河北省五个一名校联盟2019届高三摸底考试物理试题
河北省“五个一”名校联盟2023届高三年级摸底考数学卷及答案
河北省“五个一”名校联盟2023届高三年级摸底考试数学试卷命题单位：邯郸市第一中学（满分：150分,测试时间：120分钟）第I 卷（选择题,共60分）一、选择题：本题共8小题,每小题5分,共40分,在每小题给出的四个选项中,只有一项是符合题目要求的.1.设集合{}2280A x x x =--<,{}2,3,4,5B =,则A B = （）A.{2}B.{}2,3 C.{}3,4 D.{}2,3,42.已知2i z =+,则()i z z -=（）A.62i- B.42i- C.62i+ D.42i+3．已知圆锥的高为1,母线长为6,则过此圆锥顶点的截面面积的最大值为（）A.2B.52D.34．设0>ω,若函数()2cos()2f x x πω=-在[,42ππ-上单调递增,则ω的取值范围是（）A.1(0,]2B.3(1,]2C.3[0,]2D.(0,1]5．如图,在底面半径为1,高为6的圆柱内放置两个球,使得两个球与圆柱侧面相切,且分别与圆柱的上下底面相切．一个与两球均相切的平面斜截圆柱侧面,得到的截线是一个椭圆．则该椭圆的离心率为()A.226．已知82βαππ<<<,且5sin 2sin cos 2sin 4413πααπ-=，sin 2cos 4πβ+cos 2sin4πβ33=,则()βα22sin -的值为（）B.96 C. D.96-7．若过点(,)m n 可以作曲线2log y x =的两条切线,则（）A.2log m n> B.2log n m> C.2log m n< D.2log n m<8．先后抛掷两枚质地均匀的骰子,甲表示事件“第一枚骰子掷出的点数是1”,乙表示事件“第二枚骰子掷出的点数是2”,丙表示事件“两枚骰子掷出的点数之和是8”,丁表示事件“两枚骰子掷出的点数之和是7”,则下列说法正确的有（）①甲与乙相互独立②乙与丁相互独立③乙与丙不互斥但相互独立④甲与丙互斥但不相互独立A.1个B.2个C.3个D.4个二、选择题：本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,部分选对的得2分,有选错的得0分.9．有6个相同的球,分别标有数字1,2,3,4,5,6,现从中有放回的取出5个球并记录取球结果,则下列统计结果中可能取出6号球的是（）A.平均数为3,中位数为2B.中位数为3,众数为2C.平均数为2,方差为2.4D.中位数为3,极差为210．已知(cos ,sin ),(cos )a x x b x x ==r r ,函数()f x a b =⋅r r,则下列选项正确的是()A.函数f (x )的值域为13[,]22-．B.将函数1sin 2y x =+图像上各点横坐标变为原来的12（纵坐标不变）,再将所得图像向左平移12π个单位长度,可得函数()f x 的图像．C.函数f (x )是奇函数．D.函数f (x )在区间[]π20，内所有零点之和为143π．11．如图,正方体ABCD -A 1B 1C 1D 1棱长为1,P 是1A D 上的一个动点,下列结论中正确的是（）A.BP 的最小值为23B.PA PC +C.当P 在直线1A D 上运动时,三棱锥1B ACP -的体积不变D.以点B 为球心,2为半径的球面与面AB 1C 的交线长为π312．已知圆221:(12C x y +-=上两点A 、B 满足AB 点()0,0M x 满足:MA MB =,则下列结论中正确的是（）A.当AB =,012x =B.当00x =时,过M 点的圆C 的最短弦长是C.线段AB 的中点纵坐标最小值是12D.过M 点作圆C 的切线且切点为A,B,则0x 的取值范围是(,)-∞⋃+∞第II 卷（非选择题,共90分）三､填空题:本题共4小题,每小题5分,共20分.13．已知函数()3(xxa e f x e x -=是偶函数,则=a ______.14．设抛物线2y =的焦点为F ,准线为l ,过抛物线上一点A 作l 的垂线,垂足为B ．设0C (),AF 与BC 相交于点D ．若CF AF =,则△ACD 的面积为_____．15.,212xx R e x a ∀∈-≥+,则a 的最大值为______.16.德国大数学家高斯年少成名,被誉为数学届的王子,19岁的高斯得到了一个数学史上非常重要的结论,就是《正十七边形尺规作图之理论与方法》.在其年幼时,对1+2+3+……+100的求和运算中,提出了倒序相加法的原理,该原理基于所给数据前后对应项的和呈现一定的规律生成,因此,此方法也称之为高斯算法,现有函数()xf x =设数列{}n a 满足*121(0)()()()(1)()n n a f f f f f n N n n n-=+++++∈ ，若12,{}n n n n b a b n +=则的前项_________.n S =和四､解答题：本题共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10分)已知正项数列{}n a 满足11a =,且112++=-n n n n a a a a .(1)求数列{}n a 的通项公式;(2)记21n n a b n =+,求数列{}n b 的前n 项和为n S ,求证:11.32n S ≤<18.(本小题满分12分)某学校组织“纪念共青团成立100周年”知识竞赛,有A ,B,C 三类问题,每位参加比赛的同学需要先选择一类并从中随机抽取一个问题回答,只有答对当前的问题才有资格从下一类问题中再随机抽取一个问题回答.A 类问题中的每个问题回答正确得10分,否则得0分;B 类问题中的每个问题回答正确得20分,否则得0分,C 类问题中的每个问题回答正确得30分,否则得0分.已知小康同学能正确回答A 类问题的概率为0.8,能正确回答B 类问题的概率为0.6,能正确回答C 类问题的概率为0.4,且能正确回答问题的概率与回答次序无关.（1）若小康按照CBA 的顺序答题,记X 为小康的累计得分,求X 的分布列;（2）相比较小康自选的CBA 的答题顺序,小康的朋友小乐认为按照ABC 的顺序答题累计得分期望更大,小乐的判断正确吗？并说明理由.19.(本小题满分12分)已知ABC ∆的内角C B A ,,的对边分别为c b a ,,,若4,b =在①()(sin sin )(sin sin )b c B C A C a +-=-,②1cos 3)(2cos =++B C A 两个条件中任选一个完成以下问题：（1）求;B （2）若D 在AC 上,且,AC BD ⊥求BD 的最大值.20.(本小题满分12分)如图,ABCD 为圆柱OO '的轴截面,EF 是圆柱上异于AD ,BC 的母线.（1）证明:BE ⊥平面DEF ;（2）若6==BC AB ,当三棱锥B DEF -的体积最大时,求二面角B DF E --的正弦值.21.(本小题满分12分)已知双曲线C :22221x y a b-=的离心率为2,1F 、2F 为它的左、右焦点,点P 为双曲线在第一象限上的一点,且满足120PF PF ⋅=uuu r uuu r,126PF PF =.（1）求C 的方程;（2）过点2F 作直线l 交双曲线于,A B 两点,在x 轴上是否存在定点(),0Q m ,使得⋅uur uuu rQA QB 为定值,若存在,求出m 的值和该定值;若不存在,请说明理由.2212012.()()ln ().();():(本小题满分分已知函数（）讨论的零点个数（）证明x f x x ax a f x f e xf x a=+≠≤-河北省“五个一”名校联盟2023届高三年级摸底考试数学参考答案一、单选题1——4:BADD 5——8:BBBC 二、多选题9.AB10.ABD 11.BCD12.CD三、填空题13.1-14.15.116.12n n +⋅四、解答题17.【解析】(1)数列{}n a 中,0n a >,由112++=-n n n n a a a a ,可得2111=-+nn a a .…………………………………………………………………………2分又11111a ==,则数列1n a ⎧⎫⎨⎬⎩⎭是首项为1公差为2的等差数列,则12)1(211-=-+=n n a n,则数列{}n a 的通项公式为121-=n a n .…………………………………………………4分(2)由(1)知121-=n a n ,则1111(21(21)(21)22121n n a b n n n n n ===-+-+-+,…………………………………6分则数列{}n b 的前n 项和111111111123352121221()()n S n n n =-+-++-=--++L ,………………………8分,012131,311210,312,*<+-≤-∴≤+<∴≥+∴∈n n n N n .2131,1121132<≤∴<+-≤∴n S n …………………………………………………10分18.【解析】(1)由题可知,X 的所有可能取值为0,30,50,60……………………………1分()010.40.6P X ==-=()()300.410.60.16P X ==⨯-=()500.40.6(10.8)0.048P X ==⨯⨯-=()600.40.60.80.192P X ==⨯⨯= (5)分所以X 的分布列为X0305060P0.60.160.0480.192………………………………………………………………………………………………6分(2)由(1)知,()00.6300.16500.048600.19218.72E X =⨯+⨯+⨯+⨯=．