湖南四大名校内部资料试卷-2018-2019-1师大附中高一期末考
湖南师大附中2018-2019学年高一下学期期末考试数学含答案
湖南师大附中2018-2019学年度高一第二学期期末考试命题:柳 叶 审题:谭泽仁时量:120分钟 满分:150分得分: ____________第I 卷(满分100分)一、选择题:本大题共11个小题,每小题5分,共55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. 若a , b , c 是平面内任意三个向量,入€ R ,下列关系式中,不一定成立的是A . a + b = b + aB .入(a + b )=扫+AbC . (a + b ) + c = a + (b + c )D . b =扫2. 下列命题正确的是A .若a 、b 都是单位向量,贝U a = bB .若AB = DC ,则A 、B 、C 、D 四点构成平行四边形C. 若两向量a 、b 相等,则它们是起点、终点都相同的向量D. AB 与BA 是两平行向量3. cos 12° cos 18°— sin 12° sin 18°的值等于3 1 1 3A .~2B .2C . — 2D •— ~2" 4. 函数f(x) = tan X 2的最小正周期为 1 + tan x n nA.~B.2 C . n D . 2 n5. 设a , b 是非零向量,则下列不等式中不恒成立的是A . |a + b |w |a 汁 |b |B . |a |— |b |< |a + b |C . |a |— |b |< |a |+ |b |D . |a |< |a + b |6. 函数 f(x) = Asi n( 3X +$)A ,3, $为常数,A > 0, 3> 0, W |< 专 I示,贝 U f( n )=C. 7. 如图,角a B 均以Ox 为始边,终边与单位圆O 分别交于点A 、B ,贝U OA OB =1 zS B.(n y(n \ 19. 已知 a€ 0, "2,cos "6 + a = 3,则 Sin a 的值等于 2 [2— \ 3 2 j'2 + ... 3 2 .:6 — 12 ,'6 — 1 A. 6 B. 6 C.6D . — 6 10. 将函数y = 3sin 2x +才的图象向右平移 寺个单位长度,所得图象对应的函数 A .在区间12,7i2上单调递减B .在区间n 号上单调递增11 .设O 是平面上一定点,A 、B 、C 是该平面上不共线的三点 ,动点P 满足OP = OA + AB AC入 ------- + ------ ,入€ [0,+^),则点P 的轨迹必经过△ ABC 的(AB | c os B |A C I C OS C 丿A .夕卜心B .内心C .重心D .垂心答题卡题号1 2 3 4 5 6 7 8 9 10 11 得分答案、填空题:本大题共3个小题,每小题5分,共15分.n12. __ 已知直线x =匸是函数f(x) = sin(2x +柏的图象上的一条对称轴,则实数$的最小正值 为 __________ .13. 已知 sin a + cos 3 = 1, cos a + sin 3 = 0,贝U sin( a+ 3= ____________ .A . sin( a — 3B . sin(a+ 3C . cos(a — 3D . cos(a+ 3nn 已知 ~4v a< 2, 且 sin a •s a 3 小 =10,贝y sina — cos a 的值是 C. 在区间D. 在区间上上 n 3n3 冗一 6 冗-6 单调递减 单调递增14. 已知AB丄AC,|AB||AC|= 1.点P为线段BC上一点,满足AP = 巫 +」A^.若点I I|AC||A B|4|AC|Q ABC 外接圆上一点,则AQ AP 的最大值等于三、解答题:本大题共3个小题,共30分.15. (本小题满分8分) (1)求tan a 的值;⑵求tan 2a + 4的值.已知 5sin a — COS a COS a + Sin a1.16. (本小题满分10分)已知向量 a = ( 2sin a , 1), b= 1, sin a + —.(1)若角a的终边过点(3, 4),求a b的值;⑵若a// b,求锐角a的大小.17. (本小题满分12分)已知函数f(x) = sin =■— x sin x —^3cos2x.(1)求f(x)的最小正周期和最大值;⑵讨论f(x)在n,务上的单调性.第n卷(满分50分)一、填空题:本大题共2个小题,每小题6分.18. 两等差数列{a n}和{b n},其前n项和分别为S n、T n,且Sn= 7n+j ,则严20等于T nn+ 3 b7+ b i5(x+ 1) + sin x19. ------------------------------ 设函数f(x) = _________________________________________----------------------------------------- 的最大值为M ,最小值为m,贝V M + m = __________________ .x十1二、解答题:本大题共3个小题,共38分,解答应写出文字说明,证明过程或演算步骤.20. (本小题满分12分)如图,在四棱锥P—ABCD 中,PA丄底面ABCD , AD 丄AB , AB // DC , AD = DC = AP = 2, AB = 1,点E为棱PC的中点.(1) 证明:BE丄DC ;(2) 求直线BE与平面PBD所成角的正弦值.21. (本小题满分13分)在四边形ABCD中,AD // BC ,(1)求AD的长;⑵若/ BCD = 105° ,求四边形7122.(本小题满分13分) 已知函数 f(x) = x|x — a|+ bx(a , b € R ). (1)当b =— 1时,函数f(x)恰有两个不同的零点,求实数a 的值; ⑵当b = 1时, ① 若对于任意x € [1 , 3],恒有f (x x )三2 x + 1,求a 的取值范围; ② 若a >0,求函数f(x)在区间[0, 2]上的最大值g(a). 湖南师大附中2018-2019学年度高一第二学期期末考试 数学参考答案 、选择题1. D 【解析】 选项A ,根据向量的交换律可知正确;选项 B ,向量具有数乘的分配律, 可知正确;选项C ,根据向量的结合律可知正确; 选项D , a , b 不一定共线,故D 不正确.故 选D.2. D 【解析】A.单位向量长度相等,但方向不一定相同,故A 不对;B.A 、B 、C 、D 四点可能共线,故B 不对;C.只要方向相同且长度相等,则这两个向量就相等,与始点、终 点无关,故C 不对; D.因AB 和BA 方向相反,是平行向量,故D 对.故选D. 【解析】 cos 12° cos 18°— sin 12 ° sin 18 ° = cos (12°+ 18° )= cos 30° 故选A. 【解析】 tan x sin xcos x 1 2 n 函数f(x)=齐岳=cos 2x +sin 2x = 1sin 2x 的最小正周期为 选C. 由向量模的不等关系可得: |a |一 |b |w |a + b |w |a |+ |b |. 5. |a + b |< |a + |b |,故 A 恒成立. |a |—|b |w |a + b |,故 B 恒成立. |a |—|b |w |a + b |w |a |+ |b |,故 C 恒成立. 令 a = (2, 0), b = (— 2, 0),则 |a |= 2, |a + b |= 0,贝U D 不成立.故选 D. 6. B 【解析】根据函数的图象 A = ,2. 【解析】 由图象得:T = 4 所以 3= 2|7= 2. 号+A k n ,—号 + k n .k €z . 由于|『2,取k = 1,解得:片n ,所以f(x) = -2sin : 2x + 3 .则:f( n )=扌,故选B.7. C 【解析】根据题意,角a, B 均以Ox 为始边,终边与单位圆O 分别交于点A , B , 则 A(cos a ,sin a ), B(cos 3 , sin 3 ),故选C. 则有OAOB = cos a cos 3+ sina sin 3 = cos ( a —3);8. B=(sin 2 【解析】T (sin a — cos 2a+ cos a ) — 2sin a COSa )2= sin 2 a — 2sin a cos a+ cos 2aa ;又■/ sin2i 2 “ ・a + cos a= 1 , sin a cos a 10'/• (si n—cos得sin—cos a=±10; 由才V n aV"2, 2 知2<sin a <1, 0<cos故有 sin a — cos a> 0,则sin=sin —+ \6a cos 丁 cos n + ;sin n =寧半—2 x 1=吟故选 C ./ 6\6 y 6 32 3 2 6 •故选7t10 . Bn图象向右平移勺个单位长度后得到y =7t3sin 2 lx — y = 3sin 2x2 n 〜 n 2 n n才的图象,令一—+ 2k n W 2x — 丁+ 2k n , k € Z , 化简可得n x € —+ k nJ2 ,k € Z ,即函数 y = 3sin 2x —号 的单调递增区间为n五+k n故选B.7 n , 12 + k nk € Z ,令i 2 nk = 0,可得 y = 3sin 2x —三区间n 7 nJ2, 12单调递增,11. 【解析】由题意可得 OP — OA = AP =AC入+ T<|ABI COS B|AC I C OS CAB BC + (AB| cos B |AC|cos C 经过△ ABC的垂心,故选D.二、填空题12. n 【解析】(略)113. —2 【解析】sin a + cos 3= 1,两边平方可得:sin 2a + 2sin a cos 3 + cos 23 = 1,①,cos a + sin 3= 0,两边平方可得:cos a + 2cos a sin 3+ sin 3 = 0,②,由① + ②得:2 + 2(sin a cos 3 + cos a sin 3 ) = 1,即 2 + 2sin( a+ 3) = 1, 2sin( a+ 3) = —1.••/小 1--si n( a+ 3 = —217 打r f f f f 114.y 【解析】T AB丄AC , |AB||AC|= 1 ,建立如图所示坐标系,设B \, 0 , C(0, t),f 1 f f AB AC 丄1 1 1 1AB = 7,0, AC = (0, t), AP= i + ^ = t[ , 0+ 4t(0 , t) = (1, 4, • P••• P为线段BC上一点,•可设PC = XPB ,从而有—1 , t—:=入勺BC中点,•点P ABC外接圆圆心.Q在厶ABC外接圆上,又当AQ过点P时|A Q|有最大值为2|A P|=—夢,此时A P与A Q夹角为9= 0° ° cos 9 = 1.^ (A P A Q ) =^T7^^T^=17max 2 4 8、解答题即5tan a — 1 = 1 + tan a ,解得tan. 心2tan a 4AC BC所以AP BC =入=〈—|BC|+ |BC|)= 0,所以A P丄B C,即点P在BC边的高所在直线上,即点P的轨迹|AB| 4|AC|1 1解之得t=2.t—4 =— 4 人15.【解析】(1)由题意,cos a^ 0 ,5sin a —cos acos a + sin a可得5tan a —11 + tan a a= ;(4 分)• B(2, o) , C 0 , 2 •显然P(2)由(1)得tan 2 a= 厂=1—tan a16.【解析】(1)角a的终边过点(3 , 4) , • r= ,32+ 42= 5 ,• sin a = — 4, cos a =兰=|;r 5r 5• a • b = 2sin a + sin a + —•—n2sin a + sin a cos^ + cos a =2 X 5+ 4 冷 + 5 撐=孚.(5 分)⑵若 a // b ,贝2sin a sin=cos xsin x -于(1 + cos 2x)2x + sin x =++1 ,易知函数g(x)为奇函数关于原点对称,•函数f(x)图象关于点(0, 1)对称•若x x ~+ =x 0时,函数f(x)取得最大值 M ,则由对称性可知,当x =— x 0时,函数f(x)取得最小值m ,因此,M + m = f(x O ) + f( — x o )= 2.20.【解析】(1)如图,取PD 中点M ,连接EM 、AM.由于E 、M 分别为PC 、PD 的中点,1故EM // DC ,且EM = 2DC ,又由已知,可得EM // AB 且EM = AB ,故四边形ABEM 为平行 四边形,所以BE // AM..n sin 4即 2sin asin a cos -^ + cos• sin a+ sin a cos a= 1 ,2--Sin a cos a= 1 — Sin a = 对锐角a 有 cos a^ 0,na sin = 1,4丿2COS a ,二 tan •••锐na= 7.(10 分)cos 2x1 3 =^sin 2x — —cos 2x — ?因此f(x)的最小正周期为n 亞=sin : 亚=sin2x —3 — 2, 2—、3八,最大值为⑵当x €2n 牛时,O W 2x —亍W n n n n 5 n,从而当O W 2x —三W —,即$ W x w 〒2时,f(x)n 单调递增;■— W 2x — — W n 237t即鳥冗W x < 12 3时,f(x)单调递减.n综上可知,f(x)在〒, 5 n 12单调递增;在5 n 2 n -12, 3 单调递减.(12分)14918药 19. 2【解析】生±他=型±鱼=S 21 =他.b 7+ b 15 b 1+ b 21 T 21242 .x + 1 + 2x + sin x 2x + sin x ,、人【解析】可以将函数式整理为 f(x) = - 2 = 1 + x ++1 ,不妨令g(x)x 2 + 1 17.【解析】 ⑴f(x) = sin因为PA丄底面ABCD ,故PA丄CD ,而CD丄DA ,从而CD丄平面PAD,因为AM祚平面PAD,于是CD丄AM ,又BE // AM ,所以BE丄CD.(5分)(2)连接BM ,由⑴有CD丄平面PAD,得CD丄PD,而EM // CD,故PD丄EM ,又因为AD = AP, M为PD的中点,故PD丄AM , 可得PD 丄BE,所以PD丄平面BEM ,故平面BEM丄平面PBD.所以直线BE在平面PBD内的射影为直线BM,而BE丄EM,可得/ EBM为锐角,故/ EBM为直线BE与平面PBD所成的角.依题意,有PD = 2 2,而M为PD中点,可得AM = 2,进而BE =-, 2•故在直角三角形BEM 中,tan/ EBM =器=ABE ,因此sin / EBM =申.所以直线BE与平面PBD所成角的正弦值为 f.(13分)21.【解析】⑴•••在四边形ABCD中,AD // BC, AB = .3, /A = 120° , BD = 3.•由余弦定理得cos 120 ° =2 c3 + AD —9 -2 x :3x AD ,解得AD = ,3(舍去AD = - 2 3),••• AD的长为3.(5分)(2) •/ AB = AD = .3, / A = 120° , ADB = £(180 ° - 120 ° ) = 30° ,又AD // BC , / DBC = / ADB = 30° .•// BCD = 105 ° , / DBC = 30 ° , BDC = 180°- 105从而S A BDC = ^BC-BDsin / DBC = (3 3 —3)x 3x sin 30°= 4( . 3—1). (10 分)S A ABD = 2AB x ADsin A = ^x ^x 羽x sin 120 ° =泅.(11 分)22.【解析】(1)当b =—1 时,f(x) = x|x —a|— x = x(|x —a|—1), 由f(x) = 0,解得x= 0 或|x —a|= 1,由|x —a|= 1,解得x = a+ 1 或x = a— 1.•/f(x)恰有两个不同的零点且a+1丰a—1,•- a + 1 = 0 或a—1 = 0,得a = ±1.(4 分)(2)当b= 1 时,f(x) = x|x —a|+ x,①•••对于任意x € [1 , 3],恒有30°= 45° , △ BCD 中,由正弦定理得BCsin 453sin 105,解得BC = 3,3 — 3.(9 分)…S = S A ABD+ S A BDC = .(13 分)xy = f(x)在 0, a + 1 2 「单调递增,在 a + 1 2 , 单调递减,在[a , 2]上单调递增,(a + 1) 24, f(2) = 6 — 2a ,(a + 1) 2(a + 5) 2— 48 —(6 — 2a)= ----- : ------ ,• g(a) = max f宁-f(2)=4当 1 v a v 4 3 — 5 时,g(a)= f(2) = 6— 2a ;a + C (a + 1) 2 T = 4;当 4 .3— 5W a v 2 时,g(a)= fa — 1 a + 1当 2w a v 3 时,-^<—^<2 w a ,这时 y = f(x)在 ||0,(a +1) 2 =;「单调递增,在 当a > 3时,专 > 2, y = f(x)在[0 , 2]上单调递增, 此时 g(a)= f(2) = 2a — 2.单调递减, 即x|x - a|+ X w 2 x + 1,即 |x — a|< 2 x + 1 — 1,•/x € [1 , 3]时,2 x + 1 —1>0 , ••• 1 — 2 x + 1 <x — a <2 x + 1 — 1, a w x + 2\/x + 1 — 1,即x € [1 , 3]时恒有丿 ___ 成立.^a > x — 2p x + 1 + 1,令 t = x + 1,当 x € [1 , 3]时,t € [2, 2], x = t 2— 1. ••• x + 2 x + 1 — 1 = t 2+ 2t — 2 = (t + 1)2— 3> ( 2 + 1)2— 3= 2 2,• x — 2 x + 1 + 1 = t 2— 2t = (t — 1)2— 1w 0,当0 v a w 1时,号w 0,旦+」》a , 这时y = f(x)在[0, 2]上单调递增, 此时 g(a)= f(2) = 6— 2a ; 当 1 v a v 2 时,O v <a+^ v a v 2,2 2综上,a 的取值范围是[0 , 2 2]. (8分)(a + 1 F (a +1) -x — -2-' ② f(x)=2—x + ax + x , x w a x 2 — ax + x , x>a)2+4 , x w a ,(a — 1) 24, x >a.f (2) 此时g(a)= f综上所述,x€ [0, 2]时,g(a) = 6—2a, 0<a<4 3—5(叮[4 3 —5< a<3(13 分)2a—2, a>3.。
