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2024-2025学年福建省福州市第一中学九年级上学期开学考试语文试题
2024-2025学年福建省福州市第一中学九年级上学期开学考试语文试题1. 根据语境,补写出古代诗文名句。
(1)蒹葭采采,______。
(《蒹葭》)(2)安得广厦千万间,______!(杜甫《茅屋为秋风所破歌》)(3)缺月挂疏桐,______。
(苏轼《卜算子·黄州定慧院寓居作》)(4)一封朝奏九重天,______。
(韩愈《左迁至蓝关示侄孙湘》)(5)______,连月不开。
(范仲淹《岳阳楼记》)(6)杜甫在《石壕吏》中以“______,______”两句显示出诗人极为关切劳动人民,通夜倾耳细听,未能入睡。
(7)李白《行路难》中,表达诗人坚信理想终会实现的诗句是:______,______。
2. 阅读下面的文字,按要求作答。
中华民族有着优良的读书传统。
闲来读书是一种良好的习惯,勤学善思之风更是A ()。
孔子“敏而好学,不耻下问”,董遇“读书百遍,而义自见”,董其昌“读万卷书,行万里路”,于谦“书卷多情似故人,晨昏忧乐每相亲”……这些无不体现了中国人对阅读的B()。
书籍,能描绘hào hàn()天地间千姿百态的风俗和人情,能传递漫长岁月里灿烂美好的理想和智慧。
人生几十年光阴,居然能chí chěng()古今,经天纬地,这种奇迹的产生,至少有一半左右要归功于阅读。
(1)根据拼音在文中括号内填写正确的词语。
(2)选择恰当的词语填在文中A、B处。
一脉相承一脉相通推重推崇(3)文中画线句有语病,请把正确的句子写下来。
3. 班级开展名著《水浒传》阅读活动,完成相关阅读记录。
水调歌头苏轼丙辰中秋,欢饮达旦,大醉,作此篇,兼怀子由。
明月几时有?把酒问青天。
不知天上宫阙,今夕是何年。
我欲乘风归去,又恐琼楼玉宇,高处不胜寒。
起舞弄清影,何似在人间。
转朱阁,低绮户,照无眠。
不应有恨,何事长向别时圆?人有悲欢离合,月有阴晴圆缺,此事古难全。
但愿人长久,千里共婵娟。
4. 下列对这首词的理解和分析,不正确...的一项是()A.词人开篇陡然发问,表现出对宇宙和人生的疑惑,显露出对明月的怨恨之情。
福州一中2023-2024学年第一学期初一数学期中考试附详细答案
福州一中2023-2024学年第一学期期中考试初一 数学试卷(完卷120分钟 满分150分)注意事项:1.答题前,考生务必在答题卡规定位置填写本人准考证号、姓名等信息,考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名”与考生本人准考证号、姓名是否一致.2.选择题每小题选出答案后,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.非选择题答案用0.5毫米黑色墨水签字笔在答题卡上相应位置书写作答,在试题卷上答题无效.第I 卷一、选择题(每小题3分,共36分,请将正确的答案涂在答题卡上) 1.−2023的相反数是( )A.−2023B.2023C.−12023D.120232.节约是一种美德,节约是一种智慧,据不完全统计,全国每年浪费食物总量折合粮食可养活约3亿5千万人,350 000 000这个数用科学记数法表示为( ) A.3.5×107 B.35×107 C.3.5×108 D.0.35×1093.如图,检测4个足球,其中超过标准质量的克数记为正数,不足标准质量的克数记为负数,从轻重的角度看,最接近标准的是( )+0.9−3.6−0.8+2.5A. B. C. D.4.下列各式是整式的是( ) A.2a −b ,2ab 35,a bB.2,25a+3ab ,5πa 2C.a−b 3,−2b 7,3a −5b 2D.a 2,−43a−1,(3a+b)25.3.598精确到百分位是( )A.3.6B.3.600C.3.60D.3.596.实数a ,b 在数轴上的对应点的位置如图所示,则下列不等关系正确的是( )A.a >bB.ab >0C.|a|<|b|D.a <−b7.下列说法:①正整数和负整数统称整数;②正分数和负分数统称分数;③整数和分数统称有理数;④单项式和多项式统称整式;⑤零既不是正数,也不是非负数.其中正确的有( )A.2个B.3个C.4个D.5个 8.下列运算正确的是( )A.−3(a+b)=−3a+3bB.−3(−a −b)=3a+3bC.−3(−a+b)=−3a+3bD.−3(a −b)=−3a+b 9.在数轴上距−3有2个单位长度的点所表示的数是( )A.−5B.−1C.5或−1D.−5或−1 10.已知|a|=2,|b|=3,|a −b|=−(a −b),则a+b 的值为( ) A.5或1 B.−1或1 C.5或−1 D.−5或1 11.对于正数x ,规定f(x )=2x+1,例如f(4)=24+1=25,f(14)=21+14=84+1=85,则f(100)+f(99)+f(98)+…+f(2)+f(1)+f(12)+...+f(198)+f(199)+f(1100)=( ) A.198 B.199 C.200 D.201212.定义运算a ○×b=a(1−b),下面给出了关于这种运算的四个结论:①12○×(−2)=6;②a ○×b=b ○×a ;③(5○×a)+(6○×a)=11○×a ;④若3○×b=3,则b=1.其中正确结论的个数( )A.1个B.2个C.3个D.4个第Ⅱ卷12 340 -1 -2 -3 -4二、填空题(每小题4分,共24分,请将正确的答案写在答题卡上)13.我国古代数学名著《九章算术》中对正负数的概念注有“今两算得失相反,要令正负以名之”,如:向东走30米记为“+30米”,则“−50米”表示_______. 14.倒数等于本身的数是_______,平方等于本身的数是_______. 15.−3a 2bc 5系数是______,次数是_______.16.若(x +2)2+|y −3|=0,则x y 的值为_______.17.若多项式4x 2y m +x y 2−(n −2)x 2y 2+3是关于x ,y 的五次三项式,则m −n=_______. 18.如用形状和大小相同的黑色棋子按下图所示的方式排列,按照这样的规律,第n 个图形需要棋子______枚.(用含n 的代数式表示)三、解答题(本大题共10小题,合计90分,请将答案写在答题卡上) 19.(每小题4分,共16分)计算(1)|−6|−7+(−3) (2)(12−59+23)×(−18)(3)4÷(−2)×(−112)−(−22) (4)−57×[(−3)2×(−223)−2]20.(每小题4分,共8分)化简(1)2x 3−4x 2−3x −2x 2−x 3+5x −7; (2)已知A=2m 2−mn ,B=m 2+2mn −5,求4A −2B. 21.(本题6分)在数轴上表示下列有理数:−12,0,−|−2|,(−2)2,并用“<”把它们连接起来.22.(本题6分)a 、b 、c 在数轴上的位置如图所示,化简|a+b|−2|a+c|−3|c −b|+|a|.23.(本题10分)商店出售茶壶每只定价25元,茶杯每只定价5元,该店制定了两种……①②③④优惠方案,方案一:买一只茶壶赠送一只茶杯;方案二:按总价的94%付款.某顾客需购茶壶4只,茶杯x只(x≥4).(1)分别求出两种优惠办法分别付多少钱.(2)当x=47时,两种方案哪一种更省钱?24.(本题8分)已知−2a2b y+3与4a x b2的和是单项式.(1)x=______,y=______.(2)在(1)的条件下,先化简再求值:2(x2y−3y3+2x)−3(x+x2y−2y3)−x.25.(本题6分)若多项式4m x2+5x−2y2+8x2−n x+y−1的值与x的取值无关,求(m+n)2的值.26.(本题8分)阅读材料:我们知道,4x−2x+x=(4−2+1)x=3x,类似地,我们把(a+b)看成一个整体,则4(a+b)−2(a+b)+(a+b)=(4−2+1)(a+b)=3(a+b).“整体思想”是中学教学解题中的一种重要的思想方法,它在多项式的化简与求值中应用极为广泛.(1)把(m−n)2看成一个整体,合并3(m−n)2−4(m−n)2+3(m−n)2的结果是_______.(2)已知x2+2y=4,则3x2+6y−2的值是_________.(3)已知x2+x y=2,2y2+3x y=5,求2x2+11x y+6y2的值.27.(本题10分)探索规律,观察下面的算式,解答问题.1+3=4=22;1+3+5=9=32;1+3+5+7=16=42;1+3+5+7+9=25=52…(1)请猜想:1+3+5+7+9+…+99=_______.(2)请猜想:1+3+5+7+9+…+(2n−1)=_______(n是正整数且n≥l).(3)计算:201+203+205+…+395+397+399.28.(本题12分)如图1,有P、Q两动点在线段AB上各自做不间断的往返匀速运动(即只要动点与线段AB的某一端点重合则立即转身以同样的速度向AB的另一端点运动与端点重合之前动点运动方向、速度均不改变),已知点P的速度为3米/秒,点Q 的速度为5米/秒.(1)已知AB=120米,若点P 先从点A 出发,当AP=12米时,点Q 从点A 出发,点Q 出发后经过____秒与点P 第一次重合.(2)已知AB=120米,若P 、Q 两点同时从点A 出发,经过几秒P 、Q 两点第一次重合. (3)如图2,若P 、Q 两点同时从点A 出发,点P 与点Q 第一次重合于点M ,第二次重合于点N ,且MN=50米,求AB 的长.福州一中2023-2024学年第一学期期中考试参考答案初一 数学试卷(完卷120分钟 满分150分)注意事项:1.答题前,考生务必在答题卡规定位置填写本人准考证号、姓名等信息,考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名”与考生本人准考证号、姓名是否一致.2.选择题每小题选出答案后,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.非选择题答案用0.5毫米黑色墨水签字笔在答题卡上相应位置书写作答,在试题卷上答题无效.第I 卷一、选择题(每小题3分,共36分,请将正确的答案涂在答题卡上) 1.−2023的相反数是( )A.−2023B.2023C.−12023D.120231.解:负数的相反数是正数,绝对值不变,故选B 。
2023-2024学年福建省福州一中高三(上)期中数学试卷【答案版】
2023-2024学年福建省福州一中高三(上)期中数学试卷一、单项选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个1.命题“∀x∈(1,+∞),sin x﹣2x<0”的否定是()A.∃x0∈(1,+∞),sinx0−2x0≥0B.∀x∈(﹣∞,1],sin x﹣2x≥0C.∀x∈(1,+∞),sin x﹣2x≥0D.∃x∈(﹣∞,1],sin x0−2x0≥02.已知集合A={x|x−2x≤0,x∈N},B={x|log2√x<1,x∈Z},则A∩B=()A.{0,1,2,3,4}B.{1,2,3,4}C.{1,2}D.{0,1,2}3.已知a=cos2,b=sin3,c=tan4,则a,b,c的大小关系是()A.a<b<c B.b<a<c C.c<a<b D.a<c<b4.设函数y=f(x)的定义域为D,∀x1,x2∈D,当x1+x2=2a时,恒有f(x1)+f(x2)=2b,则称点(a,b)为函数y=f(x)图象的对称中心.利用对称中心的上述定义,研究函数f(x)=e x﹣e﹣x+sin x+1,可得到f(﹣2023)+f(﹣2022)+…+f(2022)+f(2023)=()A.0B.2023C.4046D.40475.设a,b,c都是正数,且3a=4b=6c,那么下列关系正确的是()A.a+2b=c B.ac+bc=2ab C.1a+12b=1cD.1a+1b=2c6.设a,b,c分别是△ABC中内角A,B,C的对边,且ab+ba=4cosC,则tanCtanA+tanCtanB=()A.1B.2C.3D.47.若(3x+y)2023+x2023+4x+y=0,则4x+y=()A.0B.1C.2D.38.函数f(x)=a sin x+cos x(a<0)的图象向右平移φ个单位长度后,所得的函数为偶函数,则2sin2φ−a−1a的最小值为()A.2B.4C.6D.8二、多项选择题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,部分选对的得2分,有选错的得0分9.已知函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<π2)的部分图象如图所示,下列说法正确的是()A.该图象对应的函数解析式为f(x)=2sin(2x+π3)B.函数y=f(x)的图象关于点(−5π12,0)对称C.函数y=f(x)的图象关于直线x=−π6对称D.函数y=f(x)在[π12,7π12]上单调递减10.已知函数f(x)的定义域为R,且函数f(2x+1)是周期为2的奇函数,则()A.函数f(x)的图像关于点(12,0)中心对称B.函数f(x)的图像关于点(1,0)中心对称C.2是函数y=f(x)的一个周期D.4是函数y=f(x)的一个周期11.已知函数f(x)=x cos x﹣sin x﹣x的定义域为[﹣2π,2π],则()A.f(x)为奇函数B.f(x)在[0,π)上单调递减C.f(x)恰有2个极值点D.f(x)有且仅有2个极大值点12.函数f(x)=x3﹣ax2﹣x+1,则下列结论正确的是()A.若函数f(x)在(−12,13)上为减函数,则−1≤a≤14B.若函数f(x)的对称中心为(1,﹣2),则a=32C.当a=3时,若f(x)=﹣2有三个根x1,x2,x3,且x1+x2+x3=3D.当a=1时,若过点(﹣1,n)可作曲线y=f(x)的三条切线,则0<n<6427三、填空题:本题共4小题,每小题5分,共20分.13.已知角θ的终边过点P(﹣3,4),则sinθ+cosθ的值为.14.已知正实数x,y满足1x+2y=1,则2xy﹣2x﹣y的最小值为.15.已知函数f(x)=sin(2ωx +2π3)−√32(x ∈[0,π2],ω>0)有且仅有两个零点,则实数ω的取值范围是 .16.已知在函数f (x )=lnx +3x 2与函数g (x )=4x 2﹣ax (a ≠0)的图象上存在关于y 轴对称的点,则实数a 的取值范围是 .四、解答题:本题共6小题,共70分,17题10分,其他小题各12分.解答应写出文字说明、证明过程或演算步骤.17.(10分)已知S n 是数列{a n }的前n 项和,S n =n 2+3n2.(1)求数列{a n }的通项公式; (2)设T n 为数列{1a n a n+1}前n 项的和,若λT n ≤a n +1对一切n ∈N *恒成立,求实数λ的最大值. 18.(12分)在图1中,∠B =90°,AB =BC =√2,△ACD 为等边三角形,O 为AC 边的中点,E 在BC 边上,且EC =2BE ,沿AC 将△ACD 进行折叠,使点D 运动到点F 的位置,如图2,连接FO ,FB ,FE ,FB =2.(1)证明:FO ⊥平面ABC . (2)求二面角E ﹣F A ﹣C 的余弦值.19.(12分)已知函数f(x)=cos 2x +2√3sinxcosx −sin 2x . (1)若f(x 02)=15,x 0∈[0,π3],求cos2x 0的值.(2)A 、B 、C 是△ABC 的三个内角,f (B )=2;若D 是AC 边上的点,且AD =3DC =3,∠A =∠ABD =θ,求sin θ的值.20.