2019年10月浙江省学考选考七彩阳光联盟二联期中高2020届高2017级高三英语试题原稿

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浙江省七彩阳光新高考研究联盟2019-2020学年高三上学期期中物理试卷 (含解析)

浙江省七彩阳光新高考研究联盟2019-2020学年高三上学期期中物理试卷 (含解析)

浙江省七彩阳光新高考研究联盟2019-2020学年高三上学期期中物理试卷一、单选题(本大题共11小题,共34.0分)1.一物体静止在斜面上,下列说法正确的是()A. 物体所受重力与斜面对物体弹力的合力,就是物体对斜面的静摩擦力B. 物体所受重力与斜面对物体静摩擦力的合力,就是物体对斜面的正压力C. 斜面对物体弹力与斜面对物体静摩擦力的合力,就是物体所受的重力D. 斜面对物体弹力与斜面对物体静摩擦的合力,就是物体所受重力的平衡力2.设地球半径为R,a为静止在地球赤道上的一个物体,b为一颗近地绕地球做匀速圆周运动的人造卫星,c为地球的一颗同步卫星,其轨道半径为r.下列说法中正确的是()A. a与c的线速度大小之比为√rR B. a与c的线速度大小之比为√RrC. b与c的周期之比为√rR D. b与c的周期之比为Rr√Rr3.如图所示,位于水平地面上的质量为M的小木块,在大小为F、方向与水平成α角的拉力作用下沿地面做加速运动,若木块与地面间的动摩擦因数为μ,则木块的加速度为()A. FM B. (Fcos α−μMg)MC. FcosαM D. Fcos α−μ(Mg−Fsin α)M4.做匀速圆周运动的物体,在运动过程中,保持不变的物理量是()A. 速度B. 加速度C. 速率D. 所受的合力5.如图所示,导体杆OP在作用于OP中点且垂直于OP的力作用下,绕O轴沿半径为r的光滑的半圆形框架在匀强磁场中以一定的角速度转动,磁场的磁感应强度为B,AO间接有电阻R,杆和框架电阻不计,回路中的总电功率为P,则()A. 外力的大小为2Br√PRB. 外力的大小为Br√PRC. 导体杆旋转的角速度为2√PRBr2D. 导体杆旋转的角速度为2Br2√PR6.如图所示,一通电圆环质量为m,电流方向为逆时针方向放在水平桌面上,一条形磁铁竖直放在环的中心处,N极在下端,下面判断正确是()A. 环对桌面的压力仍为mgB. 环对桌面的压力小于mgC. 环对桌面的压力大于mgD. 环所受桌面对它的摩擦力向左7.如图所示,OA、OB是两根轻绳,AB是轻杆,它们构成一个正三角形.在A、B处分别固定着质量均为m的小球,此装置悬挂在O点.现对B处小球施加水平外力F,让绳OA位于竖直位置.设此状态下OB绳中张力大小为T,已知当地重力加速度为g,则()A. T=2mgB. T>2mgC. T<2mgD. 三种情况皆有可能8.如图所示是电磁流量计的示意图。

2019年10月浙江省学考选考2019学年第一学期浙江七彩阳光新高考研究联盟期中联考高三数学答案

2019年10月浙江省学考选考2019学年第一学期浙江七彩阳光新高考研究联盟期中联考高三数学答案

2019学年第一学期浙江“七彩阳光”新高考研究联盟期中联考高三年级数学学科参考答案(命题:海盐高级中学)一、选择题:(本大题共10小题, 每小题4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.D2.B3.B4.C5.B6.A7.C8.A9.C 10.D二、填空题:(本大题共7小题, 双空题每小题6分, 单空题每小题4分, 共36分)11. {}10<<x x , {}21≥<x x x 或 12. 32- ,10 13. )0,3(, 1339± 14. 23-n , ()2209)1(++n n 15. 5 16. 40 17. 9 三、解答题:(本大题共5小题, 共74分)18.解:()1由题意得()2sin 6f x x π⎛⎫=- ⎪⎝⎭, ------------------------3分 5366x πππ≤-≤, 所以()[]1,2f x ∈. ------------------------6分 ()2由78,663f A f B ππ⎛⎫⎛⎫+=+- ⎪ ⎪⎝⎭⎝⎭化简得4sin sin 3A B +=, ------------------------.8分 4sin sin 3sinB sin B a A b B-==413sin B =-,而1sin 13B ≤≤, ...............12分 所以1,33a b ⎡⎤∈⎢⎥⎣⎦. ------------------------14分 19. ()1证明:分别取线段AC 、AB 的中点记为O 、D , 连接SO 、OD ,因为 SAC ∆为等边三角形, 则AC SO ⊥,又OD //BC , 则AC OD ⊥, O OD SO = ,则AC 平面SOD ⊥, 所以AC SD ⊥. ------------------------6分()2延长SO , 过D 做SO 延长线的垂线, 垂足记为H ,易知DH ⊥平面SAC ,所以DSH ∠为直线SD 与平面SAC 所成角. ------------------------10分在SBC ∆中,因为cos SDA+cos SDB=0∠∠, 求得=6SD , ------------------------12分又1OD=2且, 则DSH=6π∠, 故直线SD 与平面SAC 所成的角为6π. .------------------------15分20.解:(1)na d a a a a a a a a n =∴=∴==∴=++=++14,312,943642531------2分208,20311342=+∴==+q b q b b b b ① 821=q b ②由①②得2=q 或21=q (舍)21=bnn b 2=∴ ------------------------5分(2)n nn c 24-=3224341+-⨯=∴+n n n B ---------------------9分 )121121(23)12)(12(32211---=--=∴++n n n n nn nB b -----------------13分23)1211(231<--=∴+n n T ----------------------------------15分21.解(1)直线AB 方程:1+=x y , 设),(),,2211y x B y x A (联立方程⎩⎨⎧=+=y x x y 4120442=--⇒x x4,42121-==+x x x x . .....................2分 ⋅--=⋅∴2111x y K K BP AP ⋅--2122x y =424221+⋅+x x =164)(22121+++x x x x2116484=++-= ....................5分(2)设),(),,2211y x B y x A (),22t t P (,, 设直线BP 斜率为K设直线BP 方程)2(-2t x k t y -= 不妨)0(>k联立方程⎩⎨⎧==y x 42t)-k(x t -y 22048422=-+-⇒t kt kx x211482,42t kt t x k t x -=⋅=+ ....................7分 =-+=∴t x k BP 2112t k k -+214 同理可得t k k AP ++=∴11142 ....................9分 由BP AP =得k k k t +-=231....................11分 故:222)1(821t k k BP AP S ABP -+==∆16)1(2)1()2(8)1()1()1(8222222222=++≥+++=k k k k k k k k 当且仅当1=k 时取等号, 所以ABP ∆面积最小值为16.....................15分22.解:(1)22)(ex e x f x += 0222>+=ex e ex x ......................3分又 0≠x)(x f ∴增区间为()0-,∞, ()∞+,0......................5分(2)由题得2)2(x me ex e x x =⋅-有三个实根所以m e exe x x x =⋅-)2(2有三个非零实根 即m e xe xe x x =-)2(有三个非零实根......................7分令)0)(≠⋅==x e x x g t x ( )01)('≠⋅+=x e x x g x ()( )(x g ∴在()1--,∞单调递减, ),(∞+1-单调递增......................9分 022=--∴m t e t 一个根在⎪⎭⎫ ⎝⎛0,e 1-, 另一个根在()∞+,0;或者一个根等于e 1-, 另一个根在⎪⎭⎫⎝⎛0,e 1-内(舍) ......................12分 令=)(t h m t et --22 由⎪⎩⎪⎨⎧<=>-0)0()2(0)1(h eh e h 230e m <<⇔ ......................15分。

