2020年1月9日福建省泉州市高2020届高2017级高三高中毕业班期末质量检测理科数学试题及参考答案解析
2017年泉州市普通高中毕业班质量检查理科数学含答案
2017年泉州市普通高中毕业班质量检查理科数学第Ⅰ卷一、选择题:本大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知z 为复数z 的共轭复数,且()11i z i -=+,则z 为( ) A .i - B . i C .1i - D .1i +2.已知集合11|<22,|ln 022x A x B x x ⎧⎫⎧⎫⎛⎫=≤=-≤⎨⎬⎨⎬ ⎪⎩⎭⎝⎭⎩⎭,则()R A C B = ( ) A . ∅ B .11,2⎛⎤- ⎥⎝⎦C .1,12⎡⎫⎪⎢⎣⎭D .(]1,1-3. 若实数,x y 满足约束条件1222x y x y ≤⎧⎪≤⎨⎪+≥⎩,则22z x y =+的最小值是( )AB .45C .1D . 44.已知向量,a b满足()1,0a a b a a b =-=-= ,则2b a -= ( ) A . 2 B..5. 已知n S 为数列{}n a 的前n 项和且22n n S a =-,则54S S -的值为( ) A . 8 B .10 C. 16 D .32 6.已知函数()2sin cos 222x x f x ϕϕπϕ++⎛⎫⎛⎫⎛⎫=<⎪ ⎪⎪⎝⎭⎝⎭⎝⎭,且对于任意的x R ∈,()6f x f π⎛⎫≤ ⎪⎝⎭.则 ( )A .()()f x f x π=+B .()2f x f x π⎛⎫=+⎪⎝⎭C. ()3f x f x π⎛⎫=-⎪⎝⎭ D .()6f x f x π⎛⎫=- ⎪⎝⎭7. 函数()()ln sin 0f x x x x x ππ=+-≤≤≠且的图象大致是( )A .B .C. D .8.关于x 的方程ln 10x x kx -+=在区间1,e e ⎡⎤⎢⎥⎣⎦上有两个不等实根,则实数k 的取值范围是( )A .11,1e ⎛⎤+ ⎥⎝⎦ B .(]1,1e - C. 11,1e e⎡⎤+-⎢⎥⎣⎦D .()1,+∞9.机器人AlphaGo (阿法狗)在下围棋时,令人称道的算法策略是:每一手棋都能保证在接下来的十几步后,局面依然是满意的.这种策略给了我们启示:每一步相对完美的决策,对最后的胜利都会产生积极的影响.下面的算法是寻找“1210,,,a a a ”中“比较大的数t ”,现输入正整数“42,61,80,12,79,18,82,57,31,18“,从左到右依次为1210,,,a a a ,其中最大的数记为T ,则T t -= ( )A .0B . 1 C. 2 D .310.某几何体的三视图如图所示,则该几何体的侧视图中的虚线部分是 ( )A .圆弧B .抛物线的一部分 C. 椭圆的一部分 D .双曲线的一部分 11.已知抛物线E 的焦点为F ,准线为l 过F 的直线m 与E 交于,A B 两点,,CD 分别为,A B 在l 上的射影,M 为AB 的中点,若m 与l 不平行,则CMD ∆是( )A .等腰三角形且为锐角三角形B .等腰三角形且为钝角三角形 C.等腰直角三角形 D .非等腰的直角三角形 12. 数列{}n a 满足12sin122n n n a a n π+⎛⎫=-+ ⎪⎝⎭,则数列{}n a 的前100项和为( ) A . 5050 B .5100 C.9800 D .9850第Ⅱ卷二、填空题:本大题共4小题,每题5分,满分20分,将答案填在答题纸上13.某厂在生产甲产品的过程中,产量x (吨)与生产能耗y (吨)的对应数据如下表:根据最小二乘法求得回归直线方程为ˆ0.65yx a =+.当产量为80吨时,预计需要生产能耗为 吨.14. ()()4121x x -+的展开式中,3x 的系数为 .15.已知l 为双曲线()2222:10,0x y C a b a b-=>>的一条渐近线,l 与圆()222x c y a-+=(其中222c a b =+)相交于,A B 两点,若AB a =,则C 的离心率为 .16.如图,一张4A 纸的长、宽分别为,2a .,,,A B C D 分别是其四条边的中点.现将其沿图中虚线掀折起,使得1234,,,P P P P 四点重合为一点P ,从而得到一个多面体.关于该多面体的下列命题,正确的是 .(写出所有正确命题的序号) ①该多面体是三棱锥; ②平面BAD ⊥平面BCD ;③平面BAC ⊥平面ACD ; ④该多面体外接球的表面积为25a π三、解答题 (本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤.)17. ABC ∆的内角,,A B C 的对边分别为,,a b c ,且()2cos cos cos sin A C A C B -+= .(1)证明:,,a b c 成等比数列;(2)若角B 的平分线BD 交AC 于点D ,且6,2BAD BCD b S S ∆∆==,求BD .18.如图,在以,,,,,A B C D E F 为顶点的多面体中,AF ⊥平面ABCD ,DE ⊥平面ABCD ,0//,,60,244AD BC AB CD ABC BC AF AD DE =∠=====.(1)请在图中作出平面α,使得DE α⊂,且//BF α,并说明理由; (2)求直线EF 和平面BCE 所成角的正弦值.19.某校为了解校园安全教育系列活动的成效,对全校学生进行一次安全意识测试,根据测试成绩评定“合格”、“不合格”两个等级,同时对相应等级进行量化:“合格”记5分,“不合格”记为0分.现随机抽取部分学生的答卷,统计结果及对应的频率分布直方图如下所示.(1)求,,a b c 的值;(2)用分层抽样的方法,从评定等级为“合格”和“不合格”的学生中选取10人进行座谈.现再从这10人中任选4人,记所选4人的量化总分为ξ,求ξ的分布列及数学期望()E ξ; (3)某评估机构以指标M (()()E M D ξξ=,其中()D ξ表示ξ的方差)来评估该校安全教育活动的成效.若0.7M ≥,则认定教育活动是有效的;否则认定教育活动无效,应调整安全教育方案.在(2)的条件下,判断该校是否应调整安全教育方案?20. ABC ∆中,O 是BC 的中点,BC =,其周长为6+,若点T 在线段AO 上,且2AT TO =.(1)建立合适的平面直角坐标系,求点T 的轨迹E 的方程;(2)若,M N 是射线OC 上不同两点,1OM ON = ,过点M 的直线与E 交于,P Q ,直线QN 与E 交于另一点R .证明:MPR ∆是等腰三角形. 21. 已知函数()()ln 11,f x mx x x m R =+++∈.(1)若直线l 与曲线()y f x =恒相切于同一定点,求l 的方程; (2)当0x ≥时,()xf x e ≤,求实数m 的取值范围.请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分.22.选修4-4:坐标系与参数方程在平面直角坐标系xOy 中,直线l 的参数方程为3cos 1sin x t y t ϕϕ=+⎧⎨=+⎩(t 为参数),在以坐标原点为极点,x 轴的正半轴为极轴的极坐标系中,圆C 的方程为4cos ρθ=. (1)求l 的普通方程和C 的直角坐标方程;(2)当()0,ϕπ∈时,l 与C 相交于,P Q 两点,求PQ 的最小值. 23.选修4-5:不等式选讲 已知函数()124f x x x =++-. (1)解关于x 的不等式()9f x <;(2)若直线y m =与曲线()y f x =围成一个三角形,求实数m 的取值范围,并求所围成的三角形面积的最大值.试卷答案一、选择题1-5: ABBAD 6-10: CDADD 11、12:AB二、填空题16. ①②③④ 三、解答题17.解法一:(1)因为()2cos cos cos sin A C A C B -+= ,所以()2cos cos cos cos sin sin sin A C A C A C B --= ,化简可得2sin sin sin A C B =,由正弦定理得,2b ac =,故,,a b c 成等比数列.(2)由题意2BAD BCD S S ∆∆=,得11sin 2sin 22BA BD ABD BC BD CBD ∠=⨯∠ , 又因为BD 是角平分线,所以ABD CBD ∠=∠,即sin sin ABD CBD ∠=∠, 化简得,2BA BC =,即2c a =.由(1)知,2ac b =,解得a c == 再由2BAD BCD S S ∆∆=得,11222AD h CD h ⎛⎫=⨯ ⎪⎝⎭(h 为ABC ∆中AC 边上的高), 即2AD CD =,又因为6AC =,所以4,2AD CD ==. 【注】利用角平分线定理得到4,2AD CD ==同样得分,在ABC ∆中由余弦定理可得,222cos2b c a A bc +-===在ABD ∆中由余弦定理可得,2222cos BD AD AB AD AB A =+-,即(22242428BD =+-⨯⨯=,求得BD =解法二:(1)同解法一.(2)同解法一,4,2AD CD ==.在ABC ∆中由余弦定理可得,222cos 2b a c C ab +-==, 在BCD ∆中由余弦定理可得,2222cos BD CD BC CD BC C =+- ,即(22222228BD =+-⨯⨯=,求得BD =解法三: (1)同解法一.(2)同解法二,4,2AD CD ==.在ABC ∆中由余弦定理可得,222543cos 2724a cb B ac +-===, 由于2cos 12sin2BB =-,从而可得sin 2B =, 在ABC ∆中由余弦定理可得,222cos 2b a c C ab +-==,求得sin C = 在BCD ∆中由正弦定理可得,sin sin CD BD CBD C =∠,即sin sin CD CBD CBD==∠ . 【注】若求得sin A 的值后,在BDA ∆中应用正弦定理求得BD 的,请类比得分. 解法四: (1)同解法一.(2)同解法一,4,2AD CD ==.在BCD ∆中由余弦定理得,(2222214cos 224BD BD BDC BD BD +--∠==⨯⨯,在BDA ∆中由余弦定理得,(2222456cos 248BD BD BDA BDBD+--∠==⨯⨯,因为BDA BDC π∠+∠=,所以有cos cos 0BDC BDA ∠+∠=,故221456048BD BD BD BD--+=,整理得,2384BD =,即BD =18.解:(1)如图,取BC 中点P ,连接,PD PE ,则平面PDE 即为所求的平面α. 显然,以下只需证明//BF 平面α; ∵2,//BC AD AD BC =, ∴//AD BP 且AD BP =, ∴四边形ABPD 为平行四边形, ∴//AB DP .又AB ⊄平面PDE ,PD ⊂平面PDE , ∴//AB 平面PDE .∵AF ⊥平面ABCD ,DE ⊥平面ABCD , ∴//AF DE .又AF ⊄平面PDE ,DE ⊂平面PDE , ∴//AF 平面PDE ,又AF ⊂平面,ABF AB ⊂平面,ABF AB AF A ⋂=, ∴平面//ABF 平面PDE . 又BF ⊂平面ABF ,∴//BF 平面PDE ,即//BF 平面α.(2)过点A 作AG AD ⊥并交BC 于G , ∵AF ⊥平面ABCD ,∴,AF AG AF AD ⊥⊥,即,,AG AD AF 两两垂直,以A 为原点,以,,AG AD AF 所在直线分别为,,x y z 轴,建立如图所示空间直角坐标系A xyz -.在等腰梯形ABCD 中,∵060,24ABG BC AD ∠===,∴1,BG AG ==则))1,0,BC-.∵44AF DE ==,∴()()0,2,1,0,0,4E F ,∴()()0,4,0,BC BE ==.设平面BCE 的法向量(),,n x y z =,由00n BC n BE ⎧=⎪⎨=⎪⎩,得4030y y z =⎧⎪⎨++=⎪⎩,取x =BCE的一个法向量)n =.设直线EF 和平面BCE 所成角为θ,又∵()0,2,3EF =-,∴sin cos ,n EF θ===,故直线EF 和平面BCE所成角的正弦值为26. 19.解:(1)由频率分布直方图可知,得分在[)20,40的频率为0.005200.1⨯=, 故抽取的学生答卷数为:6600.1=, 又由频率分布直方图可知,得分在[]80,100的频率为0.2, 所以600.212b =⨯=,又2460b a b +++=,得30a b +=, 所以18a =.180.0156020c ==⨯.(2)“不合格”与“合格”的人数比例为24:36=2:3, 因此抽取的10人中“不合格”有4人,“合格”有6人. 所以ξ有20,15,10,5,0共5种可能的取值.ξ的分布列为:()()()431226646444410101018320,15,1014217C C C C C P P P C C C ξξξ=========,()()134644441010415,035210C C C P P C C ξξ======. ξ的分布列为:所以()20151050121421735210E ξ=⨯+⨯+⨯+⨯+⨯=. (3)由(2)可得()()()()()()2222218341201215121012512012161421735210D ξ=-⨯+-⨯+-⨯+-⨯+-⨯=,所以()()120.750.716E M D ξξ===>,故我们认为该校的安全教育活动是有效的,不需要调整安全教育方案. 20.解法一:(1)以O 为坐标原点,以BC的方向为x 轴的正方向,建立平面直角坐标系xOy .依题意得,B C ⎛⎫⎫⎪⎪ ⎪⎪⎝⎭⎝⎭.由6AB AC BC ++=+6AB AC +=, 因为故6AB AC BC +=>,所以点A 的轨迹是以,B C 为焦点,长轴长为6的椭圆(除去长轴端点),所以A 的轨迹方程为()2221399x y x +=≠±. 设()()00,,,A x y T x y ,依题意13OT OA =,所以()()001,,3x y x y =,即0033x x y y =⎧⎨=⎩,代入A 的轨迹方程222199x y +=得,()()22323199x y +=,所以点T 的轨迹E 的方程为()22211x y x +=≠±.(2)设()()()()()1122331,0,,0,1,,,,,,M m N m Q x y P x y R x y m ⎛⎫≠⎪⎝⎭. 由题意得直线QM 不与坐标轴平行, 因为11QM y k x m =-,所以直线QM 为()11y y x m x m=--, 与2221x y +=联立得,()()()22222211111122120mmx x m x x mx x m x +---+--=,由韦达定理2221111221212mx x m x x x m mx --=+-,同理222222111*********111122121112x x x mx m x x m m x x x x m mx x m m ⎛⎫⎛⎫-- ⎪ ⎪--⎝⎭⎝⎭===+-⎛⎫⎛⎫+- ⎪ ⎪⎝⎭⎝⎭, 所以23x x =或10x =, 当23x x =时,PR x ⊥轴, 当10x =时,由()()2112212112m x x x mmx -+=+-,得2221mx m =+,同理3222122111m m x x m m ⎛⎫ ⎪⎝⎭===+⎛⎫+ ⎪⎝⎭,PR x ⊥轴.因此MP MR =,故MPR ∆是等腰三角形. 解法二:(1)以O 为坐标原点,以BC的方向为x 轴的正方向,建立平面直角坐标系xOy .依题意得,22B C ⎛⎫⎛⎫-⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭. 在x轴上取12,F F ⎛⎫⎫ ⎪⎪ ⎪⎪⎝⎭⎝⎭,因为点T 在线段AO 上,且2AT TO =, 所以12//,//FT AB F T AC ,则()1212116233FT F T AB AC F F +=+=⨯=>= 故T 的轨迹是以12,F F 为焦点,长轴长为2的椭圆(除去长轴端点), 所以点T 的轨迹E 的方程为()22211x y x +=≠±.(2)设()()()1,0,,0,1,M m N n m n m ⎛⎫≠=⎪⎝⎭,()()()112233,,,,,Q x y P x y R x y , 由题意得,直线QM 斜率不为0,且()01,2,3i y i ≠=,故设直线QM 的方程为:x t y m =+ ,其中11x mt y -=, 与椭圆方程2221x y +=联立得,()2222210t y mty m +++-=,由韦达定理可知,212212m y y t -=+ ,其中()22221211122112222x m x mx m y t y y --+++=+=,因为()11,Q x y 满足椭圆方程,故有221121x y +=,所以22121122mx m t y -++=. 设直线RN 的方程为:x sy n =+,其中11x ns y -=, 同理222113221121,22nx n n y y s s y -+-=+=+ , 故()()()()()()222222212222231321122211222m m s m s y y y t n y y y n t t s --+++====---+++ 222121212211211221111212nx n m m x y m m mx m mx my -+⎛⎫-+ ⎪⎝⎭=-=-=--+-+ , 所以23y y =-,即PR x ⊥轴,因此MP MR =,故MPR ∆是等腰三角形.21.解:(1)因为直线l 与曲线()y f x =恒相切于同一定点, 所以曲线()y f x =必恒过定点,由()()ln 11f x mx x x '=+++,令()ln 10x x +=,得0x =, 故得曲线()y f x =恒过的定点为()0,1.因为()()ln 111x f x m x x ⎛⎫'=+++ ⎪+⎝⎭,所以切线l 的斜率()01k f '==, 故切线l 的方程为1y x =+,即10x y -+=.(2)令()()()[)ln 11,0,xxg x e f x e x mx x x =-=--+-∈+∞,()()[)1ln 1,0,1x xg x e m x mx x '=--+-∈+∞+. 令()()[)1ln 1,0,1xx h x e m x mx x =--+-∈+∞+, ()()[)()211,0,,01211xh x e m x h m x x ⎡⎤''=-+∈+∞=-⎢⎥++⎢⎥⎣⎦. ① 当0m ≤时,因为()0h x '>,所以()h x 在[)0,+∞上单调递增,故()()()00h x g x h '=≥=, 因为当[)0,x ∈+∞时,()0g x '≥,所以()g x 在[)0,+∞上单调递增,故()()00g x g ≥=. 从而,当0x ≥时,()xe f x ≥恒成立.② 当102m <≤时, 因为()h x '在[)0,+∞上单调递增,所以()()0120h x h m ''≥=-≥, 故与①同理,可得当0x ≥时,()xe f x ≥恒成立.③ 当12m >时,()h x '在[)0,+∞上单调递增, 所以当0x =时,()h x '在[)0,x ∈+∞内取得最小值()0120h m '=-<. 取410x m =->,因为()()()22111111111xh x e m x m x x x x ⎡⎤⎡⎤'=-+≥+-+⎢⎥⎢⎥++++⎢⎥⎢⎥⎣⎦⎣⎦, 所以()1111141440164284h m m m '-≥-->⨯-->, 前述说明在()0,41m -内,存在唯一的()00,41x m ∈-,使得()00h x '=,且当[]00,x x ∈时,()0h x '≤,即()h x 在[]00,x 上单调递减,所以当[]00,x x ∈时,()()()00h x g x h '=≤=, 所以()g x 在[]00,x 上单调递减,此时存在00x x =>,使得()()000g x g <=,不符合题设要求. 综上①②③所述,得m 的取值范围是1,2⎛⎤-∞ ⎥⎝⎦.说明:③也可以按以下方式解答: 当12m >时,()h x '在[)0,+∞上单调递增, 所以当0x =时,()h x '在[)0,x ∈+∞内取得最小值()0120h m '=-<,当x →+∞时,()211,011xe m x x ⎡⎤→+∞-+→⎢⎥++⎢⎥⎣⎦,所以()h x '→+∞, 故存在()00,x ∈+∞,使得()00h x '=,且当()00,x x ∈时,()0h x '<, 下同前述③的解答.22.解一:(1)由直线l 的参数方程3cos 1sin x t y t ϕϕ=+⎧⎨=+⎩(t 为参数),消去参数t 得,()()3sin 1cos 0x y ϕϕ---=,即直线l 的普通方程为()()sin cos cos 3sin 0x y ϕϕϕϕ-+-=, 由圆C 的极坐标方程为4cos ρθ=,得()24cos 0*ρρθ-=,将222cos x x y ρθρ=⎧⎨+=⎩代入(*)得, 2240x y x +-=, 即C 的直角坐标方程为()2224x y -+=.(2)将直线l 的参数方程代入()2224x y -+=得,()22cos sin 20t t ϕϕ++-=,()24cos sin 80ϕϕ∆=++>,设,P Q 两点对应的参数分别为12,t t , 则()12122cos sin ,2t t t t ϕϕ+=-+=-,所以12PQ t t =-==因为()()0,,20,2ϕπϕπ∈∈, 所以当3,sin 214πϕϕ==-时,PQ 取得最小值【注:未能指出取得最小值的条件,扣1分】 解法二:(1)同解法一(2)由直线l 的参数方程知,直线l 过定点()3,1M , 当直线l CM ⊥时,线段PQ 长度最小. 此时()223212CM=-+=,PQ ===所以PQ 的最小值为解法三: (1)同解法一(2)圆心()2,0到直线()()sin cos cos 3sin 0x y ϕϕϕϕ-+-=的距离,cos sin 4d πϕϕϕ⎛⎫=-=- ⎪⎝⎭,又因为()0,ϕπ∈, 所以当34ϕπ=时,d又PQ == 所以当34ϕπ=时,PQ 取得最小值23.解:(1)()33,11245,1233,2x x f x x x x x x x -+≤-⎧⎪=++-=-+-<<⎨⎪-≥⎩.①当1x ≤-时,由不等式339x -+<,解得2x >-. 此时原不等式的解集是:{|21x x -<≤-.②当12x -<<时,由不等式59x -+<,解得4x >-. 此时原不等式的解集是:{}|12x x -<<.③当2x ≥时,由不等式339x -<,解得4x <, 此时原不等式的解集是:{}|24x x ≤<. 综上可得原不等式的解集为()2,4-.(2)由(1)可得,函数()f x 的图像是如下图所示的折线图. 因为()()()min 16,23f f x f -===,故当36m <≤时,直线y m =与曲线()y f x =围成一个三角形, 即m 的范围是(]3,6. 【注:范围正确,不倒扣】 且当6m =时,()()max 1316362S =+-=.。
福建泉州新世纪中学2017届高三普通高中毕业班质量检查数学(理)试题(原卷版)
2017年普通高中毕业班质量检查数学(理)试卷一、选择题:本大题共12小题,每小题5分.在每小题给出的四个选项中,只有一项是符合题目要求的.1. 已知集合,若,则实数的取值范围是A. B. C. D.2. 已知复数满足,则复数的共轭复数为A. B. C. D.3. 已知随机变量服从正态分布,若,则A. B. C. D.4. 若双曲线的渐近线方程为,则的值为A. B. C. D. 或5. 执行如图所示的程序框图,运行相应的程序,若输入的值为 2,则输出的值为......A. 64B. 84C. 340D. 13646. 已知数列的前项和为,且,则A. B. C. D.7. 已知,则A. B. C. D.8. 在区域中,若满足的区域面积占面积的,则实数的值是A. B. C. D.9. 在四面体中,若,,,则直线与所成角的余弦值为A. B. C. D.10. 如图为中国传统智力玩具鲁班锁,起源于古代汉族建筑中首创的榫卯结构,这种三维的拼插器具内部的凹凸部分(即榫卯结构)啮合,外观看是严丝合缝的十字立方体,其上下、左右、前后完全对称,六根完全相同的正四棱柱分成三组,经榫卯起来.现有一鲁班锁的正四棱柱的底面正方形边长为,欲将其放入球形容器内(容器壁的厚度忽略不计),若球形容器表面积的最小值为30,则正四棱柱的高为A. B. C. D. 511. 已知,是椭圆的左、右焦点,点在椭圆上,线段与圆相切于点,且点为线段的中点,则(其中为椭圆的离心率)的最小值为A. B. C. D.12. 曲线是平面内与两个定点,的距离之积等于的点的轨迹.给出下列命题:①曲线过坐标原点;②曲线关于坐标轴对称;③若点在曲线上,则的周长有最小值;④若点在曲线上,则面积有最大值.其中正确命题的个数为A. B. C. D.二、填空题:本大题共4小题,每小题5分.13. 已知向量,满足,,且,则实数__________.14. 甲、乙、丙三位同学获得某项竞赛活动的前三名,但具体名次未知.3人作出如下预测:甲说:我不是第三名;乙说:我是第三名;丙说:我不是第一名.若甲、乙、丙3人的预测结果有且只有一个正确,由此判断获得第一名的是______.15. 已知函数,数列满足,则________.16. 在中,,,延长线段至点,使得,若,则______.三、解答题:解答应写出文字说明、证明过程或演算步骤.17. 已知等差数列前5项和为50,,数列的前项和为,,. (Ⅰ)求数列,的通项公式;(Ⅱ)若数列满足,,求的值.18. 如图,在四棱锥中,侧面底面,底面是平行四边形,,,,为的中点,点在线段上.(Ⅰ)求证:;(Ⅱ)试确定点的位置,使得直线与平面所成的角和直线与平面所成的角相等.19. 某市政府为了引导居民合理用水,决定全面实施阶梯水价,阶梯水价原则上以住宅(一套住宅为一户)的月用水量为基准定价:若用水量不超过12吨时,按4元/吨计算水费;若用水量超过12吨且不超过14吨时,超过12吨部分按6.60元/吨计算水费;若用水量超过14吨时,超过14吨部分按7.80元/吨计算水费.为了了解全市居民月用水量的分布情况,通过抽样,获得了100户居民的月用水量(单位:吨),将数据按照,,…,分成8组,制成了如图1所示的频率分布直方图.(图1)(图2)(Ⅰ)假设用抽到的100户居民月用水量作为样本估计全市的居民用水情况.(i)现从全市居民中依次随机抽取5户,求这5户居民恰好3户居民的月用水用量都超过12吨的概率;(ⅱ)试估计全市居民用水价格的期望(精确到0.01);(Ⅱ)如图2是该市居民李某2016年1~6月份的月用水费(元)与月份的散点图,其拟合的线性回归方程是. 若李某2016年1~7月份水费总支出为294.6元,试估计李某7月份的用水吨数.20. 已知椭圆的离心率为,短轴长为.(Ⅰ)求椭圆的标准方程;(Ⅱ)若圆的切线与曲线相交于、两点,线段的中点为,求的最大值.21. 已知函数,.(Ⅰ)当时,求证:过点有三条直线与曲线相切;(Ⅱ)当时,,求实数的取值范围.请考生在(22)、(23)两题中任选一题作答.注意:只能做所选定的题目.如果多做,则按所做第一个题目计分,做答时请用2B铅笔在答题卡上将所选题号后的方框涂黑.22. 选修4-4:坐标系与参数方程在平面直角坐标系中,以为极点,轴的正半轴为极轴建立极坐标系.若直线的极坐标方程为,曲线的极坐标方程为:,将曲线上所有点的横坐标缩短为原来的一半,纵坐标不变,然后再向右平移一个单位得到曲线.(Ⅰ)求曲线的直角坐标方程;(Ⅱ)已知直线与曲线交于两点,点,求的值.23. 选修4-5:不等式选讲已知函数,.(Ⅰ)若,求函数的最小值;(Ⅱ)若不等式的解集为,且,求的取值范围.。
