09高二上期末考试

2022-2023学年江苏省泰州市高二上学期期末数学试题一、单选题1．已知集合，则（ ）{21},{12}A x x B x x =-<≤=-<≤∣∣A B = A ．B ．C ．D ．(]1,1-(2,2]-(2,1]-(1,2]-【答案】A【分析】根据给定条件，利用交集的定义直接求解作答.【详解】集合，所以.{21},{12}A x x B x x =-<≤=-<≤∣∣(1,1]A B ⋂=-故选：A 2．复数，则（ ）1i1i z +=-3z =A ．B ．C ．-1D ．1i -i【答案】A【分析】直接利用复数代数形式的乘除运算化简，然后利用即可求出结果．2i 1=-【详解】解：，21i (1i)i 1i (1i)(1i)z ++===--+ ，33i i z ∴==-故选：A ．3．已知点，，若直线与直线垂直，则（ ）()1,0A ()3,1B AB 10x my -+=m =A ．B ．C ．D ．2-12-122【答案】B【分析】先求出直线的斜率，再根据两直线垂直斜率乘积为即可求的值．AB 1-m 【详解】依题意可得直线的斜率为，AB 101312-=-因为直线与直线垂直，AB 10x my -+=且直线的斜率为，10x my -+=2-所以，解得．12m =-12m =-故选：B ．4．数学家斐波那契在研究兔子繁殖问题时，发现有这样一个数列，，，，，，{}:1n a 12358其中从第项起，每一项都等于它前面两项之和，即，，这样的数列称 3121a a ==21n n n a a a ++=+为“斐波那契数列”若，则（ ）.()36912621m a a a a a =+++++m =A ．B ．C ．D ．126127128129【答案】C【分析】根据数列的特点，每个数等于它前面两个数的和，移项得： ，使用累加法2n n a a +=-1n a +求得，然后将的系数倍展开即可求解．21n n S a +=-()36912621a a a a +++++2【详解】由从第三项起，每个数等于它前面两个数的和，，121a a ==由，得 ，所以，，，()*21n n n a a a n ++=+∈N 2n n a a +=-1n a +132a a a =-243a a a =-⋯2n n a a +=-，1n a +将这个式子左右两边分别相加可得：，所以.n 1221n n n S a a a a +=+++=-21n n S a ++=所以.()3691261234567891241251261261282111a a a a a a a a a a a a a a a a S a +++++=++++++++++++=+=故选：C ．5．已知双曲线的焦点在轴上，渐近线方程为，则的离心率为（ ）C y 2y x =±C AB ．CD2【答案】A【分析】根据已知条件求得，从而求得双曲线的离心率.ab 【详解】由题意，双曲线的焦点在轴上，y 由于双曲线的渐近线方程为，2y x =±所以，即，2ab =12ba =所以c e a =====故选：A6．已知函数的导函数为，且，则（ ）()f x ()'f x ()2cos 6f x xf xπ⎛⎫'=+ ⎪⎝⎭6f π⎛⎫=⎪⎝⎭A ．B ．CD12-126π6π【答案】D【分析】将求导并代入即可得出，即可得到的具体解析式，再代入()f x 6x π=6f π⎛⎫' ⎪⎝⎭()f x 即可得出答案.6x π=【详解】，()2cos 6f x xf xπ⎛⎫'=+ ⎪⎝⎭ ，()2sin 6f x f xπ⎛⎫''∴=- ⎪⎝⎭令，则，6x π=2sin666f f πππ⎛⎫⎛⎫''=- ⎪ ⎪⎝⎭⎝⎭，162f π⎛⎫'=⎪⎭∴⎝则，()cos f x x x=+ cos 6666f ππππ⎛⎫=+= ⎪⎭⎝∴故选：D.7．已知等差数列中， 记，，则数列的前项和为（ ）{}n a 452a a +=11n n n a b a +=-*N n ∈{}n b 8A ．B ．C ．D ．04816【答案】C【分析】分离常数可得，设，当，时，可得，故21,1n n b a =+-21n n c a =-18n ≤≤*N n ∈90n n c c -+=可得数列的前项和．{}n b 8【详解】由等差数列性质得945n n a a a a -++＝11221,111n n n n n n a a b a a a +-+===+---设，当，时，21n n c a =-18n ≤≤*N n ∈()()()()94599992222220,111111n n n n n n n n n n a a a a c c a a a a a a -----+-+-+=+=⋅=⋅=------故1238b b b b ++++1281282221118111c c c a a a =++++++=++++---()()()()1827364588c c c c c c c c =++++++++=故选：C 8．已知函数及其导函数的定义域均为，且是奇函数，记，若()f x ()f x 'R ()1f x +()()g x f x '=是奇函数，则（ ）()g x ()10g =A ．B ．C ．D ．201-2-【答案】B 【分析】根据是奇函数，可得，两边求导推得，()1f x +()()11f x f x -+=-+()()2g x g x =-+，再结合题意可得4是函数的一个周期，且，进而可求解．()()20g g =()g x ()00g =【详解】因为 是奇函数，所以，()1f x +()()11f x f x -+=-+两边求导得 ，()()11f x f x ''--+=-+即，()()11f x f x ''-+=+又，()()g x f x '=所以，即，()()11g x g x -+=+()()2g x g x =-+令 ，可得 ，2x =()()20g g =因为是定义域为的奇函数，所以，即．()g x R ()00g =()20g =因为是奇函数，()g x 所以 ，又，()()g x g x -=-()()2g x g x =-+所以，则，，()()2g x g x -+=--()()2g x g x +=-()()()42g x g x g x +=-+=所以4是函数的一个周期，()g x 所以．()()1020g g ==故选：B ．二、多选题9．已知圆，点，，则（ ）:C ()()225516x y -+-=()4,0A ()0,2B A ．点在圆外B ．直线与圆相切A C 1x =C C ．直线与圆相切D ．圆与圆相离AB C 2249x y +=C 【答案】AB【分析】根据已知写出圆心、半径.代入点坐标，即可判断A 项；分别求出圆心到直线的距离，A 比较它们与半径的关系，即可判断B 、C 项；求出圆心距，根据与两圆半径的关系即可判OCOC断D 项.【详解】解：由题，圆的圆心坐标为，半径为，C ()5,5C 4r =对于A 项，因为，所以点在圆外，故A 正确；()()2245052616-+-=>A C 对于B 项，圆心到直线的距离为，故直线与圆相切，故B 项正确；1x =1514d r=-==1x =C 对于C 项，直线的方程为，整理得，则圆心到直线的距离为AB 142x y+=240x y +-=C AB，24d r >=所以直线与圆相离，故C 错误；AB C 对于D 项，圆的圆心坐标为，半径为，则圆心间的距离为2249x y +=()0,0O 7R ==因为，所以圆与圆相交，故D 错误.310R r R r -=<<=+2249x y +=C 故选：AB ．10．已知等差数列的前项和为，当且仅当时取得最大值，则满足的最大的{}n a n n S 7n =n S 0k S >正整数可能为（ ）k A ．B ．C ．D ．12131415【答案】BC 【分析】由题意可得，公差，且，，分别求出，讨论的10a >0d <70a >80a <131415S S S ，，78a a +符号即可求解．【详解】因为当且仅当时，取得最大值，7n =n S 所以，公差，且，．10a >0d <70a >80a <所以，，，()113137131302a a S a ⨯+==>()()11414781472a a S a a ⨯+==+()115158151502a a S a ⨯+==<故时，．15n ≥0n S <当时，，则满足的最大的正整数为；780a a +>140S >0k S >k 14当时，，则满足的最大的正整数为，780a a +≤140S ≤0k S >k 13故满足的最大的正整数可能为与．0k S >k 1314故选：BC ．11．已知抛物线的焦点为，为上一动点，点，则（ ）2:4C y x =F ()00P x y ，C ()21A ，A ．当时，02x =3PF =B ．当时，在点处的切线方程为01y =C P 2210x y -+=C ．的最小值为PA PF+3D ．的最大值为PA PF-【答案】ACD 【分析】当时，求出判断A ；02x =PF 设切线与抛物线联立使求出切线方程判断B ；Δ0=利用抛物线的定义转化求解的最小值可判断C ；PA PF+根据三角形两边之差小于第三边判断D ．【详解】因为抛物线，所以准线的方程是.2:4C y x =l =1x -对于，当时，，此时，故A 正确;A 02x =24p =0||2132pPF x =+=+=对于B ，当时，，令切线方程为:，与联立得01y =014x =1(1)4m y x -=-24y x =2y -，4410my m +-=令，解得，即切线方程为:，即，故B 错2161640m m ∆=-+=12m =11(1)24y x -=-4210x y -+=误;对于C ，过点分别作准线的垂线，垂足为,P A l ,,Q B则，所以的最小值为故C 正确.||||||||||3PA PF PA PQ AB +=+≥=||||PA PF +3,对于D ，因为焦点，所以(1,0)F ||||||PA PF AF -≤==所以故D 正确.||||PA PF -故选：ACD12．已知 ，则（ ）22e e x yx y --<-A ．B ． ()ln 10x y ++<2()1e x yx y +++<C ．D ．sin sin x y x y +>--22cos cos x y y x ->-【答案】BC【分析】根据条件构造函数，求导，计算出x 与y 的关系，再根据函数的性质逐项分析.【详解】因为 ，即 ．22e e x y x y --<-()22e e x y x y --<--令 ，则有，()2e xf x x =-()()f x f y <-则 ，令，则，()'2e x f x x =-()2e xg x x =-()'2e xg x =-令 ，可得，()'2e 0x g x =-=ln2x =当时，，函数单调递增，()ln2x ∈-∞，()'0g x >()g x 当时，，函数单调递减，()ln2x ∈+∞，()'0g x <()g x 故，()()max ln22ln220g x g ==-<所以总有 ，故单调递减；所以，即；()'0f x <()f x x y >-0x y +>对于A ，，故A 错误；()ln 1ln10x y ++>=对于B ，设 ，则，()()2e 10x h x x x =-->()()''e 20x h x x f x =-=->故在上单调递增，所以，()h x ()0+∞，()()00h x h >=所以，因为，所以，故B 正确；()21e 0x x x +<>0x y +>()21ex yx y +++<对于C ，，即．sin sin x y x y +>--()()sin sin x x y y +>-+-设，则，()sin u x x x=+()()u x u y >-则，所以单调递增．()1cos 0u x x ='+≥()u x因为，所以，故C 正确；x y >-()()u x u y >-对于D ，，即，22cos cos x y y x ->-22cos cos x x y y +>+令，则，()2cos t x x x=+()()t x t y >因为，所以为偶函数，()()()()22cos cos t x x x x x t x -=-+-=+=()2cos t x x x=+所以即为．()()t x t y >()()t x t y >-则 ，令，则，所以单调递增．()'2sin t x x x=-()2sin m x x x =-()'2cos 0m x x =->()m x 又，()00m =所以当时，，，函数单调递减；()0x ∈-∞，()0m x <()'0t x <()t x 当时，，，函数单调递增，()0x ∈+∞，()0m x >()'0t x >()t x 当时，，故D 错误；0y x -<<()()t y t x ->故选：BC.三、填空题13．已知等比数列的公比不为，，且，，成等差数列，则__________．{}n a 111a =2a 4a 3a 5a =【答案】/0.0625116【分析】根据条件求出公比q ，再运用等比数列通项公式求出 .5a 【详解】根据题意得，， 且，32420a a a +-=2311120a q a q a q ∴+-=2320q q q +-=1q ≠解得，，；12q =-11a = 445111216a a q ⎛⎫∴==-= ⎪⎝⎭故答案为： .11614．已知点，，点满足直线，的斜率之积为，则的面积的最大()50A -，()50B ，P PA PB 1625-PAB 值为__________．【答案】20【分析】根据条件，运用斜率公式求出P 点的轨迹方程，再根据轨迹确定 面积的最大值.PAB 【详解】设，由题意可知，，()P m n ，2216552525PA PBn n n k k m m m ⋅=⋅==-+--整理得；()22152516m n m +=≠±得动点的轨迹为以，为长轴顶点的椭圆除去，两点，P A B (A B )显然当点位于上下顶点时面积取得最大值，P PAB 因为，，5a =4b =所以；()max 12202PAB S a b =⨯⨯= 故答案为：20.15．已知函数及其导函数的定义域均为，为奇函数，且则不等()f x ()f x 'R ()f x ()()0.f x f x '->式的解集为__________．()2320f x x -+>【答案】()1,2【分析】设，由导数法可得单调递减，可转化为()()x f x g x =e ()g x ()2320f x x -+>，根据单调性即可求解．()()2320g x x g -+>【详解】设，则，故单调递减．()()x f x g x =e ()()()0xf x f xg x e '-'=<()g x 因为为奇函数，定义域为，所以，故．()f x R ()00f =()()0000e f g ==可转化为，即．()2320f x x -+>()223232ex x f x x -+-+>()()2320g x x g -+>因为单调递减，所以，解得．