2019年10月浙江省学考选考七彩阳光联盟二联期中高2020届高2017级高三生物试题参考答案

2019年10月浙江省学考选考七彩阳光联盟二联期中高2020届高2017级高三生物试题参考答案

2019学年第一学期浙江“七彩阳光”新高考研究联盟期中联考高三年级生物学科参考答案命题:景宁中学张晓芳李郭联系电话审稿:景宁中学张玲玲联系电话:一、选择题(每题2分,共25×2=50分)1-5 DBAAD 6-10 DADCD11-15 DCCCC 16-20 DBCDD21-25 BDBDC二、非选择题(共50分)26.(8分,每空1分)(1)生产者因含N、P元素的污染物大量流入(2)鲢鱼鲢鱼以藻类等浮游植物为食物(3)生物量捕食者/天敌(4)通过增加生物多样性来提高生态系统的自我调节能力(5)定期合理投放鲢鱼;定期适度捕捞鲢鱼;控制性捕杀鲢鱼的捕食者;控制人类活动(工业、农业、旅游等)对该水库生态环境的负面影响(答出任意一条给分)27.(8分,每空1分)(1)光强度、温度、植株类型(答全给分)单位时间单位叶面积内叶绿体和线粒体(2)大于左下(3)增多光强度和温度(4)C1802经光合作用产生H2180,然后光合作用利用H2180产生含180的氧气28.(共8分,每空1分)(1)遵循两对等位基因位于非同源染色体上(2)bbZ A Z A或bbZ A Z a 6(3)绿羽色:黄羽色:蓝羽色:白羽色= 3 :3 :1 :1 1/6(4)遗传图解(2分)(亲本基因型、表现型和子代基因型、表现型写对1分,配子、符号写对1分)绿羽色雄鸟白羽色雌鸟P BBZ A Z A×bbZ a W配子BZ A bZ a bWF1 BbZ A Z a BbZ A W绿羽色雄鸟绿羽色雌鸟比例 1 : 129.(8分,每空1分)(1)高压蒸汽灭菌用牛皮纸(或报纸)包扎器皿(2)苯酚涂布分离法(稀释涂布法)(3)扩大血细胞计数板大4℃冰箱30.(8分,每空1分)(1)受精卵显微注射胚胎体外同期发情处理(2)骨髓瘤细胞克隆(3)抗原抗体杂交 C31.(10分)(1)①取4个广口瓶,标号为甲、乙、丙、丁(1分)②甲广口瓶内加入一定量的池塘水和饲料;乙广口瓶内加入等量池塘水、含牛甲状腺磨碎物的饲料;丙广口瓶内加入等量池塘水+甲状腺激素溶液和饲料;丁广口瓶内加入等量池塘水+碘液和饲料(2分)③将生长发育状况一致的同种蝌蚪随机均分为4组,用尺子测量蝌蚪长度后,分别放入4个广口瓶中,将各广口瓶放置在相同且适宜的条件下培养(2分)④统计分析每组数据的平均值(1分)(2)(3分)验证甲状腺中的甲状腺激素对蝌蚪生长的影响实验结果记录表。