湖南师大附中2018~2019学年度高一第一学期期末考试物理试题卷及答案
湖南师大附中2018~2019学年度高一第一学期期末考试物理时量:90分钟满分:100分一、选择题(本大题共12小题,1~8题为单选题,9~12题为多选题,每题4分,多选题少选计2分,有错选不计分,共48分)1.2019年1月1日,为庆祝元旦佳节,长沙市在橘子洲举行了烟花燃放活动.燃放时先用专用的发射炮筒将礼花弹发射到空中.假设礼花弹沿竖直方向升空,以下说法正确的是A.若不计空气阻力,礼花弹的速度越大,加速度越小B.若不计空气阻力,礼花弹的速度越来越小,加速度越来越小C.礼花弹的速度变化越快,加速度一定越大D.某时刻礼花弹上升到最高点,爆炸之前速度为零,其加速度也为零2.下列各图中,所有接触面都是光滑的,所有P、Q两球都处于静止状态.P、Q两球之间不存在弹力的是A.B.C.D.3.在强冷空气影响下,2018年12月,湖南遭遇强降雪.为防止道路结冰,长沙市公路局对各湘江大桥桥面进行撒盐除冰.假设撒盐车的牵引力不变,车所受阻力与车重成正比,若未撒盐时,撒盐车匀速直线行驶,则撒盐过程中它的运动将是A.做变加速运动B.做初速度不为零的匀加速直线运动C.做匀减速运动D.继续保持匀速直线运动4.如图所示,关于静止于水平桌面上的物体受力的说法中正确的是A.桌面对物体的支持力的大小大于物体的重力B.物体所受的重力和桌面对它的支持力是一对作用力与反作用力C.物体对桌面的压力就是物体的重力D.物体对桌面的压力和桌面对物体的支持力是一对相互作用力5.物体在足够大的水平面上运动,其v―t图象如图所示,设向右为正方向,关于前4s内物体运动情况的判断,下列说法正确的是A.物体始终向右运动B.物体先向右运动,第2s末开始向左运动C.第3s末物体在出发点的左侧D.第4s末物体距出发点最远6.关于平抛运动,下列说法正确的是A.平抛运动是加速度大小不变、方向改变的曲线运动B.做平抛运动的物体,在任何相等的时间内,位移的增量都是相等的C.平抛运动可以分解为水平方向的匀速直线运动和竖直方向的自由落体运动D.落地时间和落地时的速度只与抛出点的高度有关7.质量为60kg的某同学站在升降机中的磅秤上,某一时刻该同学发现磅秤的示数为50kg,则在该时刻升降机可能是以下列哪种方式运动A.匀速上升B.匀速下降C.加速上升D.减速上升8.停在水平地面上的小车内,用绳子AB、BC栓住一个重球,绳BC呈水平状态,绳AB的拉力为T1,绳BC的拉力为T2.若小车由静止开始加速向左运动,但重球相对小车的位置不发生变化,则两绳的拉力的变化情况是A.T1变小,T2变小B.T1变小,T2变大C.T1不变,T2变小D.T1变大,T2不变9.如图所示,A、B是两根竖直立在地上的木桩,轻绳系在两木桩上不等高的P、Q 两点,C为光滑的质量不计的滑轮,下面悬挂重物G、现保持结点P的位置不变,当Q和A杆的位置变化时,轻绳的张力大小变化情况是A.Q点上下移动时,张力不变B.Q点向上移动时,张力变大C.A杆向右移动时,张力变小D.A杆向左移动时,张力不变10.做曲线运动的物体,在运动过程中,下列物理量可能不变的是A.速率B.速度C.加速度D.合外力11.关于运动的合成与分解,以下说法正确的是A.两个直线运动的合运动一定是直线运动B.运动的合成与分解都遵循平行四边形定则C.两个分运动总是同时进行着的D.某个分运动的规律会因另一个分运动而改变12.如图所示,在倾角为α的斜面上,放一个质量为m的小球,小球和斜坡及挡板间均无摩擦,在挡板绕O点逆时针缓慢地转向水平位置的过程中A.小球对斜面的压力逐渐减小B.小球对斜面的压力逐渐增大C.小球对挡板的弹力先减小后增大D.小球对挡板的弹力先增大后减小二、实验题填空题(本大题共2小题,每空2分,共20分)13.目前实验室用的打点计时器有电磁打点计时器和________计时器两种,它们的原理基本一样,所接电源均为频率为50Hz的电源,每隔________打一次点,其中电磁打点计时器使用的是________(“直流”或“交流”)________(填电压大小)的电源.14.某实验小组“研究平抛物体的运动”实验的装置如图甲所示.钢球从斜槽上滚下,经过水平槽飞出后做平抛运动.每次都使钢球从斜槽上同一位置由静止滚下,把笔尖放在小球可能经过的位置,如果小球能够碰到笔尖,就说明位置找对了.通过多次实验,在竖直白纸上记录钢球所经过的多个位置,用平滑曲线连起来就得到钢球做平抛运动的轨迹.(1)实验所需的器材有:白纸、图钉、平板、铅笔、斜槽、小球、重垂线,除此之外还需要的一项器材是________.A.天平B.秒表C.刻度尺(2)在此实验中,小球与斜槽间有摩擦________(选填“会”或“不会”)使实验的误差增大;如果斜槽末端点到小球落地点的高度相同,小球每次从斜槽滚下的初始位置不同,那么小球每次在空中运动的时间________(选填“相同”或“不同”).(3)如图乙所示是在实验中记录的一段轨迹.已知小球是从原点O水平抛出的,经测量A点的坐标为(40cm,20cm),g取10m/s2,则小球平抛的初速度v0=________m/s,若B点的横坐标为x B=60cm,则B点纵坐标为y B=________m.(4)一同学在实验中采用了如下方法:如图丙所示,斜槽末端的正下方为O点.用一块平木板附上复写纸和白纸,竖直立于正对槽口前的O1处,使小球从斜槽上某一位置由静止滚下,小球撞在木板上留下痕迹A将木板向后平移至O2处,再使小球从斜槽上同一位置由静止滚下,小球撞在木板上留下痕迹B.O、O1间的距离为x1,O、O2间的距离为x2,A、B间的高度差为y.则小球抛出时的初速度v0′为________.A.2221()2x x gy-B.2221()2x x gy+C.2122x x gy+D.2122x x gy-三、计算题(本大题共4小题,第15题8分,16题8分,17题8分,18题8分,共32分)15.在平直的公路上,以速度v0=12m/s匀速前进的汽车,遇紧急情况刹车后,轮胎停止转动在地面上滑行,经过时间t=1.5s汽车停止,求:(1)刹车时,汽车的加速度a;(2)从开始刹车到停止,汽车行驶的位移x;(3)开始刹车后,1s末的速度.16.如图所示,水平传送带以不变的速度v=10m/s向右运动,将工件轻轻放在传送带的左端,由于摩擦力的作用,工件做匀加速运动,经过时间t=2s,速度达到v;再经过时间t′=4s,工件到达传送带的右端,g取10m/s2,求:(1)工件在水平传送带上滑动时的加速度的大小;(2)工件与水平传送带间的动摩擦因数;(3)工件从水平传送带的左端至到达右端通过的距离.17.如图所示,可视为质点的质量m=2.0kg的木块静止在高h=1.8m的水平台上,木块距平台右边缘x=4m,木块与平台间的动摩擦因数μ=0.2.用F=20N的水平拉力拉动木块,木块向右运动到平台边缘时撤去F.不计空气阻力,g取10m/s2,求:(1)木块在平台上运动的时间;(2)木块离开平台时的速度大小;(3)木块落地时距平台边缘的水平距离.18.如图所示,水平面与倾角θ=37°的斜面在B处平滑相连(物体经过B处时速率保持不变),水平面上A、B两点间距离s0=8m.质量m=1kg的物体(可视为质点)在F=6.5N的水平拉力作用下由A点从静止开始运动,到达B点时立即撤去F,物体将沿粗糙斜面继续上滑.已知物体与水平面及斜面间的动摩擦因数μ均为0.25.(g取10m/s2,sin37°=0.6,cos37°=0.8)求:(1)物体从A运动到B处时的速度大小v B;(2)物体沿斜面上滑过程中的加速度大小;(3)物体停止运动后距离B点的距离.。
2018-2019学年湖南省师范大学附属中学高一上学期期末考试数学试题(答案+解析)
湖南省师范大学附属中学2018-2019学年高一上学期期末考试数学试题第Ⅰ卷一、选择题1.若直线过点(1, 2),(2, 2+3),则此直线的倾斜角是() A .30° B .45° C .60° D .90°2.已知直线l 1:ax -y -2=0和直线l 2: (a +2)x -y +1=0,若l 1⊥l 2,则a 的值为() A .2 B .1 C .0 D .-13.若a 、b 表示直线,α表示平面,下列命题中正确的个数为() ①a ⊥α,b ∥αa ⊥b ;②a ⊥α,a ⊥b b ∥α;③a ∥α,a ⊥b b ⊥α. A .1 B .2 C .3 D .04.在空间直角坐标系中,点B 是A (1,2,3)在xOz 坐标平面内的射影,O 为坐标原点,则|OB |等于()A.14B.13C. 5D.105.两圆x 2+y 2-1=0和x 2+y 2-4x +2y -4=0的位置关系是() A .内切 B .相交 C .外切 D .外离6.如图,某几何体的正视图与侧视图都是边长为1的正方形,且体积为12,则该几何体的俯视图可以是()7.已知圆C :x 2+y 2-4x -5=0,则过点P (1,2)的最短弦所在直线l 的方程是() A .3x +2y -7=0 B .2x +y -4=0 C .x -2y -3=0 D .x -2y +3=08.直三棱柱ABC —A 1B 1C 1中,若∠BAC =90°,AB =AC =AA 1,则异面直线BA 1与AC 1所成的角等于()A .30°B .45°C .60°D .90°9.从直线x -y +3=0上的点向圆x 2+y 2-4x -4y +7=0引切线,则切线长的最小值为() A.322 B.142 C.324 D.322-110.如图,等边三角形ABC 的中线AF 与中位线DE 相交于G ,已知△A ′ED 是△AED 绕DE 旋转过程中的一个图形,下列命题中,错误的是() A .恒有DE ⊥A ′FB .异面直线A ′E 与BD 不可能垂直C .恒有平面A ′GF ⊥平面BCDED .动点A ′在平面ABC 上的射影在线段AF 上 二、填空题11.如图,△O ′A ′B ′是水平放置的△OAB 的直观图,O ′A ′=3, O ′B ′=4,则△AOB 的面积是________.12.在三棱锥A -BCD 中,AB ⊥AC ,AB ⊥AD ,AC ⊥AD ,若AB =3,AC =4,AD =5,则三棱锥A -BCD 的外接球的表面积为________.13.如图所示,已知矩形ABCD 中,AB =3,BC =a ,若P A ⊥平面AC ,在BC 边上取点E ,使PE ⊥DE ,则满足条件的E 点有两个时,a 的取值范围是________. 三、解答题14.已知直线l 经过点P (-2,5),且斜率为-34.(Ⅰ)求直线l 的方程;(Ⅱ)求与直线l 切于点(2,2),圆心在直线x +y -11=0上的圆的方程.15.已知坐标平面上动点M (x ,y )与两个定点A (26,1),B (2,1)的距离之比等于5. (Ⅰ)求动点M 的轨迹方程,并说明轨迹是什么图形;(Ⅱ)记(Ⅰ)中的轨迹为C ,过点P (-2,3)的直线l 被C 所截得的线段的长为8,求直线l 的方程.16.如图所示,ABCD 是正方形,O 是正方形的中心,PO ⊥底面ABCD ,底面边长为a ,E 是PC 的中点.(Ⅰ)求证:P A ∥平面BDE ; (Ⅱ)求证:平面P AC ⊥平面BDE ;(Ⅲ)若二面角E -BD -C 为30°,求四棱锥P -ABCD 的体积.第Ⅱ卷一、选择题17.中国古代数学著作《孙子算经》中有这样一道算术题:“今有物不知其数,三三数之余二,五五数之余三,问物几何?”人们把此类题目称为“中国剩余定理”问题,若正整数N 除以正整数m 后的余数为n ,则记为N =n (mod m ),例如11=2(mod 3).现将该问题以程序框图的算法给出,执行该程序框图,则输出的n 等于()A .21B .22C .23D .2418.在四棱锥P -ABCD 中,AD ⊥面P AB ,BC ⊥面P AB ,底面ABCD 为梯形,AD =4,BC =8,AB =6,∠APD =∠CPB ,满足上述条件的四棱锥的顶点P 的轨迹是 A .直线的一部分 B .半圆的一部分C .圆的一部分D .球的一部分 二、填空题19.定义在R 上的奇函数f (x ),当x ≥0时,f (x )=⎩⎪⎨⎪⎧log 12(x +1),x ∈[0,1),1-|x -3|,x ∈[1,+∞),则关于x 的函数F (x )=f (x )-20172018的所有零点之和为________.三、解答题20.如图,在正方体ABCD -A 1B 1C 1D 1中.(Ⅰ)求证:AC⊥BD1;(Ⅱ)是否存在直线与直线AA1,CC1,BD1都相交?若存在,请你在图中画出两条满足条件的直线(不必说明画法及理由);若不存在,请说明理由.21.平面直角坐标系中,在x轴的上方作半径为1的圆Γ,与x轴相切于坐标原点O.平行于x 轴的直线l1与y轴交点的纵坐标为-1,A(x,y)是圆Γ外一动点,A与圆Γ上的点的最小距离比A到l1的距离小1.(Ⅰ)求动点A的轨迹方程;(Ⅱ)设l2是圆Γ平行于x轴的切线,试探究在y轴上是否存在一定点B,使得以AB为直径的圆截直线l2所得的弦长不变.22.已知函数f (x )=log 2(x +1).(Ⅰ)若f (x )+f (x -1)>0成立,求x 的取值范围;(Ⅱ)若定义在R 上奇函数g (x )满足g (x +2)=-g (x ),且当0≤x ≤1时,g (x )=f (x ),求g (x )在[-3,-1]上的解析式,并写出g (x )在[-3,3]上的单调区间(不必证明);(Ⅲ)对于(Ⅱ)中的g (x ),若关于x 的不等式g ⎝ ⎛⎭⎪⎫t -2x8+2x +3≥g ⎝⎛⎭⎫-12在R 上恒成立,求实数t 的取值范围.【参考答案】第Ⅰ卷一、选择题 1.C【解析】利用斜率公式k =3=tan θ,可求倾斜角为60°. 2.D【解析】由题知(a +2)a +1=a 2+2a +1=(a +1)2=0,∴a =-1.也可以代入检验. 3.A【解析】①正确. 4.D【解析】点A (1,2,3)在xOz 坐标平面内的射影为B (1,0,3), ∴|OB |=12+02+32=10. 5.B【解析】将两圆化成标准方程分别为x 2+y 2=1,(x -2)2+(y +1)2=9,可知圆心距d =5,由于2<d <4,所以两圆相交. 6.C【解析】当俯视图为A 中正方形时,几何体为边长为1的正方体,体积为1;当俯视图为B 中圆时,几何体为底面半径为12,高为1的圆柱,体积为π4;当俯视图为C 中三角形时,几何体为三棱柱,且底面为直角边长为1的等腰直角三角形,高为1,体积为12;当俯视图为D 中扇形时,几何体为圆柱的14,且体积为π4.7.D【解析】化成标准方程(x -2)2+y 2=9,过点P (1,2)的最短弦所在直线l 应与PC 垂直,故有k l ·k PC =-1,由k PC =-2得k l =12,进而得直线l 的方程为x -2y +3=0.8.C【解析】将直三棱柱ABC -A 1B 1C 1补形为正方体ABDC -A 1B 1D 1C 1, 则异面直线BA 1与AC 1所成的角等于BA 1与BD 1所成的角,为60°. 9.B【解析】当圆心到直线距离最短时,可得此时切线长最短.d =322,切线长=⎝⎛⎭⎫3222-12)=142.10.B【解析】对A 来说,DE ⊥平面A ′GF ,∴DE ⊥A ′F ;对B 来说,∵E 、F 为线段AC 、BC 的中点,∴EF ∥AB ,∴∠A ′EF 就是异面直线A ′E 与BD 所成的角,当(A ′E )2+EF 2=(A ′F )2时,直线A ′E 与BD 垂直,故B 不正确;对C 来说,因为DE ⊥平面A ′GF ,DE 平面BCDE ,∴平面A ′GF ⊥平面BCDE ,故C 正确; 对D 来说,∵A ′D =A ′E ,∴DE ⊥A ′G ,∵△ABC 是正三角形,∴DE ⊥AG ,又A ′G ∩AG =G ,∴DE ⊥平面A ′GF ,从而平面ABC ⊥平面A ′AF ,且两平面的交线为AF ,∴A ′在平面ABC 上的射影在线段AF 上,正确. 二、填空题 11.12【解析】△OAB 为直角三角形,两直角边分别为4和6,S =12. 12.50π【解析】三棱锥A -BCD 的外接球就是长宽高分别为3、4、5的长方体的外接球,所以外接球的半径R 满足:2R =32+42+52=5 2.所以三棱锥A -BCD 的外接球的表面积S =4 πR 2=50 π. 13. a >6【解析】由P A ⊥平面AC ,PE ⊥DE ,得AE ⊥DE .问题转化为以AD 为直径的圆与BC 有两个交点,所以a2>3,解得a >6.三、解答题14.(Ⅰ)3x +4y -14=0 (Ⅱ)(x -5)2+(y -6)2=2515.解:(Ⅰ)由题意,得|MA ||MB |=5.(x -26)2+(y -1)2(x -2)2+(y -1)2=5,化简,得x 2+y 2-2x -2y -23=0. 即(x -1)2+(y -1)2=25.∴点M 的轨迹方程是(x -1)2+(y -1)2=25,轨迹是以(1,1)为圆心,以5为半径的圆. (Ⅱ)当直线l 的斜率不存在时,l :x =-2, 此时所截得的线段的长为252-32=8, ∴l :x =-2符合题意.当直线l 的斜率存在时,设l 的方程为 y -3=k (x +2),即kx -y +2k +3=0, 圆心到l 的距离d =|3k +2|k 2+1, 由题意,得⎝ ⎛⎭⎪⎫|3k +2|k 2+12+42=52,解得k =512.∴直线l 的方程为512x -y +236=0.即5x -12y +46=0.