2020-2021学年广东省实验中学高三(上)第一次段考数学试卷

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高中数学压轴题题型名校模考题汇总

高中数学压轴题题型名校模考题汇总

专题10压轴题题型汇总压轴题型一、保值函数型“保值函数”,又称为“k 倍值函数”,“和谐函数”,“美好区间”等等。

1、现阶段主要是一元二次函数为主的。

核心思路是转化为“根的分布”。

2、函数单调性是解决问题的入口之一。

3、方程和函数思想。

特别是通过两个端点值构造对应的方程,再提炼出对应的方程的根的关系。

如第1题1.(江苏省连云港市市区三星普通高中2020-2021学年高一上学期期中联考)对于区间[,]a b 和函数()y f x =,若同时满足:①()f x 在[,]a b 上是单调函数;②函数(),[,]y f x x a b =∈的值域还是[,]a b ,则称区间[,]a b 为函数()f x 的“不变”区间.(1)求函数2(0)y x x =≥的所有“不变”区间;(2)函数2(0)y x m x =+≥是否存在“不变”区间?若存在,求出实数m 的取值范围;若不存在,请说明理由.2.(北京市昌平区2020-2021学年高一上学期期中质量抽测)已知函数2()f x x k =-.若存在实数,m n ,使得函数()f x 在区间上的值域为,则实数k 的取值范围为()A .(1,0]-B .(1,)-+∞C .2,0]D .(2,)-+∞3.(广东省广州市第一中学2020-2021学年高一上学期11月考试)已知函数221()x f x x-=.(1)判断函数()f x 的奇偶性并证明;(2)若不等式23()1x f x kx x +-≥在1,14x ⎡⎤∈⎢⎥⎣⎦上恒成立,求实数k 的取值范围;(3)当11,(0,0)x m n m n ⎡⎤∈>>⎢⎥⎣⎦时,函数()()1(0)g x tf x t =+>的值域为[23,23]m n --,求实数t 的取值范围.4.(江苏省盐城市实验高级中学2020-2021学年高一上学期期中)一般地,若()f x 的定义域为[],a b ,值域为[],ka kb ,则称[],a b 为()f x 的“k 倍跟随区间”;特别地,若()f x 的定义域为[],a b ,值域也为[],a b ,则称[],a b 为()f x 的“跟随区间”,(1)若[]1,b 为2()22f x x x =-+的跟随区间,则b =______;(2)若函数()f x m =m的取值范围是______.压轴题型二、方程根的个数1.一元二次型“根的分布”是期中考试的一个难点和热点。

实验中学2021届高三上学期第一次阶段考试数学试卷含答案

实验中学2021届高三上学期第一次阶段考试数学试卷含答案

广东省实验中学2021届高三年级第一次阶段考试数 学本试卷分选择题和非选择题两部分,共4页,满分150分,考试用时120分钟。

注意事项:1.答卷前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名、考号填写在答题卷上。

2.选择题每小题选出答案后,用2B 铅笔把答题卷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;不能答在试卷上.3.非选择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答题卷各题目指定区域内的相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改液.不按以上要求作答的答案无效。