浙江省“七彩阳光”新高考研究联盟2019-2020学年高一上学期10月联考数学试题 PDF版含答案

浙江省“七彩阳光”新高考研究联盟2019-2020学年高一上学期10月联考数学试题 PDF版含答案
f (x)是奇函数 x (-4,0)时,f (x) (- ,0) f(0) 0 f (x)的值域R
,(8 分) (12 分)
19.解:(1)令 t=ax,则 t>0,∴g(t)=﹣t2﹣2t+3=﹣(t+1)2+4 ∵t>0,∴g(t)<3,即函数 f(x)的值域为(﹣∞,3);
D. a b c
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y
=
f
(x) 在定义域 (0,+) 上是单调函数,当
x (0,+) 时,都有
f

f
(x) −
1 x
=
2,则
f
(3)
的值为( )
A. 2
B. 3
C. 3
D. 4
2
3
二、填空题:本大题共 6 小题,多空题每题 6 分,单空题每题 4 分,共 30 分.
2.给定下列函数,其中在区间 (0,1) 上单调递增的函数是( )
A. y = −x2
B. y =| x2 − 2x |
C. y = ( 1 )x+1 2
3.设函数 f (x) =
x2 −1, x 2
,则 f ( f (2)) 的值为(

f (x − 2), x 2
A.0
B.3
C. −1
C.有最大值-3
D.有最小值-3
7.函数 f (x) = 2x ,对任意的 x1, x2 ,且 x1 x2 ,则下列四个结论不一定正确的是(
)
A. f (x1 + x2 ) = f (x1) f (x2 )
B. (x1 − x2 )[ f (x1) − f (x2 )] 0
C. f (x1 x2 ) = f (x1) + f (x2 )

浙江省“七彩阳光”新高考研究联盟2019-2020学年高一上学期10月联考数学试题 PDF版含答案

浙江省“七彩阳光”新高考研究联盟2019-2020学年高一上学期10月联考数学试题 PDF版含答案

2.给定下列函数,其中在区间 (0,1) 上单调递增的函数是( )
A. y = −x2
B. y =| x2 − 2x |
C. y = ( 1 )x+1 2
3.设函数 f (x) =
x2 −1, x 2
,则 f ( f (2)) 的值为(

f (x − 2), x 2
A.0
B.3
C. −1

A. a c b B. a b c
C. b a c
D. b c a
6.已知函数 f (x) 是奇函数, f (x) 在(0,+∞)上是减函数,且在区间[a, b](a b 0) 上的值域
为[-3,4],则在区间[−b, −a] 上( )
A.有最大值 4 B.有最小值-4
D. f x1 + x2 f (x1)+ f (x2 )
2
2
高一数学学科 试题 第1页(共 4 页)
8.设函数
g(x)
=
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2

f
(x)
=
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x g(x) g(x)
,则
f
(x)
的值域是(

A.[− 9 , 0] (1, +) B.[0, +) 4
C.[− 9 , +) 4
D.[− 9 , 0] (2, +) 4
9.设
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102016 102017
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b
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102017 102018
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c

浙江省七彩阳光新高考研究联盟2019-2020学年高二上学期期中物理试卷(有解析)

浙江省七彩阳光新高考研究联盟2019-2020学年高二上学期期中物理试卷(有解析)

浙江省七彩阳光新高考研究联盟2019-2020学年高二上学期期中物理试卷一、单选题(本大题共10小题,共30.0分)1.下列哪种情况是不可能出现的()A. 物体的加速度增大时,速度反而减小B. 物体的速度为零时,加速度却不为零C. 物体的加速度不为零且始终不变,速度也始终不变D. 物体的加速度大小和速度大小均保持恒定且均不为零2.在某次户外风洞体验活动中,体验者在风力的作用下漂浮在半空.若增加风力,则该体验者在加速上升的过程中()A. 处于超重状态且机械能增加B. 处于超重状态且机械能减少C. 处于失重状态且机械能增加D. 处于失重状态且机械能减少3.如图所示,两个质量均为m的物块A和B通过一轻弹簧连接在一起,并放置于水平传送带上,水平轻绳一端连接A,另一端固定在墙上,A、B与传送带间的动摩擦因数均为μ。

传送带沿顺时针方向匀速转动,系统达到稳定后,突然剪断轻绳的瞬间,设A、B的加速度大小分别为a A和a B(弹簧在弹性限度内),重力加速度为g,则()A. a A=2μg,a B=0B. a A=2μg,a B=μgC. a A=μg,a B=0D. a A=μg,a B=μg4.如图所示,当小车A以恒定的速度v向左运动时,则对于B物体来说,下列说法正确的是()A. 匀加速上升B. 匀速上升C. B物体受到的拉力大于B物体受到的重力D. B物体受到的拉力等于B物体受到的重力5.“太空涂鸦”技术的基本物理模型是:原来在较低圆轨道运行的攻击卫星在变轨后接近在较高圆轨道上运行的侦察卫星时,向其发射“漆雾”弹,“漆雾”弹在临近侦察卫星时,压爆弹囊,让“漆雾”散开并喷向侦察卫星,喷散后强力吸附在侦察卫星的侦察镜头、太阳能板、电子侦察传感器等关键设备上,使之暂时失效。