2020年福建高三毕业班质量检查测试
2020年福建高三毕业班质量检查测试(共8页)--本页仅作为文档封面,使用时请直接删除即可----内页可以根据需求调整合适字体及大小--2020年福建高三毕业班质量检查测试本试卷满分150分,考试时间120分钟注意事项:1.答卷前,考生务必将自己的姓名、准考证号等信息填写在答题卡和试卷指定位置上。
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第二部分阅读理解(共两节,满分40分)第一节 (共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C和D四个选项中,选出最佳选项。
AImproving your lifestyle through sportsDo you want to lead an active lifestyle Are you passionate about sports Have youthought about making new friends Come on down and sign up for any of ourclasses at our exclusive launch !We offer classes like badminton, tennis, basketball and Array volleyball for everyone from the age of 10 to 40. Comeon down to learn more about our classes as well as ourspecial rates. All our classes are conducted by certifiedcoaches.Highlights of our launch event:● 1 p. m. : Talk on balancing sports and studies by Dr Claire Leow● 3 p. m. : Autograph signing session by professional badminton player Kate Wee , winner of therecent Singapore Open● 5 p. m. : Talk on how sports can benefit one's lifestyle by Mr Ryan Tan● 6 p. m. : Free tennis clinic for children conducted by Michael Ismail, a former professionaltennis playerTo register for the above events , please contact Michelle at 6234 6226 or emailsports@edufit. comLimited places available on a first-come-first-served basis.Take part in a sure-win lucky draw when you enter for any sports class on the dayof our launch! Prizes include VibraSquare Mall vouchers (票券). Wellness &Fitness sports clothing and many more !Official Sponsors :2VibraSquare Mall Wellness & Fitness Glizard Drinks1. What can we know about the Mystery GiftA. It is available anytime during April.B. It is given to the first fifty class applicants.C. You may choose vouchers or clothing.D. Each participant of the launch event can get one.2. When will kids attend the launch event if they are fond of tennis?A. At 1 p. m. .B. At 3 p. m. .C. At 5 p. m. .D. At 6 p. m. .is the main purpose of the text?report the sports events. B. To introduce healthy lifestyles.C. To advertize the sports classes.D. To give advice on making friends.BAt England's University of Plymouth, Professor Eduardo Miranda has been programming pairs of robots to compose music. Miranda's robots have simple "vocal cords" (声带) and are programmed to sing and to listen to each other. The robots' unique warbling sounds ( 颤音 ) do not perfectly match the human voice, but each machine is exactly sharing music with the other in a new and unique way.Each robot is equipped with speakers, software that mimics the human voice, a mouth that opens as it "sings," a microphone for ears, and a camera for eyes. The robots also move. Miranda hopes that by studying his robot vocalists, he can discover something about how and why humans create, perform, and listen to music.When the robots sing, first one robot makes six random sounds. Its partner responds with more sounds. The first robot analyzes the sounds to see if their sequences (序列)are similar. If they are, it nods its head and commits the sounds to memory, and the second robot notices and "memorizes" the musical sequence, too. If the first robot thinks the sounds are too different, it shakes its head and both robots ignore the sounds. Then the process continues.Miranda set up an experiment in which he left the two robots alone in his study for two weeks. When he returned, his little warblers had, by imitating each other, not only shared notes but combined them. The product of their cooperation was far from symphonic, but the robots had begun to combine the notes into their own self-developed " songs".With the help of his warbling robots, one of Miranda's goals is to create music that no human would ever compose. Miranda believes the robots are ideal for this purpose because they would not be influenced by any existing musical styles or rules.is closest in meaning to the underlined word " mimics " in Paragraph 23A. Substitutes.B. Interrupts.C. Controls.D. Copies.5. What did the two robots do during Miranda's experiment?A. They interacted with each other.B. They ignored the unique sound.C. They learned to sing better than humans.D. They committed random sounds to memory.6. What does Miranda want his robots to do?A. Sing as well as human s do.B. Create new styles of music.C. Memorize a variety of music.D. Promote traditional musical from7. What is the text mainly about?A. Future robots.B. Special songs.C. Music by robots.D. Experiments by Miranda.CGeorge Nakashima always insisted that he was a simple woodworker, not ancartist. Even though major museums exhibited hi s works and the director of the American Craft Museum called him a national treasure, Mr Nakashima rejected th e lab e l of artist. For almost fifty years he simp l y wen t on shaping wood into beautiful chairs, tables, and ca binets.Nakashima had a clear goal. He intended each piece of furniture he made to be as perfect as possible. Even making a box was an act of creation, because it produced an object that had never exis ted b efo re. Initi ally Nakashima us e d lo cal woo d, sometimes from his ow n prop ter, he traveled to seek out English oak, Persian walnut, African zebra w ood and Indian teak. He especially lik ed to find giant roots that had been dug out of the ground after a t r ee was ta k en down.Na k ash ima felt that m a king this wood into furniture was a way of allowing the tree to live again.4Mo s t furniture makers prefer perfe c t board s , but Nakashima took plea s ure in using wood with interesting knots (节疤)and cracks. These irr eg ulariti es gave the woo d personalit y and s h owed th a t the tree had lived a happy life.He n ever failed to c r eate a n object th a t was both u seful a nd beautiful. One early p i ece Nakashima designed was a three-legged c hair for his small daughter, Mira, to u se when s h e sat at the t ab le for meals. The Mira c hair be came so popul ar that Nakashima l ater made both low a nd high versions. Another famous pi ece, th e Conoid chair, has two l egs supported by bladelik e feet. A l ways, Nakashima 's designs were pr ecise an d gracefu l, marked b y a simpl i c it y that re vealed his love for the wood.As th e years passed, Nakashima 's reputation grew and hi s wor k received man y awards. H is children Mira a nd Kevin, now ad ult s, join ed the t eam of crafts-p eop l e in th eir fa th e r's s tudio. Nakash im a's dream of integr a tin g work a nd family h a d come true.8.Which of th e following best describes NakashimaA.Generous and outgoing.C. Capable and friendl y.9. Why wa s Naka s him a c alled a nationaltr e a s ureA.Hi s a rt wo rk made tr ees liv e agai n.B.He u sed pr ec iou s wood mat e ri als.C.Hi s c hair s were beautifully designed.D.H e wa s d ev oted t o makin g furniture. B. Honest and simple.D. Creative a nd modest.10.What ca n we l ear n abou t Nakashima from the l ast t wo paragraphs?11.12.A.H e loved his work and famil y.B.He made c h a ir s of the sa m e s tyle.C.He so u g ht for a s impl e li fe an d a rt.D.H e w as l o s t in r es earch in g th e wood.513.What c an be inf e rred abou t Mira and Ke v inA.The y had an art s tudio of their own.B.The y still lacked the ability to cr eate art works.C.The y had a co mmon interest w ith their father.D.The y enjo ye d the same reputation with their father.DA drug designed entirely by artificial intelligence is about to enter clinical human trials for the first time. The drug , which is intended to treat obsessive-compulsive disorder ( OCD) ( 强迫症),was discovered using AI systems from Oxford-based biotech company Exscientia. While it would usually take around four and a half years to get a drug to this stage of development, Exscientia says that by using the AI tools it's taken less than 12 months.The drug, known as DSP-1181 , was created by using algorithms ( 算法)to examine potential compounds ( 化合物), checking them against a huge database of parameters , including a patient's genetic factors. Speaking to the BBC, Exscientia chief executive Professor Andrew Hopkins described the trials as a " key milestone in drug discovery" and noted that there are " billions " of decisions needed to find the right molecules ( 分子) for a drug, making their eventual creation a " huge decision. " With Al, however , "the beauty of the algorithms is that they are unknowable , so can be applied to any disease. "We've already seen multiple examples of AI being used to diagnose illness and analyze patient data, so using it to engineer drug treatment is an obvious progression of its place in medicine. But the Al-created drugs do bring about some relevant questions. Will patients be comfortable taking medicine designed by a machine How will these drugs differ from those developed by humans alone Who will make the rules for the use of AI in drug research Hopkins and his team hope that these and a great many other questions will be explored in the trials, which will begin in March.12. What is special about the drug designed by Al?A. It's a better cure for OCD.B. It has no side effect on humans.C. Its development takes less time.D. It doesn't need clinical human trials.13. Which is a key factor in creating the drug according to Paragraph 2?A. Trials.B. Algorithms.C. Compounds.D. Molecules.14. How does Hopkins feel about the way of drug creation?A. Optimistic.B. Doubtful.C. Disappointed.D. Puzzled.15. What ca n be the best title for the text?A. Medical Trials by AIB. An Example in Medical TrialsC. A Creation in AI DevelopmentD. AI-designed Drugs to Be onTrial第二节(共5小题;每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。
2020年1月福建省泉州市普通高中2020届高三毕业班上学期期末质量监测数学(文)试题(解析版)
绝密★启用前福建省泉州市普通高中2020届高三毕业班上学期期末单科教学质量监测数学(文)试题(解析版)2020年1月注意事项:1.答卷前,考生务必将自己的姓名和准考证号填写在答题卡上.2.回答选择题时,选出每小题答案后,用2B 铅笔把答题卡对应题目的答案标号涂黑.如需改动,用橡皮擦干净后,再选涂其它答案标号.回答非选择题时,将答案写在答题卡上.写在本试卷上无效.3.考试结束后,将本试卷和答题卡一并收回.一、选择题:本题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知集合{}220A x x x =-≥,则A =R ( ) A. (,0]-∞ B. [2,)+∞C. (0,2)D. (,0][2,)-∞+∞【答案】C【解析】【分析】解一元二次不等式化简集合A ,再进行集合的补运算,即可得答案. 【详解】因为{}220A x x x =-≥,所以(,0][2,)A =-∞+∞,则(0,2)A =R . 故选:C .【点睛】本题考查一元二次不等式的求解、补集的运算,考查基本运算求解能力,属于基础题.2.若复数(2i)(1i)a -+为纯虚数,则实数a 的值是( )A. 2-B. 1-C. 1D. 2 【答案】A【解析】【分析】先对得数进行乘法运算,再利用纯虚数的概念,得到实数a 的值.【详解】复数(2i)(1i)a -+(2)(2)i a a =++-为纯虚数,所以20a +=,解得2a =-.故选:A.【点睛】本题考查复数的四则运算、纯虚数的概念,考查基本运算求解能力,属于基础题.3.若ππ2sin()cos()445αα--=-,则cos2=α( ) A. 45- B. 25- C. 25 D. 45【答案】A【解析】【分析】 先利用倍角公式得到π4sin(2)25α-=-,再利用诱导公式求得cos2α的值. 【详解】因为ππ1ππ2sin()cos()2sin()cos()442445αααα--=⋅⋅--=-, 所以π4sin(2)25α-=-,即4cos 25α=-. 故选:A .【点睛】本题考查倍角公式、诱导公式的应用,考查基本运算求解能力,求解时注意符号的正负.4.新中国成立70周年,社会各界以多种形式的庆祝活动祝福祖国,其中,“快闪”因其独特新颖的传播方式吸引大众眼球.根据腾讯指数大数据,关注“快闪”系列活动的网民群体年龄比例构成,及男女比例构成如图所示,则下面相关结论中不正确的是( )。
2020届泉州市普通高中毕业班质量检查语文参考答案-定稿
★★★本材料仅限我市内部交流,不得上传网络。
泉州市2020届普通高中毕业班质量检查语文参考答案一、现代文阅读(36分)(一)论述类文本阅读(本题共3小题,9分)1.(3分)B2.(3分)C3.(3分)C(二)实用类文本阅读(本题共3小题,12分)4.(3分)C5.(3分)D6.(6分)①利用举办国际活动的契机,加大信息化基础设施建设力度,为提升社会服务与管理水平奠定物质基础。
②依托城市数据大脑,构建各种智能平台,提高社会服务与管理的效率、安全性和智能化水平。
③利用北斗—羲和系统,推进室内外定位、物联网新发展,提高社会服务与管理的精度。
答对一点给2分。
意思答对即可。
其他答案,只要言之成理,可酌情给分。
(三)文学类文本阅读(本题共3小题,15分)7.(3分)C8.(6分)(第一问)①多为短句,简洁明快;②常化用熟语,通俗生动;③语带情感,富有生活气息。
(第二问)①为人坦诚,个性爽直;②阅历丰富,善于交谈;③坚强乐观,热爱生活。
每一问3分,答对一点给1分。
意思答对即可。
其他答案,只要言之成理,可酌情给分。
9.(6分)①可以节省不必要的叙述交代,使文章内容更集中,行文更紧凑;②可以更大限度地坦露养蜂人的心灵,使贺福平的个性得以更真实的表现,赞美养蜂人勤劳进取的主题更明晰;③富有生活气息和现场亲切感,使读者如临其境、如闻其声,从而增强文章的感染力。
答对一点给2分。
意思答对即可。
其他答案,可根据观点明确、分析合理的程度,酌情给分。
二、古代诗文阅读(34分)(一)文言文阅读(本题共4小题,19分)10.(3分)C11.(3分)D12.(3分)B13.(10分)(1)(5分)那些曾被弹劾的大官,全部命令他们自己陈述后选择剔除他们,用以警戒在位官吏。
译对大意给3分;“简去”“儆”两处,译对一处给1分。
(2)(5分)见朝廷政事大变,敢于进言的人都被贬斥,而刘瑾更加肆虐,许天锡特别愤怒。