()g x 2320x x -+<12x <<故答案为：．()1,216．已知实数，，，满足，，，则1x 2x 1y 2y 22114x y +=22229x y +=12120x x y y +=的最大值是___________．112249x y x y +-++-【答案】1313+【分析】由已知得分别在圆和圆上，利用数形结合法，将所求问题转化,A B 224x y +=229x y +=为两点到直线和倍，再利用三角函数求出其最大值即,A B 40x y +-=90x y +-=可．【详解】解：由，可知，22114x y +=22229x y +=点，分别在圆和圆上，()11,A x y ()22,B x y 224x y +=229x y +=如图，作直线，过作于，过A 作于，:l y x =-B BD l ⊥D AE l ⊥E而，1122|4||9|x y x y +-++-表示A 到直线的距离，40x y +-=1d表示到直线的距离，B 90x y +-=2d 因为与，平行，y x =-40x y +-=90x y +-=且与的距离为，y x =-40x y +-=3d与的距离为y x =-90x y +-=4d 要使的取最大值，则需在直线的左下角这一侧，112249x y x y +-++-,A B :l y x =-所以1||d AE =+2||d BD =由得，12120x x y y +=OA OB ⊥设，因为，所以，π,0,2DOB θθ⎛⎫∠=∈ ⎪⎝⎭OA OB ⊥π2AOE θ∠=-从而，π||||sin 3sin ,||||sin 2cos 2BD BO AE AO θθθθ⎛⎫=⋅==⋅-= ⎪⎝⎭故，()||||3sin 2cos BD AE θθθθθϕ⎫+=+=+⎪⎭其中，π20,,tan 23ϕϕ⎛⎫∈=⎪⎝⎭故当时，π2θϕ=-||||BD AE +从而，)1122124913x y x y d d +-++-=+=≤即．1122|4||9|x y x y +-++-13．13【点睛】关键点睛：本题解答的关键是将代数问题转化为几何问题，数形结合，再借助三角函数的性质求出最值.四、解答题17．已知中，．ABC )222sin sin sin 2sin sin cos A B C A BC +-=-(1)求；C (2)若，，求的面积．45A =2BC =ABC 【答案】(1)π6【分析】由正弦定理得，再由余弦定理得，可得()1)2222cos a b c abC +-=cos cos C C =；cos C =C 由正弦定理得，得出，再得出，由三角形面积公式可得的面积．()22sin45sin30BA= BA sin B ABC 【详解】（1）设，，对边长，，A B C a b c因为)222sin sin sin 2sin sin cos A B C A BC +-=由正弦定理，2sin sin sin a b cR A B C ===所以，)2222cos a b c abC +-=所以，222cos 2a b c Cab +-=即，cos cos C C =-所以，cos C =因为，()0,πC ∈所以；π6C =（2）中，，，，ABC 45A =2BC =30C =因为，sin sin BC BAA C =所以，2sin45sin30BA=所以，BA =因为，()sin sin sin45cos30cos45sin30B A C =+=+= 所以1sin 2ABC S BA BC B =⋅⋅122=⨯．=18．已知数列中，，当时，记，．{}n a 15a =2n ≥1221.nn n a a -=+-12n n n a b -=*n N ∈(1)求证：数列是等差数列，并求数列的通项公式{}n b {}n a ;(2)求数列的前项和．{}1n a -n n T 【答案】(1)证明见解析，()121n n a n =++(2)．12n n T n +=⋅【分析】（1）对递推公式变形，求出 的通项公式，再求出 的通项公式；{}n b {}n a （2）运用错位相减法求和.【详解】（1）因为且当时，，15a =2n ≥1221nn n a a -=+-所以当时，，2n ≥()11212nn n a a --=-+所以，因为，即，1111122n n nn a a ----=+12n n n a b -=11n n b b --=所以是以为首项，为公差的等差数列，{}n b 11122a b -==1所以，()121112n na n n -=+-⨯=+所以；()121n n a n =++（2）由知，()2()1()112n n a n -=+则…①…②，()12223212nn T n =⨯+⨯+++⨯ ()2312223212n n T n +=⨯+⨯+++⨯ ①-②得()12312222212n n n T n +-=⨯++++-+⨯ 所以；()()1141241212n n n -+-=+-+-()111442122n n n n T n n +++=-+-++=⋅综上，， .()121n n a n =++12n n T n += 19．已知函数．()()ln 1=1x x f x x +--(1)求函数的最大值；()f x (2)记，．若函数既有极大值，又有极小值，求()()()()21211g x x f x x a x =+++--+a ∈R ()g x 的取值范围．a 【答案】(1)2(2)()+∞【分析】（1）对函数求导，研究函数的单调性，从而可得函数的最值；（2）条件等价于方程在区间上有两个不相等的实数根，列关于()22430x a x a +-+-=()1,-+∞的不等式，求解即可．a 【详解】（1）由函数，则其定义域为，，()()ln 1=1x x f x x +--()1,∞+()()()2ln 1=1x f x x '---当时，；当时，，()1,2x ∈()0f x ¢>()2,x ∈+∞()0f x '<所以函数在区间上为增函数；在区间为减函数，()f x ()1,2()2,+∞所以；()()max 22f x f ==（2）由，()()()()()()221211ln 123g x x f x x a x x x a x =+++--+=++--+(1)x >-则，()()()224312211x a x a g x x a x x +-+-=+--='++因为既有极大值，又有极小值，()g x 即等价于方程在区间上有两个不相等的实数根，()22430x a x a +-+-=()1,-+∞即，解得()()()22430414Δ4830a a a a a ⎧--+->⎪-⎪>-⎨⎪⎪=--->⎩a >所以所求实数的取值范围是．a ()+∞20．设数列的前项积为，且．{}n a n nT22nn T =(1)求数列的通项公式{}n a ;(2)记区间内整数的个数为，数列的前项和为，求使得的最[]()*1,m m a a m N +∈m b {}m b m m S 2023m S >小正整数．m 【答案】(1)212n n a -=(2)5【分析】（1）根据与的关系，类比与的关系求通项即可；n T n a n S n a （2）根据定义求出的通项，再由公式法求和，最后解不等式即可.m b 【详解】（1）因为数列的前项积，{}n a n 22n n T =①当时，，1n =12a =当时，，2n ≥2(1)1212n n a a a --⋅=②除以得，①②212n n a -=又时，满足，1n =12a =212n na -=所以.212n n a -=（2）因为区间内整数的个数为，[]()*1,m m a a m N +∈m b 所以，212121221321m m m m b +--=-+=⋅+所以.()214324214m mm S m m -=⨯+=⨯+--由，得，即，2023m S >2422023m m ⨯+->242025m m ⨯+>当时，，4m =424451245162025⨯+=+=<当时，，5m =52452048520532025⨯+=+=>因为随的增大而增大，24mm ⨯+m 所以的最小整数为．2023m S >521．已知椭圆，，左、右顶点分2222:1(0)x y C a b a b +=>>1F 2F 别为，，上顶点为，的周长为点，异于两点且在上，直线，1A 2A B 12BF F △ 4.P Q 12,A A C 1A P ，的斜率分别为，，，且2A Q 2A P 1k 2k3k 123k k =(1)证明为定值13k k ;(2)求点到直线距离的最大值．B PQ 【答案】(1)证明见解析【分析】（1）利用题意得到关于的等式，联立方程组即可求得，设，代入椭圆,,a b c ,,a b c ()11,P x y 方程可得到，然后利用两点斜率公式即可求证；221114x y +=（2）先推断出直线斜率必不为，设其方程为，与椭圆进行联立得到二次方PQ 0()2x ty n n =+≠±程，可得到代入即可算出答案12221222444tn y y t n y y t ⎧+=-⎪⎪+⎨-⎪=⎪+⎩，()()1223121212122y y kk x x ⋅-==--【详解】（1）设椭圆焦距为，2c 由题知，解得，222224c a a c a b c ⎧=⎪⎪⎪+=⎨⎪=+⎪⎪⎩21a b c ⎧=⎪=⎨⎪=⎩所以椭圆的标准方程为，2214x y +=依题意，，设椭圆上任一点，则，()12,0A -()22,0A ()11,P x y 221114x y +=所以；21211113221111114·22444x y y y k k x x x x -⋅====-+---（2）设，若直线的斜率为，则，关于轴对称，必有，不合题意，()22,Q x y PQ 0P Q y 12k k =-所以直线斜率必不为，设其方程为，PQ 0()2x ty n n =+≠±与椭圆联立，整理得：，C 2244x y x ty n⎧+=⎨=+⎩()2224240t y tny n +++-=所以，且()221640t n ∆=+->12221222444tn y y t n y y t ⎧+=-⎪⎪+⎨-⎪=⎪+⎩，由（1）知，即，123134k k k =-=23121k k ⋅=-即，即，()()121212122y y x x =---()()()121212121224y y ty n ty n t y y n =-⎡⎤++-+++⎣⎦即，()()122212121212(2)y y t y y t n y y n =-+-++-即，()()()()22212212224n t n t n n t +=-+-+-+所以，此时，1n =-()()2221641630t n t ∆=+-=+>故直线恒过轴上一定点，:1PQ x ty =-x ()1,0D -所以点到直线B PQ 【点睛】方法点睛：利用韦达定理法解决直线与圆锥曲线相交问题的基本步骤如下：（1）设直线方程，设交点坐标为；()()1122,,,x y x y （2）联立直线与圆锥曲线的方程，得到关于（或）的一元二次方程，必要时计算；x y ∆（3）列出韦达定理；（4）将所求问题或题中的关系转化为、（或、）的形式；12x x +12x x 12y y +12y y （5）代入韦达定理求解.22．已知函数，其中，()ln f x a x bx=-a .b R ∈(1)若，求函数的单调区间1a =()f x ;(2)若，函数有两个相异的零点，，求证：．1b =()f x 1x 2x 212e x x >【答案】(1)答案见解析(2)证明见解析【分析】（1）求出函数的导数，解关于导函数的不等式，求出函数的单调区间即可；（2）不妨令，用分析法对进行等价转化，最后可构造函数即可证明结论成立．120x x >>212e x x >【详解】（1）当时，，定义域为，1a =()ln f x x bx=-()0+∞，所以，，()11bx f x b x x -'=-=所以，时，在上恒成立，0b ≤()0f x '≥()0+∞，故在上单调递增，()f x ()0+∞，当时，令得，0b >()0f x '=1x b =所以，当时，，单调递增，11x b ⎛⎫∈ ⎪⎝⎭，()0f x ¢>()f x 时，，单调递减，1x b ⎛⎫∈+∞ ⎪⎝⎭，()'0f x <()f x 综上，时，在上单调递增，0b ≤()f x ()0+∞，时，在上单调递增，在上单调递减；0b >()f x 10b ⎛⎫ ⎪⎝⎭，1b ⎛⎫+∞⎪⎝⎭，（2）由题知，，1b =()ln f x a x x=-因为函数有两个相异零点，，且，()y f x =1x 2x ()11f =-所以且，，即，11x ≠21x ≠1122ln 0ln 0a x x a x x -=⎧⎨-=⎩1122ln ln x a x x a x ⎧=⎪⎪⎨⎪=⎪⎩所以，方程有两个不相等的实数根，ln xa x =令，则，()ln x g x x =()2ln 1(ln )x g x x -'=故当时，，时，，()()011e x ∈⋃，，()0g x '<()e x ∈+∞，()0g x '>所以，在，上单调递减，在上单调递增，()g x ()01，()1e ，()e ，+∞因为，，，，()01x ∈，()0g x <()1x ∈+∞，()0g x >所以，要使方程有两个不相等的实数根，ln xa x =则，()e ea g >=不妨令，则，，120x x >>11ln a x x =22ln a x x =所以，()()1212121212ln ln ln ln ln a x x a x x x x a x x x x +==+-=-，要证，只需证，即证：，212ex x >12ln 2x x >1212ln 2x x x x a +=>因为，1212ln ln 1x x a x x -=-所以，只需证，()121211212122ln ln ln 2x x x x x x x x x x x x -++=>--只需证，即，1211221ln 21x x xx x x +>-1112221ln 21x x x x x x ⎛⎫⎛⎫+>- ⎪ ⎪⎝⎭⎝⎭故令，121x t x =>故只需证，成立，()()1ln 21t t t +>-1t >令，，()()()1ln 21h t t t t =+--1t >则，()11ln 2ln 1+'=+-=+-t h t t t t t 令，()1ln 1g t t t =+-在恒成立，()221110t t t g t t '-=-=>()1+∞，所以，在上单调递增，()h t '()1+∞，因为，()10h '=所以在恒成立，()0h t '>()1+∞，所以，在上单调递增，()h t ()1+∞，所以，，即，()()10h t h >=()()1ln 21t t t +>-所以，成立．212e x x >【点睛】思路点睛：本题第二问令，用分析法对进行等价转化，最后可构造函数120x x >>212e x x >即可证明结，本题考查了函数的零点、应用导数研究函数的单调性、最值，对于恒成立问题往往转化为函数最值解决，属于难题．。

