2019~2020学年度学年度吉林省榆树市第一高级中学高一第1学期尖子生第二次考试数学理试卷及参考答案解析
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语文试题(1)一、现代文阅读(36分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1~3题。
怎样读中国书余英时中国传统的读书法,讲得最亲切有味的无过于朱熹。
古今中外论读书,大致都不外专精和博览两途。
“专精”是指对古代经典之作必须下基础工夫。
古代经典很多,今天已不能人人尽读。
像清代戴震,不但“十三经”文本能背诵,而且“注”也能背诵,只有“疏”不尽记得,这种工夫今天已不可能。
因为我们的知识范围扩大了无数倍,无法集中在几部经、史上面。
但是我们若有志治中国学问,还是要选几部经典,反复阅读,虽不必记诵,但至少要熟。
近人余嘉锡在他的《四库提要辩证》的序录中说:“董遇谓读书百遍,而义自见,固是不易之论。
百遍纵或未能,三复必不可少。
”至少我们必须在自己想进行专门研究的范围之内,作这样的努力。
不但中国传统如此,西方现代的人文研究也还是如此。
精读的书给我们建立了作学问的基地;有了基地,我们才能扩展,这就是博览了。
博览也须要有重点,不是漫无目的地乱翻。
现代是知识爆炸的时代,古人所谓“一物不知,儒者之耻”,已不合时宜了。
所以我们必须配合着自己专业去逐步扩大知识的范围。
博览之书虽不必“三复”,但也还是要择其精者作有系统的阅读,至少要一字不遗细读一遍。
稍稍熟悉之后,才能“快读”、“跳读”。
朱子曾说过:读书先要花十分气力才能毕一书,第二本书只用花七八分功夫便可完成了,以后越来越省力,也越来越快。
这是从“十目一行”到“一目十行”的过程,无论专精和博览都无例外。
读书要“虚心”,这是中国自古相传的不二法门。
朱子说得好:“读书别无法,只管看,便是法。
正如呆人相似,捱来捱去,自己却未先要立意见,且虚心,只管看。
看来看去,自然晓得。
”这似乎是最笨的方法,但其实是最聪明的方法。
我劝青年朋友们暂且不要信今天从西方搬来的许多意见,说甚么我们的脑子已不是一张白纸,我们必然带着许多“先入之见”来读古人的书,“客观”是不可能的等等昏话。
吉林省榆树市第一高级中学2019-2020学年高一数学上学期尖子生考试试题 理总分150分 时间120分一、选择题(本题共12个小题,每题5分,共60分) 1.已知集合2{|4},{|25}A x x x B x x =<=<<,则A B =( )A.{|02}x x <<B.{|45}x x <<C.{|24}x x <<D.{|05}x x <<2.与函数()lg 110x y -=的图象相同的函数是( )A.1y x =-B.1y x =-C.211x y x -=+ D.2y =3.tan 3α=则sin cos sin cos αααα+=-( )A. 2B. 1C. 3D. 44.若()()314,1,,1a x a x f x ax x ⎧-+<⎪=⎨-≥⎪⎩是定义在(),-∞+∞上的减函数,则a 的取值范围是( )A.11,83⎡⎫⎪⎢⎣⎭ B.11,83⎛⎤ ⎥⎝⎦ C.10,3⎛⎫ ⎪⎝⎭D.1,3⎛⎤-∞ ⎥⎝⎦5.函数()πsin 23f x x =+⎛⎫ ⎪⎝⎭的最小正周期为( )A.4πB.2πC.πD.π26.函数2sin(2)π3y x =- ([0,π])x ∈为增函数的区间是( )A .5π[0,]12 B .π[0,]2C .5π11π[,]1212 D .11π[,π]127.23log 9log 4⋅= ( )A.14 B. 12C. 2D. 4 8.设D 为ABC △所在平面内一点,3BC BD =,则( ) A .23AC AB AD =-+ B .32AC AB AD =- C .34AC AB AD =-+D .43AC AB AD =-9.已知向量(),6a x =,()3,4b =,且a 与b 的夹角为锐角,则实数x 的取值范围为( ) A .[)8,-+∞B .998,,22⎛⎫⎛⎫-+∞ ⎪⎪⎝⎭⎝⎭C .998,,22⎡⎫⎛⎫-+∞⎪ ⎪⎢⎣⎭⎝⎭D .()8,-+∞10.如果向量如果向量()()14a k b k ==,与,共线且方向相反,则k =( ) A.2±B.2-C.2D.011.已知.2230log 7,log 0.8,3c a b === ,则,,a b c 的大小关系为( ) A. c b a <<B. a b c <<C. b c a <<D. c a b <<12.已知函数2log ,0()21,0x x f x x x ⎧>⎪=⎨+-≤⎪⎩,若函数()1y f x m =-+有四个零点,零点从小到大依次为,,,a b c d ,则a b cd ++的值为( )A.2B.-2C.-3D.3二、填空题(本题共4个小题,每个小题5分,共20分)13.设α是第三象限角,5tan 12α=,则cos(π)α-=___________ 14.已知()f x 是奇函数,且当0x <时,()e axf x =-.若(ln 2)8f =,则a =__________.15.要使函数124xxy a =++⋅在(],1x ∈-∞时恒大于0,则实数a 的取值范围是.16.给出下列命题:①函数2cos 32y x π⎛⎫=+ ⎪⎝⎭是奇函数;②将函数cos 23y x π⎛⎫=- ⎪⎝⎭的图象向左平移3π个单位长度,得到函数cos2y x =的图象;③若,αβ是第一象限角且αβ<,则tan tan αβ<; ④8x π=是函数5sin 24y x π⎛⎫=+ ⎪⎝⎭的图象的一条对称轴;⑤函数sin 23y x π⎛⎫=+ ⎪⎝⎭的图象关于点,012π⎛⎫⎪⎝⎭中心对称,其中,正确命题的序号是__________.三、解答题(本题共6个题,满分70分) 17.(本题满分12分)已知,i j 是互相垂直的两个单位向量,3a i j =+,3b i j =--. (1)求a 和b 的夹角;(2)若()a a b λ⊥+,求λ的值. 18(本题满分12分)如图是函数在一个周期内的图像,试确定的值。
2019-2020学年吉林省榆树市第一高级中学高一上学期尖子生考试数学(理)试卷总分150分 时间120分一、选择题(本题共12个小题,每题5分,共60分) 1.已知集合2{|4},{|25}A x x x B x x =<=<<,则A B =U ( ) A.{|02}x x << B.{|45}x x << C.{|24}x x << D.{|05}x x <<2.与函数()lg 110x y -=的图象相同的函数是( )A.1y x =-B.1y x =-C.211x y x -=+ D.2y =3.tan 3α=则sin cos sin cos αααα+=-( )A. 2B. 1C. 3D. 44.若()()314,1,,1a x a x f x ax x ⎧-+<⎪=⎨-≥⎪⎩是定义在(),-∞+∞上的减函数,则a 的取值范围是( )A.11,83⎡⎫⎪⎢⎣⎭ B.11,83⎛⎤ ⎥⎝⎦ C.10,3⎛⎫ ⎪⎝⎭D.1,3⎛⎤-∞ ⎥⎝⎦5.函数()πsin 23f x x =+⎛⎫ ⎪⎝⎭的最小正周期为( )A.4πB.2πC.πD.π26.函数2sin(2)π3y x =- ([0,π])x ∈为增函数的区间是( )A .5π[0,]12 B .π[0,]2C .5π11π[,]1212 D .11π[,π]127.23log 9log 4⋅= ( )A.14 B. 12C. 2D. 4 8.设D 为ABC △所在平面内一点,3BC BD =u u u r u u u r,则( ) A .23AC AB AD =-+u u u r u u u r u u u rB .32AC AB AD =-u u u r u u u r u u u rC .34AC AB AD =-+u u u r u u u r u u u r D .43AC AB AD =-u u u r u u u r u u u r9.