.It focuses more on outdoor activitiesC.It gives teachers a chance to receive trainingD.It provides activities about the Spanish culture2.What is the best way to learn a language according to the text?A.Memorizing a larger vocabularyB.Speaking more in the new language.C.Mastering more grammar structuresD.Writing stories to share with others3.What is the purpose of the text?A.To employexperienced Spanish teacherB.To hire foreign volunteers for a programC.To attract teen foreigners to a programD.To introduce language learning methodsBGerardo Ixcoy is a 27-year-old teacher in Guatemala. He teaches in the area where having electricity is something of a luxury and students have no access to mobile apps or computers, which became a problem when Guatemalan government had to choosedistance education for students at home because of COVID-19 inmid-March.Gerardo Ixcoy felt that he had to teach those children living far from the digital world. After all, education is a universal right. But what was the solution?Gerardo bought himself a secondhand tricycle with his savings. Once he had the tricycle, the next step was to change it into a classroom on wheels. He put a roof with a solar panel（太阳能电池板）on it, along with a screen to avoid the spread of COVID-19. He fixed a whiteboard on it so he could explain to the students the basics of primary education：math, the local language and the national language Spanish.The purpose of the solar panelis to provide constant power for a small loudspeaker so he can teach from a distance. It's a matter of respecting the healthy distance required to be safe from the virus. He, the teacher, stays outside, and the student participates from his or her house.He tries to visit his students at least twice a week, riding his classroom — cycle. The children he serves say that, although he visits them for only a few hours a week, they look forward to it. They appreciate his visit because they know he has limited time he can devote to them. They must take advantage of it to learn.4. What happened to Gerardo Ixcoy's students because of COVID-19?A. They chose to have classes in groups.B. They had to stop all the classes.C. They continued to have classes online.D. They started to learn about computers.5. Why does the teacher stay from a distance while teaching?A. Because healthy distance is reqired.B. Because he doesn't want to see his students.C. Because he wants to show his authority.D. Because the local custom reqiries him to.6. Where did Gerardo Ixcoy teach his students?A. In his own house.B. In his student's house.C. On the school playground.D. Outside his student's house.7. What can be the best title of the story?A. Love breaks down barriers.B. Hard work pays off.C. Time is money.D. Education is the key to success.CMy mother is a teacher. I often asked her, “ Why do you teach? What keeps you teaching?” The answer was always the same. “There is always that one child, that one moment that is worth it.”Now, I am a teacher, too. But different from my mother, I teach students challenging activities outdoors. And when my mother asked me the similar questions, my answer was, “It’s that one child, that one special moment.”One of those moments happened recently. I was working with a group of girls in a four-week programme. Everything went on well through the “Team” events and we were moving on to a“High” one called the Wire Walk.In the Wire Walk, each girl had to climb up the pegs in a tree to a wire, 8 meters high, and then walk across it. Of course, everything was safe for sure.When it was Susie's turn, it seemed that she didn't really want to. I asked Susie if she was ready. She answered softly, “I suppose.”Now, Susie was at the foot of the tree. She started to make the long reach. Then I found Susie was becoming nervous with every step. I realized she would not go much further.Susie was halfway up. Then she stopped and held the tree tight, looking very afraid. Her eyes were closed. With her face against the tree, she cried, “I can't.”The other girls sat in silence. I talked quietly to Susie, trying to relax her. I talked for a long time till I ran out of words.