译对大意给3分;“加甚”“大”两处,译对一处给1分。
2020届福建省泉州市高三普通高中毕业班单科质量检查数学试题(文)(解析版)
福建省泉州市2020届高三普通高中毕业班单科质量检查数学试题(文)一、选择题:本题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知集合{}220A x x x =-≥,则A =R( )A. (,0]-∞B. [2,)+∞C. (0,2)D. (,0][2,)-∞+∞『答案』C『解析』因为{}220A x x x =-≥,所以(,0][2,)A =-∞+∞,则(0,2)A =R.故选:C .2.若复数(2i)(1i)a -+为纯虚数,则实数a 的值是( ) A. 2-B. 1-C. 1D. 2『答案』A『解析』复数(2i)(1i)a -+(2)(2)i a a =++-为纯虚数,所以20a +=,解得2a =-. 故选:A.3.若ππ2sin()cos()445αα--=-,则cos2=α( ) A. 45-B. 25-C. 25D.45『答案』A『解析』因ππ1ππ2sin()cos()2sin()cos()442445αααα--=⋅⋅--=-,所以π4sin(2)25α-=-,即4cos 25α=-.故选:A .4.新中国成立70周年,社会各界以多种形式的庆祝活动祝福祖国,其中,“快闪”因其独特新颖的传播方式吸引大众眼球.根据腾讯指数大数据,关注“快闪”系列活动的网民群体年龄比例构成,及男女比例构成如图所示,则下面相关结论中不正确的是( )A. 35岁以下网民群体超过70%B. 男性网民人数多于女性网民人数C. 该网民群体年龄的中位数在15~25之间D. 25~35岁网民中的女性人数一定比35~45岁网民中的男性人数多『答案』D『解析』对A ,依题意可得,35岁以下网民群体所占比例为74%,故A 正确;对B ,由男女比例构成图可得男性所占比例55%,故B 正确;对C ,因为15岁以下所占比例为23%,35岁以下所占比例为54%,故该网民群体年龄的中位数在15~25之间,故C 正确; 对D ,『答案』无法判断,故D 错误. 故选:D.5.设E 是中心在坐标原点的双曲线.若(2,0)A 是E 的一个顶点,(4,0)F -是E 的一个焦点,则E 的一条渐近线方程为( )A. 13y x =B. y x =C. y =D. 3y x =『答案』C『解析』由已知双曲线的焦点在x 轴上,2a =,4c =,所以b ==,所以双曲线E 的渐近线方程为by x a=±,即y =. 故选:C .6.已知等差数列{}n a 中,36+8a a =,则475+a a =( ) A. 32B. 27C. 24D. 16『答案』C『解析』设等差数列{}n a 公差d ,则361278a a a d +=+=,所以471156213(27)24a a a d a d +=+=+=. 故选:C .7.“堑堵”是中国古代数学名著《九章算术》中记载着的一种多面体.如图,网格纸上小正方形的边长为1,粗实线画出的是某“堑堵”的三视图,则该“堑堵”的体积等于( )A. 12B. 8C. 6D. 4『答案』C『解析』由已知可得该“堑堵”是直三棱柱,所以体积123262V =⨯⨯⨯=, 故选:C .8.函数()sin f x x x =的图象大致为( )A. B.C. D.『答案』B『解析』因为()sin f x x x =的定义域为R 关于原点对称,且()()[sin()]()f x x x f x -=--=,所以()f x 为偶函数,故排除C ,D ;因为'()sin cos 0f x x x x =+≥在π[0,]2x ∈恒成立,所以()sin f x x x =在π[0,]2上单调递增. 故选:B .9.明代朱载堉创造了音乐学上极为重要的“等程律”.在创造律制的过程中,他不仅给出了求解三项等比数列的等比中项的方法,还给出了求解四项等比数列的中间两项的方法.比如,若已知黄钟、大吕、太簇、夹钟四个音律值成等比数列,则有大吕大吕太簇据此,可得正项等比数列{}n a 中,k a =( )A. n k -B. n -C.D.『答案』C『解析』因为三项等比数列的中项可由首项和末项表示,四项等比数列的第2、第3项均可由首项和末项表示, 所以正项等比数列{}n a 中的k a 可由首项1a 和末项n a 表示,因为11n n a a q -=,所以=q所以11=k k a a -⎛ ⎝1111=k n n a a a --⎛⎫ ⎪⎝⎭1111=n k k n n na a ----⋅=故选:C.10.若直线0x ay -=与函数e ()xf x x=的图象有且只有一个公共点,则a 的取值范围为( )A. 24(,0)(,)e-∞+∞B. 24(,)e +∞C. 2e (,)4+∞D. 21e (,)e 4『答案』B『解析』由0ex x ay y x -=⎧⎪⎨=⎪⎩,消去y ,得2(0)e x x a x =≠,则依题意,可知直线y a =与曲线2(0)ex x y x =≠恰有一个公共点,令2()(0)ex x f x x =≠,则22'22e e 2()(e )e x x x xx x x x f x ⋅-⋅-==, 当'()002f x x >⇒<<;当'()00f x x <⇒<或2x >; 所以()f x 在(,0),(2,)-∞+∞单调递减,在(0,2)单调递增,如图所示,作出y a =与2(0)ex x y x =≠的图象,因为两个函数图象有且只有一个公共点, 所以24ea >.故选:B11.若椭圆E 的顶点和焦点中,存在不共线的三点恰为菱形的中心和顶点,则E 的离心率等于( )A.2B.C. 12或2D.2『答案』D『解析』依题意,由菱形对角线互相垂直可转化为,在椭圆的顶点和焦点中找到不共线的三点能构成一个直角三角形,结合椭圆的对称性,只须考虑三种情况:(1)如图1,若以顶点D 焦点B 为菱形顶点,C 为中心,则DC BC ⊥,由勾股定理得,2222()()a b a a c ++=+,由222b a c =-化简得220c ac a +-=,两边同除以2a ,得210e e +-=,又因为01e <<,可得e =. (2)如图2,若以焦点A ,B 为菱形顶点,C 为中心,则AC BC ⊥,故45OCB ∠=,易得c e a ==;(3)如图3,若以焦点B 为菱形的中心,C ,E 为顶点,则CB EB ⊥,易得2c e a ==,故选D.12.已知函数π()sin()(0,02f x x ωϕωϕ=+><<).若π()8f x -为奇函数,π()8f x +为偶函数,且()2f x =在π(0,) 6至多有2个实根,则ω的最大值为( )A. 10B. 14C. 15D. 18『答案』A『解析』由题意,得π(0)8-,为()f x 的图象的对称中心,直线π8x =为()f x 的图象的一条对称轴,所以1122ππ8,ππ+π82k k k k ωϕωϕ⎧-+=⎪⎪∈⎨⎪+=⎪⎩Ζ(),两式相加得12ππ42k k ϕ+=+, 又因为π02ϕ<<,所以π4ϕ=,代入2ππ+π82k ωϕ+=,得82()k k ω=+∈Ζ, 因为π(0,) 6x ∈时,ππππ(,) 4464t x ωω=+∈+,即由已知可得sin 2t =,πππ(,) 464t ω∈+至多有2个实根,即ππ11π644ω+≤,由此可得015ω<≤, 又因为82()k k ω=+∈Ζ,所以1k =时ω的最大值为10, 故选:A .二、填空题:本题共4小题,每小题5分,共20分.13.已知向量,a b ,且()=3,2a -,()=52a b +,,则b =____________.『答案』『解析』因为()(2,4)b a b a =+-=,所以22242b =+=故『答案』为:14.若,x y 满足约束条件0,10,220,x x y x y ≤⎧⎪--≤⎨⎪-+≥⎩则32z x y =+的最大值为___________.『答案』4『解析』不等式组表示的可行域如图所示:由32z x y =+得322zy x =-+在y 轴上的截距越大,z 越大, 所以当直线322zy x =-+过点()0,2A 时,z 取得最大值,所以z 的最大值为4. 故『答案』为:415.已知直线:(1)10()l mx m y m R +--=∈与圆22:8O x y +=交于,A B 两点,,C D 分别为,OA AB 的中点,则AB CD ⋅的最小值为____________.『答案』『解析』直线l 的方程可化为()10m x y y -+-=,由0,10,x y y -=⎧⎨-=⎩解得1,1,x y =⎧⎨=⎩,直线l 恒过点(1,1)P ,因为,C D 分别为,OA AB的中点,所以12CD OB =. 当OP AB ⊥时,AB最小,此时AB =所以AB CD ⋅ 故『答案』为:16.已知正方体1111ABCD A B C D -的棱长为1,动点P 在棱1AA 上,四棱锥11P BDD B -的顶点都在球O 的球面上,则球O 的表面积取值范围是_____________.『答案』25π[3π,]8『解析』如图,设球O 的球心为G ,1AA 的中点1O ,1CC 的中点2O ,12O O 的中点O ,且1OO =,2OA OB ==因11,,,B D D B 在球面上,所以球心在线段2OO 上,点P 也在球面上,GP GB R ==.设11,O P x O G y ==.则2OG y =-. 在1Rt O PG △中,222R x y =+…………①在Rt BOG △中,22222R y ⎛⎛=+- ⎝⎭⎝⎭…………②,联立①②,得254x =-,因为102x ≤≤y .所以2222253325([,]44432R x y y y=+=+=+∈,所以球O的表面积取值范围为25[3π,π]8.故『答案』为:25π[3π,]8三、解答题:共70分,解答应写出文字说明、证明过程或演算步骤.第17~21题为必考题,每个试题考生都必须作答,第22、23题为选考题,考生根据要求作答.(一)必考题:共60分.17.记数列{}n a的前n项和为n S.若233n nS a=-.(1)证明:{}n a为等比数列;(2)设9logn nb a=,求数列11n nb b+⎧⎫⎨⎬⎩⎭前n项和n T.解:(1)由已知,得233n nS a=-,……①当2n≥时,11233n nS a--=-,……②①—②,得-1-122(33)(33)n n n nS S a a-=---,即-1233n n na a a=-,整理,得()132nnana-=≥,又由11233S a=-,得1=30a≠,所以{}n a是以3为首项,3为公比的等比数列.(2)由(1)得=3nna,所以9log32nnnb==,所以()11411=4()11n nb b n n n n+=-++,故nT=111111144()+4()++4()4(1)1223111nn n n n---=-=+++.18.ABC的内角,,A B C所对的边分别为,,a b c.已知cos)cosa C c A=.(1)求ba;(2)求cos A的最小值.解:(1)在ABC中,由正弦定理,得sin sin sina b cA B C==,的从而由cos )cos a C c A =,可得sin cos )sin cos A C C A =,sin cos cos sin A A C A C =+sin()A A C =+,又因为πA B C ++=,所以sin B A =,所以b a=(2)由(1)不妨设b ==11c <<, 在ABC 中,由余弦定理,得222cos 2b c a A bc+-=,所以22222cos ()3A c c ===+,当2c c =即c =cos A 取到最小值为3. 19.如图,MA ⊥平面ABCD ,CN ⊥平面ABCD ,四边形ABCD 是边长为2的菱形,60BAD ∠=,1CN =,3AM =.(1)证明://BN 平面ADM ;(2)求三棱锥N ADM -的体积.(1)证明:因为MA ⊥平面ABCD ,CN ⊥平面ABCD ,所以//MA NC ,又MA ⊂平面ADM ,NC ⊄平面ADM ,所以//NC 平面ADM .因为四边形ABCD 是菱形,所以//BC AD ,又AD ⊂平面ADM ,BC ⊄平面ADM ,所以//BC 平面ADM ,又BC NC C =,BC ⊂平面BCN ,CN ⊂平面BCN ,所以平面//BCN 平面ADM ,又BN ⊂平面BCN ,所以//BN 平面ADM .(2)解:由(1)知,//BN 平面ADM ,所以点N 到平面ADM 的距离等于点B 到平面ADM 的距离.取AD 的中点E ,连接BE ,BD .因为四边形ABCD 是边长为2的菱形,60BAD ∠=,所以ABD △是边长为2的等边三角形,所以BE AD ⊥,且BE =又因为MA ⊥平面ABCD ,BE ⊂平面ABCD ,所以MA BE ⊥,又MA AD A =,MA ⊂平面ADM ,AD 平面ADM ,所以BE ⊥平面ADM ,故所以点N 到平面ADM 的距离为BE .所以三棱锥N ADM -的体积11123332N ADM ADM V S BE -∆=⨯⨯=⨯⨯⨯= 20.已知抛物线E 的顶点在原点,焦点在y 轴上,过点(1,0)A 且斜率为2的直线与E 相切. (1)求E 的标准方程;(2)过A 的直线l 与E 交于,P Q 两点,与y 轴交于点R ,证明:2AR AP AQ =⋅. 解:(1)过点(1,0)A 且斜率为2的直线方程为2(1)y x =-,即22y x =-,设E 的方程为2(0)x ay a =≠, 由222,,y x x ay =-⎧⎨=⎩消去y ,得2220x ax a -+=, 因为直线与E 相切,所以2480a a ∆=-=,解得0a =(舍去)或2a =,所以E 的标准方程为22x y =.(2)设l 的方程为(1)y k x =-,1122(,),(,)P x y Q x y .令0x =,得y k =-,即(0,)R k -,由2(1),2,y k x x y =-⎧⎨=⎩消去y ,得2220x kx k -+=, 因为l 与E 相交,所以2480k k ∆=->,解得2k >或k 0<,设,则12122,2x x k x x k +==,11AP =-,21AQ =-, 从而222211212(1)11(1)()11AP AQ k x x k x x x x k ⋅=+--=+-++=+, 又221AR k =+, 所以2AR AP AQ =⋅.21.已知函数2()(1)ln 2a f x x a x x =+--. (1)讨论()f x 的单调性;(2)当1x ≥时,e ()2f x -≥,求a 的取值范围. 解:(1)函数()f x 的定义域为(0,)+∞,21(1)1(1)(1)()1ax a x ax x f x ax a x x x+---+'=+--==, 当0a ≤时,10ax ,则()0f x '<,故()f x 在(0,)+∞单调递减;当0a >时,令()0f x '>,得1x a >;令()0f x '<,得10x a<<, 故()f x 在1(0,)a 上单调递减,在1()a +∞,单调递增.综上,可得当0a ≤时,()f x 在(0,)+∞单调递减;当0a >时,()f x 在1(0,)a 单调递减,在1()a +∞,单调递增. (2)①当0a ≤时,因为242e (e )e (1)e 2222a f a =+--<-<-,所以0a ≤不符合题意; ②当0a >时,由(1),知()f x 在1(0,)a 单调递减,在1()a+∞,单调递增. (ⅰ)当11a ≤即1a ≥时,1()(1)()0a x x a f x x-+'=≥,所以()f x 在[1,)+∞单调递增, 故31e ()(1)1222f x f a =->-≥≥,故1a ≥满足题意. (ⅱ)当11a >即01a <<时,()f x 在1[1,)a 单调递减,在1(,)a+∞单调递增, 故min 111()()1ln 2f x f a a a ==--, 当1x ≥时,e ()2f x -≥,当且仅当1e ()2f a -≥, 令1()1ln (0)2g t t t t =-->,则11()02g t t'=--<,故()g t 在(0,)+∞单调递减, 又e (e)2g =-,从而由1e ()2f a -≥即1()(e)g g a ≥,可得1e a ≤,解得11ea <≤, 综上,可得a 的取值范围为1[,)e+∞. (二)选考题:共10分.请考生在第22,23两题中任选一题作答.如果多做,则按所做的第一题计分.22.在同一平面直角坐标系xOy 中,经过伸缩变换2,x x y y''=⎧⎨=⎩后,曲线221:1C x y +=变为曲线2C .(1)求2C 的参数方程;(2)设()2,1A ,点P 是2C 上的动点,求OAP △面积的最大值,及此时P 的坐标.解:(1)由伸缩变换2,x x y y ''=⎧⎨=⎩得到1,2.x x y y ⎧=⎪⎨⎪='⎩'……①将①代入221x y +=,得到221+=12x y ''(),整理得222:+=14x C y ''. 所以2C 的参数方程为2cos ,sin x y αα=⎧⎨=⎩(α为参数).(2)设()()2cos ,sin 02πP ααα<≤,直线:20OA x y -=,则P 到直线OA的距离为d ==,所以111222OAP S OA d d =⋅==△≤ 当3π=4α或7π=4α时,OAP △, 此时P的坐标为)2-或(2. 23.已知函数1()||||f x x a x a =++-. (1)证明:()2f x ≥;(2)当12a =时,()f x xb +≥,求b 的取值范围. (1)证明:111()||||||||||2f x x a x a a a a a =++-+=+=≥≥; (2)解:312,,22151()2=,2,22232,2,2x x f x x x x x x ⎧-+≤-⎪⎪⎪=++--<<⎨⎪⎪-≥⎪⎩ 作出()f x 的图象,如图由图,可知()f x x b +≥,当且仅当(2)2f b +≥,解得12b =, 故b 的取值范围为1(,]2-∞.。
福建省泉州市2017届高三第二次质量检查数学理试题含答案-精选
2017年泉州市普通高中毕业第二次质量检查理 科 数 学第 Ⅰ 卷一、选择题:本大题共12小题,每小题5分,在每个小题给出的四个选项中,只有一项是符合题目要求的.(1)已知集合{}21xA x =>,{}2560B x x x =-+<,则A B =ð(A )(2,3) (B )(,2][3,)-∞+∞ (C )(0,2][3,)+∞ (D )[3,)+∞(2)已知复数i()z a a =+∈R .若z <2i z +在复平面内对应的点位于 (A )第一象限 (B )第二象限 (C )第三象限 (D )第四象限(3)公差为2的等差数列{}n a 的前n 项和为n S .若312S =,则3a =(A )4 (B )6 (C )8 (D )14(4)已知实数,x y 满足约束条件,220,y x x y ≤⎧⎨--≤⎩z x y =+,则满足1z ≥的点(,)x y 所构成的区域面积等于(A )14 (B )12 (C )34(D )1(5)榫卯(s ǔn m ǎo )是古代中国建筑、家具及其它器械的主要结构方式,是在两个构件上采用凹凸部位相结合的一种连接方式,凸出部分叫做“榫头”.某“榫头”的三视图及其部分尺寸如图所示,则该“榫头”体积等于(A )12 (B )13 (C )14 (D )15(6)执行一次如图所示的程序框图,若输出i 的值为0,则下列关于框图中函数()()f x x ∈R 的表述,正确的是 (A )()f x 是奇函数,且为减函数(B )()f x 是偶函数,且为增函数(C )()f x 不是奇函数,也不为减函数(D )()f x 不是偶函数,也不为增函数(7)已知以O 为中心的双曲线C 的一个焦点为F ,P 为C 上一点,M 为PF 的中点.若OMF ∆为等腰直角三角形,则C 的离心率等于(A 1 (B )1 (C )2 (D (8)已知曲线π:sin(2)()2C y x ϕϕ=+<的一条对称轴方程为π6x =,曲线C 向左平移θ(0θ>)个单位长度,得到的曲线E 的一个对称中心为π(,0)6,则ϕθ-的最小值是(A )π12 (B )π4(C )π3 (D)5π12(9)在梯形ABCD 中,ABCD,1AB =,2AC =,BD =60ACD ∠=,则AD =(A )2 (B (C (D )13-(10)某密码锁共设四个数位,每个数位的数字都可以是1,2,3,4中的任一个.现密码破译者得知:甲所设的四个数字有且仅有三个相同;乙所设的四个数字有两个相同,另两个也相同;丙所设的四个数字有且仅有两个相同;丁所设的四个数字互不相同.则上述四人所设密码最安全的是(A )甲 (B )乙 (C )丙 (D )丁(11)已知直线,PA PB 分别与半径为1的圆O 相切于点,A B ,2PO =,2(1)PM PA PB λλ=+-.若点M 在圆O 的内部(不包括边界),则实数λ的取值范围是(A )(1,1)- (B )2(0,)3 (C )1(,1)3(D )(0,1) (12)已知函数()e xf x =,2()g x ax ax =-.若曲线()y f x =上存在两点关于直线y x =的对称点在曲线()y g x =上,则实数a 的取值范围是(A )(0,1) (B )(1,)+∞ (C )(0,)+∞ (D )(0,1)(1,)+∞第 Ⅱ 卷本卷包括必考题和选考题两个部分.第(13)题~第(21)题为必考题,每个试题考生都必须做答.第(22)、(23)题为选考题,考生根据要求作答. 二、填空题:本大题共4小题,每小题5分.(13)已知椭圆22:143x y C +=的左顶点、上顶点、右焦点分别为,,A B F ,则A B A F ⋅=_________.(14)已知曲线2:2C y x x =+在点(0,0)处的切线为l ,则由,C l 以及直线1x =围成的区域面积等于__________.(15)在平面直角坐标系xOy 中,角θ的终边经过点(,1)(1)P x x ≥,则cos sin θθ+的取值范围是_____.(16)已知在体积为12π的圆柱中,,AB CD 分别是上、下底面两条不平行的直径,则三棱锥A BCD -的体积最大值等于_________.三、解答题:解答应写出文字说明,证明过程或演算步骤. (17)(本小题满分12分)在数列{}n a 中,14a =,21(1)22n n na n a n n +-+=+.(Ⅰ)求证:数列n a n ⎧⎫⎨⎬⎩⎭是等差数列; (Ⅱ)求数列1n a ⎧⎫⎨⎬⎩⎭的前n 项和n S .(18)(本小题满分12分)某测试团队为了研究“饮酒”对“驾车安全”的影响,随机选取100名驾驶员先后在无酒状态、酒后状态下进行“停车距离”测试. 测试的方案:电脑模拟驾驶,以某速度匀速行驶,记录下驾驶员的“停车距离”(驾驶员从看到意外情况到车子完全停下所需要的距离).无酒状态与酒后状态下的试验数据分别列于表1和表2.表1(Ⅰ)求,a b 的值,并估计驾驶员无酒状态下停车距离的平均数;(Ⅱ)根据最小二乘法,由表2的数据计算y 关于x 的回归方程ˆˆˆy bx a =+;(Ⅲ)该测试团队认为:驾驶员酒后驾车的平均“停车距离”y 大于(Ⅰ)中无酒状态下的停车距离平均数的3倍,则认定驾驶员是“醉驾”.请根据(Ⅱ)中的回归方程,预测当每毫升血液酒精含量大于多少毫克时为“醉驾”?(附:对于一组数据1122(,),(,),,(,)n n x y x y x y ,其回归直线ˆˆˆy bx a =+的斜率和截距的最小二乘估计分别为1221ˆni ii nii x y nxybxnx ==-=-∑∑,ˆˆa y bx=-.)(19) (本小题满分12分)如图,在三棱锥A BCD -中,平面ABD ⊥平面BCD ,AB AD =,60CBD ∠=,24BD BC ==,点E 在CD 上,2DE EC =.(Ⅰ)求证AC BE ⊥;(Ⅱ)若二面角E BA D --的余弦值为5,求三棱锥A BCD -的体积.(20) (本小题满分12分)在平面直角坐标系xOy 中,抛物线2:2(0)C x py p =>的焦点为F ,过F 的直线l 交C 于,A B 两点,交x 轴于点D ,B 到x 轴的距离比BF 小1.(Ⅰ)求C 的方程;(Ⅱ)若BOF AOD S S ∆∆=,求l 的方程.(21) (本小题满分12分)已知函数()ln f x x kx k =-+.(Ⅰ)若()0f x ≥有唯一解,求实数k 的值;(Ⅱ)证明:当1a ≤时,2(())e 1xx f x kx k ax +-<--.(附:ln 20.69≈,ln3 1.10≈,32e 4.48≈,2e 7.39≈)请考生在第(22)、(23)两题中任选一题作答.注意:只能做所选定的题目.如果多做,则按所做的第一个题目计分,作答时请用2B 铅笔在答题卡上将所选题号后的方框涂黑. (22)(本小题满分10分)选修4—4:坐标系与参数方程在平面直角坐标系xOy 中,曲线1C 的参数方程为1cos ,sin x y αα=+⎧⎨=⎩(α为参数);在以O为极点,x 轴正半轴为极轴的极坐标系中,曲线2C 的极坐标方程为2cos sin ρθθ=.(Ⅰ)求1C 的普通方程和2C 的直角坐标方程;(Ⅱ)若射线l :y kx =(0)x ≥分别交1C ,2C 于,A B 两点(,A B 异于原点).当k ∈时,求OA OB ⋅的取值范围.(23)(本小题满分10分)选修4—5:不等式选讲已知函数()f x x a x a =-++. (Ⅰ)当2a =时,解不等式()6f x >;(Ⅱ)若关于x 的不等式()21f x a <-有解,求实数a 的取值范围.2017年泉州市普通高中毕业班质量检查理科数学试题答案及评分参考评分说明:1.本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的主要考查内容比照评分标准制定相应的评分细则.2.对计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应给分数的一半;如果后继部分的解答有较严重的错误,就不再给分. 3.解答右端所注分数,表示考生正确做到这一步应得的累加分数. 4.只给整数分数.选择题和填空题不给中间分.一、选择题:本大题考查基础知识和基本运算.每小题5分,满分60分.(1)C (2)B (3)B (4)C (5)C (6)D (7)B(8)A(9)B(10)C(11)B(12)D(11)解法一:以圆心O 为原点,OP 的方向为x 轴的正方向建立平面直角坐标系,则有()2,0P ,1(,)22A ,1(,22B -.设()00,M x y ,可解得()01132x λ=-,)0312y λ=-,因为()00,M x y 在圆内,所以()()22131331144λλ-+-<,整理,得311λ-<,解得2(0,)3λ∈,故答案选(B ).解法二:如图,在线段PA 的延长线上取点Q ,使得PA AQ =.连结OQ ,交圆O 于C .可求得60BOP AOP AOQ ∠=∠=∠=,故,,B O Q 三点共线.因为2PA PQ =,所以2(1)(1)PM PA PB PQ PB λλλλ=+-=+-,故BM B Q λ=.又因为点M 在圆O 的内部(不包括边界),所以2(0,)3λ∈,答案选(B ).