综上,直线l 的方程为x =-2,或5x -12y +46=0. 16.(Ⅰ)证明:连接OE . ∵O 、E 分别为AC 、PC 中点, ∴OE ∥P A .∵OE 面BDE ,P A 平面BDE , ∴P A ∥平面BDE .(Ⅱ)证明:∵PO ⊥平面ABCD ,∴PO ⊥BD . 在正方形ABCD 中,BD ⊥AC , 又∵PO ∩AC =O ,∴BD ⊥平面P AC .又∵BD 平面BDE ,∴平面P AC ⊥平面BDE . (Ⅲ)解:取OC 中点F ,连接EF . ∵E 为PC 中点,∴EF 为△POC 的中位线,∴EF ∥PO . 又∵PO ⊥平面ABCD , ∴EF ⊥平面ABCD , ∵OF ⊥BD ,∴OE ⊥BD .∴∠EOF 为二面角E -BD -C 的平面角,∴∠EOF =30°.在Rt △OEF 中,OF =12OC =14AC =24a ,∴EF =OF ·tan 30°=612a ,∴OP =2EF =66a . ∴V P -ABCD =13×a 2×66a =618a 3.第Ⅱ卷17.C 18.C【解析】因为AD ⊥平面P AB ,BC ⊥平面P AB ,所以AD ∥BC ,且∠DAP =∠CBP =90°.又∠APD =∠CPB ,AD =4,BC =8,可得tan ∠APD =AD P A =CB PB =tan ∠CPB ,即得PB P A =CBAD =2,在平面P AB 内,以AB 所在直线为x 轴,AB 中点O 为坐标原点,建立平面直角坐标系,则A (-3,0)、B (3,0).设点P (x ,y ),则有|PB ||P A |=(x -3)2+y 2(x +3)2+y 2=2,整理得x 2+y 2+10x +9=0.由于点P 不在直线AB 上,故此轨迹为一个圆,但要去掉二个点,选C. 19.1-220172018【解析】∵当x ≥0时,f (x )=⎩⎪⎨⎪⎧log 12(x +1),x ∈[0,1);1-|x -3|,x ∈[1,+∞);即x ∈[0,1)时,f (x )=log 12(x +1)∈(-1,0];x ∈[1,3]时,f (x )=x -2∈[-1,1]; x ∈(3,+∞)时,f (x )=4-x ∈(-∞,-1); 画出x ≥0时f (x )的图象,再利用奇函数的对称性,画出x <0时f (x )的图象,如图所示;则直线y =20172018,与y =f (x )的图象有5个交点,则方程f (x )-20172018=0共五个实根, 最左边两根之和为-6,最右边两根之和为6,∵x ∈(-1,0)时,-x ∈(0,1),∴f (-x )=log 12(-x +1), 又f (-x )=-f (x ),∴f (x )=-log 12(-x +1)=log 12(1-x )-1=log 2(1-x ),∴中间的一个根满足log 2(1-x )=20172018, 即1-x =220172018,解得x =1-220172018, ∴所有根的和为1-220172018. 20.(Ⅰ)证明:如图,连结BD .∵正方体ABCD -A 1B 1C 1D 1,∴D 1D ⊥平面ABCD .∵AC 平面ABCD ,∴D 1D ⊥AC .∵四边形ABCD 是正方形,∴AC ⊥BD .∵BD ∩D 1D =D ,∴AC ⊥平面BDD 1.∵BD 1平面BDD 1,∴AC ⊥BD 1.(5分)(Ⅱ)解:存在.答案不唯一,作出满足条件的直线一定在平面ACC 1A 1中,且过BD 1的中点并与直线A 1A ,C 1C 相交.下面给出答案中的两种情况,其他答案只要合理就可以给满分.21.解:(Ⅰ)设圆Γ的圆心为O 1,显然圆Γ上距A 距离最小的点在AO 1上,于是依题意知AO 1的长度等于A 到l 1的距离.显然A 不能在l 1的下方,若不然A 到l 1的距离小于AO 1的长度, 故有(y -1)2+x 2=y -(-1),即y =14x 2 (x ≠0). (Ⅱ)若存在这样的点B ,设其坐标为(0,t ),以AB 为直径的圆的圆心为C ,过C 作l 2的垂线,垂足为D .则C 点坐标为⎝⎛⎭⎫x 2,y +t 2,于是CD =|y +t -4|2, AB =x 2+(y -t )2=4y +(y -t )2设所截弦长为l ,则l 24=⎝⎛⎭⎫AB 22-CD 2=4y +(y -t )24-(y +t )2-8(y +t )+164, 于是l 2=(12-4t )y +8t -16,弦长不变即l 不随y 的变化而变化,故12-4t =0,即t =3.即存在点B (0,3),满足以AB 为直径的圆截直线l 2所得的弦长不变.22.解:(Ⅰ)由f (x )+f (x -1)>0得log 2(x +1)+log 2x >0,得⎩⎪⎨⎪⎧x 2+x >1x >0x +1>0,解得x >5-12,所以x 的取值范围是x ∈⎩⎨⎧⎭⎬⎫x ⎪⎪⎪x >5-12(5分); (Ⅱ)当-3≤x ≤-2时,g (x )=-g (x +2)=g (-x -2)=f (-x -2)=log 2(-x -2+1)=log 2(-x -1), 当-2<x ≤-1时,g (x )=-g (x +2)=-f (x +2)=-log 2(x +3),综上可得g (x )=⎩⎪⎨⎪⎧log 2(-1-x ),(-3≤x ≤-2)-log 2(3+x ),(-2<x ≤-1), g (x )在[-3,-1]和[1,3]上递减;g (x )在[-1,1]上递增;(Ⅲ)因为g ⎝⎛⎭⎫-12=-g ⎝⎛⎭⎫12=-f ⎝⎛⎭⎫12=-log 232, 由(Ⅱ)知,若g (x )=-log 232,得x =-32或x =52, 由函数g (x )的图象可知若g ⎝ ⎛⎭⎪⎫t -2x8+2x +3≥g ⎝⎛⎭⎫-12在R 上恒成立. 设u =t -2x 8+2x +3=-18+t +18(1+2x ), 当t +1≥0时,u =-18+t +18(1+2x )∈⎝⎛⎭⎫-18,-18+t +18, 则u ∈⎝⎛⎭⎫-18,-18+t +18⎝⎛⎭⎫-12,52,则-18+t +18≤52, 解得-1≤t ≤20.当t +1<0时,u =18+t +18(1+2x )∈⎝⎛⎭⎫-18+t +18,-18, 则u ∈⎝⎛⎭⎫-18+t +18,-18⎝⎛⎭⎫-12,52,则-18+t +18≥-12, 解得-4≤t <-1.综上,故-4≤t ≤20.。
湖南四大名校内部资料高一年级数学2018—2019—2师大附中高一第二次质量检测数学试卷
湖南师大附中20182019-学年度高一第二学期第二次阶段性检测数学命题:苏萍 苏林 赵优良审题:赵优良时量:120分钟 满分:150分一、选择题:本大题共12小题,每小题5分,共60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. 若点()cos ,sin P θθ在直线20x y -=上,则tan 2θ=( ) A. 45-B.43C. 43-D.452. 已知α是第二象限角,1sin cos 5αα=-,则cos sin αα-=( )A.B. C.或 D.753. 已知等差数列{}n a 前9项的和为27,108a =,则100a =( ) A. 100B. 99C. 98D. 974. 函数22cos sin 44y x x ππ⎛⎫⎛⎫=+-+ ⎪ ⎪⎝⎭⎝⎭的最小正周期为( )A. 2πB. πC.2πD.4π 5. 在ABC ∆中,若22AB BC AB AC -=⋅,则ABC ∆是( ) A. 等腰三角形B. 直角三角形C. 等腰直角三角形D. 等边三角形6. 已知1e 、2e 是两个单位向量,且夹角为3π,则12e te +与12te e +数量积的最小值为( )A. 32-B. 6-C.12D.37. 如图,已知OAB ∆,若点C 满足3AC CB =,OC xOA yOB =+(),x y R ∈,则11x y+=( ) A.14B.34C.316D.1638. 将函数3y cos x π⎛⎫=-⎪⎝⎭的图象上各点的横坐标伸长到原来的2倍(纵坐标不变),再向左平移6π个单位,所得函数图象的一条对称轴是( ) A. 4x π=B. 6x π=C. x π=D. 2x π=9. 已知3sin 2252πααπ⎛⎫=<< ⎪⎝⎭,()1tan 2αβ-=-,则()tan αβ+等于( ) A. 2-B. 1-C. 211-D.21110. 已知{}n a 为无穷等比数列,且公比1q >,记n S 为{}n a 的前n 项和,则下列结论正确的是( ) A. 21a a >B. 120a a +>C. {}2n a 是递增数列D. n S 存在最小值11. 已知ABC ∆中,a 、b 分别是角A 、B 所对的边,且()0a x x =>,4b =,60A =︒,若三角形有两解,则x 的取值范围是( )A. x >B. 4x ≤≤C. 4x <<D. 4x <≤12. 已知O 为ABC ∆的外心,角A 、B 、C 的对边分别为a 、b 、c ,若CO AB BO CA ⋅=⋅,则222a b c+的值是( ) A. 13B.12C. 1D. 2二、填空题:本大题共4个小题,每小题5分,共20分.13. = ; 14. 函数()()sin f x A x ωϕ=+0,0,2A πωϕ⎛⎫>><⎪⎝⎭的部分图象如图所示,现将此图象向左平移12π个单位长度得到函数()g x 的图象,则函数()g x 的解析式为 ;15. 已知函数()1xf x x=+,则()()()()111112320192342019f f f f ff f f ⎛⎫⎛⎫⎛⎫⎛⎫+++++++++= ⎪ ⎪ ⎪⎪⎝⎭⎝⎭⎝⎭⎝⎭;16. 对于数列{}n a ,定义1123242n nn a a a a T n-++++=为数列{}n a 的“好数”,已知某数列{}n a 的“好数”为12n n T +=,记数列{}n a pn -的前n 项和为n S ,若8n S S ≤对任意的n N *∈恒成立,则实数p 的取值范围为 .三、解答题:本大题共6个小题,共70分,解答应写出文字说明,证明过程或演算步骤. 17. (本小题满分10分)已知在ABC ∆中,A 、B 、C 所对的边分别为a 、b 、c ,若2228a b c +=+,3C π=.(1)求ABC ∆的面积;(2)若c =,求sin sin A B +的值.在数列{}n a 中,12a =,1122n n n a a ++=+,设2nn n a b =. (1)证明:数列{}n b 是等差数列并求数列{}n a 的通项公式; (2)求数列{}n a 的前n 项和.19. (本小题满分12分)ABC ∆中,记角A 、B 、C 所对的边分别为a 、b 、c ,已知a 、b 、c 依次成等比数列且2a 、2b 、2c 依次成等差数列.(1)求B 的大小;(2)若a c b λ+=,求λ的取值范围.已知函数()22sin 214f x x x π⎛⎫=+-⎪⎝⎭,x R ∈. (1)求函数()f x 的单调递增区间;(2)在ABC ∆中,角A 、B 、C 所对的边分别为a 、b 、c ,且满足()2cos cos a c B b C -=,若方程()1f A m +=恰有两个不同的解,求实数m 的取值范围.21. (本小题满分12分)如图所示,在xOy 平面上,点()1,0A ,点B 在单位圆上且()0AOB ααπ∠=<<. (1)若点34,55B ⎛⎫- ⎪⎝⎭,求tan 24πα⎛⎫- ⎪⎝⎭的值;(2)若OA OB OC +=,四边形OACB 的面积用S 表示,求S OA OC +⋅的最大值.已知数列{}n a 满足2112n n n a a a +=+,n N *∈. (1)若120a -<<,证明:110n a a +<<; (2)若10a >,记12111222n n S a a a =++++++,问:是否存在常数M ,使得n S M <对n N *∈均成立.。
湖南师大附中2018~2019学年度高一第一学期期末考试语文试题卷及答案
湖南师大附中2018~2019学年度高一第一学期期末考试语文时量:120分钟满分:150分一、语言基础知识及运用(16分,每小题4分)1.下列词语中,加点字的读音全都正确的一项是A.袅.娜(niǎo)流觞.(shāng)消弭.(mǐ)镣.铐(liào)B.椽.子(chuán)愀.然(qiǎo)肄.业(sì)悼.念(dào)C.涸.辙(hé)谬.误(miào)邮戳.(chuō)祈.祷(qí)D.诟.病(gòu)相勖.(mào)蜷.缩(quán)卓.越(zhuó)2.下列词语的字形完全正确的一项是A.斑驳琴瑟抵砺德行唯物辩证B.驯鸽枕藉义愤填膺坚忍不拔C.倩影庐冢浅尝辄止蘩芜丛杂D.伶俜赎罪了草塞责孜孜以求3.下列各句中,加点的词语使用正确的一句是A.青年学子要以天下为己任,爱国为民,把匡正流俗当作自己责无旁贷....的职责。
B.长沙市和常德市在众多的竞争者中独占鳌头....,该年光荣入选“全国文明城市”。
C.鲁迅先生对友人尤其是青年朋友的爱护无所不至....,哪怕对犯错的青年也是如此。
D.这部造势强劲的3D电影上映后,观者如堵,好评如潮,更让部分影迷叹为观止....。
4.下列各句中,有语病...的一句是A.人们普遍担心,迟早要出现的克隆人将深刻影响人类生活,使人类在伦理道德等方面面临巨大挑战。
B.调查显示,不少时尚的网络游戏带有暴力情节和血腥场面,这无疑会严重影响青少年的身心健康。
C.我国残疾人公共服务,主要是由康复、特殊教育、就业、无障碍设施、文体服务等专项内容构成的。
D.大选前18个月,他就开始接受特勤局保护,部分原因在于许多带有种族色彩的言论都是针对他引发的。
二、古诗文默写(8分)5.依据课文默写。
(1)越陌度阡,________________。
________________,心念旧恩。
【英语】湖南师范大学附属中学2018-2019学年高一上学期期末考试试题 (解析版)
湖南师范大学附属中学2018-2019学年高一上学期期末考试英语试题时量:120分钟满分:50分第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.Where does the conversation probably take place?A. In a clinic.B. In a hotel.C. In a store.【答案】C2.What did the woman think they would do?A. See an exhibition.B. Have a meeting.C. Attend a lecture.【答案】A3.What will the man probably do next?A. Go back to his work.B. Eat out for lunch.C. Pick up Jenny.【答案】B4.What is the probable relationship between the two speakers?A. Professor and student.B. Hotel manager and tourist.C. Salesman and customer.【答案】A5.How much will the woman pay for one chair?A. $ 59.B. $62.C. $65.【答案】C第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
每段对话或独白后有几个小题。
从题中所给的A、B、C三个选项中选出最佳选项。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
湖南师范大学附属中学2018-2019学年高一上学期期末考试英语试题附答案解析
湖南师大附中2018—2019学年度高一第一学期期末考试英语时量:120分钟满分:50分第一部分听力(共两节,满分30分)做题时,先将答案标在试题卷上。
录音内容结束后,你将有两分钟的时间将试题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。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.Where does the conversation probably take place?A. In a clinic.B. In a hotel.C. In a store.【答案】C【解析】【详解】此题为听力题,解析略。
2.What did the woman think they would do?A. See an exhibition.B. Have a meeting.C. Attend a lecture.【答案】A【解析】【详解】此题为听力题,解析略。
3.What will the man probably do next?A. Go back to his work.B. Eat out for lunch.C. Pick up Jenny.【答案】B【解析】【详解】此题为听力题,解析略。
4.What is the probable relationship between the two speakers?A. Professor and student.B. Hotel manager and tourist.C. Salesman and customer.【答案】A【解析】【详解】此题为听力题,解析略。
5.How much will the woman pay for one chair?A. $ 59.B. $62.C. $65.【答案】C【解析】【详解】此题为听力题,解析略。
湖南省师大附中2018-2019学年高一上学期期末考试语文试卷PDF版含答案
有以合乎世必违乎古有以同乎俗必离乎道矣( 生其 无急于解里人之惑" 则于是焉" 必能 择而取之 ( 遂书以赠二生" 并示苏君" 以为何如也$ 对下列句子中加点的词的解释! 正确的一项是 # " * 予之同年友也 +* 而苏君固可谓善知人者也 , * 同年, 年龄相同 善知, 善良而智慧
湖南师大附中! 共# " # $ % ! " # &学年度高一第一学期期末考试语文 第 # 页 "页
考场号 年级 班级 姓名 学号
密封线内不要答题