“I will still be your friend no matter what, Susie!” Mary broke the silence.To my surprise, Susie lifted her head and looked up to the wire. Then she turned to look down at Mary and smiled. Mary smiled back.Finally, Susie made it all the way across that wire. When she returned to the ground, the first hug she looked for was from Mary. We all cheered. Moments like this keep me doing what I do.8. Why did Susie stop halfway up to the wire?A. It was too high for her to reach.B. She was too afraid to move on.C. Something was wrong with it.D. The teacher asked her to do it.9. Why did Mary break the silence?A. To try again herself.B. To relax the writer.C. To encourage Susie.D. To draw Susie's attention.10. What can we learn from the story of Susie?A. Helping others is helping ourselves.B. A word of encouragement matters a lot.C. Outdoor activities bring peoplecloser than ever.D. A good method plays an important role in teaching.11. Which of the following can be the best title for the passage?A. The Wire WalkB. That One MomentC. Fear and CourageD. The Young HeartsDI am a part of the Windward Robotics team known as the Omnicats, one of the many clubs offered at my school. My eight teammates and I started out by watching the robotics qualifiers and finals of Chinese teams. We used these robotics videos for inspiration and drew up mind maps of what our team wanted to do.We meet every day for two hours after school in order to build our robots. Typically, adultsoverseewhat students are building, but my school lets teens take control. It leaves room for more errors, but we're prouder of our final product.Before our first real match, a practice was offered at the actual site. My sub-team had been further along than other teams but quickly fell behind. When we arrived on that big day, I immediately felt my teammates' nervousness. Our first match was in three minutes. All of us glanced around at each other since we were not prepared at all. Our robot was still sitting in a box. Two of our teammates rushed to take the robot for an officialexamination as the rest of us discussed our strategy (策略).Our team captain explained that we all should try driving the robot now, while it was early in the season. She gave me the remote. I had only driven for a limited amount of time in practice so I was extremely nervous. Our autonomous code (自动代码) beganto run almost immediately after our short discussion. Despite all the drama, we ended up winning the match.Not too bad for the first match! I was so proud. I did not enjoy the pressure of driving but liked building the robot behind the scenes instead.We have transferred to a different type of robotics at the end of the semester, which involves the whole team designing one giant robot with large tools. I am a part of the electrical team now, so I make sure the motherboard (母板) can supply the robot's energy needs. We have all enjoyed the rest of the season and learned that robotics it never dull!12. What does the underlined word in paragraph 2 probably mean?A. InspectB. AdmireC. OverlookD. Assume13. Why did her teammatesfed so nervous before first match?A. Because it was really too difficult for them.B. Because they didn't prepare for it at all.C. Because they didn't know what the match would be like.D. Because they didn't make full preparations before the match.14. How did the author feel about their first robotics match?A. It was tiring but exciting.B. It was stressful but wonderful.C. It was terrible but powerful.D. It was boring but professional15. What is the best title for the text?A. A Girl Who Loves RobotsB. My First Robotics MatchC. Teamwork for RobotsD. Robotics Is Never Dull第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2021届宁波市镇海中学高三语文上学期期中考试试题及答案