已知向量(),6a x =r ,()3,4b =r ,且a r 与b r的夹角为锐角,则实数x 的取值范围为( )A .[)8,-+∞B .998,,22⎛⎫⎛⎫-+∞ ⎪ ⎪⎝⎭⎝⎭UC .998,,22⎡⎫⎛⎫-+∞⎪ ⎪⎢⎣⎭⎝⎭UD .()8,-+∞10.如果向量如果向量()()14a k b k ==,与,共线且方向相反,则k =( ) A.2±B.2-C.2D.011.已知.2230log 7,log 0.8,3c a b === ,则,,a b c 的大小关系为( ) A. c b a <<B. a b c <<C. b c a <<D. c a b <<12.已知函数2log ,0()21,0x x f x x x ⎧>⎪=⎨+-≤⎪⎩,若函数()1y f x m =-+有四个零点,零点从小到大依次为,,,a b c d ,则a b cd ++的值为( ) A.2B.-2C.-3D.3二、填空题(本题共4个小题,每个小题5分,共20分)13.设α是第三象限角,5tan 12α=,则cos(π)α-=___________ 14.已知()f x 是奇函数,且当0x <时,()e axf x =-.若(ln 2)8f =,则a =__________.15.要使函数124xxy a =++⋅在(],1x ∈-∞时恒大于0,则实数a 的取值范围是.16.给出下列命题:①函数2cos 32y x π⎛⎫=+ ⎪⎝⎭是奇函数;②将函数cos 23y x π⎛⎫=- ⎪⎝⎭的图象向左平移3π个单位长度,得到函数cos2y x =的图象;③若,αβ是第一象限角且αβ<,则tan tan αβ<; ④8x π=是函数5sin 24y x π⎛⎫=+ ⎪⎝⎭的图象的一条对称轴;⑤函数sin 23y x π⎛⎫=+ ⎪⎝⎭的图象关于点,012π⎛⎫⎪⎝⎭中心对称,其中,正确命题的序号是__________.三、解答题(本题共6个题,满分70分) 17.(本题满分12分)已知,i j r r是互相垂直的两个单位向量,3a i j =+r r r,3b i j =--r r r . (1)求a r 和b r的夹角;(2)若()a a b λ⊥+r r r,求λ的值.18(本题满分12分) 如图是函数在一个周期内的图像,试确定的值。
数学(理)试题总分150分 时间120分 一、选择题(本题共12个小题,每题5分,共60分) 1.已知集合2{|4}A x x x =<,{|25}B x x =<<,则A B =U ( ) A.{|02}x x <<B.{|45}x x <<C.{|24}x x <<D.{|05}x x <<2.已知函数2()2(21)Z f x x x x x =+-≤≤∈且,则()f x 的值域是( ) A .[0,3]B .{}1,0,3-C .{}0,1,3D .[1,3]-3.若向量(1,2),(3,4)AB BC ==u u u r u u u r,则AC =u u u r ( )A. ()4,6B. ()4,6--C. (2,2)--D. ()2,24.已知tan 3α=,则222sin 2cos sin cos sin ααααα+=+( ). A.38 B.916 C.1112D.795.若将函数1()cos22f x x =的图像向左平移π6个单位长度,则平移后图像的一个对称中心可以为( )A. π ,012⎛⎫ ⎪⎝⎭B. π,06⎛⎫ ⎪⎝⎭C. π,03⎛⎫ ⎪⎝⎭D. π,02⎛⎫⎪⎝⎭6.设R m ∈,向量(1,2),(,2)a b m m =-=-r r若a b ⊥r r ,则m 等于( ) A .23- B .23 C .4- D .47.将函数πsin 3y x ⎛⎫=- ⎪⎝⎭的图象上所有点的横坐标伸长到原来的2倍(纵坐标不变),再将所得的图象向左平移π3个单位,得到的图象对应的解析式是( ) A. 1sin 2y x =B. 1πsin 26y x ⎛⎫=- ⎪⎝⎭C. 1πsin 22y x ⎛⎫=- ⎪⎝⎭D. πsin 26y x ⎛⎫=- ⎪⎝⎭8.函数()()2ln 28f x x x =--的单调递增区间是( ) A.(),2-∞-B.(1),-∞C.(1,)+∞D.(4,)+∞9.函数2sin 2x y x =的图象可能是( )A.B.C. D.10.如果角θ满足sin cos 2θθ+=那么1tan tan θθ+的值是( ) A.-1B.-2C.1D.211.对于幂函数()45f x x =,若120x x <<,则()()1212,22f x f x x x f ++⎛⎫ ⎪⎝⎭的大小关系是( ) A. ()()121222f x f x x x f ++⎛⎫>⎪⎝⎭ B. ()()121222f x f x x x f ++⎛⎫< ⎪⎝⎭C. ()()121222f x f x x x f ++⎛⎫=⎪⎝⎭D.无法确定 12.已知偶函数()f x 在区间[0,)+∞上单调递增,则满足(21)(1)f x f -<的x 取值范围是( ) A.(1,0)-B.(0,1)C.(1,2)D.(1,1)-二、填空题(本题共4个小题,每个小题5分,共20分)13.若关于x 的方程245x x m -+=有4个互不相等的实数根,则实数m 的取值范围是______. 14.平面内有三点(0,3),(3,3),(,1)A B C x --,且//AB AC u u u r u u u r,则x 为______.15.将函数sin y x =的图象上所有的点向右平行移动10π个单位长度,再把所得各点的横坐标伸长到原来的2倍(纵坐标不变),所得图象的函数解析式是______________.16.已知5122x x -⎛⎫≤ ⎪⎝⎭,则函数241y x x =-+的值域为______.三、解答题(本题共6个题,共70分) 17.(本题满分12分)已知向量()3,2a r=,),3(1b r =-,()5,2c r =. (1)求62a b c r r r +-; (2)求满足a mb nc r r r=+的实数m ,n ;(3)若//()(2)a kc b a r r r r +-,求实数k .18.(本题满分12分)已知三个点()()()2,1,3,2,1,4A B D -.1.求证: AB AD ⊥u u u r u u u r;2.要使四边形ABCD 为矩形,求点C 的坐标,并求矩形ABCD 两对角线所夹锐角的余值。
吉林省榆树市第一高级中学2019-2020学年高一化学上学期尖子生考试试题本试卷满分102分,考试时间70分钟可能用到的相对原子质量:H -1 C -12 N -14 O -16 Na -23 Mg -24 Al -27 S -32Cl -35.5 Fe -56 Cu- 64一、选择题(本题共18小题,每小题3分,共54分,每小题只有一个正确答案)1.下列有关物质组成的说法正确的是()A. 物质均是由分子构成,分子均是由原子构成B. 只由一种元素组成的物质一定是单质C. 金属氧化物都是碱性氧化物D. 硫酸是纯净物,盐酸是混合物2.下列实验方案中可行的是()A.用澄清石灰水检验CO中是否含有CO2B.用BaCl2除去NaOH溶液中混有的少量Na2SO4C.用酒精把碘水中的碘萃取出来D.用溶解、过滤的方法分离CaCl2和NaCl固体混合物3.过滤后的食盐水仍含有可溶性的CaCl2、MgCl2、Na2SO4等杂质,通过如下几个实验步骤,可制得纯净的食盐水,则下列操作顺序正确的是()①加入稍过量的Na2CO3溶液;②加入稍过量的NaOH溶液;③加入稍过量的BaCl2溶液;④滴入稀盐酸至无气泡产生;⑤过滤。
