湖南省长望浏宁四县2018届高三联合调研考试数学(文)试卷(含答案)
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2018届高三理综联合调研试卷(长望浏宁四县有答案)
5 Pa~Pβ~Pγ)。
下列说法错误的
A.已知某种酶可以催化ATP的一个磷酸基团转移到DNA末端,若要使已有DNA末端被32P标记上,则带有32P的磷酸基团应在ATP 的“γ”位上
B.若用带有32P标记的dATP作为DNA生物合成的原料将32P标记到新合成的DNA分子上,则带有32P的磷酸基团应在dATP的“γ”位上
c.ATP中两个磷酸基团被水解掉后所得的物质是合成RNA分子的单体之一
D.dATP中的两个高能磷酸键储存的能量基本相同,但稳定性不同
4.某雌雄同株的植物的花色由位于2号染色体上的一对等位基因控制,A控制红色,a控制白色。
某杂合的红花植株经射线处理后2号染色体缺失了一个片段,含有片段缺失染色体的雄配子不能正常发育,该植株自交产生的后代中红花白花=11。
下列有关说法正确的是
A.A基因所在染色体发生了片段缺失,且A基因位于缺失的片段上
B.A基因所在染色体发生了片段缺失,且A基因不位于缺失的片段上
c.a基因所在染色体发生了片段缺失,且a基因位于缺失的片段上
D.a基因所在染色体发生了片段缺失,且a基因不位于缺失的片段上
5.豌豆种子刚萌发时,只有一个主茎,没有侧枝;切除主茎顶端,常会长出甲、乙两个侧枝(图①)。
若将甲的顶芽切除后,甲、乙。
2018届高三十四校联考第一次考试试卷数学(文科) 第Ⅰ卷(共60分)一、选择题:本大题共12个小题,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.已知集合{}1,2,3A =,{}2,4B =,则A B =( )A .{}2B .{}4C .{}1,2D .{}1,2,3,42.已知θ的始边与x 轴非负半轴重合,终边上存在点(1,)P a -且sin 2θ=,则a =( )A .1-B .1C .D 3.复数z 满足23i z i ⋅=+,则||z =( )A BC D 4.若三棱锥的三视图如图所示,则该三棱锥的四个面中直角三角形的个数是( )A .1B .2C .3D .45.在区间[]2,3-上随机取一个数x ,则满足|1|1x -≤的概率是( ) A .15B .25C .35D .456.我国古代数学著作《九章算术》有如下问题:“今有金箠,长五尺,斩本一尺,重四斤,斩末一尺,重二斤,问次一尺各重几何?”意思是:“现有一根金杖,长5尺,一头粗,一头细,在最粗的一端截下1尺,重4斤;在最细的一端截下1尺,重2斤;问依次每一尺各重多少斤?”设该金杖由粗到细是均匀变化的,其总重量为W ,则W 的值为( )A .4B .12C .15D .187.已知双曲线方程为2212015x y -=,则该双曲线的渐近线方程为( )A .34y x =±B .43y x =±C .y x =D .y x = 8.某程序框图如图所示,该程序运行后输出的值是( )A .1011B .511C .89D .499.已知定义在R 上的奇函数()f x 满足当0x >时,()224xf x x =+-,则()f x 的零点个数是( ) A .2B .3C .4D .510.如图,已知边长为2的正方体1111ABCD A BC D -,点E 为线段1CD 的中点,则直线AE 与平面11A BCD 所成角的正切值为( )A.2B .12C.2D11.已知函数()2sin cos (0)f x x x ωωω=->,若()f x 的两个零点1x ,2x 满足12min ||2x x -=,则(1)f 的值为( )AB. C .2 D .2-12.已知函数()f x 是定义在R 上的奇函数,其导函数为'()f x ,若对任意的正实数x ,都有'()2()0xf x f x +>恒成立,且1f =,则使2()2x f x <成立的实数x 的集合为( )A.(,(2,)-∞+∞B.(C .(-∞D .)+∞第Ⅱ卷(共90分)二、填空题(每题5分,满分20分,将答案填在答题纸上)13.已知矩形ABCD 的边2AB =,1AD =,则BD CD ⋅= .14.若实数x ,y 满足约束条件2,6,0,x x y x y ≥⎧⎪+≤⎨⎪-≤⎩则目标函数23z x y =-的最大值是 .15.在ABC ∆中,a ,b ,c 分别是内角A ,B ,C 的对边,sin cos (cos )sin 0A B c A B --⋅=,则边b = .16.已知在三棱锥P ABC -中,90BAC ∠=︒,2AB AC ==,BC的中点为M 且PM =,当该三棱锥体积最大时,它的内切球半径为 .三、解答题 (本大题共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.已知等比数列{}n a 满足12a =且235a a a ⋅=. (1)求{}n a 的通项公式;(2)设n n b a n =-,求{}n b 的前n 项和n S .18.已知某班的50名学生进行不记名问卷调查,内容为本周使用手机的时间长,如表:(1)求这50名学生本周使用手机的平均时间长;(2)时间长为[0,5)的7名同学中,从中抽取两名,求其中恰有一个女生的概率; (3)若时间长为[0,10)被认定“不依赖手机”,[]10,25被认定“依赖手机”,根据以上数据完成22⨯列联表:能否在犯错概率不超过0.15的前提下,认为学生的性别与依赖手机有关系?(参考公式:22()()()()()n ad bc K a b c d a c b d -=++++,n a b c d =+++)19.在四棱锥P ABCD -中,//AB CD ,24CD AB ==,60ADC ∠=︒,PAD ∆是一个边长为2的等边三角形,且平面PAD ⊥平面ABCD ,M 为PC 的中点.(1)求证://BM 平面PAD ; (2)求点M 到平面PAD 的距离.20.在平面直角坐标系中,动点(,)M x y (0x ≥)到点(1,0)F 的距离与到y 轴的距离之差为1.(1)求点M 的轨迹C 的方程;(2)若(4,2)Q -,过点(4,0)N 作任意一条直线交曲线C 于A ,B 两点,试证明QA QB k k +是一个定值. 21.已知函数3211()332f x ax x x =+--(a 为实数). (1)当()f x 与3y =-切于00(,())A x f x ,求a ,0x 的值;(2)设()'()x F x f x e =⋅,如果()1F x >-在(0,)+∞上恒成立,求a 的范围. 请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分. 22.选修4-4:坐标系与参数方程以原点O 为极点,x 轴的非负半轴为极轴建立极坐标系,已知曲线C 的极坐标方程为:4sin ρθ=,在平面直角坐标系xOy 中,直线l的方程为1,22x y ⎧=-+⎪⎪⎨⎪=⎪⎩(t为参数). (1)求曲线C 和直线l 的直角坐标方程;(2)已知直线l 交曲线C 于A ,B 两点,求A ,B 两点的距离. 23.选修4-5:不等式选讲已知函数()|2||1|f x x x =++-. (1)求证:()3f x ≥;(2)求不等式2()f x x ≥的解集.2018届高三十四校联考第一次考试数学(文科)试卷答案一、选择题1-5:ABADB 6-10:CCBBA 11、12:CC 二、填空题13.4 14.2- 15.1 16.三、解答题17.解:(1)因为12a =且235a a a ⋅=,所以2q =, 从而2n n a =.(2)由(1)得2n n n b a n n =-=-, ∴23(2222)(123)nn S n =++++-++++……2(12)(1)(1)2(21)1222n n n n n n -++=-=---. 18.解:(1)1(2.577.52812.5917.5522.51)950⨯+⨯+⨯+⨯+⨯=, 所以,这50名学生本周使用手机的平均时间长为9小时.(2)时间长为[0,5)的有7人,记为A 、B 、C 、D 、E 、F 、G ,其中女生记为A 、B 、C 、D ,从这7名学生中随机抽取两名的基本事件有:{},A B ,{},A C ,{},A D ,{},AE ,{},A F ,{},A G ,{},B C ,{},B D ,{},B E ,{},B F ,{},B G ,{},C D ,{},C E ,{},C F ,{},C G ,{},D E ,{},D F ,{},D G ,{},E F ,{},E G ,{},F G 共21个.设事件M 表示恰有一位女生符合要求的事件有:{},A E ,{},A F ,{},A G ,{},B E ,{},B F ,{},B G ,{},C E ,{},C F ,{},C G ,{},D E ,{},D F ,{},D G 共12个.所以恰有一个女生的概率为24()217P M 1==. (3)2250(1510520)0.397 2.07215352030K ⨯-⨯=≈<⨯⨯⨯,不能在犯错概率不超过0.15的前提下,认为学生的性别与依赖手机有关系. 19.(1)证明:过M 作//MN CD ,交PD 于点N ,连接AN , 可知1//2MN CD ,而1//2AB CD , 所以//MN AB ,从而四边形ABMN 为平行四边形, 所以//AN BM ,又AN ⊂平面PAD ,BM ⊄平面PAD, 所以//BM 平面PAD .(2)由(1)可知M 到平面PAD 的距离等于B 到平面PAD 的距离, 设B 到平面PAD 的距离为h , 由B PAD PABD V V --=,∴1133PAD ABD S h S ∆∆⋅⋅=⋅h = 故M 到平面PAD20.解:(1)M 到定点(1,0)F 的距离与到定直线1x =-的距离相等, ∴M 的轨迹C 是一个开口向右的抛物线,且2p =, ∴M 的轨迹方程为24y x =.(2)设过(4,0)N 的直线的方程为4x my =+,联立方程组24,4,y x x my ⎧=⎨=+⎩整理得24160y my --=,设直线l 与抛物线的交点为11(,)A x y ,22(,)B x y , 则有124y y m +=,1216y y =-, 又212122121222228321448816642QA QBy y y y m k k x x my my m ------+=+=+==-+++++,因此QA QB k k +是一个定值为12-. 21.解:(1)2'()1f x ax x =+-, 由()f x 与3y =-切于点00(,())A x f x ,则320000200011()33,32'()10,f x ax x x f x ax x ⎧=+--=-⎪⎨⎪=+-=⎩解得316a =-,04x =. (2)2()(1)x F x ax x e =+-⋅,∴2'()((21))x F x e ax a x =⋅++,且(0)1F =-.①当0a =时,'()x F x x e =⋅,可知()F x 在(0,)+∞递增,此时()1F x >-成立; ②当102a -<<时,21'()()x a F x e ax x a +=⋅+,可知()F x 在21(0,)a a+-递增,在21(,)a a +-+∞递减,此时11()1a F e a--=-<-,不符合条件;③当12a =-时,21'()()02xF x e x =⋅-<恒成立,可知()F x 在(0,)+∞递减,此时()1F x <-成立,不符合条件; ④当12a <-时,21'()()xa F x e ax x a+=⋅+,可知()F x 在(0,)+∞递减,此时()1F x <-成立,不符合条件;⑤当0a >时,21'()()xa F x e ax x a+=⋅+,可知()F x 在(0,)+∞递增,此时()1F x >-成立. 综上所述,0a ≥.22.解:(1)由题知,曲线C 化为普通方程为22(2)4x y +-=, 直线l 的直角坐标方程为10x y -+=.(2)由题知,直线l的参数方程为1,22x y t ⎧=-+⎪⎪⎨⎪=⎪⎩(t 为参数), 代入曲线C :22(2)4x y +-=中,化简,得210t -+=,设A ,B 两点所对应的参数分别为1t ,2t,则12121,t t t t ⎧+=⎪⎨⋅=⎪⎩所以21||t t -A ,B23.解:(1)证明:()|2||1||(2)(1)|3f x x x x x =++-≥+--=.(2)21,2,()3,21,21,1,x x f x x x x --≤-⎧⎪=-<<⎨⎪+≥⎩所以22,21,x x x ≤-⎧⎨--≥⎩或221,3,x x -<<⎧⎨≥⎩或21,21,x x x ≥⎧⎨+≥⎩解得1x ≤≤故解集为{|1x x ≤.。