若小康按照ABC 顺序答题,记Y 为小康答题的累计得分,则Y 的所有可能取值为0,10,30,60()010.80.2P Y ==-=()()100.810.60.32P Y ==-=()300.80.6(10.4)0.288P X ==⨯⨯-=()600.80.60.40.192P X ==⨯⨯=………………………………………………………10分所以()00.2100.32300.288600.19223.36E Y =⨯+⨯+⨯+⨯=故小乐的判断正确…………………………………………………………………………12分19.【解析】(1)若选①,由正弦定理得,(),)()(a c a c b c b -=-+………………………2分即,222ac a c b -=-即,222ac b c a =-+2221cos ,222a cb ac B ac ac +-∴===……4分(0,),,3B B ππ∈∴=Q ……………………………………………………………………5分若选②cos2()3cos cos2()3cos cos23cos 1,A C B B B B B π++=-+=+=Q …………………2分,1cos 31cos 22=+-∴B B 即22cos 3cos 20,B B +-=即2cos -=B (舍)或21cos =B ,…………………………………………………………4分(0,),,3ππ∈∴=Q B B ……………………………………………………………………5分(2)BD AC ⊥Q ,BD 为AC 边上的高,当面积最大时,高取得最大值.…………………6分法一：由余弦定理得,B ac c a b cos 216222-+==,由重要不等式得162ac ac ac ≥-=,当且仅当a=c 时取等,……………….…….…….…….…….……….…………………9分所以34sin 21≤=∆B ac S ABC .…….…….…….…….…….…….………………10分所以AC 边上的高的最大值为4312b =..…….…….…….…….………………12分法二：由正弦定理得ABC ∆外接圆的直径为2sin b R B ==,.……………………7分利用正弦定理表示面积得:11sin sin 2233ABC S ac B A C B ∆==⋅122sin()sin()233A A A A ππ=-=-)363A π=-+≤……………………………………………………10分所以AC 边上的高的最大值为322134=b ..…….…….…….…….………………12分20.【解析】(1)证明：如右图,连接AE ,由题意知AB 为O 的直径,所以AE BE ⊥.因为AD ,EF 是圆柱的母线,所以AD EF ∥且AD EF =,所以四边形AEFD 是平行四边形.所以AE DF ∥,所以BE DF ⊥.因为EF 是圆柱的母线,所以EF ⊥平面ABE ,又因为BE ⊂平面ABE ,所以EF BE ⊥.又因为DF EF F = ,DF 、EF ⊂平面DEF ,所以BE ⊥平面DEF .………………………………………4分(2)由(1)知BE 是三棱锥B DEF -底面DEF 上的高,由(1)知EF AE ⊥,AE DF ∥,所以EF DF ⊥,即底面三角形DEF 是直角三角形.设DF AE x ==,BE y =,则22:6Rt ABE x y+=在中有,………………………………………………………………5分所以221113326622B DEF DEFx yV S BE x y-∆+⎛=⋅=⋅⋅⋅=≤=⎝,当且仅当3==yx时等号成立,即点E,F分别是»AB,»CD的中点时,三棱锥B DEF-的体积最大,…………………………………………………………………………………7分(:另解等积转化法:1.3B DEF D BEF D BCF B CDF CDFV V V V S BC----∆====⋅,)F CD E F AB CD易得当与距离最远时取到最大值此时、分别为 、 中点下面求二面角B DF E--的正弦值：法一：由(1)得BE⊥平面DEF,因为DF⊂平面DEF,所以BE DF⊥.又因为EF DF⊥,EF BE E⋂=,所以DF⊥平面BEF.因为BF⊂平面BEF,所以BF DF⊥,所以BFE∠是二面角B DF E--的平面角,……9分由(1)知BEF为直角三角形,则3BF==.故3sin3BEBFEBF∠==,所以二面角B DF E--的正弦值为分法二：由(1)知EA,EB,EF两两相互垂直,如图,以点E为原点,EA,EB,EF所在直线为x,y,z轴建立空间直角坐标系E xyz-,则00000000(),(,,),(,B D E F.由(1)知BE⊥平面DEF,故平面DEF的法向量可取为00()EB=uuu r.设平面BDF的法向量为(,,)n x y z=,由((0,DF BF==,……………………………………………………8分得n DFn BF⎧⋅=⎨⋅=⎩,即⎧=⎪⎨+=⎪⎩,即xy=⎧⎪⎨=⎪⎩,取1z=,得n= (10)分设二面角B DF E --的平面角为θ,cos cos ,n EB n EB n EBθ⋅=<>==⋅r uur r uurr uur ,所以二面角B DF E --的正弦值为33.………………………………………………12分21.【解析】(1)解法一：由2ce a==得：2c a =,b ∴=,120PF PF ⋅=uuu r uuu rQ ,∴12PF PF ⊥,在12Rt F PF V 中,由122PF PF a -=得：222121224PF PF PF PF a +-=,代入222124PF PF c +=,126PF PF =得:224124c a -=解得:23b =,21a =,∴双曲线方程为：2213y x -=.………………………………………4分解法二：由2ce a==得：2c a =,b ∴==,设点()(),0P x y y >,则点P满足22221x y a b-=…①,120PF PF ⋅=uuu r uuu r Q ,()()222,,0c x y c x y x c y ∴---⋅--=-+=,即222x y c +=…②,121211222F PF S PF P y c F ⋅==,即3y c ⋅=…③,则由①②得：2b y c =,代入③得：23b =,21a =,∴双曲线方程为:2213y x -=.…………4分(2)解法一:当l 斜率不存在时,:2l x =,此时()2,3A ,()2,3B -,2(2)9QA QB m ⋅=--，uur uuu r当l 斜率为0时,:0l y =,此时()1,0A -,()10B ,,21QA QB m ⋅=-uur uuu r；QA QB ⋅若为定值，uur uuu r 22:(2)91.,0,1m m m QA QB ⋅=--=-=-则有解得uur uuu r:(10),:0.QA QB Q ⋅=-uur uuu r下证当为，时恒有；………………………………………………6分当l 斜率存在时,设():2l y k x =-,()11,A x y ,()22,B x y ,联立()22233y k x x y ⎧=-⎨-=⎩得()222234430k x k x k -+--=,则236360k ∆=+>,212243k x x k -∴+=-,2122433k x x k --=-,…………………………………8分()()121211QA QB x x y y ∴⋅=+++uur uuu r ()()212121212124x x x x k x x x x =++++-++⎡⎤⎣⎦()()()222121212114k x x k x x k =+--+++………………………………………………10分()()22222224341211433k k k k k k k ---=+--++--()222241(3)410.3k k k k +-=++=-综上所述：存在1m =-,使得0QA QB ⋅=uur uuu r ；……………………………………………12分解法二：当l 斜率为0时,:0l y =,此时()1,0A -,()10B ,,由(),0Q m 得：21QA QB m ⋅=-uur uuu r ；………………………………………………………………………6分当l 斜率不为0时,设:2l x ty =+,()11,A x y ,()22,B x y ,联立22233x ty x y =+⎧⎨-=⎩得：()22311290t y ty -++=,则236360t ∆=+>,1221231t y y t -∴+=-,122931y y t =-,…………………………………………………………8分()()()()11221212,,QA QB x m y x m y x m x m y y ∴⋅=-⋅-=--+uur uuu r 2212121212(2)(2)(1)(2)()(2)ty m ty m y y t y y m t y y m =+-+-+=+⋅+-++-()2222222129(1215)9(1)(2)(2)(2)313131t m t t m t m m t t t --+=++-+-=+----,………………………10分若⋅uur uuu r QA QB 为定值,则1215931m -=-,1m ∴=-,()1,0Q ∴-,此时0QA QB ⋅=uur uuu r ；当1m =-,l 斜率为0时,210QA QB m ⋅=-=uur uuu r ；综上所述,存在1m =-,使得0QA QB ⋅=uur uuu r ；………………………………………………………………………………12分2min ln ln ln 122.