二 古诗文默写 $分 依据课文默写$ ' * # 越陌度阡 ! 心念旧恩$" 曹操( 短歌行) # " # $! " # 或取诸怀抱 ! 或因寄所托 ! 王 ! * $" 羲之( 兰亭集序) # " # 而! 常在于险远! 而人之所罕至焉! / $ 王安石( 游褒禅山记) # " " # ( 孔雀东南飞) 中% 便利此月内! 六合正相应& 所写婚俗与( 诗经 + 氓) 中% ) 两句所写的做法相似$ ! & 三 课内文言文阅读 ! !分 阅读下面的文言文! 完成( 选择题每小题)分 & 题 $ 正襟危坐而问客曰! # 何为其然也$% 客曰 ! # & 月明星稀" 乌鹊南飞 ' " 此 苏子愀然" 非曹孟德之诗乎 $ 西望夏口" 东望武昌" 山川相 缪" 郁乎苍苍" 此非孟德之困 于 周郎者 乎$ 方其破荆州 " 下江陵" 顺流而 东也" 舳舻千里" 旌旗蔽空" 酾酒临江" 横槊赋诗" 固一 世之雄也 " 而今安在哉$ 况吾与子渔樵于江渚之上" 侣鱼虾而友麋鹿" 驾一叶之扁舟" 举 匏樽以相属 (寄蜉蝣于天地" 渺沧海之一粟( 哀吾生之须臾" 羡长江之无穷( 挟飞仙以 遨游" 抱明月而长终(知不可乎骤得" 托遗响于 悲风( % # 客亦知夫水与月乎$ 逝者如斯" 而未尝往也 ) 盈虚者如彼" 而卒莫消长 苏子曰! 则天地曾不能以一瞬) 自其不变者而观之" 则物与我皆无尽 也(盖将自其变者而观之 " 也" 而又何羡乎* 且夫天地之间" 物各有主" 苟非吾之 所有" 虽一毫而莫取( 惟江上之清 风" 与山间之明月 " 耳得之而为声" 目遇之而成色 " 取之无禁" 用之不竭" 是造物者之无尽 藏也" 而吾与子之所共适( % 节选自苏轼 赤壁赋 对下列句子中加点词语的解释! 不正确的一项是 ( * 苏子愀然 ! 正襟危坐愀然, 容色改变状 +* 方其破荆州 ! 下江陵 下, 顺江而下 , * 山川相缪! 郁乎苍苍 缪, 盘绕 -* 而吾与子之所共适 适, 享有 .* 下列各组句子中! 加点词的意义和用法相同的一组是 0 * +* 此非孟德之困于周郎者乎 , * 顺流而 东也 -* 知不可乎 骤得 .* 苟非吾之所有 托遗响于 悲风 抱明月而 长终 而又何羡乎 目遇之 而成色
湖南师大附中2018-2019学年高一下学期期末数学试题
湖南师大附中2018-2019学年高一下学期期末数学试题学校:___________姓名:___________班级:___________考号:___________1.不等式2450x x -->的解集为( )A .{|5x x …或1}x -… B .{|5x x >或1}x <-C .{|15}x x -≤≤D .{|15}x x -<<2.△ABC 中, 如果cos A cos B cosCa b c==, 那么△ABC 是( ) A .直角三角形 B .等边三角形 C .等腰直角三角形 D .钝角三角形3.已知数列{a n }中,,*11()2n n a a n N +=+∈,则的值为 ( )A .49B .50C .51D .524.在ABC V 中,已知222b c a bc +-=,则(A ∠= ) A .6π B .3π C .23π D .3π或23π5.已知,2παπ⎛⎫∈ ⎪⎝⎭,3sin 5α=,则tan 4πα⎛⎫+= ⎪⎝⎭( )A .17B .7C .17-D .-76.设110a b<<,则( )A .22a b >B .a b +>C .2ab b <D .22a b a b +>+7.关于简单随机抽样,下列说法正确的是( )①它要求被抽取样本的总体的个数有限;②它是从总体中逐个地进行抽取;③不做特殊说明时它是一种不放回抽样;④它是一种等可能性抽样 A .①②③④B .③④C .①②③D .①③④811的等比中项是( ) A .1B .-1C .±1D .129.已知11a =,1()n n n a n a a +=-(*n N ∈),则数列{}n a 的通项公式是 ( )A .21n -B .11()n n n-+ C .nD .2n10.若1a b >>,P =,()1lg lg 2Q a b =+,lg 2a b R +⎛⎫= ⎪⎝⎭,则( ) A .R P Q <<B .P Q R <<C .Q P R <<D .P R Q <<11.在ABC V 中,D 为AB 的中点,60A ∠=︒且2AB AC AB CD ⋅=⋅u u u r u u u r u u u r u u u r,若ABC V 的面积为AC 的长为( )A .BC .3D .12.已知函数22sin sin ,[1,1]()22,(1,)x x a a x f x x ax a x ⎧++-∈-=⎨-+∈+∞⎩若关于x 的不等式()0f x …对任意[1,)x ∈-+∞恒成立,则实数a 的范围是( ) A .[0,2] B .(,0][2,)-∞+∞U C .(,0][1,2]-∞UD .[0,1][2,)⋃+∞13.已知(4,3),(5,6)a b =-=r r,则23||4a a b -⋅=r r r _________.14.等差数列{a n }的前m 项和为30,前2m 项和为100,则它的前3m 项和为 . 15.已知函数245x y a +=-(0a >,且1a ≠)的图像横过定点P ,若点P 在直线20Ax By ++=上,且0AB >,则12A B+的最小值为_________. 16.定义:如果一个数列从第二项起,后一项与前一项的和相等且为同一常数,这样的数列叫“等和数列”,这个常数叫公和.给出下列命题: ①“等和数列”一定是常数数列;②如果一个数列既是等差数列又是“等和数列”,则这个数列一定是常数列; ③如果一个数列既是等比数列又是“等和数列”,则这个数列一定是常数列; ④数列{}n a 是“等和数列”且公和100h =,则其前n 项之和50n S n =; 其中,正确的命题为__________.(请填出所有正确命题的序号) 17.已知(sin cos ,3),(sin cos ,1)a b αααα=+=-r r,且//a b r r. (1)求tan α值;(2)求1tan 2cos 2αα+的值.18.如图,在圆内接ABC V 中,A 、B 、C 所对的边分别为a 、b 、c ,满足cos cos 2cos a C c A b B +=(1)求B 的大小;(2)若点D 是劣弧AC 上一点,2AB =,3BC =,1AD =,求ABCD 四边形的面积.19.设数列{}n a 满足10a =,*1111,11n nn a a +-=∈--N (1)求{}n a 的通项公式; (2)设n b =,记12n n S b b b =++⋯+,求n S .20.某工厂生产某种产品,每日的成本C (单位:万元)与日产量(单位:吨)满足函数关系式3C x =+,每日的销售额S (单位:万元)与日产量x 的函数关系式27,(06)813,(6)k x x S x x ⎧++<<⎪=-⎨⎪≥⎩,已知每日的利润L S C =-,且当2x =时,92L =. (1)求k 的值;(2)当日产量为多少吨时,每日的利润可以达到最大,并求出最大值. 21.已知函数213()22f x x x =+,数列{}n a 的前n 项和为n S ,点()()*,n n S n N ∈()y f x =的图像上.(1)求数列{}n a 的通项公式; (2)令11n n n n n a a c a a ++=+,证明:121222n n c c c n <++⋯+<+;(3)设()ln 1n n b a =-,是否存在(,2)k k N k ∈…,使得12,,k k k b b b ++成等比数列,若存在,求出所有的k,若不存在,请说明理由.参考答案1.B 【解析】 【分析】根据一元二次不等式的解法,求得原不等式的解集. 【详解】依题意()()245150x x x x --=+->,解得1x <-或5x >.所以不等式的解集为{|5x x >或1}x <-.故选:B 【点睛】本小题主要考查一元二次不等式的解法,属于基础题. 2.B 【解析】试题分析:由题意得,由正弦定理得,所以,,所以,同理可得,所以三角形是等边三角形.考点:正弦定理在三角形中的应用. 3.D 【解析】试题分析:∵*11()2n n a a n N +=+∈,∴a n+1−a n =12, 则数列{a n }构成以12为公差的等差数列,又a 1=2,∴a 101=a 1+(101−1)×12=2+100×12=52. 故选D .考点:数列递推关系式;等差数列的通项公式. 4.B 【解析】 【分析】由已知直接利用余弦定理求得cos A ,则A ∠可求. 【详解】由222b c a bc +-=,得2221cos 222b c a bc A bc bc +-===,0A Q π<<,3A π∴=.故选B . 【点睛】本题考查三角形的解法,考查余弦定理的应用,是基础题. 5.A 【解析】 【分析】先求出tan α的值,再利用和角的正切求tan 4πα⎛⎫+ ⎪⎝⎭的值. 【详解】 因为,2παπ⎛⎫∈⎪⎝⎭,3sin 5α=,所以3tan 4α=-,所以tan 4πα⎛⎫+ ⎪⎝⎭=3114371()14-+=--⋅. 故选A 【点睛】本题主要考查同角的三角函数关系,考查和角的正切的计算,意在考查学生对这些知识的掌握水平和分析推理能力. 6.C 【解析】 试题分析:110a b<<∴Q设1,2a b =-=-,代入四个不等式验证可知2ab b <正确 考点:不等式性质 7.A 【解析】 【分析】根据简单随机抽样的定义和性质得到答案.根据简单随机抽样的定义和性质知:①它要求被抽取样本的总体的个数有限,正确; ②它是从总体中逐个地进行抽取,正确; ③不作特殊说明时它是一种不放回抽样,正确; ④它是一种等可能性抽样,正确; 故选:A 【点睛】本题考查了简单随机抽样的定义和性质,属于简单题. 8.C 【解析】试题分析:设两数的等比中项为)21111x x x ∴==∴=±,等比中项为-1或1考点:等比中项 9.C 【解析】由()1n n n a n a a +=-,得:()11n n n n a a ++=,11n na a n n+=+ ∴n a n ⎧⎫⎨⎬⎩⎭为常数列,即111n a a n ==,故n a n =故选C 10.B 【解析】 【分析】由基本不等式以及对数函数的单调性可得出三个数P 、Q 、R 的大小关系. 【详解】由于函数lg y x =在()0,∞+上是增函数,1a b >>Q ,则lg lg 0a b >>,由基本不等式可得()()11lg lg lg lg222a bP a b ab R +=<+===, 因此,P Q R <<,故选B .本题考查利用基本不等式比较大小,在利用基本不等式比较各数的大小关系时,要注意“一正、二定、三相等”这些条件的应用,考查推理能力,属于中等题. 11.B 【解析】 【分析】设,,AB c AC b ==先化简2AB AC AB CD ⋅=⋅u u u r u u u r u u u r u u u r得3c b =,由ABC V 的面积为16bc =,即得AC 的长.【详解】设,,AB c AC b ==由题得2AB AC AB CD ⋅=⋅u u u r u u u r u u u r u u u r,所以2()AB AC AB AD AC AB AD AB AC ⋅=⋅-=⋅-⋅u u u r u u u r u u u r u u u r u u u r u u u r u u u r u u u r u u u r,所以3,3cos cos0,332cAB AC AB AD c b c c b π⋅=⋅∴⨯⨯⨯=⨯⨯∴=u u u r u u u r u u u r u u u r .因为ABC V 的面积为1sin 1623b c bc π⨯⨯⨯=∴=.所以2316,b b =∴=所以3AC =. 故选:B 【点睛】本题主要考查平面向量的数量积运算,考查三角形的面积的应用,意在考查学生对这些知识的理解掌握水平. 12.C 【解析】 【分析】先通过分析[1,1]x ∈-的函数得到1a ≥或0a ≤,再通过分析(1,)x ∈+∞的函数得到2a ≤,综合即得解. 【详解】(1)22()sin |||sin |,()sin |||sin |()f x x x a a f x x x a a f x =++-∴-=++-=,所以函数2()sin |||sin |,(11)f x x x a a x =++--≤≤是一个偶函数, 又因为sin |||sin |2sin ,(01)y x x x x =+=≤≤,是一个增函数, 所以sin |||sin |2sin 0y x x x =+=≥,所以22()sin |||sin |01f x x x a a a a a =++-≥-≥∴≥,或0a ≤.(2)222()=22()2(1)f x x ax a x a a a x -+=--+>,当1a ≤时,()f x 在(1,)+∞单调递增,所以(1)=10f ≥成立,符合题意. 当1a >时,2min ()()20,02,12f x f a a a a a ==-≥∴≤≤∴<≤. 所以2a ≤.综合(1)(2)得实数a 的范围是(,0][1,2]-∞U . 故选:C 【点睛】本题主要考查分段函数的图象和性质,考查不等式的恒成立问题的求解方法,意在考查学生对这些知识的理解掌握水平和分析推理能力. 13.83 【解析】 【分析】先求出2||a r ,再求出a b ⋅r r,即得解. 【详解】由题得22||25a ==r ,(4)5362a b ⋅=-⨯+⨯=-r r , 所以23||43254(2)83a a b -⋅=⨯-⨯-=rr r .故答案为:83 【点睛】本题主要考查向量模的坐标计算,考查数量积的坐标表示,意在考查学生对这些知识的理解掌握水平. 14.210【解析】试题分析:设前3m 项和为 x ,则 30,100﹣30,x ﹣100 成等差数列,解出 x 的值,即为所求.解:等差数列{a n }的每m 项的和成等差数列,设前3m 项和为 x ,则 30,100﹣30,x ﹣100 成等差数列,故 2×70=30+(x ﹣100 ),x=210, 故答案为210.考点:等差数列的性质. 15.4 【解析】 【分析】先求出定点P 的坐标,由题得22A B +=,再利用基本不等式求12A B+的最小值得解. 【详解】令020,2,451x x y a +=∴=-∴=⨯-=-,所以定点P 的坐标为(2,1)--. 所以(2)20,22,0,0,0A B A B A B A B ⨯--+=∴+=⋅>∴>>Q .所以12112141(2)()(4)[44222A B A B A B A B B A +=⨯+⨯+=++≥+=. 当且仅当1,12A B ==时取“等号”. 所以12A B+的最小值为4. 故答案为:4 【点睛】本题主要考查指数型函数的定点问题,考查基本不等式求最值,意在考查学生对这些知识的理解掌握水平. 16.② 【解析】 【分析】利用“等和数列”的定义对每一个命题逐一分析判断得解.【详解】①“等和数列”不一定是常数数列,如数列1,0,1,0,1,0,1,0,1,0,L 是“等和数列”,但是不是常数数列,所以该命题错误;②如果一个数列既是等差数列又是“等和数列”,则这个数列一定是常数列.如果数列{}n a 是等差数列,所以112(2)n n n a a a n +-+=≥,如果数列{}n a 是“等和数列”,所以11+(2),n n n n a a a a n -+=+≥所以11(2),n n a a n -+=≥所以122(2)n n a a n -=≥,所以1(2)n n a a n -=≥,所以这个数列一定是常数列,所以该命题是正确的.③如果一个数列既是等比数列又是“等和数列”,则这个数列一定是常数列. 如果数列{}n a 是等比数列,所以211(2)n n n a a a n +-⋅=≥,如果数列{}n a 是“等和数列”,所以 11+(2),n n n n a a a a n -+=+≥所以11(2),n n a a n -+=≥所以221(2)n n a a n -=≥,所以1(2)n n a a n -=±≥,所以这个数列不一定是常数列,所以该命题是错误的.④数列{}n a 是“等和数列”且公和100h =,则其前n 项之和50n S n =,是错误的.举例“等和数列”1,99,1,99,1,其5201505S =≠⨯,所以该命题是错误的.故答案为:②【点睛】本题主要考查数列的新定义的理解和应用,考查等差数列和等比数列的应用,意在考查学生对这些知识的理解掌握水平和分析推理能力.17.(1)2;(2)3-.【解析】【分析】(1)由题得sin +cos 3(sin cos )0αααα--=化简即得解;(2)求出tan 2,cos 2αα即得解.【详解】(1)因为//a b r r ,所以sin +cos 3(sin cos )0,sin 2cos ,tan 2ααααααα--=∴=∴=.(2)由题得22tan 44tan 21tan 143ααα===---.因为tan 2,cosααα=∴==或cos αα== 所以22143cos 2cossin 555ααα=-=-=-. 所以154tan 23cos 233αα+=--=-. 【点睛】本题主要考查向量平行的坐标表示,考查同角的商数关系和二倍角公式,意在考查学生对这些知识的理解掌握水平.18.(1)3B π=;(2)【解析】【分析】(1)根据正弦定理化简即可;(2)在ABC ∆,利用余弦定理求出AC ,已知B ,可得ADC ∠,再余弦定理求出DC ,即可ABC ∆和ADC ∆面积,可得四边形ABCD 的面积.【详解】(1)cos cos 2cos a C c A b B +=Q .由正弦定理,可得sin cos sin cos 2sin cos A C C A B B +=.得sin 2sin cos B B B =. 0B Q π<<,sin 0B ≠,1cos 2B ∴=, 即3B π=.(2)在ABC ∆中,2AB =,3BC =,3B π=. 由余弦定理,222249cos 3212AB BC AC AC AB BC π+-+-==g ,可得:AC =.在ADC ∆中,AC =,1AD =,,,,A B C D 在圆上, 3B π=Q . 23ADC π∴∠=.