关于这一过程下列正确说法是()A. 攻击卫星接近侦察卫星的过程中受到地球的万有引力一直在增大B. 攻击卫星在原轨道上运行的周期比侦察卫星的周期大C. 攻击卫星在原轨道上运行的线速度比侦察卫星的线速度大D. 攻击卫星在原轨道需要减速才能变轨接近侦察卫星6.有一只风扇,标有“U、P”,电动机线圈电阻为R,把它接入电压为U的电路中,以下几种计算电风扇所发热量的方法,正确的是()A. Q=U2⋅tRB. Q=P⋅tC. Q=(PU)2⋅Rt D. 以上三种都正确7.直角坐标系xOy中,A、B两点位于x轴上,坐标如图所示,C、D位于y轴上,C、D两点各固定一等量正点电荷,另一电量为Q的负点电荷置于O点时,B点处的电场强度恰好为零.若将该负点电荷移到A点,则B点处场强的大小和方向分别为(静电力常量为A)()A. 5kQ4l2,沿x轴正方向 B. 5kQ4l2,沿x轴负方向C. 3kQ4l2,沿x轴负方向 D. 3kQ4l2,沿x轴正方向8.利用霍尔效应制作的霍尔元件,广泛应用于测量和自动控制等领域,如图是霍尔元件的工作原理示意图,磁感应强度B垂直于霍尔元件的工作面向下,通入图示方向的电流I,C、D两侧面会形成电压U,下列说法中正确的是()A. 电压U仅与材料有关B. 若元件的载流子是自由电子,则C侧面电势高于D侧面电势C. 在测地球赤道上方的地磁场强弱时,元件的工作面应保持垂直D. 仅增大磁感应强度时,电势差U CD变大9.如图所示,当电流通过线圈时,磁针将发生偏转,以下的判断正确的是()A. 当线圈通以沿顺时针方向的电流时,磁针N极将指向读者B. 当线圈通以沿逆时针方向的电流时,磁针S极将指向读者C. 当磁针N极指向读者,线圈中电流沿逆时针方向D. 不管磁针如何偏转,线圈中的电流总是沿顺时针方向10.如图所示,用粗细均匀的电阻丝折成平面梯形框架,ab、cd边均与ad边成60°角,ab=bc=cd=L,长度为L的电阻丝电阻为r,框架与一电动势为E,内阻为r的电源相连接,垂直于框架平面有磁感应强度为B的匀强磁场,则框架受到的安培力的合力大小为()A. 0B. 5BEL11r C. 10BEL11rD. BELr二、多选题(本大题共3小题,共15.0分)11.如图所示,用水平力F推静止在斜面上的物块,当力F由零开始逐渐增大而物块仍保持静止状态,则物块()A. 所受合力逐渐增大B. 所受斜面摩擦力逐渐增大C. 所受斜面摩擦力先减小后增大D. 所受斜面弹力逐渐增大12.如图为某电场中的一条电场线,M、N是这条电场线上的两点。

浙江省“七彩阳光”新高考研究联盟2019-2020学年高一上学期10月联考数学试题(解析版)

浙江省“七彩阳光”新高考研究联盟2019-2020学年高一上学期10月联考数学试题(解析版)