湖南省2018年高考文科数学试题及答案(Word 版)(试卷满分150分,考试时间120分钟)注意事项:1.答卷前,考生务必将自己的姓名和准考证号填写在答题卡上。
2.回答选择题时,选出每小题答案后,用铅笔把答题卡对应题目的答案标号涂黑。
如需改动,用橡皮擦干净后,再选涂其它答案标号。
回答非选择题时,将答案写在答题卡上。
写在本试卷上无效。
3.考试结束后,将本试卷和答题卡一并交回。
一、选择题:本题共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.已知集合{}02A =,,{}21012B =--,,,,,则A B =A .{}02,B .{}12,C .{}0D .{}21012--,,,, 2.设1i2i 1iz -=++,则z = A .0B .12C .1D .23.某地区经过一年的新农村建设,农村的经济收入增加了一倍.实现翻番.为更好地了解该地区农村的经济收入变化情况,统计了该地区新农村建设前后农村的经济收入构成比例.得到如下饼图:则下面结论中不正确的是 A .新农村建设后,种植收入减少B .新农村建设后,其他收入增加了一倍以上C .新农村建设后,养殖收入增加了一倍D .新农村建设后,养殖收入与第三产业收入的总和超过了经济收入的一半4.已知椭圆C :22214x y a +=的一个焦点为(20),,则C 的离心率为 A .13B .12C .22D .2235.已知圆柱的上、下底面的中心分别为1O ,2O ,过直线12O O 的平面截该圆柱所得的截面是面积为8的正方形,则该圆柱的表面积为 A .122πB .12πC .82πD .10π6.设函数()()321f x x a x ax =+-+.若()f x 为奇函数,则曲线()y f x =在点()00,处的切线方程为A .2y x =-B .y x =-C .2y x =D .y x =7.在△ABC 中,AD 为BC 边上的中线,E 为AD 的中点,则EB = A .3144AB AC - B .1344AB AC - C .3144AB AC +D .1344AB AC + 8.已知函数()222cos sin 2f x x x =-+,则 A .()f x 的最小正周期为π,最大值为3 B .()f x 的最小正周期为π,最大值为4 C .()f x 的最小正周期为2π,最大值为3 D .()f x 的最小正周期为2π,最大值为49.某圆柱的高为2,底面周长为16,其三视图如右图.圆柱表面上的点M 在正视图上的对应点为A ,圆柱表面上的点N 在左视图上的对应点为B ,则在此圆柱侧面上,从M 到N 的路径中,最短路径的长度为A .217B .25C .3D .210.在长方体1111ABCD A B C D -中,2AB BC ==,1AC 与平面11BB C C 所成的角为30︒,则该长方体的体积为 A .8B.C.D.11.已知角α的顶点为坐标原点,始边与x 轴的非负半轴重合,终边上有两点()1A a ,,()2B b ,,且2cos 23α=,则a b -= A .15B.5C.5D .112.设函数()201 0x x f x x -⎧=⎨>⎩,≤,,则满足()()12f x f x +<的x 的取值范围是A .(]1-∞-,B .()0+∞,C .()10-,D .()0-∞,二、填空题(本题共4小题,每小题5分,共20分)13.已知函数()()22log f x x a =+,若()31f =,则a =________.14.若x y ,满足约束条件220100x y x y y --⎧⎪-+⎨⎪⎩≤≥≤,则32z x y =+的最大值为________.15.直线1y x =+与圆22230x y y ++-=交于A B ,两点,则AB =________.16.△ABC 的内角A B C ,,的对边分别为a b c ,,,已知sin sin 4sin sin b C c B a B C +=,2228b c a +-=,则△ABC 的面积为________.三、解答题:共70分。
2018年普通高等学校招生全国统一考试4月调研测试卷 文科数学文科数学测试卷共4页。
满分150分。
考试时间120分钟。
第Ⅰ卷一、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备选项中,只有一项是符合题目要求的.(1)设全集U R =,集合{1012}A =-, , , ,2{|log 1}B x x =<,则()U A B =I (A ){12},(B ){102}-, ,(C ){2}(D ){10}-,(2)复数z 满足(12i)3i z +=+,则=z(A )1i - (B )1i +(C )1i 5- (D )1i 5+ (3)设等差数列{}n a 的前n 项和为n S ,若73=a ,123=S ,则=10a(A )10(B )28(C )30(D )145(4)已知两个非零向量a r ,b r 互相垂直,若向量45m a b =+u r r r 与2n a b λ=+r r r共线,则实数λ的值为(A )5 (B )3(C )2.5 (D )2(5)“1cos 22α=”是“ππ()6k k Z α=+∈”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(6)执行如图所示的程序框图,如果输入的[22]x ∈-, ,则输出的y 值的取值范围是(A )52y -≤或0y ≥ (B )223y -≤≤(C )2y -≤或203y ≤≤(D )2y -≤或23y ≥(7)曲线250xy x y -+-=在点(12)A , 处的切线与两坐标轴所围成的三角形的面积为(A )9(B )496(C )92(D )113(8)已知定义在R 上的奇函数()y f x =满足(2)()f x f x +=-,且(1)2f =,则(2018)(2019)f f +的值为(A )2-(B )0(C )2 (D )4CA BD(9)如图,在矩形ABCD 中,2AB =,3AD =,两个圆的半径都是1,且圆心12O O ,均在对方的圆周上,在矩形ABCD 内随机取一点,则此点取自阴影部分的概率为 (A(B(C (D (10)设函数6cos y x =与5tan y x =的图象在y轴右侧的第一个交点为A ,过点A 作y 轴的平行线交函数sin 2y x =的图象于点B ,则线段AB 的长度为(A(B )2(C(D )(11)某几何体的三视图如图所示,其正视图为等腰梯形,则该几何体的表面积是(A )18(B )8+(C )24(D )12+(12)设集合22{()|(3sin )(3cos )1}A x y x y R ααα=+++=∈, , ,{()|34100}B x y x y =++=, ,记P A B =I ,则点集P 所表示的轨迹长度为 (A )(B )(C )(D )第Ⅱ卷本卷包括必考题和选考题两部分。
绝密★启用前2018年普通高等学校招生全国统一考试文科数学注意事项:1.答卷前,考生务必将自己的姓名、准考证号填写在答题卡上。
2.回答选择题时,选出每小题答案后,用铅笔把答题卡上对应题目的答案标号涂黑。
如需改动,用橡皮擦干净后,再选涂其它答案标号。
回答非选择题时,将答案写在答题卡上。
写在本试卷上无效。
3.考试结束后,将本试卷和答题卡一并交回。
一、选择题:本题共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.已知集合{0,2}A =,{2,1,0,1,2}B =--,则A B =I A .{0,2} B .{1,2}C .{0}D .{2,1,0,1,2}--2.设1i2i 1iz -=++,则||z =A .0B .12C .1D 3.某地区经过一年的新农村建设,农村的经济收入增加了一倍,实现翻番. 为更好地了解该地区农村的经济收入变化情况,统计了该地区新农村建设前后农村的经济收入构成比例,得到如下饼图:则下面结论中不正确的是 A .新农村建设后,种植收入减少B .新农村建设后,其他收入增加了一倍以上C .新农村建设后,养殖收入增加了一倍D .新农村建设后,养殖收入与第三产业收入的总和超过了经济收入的一半4.已知椭圆22214x y C a +=:的一个焦点为(2,0),则C 的离心率为A .13B .12C .2D .35.已知圆柱的上、下底面的中心分别为1O ,2O ,过直线12O O 的平面截该圆柱所得的截面是面积为8的正方形,则该圆柱的表面积为A .B .12πC .D .10π6.设函数32()(1)f x x a x ax =+-+. 若()f x 为奇函数,则曲线()y f x =在点(0,0)处的切线方程为 A .2y x =-B .y x =-C .2y x =D .y x =7.在ABC △中,AD 为BC 边上的中线,E 为AD 的中点,则EB =u u u ru u u r u u u r u u ur u u u rC .3144AB AC +u u ur u u u r D .1344AB AC +u u ur u u u r 8.已知函数22()2cos sin 2f x x x =-+,则 A .()f x 的最小正周期为π,最大值为3 B .()f x 的最小正周期为π,最大值为4 C .()f x 的最小正周期为2π,最大值为3 D .()f x 的最小正周期为2π,最大值为49.某圆柱的高为2,底面周长为16,其三视图如右图.圆柱表面上的点M 在正视图上的对应点为A ,圆柱表面上的点N 在左视图上的对应点为B ,则在此圆柱侧面上,从M 到N 的路径中,最短路径的长度为A .217B .25C .3D .210.在长方体1111ABCD A B C D -中,2AB BC ==,1AC 与平面11BB C C 所成的角为30︒,则该长方体的体积为A .8B .62C .82D .8311.已知角α的顶点为坐标原点,始边与x 轴的非负半轴重合,终边上有两点(1,)A a ,(2,)B b ,且2cos23α=,则||a b -=A .15B .5 C .25D .112.设函数2,0,()1,0,x x f x x -⎧=⎨>⎩≤ 则满足(1)(2)f x f x +<的x 的取值范围是A .(,1]-∞-B .(0,)+∞C .(1,0)-D .(,0)-∞二、填空题:本题共4小题,每小题5分,共20分。
绝密★启用前2018年长望浏宁高三调研考试语文试卷总分:150分时量:150分钟长望浏宁四县(区、市)联合命制第Ⅰ卷阅读题一、现代文阅读(35分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1~3题。
藏匿在《人骑图》里的“画外音”42岁的赵孟頫在湖州老家画出了《人骑图》,左边有两则赵孟頫的自题,第一则说北上后见到三卷唐代画马高手韩干的真迹,才开始明白韩干在画中的用意。
三年后他再次题跋说画好画是难事,但能够“识画”更难。
自己能把马画得如此好,主要是天分使然,这一幅画自认为不愧唐人。
赵孟頫的次子赵雍题跋中说这匹马吃的是与三品官俸相等的精美饲料,装饰的是价值千金的辔环。
谁能给它解脱束缚的缰绳,让他回归“汧渭中间”纵情奔跑呢?《人骑图》中的“骑”字是由“马”和“奇”组成的会意字,“奇”字古义为“异”和“大”,作为量词与“偶”相反,而“偶”就是俗的意思,《后汉书·独行列传》中对独行之士的描绘就为历代文人推崇。
远古时期的马也是以多数出现,或是被组成骑兵团参战,或是作为天子贵族出行的马车,孔颖达《左传正义》说至六国时始有单骑。
不同凡俗之人才能骑马独行,“千里走单骑”的关羽就是一例。
赵孟頫把自己画成唐代骑马的独行之士,正是一种内心独白:因为秉持道义,所以具有一意孤行的勇气。
赵雍的题跋正是以说马来暗喻自己父亲。
学者赵华《赵孟頫同知济南考》一文也认为《人骑图》是赵孟頫的自画像,他将赵孟頫侄儿赵由辰的题跋和1295年赵孟頫好友戴表元的《史廉访自济南来江东,时赵子昂同知府事,画其所乘玉鼻骍以为赠》二首对比,认为戴表元描绘的一身“唐妆”的赵孟頫就是画中这样,而这匹马也就是赵孟頫的骑乘玉鼻骍。
父子二人都在画自己的画里扮成唐人,这种艺术的复古风潮背后正是对古人道德的推崇。
《人骑图》和唐人韩干的《牧马图》有许多共同之处:二者所画马鬃都很薄很少,修剪马鬃是为了去除兽性,唐代宫廷贵族把马鬃梳成几簇的流行样式,称为“三花马”和“五花马”,多见于唐三彩的马。
长望浏宁四县市高三模拟考试英语试卷本试卷分为四个部分,包括听力、语言知识运用、阅读理解和书面表达。
考试结束后,将本试卷和答题卡一并交回。