海拉尔第二中学高二年级阶段考试历史试题2009-10-09一、单项选择题（共35题，每题2分，共70分）1、关于迪亚士和麦哲伦远洋航行的相同点的表述，正确的是A．使欧洲与非洲的联系加强B．都因为干预当地内政而被杀C．获得了葡萄牙王室的支持D．航线先后经过大西洋、太平洋2、新航路的开辟使欧洲商业发生了重大变化，其主要表现有①大西洋沿岸成为世界贸易中心②葡萄牙和西班牙资本主义经济发展③世界贸易量大增④意大利商业地位得到提高A．①③B．①②④C．①②D．①②③3、新航路开辟后，世界各民族的历史逐渐融合为一部统一的人类历史。

这种“融合”和“统一”实现的主要方式是A．商品经济的发展B．东方的借鉴学习C．正常的文化交流D．西方的殖民掠夺4、1524年，西班牙人说：“以前我们在世界的边缘，现在在它的中央了，这给我们的命运带来了前所未有的改变。

”对此理解正确的是A．西班牙已经成为世界商贸中心B．西班牙走上发展资本主义的道路C．西班牙成为殖民大帝国，掠得巨额财富D．荷兰打败西班牙，掌握世界殖民霸权5、人文主义追求个性解放，推动了资本主义的发展，但对社会道德规范的不利影响是A．其所倡导的冒险精神，不利于社会的稳定B．倡导个性自由，不利于建立法制社会C．主张建立君主立宪政体D．个人私欲的膨胀、泛滥和社会混乱6、文艺复兴时期，自然科学的兴起，为社会进步提供了巨大的科技动力。

下列对文艺复兴时期自然科学的叙述，错误的是A．布鲁诺和弗兰西斯·培根提倡唯物主义研究方法B．哥白尼确立了“太阳中心说”C．开普勒提出行星沿椭圆形轨道绕太阳运行D．伽利略提出宇宙无固定中心说7、下列关于都铎王朝统治的评价最全面、最正确的是A．加强了封建专制统治，激化了与资产阶级的矛盾B．促进了资本主义经济的发展C．加强了专制统治，但客观上有利于资本主义的发展D．竭力削弱旧贵族的实力，发展资本主义8、英国资产阶级革命时期，有一位激进人物说：“国家本来可以在一个短时期达到人类所向往的幸福，然而却由于他（克伦威尔）一个人的野心而使所有善良的人所有希望破灭了。

2009——2010第一学期期末考试考情分析莱州市高级职业学校高中部2010年3月09级教师成绩比较表（表一）分析：一、在本次全市期末统考中，09级取得了喜人的成绩，有多位老师，邱培桐、尹红红、张军、原树峰、刘洪涛、高月丽、张巧玲等等，所授课程的成绩高于兄弟院校，其中英语和数学的平均成绩远远高出我们的兄弟院校成绩，（数学平均成绩高于兄弟院校11.4分；英语平均成绩高于兄弟院校10.5分），政治和地理的成绩也比较突出（政治平均成绩超7.9分，地理平均成绩超8.0分），这些优异成绩的取得和各科老师的辛勤付出是分不开的。

二、我们在看到优异成绩的同时，也要看到我们的不足，在本次统考成绩中，化学、物理、生物学科的平均成绩低于兄弟院校的平均成绩。

（化学平均成绩落后8.5分物理平均成绩落后3.0分；生物平均成绩落后7.7分）措施：各位老师经过这次统考，要查找到学科中存在的薄弱环节，研究解决问题的对策，也希望各位老师在今后讲课时能更新教学观念，采用先进的理念指导教学，运用启发式的教学，提高学生学习的主动性和积极性，研究学习方法，因材施教，针对不同水平的学生设置不同的学习方法，使每一位学生都能积极、主动地学习，不断提高课堂教学效果。

要根据学生对知识掌握的实际情况来决定讲课的起点，加强平日的课堂测试和单元测试，及时的反馈学习信息，以便对学生掌握知识的实际情况进行了解和查找不足。

09级班级情况对照表（表二）09级班级各科名次对照表（表三）分析：09级的新生，经过一个学期的努力，并取得优异的成绩，这除了和各任课老师的努力外，更不分开的是各班主任的辛勤劳苦。

在这次的期末考试中，徐美丽老师带领的09高三取得了高中部总分成绩第一（9门课程总分平均570.8分）。

总体来看，各班的成绩比较平衡，各科之间的差距不是很大。

措施：1，也希望各班主任在今后的工作中：以教学为主，学生为主，协调与各任课老师的关系，做好桥梁纽带的工作，在对学生的管理中兼顾过程与方法，情感态度与价值观。

贵州省安顺市2024年数学（高考）统编版真题(评估卷)模拟试卷一、单项选择题（本题包含8小题，每小题5分，共40分。

在每小题给出的四个选项中，只有一项是符合题目要求的）(共8题)第(1)题中国古代数学家用圆内接正边形的周长来近似计算圆周长，以估计圆周率的值.若据此证明，则正整数至少等于（）A．B．C．D．第(2)题函数是定义在R上奇函数，且，，则（）A．0B．C．2D．1第(3)题技术的数学原理之一是著名的香农公式:.它表示:在受噪声干扰的信道中，最大信息传递速度取决于信道带宽，信道内信号的平均功率，信道内部的高斯噪声功率的大小，其中叫做信噪比.当信噪比较大时，公式中真数中的可以忽略不计.假设目前信噪比为若不改变带宽，而将最大信息传播速度提升那么信噪比要扩大到原来的约（）A．倍B．倍C．倍D．倍第(4)题在△ABC中，内角A，B，C所对的边分别是a，b，c，若a＝4，A＝，则该三角形面积的最大值是A ．2B．3C．4D．4第(5)题欧拉恒等式（为虚数单位，为自然对数的底数）被称为数学中最奇妙的公式.它是复分析中欧拉公式的特例：当自变量时，.得.根据欧拉公式，复数在复平面上所对应的点在（）A．第一象限B．第二象限C．第三象限D．第四象限第(6)题如图所示，单位圆中弧AB的长为x，f(x)表示弧AB与弦AB所围成的弓形面积的2倍，则函数y＝f(x)的图像是（）A．B．C．D．第(7)题设，已知直线与圆，则“”是“直线与圆相交”的（）A．充分不必要条件B．必要不充分条件C．充要条件D．既不充分也不必要条件第(8)题若（为虚数单位），则（）A．5B．C．D．二、多项选择题（本题包含3小题，每小题6分，共18分。

在每小题给出的四个选项中，至少有两个选项正确。

全部选对的得6分，选对但不全的得3分，有选错或不答的得0分） (共3题)第(1)题将函数的图象向右平移个单位，再把所得图象上各点的横坐标缩短为原来的一半，纵坐标不变，得到函数的图象，则关于的说法正确的是（）A．最小正周期为B．奇函数C．在上单调递增D．关于中心对称第(2)题已知复数，，则下列结论中正确的是（）A．若，则B．若，则或C．若且，则D．若，则第(3)题双曲线：，左、右顶点分别为，，为坐标原点，如图，已知动直线与双曲线左、右两支分别交于，两点,与其两条渐近线分别交于，两点,则下列命题正确的是（）A．存在直线，使得B．在运动的过程中，始终有C．若直线的方程为，存在，使得取到最大值D．若直线的方程为，，则双曲线的离心率为三、填空（本题包含3个小题，每小题5分，共15分。

高二化学期末试卷带答案考试范围：xxx ；考试时间：xxx 分钟；出题人：xxx 姓名：___________班级：___________考号：___________1．答题前填写好自己的姓名、班级、考号等信息 2．请将答案正确填写在答题卡上一、选择题1.下列离子组一定能大量共存的是（ ） A ．酚酞呈无色的溶液中：I -、CH 3COO －、NO 3-、Na + B ．甲基橙呈黄色的溶液中： Na +、CO 32－、NO 3-、NH 4＋ C ．含大量Al 3+的溶液中： K +、Na +、NO 3-、S 2－ D ．pH>7-的溶液中：CO 32-、Cl -、F -、K +2.已知下列实验事实：①Cr 2O 3固体既能溶于KOH 溶液得到KCrO 2溶液，又能溶于硫酸得到Cr 2(SO 4)3溶液；②向KCrO 2溶液中滴加H 2O 2溶液，再酸化，可得K 2Cr 2O 7溶液； ③将K 2Cr 2O 7溶液滴加到淀粉和KI 的混合溶液中，溶液变蓝。