(12分)福州某公园有一个半圆形荷花池(如图所示),为了让游客深入花丛中体验荷花美景,公园管理处计划在半圆形荷花池中设计栈道观景台P 和栈道P A 、PB 、PC 、AB ,观景台P 在半圆形的中轴线OC 上(如图,OC 与直径AB 垂直,P 与O ,C 不重合),通过栈道把荷花池连接起来,使人行其中有置身花海之感.已知AB =200米,∠P AB =θ,栈道总长度为L .(1)求L 关于θ的函数关系式.(2)若栈道的造价为每米5千元,问:栈道PC 长度是多少时,栈道的建设费用最小?并求出该最小值.21.(12分)已知△ABC 内角A ,B ,C 的对边分别为a ,b ,c ,点G 是△ABC 的重心,且AG →•BG →=0. (1)若∠GAB =π6,①直接写出AG CG=_____;②设∠CAG =α,求tan α的值.(2)求cos ∠ACB 的取值范围.22.(12分)已知函数f(x)=alnx −bx −1x(x >0,a >0),f ′(x )为f (x )的导函数.(1)当a =1时,讨论函数f (x )的单调性;(2)已知x 1,x 2∈(0,+∞)(x 1≠x 2),若存在b ∈R ,使得f (x 1)=f (x 2)成立,求证:f ′(x 1)+f ′(x 2)>0.2023-2024学年福建省福州一中高三(上)期中数学试卷参考答案与试题解析一、单项选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个1.命题“∀x∈(1,+∞),sin x﹣2x<0”的否定是()A.∃x0∈(1,+∞),sinx0−2x0≥0B.∀x∈(﹣∞,1],sin x﹣2x≥0C.∀x∈(1,+∞),sin x﹣2x≥0D.∃x∈(﹣∞,1],sin x0−2x0≥0解:“∀x∈(1,+∞),sin x﹣2x<0”的否定是:∃x0∈(1,+∞),sinx0−2x0≥0.故选:A.2.已知集合A={x|x−2x≤0,x∈N},B={x|log2√x<1,x∈Z},则A∩B=()A.{0,1,2,3,4}B.{1,2,3,4}C.{1,2}D.{0,1,2}解:∵A={x|x−2x≤0,x∈N},∴A={1,2},∵B={x|log2√x<1,x∈Z},∴B={1,2,3},∴A∩B={1,2}.故选:C.3.已知a=cos2,b=sin3,c=tan4,则a,b,c的大小关系是()A.a<b<c B.b<a<c C.c<a<b D.a<c<b解:令f(x)=x﹣sin x,则f′(x)=1﹣cos x≥0,所以函数f(x)在(0,π2)上单调递增,所以当x>0时,f(x)>f(0)=0,即有x>sin x成立,所以b=sin3=sin(π﹣3)<π﹣3;令g(x)=tanx−x(0<x<π2),则g′(x)=cos2x+sin2xcos2x−1=1−cos2xcos2x>0,所以函数g(x)在(0,π2)上单调递增,所以当0<x<π2时,g(x)>g(0)=0,即有tan x>x成立,所以c=tan4=tan(4﹣π)>4﹣π,因为4﹣π>π﹣3,所以c>b,又a=cos2<cos π2=0,所以a<b<c.故选:A.4.设函数y=f(x)的定义域为D,∀x1,x2∈D,当x1+x2=2a时,恒有f(x1)+f(x2)=2b,则称点(a,b)为函数y=f(x)图象的对称中心.利用对称中心的上述定义,研究函数f(x)=e x﹣e﹣x+sin x+1,可得到f(﹣2023)+f(﹣2022)+…+f(2022)+f(2023)=()A.0B.2023C.4046D.4047解:f(x)=e x﹣e﹣x+sin x+1的定义域为R.因为f(x)+f(﹣x)=(e x﹣e﹣x+sin x+1)+[e﹣x﹣e x+sin(﹣x)+1]=2,所以f(x)的图象关于点(0,1)对称.所以f(﹣2023)+f(﹣2022)+…+f(2022)+f(2023)=2×2023+f(0)=4047.故选:D.5.设a,b,c都是正数,且3a=4b=6c,那么下列关系正确的是()A.a+2b=c B.ac+bc=2ab C.1a+12b=1cD.1a+1b=2c解:由3a=4b=6c=k,得a=log3k,b=log4k,c=log6k,1 a =logk3,1b=logk4,1c=logk6,则12b=12logk4=logk2,根据log k3+log k2=log k6可知,1a+12b=1c.故选:C.6.设a,b,c分别是△ABC中内角A,B,C的对边,且ab+ba=4cosC,则tanCtanA+tanCtanB=()A.1B.2C.3D.4解:由ab+ba=4cosC得a2+b2=4ab cos C,所以c2=a2+b2﹣2ab cos C=2ab cos C,由正弦定理得sin2C=2sin A sin B cos C,tanC=sinCcosC=2sinAsinBsinC=2sinAsinBsin(A+B)=2sinAsinBsinAcosB+cosAsinB=21tanB+1tanA,所以tanCtanA+tanCtanB=2.故选:B.7.若(3x+y)2023+x2023+4x+y=0,则4x+y=()A.0B.1C.2D.3解:设f(x)=x2023+x,x∈R,因为f(﹣x)=(﹣x)2023+(﹣x)=﹣(x2023+x)=﹣f(x),所以f(x)为奇函数,又因为y=x2023和y=x在R上都为增函数,所以f(x)在R上为增函数.由(3x+y)2023+x2023+4x+y=0得(3x+y)2023+3x+y=﹣(x2023+x),即f(3x+y)=﹣f(x)=f(﹣x),所以3x+y=﹣x,即4x+y=0.故选:A.8.函数f(x)=a sin x+cos x(a<0)的图象向右平移φ个单位长度后,所得的函数为偶函数,则2sin2φ−a−1a的最小值为()A.2B.4C.6D.8解:f(x)=asinx+cosx=√a2+1sin(x+α),其中tanα=1 a ,函数的图象向右平移φ个单位长度后,得到函数y=√a2+1sin(x+α−φ)为偶函数,则当x=0时,α−φ=π2+kπ,k∈Z,即φ=α−π2−kπ,则2φ=2α﹣π﹣2kπ,k∈Z,sin2φ=sin(2α−π−2kπ)=−sin2α=−2sinαcosαsin2α+cos2α=−2tanα1+tan2α=−2a1+1a2=−2aa2+1,即2sin2φ−a−1a=−4aa2+1−a2+1a,因为a<0,所以−4aa2+1>0,−a2+1a>0,所以−4aa2+1−a2+1a≥2√(−4aa2+1)×(−a2+1a)=4,当4aa2+1=a2+1a,即a=﹣1时,等号成立,所以2sin2φ−a−1a的最小值为4.故选:B.二、多项选择题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,部分选对的得2分,有选错的得0分9.已知函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<π2)的部分图象如图所示,下列说法正确的是()A.该图象对应的函数解析式为f(x)=2sin(2x+π3)B .函数y =f (x )的图象关于点(−5π12,0)对称 C .函数y =f (x )的图象关于直线x =−π6对称D .函数y =f (x )在[π12,7π12]上单调递减解:由函数f (x )=A sin (ωx +φ)的部分图象知,A =2, 所以T =4×(π3−π12)=π,所以ω=2πT=2, 又因为f (π12)=2sin (2×π12+φ)=2,所以π6+φ=π2+2k π,k ∈Z ,解得φ=π3+2k π,k ∈Z , 又因为|φ|<π2,所以φ=π3,所以f (x )=2sin (2x +π3),故A 正确;因为x =−5π12时,f (−5π12)=2sin (−5π6+π3)=﹣2≠0, 所以函数y =f (x )的图象不关于点(−5π12,0)对称,选项B 错误; 因为x =−π6时,f (﹣)=2sin (−π3+π3)=0≠±2,所以f (x )的图象不关于x =−π6对称,选项C 错误;因为x ∈[π12,7π12]时,2x +π3∈[π2,3π2],所以函数f (x )=2sin (2x +π3)单调递减,选项D 正确.故选:AD .10.已知函数f (x )的定义域为R ,且函数f (2x +1)是周期为2的奇函数,则( ) A .函数f (x )的图像关于点(12,0)中心对称B .函数f (x )的图像关于点(1,0)中心对称C .2是函数y =f (x )的一个周期D .4是函数y =f (x )的一个周期解:根据题意,因为f(2x+1)是奇函数,所以f(2x+1)+f(﹣2x+1)=0,则f(x+1)+f(﹣x+1)=0,所以f(x)的图像关于点(1,0)中心对称,故B正确;因为f(2x+1)的周期是2,所以f(2x+1)=f(2(x+2)+1)=f(2x+5),即f(x+1)=f(x+5)⇒f(x)=f(x+4),即4是函数y=f(x)的一个周期,故D正确;不妨令f(2x+1)=sin(πx),显然符合要求,此时f(x)=sin(π2x−π2),显然其周期为T=2ππ2=4≠2,且f(12)≠0,故A、C错误.故选:BD.11.已知函数f(x)=x cos x﹣sin x﹣x的定义域为[﹣2π,2π],则()A.f(x)为奇函数B.f(x)在[0,π)上单调递减C.f(x)恰有2个极值点D.f(x)有且仅有2个极大值点解:A.函数的定义域为[﹣2π,2π],关于原点对称,又f(﹣x)=﹣x cos(﹣x)﹣sin(﹣x)+x=﹣x cos x+sin x+x=﹣f(x),所以函数f(x)为奇函数,故A正确;B.由f(x)=x cos x﹣sin x﹣x,得f′(x)=cos x﹣x sin x﹣cos x﹣1=﹣x sin x﹣1,当x∈[0,π)时,f′(x)<0,所以函数f(x)在[0,π)上单调递减,故B正确;C.显然f′(0)≠0,当x≠0时,令f′(x)=0,则﹣x sin x﹣1=0,所以sinx=−1x,分别作出y=sin x和y=−1x在[﹣2π,2π]的图象,由图可知,这两个函数的图象在区间[﹣2π,2π]上共有4个公共点,且图象在这些公共点处都不相切,故f(x)在区间[﹣2π,2π]上的极值点的个数为4,有2个极大值点,故C错误,D正确.故选:ABD.12.函数f(x)=x3﹣ax2﹣x+1,则下列结论正确的是()A .若函数f (x )在(−12,13)上为减函数,则−1≤a ≤14B .若函数f (x )的对称中心为(1,﹣2),则a =32C .当a =3时,若f (x )=﹣2有三个根x 1,x 2,x 3,且x 1+x 2+x 3=3D .当a =1时,若过点(﹣1,n )可作曲线y =f (x )的三条切线,则0<n <6427解:对于A ,f (x )=x 3﹣ax 2﹣x +1,∴f ′(x )=3x 2﹣2ax ﹣1, 函数f (x )在(−12,13)上为减函数,则f ′(x )≤0,对∀x ∈(−12,13),所以{f ′(−12)=34+a −1≤0f′(13)=13−23a −1≤0,解得−1≤a ≤14,故A 正确;对于B ,函数f (x )的对称中心为(1,﹣2),则f (0)+f (2)=﹣4, 即1+8﹣4a ﹣2+1=﹣4,解得a =3,故B 错误;对于C ,当a =3时,f (x )=x 3﹣3x 2﹣x +1,则f (x )=﹣2,即x 3﹣3x 2﹣x +3=0,化简得(x +1)(x ﹣1)(x ﹣3)=0,其3个根为x 1=﹣1,x 2=1,x 3=3,所以x 1+x 2+x 3=3,故C 正确; 对于D ,当a =1时,f (x )=x 3﹣x 2﹣x +1,设切点为(x 0,y 0),则y 0=x 03−x 02−x 0+1,切线的斜率k =f ′(x 0)=3x 02−2x 0−1, 则切线方程为y −(x 03−x 02−x 0+1)=(3x 02−2x 0−1)(x −x 0), 将点(﹣1,n )代入上式,整理得n =﹣2x 03−2x 02+2x 0+2,过点(﹣1,n )可作曲线y =f (x )的三条切线,即方程n =﹣2x 03−2x 02+2x 0+2有三个不同的解,令g (x )=﹣2x 3﹣2x 2+2x +2,则g ′(x )=﹣6x 2﹣4x +2=﹣2(x +1)(3x ﹣1), 当x ∈(﹣∞,﹣1)时,g ′(x )<0,函数g (x )单调递减, 当x ∈(−1,13)时,g ′(x )>0,函数g (x )单调递增,当x ∈(13,+∞)时,g ′(x )<0,函数g (x )单调递减,所以函数g (x )在x =﹣1处取得极小值,极小值为g (﹣1)=0, 在x =13处取得极大值,极大值为g(13)=6427,由方程n =﹣2x 03−2x 02+2x 0+2有三个不同的解,所以0<n <6427,故D 正确. 故选:ACD .三、填空题:本题共4小题,每小题5分,共20分.13.已知角θ的终边过点P (﹣3,4),则sin θ+cos θ的值为15 . 解:∵θ的终边过点P (﹣3,4),∴cos θ=−3√(−3)+4=−35, sin θ=√(−3)+4=45, ∴sin θ+cos θ=45+(−35)=15. 故答案为:15. 14.已知正实数x ,y 满足1x +2y=1,则2xy ﹣2x ﹣y 的最小值为 8 . 解:因为x >0,y >0,所以1=1x +2y ≥2√2xy,即xy ≥8,当且仅当x =2,y =4时,等号成立, 所以2xy ﹣2x ﹣y =xy(2−2y −1x)=xy(2−1)=xy ≥8. 即2xy ﹣2x ﹣y 的最小值为8.故答案为:8.15.已知函数f(x)=sin(2ωx +2π3)−√32(x ∈[0,π2],ω>0)有且仅有两个零点,则实数ω的取值范围是 [53,2) . 解:令f(x)=sin(2ωx +2π3)−√32=0,得sin(2ωx +2π3)=√32, 由题意方程sin(2ωx +2π3)=√32在x ∈[0,π2]上有且仅有两个实根, 由x ∈[0,π2],得2ωx +2π3∈[2π3,ωπ+2π3], 所以7π3≤ωπ+2π3<8π3,解得53≤ω<2, 所以实数ω的取值范围是53≤ω<2. 故答案为:[53,2). 16.已知在函数f (x )=lnx +3x 2与函数g (x )=4x 2﹣ax (a ≠0)的图象上存在关于y 轴对称的点,则实数a 的取值范围是 (﹣∞,﹣1] .解:设f (x )=lnx +3x 2上一点坐标(x 0,f (x 0))(x 0>0),则其关于y 轴对称的点为(﹣x 0,f (x 0)),若该点在函数g(x)=4x2﹣ax(a≠0)上,则有g(﹣x0)=f(x0),故有4x02+ax0=lnx0+3x02⇒a=lnx0x0−x0,令ℎ(x)=lnxx−x,则ℎ′(x)=1−x2−lnxx2,令m(x)=1﹣x2﹣lnx,m′(x)=−2x−1x<0,所以m(x)在(0,+∞)上单调递减,又m(1)=0⇒x>1时,m(x)<0,h′(x)<0,即此时h(x)单调递减,0<x<1时,m(x)>0,h′(x)>0,即此时h(x)单调递增,所以h(x)≤h(1)=﹣1,所以a≤﹣1.故答案为:(﹣∞,﹣1].四、解答题:本题共6小题,共70分,17题10分,其他小题各12分.解答应写出文字说明、证明过程或演算步骤.17.(10分)已知S n是数列{a n}的前n项和,S n=n 2+3n 2.(1)求数列{a n}的通项公式;(2)设T n为数列{1a n a n+1}前n项的和,若λT n≤a n+1对一切n∈N*恒成立,求实数λ的最大值.