2015年高考模拟试卷及参考答案 全国卷1
2015年高考理综试题及答案(新课标全国卷1)
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2015年全国卷1理综第Ⅰ卷(选择题共126分)可能用到的相对原子质量:H 1 C 12 N 14 O 16 Cl 35.5 K 39Cr 52 Fe 56 Cu 64 Br 80 Ag 108 I 127一、选择题:本题共13小题,每小题6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.下列叙述错误的是A.DNA与A TP中所含元素的种类相同B.一个tRNA分子中只有一个反密码子C.T2噬菌体的核酸由脱氧核糖苷酸组成D.控制细菌性状的基因位于拟核和线粒体中的DNA上2.下列关于植物生长素的叙述,错误的是A.植物幼嫩叶片中的色氨酸可转变为生长素B.成熟茎韧皮部中的生长素可以进行非极性运输C.幼嫩细胞和成熟细胞对生长素的敏感程度相同D.豌豆幼苗切段中乙烯的合成受生长素浓度的影响3.某同学给健康实验兔静脉滴注0.9%的NaCl溶液(生理盐水)20 mL后,会出现的现象是A.输入的溶液会从血浆进入组织液B.细胞内液和细胞外液分别增加10 mLC.细胞内液Na+的增加远大于细胞外液Na+的增加D.输入的Na+中50%进入细胞内液,50%分布在细胞外液4.下列关于初生演替中草本阶段和灌木阶段的叙述,正确的是A.草本阶段与灌木阶段群落的丰富度相同B.草本阶段比灌木阶段的群落空间结构复杂C.草本阶段比灌木阶段的群落自我调节能力强D.草本阶段为灌木阶段的群落形成创造了适宜环境5.人或动物PrP基因编码一种蛋白(PrP°),该蛋白无致病性。
PrP°的空间结构改变后成为PrP°°(朊粒),就具有了致病性。
PrP°°可以诱导更多的PrP°转变为PrP°°,实现朊粒的增殖,可以引起疯牛病。
据此判断,下列叙述正确的是A.朊粒侵入机体后可整合到宿主的基因组中B.朊粒的增殖方式与肺炎双球菌的增殖方式相同C.蛋白质空间结构的改变可以使其功能发生变化D.PrP°转变为PrP°°的过程属于遗传信息的翻译过程6.抗维生素D佝偻病为X染色体显性遗传病,短指为常染色体显性遗传病,红绿色盲为X染色体隐性遗传病,白化病为常染色体隐性遗传病。
2015年高考语文试卷全国卷(含答案)
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2015年普通高等学校全国统一考试全国卷语文第I卷阅读题甲必考题一、现代文阅读(9分,毎小题3分)阅读下面的文字,完成1〜3题宋代的农业、手工业、商业在唐代的基础上又有了新的发展,特别是商品经济出现了空前的繁荣,在此背景下,宋代的货巾流通和信用进入迅速发展时期,开创了古代金融的新篇章.宋代在信用形式和信用工具方面都呈现出新的特点.信用形式有借贷、质、押、典、赊买赊卖等多种形式.借贷分为政府借贷和私人借贷.政府借贷主要表现为赈贷的形式,在紧急情况下通过贷给百姓粮食或种子的方式,帮助他们度过困境。
私人借贷多为高利贷,它可以解决社会分化和“钱荒"带来的平民百姓资金严重不足的问题,满足特殊支付和燃眉之急的需要.质、押是借贷的担保形式,由质库、解库等机构经营.质属于动产担保,它必须转移动产的占有;押属于不动产担保,通常将抵押物的契约交付债权人即可。
债务人违约时,债权人可用变卖价款优先受偿。
典作为不动产转移的一种形式是在宋代形成和发展起来的。
其特点是典权人向出典人支付典价后,在典期内就占有了出典人典产的使用权和收益支配权,出典人也不必向典权人支付利息。
宋代的商业贸易非常发达,但存在着通货紧缩现象,故赊买赊卖行为也很普遍,几乎生产、流通、消费领域的所有物品都能进行赊买赊卖.从实际效果看,它解决了军需、加强了流通,更重要的一点,它对束缚生产流通扩大和发展的高利贷构成了冲击。
随着社会经济的发展,宋代商业贸易对货币的要求越来越高,但是社会中货币供给和流通状况不尽理想,表现为货币流通区域的割据性、货币供给数量的有限性,以及大量流通的铜铁钱细碎和不便携带的特性,其结果是抑制了经济发展.为了解决这类问题,在高度发达的造纸和印刷技术保障下,通过民间自发力量的作用和官府的强制推行,宋语文试题第1页(共10页)代社会陆续出如现了诸茶引、盐引、交子、关子和会子等新型纸质信用工具。
茶引、盐引要求相关人员先用粮草或现钱的付出作为取得的条件,然后凭此类纸质信用工具异地兑取现钱或政府专卖货物.这些信用工具的使用,除了可发挥信用功能外,也使得政府和商人在专卖货物领域能够共同获利,既有利于商人从政府专卖的货物中分得一份利益,又有利于政府实现增加收入、补给军需等目标。
2015年高考物理全国卷1及答案解析
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计重力),从较强磁场区域进入到较弱磁场区域后,粒子的
A.轨道半径减小,角速度增大
B.轨道半径减小,角速度减小
C.轨道半径增大,角速度增大
D.轨道半径增大,角速度减小
15.如图,直线 a、b 和 c、d 是处于匀强电场中的两组平行线,M、N、P、Q 是它们的交点,四点处的电势分
别为 M 、 N 、 P 、 Q 。一电子由 M 点分别运动到 N 点和 P 点的过程中,电场力所做的负功相等。则
两端的电压为 U,原、副线圈回路中电阻消耗的功率的比值为 k。则
A.U=66V,k=1 9 1
C.U=66V,k=3
1 B.U=22V,k=9
1 D.U=22V,k=3
17.如图,一半径为 R、粗糙程度处处相同的半圆形轨道竖直固定放置,直径 POQ 水平。一质量为 m 的质点 自 P 点上方高度 R 处由静止开始下落,恰好从 P 点进入轨道。质点滑到轨道最低点 N 时,对轨道的压力为
4 mg,g 为重力加速度的大小。用 W 表示质点从 P 点运动到 N 点的过程中克服摩擦力所做的功。则
1 A.W=2mgR,质点恰好可以到达 Q 点
1 B.W>2mgR,质点不能到达 Q 点
C.W=12mgR,质点到达 Q 点后,继续上升一段距离
D.W<12mgR,质点到达 Q 点后,继续上升一段距离
撞(碰撞时间极短)。碰撞前后木板速度大小不变,方向相反;运动过程中小物块始终未离开木板。已知碰撞 后 1 s 时间内小物块的 -t 图线如图(b)所示。木板的质量是小物块质量的 l5 倍,重力加速度大小 g 取 10 m
/s2 。
求 (1)木板与地面间的动摩擦因数μ1 及小物块与木板间的动摩擦因数μ2; (2)木板的最小长度; (3)木板右端离墙壁的最终距离。 (二)选考题:共 45 分。请考生从给出的 3 道物理题、3 道化学题、2 道生物题中每科任选一题做答,并用 2B 铅 笔在答题卡上把所选题目题号后的方框涂黑。注意所做题目的题号必须与所涂题目的题号一致。在答题卡
2015年高考语文试卷全国一卷(含答案)
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2015年高考语文试卷新课标全国1卷(解析版)甲必考题一、现代文阅读(9分,毎小题3分)阅读下面的文字,完成1-3题宋代的农业、手工业、商业在唐代的基础上又有了新的发展,特别是商品经济出现了空前的繁荣。
在此背景下,宋代的货币流通和信用进入迅速发展时期,开创了古代金融的新篇章。
宋代在信用形式和信用工具方面都呈现出新的特点。
信用形式有借贷、质、押、典、赊买赊卖等多种形式。
借贷分为政府借贷和私人借贷。
政府借贷主要表现为赈贷的形式,在紧急情况下通过贷给百姓粮食或种子的方式,帮助他们度过困境。
私人借贷多为高利贷,它可以解决社会分化和“钱荒”带来的平民百姓资金严重不足的问题,满足特殊支付和燃眉之急的需要。
质、押是借贷的担保形式,由质库、解库等机构经营。
质属于动产担保,它必须转移动产的占有;押属于不动产担保,通常将抵押物的契约交付债权人即可。
债务人违约时,债权人可用变卖价款优先受偿。
典作为不动产转移的一种形式是在宋代形成和发展起来的。
其特点是典权人向出典人支付典价后,在典期内就占有了出典人典产的使用权和收益支配权,出典人也不必向典权人支付利息。
宋代的商业贸易非常发达,但存在着通货紧缩现象,故赊买赊卖行为也很普遍,几乎生产、流通、消费领域的所有物品都能进行赊买赊卖。
从实际效果看,它解决了军需、加强了流通,更重要的一点,它对束缚生产流通扩大和发展的高利贷构成了冲击。
随着社会经济的发展,宋代商业贸易对货币的要求越来越高,但是社会中货币供给和流通状况不尽理想,表现为货币流通区域的割据性、货币供给数量的有限性,以及大量流通的钢铁钱细碎和不便携带的特性,其结果是抑制了经济发展。
为了解决这类问题,在高度发达的纸币和印刷技术保障下,通过民间自发力量的作用和官府的强制推行,宋代社会陆续出现了诸如茶引、盐引、交子、关子和会子等新型纸质信用工具。
茶引、盐引要求相关人员先用粮草或现钱的付出作为取得的条件,然后凭此类纸质信用工具异地兑取现钱或政府专卖货物。
2015年高考理综全国卷1(含详细答案)
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理科综合能力测试试卷 第1页(共52页)理科综合能力测试试卷 第2页(共52页)绝密★启用前 2015年普通高等学校招生全国统一考试(全国新课标卷1)理科综合能力测试使用地区:陕西、山西、河南、河北、湖南、湖北、江西本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。
第Ⅰ卷1至5页,第Ⅱ卷6至16页,共300分。
考生注意:1. 答题前,考生务必将自己的准考证号、姓名填写在答题卡上。
考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名、考试科目”与考试本人准考证号、姓名是否一致。
2. 第Ⅰ卷每小题选出答案后,用2B 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,再选涂其他答案标号。
第Ⅱ卷用黑色墨水签字笔在答题卡上书写作答,在试题卷上作答,答案无效。
3. 考试结束,监考员将试题卷、答题卡一并收回。
可能用到的相对原子质量:H —1 C —12 N —14 O —16 Cl —35.5 K —39Cr —52 Fe —56 Cu —64 Br —80 Ag —108 I —127第Ⅰ卷(选择题 共126分)本卷共21小题,每小题6分,共126分。
一、选择题(本大题共13小题,每小题6分,共78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. 下列叙述错误的是( )A. DNA 与ATP 中所含元素的种类相同B. 一个tRNA 分子中只有一个反密码子C. T 2噬菌体的核酸由脱氧核糖苷酸组成D. 控制细菌性状的基因位于拟核和线粒体中的DNA 上 2. 下列关于植物生长素的叙述,错误的是( )A. 植物幼嫩叶片中的色氨酸可转变为生长素B. 成熟茎韧皮部中的生长素可以进行非极性运输C. 幼嫩细胞和成熟细胞对生长素的敏感程度相同D. 豌豆幼苗切段中乙烯的合成受生长素浓度的影响3. 某同学给健康实验兔静脉滴注0.9%的Na Cl 溶液(生理盐水)20 mL 后,会出现的现象是 ( )A. 输入的溶液会从血浆进入组织液B. 细胞内液和细胞外液分别增加10 mLC. 细胞内液Na +的增加远大于细胞外液Na +的增加D. 输入的Na +中50%进入细胞内液,50%分布在细胞外液 4. 下列关于初生演替中草本阶段和灌木阶段的叙述,正确的是( )A. 草本阶段与灌木阶段群落的丰富度相同B. 草本阶段比灌木阶段的群落空间结构复杂C. 草本阶段比灌木阶段的群落自我调节能力强D. 草本阶段为灌木阶段的群落形成创造了适宜环境5. 人或动物PrP 基因编码一种蛋白(c PrP ),该蛋白无致病性。
(完整)2015年高考理科数学试卷全国卷1含答案),推荐文档
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2015年高考理科数学试卷全国卷11.设复数z 满足11zz+-=i ,则|z|=( ) (A )1 (B )2 (C )3 (D )2 2.o o o o sin 20cos10cos160sin10- =( ) (A )3-(B )3 (C )12- (D )123.设命题p :2,2nn N n ∃∈>,则p ⌝为( )(A )2,2nn N n ∀∈> (B )2,2nn N n ∃∈≤(C )2,2nn N n ∀∈≤ (D )2,=2nn N n ∃∈4.投篮测试中,每人投3次,至少投中2次才能通过测试。
已知某同学每次投篮投中的概率为0.6,且各次投篮是否投中相互独立,则该同学通过测试的概率为( ) (A )0.648 (B )0.432 (C )0.36 (D )0.3125.已知M (00,x y )是双曲线C :2212x y -=上的一点,12,F F 是C 上的两个焦点,若120MF MF •<u u u u r u u u u r,则0y 的取值范围是( )(A )(-33,33) (B )(-36,36) (C )(223-,223) (D )(23-,23)6.《九章算术》是我国古代内容极为丰富的数学名著,书中有如下问题:“今有委米依垣内角,下周八尺,高五尺。
问:积及为米几何?”其意思为:“在屋内墙角处堆放米(如图,米堆为一个圆锥的四分之一),米堆为一个圆锥的四分之一),米堆底部的弧长为8尺,米堆的高为5尺,问米堆的体积和堆放的米各为多少?”已知1斛米的体积约为1.62立方尺,圆周率约为3,估算出堆放斛的米约有( ) (A )14斛 (B )22斛 (C )36斛 (D )66斛7.设D 为ABC ∆所在平面内一点3BC CD =u u u r u u u r,则( )(A )1433AD AB AC =-+u u u r u u ur u u u r (B )1433AD AB AC =-u u u r u u u r u u u r(C )4133AD AB AC =+u u u u u r u u u r u u u r (D )4133AD AB AC =-u u u u u u u ru u u r u u u r8.函数()f x =cos()x ωϕ+的部分图像如图所示,则()f x 的单调递减区间为( )(A)13 (,),44 kk k Zππ-+∈(B)13(2,2),44k k k Zππ-+∈(C)13(,),44k k k Z-+∈(D)13(2,2),44k k k Z-+∈9.执行右面的程序框图,如果输入的t=0.01,则输出的n=()(A)5 (B)6 (C)7 (D)810.25()x x y++的展开式中,52x y的系数为()(A)10 (B)20 (C)30 (D)6011.圆柱被一个平面截去一部分后与半球(半径为r)组成一个几何体,该几何体三视图中的正视图和俯视图如图所示.若该几何体的表面积为16 + 20π,则r=()(A)1 (B)2 (C)4 (D)812.设函数()f x=(21)xe x ax a--+,其中a1,若存在唯一的整数x,使得()f x 0,则a的取值范围是()(A)[-32e,1)(B)[-32e,34)(C)[32e,34)(D)[32e,1)13.若函数f(x)=2ln()x x a x+为偶函数,则a=14.一个圆经过椭圆221164x y+=的三个顶点,且圆心在x轴的正半轴上,则该圆的标准方程为 .15.若,x y满足约束条件1040xx yx y-≥⎧⎪-≤⎨⎪+-≤⎩,则yx的最大值为 .16.在平面四边形ABCD中,∠A=∠B=∠C=75°,BC=2,则AB的取值范围是 .17.(本小题满分12分)n S 为数列{n a }的前n 项和.已知n a >0,2n n a a +=43n S +.(Ⅰ)求{n a }的通项公式; (Ⅱ)设11n n n b a a +=,求数列{n b }的前n 项和. 18.如图,四边形ABCD 为菱形,∠ABC=120°,E ,F 是平面ABCD 同一侧的两点,BE ⊥平面ABCD ,DF ⊥平面ABCD ,BE=2DF ,AE ⊥EC.(Ⅰ)证明:平面AEC ⊥平面AFC ;(Ⅱ)求直线AE 与直线CF 所成角的余弦值.19.某公司为确定下一年度投入某种产品的宣传费,需了解年宣传费x (单位:千元)对年销售量y (单位:t )和年利润z (单位:千元)的影响,对近8年的年宣传费i x 和年销售量i y (i =1,2,···,8)数据作了初步处理,得到下面的散点图及一些统计量的值.x ry u rw u r821()ii x x =-∑821()ii w w =-∑81()()iii x x y y =--∑ 81()()iii w w yy =--∑46.6 56.3 6.8 289.8 1.6 1469 108.8表中i i w x =,w u r =1881i i w =∑(Ⅰ)根据散点图判断,y=a+bx 与x y 关于年宣传费x 的回归方程类型?(给出判断即可,不必说明理由)(Ⅱ)根据(Ⅰ)的判断结果及表中数据,建立y 关于x 的回归方程;(Ⅲ)已知这种产品的年利率z 与x 、y 的关系为z=0.2y-x.根据(Ⅱ)的结果回答下列问题:(ⅰ)年宣传费x=49时,年销售量及年利润的预报值是多少? (ⅱ)年宣传费x 为何值时,年利率的预报值最大?附:对于一组数据11(,)u v ,22(,)u v ,……,(,)n n u v ,其回归线v u αβ=+的斜率和截距的最小二乘估计分别为:20.(本小题满分12分)在直角坐标系xoy 中,曲线C :y=24x 与直线y kx a =+(a >0)交与M,N 两点,(Ⅰ)当k=0时,分别求C 在点M 和N 处的切线方程;(Ⅱ)y 轴上是否存在点P ,使得当k 变动时,总有∠OPM=∠OPN ?