浙江省“七彩阳光”新高考研究联盟2019-2020学年高一上学期10月联考试题一、选择题1.已知全集{|06}U x N x =∈≤≤,集合{4,5,6}A =,则U C A =( )A. {1,2,3}B. {0,1,2,3}C. {|03}x x ≤≤D. {|03}U x N x =∈<≤『答案』B 『解析』因为{}0,1,2,3,4,5,6U =,所以{}0,1,2,3U C A =,选B.2.给定下列函数,其中在区间(0,1)上单调递增的函数是( )A. 22y x =-B.22y x x=-C.112x y +⎛⎫= ⎪⎝⎭D. 1y x x=+『答案』B『解析』A. 212y x =-为二次函数,对称轴是0x =,开口向下,所以在区间(0,1)上单调递减;B. 当(0,1)x ∈时,2222y x x x x --+==,对称轴是1x =,开口向下,所以在区间(0,1)上单调递增;C.112x y +⎛⎫= ⎪⎝⎭中,()10,12∈,所以在区间(0,1)上单调递减;D.当11x x x =⇒=时,1y x x =+在()0,∞+上有最低点,所以在区间(0,1)上单调递减. 故选:A.3.设函数21,2()(2),2x x f x f x x ⎧-≤=⎨->⎩,则((2))f f 的值为( ) A. 0B. 3C. -1D. 2『答案』A『解析』()22213f =-=,2((2))(3)(32)(1)110f f f f f ==-==-=.故选:A.4.已知集合{1,2}A =,{4,5,6}B =,:f A B →为集合A 到集合B 的一个函数,那么该函数的值域C 的不同情况有( )种. A. 2B. 3C. 6D. 7『答案』C『解析』由函数的定义可知,函数的值域C 是集合B 的一个子集.{4,5,6}B =,非空子集共有3217-=个;而定义域A 中至多有2个元素,所以值域C 中也至多有2个元素; 所以集合B 的子集{4,5,6}不能作为值域C ,值域C 的不同情况只能有6种. 故选:C.5.三个数20.3a =,0.3(1.9)b =,0.32c =之间的大小关系是( )A. a c b <<B. a b c <<C. b a c <<D. b c a <<『答案』B 『解析』因为()0.320.31,? 1.91a b ==,0.321c =>, 又0.3y x =为(0,)+∞上单调递增函数,所以()0.30.31.92<,综上a b c <<,选B.6.已知函数()f x 是奇函数,()f x 在(0,)+∞上是减函数,且在区间[,](0)a b a b <<上的值域为[3,4]-,则在区间[,]b a --上( ) A. 有最大值4 B. 有最小值-4C. 有最大值-3D. 有最小值-3『答案』B『解析』∵()f x 是奇函数,在(0,)+∞上是减函数,∴()f x 在(,0)-∞上也是减函数,即在区间[,](0)a b a b <<上递减. 又∵()f x 在区间[,](0)a b a b <<上的值域为[3,4]-, ∴()()4,3,f a f b ==-根据奇函数性质可知()()4,3,f a f b -=--=且在区间[,]b a --上单调递减,∴()f x 在区间[,]b a --上有最大值3,有最小值-4. 故选:B.7.函数()2xf x =,对任意的1x ,2x ,且12x x <,则下列四个结论不一定正确的是( )A.()()()1212f x x f x f x +=⋅B. ()()()12120x x f x f x -->⎡⎤⎣⎦C.()()()1212f x x f x f x ⋅=+D. ()()121222f x f x x x f ++⎛⎫<⎪⎝⎭『答案』C『解析』A. 1212222x x x x +=⨯,正确;B. 函数()2xf x =在R 上递增,若12x x <,则()()12f x <f x ,()()()12120x x f x f x -->⎡⎤⎣⎦正确;C. 1212222xx x x +≠,不正确;D. 由基本不等式,当12x x <时,1212222222x x x x ++>==,即()()121222f x f x x x f ++⎛⎫> ⎪⎝⎭,正确. 故选:C.8.设函数2()2()g x x x R =-∈,()4,(),(),().(){g x x x g x g x x x g x f x ++<-≥=则()f x 的值域是( )A. 9,0(1,)4⎡⎤-⋃+∞⎢⎥⎣⎦B. [0,)+∞C. 9[,)4-+∞D. 9,0(2,)4⎡⎤-⋃+∞⎢⎥⎣⎦『答案』D的『解析』当()x g x <,即22,(2)(1)0x x x x <--+>时,2x >或1x <-, 222()()4242(0.5) 1.75f x g x x x x x x x =++=-++=++=++,其最小值为(1)2f -=,无最大值, 因此这个区间的值域为:(2,)+∞;当()x g x ≥时,12x -≤≤,22()()2(0.5) 2.25f x g x x x x x =-=--=--, 其最小值为(0.5) 2.25f =-,其最大值为(2)0f =,因此这区间的值域为:9[,0]4-, 综合得函数值域为:9[,0](2,)4-⋃+∞ ,故选D .9.设20162017110011a +=+,20172018110011b +=+,20182019110011c +=+,则a ,b ,c 的大小关系( ) A. b c a << B. a c b << C. c b a <<D. a b c <<『答案』C『解析』由题意可得,()20172016201720172017111111011991010101010101010a ++===+++++, 同理,201720182018111101910101010b +==+++,201820192019111101910101010c +==+++,∵201920182017111999101010101010<<+++,∴c b a << 故选:C.10.设()y f x =在定义域(0,)+∞上是单调函数,当()0,x ∈+∞时,都有1()2f f x x ⎡⎤-=⎢⎥⎣⎦,则(3)f 的为( )A. 2B. 3C. 32D. 43『答案』D『解析』设1()f x t x -=,则()2f t =,1()f x t x =+ ∵()y f x =在定义域(0,)+∞上是单调函数 ∴方程()2f t =只有一解,即t 为定值.又∵()12f t t t =+=,∴1t =,即()14333f t =+=故选:D. 二、填空题11.(1)12.55(0.64)-=_________;(2)7log 22lg5lg 47++=_________. 『答案』 (1).14-(2). 4『解析』(1)()2.510.51535310.640.640.82424---⎛⎫-==-=-=- ⎪⎝⎭,()7log 222lg5lg472lg52lg222102 4.lg ++=++=+=12.函数()f x ,()g x 分别由下表给出,则((1))f g 的值为________;满足(())(())f g x g f x >的x 的值为________.『答案』 (1). 1 (2). 2 『解析』(1). ()()()131f g f ==;故答案为:1.(2).()()()131f g f ==,()()()113g f g ==;()()()223f g f ==,()()()231g f g ==;()()()311f g f ==,()()()113g f g ==;∴当(())(())f g x g f x >时,2x =.故答案为:2.13.