2021届宁波市镇海中学高三语文上学期期中考试试题及答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。
天湖海桀①走进西部，你不可以不去看湖，看那高天大地上独有的液态的翡翠，看那独属于世界第三极的纯粹的原始，看那浑古苍茫中的生命的鲜活。
②那是可以触摸到的真实的梦啊！视觉里，湖天一色，长云横空。
倒映在湖中的雪山冰肌玉骨，水面莹波荡漾。
满眼都是湿漉漉的清逸，一如天然的水墨。
③你向她走去。
④天地安详，净若装饰，却又毫不虚玄，那实实在在的鲜美质感，那青春弥漫的天灵之光，使人在茫茫然然、浩渺无涯的醺然里，充满生命的自豪。
心灵里清辉熠熠，通明如澈。
恍恍惚惚间，似有一条曲曲弯弯的小路，在洒满光斑的湖面上诱惑着、浮迤着——它伸向太阳，伸向尽头，伸向谜一般的深处，却又分分明明铺展在脚下……⑤地球上的喜马拉雅海消失了。
而这片湖水却更加年轻，更加美丽，更加温情，像一位默默无语的仙女，静静偎依在雪山的怀抱；又像是一轮永远皎洁、永远安宁的满月，在那幽远而又孤独的清澈里，在那只会消失不会变老的诗意中，守望着那个终极的谶语……⑥天湖的独异举世无双。
她的周围没有亭台，没有飞檐，没有园艺，没有楼阁；不见帆影，不见闹嚷，不见气象万千的云雾缠绵，连最最普通的一棵小树都不长……所有的只是云，只是雪，只是湖，只是草显示出的坦然。
所有的只是自然叠积出的圣洁的山，仙灵的水。
⑦无限静卧于斯。
面对这样的境域，你生命的孤独，本能的渴望，会就此蜕变。
距离消失了，坦途历历。
⑧你只想再看一眼真切的雪山、冰川、草原；你只想再领略一次三伏的清凉，超凡的意味，归宿的自然；你只想在荒野的深处，看着瀚海落日的苍凉，走向真朴，走向纯粹……继而，在突如其来的孤寂中，在恍如隔世的空寂里，在幽梦般的清逸和爱悦般的境界中，忘记尘世的喧嚣和魅惑，忘记所有的烦闷和苦恼。
心啊，如汪洋中的船，高扬着鼓胀的帆，全部的目的和向往都只会是那可以真正落脚的弥漫着人烟和爱意的村庄与大地。
2021届宁波市镇海中学高三英语上学期期中考试试题及答案
2021届宁波市镇海中学高三英语上学期期中考试试题及答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项AIt looks like 2017 is shaping up to be a record-breaking year in movie History. Here is a list of some of the year’s biggest blockbusters so far.Kong: Skull IslandA reboot (重启) of King Kong would normally get laughed at in this day and age, but it looks like this modem version of the story will be worth watching. With US actress Brie Larson and UK actor Tom Hiddleston in the mix, this film is set to be this year’s biggest monster tale.Release Date:3/10/17Beauty and the BeastDirector Bill Condon is bringing back a part of our childhood in live action, complete with the songs we all remember and love. With British stars Emma Watson and Dan Stevens leading thecast, the classic Walt Disney story already sounds like it’ll be a delight for both children and adults.Release Date:3/17/17The Fate of the FuriousThere was doubt that we’d even get a Fast 8, especially after the fitting ending US actor Paul Walker’s character was given at the end of Furious 7. Fans thought they’d never see Dom and the rest of the crew back in action, but thankfully, US star Vin Diesel himself confirmed that The Fate of the Furious is fueling up for another go.Release Date:4/14/17Spider-Man: HomecomingSpider Man is heading home to Marvel Studios for the first time. The movie will show us Peter Parker’s high school days, and will continue the threads we saw formed during his initial appearance in Captain America: Civil War. We know that Michael Keaton is playing The Vulture in this story, and that both Happy Hogan and Tony Stark, played by US actors Jon Favreau and Robert Downey Jr., are along for the ride.Release Date:7/7/171.Which of the following movies can’t you see on May Day?A.Kong: Skull Island.B.Beauty and the Beast.C.The Fate of the Furious.D.Spider-Man: Homecoming.2.What can we know about Beauty and the Beast?A.It is fueling up for another go.B.It is produced by Marvel StudiosC.It’ll show us Peter Parker’s school days.D.It’ll be enjoyable for both children and adults.3.What does the underlined word “cast” probably mean?A.Characters.B.Actors.C.Directors.D.Teachers.BSome people take their holiday decoration very seriously. And some take it to the next level. The Griffith family in Kenova, West Virginia, is in this camp. They put on a display that shows they’re just filled with Halloween spirit.Each year, this family displays 3,000 pumpkins (南瓜) in front of their home for the Halloween season. Yes, you readthatright. Ric Griffith puts out one jack-o’ -lantern (南瓜灯) for every person who lives in Kenova.Of course he doesn’t do it all on his own. He has a lot of help from his family, and also from members of the community who are super-proud of what has become quite the tourist attraction over the years. More than 30,000 people stop by to see the