4.考生必须保持答题卡的整洁,考试结束后,将答题卷收回。

第一部分选择题(共60分)一、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选项中,只有一项是符合要求的.1.已知}1,log {2>==x x y y A , ]2,1|{>==x xy y B ,则.B A =( )A .),21[∞+B .)21,0(C .),0(∞+D .),21[)0,(+∞-∞ 2.“542sin =α"是“2tan =α”的( )A .充分不必要条件B .必要不充分条件C .充要条件D .既不充分也不必要条件3.为了规定工时定额,需要确定加工某种零件所需的时间,为此进行了5次试验,得到5组数据:,(),,(211x y x ),(),,(),,(),5544332y x y x y x y ,由最小二乘法求得回归直线方程为.9.5467.0+=x y若已知54321x x x xx ++++=250,则54321y y y yy ++++=( )A .75B .155.4C .375D .4424.命题p :变量),(y x 满足约束条件⎪⎩⎪⎨⎧≥-+≤≤0543y x x y ,则x yz =的最小值为41,命题q :直线2=x 的倾斜角为2π,下列命题正确的是( )A .q p ∧B .)()(q p ⌝∧⌝C .q p ∧⌝)(D .)(q p ⌝∧ 5.已知两个单位向量21,e e ,若121)2(e e e ⊥-,则21,e e 的夹角为( )A .32πB .3πC .4πD .6π6.已知)2,0(πα∈,)0,2(πβ-∈,31)4cos(=+πα,33)24cos(=-βπ,则)2cos(βα+=( )A .33 B .33-C .935 D .96-7.已知长方形的四个顶点:)1,0(),1,2(C ),0,2(),0,0(D B A .一质点从点A 出发,沿与AB 夹角为θ的方向射到BC 上的点1P 后,依次反射到DA CD 、和AB上的点432P P P 、、(入射角等于反射角).设4P 的坐标为)0,(4x ,若214<<x ,则θtan 的范围是()A .)21,31( B .)52,31( C .)21,52( D .)32,52(8.设*N n ∈,函数n x x x f ln )(=,函数)0()(>=x xe x g n x.若函数)(x f y =与函数)(x g y =的图象分别位于直线y =1的两侧,则n 的取值集合为( ) A .}2,1{ B .}3,2{ C .}3,1{ D .}3,2,1{二、选择题:本题共4小题,每小题5分,共20分,在每小题给出的四个选项中,只有多项符合题目要求.全选对的得5分,有选错得得0分,部分选对得得3分.9.己知函数R x x x x x x f ∈-+=,cos cos sin 32sin )(22,则( )A .2)(2≤≤-x fB .)(x f 在区间),0(π上只有1个零点C .)(x f 的最小正周期为πD .3π=x 为)(x f 图象的一条对称轴10.已知空间中不同直线n m 、和不同平面βα、,下列命题中是真命题的是( )A .若n m 、互为异面直线,ββαα//,//,//,//n m n m ,则βα//B .若βα//,,n m n m ⊥⊥,则βα⊥C .若αα//,m n ⊥,则m n ⊥D .若m n m //,,αβα⊥⊥,则β//n11.设公差不为0的等差数列}{na 的前n 项和为nS ,若1817S S=,则下列各式的值为0的是( )A .17a B .35S C .1917a a- D .1619S S-12.1970年4月24日,我国发射了自己的第一颗人造地球卫星“东方红一号”,从此我国开始了人造卫星的新篇章.人造地球卫星绕地球运行遵循开普勒行星运动定律:卫星在以地球为焦点的椭圆轨道上绕地球运行时,其运行速度是变化的,速度的变化服从面积守恒规律,即卫星的向径(卫星与地球的连线)在相同的时间内扫过的面积相等.设椭圆的长轴长、焦距分别为c a 2,2,下列结论正确的是( ) A .卫星向径的取值范围是],[c a c a +-B .卫星在左半椭圆弧的运行时间大于其在右半椭圆弧的运行时间C .卫星向径的最小值与最大值的比值越大,椭圆轨道越扁D .卫星运行速度在近地点时最大,在远地点时最小第二部分 非选择题 (90分)三、填空题:本题共4小题,每小题5分,共20分.13.已知等腰三角形的底边长为6,一腰长为12,则它的内切圆面积为____.14.大学在高考录取时采取专业志愿优先的录取原则.一考生从某大学所给的8个专业中,选择3个作为自己的第一、二、三专业志愿,其中甲、乙两个专业不能同时兼报,则该考生有____种不同的填报专业志愿的方法(用数字作答).15.如图,在四棱锥ABCD S -中,⊥SA 平面ABCD ,底面ABCD是菱形,且DAB ∠1,60===AB SA,则异面直线SD 与BC 所成的角的余弦值为 ,点C 到平面SAD 的距离等于 .16.点P 在双曲线)0,0(12222>>=-b a b y a x 的右支上,其左、右焦点分别为、1F 2F ,直线1PF 与以坐标原点O 为圆心、a 为半径的圆相切于点A ,线段1PF 的垂直平分线恰好过点2F ,则该双曲线的离心率为____.四、解答题:本题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤. 17.(本小题10分)设}{na 是公差大于零的等差数列,己知.27,32231-==a a a (1)求}{na 的通项公式;(2)设}{nb 是以函数x y π2sin 4=的最小正周期为首项,以2为公比的等比数列,求数列}{n n b a ⋅的前n 项和nS .18.(本小题12分)在①41sin sin =C B ;②332tan tan =+C B 这两个条件中任选一个,补充到下面问题中,并进行作答.在ABC ∆中,内角C B A ,,的对边分别为32,31tan tan ,,,==a C B c b a , .(1)求角C B A ,,的大小; (2)求ABC ∆的周长和面积.19.(本小题12分)在多面体111B A ABCC 中,四边形11A ABB 为菱形,601=∠BA B ,平面⊥11A ABB 平面,21,11C B BC ABC =.,1C B AB BC AC ⊥⊥(1)若O 是线段AB 的中点,证明:平面⊥ABC 平面;1OC B(2)求二面角B AC C--1的正弦值.20.(本小题12分)为了了解居民的家庭收入情况,某社区组织工作人员从该社区的居民中随机抽取了100户家庭进行问卷调查,经调查发现,这些家庭的月收入在3000元到10000元之间,根据统计数据作出:(1)经统计发现,该社区居民的家庭月收入Z (单位:百元)近似地服从正态分布)196,(μN ,其中μ近似为样本平均数.若Z 落在区间)2,2(σμσμ+-的左侧,则可认为该家庭属“收入较低家庭”,社区将联系该家庭,咨询收入过低的原因,并采取相应措施为该家庭提供创收途径.若该社区A 家 庭月收入为4100元,试判断A 家庭是否属于“收入较 低家庭”,并说明原因;(2)将样本的频率视为总体的概率;①从该社区所有家庭中随机抽取n 户家庭,若这n户家庭月收入均低于8000元的概率不小于50%, 求n 的最大值;②在①的条件下,某生活超市赞助了该社区的这次调查活动,并为这次参与调查的家庭制定了赠送购物卡的活动,赠送方式为:家庭月收入低于μ的获赠两次随机购物卡,家庭月收入不低于μ的获赠一次随机购物卡;每次赠送的购物卡金额及对应的概率分别为:则A 家庭预期获得的购物卡金额为多少元?(结果保留整数)21.(本小题12分)已知曲线C 上的动点M 到y 轴的距离比到点F (1,0)的距离小1, (1)求曲线C 的方程;(2)过F 作弦RS PQ 、,设RS PQ 、的中点分别为B A 、,若0=⋅RS PQ ,求||AB 最小时,弦RS PQ 、所在直线的方程;(3)在(2)条件下,是否存在一定点T ,使得FT TB AF -=λ?若存在,求出P 的坐标,若不存在,试说明理由.22.(本小题12分)已知函数22ln )42()(x x ax xx f +-=。