Part I Listening Comprehension (30 marks)Section A (22.5 marks)Directions: In this section you’ll hear six conversations between two speakers. For each conversation, there are several questions and each question is followed by 3 choices marked A, B and C. Listen carefully and then choose the best answer for each question.You will hear each conversation TWICE.Example:When will the magazine probably arrive?A. Wednesday.B. Thursday.C. Friday.The answer is B.Conversation 11. Who made the phone call?A. An air hostess.B. The boy’s father.C. A doctor.2. What caused the delay of the plane?A. A strike.B. The bad weather.C. The engine.Conversation 23. When will the speakers meet on Friday?A. At 10:00 am.B. At 11:00 am.C.At 12:00.4. Where will the speakers meet?A. At the man’s.B. In the office.C. At the restaurant.Conversation 35. Why does the woman want to find Mr. Green?A. To tell him about the meeting.B. To askfor a day’s leave.C. To make an appointment.6. Who will the woman visit tomorrow?A. Tom.B. Mr. Green.C. Her grandma.Conversation 47. How many coats does the man mention?A. Two.B. Three.C.Four.8. Where was the second coat lost according to the man?A. On the bus.B. In the park.C. On the underground.9. How does the woman sound at last?A. Confused.B. Patient.C. Annoyed.Conversation 510. Why does the company fire Jim?A. There’s a crisis and a new machine.B.He is always late for work.C. He’s not up to the work.11. What’s the probable relationship between the speakers?A. Headmaster and teacher.B. Manager andsecretary. C. Director and actress.12. What do we know about the man speaker?A. He has ever been dismissed.B. He needn’ttalk to Jim himself.C. He asked Sara to work here.Conversation 613. What is the survey about?A. Family life.B. Unusual hobbies.C. Holiday activities.14. What is the man interested in?A. Special exhibition.B. Bird shooting.C. Mountain climbing.15. Where does the man’ wife prefer to go?A. The seaside.B. Castle.C. Ski resorts.Section B (7.5 marks)Directions: In this section, you will hear a short passage. Listen carefully and then fill in the numberedblanks with the information you’ve heard. Fill in eachblank with NO MORE THAN THREE WORDS.You will hear the short passage TWICE.Part П Language Knowledge (45 marks)Section A (15 marks)Directions: Beneath each of the following sentencesthere are four choices marked A, B, C and D. Choose theone answer that best completes the sentence.Example:The wild flowers looked like a soft orange blanket the desert.A. coveringB. coveredC. coverD.to coverThe answer is A.21. Although nuclear power may lead to disasters, ______the nuclear power industry has been operating safelyfor three decades in China.A. andB. butC. soD. yet22. President Xi has advised young people to avoid staying up late, ______ heated response from Internetusers on microblog.A. having drawnB. drawingC. to drawD. drawn23. The silence of the library ______only by the soundof pages being turned over.A had been broken B. breaks C. brokeD. was broken24. When Mary arrived she found all her children ______for nearly two hours.A. have gone to sleepB. fell asleepC. was falling asleepD. had been asleep25. —Excuse me, can you show me the way to the nearestbus stop?—Sorry. I’m a stranger here. I ______ here untilmy guide arrives.A. standB. have stoodC. am standingD. will stand26. Thailand has decided to begin building its first standard-gauge railways in cooperation with China,______may cost $12.2 billion.A. whenB. thatC. itD. which27. Suddenly a good idea occurred to her, but shecouldn’t find any paper ______.A. to writeB. to write onC. writtenD. writing on28. Thank you for your hard work last week. I don’t thinkwe _______ it without you.A. could have managedB. can manageC. managedD. could manage29. Well, I really don’t know what you mean, ______ youwant me to do?A what it is that B. what is it that C. how itis that D. how is it that30. You mean I was absent? No! I ______at noon and I have been in the office since.A. came backB. have come backC. was backD. have been back31. In this socialist country led by the Communist Partyof China, every corrupt official _______ be dealt withonce evidence is found.A. mustB. shouldC. canD.might32. ______the civil servants want a pay increase, theywill have got promoted as well.A. Not do onlyB. Do not onlyC. Only not doD. Not only do33. _______ by a great demand for environmental-friendly cars, those factories has produced more green ones.A. Driven B Being driven C Having drivenD To drive34. What our parents expect from us is going home and greeting them frequently, ________ giving them much money.A. less thanB. rather thanC. as well asD. as much as35. _______ China views US-Japan relationship will have an effect on China’s military policy.A. WhenB. WhyC. WhatD. HowSection B (18 marks)Directions: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.From my second grade on, there was one event I fearedevery year: the piano recital(独奏演唱会). A recital 36 I had to practice a boring piece of music and perform before strangers. Each year I would ask my father if I could skip the recital “just this once”. And each year he would shake his head, saying something about building 37 and working toward a goal.One recent Sunday I stood in church, video camera in hand, and 38 my 68-year-old father play the pianoin his very first recital.My father had longed to play music since childhood, but his family was poor and couldn’t 39 lessons. He could have gone on regretting it, 40 too many of us do. But he wasn’t stuck in the past. When he retired three years ago, he 41 his church music director to take him as 42 .For a moment after my father sat down at the keyboard, he stared down at his fingers. Has he forgotten the 43 ? I worried. But then came the beautiful melody (旋律). And I 44 he had been doing what music teachers always stress: 45 the notes and pretend the others aren’t there.