说明理由.21.(本小题满分12分)已知函数f (x )=31,()ln 4x ax g x x ++=-. (Ⅰ)当a 为何值时,x 轴为曲线()y f x = 的切线;(Ⅱ)用min {},m n 表示m,n 中的最小值,设函数}{()min (),()(0)h x f x g x x => ,讨论h (x )零点的个数. 22.(本题满分10分)选修4-1:几何证明选讲 如图,AB 是的直径,AC 是的切线,BC 交于E.(Ⅰ)若D 为AC 的中点,证明:DE 是的切线;(Ⅱ)若3OA CE =,求∠ACB 的大小.23.(本小题满分10分)选修4-4:坐标系与参数方程 在直角坐标系xOy 中,直线1C :x =-2,圆2C :()()22121x y -+-=,以坐标原点为极点, x 轴的正半轴为极轴建立极坐标系. (Ⅰ)求1C ,2C 的极坐标方程; (Ⅱ)若直线3C 的极坐标方程为()4R πθρ=∈,设2C 与3C 的交点为M ,N ,求2C MN ∆的面积.24.(本小题满分10分)选修4—5:不等式选讲 已知函数=|x+1|-2|x-a|,a>0.(Ⅰ)当a=1时,求不等式f (x )>1的解集;(Ⅱ)若f (x )的图像与x 轴围成的三角形面积大于6,求a 的取值范围.【答案解析】 1.【答案】A 【解析】由11z i z +=-得,11i z i-+=+=(1)(1)(1)(1)i i i i -+-+-=i ,故|z|=1,故选A. 考点:本题主要考查复数的运算和复数的模等.2.【答案】D【解析】原式=o o o o sin 20cos10cos 20sin10+ =o sin30=12,故选D. 考点:本题主要考查诱导公式与两角和与差的正余弦公式. 3.【答案】C【解析】p ⌝:2,2nn N n ∀∈≤,故选C. 考点:本题主要考查特称命题的否定 4.【答案】A【解析】根据独立重复试验公式得,该同学通过测试的概率为22330.60.40.6C ⨯+=0.648,故选A.考点:本题主要考查独立重复试验的概率公式与互斥事件和概率公式 5.【答案】A【解析】由题知12(3,0),(3,0)F F -,220012x y -=,所以12MF MF •u u u u r u u u u r = 0000(3,)(3,)x y x y --•- =2220003310x y y +-=-<,解得033y <<,故选A.考点:双曲线的标准方程;向量数量积坐标表示;一元二次不等式解法. 6.【答案】B【解析】设圆锥底面半径为r ,则12384r ⨯⨯==163r =,所以米堆的体积为211163()5433⨯⨯⨯⨯=3209,故堆放的米约为3209÷1.62≈22,故选B.考点:圆锥的性质与圆锥的体积公式 7.【答案】A【解析】由题知11()33AD AC CD AC BC AC AC AB =+=+=+-=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=1433AB AC -+u u ur u u u r ,故选A. 考点:平面向量的线性运算 8.【答案】D【解析】由五点作图知,1+4253+42πωϕπωϕ⎧=⎪⎪⎨⎪=⎪⎩,解得=ωπ,=4πϕ,所以()cos()4f x x ππ=+,令22,4k x k k Z πππππ<+<+∈,解得124k -<x <324k +,k Z ∈,故单调减区间为(124k -,324k +),k Z ∈,故选D. 考点:三角函数图像与性质9.【答案】C【解析】执行第1次,t=0.01,S=1,n=0,m=12=0.5,S=S-m=0.5,2mm ==0.25,n=1,S=0.5>t=0.01,是,循环,执行第2次,S=S-m=0.25,2mm ==0.125,n=2,S=0.25>t=0.01,是,循环, 执行第3次,S=S-m=0.125,2mm ==0.0625,n=3,S=0.125>t=0.01,是,循环,执行第4次,S=S-m=0.0625,2mm ==0.03125,n=4,S=0.0625>t=0.01,是,循环,执行第5次,S=S-m=0.03125,2mm ==0.015625,n=5,S=0.03125>t=0.01,是,循环,执行第6次,S=S-m=0.015625,2mm ==0.0078125,n=6,S=0.015625>t=0.01,是,循环,执行第7次,S=S-m=0.0078125,2mm ==0.00390625,n=7,S=0.0078125>t=0.01,否,输出n=7,故选C.考点:本题注意考查程序框图 10.【答案】C【解析】在25()x x y ++的5个因式中,2个取因式中2x 剩余的3个因式中1个取x ,其余因式取y,故52x y 的系数为212532C C C =30,故选 C.考点:本题主要考查利用排列组合知识计算二项式展开式某一项的系数.【名师点睛】本题利用排列组合求多项展开式式某一项的系数,试题形式新颖,是中档题,求多项展开式式某一项的系数问题,先分析该项的构成,结合所给多项式,分析如何得到该项,再利用排列组知识求解. 11.【答案】B【解析】由正视图和俯视图知,该几何体是半球与半个圆柱的组合体,圆柱的半径与球的半径都为r ,圆柱的高为2r ,其表面积为22142222r r r r r r πππ⨯+⨯++⨯=2254r r π+=16 + 20π,解得r=2,故选B.考点:简单几何体的三视图;球的表面积公式、圆柱的测面积公式 12.【答案】D【解析】设()g x =(21)xe x -,y ax a =-,由题知存在唯一的整数0x ,使得0()g x 在直线y ax a =-的下方.因为()(21)xg x e x '=+,所以当12x <-时,()g x '<0,当12x >-时,()g x '>0,所以当12x =-时,max [()]g x =12-2e -,当0x =时,(0)g =-1,(1)30g e =>,直线y ax a =-恒过(1,0)斜率且a ,故(0)1a g ->=-,且1(1)3g e a a --=-≥--,解得32e≤a <1,故选D.考点:本题主要通过利用导数研究函数的图像与性质解决不等式成立问题13.【答案】1【解析】由题知ln(y x =是奇函数,所以ln(ln(x x +- =22ln()ln 0a x x a +-==,解得a =1. 考点:函数的奇偶性 14.【答案】22325()24x y -+=【解析】设圆心为(a ,0),则半径为4a -,则222(4)2a a -=+,解得32a =,故圆的方程为22325()24x y -+=. 考点:椭圆的几何性质;圆的标准方程 15.【答案】3【解析】作出可行域如图中阴影部分所示,由斜率的意义知,yx是可行域内一点与原点连线的斜率,由图可知,点A (1,3)与原点连线的斜率最大,故yx的最大值为3.考点:线性规划解法16.【答案】【解析】如图所示,延长BA ,CD 交于E ,平移AD ,当A 与D 重合与E 点时,AB 最长,在△BCE 中,∠B=∠C=75°,∠E=30°,BC=2,由正弦定理可得sin sin BC BEE C=∠∠,即o o2sin 30sin 75BE=,解得BE AD ,当D 与C 重合时,AB 最短,此时与AB 交于F ,在△BCF 中,∠B=∠BFC=75°,∠FCB=30°,由正弦定理知,sin sin BF BC FCB BFC =∠∠,即o o2sin 30sin 75BF =,解得AB 的取值-).考点:正余弦定理;数形结合思想17.【答案】(Ⅰ)21n +(Ⅱ)11646n -+ 【解析】试题分析:(Ⅰ)先用数列第n 项与前n 项和的关系求出数列{n a }的递推公式,可以判断数列{n a }是等差数列,利用等差数列的通项公式即可写出数列{n a }的通项公式;(Ⅱ)根据(Ⅰ)数列{n b }的通项公式,再用拆项消去法求其前n 项和.试题解析:(Ⅰ)当1n =时,211112434+3a a S a +=+=,因为0n a >,所以1a =3,当2n ≥时,2211n n n n a a a a --+--=14343n n S S -+--=4n a ,即111()()2()n n n n n n a a a a a a ---+-=+,因为0n a >,所以1n n a a --=2,所以数列{n a }是首项为3,公差为2的等差数列, 所以n a =21n +; (Ⅱ)由(Ⅰ)知,n b =1111()(21)(23)22123n n n n =-++++,所以数列{nb }前n项和为12nb b b +++L =1111111[()()()]235572123n n -+-++-++L =11646n -+. 考点:数列前n 项和与第n 项的关系;等差数列定义与通项公式;拆项消去法18.【答案】(Ⅰ)见解析(Ⅱ)3【解析】 试题分析:(Ⅰ)连接BD ,设BD∩AC=G,连接EG ,FG ,EF ,在菱形ABCD 中,不妨设GB=1易证EG ⊥AC ,通过计算可证EG ⊥FG ,根据线面垂直判定定理可知EG ⊥平面AFC ,由面面垂直判定定理知平面AFC ⊥平面AEC ;(Ⅱ)以G 为坐标原点,分别以,GB GC u u u r u u u r的方向为x 轴,y 轴正方向,||GB u u u r为单位长度,建立空间直角坐标系G-xyz ,利用向量法可求出异面直线AE 与CF 所成角的余弦值. 试题解析:(Ⅰ)连接BD ,设BD∩AC=G,连接EG ,FG ,EF ,在菱形ABCD 中,不妨设GB=1,由∠ABC=120°,可得 由BE ⊥平面ABCD ,AB=BC 可知,AE=EC ,又∵AE ⊥EC ,∴EG ⊥AC ,在Rt △EBG 中,可得,故DF=2.在Rt △FDG 中,可得FG=2在直角梯形BDFE 中,由BD=2,,DF=2可得EF=2, ∴222EG FG EF +=,∴EG ⊥FG ,∵AC∩FG=G,∴EG ⊥平面AFC ,∵EG ⊂面AEC ,∴平面AFC ⊥平面AEC.(Ⅱ)如图,以G 为坐标原点,分别以,GB GC u u u r u u u r 的方向为x 轴,y 轴正方向,||GB u u u r为单位长度,建立空间直角坐标系G-xyz ,由(Ⅰ)可得A (0,,0),E (),F (-1,0,2),C (00),∴AE u u u r =(1),CF uuu r =(-1,,2) (10)分故cos ,||||AE CF AE CF AE CF ⋅<>==u u u r u u u r u u u r u u u r u u u r u u u r .所以直线AE 与CF. 考点:空间垂直判定与性质;异面直线所成角的计算;空间想象能力,推理论证能力19.【答案】(Ⅰ)y c =+适合作为年销售y 关于年宣传费用x 的回归方程类型;(Ⅱ)$100.6y =+46.24【解析】 试题分析:(Ⅰ)由散点图及所给函数图像即可选出适合作为拟合的函数;(Ⅱ)令w =先求出建立y 关于w 的线性回归方程,即可y 关于x 的回归方程;(Ⅲ)(ⅰ)利用y 关于x 的回归方程先求出年销售量y 的预报值,再根据年利率z 与x 、y 的关系为z=0.2y-x 即可年利润z 的预报值;(ⅱ)根据(Ⅱ)的结果知,年利润z 的预报值,列出关于x 的方程,利用二次函数求最值的方法即可求出年利润取最大值时的年宣传费用.试题解析:(Ⅰ)由散点图可以判断,y c =+适合作为年销售y 关于年宣传费用x 的回归方程类型.(Ⅱ)令w =y 关于w 的线性回归方程,由于$81821()()()iii ii w w yy dw w ==--=-∑∑=108.8=6816, ∴$cy dw =-$=563-68×6.8=100.6. ∴y 关于w 的线性回归方程为$100.668y w =+,∴y 关于x 的回归方程为$100.6y =+(Ⅲ)(ⅰ)由(Ⅱ)知,当x =49时,年销售量y 的预报值 $100.6y =+,576.60.24966.32z=⨯-=$. (ⅱ)根据(Ⅱ)的结果知,年利润z 的预报值0.2(100.620.12zx x =+-=-+$,=13.6=6.82,即46.24x =时,z $取得最大值. 故宣传费用为46.24千元时,年利润的预报值最大.……12分考点:非线性拟合;线性回归方程求法;利用回归方程进行预报预测;应用意识20.【答案】0y a --=0y a ++=(Ⅱ)存在【解析】试题分析:(Ⅰ)先求出M,N 的坐标,再利用导数求出M,N.(Ⅱ)先作出判定,再利用设而不求思想即将y kx a =+代入曲线C 的方程整理成关于x 的一元二次方程,设出M,N 的坐标和P 点坐标,利用设而不求思想,将直线PM ,PN 的斜率之和用a 表示出来,利用直线PM ,PN 的斜率为0,即可求出,a b 关系,从而找出适合条件的P 点坐标.试题解析:(Ⅰ)由题设可得)M a ,()N a -,或()M a -,)N a .∵12y x '=,故24x y =在x =,C 在,)a 处的切线方程为y a x -=-0y a --=.故24x y =在x =-处的到数值为,C 在(,)a -处的切线方程为y a x -=+0y a ++=.0y a --=0y a ++=.(Ⅱ)存在符合题意的点,证明如下:设P (0,b )为复合题意得点,11(,)M x y ,22(,)N x y ,直线PM ,PN 的斜率分别为12,k k .将y kx a =+代入C 得方程整理得2440x kx a --=.∴12124,4x x k x x a +==-. ∴121212y b y b k k x x --+=+=1212122()()kx x a b x x x x +-+=()k a b a+. 当b a =-时,有12k k +=0,则直线PM 的倾斜角与直线PN 的倾斜角互补,故∠OPM=∠OPN ,所以(0,)P a -符合题意.考点:抛物线的切线;直线与抛物线位置关系;探索新问题;运算求解能力21..【答案】(Ⅰ)34a =;(Ⅱ)当34a >-或54a <-时,()h x 由一个零点;当34a =-或54a =-时,()h x 有两个零点;当5344a -<<-时,()h x 有三个零点. 【解析】试题分析:(Ⅰ)先利用导数的几何意义列出关于切点的方程组,解出切点坐标与对应的a 值;(Ⅱ)根据对数函数的图像与性质将x 分为1,1,01x x x >=<<研究()h x 的零点个数,若零点不容易求解,则对a 再分类讨论.试题解析:(Ⅰ)设曲线()y f x =与x 轴相切于点0(,0)x ,则0()0f x =,0()0f x '=,即3002010430x ax x a ⎧++=⎪⎨⎪+=⎩,解得013,24x a ==. 因此,当34a =时,x 轴是曲线()y f x =的切线. (Ⅱ)当(1,)x ∈+∞时,()ln 0g x x =-<,从而()min{(),()}()0h x f x g x g x =≤<, ∴()h x 在(1,+∞)无零点.当x =1时,若54a ≥-,则5(1)04f a =+≥,(1)min{(1),(1)}(1)0h fg g ===,故x =1是()h x 的零点;若54a <-,则5(1)04f a =+<,(1)min{(1),(1)}(1)0h f g f ==<,故x =1不是()h x 的零点.当(0,1)x ∈时,()ln 0g x x =->,所以只需考虑()f x 在(0,1)的零点个数.(ⅰ)若3a ≤-或0a ≥,则2()3f x x a '=+在(0,1)无零点,故()f x 在(0,1)单调,而1(0)4f =,5(1)4f a =+,所以当3a ≤-时,()f x 在(0,1)有一个零点;当a ≥0时,()f x 在(0,1)无零点.(ⅱ)若30a -<<,则()f x 在(01)单调递增,故当x ()f x 取的最小值,最小值为f 14.①若f >0,即34-<a <0,()f x 在(0,1)无零点.②若f =0,即34a =-,则()f x 在(0,1)有唯一零点;③若f <0,即334a -<<-,由于1(0)4f =,5(1)4f a =+,所以当5344a -<<-时,()f x 在(0,1)有两个零点;当534a -<≤-时,()f x 在(0,1)有一个零点.…10分 综上,当34a >-或54a <-时,()h x 由一个零点;当34a =-或54a =-时,()h x 有两个零点;当5344a -<<-时,()h x 有三个零点. 考点:利用导数研究曲线的切线;对新概念的理解;分段函数的零点;分类整合思想22.【答案】(Ⅰ)见解析(Ⅱ)60°【解析】试题分析:(Ⅰ)由圆的切线性质及圆周角定理知,AE ⊥BC ,AC ⊥AB ,由直角三角形中线性质知DE=DC ,OE=OB ,利用等量代换可证∠DEC+∠OEB=90°,即∠OED=90°,所以DE 是圆O 的切线;(Ⅱ)设CE=1,由OA =得,AB=AE=x ,由勾股定理得BE ,由直角三角形射影定理可得2AE CE BE =⋅,列出关于x 的方程,解出x ,即可求出∠ACB 的大小.试题解析:(Ⅰ)连结AE ,由已知得,AE ⊥BC ,AC ⊥AB ,在Rt △AEC 中,由已知得DE=DC ,∴∠DEC=∠DCE ,连结OE ,∠OBE=∠OEB ,∵∠ACB+∠ABC=90°,∴∠DEC+∠OEB=90°,∴∠OED=90°,∴DE 是圆O 的切线.(Ⅱ)设CE=1,AE=x ,由已知得AB=BE ,由射影定理可得,2AE CE BE =⋅,∴2x =,解得x考点:圆的切线判定与性质;圆周角定理;直角三角形射影定理23.【答案】(Ⅰ)cos 2ρθ=-,22cos 4sin 40ρρθρθ--+=(Ⅱ)12【解析】 试题分析:(Ⅰ)用直角坐标方程与极坐标互化公式即可求得1C ,2C 的极坐标方程;(Ⅱ)将将=4πθ代入22cos 4sin 40ρρθρθ--+=即可求出|MN|,利用三角形面积公式即可求出2C MN V的面积. 试题解析:(Ⅰ)因为cos ,sin x y ρθρθ==,∴1C 的极坐标方程为cos 2ρθ=-,2C 的极坐标方程为22cos 4sin 40ρρθρθ--+=.……5分(Ⅱ)将=4πθ代入22cos 4sin 40ρρθρθ--+=,得240ρ-+=,解得1ρ=2ρ|MN|=1ρ-2ρ因为2C 的半径为1,则2C MN V 的面积o 11sin 452⨯=12. 考点:直角坐标方程与极坐标互化;直线与圆的位置关系24.【答案】(Ⅰ)2{|2}3x x <<(Ⅱ)(2,+∞) 【解析】试题分析:(Ⅰ)利用零点分析法将不等式f (x )>1化为一元一次不等式组来解;(Ⅱ)将()f x 化为分段函数,求出()f x 与x 轴围成三角形的顶点坐标,即可求出三角形的面积,根据题意列出关于a 的不等式,即可解出a 的取值范围.试题解析:(Ⅰ)当a=1时,不等式f (x )>1化为|x+1|-2|x-1|>1, 等价于11221x x x ≤-⎧⎨--+->⎩或111221x x x -<<⎧⎨++->⎩或11221x x x ≥⎧⎨+-+>⎩,解得223x <<, 所以不等式f (x )>1的解集为2{|2}3x x <<.(Ⅱ)由题设可得,12,1()312,112,x a x f x x a x a x a x a --<-⎧⎪=+--≤≤⎨⎪-++>⎩,所以函数()f x 的图像与x 轴围成的三角形的三个顶点分别为21(,0)3a A -,(21,0)B a +,(,+1)C a a ,所以△ABC 的面积为22(1)3a +. 由题设得22(1)3a +>6,解得2a >. 所以a 的取值范围为(2,+∞).考点:含绝对值不等式解法;分段函数;一元二次不等式解法。