广东省广州市海珠外国语实验中学2024-2025学年高一上学期段考(一)数学试题

广东省广州市海珠外国语实验中学2024-2025学年高一上学期段考(一)数学试题

广东省广州市海珠外国语实验中学2024-2025学年高一上学期段考(一)数学试题一、单选题1.设全集{}1,2,3,4,5U =,集合{}{}1,4,2,5M N ==,则U N M =U ð( ) A .{}2,3,5B .{}1,3,4C .{}1,2,4,5D .{}2,3,4,52.“0a b >>”是“1a b +>-”的( ) A .充分不必要条件 B .必要不充分条件 C .充要条件D .即不充分也不必要3.已知a ∈R ,b ∈R ,若集合{}2,,1,,0b a a a b a ⎧⎫=+⎨⎬⎩⎭,则20222023b a -的值为( )A .2-B .1-C .1D .24.移动支付、高铁、网购与共享单车被称为中国的新“四大发明”.某中学为了解本校学生中新“四大发明”的普及情况,随机调查了100位学生,其中使用过移动支付或共享单车的学生共90位,使用过移动支付的学生共有80位,使用过共享单车且使用过移动支付的学生共有60位,则该校使用共享单车的学生人数为( ) A .50B .60C .70D .805.若“0(0,2)x ∃∈,使得20210x x λ-+≤成立”是假命题,则实数λ可能的值是( )A .1B .C .3D .6.已知函数()f x 的定义域为[)0,+∞,则函数()25f x y x -=-的定义域为( )A .()()2,55,-+∞UB .[)()2,55,-+∞UC .()()2,55,⋃+∞D .[)()2,55,+∞U7.两次购买同一物品,可以用两种不同的策略,第一种不考虑物品价格的升降,每次购买这种物品所花的钱一定;第二种是不考虑物品价格的升降,每次购买这种物品数量一定.设第一种方式购买的平均价格为a 元,第二种方式购买的平均价格为b 元,下列说法正确的是( ) A .a b ≤ B .a b < C .a b ≥D .a b >8.若关于x 的不等式()22120x a x a -++<恰有两个整数解,则a 的取值范围是( )A .322a a ⎧⎫<≤⎨⎬⎩⎭∣ B .112aa ⎧⎫-<≤-⎨⎬⎩⎭∣ C .112aa ⎧-<≤-⎨⎩∣或322a ⎫≤<⎬⎭ D .112aa ⎧-≤<-⎨⎩∣或322a ⎫<≤⎬⎭二、多选题9.下列说法正确的有( ) A .不等式21131x x +>-的解集是123x x ⎧⎫-<<⎨⎬⎩⎭B .命题“[1,2]x ∀∈,20x a -≤”为真命题的一个充分不必要条件是5a ≥C .命题:R p x ∀∈,20x >,则:p x ⌝∃∈R ,20x <D .{}2210M x x x =-+=,{}1N =表示同一集合10.已知正实数x ,y 满足2x y xy +=,则( )A .8xy ≥B .6x y +>C .29x y +≥D .1831x y+≥- 11.下列说法正确的是( )A .()xf x x =与()1,01,0x g x x ≥⎧=⎨-<⎩表示同一函数 B .函数()y f x =的图象与直线1x =的交点至多有1个C .若()1f x x x =--,则112f f ⎛⎫⎛⎫= ⎪⎪⎝⎭⎝⎭D .关于x 的方程()230x m x m +-+=有一个正根,一个负根的充要条件是()0,m ∈+∞三、填空题12.若不等式20ax bx c --<的解集是{23}xx <<∣,则不等式20cx bx a -->的解集为. 13.设集合{}28150A x x x =-+=,{}10B x ax =+=.若A B A =U ,求实数a 的取值集合是.14.已知函数()(0)f x ax b a =->,(())43f f x x =-,则(2)f =.四、解答题15.如图,某广场要划定一矩形区域ABCD ,并在该区域内开辟出三块形状大小相同的矩形绿化区,这三块绿化区四周和绿化区之间设有1米宽的走道.已知三块绿化区的总面积为800平方米,求该矩形区域ABCD 占地面积的最小值.16.已知函数()f x 的解析式为()22,1,126,2x x f x x x x x +≤-⎧⎪=-<≤⎨⎪-+>⎩(1)求()1f ,()()2f f -的值;(2)画出这个函数的图象,并写出()f x 的最大值; (3)解不等式()2f x <. 17.(1)已知54x <,求14245x x -+-的最大值; (2)若正数x ,y 满足220x xy +-=,求3x y +的最小值.18.设集合{A x y =,{}521B x m x m =-≤≤+. (1)若1m =时,求A B U ;(2)若x A ∈是x B ∈的充分不必要条件,求实数m 的取值范围.19.已知函数()()()2111f x m x m x m =+--+-.(1)当0m <时,解关于x 的不等式()32f x x m ≥+-;(2)若存在x∈0,2,使得不等式2≤+-成立,求实数m的取值范围.()21f x x x。

2020-2021学年辽宁省实验中学高三(上)月考数学试卷(10月份)(附答案详解)

2020-2021学年辽宁省实验中学高三(上)月考数学试卷(10月份)(附答案详解)