(12)解法一:可以看出,(1,0)是曲线(1)y ax x =-与曲线ln y x =的一个公共点,且当1a =时,两曲线在点(1,0)处的切线方程均为1y x =-.由导数的概念,可知当01a <<或1a >时,曲线(1)y ax x =-与直线1y x =-交于两点,必与曲线ln y x =交于两点,故答案为(D ).解法二:方程2ln ax ax x -=显然有一个根1x =.若满足在去心邻域(1,1)δδ-+存在非1的根则符合题意.又因为对于区间(1,1)δδ-+(其中δ为任意充分小正数),1ln x x -(表示等价无穷小 ),故去心邻域(1,1)δδ-+中,方程等价为1ax =,所以a 取遍去心邻域11(,)11δδ+-,所以排除选项(A )(B )(C ),答案为(D ).解法三:2ln ax ax x -=有两个不同根,由于两者都是连续函数,令特殊值1a =,不合题意;令特殊值2a =,符合题意;令特殊值12a =,符合题意.故选项(D ). 解法四:依题意,可知()ln 1x a x x =-有两个不同实根.设()ln x F x x =,则()21l n 'xF x x-=. 当(0,1)x ∈时,()F x 单调递增;当(1,)x ∈+∞时,()F x 单调递减;当1a =时,()()1F x a x ≤-恒成立,当且仅当1x =取到等号,即只有一个根,与题意不合.当1a <时,显然符合题意.当1a >时,可以发现0x +→时,()()1F x a x <-;(或者()()111F a a a --<-)21x a =当时,()211F x a a ⎛⎫>- ⎪⎝⎭(证明后补).根据零点存在性定理可得在(0,1)必有一根.故两图象有两个公共点.故a 的取值范围是(0,1)(1,)+∞.补证:21x a =时,()()1F x a x >-,即证2221ln 1a a a a ⎛⎫>- ⎪⎝⎭,即证221ln a a a a >-, 这是显然的22ln 0a a >,而10a a-<.得证解法五:方程2ln ax ax x -=显然有一个实根1x =,故当1x ≠时方程()ln 1xa x x =-还有另一个实根,当0x +→时,()ln 1x x x →+∞-;当x →+∞时,()ln 01xx x +→-;且()()()()()2111111ln 'ln 'ln 1lim lim lim lim lim 112121'1'x x x x x x x xx x x x x x x x x x -----+→→→→→=====-----⎡⎤⎡⎤⎣⎦⎣⎦, ()()()()()2111111ln 'ln 'ln 1lim lim lim lim lim 112121'1'x x x x x x x x x x x x x x x x x x +++++-→→→→→=====-----⎡⎤⎡⎤⎣⎦⎣⎦; 显然,0a >,且1a ≠都是符合题意.二、填空题:本大题考查基础知识和基本运算.每小题5分,满分20分.(13)6 (14)13(15) (16)8解析:(15)解法一:依题意,可知π(0,]4θ∈,所以ππ(,]442πθ+∈,故πsin(),1]42θ+∈,所以πcos sin )4θθθ+=+∈,故答案为.解法二:由三角函数定义,得cos θ=,sin θ=所以cos sin θθ+===== 因为1y x x=+在[1,)+∞单调递增,所以[2,)y ∈+∞, 所以2(0,1]1x x∈+,从而cos sin θθ+∈,故答案为.(16)解:设上、下底面圆的圆心分别为1,O O ,圆的半径为r ,由已知21π12πV r OO =⋅=圆柱,所以2112r OO ⋅=,则A BCD C OAB D OAB V V V ---=+,因为O 是CD 中点,所以C 到平面OAB 的距离与D 到平面OAB 的距离相等,故C OABD OAB V V --=,从而2A BCD C OAB V V --=.设三棱锥C OAB -的高为h ,则h r ≤,所以11221223323A BCD D OAB OAB V V S h AB OO h r OO h --∆===⋅⋅=⋅212212833r OO ≤⋅=⨯=, 故三棱锥A BCD -的体积最大值等于8.三、解答题:本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤. (17)(本小题满分12分)解法一:(Ⅰ)21(1)22n n na n a n n +-+=+的两边同时除以(1)n n +,得*12()1n na a n n n+-=∈+N , ····························· 3分 所以数列n a n ⎧⎫⎨⎬⎩⎭是首项为4,公差为2的等差数列. ················· 6分 (Ⅱ)由(Ⅰ),得22na n n=+, ························· 7分 所以222n a n n =+,故2111(1)111()222(1)21n n n a n n n n n n +-==⋅=⋅-+++, ········ 8分 所以111111[(1)()()]22231n S n n =-+-++-+, 1111111[(1)()]223231n n =++++-++++, 11(1)212(1)n n n =-=++. ····························· 12分 解法二:依题意,可得1(1)22nn n a a n n++=++, ··················· 1分 所以1(1)222211nn n n n n n a n a a a a a n n n n n n n++++-=-=+-=++, 即*12()1n na a n n n+-=∈+N , ······························ 3分所以数列n a n ⎧⎫⎨⎬⎩⎭是首项为4,公差为2的等差数列. ··················· 6分 (Ⅱ)同解法一. ······························· 12分(18)(本小题满分12分)本小题主要考查频率分布直方图、数学期望等基础知识;考查抽象概括能力、数据处理能力、运算求解能力、应用意识;考查统计与概率思想、分类与整合思想. 解:(Ⅰ)依题意,得6502610a =-,解得40a =, ··················· 1分 又36100ab ++=,解得24b =; ·························· 2分 故停车距离的平均数为26402482152535455527100100100100100⨯+⨯+⨯+⨯+⨯=. ······ 4分 (Ⅱ)依题意,可知50,60x y ==, ························· 5分2222221030305050607070909055060ˆ1030507090550b⨯+⨯+⨯+⨯+⨯-⨯⨯=++++-⨯, ·············· 6分 710=, ····································· 7分 7ˆ60502510a=-⨯=, 所以回归直线为ˆ0.725yx =+. ···························· 8分 (Ⅲ)由(I )知当81y >时认定驾驶员是“醉驾”. ·················· 9分令ˆ81y>,得0.72581x +>,解得80x >, ······················ 11分 当每毫升血液酒精含量大于80毫克时认定为“醉驾”. ·················· 12分(19) (本小题满分12分)解法一(Ⅰ)取BD 的中点O ,连结,,AO CO EO .因为AB AD =,BO OD =,所以AO BD ⊥, ····················· 1分 又平面ABD ⊥平面BCD ,平面ABD平面BCD BD =,AO ⊂平面ABD ,所以AO ⊥平面BCD , ······························· 2分 又BE ⊂平面BCD ,所以AO BE ⊥. 在BCD ∆中,2BD BC =,2DE EC =,所以2BD DEBC EC==,由角平分线定理,得CBE DBE ∠=∠, ························ 3分 又2BC BO ==,所以BE CO ⊥, ·························· 4分 又因为AOCO O =,AO ⊂平面ACO ,CO ⊂平面ACO ,所以BE ⊥平面ACO , ······························· 5分 又AC ⊂平面ACO ,所以AC BE ⊥. ························· 6分 (Ⅱ)在BCD ∆中,24BD BC ==,60CBD ∠=,由余弦定理得CD =222BC CD BD +=,即90BCD ∠=,所以30EBD EDB ∠=∠=,BE DE =,所以EO BD ⊥, ··············· 7分 结合(Ⅰ)知,,,OE OD OA 两两垂直.以O 为原点,分别以向量,,OE OD OA 的方向为x 轴、y 轴、z 轴的正方向建立空间直角坐标系O xyz -(如图),设(0)AO t t =>,则()0,0,A t ,()0,2,0B -,E , 所以()0,2,BA t =,2(2,0)BE =, ······················ 8分 设(),,x y z =n 是平面ABE 的一个法向量,则0,0,BA BE ⎧⋅=⎪⎨⋅=⎪⎩n n 即20,20,y tz x y +=⎧+=,整理,得,2,x z y t ⎧=⎪⎨=-⎪⎩令1y =-,得21,)t=-n . ···························· 9分 因为OE ⊥平面ACD ,所以(1,0,0)=m 是平面ABD 的一个法向量. ··········· 10分 又因为二面角E BAD --,所以cos,5<>==m n,解得2t=或2t=-(舍去),···········11分又AO⊥平面BCD,所以AO是三棱锥A BCD-的高,故111223323A BCD BCDV AO S-∆=⋅⋅=⨯⨯⨯⨯=. ·················12分解法二:(Ⅰ)取BD中点O,连结,,OA OC OE.因为AB AD=,BO DO=,所以AO BD⊥,···················1分又因为平面ABD⊥平面BCD,平面ABD平面BCD BD=,AO⊂平面ABD,所以AO⊥平面BCD,·······························2分在平面BCD内,过O作OF OD⊥(如图),则OF,OD,OA两两垂直.以O为原点,分别以向量,,OF OD OA的方向为x轴、y轴、z轴的正方向建立空间直角坐标系O xyz-(如图),设()0AO t t=>,······················3分在BCD∆中,24BD BC==,60CBD∠=,由余弦定理得CD=因为222BC CD BD+=,所以90BCD∠=,故30CDB∠=,·············4分则有()0,0,A t,()0,2,0B-,1,0)C-,E,·············5分所以(3,1,)AC t=--,2(2,0)BE =,所以()()312003AC BE t⋅=⨯+-⨯+-⨯=,所以AC BE⊥. ··································7分(Ⅱ)由(Ⅰ)可得()0,2,BA t=.设(),,x y z=n是平面ABE的法向量,则0,0,BA BE ⎧⋅=⎪⎨⋅=⎪⎩n n即20,20,y tz x y +=⎧+=整理,得,2,x z y t ⎧=⎪⎨=-⎪⎩令1y =-,得21,)t=-n . ···························· 9分 因为OE ⊥平面ACD ,所以(1,0,0)=m 是平面ABD 的一个法向量. ··········· 10分 又因为二面角E BA D --的余弦值为5,所以cos ,5<>==m n ,解得2t =或2-(不合,舍去), ········· 11分 又AO ⊥平面BCD ,所以AO 是三棱锥A BCD -的高,故11122332A BCD BCD V AO S -∆=⋅⋅=⨯⨯⨯⨯=. ················· 12分 解法三(Ⅰ)同解法一. ······························ 6分 (Ⅱ)过点O 作OF AB ⊥于点F ,连结EF.在BCD ∆中,24BD BC ==,60CBD ∠=,由余弦定理可得CD =因为222BC CD BD +=,所以90BCD ∠=,故30EBD EDB ∠=∠=,BE DE =,所以EO BD ⊥, ··············· 7分 又平面ABD ⊥平面BCD ,平面ABD平面BCD BD =,EO ⊂平面BCD ,所以EO ⊥平面ABD ,又AB ⊂平面ABD ,所以EO AB ⊥, ·············· 8分 又因为EOOF O =,所以AB ⊥平面EOF ,又EF ⊂平面EOF ,所以AB EF ⊥,所以EFO ∠为二面角E BA D --的平面角, ·············· 9分所以cos EFO ∠=,所以3tan EO EFO FO FO ∠===,解得FO = ······ 10分设()0AO t t =>,则2t =2t =或2-(不合,舍去), ········· 11分 又AO ⊥平面BCD ,所以AO 是三棱锥A BCD -的高,所以111223323A BCD BCD V AO S -∆=⋅⋅=⨯⨯⨯⨯=. ················ 12分(20) (本小题满分12分) 解法一:(Ⅰ)C 的准线方程为2px =-, ························ 1分 由抛物线的定义,可知BF 等于点B 到C 的准线的距离. ················ 2分 又因为点B 到x 轴的距离比BF 小1,所以点B 到x 轴的距离比点B 到抛物线准线的距离小1, ················· 3分 故12p=,解得2p =, 所以C 的方程为24x y =. ······························ 4分 (Ⅱ)由(Ⅰ)得C 的焦点为(0,1)F ,设直线l 的方程为()10y kx k =+≠,11(,)A x y ,22(,)B x y .则1(,0)D k-. ·························· 5分联立方程组24,1,x y y kx ⎧=⎨=+⎩消去y ,得2440x kx --=. ·················· 6分22(4)41(4)16160k k ∆=--⨯⨯-=+>,由韦达定理,得12124,4x x k x x +==-. ························ 7分 设点O 到直线l 的距离为d ,则12BOF S d BF ∆=⋅,12AOD S d AD ∆=⋅. 又BOF AOD S S ∆∆=,所以BF AD =. ························· 8分 又,,,A B D F 在同一直线上,所以121()x x k --=,即211x x k-=, ············· 9分 因为222211212()()4(4)4(4)x x x x x x k -=+-=-⨯-, ················· 10分所以221(4)4(4)()k k-⨯-=,整理,得42161610k k +-=,故2k =k =, ························ 11分所以l 的方程为1y x =+. ························· 12分 解法二:(Ⅰ)C 的焦点为(0,)2pF , ························· 1分 将2p y =代入22x py =,得x p =或x p =-,故2p BF =,因为点B 到x 轴的距离比BF 小1,12p BF =+,即12pp =+, ············· 2分 解得2p =,所以C 的方程为24x y =, ························ 3分 经检验,抛物线的方程24x y =满足题意. ······················· 4分 (Ⅱ)同解法一. ·································· 12分(21) (本小题满分12分)解法一:(Ⅰ)函数()f x 的定义域为(0,)+∞.要使()0f x ≥有唯一解,只需满足()max 0f x =,且()max 0f x =的解唯一, ······· 1分()1kxf x x-'=, ································· 2分 ①当0k ≤时,()0f x '≥,()f x 在(0,)+∞上单调递增,且()10f =,所以()0f x ≥的解集为[1,)+∞,不符合题意; ···················· 4分 ②当0k >时,且1(0,]x k ∈时,()0f x '≥,()f x 单调递增;当,)(1kx +∞∈时,()0f x '<,()f x 单调递减,所以()f x 有唯一的一个最大值为1()f k,令1()ln 10f k k k=--=,得1k =,此时()f x 有唯一的一个最大值为()1f ,且()10f =,故()0f x ≥的解集是{}1,符合题意;综上,可得1k =. ································· 6分 (Ⅱ)要证当1a ≤时,2(())e 1xx f x kx k ax +-<--,即证当1a ≤时,2e ln 10x ax x x --->,即证2e ln 10x x x x --->. ···························· 7分由(Ⅰ)得,当1k =时,()0f x ≤,即ln 1x x ≤-,从而ln (1)x x x x ≤-,故只需证2e 210x x x -+->,当0x >时成立; ···················· 8分 令2()e 21(0)xh x x x x =-+-≥,则()e 41xh x x '=-+, ················ 9分 令()()F x h x '=,则()F x '=e 4x -,令()0F x '=,得2ln 2x =.因为()F x '单调递增,所以当(]0,2ln 2x ∈时,()0F x '≤,()F x 单调递减,即()h x '单调递减,当()2ln 2,x ∈+∞时,()0F x '>,()F x 单调递增,即()h x '单调递增, 所以(ln 4)58ln 20h '=-<,(0)20h '=>,2(2)e 810h '=-+>,由零点存在定理,可知1(0,2ln 2)x ∃∈,2(2ln 2,2)x ∃∈,使得()()120h x h x ''==, 故当10x x <<或2x x >时,()0h x '>,()h x 单调递增;当12x x x <<时,()0h x '<,()h x 单调递减,所以()h x 的最小值是(0)0h =或2()h x . 由()20h x '=,得22e41x x =-,2()h x ()()222222222e 21252221x x x x x x x =-+-=-+-=---,因为()22ln 2,2x ∈,所以2()0h x >,故当0x >时,()0h x >,所以原不等式成立. ····················· 12分解法二:(Ⅰ)函数()f x 的定义域为(0,)+∞.1()kxf x x-'=, ································· 1分 ①当0k ≤时,()0f x '≥,()f x 在()0,+∞上单调递增,且(1)0f =,所以()0f x ≥的解为[1,)+∞,此时不符合题意; ···························· 2分 ②当0k >时,11()()kx k f x x x x k-'==--, 所以当1(0,]x k ∈时,()0f x '≥,()f x 单调递增;当,)(1kx +∞∈时,()0f x '<,()f x 单调递减,所以1()()f x f k ≤,1()ln 1f k k k=--, ··················· 3分 令()ln 1g k k k =--,11()1k g k k k-'=-=, ····················· 4分 当(]0,1k ∈时,()0g k '≤,()g k 单调递减,当()1,k ∈+∞时,()0g k '>,()g k 单调递增,所以()(1)0g k g ≥= ,由此可得当0k >且1k ≠时,1()0f k>,且当0,x x +→→+∞时,()f x →-∞,由零点存在定理,1211(0,),(,)x x k k∃∈∈+∞,使得()()120f x f x ==,当12x x x ≤≤时,()0f x ≥,解集不唯一,不符合题意; 当1k =时,()f x ≤()10f =,所以()0f x ≥的解集是{}1,符合题意;综上可得,当1k =时,()0f x ≥有唯一解; ····················· 6分 (Ⅱ)要证明当1a ≤时,2(())e 1xx f x kx k ax +-<--,即证当1a ≤时,2e ln 10x ax x x --->,(因为22ax x ≤)即证2e ln 10x x x x --->, ···························· 7分 令2()e ln 1(0)xF x x x x x =--->,则()e 2ln 1xF x x x '=---, ············ 8分 令()()G x F x '=,则1()e 2x G x x'=--在(0,)+∞上单调递增,且(1)0G '<,(2)0G '>, 所以0(1,2)x ∃∈使得0()0G x '=,即01e2x x =+, 所以当0x x >时,()0G x '>,()G x 单调递增,即()F x '递增; 当00x x <<时,()0G x '<,()G x 单调递减,即()F x '递减, 所以00min 000001()e 2ln 12ln 1xF x x x x x x '=---=--+,1()2ln 1H x x x x=--+, 当(1,2)x ∈时递减,0min ()(1)0F x H '<=,当0x →时,()F x '→+∞,3233()e 3ln 1022F '=--->,由零点存在定理,可得10(0,)x x ∃∈,203(,)2x x ∈, 12()()0F x F x ''==, 故当10x x <<或2x x >时,()0F x '>,()F x 单调递增, 当12x x x <<时,()0F x '<,()F x 单调递减, 当0x +→时,()0F x →,由2()0F x '=得,222e2ln 1x x x =++,02312x x <<<, 又2()F x 22222222222e ln 12ln ln xx x x x x x x x =---=-++-,令2()2ln ln M x x x x x x =-++-(312x <<), 则1()22ln 1M x x x x '=-++--在3(1,)2递减,且(1)0M '=,所以()0M x '<,所以()M x 在3(1,)2递减,393331()3ln ln 0.75(ln 3ln 2)0242222M =-++-=-->, 所以当312x <<,()0M x >,即2()0F x >, 所以()0F x >,即原不等式成立. ·························· 12分请考生在第(22),(23)题中任选一题作答,如果多做,则按所做的第一题计分,作答时请写清题号.(22)选修44-;坐标系与参数方程本小题主要考查参数方程、极坐标方程等基础知识,考查运算求解能力,考查数形结合思想、化归与转化思想等.满分10分.解:(Ⅰ)由题意得,由1cos ,sin x y αα=+⎧⎨=⎩可得2222(1)cos sin x y αα-+=+,即1C 的普通方程为22(1)1x y -+=. ·························· 2分 方程2cos sin ρθθ=可化为22cos sin ρθρθ= ……(*), 将cos ,sin ,x y ρθρθ=⎧⎨=⎩代入方程(*),可得2x y =. ····················· 5分(Ⅱ)联立方程22(1)1,x y y kx ⎧-+=⎨=⎩ 得2222(,)11kA k k ++. ················· 7分联立方程组2y kx y x=⎧⎨=⎩,可得2(,k )B k ,所以2221OA OB k k k ⋅==+. ···················· 9分又k ∈,所以OA OB ⋅∈. ······················· 10分(23)选修45-:不等式选讲本小题主要考查绝对值不等式等基础知识,考查运算求解能力、推理论证能力,考查化归与转化思想、分类与整合思想等.满分10分.解:(Ⅰ)当2a =时,()22224,222,2x x f x x x x x x , >⎧⎪=-++= -≤≤⎨⎪- <- ⎩. ········· 1分当2x >时,可得26x >,解得3x >. ······················· 2分 当22x -≤≤时,因为46>不成立,故此时无解; ················· 3分 当2x <-时,由26x ->得,3x <-,故此时3x <-. ················ 4分 综上所述,不等式()6f x >的解集为(,3)(3,)-∞-+∞. ··············· 5分(Ⅱ)因为()2f x x a x a x a x a a =-++≥---=, ·············· 6分 要使关于x 的不等式()21f x a <-有解,只需221a a <-成立即可. ·········· 7分当0a ≥时,221a a <-即221a a <-,解得1a >1a < ····················· 8分 当0a <时,221a a <-,即221a a -<-,解得1a >-+1a <- ···················· 9分 所以,a的取值范围为(,1(12,)-∞--++∞. ················· 10分。
2017年泉州市普通高中毕业班质量检测语文
2017年泉州市普通高中毕业班质量检查语文答案及评分说明一、现代文阅读(35分)1.(3分)D2.(3分)B3.(3分)A(二)实用类文本阅读(12分)4.(3分)B5.(5分)A D答对一项给3分,答对两项给5分。
回答三项或三项以上,不给分。
6.(4分)①谭云山和玄奘一样,都为中印文化交流做出卓越贡献,因此可称为“现代玄奘”。
②谭云山在印度“归西”而没有像玄奘一样归国,他的贡献主要是“东土送经”而不是玄奘般的“西天取经”,所以说他不是“现代玄奘”。
第①点1分,第②点3分。
意思答对即可。
【评分说明】■关于谭云山是“现代玄奘”根据考生分析概括的准确性和完整性给分,没有答出“中印文化交流”不给分。
■关于谭云山不是“现代玄奘”①答出谭云山在印度“归西”而没有像玄奘一样归国,给1分。