amazing display, which includes jack‑o’‑lanterns cut to look like the faces of famous people, animals, cartoon characters, and other creative designs. It’s certainly a must-see as part of the area’s Ceredo-Kenova Autumnfest.To fit in all 3,000 pumpkins, Griffith and his helpers spread them across the home’s garden and front porch (门廊), as well as on the roof!Griffith began the tradition back in 1978 with just fivepumpkins and, many years later, it’s much bigger and better. People can’t help but come to the area, walking along the sidewalk outside the house to get that perfect Halloween experience.“Locals take great pride in it, and then there are people from around the country who plan fall trips and include it in their trips so they can see it,” said Tyson Compton, president of the Cabell‑Huntington Convention and Visitors Bureau. “It’s really something.”“It’s become a tradition for many people in our area, and it feels good to keep that going,” Griffith said.4. What does the underlined word “that” in Paragraph 2 refer to?A. The Griffiths selling 3,000 pumpkins a day.B. Ric Griffith giving 3,000 jack-o’‑lanterns to tourists.C. Ric Griffith teaching 3,000 people how to display jack-o’‑lanterns.D. The Griffiths decorating their house with 3,000 pumpkins.5. What can we learn from Paragraph 3?A. Ric Griffith’s idea has received support from his community.B. There are only two kinds of jack-o’‑lanterns.C. Ric Griffith dislikes asking for help from others.D. There are 30,000 people in Kenova.6. What do we know about the tradition?A. It began with 1,978 pumpkins.B. It has lasted more than 40 years.C. It includes five pumpkin competitions.D. It encourages people to do more exercise.7. What’s Tyson Compton’s attitude towards the tradition?A. Worried.B. Uncaring.C. Favorable.D. Uncertain.CThe British poet Matthew Byrne moved to Beijing in 2013 and felt that the capital city's poetry scene was lacking．His obsession（痴迷）for starting poetryevents led to the foundation of the Spittoon Collective in May 2015．“At that time，the literary activity in Beijing was The Bookworm based in Sanlitun，”Byrne says．While some of Beijing's literary institutions would go on to close in the fall of 2019，Spittoon would continue to grow as a community for poets and writers，as well as musicians and others in the creative scene．Byrne describes the Spittoon Collective as a platform for people to share ideas，from literary works to different forms of art，with projects developing from the creative energy within the community．Spittoon originally started as a poetry night at the Mado Bar in Dongcheng District 's Baochao Hutong．Byrne says，“In Beijing，you have these wonderful hutongs，ancient structures where you can walk down and visit cool bars，so I thought it would be good to have a poetry event as it seemed like poetry belonged naturally to this area．”He adds，“The objective now is to discover Chinese voices and broadcast them to the rest of the world．We create a kindof theme park-like atmosphere where every Thursday is occupied by a different literary style or art form．”The readings would mainly be in English，but with an international community，a new section called“Poetry-in-Translation”was started，which featured works in Chinese，French，Arabic，Russian，Spanish and other languages．Joining organized activities like Spittoon can be a major help for those caught up in a boring life．And it's especially important for the people who have moved to China as they need to buildnew relationships while living in a different country．8. What can we know about the Spittoon Collective?A. It was closed in 2019．B. It was set up in 2013 in Beijing．C. It's popular with literature lovers．D. It's a community just for foreigners．9. Why did Byrne start the Spittoon Collective in Baochao Hutong?A. To attract students' love of poetry．B. To expandChina's literature globally．