由余弦定理,2222217cos 322AD DC AC DC AD DC DCπ+-+-==g . 解得:2DC =四边形ABCD的面积121sin sin 2323ABC ADC S S S AD DC AB BC ππ∆∆=+=+=g g g g . 【点睛】本题考查三角形的面积的求法,正弦余弦定理的合理运用.圆内角四边形的角的关系,意在考查学生对这些知识的理解掌握水平.19.(1)1n n a n -=;(2)=1n S . 【解析】【分析】(1)由题得数列1{}1n a -是一个以1111a =-为首项,以1为公差的等差数列,再利用等差数列的通项求解即可;(2)先求出n b =. 【详解】(1)因为*1111,11n nn a a +-=∈--N , 所以数列1{}1n a -是一个以1111a =-为首项,以1为公差的等差数列, 所以11=1+(1)1,1n n n n n a a n--⨯=∴=-. 所以{}n a 的通项公式为1n n a n -=. (2)n b =,所以1=11n S ++=L 【点睛】 本题主要考查等差数列的通项的求法,考查裂项相消法求和,意在考查学生对这些知识的理解掌握水平.20.(1)9k =;(2)当日产量为5吨时,日利润达到最大6万元.【解析】【分析】(1)利用每日的利润L S C =-,且当2x =时,92L =,可求k 的值;(2)利用分段函数,分别求出相应的最值,即可得出函数的最大值.【详解】 由题意,每日利润L 与日产量x 的函数关系式为4,06810,6k x x L x x x ⎧++<<⎪=-⎨⎪-⎩…(1)当2x =时,92L =,即:924228k =++- 9k ∴= (2)当6x …时,10L x =-为单调递减函数, 故当6x =时,4max L =;当06x <<时,9(8)121268L x x =-++-=-… 当且仅当5x =时,6max L =综合上述情况,当日产量为5吨时,日利润达到最大6万元.【点睛】本题考查函数解析式的确定,考查分段函数的最值的求法,考查基本不等式求函数的最值,意在考查学生对这些知识的理解掌握水平,确定函数的解析式是关键.21.(1)=1n a n +;(2)证明见解析;(3)不存在,证明见解析.【解析】【分析】(1)由题得213,22n S n n =+再利用项和公式1=S (2)n n n a S n --≥即可求出数列{}n a 的通项公式;(2)求出1211222n c c c n n =++⋯++-+,不等式即得证;(3)假设存在,由题得ln(1)ln(2),ln ln(1)k k k k ++=+再证明ln(1)ln(2),ln ln(1)k k k k ++>+即得解.【详解】(1)由题得221313,(1)(1),(2)2222n n S n n S n n n =+∴=-+-≥, 两式相减得1=S 1n n n a S n --=+,1n =时,112,a S ==适合,1n a n =+,所以1n a n =+.所以数列{}n a 的通项公式为1n a n =+.(2)1(22)111=22(1)121112n n n c n n n n n n n n ++-=++=+-+++++++++, 1211111122223341122122n n n c c c n n =+-++-∴++⋯+=+-+++-+++L , 因为110,(1)22n n ->≥+,所以121222n n c c c n <++⋯+<+. (3)假设存在,因为12,,k k k b b b ++成等比数列,所以211222ln (1)ln(1)l ),=n(1k k k k k k b b b a a a ++++∴-=--gg , 所以2ln (1)ln()ln(2),k k k +=+g 所以ln(1)ln(2),ln ln(1)k k k k ++=+ 设2ln(1)ln (1)ln(1)(),(2),(=ln (ln )(1)x x x x x g x x g x x x x x +-++'=≥∴+g g ), 设()ln (1)ln(1),(2),()ln ln(1)0h x x x x x x h x x x '=-++≥∴=-+<,所以()h x 在[2,)+∞上单调递减,所以()(2)2ln 23ln3ln 4ln 270h x h ≤=-=-<,所以()0g x '<,所以()(1)g k g k >+, 所以ln(1)ln(2),ln ln(1)k k k k ++>+所以ln(1)ln(2)ln ln(1)k k k k ++=+不成立. 所以不存在(,2)k k N k ∈…,使得12,,k k k b b b ++成等比数列. 【点睛】本题主要考查数列通项的求法,考查数列求和和数列不等式的证明,考查数列的存在性问题的求解,意在考查学生对这些知识的理解掌握水平和分析推理能力.。
湖南师大附中18-19学度高一下年末考试-数学
y O6π 2 512π湖南师大附中18-19学度高一下年末考试-数学2017—2018学年度下学期期末考试高一数学试题时 量:120分钟 满 分:150 分〔必考I 部分100分,必考II 部分50分〕必考I 部分【一】选择题:本大题共8个小题,每题4分,共32分,在每个小题给出的四个选项中,只有一项为哪一项符合题目要求的、 1.以下命题正确的选项是〔B 〕A.第一象限角是锐角B.相等向量一定共线C.终边相同的角一定相等D.小于90的角是锐角 2、sin(-3300)的值为〔C 〕A 、23-B 、21-C 、21D 、233.假设)4,3(-P 为角α终边上一点,那么cos a =〔A 〕A 、35-B 、45C 、34-D 、34- 4.函数)12tan(3+=xy 的最小正周期是〔B 〕A 、πB 、2πC 、4πD 、4π5、如图,1e ,2e 为互相垂直的单位向量,那么向量c b a ++可表示为(C)A 、-13e 22eB 、--13e 23eC 、+13e 22eD 、+12e 23e6.平面向量)2,(),1,2(-==x ,且⊥,那么=x 〔D 〕A 、-3B 、3C 、-1D 、17.在四边形ABCD 中,假如=,且BD AC =,那么四边形ABCD 的形状为〔C 〕A.梯形B.菱形C.长方形D.正方形8.函数()sin()(0,0,||)2f x A x A ωϕωϕπ=+>><的部分图象如下图所示,那么函数()f x 的解析式为(D)A.1()2sin()26f x x π=+B.1()2sin()26f x x π=-C.()2sin(2)6f x x π=- D.()2sin(2)6f x x π=+【二】填空题:本大题共6个小题,每题4分,共24分,请把答案的最简形式填在横线上.9.sin 27cos63cos27sin63︒︒+︒︒=1.10、设一扇形的弧长为4cm ,面积为4cm 2,那么那个扇形的圆心角的弧度数是2. 11.将函数x y 4sin =的图象向左平移12π个单位,得到函数sin(4)y x φ=+的图象,那么φ=3π.53==且12=⋅b a ,那么a 在b 方向上的投影为125. 13.)cos ,(sin ),1,3(αα==,且//,那么4sin 2cos 5cos 3sin αααα-+=57.14.设函数()3sin(2)3f x x π=-的图象为C ,给出以下命题: ①图象C 关于直线1112x =π对称;②函数)(x f 在区间5(,)1212ππ-内是增函数;③函数()f x 是奇函数;④图象C 关于点(,0)3π对称.其中,正确命题的编号是①②.〔写出所有正确命题的编号〕【三】解答题:本大题共4个小题,共44分,解承诺写出文字说明,证明过程或演算步骤 15.〔此题总分值10分〕3sin 5θ=,(,)2θπ∈π,求tan θ,cos()4θπ-的值.解:∵3sin 5θ=,(,)2πθπ∈,∴54sin 1cos 2-=--=θθ,∴43cos sin tan -==θθθ,∴cos()cos cossin sin444πππθθθ-=+435252=-⨯+⨯=10-.16.〔此题总分值10分〕1,2==.〔1〕假设,的夹角θ为45; 〔2〕假设b b a ⊥-)(,求与的夹角θ. 解:〔1〕22221a b a ab b -=-+=- 〔2〕()a b b -⊥,2()21cos 10a b b a b b θ∴-⋅=⋅-=⨯⨯-=,cos )θθπ∴=≤≤,4πθ∴=. 17.〔本小题总分值12分〕函数()22sin cos cos f x x x x x =+-. 〔1〕求函数()f x 的单调递增区间;〔2〕求()f x 的最大值及取最大值时x 的集合.解:由,()2cos 22sin(2)6f x x x x π=-=-.〔1〕由222262k x k πππππ-≤-≤+,k Z ∈,得增区间为[,]()63k k k Z ππππ-+∈.〔2〕当2262x k πππ-=+,k Z ∈,即sin(2)16x π-=时,()f x 取最大值2,如今x 的集合为{|,}3x x k k Z ππ=+∈.18.〔本小题总分值12分〕向量)0(),cos ,(cos ),cos ,sin 3(>-==ωωωωωx x b x x a ,函数21)(+⋅=b a x f 的图象的两相邻对称轴间的距离为4π. 〔1〕求ω的值;〔2〕假设75(,)2412x ππ∈,53)(-=x f ,求x 4cos 的值; 〔3〕假设1cos ,(0,)2x x ≥∈π,且m x f =)(有且仅有一个实根,求实数m 的值.解:由题意,21cos cos sin 3)(2+-⋅=x x x x f ωωω2122cos 12sin 23++-=x x ωω x x ωω2cos 212sin 23-=)62sin(πω-=x , 〔1〕∵两相邻对称轴间的距离为4π, ∴222πωπ==T ,∴2=ω. 〔2〕由〔1〕得,53)64sin()(-=-=πx x f , ∵75(,)2412x ππ∈,∴)23,(64πππ∈-x , ∴54)64cos(-=-πx ,∴)664cos(4cos ππ+-=x x 6sin )64sin(6cos )64cos(ππππ---=x x 21)53(23)54(⨯--⨯-=103532+-=.〔3〕21cos ≥x ,且余弦函数在),0(π上是减函数,∴]3,0(π∈x , 令21)(+⋅=b a x f =)64sin(π-x ,m x g =)(,在同一直角坐标系中作出两个函数的图象,可知211-==m m 或.必考II 部分19、(本小题总分值12分)在等差数列{}n a 和等比数列{}n b 中,1141,8a b b ===,{}n a 的前10项和1055S =. 〔1〕求n a 和n b ;〔2〕现分别从{}n a 和{}n b 的前3项中各随机抽取一项,写出相应的差不多事件,并求这两项的值相等的概率.解:〔1〕由{}n a 是等差数列得,d a S 291010110⨯+=,又55,1101==S a , ∴1=d ,n d n a a n =-+=∴)1(1.又{}n b 是等比数列,且8,141==b b ,38=1q ∴⨯,2q ∴=,12-=∴n n b .〔2〕因为{}n a 前3项为1,2,3,{}n b 前3项为1,2,4,因此差不多事件为: 〔1,1〕,〔1,2〕,〔1,4〕,〔2,1〕,〔2,2〕,〔2,4〕,〔3,1〕,〔3,2〕,〔3,4〕 共9种,设两项值相等为事件A ,那么事件A 包括2种:〔1,1〕,〔2,2〕,92)(=∴A P . 20、(本小题总分值12分)()lg(1)f x x =+,〔1〕假设0(12)()1f x f x <--<,求x 的取值范围;〔2〕假设()g x 是以2为周期的偶函数,且当[]0,1x ∈时,()()g x f x =,当[]1,2x ∈时,求函数()y g x =的取值范围.解:〔1〕由⎩⎨⎧>+>-01022x x ,得11<<-x .由1012211122lg)1lg()22lg(0<+-<<+-=+--<x xx x x x 得. 因为01>+x ,因此211221010,33x x x x +<-<+-<<即.由⎪⎩⎪⎨⎧<<-<<-313211x x ,得3132<<-x . 〔2〕当]2,1[∈x 时,]1,0[2∈-x ,因此)3lg()2()2()2()(x x f x g x g x g y -=-=-=-==. 由单调性可得,函数()y g x =的取值范围为[0,lg 2]. 21、(本小题总分值13分)如图,在四棱锥P-ABCD 中,底面ABCD 是矩形,AD ⊥PD ,BC=1,PD=CD=2. 〔1〕求异面直线PA 与BC 所成角的正切值;〔2〕证明平面PDC ⊥平面ABCD ;〔3〕求直线PB 与平面ABCD 所成角的正弦值. 解:〔1〕如图,在四棱锥ABCD P -中,因为底面ABCD 是矩形, 因此BC AD =且BC AD //,又因为PD AD ⊥,故PAD ∠为异面直 线PA 与BC 所成的角.在PDA Rt ∆中,2tan ==∠ADPDPAD ,因此,异面直线PA 与BC 所成角的正切值为2.〔2〕证明:由于底面ABCD 是矩形,故CD AD ⊥,又由于D PD CD PD AD =⊥ ,, 因此⊥AD 平面PDC,而⊂AD 平面ABCD ,因此平面⊥PCD 平面ABCD . 〔3〕在平面PCD 内,过点P 作CD PE ⊥交直线CD 于点E ,连接EB.由于平面⊥PCD 平面ABCD ,而直线CD 是平面PCD 与平面ABCD 的交线, 故⊥PE 平面ABCD ,由此得PBE ∠为直线PB 与平面ABCD 所成的角.在PDC ∆中,由于,32,2===PC CD PD 可得=∠PCD 在PEC Rt ∆中,330sin ==PC PE ,由⊥AD BC AD ,//平面PDC ,得⊥BC 平面PDC , 因此PC BC ⊥,在PCB Rt ∆中,1322=+=BC PC PB .P A BCD在PEB Rt ∆中,1339sin ==∠PB PE PBE . 因此直线PB 与平面ABCD所成的角的正弦值为13. 22、(本小题总分值13分)在某海滨城市附近海面有一台风,据监测,当前台风中心位于城市O 〔如图〕的东偏南)102(cos =θθ方向300km 的海面P 处,并以20km/h 的速度向西偏北45°方向移动,台风侵袭的范围为圆形区域,当前半径为60km ,并以10km/h 的速度不断增大,问几小时后该城市开始受到台风的侵袭?受到台风侵袭的时间有多少小时? 解:设在t 时刻台风中心位于点Q ,如今300OP =,20PQ t =,台风侵袭范围的圆形区域半径为()1060r t t =+,由102cos =θ,可知1027cos 1sin 2=-=θθ, cos cos(45)cos cos 45sin sin 45OPQ θθθ∠=-︒=︒+︒=5422102722102=⨯+⨯, 在△OPQ 中,由余弦定理,得OPQ PQ OP PQ OP OQ ∠⋅-+=cos 2222=54203002)20(30022⨯⨯⨯-+t t =9000096004002+-t t假设城市O 受到台风的侵袭,那么有()OQ r t ≤,即22)6010(900009600400+≤+-t t t ,整理,得0288362≤+-t t ,解得12≤t ≤24,答:12小时后该城市开始受到台风的侵袭,受到台风侵袭的时间有12小时.O Pθ45°东西北东。
湖南师范大学附属中学2018-2019学年高一下学期期末数学试卷
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湖南师大附中2018—2019学年度高一第一学期期末考试生物
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湖南师大附中高一第一学期期末考试生物试题 第 共 "页 $ 页
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湖南师大附中18-19学度高一上年末考试卷—英语