A.③⑤②①④B.②③①④⑤C.①②③⑤④D.③②①⑤④4.下列变化,需要加入适当的氧化剂才能完成的是()A.SO2→SO32ˉB.PCl3→PCl5C.MnO4ˉ→Mn2+D.Cl2→HClO5.Na2FeO4是一种高效多功能水处理剂,应用前景十分看好.一种制备Na2FeO4的方法可用化学方程式表示如下:2FeSO4+6Na2O2=2Na2FeO4+2Na2O+2Na2SO4+O2↑,对此反应下列说法中正确的是( )A.Na2O既不是氧化产物又不是还原产物B.FeSO4是氧化剂C.若有2molFeSO4参与反应,则该反应中共有8mol电子转移D.Na2O2既是氧化剂又是还原剂6.已知M2O7X-+3S2-+14H+=2M3++3S↓+7H2O,则M2O7X-中的M的化合价为()A 、+2价B 、 +3价C 、+4价D 、 +6价7.为除去某物质中所含的杂质,所选用的除杂试剂或操作方法正确的是( )序号物 质杂质除杂试剂或操作方法①NaCl 溶液Na 2CO 3加入盐酸,蒸发②FeSO 4溶液CuSO 4加入过量铁粉并过滤③H 2CO 2依次通过盛有NaOH 溶液和浓硫酸的洗气瓶④NaNO 3CaCO 3加稀盐酸、过滤、蒸发结晶A .①②③B .②③④C .①③④D .①②③④8.据报道,科学家研制了一种间距为50 nm 的“镊子”,利用它可以操控活细胞中的DNA 分子等。
吉林省榆树市第一高级中学2019-2020学年高一上学期尖子生第二次考试数学(理)试卷总分150分 时间120分 一、选择题(本题共12个小题,每题5分,共60分) 1.已知集合2{|4}A x x x =<,{|25}B x x =<<,则A B =U ( ) A.{|02}x x <<B.{|45}x x <<C.{|24}x x <<D.{|05}x x <<2.已知函数2()2(21)Z f x x x x x =+-≤≤∈且,则()f x 的值域是( ) A .[0,3]B .{}1,0,3-C .{}0,1,3D .[1,3]-3.若向量(1,2),(3,4)AB BC ==u u u r u u u r,则AC =u u u r ( )A. ()4,6B. ()4,6--C. (2,2)--D. ()2,24.已知tan 3α=,则222sin 2cos sin cos sin ααααα+=+( ).A.38B.916C.1112D.795.若将函数1()cos22f x x =的图像向左平移π6个单位长度,则平移后图像的一个对称中心可以为( )A. π ,012⎛⎫ ⎪⎝⎭B. π,06⎛⎫ ⎪⎝⎭C. π,03⎛⎫ ⎪⎝⎭D. π,02⎛⎫⎪⎝⎭6.设R m ∈,向量(1,2),(,2)a b m m =-=-r r若a b ⊥r r ,则m 等于( ) A .23- B .23 C .4- D .47.将函数πsin 3y x ⎛⎫=- ⎪⎝⎭的图象上所有点的横坐标伸长到原来的2倍(纵坐标不变),再将所得的图象向左平移π3个单位,得到的图象对应的解析式是( ) A. 1sin 2y x =B. 1πsin 26y x ⎛⎫=- ⎪⎝⎭C. 1πsin 22y x ⎛⎫=- ⎪⎝⎭D. πsin 26y x ⎛⎫=- ⎪⎝⎭8.函数()()2ln 28f x x x =--的单调递增区间是( ) A.(),2-∞-B.(1),-∞C.(1,)+∞D.(4,)+∞9.函数2sin 2x y x =的图象可能是( )A.B.C. D.10.如果角θ满足sin cos 2θθ+=那么1tan tan θθ+的值是( ) A.-1B.-2C.1D.211.对于幂函数()45f x x =,若120x x <<,则()()1212,22f x f x x x f ++⎛⎫ ⎪⎝⎭的大小关系是( ) A. ()()121222f x f x x x f ++⎛⎫>⎪⎝⎭ B. ()()121222f x f x x x f ++⎛⎫< ⎪⎝⎭C. ()()121222f x f x x x f ++⎛⎫=⎪⎝⎭D.无法确定 12.已知偶函数()f x 在区间[0,)+∞上单调递增,则满足(21)(1)f x f -<的x 取值范围是( ) A.(1,0)-B.(0,1)C.(1,2)D.(1,1)-二、填空题(本题共4个小题,每个小题5分,共20分)13.若关于x 的方程245x x m -+=有4个互不相等的实数根,则实数m 的取值范围是______. 14.平面内有三点(0,3),(3,3),(,1)A B C x --,且//AB AC u u u r u u u r,则x 为______.15.将函数sin y x =的图象上所有的点向右平行移动10π个单位长度,再把所得各点的横坐标伸长到原来的2倍(纵坐标不变),所得图象的函数解析式是______________.16.已知5122x x -⎛⎫≤ ⎪⎝⎭,则函数241y x x =-+的值域为______.三、解答题(本题共6个题,共70分) 17.(本题满分12分)已知向量()3,2a r=,),3(1b r =-,()5,2c r =. (1)求62a b c r r r +-; (2)求满足a mb nc r r r=+的实数m ,n ;(3)若//()(2)a kc b a r r r r +-,求实数k .18.(本题满分12分)已知三个点()()()2,1,3,2,1,4A B D -.1.求证: AB AD ⊥u u u r u u u r;2.要使四边形ABCD 为矩形,求点C 的坐标,并求矩形ABCD 两对角线所夹锐角的余值。
2019-2020学年吉林省长春市榆树一中高一上学期尖子生第二次考试数学(文)试题一、单选题1.已知集合2{|4}A x x x =<,{|25}B x x =<<,则A B =( )A .{|02}x x <<B .{|45}x x <<C .{|24}x x <<D .{|05}x x <<【答案】D【解析】解出A 集合,再由并集的定义写出A B 即可。
【详解】由2{|4}A x x x =<⇒{|04}A x x =<<,则{|05}A B x x ⋃=<<.故选D . 【点睛】本题主要考查集合的并集,正确求解一元二次不等式,是首要条件。
属于基础题 2.已知函数()()2221f x x x x x Z =+-≤≤∈且,则()f x 的值域是( )A .[]0,3B .{}1,0,3-C .{}0,1,3D .[]1,3-【答案】B【解析】试题分析:求出函数的定义域,然后求解对应的函数值即可.函数()()2221f x x x x x Z =+-≤≤∈且,所以2101x =--,,,;对应的函数值分别为:0103-,,,;所以函数的值域为:{}1,0,3-故答案为B .【考点】函数值域 3.若向量=(1,2),=(3,4),则=A .(4,6)B .(-4,-6)C .(-2,-2)D .(2,2) 【答案】A 【解析】.4.已知tan 3α=,则222sin 2cos sin cos sin ααααα+=+( ).A .38B .916C .1112D .79【答案】C【解析】分子分母同时除以2cos α,利用同角三角函数的商关系化简求值即可. 【详解】因为tan 3α=,所以2cos 0α≠,于是有2222222222sin 2cos sin 2cos 211sin cos sin sin cos sin tan tan 1tan cos cos 2ααααααααααααααα+++===+++,故本题选C.【点睛】本题考查了同角三角函数的商关系,考查了数学运算能力. 5.若将函数1()cos 22f x x =的图像向左平移6π个单位长度,则平移后图像的一个对称中心可以为( ) A .(,0)12πB .(,0)6πC .(,0)3πD .(,0)2π【答案】A【解析】通过平移得到1cos(2)23y x π=+,即可求得函数的对称中心的坐标,得到答案. 