C. To makeBeijing's hutongs famous．D. To combine poetry with the hutong．10. What effect would the Spittoon Collective have on Chinese culture?A. Beneficial．B. Negative．C. Challenging．D. Controversial．11. What is the author's attitude to the Spittoon Collective?A. Unclear．B. Intolerant．C. Doubtful．D. Favorable．DThe World Wildlife Fund (WWF) says more than half of the world’s wildlife population has been lost, whichthe conservation group says has placed the health of the planet at risk.The WWF recently released its 10th Flagship Living Planet Report. The group warns the condition of the world's animals is worse than its earlier reports showed, indicating worldwide action is needed.The WWF is worried about the loss of and damage to Earth’s environment. The report provides information about more than 10,000 animal populations from 1970 to 2010. These populations are called “vertebrate species,” oranimals with backbones — like fish, birds, mammals, amphibians and reptiles. The report shows these populations have dropped by 52 percent in just 40 years. It warns freshwater species have fallen by 76 percent, which is almost twice the loss of land and ocean species. Most of these losses are in the tropics, with the biggest drop inLatin America.Marco Lambertini, the WWF’s International Director-General said, “This is about losing natural habitats. This is aboutconvertingforests, grasslands, and wetlands into agriculture mainly, and it is about unsustainable use of wildlife. Illegal hunting has been actually increasing over the last 10 years, which definitely a driving force for extinction, particularly of large species.”The report also notes what it calls the world’s “Ecological Footprint”, that is, the effect of human activities on the planet. Mr. Lambertini says there has been an increase in carbon dioxide gases and the pouring of nitrogen into oceans and rivers from fertilizers used in agriculture, which certainly cannot continue.“We are consuming on average every year about the equivalent of about 1.5, one and a half times the resources available to the planet. That means we are cutting trees more quickly than they can be restored. We are fishing the oceans more quickly than fishing stocks can reproduce, and we are emitting in the atmosphere more CO2than the natural systems can actually absorb, which is clearly not sustainable.”Mr. Lambertini warns climate change affects almost everyone on the planet and that whole species may disappear if the world does not reduce the effects of humans on the climate.12. According to the passage, what kind of species faces the biggest drop in population?A. Land and ocean species.B. Animals with backbones.C. Freshwater species inLatin America.D. Freshwater species in the tropics.13. All the following can contribute to the loss of world’s wildlife population EXCEPT ________.A. turning wildlife habitats into agriculture land.B. making sustainable use of wildlife.C. hunting illegally.D. emitting CO2 gases and pouring nitrogen.14. Which does the underlined word “converting” in paragraph 4 mean?A. Conserving.B. Conveying.C. Exchanging.D. Transforming.15. It can be inferred from the passage that _______.A. Marco is much concerned about human’s current behaviors towards wildlife.B. what the planet provides now can satisfy human’s sustainable development.C. more than half of the world’s wildlife population has been lost.D. if humans reduce the effects on the climate, the whole species will not disappear.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2021届宁波市镇海中学高三英语期中考试试题及答案解析