下列判断不正确的是( )A ．化合物KCrO 2中Cr 元素为+3价B ．实验①证明Cr 2O 3是两性氧化物C ．实验②证明H 2O 2既有氧化性又有还原性D ．实验③证明氧化性：Cr 2O >I 2 3.下列说法错误的是A ．萃取操作时，选择有机萃取剂，则溶质在萃取剂的溶解度必须比水大B ．分液操作时，分液漏斗中下层液体从下口放出，上层液体从上口倒出C ．蒸馏操作时，应使温度计水银球插入混合溶液的液面下D ．配制一定物质的量浓度的溶液时，洗涤烧杯和玻璃棒的溶液必须转入容量瓶中4.化学在生产和日常生活中有着重要的应用。

下列说法不正确的是( )。

A ．铵态氮肥不能与草木灰混合施用B．工业上用石灰乳对煤燃烧形成的烟气进行脱硫，最终能制得石膏C．镁合金的硬度和强度均高于纯镁，工业上采用电解MgCl2饱和溶液制得镁D．MgO的熔点很高，可用于制作耐高温材料，但工业上不用MgO制镁5.根据下表中烃的分子式排列规律，判断空格中烃的同分异构体数目是A．①④B．③④C．②③D．①②9.已知充分燃烧a g 乙炔气体时生成1mol 二氧化碳气体和液态水，并放出热量bkJ，则乙炔燃烧的热化学方程式正确的是A．2C2H2(g) + 5O2(g) ="=" 4CO2(g) + 2H2O(l)；△H = －4b KJ/molB．C2H2(g) + 5/2O2(g) ="=" 2CO2(g) + H2O(l)；△H =" 2b" KJ/molC．2C2H2(g) + 5O2(g) ="=" 4CO2(g) + 2H2O(l)；△H = －2b KJ/molD．2C2H2(g) + 5O2(g) ="=" 4CO2(g) + 2H2O(l)；△H =" b" KJ/mol10.在密闭容器中发生如下反应：mA（g）+nB（g） pC（g）达到平衡后，保持温度不变，将气体体积缩小到原来的1/2，当达到新平衡时，C的浓度为原来的1.9倍。