2015届普通高等学校招生全国统一考试(新课标全国卷Ⅰ)模拟英语试题(河南卷)word版版 含答案
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2015年普通高等学校招生全国统一考试(新课标全国卷Ⅰ)模拟卷英语试题第二部分阅读理解(共两节,满分40分)第一节(共15小题:每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑。
ABefore he sailed round the world alone, Francis Chichester had already surprised his friends several times. He had tried to fly round the world but failed. That was in 1931.The years passed. He gave up flying and began sailing. He enjoyed it greatly. Chichester was already 58 years old when he won the first solo transatlantic sailing race. His old dream of going round the world came back, but this time he would sail.His friends and doctors did not think he could do it, as he had lung cancer. But Chichester was determined to carry out his plan. In August 1966, at the age of nearly 65, an age when many men retire, he began the greatest voyage of his life.Chichester covered 14100 miles before stopping in Sydney, Australia. This was more than twice the distance anyone had previously sailed alone. He arrived in Australia on 12 December, just 107 days out from England. He received a warm we lcome from the Australians and from his family who had flown there to meet him. On shore, Chichester could not walk without help. Everybody said the same thing: he had done enough; he must not go any further. But he did not listen.After resting in Sydney for a few weeks, Chichester set off once more in spite of his friends' attempts to dissuade him. The second half of his voyage was by far the more dangerous part, during which he sailed round the treacherous Cape Horn.After succeeding in sailing round Cape Horn, Chichester sent the following radio message to London: "I feel as if I had wakened from a nightmare. Wild horses could not drag me down to Cape Horn and that sinister Southern Ocean again."Just before 9 o'clock on Sunday evening 28 May 1967, he arrived back in England, where a quarter of a million people were waiting to welcome him.Queen Elizabeth II knighted(授以爵位) him with the very sword that Queen Elizabeth I had used almost 400 years earlier to knight Sir Francis Drake after he had sailed round the world for the first time.The whole voyage from England and back had covered 28,500 miles. It had taken him nine months, of which the sailing time was 226 days. He had done what he wanted to accomplish.21.What can we learn about Chichester?A.He failed the solo transatlantic sailing race in 1959.B.He was a brave and determined man.C.The second half of his voyage was not as dangerous as the first half.D.The radio message expressed his concern about the sailing.22.What did Queen Elizabeth II do after Chichester arrived back in England?A.She called on the English to learn form him.B.She was waiting to congratulate on his success in sailingC.She thought poorly of his achievements.D.She knighted him for praising him.23.We can infer from the text that ____A.Anyone who had sailed alone traveled less than 7050miles before 1966B.Chichester sailed round the Atlantic in 1931C.Most of the English retire at the age of 65D.Chichester died of lung cancer in 1967 after he went back to EnglandBBanquets are usually held in restaurants in private rooms that have been reserved for the purpose. You will be met at the door and led to the banquet room. Traditionally, the head of your delegation should enter the room first. Do not be surprised if your hosts greet you with a loud round of applause. The proper response is to applaud back.Seating arrangements are stricter than in the West. Guests should never assume that they may sit where they please and should wait for hosts to guide them to their places. Traditionally, the Chinese regard the right side as the superior and the left side as the inferior. Therefore on formal occasions, the host invariably arranges for the main guests to sit on his right side.It is the host's responsibility to serve the guests, and at very formal banquets people do not begin to eat until the host has served a portion to the principal guest. Or, the host may simply raise his chopsticks and announce that eating has begun. After this point, one may serve oneself any food in any amount. Remember to go slow on eating. Don't fill yourself up when five courses are left to go. To stop eating in the middle of a banquet is rude, and your host may incorrectly think that something has been done to offend you.Drinking takes an important place in Chinese banquets. It is likely that the host will stand and hold his glass out with both hands while saying a few words. When he says the words "gan bei", which means bottoms up, all present should drai n their glasses. After this initial toast, drinking and toasting are open to all. No words are needed to make a toast, and it is not necessary to drain your glass, although to do so is more respectful. When filling another glass, it is polite to fill it as full as you can. This symbolizes full respect and friendship.When the last dish is finished, the banquet has officially ended. There is little ceremony involved with its conclusion. The host may ask if you have eaten your fill. Then the principal host will rise, signaling that the banquet has ended. Generally, the principal host will bid good evening to everyone at the door and stay behind to settle the bill with the restaurateur. Other hosts usually accompany guests to their vehicles and remain outside waving until the cars have left the premises.24.To attend a formal banquet,you_______.A.may enter the banquet room directly when you arrive .B.may stop eating in the middle of the banquet.C.may help yourself to any food immediately the dishes are served.D.should applaud back when greeted with a loud round of appla use .25.What is the host …s responsibility in the author‟s eyes?A.Announcing that eating ends.B.Driving guests to their home.C.Filling the guests‟ glasses.D.Arranging for guests to go to their place..26.We can conclude form the passage that____A.To stop eating halfway means that the dishes taste bad.B.Guests should drain their g1ass es the instant they are filledC.Important guests are arranged to sit on the host‟s right sideD.The more you drink , the more you respect the host.27.The passage mainly tells us ____A.Chinese food is delicious.B.customs and traditions at banquets in ChinaC.what food guests should eat at banquetsD.different customs between China and western countries.CA machine that takes sweat-laden clothes and turns the moisture into drinking water is in use in Sweden.The device spins and heats the material to remove the sweat, and then passes the vapor through a special membrane designed to only let water molecules get through.Since its Monday launch, its creators say more than 1,000 people have "drunk other's sweat" in Gothenburg.They add the liquid is cleaner than local tap water.The device was built for the United Nation's child-focused charity