【详解】 向左平移6π个单位长度后得到1cos 223y x π⎛⎫=+ ⎪⎝⎭的图像,则其对称中心为(),0122k k Z ππ⎛⎫+∈ ⎪⎝⎭,或将选项进行逐个验证,选A. 【点睛】本题主要考查了三角函数的图象变换,以及三角函数的图象与性质的应用,其中解答中根据三角函数的图象变换,以及熟记三角函数的图象与性质是解答的关键,着重考查了推理与运算能力.6.设m R ∈,向量(1,2),(,2)b a m m =-=-,若a b ⊥,则m 等于( )A .23-B .23C .-4D .4【答案】D【解析】直接利用向量垂直的充要条件列方程求解即可. 【详解】因为(1,2),(,2)b a m m =-=-,且a b ⊥,所以()(1,2)(,2)220a m m m m b ⋅=-⋅-=--=, 化为40m -=,解得4m =,故选D. 【点睛】利用向量的位置关系求参数是命题的热点,主要命题方式有两个:(1)两向量平行,利用12210x y x y -=解答;(2)两向量垂直,利用12120x x y y +=解答. 7.将函数sin 3y x π⎛⎫=-⎪⎝⎭的图象上所有点的横坐标伸长到原来的2倍(纵坐标不变),再将所得的图象向左平移3π个单位,得到的图象对应的解析式是( ) A .1sin2y x = B .1sin 22y x π⎛⎫=-⎪⎝⎭C .1sin 26y x π⎛⎫=- ⎪⎝⎭D .sin 26y x π⎛⎫=-⎪⎝⎭【答案】C【解析】【详解】将函数y=sin(x -3π)的图象上所有点的横坐标伸长到原来的2倍(纵坐标不变)得到y=sin(12x -3π),再向左平移3π个单位得到的解析式为y=sin(12(x+3π)-3π)=y=sin(12x -6π),故选C8.函数2()ln(28)f x x x =--的单调递增区间是 A .(,2)-∞- B .(,1)-∞ C .(1,)+∞ D .(4,)+∞【答案】D【解析】由228x x -->0得:x ∈(−∞,−2)∪(4,+∞), 令t =228x x --,则y =ln t ,∵x ∈(−∞,−2)时,t =228x x --为减函数; x ∈(4,+∞)时,t =228x x --为增函数; y =ln t 为增函数,故函数f (x )=ln(228x x --)的单调递增区间是(4,+∞), 故选:D.点睛:形如()()y f g x =的函数为()y g x =,() y f x =的复合函数,() y g x =为内层函数,()y f x =为外层函数. 当内层函数()y g x =单增,外层函数()y f x =单增时,函数()()y f g x =也单增; 当内层函数()y g x =单增,外层函数()y f x =单减时,函数()()y f g x =也单减; 当内层函数()y g x =单减,外层函数()y f x =单增时,函数()()y f g x =也单减; 当内层函数()y g x =单减,外层函数()y f x =单减时,函数()()y f g x =也单增. 简称为“同增异减”.9.函数y =2x sin2x 的图象可能是A .B .C .D .【答案】D【解析】分析:先研究函数的奇偶性,再研究函数在π(,π)2上的符号,即可判断选择.详解:令()2sin 2xf x x =, 因为,()2sin 2()2sin 2()x x x R f x x x f x -∈-=-=-=-,所以()2sin 2xf x x =为奇函数,排除选项A,B;因为π(,π)2x ∈时,()0f x <,所以排除选项C ,选D.点睛:有关函数图象的识别问题的常见题型及解题思路:(1)由函数的定义域,判断图象的左、右位置,由函数的值域,判断图象的上、下位置;(2)由函数的单调性,判断图象的变化趋势;(3)由函数的奇偶性,判断图象的对称性;(4)由函数的周期性,判断图象的循环往复.10.如果角θ满足sin cos θθ+=1tan tan θθ+的值是( ) A .-1 B .-2 C .1 D .2 【答案】D【解析】试题分析:sin cos θθ+=()2sin cos 12sin cos 2θθθθ∴+=+=,1sin cos 2θθ∴=. 221sin cos sin cos 1tan 21tan cos sin sin cos 2θθθθθθθθθθ+∴+=+===.故D 正确.【考点】同角三角函数基本关系式.11.对于幂函数f(x)=45x ,若0<x 1<x 2,则12()2x x f +,12()()2f x f x +的大小关系是( ) A .12()2x x f +>12()()2f x f x + B .12()2x x f +<12()()2f x f x + C .12()2x x f +=12()()2f x f x + D .无法确定【答案】A【解析】本题考查幂函数图象及性质。
吉林省榆树市第一高级中学2019-2020学年高一英语上学期尖子生第二次考试试题第I卷选择题(共7分)第二部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、 B、C和D)中,选出最佳选项。
AAccommodation at University College Cork, IrelandOn-campus accommodationUCC Campus accommodation is more than just a place to eat and sleep---it provides social, cultural and sporting shops and the chance to develop friendships that will last a lifetime. Our accommodation services are divided into our two areas of operation, one being the safe, modern accommodation for those studying during the academic year. The other one is the excellent summer accommodation for holiday or summer course visitors who are seeking what is a very special environment to stay in with every modern convenience but dipped in history and the atmosphere of one of Europe's oldest and most highly regarded universities.Rental HousingThis is where the student lives with a family and 1~5 other students and you get your breakfast, evening meal and light supper. Rates are available for both single and shared rooms, five days or seven-day board. Prices are set and no landlady can charge over this agreed rate. This is a very economical and safe accommodation option and particularly well suited to freshmen. Lists of landladies are available free from the Accommodation Office by post, fax and email.Shared Student