解:(1)因为S n是数列{a n}的前n项和,且S n=n2+3n2,则S n−1=(n−1)2+3(n−1)2,当n≥2时,a n=S n−S n−1=n2+3n2−(n−1)2+3(n−1)2=n+1,当n=1时,a1=S1=2,满足通项公式,所以{a n}的通项公式为a n=n+1.(2)因为T n为数列{1a n a n+1}前n项的和,令b n=1a n a n+1=1(n+1)(n+2)=1n+1−1n+2,则T n=b1+b2+b3+⋯+b n−1+b n=(12−13)+(13−14)+(14−15)+⋯+(1n−1n+1)+(1n+1−1n+2),=12−1n+2=n2(n+2),因为λT n≤a n+1对一切n∈N*恒成立,则λ≤a n+1T n=2(n+4n)+8,因为n+4n≥2√n⋅4n=4,当且仅当n=2时,等号成立.所以2(n+4n)+8≥16,所以实数λ的最大值为16.18.(12分)在图1中,∠B=90°,AB=BC=√2,△ACD为等边三角形,O为AC边的中点,E在BC 边上,且EC=2BE,沿AC将△ACD进行折叠,使点D运动到点F的位置,如图2,连接FO,FB,FE,FB=2.(1)证明:FO⊥平面ABC.(2)求二面角E﹣F A﹣C的余弦值.(1)证明:由题意FO⊥AC,又AC=2,FO=√3,OB=1,而FB=2,所以FO2+BO2=FB2,所以FO⊥BO,因为AC∩BO=O,AC,BO⊂平面ABC,所以FO⊥平面ABC;(2)解:以O为坐标原点,分别以OB,OC,OF所在直线为x轴,y轴,z轴,建立空间直角坐标系,如图,则A(0,﹣1,0),F(0,0,√3),E(23,13,0),AF →=(0,1,√3),AE →=(23,43,0), 设平面AEF 的一个法向量是m →=(x ,y ,z ),则{m →⋅AF →=y +√3z =0m →⋅AE →=23x +43y =0,取y =−√3,则x =2√3,z =1,则m →=(2√3,−√3,1), 显然n →=(1,0,0)是平面AFC 的一个法向量,cos <m →,n →>=m →⋅n →|m →||n →|=√312+3+1×1=√32, 所以二面角E ﹣F A ﹣C 的余弦值为√32. 19.(12分)已知函数f(x)=cos 2x +2√3sinxcosx −sin 2x . (1)若f(x 02)=15,x 0∈[0,π3],求cos2x 0的值. (2)A 、B 、C 是△ABC 的三个内角,f (B )=2;若D 是AC 边上的点,且AD =3DC =3,∠A =∠ABD =θ,求sin θ的值.解:(1)因为f(x)=cos 2x +2√3sinxcosx −sin 2x =√3sin2x +cos2x =2sin(2x +π6), 所以f(x 02)=√3sinx 0+cosx 0=15, 将cosx 0=15−√3sinx 0代入sin 2x 0+cos 2x 0=1, 得4sin 2x 0−2√35sinx 0−2425=0, 解得sinx 0=√3+3√1120或sinx 0=√3−3√1120又x 0∈[0,π3],所议sinx 0=√3+3√1120, 所以cos2x 0=1−2sin 2x 0=1−2(√3+3√1120)2=49−3√33100. (2)f(B)=2sin(2B +π6)=2,因为B ∈(0,π),所以2B +π6∈(π6,13π6), 所以2B +π6=π2,即B =π6, 因为∠A =∠ABD =θ,所以BD =AD =3, 在△ABC 中,由正弦定理可得BC =ACsinθsin30°=8sinθ, 因为∠BCD =2θ,CD =1,所以,在△BCD 中,由余弦定理有32+12−2×3×1×cos2θ=(8sinθ)2,整理得10﹣6cos2θ=64sin 2θ,即10﹣6+12sin 2θ=64sin 2θ,解得sin 2θ=113, 因为0<θ<π6, 所以sinθ=√1313.20.(12分)福州某公园有一个半圆形荷花池(如图所示),为了让游客深入花丛中体验荷花美景,公园管理处计划在半圆形荷花池中设计栈道观景台P 和栈道P A 、PB 、PC 、AB ,观景台P 在半圆形的中轴线OC 上(如图,OC 与直径AB 垂直,P 与O ,C 不重合),通过栈道把荷花池连接起来,使人行其中有置身花海之感.已知AB =200米,∠P AB =θ,栈道总长度为L .(1)求L 关于θ的函数关系式.(2)若栈道的造价为每米5千元,问:栈道PC 长度是多少时,栈道的建设费用最小?并求出该最小值.解:(1)因为P 在半圆形的中轴线OC 上,OC ⊥AB ,AB =200米,∠P AB =θ,所以PA =PB =12OA cosθ=100cosθ,PO =12ABtanθ=100tanθ, 所以PC =OC ﹣PO =100﹣100tan θ,所以栈道总长度L =PA +PB +PC +AB =100cosθ+100cosθ+100−100tanθ+200=200cosθ−100tanθ+300,θ∈(0,π4). (2)由(1)得L ′(θ)=100(2sinθ−1)cos 2θ,θ∈(0,π4), 所以当0<θ<π6时,L ′(θ)<0,L 单调递减,当π6<θ<π4时,L ′(θ)>0,L 单调递增, 所以当θ=π6,即PC =100−100tan π6=100(3−√3)3时,栈道的建设费用最小, 建设费用最小值为5×(200cos π6−100tan π6+300)=500√3+1500千元. 21.(12分)已知△ABC 内角A ,B ,C 的对边分别为a ,b ,c ,点G 是△ABC 的重心,且AG →•BG →=0.(1)若∠GAB =π6,①直接写出AG CG=_____;②设∠CAG =α,求tan α的值. (2)求cos ∠ACB 的取值范围.解:(1)①设AB 的中点为D ,则D ,G ,C 三点共线且CG =2DG ,因为AG →⋅BG →=0,所以AG ⊥BG ,所以AD =DG ,因为∠GAB =π6,所以∠GAB =∠AGD =π6,∠ADG =2π3, 所以在△ADG 中,由余弦定理得AG =√AD 2+DG 2−2AD ⋅DGcos∠ADG =√3DG ,所以AG CG =√32. ②以A 为原点,AB 所在直线为x 轴建立如图平面直角坐标系,设AB =2,则B (2,0),G(32,√32),D (1,0), DG →=(12,√32),GC →=2DG →=(1,√3),故C(52,3√32), 所以tan ∠BAC =3√35, 所以tanα=tan(∠BAC −π6)=tan∠BAC−tan π61+tan∠BAC⋅tan π6=3√35−√331+3√35×√33=√36. (2)设∠GAB =θ,θ∈(0,π2),则G (2cos 2θ,2cos θsin θ),DG →=(2cos 2θ−1,2cosθsinθ)=(cos2θ,sin2θ),GC →=2DG →=(2cos2θ,2sin2θ),故C (2cos2θ+2cos 2θ,3sin2θ),即C (3cos2θ+1,3sin2θ)所以CA →=(−3cos2θ−1,−3sin2θ),CB →=(1−3cos2θ,−3sin2θ),|CA → |=√9cos 22θ+1+6cos2θ+9sin 22θ=√10+6cos2θ,|CB → |=√1−6cos2θ+9cos 22θ+9sin 22θ=√10−6cos2θ,所以cos ∠ACB =CA →⋅CB→|CA →||CB →|=8√100−36cos 2θ,因为θ∈(0,π2),所以2θ∈(0,π),所以﹣1<cos2θ<1,所以45≤√100−36cos 22θ1, 即cos ∠ACB ∈[45,1).22.(12分)已知函数f(x)=alnx −bx −1x(x >0,a >0),f ′(x )为f (x )的导函数. (1)当a =1时,讨论函数f (x )的单调性;(2)已知x 1,x 2∈(0,+∞)(x 1≠x 2),若存在b ∈R ,使得f (x 1)=f (x 2)成立,求证:f ′(x 1)+f ′(x 2)>0.解:(1)当a =1时,f(x)=lnx −bx −1x,x >0, f ′(x)=1x −b +1x 2=−bx 2+x+1x 2, 当b =0时,f ′(x )在区间(0,+∞)上恒大于0,此时函数的单调递增区间是(0,+∞); 当b >0时,设g (x )=﹣bx 2+x +1=0,其中Δ=1+4b >0,当x ∈(0,1+√1+4b 2b ),f ′(x )>0,函数单调递增, 当x ∈(1+√1+4b 2b,+∞),f ′(x )<0,函数单调递减, 当b <0时,Δ=1+4b ,当b ≤−14时,Δ≤0,此时f ′(x )≥0恒成立,函数的单调递增区间是(0,+∞), 当−14<b <0时,Δ=1+4b >0, 当1−√1+4b 2b <0且1+√1+4b 2b<0, 所以f ′(x )在区间(0,+∞)上恒大于0,即函数的单调递增区间是(0,+∞),综上可知,b ≤0时,函数的单调递增区间是(0,+∞),当b >0时,函数的单调递减区间是(0,1+√1+4b 2b ),函数的单调递减区间是(1+√1+4b 2b,+∞); (2)证明:不妨设x 1>x 2,因为f (x 1)=f (x 2),则alnx1−bx1−1x1=alnx2−bx2−1x2,即alnx1−alnx2−1x1+1x2=b(x1−x2),得aln x1x2x1−x2+1x1x2=b,由f′(x)=1x2+ax−b,则f′(x1)+f′(x2)=(1x12+1x22)+a(1x1+1x2)−2(aln x1x2x1−x2+1x1x2),f′(x1)+f′(x2)=(1x12+1x22−2x1x2)+a(1x1+1x2−2ln x1x2x1−x2),所以f′(x1)+f′(x2)=(x1−x2)2x12x22+ax1−x2(x12−x22x1x2−2lnx1x2),f′(x1)+f′(x2)=(x1−x2)2x12x22+ax1−x2(x1x2−x2x1−2lnx1x2),设t=x1x2∈(1,+∞),构造函数φ(t)=t−1t−2lnt(t>1),φ′(t)=1+1t2−2t=(t−1)2t2>0,所以φ(t)在(1,+∞)上为增函数,所以φ(t)>φ(1)=0,即x1x2−x2x1−2lnx1x2>0,又(x1−x2)2x12x22>0,a>0,x1﹣x2>0,所以f′(x1)+f′(x2)>0.。
2024-2025学年福建省福州一中高三(上)开学数学试卷(含答案)
2024-2025学年福建省福州一中高三(上)开学数学试卷一、单选题:本题共8小题,每小题5分,共40分。
在每小题给出的选项中,只有一项是符合题目要求的。
1.已知复数z 满足z +−z =4,且z−−z =2i ,则|z|=( )A. 2 B. 3 C. 2 D. 52.已知集合A ={x|2<x <4},B ={x||x−4|>1},则A ∩(∁R B)=( )A. (2,3)B. (3,4)C. [3,4)D. (−∞,4)∪(5,+∞)3.已知向量a ,b ,满足|a +2b |=2 7,|a |=2,|b |= 3,则向量a 与b 的夹角为( )A. π6 B. π3 C. 2π3 D. 5π64.已知函数f(x)=2x2−ax−1,在(1,2)上单调递增,则a 的取值范围为( )A. (0,2] B. (−∞,2] C. (2,4)D. [4,+∞)5.已知α,β为锐角,且cosα=35,sin (α−β)=513,则cosβ=( )A. −1665B. 5665C. 1665D. −56656.设四棱台ABCD−A 1B 1C 1D 1的上、下底面积分别为S 1,S 2,侧面积为S ,若一个小球与该四棱台的每个面都相切,则( )A. S 2=S 1S 2B. S =S 1+S 2C. S =2 S 1S 2D. S = S 1+ S 27.如图,将绘有函数f (x )=M sin (π3x +φ)(M >0,0<φ<π)部分图像的纸片沿x 轴折成直二面角,此时A,B 之间的距离为 15,则φ=( )A. π6B. π3C. 2π3D. 5π68.已知f(x)=2x +2−x +cosx +x 2,若a =f(4lnπ3),b =f(πln 43),c =f(4ln 3π),则( )A. a <b <cB. b <c <aC. c <a <bD. b <a <c二、多选题:本题共3小题,共15分。
福州第一中学2025届高三第四次模拟考试语文试卷含解析2
福州第一中学2025届高三第四次模拟考试语文试卷注意事项1.考试结束后,请将本试卷和答题卡一并交回.2.答题前,请务必将自己的姓名、准考证号用0.5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置.3.请认真核对监考员在答题卡上所粘贴的条形码上的姓名、准考证号与本人是否相符.4.作答选择题,必须用2B铅笔将答题卡上对应选项的方框涂满、涂黑;如需改动,请用橡皮擦干净后,再选涂其他答案.作答非选择题,必须用05毫米黑色墨水的签字笔在答题卡上的指定位置作答,在其他位置作答一律无效.5.如需作图,须用2B铅笔绘、写清楚,线条、符号等须加黑、加粗.1、下列各句中,没有错别字且加点字的注音全都正确的一项是A.中国质检总局宣布,自9 月15 日起,中国暂停进口蒙.(měng)古国牛羊肉及其制品。
有网友发帖子评论:“怎么办?我每天都要吃土豆牛腩.(nǎn)饭。
”B.嘉庆对内政的整饬.(chì)是从诛杀和珅开始的。
在清除飞扬跋扈、贪赃妄法的和珅后,嘉庆即打出“咸与维新”的旗号,诏求直言,广开言路,祛.(qù)邪扶正,以期光风霁月。
C.网络社会中,一个小事件容易在论坛中被围观,被热炒,以致不断发酵,上升为论坛主题帖.(tiè),并被更多的论坛转载.(zǎi),成为网络社会的一时热点。
D.作为鲁迅先生的扛.(gāng)鼎之作,《阿Q正传》是一部惊世骇俗、震聋发聩的中篇力作,它借用小说的外型却以犀利的杂文笔法直刺国民的灵魂深处,催人反省。
2、补写出下列名篇名句的空缺部分(只选3小题)(1)______,钻之弥坚。
_____,忽焉在后。
(《论语》)(2)三岁为妇,_____。
_____,靡有朝矣。
(《诗经·氓》)(3)渔舟唱晚,_____;雁阵惊寒,_____。
(王勃《滕王阁序》)(4)爱其子,_____,于其身也,_____,惑矣。
(韩愈《师说》)(5)_____。
一种相思,_____。
福建省福州第一中学2023-2024学年高一下学期期末考试英语试题(含答案)
福州第一中学2023-2024学年高一下学期期末考试英语试卷(完卷120分钟满分150分)第一部分听力(略)第二部分阅读(共两节,满分50分)第一节(共15小题;每小题2.5分,满分37.5分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项。