高二2023-2024学年度上期期末能力测评数学（答案在最后）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、准考证号填写在答题卡指定位置；2.回答选择题时，选出每小题答案后，用铅笔把答题卡上相应题目答案标号涂黑.如需改动，请用橡皮擦干净；3.回答非选择题时，在答题卡上作答.写在本试卷上无效；4.考试结束后，将本试卷和答题卡一并交回.一、选择题：本题共8小题，每小题5分，共40分.每小题的四个选项中，只有一个选项符合题目要求.1.直线:l 2310x y +-=的一个方向向量为（）A.()2,3- B.()3,2- C.()2,3 D.()3,2【答案】B 【解析】【分析】利用直线方向向量的定义和直线斜率与方向向量的关系直接求解即可.【详解】由2310x y +-=得，2133y x -+，所以直线的一个方向向量为2(1,)3-，而2(3,2)3(1,)3-=--，所以(3,2)-也是直线的一个方向向量.故选：B.2.对于变量x ，条件:p Q x ∈，条件:q R ，则p 是q 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【答案】D 【解析】【分析】根据充分必要条件的要求，分别判断p 能否推出q ，以及q 能否推出p 即得.【详解】由Q x ∈，若取=1x -R ，即p 不是q 的充分条件；R ，若取πx =，显然不满足Q x ∈，即p 不是q 的必要条件.3.对某社团进行系统抽样，编号为001，002，⋯，120，则抽取的序号不可能是（）A.001，004，⋯，117B.008，020，⋯，116C.005，015，⋯，115D.014，034，⋯，114【答案】A 【解析】【分析】根据系统抽样的要求抽取的序号的间隔相同，序号构成等差数列，逐项验证.【详解】根据系统抽样的要求抽取的序号的间隔相同,序号构成等差数列，对A ：121,4,3,32n a a d a n ====-，令32117n -=此方程没有正整数解，故A 不可能；对B ：128,20,12,124n a a d a n ====-，令124116n -=得10n =满足要求，故B 可能；对C ：125,15,10,105n a a d a n ====-，令105115n -=得12n =满足要求，故C 可能；对D ：1214,34,20,206n a a d a n ====-，令206114n -=得6n =满足要求，故D 可能；故选：A4.若直线:l 260x y m -+-=平分圆:C 22240x mx y +++=，则实数m 的值为（）A .2- B.2 C.3 D.2-或3【答案】C 【解析】【分析】列出22240x mx y +++=所满足的条件，由直线l 过圆心求得m 的值.【详解】22240x mx y +++=可化为()2224x m y m ++=-，则240m ->，直线260x y m -+-=始终平分圆22240x mx y +++=的周长,则直线l 经过圆心(,0)m -.代入直线得260m m --=，解得3m =或2m =-.因为2m =-不满足240m ->，故3m =故选：C.5.若数列{}n a 满足12a =，1123n nn S S n a +++=+，则88S a +的值为（）A.9B.10C.11D.12【解析】【分析】由n S 与n a 的关系求得()()112n n S n S n +=++，从而1n S n ⎧⎫⎨⎬+⎩⎭为常数列，得到1n S n =+，即可求88S a +的值.【详解】由11n n n S S a ++-=及1123n nn S S n a +++=+得()()1123n n n n S S n S S +++=+-，即()()112323n n n n S S n S n S ++-+=++，即()()112n n S n S n +=++，所以112n n S S n n +=++，即1n S n ⎧⎫⎨⎬+⎩⎭为常数列，又11221S a ==，所以11n Sn =+，即1n S n =+，所以878879,81,S S a S S ===-=，所以8810S a +=.故选：B6.已知实数,x y28x y =+-，则点(),P x y 的轨迹为（）A.抛物线B.双曲线C.一条直线D.两条直线【答案】D 【解析】【分析】将已知方程等价变形为()()334170x x y -⋅+-=，即可判断点(),P x y 的轨迹.28x y =+-，所以两边平方得()()22223246443216x y x y xy x y -+-=+++--，化简整理得2351426120x xy x y ++--=，所以()()334170x x y -⋅+-=，所以30x -=或34170x y +-=，即点(),P x y 的轨迹方程为30x -=或34170x y +-=，所以点(),P x y 的轨迹为两条相交直线.故选：D7.若复数z 满足()24z z z ⋅+=，则23z z +的最小值为（）A .16B. C. D.【答案】C 【解析】【分析】设i z x y =+，利用复数的乘法运算及模的公式得422491016x x y y ++=，所求式子为()2244x y +，令224t x y =+，则利用422152160x tx t --+=有解求得t ≥，即可得解.【详解】设i z x y =+，则()()()()222i 3i 34i 4z z z x y x y x yxy ⋅+=+⋅+=-+=，所以()()22223416x y xy -+=，即422491016x x y y ++=，而()()()2222222333i i 42i 16444z zx y x y x y x y x y +=++-=+=+=+，令224t x y =+，则224y t x =-，所以()()242229104416x x t x t x +-+-=，即422152160x tx t --+=，记20m x =≥，则22152160m tm t --+=，由题意，该方程存在非负根，且二次函数对称轴015tm =>，所以()()22Δ2415160t t =-⨯⨯-+≥，所以215t ≥，又0t >，所以t ≥，所以234z z t +=≥，即23z +的最小值为.故选：C8.计算：cos 20cos 40cos 40cos80cos80cos 20-+= （）A.12B.23C.34D.2【答案】C 【解析】【分析】根据和差角公式以及积化和差公式即可求解.【详解】()()()()11cos 20cos 40cos 40cos80cos80cos 20cos 4020cos 4020cos 8040cos 804022⎡⎤⎡⎤-+=++--++-⎣⎦⎣⎦()()1cos 8020cos 80202⎡⎤+++-⎣⎦111131cos 20cos 40cos100cos 202cos 40cos100222242112⎡⎤⎡⎤⎡⎤⎡⎤=+-+++=+⎣⎦-⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣-⎦+()()3131cos 20cos 40cos100cos 3010cos 3010sin104242⎡⎤⎡⎤=+=+--+-⎣⎦-+⎦⎣3132sin 30sin10sin10424⎡⎤=+-=⎣⎦，故选：C二、选择题：本题共4小题，每小题5分，共20分.在每小题的四个选项中，有多个选项符合题目要求.全部选对得5分，部分选对得2分，有选错的得0分.9.设集合A ={|αα为两个非零向量可能的夹角}，集合B ={|ββ为两条异面直线可能的夹角}，则下列说法错误的是（）A.4π3A ∉ B.2π3B ∈C.ππ2A B θθ⎧⎫⊆≤≤⎨⎬⎩⎭ð D.ππ2A B θθ⎧⎫⊇≤≤⎨⎬⎩⎭ð【答案】BCD 【解析】【分析】由向量夹角定义和异面直线所成角取值范围求出集合A ，B ，再结合集合相关概念即可求解.【详解】由题集合[]0,πA =，π0,2B ⎛⎤= ⎥⎝⎦，所以4π3A ∉，2π3B ∈，故A 对，B 错；由上{}π0,π2A B ⎛⎤=⋃ ⎥⎝⎦ð，故C 、D 错.故选：BCD.10.已知曲线:Γ1x x y y +=-，将曲线Γ用函数()f x 表示，则下列说法正确的是（）A.()f x 在R 上单调递减；B.()y f x =的图象关于34y x =对称；C.()22fx x +的最小值为9；D.若直线:l y kx b =+()0b <与()y f x =的图象没有交点，则实数k 为定值.【答案】ACD 【解析】【分析】分段讨论确定Γ所表示的曲线方程作出图象，由图象判断A ，B ，D 选项；求出()22f x x +的表达式求其最小值判断C 选项；【详解】当0,0x y >≥时，221916x y+=-不存在，故在第一象限内无图象；当0,0x y <≥时，221916x y-+=-，在第二象限内为双曲线的一部分，其渐近线为43y x =-，此时2216169x y =-，即()()221616,39x f x x =-≤-，所以()222251699x f x x +=-≥；当0,0x y ≤<时，221916x y +=，在第三象限内为椭圆的一部分；此时2216169x y =-，即()()221616,309x f x x =--<≤，所以()22271699x f x x +=->当0,0x y ><时，22916x y -=-，在第四象限内为双曲线的一部分，其渐近线为43y x =-；此时2216169x y =+，即()()221616,09x f x x =+>，所以()2222516169x f x x +=+>；综上：()22fx x +的最小值为9，故C 正确；()y f x =图象如图所示：对于A ：由图象可得()f x 在R 上单调递减，故A 正确；对于B ，由图象可得()f x 图象不关于直线34y x =成轴对称图形，也可以求得()3,0-关于直线34y x =对称的点2172,2525⎛⎫-- ⎪⎝⎭不在()f x 图象上，故B 错误；对D ：若直线:l y kx b =+()0b <与()y f x =的图象没有交点，则直线l 与渐近线平行，即43k =-为定值，否则直线l 与渐近线相交，则一定会与()y f x =的图象相交，故D 正确.故选：ACD【点睛】关键点点睛：本题关键是能根据,x y 的正负去掉绝对值符号得到曲线方程，作出图象，数形结合分析.11.已知独立的事件A 、B 满足()()0P A P B <<，则下列说法错误的是（）A.()()P A P AB +一定小于()2P B ；B.()()P A B P AB +可能等于()2PB ；C.事件AB 和事件AB 不可能相互独立；D.事件AB 和事件A B +可以相互独立.【答案】BC 【解析】【分析】利用独立事件的定义和性质可判断A 正确，B 错误；根据事件A 与B ，A 与B ，A 与B ，A 与B 都相互独立，利用相互独立事件概率公式计算即可.【详解】()()P A P B <且,A B 相互独立，则()()P AB P B <，()()2()P A P AB P B +<,A 正确.∵A B +表示事件,A B 至少发生一个，AB 表示事件,A B 同时发生，∴()(),()()()()P A B P B P AB P A P B P B +>=<，∴()()P A B P AB +不能等于()2P B ，B 错误.若1()2P B =，则1()2P B =，此时()()P AB P AB =，∵AB AB A = .∴()(()(()()()P A P AB AB P AB P AB P A P B P AB ==+=+ .∴移项得(()()()()()()(1())()()P AB P A P AB P A P A P B P A P B P A P B =-=-=-=.∴事件A 与B 相互独立，同理可知事件A 与B ，A 与B 也都相互独立.∴事件AB 和AB 可能相互独立，事件AB 和A B +可能相互独立，C 错误，D 正确.故选：BC【点睛】关键点点睛：解题的关键是已知独立事件A 、B ，可推出事件A 与B ，A 与B ，A 与B ，A 与B 都相互独立.12.如图，在棱长为6的正方体1111ABCD A B C D -上，点M 为体对角线1BD 靠近1D 点的三等分点，点E F 、为棱AB 、1CC 的中点，点P 在平面MEF 上，且在该平面与正方体表面的交线所组成的封闭图形中（含边界），则下列说法正确的是（）A.平面MEF 与底面ABCD 的夹角余弦值为77；B.点D 到平面MEF 的距离为11； C.点D 到点P 的距离最大值为6345；D.设平面MEF 与正方体棱的交点为1T 、…、n T ，则n 边形1n T T ⋯最长的对角线的长度大于172.【答案】BCD 【解析】【分析】建立空间直角坐标系，即可利用法向量的夹角求解A ，根据点面距离的向量法即可求解B ，根据面面平行的性质可得截面为六边形EQFNKT ，即可根据点点距离公式求解CD.【详解】建立如图所示的空间直角坐标系，则()()()2,2,4,6,3,0,0,6,3M E F ,()()4,1,4,2,4,1ME MF =-=--,设平面MEF 法向量为(),,m x y z =，440240ME m x y z MF m x y z ⎧⋅=+-=⎪⎨⋅=-+-=⎪⎩，取4y =，则()5,4,6m = ，而平面ABCD 的一个法向量为()10,0,6AA =，所以平面MEF 与底面ABCD的夹角余弦值为1677cos ,77m AA ==.故A 错误，()2,2,4,DM = 所以点D 到平面MEF的距离为11DM m m ⋅==，故B正确，延长EM 交11D C 于点N ,连接NF 交DC 延长线于点H ,连接EH 交BC 于Q ，由于点M 为体对角线1BD 靠近1D 点的三等分点，所以1111322D M D N D N MB EB ==⇒=,11912C N C F CH CH CF ==⇒=,9612235CH CQ BQ BQ EB BQ BQ -=⇒=⇒=,在棱11A D 上取K ，使得165D K =,由于11116124455,35352D K D KBQ BQ D N EB EB D N==⇒=⇒=,故//KN EQ ，连接,,TE TK FQ ,故六边形EQFNKT 即为平面MEF 上与正方体所截得的截面，由于1121863,6,555FC AE CQ D K ===-==113//,2932C F AT ATNF TE AT NC AE ∴=⇒=⇒= ,由于CQ 最大，故DQ为最大值5DQ =，故当P 在Q 处时，DP最大为5,C正确，由于()()()1863,6,0,6,3,0,0,6,3,6,0,2,,0,6,0,,6,552Q E F T K N ⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭172NE ==>,因此六边形EQFNKT 的最长对角线的长度不小于NE 的长度，因此六边形EQFNKT 的最长对角线的长度大于172，故D 正确，故选：BCD【点睛】方法点睛：作截面的常用三种方法：直接法，截面的定点在几何体的棱上；平行线法，截面与几何体的两个平行平面相交，或者截面上有一条直线与几何体的某个面平行；延长交线得交点，截面上的点中至少有两个点在几何体的同一平面上.三、填空题：本题共4小题，每小题5分，共20分.13.函数()f x =的定义域为______.【答案】()11,2∞⎧⎫+⋃⎨⎬⎩⎭【解析】【分析】根据根式函数和对数函数及分式函数定义域法则列不等式求解即可.【详解】由题意2100ln 0x x x -≥⎧⎪>⎨⎪>⎩或2100ln 0x x x -=⎧⎪>⎨⎪≠⎩，解得1x >或12x =，所以函数()f x =的定义域为()11,2∞⎧⎫+⋃⎨⎬⎩⎭.故答案为：()11,2∞⎧⎫+⋃⎨⎬⎩⎭14.已知某平面内三角形ABC 为等腰三角形，AB AC =，点D 为AC 中点，且3BD =，则ABC 面积的最大值为____________.【答案】6【解析】【分析】根据向量的模长公式可得259cos 4A x=-，即可利用面积公式得()()2229203664ABC S x =--+ ，利用二次函数的性质即可求解.【详解】设AB AC x==由于12BD AC AB =- ,所以2222215cos 44BD AC AB AC AB x x A =+-⋅=- ,故259cos 4A x=-,()()222424211159sin 1cos 12444ABC S AB AC A x A x x ⎡⎤⎛⎫⎛⎫==-=--⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦()24229458192036648464x x x =-+-=--+故当220x =时，此时()2ABC S 取最大值36，故面积的最大值为6，故答案为：615.已知锐角α，β满足2tan cos αβ=，2tan tan2αβ=，则sin sin βα的值为______.