江苏省南通市如东高级中学、沭阳如东中学2020-2021学年高三10月联考生物试题
江苏省如东高级中学2020-2021学年高三上学期10月调研物理试题
江苏省如东高级中学2020-2021学年高三上学期10月调研物理试题学校:___________姓名：___________班级：___________考号：___________一、单选题1．如图，滑块以初速度v0沿表面粗糙且足够长的固定斜面，从顶端下滑，直至速度为零．对于该运动过程，若用h、s、v、a分别表示滑块的下降高度、位移、速度和加速度的大小，t表示时间，则下列图象最能正确描述这一运动规律的是（）A．B．C．D．2．如图所示，粗糙水平面上的物体在水平拉力F作用下做匀加速直线运动，现使F不断变小到零，则在滑动过程中（）A．物体的加速度不断减小，速度不断增大B．物体的加速度不断增大，速度不断减小C．物体的加速度先增大后减小，速度先减小后增大D．物体的加速度先减小后增大，速度先增大后减小3．如图所示，曲线I是一颗绕地球做圆周运动卫星轨道的示意图，其半径为R，曲线II是一颗绕地球做椭圆运动卫星轨道的示意图，O点为地球球心，AB为椭圆的长轴，两轨道和地心都在同一平面内，已知在两轨道上运动的卫星的周期相等，万有引力常量为G，地球质量为M，下列说法正确的是（）A．椭圆轨道的长轴长度为1.5RB．卫星在I轨道的速率为v0，卫星在II轨道B点的速率为v B，则v0>v BC．卫星在I轨道的加速度大小为a0，卫星在II轨道A点加速度大小为a A，则a0>a AD．若OA= 0.5R，则卫星在B点的速率v B4．如图所示物块A和B的质量分别为4m和m，开始A、B均静止，细绳拉直，在竖直向上拉力F＝6mg作用下，动滑轮竖直向上运动．已知动滑轮质量忽略不计，动滑轮半径很小，不考虑绳与滑轮之间的摩擦，细绳足够长，在滑轮向上运动中，物块A和B 的加速度分别为A．a A＝12g，a B＝5g B．a A＝a B＝15gC．a A＝14g，a B＝3g D．a A＝0，a B＝2g5．如图所示,验电器带有少量正电荷,将一带负电的小球从远处逐渐靠近验电器的金属球.此过程中,可能看到金属箔片张开的角度A．不断增大B．先减小至零,后逐渐增大C．先增大,后减小D．先增大,后不变6．如图所示，质量均可忽的轻绳与轻杆所能承受的弹力的最大值一定，A 端用铰链固定，滑轮在A 点正上方（滑轮大小及摩擦均可不计），B 端吊一重力为G 的重物． 现将绳的一端挂在杆的B 端，用拉力F 将B 端缓慢向上拉（均未断），在仙杆转到竖直方向前，以下分析正确的是（）A ．绳子拉力越来越大B ．绳子拉力越来越小C ．AB 杆的弹力越来越大D ．AB 杆的弹力越来越小二、多选题7．在研究落体运动规律时，伽利略让小球从静止开始沿斜面滚下，则 （ ） A ．实验中测量小球运动的速度v 与时间tB ．实验中测量小球运动的位移x 与时间tC ．用斜面实验来“冲淡”重力，是为了方便测量时间D ．用斜面实验来“冲淡”重力，是为了方便测量速度8．如图，质量为m 2的物体2放在正沿平直轨道向右行驶的车厢底板上，并用竖直细绳通过光滑定滑轮连接质量为m 1的物体1，与物体1相连接的绳与竖直方向成θ角，则（ ）A ．车厢的加速度为gsinθB ．绳对物体1的拉力为1cos m g θC ．物体2所受底板的摩擦力为m 2gtanθD ．底板对物体2的支持力为(m 2－m 1)g 9．质量为m 的物体P 置于倾角为θ1的固定光滑斜面上，轻细绳跨过光滑定滑轮分别连接着P 与小车，P 与滑轮间的细绳平行于斜面，小车以速率v 水平向右做匀速直线运动．当小车与滑轮间的细绳和水平方向成夹角θ2时，下列判断正确的是（）A ．P 的速率为2cos v θ B ．P 的速率为v cos θ2C ．绳的拉力等于mg sin θ1D ．绳的拉力大于mg sin θ110．2021年6月14日.承担嫦娥四号中继通信任务的“鹊桥”中继星抵达绕地月第二拉格朗日点的轨道，第二拉格朗日点是地月连线延长线上的一点，处于该位置上的卫星与月球同步绕地球公转，则该卫星的A ．向心力仅来自于地球引力B ．线速度大于月球的线速度C ．角速度大于月球的角速度D ．向心加速度大于月球的向心加速度11．如图所示，质量为m 1的木块和质量为m 2的长木板叠放在水平地面上．现对木块施加一水平向右的拉力F ，木块在长木板上滑行，长木板始终静止．已知木块与长木板间的动摩擦因数为μ1，长木板与地面间的动摩擦因数为μ2，且最大静摩擦力与滑动摩擦力相等．则( )A ．μ1一定小于μ2B ．μ1可能大于μ2C ．改变F 的大小，F>μ2(m 1＋m 2)g 时，长木板将开始运动D．改F作用于长木板，F>(μ1＋μ2)(m1＋m2)g时，长木板与木块将开始相对滑动三、实验题12．用等效替代法验证力的平行四边形定则的实验情况如下图甲所示，其中A为固定橡皮筋的图钉，O为橡皮筋与细绳的结点，OB和OC为细绳，图乙是白纸上根据实验结果画出的图．(1)本实验中“等效代替”的含义是________．A．橡皮筋可以用细绳替代B．左侧弹簧测力计的作用效果可以替代右侧弹簧测力计的作用效果C．右侧弹簧测力计的作用效果可以替代左侧弹簧测力计的作用效果D．两弹簧测力计共同作用的效果可以用一个弹簧测力计的作用效果替代(2)图乙中的F与F′两力中，方向一定沿着AO方向的是________，图中________是F1、F2合力的理论值．(3)完成该实验的下列措施中，能够减小实验误差的是________．A．作图要用细芯铅笔，图的比例要尽量大些，要用严格的几何作图法作出平行四边形，图旁要画出表示力的比例线段，且注明每个力的大小和方向B．拉橡皮筋时，弹簧秤、橡皮筋、细绳应贴近木板且与木板面平行C．拉橡皮筋的细绳要长些，标记同一细绳方向的两点要远些D．使拉力F1和F2的夹角很大13．用图甲所示的装置“探究加速度与力、质量的关系”，实验中，保持小车质量一定时，探究小车加速度与合力F的关系。
江苏省如东高级中学2021届高三10月调研数学试题
度第一学期高三数 学 试 卷一、填空题：本大题共14小题，每小题5分，共70分．不需写出解答过程，请把答案直接填写在答题卡相应位置上．1. 设集合{1,2,3,4}P =，{|22,}Q x x x R =-≤≤∈，则P Q = ▲ .2. 函数()2ln(3)=-f x x x 的定义域是 ▲ .3. 命题“20,0x x ∀<>都有”的否定是 ▲ .4. 已知1tan()42πα+=，且02πα-<<，则=αsin ▲ .5. 若直线1l ：30++=x y m (0>m )与直线2l ：2630x y +-=的距离为10，则m = ▲ .6. 已知函数()sin 26f x x π⎛⎫=+⎪⎝⎭，若函数()(0)2y f x πϕϕ=-<<是偶函数，则ϕ= ▲ .7. 设函数()=f x 2-10310.x x x x⎧≥⎪⎪⎨⎪<⎪⎩，，，若f (a )>a ，则实数a 的取值范围为 ▲ .8. 定义在R 上的奇函数()f x 满足：当0>x 时，x x f x2019log 2019)(+=，则在R 上方程()0f x =的实根个数为▲ .9. 若{}1,∈-x m 是不等式2230--≤x x 成立的充分不必要条件,则实数m 的范围是 ▲ .10. 已知直线l 的方程是60x y +-=，,A B 是直线l 上的两点，且∆OAB 是正三角形（O 为坐标原点），则∆OAB 外接圆的方程是 ▲ .11. 在平面直角坐标系xOy 中，若曲线2x bax y +=(b a ,为常数)过点)4,1(P ，且该曲线在点P 处的切线与直线03=++y x 垂直，则b a 2+的值是 ▲ .12. 在三角形ABC ∆中，=4AB ，0AC λλ=>（），若2CA CB ⋅≥-对任意的0λ>恒成立，则角A 的取值范围为注 意 事 项考生在答题前请认真阅读本注意事项及各题答题要求 1．本试卷共4页，包含［填空题（第1题～第14题，共70分）、解答题（第15～20题，共90分）。
江苏省如东高级中学2020届高三数学10月调研试题(扫描版)
1如东高级中学2019-2020学年度第一学期高三年级10月调研测试数 学 试 卷一、填空题：本大题共14小题，每小题5分，共70分．不需写出解答过程，请把答案直接填写在答题卡相应位置上．1. 设集合{1,2,3,4}P =，{|22,}Q x x x R =-≤≤∈，则P Q = ▲ .2. 函数()2ln(3)=-f x x x 的定义域是 ▲ .3. 命题“20,0x x ∀<>都有”的否定是 ▲ .4. 已知1tan()42πα+=，且02πα-<<，则=αsin ▲ .5. 若直线1l ：30++=x y m (0>m )与直线2l ：2630x y +-=m = ▲ .6. 已知函数()sin 26f x x π⎛⎫=+ ⎪⎝⎭，若函数()(0)2y f xπϕϕ=-<<是偶函数，则ϕ= ▲ .7. 设函数()=f x 2-10310.x x x x⎧≥⎪⎪⎨⎪<⎪⎩，，，若f (a )>a ，则实数a 的取值范围为 ▲ .8. 定义在R 上的奇函数()f x 满足：当0>x 时，x x f x2019log 2019)(+=，则在R 上方程()0f x =的实根个数为 ▲ .29. 若{}1,∈-x m 是不等式2230--≤x x 成立的充分不必要条件,则实数m 的范围是 ▲ . 10. 已知直线l 的方程是60x y +-=，,A B 是直线l 上的两点，且∆OAB 是正三角形（O 为坐标原点），则∆OAB 外接圆的方程是 ▲ . 11. 在平面直角坐标系xOy 中，若曲线2xbax y +=(b a ,为常数)过点)4,1(P ，且该曲线在点P 处的切线与直线03=++y x 垂直，则b a 2+的值是 ▲ .12. 在三角形ABC ∆中，=4AB ，0AC λλ=>（），若2CA CB ⋅≥-对任意的0λ>恒成立，则角A 的取值范围为 ▲ . 13. 已知函数()⎪⎭⎫⎝⎛≤≤=2211x x x f ，记()m k d ,为函数()x f y =图像上的点到直线m kx y +=的距离的最大值，那么()m k d ,的最小值为 ▲ .14. 若存在[]1,2a ∈，使得关于x 的方程22()()+-=a a tx a x有四个不等的实数根，则实数t 的取值范围是 ▲ .二、解答题: 本大题共6小题．共90分．请在答题卡指定区域内作答，解答时应写出文字说明、证明过程或演算步骤． 15. (本小题满分14分)已知函数22()2sincos cos )4444x x x x f x =-． （1）求函数()f x 的最小正周期；（2）若[],x ππ∈-，[]()lg ()1g x f x =-，求函数()g x 的单调递增区间．16. (本小题满分14分)在ABC ∆中，=BC 2=⋅AC AB ．（1）求ABC ∆三边的平方和；（2）当ABC ∆的面积最大时，求cos B 的值．317．(本小题满分14分)已知直线l ：120kx y k -++= (k R ∈). （1）证明：直线l 过定点；（2）若直线l 交x 轴负半轴于A ，交y 轴正半轴于B ，AOB ∆的面积为S (O 为坐标原点)，求S 的最小值并求此时直线l 的方程．18. (本小题满分16分)如图,某市有一条东西走向的公路l ，现欲经过公路l 上的O 处铺设一条南北走向的公路m .在施工过程中发现在O 处的正北方向1百米的A 处有一汉代古迹.为了保护古迹,该市决定以A 为圆心、1百米为半径设立一个圆形保护区.为了连通公路l ，m ，欲再新建一条公路PQ ，点,P Q 分别在公路l ，m 上（点,P Q 分别在点O 的正东、正北方向），且要求PQ 与圆A 相切. （1） 当点P 距O 处2百米时，求OQ 的长； （2）当公路PQ 的长最短时, 求OQ 的长419．(本小题满分16分)已知a ∈R ，函数1()e x f x ax -=-的图象与x 轴相切． （1）求实数a 的值； （2）求()f x 的单调区间；（3）当1x >时，恒有()(1)ln f x m x x >-，求实数m 的取值范围．20. (本小题满分16分)已知函数f (x )＝x ln x －x ．（1）设g (x )＝f (x )＋|x －a |，a ∈R ．e 为自然对数的底数． ①当a ＝－2e 3时，判断函数g (x )零点的个数；②当x ∈ [1e ，e]时，求函数g (x )的最小值．（2）设0＜m ＜n ＜1，求证：f (n )＋2mm 2＋1＜0．5如东高级中学2019-2020学年度第一学期高三年级10月调研测试数学加试试卷（物理方向考生作答）解答题（共4小题，每小题10分共40分，解答时应写出文字说明，证明过程或演算步骤） 1. 设:p 实数x 满足22430x ax a -+<（其中0>a ），:q 实数x 满足302x x -≤-． 若p 是q 的必要不充分条件，求实数a 的取值范围．2. 已知函数()2sin cos 3f x x x π⎛⎫=+ ⎪⎝⎭．若02x π≤≤，求函数()f x 的值域.3. 二次函数2(0)y x bx b =+≠图像与x 轴交于O ，A 两点，交直线:l y x =于O ，B 两点，经过三点O ，A ，B 作圆C ．（1）求证：当b 变化时，圆C 的圆心在一条定直线上； （2）求证：圆C 经过除原点外的一个定点．4. 已知函数x e a x f x +-=2)21()(.（R a ∈） （Ⅰ）若)(x f 在区间)0(∞+，上单调递减，求实数a 的取值范围；（Ⅱ）若在区间),0(+∞上，函数)(x f 的图象恒在曲线xae y 2=下方，求a 的取值范围．6如东高级中学2019-2020学年度第一学期高三年级10月调研测试数学参考答案一、填空题: 本大题共14小题，每小题5分，共70分．不需写出解答过程，请把答案直接填写在答题卡相应位置上1. {}1,22. (0,3)3. 20,0x x ∃<≤有4. 1010- 5. 172 6. 3π 7. (-∞，-1)8. 3 9. ]23,1(- 10. 8)2()2(22=-+-y x 11. 5 12. [,]4ππ 13.8214. ( 二、解答题: 本大题共6小题．共90分．请在答题卡指定区域内作答，解答时应写出文字说明、证明过程或演算步骤． 15.(本小题满分14分) 解：（Ⅰ）()fx sin 22x x =π2sin 23x ⎛⎫=+ ⎪⎝⎭， ……………4分()f x ∴的最小正周期2π4π12T ==． ……………6分 （Ⅱ）由（Ⅰ）知[]()lg ()1lg 2sin()123x g x f x π⎡⎤=-=+-⎢⎥⎣⎦， ……………8分 故1sin()232x π+>，得5226236x k k πππππ+<+<+， 结合[]()lg ()1g x f x =-单调递增得226232x k k πππππ+<+≤+， ……………10分4433k x k k Z ππππ∴-<≤+∈，， ……………12分[],x ππ∈-，∴函数()g x 的单调递增区间为,33ππ⎛⎤- ⎥⎝⎦． ……………14分716. (本小题满分14分)解：（Ⅰ）（1）因为2=⋅AC AB ，所以cos 2AB AC A ∙∙=． ……………2分 在ABC ∆中，由余弦定理得：2222cos BC AB AC AB AC A =+-∙∙，即2224AB AC =+-，于是2210AB AC +=， ……………4分故22210616AB BC AC ++=+=为定值． ……………6分 （2）由（1）知：2210AB AC +=，所以2252AB AC AB AC +∙≤=，当且仅当AB AC =时取“=”号， ……………8分 因为cos 2AB AC A ∙∙=，所以2cos A AB AC=∙，从而sin A == ……………10分ABC ∆的面积11sin 22S AB AC A AB AC =∙∙=∙2=≤=， ……………12分 当且仅当AB AC =时取“=”号．因为2210AB AC +=，所以当AB AC =时，AB AC ==故2cos BCB AB ===． ……………14分17. 解:(1)证明：∵直线l 的方程可化为(2)(1)0k x y ++-=， ……………2分令2010x y +=⎧⎨-=⎩，解得：21x y =-⎧⎨=⎩， ……………4分∴无论k 取何值，直线总经过定点(2,1)-. ……………6分 (2)解：由题意可知0k ≠，再由l 的方程，得12(,0)kA k+-，(012)B k +，．