2020-2021学年辽宁省实验中学高三(上)月考数学试卷(10月份)一、单选题(本大题共10小题,共50.0分)1. 集合A ={x|y =√2x −1},B ={x|x 2−5x −6<0},则∁R (A ∩B)=( )A. {x|x <2或x >3}B. {x|x ≤2或x ≥3}C. {x|x <12或x ≥6}D. {x|x ≤12或x >6}2. 下列命题正确的是( )A. 若a <b ,则ac 2<bc 2B. 若a >b ,则1a <1b C. 若a >b ,c >d ,则ac >bdD. 若1ab 2<1a 2b ,则a <b3. 已知q :∀x ∈[−2,3),x 2<9,则¬q 为( )A. ∃x ∈[−2,3),x 2<9B. ∃x ∉[−2,3),x 2<9C. ∃x ∈[−2,3),x 2≥9D. ∃x ∉[−2,3),x 2≥94. 已知函数f(x)={(13)x ,x ≥3f(x +1),x <3,则f(2+log 32)的值为( )A. −227B. 154C. 227D. −545. 函数y =f(x +1)为偶函数且满足f(x)+f(−x)=0,x ∈[0,1]时,f(x)=x 3,则f(985)=( )A. 1B. −1C. 9853D. −98536. 甲、乙、丙三位同学被调查是否去过A 、B 、C 三个城市,甲说:我去过的城市比乙多,但没去过B 城市;乙说:我没去过C 城市;丙说:我们三人去过同一城市;由此可判断乙去过的城市为( )A. AB. BC. CD. A 和B7. 已知函数f(x)=ln(e x +1)−12x ,下列选项正确的是( )A. 奇函数,在(−1,1)上有零点B. 奇函数,在(−1,1)上无零点C. 偶函数,在(−1,1)上有零点D. 偶函数,在(−1,1)上无零点8. 如图,《九章算术》中记载了一个“折竹抵地”问题:今有竹高一丈,末折抵地,去本三尺,问折者高几何?意思是:有一根竹子,原高一丈(1丈=10尺),现被风折断,尖端落在地上,竹尖与竹根的距离三尺,问折断处离地面的高为( )尺.A. 5.45B. 4.55C. 4.2D. 5.89.下列命题正确的是()A. x+1x≥2恒成立B. √a2+4+1√a2+4的最小值为2C. m,n都是正数时,(m+1m )(n+1n)最小值为4D. a>0,b>0是b3a +3ab≥2的充要条件10.函数y=lncosx(−π2<x<π2)的图象是()A. B.C. D.二、多选题(本大题共2小题,共10.0分)11.为了了解市民对各种垃圾进行分类的情况,加强垃圾分类宣传的针对性,指导市民尽快掌握垃圾分类的方法,某市垃圾处理厂连续8周对有害垃圾错误分类情况进行了调查.经整理绘制了有害垃圾错误分类重量累积统计图,图中横轴表示时间(单位:周),纵轴表示有害垃圾错误分类的累积重量(单位:吨).根据图形分析,下列结论正确的是()A. 第1周和第2周有害垃圾错误分类的重量加速增长B. 第3周和第4周有害垃圾错误分类的重量匀速增长C. 第5周和第6周有害垃圾错误分类的重量相对第3周和第4周增长了30%D. 第7周和第8周有害垃圾错误分类的重量相对第1周和第2周减少了1.8吨12.已知当x>0时,f(x)=−2x2+4x,x≤0时,y=f(x+2),以下结论正确的是()A. f(x)在区间[−6,−4]上是增函数B. f(−2)+f(−2021)=2C. 函数y=f(x)周期函数,且最小正周期为2<k<4−2√2或k=2√2−4D. 若方程f(x)=kx+1恰有3个实根,则12三、单空题(本大题共4小题,共20.0分)13.命题“∃x∈R,2x2−3ax+9<0”为假命题,则实数a的取值范围为______.14.函数f(x)=x2sinx−2,则f(2021)+f(−2021)=______ .15.有一支队伍长L米,以一定的速度匀速前进,排尾的传令兵因传达命令赶赴排头,到达排头后立即返回,且往返速度不变,如果传令兵回到排尾后,整个队伍正好前进了L米,则传令兵所走的路程为______ .16.若集合A1,A2满足A1∪A2=A,则称(A1,A2)为集合A的一个分拆,并规定:当且仅当A1=A2时,(A1,A2)与(A2,A1)为集合A的同一种分拆,则集合A={−1,0,2}的不同分拆种数是______ .四、解答题(本大题共6小题,共70.0分)+a,x>−1}.17.已知集合A={x|y=log2(4−2x)+1},B={y|y=x+1x+1(1)求集合A和集合B;(2)若“x∈∁R B”是“x∈A”的必要不充分条件,求a的取值范围.18.已知函数f(x)=2(m+1)x2+4mx+2m−1.(Ⅰ)若m=0,求f(x)在[−3,0]上的最大值和最小值;(Ⅱ)若关于x的方程f(x)在[0,1]上有一个零点,求实数m的取值范围.19.已知函数f(x)为偶函数,x≥0时,f(x)=x2+4x.(1)求f(x)解析式;(2)若f(2a)<f(1−a),求a的取值范围.20.新冠肺炎疫情造成医用防护服短缺,某地政府决定为防护服生产企业A公司扩大生产提供x(x∈[0,10])(万元)的专项补贴,并以每套80元的价格收购其生产的全部防)(万护服.A公司在收到政府x(万元)补贴后,防护服产量将增加到t=k⋅(6−12x+4件),其中k为工厂工人的复工率(k∈[0.5,1]).A公司生产t万件防护服还需投入成本(20+9x+50t)(万元).(1)将A公司生产防护服的利润y(万元)表示为补贴x(万元)的函数(政府补贴x万元计入公司收入);(2)在复工率为k时,政府补贴多少万元才能使A公司的防护服利润达到最大?(3)对任意的x∈[0,10](万元),当复工率k达到多少时,A公司才能不产生亏损?(精确到0.01).21.已知函数f(x)=−x|x−2a|+1(x∈R).(1)当a=1时,求函数y=f(x)的零点;),求函数y=f(x)在x∈[1,2]上的最大值.(2)当a∈(0,3222.若存在常数k(k>0),使得对定义域D内的任意x1,x2(x1≠x2),都有|f(x1)−f(x2)|≤k|x1−x2|成立,则称函数f(x)在其定义域D上是“k−利普希兹条件函数”﹒(1)举例说明函数f(x)=log2x不是“2−利普希兹条件函数”;(2)若函数f(x)=√x(1≤x≤4)是“k−利普希兹条件函数”,求常数k的最小值;(3)若存在常数k(k>0),使得对定义域D内的任意x1,x2(x1≠x2),都有|f(x1)−f(x2)|>k|x1−x2|成立,则称函数f(x)在其定义域D上是“非k−利普希兹条件函数”.