【答案】56【解析】【分析】根据已知结合同角关系消去β得1tan tan2tan ααα-=，再根据二倍角公式化弦为切得1sin 2cos αα+=，然后利用同角三角函数关系求得33sin ,tan 54αα==，然后代入sin sin βα==计算可得.【详解】因为2tan cos αβ=，2tan tan 2αβ=，所以22sin 1tan tan 2cos tan αβαβα-==，又2sin2sin 1cos 22tan 2sin cos 2sin cos 222αααααααα-===，所以1cos 1tan cos sin sin tan sin ααααααα---==，所以1cos cos sin ααα-=-，即1sin 2cos αα+=，又22sin cos 1αα+=，所以25sin 2sin 30αα+-=，又α为锐角，解得3sin 5α=，或sin 1α=-（舍去），所以43cos ,tan 54αα==，所以sin 5sin 6βα==.故答案为：5616.假设视网膜为一个平面，光在空气中不折射，眼球的成像原理为小孔成像.思考如下成像原理:如图，地面内有圆1O ，其圆心在线段MB 上，且与线段MB 交于不与,M B 重合的点A ，PM ⊥地面，且24BM PM ==，P 点为人眼所在处，视网膜平面与直线BM 垂直.过A 点作平面α平行于视网膜平面.科学家已经证明，这种情况下圆1O 上任意一点到P 点的直线与平面α交点的轨迹（令为曲线C ）为椭圆或圆，且由于小孔成像，曲线C 与圆1O 在视网膜平面上的影像是相似的，则当视网膜平面上的圆1O 的影像为圆时，圆1O 的半径r 为____________.当圆1O 的半径r 满足112r ≤≤时，视网膜平面上的圆1O 的影像的离心率的取值范围为____________.【答案】①.32②.26,23⎣⎦【解析】【分析】使用空间向量方法可以验证曲线C 的两条半轴（半长轴和半短轴，但顺序可能不对应）的长分别为2r和，然后根据题设求解.【详解】由于视网膜平面与直线BM 垂直，平面α平行于视网膜平面，故平面α与直线BM 垂直.设地面平面为β，则据已知条件有PM β⊥.从而在β内可过M 作MA 的垂线MD ，使得,,MA MD MP 可分别作为以M为原点的一个右手坐标系的,, x y z轴正方向.由已知有4BM=，2PM=，故()0,0,0M，()4,0,0B，()0,0,2P.而42MA MB AB r=-=-，故()42,0,0A r-.再由1O A r=，知()14,0,0O r-.由于平面α与直线BM垂直，即平面α与x轴垂直，从而平面α上每一点的坐标的x轴分量都是定值42r-.再根据点A在线段MB内部及4BM=，又有0424r<-<，得02r<<.此时，地面平面即平面xOy，故圆1O的方程为()2224x r y rz⎧+-+=⎪⎨=⎪⎩.据此可设圆1O上的一点Q的坐标为()4cos,sin,0r r t r t-+，故()4cos,sin,2PQ r r t r t=-+-.设直线PQ和平面的交点为R，则,,P Q R三点共线，且R的坐标的x轴分量是42r-.故()22sin424842,,4cos4cos4cosr r tr rPR PQ rr r t r r t r r t⎛⎫---==-⎪-+-+-+⎝⎭，这得到R的坐标为()()22sin21cos42,,4cos4cosr r t r trr r t r r t⎛⎫-+-⎪-+-+⎝⎭.设()22sin4cosr r tyr r t-=-+，()21cos4cosr tzr r t+=-+，则()222221682242r ry zrr r-⎛⎫⎛⎫⋅+-⎪ ⎪⎝⎭⎝⎭-()()22222242142r ry zrr r--⎛⎫=⋅+-⎪⎝⎭-()()()222168sin41cos14cos4cosr t tr r tr r t⎛⎫-+=+-⎪-+-+⎝⎭()()()()()()22221681cos4cos4cos4cosr t r t rr r t r r t---+=+-+-+()()()()()2221681cos 4cos 4cos r t r t r r r t --+-+=-+()()()()()22222168168cos 168cos 24cos 4cos r r t r r t r r t r r r t ---+-++-+=-+()()()222216824cos cos 4cos r r r r t r tr r t -++-+=-+()()224cos 4cos r r t r r t -+=-+1=.所以我们得到点R 的轨迹为()222224216821242x r r r y z r r r =-⎧⎪-⎛⎫⎛⎫⎨⋅+-= ⎪ ⎪⎪⎝⎭⎝⎭-⎩.由此可知，曲线C 是位于平面α内，以42,0,2r r ⎛⎫- ⎪⎝⎭为中心，半长轴和半短轴分别（顺序可能不对应）为2r22-=的椭圆（或者是圆，因为在二者相等时是圆）.而曲线C 和视网膜平面上的圆1O 的影像相似，故其中一个是圆当且仅当另一个是圆，且二者离心率相等.当曲线C 是圆时，有2r=12=，两边平方可得32r =.当112r ≤≤时，2r>=>，故和2r分别（顺序对应）是半长轴和半短轴的长，从而离心率e =再由112r≤≤,23⎣⎦.故答案为：32，26,23⎣⎦.【点睛】关键点点睛：本题的关键点在于，利用已知的坐标，采取适当的配凑得到类似椭圆的方程，从而得到相应曲线的性质.四、解答题：本题共5小题，共70分.解答题应写出文字说明、证明过程或演算步骤.17.已知抛物线C 的顶点是坐标原点O ，焦点是双曲线2241x y -=的右顶点.（1）求抛物线C 的方程；（2）若直线:l 2x y +=与抛物线相交于A 、B 两点，解决下列问题：（i ）求弦长AB ；（ii ）求证：OA OB ⊥.【答案】（1）22y x =；（2）（i）；（ii ）证明见解析.【解析】【分析】（1）求出双曲线右顶点，再求出抛物线的方程即得.（2）把直线l 的方程与抛物线方程联立，利用韦达定理，结合弦长公式及数量积的坐标表示求解即得.【小问1详解】双曲线2241x y -=，即22114x y -=，其右顶点为1(,0)2，则抛物线C 的焦点为1(,0)2，而抛物线C 的顶点是坐标原点O ，所以抛物线C 的方程：22y x =.【小问2详解】（i ）设211)1(,2A y y ，222)1(,2B y y ，由222y xx y ⎧=⎨=-+⎩消去x 得：2240y y +-=，则122y y +=-，124y y =-，于是12y y -==所以12AB y y =-==.（ii ）显然211)1(,2OA y y = ，222)1(,2OB y y = ，则221212121211(1)044OA OB y y y y y y y y ⋅=+=+= ，显然0,0OA OB ≠≠ ，即OA OB ⊥ ，所以OA OB ⊥.18.已知递增数列{}n a 和{}n b 分别为等差数列和等比数列，且113=a b ，422a b =，73a b =，126a b +=（1）求数列{}n a 和{}n b 的通项公式；（2）若ln ln n nb n a ac b =，证明：1211nc c c n 迹+.【答案】（1）2n a n =+，13n n b -=（2）证明见解析【解析】【分析】（1）由等差和等比数列的性质结合题意列方程组，解出11,,,a d q b ，再由基本量法求出通项即可；（2）由对数的运算性质化简再简单放缩可得()11133log 32log 31n n n n n nc n ++-=+≤=+，最后利用累乘法可证明.【小问1详解】设等差数列{}n a 的公差为d ，等比数列{}n b 的公比为q ，由题意可得：11112111133266a b a d b q a d b q a b q =⎧⎪+=⎪⎨+=⎪⎪+=⎩，前两式化简后有1111131322a b a d b q ⎧=⎪⎪⎨⎪+=⎪⎩，由上述式子可得：()21111136322a a d a d ⎛⎫+=+ ⎪⎝⎭，化简得：()()11930a d a d +-=，则19a d =-或13a d =，若19a d =-，可得1233b b b d ===-，数列{}n b 为常数列，故舍去；若13a d =，带入得3q =，又由116a b q +=，解得1d =，13a =，11b =，于是得到数列{}n a 的通项公式为2n a n =+，数列{}n b 的通项公式为13n n b -=.【小问2详解】由题可得()113ln log log 32ln n n a n nnb n n b b a ac a b +-===+，由于N n *∈时，()()113322310nn n ---+=-≥，则1332n n -³+（当且仅当1n =时取等号），所以()11133log 32log 31n n n n n nc n ++-=+≤=+，则121212311nn c c c n n 迹创即=++（当且仅当1n =时取等号）.所以1211n c c c n 迹+.19.如图，1111ABCD A B C D -为一个平行六面体，且12AB AD AA ===，1BAA ∠=23πBAD ∠=，13DAA π∠=.（1）证明：直线AB 与直线1AC 垂直；（2）求点1B 到平面ABCD 的距离；（3）求直线1AC 与平面ABCD 的夹角的余弦值.【答案】（1）证明见解析（2）3（3）3【解析】【分析】（1）利用垂直关系的向量表示求1AB AC即可证明.（2）由已知条件得三棱锥1B ABC -为正四面体，再利用正四面体结构特征即可求解得到点1B 到平面ABCD 的距离.（3）由（1）可得1AC，再由（2）得点1C 到平面ABCD 的距离，进而可求出线面角的正弦值，再结合同角三角函数平方和为1求解余弦值即可.【小问1详解】由题可得111AC AC CC AB AD AA =+=++，所以()2111····AB AC AB AB AD AA AB AB AD AB AA =++=++ 2π2π422cos 22cos 033=+⨯+⨯=，则1AB AC ⊥，于是得证：1AB AC ⊥.【小问2详解】连接11,,AB CB AC ，则由题意可知1113DAA CBB ABC ABB π∠=∠=∠=∠=，且1AB BB BC ==，所以三棱锥1B ABC -为正四面体，所以由正四面体结构性质1B 在底面ABC 的投影O 在BG （G 为AC 中点）上，且1112333GO BO BG ====，所以1B O ⊥平面ABC ，且1263B O ==，即点1B 到平面ABCD 的距离为3.【小问3详解】设直线1AC 与平面ABCD 的夹角为θ，由于1111ABCD A B C D -为一个平行六面体，则点1C 到平面ABCD 的距离等于点1B 到平面ABCD 的距离为3d =，由（1）中11AC AB AD AA =++，得到：1AC === ，则1sin 3d AC θ== ，显然π0,2θ⎛⎫∈ ⎪⎝⎭，则cos 3θ==.20.已知圆1:O 224x y +=，圆2:O ()221x y m +-=()01m ≤<，点P 为圆2O 上的一点.（1）若过P 点作圆2O 的切线l 交圆1O 于A 、B 两点，且弦AB长度最大值与最小值之积为m 的值；（2）当0m =时，圆1O 上有C 、D 两点满足PC PD ⊥，求线段CD 长度的最大值.【答案】（1）12（21【解析】【分析】（1）画出图形，得出AB =，进一步由三角形三边关系得出1O Q 的最值，由此即可顺利得解.（2）由三角形三边关系、直角三角形性质可得关于CD 的不等式，解不等式即可得解.【小问1详解】设AB 中点为Q 点，连接12O O 、1O Q 、2O Q 、2O P ，由01m ≤<，得12211O O <-=，则圆1O 内含圆2O ，由垂径定理得：AB =，1AB O Q ⊥，由切线l 可得2AB O P ⊥，可得112121O Q O P O P O O m ≤≤+=+（当且仅当直线AB 为1y m =+时都取等），12121121O Q O P O O O P O O m ≥-≥-=-（当且仅当直线AB 为1y m =-+时都取等），所以111m O Q m -≤≤+，于是=，解得12m =.【小问2详解】取CD 中点T ，连接1O T 、TP 、1O P .当0m =时，1O 和2O 重合，由于PC PD ⊥，则12PT CD =，而11112O T PT O P CD ≥-=-，221144O T CD +=，则22114142CD CD ⎛⎫-≥- ⎪⎝⎭，解得：1CD ≤，当且仅当1O 在线段TP 上时取等，所以CD 1.21.请解决以下两道关于圆锥曲线的题目.（1）已知圆:M ()22224x y a ++=()02a <<，圆P 过点()2,0N 且与圆M 外切.设P 点的轨迹为曲线E .①已知曲线Γ:x yλ=()R λ∈与曲线E 无交点，求λ的最大值（用a 表示）；②若记（2）中题①的λ最大值为0λ，圆:Q ()2211x y -+=和曲线00Γ:x y λ=相交于A 、B 两点，曲线E 与x 轴交于K 点，求四边形OAKB 的面积的最大值，并求出此时a 的值.（参考公式：322223a b c abc ⎛⎫++≤ ⎪⎝⎭，其中,,0a b c >，当且仅当a b c ==时取等号）（2）如图，椭圆:C 22221x y a b+=()0a b >>的左右焦点分别为1F 、2F ，其上动点M 到1F 的距离最大值和最小值之积为1，且椭圆C 的离心率为2.①求椭圆C 的标准方程；②已知椭圆C 外有一点P ，过P 点作椭圆C 的两条切线，且两切线斜率之积为12-.是否存在合适的P 点，使得123F PF π∠=？若存在，请写出P 点的坐标；若不存在，请说明理由.【答案】（1；②四边形OAKB 的面积的最大值为839，实数a的值为3（2）①2214x y +=；②不存在P 点使得123F PF π∠=，理由见解析【解析】【分析】（1）①根据已知条件求出点P 的轨迹方程E ，再将两个曲线无交点转化为对应的方程组无解即可.②根据已知条件求出,A B 两点坐标，表示出所求四边形的面积结合参考的不等式求解即可.（2）①根据焦点弦的范围和离心率列方程组求解即可.②由点P 和椭圆关系可以求出点P 的轨迹方程；再根据123F PF π∠=也以确定点所在圆弧的轨迹方程；根据联立两个方程有没有解来判断是否存在这样的点P 即可.【小问1详解】由圆P 过点()2,0N 且与圆M 外切可得：2P P M P ON R OM R R R a ⎧=⎪⎨=+=+⎪⎩，所以有24OM ON a MN -=<=，则点P 的轨迹为以M 、N 为左右焦点，实轴长为2a 的双曲线右支，所以曲线:E 222214x y a a-=-()0x >.①显然，当0λ≤时，曲线Γ与曲线E 无交点，当0λ>时，()222Γ:Γ:0x y x y x λλ=⇔=≥，于是令2222222014x x y a a x y λ>⎧⎪⎪-=⎨-⎪=⎪⎩，得222241a a x λ⎛⎫--= ⎪⎝⎭，若该方程在()0,∞+上无实数解，则22240a a λ--≤，解得λ≤所以λ.②将0λ=曲线00Γ:x y λ=得：曲线0Γ:x =22224a x y a ⇔=-()0x ≥，不妨令()222222411a x y a x y ⎧=⎪-⎨⎪-+=⎩，得0x =或212a ，于是212A B x x a ==，则四边形OAKB的面积12OAKB S a ==根据参考公式将该式化为32222228283269OAKB a a a S a ⎛⎫⎛⎫++-=≤= ⎪ ⎪ ⎪⎝⎭⎝⎭，2a =取等号，解得263a =或3-，负值舍去）所以四边形OAKB 的面积的最大值为839，此时实数a 的值为263.【小问2详解】①由焦点弦取值范围1a c MF a c -≤≤+，离心率c e a =得：()()21c a a c a c ⎧=⎪⎨⎪-+=⎩，解得：21a b c ⎧=⎪=⎨⎪=⎩，所以椭圆C 的标准方程为2214x y +=.②设00(,)P x y ，过点P 的切线方程为()00y y k x x -=-，由对称性不妨令00≥y ，()220014x y y y k x x ⎧+=⎪⎨⎪-=-⎩，消元得()()()2220000418440k x k y kx x y kx ++-+--=，令Δ0=，化简得：()()22200004210x k x y k y --+-=，由于两切线斜率之积为12-，则202020401142x y x ⎧-≠⎪-⎨=-⎪-⎩，化简得：2200163x y +=()02x ≠±，由于123F PF π∠=，则点P 在以12F F 为弦所对圆心角为23π的圆的优弧 12F F 上，当00≥y 时，易得该圆的方程为()2214x y +-=，不妨令()22221631420x y x y x y ⎧+=⎪⎪⎪+-=⎨⎪≠±⎪⎪≥⎩，解得该方程组无实数解，则当00≥y 时，不存在P 点使得123F PF π∠=，由对称性可知，当00≤y 时也不存在P 点使得123F PF π∠=，综上，不存在P 点使得123F PF π∠=.。