2021届宁波市镇海中学高三英语期中考试试题及答案解析第一部分阅读（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A、B、C、D四个选项中选出最佳选项AThe Internet can provide a wealth of educational resources for small children, if you know where to look.Enchanted LearningEnchanted learning is a great website for children, and I know several teachers who rely on it for materials for their classrooms. Most of the information is free, but for $ 20.00 per year, you can purchase a membership that allows you to have access to the site without the advertising. The pages here are great. There are color1 ed pages and worksheets for toddlers (学步儿童) and school-age children.StarfallStarfall is another good educational website for small children. Teachers often use this website in classrooms. It emphasizes reading skills for early learners. The website is aimed at first-grade-level learners, but it has tools that can help all children from pre-K to second grade. The website has reading activities, worksheets to download, and a store where you can purchase educational materials for your children.Sesame Street WorkshopKids loveSesame Street, and it is always educational. This is one of the funniest websites online for children because it is very well animated and has great sound effects. The website has games and stories for small kids, and they can create letters at the post office and mail them to their favorite characters.1. Why does Enchanted Learning offer purchasing memberships?A. Charging for the information.B. Charging for downloading materials.C. Giving access to the site without the ads.D. Providing unlimited search for more pages.2. Where can you buy educational materials for your children?A. Enchanted LearningB. StarfallC. Sesame Street WorkshopD. Colored pages3. What is the purpose of this text?A. To introduce.B. To discuss.C. To persuade.D. To educate.BNaomi Cooke was walking with a friend and their dogs through her local park in Burnside, on Tuesday when she heard someone shout to watch out. Cooke turned and hardly had time to react before a flying disc hit her in the face with a "big bang”, leaving her right cheek swollen almost to the size of a golfball.Two men playing disc golf at the course in Jellie Park were about 20 metres fromthe pairwhen one of them threw the disc hard, aiming for a nearby goal.After being hit Cooke immediately went to the emergency department, where two CT scans on her face and cheek found she had escaped any broken bones. "I'm lucky it didn't hit my eye because I think I would have lost it." Cooke said.Cooke often walks her dog at the park and said it was always busy with people playing disc golf, but it was not until after Tuesday that she became concerned about public safety there.There were no signs about the disc golf course in the park, she said, and the area is shared with children and people walking their dogs.“If it had hit one of the kids in the head, it could have killed them.” Cooke did not think she was the only person who had been hit before, and said there would be others who share her concerns.Cooke planned to go to the council, saying it needed to realise how dangerous it was for the space to be shared by everyone and to provide disc golfers with a space where they can play safely. "There should be rules about how it's done, making it safe for everyone.”4. What happened to Cooke on Tuesday?A. She was struck by a golf ball.B. She was hit by a flying disc.C. She was beaten by two men.D. She was frightened by a mad dog.5. What do the underlined words "the pair" in Paragraph 2 refer to?A. Cooke and her friend.B. Cooke and her dog.C. The two disc golfers.D. The two CT scans.6. How did Cooke feel about people playing disc golf in the park?A. Acceptable.B. Shocked.C. Angry.D. Worried.7. Why did Cooke plan to go to the council?A. To get the two men in trouble.B. To call for a ban on disc golf.C. To ask for personal protection.D. To call for safer places for disc golf.COnce a rich and clever boy had practically everything a boy could want, so he was not interested in most toys. But he couldn't get a very old mirror, and heconvinced his parents to buy it from mysterious（神秘的）old man. When the mirror arrived home, the boy