AIf you are a fruit grower — or would like to become one — take advantage of Apple Day to see what’s around. It’s called Apple Day but in practice it’s more like Apple Month. The day itself is on October 21, but since it has caught on, events now spread out over most of October around Britain.Visiting an apple event is a good chance to see, and often taste, a wide variety of apples. To people who are used to the limited choice of apples such as Golden Delicious and Royal Gala in supermarkets, it can be quite an eye opener to see the range of classical apples still in existence, such as Decio which was grown by the Rom ans. Although it doesn’t taste of anything special, it’s still worth a try, as is the knobbly(多疙瘩的) Cat’s Head which is more of a curiosity than anything else.There are also varieties developed to suit specific local conditions.One of the very best varieties for eating quality is Orleans Reinette, but you’ll need a warm, sheltered place with perfect soil to grow it, so it’s a pipe dream for most apple lovers who fall for it.At the events, you can meet expert growers and discuss which ones will best suit your conditions, and because these are family affairs, children are well catered for with apple-themed fun and games. Apple Days are being held at all sorts of places with an interest in fruit, including stately gardens and commercial orchards(果园). If you want to have a real orchard experience, try visiting the National Fruit Collection at Brogdale, near Faversham in Kent.1. What can people do at the apple eventsA. Attend experts’ lectures.B. Visit fruit-loving families.C. Plant fruit trees in an orchard.D. Taste many kinds of apples.2. What can we learn about DecioA. It is a new variety.B. It has a strange look.C. It is rarely seen now.D. It has a special taste.3. What does the underlined phrase “a pipe dream” in Paragraph 3 meanA. A practical idea.B. A vain hope.C. A brilliant plan.D. A selfish desire.4. What is the author’s purpose in writing the textA. To show how to grow apples.B. To introduce an apple festival.C. To help people select apples.D. To promote apple research.BAnyone who experienced wet weather this summer will not be surprised to learn that this marks the fourth straight year the UK has endured a wet summer, a first in modern rainfall records dating back to 1914. This summer was only slightly wetter than usual, with the UK receiving 248.5 mm (9.78 inches) of rainfall —10 percent above average, according to Met Office figures.The national rainfall average painted an interesting weather picture. Rain tended to fall in heavy showers, triggering flash floods. As a result, East Anglia in eastern UK experienced its third wettest August with more than double the normal rainfall, while the South East had one of its dullest Augusts, with only two-thirds of its usual sunshine. Yet, both areas had been dry and sunny for the previousfive months and seemed to be heading for a long-lasting drought. In contrast, western and northwestern areas had the opposite experience —extremely wet in July, followed by a dry August. Remarkably, the North West had suffered from such a prolonged drought that it led to hosepipe (软水管) bans.Although some weather companies predicted an extremely hot barbecue summer, temperatures were only 0.6℃above average across the UK. Even so, the first half of this summer saw Glastonbury enjoy a mud-free festival and Wimbledon remain hot and dry. While it didn’t break any temperature records, there were plenty of opportunities for barbecues this summer.5. What was unusual about the weather in the UK this summerA. It was rainy for the fourth summer in a row.B. It was the hottest summer on record.C. It had the least amount of sunshine ever.D. It had the biggest imbalance across the country.6. What contributed to the weather patterns this yearA. The long-lasting drought.B. The geographic factors.C. Sudden heavy rain.D. Small amounts of sunshine.7. What can be inferred from the last paragraph of the passageA. People in Glastonbury celebrated a mud festival freely.B. Some weather companies did not give an accurate forecast.C. The western part of the UK was extremely wet in August.D. The extreme hot weather this summer led to fewer barbecues.CAfter a sleepy Saturday morning on my son Andrew’s 15th birthday, I drive him to a shoe shop near our home in Toronto to get a pair of sandals. We time the trip to arrive right when the store opens. Andrew has difficulties speaking and prefers to go shopping when it’s not busy.“Size 41 of those black slip-on sandals, please,” I tell the two clerks at the shop when we arrive. Andrew slips his socked feet into the shoes. A perfect fit. We box them up and pay. Then I hold up his letter board.For 10 years, Andrew has used a picture chart to communicate. The images represent important and common words: people, places, food, greetings and activities. Years ago, we discovered that Andrew could communicate more than just his basic needs with a letter board. He can point to the letters and spell out words, statements, thoughts.Therefore, instead of using the “thank you” symbol on his picture chart, Andrew points to each letter, o ne by one: “Thank you.” When we look up, they are wide-eyed. One of the staff speaks up, “Can I ask you what that is Because I have a brother-in-law who doesn’ttalk.”This is what happens when we show up —by showing our true selves —in our strengths and our “scantiness”. We invite others into our humanness, and we allow them to share theirs. “Let me give you my contact information. I will teach you this method,” I continue. I ask the shop assistant about his brother-in-law, Jason is 30 and doesn’t speak, but he can do a lot for himself. Still, no one really knows him.“Amazing!” I say. “We’ve met people who started using this method when they were 50 or 15 or five! Andrew, what do you think ” Andrew starts pointing to letters: “Tell Jason he will change e veryone’s opinion of him in 26 letters.”We are all moved, inspired, and thrilled. “Wow,” the clerk says, “Thank you.” Andrew smiles. This is how it happens, how we impact another person’s life in a split second: by delicately leading with our own.8. What do we know about Andrew from the first two paragraphsA. He enjoys shopping on weekends.B. He has a habit of arriving on time.C. He likes less crowded places better.D. He prefers wearing sandals with socks on.9. What does the author intend to do in paragraph 3A. Introduce a new topic.B. Give an example.C. Show a solution.D. Provide a reason.10. What does the underlined word “scantiness” in paragraph 5 probably meanA. Advantages.B. Weaknesses.C. Characters.D. Embarrassments.11. What does the author want to convey in the last paragraphA. Accept disabled people as they are.B Respect a man and he will do the more.C. You will succeed as long as you keep doing it.D. A little kindness goes a long way.DIn the 1950s, British historian Northeaster Parkinson came up with a concept which was later known as Parkinson’s