“I’m 46 of him for starting something new athis age,” I said to my son J eff.“Yeah, and doing it so 47 ,” Jeff added.With his first recital, my father taught me more about self-confidence and the life goal than all the words he used those 30-plus years ago.36.A. reflected B. meant C. explainedD. proved37. A. self-confidence B. self-control C.self-defense D. self-discipline38. A kept B. sent C. watched D.felt39. A. miss B. afford C. selectD. understand40. A. as B. once C. ifD. while41. A. allowed B. invited C. inspiredD. persuaded42. A. a teacher B. an old man C. a studentD. a singer43. A. words B. videos C. notesD. lessons44. A. predicted B. realized C. imaginedD. insisted45. A. pass over B . turn up C. bring inD. concentrate on46. A. ashamed B. aware C. tiredD. proud47. A. nicely B. anxiously C. casuallyD. frequentlySection C (12 marks)Directions: Complete the following passage by filling in each blank with one word that best fits the context.A report found that over half of the high school boys and two-thirds of the girls never shower after sports at school. Researchers suggest students don't want to sweat 48 take a shower, so they are less active. The researchers questioned almost 4,000 children. Lead researcher Dr Gavin said he was surprised 49 how rarely students showered. He said children were getting poor health because of less exercise. He said: " 50 the unwillingness to shower is a barrier(障碍)to playing sport, we need to do something to promote activity at schools."51 , the BBC says the study did not look at theexact reasons 52 students do not shower. Maybe there are some other reasons. Undressing in front of 53 may be too much for some children. A spokeswoman for a health organisation said children worry about their body image. She said schools had 54 role to play in changing attitudes. She said schools should encourage students to do physical activity and let 55 know they need to shower after.Part Ш Reading Comprehension (30 marks)Directions: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage.ARegarded as one of the English language's most gifted poets, John Keats wrote poetry that concentrated on imagery, human nature, and philosophy. Although Keats didn't receive much formal literary education, his own studies and passion brought him much success. Additionally, his own life situation influenced his poetry greatly.Growing up as a young boy in London in a lower, middle-class family, the young John didn't attend a private school, but went to a public one. His teachers and his family's friends regarded him as an optimistic boy who favored playing and fighting much more than minding his studies. After his father's death in the early 1800s, followed by his mother's passing due to tuberculosis (肺结核), he began viewing life differently. He wanted to escape the world and did so by reading anything he could get his hands on.At around the age of 16, the teenage John Keats began studying under a surgeon so that he too might become a doctor. However, his literary appetite had taken too much of his fancy, especially with his addiction to the poetry of Ehmund Spenser. He was able to have his first full poem published in the Examiner in 1816, entitled O Solitude! If I Must With Thee Dwell. Within two months in 1817, Keats had written an entire volume of poetry, but was sharply criticized by a magazine. However, the negative response didn't stop his pursuit of rhythm.John Keats' next work was Endymion, which waspublished in May 1818. The story involves a shepherd who falls in love with the moon goddess and leads him on an adventure of one boy's hope to overcome the limitations of being human. Following Engymion, however, he tried something more narrative-based and wrote Isabella. During this time, John Keats began seeing his limitations in poetry due to his own limit in life experiences. He would have to have the "knowledge" associated with his poems. His next work was Hyperion that would attempt to combine all that he learned. However, a bout (发作) with tuberculosis while visiting Italy would keep him from his work and eventually take his life in 1821.56. John Keats' attitude towards life changed because of________.A. his early education from schoolB. the criticism of a magazineC. Edmund Spenser's poetryD. the deaths of his parents57. What is the common thing between John Keats and his mother?A. They read many books.B. They died ofthe same disease.C. They had a bad childhood.D. They showed strong interest in poetry.58. What do we know from the passage?A. Keats once had a chance of becoming a doctor.B. Keats received little education at school.C. In 1816 Keats spent two months writing a poem.D. Endymion was about a real love story.59. While pursuing his dream of becoming a poet at first, John Keats was________.A. knowledgeableB. experiencedC. determinedD. impatient60. What can we infer from the passage?A. The poem Hyperion wasn't completed by Keats.B. Edmund Spenser was the greatest poet in Keats' time.C. It is likely that Keats rewrote his poem Isabella.D. Keats' family must have been very poor when he was young.BWhen Jeff Sparkman draws his cartoon superheroes with colored pencils, he often has to ask other people to tellhim what color his masked men turned out to be because he's color-blind. Now, a new smart phone application (app) can help him figure out what colors he's using and how the picture looks to most everyone else.The DanKam app, available for iPhone and Android for $2.99, is an application that turns the vague colors that one percent of the population with color-blindness sees into the "true" colors as everyone else sees them. In America, an estimated 32 million color-blind Americans —95% are males—can soon have their life improved.