(1)()ln 0,,(),()(0,),()0,(,),()0,()(0,)1(,),()(),20,();,()0,()x x x f x x ax a g x g x x x x x e g x x e g x g x e e g x g e ex g x x e g x x g x -'=+==-=-=''∈<∈+∞>∴+∞∴==-→→+∞><→+∞→【解析】令则设当时时在上单调递减，在上单调递增分时当时且时L L L L L L L L L L L L L L L L L Q 0,311,(),0,(),a f x a a f x e e∴<-=->分当时无零点当或时有一个零点L L L L L L L10,().5L L L L L L L L L L L L L L L L L L L L L L L L L a f x e-<<当时有两个零点分ln ()()()(2),((),7ln 10(0)ln 10(0),:()10(0)8()1,()1,(,0)x at atat t f x x x x f e x f e t f f t a x a ate t at t t at e t tf x e x h x x e h x e x --------=≤-⇔≤-++-≥>++-≥>+-≥>'=+-=-∈-∞设则分即证，即证即证，分设则当时L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L 00,()0,(0,),()0,()(,0),()(0),()(0)010110,0"",(1),,,()0x h x x h x h x h x h x h x e x a x ef x -'<∈+∞'>∴-∞+∞∴≥=∴+-≥==>-=当时在单调递减在，单调递增，分当且仅当时成立由知当时存在使得L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L ()11()()10,().12x f x f e x f x e f x a-∴+-≥∴≤-分分L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L。
河北省“五个一”名校联盟2025届高三上学期第一次联考英语试卷 (含答案,无听力音频无听力原文)
绝密★启用前河北省“五个一”名校联盟2025届高三第一次联考英语本试卷共8页，满分150分，考试用时120分钟。
注意事项：1．本试卷由四个部分组成，其中，第一、二部分和第三部分的第一节为选择题，第三部分的第二节和第四部分为非选择题。
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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。
例：How much is the shirt?A．£ 19.15.B．£9.18.C．£9.15.答案：C。
1．What day is it today?A．Thursday.B．Friday.C．Saturday.2．What does the man like least about the movie?A．The action scenes.B．The special effects.C．The main character.3．What is happening with the girl?A．She is looking for a job.B．She is visiting a new city.C．She is distancing herself from her father.4．What difficulty does the man have in learning French?A．Pronunciation.B．Spelling.C．Grammar.5．What are the speakers mainly talking about?A．Mike’s good manners.B．Mike’s eye problem.C．Mike’s meeting with his boss.第二节（共15小题，每小题1.5分，满分22.5分）听下面5段对话或独白。
河北省“五个一”名校联盟 2023 届高三年级英语考试试题
河北省“五个一”名校联盟2023届高三年级摸底考试英语试卷命题单位：石家庄市第一中学（考试时间：120分钟试卷满分：150分）注意事项：1.答卷前，考生务必将自己的姓名、考生号等填写在答题卡上。
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答案标号涂黑。
如需改动，用橡皮擦干净，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.考试结束后，将答题卡交回。
第一部分听力（共两节，满分30分）第一节（共5小题；每小题1.5分，满分7.5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。
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.Where are the speakers?A.In Singapore.B.In China.C.In Russia.2.Who has probably arrived?A.The man’s dad.B.The man’s sister.C.The man’s friend.3.What does the man suggest the woman do?A.Wear warmer clothing.B.Follow him.C.Have lunch first.4.Which is the most popular job nowadays?A.A surgeon.B.A general doctor.C.A dentist.5.What are the speakers talking about?A.Weather.B.Fashion.C.Shopping.第二节（共15小题；每小题1.5分，满分22.5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。
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。
河北省“五个一”名校联盟2020届高三上学期一轮复习收官考试(理综)
理综试卷第Ⅰ卷(选择题)可能用到的相对原子质量：H 1 C 12 N 14 O 16 S 32 K 39 Cr 52 Mn 55 Fe 56一、选择题1．细胞骨架是真核细胞借以维持其基本形态的重要结构，它通常也被认为是广义上细胞器的一种，下列相关叙述错误的是( )A．如果破除细胞骨架，动物蛋白分泌速度明显降低，说明细胞骨架是一种膜性细胞器B．细胞骨架与物质运输、能量转换、信息传递等生命活动密切相关C．细胞骨架是由蛋白质纤维组成的网架结构D．细胞骨架有保持细胞内部结构有序性的功能2．多细胞生物，可以看作是由一个受精卵经过多次分裂、分化形成的细胞社会，下列相关叙述错误的是( )A．不同细胞的遗传信息执行情况不同，某一细胞的很多信息处于关闭状态B．细胞在分裂间期进行活跃的物质准备，同时细胞有适度的生长C．细胞凋亡对抵御外界各种因素的干扰都起着非常关键的作用D．不同细胞寿命有很大区别，寿命长的细胞分裂能力较强3．人感染乳头瘤病毒(HPV)可诱发宫颈癌等恶性肿瘤。
国际上至今已经有预防性的四价疫苗(HPV6，11，16，18)等可以预防这四种病毒类型感染，下列说法错误的是( )A．为制备该疫苗，将HPV外壳蛋白L1基因与载体连接，导入受体细胞。
受体细胞将目的基因转录，再翻译出L1蛋白，这样就获得了疫苗的有效成分。
B．人体接种该疫苗后，疫苗作为抗原刺激机体产生特异性抗体。
C．免疫过的机体，再次接触HPV，浆细胞会迅速繁殖、分化，产生大量抗体。
D．HPV感染者接种疫苗后病变的比例没有明显差异，原因可能是该疫苗未能明显诱导细胞免疫清除体内HPV。
4．为研究蛋白A的功能，选用细胞膜中缺乏此蛋白的非洲爪蟾卵母细胞进行实验，处理及结果见下表。
下列叙述错误的是( )实验组号在等渗溶液中进行的处理在低渗溶液中测定卵细胞的水通透速率(cm／s×10-4)Ⅰ向卵母细胞注入微量水(对照) 27.9Ⅱ向卵母细胞注入蛋白A的mRNA 210.0Ⅲ将部分Ⅱ细胞放入含HgCl2的等渗溶液中80.7A．将Ⅰ组卵母细胞放入低渗溶液后，水分子经协助扩散穿过膜的磷脂双分子层进入卵母细胞。
2023届河北省五个一名校联盟高三年级摸底考试英语试题听力