UNICEF to promote a campaign highlighting the fact that 780 million people in the world la ck access to clean water. The machine was designed and built by engineer Andreas Hammar, known locally for his appearances on TV tech show Mekatronik.He said the critical part of the sweat machine was a new water purification component developed by a company named HVR in collaboration with Sweden's Royal Institute of Technology."It uses a technique called membrane distillation(膜蒸馏)," he told the BBC."We use a substance that's a bit like Gortex that only lets steam through but keeps bacteria, salts, clothing fibers and other substances out."They have something similar on the International Space Station to treat astronaut's urine - but our machine was cheaper to build.V olunteers have been sampling the treated sweat since the start of the week in Gothenburg"The amount of water it produces depends on how sweaty the person is - but one person's T-shirt typically produces 10ml , roughly a mouthful."The equipment has been put on show at the Gothia Cup - the world's largest international youth football tournament.MattiasRonge, chief executive of Stockholm-based advertising agency Deportivo - which organized the stunt(惊人的表演) - said the machine had helped raise awarenessfor UNICEF, but in reality had its limitations."People haven't produced as much sweat as we hoped - right now the weather in Gothenburg is lousy," he said."So we've installed exercise bikes alongside the machine and volunteers are cycling like crazy."Even so, the demand for sweat is greater than the supply. And the machine will never be mass produced - there are better solutions out there such as water purifying pills."28.Which of the following is the disadvantage of the machine?_______.A.The water processed by the machine is cleaner than local tap water .B.The amount of water the machine produces is too small.C.the machine is cheaper than the similar one on the International Space StationD.The machine can help raise awareness of lacking water in the world .29.What is the main idea of the fifth paragraph ?A.How the machine works.B.Who developed the machine.C.How the machine was invented.D.Why the machine was invented..30.What does Mttias Ronge think of the machine?____A.The machine can solve the severe water shortage in the world..B.People do not like the water the machine processesC.The machine should be used in rathe r hot areas in summer.D.The machine is not worth popularizing..31.The passage is most likely to be____A.a tourist brochureB.a book reviewC.a news reportD.a blog.DMany people rely on a cup of coffee or two to wake them up in the morning or pick them up during the working day, but now a chemist has come up with a speedy alternative to crafting a cup of coffee.U.S. biochemist Ben Y u has created `Sprayable Energy,` which claims to be the world`s first caffeine - based topical energy spray.He said tired workers can spray a `shot` of caffeine onto their skin without experiencing a strong buzz, loading up on unnecessary calories or being stuck with a nasty aftertaste like they might get from drinking energy drinks or coffee.The patent-pending caffeine spray is an odourless liquid that is absorbed through the skin and distributed through the body over a number of hours to deliver a caffeine hit that apparently lastslonger than guzzling a cup of coffee.Each small aluminum bottle of Sprayable Energy contains around 160 sprays - the equivalent amount of caffeine to 40 cups of coffee and the creators say it is a much cheaper way of getting a caffeine fix than popping to a cafe.The only active ingredient in the spray is caffeine, which can naturally enter the human body through the skin by passing through cell membranes as it is very similar to nicotine in structure.Each spritz of Sprayable Energy contains around a quarter of the amount of caffeine found in a cup of coffee, but apparently has the same effect as a full cup.The website said: `The reason for this is our product not being ingested, isn`t almost entirely metabolised(新陈代谢)by the liver before entering your system and becoming available to your body.``Thus, a smaller amount of caffeine can have just the same effect as a very large amount of caffeine ingested through an energy drink or cup of coffee.`c`s website recommends that users apply the spray in places where they normally spritz perfume, such as the neck or wrists, but warns users not to exceed 20 sprays a day.It claims that after spraying the product on the skin, users will feel `awake and focused without being over-stimulated,` which is common with coffee and energy drinks.32.What can we learn about Sprayable Energy from the passage ?_______.A.A bottle of Sprayable Energyis cheaper than 40 cups of coffee. .B.Sprayable Energy lets users not worry about taking in unecessary calories..C.Sprayable Energy can be used at least 20 sprays a dayD.The caffeine spray is a colorless liquid that is absorbed through the skin..33.What does the fifth paragraph mainly tell us ?____A,Sprayable Energy is taken by mouth...B.Before Sprayable Energy is absorbed by body, the liver entirely metabolizes itC.How Sprayable Energy acts on user‟s body and how users use the product.D.Sprayable Energy is convenient but expensive.34.We can conclude from the passage that ____A.Sprayable Energy can not be permitted to be used by people now.B. Sprayable Energy has not already gone into mass productionC.each spritz of Sprayable Energy has the same effect as a full cup of coffeeD.Sprayable Energy is well received by flagging workers.35.The writer‟s attitude towards Sprayable Energy ____A.negative.B. positiveC.uncertainD.indifferent第二节(共5小题;每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。
2015年高考全国卷1英语试题及答案
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2015年普通高等学校招生全国统一考试(全国卷1)英语本试卷分第I卷(选择题)和第II卷(非选择题)两部分。
考试结束后,将本试卷和答题卡一并交回。
第I卷第一部分听力(共两节,满分30分)做题时, 先将答案标在试卷上。
录音内容结束后, 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, 并标在试卷的相应位置。
听完每段对话后, 你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
例:How much is the shirt?A. £19。
15.B. £9.18.C. £9.15。
答案是C。
( )1。
What time is it now?A。
9:10 B。
9:50 C。
10:00()2. What does the woman think of the weather?A。
It's nice. B。
It’s warm。
C. It's cold.( )3。
What will the man do?A。
Attend a meeting. B. Give a lecture. C. Leave his office。
( )4。
What is the woman’s opinion about the course?A. Too hard。
B. Worth taking.C. Very easy。
()5. What does the woman want the man to do?A. Speak louder。
B. Apologize to her.C. Turn off the radio.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。
2015年高考文科数学全国卷1(含详细答案)