高二物理期末试卷参考答案
一、二 选择题
四．计算题
15． 22D B FR 16． 略
17．解：（1A 点射入，由C 2分
又
2分
则粒子的比荷 2分
（2）粒子从D 点飞出磁场速度方向改变了60°角，故AD 弧所对圆心角60°，粒子做圆周运动的半径
'cot 30R r ==
１分 又 ''
mv
R qB =
１分
所以
'B B =
2分
粒子在磁场中飞行时间
11266'3m r
t T qB v
π==⨯=
2分
18．(1) 棒匀速向左运动，感应电流为顺时针方向，电容器上板带正电．
∵ 微粒受力平衡，电场力向上，场强方向向下。

∴微粒带负电。

设微粒带电量大小为 q ，由平衡条件知：mg = qU C /d
对R 1、R 2和金属棒构成的回路，由欧姆定律可得I = E /3R ，U C = IR 2 = IR 由法拉第电磁感应定律可得E = Blv 0
由以上各式求得0
3υBL mgd
q =
(2) 因带电微粒从极板中间开始向下作初速度为零的匀加速运动，
U C Blv 0/6
设棒ab ∴v x = v 0v 0/2.。

高二英语试题The document was prepared on January 2, 2021江苏省大丰高级中学09—10学年度高二年级期终考试英语试卷考试时间：120分钟分值：120分本试卷分为第一卷选择题和第二卷非选择题两部分.考试时间120分钟.第一卷第一部分：听力共两节,满分20分第一节共5小题,每题1分,满分5分听下面5段对话.每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置.听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题.每段对话仅读一遍.1．What does the woman meanA．The man is a hard-working student．B．The man is usually the last to hand in his test paper.C．The man has bad study habits．2．Where does the conversation probably happenA．In the dining room. B. In the yard. C. In the living room.3．What does the woman think about the manA．He will be late for the meeting．B．He will not be late if he leaves at once．C．He will not go for the meeting．4．What is Frank’s telephone numberA. ．B. ．C. ．5．How will the woman get to LondonA. By car.B. By train.C. By plane.第二节共15小题；每小题1分,满分15分听下面5段对话或独白.每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置.听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟；听完后,各小题将给出5秒钟的作答时间,每段对话或独白读两遍.听第6段材料,回答第6至7题.6．What are the two speakers talking aboutA．About a friend’s address．B．About the software．C．About how to download software from the Internet．7．Which of the following is mentioned in the dialogueA．EPT． B．FPT． C．FTP.听第7段材料,回答第8至9题.8．What is the price of the shoesA. 354 yuan.B. 435 yuan.C. 345 yuan. 9．What does the man feel about the shoes at lastA．The shoes are too small.B．He does not like the color.C．The shoes are too dear.听第8段材料,回答第10至12题.10．What does the man want to doA．Have his hair cut. B．Go to the cinema. C．Go to work．11．What does the woman complain about the manA．He’s a lit tle wasteful． B. He is forgetful. C. He is lazy.12．Who has made the mistakeA. The man speaker.B. Not sure. C．Edison.听第9段材料,回答第13至16题.13. Where did the woman celebrate the last New YearA．In Bridport, a small town in East England．B．In a pub酒吧 in London．C．In Bridport, a small town in West England．14．How did the man like his last year’s celebrationA．He enjoyed it very much,B．He thought it was too crowded to get a drink in the pub．C．He enjoyed the drink brought by himself.15．What did people do when the clock struck twelve in Bridport last yearA．They drank, danced and sang．B．They set off fireworks．C．They wore fancy dress costumes服装 and drank．16．What is the woman’s favorite part of every New YearA．The kissing that people give her．B．The fireworks and the high spirit of the people．C．The drinks, music and fireworks．听第10段材料,回答第17至20题.17．What is the speakerA．A radio announcer. B．A doctor. C．An actor.18．What will be shown on TV on Oct 10thA．A program about the plane．B．A program about the grains．C．A program about the brain．19．What will be the main purpose of the programA．To explain the working of the brain．B．To show the latest use of computer pictures.C．To increase people’s knowledge of farming．20．Why is the program easy to followA．Because it’s for children．B．Because it’s for scientists．C．Because it’s a public show．第二部分：英语知识运用共两节,满分35分第一节：单项选择共15小题；每小题1分,满分15分请认真阅读下面各题,从题中所给的A、B、C、D四个选项中,选出最佳选项,并在答题卡上将该项涂黑.21. We have every reason to believe that_________2010 World Expo inShanghai will be ____ success.A. /; aB. the; /C. the; aD. a; a22. The company is _________ at training each employee by the end of this year.A. aidingB. aimingC. attemptingD. pointing23. His health _________ under the pressure of work and he had to stay in hospital for a month.A. broke upB. broke awayC. broke outD. broke down24. – How was the football match last night-- I wish I _________ it. It was quite a disappointment.A. had watchedB. hadn’t watchedC. watchedD. didn’t watch25. －How do you find the film－Oh, excellent. It’s well _________once again.A. worth being seenB. worthy to seeC. worthy of seeingD. worth seeing26. –Why did Susan leave so early without saying bye-- Luckily she did. Otherwise she _________something much more unpleasant.A. saidB. would sayC. would have saidD. had said27. With so many dishes to _________, customers can enjoy theirfavorite food in the newly-opened restaurant.A. be chosenB. be chosen fromC. chooseD. choose from28. I don’t want to buy the dictionary. _________, it’s tooexpensive; _________I don’t have enough money with me at present.A. Because; andB. At first; thenC. On one hand; on the other handD. On one hand; on other hand29. _________was known to all, he had broken his promise _________he would give us a rise.A. As; whichB. As; thatC. It;that D. It; which30. All the money he had had _________,so he had to make a living by begging.A. run outB. used upC. taken upD. stayed up31. How was _________ that he missed such a filmA. itB. thisC. thatD. what32. ——Your flight is boarding now. We’ll have to part.—— Don’t feel blue. _________.A. A still tongue makes a wise headB. A single flower does not make a springC. All good things come to an endD. All that glittersis not gold33. It is _________for me to finish reading in _________.A. too difficult a book; a so short timeB. a toodifficult book; so short a timeC. too difficult a book; so short a timeD. a toodifficult book; a so short time34. The doctor advised Vera strongly that she should take a holiday, but _________didn’t help.A. itB. sheC. whichD. he35. On October 1st every year all the Chinese people hold greatparties _________the birthday of New China.A. in celebration ofB. in memory ofC. in favor ofD. in praise of第二节：完形填空共20小题,每小题1分, 满分20分阅读下面短文,从短文后所给各题的四个选项A、B、C、D中选出能填入空白处的最佳选项.You've probably heard the expression, "What you see is what you get." My grandfather used to say, "If you 36 a tree long enough, it will move." We see 37 we want to see. Psychologists tell us that 38 controls our life more than our self-image. We live like the person we see in the 39 . We are what wethink we are. If you don't think you'll be successful,you 40 . You can't be it if you can't see it. Your lifeis 41 to your vision. If you want to change your 42 , you must change your vision of life.Arnold Schwarzenegger was not that famous whenhe 43 a newspaper reporter. The reporter asked Schwarzenegger,“ 44 you've retired from body building, what do you plan to do next” Schwarzenegger answered very calmlyand 45 : "I’m going to become the No. 1 movie star inHollywood. ”The reporter was 46 and amused at Schwarzenegger's plan. At that time, it was very hard to 47 how this muscle-bound bodybuilder, who was not a 48 actor and who spoke poor English with a strong Australian accent, could ever 49 to be Hollywood's No. 1 movie starSo the reporter asked Schwarzenegger 50 he planned to make his dream come true, Schwarzenegger said, “I'll do it the same 51 I became the bodybuilder in the world. What I do is create a vision of who I want to be, then I start living like that person in my 52 as if it were already true.” Sounds alm ost childishly 53 ,doesn't it But it 54 Schwarzenegger did become the No. 1 highest paid movie star in Hollywood Remember: “If you can see it, you can 55 it.”36. A. look after B. livewith C. look at D. care for37. A. what B.where C. thatD. why38. A. something B.everything C. nothing D. anything39. A. water B.picture C. novelD. mirror40. A. don't B.won't C. can'tD. will41. A. limited B.contributed C. devoted D. offered42. A. idea B.image C. lifeD. vision43. A. turned into B. met withC. acted asD. worked as44. A. Now that B. Evenif C. In case D. Only if45. A. proudly B.anxiously C. confidently D. happily46. A. surprised B.excited C. disappointed D. scared47. A. report B.imagine C. findD. judge48. A. famous B.good C. professionalD. popular49. A. hope B.have C. failD. happen50. A. when B.why C. whatD. how51. A. chance B.method C. wayD. effort52. A. film B.play C. familyD. mind53. A. foolish B.simple C. funnyD. clever54. A. succeeded B. workedC. didD. completed55. A. find B.leave C.get D. touch第三部分：阅读理解共15小题；每小题2分,满分30分请认真阅读下面短文,从短文后各题所给的A、B、C、D四个选项中,选出最佳选项,并在答题卡上将该项涂黑.AI was in the middle of coding为……编程a web page when my wife emailed me these questions: Ever wonder what it would be like to have the face, the brains, the personality and the body What it would be like to have everyone stop when you walk in the room What it would be like to be able to get anything or anyone you want I stopped for a moment and thought about it because my wife wouldn’t email me this unless something had driven her to do so. I emailed her back with what I thought was a pretty good answer. Here is what I wrote her back.Yes, I had thought many times about what it would be like to be one of the beautiful people. To be able to take your breath away when I walked into a room, or to be the life of the party as I wore only the finest clothes and sported the perfect body. But then I alwayscame back to the realization that a lifestyle like that is so easily broken. As you get older, your body changes; as you get older, the money changes. Your body never looks the same, the clothes become more and more expensive to continue. And once you have crossed the line, suddenly you are out. The next fresh face comes in, and you are quickly forgotten.All through growing up I was never an attractive person. I was overweight and laughed at. But that didn’t stop me from being a nice person. A good, clean, funny, helpful person. I was the person who you came to when you needed a friend after a breakup. I was the one you came to when you needed a joke to brighten up your day. And in the long run, I will be the one you remember, not the new face, or the fresh style.In closing, I would like to say that we are, as people, have developed into looking for things that are bigger and better instead of what will last. I don’t know about you, but I will remember the friend who helped me when I was down, more than the beauty I just saw walking down the street.56. Why does the writer regard outside looks as less importantA. Because it won’t last lo ng.B. Because it costs too little to remain.C. Because others don’t value it.D. Because he thinkshis wife less values it.57. Which of the following has the same meaning as the underlinedphrase “to take your breath away”A. To impress you very much.B. To prevent you from breathing.C. To make you take a deep breath.D. To make you curious.58. Which of the following is the best title of the passageA. Wisdom.B. Beauty or the Brains.C. Three Questions to Answer.D. My W ife’s Questions.BIn a traditional Chinese family, women are expected to do the housekeeping and leave the “other business” to men. However, the appearance of the full-time “househusband” is changing traditional family.A survey in Beijing, Shanghai, Guangzhou and Shenzhen shows that22 percent, 73 percent, 34 percent and 32 percent of white-collar male workers, aged between 28 and 33, would be prepared to do the housekeeping if the conditions were right.Yang Wenhui, 32 years old, worked at the office of a companyprefer to quit离开 the job. “My job was dull and steady. I was not promoted提拔. My wife, in contrast, really likes her job. So, after our baby was born, I chose to stay at home and take care of the family while my wife works full-time outside the home,” said Yang.Sociologists have found the full-time househusband emerges inthree main situations.Firstly, if the wife is ambitious有雄心的, well-paid and has good job prospects, while her husband is paid poorly and has no job prospects, it makes economic sense for the woman to become the main income earner for the household. Secondly, if the wife is tired of household chores and eager to work outside the home, her husband may forfeit his job. Thirdly, if the husband can do his work at home, he may take this choice as it allows him more time to take care of the family.Influenced by traditional ideas, some families with full-time househusbands prefer others not to know about their arrangement, concerned people wou ld laugh at a husband with “no prospects” orwife who is “too strong”.Zhou Wei said he has become usual to being a full-time househusband although his relatives doubted this when they gathered during the holidays. “A happy life is the most important thin g, not other peoples’ opinions,” added Zhou.59. The reason for Yang Wenhui quitting his job is that ________.A. it is too difficult for him to do itB. it is too boringfor him to do itC. he is too old to go on doing itD. his wife wantedhim to quit it60. In which situation is a man unlikely to become a full-time househusbandA. He can earn enough money to keep family and has a good job prospects.B. He can earn much less than his wife and will never get promoted.C. His wife hates housework and is busy with her work outside.D. His job can be done at home and he would like stay at home.61. Which of the following is TRUE according to the passageA. A full-time househusband is a man without prospects in life.B. A full-time househusband is much weaker than his wife.C. A full-time househusband is willing to share his experience with his relatives.D. A full-time househusband can also enjoy happiness from housework.62. The underlined word “forfeit” probably means ________.A. appreciateB. quitC. continueD. escapeCHaving finished her homework, Ma Li wants some music for relaxation. As usual, she starts her computer and goes to to download music files. But this time she is surprised when an announcement about protecting songs’copyright版权 bursts onto the screen. The age of free music and movie downloads may have come to anend as Web companies like Baidu are accused 控告 of pirating 盗用copyright. Lawsuits 诉讼 have been filed against four websites offering free downloads. In September, a Beijing court 法院 ordered Baidu to pay recording company Shanghai Push compensation for their losses. Baidu was also told to block links to the pirated music on the website. This caused a heated discussion on Internet file sharing.“Baidu’s defeat in the lawsuit shows it is not right to get copyrighted songs without paying. Downloaders may face lawsuits or fines,” said an official.Like many teens, Huang Ruoru, an 18-year-old girl from Puning in Guangdong Province , doesn’t think that getting music from websiteis wrong. She always shares her favorite songs downloaded from Baidu with her friends. When told about the lawsuit, she began to feel a little guilty about obtaining others’ work without paying.However, other teenagers have different ideas. Wang Yafei, a Senior 2 girl from Jinan, Shandong Province, pointed out that file sharing is a good way to promote pop singers. “If I download a song and really like it, I will buy the CD,” She said. “So what the recording companies really should concentrate on is improving their music, rather that pursuing file-sharers.”63. Which of the following best describes the passageA. Music on the Internet is of better quality.B. Downloading material can be illegalC. It’s good to get free music on the Internet.D. Baidu is a popular web company.64. The four web companies were put to court because __________.A. they got copyrighted songs without payingB. they downloaded copyrighted music for peopleC. they make copyrighted files for free downloads.D. they offer free music on line65. How do some of the teenagers feel while downloading free music after the lawsuitA. A bit guiltyB. A little sadC. very angryD. awfully sorry66. What’s the advantage of file sharing for recording companiesA. Getting more money from web companiesB. Enabling people to download favorite songsC. Helping to improve the musicD. Making pop singers more popularDSometime today—perhaps several times —Dick Winter will think about the 19-year-old who saved his life.Because of this young man, Winter enjoys things like friendships, colours and laughter every day.The young man saved Winter’s life by signing an organ donor card 器官捐献卡.“I can’t say thank you enough,” Winter said yesterday at a news conference marking the tenth anniversary of the Multi Organ Transplant program at Toronto General Hospital.What Winter knows of the 19-year-old who saved his life is only that he died in a car accident and that his family was willing to honour his wishes and donate his organs for transplantation.His liver 肝脏 went to Winter, who was dying from liver trouble. “Not a day goes by that I don’t think of what a painful thing it must have been for them,” Winter said yesterday.“They are very, very special people.”Winter, 63, is fitter now than he was 10 years ago, when he got the transplant. He has five modals from the 1995 World Transplant Games in swimming and hopes to collect some more next year in Japan.“At one time, we were probably strange people in the eyes of other people. Now it’s expected you should be able to go back and do everything you did before. Only better.”The biggest change for Winter, however, isn’t that he has become a competitive athlete. The biggest change is how deeply he appreciates every little thing about his now.“ I have no time for arguments,” said Winter.“ You change everything. Material things don’t mean as much. Friendships mean a lot.”Also at yesterday’s news conference was Dr. Gray Levy, Winter’s doctor.Levy said he has bitter-sweet feelings when he looks at Winter and hears of his athletic exploits.Levy knows that for every recipient 接受者 like Winter, there are several others who die even though they could be saved because there aren’t enough donated organs.“For every Mr. Winter, we have 5 to 10 people that will never be given the chance that Mr. Winter was given,” Levy said.Levy said greater public awareness and more resources are needed. He noted that in Spain and the United States, hospitals receive $10,000 per donor to cover the costs of the operating room, doctors, nurses and teams to work with the donors’ families.67. Which of the following is true about the 19-year-oldA. He died of liver troubleB. He got wounded in a battle.C. He was willing to donate his organs.D. He became a recipient of a prize.68. What do we learn about Dick WinterA. He is becoming less competitive now.B. He is always thinking about his early life.C. He knows all about the young man and his family.D. He values friendships more than material things.69. Dr. Levy would agree that .A. Spanish hospitals have more favorable conditions for organ transplantB. the Canadian public have realized the importance of organ donationC. there are enough donated organsD. Canadian hospitals now have enough donated organs70. What’s the author’s purpose in writing this articleA. The public should give more support to organ transplant.B. Transplant patients are thankful for the help they receive.C. Transplant can change a patent ’s life greatly.D. It is easy to get organs for transplant.第二卷非选择题共35分第四部分单词拼写10分situation is ___________令人为难的, and he doesn’t know how to deal with it.more people live ___________海外these days.73. ___________消费者in the USA had access to 200 channels.74. My brother has a large ____________ 收藏 of stamps and old coins. remote desert area is ___________可接近only by helicopter.76. Let's _________ 回顾 what has happened so far.tried to ___________安慰Jane after her mother's death.model is technically ___________优越于to its competitors.79. John was doing his homework and ___________同时 Cathy waswatching TV.house looked strangely ___________熟悉的, though she knew she'd never been there before.第五部分任务型阅读共10小题；每小题1分,满分10分请认真阅读下列短文,并根据所读内容在文章后表格中的空格里填入最恰当的单词.注意：每空格1个单词.请将答案写在答题纸上相应题号的横线上.Smoking is one of the worst things kids can do to their bodies. Every single day, about 4,000 kids between the ages of 12 and 17start smoking. Most junior school students don’t smoke—only about 1 in 10 does. Most senior school students don’t smoke either—about 1 in 4 does.But why do those who smoke ever begin There’s more than just one simple answer. Some kids may start smoking just because they’re curious. Others may like the idea of doing something that grown-ups don’t want them to do. Still others might think smoking is a way to act or smoking makes them look like an adult.Luckily, fewer people are starting to smoke than a few years ago. Maybe that’s because more and more people have learned that smokingcan cause cancer and heart disease. Sometimes kids don’t worry about what future illness they might get.Nicotine and other poisonous chemicals in tobacco cause lots of diseases, like heart problems and some kinds of cancer. If kids smoke, it will hurt their lungs and hearts each time they light up. It can also make it more difficult for blood to flow in the body, so smokers may feel tired. The longer they smoke, the worse the damage becomes.The human body is smart, and it knows when it’s being poisoned. When kids try smoking for the first time, they often cough a lot and feel pain or burnt in their throats and lungs. This is their lungs’way of trying to protect them. Also, many kids say that they feel第六部分书面表达满分15分最近,你校同学针对是否应该每天留家庭作业进行了讨论.讨论详情见下表.请根据2．根据内容要点适当增加细节,以使行文连贯；3．文章题目和开头已给出不计入词数.Is daily homework necessaryRecently our school students have had a discussion aboutwhether it is necessary to have daily homework. Different studentshave different opinions.______________________________________________________________________________________________________________________________________________________________大丰高级中学高二年级期终考试英语答题纸班级姓名学号单词拼写满分10分71. 72. 73. 74. 75.76. 77. 78. 79. 80.任务型阅读满分10分81. 82. 83. 84. 85.86. 87. 88. 89. 90.书面表达满分15分Is daily homework necessaryRecently our school students have had a discussion aboutwhether it is necessary to have daily homework. Different studentshave different opinions.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ _____________________________________________________________________ _________________________________________________________________________________________ _______________________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ ____________________________________________________________。