8依题意得：120120k k k +⎧-<⎪⎨⎪+>⎩，解得0k >． ……………8分 ∵21112(12)11112(44)(224)422222k k S OA OB k k k k k ++=⋅⋅=⋅+==++≥⨯⨯+=，……10分当且仅当 140k k =>，即12k =，取“＝” ……………12分 ∴min 4S =，此时直线l 的方程为240x y -+=． ……………14分 18. 解：以O 为原点，直线l 、m 分别为,x y 轴建立平面直角坐标系．设PQ 与圆A 相切于点B ，连结AB ，以1百米为单位长度，则圆A 的方程为22(1)1x y +-=，……………2分(1)由题意可设直线PQ 的方程为12x yq+=，即220qx y q +-=，(2)q >， PQ与圆A 相切，∴1=，解得83q =， ……………4分故当P 距O 处2百米时，OQ 的长为83百米． ……………6分 （2）设直线PQ 的方程为1x yp q+=，即0qx py pq +-=，(1,2)p q >>， ∵PQ 与圆A1=，化简得22q p q =-，则22222q PQ p q q q =+=+-， ……………9分令2()(2)2q f q q q q =+>-，∴22222(1)(31)()2(2)(2)q q q f q q q q --+'=-=--(2)q >，…11分当322q +<<时，()0f q '<，即()f q在3(2,2+上单调递减；当q >()0f q '>，即()f q在)+∞上单调递增， …………13分9∴()f q在q =PQ 长最短时，OQ14分答：（1)当P 距O 处2百米时，OQ 的长为83百米；（2）当公路PQ 长最短时，OQ的长为32+百米 ……………16分19. 解：（1）()1ex f x a -'=-，设切点为0(,0)x ，依题意，00()0,()0,f x f x =⎧⎨'=⎩即00101e 0,e 0,x x ax a --⎧-=⎪⎨-=⎪⎩解得01,1,x a =⎧⎨=⎩所以()1e 1x f x -'=-． ……………4分 (2)当1x <时，()0f x '<；当1x >时，()0f x '>．故()f x 的单调递减区间为(,1)-∞，单调递增区间为(1,)+∞． ……………6分（3）令()()(1)ln g x f x m x x =--，0x >． 则11()e (ln )1x x g x m x x --'=-+-，令()()h x g x '=，则1211()e ()x h x m x x-'=-+， （ⅰ）若21m ≤，因为当1x >时，1e 1x ->，211()1m x x +<，所以()0h x '>，所以()h x 即()g x '在(1,)+∞上单调递增． ……………8分又因为(1)0g '=，所以当1x >时，()0g x '>，从而()g x 在[1,)+∞上单调递增，而(1)0g =，所以()0g x >，即()(1)ln f x m x x >-成立． ……………10分（ⅱ）若12m >，可得1211()e ()x h x m x x -'=-+在(0,)+∞上单调递增．因为(1)120h m '=-<，211(1ln(2))201ln(2)[1ln(2)]h m m m m m ⎡⎤'+=-+>⎢⎥++⎣⎦，……………12分 所以存在1(1,1ln(2))x m ∈+，使得1()0h x '=，且当1(1,)x x ∈时，()0h x '<，所以()h x 即()g x '在1(1,)x 上单调递减，又因为(1)0g '=，所以当1(1,)x x ∈时，()0g x '<， 从而()g x 在1(1,)x 上单调递减，而(1)0g =，所以当1(1,)x x ∈时，()0g x <，即()(1)ln f x m x x >-不成立综上所述，k 的取值范围是1(,]2-∞……………16分1020. 解：（1）①当a ＝－2e 3时，g (x )＝x ln x －x ＋|x ＋2e 3|＝x ln x ＋2e3，g′(x )＝1＋ln x ，当0＜x ＜1e 时，g′(x )＜0；当x ＞1e时，g′(x )＞0；因此g (x )在 (0，1e )上单调递减，在(1e ，＋∞)上单调递增，又g (1e 4)＝2e 3－4e 4＝2e －4e 4＞0，g (1e )＝－1e ＋2e 3＝2－e 2e 3＜0，g (1)＝2e3＞0，所以g (x )有且仅有两个零点． ……………2分 ②（i ）当a ≤1e时，g (x )＝x ln x －x ＋x －a ＝x ln x －a ，因为x ∈[1e，e]，g′(x )＝1＋ln x ≥0恒成立，所以g (x )在[1e ，e]上单调递增，所以此时g (x )的最小值为g (1e )＝－1e －a ．……………4分（ii ）当a ≥e 时，g (x )＝x ln x －x ＋a －x ＝x ln x －2x ＋a ，因为x ∈[1e，e]，g′(x )＝ln x －1≤0恒成立，所以g (x )在[1e ，e]上单调递减，所以此时g (x )的最小值为g (e)＝a －e ．……………6分（iii ）当1e＜a ＜e 时，若1e ≤x ≤a ，则g (x )＝x ln x －x ＋a －x ＝x ln x －2x ＋a ， 若a ≤x ≤e，则g (x )＝x ln x －x ＋x －a ＝x ln x －a ，由（i ），（ii ）知g (x )在[1e，a ]上单调递减，在[a ，e]上单调递增，所以此时g (x )的最小值为g (a )＝a ln a －a ， ……………8分 综上有：当a ≤1e 时，g (x )的最小值为－1e－a ；当1e＜a ＜e 时，g (x )的最小值为a ln a －a ； 当a ≥e 时，g (x )的最小值为a －e ． ……………10分 （2）设h (x )＝2xx 2＋1， 则当x ∈(0，1)时，h′(x )＝2(1－x 2)(1＋x 2)2＞0，于是h (x )在(0，1)单调递增，又0＜m ＜n ＜1，所以h (m )＜h (n )，11 从而有f (n )＋2m m 2＋1＜f (n )＋h (n )＝n (ln n －1＋2n 2＋1) ……………12分 设φ(x )＝ln x －1＋2x 2＋1，x ＞0 则φ′(x )＝1x －4x (1＋x 2)2＝(x 2－1)2x (1＋x 2)2≥0， 因此φ(x )在(0，＋∞)上单调递增，因为0＜n ＜1，所以φ(n )＜φ(1)＝0，即ln n －1＋2n 2＋1＜0， 因此f (n )＋2m m 2＋1＜n (ln n －1＋2n 2＋1)＜0， 即原不等式得证． ……………16分如东高级中学2019-2020学年度第一学期高三年级10月调研测试数学（加试）参考答案1.解： 设{}3A x a x a =<<， {}23B x x =<≤，p 是q 的必要不充分条件，则B A Ö； 则02 33a a <≤⎧⎨>⎩，所以实数a 的取值范围是12a <≤. ……………10分 2.解： ()()2sin cos sin cos f x x x x x x x ==+1sin 22sin 223x x x π⎛⎫==+ ⎪⎝⎭ ……………5分 由02x π≤≤得，42333x πππ≤+≤，sin 213x π⎛⎫≤+≤ ⎪⎝⎭．∴0sin 21322x π⎛⎫≤++≤+ ⎪⎝⎭，即函数()f x的值域为0,12⎡+⎢⎣⎦． ……………10分 3.解：（I ）在方程bx x y +=2中．令x y y ==,0，易得()()b b B b A ---1,1,0,设圆C 的方程为022=+++Ey Dx y x 则()()()()⎩⎨⎧=-+-+-+-=-011110222E b D b b b bD b ⇒⎩⎨⎧-==2b E b D ， 故经过三点O ，A ，B 的圆C 的方程为x 2+y 2+bx+（b ﹣2）y=0，设圆C 的圆心坐标为（x 0，y 0），12 则x 0=﹣，y 0=﹣，∴y 0=x 0+1，这说明当b 变化时，（I ）中的圆C 的圆心在定直线y=x+1上． ……………5分（II ）设圆C 过定点（m ，n ），则m 2+n 2+bm+（b ﹣2）n=0，整理得（m+n ）b+m 2+n 2﹣2n=0，它对任意b≠0恒成立，∴⇒或故当b 变化时，（I ）中的圆C 经过除原点外的一个定点坐标为（﹣1，1）． ……………10分4.解：（Ⅰ）)(x f 在区间)0(∞+，上单调递减，则01)12()(2≤+-='x e a x f 在区间)0(∞+，上恒成立． …………1分 即x e a 2121≥-，而当)0(∞+∈，x 时，112<x e，故121≥-a ． 所以0≤a ． …………3分 （Ⅱ）令x ae e a ae x f x g x x x +--=-=2)21(2)()(2，定义域为R .在区间),0(+∞上，函数)(x f 的图象恒在曲线x ae y 2=下方等价于0)(<x g 在区间),0(+∞上恒成立. ∵]1)12)[(1(12)12()(2---=+--='x x x xe a e ae e a x g …………4分 ①若21>a ，令0)(='x g ，得极值点01=x ，121ln 2-=a x ， 当012=>x x ，即121<<a 时，在（2x ，+∞)上有0)(>'x g ，此时)(x g 在区间),(2+∞x 上是增函数，并且在该区间上有)),(()(2+∞∈x g x g ，不合题意；当012=≤x x ，即1≥a 时，同理可知，)(x g 在区间),0(+∞上，有)),0(()(+∞∈g x g ，也不合题意； …………6分 ②若21≤a ，则有012≤-a ，此时在区间),0(+∞上恒有0)(<'x g ，从而)(x g 在区间),0(+∞上是减函数；要使0)(<x g 在此区间上恒成立，只须满足021)0(≤--=a g 21-≥⇒a ， 由此求得a 的范围是]21,21[-． …………8分 综合①②可知，当]21,21[-∈a 时，函数)(x f 的图象恒在直线x ae y 2=下方.………10分。
2021届江苏省南通市如东中学高三10月月考数学试题
2020-2021学年如东中学10月高三月考数学试题一、选择题（本大题共8小题，共40分）1.已知集合0，1，2，，0，，，则A. B. 2，C. 0，D. 0，2，2.设，则“”是“”的A. 充分不必要条件B. 必要不充分条件C. 充要条件D. 既不充分也不必要条件3.若，则下列结论正确的是A. B.C. D.4.已知，且，则A. B. C. D.5.已知，向量，，若，则实数A. B. C. D. 26.若抛物线的焦点是椭圆的一个焦点，则A. 2B. 3C. 4D. 87.“蒙日圆”涉及几何学中的一个著名定理，该定理的内容为：椭圆上任意两条互相垂直的切线的交点必在一个与椭圆同心的圆上，该圆称为原椭圆的蒙日圆．若椭圆：的离心率为，则椭圆的蒙日圆方程为A. B. C. D.8.已知M为函数的图像上任意一点，过M作直线MA，MB分别与圆相切于A，B 两点，则原点O到直线AB得距离的最大值为A. B. C. D.二、不定项选择题（本大题共4小题，共20分）9.设M，N是抛物线上的两个不同的点，O是坐标原点，若直线OM与ON的斜率之积为，则下列说法错误的是A. B. 以MN为直径的圆的面积大于C. 直线MN过抛物线的焦点D. O到直线MN的距离不大于210、已知是定义域为R的函数，满足，，当时，，则下列说法正确的是A. 的最小正周期为4B. 的图像关于直线对称C. 当时，函数的最大值为2D. 当时，函数的最小值为11.已知函数的图象如图所示，令，则下列关于函数的说法中正确的是A. 函数图象的对称轴方程为B. 函数的最大值为C. 函数的图象上存在点P，使得在P点处的切线与直线平行D. 方程的两个不同的解分别为，，则最小值为12.把方程表示的曲线作为函数的图象，则下列结论正确的有A. 的图象不经过第三象限B. 在R上单调递增C. 的图象上的点到坐标原点的距离的最小值为1D. 函数不存在零点三、填空题（本大题共4小题，共20分）13.已知双曲线C：，O为坐标原点，F为C的右焦点，过F的直线与C的两条渐近线的交点分别为M、若为直角三角形，则______________．14.将函数的图象向右平移个单位长度，则平移后的图象中与y轴最近的对称轴的方程是______________．15.在中，，，，M、N分别为BC、AM的中点，则______________．16.已知，，记，则M 的最小值为_____；当M最小时，________．四、解答题（本大题共6小题，共70分）17.在中，a，b，c分别为内角A，B，C的对边，且满足．求A的大小；若，，求的面积．18.为调查某校学生的课外阅读情况，随机抽取了该校100名学生男生60人，女生40人，统计了他们的课外阅读达标情况一个学期中课外阅读是否达到规定时间，结果如下：是否达标不达标达标性别男生3624女生1030是否有的把握认为课外阅读达标与性别有关？附：，k如果用这100名学生生中男生和女生课外阅读“达标”的频率分别代替该校男生和女生课外阅读“达标”的概率，且每位学生是否“达标”相互独立．现从该校学生中随机抽取3人男1女，设随机变量X表示“3人中课外阅读达标的人数”，试求X的分布列和数学期望．19.已知圆．求过点且和圆C相切的直线方程；若斜率为1的直线n与圆交于两点，求面积的最大值及此时直线n 的方程．20.在平面直角坐标系xOy中，椭圆E：的离心率为，焦距为2．求椭圆E的方程；如图，动直线l：交椭圆E于两点，C是椭圆E上一点，直线OC的斜率为，且，M是线段OC延长线上一点，且，的半径为，是的两条切线，切点分别为求的最大值，并求取得最大值时直线l的斜率．21已知函数．讨论的单调性；当时，记在区间的最大值为M，最小值为m，求的取值范围．22.已知函数有两个极值点．求a的取值范围设，是的两个极值点，证明：．2020-2021学年如东中学10月高三月考高一试题一、选择题（本大题共8小题，共40分）1.已知集合0，1，2，，0，，，则A. B. 2，C. 0，D. 0，2，【答案】A2.设，则“”是“”的A. 充分不必要条件B. 必要不充分条件C. 充要条件D. 既不充分也不必要条件【答案】A3.若，则下列结论正确的是A. B.C. D.【答案】A4.已知，且，则A. B. C. D.【答案】A5.若抛物线的焦点是椭圆的一个焦点，则A. 2B. 3C. 4D. 8【答案】D6.已知，向量，，若，则实数A. B. C. D. 2【答案】D7.“蒙日圆”涉及几何学中的一个著名定理，该定理的内容为：椭圆上任意两条互相垂直的切线的交点必在一个与椭圆同心的圆上，该圆称为原椭圆的蒙日圆．若椭圆：的离心率为，则椭圆的蒙日圆方程为A. B. C. D.【答案】B8.已知M为函数的图像上任意一点，过M作直线MA，MB分别与圆相切于A，B两点，则原点O到直线AB得距离的最大值为A. B. C. D.【答案】B二、不定项选择题（本大题共4小题，共20分）9.设M，N是抛物线上的两个不同的点，O是坐标原点，若直线OM与ON的斜率之积为，则下列说法错误的是A. B. 以MN为直径的圆的面积大于C. 直线MN过抛物线的焦点D. O到直线MN的距离不大于2【答案】ABC10.已知是定义域为R的函数，满足，，当时，，则下列说法正确的是A. 的最小正周期为4B. 的图像关于直线对称C. 当时，函数的最大值为2D. 当时，函数的最小值为【答案】ABC11.已知函数的图象如图所示，令，则下列关于函数的说法中正确的是A. 函数图象的对称轴方程为B. 函数的最大值为C. 函数的图象上存在点P，使得在P点处的切线与直线平行D. 方程的两个不同的解分别为，，则最小值为【答案】ABD12.把方程表示的曲线作为函数的图象，则下列结论正确的有A. 的图象不经过第三象限B. 在R上单调递增C. 的图象上的点到坐标原点的距离的最小值为1D. 函数不存在零点【答案】ACD三、填空题（本大题共4小题，共20分）13.将函数的图象向右平移个单位长度，则平移后的图象中与y轴最近的对称轴的方程是______________．【答案】14.在中，，，，M、N分别为BC、AM的中点，则______________．【答案】15已知双曲线C：，O为坐标原点，F为C的右焦点，过F的直线与C的两条渐近线的交点分别为M、若为直角三角形，则______________．【答案】316.