河南省驻马店市正阳县高级中学2019-2020学年高一上学期第三次素质检测英语试卷+PDF版含答案
河南省驻马店市正阳县高级中学2019-2020学年高一上学期第三次素质检测数学(理)试卷

数学试题(理科)一、单选题(每小题5分,共60分) 1.已知集合A ={x |x <1},B ={x |31x <},则A .{|0}AB x x =< B .A B R =C .{|1}A B x x =>D .A B =∅2.函数22y x x =-+,[]0,3x ∈的值域为( )A .[]0,3B .[]3,0-C .[]3,1-D .[]0,13.过两点A (4,y ),B (2,-3)的直线的倾斜角是135°,则y 等于 ( )A .1B .5C .-1D .-54.函数()12x f x a -=+(0a >且1a ≠)的图象恒过定点( )A .()0,3B .()1,3C .()1,2-D .()1,3-5.若奇函数()f x 在[1,3]上为增函数,且有最小值-1,则它在[3,1]--上 ( ) A .是减函数,有最小值-1 B .是增函数,有最小值-1 C .是减函数,有最大值1 D .是增函数,有最大值16.函数()23log f x x x=-的零点所在的大致区间是( )A .(0,1)B .(1,2)C .(2,3)D .(3,4)7.已知圆锥的表面积为6,且它的侧面展开图是一个半圆,则这个圆锥的底面半径为A B C D 8.设函数()f x 是R 上的奇函数,当0x >时,()3x f x e x =+-,则()f x 的零点个数是A .1B .2C .3D .49.若直线1:220l ax y ++=与直线2:(1)10l x a y +-+=平行,则实数a 的值是( )A .2B .1-或2C .1-D .010.已知()f x 是定义在(,)-∞+∞上的偶函数,且在(,0]-∞上是增函数,设4(log 7)a f =,12(log 3)b f =, 1.6(2)c f =,则,,a b c 的大小关系是( )A .c a b <<B .b c a <<C .c b a <<D .a b c <<11.已知,l m 是两条不同的直线,,αβ是两个不同的平面,则下列命题正确的是( )A .若,l l m α⊥,则m α⊥B .若,l l αβ,则αβ∥C .若,l ααβ⊥⊥,则l β∥D .若,l l αβ⊥⊥,则αβ∥12.函数()f x 的定义域为R ,其图像上任意两点111222(,),(,)P x y P x y 满足2121()()0x x y y --<, 若不等式(22)(4)x x f m f m -<-恒成立,则m 的取值范围是( )A .[)0+∞,B .(],0-∞C .14⎡⎫-+∞⎪⎢⎣⎭, D .14⎛⎤-∞ ⎥⎝⎦,-二、填空题(每小题5分,共20分)13.021.10.5lg252lg2-++=__________.14.点(5,2)到直线5121-=-+-m y m x m )()(的距离的最大值为________。
2019-2020学年河南省驻马店市正阳县高级中学高一上学期第三次素质检测化学试题(解析版)

C.排水法收集满KMnO4分解产生的O2后,先移出导管,后熄灭酒精灯
D.使用分液漏斗前首先检查其气密性是否良好
『答案』C
『解析』
『详解』A.用试管加热固体时,试管口应向下倾斜,防止水蒸气在试管口冷凝,流回试管底,使试管底因受热不均而破裂,A错误;
『答案』D
『解析』
『详解』A. NaHCO3溶液中加入NaOH溶液,反应生成碳酸钠和水,正确的离子方程式为:HCO3-+OH-═CO32-+H2O,故A错误;
B.铁和稀硫酸反应生氢气和二价铁,正确的离子方程式为:Fe+2H+=Fe2++H2 ,故B错误;
C.少量金属钠加到冷水中的反应实质是:2Na+2H2O═2Na++2OH-+H2↑,故C错误;
河南省驻马店市正阳县高级中学2019-2020学年高一上学期第三次素质检测试题
可能用到的相对原子质量:H 1 He 4 C 12 N 14 O 16 Na 23 Al 27 S 32 Cl 35.5 K 39 P 31 Ba 137 Fe 56 Cu 64 Zn 65
第Ⅰ卷(选择题)
一、选择题(每小题3分,共16小题,共48分。在四个选项中,只有一项是符合题目要求的)
B、②符合,复分解反应中各物质反应前后化合价不变,③中元素化合价有变化,所以不是复分解反应,B错误;
C、氯化镍与氢氧化钠能发生复分解反应,因交换成分有沉淀氢氧化镍生成,C正确;
D、①中有单质形式变成了化合物形式,化合价一定改变;③中根据化合物中元素化合价代数和为零,反应物中镍的化合价为+4价,生成物中镍的化合价为+2价,也发生了改变,D正确;
河南省驻马店市正阳县高级中学2019-2020学年高一数学上学期第三次素质检测试题理【含答案】

三次素质检测试题 理
一、单选题(每小题 5 分,共 60 分)
1.已知集合 A={x|x<1},B={x| 3x 1 },则
A. A B {x | x 0}
B. A B R
C. A B {x | x 1}
D. A B
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2 A.