Law of Triviality. It says that the amount of time spent discussing an issue in an organization is oppositely connected with its actual importance. Parkinson’s Law of Triviality is also known as “bike-shedding (车棚)”, after the story Parkinson uses to illustrate it. He asks readers to imagine a financial committee meeting to discuss a three-point agenda. The points are as follows: A proposal for a 10 million nuclear power plant; A proposal for a 350 bike shed; A proposal for a 21 annual coffee budget.What happens The committee ends up running through thenuclear power plant proposal in little time. It’s too advanced for anyone to really dig into the details. The discussion soon moves to t he bike shed. Here, everyone’s an expert. In the end, the committee runs out of time and decides to meet again to complete their analysis.Bike-shedding happens because the smaller a matter is, the more people will have an opinion on it, even when there is no genuine value to add. When something is outside of our circle of capability, like a nuclear power plant, we don’t even try to express an opinion. But when something is understandable, everyone wants to show that they know about the topic at hand.How can we avoid bike-shedding The main thing you can do is to have a clear purpose. Priya Parker, the author of The Art of Gathering: How We Meet and Why It Matters, says that any successful gathering needs to have a focused purpose. “Specificity,” she says, “is a key element.”When it comes to choosing your list of invitees, Parker writes, “if the purpose of your meeting is to make a decision, you may want to consider having fewer cooks in the kitchen.” Getting the result you want —a thoughtful, educated discussion about that power plant — depends on having the right people in the room.12. Which one actually takes up the most time in Parkinson’simaginary financial meetingA The story used to illustrate the solution.B. The nuclear power plant proposal.C. The bike shed proposal.D. The coffee budget proposal.13. What can be inferred from Paragraph 4A. What requires more effort may get less attention.B. The more you know, the less you speak.C. What is simple for you may be tough for others.D. The more you put in, the better your result will be.14. What does the author suggest for a successful meetingA. Taking different opinions into account.B. Setting a specific time limit before the meeting.C. Spending less time in discussing the minor issues.D. Avoiding inviting people with little relevant knowledge.15. What does the text mainly talk aboutA. An important element in decision-making.B. A nuclear power plant dilemma.C. Why we call for an effective meeting.D. Why we tend to focus on small matters.第二节(共5小题;每小题2.5分,满分12.5分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
福建省福州第一中学2022-2023学年高三上学期第一次调研测试英语试题及参考答案
福建省福州第一中学2023届高三第一次调研测试英语第一部分阅读(共两节,满分50分)第一节(共15小题;每小题 2.5分,满分37.5分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项。
ACool InventionsFor Light SleepersMillions of people have trouble falling asleep each night,which is why LightSleeper was created.Developed in the United Kingdom,LightSleeper’s makers say it can help a person fall asleep and also help a person fall back to sleep after waking up in the middle of the night.LightSleeper works by projecting(投射)a light onto the ceiling while the person is lying down.The idea is to watch the light,because the light moving in a circle has a relaxing effect.One of the advantages of the product is that it won’t disturb other people in the room.The product can be easily moved and will switch itself off after30minutes to save power.Stop the Snoring(打鼾)Here’s a new product designed to slow down the frequency and level of your snoring.The Snore Stopper has a sensitive instrument that will detect when you’re snoring.Once it does that,it will send a very mild electronic signal to your wrist.This consequently changes your sleeping position to stop you from snoring.The Snore Stopper won’t cause any damage to your body.Get Your True ColorsPhotoshop features a pen that will copy the color settings of what you are monitoring on screen to use as a color in your program.Now an offline product allows you to pick a color when not using the computer software.The Color Picker is a concept pen.After placing the pen against an object,the user just presses the scan button.Color sensors(感应器)inside the pen send information to a box,which then mixes red,green and blue to produce the desired ink color.Keep It FreshMore high technology products are showing up in the kitchen.A new Bagel dome(圆顶形物)advertises that it will keep your baked goods fresher for longer periods of time.The product,which is able to remove air,works for bread rolls,cakes and other perishables.To use it,you just push a button at the top.1.How can LightSleeper help a person fall asleep?A.Its light makes people feel calm.B.It doesn’t make any noise.C.It is fixed on the ceiling.D.Its light moves fast.2.What do we know about the Snore Stopper?A.It will cause slight damage to a person’s body.B.It will wake a person up once he starts snoring.C.It can make sure a person doesn’t snore all night.D.It can make a person snore less by changing his sleeping position.3.The Color Picker will probably be used while you’re________.A.using the computerB.drawing picturesC.collecting informationD.designing a program4.The underlined word“perishables”in the last paragraph refers to foods which________. A.are easy to keep B.go bad easilyC.are rich in vitamins D.keep you thinBAn artist in Oakland,California is using his skills to help the homeless.Greg Kloehn builds very small shelters that make life on the streets a little more comfortable.The structures offer the homeless some safety and protection from bad weather.Each little house also has wheels on the bottom so it can go wherever its owner goes.Greg Kloehn has given away at least20tiny houses.Several are on the roadside near anactive railroad.On a recent day,Mr.Kloehn stops at one to visit Oscar Young.The two men hug.Inside his little shelter Mr.Young gets relief from cold nights on the streets.Mr.Kloehn also visits SweetPea,another friend who also lives in one of the little homes the artist built. She says it keeps her safe and protects her belongings.In the mornings,Mr.Kloehn searches the streets for building materials.He gathers what he can and takes it to his studio.There,he puts the houses together.Empty coffee bags become roof material.A washing machine door and refrigerator part become windows. Nails,screws and the sticky glue hold all the pieces together.The artist also attaches a small electrical device to the house.The device is powered by the sun.Some of the people living on the streets once had normal houses of their own.But some of the people say they have learned to live with less and they are thankful to that man.Mr.Kloehn says his work is not a social project.He says he is just someone using his skills to help his homeless neighbors.5.The following are the advantages of the small shelters EXCEPT________. A.saving power and energyB.protecting possessions of the homelessC.decorating the streets where they areD.keeping the homeless safe and comfortable6.We can learn from the passage that________.A.Greg Kloehn has sold at least20tiny housesB.an act of kindness has made people simple and gratefulC.the shelters are immovableD.the government has got involved in the action7.The passage is likely taken from________.A.an official reportB.an art reviewC.a science magazineD.a news report8.What’s the main idea of the passage?A.An artist creates homes for the homeless.B.A more comfortable shelter on the streets.C.A successful social project in Oakland.