“DanKam takes the stream of data coming in through the phone's camera and changes the colors slightly so they fall within the range that people who are color-blind see,”developer Dan Kaminsky told CNET. He came up with the idea after watching the 2009 film Star Trek with a color-blind friend.It was then that he got to know more about colorblindness like its varying types and degrees. A vast majority, for instance, have trouble seeing red or green due to a genetic defect(遗传缺陷). Blue-yellow colorblindness, however, is rarer and develops later in life because of aging, illness or head injuries, etc.He started experimenting with one of the most common representations of points in the RGB color model. What the DanKam app attempts to do is to clean up the color space of the image or video signal so that colors can be visible to those suffering from viewing problems. “You can adjust the app to fit y our needs. There is a range and not everyone who is color-blind sees things the same.” Says Kaminsky.Sparkman, a copy editor at CNET, tried out the app and was pleased with the results. "It would be useful for dressing for a job interview," he said. But using it for his art is “the most practical application." It worked well on LED and other lights on electronic gadgets, which means Sparkman can now identify the power light on his computer display as green.61. According to the first two paragraphs, we can know that DanKam ___________.A. is designed to help people with colorblindnessB. can turn vague colors into real onesC. is a phone used to help drawing picturesD. appeared in the movie Star Trek62. How does DanKam’s app work?A. It puts LED and other lights on electronic gadgets.B. It shows common representations of points in the RGB color modelC. It checks color-blind people’s types of degrees of colorblindness.D. It changes the colors so that color-blind people can see them.63. It can be inferred from the passage that colorblindness __________.A. cannot be cured by any methodsB. is not necessarily inborn diseaseC. is more commonly seen in womenD. makes people unable to tell any colors64. The underlined word visible in Paragraph 4 is closest in meaning to _________.A. recognizableB. enjoyableC. adjustableD. portable65. Which of the following is NOT included in the things that DanKam helps Sparkman with?A. Choosing clothes.B. Playing computer gamesC. Drawing his pictures.D. Handling electronic gadgets.CBicycles, roller skates and skateboards are dangerous.I still have scars (伤疤) on my knees from my childhood run-ins with various wheeled devices. Admittedly, I was a foolish kid, but I’m glad I didn’t spend my childhood trapped indoors to protect me from any injury.“That which does not kill us makes us stronger.” But parents can’t handle it when teenagers put this theory into practice. And now technology has become the new field for the age-old battle between adults and their freedom-seeking kids.Locked indoors, unable to get on their bicycles and hang out with their friends, teens have turned to social media and their mobile phones to gossip and socialize with their friends. What they do online often mirrors what they might otherwise do if their mobility weren’t so heavily restricted (限制) in the age of helicopter parenting. Social media and smartphones have become so popular in recent years because teens need a place to call their own. They want the freedom to explore theiridentity and the world around them. Instead of climbing out of windows, they jump online.As teens have moved online, parents have projected their fears onto the Internet, imagining all the potential dangers that youth might face.Rather than helping teens develop strategies (策略) to deal with public life and the potential risks of interacting with others, fearful parents have focused on tracking, monitoring and blocking. These approaches don’t help teens develop the skills they need to manage complex social situations. “Protecting” kids may feel like the right thing to do, but it denies teens the chances of learning as they come of age in a technology-soaked world.The key to helping youth in the modern digital life isn’t more restrictions. It’s freedom —plus communication. Urban theorist Jane Jacobs used to argue that the safest neighborhoods were those where communities collectively took interest in and paid attention to what happened on the streets. Safety didn’t come from surveillance (监视) cameras or keeping everyone indoors but from a collective willingness towatch out for one another and be present as people struggled. The same is true online.What makes the digital street safe is when teens and adults collectively agree to open their eyes and pay attention, communicate and work together to deal with difficult situations. Teens need the freedom to wander the digital street, but they also need to know that caring adults are behind them and supporting them wherever they go. The first step is to turn off the tracking software. Then ask your kids what