1 B.
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)
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C. c b a
D. a b c
数学试题驻马店市2019_2020学年高一数学上学期第三次素质检测试题文

河南省驻马店市正阳县高级中学2019-2020学年高一数学上学期第三次素质检测试题 文一、单选题(每小题5分,共60分)1.设集合,,则A .B .C .D .2.函数22y x x =-+,[]0,3x ∈的值域为( )A .[]0,3B .[]3,0-C .[]3,1-D .[]0,13.过两点A (4,y ),B (2,-3)的直线的倾斜角是135°,则y 等于 ( )A .1B .5C .-1D .-5 4.函数()12x f x a-=+(0a >且1a ≠)的图象恒过定点( )A .()0,3B .()1,3C .()1,2-D .()1,3-5.若奇函数()f x 在[1,3]上为增函数,且有最小值1,则它在[3,1]--上 ( )A .是减函数,有最小值1B .是增函数,有最小值-1C .是减函数,有最大值1D .是增函数,有最大值-1 6.函数()23log f x x x=-的零点所在的大致区间是( ) A .(0,1)B .(1,2)C .(2,3)D .(3,4)7.若直线1:220l ax y ++=与直线2:(1)10l x a y +-+=平行,则实数a 的值是( ) A .2 B .1-或2 C .1-D .08.下列函数中,在其定义域既是奇函数又是减函数的是( )A .y=|x|B .y=﹣3xC .1y x x=+D .y=9.已知a =log 20.3,b =20.1,c =0.21.3,则a ,b ,c 的大小关系是( )A .a b c <<B .c a b <<C .b c a <<D .a c b << 10.已知圆锥的表面积为6,且它的侧面展开图是一个半圆,则这个圆锥的底面半径为A B C D 11.已知,l m 是两条不同的直线,,αβ是两个不同的平面,则下列命题正确的是( )A .若,l l m α⊥,则m α⊥B .若,l l αβ,则αβ∥C .若,l ααβ⊥⊥,则l β∥D .若,l l αβ⊥⊥,则αβ∥12.设函数()f x 是R 上的奇函数,当0x >时,()3xf x e x =+-,则()f x 的零点个数是A .1B .2C .3D .4二、填空题(每小题5分,共20分)13.021.10.5lg252lg2-+++=__________.14.过点A (-1,0)且与直线2x -y +1=0平行的直线方程为________.15.已知函数()f x 是定义在R 上的偶函数,且在[)0,+∞上单调递增,若()30f -=,实数a 满足()250f a -≤,则a 的取值范围为________.16.已知正三棱柱的各条棱长都相等,且内接于球,若正三棱柱的体积是,则球的表面积为_____. 三、解答题17.(10分)已知全集{}|0U x x =>,集合{}{}{}|37|210|5A x x B x x C x a x a =≤==-<,<<,<<.(1)求()U A B C A B ⋃⋂,;(2)若()C A B ⊆⋃,求a 的取值范围.18.(12分)(1)求过直线220x y -+=与220x y --=的交点,且与直线3+410x y +=垂直的直线方程.(2)求经过点(1,2)且在x 轴上截距等于y 轴上截距的直线方程;19.(12分)已知函数()2f x x ax b =++为偶函数,且有一个零点为2.(1)求实数a ,b 的值.(2)若()()g x f x kx =-在[]0,3上的最小值为-5,求实数k 的值.20.(12分)已知函数()20,log 0,ax x f x x x +≤⎧=⎨>⎩,,且点(4,2)在函数f (x )的图象上.(1)求函数f (x )的解析式,并在图中的直角坐标系中画出函数f (x )的图象;(2)求不等式f (x )<1的解集;21.(12分)如图,矩形ABCD 所在平面与半圆弧CD 所在平面垂直,M 是CD 上异于C ,D 的点.(1)证明:平面AMD ⊥平面BMC ;(2)在线段AM 上是否存在点P ,使得MC ∥平面PBD ?说明理由.22.(12分)已知函数()21ax bf x x +=+定义在()1,1-上的奇函数,且1225f ⎛⎫= ⎪⎝⎭. (1)求函数()f x 的解析式;(2)判断函数()f x 的单调性,并证明; (3)解关于x 的不等式()()210f x f x -+<.数学(文科)参考答案1.D 2.C 3.D 4.B 5.D 6.C 7.C 8.B 9.D 10.A 11.D 12.C 13.5 14.220x y -+= 15.14a ≤≤ 16.17.(1){|210}x x <<, {|23x x <<或710}x ≤<。
河南省正阳高中2020学年度高三英语上期第三次素质检测试题

河南省正阳高中2020学年度高三英语上期第三次素质检测试题第一卷(共115分)第一部分:听力(共两节,满分30分)听下面5段对话,每段对话有一个小题,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 are they talking about?A. Diving in EnglandB. How to drive a car.C. Whether to have the right to drive a car in England.2. What does the woman mean?A. Mary is illB. Mary thinks well of the concert.C. She has no chance to talk to Mary.3. Does Jane know Arvid Brown?A. She doesn’t know himB. She knows something about himC. She knows him very well.4.What can we learn from the conversation?A. They are neighborsB. They are classmates.C. They are not from the same country.5. Where does he conversation probably take place?A. In a clothing store.B. At the woman’s home.C. In the sitting room.听下面几段对话或独白。
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,你交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。
河南省驻马店市正阳县高级中学2019-2020学年高一上学期第三次素质检测数学(文)试题

一、单选题
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,则 ,则
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,则
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二、填空题