若函数f(x)=log2(2x−a)为[1,2]上的“非1−利普希兹条件函数”,求实数a的取值范围.答案和解析1.【答案】C【解析】解:集合A={x|y=√2x−1}={x|x≥12},B={x|x2−5x−6<0}={x|−1< x<6},所以A∩B={x|12≤x<6},则∁R(A∩B)={x|x<12或x≥6}.故选:C.先求出集合A,B,然后利用集合交集与补集的定义求解即可.本题考查了集合的运算,主要考查了集合交集与补集定义的运用,涉及了函数定义域的求解以及一元二次不等式的解法,属于基础题.2.【答案】D【解析】解:对于A,若c=0,则ac2=bc2,故A错误;对于B,若a>0>b,则1a >1b,故B错误;对于C,若a>b,c>d,取a=2,b=1,c=−1,d=−2,此时ac=bd,故C错误;对于D,若1ab2<1a2b,则a2b2>0,所以a2b2⋅1ab2<a2b2⋅1a2b,即a<b,故D正确.故选:D.由不等式的性质逐一判断即可.本题主要考查不等式的基本性质,考查逻辑推理能力,属于基础题.3.【答案】C【解析】解:命题q:∀x∈[−2,3),x2<9,则¬q:∃x∈[−2,3),x2≥9.故选:C.根据全称命题的否定是存在量词命题,写出对应的命题即可.本题考查了全称命题的否定是存在量词命题应用问题,是基础题.4.【答案】B【解析】解:∵2+log 31<2+log 32<2+log 33,即2<2+log 32<3 ∴f(2+log 32)=f(2+log 32+1)=f(3+log 32) 又3<3+log 32<4∴f(3+log 32)=(13)3+log 32=(13)3×(13)log 32=127×(3−1)log 32=127×3−log 32=127×3log 312=127×12=154∴f(2+log 32)=154故选B先确定2+log 32的范围,从而确定f(2+log 32)的值本题考查指数运算和对数运算,要求能熟练应用指数运算法则和对数运算法则.属简单题5.【答案】A【解析】解:根据题意,函数y =f(x +1)为偶函数,则f(x)的图象关于直线x =1对称,则有f(−x)=f(x +2),又由f(x)满足f(x)+f(−x)=0,即f(−x)=f(x +2), 则有f(x +2)=−f(x),综合可得:f(x +4)=−f(x +2)=f(x),f(x)是周期为4的函数, 则f(985)=f(1+4×246)=f(1)=1, 故选:A .根据题意,分析可得f(x +4)=f(x),则f(x)是周期为4的函数,据此可得f(985)=f(1),结合函数的解析式计算可得答案.本题考查函数的奇偶性的性质以及应用,涉及函数的周期性,属于基础题.6.【答案】A【解析】解:由乙说:我没去过C 城市,则乙可能去过A 城市或B 城市,但甲说:我去过的城市比乙多,但没去过B 城市,则乙只能是去过A ,B 中的任一个, 再由丙说:我们三人去过同一城市, 则由此可判断乙去过的城市为A . 故选:A .可先由乙推出,可能去过A 城市或B 城市,再由甲推出只能是A ,B 中的一个,再由丙即可推出结论.本题主要考查简单的合情推理,要抓住关键,逐步推断,是一道基础题.7.【答案】D【解析】解:根据题意,函数f(x)=ln(e x +1)−12x =ln(√e x+1√ex),其定义域为R ,有f(−x)=ln(√e x+1√ex)=f(x),即函数f(x)为偶函数,设t =√e x+1√ex ,在区间[0,1)上,t =√e x+1√ex>2且是增函数,而y =lnt ,在(2,+∞)上为增函数,则f(x)在区间[0,1)上为增函数,又由f(0)=ln2>0,则在区间[0,1)上,f(x)≥f(0)>0恒成立,故f(x)在区间[0,1)上没有零点,又由f(x)为偶函数,则f(x)在(−1,1)上无零点; 故选:D .根据题意,先分析函数的奇偶性,再设t =√e x+1√ex,则y =lnt ,利用复合函数的单调性判断方法可得f(x)在区间[0,1)上为减函数,求出f(1)的值,分析可得区间[0,1)上没有零点,结合函数的奇偶性分析可得答案.本题考查函数奇偶性的性质以及应用,涉及函数零点的判断,属于基础题、8.【答案】B【解析】解:如图,已知AC +AB =10(尺),BC =3(尺),AB 2−AC 2=BC 2=9,所以(AB +AC)(AB −AC)=9,解得AB −AC =0.9, 因此{AB +AC =10AB −AC =0.9,解得{AB =5.45AC =4.55,故折断后的竹干高为4.55尺,故选:B.由题意可得AC+AB=10(尺),BC=3(尺),运用勾股定理和解方程可得AB,AC,即可得到所求值.本题考查三角形的勾股定理的运用,考查方程思想和运算能力,属于基础题.9.【答案】C【解析】解:x+1x≥2恒成立,不成立,因为x可以小于0,所以A不正确;√a2+4√a2+4的最小值大于2,所以B不正确;m,n都是正数时,(m+1m )(n+1n)≥2√m⋅1m⋅2⋅√n⋅1n=4,当且仅当m=n=1,表达式取得最小值为4,所以C正确;a>0,b>0是b3a +3ab≥2的充分不必要条件,所以D不正确;故选:C.利用基本不等式,判断选项的正误即可.本题考查命题的真假的判断与应用,基本不等式的应用,是基础题.10.【答案】A【解析】【分析】本题主要考查复合函数的图象识别.属于基础题.利用函数y=lncosx(−π2<x<π2)的奇偶性可排除一些选项,利用函数值与0的关系可排除一些选项.从而得以解决.【解答】解:∵cos(−x)=cosx,∴y=lncosx(−π2<x<π2)是偶函数,可排除B、D,由cosx≤1⇒lncosx≤0排除C,故选:A.11.【答案】ABD【解析】对于A ,第1周和第2周有害垃圾错误分类的重量明显增多,是加速增长,故A 正确;对于B ,第3周和第4周有害垃圾错误分类的重量图象是线段,是匀速增长,故B 正确; 对于C ,第5周和第6周有害垃圾错误分类的重量相对第3周和第4周是减少,故C 错误;对于D ,第7周和第8周有害垃圾错误分类的重量增长0.6吨, 第1周和第2周有害垃圾错误分类的重量增长2.4吨,∴第7周和第8周有害垃圾错误分类的重量相对第1周和第2周减少了1.8吨,故D 正确. 