函数y =________;值域是________.『答案』 (1). (2,3) (2). [1,2]『解析』(1). 2430x x -+-≥解得函数的定义域为[]1,3,设t =,对称轴为()4221x =-=⨯-,得出t =()1,2上递增,()2,3上递减; 又∵2t y =恒单调递增,∴根据复合函数单调性同增异减,可得y =()1,2上递增,()2,3上递减;故答案为:(2,3) . (2). 由(1)得,[]1,3x ∈,所以[]2430,1x x -+-∈[]0,1,[]1,2y =∈,即函数y =[1,2].故答案为:[1,2].14.已知函数22y x x =+在闭区间[,]a b 上的值域为[1,3]-,则⋅a b 的最大值为________. 『答案』3 『解析』画出函数()22f x x x=+的图像可知,要使其在闭区间[,]a b 上的值域为[1,3]-, 由于有且仅有()11f -=-,所以1[,]1a b a b -∈⇒≤-≤,而()()313f f -==,所以有[][,]3,1a b ⊆-,3a =-或1b =,又∵0a <,⋅a b 的最大值为正值时,0b <,∴1,3b a ≠=-, 所以3a b b ⋅=-,当b 取最小值时,⋅a b 有最大值. 又∵1b ≥-,∴⋅a b 的最大值为()()313-⨯-=;故答案为:3.15.函数()y f x =是定义在R 上的增函数,函数(2)y f x =-的图像关于点(2,0)对称,则满足()2(4)40x f f x x -+-<的实数x 的取值范围为________.『答案』(4,1)-『解析』函数(2)y f x =-的图像关于点(2,0)对称,则函数()y f x =的图像关于点(0,0)对称,即()y f x =为奇函数, 满足()()f x f x -=-.所以()2(4)40x f f x x -+-<,()()22(4)4)4(4f x f x f x f x x x -<---<-+⇒,又∵()y f x =是定义在R 上的增函数,∴21444x x x x +-<-⇒-<< 故答案为:(4,1)-16.已知0a >时,对任意0x >,有2()()0x a x bx a -+-≥恒成立,则ab 的取值范围是 _________________. 『答案』()(),10,-∞-+∞『解析』因为对任意0x >,有()()20x a x bx a -+-≥恒成立, 所以x a =为方程20x bx a +-=的根,即210,?10,?1,?111a a a ba a a b b a b a a +-=+-==-==-+--, 因为0a >,所以11a 1,11a -∴-或101a <-,即1a b <-或0a b >.三、解答题 17.已知集合{|A x y ==,{}22|60B x x ax a =--<,其中0a ≥.(1)当1a =时,求集合A B ⋃,()R C A B⋂;(2)若()R C A B B⋂=,求实数a 的取值范围.『解』(1){()(){}[]||3103,1A x y x x x ===+-≥=-当1a =时,{}{}()222|60|602,3B x x ax a x x x =--<=--<=-,所以[)3,3,A B ⋃=-因为()()(),31,R C A =-∞-⋃+∞,所以()()1,3R C A B ⋂=(2)因为()R C A B B⋂=,所以R B C A⊆,当B =∅时,0a =,满足条件,{}()220|602,3a B x x ax a a a >=--<=-当时,不满足条件,因此0a =.18.已知函数()f x 是定义在(4,4)-上的奇函数,满足(2)1f =,当40x -<≤时,有()4ax bf x x +=+.(1)求实数a ,b 的值;(2)求函数()f x 在区间(0,4)上的解析式; (3)求函数()f x 在区间(4,4)-上的值域.『解』(1)由题可知,2(2)12(0)04a b f b f -+⎧-==-⎪⎪⎨⎪==⎪⎩,解得10a b =⎧⎨=⎩; (2)由(1)可知当(4,0)∈-x 时,()4xf x x =+,当(0,4)x ∈时,(4,0)-∈-x ,()()44x xf x f x x x -=--=-=-+-+.(3)4()14f x x =---,当(0,4)x ∈时,4(,1)4x ∈-∞--,4()1(0,)4f x x =--∈+∞-,∵()f x 是奇函数,∴(4,0)∈-x 时,()(,0)f x ∈-∞, 又∵(0)0f =,∴()f x 的值域为R .19.已知函数()(1)(3)x xf x a a =-+(1a >) (1)求函数()f x 的值域;(2)若[2,1]x ∈-时,函数()f x 的最小值为5-,求a 的值和函数()f x 的最大值.『解』(1)设xa t =,则0t >,()()()()22132314143x x f x a a t t t =-+=--+=-++<-+=,即(),3-∞值域为,(2) 设xa t =,则2,t a a -⎡⎤∈⎣⎦,而()()()()22132314x x f x a a t t t =-+=--+=-++,所以当t a =时, 函数()f x 取最小值,即2235a a --+=-,因为1a >,所以2a =,当214t a -==时函数()f x 取最大值,为1139316216--+=.20.已知函数22221,0()2,0x ax a x f x x a x x ⎧-++≤⎪=⎨+->⎪⎩.(1)证明:()f x 在(0,1)上单调递减,在(1,)+∞上单调递增; (2)记函数()f x 最小值为()M a ,求()M a 的最大值.『解』(1)设120x x <<,()()2212121222f x x f x a x x a x ⎛⎫⎛⎫+--+- ⎪ ⎪=⎝⎭-⎝⎭ ()()331212121212211212222x x x x x x x x x x x x x x x x -+-⎡⎤-+-⎣⎦==又∵120x x <<,∴12120,0x x x x >-<.当1201x x 时,()1212121201,0220x x x x x x x x <<<+<⇒+-<,∴()()()()()()21212121122120x x x x x x f x f x f x f x x x -+-⎡⎤⎦>⇒>⎣-=.当121x x <<时,()121212121,220x x x x x x x x >+>⇒+->,∴()()()()()()21212121122120x x x x x x f x f x f x f x x x -+-⎡⎤⎦<⇒<⎣-=.即()f x 在(0,1)上单调递减,在(1,)+∞上单调递增. (2)由(1)得,()f x 在0x >时的最小值为()13f a=-.由∵当x ∈R 时,二次函数2221y x ax a =-++的对称轴为x a =, 由题意可得,22()21f x x ax a =-++时,0x ≤. ∴当a ≥0时, 2()()1f x x a =-+在(-∞,0』上递减,故在(-∞,0』上的最小值为2(0)1f a =+, f (x )在(0,+∞)上的最小值为f (1)=3-a ;∵213a a +-,∴01a ≤≤.当a <0时,f (x )在(-∞,0』上的最小值为f (a )=1,f (x )在(0,+∞)上的最小值为f (1)=3-a ;∵13a -,∴0a <,即21,01()1,03,1a a M a a a a ⎧+⎪=<⎨⎪-⎩,所以M (a )在(-∞,0)上为常数函数,在(0,1)上是增函数,在(1,+∞)上是减函数, 作出M (a )的函数图象如图所示:所以M (a )的最大值为2.。