13.
__________.
D.4
14. 过点A(-1,0)且与直线2x-y+1=0平行的直线方程为________.
8. 下列函数中,在其定义域既是奇函数又是减函数的是( )
A.y=|x|
B.y=﹣
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9. 已知a=log20.3,b=20.1,c=0.21.3,则a,b,c的大小关系是( )
A.
B.
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D.
10. 已知圆锥的表面积为6,且它的侧面展开图是一个半圆,则这个圆锥的底面半径为
A.
A.
B.
C.
D.
5. 若奇函数 在 上为增函数,且有最小值1,则它在 A.是减函数,有最小值1 B.是增函数,有最小值-1
河南省驻马店市正阳县高级中学2020学年高一英语上学期第三次素质检测试题(最新整理)

河南省驻马店市正阳县高级中学2019—2020学年高一英语上学期第三次素质检测试题第一部分听力(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,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。
Where will they meet?A. At the underground。
B. At the cinema.C. At the museum.2. Which bus will the woman take?A。
Bus number 4。
B. Bus number 6.C。
Bus number 2.3。
What is the man doing?A。
Saying good—bye to a friend. B。
Arranging a plane trip. C. Paying a bill at the bank。
4。
How will the woman pay for the meal?A。
With Master Card. B. With Visa Card。
C。
With cash。
5。
When will the woman get the TV?A。
In about an hour. B. A long time。
C. Several days。
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,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8题.6。
Where does the conversation probably take place?A。
河南省驻马店市正阳县高级中学2019-2020学年高一上学期第三次素质检测(理)数学试题(解析版)

河南省驻马店市正阳县高级中学2019-2020学年高一上学期第三次素质检测(理)试题一、单选题(每小题5分,共60分)1.已知集合A ={x |x <1},B ={x |31x<},则( )A. {|0}A B x x =<B. A B =RC. {|1}AB x x =>D. AB =∅『答案』A『解析』∵集合{|31}xB x =<,∴{}|0B x x =< ∵集合{|1}A x x =<,∴{}|0A B x x ⋂=<,{}|1A B x x ⋃=<故选A2.函数22y x x =-+,[]0,3x ∈的值域为( )A.[]0,3B.[]3,0-C.[]3,1-D.[]0,1『答案』C『解析』二次函数22y x x =-+图象开口向下,对称轴为直线1x =,该函数在区间[]0,1上单调递增,在区间[]1,3上单调递减,所以,当1x =时,函数22y x x =-+取得最大值,即max 121y =-+=. 当0x =时,0y =,当3x =时,23233y =-+⨯=-,该函数的最小值为min 3y =-. 因此,函数22y x x =-+,[]0,3x ∈的值域为[]3,1-.故选C.3.过两点A (4,y ),B (2,-3)的直线的倾斜角是135°,则y 等于 ( ) A. 1B. 5C. -1D. -5『答案』D『解析』∵过两点A (4,y ),B (2,-3)的直线的倾斜角是135°, ∴3tan135142y +=︒=--,解得5y =-.选D . 的4.函数()12x f x a -=+(0a >且1a ≠)的图象恒过定点( )A.()0,3 B.()1,3 C.()1,2-D.()1,3-『答案』B 『解析』()12x f x a -=+,当1x =时,()13f =,即函数图像恒过定点()1,3故选B.5.若奇函数()f x 在[1,3]上为增函数,且有最小值-1,则它在[3,1]--上( ) A. 是减函数,有最小值-1 B. 是增函数,有最小值-1 C. 是减函数,有最大值1 D. 是增函数,有最大值1『答案』D『解析』因为奇函数()f x 在[1,3]上为增函数, 且有最小值-1 所以函数()f x 在[3,1]--上为增函数,且有最大值1. 故选:D6.函数()23log f x x x =-的零点所在的大致区间是( )A. (0,1)B. (1,2)C. (2,3)D. (3,4)『答案』C『解析』由题意,()3121022f =-=-<,()23log 310f =->,所以,所以函数()23log f x x x=-的零点所在的大致区间是()2,3,故选C. 7.已知圆锥的表面积为6,且它的侧面展开图是一个半圆,则这个圆锥的底面半径为A.B. C. D. 『答案』A『解析』设底面半径为R ,侧面展开图半径为r ; 底面周长等于侧面半圆周长,即2ππ,2R r r R ==2221ππ3π6,2S R r R R =+===表A 8.设函数()f x 是R 上的奇函数,当0x >时,()e 3xf x x =+-,则()f x 的零点个数是( )A. 1B. 2C. 3D. 4『答案』C『解析』∵函数f (x )是定义域为R 的奇函数,∴f (0)=0,所以0是函数f (x )的一个零点;当x >0时,令f (x )=e x +x -3=0,则e x =-x +3,分别画出函数y =e x ,和y =-x +3的图象,如图所示,有一个交点,所以函数f (x )有一个零点,又根据对称性知,当x <0时函数f (x )也有一个零点.综上所述,f (x )的零点个数为3个, 故选C . 9.若直线1:220l ax y ++=与直线2:(1)10l x a y +-+=平行,则实数a 的值是( )A. 2B. 1-或2C. 1-D. 0『答案』C『解析』∵已知两直线平行,∴(1)20a a --=,解得1a =-或2a =,2a =时,两直线重合,舍去,1a =-时两直线平行.故选C .10.已知()f x 是定义在(,)-∞+∞上的偶函数,且在(,0]-∞上是增函数,设4(log 7)a f =,12(log 3)b f =,1.6(2)c f =,则,,a b c 的大小关系是( ) A. c a b << B. b c a <<C. c b a <<D. a b c <<『答案』C 『解析』()f x 是定义在(,)-∞+∞上的偶函数,1222(log 3)(log 3)(log 3)b f f f ∴==-=, 22442log 4log 3log 9log 71=>=>>, 1.6122>,1.6420log 7log 32<<<,在(-∞,0]上是增函数,∴在[0,)+∞上为减函数, 则1.642(log 7)(log 3)(2)f f f >>,即c b a <<,故选C .11.