只答“归西”不给分。
②答出谭云山的贡献主要是“东土送经”而不是玄奘般的“西天取经”给2分,只答其中一方面,给1分。
③分析的要点不是谭云山不是“现代玄奘”的原因,不给分。
④考生答案的意思与本答案相同或很接近均可。
(三)文学类文本阅读(14分)7.(3分)C8.(5分)①“捶背”将两人的命运连在一起,强化了情节的戏剧性,也使情节更集中。
②借助“捶背”来呈现两人的言行,使人物形象的刻画具体集中。
③通过“捶背”这一小事来表现军营练兵的主题,体现出了以小见大的写作特点。
答对一点给2分,答对三点给满分。
意思对即可。
【评分说明】①分析要针对情节、人物、主题的作用进行,没有针对性,不给分。
②考生答案的意思与本答案相同或很接近均可。
③考生答案要点不是本答案所提供的,但言之有据,析之成理,可酌情给分。
给满5分为止。
9.(6分)观点一:辉是主人公。
①小说借写“捶背”来刻画人物形象,而辉是“捶背”的实施者。
②小说的主要情节都是围绕辉而展开的,小说着重写辉从一个“新兵”到一名“训练标兵”的成长过程。
③小说着力要表现的主题,是新一代青年士兵挑战自我、刻苦拼搏、不断成熟的精神风貌。
福建省泉州市2020届普通高中毕业班质量检查语文试题及答案解析
福建省泉州市2020届普通高中毕业班质量检查语文试题及答案解析注意事项:1.答题前,考生务必将自己的姓名、准考证号填写在答题卡上。
2.回答选择题时,选出每小题答案后,用铅笔把答题卡上对应题目的答案标号涂黑。
如需改动,用橡皮擦干净后,再选涂其他答案标号回答非选择题时,将答案写在答题卡上。
写在本试卷上无效。
3.考试结束,将本试卷和答题卡一并交回。
一、现代文阅读(36分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1~3题。
谈到技术的起源时,人们往往会强调技术是人类肢体能力的延伸。
时至今日,在看到科学技术给我们带来巨大便利的同时,更应该看到,它们作为一种力量,介入到人类社会之后所带来的秩序改变甚至重组。
正是在此意义上,恩格斯在强调劳动在人诞生过程中的重要性时,实际上也就是强调了技术对人自身以及人所存在于其中的社会关系的重要性。
这种重要性在当下更是体现为人类本身的技术化趋势,这种趋势一方面表现为人类身体的技术化;另一方面,在更广泛的层面上也表现为人类生存的技术化,人类生存的各个方面都被技术深刻影响。
正是看到了技术对人类身体乃至人类社会的这种重大影响,自古以来思想家们就非常强调人类伦理秩序与技术的和谐发展。
不管是在日常生活之中,还是在前沿的科技领域,科学技术都参与到了伦理秩序的维持或建构之中。
为了做到这一点,人们通常会将某种伦理秩序内嵌到科学技术特别是以人工制品形式呈现的技术物之中,于是,技术物就成了伦理秩序的一种制度化呈现方式。
相较于传统强调以道德约束作为主要手段维持伦理秩序,这种制度化的优势在于其强制性程度更刚性、执行效率更高,就如同与纯粹的道德呼吁相比,红绿灯和交通监控设施更能维持交通秩序一样。
然而,科学技术对伦理秩序的建构方式并不是决定论式的,具有偶然性。
偶然性是指科学技术及其人工制品具有多样实现性的特征,例如核技术既可以用于制造武器,亦可用于增加能源。
这种偶然性给科技管控带来了很大困难,进而也就给伦理秩序的维持带来了较大风险。
泉州市2017届高中毕业班质量检查试卷
泉州市2017届高中毕业班质量检查试卷文科综合能力测试地理2017.03第I卷本卷共35小题。
每小题4分,共140分。
在每个小题给出的四个选项中,只有一项是符合题目要求的。
相对于修筑堤防、改迁河道等耗资巨大的主动防洪工程,在人力、资金相对不足的古代, 珠江三角洲西部高要地区有30多个村落利用当地有利的自然条件进行被动防洪,形成独特有趣的八卦形态。
图1示意高要地区八卦村落分布区,图2遥感图片示意某“八卦村”的道路和排水系统。
据此完成1〜3题。
0城审•入卄村CZ? ter乜o商地尹=河i*图1 图21•与北岸相比,南岸的村落多呈八卦形态主要是因为这里A .水源丰富B .水灾多发C .水运便利D .耕地充足2.根据“八卦村”排水系统的形态可以推断A •池塘位于村中心方便蓄水B •道路都与排水系统并行方便出行C.村落选址在近似圆形的小山岗上 D •村落选址在近似圆形的小盆地里3•近20年来,高要地区许多“八卦村”的形态逐渐瓦解,可能是由于该地区A .年降水量减少B .台风登陆减少C.防灾意识增强 D •堤防趋于完备图3为中国人口老龄化的全局趋势三维透视图,该图揭示中国县域人口老龄化空间分布的总体格局及特定方向的变化趋势。
人口老龄化系数与人口老龄化程度正相关。
读图完成4〜6题。
4.2000〜2010年,中国人口老龄化程度空间分布变化的趋势是A.整体升高,东西差异扩大 C.西北升高,东西差异缩小5. 2000〜2010年,东北地区老龄化程度变化的主要原因是低 D.城市化水平提高 6.该时期南部沿海地区人口老龄化程度降低会A.增加青壮年的社会负担 降低C.促进社会养老体系完善 升级从秦岭第二高峰鳌山(海拔 3475米)沿山脊徒步至第一高峰太白山(海拔 3767米)的户外徒步线路一一鳌太线,以山水形胜而出名。
但其积雪多、难度大、危险性高对户外示意“石海”景观与秦岭太白山北坡植被分布。
据此完成口老龄化系数图32000年 201()年A.人口迁入B.放开二胎C.出生率持续较 B.西北升高,东部降幅小D.整体升高,东部升幅小B.使劳动力成本增幅D.制约产业结构调整 爱好者提出挑战,尤其以称为“石海”段的路段最难,基本上是在碎石上攀爬。
2020年1月福建省泉州市普通高中2020届高三毕业班上学期期末质量监测英语试题(解析版)
绝密★启用前福建省泉州市普通高中2020届高三毕业班上学期期末单科教学质量监测英语试题(解析版)2020年1月第二部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑。
AReasons Every Teen Should Go To Summer Camp1 Improve Interpersonal Skills & Form Close FriendshipsIn a world where anyone can look up a fact and where machines are replacing even complex workplace tasks, employers need employees who can interact effectively with other people. This is one of the most important skills teens learn at camp. In the non-competitive camp culture,teens build up their “emotional intelligence” (EQ),their face-to-face communication and relationship skills.2 Experience Character Development and Develop Life SkillsTeens develop other important life skills at camp, including independence,responsibility, and decision-making. Teens grow considerably in environment away from their parents where they are forced to live on their own and find their own resources.3 Meet Positive Role ModelsWalk into any well-run summer camp and you’ll be surrounded by wh olesome,outdoorsy young people. Camp offers teens the opportunity to be among young adults who are positive role models and to form close relationships with them. Most campcounselors are hard-working college students who want to serve others. Aren’t they just the kind of young adults you want your teen to become?4 Discover Their Best SelfCamp experiences offer teens the chance to step back from the tiring task of academic and competitive sports and instead think about what’s important to them. Many campers become less self-absorbed after spending a few weeks at camp, learning to train their focus on others. They discover new hobbies and avenues to pursue in education and their future careers.1. What can teens acquire in the camp to meet their future career?A. Computer competence.B. Communication skills.C. Adventurous spirit.D. Academic quality.2. Which of the following best describes camp counselors?A. Committed.B. Ambitious.C. Humorous.D. Demanding.3. What change can camp experiences bring to many campers?A. Preferring non-competitive culture.B. Becoming positive role models.C. Focusing more on academics competition.D. Finding more suitable future career choices.【答案】1. B 2. A 3. D【解析】本文是一篇应用文。
2020年1月福建省高2020届高2017级高三泉州市单科质检理科数学试题答题全析解答题部分
保密★启用前泉州市2020届高中毕业班单科质量检查理科数学2020.1注意事项:1.答题前,考生先将自己的姓名、准考证号填写在答题卡上.2.考生作答时,将答案答在答题卡上.请按照题号在各题的答题区域(黑色线框)内作答,超出答题区域书写的答案无效.在草稿纸、试题卷上答题无效.3.选择题答案使用2B 铅笔填涂,如需改动,用橡皮擦干净后,再选涂其它答案标号;非选择题答案使用5.0毫米的黑色中性(签字)笔或碳素笔书写,字体工整、笔迹清楚.4.保持答题卡卡面清洁,不折叠、不破损.考试结束后,将本试卷和答题卡一并交回.三、解答题:共70分.解答应写出文字说明、证明过程或演算步骤.第17~21题为必考题,每个试题考生都必须作答.第22、23题为选考题,考生根据要求作答.(一)必考题:共60分.17.(12分)如图,四棱锥ABCD P -的底面是正方形,⊥PA 平面ABCD ,AE PD ⊥.(1)证明:AE ⊥平面PCD ;(2)若AP AB =,求二面角D PC B --的余弦值.【命题意图】本小题考查线面垂直的判定与性质、二面角的求解及空间向量的坐标运算等基础知识,考查空间想象能力、逻辑推理及运算求解能力,考查化归与转化思想、函数与方程思想等,体现基础性、综合性与应用性,导向对发展数学抽象、逻辑推理、直观想象等核心素养的关注.【试题解析】解法一:(1)因为PA ⊥平面ABCD ,CD ⊂平面ABCD ,所以PA CD ⊥.·········································································································1分又底面ABCD 是正方形,所以AD CD ⊥.·····································································2分又PA AD A = ,所以CD ⊥平面PAD .······································································3分又AE ⊂平面PAD ,所以CD AE ⊥.···········································································4分又因为AE PD ⊥,CD PD D = ,,CD PD ⊂平面PCD ,·············································5分所以AE ⊥平面PCD .·······························································································6分(2)因为PA ⊥平面ABCD ,底面ABCD 为正方形,所以PA AB ⊥,PA AD ⊥,AB AD ⊥,分别以AB 、AD 、AP 所在的直线为x 轴、y 轴、z 轴建立空间直角坐标系A xyz -(如图所示).······································································7分设1PA AB ==,则A 0,0,0(),B 1,0,0(),C 1,1,0(),D 0,1,0(),(0,0,1)P ,11(0,,)22E ,1,0,1PB =- (),1,1,1PC =- (),11(0,,22AE = .··························································8分由(1)得11(0,,)22AE = 为平面PCD 的一个法向量.·······················································9分设平面PBC 的一个法向量为111()m x ,y ,z =.由0,0,PB m PC m ⎧⋅=⎪⎨⋅=⎪⎩得111110,0,x z x y z -=⎧⎨+-=⎩令11x =,解得11z =,10y =.所以(1,0,1)m =.·····································································································10分因此112cos ,2m AEm AE m AE⋅===⋅.·······························································11分由图可知二面角B PC D --的大小为钝角.故二面角B PC D --的余弦值为12-.·········································································12分解法二:(1)同解法一.·····································································································6分(2)过点B 作BF 垂直于PC 于点F ,连接DF 、BD .因为PB PD =,BC CD =,PC PC =,所以PBC PDC △≌△.······························································································7分因此易得090DFC BFC ∠=∠=,BF DF =.································································8分所以BFD ∠为二面角B PC D --的平面角.···································································9分设1PA AB ==,则BD =3BF DF ==.·························································10分在BDF △中,由余弦定理,得222222)133cos 2263BF DF BDBFD BF DF+-+-∠==-⋅.故二面角B PC D --的余弦值为12-.·········································································12分18.(12分)记n S 为数列{}n a 的前n 项和.已知0n a >,2634n n n S a a =+-.(1)求{}n a 的通项公式;(2)设2211n n n n n a a b a a +++=,求数列{}n b 的前n 项和n T .【命题意图】本小题主要考查递推数列、等差数列的通项公式与数列求和等基础知识,考查推理论证能力与运算求解能力等,考查化归与转化思想、特殊与一般思想等,体现基础性,导向对发展逻辑推理、数学运算等核心素养的关注.【试题解析】解:(1)当1n =时,2111634S a a =+-,所以14a =或1-(不合,舍去).································1分因为2634n n n S a a =+-①,所以当2≥n 时,2111634n n n S a a ---=+-②,由①-②得2211633n n n n n a a a a a --=+--,······································································2分所以()()1130n n n n a a a a --+--=.················································································3分又0n a >,所以13n n a a --=.······················································································4分因此{}n a 是首项为4,公差为3的等差数列.···································································5分故()43131n a n n =+-=+.························································································6分(2)由(1)得()()()()22313433231343134n n n b n n n n +++==+-++++,········································9分所以()33333392()2477103134434n nT n n n n n =+-+-+⋅⋅⋅+-=++++.····························12分19.(12分)ABC △中,60B =︒,2AB =,ABC △的面积为(1)求AC ;(2)若D 为BC 的中点,,E F 分别为,AB AC 边上的点(不包括端点),且120EDF ∠=︒,求DEF △面积的最小值.【命题意图】本小题主要考查解三角形、三角恒等变换等基础知识,考查推理论证能力和运算求解能力等,考查数形结合思想和化归与转化思想等,体现综合性与应用性,导向对发展直观想象、逻辑推理、数学运算及数学建模等核心素养的关注.【试题解析】解法一:(1)因为60B =︒,2AB =,所以1sin 2ABC S AB BC B =⋅⋅⋅△13222BC =⨯⨯32BC =,·············································2分又ABC S =△,所以4BC =.···················································································3分由余弦定理,得2222cos AC AB BC AB BC B =+-⋅⋅·······················································4分221242242=+-⨯⨯⨯12=,·························································································5分所以AC =········································································································6分(2)设BDE θ∠=,[]0,60θ∈︒︒,则60CDF θ∠=︒-.在BDE △中,由正弦定理,得sin sin BD DEBED B=∠,·························································7分即2sin(60)32θ=︒+,所以3sin(60)DE θ=︒+;···························································8分在CDF △中,由正弦定理,得sin sin CD DFCFD C=∠,由(1)可得30C =︒,即21sin(90)2DF θ=︒-,所以1cos DF θ=;·····································9分所以1sin 2DEF S DE DF EDF =⋅⋅⋅∠△34sin(60)cos θθ=︒+⋅=········································································10分=,············································································11分当15θ=︒时,sin(260)1θ+︒=,min ()6DEF S ==-△故DEF △面积的最小值为6-.············································································12分解法二:(1)同解法一.·····································································································6分(2)设CDF θ∠=,[]0,60θ∈︒︒,则60BDE θ∠=︒-.在CDF △中,由正弦定理,得sin sin CD DFCFD C=∠,························································7分由(1)可得30C =︒,即21sin(30)2DFθ=︒+,所以()1sin 30DF θ=︒+;···························8分在BDE △中,由正弦定理,得sin sin BD DEBED B=∠,即2sin(120)32θ=︒-,所以sin(120)DE θ=︒-;·························································9分所以1sin 2DEF S DE DF EDF =⋅⋅⋅∠△()334sin 30sin(120)θθ=⋅︒+⋅︒-13312222=⎝⎭⎝⎭ (10)分=······················································································11分当45θ=︒时,sin 21θ=,min ()6DEF S ==-△故DEF △面积的最小值为6-.············································································12分20.(12分)已知椭圆2222:1(0)x y E a b a b+=>>的离心率为12,点32A 在E 上.(1)求E 的方程;(2)斜率不为0的直线l 经过点1(,0)2B ,且与E 交于Q P ,两点,试问:是否存在定点C ,使得QCB PCB ∠=∠?若存在,求C 的坐标;若不存在,请说明理由.【命题意图】本小题主要考查椭圆的几何性质、直线与椭圆的位置关系等基础知识,考查推理论证能力、运算求解能力等,考查化归与转化思想、数形结合思想、函数与方程思想等,体现基础性、综合性与创新性,导向对发展逻辑推理、直观想象、数学运算、数学建模等核心素养的关注.【试题解析】解法一:(1)因为椭圆E的离心率12e a ==,所以2234a b =①,··································1分点)23,3(A 在椭圆上,所以143322=+ba ②,·······························································2分由①②解得42=a ,32=b .························································································3分故E 的方程为13422=+y x .··························································································4分(2)假设存在定点C ,使得PCB QCB ∠=∠.由对称性可知,点C 必在x 轴上,故可设(,0)C m .··························································5分因为PCB QCB ∠=∠,所以直线PC 与直线QC 的倾斜角互补,因此0PC QC k k +=.·············6分设直线l 的方程为:21+=ty x ,),(11y x P ,),(22y x Q .