they’re doing when they’re online —and why it’s so important to them.66. When he was a child, the writer ______.A. became disabledB. spent much time outdoorC. always stayed at homeD. was ignored by his parents67. Teens go online mainly because ______.A. online games mirror real lifeB. they want to fight against their parentsC. online experiences make them strongD. they need a space of their own68. By mentioning “helicopter parenting” (Paragraph3), the writer means parents ______.A. remove any hidden dangers their kids may faceB. use helicopters to track their kidsC. prevent their kids from going to schoolD. protect their kids too much69. According to the passage, helicopter parents may make kids ______.A. lose the chances of learningB. handle complex social situations wellC. adapt to the digital world quicklyD. develop strategies to deal with public life70. The main idea of the passage is that ______.A. kids should be given freedom to deal with online risksB. safe neighborhoods come from joint efforts of allC. the digital street is a threat to kids’ safetyD. kids should be warned against potential dangers in societyPart IV Writing (45 marks)Section A (10 marks)Directions: Read the following passage. Fill in the numbered blanks by using the information from the passage.Write NO MORE THAN THREE WORDS for each answer. Being active in your local community is important because it is an essential part of living a satisfying life.Humans by nature are social beings. We need to be around other people to feel happy, secure and safe. Hanging out with friends helps, but that isn’t qu ite enough. We need to be active in our local community. Everyone benefits from a strong community. Being active within our society helps individuals feel less lonely, have a more vital and interesting life, stay healthy and feel happier. Being active within our society brings us inspiration, helps us succeed in business and helps us find our way in life. Communities thrive﹙兴旺﹚ when people are better connected. Local economies expand, businesses succeed, education grows, support systems become more efficient, and so on.There may be some obstacles on the way to being active in the local community. People may become disconnected from their community. It is a modern day reality that people turn to their computers for connection with others. This can be a great thing, but it can also go too far and we can become separated from our local, physical world society. Computers are usually at their best when used as a tool to accomplish tasks and not as a portal (传送门) into an online existence that controls our life. Online communities have become very real and they help millions feel more connected, but they must not take the place of our physical life community.It is also common for people to find jobs online and to migrate to different locations around the world. This uproots﹙使人迁移他处居住﹚ us and takes us away from the local society we grew up in, forces us to make new friends and establish ourselves in a new community. Becoming active in our community does require effort, but it’s well worth it. Understanding the importa nce of community and becoming more active within yours will greatly increase your quality of life.Section B (10 marks)Directions: Read the following passage. Answer the questions according to the information given in the passage.Fast food and too much TV time should not take all the blame for the weight problems. A group of researchers say that a number of aspects of modern living from lack of sleep to environmental chemicals and to living with air conditioning—may be closely related to the weight problems of many people.Lack of sleep is one of factors. Researches in animals and humans suggest that a lack of sleep over a long time can increase appetite. Studies also show that many adults and children are sleeping less than they used to. In recent decades, adults have gone from sleeping for all average of nine hours to about seven hours. There is also evidence that industrial chemicals may increase body fat. These chemicals, which are used in products, change hormonal(荷尔蒙的) activity when they get into the body. Studies suggest that people have been getting more and more of these chemicals through the foodchain in recent decades.Another factor that may be making people overweightis air conditioning. The body burns calories when forcedto eat less in hot and wet weather, but air conditioningmakes it unnecessary for the body to make any adjustment.Researchers also list other possible risk factors for overweight, including more older mothers, whosechildren may become overweight more easily; a number of medicines which can lead to weight gain; and fewer peoplewho smoke, since people often gain weight when they stopsmoking because nicotine can make people eat less.No one is suggesting that people should stop working andsleep more, or keep smoking. When it comes to saying anyperson’s weight, what really matters is how much theyeat and how much of the food they use every day. Thatmeans diet and exercise are still the key.81. Why does lack of sleep will result in weight problem? (No more than 5 words)____________________________________________________.82. How long do adults averagely sleep nowadays? (No more than 2 words)_____________________________________________________.83. How do the industrial chemicals affecting our weightcome into our body?(No more than 4 words)______________________________________________________.84. What is the main idea of this passage? (No more than 18 words)_______________________________________________________.Section C (25 marks)Directions: Write an English composition according tothe instructions given below in Chinese.随着老龄化问题的日趋严重,空巢老人越来越多,有的老人呆在家里无所事事,感到空虚寂寞。