浙江省“七彩阳光”新高考研究联盟2019-2020学年高一10月联考英语试题(pdf版)

浙江省“七彩阳光”新高考研究联盟2019-2020学年高一10月联考英语试题(pdf版)

2019学年第一学期浙江“七彩阳光”新高考研究联盟10月联考高一年级英语学科试题命题:杭州市余杭第二高级中学考生须知:1.本卷共7页满分120分,考试时间90分钟;2.答题前,在答题卷指定区域填写班级、姓名、考场号、座位号及准考证号并填涂相应数字。

3.所有答案必须写在答题纸上,写在试卷上无效;4.考试结束后,只需上交答题纸。

第I卷第一部分:听力(共两节,满分15分)第一节(共5 小题;每小题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 color is the man’s umbrella?A.BrownB. BlackC. Blue2.Who is the man?A.Doctor CookB. Mr. SmithC. Mr. Smith’s secretary3.Where does the conversation take place?A.In a restaurantB. In a hotelC. In a shop4.Why does the woman want to live in the city?A.To save moneyB.To take care of her parentsC.To go to work conveniently.5.What are the speakers talking about in general?A.SpeechB. New classroomsC. Screens第二节(共10 小题;每小题 1 分,满分10 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。

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2019学年第一学期浙江“七彩阳光”新高考研究联盟期中联考高三年级英语试题命题:杭州市富阳二中考生须知:1.本卷共8页,满分150分,考试时间120分钟;2.答题前,在答题卷指定区域填写班级、姓名、考场号、座位号及准考证号;3.所有答案必须写在答题卷上,写在试卷上无效;4.考试结束后,只需上交答题卷。

第I卷第一部分:听力(共两节,满分30分)做题时,先将答案标在试卷上。

录音内容结朿后,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。

1. Why did the man get a new phone?A. His old one was broken.B. He wanted to take good photos.C. He gave his old one to his nephew.2. What is the woman best at?A. Cooking.B. Gardening.C. Doing magic.3. What does the woman want for her birthday?A. Time with her loved ones.B. Something cheap.C. A new car.4. What’s the probable relationship between the speakers?A. Bus driver and passenger.B. Co-workers.C. A couple.5. Why is the speakers’ cl ass size smaller this semester?A. It was broken up into two classes.B. Some students failed the first class.C. There were too many kids last year.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、7 题。

6. What’s wrong with the woman’s gloves?A. They don’t fit.B. They are too thin.C. They have holes in them.7. What will the speakers do first?A. Go skating.B. Have a drink.C. Buy something.听第7段材料,回答第8至10题。

8. Who is the man?A. A doctor.B. A banker.C. A personal trainer.9. How often does the man eat meat?A. Every day.B. Twice a week.C. Three times a week.10. What does the woman suggest the man do?A. Eat no more meat.B. Change his habits.C. Come to see her more often.听第8段材料,回答第11至13题。

11. What will the man do next month?A. Find a new roommate.B. Move out of his apartment.C. Fix the windows and doors.12. What does the woman like about the apartment?A. The rent.B. The roof.C. The position.13. What will the woman probably do next?A. Call the house owner.B. Go back to her office.C. Bargain with the man.听第9段材料,回答第14至17题。

14. What is the primary reason the man wants furniture?A. To receive guests.B. To have a place to sleep.C. To make his apartment look nice.15. What does the woman offer to give the man?A. A sofa.B. Two tables.C. A television.16. Where will the man probably get a wardrobe?A. From garage sales.B. From a f urniture store.C. From his parents’ house.17. What’s the probable relationship between the speakers?A. Friends.B. Brother and sister.C. Husband and wife.听第10段材料,回答第18至20题。

18. How do French people feel about their country and culture?A. Proud.B. Modest.C. Uninterested.19. Where do French Fries come from?A. Italy.B. France.C. Spain.20. How many fashion designers are mentioned in the talk?A. Two.B. Three.C. Four.第二部分:阅读理解(共两节,满分35分)第一节:(共10小题;每小题2.5分,满分25分)阅读下列短文,从每题所给的A、B、C和D四个选项中,选出最佳选项,并在答题纸上将该项涂黑。