湖南师大附中18-19学度高一上年末考试卷—英语时量:120分钟分值150分必考I部分PART ONE: LISTENING COMPREHENSION 〔30分〕Section A 〔22.5marks〕Directions: In this section, you’ll hear 6 conversations between 2 speakers. For each conversation, there are several questions and each question is followed by 3 choices. Listen to the conversations carefully and then answer the questions by marking the corresponding letter (A, B or C) on the question booklet. You will hear each conversation TWICE.Conversation 11. What does the woman think of shopping here?A. It’s boring.B. It’s enjoyable.C. It’s inconvenient.2. What are the speakers doing now?A. Having dinner.B. Shopping.C. Cooking.Conversation 23. What kind of exercise would the woman like to do?A. Jogging.B. Swimming.C. Weight-lifting.4. How often does the man suggest the woman work out?A. Twice a week.B. Three times a week.C. Four times a week. Conversation 35. What does the woman think of the painting Les Demoise de Avignon?A. Modern.B. Strange.C. Beautiful.6. What do we know about Picasso?A. He was a strange man.B. He died at an old age.C. He was born 200 years ago.Conversation 47. How are the speakers going to Paris?A. By bike.B. By plane.C. By train.8. How much should the speakers pay for hiring bikes every day?A. £5.B. £10.C. £15.9. What do the speakers decide to do in the end?A. Take their bikes along.B. Hire bikes in Paris.C. Buy bikes in Paris. Conversation 510. Who is the woman probably?A. A waitress.B. A librarian.C. A salesgirl.11. Why does the man want to change the CD?A. It is broken.B. His mum took it wrong.C. He doesn’t like it any more.12. What does the man want to exchange the CD for?A. A computer game.B. The first CD by Chaos Theory.C. The latest CD by Chaos Theory.Conversation 613. How did Hemingway get his first job?A. With the help of his uncle.B. With the help of his teacher.C. By himself.14. Why did Hemingway fail to join the army?A. He was too young.B. He was seriously wounded.C. His eyesight was poor.15. Among the following three books, which one is the latest?A. The Sun Also RisesB. A Farewell to ArmsC. The Old Man and the Sea Section B (7.5marks)Directions: In this section, you will hear a short passage. Listen carefully and then fill in the numbered blanks with the information you have heard. Fill in each blank with NO MORE THAN THREE WORDS.You will hear the short passage TWICE.Neil Armstrong – A famous 16. _____________Personal Data Born on August 5, 1930, died on August 25, 201817. __________ AmericanChildhood Showing a great interest in flyingEducation Studying engineering at universityWork experience Joining the army and working as a Navy pilotBeing a test pilot in a flight stationBeing an 18. _____________Achievement Being the first man setting foot on the moon after four days’19. _____________ in spaceConclusion He 20. _____________ the young people around the world to be willing to explore and push the limits, and to work hard to make their dreams come true.PART TWO: LANGUAGE KNOWLEDGE 〔40 marks〕Section A 〔10 marks〕Directions: Beneath each of the following sentences there are 4 choices marked A, B, C and D. Choose one answer that best completes the sentence.21. Yang Liwei was a student at this flight school from 1983 to 1987, he studied very hard and was made chairman of the Students’ Union.A. during which timeB. for which timeC. during whose timeD. by that time22. George Gould, a friend of Carnarvon’s, went to Egypt after hearing of his strange death and visited the tomb, only ______a high fever the next day.A. catchingB. to catchC. caughtD. catch23. Although medical science________ control over several dangerous diseases, what worries us is that some of them are returning.A. achievedB. has achievedC. will achieveD. had achieved24. ----The train will arrive at the station in half an hour. Who ______for us?----I am sure Dad will be standing there when we get off the train.A. will be waitingB. was to waitC. waitD. was about to wait25. ---- What were you doing when Tony phoned you?---- I had just finished my work and _____to take a shower.A. had startedB. startedC. have startedD. was starting26. ----My car broke down yesterday on the way to the Yue Lu mountain.----I am sorry to hear that, but you _______that car every day for ten years before it broke down. Now you really need to buy a new one.A. have usedB. have been usedC. had used D had been used27. —I’m still working on my project.—Oh, you’ll miss the deadline. Time is .A. running outB. going outC. giving outD. losing out28. If you _____faults but you still want the bicycle, ask the shop assistant toreduce the price.A. come across B. care about C. look intoD. look up to29. -----Where are the children? Professor Zhang has come and his speech will start soon.------I wish they always late.A. hadn’t been C. wouldn’t be D. wouldn’t have been D. weren’t30. ------Have you packed for your adventure holiday?------Oh, almost, but I still need a flashlight _____I will be able to see in the dark.A. becauseB. so thatC. even ifD. asSection B 〔12 marks〕Directions: For each blank in the following passage there are 4 words or phrases marked A, B, C and D. Fill in each blank with a word or phrase that best fits the context.An amazing teacherYears ago a John Hopkins professor gave a group of graduate students this task: Go to the slums〔贫民窟〕. Took 200 boys, between the ages of 12 and 16, and 31 their background and environment, then 32 their chances for the future.The students, after consulting social statistics, talking to the boys, and collecting much data, concluded that 90 percent of the boys would 33 some time in prison.Twenty-five years later 34 group of graduate students was given the job of35 the prediction. They went back to the same area. Some of the boys---by then36 ---were still there, a few had died, some had moved away, but they got in touch with 180 of the 37 200.They found that only four of the group had ever been sent to 38 .Why was it that these men, who had lived in a breeding place of crime, had such a surprisingly good record? The researchers were 39 told,“Well, there was a teacher…”They pressed〔追问〕further, and found that in 75 percent of the cases it was the same woman. The researchers went to this teacher, now living in a home for retiredteachers. How had she had this brilliant 40 on that group of children? Could she give them any reason why these boys should have remembered her?“No,” she said,“No,I really couldn’t.”And then, thinking back 41 the years, she said musingly〔沉思地〕,42 to herself than to her questioners,“Iloved those boys…”31. A. look into B. make up C. show up D. result in32. A. control B. arrange C. desert D. predict33. A. cost B. spend C. take D. waste34. A. others B. the other C. another D. other35. A. cursing B. convincing C. applying D. testing36. A. teachers B. students C. men D. professors37. A. native B. rare C. recorded D. discouraged38. A. slums B. society C. prison D. school39. A. continually B. optimistically C. curiously D.officially40. A. reflection B. harmony C. effect D. existence41. A. before B. after C. in D. over.42. A. less B. more C. better D. worse SECTION C (8 points)Directions: Complete the passage below with the correct forms of the given words and phrases in the column (two are extra).Last summer holiday, Colin and Toby spent a few weeks traveling in Africa. The two British brothers chose Africa as their tourist 43.________ because they considered it to be a place of mystery and 44. ________. They made sure that they 45. ________ everything well before they started their adventure.Their first extraordinary experience was riding camels through the Sahara Desert. In the desert, they were more than pleased to see the 46. ________ stars on clear nights. Then, they traveled down the River Nile and experienced the exciting white-water rafting in the 47. ______ water. Afterwards, they went on a trip to see wild animals in Kenya. During the weeks of walking in Kenya, Colin and Toby felt lucky to have prepared a large backpack 48. ________ to carry the supplies of water and food. As they were very curious to see the wild animals up close, they brought guns with them 49. ________ the animals came too close. Then, they moved on to climb Mount Kilimanjaro and enjoyed the splendid scenery there.50. ________, Colin and Toby’s adventure in Africa lasted for about a month. Tired but delighted, they knew that they would bear it in mind forever. SECTION D (10 points)Directions: Read the passage carefully and then translate the underlined sentences into English or Chinese. Write your answer on your answer sheet.Howard Carter is one of the most famous explorers the world has ever known. He was brave and loved to visit and explore new places. During his life, he discovered many amazing things.51.