已知,l m 是两条不同的直线,,αβ是两个不同的平面,则下列命题正确的是( ) A. 若,l l m α⊥,则m α⊥ B. 若,l l αβ,则αβ∥ C. 若,l ααβ⊥⊥,则l β∥ D. 若,l l αβ⊥⊥,则αβ∥ 『答案』D『解析』当直线,l m 是相交且垂直,确定的平面与α平行时,m α,故A 错误; 当,αβ相交,直线l 与交线平行时,,l l αβ,故B 错误;当直线l 在面β内,且αβ⊥,直线l 垂直,αβ的交线时,l α⊥,故C 错误; 垂直与同一直线的两个平面平行,故D 正确. 故选D.12.函数()f x 的定义域为R ,其图像上任意两点111222(,),(,)P x y P x y 满足2121()()0x x y y --<,若不等式(22)(4)x xf m f m -<-恒成立,则m 的取值范围是( ) A. [)0+∞, B. (],0-∞ C.14⎡⎫-+∞⎪⎢⎣⎭, D. 14⎛⎤-∞ ⎥⎝⎦,-『答案』B 『解析』任意两点111222(,),(,)P x y P x y 满足2121()()0x x y y --<,则函数单调递减.(22)(4)x x f m f m -<-恒成立,即224x x m m ->-恒成立.设2(0)xt t => 故222411111()(0)3333212x x m m t t t t >∴<+=+->+ 2111()0(0)3212t t +->>恒成立,所以0m ≤故选B.二、填空题(每小题5分,共20分)13.021.10.5lg 252lg 2-+++=__________. 『答案』5 『解析』原式()2162lg 254325=+-+⨯=+=.14.点(5,2)到直线()1(21)5m x m y m -+-=-的距离的最大值为________.『答案』『解析』化简()()1215m x m y m -+-=-可得m ()()2150x y x y +--+-=,由2109504x y x x y y +-==⎧⎧⇒⎨⎨+-==-⎩⎩, 所以()()1215m x m y m -+-=-过定点()9,4-,点(5,2)到直线()()1215m x m y m -+-=-的距离的最大值就是点(5,2)与点()9,4-==故答案为 15.已知函数()f x 是定义在R 上的偶函数,且在[)0,+∞上单调递增,若()30f -=,实数a 满足()250f a -≤,则a 的取值范围为____________.『答案』14a ≤≤ 『解析』因为函数()f x 是定义在R 上的偶函数,且()30f -=所以()()()252503f a f a f -=-≤=,因为()f x 在[)0,+∞上单调递增,所以253a -≤,解得14a ≤≤.故答案为:14a ≤≤ 16.已知正三棱柱111ABC A B C -的各条棱长都相等,且内接于球O ,若正三棱柱111ABC A B C -的体积是O 的表面积为_____.『答案』28π3『解析』设111AA A B a ==,则正三棱柱111ABC A B C -的体积是34a =解得2a =,底面正三角形的外接圆半径2sin 60a r ==︒,所以球的半径R ==,所以球O 的表面积为228π4π3R =. 三、解答题 17.已知全集{}|0U x x =>,集合{}{}{}|37|210|5A x x B x x C x a x a =≤==-<,<<,<<.(1)求()U A B C A B⋃⋂,;(2)若()C A B ⊆⋃,求a的取值范围.『解』(1) {|210}A B x x ⋃=<<,{|037}U C A x x x =<<≥,或,(){|23U C A B x x ⋂=<<,或710}x ≤<.(2)①若C 为空集,则5a a ≥-,解得a52≤.②若C 不是空集,则2510a a ≤<≤-,解得53.2a <≤综上所述, 3a ≤, 即a 的取值范围是(],3-∞18.(1)求过直线220x y 与220x y --=的交点,且与直线3+410x y +=垂直的直线方程.(2)求经过点(1,2)且在x 轴上截距等于y 轴上截距的直线方程;『解』(1)由220220x y x y -+=⎧⎨--=⎩得22x y =⎧⎨=⎩,交点为(2,2).设所求直线430x y C -+= 代入点(2,2)得,C =-2 故所求直线方程为4320x y --=.(2)当直线过原点时,直线方程为20x y -=;当直线不过原点时,设直线方程为x 1(ya a +=或x )y a +=直线经过(2,1)21a +=即3a =直线方程为30x y +-=综上所述:直线方程为20x y -=或30x y +-= 19.已知函数()2f x x ax b=++为偶函数,且有一个零点为2.(1)求实数a ,b 的值. (2)若()()g x f x kx=-在[]0,3上的最小值为-5,求实数k 的值.详解』(1)因为函数()2f x x ax b=++为偶函数,所以()()22f x f x x ax b x ax b=-++=-+,,即20ax =,因此0a =,又因为零点为2,所以()2040 4.f b b =+==-,, (2)()()24g x f x kx x kx =-=--,当2k<0时,()g x 在[]0,3上的最小值为()045g =-≠-,舍去,当2k >3时,()g x 在[]0,3上的最小值为()103535,63g k k =-=-=<,舍去,当02k ≤≤3时,()g x 在[]0,3上的最小值为245,224k k g k ⎛⎫=--=-=± ⎪⎝⎭,因为02k≤≤3,所以2k =,综上2k =.20.已知函数()20,log 0,a x x f x x x +≤⎧=⎨>⎩,,且点(4,2)在函数f (x )的图象上. (1)求函数f (x )的解析式,并在图中的直角坐标系中画出函数f (x )的图象; (2)求不等式f (x )<1的解集;(3)若方程f (x )-2m =0有两个不相等的实数根,求实数m 的取值范围.『解』(1)∵点()4,2在函数的图象上,∴()4log 42a f ==,∴2a =.∴()220,log 0,x x f x x x +≤⎧=⎨>⎩,, . 画出函数的图象如下图所示.(2)不等式()1f x <等价于20,log 1,x x >⎧⎨<⎩或0,21,x x ≤⎧⎨+<⎩解得02x <<,或1x <-,所以原不等式的解集为()(),10,2-∞-⋃.(3)∵方程f (x )-2m =0有两个不相等的实数根,∴函数2y m =的图象与函数()y f x =的图象有两个不同的交点. 结合图象可得22m ≤,解得1m ≤ ∴实数m 的取值范围为(],1-∞.21.如图,矩形ABCD 所在平面与半圆弧CD 所在平面垂直,M 是CD 上异于C ,D 的点. (1)证明:平面AMD ⊥平面BMC ;(2)在线段AM 上是否存在点P ,使得MC ∥平面PBD ?说明理由.『解』(1)由题设知,平面CMD ⊥平面ABCD ,交线为CD .因为BC ⊥CD ,BC ⊂平面ABCD ,所以BC ⊥平面CMD ,故BC ⊥DM . 因为M 为CD 上异于C ,D 的点,且DC 为直径,所以DM ⊥CM . 又BC ∩CM =C ,所以DM ⊥平面BMC . 而DM ⊂平面AMD ,故平面AMD ⊥平面BMC . (2)当P 为AM 的中点时,MC ∥平面PBD .