由221,2143x ty x y ⎧=+⎪⎪⎨⎪+=⎪⎩消去x ,得04512)1612(22=-++ty y t ,···············································7分2222(12)4(1216)(45)144180(1216)0t t t t ∆=-⨯+⨯-=+⨯+>,所以t ∈R ,122121216t y y t +=-+,122451216y y t =-+,····································································8分因为0=+QC PC k k ,所以02211=-+-mx y m x y ,所以0)()(1221=-+-m x y m x y ,即0)21()21(1221=-++-+m ty y m ty y .·························9分整理得121212()()02ty y m y y +-+=,所以0161212)21()161245(222=+-⨯-++-⨯t t m t t ,即01612)12)(21(902=+--+-t t m t .·················10分所以0)21(1290=-+m t t ,即0)]21(1290[=-+t m ,对t ∈R 恒成立,即0)1296(=-t m 对t ∈R 恒成立,所以8=m .·····························································11分所以存在定点)0,8(C ,使得QCB PCB ∠=∠.·······························································12分解法二:(1)同解法一.·····································································································4分(2)若点C 存在,当直线PQ 垂直x 轴时,点C 必在x 轴上,如果直线PQ 不垂直x 轴,由对称性可知,点C 也必在x 轴上.···········································5分假设存在点)0,(m C ,使得QCB PCB ∠=∠,即直线PC 与直线QC 的倾斜角互补,所以0=+QC PC k k .····································································································6分设直线l 的方程为)21(-=x k y ,),(11y x P ,),(22y x Q .由221(2143y k x x y ⎧=-⎪⎪⎨⎪+=⎪⎩消去x ,得0124)34(2222=-+-+k x k x k ,··········································7分22222(4)4(43)(12)1801440k k k k ∆=--⨯+-=+>,所以k ∈R ,2122443k x x k +=+,34122221+-=k k x x ,··············································································8分因为0=+QC PC k k ,所以02211=-+-m x y m x y ,所以0)()(1221=-+-m x y m x y ,················9分即122111()()()022k x x m k x x m --+--=.整理得0]))(21(2[2121=+++-m x x m x x k ,··································································10分所以0]34421(34242[2222=++⨯+-+-m k k m k k k ,整理得0342432=+-⨯k m k ,对任意的k ∈R 恒成立,····························································11分所以8=m ,故存在x 轴上的定点)0,8(C ,使得QCB PCB ∠=∠.····································12分21.(12分)已知函数()2()1e xf x x ax =++.(1)讨论()f x 的单调性;(2)若函数()2()1e 1xg x x mx =+--在[)1,-+∞有两个零点,求m 的取值范围.【命题意图】本小题主要考查导数的综合应用,利用导数研究函数的单调性、最值和零点等问题,考查抽象概括、推理论证、运算求解能力,考查应用意识与创新意识,综合考查化归与转化思想、分类与整合思想、函数与方程思想、数形结合思想、有限与无限思想以及特殊与一般思想,考查数学抽象、逻辑推理、直观想象、数学运算、数学建模等核心素养.【试题解析】。
2017届福建省泉州市高三质检物理试题及答案
泉州市2017届普通中学高中毕业班质量检查
理科综合能力测试(物理)
本试卷分第I卷(选择题)和第II卷(非选择题)两部分。
第I卷1至4
页,均为必考题,
第II卷5至12页,包括必考和选考两部分。
满分300分。
可能用到的相对原子质贝:H一1 C一12 O一16 Fe --56 Ag
一10
注意事项:
1.答题前考生务必在试题卷、答题卡规定的地方填写自己的准考证
号、姓名。
考生要认真核对答题卡上粘贴的条形码的“准考证号、姓
名”与考生本人准证号、姓名是否一致。
2.第I卷每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,如
需改动用橡擦干净后。
再选涂其它答案标号。
第II卷用0.5毫米黑色签字笔在答题卡上书写作答,在试题卷上作答,答案无效。
3.考试结束后,监考员将试题卷和答题卡一并收回。
第I卷
(必考)
本卷共18小题.每题6分,共108分。
一、选择题(本题共18小题。
在每小题给出的四个选项中,只有一个选项
符合题目要求。
)
13.如图所示。
间距为d的两细束平行单色
光线以相同的人射角a射到夹角为的两平。
2017年泉州市普通高中毕业班质量检查
2017年福建省泉州市普通高中毕业班质量检查语文【注意事项】1.答卷前,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,用合乎要求的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.选考题的作答:先把所选题目的题号在答题卡上指定的位置用2B铅笔涂黑。
答案写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。
5.考试结束后,请将本试题卷和答题卡一并上交。
一、现代文阅读(35分)(一)论述类文本阅读(9分,每小题3分)阅读下面的文字,完成1-3题。
清季与民国的思想界大家辈出,派别林立,人们围绕救国救民方案、民族复兴路径、人生价值取向进行了不尽相同的思考与设计,并分为不同思想流派与社会思潮,相互间进行着激烈的论争、交锋。
就基本格局而言,主要存在着自由主义、激进主义与保守主义的思潮,三者之间呈现出既分立又并生、既交锋又交集、既对立又对话的离合关系。
《中国现代三大思潮及其相互关系》所考察的三大思潮是三个鼎立的、互相抗衡的价值系统,由于政治取向、文化取向不同,三者之间展开过激烈的交锋、论争。
书中考察重点放在诠释三大思潮的并生系统、共同观念上,着重于思潮之“合”,考察三大思潮的多元并存对整合与凝聚社会共识的启示。
书中指出三大思潮有着相近相似的思想倾向,相似的、关注与相近的关怀。
它们最大的共同点在于对实现国家富强目标的关怀、对实现民族复兴愿景的向往,都有着民族情怀与爱国情结。
民族主义是三大思潮的并生系统,是三大思潮的同源潜流,是三大思潮的共同观念,但三大思潮对民族主义诉求的表达方式却有所不同;民族复兴成为现代三大思潮的共同梦想,成为各思想流派的最大公约数与最基本共识,三大思潮均怀着实现民族复兴之梦,只是对民族复兴的实施方案、民族复兴的实现路径有着不尽一致的设计。
2017届福建省泉州市高三质量检测物理试题及答案
福建省泉州市2017届高中毕业班质量检测理综物理试题13.某校航模小组正在操场上试射火箭模型,若某次竖直向上发射时它的v-t 图象如图所示,则下列说法正确的是A. 0一2s内火箭模型匀速上升B. 2一4s内火箭模型静止不动C.6 s末火箭模型的加速度为零D.10 s末火箭模型恰好回到出发点14.为了研究乐音的振动规律,某同学用计算机录制下优美的笛音do和sol,然后在电脑上用软件播放,分别得到如图中a和b所示的两个振动图线,则下列说法正确的是A. do和sol两笛音的频率之比约为3:2B. do和sol两笛音的周期之比约为5:3C. do和sol两笛音在空气中的波长之比约为3:2D. do和sol两笛音在空气中的传播速度之比约为5:315.“太空涂鸦”技术就是使低轨运行的攻击卫星在接近高轨侦查卫星时,准确计算轨道向其发射“漆雾”弹,并在临近侦查卫星时,压爆弹囊,让“漆雾”散开并喷向侦查卫星,喷散后强力吸附在侦查卫星的侦察镜头、太阳能板、电子侦察传感器等关键设备上,使之暂时失效。
下列说法正确的是A.攻击卫星在轨运行速率大于7.9 km/sB.攻击卫星进攻前的速度比侦查卫星的速度小C.攻击卫星完成“太空涂鸦”后应减速才能返回低轨道上D.若攻击卫星周期已知,结合万有引力常量就可计算出地球质量16.如图所示,水平桌面上平放着一副扑克牌,总共54张,每一张牌的质量都相等,牌与牌之间以及牌与桌面之间的动摩擦因数都相等。
用手指竖直向下按压第一张牌,并以一定速度水平移动手指,将第一张牌从牌揉中水平向右移出(牌与手指之间无相对滑动)。
设最大静摩擦力等于滑动摩擦力,则下列说法正确的是A.手指对第1张牌的摩擦力方向与手指运动方向相反B.从第2张到第54张之间的牌不会发生相对滑动C.从第2张到第54张之间的牌可能发生相对滑动D.桌面对第54张牌的摩擦力方向与手指运动方向相同17.如图所示,以0为圆心的圆形区域内有垂直于纸面向里的匀强磁场,一个带电粒子以速度v从A点向圆心方向垂直磁场射入,经过时间t从C点射出磁场,AO与OC成900角。
2020届福建省泉州市2017级高三上学期单科质量检测数学(理)(答题全析—小题部分)
段 CF 的中点,则 AB
A. 9
B. 12
【解析】依题意得 p 4 ,焦点 F (0,2) ,
C. 18
D. 72
解法一:因为 A 为线段 CF 的中点,所以 A(2
2,1) ,kAF
2 1 0 (2
2)
2 ,所以直线 AF 的方 4
程为 y
2 x 2 , 将 其 代 入 x2 8y 得 x2 2 4
4,左焦点 F
到C
的一条渐近线的距离为 3 ,则 C
的方程为
A. x2 y2 1 23
B. x2 y2 1 43
C. x2 y2 1 49
D. x2 y2 1 16 9
【解析】因为实轴长 2a 4 ,所以 a 2 , F (c,0) ,由对称性,双曲线的一个焦点到两条渐近线的距离
log π e 2
1 2
,
a c 1 (2 ln π) 1 ln π 2 0 ,所以 b c a ,故选 B.
ln π
ln π
11.在平面直角坐标系 xOy 中,直线 l : kx y 4k 0 与曲线 y 9 x2 交于 A, B 两点,且 AOAB 2 ,
为 DMA∽DNB ,
所以
AM BN
AD DB
,即
3 k
9 12
k
,解得 k
6,
因此 AB 3 6 9 ,故选 A.
市单科质检数学(理科)试题 第 4 页(共 8 页)
10.已知
a
log π
e
,b
ln
π e
2017届福建省泉州市高三质检语文试题及答案
准考证号姓名(在此卷上答题无效)保密★启用前泉州市2017届普通中学高中毕业班质量检查语文本试卷满分150分。
注意事项:1.答题前,考生务必在试题卷、答题卡规定的地方填写自己的准考证号、姓名。
考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名”与考生本人准考证号、姓名是否一致。
2.用0.5毫米黑色签字笔在答题卡上书写作答,在试题卷上作答,答案无效。
做选做题时,用2B铅笔把答题卡上对应题目的标号涂黑。
3.考试结束,考生必须将本试卷和答题卡一并交回。
一、古代诗文阅读(27分)(一)默写常见的名句名篇(6分)1.补写出下列名句名篇中的空缺部分。
(6分)(1)芳与泽其杂糅兮,。
(屈原《离骚》)(2)日月之行,若出其中;,若出其里。
(曹操《观沧海》)(3)。
夕阳西下,断肠人在天涯。
(马致远《天净沙·秋思》)(4)外无期功强近之亲,内无应门五尺之僮,,形影相吊。
(李密《陈情表》)(5),师不必贤于弟子,闻道有先后,术业有专攻,如是而已。
(韩愈《师说》)(6)登斯楼也,则有心旷神怡,,把酒临风,其喜洋洋者矣。
(范仲淹《岳阳楼记》)(二)文言文阅读(15分)阅读下面的文言文,完成2~5题。
鲁云谷传[明]张岱会稽宝祐桥南,有小小药肆.,则吾友云谷悬壶地也。
云谷深于茶理,相知者日集试茶,纷至沓来,应接不暇。
人病其烦,而云谷乐此不为疲也。
术擅痈疽,更专痘疹。
然皆以聪明用事,医不经师,方不袭古,每以劫剂臆见,起死回生。
人终疑其游戏岐黄①,不尊不信,故凡患痘之家,非极险极逆时,医之所谢绝者,决不顾吾云谷也。
云谷也诊视灵敏,可救则救,不可救则望之却.走,未尝依回盼睐,受人一钱。
性极好洁,负米颠②之癖,恨烟,恨酒,恨人撷花;尤恨人唾洟秽地。
故非解人韵士,不得与之久交。
自小多艺,凡羌笛、胡琴、凤笙、斑管,无不精妙。
而尤喜以洞箫和人度曲。
向.与李玉成竹肉③相得,后惟王公端与之合调,余皆非其敌手也。
其密友惟陆癯庵、金尔和与余三人,非大风雨,非至不得已事,必日至其家,啜茗焚香,剧谈谑笑,十三年于此。
福建省泉州市普通中学2020届高中数学毕业班质量检查试题 理
泉州市普通高中毕业班质量检查理 科 数 学本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题),第Ⅱ卷第21题为选考题,其它题为必考题.本试卷共6页,满分150分.考试时间120分钟. 参考公式:样本数据1x 、2x 、…、n x 的标准差:s =x 为样本平均数; 柱体体积公式:V Sh =,其中S 为底面面积,h 为高;锥体体积公式:13V Sh =,其中S 为底面面积,h 为高; 球的表面积、体积公式:24S R π=,343V R π=,其中R 为球的半径.第Ⅰ卷(选择题 共50分)一、选择题:本大题共10小题,每小题5分,共50分.在每小题给出的四个选项中,只有一项是符合题意要求的.1. 复数()1i i +等于A .1i -+B .1i +C .1i --D .1i -2. 已知集合{}13A x x =<<,{}21log 2B x x =<<,则A B I 等于A.{}03x x << B.{}23x x << C.{}13x x << D.{}14x x <<3. 已知(2,1),(1,3)a b ==--r r ,则||a b -r r等于ABC .5D .254. 执行右侧框图所表达的算法,如果最后输出的S 值为12012,那么判断框中实数a 的取值范围是 A .20112012a ≤< B .20112012a <≤ C .20112012a ≤≤ D .20122013a ≤<5. 下列四个条件:①x ,y ,z 均为直线; ②x ,y 是直线,z 是平面;③x 是直线,y ,z 是平面;④x ,y ,z 均为平面. 其中,能使命题“,x y y z x z ⊥⇒⊥P ”成立的有A .1个B .2个C .3个D .4个6. 已知实数,x y 满足2220,0,4,x y x y x y ⎧-+≥⎪+≥⎨⎪+≤⎩则2z x y =+的最大值是 A .5 B .-1 C .2D.7. 已知二次函数2()f x ax bx =+,则“(2)0f ≥”是“函数()f x 在()1,+∞单调递增”的A .充要条件B .充分不必要条件C .必要不充分条件 D. 既不充分也不必要条件8. 已知12,A A 分别为椭圆2222:1(0)x y C a b a b+=>>的左右顶点,椭圆C 上异于12,A A 的点P 恒满足1249PA PA k k ⋅=-,则椭圆C 的离心率为A .49 B .23 C .59D.39. 为调查某校学生喜欢数学课的人数比例,采用如下调查方法:(1)在该校中随机抽取100名学生,并编号为1,2,3, (100)(2)在箱内放置两个白球和三个红球,让抽取的100名学生分别从箱中随机摸出一球,记住其颜色并放回;(3)请下列两类学生举手:(ⅰ)摸到白球且号数为偶数的学生;(ⅱ)摸到红球且不喜欢数学课的学生.如果总共有26名学生举手,那么用概率与统计的知识估计,该校学生中喜欢数学课的人数比例大约是A.88%B. 90%C. 92%D.94%10. 函数的图象与方程的曲线有着密切的联系,如把抛物线2y x =的图象绕原点沿逆时针方向旋转90o就得到函数2y x =的图象.若把双曲线2213x y -=绕原点按逆时针方向旋转一定角度θ后,能得到某一个函数的图象,则旋转角θ可以是A .30oB .45oC .60oD .90o第Ⅱ卷(非选择题 共100分)二、填空题:本大题共5小题,每小题4分,共20分.将答案填在答题卡的相应位置. 11. 已知等差数列}{n a 中, 51a =,322a a =+,则11S = . 12. 一个三棱锥的正视图和侧视图及其尺寸如图所示,则该三棱锥俯视图的面积为 .123侧视图正视图13. 在ABC V中,60,B AC ==oABC V 周长的最大值为 .14. 已知{}()(),min ,a b a a b a b b ≤⎧⎪=⎨>⎪⎩,设()31min ,f x x x ⎧⎫=⎨⎬⎩⎭,则由函数()f x 的图象与x 轴、直线x e =所围成的封闭图形的面积为 .15. 数学与文学之间存在着许多奇妙的联系. 诗中有回文诗,如:“云边月影沙边雁,水外天光山外树”,倒过来读,便是“树外山光天外水,雁边沙影月边云”,其意境和韵味读来真是一种享受!数学中也有回文数,如:88,454,7337,43534等都是回文数,无论从左往右读,还是从右往左读,都是同一个数,称这样的数为“回文数”,读起来还真有趣!二位的回文数有11,22,33,44,55,66,77,88,99,共9个;三位的回文数有101,111,121,131,…,969,979,989,999,共90个; 四位的回文数有1001,1111,1221,…,9669,9779,9889,9999,共90个; 由此推测:10位的回文数总共有 个.三、解答题:本大题共6小题,共80分.解答应写出文字说明,证明过程或演算步骤. 16.(本小题满分13分)已知点(1,0)F ,直线:1l x =-,动点P 到点F 的距离等于它到直线l 的距离. (Ⅰ)试判断点P 的轨迹C 的形状,并写出其方程.(Ⅱ)是否存在过(4,2)N 的直线m ,使得直线m 被截得的弦AB 恰好被点N 所平分?17.(本小题满分13分)将边长为1的正三角形ABC 按如图所示的方式放置,其中顶点A 与坐标原点重合.记边AB 所在直线的倾斜角为θ,已知0,3πθ⎡⎤∈⎢⎥⎣⎦. (Ⅰ)试用θ表示BC uuu r的坐标(要求将结果化简为形如(cos ,sin )αα的形式);(Ⅱ)定义:对于直角坐标平面内的任意两点()11,P x y 、()22,Q x y ,称1212x x y y -+-为P 、Q 两点间的“taxi 距离” ,并用符号PQ 表示.试求BC 的最大值.18.(本小题满分13分)已知12310,,,,A A A A L 等10所高校举行的自主招生考试,某同学参加每所高校的考试获得通过的概率均为12. (Ⅰ)如果该同学10所高校的考试都参加,试求恰有2所通过的概率;(Ⅱ)假设该同学参加每所高校考试所需的费用均为a 元,该同学决定按12310,,,,A A A A L 顺序参加考试,一旦通过某所高校的考试,就不再参加其它高校的考试,试求该同学参加考试所需费用ξ的分布列及数学期望.19. (本小题满分13分)如图,侧棱垂直底面的三棱柱111ABC A B C -中,AB AC ⊥,13AA AB AC ++=,(0)AB AC t t ==>,P 是侧棱1AA 上的动点.C 11C(Ⅰ)当1AA AB AC ==时,求证:11A C ABC ⊥平面; (Ⅱ)试求三棱锥1P BCC -的体积V 取得最大值时的t 值;(Ⅲ)若二面角1A BC C --的平面角的余弦值为10,试求实数t 的值. 20.(本小题满分14分)已知()0xf x x e =⋅,()()10f x f x '=,()()21f x f x '=,…,()()1n n f x f x -'=(n N *∈).(Ⅰ)请写出()n f x 的表达式(不需证明); (Ⅱ)设()n f x 的极小值点为(),n n n P x y ,求n y ;(Ⅲ)设()()22188n g x x n x n =--+-+, ()n g x 的最大值为a ,()n f x 的最小值为b ,试求a b -的最小值.21. 本题有(1)、(2)、(3)三个选答题,每题7分,请考生任选2题作答,满分14分.如果多做,则按所做的前两题记分.作答时,先用2B 铅笔在答题卡上把所选题目对应的题号涂黑,并将所选题号填入括号中.作(1)(本小题满分7分)选修4—2:矩阵与变换若二阶矩阵M 满足127103446M ⎛⎫⎛⎫= ⎪ ⎪⎝⎭⎝⎭. (Ⅰ)求二阶矩阵M ;(Ⅱ)把矩阵M 所对应的变换作用在曲线223861x xy y ++=上,求所得曲线的方程. (2)(本小题满分7分)选修4-4:坐标系与参数方程已知在直角坐标系xOy 中,曲线C 的参数方程为2cos 2sin x t y θθ=⎧⎨=⎩(t 为非零常数,θ为参数),在极坐标系(与直角坐标系xOy 取相同的长度单位,且以原点O 为极点,以x 轴正半轴为极轴)中,直线l 的方程为sin()4πρθ-=(Ⅰ)求曲线C 的普通方程并说明曲线的形状;(Ⅱ)是否存在实数t ,使得直线l 与曲线C 有两个不同的公共点A 、B ,且10OA OB ⋅=u u u r u u u r(其中O 为坐标原点)?若存在,请求出;否则,请说明理由. (3)(本小题满分7分)选修4—5:不等式选讲已知函数()24f x x x =-+-的最小值为m ,实数,,,,,a b c n p q 满足222222a b c n p q m ++=++=.(Ⅰ)求m 的值;(Ⅱ)求证:4442222n p q a b c++≥.参考解答及评分标准一、选择题:本大题考查基础知识和基本运算.每小题5分,满分50分. 1. A 2.B 3.C 4.A 5.C6. D 7.C 8.D 9.B 10.C二、填空题:本大题考查基础知识和基本运算.每小题4分,满分20分.11.33 12.1 13. .5415.90000三、解答题:本大题共6小题,共80分.解答应写出文字说明,证明过程或演算步骤. 16. 本小题考查抛物线的标准方程、直线与圆锥曲线的位置关系等基础知识,考查推理论证能力、运算求解能力,考查化归与转化思想、数形结合思想等.满分13分.解:(Ⅰ)因点P 到点F 的距离等于它到直线l 的距离,所以点P 的轨迹C 是以F 为焦点、直线1x =-为准线的抛物线, ………………2分 其方程为24y x =. ………………5分(Ⅱ)解法一:假设存在满足题设的直线m .设直线m 与轨迹C 交于1122(,),(,)A x y B x y ,依题意,得121284x x y y +=⎧⎨+=⎩. ………………6分①当直线m 的斜率不存在时,不合题意. ………………7分②当直线m 的斜率存在时,设直线m 的方程为2(4)y k x -=-,………8分联立方程组22(4)4y k x y x-=-⎧⎨=⎩,消去y ,得2222(844)(24)0k x k k x k --++-=,(*) ………………9分∴21228448k k x x k -++==,解得1k =. ………………10分 此时,方程(*)为2840x x -+=,其判别式大于零, ………………11分 ∴存在满足题设的直线m ………………12分 且直线m 的方程为:24y x -=-即20x y --=. ………………13分解法二:假设存在满足题设的直线m .设直线m 与轨迹C 交于1122(,),(,)A x y B x y ,依题意,得121284x x y y +=⎧⎨+=⎩. ………………6分易判断直线m 不可能垂直y 轴, ………………7分 ∴设直线m 的方程为4(2)x a y -=-,………8分联立方程组24(2)4x a y y x -=-⎧⎨=⎩,消去x ,得248160y ay a -+-=, ………………9分∵216(1)480a ∆=-+>,∴直线与轨迹C 必相交. ………………10分 又1244y y a +==,∴1a =. ………………11分 ∴存在满足题设的直线m ………………12分且直线m 的方程为:24y x -=-即20x y --=. ………………13分解法三:假设存在满足题设的直线m .设直线m 与轨迹C 交于1122(,),(,)A x y B x y ,依题意,得121284x x y y +=⎧⎨+=⎩. ………………6分∵1122(,),(,)A x y B x y 在轨迹C 上,∴有2112224142y x y x ⎧=⎪⎨=⎪⎩L L ()(),将(1)(2)-,得2212124()y y x x -=-. ………8分当12x x =时,弦AB 的中点不是N ,不合题意, ………9分 ∴12121241y y x x y y -==-+,即直线AB 的斜率1k =, ………10分注意到点N 在曲线C 的张口内(或:经检验,直线m 与轨迹C 相交)…11分 ∴存在满足题设的直线m ………………12分且直线m 的方程为:24y x -=-即20x y --=. ………………13分17. 本小题主要考查三角函数的定义、两角和与差的三角函数公式、平面向量等基础知识,考查运算求解能力,考查化归与转化思想.满分13分.