他天资聪颖,对家乡之外的世界充满了好奇心when he was very young.Howard Carter never went to school, but learnt to draw from his father. 52. By the 1920s, he had become an explorer, searching for the tombs of the Egyptian kings. He discovered a great fortune in jewels and gold, along with the preserved bodies of dead kings. 53. 这些通过处理的遗体确实是为人所知的木乃伊. In 1922, Howard Carter made his most amazing discovery of all, in the Valley of the Kings, in Egypt, where he found the tomb of King Tutankhamun.However, not long after the tomb had been opened, people in Carter’s team began to fall ill and die strangely. 54.Within seven years, 21 people who had something to do with the opening of the tomb died.Some people say the deaths were just coincidence. 55.Others believe that they were in connection with a mummy’s curse. However, still others believe that there is a scientific explanation, because inside the tombs there are many viruses. PART THREE :READING COMPREHENSION (30 points)Directions: Read the following three passages. Each passages is followed by several questions or unfinished statements. For each of them there are four choices marked with A, B, C and D. Choose the one that fits best according to the information given in the passage.ATwo years ago my grandmother was going to turn 75. My family discussed what the best way to celebrate was. Should we throw her a party? Should we take her on a trip? We remembered that she had touched so many people's lives, and there were so many people for her to consider. Then someone got the idea that we should include everyone in the celebration by turning it into a tribute(献礼) to my grandmother.We secretly sent out letters to the people in Grandmother's address book and asked them to send a letter with a memory that they had shared with her. People sent us letters with poems, stories and pictures. The deep feeling that was shared through the response (回应) surprised us. We compiled(编辑)these letters into a memory book and amazed her with it on the morning of her birthday.The unusual thing about my grandmother's friends was not the number that she had, but the connection they shared. In many ways this book of friendship was the greatest achievement of my grandmother's life.I believe that developing true friendships is one of the most important things that anyone can do in one's lifetime. It is not a matter of the number of friends one has, but the quality of the bonds. If one has had at least one true friendship before dying, then one can say one has lived a successful life. I have made many friends and I believe I have begun to develop the same types of friendships my grandmother kept up over her lifetime. I only hope that I will be as successful as she has been.56. How did the author's family celebrate Grandmother's birthday ?A. They took her on a trip across the country.B. They gave her a memory book of friendship.C. They invited all her friends to her birthday party.D. They asked all her friends to send her cards.57. When receiving her birthday gift, the author's grandmother probably felt _____.A. disappointed and lonelyB. sorry and sadC. surprised and pleasedD. nervous and excited58. The underlined word “bonds” in the last paragraph probably means________A. relationshipsB. worksC. successesD. celebrations59. According to the passage, the author probably agrees that _____.A. the more friends you have, the betterB. friends are more important than familyC. understanding leads to greater successD. true friendship is very important to us60. Which of the following words can best describe the author's grandmother ?A. FamousB. Great.C. PoliteD. PleasantBInternational students in Christine Rhodes’ English class in Australia share their favorite places they have visited in Australia.Canberra, the nicest city in Australia.Siggi Siebold from GermanyI went for a short holiday to Canberra because my son lives there. It’s the nicest city in Australia, a little bit similar to Cairns. There’re lots of big rivers and trees around the city.The small city of Mount IsaIndra Ekanayake from MalaysiaI came to Mount Isa six months ago. It’s a long way from Brisbane-about 2,000 km .It’s a small city, with a population of 21,000, but it has all the basic facilities(设施)such as a hospital, six schools and supermarkets. It’s a mining city. It’s hot and gets little rain. There’re two big lakes to keep the rainwater for drinking for the city.Lake Moondarra and a mineIda Robb from IndonesiaIn Mount Isa, there’s a dam called Lake Moondarra. On the weekend you can go there with your friends. People go fishing and even catch a crocodile. There’s a big mine in Mount Isa. Many people come to work in the mine and earn much money. It isn’t a beautiful green place, but many people love it.My impression(印象)of SydneyNamfon Pitaxsin from ThailandThe fist time I came to Australia, I was happy. I traveled to Sydney. Four million people live there ,and many are Asians. I saw the harbor (港口),Sydney Harbor Bridge and the Opera House and I went to see the koalas and kangaroos in Taronga Park Zoo. After that I went to the Thai restaurants. There is a revolving (旋转的) restaurant, so you can eat and see the city.61. Why did Siggi Siebold go to Canberra?A. Because it’s the nicest city in Australia.B. Because her son lives thereC. Because it is similar to her hometownD. Because there’re lots of rivers62. According to Indra Ekanayake, Mount Isa______A. is a famous harbor in AusraliaB. has a large populationC. probably doesn’t have enough drinking waterD. doesn’t have enough basic facilities63. Where is Lake Moondarra?A. In Canberra.B. In Mount Isa.C. In Brisbane.D. In Sydney64. The first time Namfon Pitaxsin went to Australia, she________A. saw some special Australian animalsB. met more Americans than AsiansC. ate in a revolving restaurantD. enjoyed the music in the Opera House65.How many cities does the passage mainly talk about?A. TwoB. ThreeC. Four.D. FiveCPublic schools in Washington, D.C. provide students with musical instruments for free. When something goes wrong with an instrument, Charles West and Larry Jernigan do the repairs. Both men approach their work with a passion(激情). For them, it’s important that students have a joyful experience with music.The two have worked together for almost 20 years. This year alone, they’ve fixed about 450 instruments. Both men are musicians and music lovers, so learning to do repairs came naturally.“I have been a musician all my life.” says West. “I played in an orchestra here in the city. I majored in music in college. I played in an army band.”Jernigan’s musical interests are varied. “I was formerly trained in the piano and guitar. The alto sax, and the flute, I picked up while working here.”In addition to fixing instruments, the two also go to schools to instruct teachers and students on how to make minor repairs on their own.West believes if children start early and stay concerned with music, it enriches other areas of their lives. “I see that in other kids. I see it in myself. I have seen it hundreds of times and it works,” he says. “They learn teamwork. They learn patience and respect.”But West has concerns about the future of music in the electronic age.“This instant age has taken away from the sit-down, the patience. And to learn to play an instrument, it takes patience, it takes diligence, it takes time.”Being able to enjoy music on the job is one of the benefits of the job. Both men agree their best rewards are the students’ performances.66. What’s the job of West and Jernigan at school?A. Teaching music.B. Writing music.C. Making musical instruments.D. Repairing musical instruments.67. They love the job because they can .A. earn more moneyB. learn repair skillsC. enjoy musicD. watch performances68. Which of the following is true of the two men?A. They have fixed 450 instruments in the past 20 years.B. They can play and repair musical instruments.C. Jernigan used to play in an army band.D. West was trained to play the piano.69. According to West, what can people learn from music?A. Teamwork and patience.B. The value of time.C. The truth of society.D. Diligence and confidence.70. What is mainly talked about in the text?A. How to repair musical instruments.B. Learning experiences of two repairmen.C. How to prepare a musical performance.D. The enjoyable job of two music lovers.