A“To be a musician, it’s like fulfilling a part of your life that's irreplaceable. It’s something that I enjoy and I love to do and I get a satisfaction from it.” That is Renald Richard. Born in 1925 in Thibodaux, Louisiana, Richard’s memories of childhood are filled with music. His father played piano by ear. His uncles and cousins all played instruments and sang. So at the age of six, Richard started piano lessons at home.At the age of 13, Renald Richard went to see a band perform at a church dance. He noticed two young trumpet players in the band. He says they were dressed “clean and sharp.” All the girls -- including his girlfriend -- fell for the trumpet players.Richard knew he had to take action. “I got to do something about this. I asked my Dad if I could playthe trumpet. And he said ‘I'll get a trumpet for you but you cannot stop the piano’. And I got the trumpet. I had lessons on the piano and trumpet. And it went on like that for a while. And I got my girl back.”Renald Richard played the trumpet in the high school band. Later, he performed while touring in a USO variety show. While playing in Houston, Texas in 1954, Richard was approached by Jeff Brown, the manager of musician Ray Charles. Renald Richard soon was playing his trumpet for Ray Charles’ band. In time, Richard became Charles’ bandleader.Renald Richard toured all over the South and West with Ray Charles. Sitting in the back seat of the car on the way to a gig, Richard wrote the song “I Got a Woman” for Charles. It became Ray Charles’ first number one radio hit.I was thrilled when music artist Kanye West used “I Got a Woman”as the basis for his hit song “Gold Digger”, Richard says. The song was number one on the Billboard Charts for 16 consecutive weeks. “It was a big, big hit,” says Richard.Living now in Southern Florida, Richard is still active at the age of 94. “I'm playing music mainly with the Marco Island Strummers and I'm playing with the Jazz Masters in Naples, so I do a lot of playing and I enjoy that.”As you can see Renald Richard’s musical days are by no means near an end.21. Why did Renald Richard begin to play the trumpet?A. His father asked him to play the trumpet.B. The trumpet players were handsomely dressed.C. He fell in love with trumpet at a church dance.D. He needed to take action to win his girlfriend back.22. Which of the following statements is true?A. Nothing can replace being a musician in Renald Richard’s life.B. Renald Richard played the piano in Ray Charles' band.C. “I Got a Woman” ranked first on the Billboard Charts for 16 weeks.D. At the age of 94, Renald Richard’s musical life is close to an end.23.What can be the best title for the text?A. A Successful PianistB. A Life Filled with MusicC. A Wonderful BandleaderD. A Hit SongBMore women and newborns survive now than ever before, says a new United Nations report. But it also found that a baby or a pregnant woman still dies every 11 seconds somewhere in the world.The UN agencies United Nations International Children’s Emergency Fund (UNICEF) and the World Health Organization (WHO) wrote the report. It says that child deaths have decreased by almost half since the year 2000. Deaths of pregnant women have dropped by more than a third. These reductions are mostly the result of wider availability of lower-cost, good quality health services.But the latest data also showed that 5.3 million children worldwide died in 2018 by the age of five. Almost half of those deaths happened before a baby turned one month. The data also showed more than 290,000 women died because of problems during pregnancy and childbirth in 2017.Henrietta Fore is UNICEF's lead director. She said in a report, “A skilled pair of hands to help mothers and newborns around the time of birth, along with clean water, adequate nutrition, basic medicines and vaccines, can make the difference between life and death.”Fore urged governments and health leaders to invest in health services to, in her words, “do all it takes ... to save these precious lives.”The data showed huge inequalities around the world. Women and young children in sub-Saharan Africa, for example, have a much higher risk of death than in other large areas.Levels of maternal(母亲的) deaths are nearly 50 times higher for women in sub-Saharan Africa than in wealthy countries. Babies born there are also 10 times more likely to die in their first month of life, the report found.The report did note improvements. Since 1990, there has been a 56-percent drop in deaths of children under 15, from 14.2 million a year to 6.2 million in 2018.Countries in eastern and southeastern Asia have made the most progress. Those areas have seen an 80 percent drop in the number of deaths among children under five.24. The writer shows the situation of deaths of women and newborns by ________.A. giving examplesB. analyzing causesC. listing figuresD. following time order25. What can we infer from the text?A. Health services are important for newborns and pregnant women.B. Most child deaths happened in their first month of life in 2018.C. Skilled doctors and expensive medicines can save lives around the time of birth.D. Levels of maternal deaths are much higher than child deaths in sub-Saharan Africa.26.The purpose of the text is to _______.A. show the reductions in women and babies’ deathB. stress the problems of women and newborns’ deathC. show huge inequalities between countries in the worldD. display the improvements in eastern and southeastern AsiaCResearchers have developed a long-lasting, environmentally safe substance that can prevent wildfires from spreading. The new fluid is the creation of scientists and engineers at Stanford University in California. A report on their work was published in Proceedings of the National Academy of Sciences.The Stanford team says the treatment could give firefighters a valuable weapon in their battle against wildfires. Wildfires destroy an approximate 4 million hectares across the United States every year. In 2018, the federal government spent more than $3 billion on firefighting efforts, with billions more lost in property damage.The new report notes that a large majority of wildfires result from human activities. It says they often start along roadsides, in campgrounds and around electrical equipment. Firefighters often use chemical fire retardants to put out fires. Fire retardants are materials that can stop or slow the spread of fires. But firefighting crews currently use them only as a kind of reactive tool. This is because the chemical treatments are not designed to stay on trees or other vegetation for long periods. They can quickly disappear because of environmental and weather conditions.The thick liquid tested by the Stanford University team was found to extend the life of the fire retardants without harming the environment. Such a treatment could open up possibilities for firefighters to take preventive actions before fires even start, notes Eric Appel. He helped to write the new report. Appel is a professor of materials science and engineering at Stanford.The treatment holds the promise “to make wildland firefighting much more proactive, rather than reactive,” Appel said. “What we do now is monitor wildfire-prone areas and wait with bated breath for fires to start, then rush to put them out.”The researchers worked with California's Department of Forestry and Fire Protection, or CalFire, to carry out experiments. The testing found that the material “provides complete fire protection even afterhalf an inch of rainfall,” the study said. Under the same conditions, commonly used fire retardan ts provided little or no protection from fire.The development team says the treatment contains only safe materials widely used in food, drug and agricultural products. Over time, the material washes away and breaks down safely in the environment, the researchers noted.Eric Appel said he hopes the new treatment “can open the door to identifying and treating high-risk areas to protect people's lives and livelihoods.”27. What can we learn from the text?A. Most of the wildfires are caused by human activities.B. Wildfires destroy more than 4 million hectares in America every year.C. The chemical fire retardants are a valuable weapon to fight against wildfires.D. The money spent on firefighting was more than the loss in property damage.28. The new treatment has the following advantages except_______.A. providing much more protection from fireB. making it possible to take preventive actionsC. making the effect of the fire retardants last longerD. opening the door to identifying and treating high-risk areas29. What’s Eric Appel’s attitude towards the new treatment?A. Confident.B. Doubtful.C. Concerned.D. Neutral.30. Where does the text probably come from?A. A research plan.B. A science magazine.C. An advertisement.D. A science fiction.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。

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