证明如下:连结AC 交BD 于O .因为ABCD 为矩形,所以O 为AC 中点. 连结OP ,因为P 为AM 中点,所以MC ∥OP .MC ⊄平面PBD ,OP ⊂平面PBD ,所以MC ∥平面PBD .22.已知定义域为R 的函数2()21x xaf x -+=+是奇函数.(1)求实数a 的值,并判断f (x )的单调性;(2)已知不等式3(log )(1)04m f f +->恒成立, 求实数m 的取值范围.『解』(1)()f x 是R 上的奇函数,()00f ∴=,()10011af -+==+ 得1a =,经检验1a =时,函数为奇函数.∴22()12112x x x a f x -+==-+++, 2x y =是R 上的增函数,()f x ∴在R 上为减函数. (2)不等式()3log 104m f f ⎛⎫+-> ⎪⎝⎭恒成立,()3log 14m ff ⎛⎫∴>-- ⎪⎝⎭()f x 是奇函数,()()11f f ∴--=,即不等式()3log 14m f f ⎛⎫> ⎪⎝⎭恒成立又()f x 在R 上是减函数,∴不等式3log 14m<恒成立当01m <<时,得34m <304m ∴<<当1m 时,得34m >1m ∴>综上,实数m 的取值范围是()30,1,4⎛⎫+∞ ⎪⎝⎭.。
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英语试题第一部分听力(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,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. Where will they meet?A. At the underground.B. At the cinema.C. At the museum.2. Which bus will the woman take?A. Bus number 4.B. Bus number 6.C. Bus number 2.3. What is the man doing?A. Saying good-bye to a friend.B. Arranging a plane trip.C. Paying a bill at the bank.4. How will the woman pay for the meal?A. With Master Card.B. With Visa Card.C. With cash.5. When will the woman get the TV?A. In about an hour.B. A long time.C. Several days. 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,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8题。
6. Where does the conversation probably take place?A. At school.B. At home.C. In the office.7. How many hours will the girl work a day?A. 2.B.4.C.6.8. How much could she get in a week?A. 24 dollars.B. 80 dollars.C. 120 dollars.听第7段材料,回答第9至11题。
9. What does the man do?A. A doctor.B. A psychologist.C. A news reporter.10. Which of the following room did the man book before he came?A. A warm room with a shower.B. A single room with a shower.C. Room 523.11. When is the man checking in?A. Saturday.B. Thursday.C. Tuesday.听第8段材料,回答第12至14题。
12. What are the speakers talking about?A. Adventure.B. Vacation.C. Weather.13. Which is the best part of the woman’s tour?A. Hiking.B. Skydiving.C. Visiting relatives.14. Why did the man stay at home and watch TV?A. It was raining.B. He was too tired.C. He was with his relatives.听第9段材料,回答第15至17题。
15. What do we learn about Ms Rowling’s first book?A. It was written for adults.B. It made her famous soon.C. It was about a kind of animal.16. Why does Ms Rowling consider herself so lucky?A. She can make a living by writing.B. She is loved by many people.C. She likes writing so much.17. Where did Ms Rowling get the ideas for the Harry Potter books?A. She read many books written by others.B. She collected them from daily life.C. She didn’t know it herself.听第10段材料,回答第18至20题。
18. How long should the participants(参与者) in Big Brother stay inside?A. 24 days.B. 100 days.C. 124 days.19. Who will be the winner of Survivor?A. The one who is left at last.B. The one who can find their own food.C. The one who wins one million dollars.20. What is the purpose of the passage?A. To introduce Survivor.B. To introduce Big Brother.C. To introduce reality shows.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,共30分)请阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该选项的标号涂黑。