必考II部分PART FOUR WRITING 〔50 marks〕Section A (10 points)Directions: Complete the following sentences according to the requirement. 71. The sunshine reflects on the lakes, which makes them shine like diamonds. (改为含分词结构的句子,不改变原意)72. His hard work wins him respect from others. (使用强调句型改写)73. People who had discovered how to stay young forever lived in this perfect world. (改倒装句)74. Don’t forget small things, such as waterproof matches. You may need them to make a fire.(使用定语从句改写句子)75. People often say a friend in need is a friend indeed. (使用it句型改写) Section B (10 points)Directions: Read the following passage. Complete the diagram by using the information for the passage. Write NO MORE THAN 3 WORDS for each answer.If we agree that the function of education is to prepare us for life, then there is very little time to waste. So, while we can, we ought to concentrate on teaching children something really useful. Here is what our schools should teach.Politeness is a mark of civilization. The sooner children learn this, the better. In any case, a lot can be accomplished by a smile and good manners.Like it or not, our adult lives will be consumed by the struggle for money, but we don’t make an effort to teach children how to manage it. So our schools have a duty to teach them this ability from the beginning.We’re likely to accept something we are told, but that’s not what educated people do. Educated people are reasonable and they look at facts. If our schools teach nothing else, they should at least teach critical〔批判性的〕thinking.Children should learn to take care of their health. They should know that if they eat junk food, they will become fat and unhealthy. They should be very clear about what happens to their bodies when they drink or smoke.All of us are part of society. We have rights and responsibilities. We ought to understand what they are. We have to know a little bit of history and geography, because we need to have an environment in which to relate to the people around us.How will we test students on these? We can’t. But that’s not a reason to avoid teaching what is important. Our schools should spend every moment they have telling this to our children: ”This is life, this is what you are going to face, and thisSection C (10 points)Directions: Read the following passage. Answer the questions according to the information given in the passage and required words limit. Write your answers on your answer sheet.Abraham Lincoln was born in Kentucky on February 12, 1809. When he was a small boy, his family moved to Indiana. Here, his mother taught him to read and write. Lincoln had very little formal education, but he became one of the best-educated men of the Great West.When Lincoln was a young man, his family moved to the new state of Illinois. Lincoln had to make a living at an early age, but in his spare time he studied law. He soon became one of the best-known lawyers in the state capital at Springfield, Illinois. It was here that Lincoln became famous for his debates with Stephen A. Douglas on the problem of slavery (奴隶制度).In 1860, Lincoln was elected President of the United States. He was the candidate of the new Republican Party. This party is against the creation of new slave states. Soon after his election, some of the Southern states quit from the Union and set up the Confederate States of America(美利坚联盟国). This action resulted in the terrible Civil War which lasted from 1861 to 1865.On January 1,1863, during the war, Lincoln published his famous Emancipation Proclamation(解放宣言). In this file, Lincoln announced that all the slaves were to be free from that day. In 1865, after the war ended, another file was added to end the slavery everywhere in the United States. Early in 1865, the Civil War came to an end with the defeat of the South by the North.Only a few days after the end of the War, Lincoln was shot by an actor named John Wilkes Booth. The President died on April 14, 1865. In his death, the world lost one of the greatest men of all time.86. Who taught Lincoln to read and write in his childhood? (no more than 10 words) (2 marks)87. When did Lincoln become the president of the U.S.? (no more than 3 words) (2 marks)88. What does the underlined phrase “this action” in Paragraph Three refer to?(no more than 20 words) (3 marks)89. How did the Civil War come to an end? (no more than 10 words) (3 marks) Section D (20 points)Directions: Write an English composition according to the instructions given below in Chinese.你的笔友Bob将从加拿大来长沙看你。
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湖南师大附中2018-2019高一第一学期数学期末考试
数 学
时量:120分钟 满分:150分
一、选择题(共12小题,每小题5分,共60分)
1.已知集合{}1,1,0M =-,{}0,1,2N =,则M N =I ( )
A.{}0,1
B.{}1,0,2-
C.{}1,0,1,2-
D.{}1,0,1- 2.若直线过点()1,2A -,()3,243B +,则此直线的倾斜角为( )
A.30o
B.45o
C.60o
D.90o 3.已知两条直线a ,b ,两个平面α,β,下面四个命题中不正确的是( )
A.a α⊥,//αβ,b a b β⊂⇒⊥
B.//αβ,//a b ,a b αβ⊥⇒⊥
C.a α⊥,//b a b α⊥⇒
D.//a b ,////a b αα⇒
4.函数21x y =-的定义域为( )
A.(),0-∞
B.[)0,+∞
C.(),1-∞-
D.()1,+∞ 5.在空间直角坐标系中,点()0,2,1A ,()2,0,2B ,点A 关于平面xOy 对称的点为A ',则A ',B 两点间的距离A B '为( )
A.32
B.17
C.4
D.3
6.如图,OAB ∆是边长为2的正三角形,记OAB ∆位于直线()0x t t =>,左侧的图形的面积为()f t .则函数()y f t =的图像为( )
A. B.
C. D.
7.如图,将一个长方体1111ABCD A B C D -由四个面的对角线截出一个棱锥11A B CD -,
则棱锥11A B CD -的体积与长方体1111ABCD A B C D -的体积之比是( )
A.2:3
B.1:2
C.1:3
D.1:6
8.设函数()()()ln 1ln 1f x x x =+--,则()f x 是( )
A.奇函数,且在()0,1上是增函数
B.奇函数,且在()0,1上是减函数
C.偶函数,且在()0,1上是增函数
D.偶函数,且在()0,1上是减函数 9.圆心在直线13y x =上的圆C 与y 轴的正半轴相切,圆C 截x 轴所得的弦长为42,则圆C 的标准方程为( )
A.()()22319x y -+-=
B.()()22319x y +++=
C.()2244163x y ⎛⎫-+-= ⎪⎝⎭
D.()()22629x y -+-= 10.底面为正方形的直四棱柱1111ABCD A B C D -中12AA AB =,则异面直线1CD 与11A C 所成角的余弦值为( )
A.2
B.12
C.3
D.10 11.若动点()11,M x y ,()22,N x y 分别在直线1:20l x y -+=,2:100l x y -+=上移动,则MN 中点Q 到原点距离的最小值为( )
A.23
B.32
C.33
D.42 12.设函数()x x x f x a b c =+-,其中0c a >>,0c b >>,若a ,b ,c 是ABC ∆的三条边长.
现有下列命题:
①任意(),1x ∈-∞,()0f x >
②若a b =,则()f x 的零点的取值范围为()0,1
③若222a b c +<,则存在()1,2x ∈,使得()0f x =.
其中所有正确命题的序号为( )
A.①
B.①③
C.②③
D.①②③
二、填空题(本大题共4个小题,每小题5分,共20分)
13.已知直线l 过点()2,1-且与直线230x y -+=垂直,则直线l 的方程为___________(请用直线方程的一般式表示).
14.已知11225x x -+=,则1x x -+=___________.
15.已知()1,2B -、()5,1C ,若CAB ∠的平分线在1y x =+上,则AC 所在直线方程是___________.
16.如图,在长方形ABCD 中,2AB =,1BC =,E 为DC 的中点,F 为线段EC (端点除外)上一动点.现将AFD 沿AF 折起,使平面ABD ⊥平面ABC .在平面ABD 内过点D 作DK AB ⊥,K 为垂足.设AK t =,则t 的取值范围是___________.
三、解答题(本大题共6个小题,共70分)
17.(本小题满分10分)
已知函数()f x 是定义域为R 的奇函数,且当0x >时,()2
2f x x x =-,现已画出函数()f x 在y 轴右侧的图象,如图所示:
(Ⅰ)请补出完整函数()y f x =的图象;
(Ⅱ)根据图象写出()f x 的单调区间.
如图,四棱锥E ABCD -中,平面EBA ⊥平面ABCD ,侧面ABE 是等腰直角三角形,EA EB =,//AB CD ,AB BC ⊥,222AB CD BC ===.
(Ⅰ)求证:AB ED ⊥;
(Ⅱ)求直线CE 与平面ABCD 的所成角的正弦值.
19.(本小题满分12分)
已知圆M 过()2,2A ,()6,0B ,且圆心在直线40x y --=上.
(Ⅰ)求此圆的方程;
(Ⅱ)求与直线350x y -+=垂直且与圆相切的直线方程;
(Ⅲ)若点P 为圆M 上任意点,求ABP ∆的面积的最大值.
20.(本小题满分12分)
如图所示,平面ABCD ⊥平面BCEF ,且四边形ABCD 为矩形,四边形BCEF 为直角梯形,//BF CE ,BC CE ⊥,4DC CE ==,2BC BF ==.
(Ⅰ)求证//AF 平面CDE ;
(Ⅱ)求二面角A EF C --的正切值.
如图,某海面上有O 、A 、B 三个小岛(面积大小忽略不计),A 岛在O 岛的东北方向202km 处,B 岛在O 岛的正东方向10km 处.
(Ⅰ)以O 为坐标原点,O 的正东方向为x 轴正方向,1km 为单位长度,建立平面直角坐标系,试写出A 、B 的坐标,并A 、B 两岛之间的距离;
(Ⅱ)已知在经过O 、A 、B 三个点的圆形区域内有未知暗礁,现有一船在O 岛的南偏西30o 方向距O 岛20km 处,正沿东北方向行驶,若不改变方向,试问该船有没有触礁的危险?
22.(本小题满分12分)
已知0a >,函数()21log f x a x ⎛⎫=+ ⎪⎝⎭
. (Ⅰ)证明:()f x 在()0,+∞上单调递减;
(Ⅱ)若对任意1,12t ⎡⎤∈⎢⎥⎣⎦,函数()f x 在区间[],1t t +上的最大值与最小值的差不超过1,求a 的取值范围;
(Ⅲ)若关于x 的方程()()2log 4250f x a x a --+-=⎡⎤⎣⎦的解集中恰好有一个元素,求a 的取值范围.。