2018年长望浏宁高三调研考试英语试卷本试卷分第Ⅰ卷﹙选择题﹚和第Ⅱ卷﹙非选择题﹚两部分。
满分150分,考试时间120分钟。
注意:所有试题均须在答题卡上作答。
第Ⅰ卷第一部分:听力技能(共两节,满分30分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项。
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。
例;How much is the shirt?A. £19.5B. £9.18C. £9.15答案是C1. What will the man do next?A. Go shopping.B. Call his friend.C. Hold a party.2. What is the weather like now?A. Sunny.B. Windy.C. Rainy.3. What's the relationship between the speakers?A. Homeowner and babysitter.B. Husband and wife.C. Colleagues.4. What is the man doing?A. Asking for advice.B. Hiring a volunteer.C. Doing a survey.5. How many junk mails does the man get a day?A. 30.B. 60.C. 90.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
2018年长望浏宁高三调研考试数学(文科)试卷时量:120分钟总分:150分一、选择题:(本大题共12个小题,每小题5分,满分60分.在每个小题给出的四个选项中,只有一项是符合题目要求的)1. 设集合,,若,则A. B. C. D.【答案】C【解析】∵集合,,∴是方程的解,即∴∴,故选C2. 在复平面内,复数(是虚数单位)对应的点位于A. 第一象限B. 第二象限C. 第三象限D. 第四象限【答案】D【解析】,对应坐标为,对应的点位于第四象限,故选D.3. 公比为的等比数列的各项都是正数,且,则A. B. C. D.【答案】B【解析】试题分析:因为,且,所以,因为公比,所以,所以.故B正确.考点:1等比数列的通项公式,及性质;2对数的运算.4. 《九章算术》中有如下问题:“今有勾八步,股一十五步,问勾中容圆,径几何?”其大意:“已知直角三角形两直角边长分别为步和步,问其内切圆的直径为多少步?”现若向此三角形内随机投一粒豆子,则豆子落在其内切圆外的概率是A. B. C. D.【答案】D【解析】由题意可知:直角三角向斜边长为17,由等面积,可得内切圆的半径为:落在内切圆内的概率为,故落在圆外的概率为5. 已知双曲线:的一条渐近线与圆相切,则双曲线C的离心率等于A. B. C. D.【答案】A【解析】双曲线:的一条渐近线,圆化为标准方程为:∵双曲线:的一条渐近线与圆相切,∴,即∴故选:A6. 若,则的值为A. B. C. D.【答案】D【解析】∵∴,.故选:D7. 某三棱锥的三视图如图所示,则该三棱锥的体积是A. B. C. D.【答案】B【解析】三视图对应的原图如下所示:,面,∴.选.8. 在等差数列中,若,则此数列的前项的和等于A. B. C. D.【答案】B【解析】试题分析:因为在等差数列中,,,解得,所以数列的前项的和,故选B.考点:1、等差数列的前项的和;2、等差数列的性质.9. 如图,给出的是计算的值的一个程序框图,则图中判断框内(1)处和执行框中的(2)处应填的语句是A. ,B. ,C. ,D. ,【答案】C【解析】此时,经第一次循环得到的结果是:;经第二次循环得到的结果是:;经第三次循环得到的结果是:;由特例归纳总结:S中最后一项的分母与i的关系是分母=令,解得:,即需要时输出,故图中判断框内(1)和执行框的(2)处应填的语句分别是故选:C点睛:算法与流程图的考查,侧重于对流程图循环结构的考查.先明晰算法及流程图的相关概念,包括顺序结构、条件结构、循环结构,其次要重视循环起点条件、循环次数、循环终止条件,更要通过循环规律,明确流程图研究的数学问题,是求和还是求项.10. 函数(其中e为自然对数的底数)的图象大致为A. B.C. D.【答案】A【解析】,所以为偶函数,图象关于轴对称,又,所以选A.11. 已知三棱柱的侧棱与底面垂直,体积为,底面是边长为的正三角形.若为底面的中心,则PA与平面所成角的大小为A. B. C. D.【答案】B【解析】如图所示,∵AA1⊥底面A1B1C1,∴∠APA1为PA与平面A1B1C1所成角,∵平面ABC∥平面A1B1C1,∴∠APA1为PA与平面ABC所成角.∵==.∴==AA1,解得.又P为底面正三角形A1B1C1的中心,∴A1P==1,在Rt△AA1P中,tan∠APA1==,∴∠APA1=60°.故选:B.点睛:求直线和平面所成角的关键是作出这个平面的垂线进而斜线和射影所成角即为所求,有时当垂线较为难找时也可以借助于三棱锥的等体积法求得垂线长,进而用垂线长比上斜线长可求得所成角的正弦值,当空间关系较为复杂时也可以建立空间直角坐标系,利用向量求解.12. 设满足,且在上是增函数,且,若函数对所有,当时都成立,则的取值范围是A. B. 或或C. 或或D.【答案】B【解析】若函数f(x)≤t2﹣2at+1对所有的x∈[﹣1,1]都成立,由已知易得f(x)的最大值是1,∴1≤t2﹣2at+1⇔2at﹣t2≤0,设g(a)=2at﹣t2(﹣1≤a≤1),欲使2at﹣t2≤0恒成立,则⇔t≥2或t=0或t≤﹣2.故选:B....二、填空题:(本大题共4小题,每小题5分,满分20分)13. 已知两个不相等的平面向量且,则_____.【答案】【解析】∵∴,又∴即,解得又∴故答案为:14. 若、满足约束条件,则的最小值为________.【答案】2【解析】由z=2x+y,得y=﹣2x+z作出不等式组对应的平面区域如图:由图象可知当直线y=﹣2x+z过点A时,直线y=﹣2x+z的在y轴的截距最小,此时z最小,由,得,即A(0,2),此时z=2×0+2=2,故答案为:2.点睛:本题考查的是线性规划问题,解决线性规划问题的实质是把代数问题几何化,即数形结合思想.需要注意的是:一,准确无误地作出可行域;二,画目标函数所对应的直线时,要注意让其斜率与约束条件中的直线的斜率进行比较,避免出错;三,一般情况下,目标函数的最大值或最小值会在可行域的端点或边界上取得.15. 已知抛物线的焦点为,准线,点在抛物线上,点在左准线上,若,且直线的斜率,则△的面积为____.【答案】【解析】抛物线的焦点为F(,0),准线方程为x=﹣,抛物线C:y2=6x点M在抛物线C上,点A在准线l上,若MA⊥l,且直线AF的斜率k AF=,准线与x轴的交点为N,则AN=3=3,A(﹣,3),则M(,3),∴S△AMN=×6×3=9.故答案为:.16. 如果一个棱锥的底面是正多边形,并且顶点在底面的射影是底面的中心,这样的棱锥叫做正棱锥.已知一个正六棱锥的各个顶点都在半径为3的球面上,则该正六棱锥的体积的最大值为_________【答案】【解析】设球心到底面的距离为x,则底面边长为,高为x+3,则V=(9﹣x2)•6(x+3)=(﹣x3﹣3x2+9x+27),其中0<x<3,V′=0,可得x2+2x﹣3=0,解得x=1或x=﹣3(舍去)∴V max=V(1)=(﹣1﹣3+9+27)=16.故答案为:16.三、解答题:(本大题共6小题,满分70分.解答应写出文字说明、证明过程或演算步骤)(一)必考题:共60分17. 在中,分别为角的对边,已知的面积为,又。
2018年长望浏宁高三调研考试数学(文科)试卷时量:120分钟 总分:150分一、选择题:(本大题共12个小题,每小题5分,满分60分.在每个小题给出的四个选项中,只有一项是符合题目要求的)1. 设集合{}1,2,4A =,{}240B x x x m =-+=,若{}1A B =I ,则B =A .{}1,3-B .{}1,0C .{}1,3D .{}1,52. 在复平面内,复数1-=i iz (i 是虚数单位)对应的点位于 A .第一象限 B .第二象限 C .第三象限 D .第四象限3.公比为2的等比数列}{n a 的各项都是正数,且16113=a a ,则=102log a A .4 B .5 C .6 D .74.《九章算术》中有如下问题:“今有勾八步,股一十五步,问勾中容圆,径几何? ”其大意:“已知直角三角形两直角边长分别为8步和15步,问其内切圆的直径为多少步?”现若向此三角形内随机投一粒豆子,则豆子落在其内切圆外的概率是 A .310π B .320π C .3110π- D .3120π- 5. 已知双曲线C :22221x y a b-=的一条渐近线与圆226290x y x y +--+=相切,则双曲线C 的离心率等于 A.54 B.53 C. 32 D. 436. 若546sin =⎪⎭⎫ ⎝⎛-x π,则⎪⎭⎫⎝⎛+x 26sin π的值为 A .2524 B .2524- C .257 D .257- 7. 某三棱锥的三视图如图所示,则该三棱锥的体积是 A .16 B .13 C .23D .1 8. 在等差数列{}n a 中,若351024a a a ++=,则此数列的前13项的和等于A .8B .13C .16D .269. 如图,给出的是计算111114710100++++L 的值的一个程序框图,则图中判断框内(1)处和执行框中的(2)处应填的语句是A .100i >,1n n =+B .34i <,3n n =+C .34i >,3n n =+D .34i ≥,3n n =+10. 函数1()(1)x x e f x x e +=-(其中e 为自然对数的底数)的图象大致为11. 已知三棱柱111ABC A B C -的侧棱与底面垂直,体积为94,底面是边长为3的正三角形.若P 为底面111A B C 的中心,则PA 与平面ABC 所成角的大小为A .512π B .3π C .4π D .6π12. 设()f x 满足()()-=f x f x -,且在[]1,1-上是增函数,且()11f -=-,若函数()221f x t at ≤-+对所有[]1,1x ∈-,当[]1,1a ∈-时都成立,则t 的取值范围是A .1122t -≤≤ B .2t ≥或2t ≤-或0t = C .12t ≥或12t ≤-或0t = D .22t -≤≤二、填空题:(本大题共4小题,每小题5分,满分20分)13.已知两个不相等的平面向量(2,1),(2,).a b x ==r r 且(2)()a b a b +⊥-r r r r,则x = .14. 若x 、y 满足约束条件20240210x y x y x y +-≥⎧⎪-+≥⎨⎪--≤⎩,则2z x y =+的最小值为 .15.已知抛物线2:2(0)C y px p =>的焦点为F ,准线3:2l x =-,点M 在抛物线C 上,点A 在左准线l 上,若MA l ⊥,且直线AF 的斜率3AF k =-,则△AFM 的面积为 .16.如果一个棱锥的底面是正多边形,并且顶点在底面的射影是底面的中心,这样的棱锥叫做正棱锥.已知一个正六棱锥的各个顶点都在半径为3的球面上,则该正六棱锥的体积的最大值为_________三、解答题:(本大题共6小题,满分70分.解答应写出文字说明、证明过程或演算步骤) (一)必考题:共60分 17. (本小题满分12分)在ABC ∆中,,,a b c 分别为角,,A B C 的对边,已知7,2c ABC =∆的面积为332, 又tan tan 3(tan tan 1)A B A B +=-。
(Ⅰ)求角C 的大小; (Ⅱ)求a b +的值。