公众号高中僧课堂D.An artist makes a living by designing small shelters.CResearchers at the University of Scotland have discovered a protein that can influence viruses developing and even can control cancer.Now the fight is on to fully understand how it works in the hope of turning the laboratory research into a treatment.The protein is called Hira.Technically it is a histone(组蛋白)complex,but it is easier to understand in terms of what it can do.Three years ago Dr Taranjit Singh Rai and colleagues at the Beatson Cancer Institute and Glasgow University reported that Hira could possibly suppress the division of cells that causes cancer.In the course of that research,Dr Rai found out something unusual.In the lab they have established that the Hira protein has a role to play in the antiviral fight,thus,making it have a fundamental role to play in fighting against cancer.The trick in using it to fight diseases may lie in increasing Hira levels in our cells.“I think what researchers might be interested in is how we can increase levels of this protein to deal with the viruses better,Dr Rai said.Dr Rai has led an international study and support has come from Cancer Research UK and the results are published in the journal Nucleic Acids Research.But there is a major concern that the research is still limited to the laboratory.It is going to take some time,probably years,before this work can move out of the lab and into clinics and hospitals.But the researchers are excited Hira will one day be the basis of a new approach in medicine.9.What does the underlined word“suppress"in Paragraph2probably mean? A.Motivate.B.Monitor.C.Control.D.Improve. 10.What can be known from the text?A.Hira has been used in the medical treatment.B.More studies should be done on Hira.C.The levels of Hira in cells are unchangeable.D.Hira can bring about side effects.11.What is the researchers'attitude to the future of Hira?A.Cautious.B.Optimistic. C.Disapproving.D.Doubtful.12.What's the text mainly about?A.A new way to use the protein.B.A new approach to improving the division of cells.C.A new medicine that can fight diseases.D.A protein that can stop viruses developing.DThere has been a lot of publicity(宣传)recently surrounding5G,the next generation of wireless technology for the world.But what is this technology and how might it change our lives?What is5G?5G stands for fifth generation,meaning the next step in the progression of technology to replace the current4G system.4G was the replacement for3G,which came after2G,and so on.These systems are wireless computer networks.Earlier“G”systems were designed to improve mobile communication operations.Each new technology brought major improvements in speed and greatly increased network capacity.The new5G system promises more of the same.It is expected to permit more users to do more thingsat a faster rate. Higher internet speeds and larger network capacity should result in better performance for device users connected to5G.However,technology experts say there is a major way that5G is different than the earlier systems.It will move well beyond mobile network technology to affect many more devices and industries than other“G”versions.When will it be available?Before we can all use5G,wireless companies and phone makers will have to complete and deploy(部署)a whole new system.New phones and communication equipment must be built.American wireless companies have been preparing for the new system for some time. They have been creating new network equipment and buying broadcasting space to carry5G signals.They have built news G antennas(天线)to serve American cities and towns.Wireless providers will invest at least$275billion n5Grelated networks in the United States,the industry group CTIA reported.The first U.S,launch of5G is expected to happen sometime this year.Industry experts expect it will take a few more year to go nationwide.It will take even longer to reach rural areas.China is expected to launch5G sometime in2020,while European nations are likely to build their systems more slowly over time.13.Which system is widely used nowadays?A.5G.B.2G.C.3G D.4G.14.Which of the following is NOT5G system’s advantage to the other Gs?A.It will allow more users to do more things.B.The internet speed will be much higher. C.The users will spend much less money on it.D.It will affect more devices and industries 15.What can we learn from the last paragraph?A.5G system has nothing to do with phone makers.B.There is much work to do before5G comes into use.C.New phones will be much smarter than those used now.D.4G system will be stopped as soon as5G comes into use.16.In which order will the following countries use5G system according to the passage? A.USA→China→European nations.B.China→USA→European nations. C.USA→European nations→China.D.European nations→China→USA.第二节(共5小题;每小题 2.5分,满分12.5分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
福州一中2024模拟试题及答案
福州一中2024模拟试题及答案一、选择题(每题3分,共30分)1. 下列关于福州一中的说法,正确的是:A. 福州一中成立于1901年B. 福州一中位于福建省福州市C. 福州一中是全国重点中学D. 以上说法都正确答案:D2. 福州一中历史上曾获得的荣誉包括:A. 全国文明校园B. 国家级示范性高中C. 福建省优秀学校D. 全国教育系统先进集体答案:D3. 福州一中在2023年的高考成绩中,以下哪项数据是正确的?A. 一本上线率超过90%B. 有超过100名学生被985高校录取C. 有超过200名学生被211高校录取D. 以上数据均正确答案:D4. 福州一中的特色课程包括:A. 科技创新B. 艺术教育C. 体育竞技D. 以上都是答案:D5. 福州一中的学生社团活动丰富,以下哪项不是福州一中的学生社团?A. 机器人社团B. 模拟联合国C. 书画社D. 烹饪社答案:D6. 福州一中在2024年模拟试题中,下列哪项是数学科目的考点?A. 函数与方程B. 几何与空间C. 概率与统计D. 以上都是答案:D7. 福州一中在2024年模拟试题中,下列哪项是英语科目的考点?A. 阅读理解B. 完形填空C. 写作D. 以上都是答案:D8. 福州一中在2024年模拟试题中,下列哪项是语文科目的考点?A. 古文阅读B. 现代文阅读C. 作文D. 以上都是答案:D9. 福州一中在2024年模拟试题中,下列哪项是物理科目的考点?A. 力学B. 电磁学C. 光学D. 以上都是答案:D10. 福州一中在2024年模拟试题中,下列哪项是化学科目的考点?A. 无机化学B. 有机化学C. 化学实验D. 以上都是答案:D二、填空题(每题4分,共20分)1. 福州一中每年的________月________日举行校庆。
答案:10月10日2. 福州一中的学生总数约为________人。
答案:30003. 福州一中在2023年获得了________项国家级奖项。