2023届河北省五个一名校联盟高三年级摸底考试英语试题一、阅读理解Space Center Houston offers a variety of age-appropriate and inspiring camp experiences with safety top of mind. In the wake of COVID-19, we aim to continue inspiring all generations through the wonders of space exploration while maintaining the highest level of safety for all. Our Explorer Camps introduce children ages 4-11 to science, technology, engineering and math through engaging and hands-on learning activities. Discover and explore our Explorer Camps for ages 6-7 and below.JOURNEY TO MARSJune 14-18, July 12-16, Aug. 9-13 | $289.95 5-DAY EXPLORER CAMPWhile demonstrating best safety practices, campers are involved in activities that arouse their interest in science along with their sense of adventure. These young engineers learn the basics of rocketry, discover coding, and explore hands-on science activities.ENGINEERING BASICSJune 7-11, July 5-9 | $289.95 5-DAY EXPLORER CAMPCampers discover what it takes to solve problems related to space exploration in Engineering Basics. Taking LEGO bricks “out of the box” and to new and exciting places, this camp gets your little builders using LEGO bricks like never before. These engineers-in-training will participate in problem-solving activities, building simple machines, art projects, and robotics.JOURNEY TO SPACEJune 28-July 2, July 26-30 | $289.95 5-DAY EXPLORER CAMPHow do astronauts work in space? Why is Venus so hot? Will people ever go to Mars? How do robots work? What do engineers do? All these questions are answered as campers explore space through art, science, robotics, and engineering activities. This camp leaves all inquiring minds inspired to continue their journey.EXPLORATION AND DISCOVERYJune 1-5 | $229.95 5-DAY EXPLORER CAMPCampers discover what it takes to send people, supplies and spacecraft beyond Earth's atmosphere. During this five-day adventure, former astronauts will deliver speeches on how they live and work in the space station. 1．Which camp provides campers with knowledge of rocket?A．JOURNEY TO MARS. B．ENGINEERING BASICS.C．JOURNEY TO SPACE. D．EXPLORA TION AND DISCOVERY.2．What can campers learn only in ENGINEERING BASICS?A．Coding. B．Creative thinking. C．Rocket design. D．Supply delivery.3．What is special about EXPLORATION AND DISCOVERY?A．It will be open in August. B．It is a 4-day Explorer Camp.C．Campers can learn problem-solving skills. D．Campers will have a chance to meet astronauts.As a college student in Boston, I formed the habit of buying used books. I enjoy the hunt, the good price and the unrecognized treasures. I find old textbooks, ex-bestsellers, and books on subjects I’ve never heard of and now must learn all about. I don’t search for rare books, first editions, or leather-bound editions but books that are worth reading.Rereading, for me, is a pleasure during retirement. The theater and the concert hall become less appealing to me, along with crowds. Staying alone and reading books have become extremely important to me. Literature needs the flesh of experience to have its full effect. Different books offer me insights and ways of expressing that stuck in my mind as grains of sand in an oyster now shine like pearls. My taste in books improves with age.While packing for a move, which occurs at intervals of five to seven years, I clear my shelves and pick my books. I abandon a few, later regret my decisions, and look for them again. Several years ago, I got rid of books related to my job-architecture. Some were design guides, reference books, product catalogs, and things that went out of date. Some were historical or centered on a period or an architect. These had given me many hours of pleasure. Will I ever open their covers again? Certainly. I held on to the red bulk of Sir Banister Fletcher’s A History of Architecture and books on Paris, Rome, and Boston.I gave away drafting equipment and instruments. I threw out rolls and rolls of paper, and old drawings of projects completed long ago, some of which had even been damaged. This time, I stay put in a cottage that suits my status and I’ve moved on in spirit. No doubt I will acquire more used books and throw away more books as passions grow and fade, like feathers changing with the seasons.4．