故选:ABD .由分段函数图象,能够读出各段上y 对于x 变化状态,由此能求出结果.本题考查命题真假的判断,考查折线图等基础知识,考查运算求解能力、数据分析能力等数学核心素养,是基础题.12.【答案】BD【解析】解:x ≤0时,y =f(x +2),∴f(x)在x ≤0时的图象以2为周期进行循环,如下图所示,由图象可知,f(x)在区间[−6,−4]上先增后减,所以A 错误; f(−2)+f(−2021)=f(0)+f(1)=0+2=2,所以B 正确;当x >0时,f(x)=−2x 2+4x ,f(3)≠f(1),所以y =f(x)不是以2为周期的周期函数,所以C 错误;y =kx +1恒过(0,1),由图象可知,直线与f(x)交点只可能在x ∈(−2,0)或x ∈(0,+∞)处取到,x ∈(−2,0)时,f(x)=−2x 2−4x ,∴{−k =2x +1x +4,−2<x <0−k =2x +1x −4,x >0,即y =−k 和g(x)={2x +1x +4,−2<x <02x +1x−4,x >0交点个数为3,画出g(x)图象,如下图所示,x ∈(−2,0)时,g(x)最大值为4−2√2,g(−2)=−12,x ∈(0,2)时,g(x)最小值为2√2−4, ∴y =−k 和y =g(x)要有3个交点,满足−k =4−2√2或2√2−4<−k <−12, 解得12<k <4−2√2或k =2√2−4,所以D 正确. 故选:BD .画出图象,即可判断A ;由x >0时,f(x)=−2x 2+4x ,x ≤0时,y =f(x +2),即可判断BC ;参变分离得{−k =2x +1x +4,−2<x <0−k =2x +1x −4,x >0,即可判断D . 本题考查了函数的图象与性质,函数零点问题,D 选项较难下手,属于难题.13.【答案】[−2√2,2√2]【解析】解:原命题的否定为“∀x ∈R ,2x 2−3ax +9≥0”,且为真命题, 则开口向上的二次函数值要想大于等于0恒成立, 只需△=9a 2−4×2×9≤0,解得:−2√2≤a ≤2√2. 故答案为:[−2√2,2√2]根据题意,原命题的否定“∀x ∈R ,2x 2−3ax +9≥0”为真命题,也就是常见的“恒成立”问题,只需△≤0.存在性问题在解决问题时一般不好掌握,若考虑不周全、或稍有不慎就会出错.所以,可以采用数学上正难则反的思想,去从它的反面即否命题去判定.注意“恒成立”条件的使用.14.【答案】−4【解析】解:根据题意,函数f(x)=x2sinx−2,则f(−x)=−x2sinx−2,则f(x)+f(−x)=−4,则有f(2021)+f(−2021)=−4,故答案为:−4.根据题意,求出f(−x)的解析式,分析可得f(x)+f(−x)=−4,据此分析可得答案.本题考查函数值的计算,涉及函数奇偶性的性质以及应用,属于基础题.15.【答案】(√2+1)L.【解析】解:设传令兵的速度为V1,队伍的速度为V2,传令兵从队尾到队头的时间为t1,从队头到队尾的时间为t2,队伍前进用时间为t.由传令兵往返总时间与队伍运动时间相等可得如下方程:t=t1+t2,即:LV2=LV1−V2+LV1+V2整理上式得:V12−2V1V2−V22=0解得:V1=(√2+1)V2;将上式等号两边同乘总时间t,即V1t=(√2+1)v2tV1t即为传令兵走过的路程S1,V2t即为队伍前进距离S2,则有S1=(√2+1)S2=(√2+1)L.故答案为:(√2+1)L.以队伍为参照物,可求传令兵从队尾往队头的速度,从队头往队尾的速度,利用速度公式求传令兵从队尾到队头的时间t1,传令兵从队头到队尾的时间为t2,队伍前进100用的时间t,而t=t1+t2,据此列方程求出V1、V2的关系,进而求出在t时间内通讯员行走的路程.本题考查路程的计算,关键是计算向前的距离和向后的距离,难点是知道向前的时候人和队伍前进方向相同,向后的时候人和队伍前进方向相反,解决此类问题常常用到相对运动的知识.16.【答案】27【解析】解:因为集合A中有三个元素,当A1=⌀时,必须A2=A,分拆种数为1;当A1有一个元素时,分拆种数为C31⋅2=6;当A1有2个元素时,分拆种数为C32⋅22=12;当A1=A时,分拆种数为C33⋅23=8.所以总的不同分拆种数为1+6+12+8=27种.故答案为:27.由题意中的定义,分A1=⌀,A1有一个元素,A1有2个元素,A1=A四种情况,分别求出分拆种数,即可得到答案.本题考查了新定义问题,解决此类问题,关键是读懂题意,理解新定义的本质,把新情境下的概念、法则、运算化归到常规的数学背景中,运用相关的数学公式、定理、性质进行解答即可,属于中档题.17.【答案】解:(1)集合A={x|y=log2(4−2x)+1}={x|4−2x>0}={x|x<2},B={y|y=x+1x+1+a,x>−1}={x|x+1+1x+1+a−1≥2√(x+1)⋅1x+1+a−1=a+1}={x|x≥a+1}.(2)∵集合A={x|x<2},B={x|x≥a+1}.∴∁U B={x|x<a+1},∵“x∈∁R B”是“x∈A“的必要不充分条件,∴x<2⇒x<a+1,∴a+1>2,解得a>1.∴a的取值范围是(1,+∞).【解析】(1)利用对数函数的定义域能求出集合A,利用均值定理能求出集合B.(2)推导出x<2⇒x<a+1,由此能求出a的取值范围.本题考查集合、实数的取值范围的求法,对数函数的定义域、均值定理、必要不充分条件等基础知识,考查运算求解能力,是基础题.18.【答案】解:(1)当m =0时,f(x)=2x 2−1,可知函数f(x)图象在[−3,0]上单调递减,∴f(x)min =f(0)=−1,f(x)max =f(−3)=17;(2)由f(0)=0得m =12.由f(1)=0得m =−18≠12,∴m =12或−18成立; 由f(0)f(1)<0得(2m −1)(8m +1)<0,解得:−18<m <12; 综上:满足条件的m 的取值范围是:[−18,12].【解析】(1)结合函数f(x)图象可求f(x)在[−3,0]上的最大值和最小值; (2)根据f(0)f(1)<0,再验证f(0)=0及f(1)=0,可求得m 范围. 