已知，，记，则M的最小值为_____；当M最小时，________．【答案】；四、解答题（本大题共6小题，共70分）17.在中，a，b，c分别为内角A，B，C的对边，且满足．求A的大小；若，，求的面积．【答案】解：因为，由正弦定理，得，即，所以，因为，所以．由正弦定理，得，由余弦定理，得，解得，所以的面积．【解析】本题主要考查了正弦定理、余弦定理以及三角形面积公式的应用，属于基础题．根据题目条件结合正弦定理可得，代入余弦公式可得，结合A的取值范围，可得A的大小．由结合正弦定理可求得b，代入余弦定理，可解得，进而代入三角形面积公式计算可得结果．18.为调查某校学生的课外阅读情况，随机抽取了该校100名学生男生60人，女生40人，统计了他们的课外阅读达标情况一个学期中课外阅读是否达到规定时间，结果如下：是否达标不达标达标性别男生3624女生1030是否有的把握认为课外阅读达标与性别有关？附：，k如果用这100名学生生中男生和女生课外阅读“达标”的频率分别代替该校男生和女生课外阅读“达标”的概率，且每位学生是否“达标”相互独立．现从该校学生中随机抽取3人男1女，设随机变量X表示“3人中课外阅读达标的人数”，试求X的分布列和数学期望．【答案】解：假设：课外阅读达标与性别无关，根据列联表，求得的观测值，因为当成立时，的概率约为，所以有以上的把握认为课外阅读达标与性别有关；记事件A为：从该校男生中随机抽取1人，课外阅读达标；事件B为：从该校女生中随机抽取1人，课外阅读达标．由题意知：，．随机变量X的取值可能为0，1，2，3．，，，．所以随机变量X的分布列为：X0123P期望．【解析】本题考查独立性检验，古典概型的计算公式，离散型随机变量的分布列与期望的求法，考查计算能力，属于中档题．根据表格数据，利用公式求出的观测值k，判断即可．求出男生、女生课外阅读“达标”的概率，列出随机变量X的分布列，求得数学期望即可．19.已知圆．求过点且和圆C相切的直线方程；若斜率为1的直线n与圆交于两点，求面积的最大值及此时直线n 的方程．【答案】解：设所求切线l的斜率为k，因为在圆上，所以，则，所以直线l的方程为，即．解法一：设直线n的方程为，则圆心到直线的距离为，弦长，则，当且仅当，即或时取“”满足直线与圆相交，故面积的最大值为2，此时直线n的方程为．解法二：，当且仅当，即时，的面积最大，此时．设直线n的方程为，则圆心到直线n的距离，由，由，即，得或，所以直线n的方程为．【解析】本题主要考查了圆与直线的位置关系，涉及圆的标准方程、点到直线的距离、弦长公式、基本不等式、与圆有关的最值问题，属于中档题．由圆的标准方程得到圆心及半径，进而可知点P在圆上，求出直线PC的斜率后即可求得圆的切线斜率，最后得到直线的方程；解法一：先求出圆心到直线的距离及弦长DE，利用基本不等式求得最值即可；解法二：由三角形的面积公式知时三角形面积最大，此时，由弦长公式求得，进而由点到直线的距离即可求解．20.在平面直角坐标系xOy中，椭圆E：的离心率为，焦距为2．求椭圆E的方程；如图，动直线l：交椭圆E于两点，C是椭圆E上一点，直线OC的斜率为，且，M是线段OC延长线上一点，且，的半径为，是的两条切线，切点分别为求的最大值，并求取得最大值时直线l的斜率．【答案】解：由题意知，，所以，，因此椭圆E的方程为．Ⅱ设，，联立方程得，由题意知，，，所以．由题意可知圆M的半径，由题设知，所以，因此直线OC的方程为联立方程得，，因此．由题意可得，而令，则，，因此，当且仅当，即时等号成立，此时．【解析】本题考查直线与圆、圆与椭圆位置关系的应用，训练了利用配方法求函数的最值，考查计算能力，是压轴题．由题意得关于a，b，c的方程组，求解方程组得a，b的值，则椭圆方程可求设，，联立直线方程与椭圆方程，利用根与系数的关系求得A，B的横坐标的和与积，由弦长公式求得，由题意可知圆M的半径r，则由题意设知得到直线OC的方程，与椭圆方程联立，求得C点坐标，可得，由题意可知，，转化为关于的函数，换元后利用配方法求得取得最大值时直线l的斜率为．21.已知函数．讨论的单调性；当时，记在区间的最大值为M，最小值为m，求的取值范围．【答案】解：．令，得或．若，则当时，当时，，故在单调递增，在单调递减；若，在单调递增；若，则当时，当时，，故在单调递增，在单调递减．当时，由知，在单调递减，在单调递增，所以在区间的最小值为，最大值为或．于是，，所以．当时，可知单调递减，所以的取值范围是．当时，单调递增，所以的取值范围是．综上，的取值范围是．【解析】本题考查利用导数研究函数的单调性，考查利用导数研究闭区间上函数的最值，是较难题．求导，分，和三类分类讨论即可求解；当时，从而在单调递减，在单调递增，所以在区间的最小值为，最大值为或分和讨论即可求解．15.已知函数有两个极值点．求a的取值范围设，是的两个极值点，证明：．【答案】解：由，，得，函数有两个极值点等价于在上有两个变号零点，等价于在上有两个变号零点，令，则，所以时，，单调递增，时，，单调递减，所以，当时，恒成立，在上单调递减，不可能有两个极值点，舍去，当时，，，，，而，由零点存在性定理得在和内分别存在一个变号零点，此时有两个极值点，综上，所求a的取值范围为证明：因为，是的两个极值点，所以，且，由知，，，令，，则，由在恒成立，当时，，单调递减，又，所以时，，即，所以，所以，由知，在单调递减，所以，即，所以，即，因为，所以，，所以，即．【解析】本题考查了利用导数研究函数的单调性，利用导数研究函数的极值，不等式的恒成立问题．首先求得，从而得出函数有两个极值点，等价于在上有两个变号零点，等价于在上有两个变号零点，令，进而可得，进而通过研究的单调性得出，分和两种情况进行讨论，从而由零点存在性定理得在和内分别存在一个变号零点，进而可得a的取值范围；由，是的两个极值点，可得出，且，由知，，令，，从而，进而可判断，单调递减，进而判断在单调递减，进而得出，进而得出，进而证明即．。
2020-2021学年江苏省南通市如东高级中学高二10月月考数学试题(解析版)
如东高级中学 2020-2021学年度第一学期10月月考高二数学试题一、单项选择题（本大题共8小题，每小题5分，共计40分．在每小题给出的四个选项中，只有一个是符合题目要求的，请把答案添涂在答题卡相应位置上）1.数列0，0，0，，0 （）A．既不是等差数列又不是等比数列B．是等比数列不是等差数列C．是等差数列不是等比数列D．是等比数列又是等差数列2. 下列不等式中与不等式同解的是（）A．B．C．D．3.已知等差数列中，则的值为（）A. B. C. D.4.已知不等式：（1）（2）（3）2，若要同时满足不等式（1）（2）的也满足不等式（3），则有（）A．B．C．D．5.已知正项数列中，，则的值为（）A．B．4C．8 D．166. 若不等式对任意恒成立，则实数a的取值范围是A. B. C. D.7. 定义“等积数列”：在一个数列中，如果每一项与它的后一项的积都为同一个常数，那么这个数列叫做等积数列，这个常数叫做该数列的公积，已知数列是等积数列且，前41项的和为103，则这个数列的公积为A. 2B. 3C. 6D. 88.南宋数学家杨辉在详解九章算法和算法通变本末中，提出了一些新的垛积公式，所讨论的高阶等差数列与一般等差数列不同，前后两项之差并不相等，但是逐项差数之差或者高次差成等差数列对这类高阶等差数列的研究，在杨辉之后一般称为“垛积术”现有高阶等差数列，其前7项分别为1，5，11，21，37，61，95，则该数列的第8项为A. 99B. 131C. 139D. 141二、多项选择题（本大题共4小题，每小题5分，共计20分．在每小题给出的四个选项中，至少有两个是符合题目要求的，请把答案添涂在答题卡相应位置上）9.已知，，则下列不等式中，正确的是A. B.C. D.10.对于数列，若存在数列满足，则称数列是的“倒差数列”，下列关于“倒差数列”描述正确的是A. 若数列是单增数列，但其“倒差数列”不一定是单增数列；B. 若，则其“倒差数列”有最大值；C. 若，则其“倒差数列”有最小值；D. 若，则其“倒差数列”有最大值．11. 已知数列的前n项和为，且满足，，则下列说法错误的是A. 数列的前n项和为B. 数列的通项公式为C. 数列为递增数列D. 数列为递增数列12. 若数列对任意满足，下面选项中关于数列的命题正确的是A. 可以是等差数列B. 可以是等比数列C. 可以既是等差又是等比数列D. 可以既不是等差又不是等比数列三、填空题.（本大题共4题，每题5分，共20分.请同学们将答案填到答题卷上对应的位置处.）13.若数列是公差不为0的等差数列，、、成等差数列，则的值为________．14.等差数列中，，则________．15. 三个同学对问题“已知m，，且，求的最小值”提出各自的解题思路：甲：，可用基本不等式求解；乙：，可用二次函数配方法求解；丙：，可用基本不等式求解；参考上述解题思路，可求得当________时，有最小值16. 定义：关于x的两个不等式和的解集分别为和，则称这两个不等式为对偶不等式如果不等式与不等式为对偶不等式，且，则________．三、解答题（本大题共有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. 设为数列的前n项和，且．（1）若，判断数列的单调性；（2）若，求数列的前n项和．18. 若数列的前n项和，求数列的通项公式．（2）若数列的前n项和，证明为等比数列．19. 已知数列的前n项和为，，满足．（1）计算，，，猜想的一个表达式（不需要证明）（2）设，数列的前n项和为，求证：．20.已知二次函数，满足，．（1）求函数的解析式；（2）若关于x的不等式在上有解，求实数t的取值范围；（3）若函数的两个零点分别在区间和内，求实数m的取值范围．21.已知公差大于0的等差数列的前n项和为，且满足，．（1）求数列的通项公式；（2）若，求的表达式；（3）若，存在非零常数c，使得数列是等差数列，存在，不等式成立，求k的取值范围．22. 已知函数b是非零实常数满足，且关于x的方程的解集中恰有一个元素．（1）求a，b的值；（2）在直角坐标系中，求定点到函数图像上任意一点的距离的最小值；（3）当时，不等式恒成立，求实数m的取值范围．2020-2021学年度第一学期10月月考高二数学试题一、单项选择题（本大题共8小题，每小题5分，共计40分．在每小题给出的四个选项中，只有一个是符合题目要求的，请把答案添涂在答题卡相应位置上）1.数列0，0，0，，0 （）A．既不是等差数列又不是等比数列B．是等比数列不是等差数列C．是等差数列不是等比数列D．是等比数列又是等差数列【答案】C【解析】数列0，0，0，，0，是无穷数列，从第二项开始起，每一项与它前一项的差都等于常数0，符合等差数列的定义，所以数列0，0，0，，0，是等差数列，根据等比数列的定义可知，等比数列中不含有为0的项，所以数列0，0，0，，0，不是等比数列，故选C．2. 下列不等式中与不等式同解的是（）A．B．C．D．【答案】D【解析】不等式等价为，即，故选：D．3.已知等差数列中，则的值为（）A. B. C. D.【答案】B【解析】设等差数列的公差为d，由知：又，，可知：，且，，。
江苏省如东高级中学2020┄2021学年高一10月阶段测试化学试题Word版 含解析
江苏省如东高级中学2020┄2021学年第一学期阶段考试10高一化学试卷试卷说明：1、本卷分I卷和II卷两部分，请将I卷答案填涂在答题卡上，II卷答案填写在答卷相应区域内，否则答题无效。
2、本卷分值100分，答题时间100分钟。
3、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-32 Cl-35.5 K-39 Ca-40 Fe-56 Cu-64 Zn-65 Ag-108 Ba-137第I卷（选择题，共40分）选择题（本题包括10小题，每小题只有一个选项符合题意，每小题2分，共20分。
）1．我国科学家在世界上第一次为一种名为“钴酞菁”的分子（直径为1.3×10-9m）恢复了磁性。
“钴酞菁”分子的结构和性质与人体内的血红素及植物休内的叶绿素非常相似。
下列说法中不正确的是（）A．“钴酞菁”分子所形成的分散系不具有丁达尔效应B．“钴酞菁”分子能透过滤纸C．此项工作可以用来改变分子的某些物理性质D．此项工作可广泛应用于光电器件、生物技术等方面2．下列电离方程式书写正确的是（）A．Al2（SO4）3 == 2 Al3 + + 3SO42— B．Ca（NO3）2 == Ca2+ + 2（NO3）2- C．Ba（OH）2 == Ba2+ + OH2— D．Na2CO3 == Na2+ + CO32—3．下列实验操作的描述中，正确的是（）A．取用Na2CO3溶液时，发现取量过多，为了不浪费，又把过量的试剂倒入试剂瓶中B．用溶解、过滤的方法分离氯化钠和硝酸钾的混合物C．称量NaOH固体，先在天平两盘上放大小相等纸片，然后NaOH放在左盘纸片上称量D .利用丁达尔效应可以区别溶液和胶体4．下列叙述正确的是（）A．1 mol的CH4质量为16g/mol B．3.01×1023个CO2分子的质量约为22g C．H2O的摩尔质量为18g D．标准状况下,1 mol任何物质体积均为22.4L5．食盐分散到酒精中形成直径在1nm~l00nm之间的超细粒子，这样的分散系是（）A．悬浊液 B．溶液 C．胶体 D．乳浊液6. 人们常根据物质各种特点对物质进行分类．下列分类中，前者包含后者的是（）A．氧化物、化合物B．溶液、分散系C．含氧酸、酸D．化合物、电解质7．下列物质属于电解质的是（）①氢氧化钾②硫酸钡③铜④蔗糖⑤氨气⑥稀硫酸⑦熔融的KNO3A．②⑤⑦ B．①②⑦ C．①⑥⑦ D．①⑤⑥8．用下列实验装置和方法进行相应实验，正确的是（）图1 图2 图3 图4A．用图1所示方法称量固体氯化钠B．用图2装置分离碘的四氯化碳溶液C．用图3所示装置和方法进行石油分馏D．用图4装置配制150 mL稀盐酸9．在下列各溶液中，离子一定能大量共存的是（）A．强碱性溶液中：K＋、Mg2＋、Cl—、SO42—B．滴加石蕊试液显红色的溶液：：K＋、Al3＋、Cl—、NO3—C．含有0.1mol·L—1Ca2＋的溶液中：Na＋、K＋、CO32—、Cl—D．无色透明的酸性溶液中：MnO4—、K+、SO42—、Na+10．实验室欲配制6.00 mol·L—1的H2SO4溶液，现有三种不同浓度的硫酸：①240 mL 1.00 mol/L的硫酸②150 mL 3.00 mol·L—1的硫酸③足量的18.00 mol·L—1的浓H2SO4。
2020-2021学年江苏如东高级中学高三英语第二次联考试卷及参考答案