What does the author consider most important when choosing used books？A．Their appealing covers. B．Their reasonable prices.C．Their excellent content. D．Their collection value.5．What does the author realize as he grows older?A．He has a small circle of friends. B．His attitude towards work changes.C．His communication skills improve. D．He has a deeper understanding of life and books.6．Why does the author mention books on architecture?A．To show some books are worth reading repeatedly.B．To recommend Sir Banister Fletcher’s books.C．To suggest we avoid making poor decisions.D．To explain his love for architecture.7．What can we learn about the author from the last paragraph?A．He will stop throwing away used books.B．He likes updating his collection of used books.C．He is bad at using advanced drafting equipment.D．He enjoys leading an active life in the countryside.Scientists today are concerned about the growing number of species in the world that may soon become extinct. The United Nations, along with many governments, are trying to save these endangered species. To help resolve this issue, scientists have been trying to learn why species become endangered.In the United States, loss of habitats is the main reason for species becoming endangered. As populations grow and cities or towns expand, nature is destroyed and the ecosystem is affected. Even small changes in one part of an ecosystem can have a big impact on another part of it. For example, deforestation may result in a change in ground temperature, which may kill off a certain kind of plant in the forest. Animals that eat this plant may suddenly find that there is no available food so they starve to death.Another possible cause of endangered or extinct species is climate change and rising sea levels. As the Earth gets warmer, this has profound effects on animal and plant life everywhere. For example, rising sea levels make coastal areas unfit for birds to live in. Warmer temperatures melt the ice where polar bears live, wrecking their habitat and making it harder for them to find food. The rising seas also make it harder for polar bears to swim from the cold waters to frozen packs of ice so many of them drown.There are several ways that people are trying to help endangered species recover. Ecologists study different species and try to find out more information about them to try and help. In Finland, for example, scientists found that a a decrease in the number of bogs(沼泽) was responsible for the falling butterfly population. Once they understood this, they could manage the butterfly population and help it recover. Governments are also passing laws to protect endangered animals. One example of this is an area near Hawaii with a seriously low fish population. The United States’ government has made it illegal for fishing boats to enter this area.8．What can we infer from paragraph 2?A．Most endangered species are located in the United States.B．Cities and towns in the United States are growing too fast.C．Ecosystem changes can result in habitat loss and animal death.D．More trees need to be planted to prevent animals from dying out.9．What does the underlined word “wrecking” in paragraph 3 mean?A．Conserving. B．Damaging. C．Providing. D．Dominating.10．Which is the best title for this text?A．Saving Endangered Species B．A List of Endangered SpeciesC．The Causes of Global Warming D．How Polar Bears Became Extinct11．What will the author probably discuss in the paragraph that follows?A．Some other ways to help endangered species recover.B．A brief discussion of polar bears and their diet habits.C．The impact of fishing laws on the development of Hawaii.D．