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数学试卷 第1页(共30页)数学试卷 第2页(共30页)数学试卷 第3页(共30页)绝密★启用前2015年普通高等学校招生全国统一考试(全国新课标卷1)数学(文科)使用地区:河南、山西、河北、江西本试卷分第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.第Ⅰ卷(选择题 共60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.已知集合{|}32,A x x n n ==+∈N ,{6,8,10,12,14}B =,则集合A B 中元素的个数为( )A .5B .4C .3D .22.已知点0,1A (),3,2B (),向量AC =43--(,),则向量BC =( )A (-7,-4)B .(7,4)C .(-1,4)D .(1,4) 3.已知复数z 满足(z -1)i=1+i ,则z=( )A .-2-iB .-2+iC .2-iD .2+i4.如果3个正整数可作为一个直角三角形三条边的边长,则称这3个数为一组勾股数.从1,2,3,4,5中任取3个不同的数,则这3个数构成一组勾股数的概率为( )A.310B .15C .110D .1205.已知椭圆E 的中心在坐标原点,离心率为12,E 的右焦点与抛物线28C y x =:的焦点重合,A ,B 是C 的准线与E 的两个交点,则|AB |=( )A .3B .6C .9D .126. 《九章算术》是我国古代内容极为丰富的数学名著,书中有如下问题:“今有委米依垣内角,下周八尺,高五尺.问:积及为米几何?”其意思为:“在屋内墙角处堆放米(如图,米堆为一个圆锥的四分之一),米堆底部的弧长为8尺,米堆的高为5尺,问米堆的体积和堆放的米各为多少?”已知1斛米的体积约为1.62立方尺,圆周率约为3,估算出堆放的米约有( )A .14斛B .22斛C .36斛D .66斛7.已知{}n a 是公差为1的等差数列,n S 为n {}a 的前n 项和.若844S S =,则10a = ( )A .172B .192C .10D .128.函数=cos(+)x f x ωϕ()的部分图象如图所示,则f x ()的单调递减区间为 ( )A .13π,π+44k k k -∈Z (),B .132π,2π+44k k k -∈Z (), C .13,+44k k k -∈Z (),D .132,2+44k k k -∈Z (),9.执行如图所示的程序框图,如果输入的0.01t =,则输出的n = ( )A .5B .6C .7D .810.已知函数1222, 1,()log (1), 1,x x f x x x -⎧-=⎨-+⎩≤>且()3f a =-,则(6)f a -= ( )A .74-B .54-C .34-D .14-11.圆柱被一个平面截去一部分后与半球(半径为r )组成一个几何体,该几何体三视图中的正视图和俯视图如图所示.若该几何体的表面积为16π20+,则r = ( )A .1B .2C .4D .812.设函数()y f x =的图象与2x a y +=的图象关于直线y x =-对称,且(2)(4)f f -+-1=,则a =( )A .1-B .1C .2D .4--------在--------------------此--------------------卷--------------------上--------------------答--------------------题--------------------无--------------------效----------------姓名________________ 准考证号_____________数学试卷 第4页(共30页)数学试卷 第5页(共30页)数学试卷 第6页(共30页)第Ⅱ卷(非选择题 共90分)本卷包括必考题和选考题两部分.第13~21题为必考题,每个试题考生都必须作答,第22~24题为选考题,考生根据要求作答.二、填空题:本大题共4小题,每小题5分,共20分.把答案填在题中的横线上. 13.在数列{}n a 中12a =,12n n a a +=,n S 为{}n a 的前n 项和.若126n S =,则n =_____.14.已知函数31f x ax x =++()的图象在点1,1f (())处的切线过点(2,7),则a =_____. 15.若x ,y 满足约束条件20,210,220,x y x y x y +-⎧⎪-+⎨⎪-+⎩≤≤≥则z 3x y =+的最大值为_____.16.已知F 是双曲线2218yC x -=:的右焦点,P 是C 的左支上一点,0,66A ().当APF △周长最小时,该三角形的面积为_____.三、解答题:本大题共6小题,共70分.解答应写出必要的文字说明、证明过程或演算步骤. 17.(本小题满分12分)已知a ,b ,c 分别是ABC △内角A ,B ,C 的对边,2sin 2sin sin B A C =. (Ⅰ)若a b =,求cos B ;(Ⅱ)若B =90°,且2a =,求ABC △的面积. 18.(本小题满分12分)如图,四边形ABCD 为菱形,G 为AC 与BD 交点,BE ABCD ⊥平面. (Ⅰ)证明:平面AEC ⊥平面BED ;(Ⅱ)若ABC ∠=120°,AE EC ⊥,三棱锥E ACD -的体积为63,求该三棱锥的侧面积.19.(本小题满分12分)某公司为确定下一年度投入某种产品的宣传费,需了解年宣传费x (单位:千元)对年销售量y (单位:t )和年利润z(单位:千元)的影响,对近8年的年宣传费i x 和年销售量i y (i =1,2,…,8)数据作了初步处理,得到下面的散点图及一些统计量的值.xyω28i=1()ixx -∑28i=1()iωω∑-8i=1()()iiy x x y-∑-8i=1()()ii y y ωω--∑46.65636.8289.8 1.6 1 469108.8表中i ω=i x ,ω=188ii=1ω∑(Ⅰ)根据散点图判断,y a bx =+与y c d x =+哪一个适宜作为年销售量y 关于年宣传费x 的回归方程类型?(给出判断即可,不必说明理由)(Ⅱ)根据(Ⅰ)的判断结果及表中数据,建立y 关于x 的回归方程;(Ⅲ)已知这种产品的年利率z 与x ,y 的关系为z=0.2y -x .根据(Ⅱ)的结果回答下列问题:(i )年宣传费x =49时,年销售量及年利润的预报值是多少? (ii )年宣传费x 为何值时,年利润的预报值最大?附:对于一组数据11()u v ,,22(,)u v ,…,(,)n n u v ,其回归直线v u αβ=+的斜率和截距的最小二乘估计分别为121()(),()nii i nii uu v v v u uu βαβ==--==--∑∑.20.(本小题满分12分)已知过点(0,1)A 且斜率为k 的直线l 与圆22 ()2(3)1C x y -+-=:交于M ,N 两点. (Ⅰ)求k 的取值范围;(Ⅱ)若12OM ON ⋅=,其中O 为坐标原点,求||MN . 21.(本小题满分12分)设函数()2ln x f x e a x =-.(Ⅰ)讨论()f x 的导函数()f x '的零点的个数; (Ⅱ)证明:当0a >时,()22ln f x a a a+≥.请考生在第22~24三题中任选一题作答,如果多做,则按所做的第一题计分. 22.(本小题满分10分)选修4—1:几何证明选讲如图,AB 是⊙O 的直径,AC 是⊙O 的切线,BC 交⊙O 于点E . (Ⅰ)若D 为AC 的中点,证明:DE 是⊙O 的切线;(Ⅱ)若OA =3CE ,求∠ACB 的大小.23.(本小题满分10分)选修4—4:坐标系与参数方程在直角坐标系xOy 中,直线1C :x =-2,圆2C :(x -1)2+(y -2)2=1,以坐标原点为极点,x 轴的正半轴为极轴建立极坐标系. (Ⅰ)求1C ,2C 的极坐标方程;(Ⅱ)若直线3C 的极坐标方程为()π4θρ=∈R ,设2C 与3C 的交点为M ,N ,求2C MN △的面积.24.(本小题满分10分)选修4—5:不等式选讲已知函数12f x =|||x |x a --+(),0a >. (Ⅰ)当=1a 时,求不等式1f x >()的解集; (Ⅱ)若f x ()的图象与x 轴围成的三角形面积大于6,求a 的取值范围.3 / 102015年普通高等学校招生全国统一考试(全国新课标卷1)数学(文科)答案解析第Ⅰ卷{8,14A B =【答案】A 【解析】(3,1)AB OB OA =-=(7,BC AC AB ∴=-=-,故选A.【考点】向量运算 【答案】C【解析】(1)i 1i z -∴=+,22i (12i)(i)2i i z +-==--∴=【解析】抛物线,1e 2c a ==的方程解得(2,3)A -数学试卷 第10页(共试卷 第11页(共30页)数学试卷 第12页(共30页)【解析】()f a =-5 / 10第Ⅱ卷【解析】12a =,64,6n ∴=【考点】等比数列定义与前【答案】1【解析】()3f x '=又(1)f a =切线过(2,7),∴【考点】利用导数的几何意义求函数的切线,常见函数的导数【答案】4平移直线l ,当直线数学试卷 第16页(共30页) 数学试卷 第17页(共30页)数学试卷 第18页(共30页)(0,66)A ∴直线AF 66y =或22APF S ∴=△22ac .,由余弦定理可得2221cos 24a cb B ac; 22ac .因为90B ,由勾股定理得222a c b ,故222a c ac ,得2c a ,所以先由正弦定理将22sin B A =化为变得关系,结合条件a b =,用其中一边把另外两边B 22ac ,根据勾股定理即可求出7 / 10ACBD ,因为BE平面ABCD AC BE ,故AC 平面BED AEC平面BED (Ⅱ)设AB x ,在菱形中,由120ABC ,可得32AG GC x ,2x GB GD . 因为AE EC ⊥,所以在可得32EG x ,由BE 平面ABCD EBG △为直角三角形,22BEx ,由已知得,E ACD 的体积3116632243E ACD V AC GD BE x,故2x ,从而AE EC ==EAC 的面积为3,EAD △的面积与ECD △的面积均为5. 故三棱锥EACD 的侧面积为5.(Ⅰ)由四边形AC BD ,由BE平面ABCD 知ACBE ,由线面垂直判定定AC平面BED ,由面面垂直判定定理知AEC 平面BED ;AB x ,通过解直角三角形将,GC ,GB ,GD 用x 表示出来,AEC △中,根据条件三菱锥EACD 的体积为x ,即可求出三菱锥E ACD 的侧面积. 【考点】线面垂直的判定与性质,面面垂直的判定,三棱锥的体积与表面积的计算(Ⅰ)由散点图可判断,关于年宣传费用c y dw ∴=-(Ⅲ)(ⅰ)由(Ⅱ)知,当576.60.2z =⨯(ⅱ)根据(Ⅱ)的结果知,年利润的预报值0.2(100.6z =13.62x =时,z 取得最大值,故宣传费用为【提示】(Ⅰ)由散点图及所给函数图像即可选出适合作为拟合的函数;关于w 的线性回归方程,即可的回归方程先求出年销售量数学试卷 第22页(共30页) 数学试卷 第23页(共30页)数学试卷 第24页(共30页)1ykx ,因为l 231|11k k,474733k,所以k 的取值范围是4747,33.(Ⅱ)设11(,)M x y ,22(,)N x y ,将1ykx 代入方程22(2)(3)1x y ,22(1)4(1)70k xk x ,所以1224(1)1k x x k ,12271x x k . 21212121224(1)y (1)()181k k OM ONx x y k x x k x x k ,由题设可得24(1)8=121k k k ,解得所以l 的方程为1y kx ,故圆心在直线l 上,所以||2MN =.【提示】(Ⅰ)设出直线l 的方程,利用圆心到直线的距离小于半径列出关于的不等式,即可求出值范围;(Ⅱ)设()M x (,)N x y 方程代入圆的方程化为关于x 的一元二次方程,利用韦达定理将表示出来,利用平面向量数量积的坐标公式及12OM ON =列出关于(0,),2()=2(0)xaf x e x x. 0a 时,()0f x ,()f x 没有零点,当0a 时,因为2e x 单调递增,ax单调递增,()f x 在(0,)单调递增,又()0f a ,当b 满足04a b 且14b 时,(b)0f ,故当0a 时,()f x 存在唯一零点;(Ⅱ)由(Ⅰ),可设()f x 在(0,)的唯一零点为0x ,当0(0,)x x 时,()0f x ,当0(,)x x 时,()0f x ,)单调递减,在0(,)x 单调递增,所以当0xx 时,()f x 取得最小值,最小值为(f 0=0a x ,所以0022()=2ln2ln2a f x ax a a a x aa ,故当0a 时,2()2ln f x a a a. 【提示】(Ⅰ)先求出导函数,分0a 与0a 考虑()f x 的单调性及性质,即可判断出零点个数;(Ⅱ)由(Ⅰ)可设()f x 在(0,)的唯一零点为0x ,根据()f x 的正负,即可判定函数的图像与性质,求出函数的最小值,即可证明其最小值不小于22lna a a,即证明了所证不等式,90ACB∠+,90∴∠,90,DE∴1=,12BE=-,由射影定理可得,CE BE,2x,解得60.90,即90∠,所以,设AE=,由勾股定理得CE BE,列出关于的方程,解出x,即可求出ACB∠【考点】圆的切线判定与性质,圆周角定理,直角三角形射影定理cosxρθ=40+=;(Ⅱ)将2=,|MN1452=.(Ⅰ)用直角坐标方程与极坐标互化公式即可求得π代入9/ 10数学试卷第28页(共30页)数学试卷第29页(共30页)数学试卷第30页(共30页)。
2015年高考全国卷1及答案(语文)
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新课标Ⅰ2015年普通高等学校全国统一考试语文注意事项:1.本试卷分第I卷(阅读题)和第II卷(表达题)两部分.2.考生务必将自己的姓名、考生号填写在答题卡上。
3.作答时,将答案写在答题卡上。
写在本试卷上无效。
4.考试结束后.将本试卷和答题卡一并交回。
第I卷阅读题甲必考题一、现代文阅读(9分,毎小题3分)阅读下面的文字,完成1〜3题宋代的农业、手工业、商业在唐代的基础上又有了新的发展,特别是商品经济出现了空前的繁荣,在此背景下,宋代的货巾流通和信用进入迅速发展时期,开创了古代金融的新篇章。
宋代在信用形式和信用工具方面都呈现出新的特点。
信用形式有借贷、质、押、典、赊买赊卖等多种形式。
借贷分为政府借贷和私人借贷。
政府借贷主要表现为赈贷的形式,在紧急情况下通过贷给百姓粮食或种子的方式,帮助他们度过困境。
私人借贷多为高利贷,它可以解决社会分化和“钱荒”带来的平民百姓资金严重不足的问题,满足特殊支付和燃眉之急的需要。
质、押是借贷的担保形式,由质库、解库等机构经营。
质属于动产担保,它必须转移动产的占有;押属于不动产担保,通常将抵押物的契约交付债权人即可。
债务人违约时,债权人可用变卖价款优先受偿。
典作为不动产转移的一种形式是在宋代形成和发展起来的。
其特点是典权人向出典人支付典价后,在典期内就占有了出典人典产的使用权和收益支配权,出典人也不必向典权人支付利息。
宋代的商业贸易非常发达,但存在着通货紧缩现象,故赊买赊卖行为也很普遍,几乎生产、流通、消费领域的所有物品都能进行赊买赊卖。
从实际效果看,它解决了军需、加强了流通,更重要的一点,它对束缚生产流通扩大和发展的高利贷构成了冲击。
随着社会经济的发展,宋代商业贸易对货币的要求越来越高,但是社会中货币供给和流通状况不尽理想,表现为货币流通区域的割据性、货币供给数量的有限性,以及大量流通的铜铁钱细碎和不便携带的特性,其结果是抑制了经济发展。
为了解决这类问题,在高度发达的造纸和印刷技术保障下,通过民间自发力量的作用和官府的强制推行,宋代社会陆续出如现了诸茶引、盐引、交子、关子和会子等新型纸质信用工具。
2015年高考模拟试卷数学卷1答案
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2015年高考模拟试卷数学(文科)卷参考答案及评分标准一、选择题(本大题共8小题,每小题5分,共40分.)1、C2、D3、C4、B5、A6、C7、B8、A二、填空题(本大题共7小题,第9题每空2分,第10,11,12题每空3分,第13,14,15题每空4分,共36分.)9、 -1 , 1, 10、①},125|{Z k k x x ∈-=ππ ② ))(0,62(Z k k ∈-ππ 11、2312、①n a n = ②2301932++n n 13、 14、3π 15、449≤a 三、解答题(共5小题,共74分.解答时应写出必要的文字说明、证明过程或演算步骤) 16、(本小题满分15分)解:(1),,-----------------2分,,---------------4分------------------------6分(2)正根据弦定理可得:,-----------8分 ,=-----------------------12分在三角形ABC 中,得到的范围:30,4π⎛⎫⎪⎝⎭,52,444B πππ⎛⎫-∈- ⎪⎝⎭,-----------14分 则范围:(0,2+------------------------------------------------------15分17、(本题满分15分)解:当时,,得. -----------------------1分3A A C A ac c a b cos sin )cos(222+=-- A A B ac B ac cos sin cos cos 2-=-∴0cos ≠∴B 1cos sin 2=∴A A 12sin =A 即4,22ππ==∴A A CcB b A a sin sin sin ==C B bc sin sin 4=∴B C -=43π)43sin(sin 4B B bc -=∴π)sin 22cos 22(sin 4B B B +)2cos 1(22sin 2B B -+=2)42sin(2+-=⇒πB bc B bc 1n =()1111S t S a =-+11a =当时,由,即,① 得,,②,即,------------------------3分 是等比数列,且公比是,. ------------------------4分(1),即,若数列为等比数列,则有,而,故,解得, ------------------------6分 再将代入,得, 由,知为等比数列,. ------------------------7分 (2)由,知,,-----------------------8分 , ------------------------10分 由不等式恒成立,得恒成立,------------------------12分设,由, 当时,,当时,, 而------------------------14分 . ------------------------15分 18、(本题满分15分)证明:(Ⅰ) 取PD 中点为M ,连MF ME , ∵ E 是PC 的中点2n ≥()1n n n S t S a =-+()1n n t S ta t -=-+()111n n t S ta t ---=-+()11n n n t a ta ta --=-+()11,2nn n n a a ta t n a --=∴=≥{}n a ∴t n n a t ∴=()()211n n nn t t b tt t-=+⋅-212121n n n n t t t b t+++-=-{}n b 2213b b b =⋅()()23421232,21,21b t b t t b t t t ==+=++()()()2324221221t t t t t t ⎡⎤+=⋅++⎣⎦12t =12t =n b 1()2n n b =112n n b b +={}n b 12t ∴=12t =1()2n n a =14()12n n c ∴=+111422441212n n nT n n ⎛⎫- ⎪⎝⎭∴=⨯+=+--12274n k n n T ≥-+-2732nn k -≥272n n n d -=111252729222n nn n n n n n d d +++---+-=-=∴4n ≤1n n d d +>4n ≥1n n d d +<454513,,1632d d d d ==∴<313,3232k k ∴≥∴≥∴ME 是PCD ∆的中位线,∴ME CD 21------------------------2分 ∵ F 是AB 中点且ABCD 是菱形,ABCD ,∴ME AB 21. ∴MEFB ∴ 四边形MEBF 是平行四边形. ------------------------5分从而 MF BE //, ---------------------6分∵ BE 平面PDF ,MF 平面PDF , ∴ BE ∥平面PDF ------------------------7分 (Ⅱ)由(Ⅰ) 得MF BE //,∴直线BE 与平面PAD 所成的线面角就是直线MF 与平面PAD 所成的线面角。