AThe convenience store in school is a place where many students enjoy a snack and a chat.However, it may become a thing of the past. China's educational and health authorities have forbidden convenience stores in all primary, middle and high schools across the country. The new rule has come into effect (开始实施)since April 1, 2019.The new rule caused heated discussions. Many st udents were sad about it. “Without convenience stores, our school life will be less interesting,” said Fang, a student from a senior high school. “We won't be able to have different delicious drinks or desserts.”However, many parents expressed different opinions about it. The father of a junior school student said that “cheap,unhealthy snacks sold at these stores are harmful to children’s health.” “Students rushed to shops to buy snacks as soon as a class was over, which influenced the normal teaching,” h e added.The rule replies to food health concerns at public schools and is to meet the health needs of students, The Beijing News reported. It also requires schools to help students follow a healthy diet.And in fact many foreign countries have their own ways to make students eat healthily. Many parents in Australia volunteer in schools, sometimes in school dining rooms. Theseparents, at least, do their best to see that their children are served healthy food. In Japan, primary and junior high school students eat lunch in their classrooms, where they learn about diet and Japan’s food culture. They also take turns serving meals to each other, cleaning up, and recycling.21. In the past, the school convenience store couldA. help parents serve studentsB. help teachers manage studentsC. provide students with a chatting placeD. provide physical exercise for students22. The new rule is mainly made toA. keep students healthyB. cut down education costsC. make parents less worried about their kidsD. let school dining rooms make more money23. What can we learn from the passage?A. Students should bring healthy lunch to school.B. Parents should work as volunteers in schools.C. Many countries make efforts to make students eat healthily.D. Other countries are facing different problems of food safety.BA Tanzanian woman named Hadhara Charles Mjeje is travelling around Africa, showing her soccer skills as a way to feed her family. Her soccer short videos have received a lot of attention on social media.Mjeje started working on her ball movement, or juggling (颠球),skills when she was growing up in Tanzania. Mjeje hits the ball with her feet, head and upper body, but not her hands. For many years, she has played for a local women’s soccer team.For the past six years, Mjeje, a 29-year-old single mother of two boys, has been using her special juggling skills to help her support them. She asks for $4 for a two-minute performance, and she earns between $45 and $50 a day.So far, Mjeje has been to Cameroon, Burundi, Gabon and Malawi to give her performances. In Malawi, her skills got the attention of the country’s National Women’s Football officials, who were interested in bringing her into their women's soccer development program, but Mjeje refused the offer. “It’s a regret,” said an official. Sugzo Ngwira, the head of the Women’s Football Committee in Central Malawi, said that if Mjeje was ready to use her skills with other players, her organization would explore ways to work with