2020-2021学年江苏如东高级中学高三英语第二次联考试卷及参考答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项AKate Humble: Books that changed my lifeKate Humble is a writer and broadcaster specializing in science, wildlife and rural affairs. Together with her husband site runs Humble by Nature, a rural skills education centre on working farm near Monmouth intheWyeValley.Winnie the Poohby A A MilneMy father used to read this to me when I was very young — he used different voices for all the animals. The characterization (角色设定) was so clever; we all know someone just like each inhabitant of the HundredAcre Wood: gloomy Eeyore; thick but loyal Pooh; enthusiastic Tigger.A A Milne was masterful in exploring the way they got along together, opening my eyes to how society really works.Last Chance to Seeby Douglas Adams and Mark CarwardineThis book tells of the authors, adventures as they set out to find the rarest of animals, those on the edge of extinction.Their travels are rather exciting and they share a wonderful humour, which really appealed to me. Yet underpinning (支撑) everything is the realization that we can't just sit back and allow species to disappear. PicturePalaceby Paul TherouxI've always loved Theroux's travel writing, but this novel took my breath away. The words aren't long or complicated but, fromthat first paragraph, his writing grabs you by the nose hairs and drags you along. I had an art teacher who told me, “You're only an artist when you've found your own style, not when you're copying someone else, and Theroux represents this.”1. Why did the author mention the characterization ofWinnie the Pooh?A. To indicate the book has realistic values.B. To show how adorable the characters are.C. To persuade people to learn from the characters.D. To prove the writer is good at creating characters.2. What didLast Chance to Seestrike into Kate's heart?A. Curiosity.B. Responsibility.C. Exploration.D. Devotion.3. Which writer does Kate Humble like for his original writing?A. A A Milne.B. Douglas Adams.C. Mark Carwardine.D. Paul Theroux.BWhen girls play with cars, they're serious. It was at primary school that Gu Huijing first became interested in cars after watching some car-themed movies. Born in2004 inShenzhen, Guangdong province, Gu decided to major in automobiles (汽车) when she was in junior high.She's driven by interest, but she is also a realist. “I think the future of the automobile industry is bright because we cannot live without food, clothes, houses and cars,” she said.In April, she won first prize at a competition for vehicle maintenance (车辆维修) in Dongguan city, which drew 33 teams from different cities across the province.“I thought it couldn't be that hard to repair cars when I started to take courses, but I was totally wrong,” Gu said. When she started learning in 2019, she was frequently confused by various problems. “And competition within our school is intense, so I had to work extra hard,” she said.Her teacher once tried to persuade her to give up as she was a sophomore (大二学生), and wasn't as knowledgeable as the seniors, and no women had ever been selected for the competition before. “But I insisted that I would carry on,” she said. Finally, her training and hard work won her the only place to stand for her school at the competition.A woman winning first prize in a vehicle maintenance competition became a hot topic on-line, causing heated discussions over gender (性别) and career choices. “Women should not be influenced by old-fashioned thinking, and should do whatever it takes to discover their interests and strengths. There are more possibilities out there,” one netizen wrote.Gu said, “Many people think vehicle maintenance is a job for men. That's wrong. I don't think gender hasanything to do with choice of jobs.” “I have a goal and I will work harder to make it happen,” she added. “I will be responsible for my choices.”4. What was a cause of Gu's choosing automobiles as her major?A. The love for automobiles.B. The guidance from her parents.C. The high popularity of automobiles.D. The determination to contribute to society.5. Why did Gu's teacher advise her to quit the competition?A. She lacked the knowledge related to it.B. She had never taken part in a competition before.C. She was unlikely to win the competition.D. Women were not allowed to enter the competition.6. Which of the following best describes Gu Huijing?A. Creative and caring.B. Determined and hard-working.C. Independent and humorous.D. Honest and courageous.7. What would be the best title for the text?A. There Is No End To LearningB. Teenage Girl Wins Car Repair ContestC. Vehicle Maintenance Catches OnD. Taking Challenge Leads To SuccessCFrom skateboarding to Fleetwood Mac, TikTok users got creative in a pandemic (流行病) year with new songs, dances and shows in 60 seconds or less.The social mediavideo app on Wednesday shared its list of top 100 videos, creators and trends in America during 2020. “These videos brought joy and inspiration to millions of Americans in the rough year,” said Kudzi Chikumbu, director of creator community at TikTok.The platform has been widely associated with Generation Z (people born after 1996), millennials (people born in the 1980s or 1990s) and influencers who have started careers based on the shared videos.The Weeknd'sBlinding Lightsand Jason Derulo's Savage Loveinspired TikTok dance challenges that made them among the top songs used on the app.For some TikTok creators, the pandemic itself became inspiration to create a connection with other users. Comedian Caitlin Reilly used the app during quarantine (隔离) to make fun of those annoying coworkers for one of the top liked videos. And singer Curtis Roach made an song for the endless days at home with his song Bored in the House.The second most popular video was an Idaho man named Nathan Apodaca who shot to fame after coolly singing Dreams by Fleetwood Mac, while skateboarding on a highway.But the most liked TikTok video was awarded to Bella Poarch with 45 million followers, who created a simple but interesting video with a few seconds of well-timed head nods and eye rolls.While the app might be designed for youth, Chikumbu said that the growth of TikTok this year has pushed it more into mainstream culture. “You're seeing everyone from the teenager to the college students with their parents and then their grandparents all making videos,” said Chikumbu. “And now TikTok trulytranscendsgenerations, particularly in this year when people had a lot of time to dig in and not only watch, but try.”8. What do we know about The Weeknd?A. He created the famous song Dreams.B. He started a career based on shared videos.C. He is probably a popular singer on TikTok.D. He organized the TikTok dance challenges.9. What did Curtis Roach think of his life during quarantine?A. Meaningful.B. Peaceful.C. Content.D. Uninteresting.10. The most popular TikTok video in 2020 was created by ______.A. Caitlin ReillyB. Curtis RoachC. Nathan ApodacaD. Bella Poarch11. What does the underlined word “transcends” in the last paragraph mean?A. Makes use of.B. Goes beyond.C. Calls on.D. Takes charge of.DAbout a billion birds die from flying into buildings each year inNorth America. Suspicions havebeen that birds may regard the open areas behind glass as safe passageways. Or they may mistake the reflected trees for the realthing.Researchers would like to reduce collisions, which requires a solid understanding about what makes a bird more or less likely to die by crashing into a building in the first place.