HousingAvailable rooms are advertised in the Accommodation Office throughout the year. Updated lists of them are provided daily. Lifetime friendship is easily developed here.Living with a LandlordUsually this provides more mature students with a high standard of accommodation at reasonable rents, usually a little below the market price. Vacancies are advertised in the Accommodation Office throughout the year.21. What is special about the environment of on-campus accommodation?A. It benefits freshmen a lot during holidays.B. It is the most highly regarded accommodation in Europe.C. It helps develop a lifelong friendship between roommates.D. It is a perfect mix of modern equipment and historical surroundings.22. Which is recommended to first-year students at their university?A. On-campus accommodation.B. Living with a Landlord.C. Shared Student Housing.D. Rental Housing.23. What do we know about the Accommodation Office?A. It doesn't deal with rents for students.B. It serves students and gets paid for what it does.C. It provides information on the accommodation choices.D. It is seen as a bridge between the university and students.BThere are only a few people who actually enjoy public speaking. Most of the population dislikes the experience and many find it fearful. What can you do? First of all remember the following information. Most of your audience would not want to trade places with you. They can't imagine how you do it. They are your allies and not your critics. They are just people like you.Second, don't make this more difficult than it needs to be. While it is good to have some eye contact with lots of people, you should try to begin by finding just one friendly face in the crowd. Pretend it's not public speaking and you are just having a conversation with that person. Now you can speak with your everyday voice and the sound equipment does the rest. So relax and talk normally.Third, have a sense of humor. Even the best of speakers has an embarrassing moment now and then. Don't take yourself too seriously. If you make a mistake it is not the end of the world. If you appropriately begin your talk with some humor, it helpsto relax you and the audience as well.Fourth, there are many different kinds of public speaking. Some can be prepared and some are extempore(即兴的). If you have time to prepare then do it. Write the content well and practice ahead of time. Use a mirror and use a friend or two. If you have no time to practice at least get the key ideas clear in your mind before you begin.Lastly, transform your nerves into passion for your topic. If you have no energy,if you aren't a bit nervous your voice is flat and uninteresting.Public speaking isn't for everyone. However, it is a skill that can be developed.If you need to, or want to have a try to be a toastmasters. It will teach you and give you lots of experience.24,The underlined word "allies" in Paragraph 2 probably refers to.A. supportersB. guardsC. teachersD. relatives25:Which of the following can help make the audience feel easy and comfortable?A. Telling some funny stories.B. Speaking with some suitable humor.C. Making a few mistakes.D. Taking yourself seriously enough.26You can make a better preparation for an extempore speech by.A. making use of a mirrorB. meeting an old friendC.writing the content well.D. clearing the main points before you