Mjeje.Mjeje’s skills have caught the interest of male soccer players, too. Samuel Zeka plays soccer in Malawi’s capital, Lilongwe. He said it was a rare talent for women to juggle the ball as she did, “I will be very grateful if she teaches me such skills,” Zeka added. Mjeje returned to Tanzania in March, after an agent traveled to Malawi to meet her. The agent wanted to share the possibility of signing an advertising contract (合向)that could earn her a lot of money in Spain with her. Mjeje said she hoped the new contract would mark the beginning of a better life.24. Mjeje is a soccer player who _________.A. has no less than 15 performances a dayB. learned soccer skills herself at a very young ageC. charges for her soccer videos on the internetD. catches others’ attention by her soccer skills25. What does the underlined word “refused” in Paragraph 4 mean?A. Agreed with.B. Turned down.C. Longed for.D. Paid no attention to.26. Sugzo Ngwira’s oganzation has _______.A. found ways to work with MjejeB. prepared to make use of Mjeje skillsC. a great interest in Mjeje’s soccer skillsD. no player who can play soccer better than Mjeje27. Which of the following can be the best title for the passage?A. Tanzanian Woman Feeds Family with Soccer SkillsB. Some Basic Skills that Help Play Soccer WellC. The Current Living Conditions of Tanzanian WomenD. The Legendary Experience of a Woman Soccer PlayerCFor most people, completing the “Explorers Grand S l am,” which requires climbing the Seven Summits —the highest mountain in each of the seven continents —and reaching the North and South Poles, would be a great enough achievement. However, it was not so for American explorer and adventurer Victor Vescbvo, who completed the challenge in 2017. Since December 2018, the 54-year-old former US naval (海军的)officer has been on a new quest (探索)—to become the first person to take a manned submersible (载人潜水艇)to the deepest-known point in each of the worl d’s five oceans.On April 28, 2019, the adventurer came one step closer to achieving his goal when he fell 10,928 meters (35,853 feet) aboard Limiting Factor, a 15-foot-long, 12-foot-high submersible, to the Earth’s deepest known point —the Challenger Deep. In addition to establishing a new world record for the deepest human dive, Vescovo also became the first person in the world to reach the highest (Mt. Everest) and lowest (Challenger Deep) points of the planet.However, that was not the adventurer’s only visit to the Challenger Deep. Vescovo returned to the area three more times in the next eight days and on May 7, 2019,he became the first human to fall to the Sirena Deep, and collect the deepest piece of mantle rock ever recovered. While the first two visits were paid alone, during the final visit, Vescovo was accompanied (陪伴)by another member of his crew (全体海员).