本题考查二次函数图象性质,考查数学运算能力,属于中档题.19.【答案】解:(1)根据题意,设x <0,则−x >0,则有f(−x)=x 2−4x ,又由f(x)为偶函数,则f(x)=f(−x)=x 2−4x , 则f(x)={x 2+4x,x ≥0x 2−4x,x <0;(2)由函数f(x)为偶函数可知f(2a)<f(1−a)⇔f(|2a|)<f(|1−a|),由(1)知函数f(x)在[0,+∞)上是增函数,∴|2a|<|1−a|,得(2a)2<(1−a)2,解得:a ∈(−1,13).【解析】(1)令x >0,则−x <0,再根据函数为偶函数可求得解析式;(2)由函数f(x)为偶函数可知f(2a)<f(1−a)⇔f(|2a|)<f(|1−a|),可求得a 的取值范围.本题考查函数奇偶性的性质以及应用、函数解析式求法、考查数学运算能力及数学抽象能力,属于中档题.20.【答案】解:(1)y =x +80t −(20+9x +50t)=30t −20−8x =30k ⋅(6−12x+4)−20−8x =180k −360k x+4−8x −20,x ∈[0,10];(2)y=180k−360kx+4−8x−20=180k+12−8[(x+4)+45kx+4],因为x∈[0,10],所以4≤x+4≤14,则(x+4)+45kx+4≥6√5√k,当且仅当x+4=45kx+4,即x=3√5√k−4时取“=”,因为k∈[0.5,1],则3√102−4≤3√5√k−4≤3√5−4,即有3√5√k−4∈[0,10],所以y≤180k+12−48√5√k,即当政府补贴为3√5√k−4万元才能使A公司的防护服利润达到最大,最大为180k+ 12−48√5√k;(3)若对任意的x∈[0,10],公司都不产生亏损,则180k−360kx+4−8x−20≥0在x∈[0,10]恒成立,即180k≥(8x+20)(x+4)x+2,记m=x+2,则m∈[2,12],此时(8x+20)(x+4)x+2=(8m+4)(m+2)m=8m2+20m+8m=8m+8m+20,由于函数f(m)=8m+8m+20在[2,12]单调递增,所以当m∈[2,12]时,f max(m)=f(12)=11623,∴k≥1162 3180≈0.65即k≥0.65,即当工厂工人的复工率达到0.65时,对任意的x∈[0,10],公司都不产生亏损.【解析】(1)利用已知条件列出函数的解析式,写出定义域即可.(2)由y的解析式得到y=180k+12−8[(x+4)+45kx+4],根据x的范围得到(x+4)+45k x+4≥6√5√k,结合k的范围可得3√102−4≤3√5√k−4≤3√5−4,即可求得答案(3)若对任意的x∈[0,10],公司都不产生亏损,得到180k−360kx+4−8x−20≥0在x∈[0,10]恒成立,利用换元法,结合函数的单调性求解函数的最值即可得到结果.本题考查实际问题的处理方法,函数的单调性以及函数的解析式的求法,考查转化思想以及计算能力,是中档题.21.【答案】解:(1)当a=1时,令−x|x−2|+1=0.当x≥2时,−x(x−2)+1=0,解得:x=1+√2;当x<2时,−x(x−2)+1=0,解得:x=1.故函数零点为:1+√2和1;(2)f(x)={−x 2+2ax +1,x ≥2ax 2−2ax +!,x <2a ,其中f(0)=f(2a)=1,于是最大值在f(1),f(2),f(2a)中取.得0<2a ≤1,即0<a ≤12时,f(x)在[1,2]上单调递减.∴f(x)max =f(1)=2a ; 当a <1<2a <2,即12<a <1时,f(x)在[1,2a]上单调递增,在[2a,2]上单调递减,故f(x)max =f(2a)=1;当1≤a <2<2a ,即1≤a <2时,f(x)在[1,a]上单调递减,在[a,2]上单调递增,故f(x)max =max{f(1),f(2)},∵f(1)−f(2)2a −3<0,故f(x)max =f(2)=5−4a .综上:f(x)max={2a,0<a ≤12,1,12<a <1,5−4a,1≤a <32..【解析】(1)求函数零点转化为解方程可解决此问题; (2)根据a 讨论函数图象,根据图象特点可求函数最大值. 本题考查函数零点与最值,考查数学运算能力,属于难题.22.【答案】解:(1)f(x)=log 2x 的定义域为(0,+∞),令x 1=12,x 2=14,则f(12)−f(14)=log 212−log 214=−1−(−2)=1, 而2|x 1−x 2|=12,∴f(x 1)−f(x 2)>2|x 1−x 2|,∴函数f(x)=log 2x 不是“2−利普希兹条件函数”;(2)若函数f(x)=√x(1≤x ≤4)是“k −利普希兹条件函数”,则对于定义域[1,4]上任意两个x 1,x 2(x 1≠x 2),均有|f(x 1)−f(x 2)|≤k|x 1−x 2|成立,不妨设x 1>x 2,则k ≥√x 1−√x 2x 1−x2=√x +√x 恒成立,∵1≤x 2<x 1≤4, ∴14<√x +√x <12,∴k 的最小值为12;(3)∵|f(x 1)−f(x 2)|>k|x 1−x 2|,f(x)=log 2(2x −a)为[1,2]上的“非1−利普希兹条件函数”,∴设x 1>x 2,则|log 2(2x 1−a)−log 2(2x 2−a)|>|x 1−x 2|,∵2x1−a>0,2x2−a>0,且2x1−a2x2−a>1,∴2x1−a2x2−a >2x1−x2=2x12x2,∴2x1+x2−a⋅2x2>2x1+x2−a⋅2x1,∴a⋅2x1>a⋅2x2,∵x1>x2,∴a>0,∵2x−a>0,∴a<2x,∵x∈[1,2],∴a<2,综上,实数a的取值范围为(0,2).【解析】(1)令x1=12,x2=14,即可说明f(x)=log2x不是“2−利普希兹条件函数”;(2)依题意,k≥√x1−√x2x1−x2=√x+√x恒成立,而14<√x+√x<12,由此可得k的最小值;(3)由题意可得,a⋅2x1>a⋅2x2,结合x1>x2,可得a>0,由2x−a>0,x∈[1,2],可得a<2,综合即得答案.本题以新定义为背景,考查函数性质的运用,考查不等式的恒成立问题,考查分离变量法以及运算求解能力,属于中档题.。