福建省福州第一中学、三牧中学2023-2024学年九年级上学期开学考试语文试题
2023-2024学年第一学期福州一中、三牧中学九年级适应性练习语文学科(时间:120分钟总分:150分)一、积累与运用(25分)奋斗的青春最美丽,奋斗的人生最精彩。
初三生活伊始,班级开展主题为“无奋斗不青春”的综合性学习活动,邀你参加。
【活动一】初生之犊,后生可畏1.下面是小牧同学制作的诗文记录卡,请你根据语境,补写出诗文名句。
(10分)诗文记录卡在历史长河中,中华民族涌现出了一批忧国忧民的仁人志士。
杜甫在《茅屋为秋风所破歌》中,借“(1),(2)”表现其崇高的理想和忧国忧民的情怀;白居易关心民生疾苦,在《卖炭翁》一文中饱含怜惜之情,用“(3),(4)”来体悟卖炭老人之忧:欧阳修在《醉翁亭记》中,借“(5),醒能述以文者”表达与民同乐的为官之道;陆游在《卜箕子·咏梅》中,以“(6),只有香如故”来写梅花纵使凋零于地,仍香气不改,精神犹在,表现了自己矢志不渝的爱国情怀。
青少年是民族复兴的新生力量,要树立远大理想,彰显青春风采。
为此,青少目年应如李白不惧“欲渡黄河冰塞川,(7)”(《行路难》)的磨难:具备庄子“水击三千里,(8)”(《北冥有鱼》)的高远志向;涵养范仲淹“先天下之忧而忧,(9)”(《岳阳楼记》)的豁达;以及培养刘禹锡“今日听君歌一曲,(10)”(《酬乐天扬州初逢席上见赠》)的坚韧不拔的品质。
【活动二】奋斗为马,激发“核能”2.小福同学在网上搜到以下材料,请你与他一起完成下列任务。
(9分)2023年1月,中核集团在微博喊话《流浪地球2》:“你们尽管想象,我们负责实现。
”①(kēng)锵有力的话语引发社会各界()关注。
人类的想象往往源于现实,高于现实。
但如果没有强大的科技实力作为基础,想象也必然只是()。
核聚变钻研的革命性进展,为艺术作品的大胆想象供给现实基础;而能源行业的稳扎稳打,则给予了科幻作品永恒的题材。
能源既是人类文明发展的重要支撑,也是②(héng)量国家科技实力的关键要素。
福建省福州第一中学2023-2024学年高二上学期第一学段(期中)考试数学试题
福建省福州第一中学2023-2024学年高二上学期第一学段(期中)考试数学试题学校:___________姓名:___________班级:___________考号:___________二、多选题9.已知,,a b c r r r 是空间的一个基底,则下列说法中正确的是( )A .若0xa yb zc ++=r r r r ,则0x y z ===B .,,a b c r r r 两两共面,但,,a b c r r r 不共面C .一定存在实数x ,y ,使得a xb yc=+r r r D .a b +r r ,b c -r r ,2c a +r r 一定能构成空间的一个基底A .当E 点运动时,1AC ^B .当E 向1D 运动时,二面角C .二面角E AB C --的最小值为D .EF uuu r 在CB uuu r 方向上的投影向量为12.经过坐标原点O 且互相垂直的两条直线(1)写出直线PQ 的方程(答案含t );(2)在OPQ △内作内接正方形ABCD ,顶点,A B 在边OA a =,当正方形ABCD 的面积最大时,求,a t 的值19.在直三棱柱111ABC A B C -中,D ,E 分别是AA 11AA =,90BCA Ð=°.(1)证明:直线//BE平面(2)若平面PBD^平面ABCD设EP t =,显然11E P t =,则(2,1P 11221cos ,cos 1145OP O P t áñ==+°=uuu r uuuu r则(1,1,0),(0,1,0),(0,0,0),A B C D 因为E ,F 在11B D 上,且2EF =(1,,1)AE t t =--uuu r ,设平面ABE 的法向量为又(1,0,0)AB =-uuu r ,所以()01x t x -=ìí-îr(2)因为平面PBD^平面ABCD22.(1)()2214x y-+=(2)2 3。
福州一中录取通知书
福州一中录取通知书
今年夏天,福州一中的所有高一新生将到福一中大学城新校区就读。
记者昨日了解到,福一中今年的招生任务全部由新校区承担,继续面向全省招收高中生,但不再自行组织命题招考,而是通过招收保送生与中招两种办法招生。
根据省教育厅昨日下发的通知,福州一中今年仍面向23个省级课改实验区和福州市所辖八县(市)招生,总体招生计划确定为16个班,共848人。
其中安排正常招生计划12个班,648人;同时利用新校区优质高中资源,增招200名择校生。
具体招生办法为:1.面向17个省级课改实验区招收85名正常缴费生(莆田、泉州、漳州、三明、龙岩各5名,南平15名、宁德45名)。
如招收不满,计划调整到福州市;2.面向福州市招生563名,其中六区招收401名学生,招生录取工作按福州市中招办法进行;八县(市)162名,招生录取工作安排在八县(市)中招提前批进行,考生如果没有被福州一中录取,也不影响他们报考其他的中学。
此外,增招的200名择校生全部面向福州六区和八县(市)招生,其中六区180名,八县(市)20名,在招生不满的情况下,允许计划调剂。
增招的择校生降分幅度最多为20分。
据悉,去年福州一中新校区率先组织面向全省单独提前招考,吸引了8000余名初三学生参加角逐,最终录取了540名学生,录取率为1∶15。
福州一中全省招生范围
福州市六城区及八县(市);三明市梅列区、三元区;宁德市蕉城区、福安市、古田县;泉州市鲤城区、丰泽区、南安市、晋江市、石狮市;漳州市芗城区、东山县、南靖县;南平市延平区、武夷山市;龙岩市新罗区;莆田市荔城区。
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福建省福州第一中学高一下学期期末考试英语试卷
福州一中2022-2023学年第二学期第四学段模块考试高一英语学科必修二模块试卷(完卷120分钟满分150分)第一部分听力(共两节,满分30分)第一节(共5小题;每小题1. 5分,满分7. 5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What does the woman want to do?A. Find a place.B. Buy a map.C. Get an address.2. What will the man do for the woman?A. Repair her car.B. Give her a ride.C. Pick up her aunt.3. Who might Mr. Peterson be?A. A new professor.B. A department head.C. A company director.4. What does the man think of the book?A. Quite difficult.B. Very interesting.C. Too simple.5. What are the speakers talking about?A. Weather.B. Clothes.C. News.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。
每段对话或独白读两遍。
听第6段材料,回答第6, 7题。
6. Why is Harry unwilling to join the woman?A. He has a pain in his knee.B. He wants to watch TV.C. He is too lazy.7. What will the woman probably do next?A. Stay at home.B. Take Harry to hospital.C. Do some exercise.听第7段材料,回答第8, 9题。
福建省福州市第一中学2025届高三(最后冲刺)英语试卷含解析
福建省福州市第一中学2025届高三(最后冲刺)英语试卷注意事项: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。
第一部分(共20小题,每小题1.5分,满分30分)1.I'd advise anyone to put aside a proportion of their earnings — ____________.A.nothing great is easy B.nothing comes from nothingC.a bad penny always turns up D.a penny saved is a penny earned2.—There aren’t any toys in our suitcase.Where have you put them?—Oh, no.That old lady sitting next to us ______ have taken ours by mistake.A.must B.could C.should D.would3.For all the failure of his business, he remains optimistic and ________ untroubled by his problems.A.hopefully B.seeminglyC.rarely D.frequently4.We should do what we can ______ people’s har d work pays off.A.be sure B.being sureC.making sure D.to make sure5.I have often thought it would be a blessing if each human being _______ blind and deaf for a few days at some time in his life.A.has been stricken B.were strickenC.had been D.would be6.—Tom, do you enjoy your present job?—________. I just do it for a living.A.Of course B.Not reallyC.Not likely D.That’s all right7.-I was afraid I would miss the important lecture.-Oh, was that why you _______asking me to speed up?A.had kept B.are keepingC.would keep D.kept8.Great pity! His illness is ________ the treatment available so far.A.within B.above C.about D.beyond9.After talking with her teacher for about an hour, she felt relieved and began to know how to this kind of situation.A.handle B.raise C.explore D.strengthen10.Molly finally agreed, _____ reluctantly, to go and see a doctor.A.afterwards B.almostC.otherwise D.somewhat11.It was when she first arrived in China _____ she developed a passion for paper-cutting.A.where B.thatC.how D.why12.The problem lies in______ we’re now giving him a fish, not teaching him how to fish.A.what B.thatC.which D.if13.—Why not talk with your parents about your willingness to attend 2017 Peking University Summer Camp?—I tried____ to get them to listen to me.A.in time B.in vainC.in need D.in case14.The new playground to be built next year will be ____________ the old one.A.as three times big as B.three times as big asC.as big as three times D.as big three times as15.---How did she lose so much weight as expected?----_____________ eating less,she jogged for several miles every day.A.In addition to B.In spite ofC.Instead of D.Thanks to16.—Why does he wear a raincoat on sunny days?—Nor do I understand. He is ________.A.a wet blanket B.a lazy boneC.a black sheep D.an odd fish17.---Where is my Chinese book? I remember I put it here yesterday.---You _________ it in the wrong place.A.must put B.should have putC.might have put D.might put18.Being happy is a skill that can be learned, and one way to ________ ourselves to be happy is to write down the little things that cheer us up each day.A.convey B.appeal C.train D.attempt19.—Where was I?—You ________ you didn’t like your job.A.had said B.saidC.were saying D.has said20.________ who are able to work through the struggle are the ________ who are going to be successful. A.Someone; one B.Anyone; oneC.He; ones D.Those; ones第二部分阅读理解(满分40分)阅读下列短文,从每题所给的A、B、C、D四个选项中,选出最佳选项。
福建省福州第一中学2023-2024学年高一下学期4月第三学段模块考试数学试题