“It is almost indescribable how excited all of us are about achieving what we just did,” said Vescovo.The American explorer has his sights set on space. He said, “No one has done the so-called trifecta (三连胜)——the top and bottom of the world and space. For me, that would mean getting to the Karman Line, 100 km above Earth. I’d love to see the sunrise from there. I think that would be lovely.”28. What did Victor Vescovo achieve in 2017?A. He completed the “Explorers Grand Slam”.B. He dived to the Ea rth’s deepest-known point.C. He discovered a new species in the oceans.D. He became a US naval officer.29. Victor Vescovo is the first person in the world to_______,A. complete the “Explorers Grand Slam”B. reach the Earth’s highest and lowest pointsC. explore the highest point in the worldD. stay for the longest time in the deepest area of the ocean30. How many times has Victor Vescovo been to the Challenger Deep?A. Twice.B. Three times.C. Four times.D. Five times.31. What does the last paragraph mainly talk about?A. Victor Vescovo’s next goal.B. Victor Vescovo’s hobbies.C. Something about trifecta.D. Difficulties Vescovo will face.DFamily-friendly Hotels in LondonAre you planning your family holiday in London? Book one of London’s best family-friendly hotels that have a number of special features to help your stay run successfully.Melia White HouseLet the kids feel grown-up with their own check-in area at Melia White House that is close to Regent’s Park and ZSL London Zoo. There are als o treats such as activities in its kids club. Stay in a family room, which houses two adults plus two children under the age of 11 for no additional charge.Le Meridien PiccadillyHunt for the secret children's book library at family-focused Le Meridien Piccadilly. You’ll find plenty of literary works for the little ones at this west end hotel, which also allows kids under 12 to eat for free (when dining with an adult), and offers flat screen TVs, fully equipped kitchens and free Wi-Fi.AthenaeumEven before you arrive at this family hotel in London, they’ll be in touch to make sure your kids have their favourite DVDs, treats and more to keep them happy. Once checked into your family room or one of its family apartments, they can arrange your perfect journey,from bike hire (租用)to cinema tickets.The RitzAllow the kids to be taken care of at The Ritz, where children can enjoy in-room age-appropriate (适龄的)books, computer games, DVDs and free ice cream. They will also be given welcome gifts. The hotel offers rooms that connect each other, while under-15s can stay for free in their parents’bedroom, depending on the room type.NovotelBook one of Novotel’s family rooms that can house two adults and two children. Accommodations and breakfast are free for under-16s for up to two children when staying in their parents’ room. You can also enjoy a late check-out at 5pm on Sundays; and while you lie in, the kids can entertain themselves with the Cartoon Network while enjoying free ice cream.32. In Le Meridien Piccadilly, guests canA. check out at 5pmB. have dinners for freeC. have free movie ticketsD. make dinners by themselves33. Which hotel can offer the service of booking movie tickets to guests?A. Le Meridien Piccadilly.B. Melia White HouseC. NovotelD. Athenaeum.34. What do The Ritz and Novotel have in common?A. Breakfast is free for kids.B. Books are presented to kids.C. Ice cream is free for kids.D. DVDs are given as a gift.35. Where can you most probably read the passage?A. In a movie poster.B. On a tourism website.C. In a fashion magazine.D. In an ice cream advertisement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。