Various opinions and debates concerning the cause of global warming.Virtual realty can improve brain activity that may be crucial for leaning, memory and even treating Alzheimer’s, a study on rats has found.After monitoring the rats’ brain activity, rese archers from the University of California Los Angeles discovered electrical activity in a region known as the hippocampus neurons (海马体神经元) differed depending on whether the rats were placed in real-world or VR environments. The new findings are significant as the hippocampus is a primary driver of learning and memory in the brain.When rats walk around in real life, electrical activity in the hippocampus appears to synchronize (同步发生) at a rate of eight heartbeats per second. Heartbeats at this frequency are generally known as “theta (Ѳ) waves”, with stronger Ѳ waves seeming to improve the brain's ability to learn and keep sensory information. When placed in a VR environment, the rat’s Ѳ waves became stronger.“It turns out that amazing things happen when the rat is in virtual reality” said Prof Mayank Mehta from UCLA．The scientists also found that VR environments could change different electrical rhythms (节律) in different parts of the neurons, which indicates that scientists may be able to control human brain rhythms.“This is a new technology that has great potential,” Mehta said. The study also indicates why VR may stimulate these unique brain waves. A big part of it, Mehta puts, may be down to the very different set of stimuli presented in VR.Imagine that you’re approaching a doorway in real life. Your eyes see the door getting larger. But how do you know that you’re moving forward and the door isn’t coming to you? The answer is that your brain uses information such as the acceleration of your head through space or the shift of weight from one foot to the other－information that may not be present during a VR experience.12．According to the passage, why are the new findings important?A．VR’s contributions to learning have finally been identified.B．VR can affect electrical activity in the hippocampus neurons.C．VR produces the same effects on brains as real environments.D．VR is likely to become another driver for learning and memory.13．What can we learn about Ѳ waves?A．They don’t respond to different brain rhythms.B．They remain stable in different environments.C．They affect how human perceive knowledge.D．They have lite to do with the rate of heartbeats.14．What does the underlined word “it” in Paragraph 5 refer to?A．The potential of the new technology.B．The control of human brain rhythms.C．The change in different parts of neurons.D．The stimulation of the unique brain waves.15．How does the author introduce VR’s different set of stimuli?A．By conducting further tests on rats. B．By comparing different environments.C．By providing a detailed analysis D．By explaining the theory of VR.二、七选五Five years ago, my husband and I bought a former farm. It’s a short ride from our Vancouver home but a world away. As we restore the land, I feel my well-being returning, but I think the farm is also working its magic on our son’s growing resilience (适应能力). Since Dev was 20, he has lived with mental illness. We have been on this journey with him from crisis to recovery.____16____ We found the farm.It has taken a couple of years to clear the land, seed the lawn (草坪) and build garden beds. But we did it even as other parts of our life had come undone. The farm was holding our family together. We come over on weekends and holidays and have been planting. ____17____Before he went to treatment, we planted a “guild” in the old apple farm. ____18____ We dug around the weaker trees and placed in garlic, wildflowers and bone meal. It keeps the moisture (水分) and nutrients in, and the garden seems to thrive (茁壮成长) as a result.____19____ The land needed clearing, and Doug, our workman, worked steadily and helped whenever Iasked. Asking for help is a principle in the recovery community, a lesson my son is learning. It truly does take a village to restore a farm and rebuild a life.The desire to recover this land has grown in me. ____20____ He’s found his own kind of guild that supports him. Owning a farm was never our plan but it came when our family needed a meaningful project. When my hands are in the dirt, I remember that intervention (干预) is necessary and so is allowing nature to take its course. Digging is an act of faith, hope and anticipation of what will appear next.A．