福建省福州第一中学2023-2024学年高一下学期4月第三学段模块考试数学试题学校:___________姓名:___________班级:___________考号:___________一、单选题1.设向量a,b的夹角为θ,且a=(−2,1),a+2b=(2,3),则cosθ=()A.−35B.35C.55D.−2552.如图,一个水平放置的平面图形的直观图是一个长为3宽为2的矩形,则该平面图形的面积为()A.322B.32C.6 D.1223.已知直线a、b和平面α、β,则下列说法正确的是()A.若a∥α,b∥α,a⊂β,b⊂β,则α∥βB.若a∥α,b∥β,α∥β,则a∥bC.若a∥α,b∥β,a∥b,则α∥βD.若a∥α,a∥β,α∩β=b,则a∥b4.《九章算术》中的“商功”篇主要讲述了以立体几何为主的各种形体体积的计算,其中堑堵是指底面为直角三角形的直棱柱.在堑堵ABC−A′B′C′中,AB⊥AC,AB=1,AC=2,BB′=3,则此堑堵的外接球半径是()A.23B.3C.142D.145.公路北侧有一幢楼,高为60米,公路与楼脚底面在同一水平面上.某人在点A处测得楼顶的仰角为45°,他在公路上自西向东行走,行走60米到点B处,测得仰角为45°,沿该方向再行走60米到点C处,测得仰角为θ,则sinθ为()A.12B.22C.33D.136.在△ABC中,点D为边AB上一点,若BC⊥CD ,AC=32 ,AD=3, sin∠ABC=33,则△ABC的面积是A.62B.1522C.922D.1227.已知圆台上下底面的圆心分别为O1,O2,母线AB=3(点B位于上底面),且满足AO2=2BO1,圆O2的周长为2π,一只蚂蚁从点A出发沿着圆台的侧面爬行一周到AB的中点C,则蚂蚁爬行的最短路程为()A.3132B.33C.37D.3728.在△ABC中,AB=AC,若点O为△ABC的垂心,且满足AO=14AB+xAC,则cos∠BAC的值为()A.12B.13C.14D.15二、多选题9.设a,b,c分别为△ABC的内角A,B,C的对边,下列条件中可以判定△ABC一定为等腰三角形的有()A.a cos A=b cos B B.a cos B=b cos AC.b sin B=c sin C D.a=2b cos C10.已知三棱台ABC−A′B′C′,上下底面边长之比为1:2,棱AB、BC、AC的中点为点M、P、N,则下列结论错误的有()A.A′N//PC′B.A′P与AC为异面直线C.AB//面A′C′P D.面A′MN//面BCC′B′11.在△ABC中,角A,B,C的对边分别为a,b,c,且满足b=1,1−2ac sin A=a2+c2,则下列说法正确的有()A.1+a tan B=0B.sin C+cos2B=0C.c+2a>2D.△ABC的面积S<14三、填空题12.如图,八面体的每个面都是正三角形,并且四边形ABCD是边长为10的正方形,则这个八面体的体积是.13.在梯形ABCD中,AB//CD,∠DAB=90∘,AB=2,CD=AD=1,若点M在线段BD上,则AM⋅CM的最小值为.14.已知e1、e2为单位向量,且e1−e2=1,若向量a满足a+2e2=1,则a+te1t∈R的最小值为.四、解答题15.已知在平面直角坐标系中,点A1,2,B4,6,C0,3.(1)若 AB+tAC⊥ AB−tAC,求t的值;(2)记AB在AC方向上的投影向量为a,求a的坐标.16.如图,四边形ABCD中,AD⊥AB,∠ADC=120°,AB=23,AD=1,CD=2,(1)求将四边形ABCD绕直线AD旋转一周所成几何体的体积;(2)求将四边形ABCD绕直线AB旋转一周所成几何体的表面积.17.正六棱柱ABCDEF−A′B′C′D′E′F′,两条相对侧棱所在的轴截面为正方形,高为4,记C′D、E′F、AA′的中点分别为P、Q、M.(1)要经过点M和对角线F′C′将六棱柱锯开,请说明在六棱柱表面该怎样划线,并求截面面积;(2)证明:PQ//面ABCDEF;(3)直线AB上是否存在一个点N,使得面D′QN//面BPF?若存在,求出BN的长度;若不存在,请说明理由.18.已知a,b,c分别是△ABC三个内角A,B,C的对边,3c sin B+b cos C=a+c,(1)求角B;(2)若点D在边AB上,BD=2,AC=23,且DA=DC,求∠DCA.19.在△ABC中,D、E为边BC上两点,且满足∠BAD=∠CAE,BD=1,DE=5,EC=3,(1)求证:ABAC =sin∠ADB3sin∠AEC;(2)求证:ABAC为定值;(3)求△ABC面积的最大值.。
福建省福州第一中学2023-2024学年八年级上学期开学语文试卷
2023-2024学年福建省福州一中八年级(上)开学语文试卷一、积累与运用。
(25分)1.(10分)补写古诗文名句.(1)刘禹锡《陋室铭》开头四句“山不在高,有仙则名。
水不在深,有龙则灵”用类比引出“ , ”的文章主旨。
(2)杜甫在《望岳》中表达不畏困难,敢于攀登绝顶,俯视一切的雄心壮志的诗句是: , 。
(3)王安石《登飞来峰》中富于哲理,借情抒景,.表达自己锐意改革的远大政治抱负的诗句是: , 。
(4) ,关山度若飞。
《木兰诗》)(5)予独爱莲之出淤泥而不染, ,中通外直……(《爱莲说》)(6)在《游山西村》中,陆游行走于山间水畔悟道于心的句子是 , 。
2.(4分)字词拼音。
①t ān 塌②废x ū③哺.乳④滑稽.3.(3分)请修改病句。
由于老师带领我们认真地复习,使我们班的成绩得到了极大的提高。
4.(3分)依次填入下面一段文字横线处的语句,衔接最恰当的一组是( ) 赤子之心是助人实现理想的正能量。
____,___,___①孟子说:“大人者,不失其赤子之心者也。
”即有大成就的人都有一颗婴儿般的心。
②他们选择职业时,仅看到行业的光鲜和收入的可观,却看不到背后的辛苦和压力。
③这样的人一旦遇到困难就会灰心丧气,停步不前,何谈实现理想?④但是,许多年轻人是从功利的角度确立目标的。
A .④②③①B .①④②③C .②④①③D .④②①③5.(5分)班级开展名著阅读分享活动,请你从备选人物中选择一位,结合故事情节,参与交流。
一个人要实现自己的梦想,最重要的是要具备以下两个条件:勇气和行动。
——俞敏洪备选人物:祥子尼摩船长唐僧二、阅读。
(65分)6.(6分)阅读下文,回答问题。
贾生李商隐宣室求贤访逐臣,贾生才调更无伦。
可怜夜半虚前席,不问苍生问鬼神。
(1)(3分)下列对诗歌赏析不正确的一项是A.标题中的“贾生”即贾谊,西汉初期著名政论家,主张加强中央集权,曾任太中大夫,后被迁为长沙王太傅。
B.一、二句叙述贾谊被迁后,过了几年汉文帝又把他召回长安,在宣室里接见了他。
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试题进一步体现 "稳步推进,适度创新"的命题原则,主要有2个特点:
试题以《课程标准》规定的主干知识为载体,不拘泥某一版本的教科书。7道选择题仍以基础题为主,重在考查基础知识、基本概念、基本技能,力图扩大知识的覆盖面。3道客观题情景真实新颖,重在落实"三基"和综合能力的考查,设问难、易编排顺序合理,容易下手,难度控制恰当。2道选考题既体现选修模块特点,又抓住化学理论的指导功能,重在应用概念、规律解决化学问题。
试题重点考查学生对生物学科知识、能力和方法的掌握情况,主要体现5个特点。
原创性强,题在书外,理在书中,点在"说明",源在"课标"。
试题难度适中,未出现让考生无从下手的偏题、怪题。命题贴近教材,重视基础,突出能力立意。要求学生在全面、系统地理解和掌握教材知识的基础上,做出正确的解答。
在情景设计上紧扣现实生活或生产中的现象或问题。
题干精炼,尽量让考生能够快速、准确获取关键信息;对于非选择题的参考答案设计,力求严谨准确,适当减少学生的书写量。
符合《考试说明》规定的范围。
在注意高中生物学各模块的大体平衡的前提下,突出了主干知识、核心知识,如细胞代谢、遗传规律及其应用、生态系统的功能、基因工程的原理和技术等。
注重数学本质,强调数学应用。
应用性试题的设计关注数学知识的合理应用,考查了数学知识在学科内的应用。如理5、文5以抽取样品为背景,考查算法的应用;理9、文16取材于考生熟悉的学习、生活实际,不仅考查了考生对相关数学知识的理解水平,而且以这些知识为载体,检测考生将知识迁移到现实情景的能力。
理17新颖的设问方式是命题很好的尝试,改变了传统立体几何的考查模式,提供四个条件,要求考生从中选择两个条件求解,考生需在过程中进行分析和论证,判断所给四个条件的等价性。理19、文21要求从一个特殊曲线所具有的性质,运用类比推理,得出另一个特殊曲线是否具有同样的性质,并且推广到同一类曲线的一般性问题,也需要考生经历尝试、归纳、猜想与推证的过程。
今后一段时间,考生要注重回归课本,按照《考试说明》的要求,把散落在课本中的各知识点再逐一梳理清楚,形成知识网络。精做练习,提升理解能力和解题能力。检测对课本上每一个知识点的理解是否准确、到位,一个知识点与另一个知识点是否连接起来了,做练习是最好的方法。选做的练习应符合《考试说明》的要求。克服丢开教材专看"教辅"的偏向;克服题量过大和做难题、偏题、怪题的偏向,要精选试题,以质取胜,归纳总结,做一道题会一类题。
第25题,从感性到理性、由浅入深、由表及里、分层次、多角度展开化学实验探究,经典实验、定量试验、探究试验有机融合。要求学生明确实验目标,理解实验装置和实验流程,抓住:气体制备→净化或干燥→主体实验→"断后"或尾气处理这条主线。穿插考查实验试剂选择、实验设计缺陷分析、实验数据处理、实验误差分析、实验过程的变量控制、实验图表解读、实验结论和目的评价。
全卷立足基础、适度开放,凸显科学素养和分析、解决问题的能力。试题不偏不怪,无过多冗长的文字叙述,答案明确简练。
试题在体现新课程理念、领会新教材意图上比去年更趋成熟。
第23题,突破元素性质与元素周期律考查的窠臼,巧妙渗透化学反应原理。该题打破以往从元素周期表的生硬推导入题,代之以实验现象和实际生产问题入手,很自然地把元素化合物性质与化学反应原理巧妙结合,把生产实际与实验现象、定性与定量、必修与选修有机融合,体现学科内综合的思想。
认真揣摩一下,我们就明白了省质检试卷中的题型均符合《考试说明》中规定的试题类型。这就给我们的总复习提供了一个信息,就是《考试说明》中的试题类型,学生都要进行相应的训练,不能只盯着"参考试卷"的题型,"参考试卷"只供"参考",不是一成不变的。复习时必须认真学习和理解《考试说明》的命题指导思想和考试内容要求的精神实质。
文化经典题是阅读《孟子》的选段,检测了考生对中国文化经典的理解能力,要求在理解的基础上填空,题型突破了近几年来只用选择题考查的固定模式,使考题有些新意,同时也降低了难度。
文学类、实用类的选考题,选文好,两组试题的难易度接近了,探究题的探究味浓了。
语言文字运用题,选取的时文是关于网络用语方面的问题,内容既切合语文学科特点,又有时代感,也是热点问题;材料有利于考生作多向度的评说。
把握数学灵魂,强调思想方法。
试题能注重学科内涵,突出学科特色,突出对数学思想与方法的考查,涉及中学阶段出现的重要数学思想和方法。理15考查概率统计最本质的内容,体现了必然与或然思想。
关注过程考查,适度探究创新。
试卷合理地设置具有一定思维量的开放性、探索性的试题,有效地考查考生的探究精神和创新意识。试卷既有条件开放的试题,如理20、文20,要求考生根据题目要求探索结论成立的条件是否存在;也有结论开放性问题,如文19(3),符合题设条件的结果不止一种。
试题突出了数学学科的特点,贴近教学实际,既注重全面,又突出重点,重视知识内在联系之余,强化对中学数学所蕴涵的数学思想和方法的考查。
试题对高中数学的主干知识进行了较为全面地考查,函数与导数、三角函数、数列、空间几何、直线与圆锥曲线、统计与概率等主干知识的占分比例在文理科中约为87%。
题目立意新颖,综合性较强。试题几乎涵盖了《考试说明》中所有Ⅱ级要求的考点,同时又较好滴兼顾对Ⅰ级要求考点的考查。试卷结构合理、难易适当、预估难度值在0.55~0.57。
物理:
题目立意新颖
综合性较强
点评人:福州三中高三物理集备组长方秀珍
试卷突出能力立意,注重对考生科学素养的考查,较好体现新课程理念,关注科学探究和物理学史的教育功能。试题着力考查考生对物理基础知识、基本技能的掌握程度,以及综合运用所学知识分析、解决问题能力。注重考查物理思想和方法的应用,关注物理与科学、技术和社会的联系,以及物理知识在生产、科研和生活中的应用。
英语:
加大词汇覆盖面
强调活用语言
点评人:福建省首届中小学中青年学科带头人、仙游一中黄聚宝
试卷令人"耳目一新",延续近年来提倡活用语言的命题原则。试题全为原创题,语料来自国外的报刊、广告、原版教材等,材料真实,许多题干、话题以时下热门事件为背景。
试卷有4个特点:
突出能力立意,关注数学素质。
注重以知识为载体检测考生的数学能力与数学素养。"五个能力"和"两个意识"在试卷中都有所涉及。理9(即理科卷第9题,下同)、文16(即文科卷第16题,下同)注重考查考生的知识迁移能力和数学素养。理19、文21需运用合情推理,寻找规律,进而运用所学知识予以验证,有效地测量出考生将知识迁移到不同情境的能力。
福州一中、三中、师大附中等名师点评2010福建省质检
11日下午,作为高考前最后一次全省性统一考试的省质检考毕。这套试卷无疑为考生提供了一套最具参考价值的样题。福州一中、福州三中、仙游一中、南安一中、宁德进修校等学校的全省各个学科的权威名师,对省质检的试卷进行了深入、细致的点评,希望对广大考生下一阶段的复习能有所帮助,根据试卷透露的高考命题信息与热点来调整最后的复习计划。
第24题,体现科学发展观在现代经济生产中的应用。该题从学生熟悉的硫酸工业生产废气和烧渣的回收处理入题,以真实的工艺流程为载体,使学生体会科学发展观在现代循环经济生产中的应用。试题巧妙地将元素化合物性质与氧化还原反应结合、将电解质溶液的定性判断与定量计算相结合,考查学生综合运用知识解决问题的能力,
语言表达新颖、地道,强调运用,注重交际,力求最大限度地保留原来韵味,符合英美人士最新发展的语言习惯。
试卷总容量和整体难度略高于去年,没有出现偏题、怪题,难、中、易比例合理。全卷侧重考查考生的综合语言运用能力,特别是运用英语获取信息、处理信息、分析问题和解决实际问题的能力。
第30题,改变以往命题套路,变死记硬背概念、结论为运用化学理论解释化学问题。亮点在于利用教材图表,巧妙增设问题,考查学生是否掌握、理解教材图表曲线
紧扣现实生活
点评:福州一中生物教研组组长、高级教师沈鸣
作文题,用三位不同时期名人的诗句提示思路,显得风趣、幽默,容易引发考生的联想与发挥。作文的具体要求似乎还是老一套的"题目自拟、立意自定、文体自选",但"关于'高峰'历来众说纷纭,对此,你有何经历或看法?"这一要求,倾向于考生能写议论文,这对作文教学有较好地导向作用。
值得注意的是:省质检试卷中的一些题目的题型与《考试说明》参考试卷中的题型有些差别,如:古代诗歌阅读,省质检试卷只设一道问答题,而参考试卷保持往年样式,设了两道问答题。文化经典阅读,省质检试卷的第一小题要求"用上面选段原文填空",而"参考试卷"是四选一的客观题。现代文阅读,省质检试卷是两道四选一的客观题,一道主观题,而"参考试卷"是三道四选一的客观题。
数学:
考查主干知识
设置开放试题
点评人:泉州市南安一中特级教师林少安
数学卷延续了2009年省质检的特点,坚持以能力检测为主导,以学科主干知识为载体,在考查基础知识、基础技能和基本方法的同时,注重对考生应用知识分析问题、解决问题能力和探究能力的考查。
语文:
题型有所突破
不死盯"参考试卷"
点评人:省普教室语文专家
刚刚进行的高三省质检,语文试卷的考试形式、试题类型、难度、结构都符合我省今年语文科《考试说明》的规定与要求。整体感觉比较好。
诗歌题突出了赏析,又有一定的指向,如:题干中"动感""画意"这些词语,指明了答题取向,注重考查具体的鉴赏分析能力,有利于区分考生的水平。
试卷有3个特点:
能力加强,不超考纲。
试卷加大了新课程词汇的覆盖面,加强对能力培养的考查力度,体现了教学的新导向、新思维。试卷考点不变,且分布广泛、更具典范性。