“There was ly little known at a broad scale. Previous studies were at one small study site.'' Jared Elmore, a graduate student in natural resource ecology and management atOklahomaStateUniversity. So he and his colleagues used a previously created data set of building collisions for birds at 40 sites throughoutMexico,Canadaand theU.S.The first finding was obvious: bigger buildings with more glass kill more birds. But the details were more remarkable. "We found that life history predicted collisions. Migrants(候鸟), insect-eaters and woodland-inhabiting species collided more than their counterparts(同类).”Most migratory species travel at night, when lights near buildings can distract or disorient(使迷失方向)them. And Elmore thinks that insect-eating birds might be attracted to buildings because their insect prey(猎物)is attracted to the lights. He suspects that woodland species get tooled by the reflections of trees and bushes in the windows. The results are in the journal Conservation Biology.By understanding which birds are more likely to collide with buildings, researchers can perhaps determine the best way to adapt buildings, or their lighting, to help prevent such accidents. And by knowing risks, along with migration timing and behavior, building managers can better predict when birds are at their greatest danger - and improve lighting strategics accordingly.Elmore's next project will use radar to help predict bird migrations. " I think that would maybe go a long way in terms of providing information to people, to the public, to building managers, on when they can get the most benefit in terms of lights-out policies."12. What is the possible reason for birds' crashing into buildings?A. They didn't see the buildings.B. They took reflections for reality.C. They assumed the windows to be open.D. They considered buildings as safe routes.13. What is Jared Elmore's study different from the previous ones?A. It created a new data set.B. It went beyond national borders.C. It covered a wider range of sites.D. I’ll studied some specific bird species.14. What was the most noticeable finding of Jared Elmore's study?A. Migratory species travel at night.B. Birds tend to be misled by glasses.C. Bigger buildings cause more collisions.D. Birds living habits give rise to collisions.15. Which of the following can help reduce bird collision?A. Adjust the lightening system.B. Attach radars to each building.C. Adopt strict lights-out policies.D. Ban using glasses on buildings.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
南通市如东县高级中学2020届高三数学上学期10月月考试题(含解析)
江苏省南通市如东县高级中学2020届高三数学上学期10月月考试题(含解析）一、填空题：本大题共14小题，每小题5分，共70分．不需写出解答过程，请把答案直接填写在答题卡相应位置上．1。
设集合{1,2,3,4}=-≤≤∈，则P Q=_______。
Q x x x RP=,{|22,}【答案】{}1,2【解析】【分析】利用集合的交集运算求解即可【详解】{}==-≤≤∴=P Q x x P Q2},21,{1,2,3,4},{|2故答案为：{}1,2【点睛】本题考查集合的交集运算,属于基础题.2.函数()2f x x x的定义域是_______.ln(3)=-【答案】(0,3)【解析】【分析】根据复合型对数函数的定义域进行求解【详解】()22=-∴->，(0,3)f x x x x xln(3),30∴∈x故答案为:(0,3)【点睛】本题考查复合函数的定义域，是基础题.3。
命题“20,0∀><都有”的否定是________.x x【答案】2有∃<≤0,0x x【解析】【分析】将全称量词改存在量词，再否定结论即可，“都有”改为“有”，大于号改成小于等于号【详解】全称量词改存在，再否定结论,即“20,0x x∀><都有”的否定是：20,0x x ∃<≤有故答案为:20,0x x∃<≤有 【点睛】本题考查全称命题的否定，全称改存在,再否定结论 4.已知1tan()42πα+=,且02πα-<<，则sin α=________。
【答案】 【解析】【分析】先由正切的和角公式求出tan α，再根据02πα-<<,利用同角三角函数基本关系求出sin α 【详解】由tan tan 1tan 114tan()tan 41tan 231tan tan 4παπαααπαα+++===⇒=---⋅，又因02πα-<<，根据同角三角函数的基本关系,可求得sin 10α=-故答案为： 【点睛】本题考查正切三角函数和角公式的求法，同角三角函数的基本求法，解题关键是正确掌握四象限对应三角函数的正负值5。
2020-2021学年江苏沭阳如东高级中学高三生物月考试题及答案
2020-2021学年江苏沭阳如东高级中学高三生物月考试题及答案一、选择题：本题共15小题，每小题2分，共30分。
每小题只有一个选项符合题目要求。
1. 析袋通常是由半透膜制成的袋状容器（蔗糖分子不能透过）。
现将质量浓度为0．3g/mL的蔗糖溶液装入透析袋，再放于清水中，实验装置如图所示。
30min后，会发现（）A.试管中的液体浓度减小B.透析袋中的液体浓度增大C.透析袋胀大D.试管液面上升2. 细胞自噬是细胞通过溶酶体与包裹细胞自身物质的膜融合，从而降解细胞自身衰老损伤的物质或结构的过程（如图所示）。
下列有关叙述中，正确的是（）A.图中自噬体的膜由两层磷脂分子组成B.图中的水解酶是在自噬溶酶体中合成的C.图中溶酶体与自噬体融合过程体现了生物膜的选择透过性D.溶酶体分解后的产物还可能被细胞再度利用3. 下图为光合作用探究历程中的相关实验，其中叙述正确的是（）A.恩格尔曼完成的是实验一，该实验设计的对照是放置在黑暗中的水绵装片和与暴露在光下的水绵装片B.萨克斯完成的是实验二，实验前植物必须进行饥饿处理以排除叶片原有淀粉的干扰C.实验三所用的方法叫同位素标记法，A、B为释放的氧气，分子质量比为9﹕8D.小球藻与实验一、二中的植物在结构上最大的区别是没有以核膜为界限的细胞核4. 下列各种病症中，不是由于人体内环境成分发生明显变化而引起的是A.组织水肿B.手足抽搐C.贫血D.尿毒症5. 两个氨基酸缩合成二肽产生一个水分子，这个水分子中的氢来自（）A.氨基B.R基C.氨基和羧基D.羧基6. 人体发育的起点是（）A. 胎儿B. 生殖细胞C. 婴儿D. 受精卵7. 在一个细胞周期中，最可能发生在同一时期的是（）A.有关蛋白质的合成和核糖体的增生B.DNA的复制和染色单体的分离C.细胞核的出现和纺锤体的出现D.染色体数目的加倍和细胞板出现8. 下列关于动物细胞生物膜上膜蛋白的叙述，错误的是（）A.不同细胞膜上的蛋白质不同，是细胞分化的结果B.细胞呼吸过程中，催化丙酮酸分解的酶分布在膜上C.激素与细胞膜上的受体蛋白结合，可改变细胞代谢D.细胞癌变使膜上的糖蛋白减少，细胞间黏着性降低9. 草履虫、衣藻、变形虫和细菌都是单细胞生物。
2020-2021学年江苏沭阳如东高级中学高三英语第一次联考试卷及答案解析
2020-2021学年江苏沭阳如东高级中学高三英语第一次联考试卷及答案解析第一部分阅读（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A、B、C、D四个选项中选出最佳选项AFour Best Hikes in the WorldThere's nothing like getting out and getting some fresh air on a hike. No matter whether your idea of a hike is a leisure walk or climbing the highest mountain on Earth, we've got you covered. Below are four best hikes inthe world.Torres del Paine W CircuitLocation (位置): Patagonia. ChileDistance: 37 + milesTime: 5~6 daysBest time to go: October to JanuaryThe W Circuit is one of the most recommended hikes you'll find. Not only will you appreciate the diverse landscapes and striking granite pillars (花岗岩柱子), but you'll probably meet some new friends along the way.Grand Canyon Rim - to - Rim HikeLocation: Arizona, the United StatesDistance: 48 milesTime: 1~3 daysBest time to go: May to June, September to OctoberThere's no better way to experience one of the greatest wonders in the world. Located in one of the USA's most beautiful parks, the views are ly appealing. Just make sure you're prepared for the challenge.Trek to PetraLocation: JordanDistance: 47 milesTime: 5~ 6 daysBest time to go: October to AprilTake the road less traveled through the Kingdom of Jordan and experience one of the seven wonders of theworld. Hike through canyons, gorges and ridges, and see tombs and temples along the way all while avoiding crowds of tourists.Yosemite Grand TraverseLocation: California, the United StatesDistance: 60 milesTime: 6~7 daysBest time to go: July to SeptemberKnown for some of the best hiking in the world, Yosemite National Park is famous for its views and huge sequoia (红杉) trees. Praised byNational Geographic, the Yosemite Grand Traverse will take you through waterfalls and green mountaintops.1.Which of the following is the best time for the hike in Patagonia, Chile?A.AprilB.MayC.AugustD.December2.Where should you go for a less crowded hike?A.JordanB.Patagonia, ChileC.Arizona, the United StatesD.California, the United States3.What can you do along the Yosemite Grand Traverse?A.Plant sequoia treesB.Appreciate waterfallsC.Visit local templesD.Climb granite pillarsBI was in the second year of my Ph. D. program when a colleague asked what I would do if I had an extra hour every day. Without much consideration, I said I would use it to help others. The question kept coming to my mind. Like many graduate students, I was overwhelmed (难以承受的) with research, teaching, coursework, and some attempt at a personal life. Still, I asked myself, "Do I really need a 25th hour to help other people-or do I need to make better use of the 24 hours I have?"I needed something to help me return to my old self. After that conversation with my colleague, I googled (谷歌搜索) “volunteer opportunities near me”. A local organization that drives people to stores or appointments was looking for volunteers. Having grown up in a rural village where everyone knew oneanother and my grandparents