解:(Ⅰ)解法一:因为()cos ,sin B θθ,cos ,sin 33C ππθθ⎛⎫⎛⎫⎛⎫++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭, ……2分 所以cos cos ,sin sin 33BC ππθθθθ⎛⎫⎛⎫⎛⎫=+-+- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭u u u r ………3分22cos ,sin 33ππθθ⎛⎫⎛⎫⎛⎫=++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭. ………7分解法二:平移BC uuu r 到AD u u u r(B 移到A ,C 移到D ),………2分由BC uuu r 的坐标与AD u u u r的坐标相等,都等于点D 的坐标. ………3分由平几知识易得直线AD 的倾斜角为23πθ+, ∵||1AD =u u u r ,∴根据三角函数的定义可得22cos ,sin 33D ππθθ⎛⎫⎛⎫⎛⎫++ ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭, 所以22cos ,sin 33BC ππθθ⎛⎫⎛⎫⎛⎫=++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭u u u r . ………7分(Ⅱ)解法一:22cos sin 33BC ππθθ⎛⎫⎛⎫=+++ ⎪ ⎪⎝⎭⎝⎭,………8分∵0,3πθ⎡⎤∈⎢⎥⎣⎦,∴22[,]33ππθπ+∈, ………9分 ∴22cos sin 33BC ππθθ⎛⎫⎛⎫=-+++ ⎪⎪⎝⎭⎝⎭………11分 512πθ⎛⎫=+ ⎪⎝⎭, ………12分所以当12πθ=时,BC . ………13分解法二: cos cos sin sin 33BC ππθθθθ⎛⎫⎛⎫=+-++- ⎪ ⎪⎝⎭⎝⎭,………8分 ∵03πθ≤≤,∴2333πππθπ≤+≤<,即03πθθπ≤<+<, ∴cos cos cos cos()33ππθθθθ⎛⎫+-=-+ ⎪⎝⎭. ………9分 ∵03πθ≤≤,∴()232πππθθ-≥+-,∴sin sin sin sin 33ππθθθθ⎛⎫⎛⎫+-=+- ⎪ ⎪⎝⎭⎝⎭, ………10分 ||||BC =cos cos()3πθθ-++sin sin 3πθθ⎛⎫+- ⎪⎝⎭5sin()cos())6612πππθθθ=+++=+, ………12分所以当12πθ=时,BC . ………13分18. 本题主要考查概率与统计的基础知识,考查数据处理能力、运算求解能力以及应用用意识,考查必然与或然思想、分类与整合思想等.满分13分.解:(Ⅰ)因为该同学通过各校考试的概率均为12,所以该同学恰好通过2所高校自主招生考试的概率为2821011122P C ⎛⎫⎛⎫=- ⎪⎪⎝⎭⎝⎭451024=. ………4分(Ⅱ)设该同学共参加了i 次考试的概率为i P (110,i i Z ≤≤∈).∵91,19,21,102i i i i Z P i ⎧≤≤∈⎪⎪=⎨⎪=⎪⎩,∴所以该同学参加考试所需费用ξ的分布列如下:………7分所以2991111(12910)2222E a ξ=⨯+⨯++⨯+⨯L , ………8分 令29111129222S =⨯+⨯++⨯L , (1)则2391011111128922222S =⨯+⨯++⨯+⨯L , …(2) 由(1)-(2)得291011111922222S =+++-⨯L ,所以2891111192222S =++++-⨯L , ………11分所以289911111191022222E a ξ⎛⎫=++++-⨯+⨯ ⎪⎝⎭L 911122a ⎛⎫=+++ ⎪⎝⎭L 10112112a -=-101212a ⎛⎫=- ⎪⎝⎭1023512a =(元). ………13分 19. 本小题主要考查直线与直线、直线与平面、平面与平面的位置关系等基础知识,考查空间想象能力、推理论证能力及运算求解能力,考查化归与转化思想、数形结合思想、函数与方程思想及应用意识. 满分13分.解:(Ⅰ)证法一:∵1AA ⊥面ABC ,∴1AA AC ⊥,1AA AB ⊥. 又∵1AA AC =,∴四边形11AAC C 是正方形, ∴11AC A C ⊥. ………1分∵11111,,,,AB AC AB AA AA AC AAC C AA AC A ⊥⊥⊂=I 平面, ∴11AB AAC C ⊥平面. ………2分又∵111AC AAC C ⊂平面, ∴1AB AC ⊥. ………3分 ∵111,,AB AC ABC AB AC A ⊂=I 平面, ∴11A C ABC ⊥平面. ………4分证法二:∵1AA ⊥面ABC ,∴1AA AC ⊥,1AA AB ⊥. 又∵AB AC ⊥,∴分别以1,,AB AC AA 所在直线为,,x y z 轴建立空间直角坐标系. ……1分 则11(0,0,0),(0,1,1),(1,0,0),(0,1,0),(0,0,1)A C B C A ,11(0,1,1),(0,1,1),(1,0,0)AC AC AB =-==u u u r u u u u r u u u r , ∴1110,0AC AC AC AB ⋅=⋅=u u u r u u u u r u u u r u u u r , …2分 ∴111,AC AC AC AB ⊥⊥u u u r u u u u r u u u r u u u r . …3分 又∵111,,AB AC ABC AB AC A ⊂=I 平面 ∴11A C ABC ⊥平面. …4分证法三:∵1AA ⊥面ABC ,∴1AA AC ⊥,1AA AB ⊥.又∵AB AC ⊥,∴分别以1,,AB AC AA 所在直线为,,x y z 轴建立空间直角坐标系. ……1分 则11(0,0,0),(0,1,1),(1,0,0),(0,1,0),(0,0,1)A C B C A ,11(0,1,1),(0,1,1),(1,0,0)AC AC AB =-==u u u r u u u u r u u u r. 设平面1ABC 的法向量(,,)n x y z =r,则100n AC y z n AB x ⎧⋅=+=⎪⎨⋅==⎪⎩r u u u u r r u u u r,解得0x y z =⎧⎨=-⎩. 令1z =,则(0,1,1)n =-r, ……3分∵1AC n =-u u u r r , ∴11A C ABC ⊥平面. ……4分 (Ⅱ)∵111AA BB C C P 平面,∴点P 到平面11BB C C 的距离等于点A 到平面11BB C C 的距离 ∴1112231113(32)(0)6232P BCC A BCC C ABC V V V V t t t t t ---====-=-<<, …5分 '(1)V t t =--,令'0V =,得0t =(舍去)或1t =,∴当1t =时,max 6V =. …8分 (Ⅲ)分别以1,,AB AC AA 所在直线为,,x y z 轴建立空间直角坐标系. 则11(0,0,0),(0,,32),(,0,0),(0,,0),(0,0,32)A C t t B t C t A t --,11(0,,23),(0,,32),(,0,0)AC t t AC t t AB t =-=-=u u u r u u u u r u u u r , 1(0,0,32)CC t =-u u u u r,(,,0)BC t t =-u u u r . ……9分设平面1ABC 的法向量1111(,,)n x y z =u r,则111111(32)00n AC ty t z n AB tx ⎧⋅=+-=⎪⎨⋅==⎪⎩u r u u u u r u r u u u r ,解得111023x t y z t =⎧⎪⎨-=⎪⎩,令1z t =,则1(0,23,)n t t =-u r. …10分 设平面1BCC 的法向量2222(,,)n x y z =u u r,则2222120(32)0n BC tx ty n CC t z ⎧⋅=-+=⎪⎨⋅=-=⎪⎩u u r u u u r u u r u u u u r . 由于302t <<,所以解得2220x y z =⎧⎨=⎩.令21y =,则2(1,1,0)n =u u r. …11分设二面角1A BC C --的平面角为θ,则有1212|||cos|||||n nn nθ⋅===⋅u r u u ru r u u r.化简得2516120t t-+=,解得2t=(舍去)或65t=.所以当65t=时,二面角1A BC C--的平面角的余弦值为10. …13分20. 本题主要考查函数、导数、数列以及合情推理等基础知识,考查推理论证能力、运算求解能力,考查数形结合思想、化归与转化思想、分类与整合思想及有限与无限思想.满分14分.解:(Ⅰ)()()xnf x x n e=+⋅(n N*∈). ……4分(Ⅱ)∵()()1xnf x x n e'=++⋅,∴当()1x n>-+时,()0nf x'>;当()1x n<-+时,()0nf x'<.∴当()1x n=-+时,()nf x取得极小值()()()11nnf n e-+-+=-,即()1nny e-+=-(n N*∈). ……8分(Ⅲ)解法一:∵()()()()2213ng x x n n=-+++-,所以()2((1))3na g n n=-+=-.……9分又()()()11nnb f n e-+=-+=-,∴()()213na b n e-+-=-+,令()()()()2130xh x x e x-+=-+≥,则()()()123xh x x e-+'=--. ……10分∵()h x'在[)0,+∞单调递增,∴()()106h x h e-''≥=--,∵()430h e-'=-<,()5420h e-'=->,∴存在()3,4x∈使得()00h x'=. ……12分∵()h x'在[)0,+∞单调递增,∴当0x x≤<时,()00h x'<;当x x>时,()00h x'>,即()h x在[),x+∞单调递增,在[)00,x单调递减,∴()()()0minh x h x =,又∵()43h e -=,()541h e -=+,()()43h h >, ∴当3n =时,a b -取得最小值4e -. ……14分 解法二: ∵()()()()2213n g x x n n =-+++-,所以()2((1))3n a g n n =-+=-.……9分又()()()11n n b f n e -+=-+=-,∴()()213n a b n e -+-=-+,令()()213n n c n e-+=-+,则1211125n n n n c c n e e +++-=-+-,……10分当3n ≥时,1211125n n n n c c n ee+++-=-+-,又因为3n ≥,所以251n -≥,210n e+>,1101n e+<<,所以2111250n n n e e ++-+->,所以1n n c c +>.……12分又1232341114,1,c c c e e e=+=+=,123c c c >>,∴当3n =时,a b -取得最小值4e -. ……14分21.(1)选修4—2:矩阵与变换本题主要考查矩阵、逆矩阵、曲线的线性变换等基础知识,考查运算求解能力及函数与方程思想.满分7分.解:(Ⅰ)记矩阵1234A ⎛⎫= ⎪⎝⎭,故2A =-,故1213122A --⎛⎫⎪= ⎪-⎝⎭. ……2分 由已知得121710710123146461122M A --⎛⎫⎛⎫⎛⎫⎛⎫ ⎪=== ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭⎝⎭. ……3分 (Ⅱ)设二阶矩阵M 所对应的变换为1211x x y y '⎛⎫⎛⎫⎛⎫= ⎪ ⎪⎪'⎝⎭⎝⎭⎝⎭,得2x x yy x y '=+⎧⎨'=+⎩, 解得2x x y y x y ''=-+⎧⎨''=-⎩, ……5分又223861x xy y ++=,故有223(2)8(2)()6()1x y x y x y x y ''''''''-++-+-+-=,化简得2221x y ''+=.故所得曲线的方程为2221x y +=. ……7分(2)选修4—4:坐标系与参数方程 本题主要考查曲线的参数方程、直线的极坐标方程等基础知识,考查运算求解能力以及化归与转化思想、分类与整合思想.满分7分.解:(Ⅰ)∵0t ≠,∴可将曲线C 的方程化为普通方程:2224x y t+=. ……1分①当1t =±时,曲线C 为圆心在原点,半径为2的圆; ……2分 ②当1t ≠±时,曲线C 为中心在原点的椭圆. ……3分 (Ⅱ)直线l 的普通方程为:40x y -+=. ……4分联立直线与曲线的方程,消y 得222(4)4x x t++=,化简得2222(1)8120t x t x t +++=.若直线l 与曲线C 有两个不同的公共点,则422644(1)120t t t ∆=-+⋅>,解得23t >.……5分又22121222812,,11t t x x x x t t+=-=++ ……6分 故12121212(4)(4)OA OB x x y y x x x x ⋅=+=+++u u u r u u u r121224()1610x x x x =+++=.解得23t =与23t >相矛盾. 故不存在满足题意的实数t . ……7分(3)选修4—5;不等式选讲 本题主要考查绝对值的几何意义、柯西不等式等基础知识,考查运算求解能力以及推理论证能力,考查函数与方程思想以及分类与整合思想.满分7分.解:(Ⅰ)法一: 26(4)()242(24)26(2)x x f x x x x x x -≥⎧⎪=-+-=<<⎨⎪-+≤⎩,……2分 可得函数的最小值为2.故2m =. ……3分法二:()24(2)(4)2f x x x x x =-+-≥---=, ……2分 当且仅当24x ≤≤时,等号成立,故2m =. ……3分(Ⅱ)Θ222222222[()()()]()n p q a b c a b c++⋅++2222()n p q a b c a b c ≥⋅+⋅+⋅ ……5分即:444222()2n p q a b c ++⨯≥2222()4n p q ++=,故4442222n p q a b c++≥. ……7分。
2020年1月9日福建省福州市高2020届高2017级高三高中毕业班期末质量检测文科数学试题参考答案
2019—2020学年度第一学期福州市高三期末质量检测数学(文科)参考答案及评分细则评分说明:1.本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的主要考查内容比照评分标准制定相应的评分细则。
2.对计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应给分数的一半;如果后继部分的解答有较严重的错误,就不再给分。
3.解答右端所注分数,表示考生正确做到这一步应得的累加分数。
4.只给整数分数。
选择题和填空题不给中间分。
一、选择题:本大题考查基础知识和基本运算.每小题5分,满分60分.1.B2.D3.C4.D5.C6.A7.D8.C9.A10.C11.B12.B二、填空题:本大题考查基础知识和基本运算.每小题5分,满分20分.13.3214.5- 16.①③④三、解答题:本大题共6小题,共70分.解答应写出文字说明,证明过程或演算步骤. 17.本题主要考查用样本估计总体、平均数、标准差、古典概型等基础知识,考查考生运用概率与统计知识解决实际问题的能力、数据处理能力,考查的核心素养主要有数学建模、数学抽象、数学运算、数据分析.满分12分.【解答】(1)参与该活动的网友单次游戏得分的平均值 ()1351045405560654075308520200x =⨯⨯+⨯+⨯+⨯+⨯+⨯ ························ 2分 60=.······························································································· 3分标准差s =················· 5分13.60=≈. ········································································· 6分 (2)用分层抽样抽取7人,其中得分在[30,40)的有1人,得分在[60,70)的有4人,得分在[80,90]的有2人, ·················································································· 7分 分别记为a ,1234,,,b b b b ,12,c c ,7人中任选2人,有21种结果, ····························· 8分 其中2人得分在同一组的有7种,分别是{}12,b b 、{}13,b b 、{}14,b b 、{}23,b b 、{}24,b b 、{}34,b b 、{}12,c c , ················································································ 10分 故2人得分不在同一组内的概率721213P =-=.·········································· 12分 18.本小题考查数列的通项与前n 项和的关系、等比数列、裂项相消法求数列的和等基础知识,考查运算求解能力,考查函数与方程思想、化归和转化思想等,考查的核心素养主要有逻辑推理、数学运算等.满分12分.【解答】(1)当1n =时,111a S p ==+, ······················································ 1分 当2n ≥时,221[(1)(1)]21n n n a S S n pn n p n n p -=-=+--+-=-+, ··················· 3分 经检验,1n =时也满足上式,所以21n a n p =-+. ··········································· 4分 因为248,,a a a 成等比数列,所以2284a a a =, ······················································································ 5分即()()()23157p p p ++=+,解得1p =. ······················································ 6分 所以2.n a n = ························································································ 7分 (2)由(1)及题设得,14111(1)1n n n b a a n n n n +===-⋅++, ······························ 9分所以12n n T b b b =++⋅⋅⋅+11111(1)()()2231n n =-+-+⋅⋅⋅+-+ ················································· 11分111n =-+ =1n n +. ··················································································· 12分 19.本小题主要考查空间直线与直线、直线与平面、平面与平面的位置关系、空间几何体的体积等基础知识,考查空间想象能力、推理论证能力、运算求解能力,考查化归与转化思想等,考查的数学素养主要有逻辑推理、直观想象等.满分12分.【解答】解法一:(1)平面AEF 与平面PBC 互相垂直, ······························· 1分 理由如下:因为PA ⊥底面ABCD ,BC ⊂平面ABCD ,所以PA BC ⊥. ··············································· 2分因为ABCD 为正方形,所以AB BC ⊥又PA AB A =,且,PA AB ⊂平面PAB ,所以BC ⊥平面PAB .因为AE ⊂平面PAB ,所以AE BC ⊥. ························································· 3分 因为PA AB =,E 为线段PB 的中点,所以AE PB ⊥, 又PBBC B =,且,PB BC ⊂平面PBC ,所以AE ⊥平面PBC , ············································································ 4分 因为AE ⊂平面AEF ,所以平面AEF ⊥平面PBC . ········································ 6分 (2)因为PA ⊥底面ABCD ,E 为线段PB 的中点,所以点E 到底面ABCD 的距离为12PA =32, ················································ 7分则133393P ABCD V -=⨯⨯⨯=, ····································································· 8分又F 为线段BC 的三等分点,PABCDEF当113BF BC ==时,1133313224E ABF V -⎛⎫=⨯⨯⨯⨯= ⎪⎝⎭,所以多面体PAEFCD 的体积为333944P ABCD E ABF V V ---=-=; ······················· 10分 当223BF BC ==时,1133323222E ABF V -⎛⎫=⨯⨯⨯⨯= ⎪⎝⎭,所以多面体PAEFCD 的体积为315922P ABCD E ABF V V ---=-=. ·························· 11分 综上,多面体PAEFCD 的体积为334或152. ················································ 12分 解法二:(1)平面AEF 与平面PBC 互相垂直, ············································ 1分 理由如下:因为PA ⊥底面ABCD ,PA ⊂平面PAB ,所以平面PAB ⊥底面ABCD , ··············· 2分 又平面PAB 底面ABCD AB =,BC AB ⊥,BC ⊂平面ABCD ,所以BC ⊥平面PAB .············································································· 3分 因为AE ⊂平面PAB ,所以AE BC ⊥, ························································· 3分 因为PA AB =,E 为线段PB 的中点,所以AE PB ⊥, 又PBBC B =,且,PB BC ⊂平面PBC ,所以AE ⊥平面PBC , ············································································ 4分 因为AE ⊂平面AEF ,所以平面AEF ⊥平面PBC . ···································································· 6分 (2)同解法一. ·················································································· 12分 20.本小题考查圆与椭圆的标准方程及其几何性质、直线与圆的位置关系、直线椭圆的位置关系等基础知识,考查推理论证能力、运算求解能力,考查数形结合思想、化归与转化的思想、分类与整合思想等,考查的数学素养主要有逻辑推理、直观想象、数学运算等.满分12分.【解答】解法一:(1)依题意,圆O·····································1分因为椭圆的短轴长等于圆O,所以2b=解得b······················································2分因为C,所以ca=①··················································3分又因为222a c b-=,所以222a c-=, ②联立①②,解得24a=, ··········································································4分所以C的方程为22142x y+=. ···································································5分(2)证明:①当直线l斜率不存在时, 直线l的方程为x=或x=当x=,A B,则4433OA OB⋅=-=,故OA OB⊥.同理可证,当x=,OA OB⊥. ··························································6分②当直线l斜率存在时,设其方程为1122,(,),(,),y kx m A x y B x y=+因为直线l与圆相切,=,即223440m k--=, ·························7分由22,142y kx mx y=+⎧⎪⎨+=⎪⎩得,222(12)4240k x kmx m+++-=, ········································8分所以()()()2222221681228420k m k m k m∆=-+-=-+>,且12221224,1224,12kmx xkmx xk⎧+=-⎪⎪+⎨-⎪=⎪+⎩ ········································································································9分所以1212()()OA OB x x kx m kx m⋅=+++221212(1)()k x x km x x m =++++ ·············································· 10分2222222(1)(24)4(12)12k m k m m k k +--++=+22234412k m k --=+0=, ················································································ 11分 所以.OA OB ⊥综上,.OA OB ⊥ ·················································································· 12分 解法二:(1)同解法一. ········································································· 5分 (2)①当直线方程为y =, (A B ,则 44033OA OB ⋅=-+=,故.OA OB ⊥同理可证,当直线方程为y =,.OA OB ⊥ ··········································· 6分 ②当直线l 不与x 轴平行时,设其方程为1122,(,),(,),x ty m A x y B x y =+ 因为直线l 与圆相切,=,即223440.m t --= ··························· 7分由22,142x ty m x y =+⎧⎪⎨+=⎪⎩得,222(2)240.t y tmy m +++-= ··········································· 8分所以()()()22222244248240t m t m t m ∆=-+-=-+>,且12221222,24.2tm y y t m y y t ⎧+=-⎪⎪+⎨-⎪=⎪+⎩······ 9分 1212()()OA OB y y ty m ty m ⋅=+++221212(1)()t y y tm y y m =++++ ····················································· 10分2222222(1)(4)2(2)2t m t m m t t +--++=+2223442m t t --=+0=, ······················································································· 11分 所以,OA OB ⊥.