start.27.Which of the following is TRUE according to this passage?A. Feeling a bit nervous during speaking gives you passion.B. Speaking in public can please most of the people.C. The sound equipment is actually useless to speaking.D. It is rather difficult to learn to speak well in public.CI went nose to nose with a mouse last week, and I'd rather not say who won.I was cooking when my friend phoned me to help her catch a mouse she had foundin the kitchen. The mouse was under the fridge when I arrived. The cat was standing guard, slowly swinging its tail. I tried to put a broom handle beneath the fridgeto force it out, but there were too many wires. My friend pulled up a chair at thetable and I pulled a chair beside the fridge, armed with a plastic box as my trap. It was a good plan, but as is usually the case, it didn't go according to plan. After a while, the mouse put its head out. I knelt down on the floor in front of the fridge and angled the box to create a no-escape trap. The mouse moved out farther and I put the box down immediately. Unfortunately, the only thing inside the box was the mouse's tail, still attached to the mouse, which was now struggling with its little legs in an attempt to take to its heels. The cat just sat.The mouse crawled (爬) up on the broom handle I'd used earlier, and we were now eyeball to eyeball. If I'd wanted to---and I didn't---I could have seized it with my teeth. I shouted I needed something else to get the mouse. My friend handed me another container. I lowered box two over the mouse's body, still on the broom handle, with its tail still in box one. Despite my two-box move, the mouse escaped and shot back under the fridge.Shortly after I left, my friend found the mouse. It was bathed in the sunlight in her bedroom. The cat was sitting next to it, enjoying the rays as well. She said they were a cute couple. Learning from my mistakes, she got a box---and a lid---and said the mouse all but jumped in the box and helped her seal the lid. Just a small change brought about an unexpected result.28. What does the last sentence in Paragraph 2 suggest?A. The plan ended in failure. .B. The plan wasn't worth carrying out.C. The plan wasn't carried out properly.D. The plan proved to be an unexpected29. What does the underlined part “take to its heels" m ean?A. Go back.B. Stand up.C. Fight back.D. Get away.30. What were the mouse and the cat doing in the bedroom?A. Enjoying the harmony,B. Comforting each other.C. Celebrating the escape.D. Appreciating the view.31. Who had the last laugh?A. The author.B. The author’s friend.C. The mouse.D. The cat.DIn a perfect world there would be no social differences. Everybody would get along no matter what their social or financial standing (地位) is. However ,there is no such thing as a perfect world, only a world filled with prejudice( 偏见) and anger. The theme of Great Expectations shows that social class and a lot of money are worth less than love,and family.A high social standing and money do not provide the necessities of a happy life. Great Expectations shows that people of a high social class are only popular among them- selves. The reason for this is that to keep their social standing, they are not to have relationships with lower class people. If it is found that they do accept these people as part of society, they are their friends. Similarly , money does not provide a happy future. Money can only buy materials. For a truly happy life , onemust have healthy relationships ,like that of Joe and Biddy. A person with money and no social skills will often become depressed due to their lack of relationships. Even the poorest person can lead a richer life than the wealthiest man in the world. Happiness comes easy to a happy family that gets along. Money isn't needed for a fun time either. A poor family can entertain each other with good conversation. Low class people get along with a variety of people. Life is much easier socially for a person who is not concerned about what others think of them.The message that social classes and money are worth less than love and family is important to remember in everyday life. The central theme of Great Expectations is not simple. It sends a good message to readers.32. What does the writer think of the present world?A. It's not perfect at all.B. There aren't many social differences.C. People get along quite well.D. Money is nothing compared with love.33. Why are Joe and Biddy mentioned in the passage?A. They belong to a high social class.B. They don't keep their social standing.C. Their relationship is healthy enough.D. Their friends are all lower class people.34. The underlined word" depressed" probably means“A. excitedB. surprisedC. frightenedD. disappointed35. The main purpose of this passage is to______A. call on people to value love and familyB. show how difficult it is to be successfulC. encourage people to work for their dreamsD. prove social differences are hard to solve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。
吉林省榆树市第一高级中学2019~2020学年度高一第一学期尖子生第二次考试数学(理)试卷总分150分 时间120分 一、选择题(本题共12个小题,每题5分,共60分)1.已知集合2{|4}A x x x =<,{|25}B x x =<<,则A B =U ( ) A.{|02}x x <<B.{|45}x x <<C.{|24}x x <<D.{|05}x x <<2.已知函数2()2(21)Z f x x x x x =+-≤≤∈且,则()f x 的值域是( ) A.[0,3]B.{}1,0,3-C.{}0,1,3D.[1,3]-3.若向量(1,2),(3,4)AB BC ==u u u r u u u r,则AC =u u u r ( )A.()4,6B.()4,6--C.(2,2)--D.()2,24.已知tan 3α=,则222sin 2cos sin cos sin ααααα+=+( ).A.38B.916C.1112D.795.若将函数1()cos22f x x =的图像向左平移π6个单位长度,则平移后图像的一个对称中心可以为( )A.π ,012⎛⎫ ⎪⎝⎭B.π,06⎛⎫ ⎪⎝⎭C.π,03⎛⎫⎪⎝⎭ D.π,02⎛⎫ ⎪⎝⎭6.设R m ∈,向量(1,2),(,2)a b m m =-=-r r若a b ⊥r r ,则m 等于( ) A.23- B.23 C.4- D.47.将函数πsin 3y x ⎛⎫=- ⎪⎝⎭的图象上所有点的横坐标伸长到原来的2倍(纵坐标不变),再将所得的图象向左平移π3个单位,得到的图象对应的解析式是( ) A.1sin 2y x = B.1πsin 26y x ⎛⎫=- ⎪⎝⎭C.1πsin 22y x ⎛⎫=- ⎪⎝⎭D.πsin 26y x ⎛⎫=- ⎪⎝⎭8.函数()()2ln 28f x x x =--的单调递增区间是( ) A.(),2-∞-B.(1),-∞C.(1,)+∞D.(4,)+∞9.函数2sin 2x y x =的图象可能是( )A.B.C. D.10.如果角θ满足sin cos 2θθ+=那么1tan tan θθ+的值是( ) A.-1B.-2C.1D.211.对于幂函数()45f x x =,若120x x <<,则()()1212,22f x f x x x f ++⎛⎫ ⎪⎝⎭的大小关系是( ) A.()()121222f x f x x x f ++⎛⎫>⎪⎝⎭ B.()()121222f x f x x x f ++⎛⎫<⎪⎝⎭C.()()121222f x f x x x f ++⎛⎫=⎪⎝⎭D.无法确定 12.已知偶函数()f x 在区间[0,)+∞上单调递增,则满足(21)(1)f x f -<的x 取值范围是( )A.(1,0)-B.(0,1)C.(1,2)D.(1,1)-二、填空题(本题共4个小题,每个小题5分,共20分)13.若关于x 的方程245x x m -+=有4个互不相等的实数根,则实数m 的取值范围是______. 14.平面内有三点(0,3),(3,3),(,1)A B C x --,且//AB AC u u u r u u u r,则x 为______.15.将函数sin y x =的图象上所有的点向右平行移动10π个单位长度,再把所得各点的横坐标伸长到原来的2倍(纵坐标不变),所得图象的函数解析式是______________.16.已知5122x x -⎛⎫≤ ⎪⎝⎭,则函数241y x x =-+的值域为______.三、解答题(本题共6个题,共70分) 17.(本题满分12分)已知向量()3,2a r=,),3(1b r =-,()5,2c r =. (1)求62a b c r r r +-; (2)求满足a mb nc r r r=+的实数m ,n ;(3)若//()(2)a kc b a r r r r +-,求实数k .18.(本题满分12分)已知三个点()()()2,1,3,2,1,4A B D -.1.求证: AB AD ⊥u u u r u u u r;2.要使四边形ABCD 为矩形,求点C 的坐标,并求矩形ABCD 两对角线所夹锐角的余值。