2015年高考模拟试卷及参考答案 全国卷1.pdf
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2015年高三模拟考试英语试题全国卷1第二部分阅读理解(共两节,满分40分) 第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑。
AIf the word researcher brings to mind someone who works in a laboratory doing experiments or reads a11 day in a librar y,then meet Dr.Sylvia Earle.Dr.Earle has carried out much of her research deep under the sea.Her outstanding work at sea ha s included leading around 70 expeditions(探索)and spending over 6,500 hours under the water. She also holds the world record for the deepest walk alone on the ocean floor at a depth of 381 metres.It’s no wonder she ha s been given the nickname“Her Deepness”.As a child,Dr.Earle was always interested in wildlife and was never afraid to find out about nature.Her parents told her to touch animals and insects,and not to be afraid of them.When she was older,she was given the opportunity to develop this int erest thanks to scho1arships to study botany at university.She also took courses in sea diving and decided to devote herself t o marine(海洋的)biology.It was this decision and her determined character that led her to have a truly remarkable career.After several years of diving,Dr.Earle took part in a research expedition in an underwater laboratory.The laboratory was called a Tektite Habitat and it allowed divers to remain under the water to carry out research for weeks at a time.In 1970 an all-female group went on the sixth expedition of the Tektite II scientists-in-the-water programme.Dr.Earle was asked to lead the expedition.Along with another four women-three scientists and an engineer-she stayed in the Tektite Habitat for two wee ks.Marine life has always been at the centre of Dr.Earle’s work and she is highly respected for her authority in this field.She i s often asked to give talks by environmental groups about marine life as well as her plan to create a global network of marin e protected areas. What makes her extra special is that she doesn’t look upon her work as a job,but as her vocation(天职)in life.21.In what way is Dr.Earle different from most researchers we know about?A.She works in a laboratory doing experiments all day.B.She reads all day in a library.C.She carries out much of her research deep under the sea.D.She has broken lots of world records.22.What’s the main idea of Paragraph 2?A.Dr.Earle’s interest in wildlife led her to a truly remarkable career.B.Dr.Earle’s success is chiefly due to her family background.C.Dr.Earle’s success is chiefly due to the scholarships to study botany at university.D.V arious factors have contributed to Dr.Earle’s success.23.Which of the following is true according to Paragraph 3?A.Five women took part in the expedition.B.Five scientists took part in the expedition.C.Dr.Earle was an ordinary member of the group.D.The group stayed under water for three weeks.24.What does Dr.Earle hope to achieve in the future?A.To keep marine life at the centre of her work.B.To create protected areas in the oceans around the world.C.To travel around the world giving talks.D.To look upon her work as her vocation in life.BA paperclip,made of steel wire bent into a looped()shape,is an instrument used to hold sheets of paper toget her.This common device is a wonder of simplicity and function.But where did this simple,cheap,and practical invention co me from?In the late 19th century,the most common way to hold papers together was by using a pin.Although the pin was an ine xpensive tool and was easily removable,it would 1eave holes in the ter,as steelvwire became more common,invent ors began to notice its elastic()feature.With this feature,it could be stretched and twisted into various clip-like objects.In the years just before 1900,quite a few paperclip designs appeared.The name most frequently associated with the paperclip invention is Johan Vaaler,a Norwegian inventor.However,Vaaler’s clips were not the same as the paperclips curr entlyin use.Specifically,they did not have the inside loop we see today.The familiar looped design was invented by Gem Manuf acturing Ltd.in England.This clip is therefore sometimes known as the Gem clip.Because of Vaaler,the paperclip played an important historical role in Norway.During World War II,Norway was occu pied by the Nazis.Norwegians were prohibited from wearing any symbol of their national unity(),such as buttons wi th the initials of their king.Thus,in protest,they started wearing paperclips to show their unity.The reason for doing this was simple:Paperclips were a Norwegian invention whose original function was to bind together.After the war,a giant papercli p statue was put up in Oslo to honour Vaaler—even though his design was never actually produced.25.According to the first paragraphthe paperclip is________.A.made of paperB.for holding clothes togetherC.shaped like a pinD.inexpensive and useful26.One way the paperclip is better than the pin is that_________A.it is cheaperB.it is simplerC.it doesn’t damage the paperD.it can be removed more easily27.Which of the following best shows what the Gem clip looks like?28.The last paragraph is mainly about_________.A.how widely used Vaaler’s clip isB.how V aalers clip became a national symbolC.how the Nazis ruled the Norwegian peopleD.why Norwegians had the initials of their king on their buttonsCI was born on the last day of February.I’ve always felt sorry for February,squeezed between the big months of January( named for the Roman god Janus,keeper of gateways)and March(after Mars,the god of war).The first Roman calendar,it is said,had 10 months and no February.Beginning at the vernal equinox()with March,i t ended with December.In an agricultural society,winter was of little importance,and thus went undivided.January and February were added about 700 B.C.by the second king of Rome,Numa Pompilius.He made all the months 29 or 31 days,but shortened February,the last month of the year,by giving it only 28.Next,it was the church’s turn.In 1582 Pope Gregory XIII announced a new calendar in Europe.Many changes were ma de,but the Pope passed up yet another chance to give February equality with the other months.It’s messy,even dangerous,changing how we measure time,but Pope Gregory was hardly the last one to try.The League o f Nations received over 150 new calendar designs,and the United Nations has considered more proposals since.Each propos al involves something that supposedly modernizes the calendar.But I have a simpler proposal that won’t lead to chaos,and will correct the historical injustices against Februarymove the last day of January and the last day of March into February to make it a normal month with 30 days,and a respecta ble 31 on leap years().This would not add or subtract a single day from the calendar year.It’s a great idea.And unlike J ulius and Augustus,I won’t even demand a month named in my honor.29.The first Roman calendar had only 10 months because_________.A.February hated to be squeezed between the big monthsB.the god of war always fought with January and FebruaryC.the Romans did not like the cold winter weatherD.it was Considered unnecessary to divide winter30.Which of the following is closest to Numa Pompilius’calendar?31.The writer