were always close by, I thought serving senior citizens in my new hometown might be just what I needed.A short time later, I started to volunteer for an organization that provides transportation for senior citizens and people with disabilities. To my surprise, adding this activity to my busy life was just what I needed to calm theconfusion I was feeling as a first-generation international graduate student.I started to volunteer about 3 hours every weekend, the time I otherwise would have wasted oversleeping or scrolling (滚屏) through social media. Sharing stories with my riders was much more rewarding. What they told me about their lives helped me realize that in every corner of the world, humans are connected with the languageof emotions. And seeing how eager my riders were to spend time out and about inspired me to think about how to spend my time, which used to slip away. My previously overwhelming schedule began to feel manageable.I'm proud of who I have become, and I continue to reflect on how I'm using the most valuable thing in life: my time.4. Why did the author ask himself the question in Paragraph 1?A. He wanted to return to normal life.B. He was busy but wanted to help others.C. He couldn't bear too much school work.D. He couldn't answer his colleague's question.5. How did the author find the volunteer job?A. A colleague recommended it.B. A local organization offered it.C. He got it from his grandparents.D. He got it by surfing on the Internet.6. What made the author feel his volunteer job was worth doing?A. Communicating with his riders.B. Improving his language learning.C. Meeting his grandparents often.D. Realizing his previous dream.7. How did volunteering influence the author?A. It helped himbecome confident and efficient.B. He found a good way to live a free and quiet life.C. He realized he had wasted too much time pursuing his Ph.D.D. It inspired him to spare more time to accompany his grandparents.CThefirst thing we notice about new people are their faces. The next time we see these people, we remember them because we remember their faces. This seems like a simple process. However, scientists found that it is not such a simple process. The section of the brain that is responsible for face recognition seems to work differently for different people. Some people have great difficulty remembering and recognizing faces, while others almostnever forget a face.Normal babies are born with a natural ability to recognize faces. In fact, their face recognition abilities are much better than their parents. Babies are most highly skilled at face recognition at six months. But by nine months, they lose this skill. By nine months, a baby’s face-recognition skills are about the same as an adult’s.Unfortunately, some people are not born with this ability to recognize faces. The part of the brain that is responsible for face recognition doesn’t work for them. This condition is called face blindness. People with very severe face blindness cannot even recognize their own faces. In fact, people with this condition can sometimes be frightened when they look in the mirror. They don’t recognize their own face, so for a second they are startled when they see this unfamiliar face.Face blindness is not always severe. Scientists believe up to 10 percent of the population may be affected by face blindness to some degree, yet many people with mild face blindness might not even know they have it. They have no reason to know they are different from anyone else until someone points it out. This is similar to people with color1 blindness. Colorblind people can’t see the difference between certain color1 s such as red and green, until someone tells them that green and red are two different color1 s.There is no cure for face blindness. So for the time being, people with face blindness need to find simple techniques to compensate for their problem. They can try to recognize people by their hairstyle, their voice, or their glasses. Hopefully, in the future as scientists learn more about this condition, they may find a cure.8. What is the first paragraph mainly about?A. The way to improve one’s face-recognition skills.B. The simple process of the brain to recognize others’ faces.C. The fact that some people have face-recognition problems.D. The importance of face recognition in human communication.9. When do children have the best face-recognition skills?A. At birth.B. Half a year old.C. Nine months old.D. In adolescence.10. What does the underlined word “startled” in the 3rd paragraph probably mean?A. Depressed.B. Confused.C. Embarrassed.D. Surprised.11. What does the author think of the problem of face blindness?A. People need to take it seriously.B. Certain techniques can make up for it.C. It will be cured in the near future.D. It has the same effect with color1 blindness.DThe idea of low material desire, low consumption and refusing to work, marry and have children, concluded as a “lying down” lifestyle, recentlystruck a chord withmany young Chinese who are eager to take pause to breathe in this fast-paced and highly-competitive society.Many millennials (千禧一代) and generation Zs complained to the Global Times that burdens, including work stress, family disputes (纷争) and financial strains, have pushed them “against the wall”. They said they hate the “involution (内卷),” joking that they would rather give up some of what they have than get trapped in an endless competition against peers.“Instead of always following the ‘virtues’ of struggle, endure and sacrifice to bear the stresses, they prefer a temporary lying down as catharsis (宣泄) and adjustment,” said a scholar. “It is no wonder that some young people, under the growing pressures from child-raising to paying the mortgage (按揭) today, would try to live in a simple way and leave the worries behind.”Interestingly, the majority of millennials and Gen Zs reached by the Global Times, who claim to be big fans of the lying down philosophy, acknowledged that they only accept a temporary lying down as a short rest. It is true that with the great improvement of living conditions, some Chinese youth have partially lost the spirit of hardship and are not willing to bear too much hard work. But in fact, lying down is not entirely comfortable. Young people who lie down always feel guilty about their constant loss of morale (士气) far beyond their reach.“Young people on campus have both aspirations and confusion about their future, but most of us have rejected setting ourselves up in chains to waste opportunities and challenges,” a postgraduate student told the Global Times. “It’s no use running away. I have to ‘stand up’ and face the reality sooner or later.”12. What does the underlined phrase in paragraph 1 mean?A. Warned.B. Criticized.C. Touched.D. Amused.13. What might have caused the “lying down” lifestyle among the young?A. Improvements in living conditions.B. Growing pressure from family and social life.C. Increasing material possessions from families.D. Temporary adjustment to failure in competitions.14. What’s the scholar’s attitude toward the “lying down” group?A. Understanding.B. Intolerant.C.Supportive.D. Unclear.15. What can be inferred about the young generation from the text?A. They never really drop their responsibilities.B. They really enjoy the “lying down” lifestyle.C. They find their dreams far beyond their reach.D. They would rather escape than take challenges.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2020-2021学年江苏沭阳如东高级中学高三生物期中考试试卷及答案