We helped ourselves by helping others.B．We also learned how to depend on others.C．My son had great difficulty making a recovery.D．I have seen this drive toward restoration in my son as well.E．Just when it couldn’t get any worse, an unexpected light occurred.F．During this time, Dev is concentrating on his own restoration process.G．Its aim is to support the health of trees by grouping other surrounding components.三、完形填空In high school Norman Greenstein was a productive doodler (涂鸦手), and created lots of paintings. When he ____21____ his works to his teacher, the teacher responded, “The problem with modern a rt is that you can____22____ on a canvas and call it art.” Norman felt so disappointed, but he ____23____ parted with art over the years.After his ____24____, Norman spared no efforts to ____25____ his wife and three children with devotion. Although life was sometimes hard for him, it didn’t ruin his ____26____ for creative expression. After being diagnosed with Parkinson’s disease, the 65-- year-old social worker, decided to turn his ____27____ diagnosis into a positive opportunity to ____28____ a lifelong dream.With the help of his son, Norman was able to start a ____29____ involving his whole family, in order to make his own works ____30____ by all people around him, as well as his ____31____ experience of fighting against Parkinson’s disease.In 2021, an online ____32____ was launched to feature the colorful canvases and he also ____33____ a memoir called Spit on a Canvas: the Journey of the Parkinson’s Painter. Meanwhile, Norman began selling his first paintings, with galleries exhibiting his works. The team is also ____34____ releasing Norman’s first digital NFT art collection. 20% of the ____35____ has been donated to two Parkinson’s research foundations. 21．A．assigned B．offered C．conveyed D．showed22．A．set B．spit C．sneeze D．skip23．A．still B．occasionally C．never D．frequently24．A．retirement B．exhibition C．graduation D．marriage25．A．support B．remove C．judge D．replace26．A．calling B．praying C．longing D．waiting27．A．unbelievable B．dark C．embarrassing D．unsolved28．A．release B．refresh C．realize D．restore29．A．project B．request C．competition D．reform30．A．spotted B．recommended C．developed D．regarded31．A．thrilling B．pleasant C．plain D．stressful32．A．drama B．contest C．programme D．gallery33．A．published B．copied C．revised D．examined34．A．ending with B．working on C．relying on D．paying for35．A．salary B．reward C．sale D．allowance四、用单词的适当形式完成短文阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。
河北省“五个一”名校联盟2020届高三上学期一轮复习收官考试语文试题(解析版)
河北省“五个一”名校联盟2020届高三一轮复习收官考试语文试卷一、现代文阅读(一)论述类文本阅读阅读下面的文字，完成1-3题。
新一代人工智能正在全球范围蓬勃发展，推动世界从互联信息时代进入智能信息时代，给人们的生产生活方式带来颠覆性影响。
人工智能与经济社会的深度融合，将给人类社会发展进步带来强大新动能，实现创新式发展。
从科学层面看，人工智能跨越认知科学、神经科学、数学和计算机科学等学科，具有高度交叉性；从技术层面看，人工智能包含计算机视觉、机器学习、知识工程、自然语言处理等多个领域，具有极强专业性：从产业层面看，人工智能在智能制造、智慧农业、智慧医疗、智慧城市等领域的应用不断扩大，具有内在融合性；从社会层面看，人工智能给社会治理、隐私保护、伦理道德等带来新的影响，具有全面渗透性。
目前，在边界清晰、规则明确、任务规范的特定应用场景下(如下围棋、人脸识别、语音识别)设计出的智能体表现出较好的专用智能。
未来，人工智能的发展将从专用人工智能、人机共存智能向通用人工智能转变。
可以预见，通过科学研究的牵引、应用技术的交叉，人工智能必将推动人类社会实现创新式发展。
人工智能为人类认识世界引入新范式，增强科学发现能力。
人工智能的发展史是相关学科不断交叉融合、遵循不同范式的发展过程。
从符号主义、逻辑推理、知识工程到连接主义，从大数据驱动小任务到小数据驱动大任务，从神经形态类脑智能到量子计算智能，人工智能的新范式不断增强人类认识世界的能力。
传统的科学研究引入新范式后，研究效能得到了极大提升。
人工智能为人类理解世界创造新工具，扩展外界认知能力。
人工智能创造各种技术帮助人类理解复杂的拥有巨量信息的世界。
计算机视觉技术利用感知世界的每一个像素，增强人们观察场景的敏锐度。
自然语言处理技术通过深度语义分析，改善人和机器的交流互动。
知识计算引擎与知识服务技术帮助我们搜集获取海量知识，进而挖掘关系，形成新的知识图谱。
自主无人系统可以利用其不怕热、不怕冷；不怕压等特性，涉足人类无法到达或难以忍受的极端环境，帮助我们探测未知世界。
【河北卷】河北省琢名小渔2025届“五个一”名校联盟高三年级第一次联考英语试卷答案
“五个一”名校联盟2025 届高三第一次联考语文答案与解析1 . 【答案】B【解析】A 项错误，“先后”表述不当。
从原文可见，“在多巴胺之前，所谓的‘快乐分子’是血清素。
”血清素在先，多巴胺在后。
C 项错误，“都是分布在大脑中的神经递质”曲解文意。
原文并没有说“肾上腺素”是“分布在大脑中的神经递质”；事实上，“肾上腺素” 由肾上腺释放；文中说“多巴胺会通过酶转化为去甲肾上腺素，而去甲肾上腺素随即通过其他酶转化为肾上腺素”。