18. (本小题满分12分)如图,多面体ABCDEF 中,//,AD BC AB AD ⊥,FA ⊥平面,//ABCD FA DE ,且222AB AD AF BC DE =====.(Ⅰ)若M 为线段EF 中点,求证://CM 平面ABF ; (Ⅱ)求多面体ABCDEF 的体积.19. (本小题满分12分)交通拥堵指数是综合反映道路网畅通或拥堵的概念,记交通拥堵指数为 ,其范围为,分别有五个级别: 畅通; 基本畅通;轻度拥堵; 中度拥堵; 严重拥堵.晚高峰时段 ,从某市交通指挥中心选取了市区 个交通路段,依据其交通拥堵指数数据绘制的直方图如图所示.(Ⅰ)求出轻度拥堵,中度拥堵,严重拥堵路段各有多少个; (Ⅱ)用分层抽样的方法从交通指数在,,的路段中共抽取6个路段,求依次抽取的三个级别路段的个数;(Ⅲ)从(Ⅱ)中抽取的6个路段中任取2个,求至少1个路段为轻度拥堵的概率.20.(本小题满分12分)已知椭圆E:22221(0)x y a b a b+=>>经过点(2,1)P ,且离心率为32.(Ⅰ)求椭圆的标准方程;(Ⅱ)设O 为坐标原点,在椭圆短轴上有两点M 、N 满足OM NO =u u u u r u u u r,直线PM 、PN 分别交椭圆于A 、B .探求直线AB 是否过定点,如果经过定点请求出定点的坐标,如果不经过定点,请说明理由.21. (本小题满分12分)已知函数()e ln 1x f x m x =--.(Ⅰ)当1m =时,求曲线()y f x =在点()()11f ,处的切线方程; (Ⅱ)当1m ≥时,证明:()1f x >.(二)选考题:共10分,考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分.选修4-4:坐标系与参数方程 22. (本小题满分10分)在直角坐标系xoy 中,曲线C 的参数方程为sin cos sin cos x y αααα=+⎧⎨=-⎩(α为参数).(Ⅰ)求曲线C 的普通方程;(Ⅱ)在以o 为极点,x 轴的非负半轴为极轴的极坐标系中,直线l π1sin()042θ-+=,已知直线l 与曲线C 相交于A 、B 两点,求AB .选修4-5:不等式选讲 23.(本小题满分10分)设()||,.f x x a a =-∈R (Ⅰ)当5=a ,解不等式3≤)(x f ;(Ⅱ)当1=a 时,若∃R x ∈,使得不等式(1)(2)12f x f x m -+≤-成立,求实数m 的取值范围.2018年长望浏宁高三调研考试数学(文科) 参考答案一、选择题:(本大题共12个小题,每小题5分,满分60分.在每个小题给出的四个选项中,只有一项是符合题目要求的)二、填空题:(本大题共4小题,每小题5分,满分20分)13. 12- 14. 2 15.三、解答题:(本大题共6小题,满分70分.解答应写出文字说明、证明过程或演算步骤)17. 解:(1)因为tan tan tan 1)A B A B +=-,所以tan tan tan()1tan tan A BA B A B++==-又因为,,A B C 为ABC ∆的内角,所以23A Bπ+=,所以3Cπ=。
6分(Ⅱ)由133sin2ABCS ab C∆==,及3Cπ=,得6ab=,又22222()217 cos,2222a b c a b c abC cab ab+-+--=====,所以112a b+=。
12分18. 解:(Ⅰ)取AD中点N,连接CN和MN,MNΘ为梯形ADEF的中位线MN∴∥AF 1分∵FA⊂平面FAB,MN⊄平面FAB∴MN∥平面FAB 2分∵四边形ABCN为矩形∴CN∥AB 3分∵FA⊂平面FAB,CN⊄平面FAB∴CN∥平面FAB 4分∵MN⋂CN=N∴平面//CMN平面ABF∵CM⊂平面CMN 6分∴//CM平面ABF(Ⅱ)8分10分12分19. 解:(Ⅰ) 由直方图可知:,,.所以这 个路段中,轻度拥堵、中度拥堵、严重拥堵路段分别为个, 个, 个. 3分 (Ⅱ) 由(1)知拥堵路段共有个,按分层抽样从 个路段中选出 个,每种情况:,,,即这三个级别路段中分别抽取的个数为 ,, 个. 6分 (Ⅲ) 记(Ⅱ)中选取的 个轻度拥堵路段为 ,,选取的 个中度拥堵路段为,,,选取的 个严重拥堵路段为,则从 个路段选取 个路段的可能情况如下:,,,,,,,,,,,,,, 共 种可能, 其中至少有 个轻度拥堵的有:,,,,,,,,共 种可能.所以所选 个路段中至少 个路段为轻度拥堵的概率为 . 12分20. 解:(Ⅰ)由椭圆的离心率e =23122=-=a b a c ,则a 2=4b 2, 2分将P (2,1)代入椭圆142222=+b y b x ,则11122=+bb ,解得:b 2=2,则a 2=8, 4分∴椭圆的方程为:12822=+y x ; 5分 (Ⅱ)当M ,N 分别是短轴的端点时,显然直线AB 为y 轴,所以若直线过定点,这个定点一点在y 轴上,当M ,N 不是短轴的端点时,设直线AB 的方程为y =kx +t ,设A (x 1,y 1)、B (x 2,y 2),由⎪⎩⎪⎨⎧+==+t kx y y x 12822消去y 得(1+4k 2)x 2+8ktx +4t 2﹣8=0,·则△=16(8k 2﹣t 2+2)>0, x 1+x 2=1482+-k kt,x 1x 2=148422+-k t , 7分又直线PA 的方程为y ﹣1=2111--x y (x ﹣2), 即y ﹣1=2111--+x t kx (x ﹣2), 8分因此M 点坐标为(0,22)21(11---x tx k ),同理可知:N (0,22)21(22---x tx k ) 9分由OM =,则22)21(11---x t x k +22)21(22---x tx k =0,化简整理得:(2﹣4k )x 1x 2﹣(2﹣4k +2t )(x 1+x 2)+8t =0,则(2﹣4k )×148422+-k t ﹣(2﹣4k +2t )(1482+-k kt)+8t =0, 10分化简整理得:(2t +4)k +(t 2+t ﹣2)=0,·当且仅当t =﹣2时,对任意的k 都成立,直线AB 过定点Q (0,﹣2). 12分21. 解:(Ⅰ)当1m =时,()e ln 1xf x x =--,所以1()e x f x x'=-1分 所以(1)e 1f =-,(1)e 1f '=-. 2分所以曲线()y f x =在点()()11f ,处的切线方程为(e 1)(e 1)(1)y x --=--. 即()e 1y x =-. 3分 (Ⅱ)证法一:当1m ≥时,()e ln 1e ln 1xxf x m x x =--≥--.要证明()1f x >,只需证明e ln 20x x -->. 4分 以下给出三种思路证明e ln 20x x -->. 思路1:设()e ln 2xg x x =--,则1()e x g x x'=-. 设1()e x h x x =-,则21()e 0x h x x'=+>, 所以函数()h x =1()e x g x x '=-在0+∞(,)上单调递增. 6分 因为121e 202g ⎛⎫'=-< ⎪⎝⎭,(1)e 10g '=->,所以函数1()e x g x x '=-在0+∞(,)上有唯一零点0x ,且01,12x ⎛⎫∈ ⎪⎝⎭. 8分 因为0()0g x '=时,所以01ex x =,即00ln x x =-. 9分 当()00,x x ∈时,()0g x '<;当()0,x x ∈+∞时,()0g x '>.所以当0x x =时,()g x 取得最小值()0g x . 10分 故()000001()=e ln 220xg x g x x x x ≥--=+->. 综上可知,当1m ≥时,()1f x >. 12分 思路2:先证明e 1xx ≥+()x ∈R . 5分 设()e 1x h x x =--,则()e 1x h x '=-.因为当0x <时,()0h x '<,当0x >时,()0h x '>, 所以当0x <时,函数()h x 单调递减,当0x >时,函数()h x 单调递增.所以()()00h x h ≥=.所以e 1xx ≥+(当且仅当0x =时取等号). 7分 所以要证明e ln 20x x -->,只需证明()1ln 20x x +-->. 8分 下面证明ln 10x x --≥.设()ln 1p x x x =--,则()111x p x x x-'=-=. 当01x <<时,()0p x '<,当1x >时,()0p x '>, 所以当01x <<时,函数()p x 单调递减,当1x >时,函数()p x 单调递增.所以()()10p x p ≥=.所以ln 10x x --≥(当且仅当1x =时取等号). 10分 由于取等号的条件不同, 所以e ln 20x x -->. 综上可知,当1m ≥时,()1f x >. 12分(若考生先放缩ln x ,或e x、ln x 同时放缩, 请参考此思路给分!) 思路3:先证明e ln 2x x ->.因为曲线e xy =与曲线ln y x =的图像关于直线y x =对称, 设直线x t =()0t >与曲线e xy =,ln y x =分别交于点A ,B , 点A ,B 到直线y x =的距离分别为1d ,2d , 则)122AB d d =+.其中12t d =22d =()0t >.①设()e t h t t =-()0t >,则()e 1t h t '=-.因为0t >,所以()e 10t h t '=->.所以()h t 在()0,+∞上单调递增,则()()01h t h >=.所以1t d =>. ②设()ln g t t t =-()0t >,则()111t g t t t-'=-=.因为当01t <<时,()0g t '<;当1t >时,()0g t '>, 所以当01t <<时,()ln g t t t =-单调递减; 当1t >时,()ln g t t t =-单调递增.所以()()11g t g ≥=.所以2d =≥所以)122AB d d =+>+=⎭. 综上可知,当1m ≥时,()1f x >. 12分 证法二:因为()e ln 1xf x m x =--,要证明()1f x >,只需证明e ln 20x m x -->. 4分 以下给出两种思路证明e ln 20x m x -->. 思路1:设()e ln 2xg x m x =--,则1()e x g x m x'=-. 设1()e x h x m x =-,则21()e 0x h x m x'=+>. 所以函数()h x =()1e x g x m x'=-在()0+∞,上单调递增. 6分因为11221e 2e 202m m g m m m m ⎛⎫⎛⎫'=-=-< ⎪ ⎪⎝⎭⎝⎭,()1e 10g m '=->, 所以函数1()e x g x m x'=-在()0+∞,上有唯一零点0x , 且01,12x m ⎛⎫∈⎪⎝⎭. 8分因为()00g x '=,所以001e xm x =,即00ln ln x x m =-- 9分 当()00,x x ∈时,()0g x '<;当()0,x x ∈+∞时,()0g x '>.所以当0x x =时,()g x 取得最小值()0g x 10分 故()()000001e ln 2ln 20xg x g x m x x m x ≥=--=++->. 综上可知,当1m ≥时,()1f x >. 12分 思路2:先证明e 1()xx x ≥+∈R ,且ln 1(0)x x x ≤+>. 5分 设()e 1xF x x =--,则()e 1xF x '=-.因为当0x <时,()0F x '<;当0x >时,()0F x '>, 所以()F x 在(,0)-∞上单调递减,在(0,)+∞上单调递增. 所以当0x =时,()F x 取得最小值(0)0F =.所以()(0)0F x F ≥=,即e 1xx ≥+(当且仅当0x =时取等号). 7分 由e 1()xx x ≥+∈R ,得1ex x -≥(当且仅当1x =时取等号). 8分所以ln 1(0)x x x ≤->(当且仅当1x =时取等号). 9分 再证明e ln 20x m x -->.因为0x >,1m ≥,且e 1xx ≥+与ln 1x x ≤-不同时取等号, 所以()()e ln 2112x m x m x x -->+---()()11m x =-+0≥.综上可知,当1m ≥时,()1f x >. 12分22. 解:(Ⅰ)由已知sin ,cos 22x y x yθθ+-==, 由1cos sin 22=+θθ,消去θ得:普通方程为12222=⎪⎭⎫ ⎝⎛-+⎪⎭⎫ ⎝⎛+y x y x , 化简得222=+y x 5分(Ⅱ)由2ρsin(4π-θ)+21=0知021)sin (cos =+-θθρ,化为普通方程为x-y+21=0圆心到直线l 的距离h =42,由垂径定理230=AB 10分23.解:解:(Ⅰ)5a =时原不等式等价于53x -≤即353,28x x -≤-≤≤≤,所以解集为{}28x x ≤≤. 3分(Ⅱ)当1a =时,|1|)(-=x x f ,令133()21()(1)(2)2211(2)233(2)x x g x f x f x x x x x x x ⎧-+≤⎪⎪⎪=-+=-+-=+<<⎨⎪-≥⎪⎪⎩,由图像知:当12x =时,()g x 取得最小值32,由题意知:3122m ≤-,所以实数m 的取值范围为14m ≤-. 10分。