期末检测卷092024-2025年人教版选择性必修1（解析版）（时间：90分钟 满分：100分）一、选择题：本题共16个小题，每小题3分。

在每小题给出的四个选项中，只有一项符合题目要求。

1．（2023·河南·高二期中）已知下列反应的热化学方程式： ()()()2222H O l 2H g O g =+ 1ΔH()()()22222H O l 2H O l O g =+ 2ΔH()()()222NO g N g O g =+ 3ΔH则反应()()()()2222H g 2NO g N g 2H O l +=+的ΔH 为 A ．31ΔH -ΔH B ．123ΔH -ΔH +ΔH C ．32ΔH +ΔH D ．132ΔH -ΔH +ΔH【答案】A 【详解】将已知反应依次编号为①②③，由盖斯定律可知，反应③—①可得反应()()()()2222H g 2NO g N g 2H O l +=+，则ΔH =31ΔH -ΔH ，故选A 。

2．（2023·辽宁·高三阶段练习）如图表示在催化剂(25Nb O )表面进行的反应：()()()()222H g CO g CO g H O g +=+。

已知下列反应：①()()()2222H g O g 2H O g += 1ΔH ②()()()21C s O g CO g 2+= 2ΔH③()()()22C s O g CO g += 3ΔHA ．23Δ<ΔH HB ．图中的能量转化方式为太阳能转化为化学能C ．反应()()()222CO g O g 2CO g +=的()32Δ=2Δ-ΔH H HD ．反应()()()()222H g CO g CO g H O g +=+的2311Δ=Δ-Δ+Δ2H H H H【答案】A 【详解】A ．2ΔH 是1molC 完不完全燃烧的焓变，3H ∆是1molC 完全燃烧的焓变，放热焓变为负，故23H H ∆>∆，A 项错误；B ．由图知，太阳能提供能量实现2CO 和2H 反应生成CO 和2H O ，B 项正确；C ．③−②再乘以2即得()()()222CO g O g 2CO g +=的()32Δ=2Δ-ΔH H H ，C 项正确；D ．2+①②−③为()()()()222H g CO g CO g H O g ++，焓变为12312H H H ∆+∆-∆，D项正确： 故选A 。