2020-2021学年江苏沭阳如东高级中学高三生物期中考试试卷及答案一、选择题：本题共15小题，每小题2分，共30分。
每小题只有一个选项符合题目要求。
1. 在利用鸡血进行“DNA的粗提取与鉴定”的实验中，下列叙述正确的是A. 用蒸馏水将NaCl溶液浓度调至0.14 mol/L，滤去析出物B. 调节NaCl溶液浓度或加入木瓜蛋白酶，都可以去除部分杂质C. 将丝状物溶解在2 mol/LNaCl溶液中，加入二苯胺试剂即呈蓝色D. 由于DNA对高温耐受性较差，故需向DNA滤液中加入冷酒精2. 下列关于生物体内激素的叙述中，正确的是（）A. 都有单一的靶器官、靶细胞B. 能够对生物体内的化学反应起催化作用C. 都是含量微少但有高效的调节功能D. 都经过内质网和高尔基体的加工和运输3. 下图表示某生物性染色体结构示意图，相关说法不正确的是A．雌、雄个体的性染色体组成不同，有同型和异型两种形式，因而基因组成有差异B．若Y染色体上携带的基因在X染色体有相应的等位基因，则性状遗传与性别无关C．若Y染色体上携带的基因在X染色体上无相应的等位基因，则只限于相应性别的个体之间遗传D．正交结果和反交结果可作为判断遗传方式的依据4. 小麦抗病（T）对易感病（t）、高秆（D）对矮秆（d）为显性，两对基因独立遗传。
现有甲（高秆抗病）与乙（高秆易感病）两株小麦杂交，子代有4种表型。
如果让甲测交、乙自交，则它们后代的表型之比应分别为（）A.9∶3∶3∶1及1∶1∶1∶1B.3∶3∶1∶1及1∶1C.1∶1∶1∶1及3∶1D.9∶3∶3∶1及1∶15. 下列化学物质中，不是植物激素的是（）A. 乙烯B. 吲哚乙酸C. 脱落酸D. 2，4﹣D6. 如图是下丘脑及其直接或间接支配的有关腺体之间的关系示意图（+表示促进，-表示抑制），有关说法正确的是（）A.a产生的某种物质可影响动物的性行为B.b与c两者的分泌物在某些生理效应上可表现为协同作用C.c不仅具有感受刺激和传导兴奋的功能，而且有分泌功能D.a、b、c中，c对内环境稳态的维持影响最小7. 南瓜所结果实中白色（A）对黄色（a）为显性，盘状（B）对球状（b）为显性，两对基因独立遗传。
2020-2021学年江苏如东高级中学高三英语第一次联考试题及参考答案
2020-2021学年江苏如东高级中学高三英语第一次联考试题及参考答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项ANo one knows when the first printing press was invented or who invented it. but the oldest known printed text originated in China during the first millennium (千年) AD.The Diamond Sutra (《金刚经》), a Buddhist book from Dunhuang, China during the Tang Dynasty, is said to be the oldest known printed book.The Diamond Sutrawas created with a method known as block printing (雕版印刷), which used boards of hand-carved wood blocks in reverse.It was said that the moveable type was developed by Bi Sheng. He was fromYingshan,Hubei,China, living from 970 to 1051 AD. His method replaced panels of printing blocks with moveable individual Chinese characters that could be reused. The first moveable Chinese Characters were carved into clay and baked into hard blocks that were then arranged onto an iron frame that was pressed against an iron plate.The earliest mention of Bi Sheng’s printing press is in the bookDream Pool Essays, written in 1086 by Shen Kuo, who noted that his nephews came into possession of Bi Sheng’s typefaces (字体) after his death. Shen Kuo explained that Bi Sheng did not use wood because the texture is inconsistent (不一致的) and absorbs wetness too easily.By the time of the Southern Song Dynasty, which ruled from 1127 to 1279 AD, books had become popular in society and helped create a scholarly class of citizens who had the capabilities to become civil servants. Large printed book collections also became a status symbol for the wealthy class.1. When was Bi Sheng’s printing press first introduced in history?A. After Bi Sheng died and his nephews owned his typefaces.B. When books became popular in the Southern Song Dynasty.C. After the block printing was replaced by the moveable type printing.D. WhenThe Diamond Sutrawas printed into a book.2. What can we infer from the passage?A. Shen Kuo made great contributions to printing.B. The moveable type printing was invented earlier than block printing.C. Printed books were hard to get in the Song Dynasty.D. By the Southern Song Dynasty, books had helped people get to higher social positions.3. Why does the author write this passage?A. To show that Buddhism was popular in the Tang Dynasty.B. To introduce the early history of printing.C. To memorize Bi Sheng, developing the moveable type printing.D. To indicate the advantages of moveable type printing.BHumans have found an easy way to tell if others are lying. Recent research shows that the best way so far is being clever at how you ask questions and listen to the answers.Much of this research is based on the idea that telling a lie is simply harder mental work than telling the truth. Making up a story takes more effort than simply recording something that happened. And like a writer, a liar has to keep all the unreal details in his memory and sound believable when he explains them.One method that seems to work is asking them to tell their story in reverse order. This is harder when the story isn't true and makes it easier for you to tell they are lying. An even more basic way that helps is to just ask more questions, especially unexpected ones. Truth-tellers can easily find more to say, but it's a challenge for a liar to come up with something that's not in his prepared story.Researchers suggest that you shouldn't lay all your cards on the table at the start, but only gradually present what proof you have. The liars' stories may not agree with that proof, making it clear that they're lying.So it looks like there are ways to increase the chances of catching a liar; we've just been basing our methods on the wrong stories. Low-tech ways of causing people to make mistakes in conversation seem to work better than any science about eye movement or machines used to recognize a liar. To find a liar, watch less and listen more.4. Why does the author mention the writer?A. To show it's hard to make up lies.B. To show it's hard to recognise a liar.C. To show writers know liars best.D. To show writers are very clever.5. What do we know about liars?A. They often have much to say.B. They often ask many questions.C. They usually prepare a made-up story.D. They usually feel good about themselves.6. What advice is given to help people catch liars?A. Asking them to set their stories down.B. Presenting your proof one by one.C. Telling different stories to them.D. Letting them ask questions.7. Which can be the best title for the text?A. Why People LieB. How to Stop People LyingC. Low-tech Ways to Find a LiarD. LiarsAre Smarter than ThoughtC“Heavy hearts, like heavy clouds in the sky, are best relieved by the letting of a little water, the French writer Antoine de Rivarol wrote. This love letter to the cleansing beauty of a good cry is a comforting thought at atime when the continuing stress of the COVID-19 has added heaviness to each of our lives.Scientifically, de Rivarol's poetic image doesn't, if you'll forgive the words used in the poem, hold water. There's limited research on crying, partly because of the difficulty of copying the behavior of real crying in a lab. But even within the previous studies, there's little evidence to suggest that crying provides a physiological cleansing of poisons in people's body.Psychologists believe the relief of a good cry connects with a different emotional process. “It seems that crying occurs just after the peak of the emotional experience, and crying is associated with this return to homeostasis: the process of maintaining a stable psychological state,” said Lauren Bylsma. He also said holding