19. (本题满分12分)函数()3(2)6f x sin x π=+的部分图象如图所示1. 写出f ()x 的最小正周期及图中00,x y 的值;2.求f ()x 在区间[,]212ππ--上的最大值和最小值. 20.(本题满分12分) 设()121 log 1axf x x -=-为奇函数, a 为常数.1.求a 的值;2.判断函数()f x 在区间()1,+∞上的单调性并证明. 21.(本题满分12分)已知函数3sin cos tan(2)22()tan()sin()f ααααααππ⎛⎫⎛⎫-+π- ⎪ ⎪⎝⎭⎝⎭=+π+π. (1)化简()f α;(2)若1()28f f ααπ⎛⎫⋅+=- ⎪⎝⎭,且5342αππ≤≤,求()2f f ααπ⎛⎫++ ⎪⎝⎭的值;(3)若2()2f f ααπ⎛⎫+= ⎪⎝⎭,求()2f f ααπ⎛⎫⋅+ ⎪⎝⎭的值.22.(本题满分10分)已知定义在区间(0,+∞)上的函数f (x )满足f ⎝⎛⎭⎫x 1x 2=f (x 1)-f (x 2),且当x >1时,f (x )<0.①求f (1)的值;②证明:f (x )为单调递减函数;③若f (3)=-1,求f (x )在[2,9]上的最小值.(理科试卷参考答案)一、选择题 1.答案:D 解析: 2.答案:B解析:由已知得函数()22f x x x =+的定义域为{}2,1,0,1--, 则()()()20,11,00f f f -=-=-=,()13f =, 所以函数的值域为{}1,0,3-. 故正确答案为B 3.答案:A解析:∵(1,2)(3,4)(4,6)AC AB BC =+=+=u u u r u u u r u u u r,故选A.4.答案:C 解析:因为tan 3α=,所以2cos 0α≠,于是有2222222222sin 2cos sin 2cos 211sin cos sin sin cos sin tan tan 1tan cos cos 2ααααααααααααααα+++===+++,故本题选C. 5.答案:A 解析:向左平移π6个单位长度后得到1πcos 223y x ⎛⎫=+ ⎪⎝⎭的图像,则其对称中心为ππ,0()122k k Z ⎛⎫+∈ ⎪⎝⎭;或将选项进行逐个验证. 6.答案:D解析:本题主要考查平面向量基本定理及坐标表示. 因为a b ⊥r r , 所以0a b ⋅=r r,所以1(2)(2)0m m ⨯+-⨯-=, 解得4m =. 7.答案:B解析:依题意得,最后得到的曲线相应的解析式是1ππ1πsin sin 23326y x x ⎡⎤⎛⎫⎛⎫=+-=- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦,故选B8.答案:D解析:由2280x x -->,得2x <-或4x >因此,函数()()2ln 28f x x x =--的定义域是(),24,()⋃-∞-+∞.注意到函数228y x x =--在(4,)+∞上单调递增,由复合函数的单调性知,()()2ln 28f x x x =--的单调递增区间是(4,)+∞,选D.9.答案:D解析:令()2sin 2x f x x =,因为,()2sin 2()2sin 2()x x x R f x x x f x -∈-=-=-=-,所以()2sin 2x f x x =为奇函数,排除选项A,B;因为π(,π)2x ∈时,()0f x <,所以排除选项C,选D.10.答案:D解析:∵sin cos θθ+∴12sin cos 2θθ+=, 即1sin cos 2θθ=, 那么1sin cos 1tan 2tan cos sin sin cos θθθθθθθθ+=+==. 11.答案:A解析:幂函数()45f x x=在()0,+∞上是增函数,大致图像如图所示,设()()12,0,,0A x C x ,其中120x x <<,则AC 的中点E 的坐标为()()121212,0,,,22x x x x AB f x CD f x EF f ++⎛⎫⎛⎫=== ⎪ ⎪⎝⎭⎝⎭∵()12EF AB CD >+∴()()121222f x f x x x f ++⎛⎫> ⎪⎝⎭故选A12.答案:B 解析:根据题意,由函数的奇偶性分析可得f(2x-1)<f(1)⇒f(|2x-1|)<f(1),进而结合单调性分析可得|2x-1|<1,解可得x 的取值范围,即可得答案.解:根据题意,f(x)为偶函数,则f(2x-1)<f(1)⇒f(|2x-1|)<f(1), 又由函数在区间[0,+∞)上单调递增, 则f(|2x-1|)<f(1)⇒|2x-1|<1, 解可得:0<x<1, 故选:B. 二、填空题 13.答案:(1,5)解析:令()245f x x x =-+,作出函数()f x 的图象,如图所示.当15m <<时,满足条件. 14.答案:1 解析:15.答案:sin 210x y π⎛⎫=- ⎪⎝⎭解析:∵向右平移10π个单位, ∴用10x π-代替sin y x =中的x ;∵各点横坐标伸长到原来的2倍, ∴用12x 代替sin 10y x π⎛⎫=- ⎪⎝⎭中的x , ∴sin 210x y π⎛⎫=-⎪⎝⎭. 16.答案:11,4⎡⎫-+∞⎪⎢⎣⎭解析:依题意,5122X X -⎛⎫≤ ⎪⎝⎭∴522x x -≤∴5x x -≤∴52x ≥ ()224123y x x x =-+=--其图象的对称轴为2x =,当52x ≥时,函数单调递增当52x =时,114y =- ∴114y ≥-,即函数的值域为11,4⎡⎫-+∞⎪⎢⎣⎭即答案为:11,4⎡⎫-+∞⎪⎢⎣⎭解析:三、解答题17.答案:(1)()()3263,(21,325,2)a b c r r r+-=+--()()()18,121,310,471(,1)=+--=(2)∵a mb nc r r r =+,∴()()3,21,35,2)(),(532m n m n m n =-+=-++. ∴53,322,m n m n -+=⎧⎨+=⎩解得4,1711.17m n ⎧=⎪⎪⎨⎪=⎪⎩(3)∵()()()//235,2225,4()a kc b a a kc k k b a r r r r r r r r+-,+=++,-=-. ∴4355()()()220k k ⨯⨯+--+= ∴1115k =-. 解析:18.答案:1.()()1,1,3,3AB AD ==-u u u r u u u r1(3)130AB AD ⋅=⨯-+⨯=u u u r u u u r∴AB u u u r ⊥AD u u u r2.设(),C x y ,()1,4DC x y =+-u u u r,由DC AB =u u u r u u u r ,得0,5?x y ==∴() 0,5C设矩形ABCD 两对角线AC ,BD 所夹锐角为θ(2,4)AC =-u u u r ,(4,2)BD =-u u u rAC =u u ur BD =u u u r 164cos 205AC BD AC BDθ⋅===⋅u u u r u u u ru u u r u u u r 解析:19.答案:1.f(x)的最小正周期为T=22ππ=,007,36x y π== 2.因为x ∈[,]212ππ--,所以2x+6π∈5[,0]6π-,于是当2x+6π=0,即x=3π-时,f(x)取得最大值0; 当2x+6π=2π-,即x=3π-时,f(x)取得最小值3-.解析:20.答案:1.因为函数()121log 1axf x x -=-是奇函数,∴()()f x f x -=-, ∴111222111log log log 111ax ax x x x ax+--=-=----, ∴1111ax x x ax+-=---,即()()()()1111ax ax x x +-=-+-。