D 项错误，“使人变得愉快”以偏概全。
原文：“而当你没有得到你期待的东西时，它们会生闷气。
”由此可见，多巴胺的增加也可能使人不愉快。
2 . 【答案】C【解析】“才会产生积极的情绪”错误。
本文核心观点为“多巴胺奖励机制”，“它与新奇和学习有关，与标记意外事件有关，它会标注这意外事件是正面还是负面，然后触发恰当的反应。
”3 . 【答案】B【解析】“多巴胺奖励机制”，即调节“想要”某种特定奖励的心理过程的预测误差机制。
A 项不符合多巴胺“与新奇和学习有关，与标记意外事件有关”的条件；C 项和D 项都提到了增加多巴胺产生量的具体做法，D 项还介绍了多巴胺改善身体健康的功能，但与“多巴胺奖励机制”无关。
4 . 【答案】A【解析】文中谈到的“快乐分子”只有多巴胺和血清素。
5 . ①最初刷短视频时，一些“新奇”的内容会使大脑释放大量的多巴胺，给人们带来愉悦、兴奋等情绪；(2 分）②然后，人们的欲望会不断增强，即使刷短视频不再像刚开始时那么令人愉悦，人们仍然渴望它们； (2 分）③大脑释放的多巴胺会使人们不断寻找更多新奇的短视频，不刷就会感到空虚寂寞，很多人刷起来停不下，最终成瘾。
(2 分）【解析】设问围绕学生的生活实际展开，让学生用材料的观点来阐释具体问题。
首先要弄清楚理论—多巴胺奖励机制的特点，然后结合刷短视频上瘾事例分三步作答。
（一点2 分，三点6 分，意思接近即可。
）6 . 【答案】A【解析】“塑造了两个工作作风迥异的领导干部形象”错误。
2025届河北省“五个一”名校联盟高三第一次联考英语试卷
河北省“五个一”名校联盟2025届高三第一次联考英语本试卷共8页，满分150分，考试用时120分钟。
注意事项：1．本试卷由四个部分组成，其中，第一、二部分和第三部分的第一节为选择题，第三部分的第二节和第四部分为非选择题。
2．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。
3．回答选择题时，选出每小题答案后，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。
4．考试结束后，将本试卷和答题卡一并交回。
第一部分听力（共两节，满分30分）做题时，先将答案标在试卷上。
录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节（共5小题；每小题1.5分，满分7.5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。
例：How much is the shirt?A．£ 19.15.B．£9.18.C．£9.15.答案：C。
1．What day is it today?A．Thursday.B．Friday.C．Saturday.2．What does the man like least about the movie?A．The action scenes.B．The special effects.C．The main character.3．What is happening with the girl?A．She is looking for a job.B．She is visiting a new city.C．She is distancing herself from her father.4．What difficulty does the man have in learning French?A．Pronunciation.B．Spelling.C．Grammar.5．What are the speakers mainly talking about?A．Mike’s good manners.B．Mike’s eye problem.C．Mike’s meeting with his boss.第二节（共15小题，每小题1.5分，满分22.5分）听下面5段对话或独白。
2023届河北省五个一名校联盟高三年级模拟底考试化学试卷(word版)
2023届河北省五个一名校联盟高三年级模拟底考试化学试卷(word版)一、单选题(★★★) 1. 化学与生产、生活密切相关，下列叙述错误的是（）A．绿色化学要求从源头上消除或减少生产活动对环境的污染B．用浸泡过高锰酸钾溶液的硅藻土吸收水果释放的乙烯，可达到水果保鲜的目的C．可溶性的铝盐和铁盐可用作净水剂D．从海水中提取物质都必须通过化学反应才能实现(★★★) 2. 下列叙述正确的是（）A．中子数为10的氧原子：B．的结构示意图：C．HClO的结构式为：H-O-Cl D．的电子式：(★★★) 3. 下列说法不正确的是（）A．钾、钠、镁等活泼金属着火时，不能用泡沫灭火器灭火B．油脂、淀粉、蔗糖和葡萄糖在一定条件都能发生水解反应C．除去干燥中混有的少量，可将混合气体依次通过盛有酸性溶液、浓硫酸的洗气瓶D．“中和滴定”实验中，锥形瓶用蒸馏水洗涤后即可使用，滴定管用蒸馏水洗涤后，需经润洗方可使用(★★★) 4. 下列离子方程式能用来解释相应实验操作或者现象的是（）A．A B．B C．C D．D(★★★) 5. 用表示阿伏加德罗常数的值。
下列叙述中正确的是（）A．盐酸与足量共热，转移的电子数为B．电解精炼铜时，若阳极质量减少64g，则阴极得到的电子数为C．0.1mol和0.1mol于密闭容器中充分反应后，其分子总数为D．反应中生成3mol转移的电子数为(★) 6. 下列操作或装置能达到实验目的的是（）配制一定浓度的NaClA．A B．B C．C D．D(★★★) 7. 化合物M(结构如图所示)是常用的还原剂之一，其中X、Y、Z是原子序数递减的不同周期短周期元素，X与W、Z与W均能形成原子个数比为1∶1和2∶1的化合物，W的最外层电子数是电子层数的3倍。
下列叙述正确的是（）A．Y的最高价氧化物对应的水化物是一种弱酸B．Z位于第三周期第ⅦA族C．X和Z形成的化合物的水溶液呈酸性D．化合物M不能与反应(★★★) 8. 下列关于溶液中所含离子检验的说法正确的是（）A．取少量某溶液于试管中，向其中先滴加稀硝酸，再滴加溶液，有白色沉淀生成，说明该溶液中含有B．取少量某溶液于试管中，向其中加入少量新制氯水，再滴加KSCN溶液，若溶液变红，说明该溶液中含有C．取少量某溶液于试管中，加入NaOH溶液并加热，用湿润的红色石蕊试纸检验产生的气体，若红色石蕊试纸变蓝，说明该溶液中含有D．将某溶液与稀盐酸反应产生的气体通入澄清石灰水中，石灰水变浑浊，说明该溶液中含有(★★★) 9. 可将气态废弃物中的硫化氢转化为可利用的硫，自身还原为。
河北省“五个一”名校联盟2025届高三8月第一次联考化学试卷+答案
河北省“五个一”名校联盟2025届高三第一次联考化学本试卷共8页，满分100分，考试用时75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上。
2．回答选择题时，选出每小题答案后，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交回。
可能用到的相对原子质量：O-16 S-32 Zn-65一、选择题：本题共14小题，每小题3分，共42分。
在每小题给出的四个选项中，只有一项符合题目要求。
1．易县燕下都遗址出土的透雕龙凤纹铜铺首（如图）是战国时期青铜器，下列说法错误的是（ ）A ．青铜是铜锡合金B ．青铜的熔点高于纯铜C ．青铜器表面的铜绿主要成分为()232Cu OH COD ．青铜器应保存在相对干燥的环境中 2．下列说法错误的是（ ）A ．氯酸钾和硫黄不能保存在同一药品柜中B ．硝酸银溶液应保存在棕色细口瓶中C ．镁条着火时应使用二氧化碳灭火器灭火D ．滴定实验中的锥形瓶不需要润洗3．化学与生产、生活、能源、环境密切相关，下列说法正确的是（ ）A ．电解氯化钾溶液可制得钾单质和氯气B ．包装上有“OTC ”标识的药物可以放心大量使用 C ．太阳能电池板的主要成分为二氧化硅D ．含酚废水可用二氧化氯进行氧化处理4．设A N 为阿伏加德罗常数的值。
下列说法正确的是（ ）A ．11L0.1mol L −⋅的3NaHSO 溶液中含有A 0.1N 个23SO −B ．218gD O 中含有A 10N 个中子C ．1mol 共价晶体2CO 中含有A 2N 个C O −健D ．标准状况下11.2LHCHO 中含有A 0.5N 个π键5．中药材当归和白芷中提取得到的紫花前胡醇（）能提高人体免疫力。
下列关于紫花前胡醇的说法错误的是（ ） A ．分子式为14144C H O B ．1mol 该物质最多能与25molH 发生反应 C ．能使酸性高锰酸钾溶液褪色D ．能发生水解反应6．关于物质结构的下列说法正确的是（ ） A ．HCN 和2SO 均为直线形分子 B ．基态Cr 和2Fe +的未成对电子数不同 C ．4CH 和3CH I 分子中C H −键的夹角相同 D ．22Na O 和24C H 中均存在极性键和非极性健7．乙酸苯酯在一定条件下可发生如下所示的Fries 重排反应，下列说法错误的是（ ）A ．X 中碳原子均为2sp 杂化B ．Y 的熔点高于ZC ．Y 在水中的溶解度大于ZD ．X 、Y 、Z 互为同分异构体8．化合物M 的结构如图所示，该物质是一种抗真菌药物，用于治疗真菌性髓膜炎、真菌性呼吸道感染及黑色真菌症。