back tears can have negative physical consequences, including headaches and muscle tension. Such restriction can also limit our experiences of joy, gratitude and other positive emotions if we avoid acknowledging our feelings.For me crying has been easier said than done during the COVID-19. Psychologists say it's normal to feel stopped up by the stresses of the past year. We should find opportunities to release and process our emotions.Watching a tear-jerking movie, having an emotional conversation with a close friend, and writing in a journal are healthy ways toelicita cry. Physical activity like light-footed walking or even dancing can also signal our bodies to release some emotional tightness. We can then open up to the flow of feelings that leave us feeling lighter and refreshed—like a clear sky after a soaking rain.8. What is the weakness of the studies ever clone on crying?A. They were clone in a laboratory setting.B. They cared little about different forms of crying.C. They were always concentrated on people's daily life.D. They showed little about the positive physical effect of crying.9. What is the function of crying according to Lauren Bylsma?A. Curing people of their diseases.B. Keeping emotionally balanced.C. Producing negative mental results.D. Expanding people's experience of joy.10. What does the underlined word “elicit” in the last paragraph mean?A. Produce.B. Postpone.C. Control.D. Repeat.11. What are people advised to do according to the text?A. Learn to hold back their tears wisely.B. Share their emotion with their colleagues.C. Have a good cry when necessary.D. Try to avoid admitting our feelings.DWhile the start of a new school year is always exciting, this year was even more so for some elementary school students inAuckland,New Zealand. They became the world’s first kids to be “taught” by a digital teacher.Before you start imagining a human-like robot walking around the classroom, Will is just an avatar that appears on the student’s desktop, or smartphone screen, when ordered to come.The autonomous animation platform has been modeled after the human brain and nervous system, allowing it to show human-like behavior. The digital teacher is assigned to teach Vector’s “Be sustainable with energy”— a free program forAucklandelementary schools.Just like the humans it replaced, Will is able to instantly react to the students’ responses to the topic. Thanks to a webcam and microphone, the avatar not only responds to questions the kids may have, but also picks up non-verbal cues. For instance, if a student smiles at Will, he responds by smiling back. This two-way interaction not only helps attract the students’ attention, but also allows the program’s developers to monitor their engagement, and make changes if needed.Nikhil Ravishankar believes that Will-like avatars could be a novel way to catch the attention of the next generation. He says, “I have a lot of hope in this technology as a means to deliver cost-effective, rich, educational experience in the future.”The program, in place since August 2018, has been a great success thus far. Ravishankar says, “ What was fascinating to me was the reaction of the children to Will. The way they look at the world is so creative and different, and Will really captured their attention.” However, regardless of how popular it becomes, Will is unlikely to replace human educators any time soon.12. What was special for some elementary school students inAuckland?A. A digital teacher taught them.B. They first saw something digital.C. This was the start of a new school year.D. They could get close to smartphone screen.13. What is the benefit of this two-way interaction?A. It can smile back.B. It can use microphone.C. It can talk any topic for free.D. It can change if necessary.14. What’s Ravishankar’s attitude to Will’s replacing Human educators soon?A. Optimistic.B. Doubtful.C. Unclear.D. Disapproving.15. What might be the best title for the passage?A. New High-tech Contributes to EducationB. The World’s First Digital Teacher Appears in Classroom.C. The World’s First Digital Teacher, a Help to StudentsD.New ZealandWill Replace Teachers in Classrooms第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
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B．肾小没有选择性，比通道直径小的物质可自由通过
D．机体可通过调节细胞膜上通道蛋白的数量或开关来调节物质的运输
2．植物受病原菌感染后，特异的蛋白水解酶被激活，从而诱导植物细胞编程性死亡，同时病原菌被消灭。激活蛋白水解酶有两条途径：①由钙离子进入细胞后启动；②由位于线粒体内膜上参与细胞呼吸的细胞色素c含量增加启动。下列叙述正确的是
A．核糖体的主要成分是蛋白质和mRNA
B．四环素与A位点结合促进了新肽键的形成
C．图中从E位点离开的tRNA可转运甲硫氨酸
D．人体细胞与细菌细胞的核糖体没有差异
7．果蝇的两条X染色体可以连在一起，成为一条并联的X染色体，带有一条并联X染色体和一条Y染色体的果蝇（品系C）表现为雌性可育，带有一条并联X染色体和一条正常X染色体的果蝇是致死的，没有X染色体的果蝇也是致死的。用化学诱变剂诱导正常的野生型雄果蝇，然后让该雄果蝇和品系C进行杂交，以下对杂交结果分析正确的是（）
A．研究人员用γ射线处理二倍体黄毛草莓选育大果型黄毛草莓属于基因工程育种
B．二倍体黄毛草莓和八倍体红颜草莓杂交得到的F1植株是单倍体，高度不育
C．一定浓度的秋水仙素处理二倍体草莓幼苗可抑制其细胞分裂时中心体形成纺锤体
D．四倍体黄毛草莓比二倍体黄毛草莓茎秆粗壮，叶片、果实大，营养物质含量增加
9．心脏的搏动受交感神经和副交感神经的控制，其中副交感神经释放乙酰胆碱，作用于心肌细胞膜上的M型受体，使心肌细胞的收缩受到抑制，心率减慢；交感神经释放的去甲肾上腺素可以和心肌细胞膜上的β-肾上腺素受体结合，使心率加快。但交感神经和副交感神经对心脏的作用强度不是等同的，利用心得安和阿托品进行如下实验（心得安是β-肾上腺素受体的阻断剂，阿托品是M型受体的阻断剂）。对两组健康青年分别注射等量的阿托品和心得安各4次，给药次序和测得的平均心率如图所示。有关叙述正确的是（）
A．离心后大肠杆菌主要分布在沉淀物中
B．沉淀物的放射性来自T2噬菌体的DNA
C．上清液具有放射性的原因是保温时间过长
D．本结果尚不能说明T2噬菌体的遗传物质是DNA
6．图表示核糖体上合成蛋白质的过程。四环素抑制tRNA与细菌核糖体的A位点结合，使蛋白质合成停止，从而阻断细菌的生长，常用于治疗一些细菌引起的疾病。下列有关叙述正确的是
B．甲蛋白可能和碳（暗）反应有关，乙蛋白可能和光反应有关
C．类囊体蛋白质由细胞质和叶绿体中的核糖体合成
D．运至叶绿体不同部位的甲、乙蛋白都需经过加工
4．细菌紫膜质是一种膜蛋白，ATP合成酶能将H+势能转化为ATP中的化学能。科学家分别将细菌紫膜质和ATP合成酶重组到脂质体（一种由磷脂双分子层组成的人工膜）上，在光照条件下，观察到如下图所示的结果。下列叙述错误的是
A．蛋白水解酶能使磷酸二酯键断开
B．钙离子通过自由扩散进入植物细胞
C．细胞色素c与有氧呼吸第一阶段有关
D．细胞编程性死亡避免了病原菌对邻近细胞的进一步感染
3．叶绿体内绝大多数蛋白质由核基因编码，少数由叶绿体基因编码，其合成、加工与转运过程如图所示。下列说法错误的是
A．甲、乙蛋白通过类似胞吞过程从细胞质进入叶绿体
A．子代雌果蝇都带有并联X染色体，子代雄果蝇X染色体来自亲代雄果蝇
B．若子代雄果蝇有某突变性状而雌果蝇没有，则该突变基因位于Y染色体上
C．若子代雌果蝇有某突变性状而雄果蝇没有，则该突变基因位于常染色体上
D．若子代果蝇都出现了某突变性状，则该突变基因最有可能位于X染色体上
8．草莓是蔷薇科草莓属的一种多年生草本植物，主要有二倍体（2n=14）、四倍体（4n=28）、八倍体（8n=56）等类型。下列相关叙述正确的有（）
A．图中由下丘脑到腺垂体，再到对应的靶腺其调节方式称为反馈调节
B．内．外环境变化如果为寒冷刺激，则图中激素增加的有A、B、C
C．女性月经周期中，雌激素分泌增加一定会使促性腺激素释放激素分泌增加
D．图中属于长反馈的序号是①和②，属于短反馈的序号是③和④
11．某班欲调查学校四个校区乔木的种类，将同学分为A．B、C、D4个小组，分别调查东、南、西、北四个校区的乔木种类。发现东区有17种，南区有14种，西区有13种，北区有16种。问：该校四个校区的乔木有多少种?（）
A．15种B．17种<该校的乔木<60种
C．60种D．17种≤该校的乔木≤60种
12．2021年7月20日，国际医学权威期刊《柳叶刀》上发表了新冠病毒疫苗II期临床试验的成果。该试验是由陈薇院士等人领衔的团队开展的研究，试验中使用的疫苗是一种腺病毒载体重组新冠病毒疫苗(基因重组疫苗)，试验采用了随机、双盲、安慰剂对照设计，508名受试者随机分配至疫苗组和安慰剂组，在更大人群中验证了疫苗的免疫原性和安全性。试验结果表明，单次接种疫苗28天后，99．5%的受试者产生了特异性抗体，95．3%受试者产生了中和抗体，89%的受试者产生了特异性T细胞免疫反应。这表明，陈薇团队研发的新冠疫苗可为健康人群提供“三重保护”，将新冠病毒“拒之门外”。下列叙述正确的是（）
A．乙酰胆碱与M型受体结合，使得心肌细胞的静息电位绝对值减小
B．注射阿托品后交感神经的作用减弱，副交感神经作用加强
C．每一组的每位健康青年共进行了8次心率的测定
D．副交感神经对心跳的抑制作用远超过交感神经对心跳的促进作用
10．下丘脑的神经内分泌细胞在神经系统与内分泌系统功能活动的调节中起着重要的桥梁作用。如图表示下丘脑-腺垂体-靶腺之间的联系，通常将靶腺分泌的激素对下丘脑和腺垂体的负反馈作用称为长反馈，腺垂体分泌的促激素对下丘脑的负反馈作用称为短反馈，下丘脑分泌的某些释放肽对下丘脑内部的负反馈称为超短反馈。下列相关叙述正确的是（）
A．甲图中H+跨膜运输的方式是主动运输
B．ATP合成酶不能将光能直接转化为ATP中的化学能
C．ATP合成酶既具有催化作用也具有运输作用
D．破坏跨膜H+浓度梯度对ATP的合成无影响
5．用32P标记的T2噬菌体侵染未被标记的大肠杆菌，侵染一段时间后搅拌、离心得到上清液和沉淀物，检测上清液中放射性32P约占初始标记T2噬菌体放射性的30%。在实验时间内，被侵染细菌的存活率接近100%。下列相关叙述不正确的是（）
江苏省南通市如东高级中学、沭阳如东中学2020-2021学年高三10月联考生物试题
学校:___________姓名：___________班级：___________考号：___________
一、单选题
1．诺贝尔化学奖曾经授予研究细胞膜通道蛋白的科学家。通道蛋白是一类跨越细胞膜磷脂双分子层的蛋白质，它包含离子通道蛋白和水通道蛋白。下列相关叙述正确的是（）