综上,.OA OB ⊥ ·················································································· 12分 21.本小题主要考查导数及其应用、函数的零点、函数的最值与值域等基础知识,考查推理论证能力、运算求解能力、抽象概括能力等,考查函数与方程思想、化归与转化思想、分类与整合思想、数形结合思想等,考查的数学素养主要有逻辑推理、直观想象、数学运算等.满分12分.【解答】(1)()()x x x a x f sin cos /-=, ························································ 1分 因为⎥⎦⎤⎢⎣⎡∈6,0πx ,所以cos sin 0x x >≥,又10x >≥,所以1cos sin x x x ⋅>,即0sin cos >-x x x . ··················································· 2分当0>a 时,()0/>x f ,所以()f x 在区间π0,6⎡⎤⎢⎥⎣⎦上递增,所以()16312366max -=-⋅⋅=⎪⎭⎫⎝⎛=πππa f x f ,解得2=a .································· 3分 当0<a 时,()0/<x f ,所以()f x 在区间π0,6⎡⎤⎢⎥⎣⎦上递减,所以()()10max -==f x f ,不合题意. ···························································· 4分 当0a =,()1f x =-,不合题意.综上,2=a . ························································································· 5分 (2)设()x x x x g sin cos -=,则()⎪⎭⎫ ⎝⎛<<<--=200cos sin 2/πx x x x x g , ····················································· 6分所以()x g 在⎪⎭⎫ ⎝⎛2,0π上单调递减,又()022,010<-=⎪⎭⎫⎝⎛>=ππg g , ·························· 7分所以存在唯一的⎪⎭⎫⎝⎛∈2,00πx ,使得()00=x g , ················································· 8分当00x x <<时,()0>x g ,即()()02/>=x g x f ,所以()()0,0x x f 在上单调递增;当20π<<x x 时,()0<x g ,即()()02/<=x g x f ,所以()()0,0x x f 在上单调递减, ············ 9分又()012,01424,010<-=⎪⎭⎫⎝⎛>-=⎪⎭⎫ ⎝⎛<-=πππf f f , ······································ 10分 所以()f x 在π0,4⎛⎫ ⎪⎝⎭与ππ,42⎛⎫⎪⎝⎭上各有一个零点, ·········································· 11分综上,函数()x f 在区间⎪⎭⎫⎝⎛2,0π上有且仅有两个零点. ····································· 12分(二)选考题:共10分.请考生在第22,23两题中任选一题作答.如果多做,则按所做第一个题目计分,作答时请用2B 铅笔在答题卡上将所选题号后的方框涂黑.22.本题考查极坐标方程和直角坐标方程的互化,直线和圆的位置关系,以及直线的参数方程的参数的几何意义等基础知识,考查学生的逻辑推理能力,化归与转化能力.考查的核心素养是直观想象、逻辑推理与数学运算.【解答】(1)由2cos ρθ=,得22cos ρρθ=. ············································· 2分 将cos ,sin x y ρθρθ=⎧⎨=⎩代入得,222x y x +=, ·························································· 4分所以C 的直角坐标方程为22(1)1x y -+=. ·················································· 5分(2)设,A B 所对应的参数分别为12,t t ,因为直线l的参数方程为5,(12x t y t⎧=⎪⎪⎨⎪=+⎪⎩为参数 ,所以M 在l 上, ············ 6分把l 的参数方程代入22(1)1x y -+=可得2180,t ++= ····························· 7分。
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保密★启用前泉州市2020届高中毕业班单科质量检查理科数学2020.1注意事项:1.答题前,考生先将自己的姓名、准考证号填写在答题卡上.2.考生作答时,将答案答在答题卡上.请按照题号在各题的答题区域(黑色线框)内作答,超出答题区域书写的答案无效.在草稿纸、试题卷上答题无效.3.选择题答案使用2B 铅笔填涂,如需改动,用橡皮擦干净后,再选涂其它答案标号;非选择题答案使用5.0毫米的黑色中性(签字)笔或碳素笔书写,字体工整、笔迹清楚.4.保持答题卡卡面清洁,不折叠、不破损.考试结束后,将本试卷和答题卡一并交回.三、解答题:共70分.解答应写出文字说明、证明过程或演算步骤.第17~21题为必考题,每个试题考生都必须作答.第22、23题为选考题,考生根据要求作答.(一)必考题:共60分.17.(12分)如图,四棱锥ABCD P -的底面是正方形,⊥PA 平面ABCD ,AE PD ⊥.(1)证明:AE ⊥平面PCD ;(2)若AP AB =,求二面角D PC B --的余弦值.【命题意图】本小题考查线面垂直的判定与性质、二面角的求解及空间向量的坐标运算等基础知识,考查空间想象能力、逻辑推理及运算求解能力,考查化归与转化思想、函数与方程思想等,体现基础性、综合性与应用性,导向对发展数学抽象、逻辑推理、直观想象等核心素养的关注.【试题解析】解法一:(1)因为PA ⊥平面ABCD ,CD ⊂平面ABCD ,所以PA CD ⊥.·········································································································1分又底面ABCD 是正方形,所以AD CD ⊥.·····································································2分又PA AD A = ,所以CD ⊥平面PAD .······································································3分又AE ⊂平面PAD ,所以CD AE ⊥.···········································································4分又因为AE PD ⊥,CD PD D = ,,CD PD ⊂平面PCD ,·············································5分所以AE ⊥平面PCD .·······························································································6分(2)因为PA ⊥平面ABCD ,底面ABCD 为正方形,所以PA AB ⊥,PA AD ⊥,AB AD ⊥,分别以AB 、AD 、AP 所在的直线为x 轴、y 轴、z 轴建立空间直角坐标系A xyz -(如图所示).······································································7分设1PA AB ==,则A 0,0,0(),B 1,0,0(),C 1,1,0(),D 0,1,0(),(0,0,1)P ,11(0,,)22E ,1,0,1PB =- (),1,1,1PC =- (),11(0,,22AE = .··························································8分由(1)得11(0,,)22AE = 为平面PCD 的一个法向量.·······················································9分设平面PBC 的一个法向量为111()m x ,y ,z = .由0,0,PB m PC m ⎧⋅=⎪⎨⋅=⎪⎩ 得111110,0,x z x y z -=⎧⎨+-=⎩令11x =,解得11z =,10y =.所以(1,0,1)m = .·····································································································10分因此112cos ,2m AE m AE m AE ⋅===⋅ .·······························································11分由图可知二面角B PC D --的大小为钝角.故二面角B PC D --的余弦值为12-.·········································································12分解法二:(1)同解法一.·····································································································6分(2)过点B 作BF 垂直于PC 于点F ,连接DF 、BD .因为PB PD =,BC CD =,PC PC =,所以PBC PDC △≌△.······························································································7分因此易得090DFC BFC ∠=∠=,BF DF =.································································8分所以BFD ∠为二面角B PC D --的平面角.···································································9分设1PA AB ==,则BD =3BF DF ==.·························································10分在BDF △中,由余弦定理,得222222)133cos 2263BF DF BD BFD BF DF +-+-∠==-⋅.故二面角B PC D --的余弦值为12-.·········································································12分18.(12分)记n S 为数列{}n a 的前n 项和.已知0n a >,2634n n n S a a =+-.(1)求{}n a 的通项公式;(2)设2211n n n n n a a b a a +++=,求数列{}n b 的前n 项和n T .【命题意图】本小题主要考查递推数列、等差数列的通项公式与数列求和等基础知识,考查推理论证能力与运算求解能力等,考查化归与转化思想、特殊与一般思想等,体现基础性,导向对发展逻辑推理、数学运算等核心素养的关注.【试题解析】解:(1)当1n =时,2111634S a a =+-,所以14a =或1-(不合,舍去).································1分因为2634n n n S a a =+-①,所以当2≥n 时,2111634n n n S a a ---=+-②,由①-②得2211633n n n n n a a a a a --=+--,······································································2分所以()()1130n n n n a a a a --+--=.················································································3分又0n a >,所以13n n a a --=.······················································································4分因此{}n a 是首项为4,公差为3的等差数列.···································································5分故()43131n a n n =+-=+.························································································6分(2)由(1)得()()()()22313433231343134n n n b n n n n +++==+-++++,········································9分所以()33333392()2477103134434n n T n n n n n =+-+-+⋅⋅⋅+-=++++.····························12分19.(12分)ABC △中,60B =︒,2AB =,ABC △的面积为(1)求AC ;(2)若D 为BC 的中点,,E F 分别为,AB AC 边上的点(不包括端点),且120EDF ∠=︒,求DEF △面积的最小值.【命题意图】本小题主要考查解三角形、三角恒等变换等基础知识,考查推理论证能力和运算求解能力等,考查数形结合思想和化归与转化思想等,体现综合性与应用性,导向对发展直观想象、逻辑推理、数学运算及数学建模等核心素养的关注.【试题解析】解法一:(1)因为60B =︒,2AB =,所以1sin 2ABC S AB BC B =⋅⋅⋅△13222BC =⨯⨯32BC =,·············································2分又ABC S =△,所以4BC =.···················································································3分由余弦定理,得2222cos AC AB BC AB BC B =+-⋅⋅·······················································4分221242242=+-⨯⨯⨯12=,·························································································5分所以AC =········································································································6分(2)设BDE θ∠=,[]0,60θ∈︒︒,则60CDF θ∠=︒-.在BDE △中,由正弦定理,得sin sin BD DE BED B=∠,·························································7分即2sin(60)32θ=︒+,所以3sin(60)DE θ=︒+;···························································8分在CDF △中,由正弦定理,得sin sin CD DF CFD C =∠,由(1)可得30C =︒,即21sin(90)2DF θ=︒-,所以1cos DF θ=;·····································9分所以1sin 2DEF S DE DF EDF =⋅⋅⋅∠△34sin(60)cos θθ=︒+⋅=········································································10分=,············································································11分当15θ=︒时,sin(260)1θ+︒=,min ()6DEF S ==-△故DEF △面积的最小值为6-.············································································12分解法二:(1)同解法一.·····································································································6分(2)设CDF θ∠=,[]0,60θ∈︒︒,则60BDE θ∠=︒-.在CDF △中,由正弦定理,得sin sin CD DFCFD C=∠,························································7分由(1)可得30C =︒,即21sin(30)2DFθ=︒+,所以()1sin 30DF θ=︒+;···························8分在BDE △中,由正弦定理,得sin sin BD DEBED B=∠,即2sin(120)32θ=︒-,所以sin(120)DE θ=︒-;·························································9分所以1sin 2DEF S DE DF EDF =⋅⋅⋅∠△()334sin 30sin(120)θθ=⋅︒+⋅︒-13312222=⎝⎭⎝⎭ (10)分=······················································································11分当45θ=︒时,sin 21θ=,min ()6DEF S ==-△故DEF △面积的最小值为6-.············································································12分20.(12分)已知椭圆2222:1(0)x y E a b a b+=>>的离心率为12,点32A 在E 上.(1)求E 的方程;(2)斜率不为0的直线l 经过点1(,0)2B ,且与E 交于Q P ,两点,试问:是否存在定点C ,使得QCB PCB ∠=∠?若存在,求C 的坐标;若不存在,请说明理由.【命题意图】本小题主要考查椭圆的几何性质、直线与椭圆的位置关系等基础知识,考查推理论证能力、运算求解能力等,考查化归与转化思想、数形结合思想、函数与方程思想等,体现基础性、综合性与创新性,导向对发展逻辑推理、直观想象、数学运算、数学建模等核心素养的关注.【试题解析】解法一:(1)因为椭圆E的离心率12e a ==,所以2234a b =①,··································1分点)23,3(A 在椭圆上,所以143322=+ba ②,·······························································2分由①②解得42=a ,32=b .························································································3分故E 的方程为13422=+y x .··························································································4分(2)假设存在定点C ,使得PCB QCB ∠=∠.由对称性可知,点C 必在x 轴上,故可设(,0)C m .··························································5分因为PCB QCB ∠=∠,所以直线PC 与直线QC 的倾斜角互补,因此0PC QC k k +=.·············6分设直线l 的方程为:21+=ty x ,),(11y x P ,),(22y x Q .由221,2143x ty x y ⎧=+⎪⎪⎨⎪+=⎪⎩消去x ,得04512)1612(22=-++ty y t ,···············································7分2222(12)4(1216)(45)144180(1216)0t t t t ∆=-⨯+⨯-=+⨯+>,所以t ∈R ,122121216t y y t +=-+,122451216y y t =-+,····································································8分因为0=+QC PC k k ,所以02211=-+-mx y m x y ,所以0)()(1221=-+-m x y m x y ,即0)21()21(1221=-++-+m ty y m ty y .·························9分整理得121212()()02ty y m y y +-+=,所以0161212)21()161245(222=+-⨯-++-⨯t t m t t ,即01612)12)(21(902=+--+-t t m t .·················10分所以0)21(1290=-+m t t ,即0)]21(1290[=-+t m ,对t ∈R 恒成立,即0)1296(=-t m 对t ∈R 恒成立,所以8=m .·····························································11分所以存在定点)0,8(C ,使得QCB PCB ∠=∠.·······························································12分解法二:(1)同解法一.·····································································································4分(2)若点C 存在,当直线PQ 垂直x 轴时,点C 必在x 轴上,如果直线PQ 不垂直x 轴,由对称性可知,点C 也必在x 轴上.···········································5分假设存在点)0,(m C ,使得QCB PCB ∠=∠,即直线PC 与直线QC 的倾斜角互补,所以0=+QC PC k k .····································································································6分设直线l 的方程为)21(-=x k y ,),(11y x P ,),(22y x Q .由221(2143y k x x y ⎧=-⎪⎪⎨⎪+=⎪⎩消去x ,得0124)34(2222=-+-+k x k x k ,··········································7分22222(4)4(43)(12)1801440k k k k ∆=--⨯+-=+>,所以k ∈R ,2122443k x x k +=+,34122221+-=k k x x ,··············································································8分因为0=+QC PC k k ,所以02211=-+-m x y m x y ,所以0)()(1221=-+-m x y m x y ,················9分即122111()()()022k x x m k x x m --+--=.整理得0]))(21(2[2121=+++-m x x m x x k ,··································································10分所以0]34421(34242[2222=++⨯+-+-m k k m k k k ,整理得0342432=+-⨯k m k ,对任意的k ∈R 恒成立,····························································11分所以8=m ,故存在x 轴上的定点)0,8(C ,使得QCB PCB ∠=∠.····································12分21.(12分)已知函数()2()1e xf x x ax =++.(1)讨论()f x 的单调性;(2)若函数()2()1e 1xg x x mx =+--在[)1,-+∞有两个零点,求m 的取值范围.【命题意图】本小题主要考查导数的综合应用,利用导数研究函数的单调性、最值和零点等问题,考查抽象概括、推理论证、运算求解能力,考查应用意识与创新意识,综合考查化归与转化思想、分类与整合思想、函数与方程思想、数形结合思想、有限与无限思想以及特殊与一般思想,考查数学抽象、逻辑推理、直观想象、数学运算、数学建模等核心素养.【试题解析】。