sounds_____in the passage.A.sadB.humorousC.happyD. angryDOne important thing during the pre-Christmas rush at our house was the arrival of my daughter’s kindergarten report car d.She got high praise for her reading,vocabulary and overall enthusiasm.On the other hand,We learnt that she has work to do on her numbers and computer skills,though the detailed handwritten report her teachers prepared is absent of any words tha t might be interpreted as negative()in describing her efforts.A number System indicates how she’s measuring up in e ach area without any mention of passing or failing.The debate over whether formal grades should be given to kids has long existed.At one level,the advantages and disadva ntages are obvious.A grade system provides a straightforward standard by which to measure how your child is progressing a t school--and how he or she is getting on compared to other children.But as writer Sue Ferguson notes,Grades can deceive. ()The aim should be”to measure learning,not simply what a student can remember on a test.”The two aren’t the same-and if you doubt that as an adult,ask yourself whether you could sit down without any preparation and still pass those high-s chool-level examinations.If you’re old enough,You’ve lived through this debate before.At one time,it was considered unfair to put children in dire ct competition with one another if it could be avoided.The intention behind that may have been good,but it ignored the fact t hat competition,and the will to come out on top,are essential components of the human condition.32.Which of the following seems to be weak on the daughter’s kindergarten report card?A.Reading.B.V ocabulary.C.Enthusiasm.D.Numbers.33.what did the teachers try to do when preparing the report card?A.Avoid using negative words.B.Describe the facts as they really were.e numbers instead of words.D.Avoid showing weak areas.34.What does the writer suggest by the underlined part in Paragraph 2?A.Most adults can pass high-school-level examinations without preparation.B.There’s no doubt that most high-school-level examinations measure learning.C.As an adult,you should not doubt your ability to pass high-school-level examinations.D.It’s difficult for most adults to pass high-school-level examinations without preparation.35.Which of the following would the writer most probably agree with?A.Schools should not give formal grades to kids.B.Schools should give formal grades to kids.C.Giving formal grades to kids has no disadvantages at a11.D.It’s unfair to put children in direct competition with one another.第二节(共5小题,每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。
2015年高考语文试卷全国一卷(含答案)
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2015年高考语文试卷全国一卷(含答案)宋代的经济在唐代基础上得到了新的发展。
特别是商品经济的繁荣,使得宋代的货币流通和信用出现了新的特点。
宋代的信用形式和信用工具也呈现出了新的特点,包括借贷、质、押、典、赊买赊卖等多种形式。
政府借贷主要是通过赈贷的形式,在紧急情况下通过贷给百姓粮食或种子的方式,帮助他们度过困境。
私人借贷则多为高利贷,解决社会分化和“钱荒”带来的平民百姓资金严重不足的问题,满足特殊支付和燃眉之急的需要。
质、押是借贷的担保形式,由质库、解库等机构经营。
典作为不动产转移的一种形式在宋代形成和发展起来,其特点是典权人向出典人支付典价后,在典期内就占有了出典人典产的使用权和收益支配权,出典人也不必向典权人支付利息。
宋代的商业贸易非常发达,但存在着通货紧缩现象,故赊买赊卖行为也很普遍,几乎所有物品都能进行赊买赊卖。
从实际效果看,它解决了军需、加强了流通,更重要的一点,它对束缚生产流通扩大和发展的高利贷构成了冲击。
正确B.宋代政府借贷的主要目的是赈济性借贷,利率较低,与私人借贷相比有优势。
原文提到政府借贷,但没有明确提到利率较低或与私人借贷相比的优势。
C.宋代债务人可以用不动产或动产作为担保,向债权人借贷,并在不偿还债务时,债权人可以用变卖价款优先受偿。
原文提到了债务人可以用担保向债权人借贷,但没有明确提到债权人可以用变卖价款优先受偿的情况。
D.宋代新型信用工具的使用,缓解了货币流通不足的问题,有利于经济发展。
原文中提到了宋代货币供给和流通状况不尽理想,新型信用工具的使用可以弥补货币不足的问题,促进经济发展。
赋得古原草送别》离离原上草,一岁一枯荣。
野火烧不尽,春风吹又生。
远芳侵古道,XXX接荒城。
又送王孙去,萋萋满别情。
4.这首诗的主题是(B)A.描写草原的美景B.表达离别之情C.赞美春天的美好D.反映自然界的变化5.下列句子中,表达了草原枯荣变化的是(A)A.“一岁一枯荣”B.“野火烧不尽”C.“远芳侵古道”D.“晴翠接荒城”6.下列句子中,表达了离别之情的是(D)A.“离离原上草”B.“野火烧不尽”C.“远芳侵古道”D.“又送王孙去,萋萋满别情”7.下列句子中,表达了自然界的力量的是(B)A.“离离原上草”B.“野火烧不尽”C.“远芳侵古道”D.“晴翠接荒城”二)近现代文学阅读(17分)阅读下面的文学作品,完成8--10题。
2015高考全国卷 语文试卷及答案
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2015年普通高等学校招生全国统一考试语文第Ⅰ卷阅读题甲必考题一、现代文阅读(9分,每小题3分)阅读下面的文宇,完成1~3题。
艺术品的接受在过去并不被看作是重要的美学问题,20世纪解释学兴起,一个名为“接受美学”的美学分支应运而生,于是研究艺术品的接受成为艺术美学中的显学。
过去,通常只是从艺术家的立场出发,将创作看作艺术家审美经验的结晶过程.作品完成就意味着创作完成.而从接受美学的角度来看,这一完成并不说明创作已经终结.它只说明创作的第一阶段告一段落,接下来是读者或现众、听众的再创作。
由于未被阅读的作品的价值包括审美价值仅仅是一种可能的存在,只有通过阅读,它才转化为现实的存在,因此对作品的接受具有艺术本体的意义.也就是说,接受者也是艺术创作的主体之一。
艺术文本即作品对于接受者来说具有什么意义呢?接受美学的创始人、德国的伊瑟尔说艺术文本是一个“召唤结构”,因为文本有“空白”“空缺”“否定”三个要素。
所谓“空白”是说它有一些东西没有表达出来,作者有意不写或不明写,要接受者用自己的生活经验与想象去补充;所谓“空缺”,是语言结构造成的各个图像间的空白,接受者在阅读文本时要把一个个句子表现的图像片断连接起来.整合成一个有机的图像系统;所谓“否定”指文本对接受者生活的现实具有否定的功能,它能引导接受者对现实进行反思和批判。
由此可见,文本的召唤性需要接受者呼应和配合,完成艺术品的第二次创作。
正如中国古典美学中的含蓄与简洁,其有限的文字常常引发出读者脑海中的丰富意象.接受者作为主体,他对文本的接受不是被动的。
海德格尔提出“前理解”,即理解前的心理文化结构,这种结构影响着理解。
理解不可能是文本意义的重现,而只能是文本与前理解”的统一。
这样,文本与接受就呈现出一种相互作用的关系:一方面文本在相当度上规定了接受者理解的范围、方向,让理解朝它的本义靠拢;另一方面,文本不可能将接受者完全制约住、规范住,接受者必然会按照自己的方式去理解作品,于是不可避免地就会出现误读或创造。
2015年普通高等学校招生全国统一考试全国卷1及参考答案解析
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2015年普通高等学校招生全国统一考试(全国卷1)英语(试题及答案)本试卷分第I卷(选择题)和第II卷(非选择题)两部分。
考试结束后, 将本试卷和答题卡一并交回。
第I卷第一部分听力(共两节, 满分30分)做题时, 先将答案标在试卷上。
录音内容结束后, 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, 并标在试卷的相应位置。
听完每段对话后, 你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
例:How much is the shirt?A. £19.15.B. £9.18.C. £9.15.答案是C。
1. What time is it now?A. 9:10B. 9:50C. 10:002. What does the woman think of the weather?A. It’s nice.B. It’s warm.C. It’s cold.3.What will the man do?A. Attend a meeting.B. Give a lecture.C. Leave his office.4. What is the woman’s opinion about the course?A. Too hard.B. Worth taking.C. Very easy.5. What does the woman want the man to do?A. Speak louder.B. Apologize to her.C. Turn off the radio.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三个项中选出最佳选项, 并标在试卷的相应位置。
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2015年高三模拟考试英语试题全国卷1第二部分阅读理解(共两节,满分40分) 第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑。
AIf the word researcher brings to mind someone who works in a laboratory doing experiments or reads a11 day in a librar y,then meet Dr.Sylvia Earle.Dr.Earle has carried out much of her research deep under the sea.Her outstanding work at sea ha s included leading around 70 expeditions(探索)and spending over 6,500 hours under the water. She also holds the world record for the deepest walk alone on the ocean floor at a depth of 381 metres.It’s no wonder she ha s been given the nickname“Her Deepness”.As a child,Dr.Earle was always interested in wildlife and was never afraid to find out about nature.Her parents told her to touch animals and insects,and not to be afraid of them.When she was older,she was given the opportunity to develop this int erest thanks to scho1arships to study botany at university.She also took courses in sea diving and decided to devote herself t o marine(海洋的)biology.It was this decision and her determined character that led her to have a truly remarkable career.After several years of diving,Dr.Earle took part in a research expedition in an underwater laboratory.The laboratory was called a Tektite Habitat and it allowed divers to remain under the water to carry out research for weeks at a time.In 1970 an all-female group went on the sixth expedition of the Tektite II scientists-in-the-water programme.Dr.Earle was asked to lead the expedition.Along with another four women-three scientists and an engineer-she stayed in the Tektite Habitat for two wee ks.Marine life has always been at the centre of Dr.Earle’s work and she is highly respected for her authority in this field.She i s often asked to give talks by environmental groups about marine life as well as her plan to create a global network of marin e protected areas. What makes her extra special is that she doesn’t look upon her work as a job,but as her vocation(天职)in life.21.In what way is Dr.Earle different from most researchers we know about?A.She works in a laboratory doing experiments all