went to see his reflection in it. His face looked very sad indeed. He tried smiling and making funny faces, but his reflection continued with its sad expression. "What a terrible mirror! It's the first time I've seen a mirror that didn't work properly!" the boy jumped violently.That same afternoon he went into the street to play and bought a few toys, but on his way to the park he saw a little girl who was crying her heart out. The girl was crying so much and looked so lonely that the rich boy went over to help her and to see what had happened. The little girl told him that she had lost her parents.Together the two set off in search of the parents. As the little girl wouldn't stop crying, the boy spent his money buying her sweets to cheer her up. Finally, after much walking, they found her parents who were much worried and were looking for her everywhere.The rich boy said goodbye to them. As it was getting late, he decided to head for home, without being able to play. At home, he went to his room, and noticed a shining light in the corner, the same corner he had left the mirror in. Seeing this, he went over to the mirror, and realised that the light was coming from his own body, so radiant（闪亮的）with happiness he had become.And so he understood the mystery of that mirror, the only mirror which could faithfully reflect the true joy of its owner. He realised it was true. He felt very happy at having helped that little girl. And since then, each morning when he looked in that mirror and failed to see a special shine, he knew what he had to do to bring it back.8. How did the boy feel when he first looked into the mirror?A. Embarrassed.B. Angry.C. Worried.D. Delighted.9. Why was the little girl crying so hard?A. She couldn't find her parents.B. She couldn't get the mysterious mirror.C. Her parents couldn't buy toys for her.D. The boy refused to give his toys to her.10. What could the boy see in the mirror after he went back from the park?A. A shining toy.B. A broken mirror.C. His happy face.D. The lovely girl.11. What is the purpose of this text?A. To tell us a horrible story.B. To introduce to us a strange mirror.C. To warn us not to be selfish.D. To encourage us to help others.DElonMusk is a symbol of passion, hard work and success. A person may work hard in his life, but when hard work goes along with passion and a new idea, then something is created that astonishes the world. As the founder of big companies like PayPal, SpaceX, and Tesla Motors, Elon Musk is a guide and inspiration for the youth around the world.Elon Musk was born on June 28,1971 inPretoria,South Africa. He was the eldest of three children. His father Errol Musk, a British-born South African, was an engineer. Musk spent his childhood inSouth Africaand at the age of 9, he got his first personal computer. This got Elon interested in programming and he started to learn on his own. At the age of 12, he made his first computer game, Blaster, which earned him $500 when he sold the source code to a magazine.When he was 17, he decided to leave home and had wanted to move to theUnited Statesafter graduating from a secondary school inPretoria. However, he was unable to move to theUnited States. In 1989, Elon Musk moved toCanada, where his mother’s s lived. After obtaining Canadian citizenship, Musk went toMontreal. Due to a lack of money, he found a low-paying job. At the age of 19, he enteredQueensUniversityinKingston,Ontario, studying there for two years. Finally, in 1992, his dream came true—he moved to theUnited States.As a teenager, Musk struggled with depression. But then he realized that the most important thing was presenting himself with the right questions. The day he started doing this, everything started to get easier.Musk thought that a human being had to expand the limits of his consciousness—to ask the right questions and get answers to