day.B.She reads all day in a library.C.She carries out much of her research deep under the sea.D.She has broken lots of world records.22.What’s the main idea of Paragraph 2?A.Dr.Earle’s interest in wildlife led her to a truly remarkable career.B.Dr.Earle’s success is chiefly due to her family background.C.Dr.Earle’s success is chiefly due to the scholarships to study botany at university.D.V arious factors have contributed to Dr.Earle’s success.23.Which of the following is true according to Paragraph 3?A.Five women took part in the expedition.B.Five scientists took part in the expedition.C.Dr.Earle was an ordinary member of the group.D.The group stayed under water for three weeks.24.What does Dr.Earle hope to achieve in the future?A.To keep marine life at the centre of her work.B.To create protected areas in the oceans around the world.C.To travel around the world giving talks.D.To look upon her work as her vocation in life.BA paperclip,made of steel wire bent into a looped()shape,is an instrument used to hold sheets of paper toget her.This common device is a wonder of simplicity and function.But where did this simple,cheap,and practical invention co me from?In the late 19th century,the most common way to hold papers together was by using a pin.Although the pin was an ine xpensive tool and was easily removable,it would 1eave holes in the ter,as steelvwire became more common,inventors began to notice its elastic()feature.With this feature,it could be stretched and twisted into various clip-like objects.In the years just before 1900,quite a few paperclip designs appeared.The name most frequently associated with the paperclip invention is Johan Vaaler,a Norwegian inventor.However,Vaaler’s clips were not the same as the paperclips curr entlyin use.Specifically,they did not have the inside loop we see today.The familiar looped design was invented by Gem Manuf acturing Ltd.in England.This clip is therefore sometimes known as the Gem clip.Because of Vaaler,the paperclip played an important historical role in Norway.During World War II,Norway was occupied by the Nazis.Norwegians were prohibited from wearing any symbol of their national unity(),such as buttons wi th the initials of their king.Thus,in protest,they started wearing paperclips to show their unity.The reason for doing this was simple:Paperclips were a Norwegian invention whose original function was to bind together.After the war,a giant papercli p statue was put up in Oslo to honour Vaaler—even though his design was never actually produced.25.According to the first paragraphthe paperclip is________.A.made of paperB.for holding clothes togetherC.shaped like a pinD.inexpensive and useful26.One way the paperclip is better than the pin is that_________A.it is cheaperB.it is simplerC.it doesn’t damage the paperD.it can be removed more easily27.Which of the following best shows what the Gem clip looks like?28.The last paragraph is mainly about_________.A.how widely used Vaaler’s clip isB.how V aalers clip became a national symbolC.how the Nazis ruled the Norwegian peopleD.why Norwegians had the initials of their king on their buttonsCI was born on the last day of February.I’ve always felt sorry for February,squeezed between the big months of January( named for the Roman god Janus,keeper of gateways)and March(after Mars,the god of war).The first Roman calendar,it is said,had 10 months and no February.Beginning at the vernal equinox()with March,i t ended with December.In an agricultural society,winter was of little importance,and thus went undivided.January and February were added about 700 B.C.by the second king of Rome,Numa Pompilius.He made all the months 29 or 31 days,but shortened February,the last month of the year,by giving it only 28.Next,it was the church’s turn.In 1582 Pope Gregory XIII announced a new calendar in Europe.Many changes were ma de,but the Pope passed up yet another chance to give February equality with the other months.It’s messy,even dangerous,changing how we measure time,but Pope Gregory was hardly the last one to try.The League o f Nations received over 150 new calendar designs,and the United Nations has considered more proposals since.Each propos al involves something that supposedly modernizes the calendar.But I have a simpler proposal that won’t lead to chaos,and will correct the historical injustices against Februarymove the last day of January and the last day of March into February to make it a normal month with 30 days,and a respectable 31 on leap years().This would not add or subtract a single day from the calendar year.It’s a great idea.And unlike J ulius and Augustus,I won’t even demand a month named in my honor.29.The first Roman calendar had only 10 months because_________.A.February hated to be squeezed between the big monthsB.the god of war always fought with January and FebruaryC.the Romans did not like the cold winter weatherD.it was Considered unnecessary to divide winter30.Which of the following is closest to Numa Pompilius’calendar?31.The writer sounds_____in the passage.A.sadB.humorousC.happyD. angryDOne important thing during the pre-Christmas rush at our house was the arrival of my daughter’s kindergarten report car d.She got high praise for her reading,vocabulary and overall enthusiasm.On the other hand,We learnt that she has work to do on her numbers and computer skills,though the detailed handwritten report her teachers prepared is absent of any words that might be interpreted as negative()in describing her efforts.A number System indicates how she’s measuring up in e ach area without any mention of passing or failing.The debate over whether formal grades should be given to kids has long existed.At one level,the advantages and disadva ntages are obvious.A grade system provides a straightforward standard by which to measure how your child is progressing a t school--and how he or she is getting on compared to other children.But as writer Sue Ferguson notes,Grades can deceive. ()The aim should be”to measure learning,not simply what a student can remember on a test.”The two aren’t the same-and if you doubt that as an adult,ask yourself whether you could sit down without any preparation and still pass those high-s chool-level examinations.If you’re old enough,You’ve lived through this debate before.At one time,it was considered unfair to put children in dire ct competition with one another if it could be avoided.The intention behind that may have been good,but it ignored the fact t hat competition,and the will to come out on top,are essential components of the human condition.32.Which of the following seems to be weak on the daughter’s kindergarten report card?A.Reading.B.V ocabulary.C.Enthusiasm.D.Numbers.33.what did the teachers try to do when preparing the report card?A.Avoid using negative words.B.Describe the facts as they really were.e numbers instead of words.D.Avoid showing weak areas.34.What does the writer suggest by the underlined part in Paragraph 2?A.Most adults can pass high-school-level examinations without preparation.B.There’s no doubt that most high-school-level examinations measure learning.C.As an adult,you should not doubt your ability to pass high-school-level examinations.D.It’s difficult for most adults to pass high-school-level examinations without preparation.35.Which of the following would the writer most probably agree with?A.Schools should not give formal grades to kids.B.Schools should give formal grades to kids.C.Giving formal grades to kids has no disadvantages at a11.D.It’s unfair to put children in direct competition with one another.第二节(共5小题,每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。