them; thus he asked the best question of all: what will have the greatest impact on the future of humans? Musk found that these things were the Internet and space exploration.The life of Elon Musk is inspiring. He found success on the Internet, without much formal learning. By reading books and using his willpower and dedication, he reached a level of knowledge that is difficult to imagine. He did not give up until he achieved his dreams. Elon Musk truly stands out among other billionaires because of his leadership abilities, positive attitude, ability to know when to take risks, and belief in futuristic technologies.12. According to the passage, Elon Musk gained knowledge of computer by ________.A. teaching himselfB. learning from his fatherC. taking online coursesD. studying inQueensUniversity13. What made things easier for Elon Musk when he suffered from depression?A. Expanding knowledge by reading books.B. Asking himself the right questions.C. Achieving success on the Internet.D. Designing computer games.14. What can we learn from the passage?A. It’s never too late to learn.B. Youth means limitless possibilities.C. Where there’s a will, there’s a way.D. He who doesn’t advance falls backward.15. Which words can best describe Elon Musk according to the passage?A. Optimistic and generous.B. Ambitious and determined.C. Independent and helpful.D. Adventurous and realistic.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
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2019学年镇海中学高三上期中
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2019学年镇海中学高三上期中
一、选择题：每小题4分，共40分
1. 已知集合{}
2450A x x x =∈--≤Z ，{}0ln 2B x x =<<,则A
B 的元素的个数（ ）
A ．2
B ．3
C ．4
D ．7
2. 若a ,b ，c ∈R ,且a b >，则下列不等式中一定成立的是( ）
A ．ac bc >
B ．()20a b c ->
C ．11
a b < D ．22a b -<-
3. 已知n S 是等差数列{}n a 的前n 项和，且24S =，418S =，则6S 等于（ ）
A ．50
B ．42
C ．38
D ．36
4. 函数()243
x
x f x =
的图象大致为（ )
B
D
5. 如图是一个几何体的三视图，则这个几何体的表面积是
（ ) A ．
76 B ．84
C ．76+
D ．84+
6. 将函数()y f x =的图象向右平移
6
π
个单位长度后，得到 ()sin 26g x x π⎛
⎫=+ ⎪⎝⎭，则()y f x =的函数解析式为( ）
A ．()cos 2f x x =-
B ．()sin 26f x x π⎛
⎫=- ⎪⎝
⎭
C ．()cos 2f x x =
D ．()cos 26f x x π⎛
⎫=- ⎪⎝
⎭
7. 设命题p ：()lg 210x -≤，命题q ：()
10x a x a
-+≤-,若q 是p 的必要不充分条件,则实数a 的取值
范围
是（
）
俯视图
侧视图
正视图
2019学年镇海中学高三上期中
A ．10,2⎡⎤⎢⎥⎣⎦
B ．10,2⎛⎫ ⎪⎝⎭
C ．10,2⎡⎫⎪⎢⎣⎭
D ．∅
8. 已知2
2
π
π
αβ-
<-<
,sin 2cos 1αβ-=
,cos 2sin αβ+=sin 6πβ⎛⎫
+= ⎪⎝
⎭
( ）
A
B
C
． D
． 9. 已知椭圆和双曲线有相同的焦点1F ，2F ,设点
P 是该椭圆和双曲线的一个公共点，且
123
F PF π
∠=
，
若椭圆和双曲线的离心率分别为1e ，2e ，则22
12e e +的最小值为（ ）
A
B
．4+C
．2 D
．1+
10. 设a ，b 为正实数，且121322a b a b ++
+=，则12
a b +的最大值与最小值之和为（ ) A ．2
B ．92
C ．132
D ．9
二、填空题：单空题每题4分,多空题每题6分
11. 抛物线22y x =的焦点坐标为 ，准线方程为 ．
12. 已知点()1,0A ，()0,2B ,点(),P a b 在线段AB 上，则直线AB 的斜率为 ， ⋅a b 的最大
值为 ．
13. 若实数(),x y 满足约束条件20
201x y x y y --≤⎧⎪
+-≥⎨⎪≤⎩
,则2x y -的最小值为 ，
的最
小值为 ．
14. 已知长方体1111ABCD A B C D -中，11
2
AA =
，
1AB =，AD =，则直线1AA 与平面1A BD 所成的角为 ,若空间的一条直线l 与直线1AA 所成的角为4
π
，则直线l 与平面1A BD 所成的最大角为 ．
15. 已知{}n a 是等比数列且0n a >，243546225a a a a a a ++=，则4a 的最大值为 ． 16. 已知圆
O ：221x y +=，设点P 是恒过点()0,4的直线l 上任意的一点，若在该圆上存在点A
使得
3
OPA π
∠≤
,则直线l 的斜率k 的取值范围为 ．
17. 已知()11,A x y ，()22,B x y 为单位圆上两点，且满足1
2
OA OB ⋅=
，则1122x y x y +++的取值范围为 ．
三、解答题：5小题，共74分
18. 已知()sin cos sin 222x x x f x a ⎛⎫=⋅++ ⎪
⎝⎭的最大值为．
（1）求实数a 的值;
（2)
若44f a f a ππ⎛⎫⎛⎫++-= ⎪ ⎪⎝⎭
⎝⎭
,求21
41tan παα⎛
⎫-+ ⎪⎝⎭+的值．
19. 在锐角ABC △中,角A ,B ，C 所对的边分别为a ,b ，c ．已知3b =,2239a c c =-+．
（1)求A ；
(2）求22sin sin B C +的取值范围．
20. 如图，在三棱锥P ABC -中，PAB △与ABC △都为等腰直角三角形，PA PB ⊥，AB AC ⊥，M 为
AC 的中点，且PM AC =．
（1）求二面角P AB C --的大小；
（2）求直线PM 与平面PBC 所成角的正弦值．
M P
C
B
A
21. 已知数列{}n a 的前n 项和为{}n S ,且满足：12a =,()
*132n n a S n +=-+∈N ．
（1）求数列{}n a 的通项公式;
（2）数列{}n b 满足12b =-，()()()2131n n n b b n n n a +-+=+,求数列{}n b 的通项公式．
22. 在平面直角坐标系中，已知()2,0F ，()2,P t -，若线段FP 的中垂线l 与抛物线()
2:20C y px p =>总是相切．
（1）求抛物线C 的方程;
（2）若过点()2,1Q 的直线l '交抛物线C 于M 、N 两点,过M ,N 分别作抛物线的切线1l ，2l 相交于点A ，1l ，2l 分别与y 轴交于点B ，C ；
（i ）证明：当l '变化时，ABC △的外接圆过定点，并求出定点的坐标； （ii ）求ABC △的外接圆面积的最小值．。