2022-2023学年广东省广州市海珠外国语实验中学高一年级上册学期段考(二)数学试题【含答案】

2022-2023学年广东省广州市海珠外国语实验中学高一年级上册学期段考(二)数学试题【含答案】

2022-2023学年广东省广州市海珠外国语实验中学高一上学期段考(二)数学试题一、单选题1.函数()()31log 4f x x x =-+-的定义域为( ) A .{}14x x << B .{}14x x ≤≤ C .{}14x x <≤ D .{}14x x ≤<【答案】D【分析】根据开偶数次方根号里的数大于等于零,对数的真数大于零列出不等式组,即可得解.【详解】解:由题意得,10,40,x x -≥⎧⎨->⎩,解得14x ≤<,所以函数的定义域为{}14x x ≤<. 故选:D .2.扇子最早称“翣”,其功能并不是纳凉,而是礼仪器具,后用于纳凉、娱乐、欣赏等.扇文化是中国传统文化的重要门类,扇子的美学也随之融人到建筑等艺术审美之中.图1为一古代扇形窗子,此窗子所在扇形的半径(图2)120cm AO =,圆心角为45,且C 为AO 的中点,则该扇形窗子的面积为( )A .215cm 2πB .21350cm πC .21350cmD .21800cm π【答案】B【分析】根据扇形面积公式进行求解即可.【详解】因为C 为AO 的中点,所以60cm,45CO =化成弧度为4π,所以此扇形窗子的面积为()22211120601350cm .2424πππ⨯⨯-⨯⨯= 故选:B3.函数()ln 26f x x x =+-的零点一定位于下列哪个区间( ) A .()1,2 B .()2,3 C .()3,4 D .()5,6【答案】B【分析】根据零点存在性定理,即可求解.【详解】由题意可知,()2ln 220f =-<,()3ln30f =>, 故()()230f f ⋅<,又因函数()ln 26f x x x =+-在()0,∞+上单调递增, 所以函数()ln 26f x x x =+-的零点一定位于区间()2,3. 故选:B. 4.函数()sin cos 2x xf x x ⋅=+的图象大致为( )A .B .C .D .【答案】A【分析】由函数的奇偶性质可知函数为偶函数,再结合2x π=时函数的符号即可得答案.【详解】解:由题知函数的定义域为R ,关于原点对称,()()()()sin sin cos 2cos 2x x x xf x x x f x -⋅-⋅===-++-,所以函数为偶函数,其图像关于y 轴对称,故排除B ,D ,当2x π=时,sin22024cos 22f πππππ⋅⎛⎫==> ⎪⎝⎭+,故排除C ,得A 为正确选项. 故选:A5.已知3sin()35x π-=,则7cos()6x π+等于 A .35B .45C .35 D .45-【答案】C【分析】由诱导公式化简后即可求值.【详解】7πcos x 6⎛⎫+ ⎪⎝⎭=-π cos x 6⎛⎫+=- ⎪⎝⎭sin[26x ππ⎛⎫-+ ⎪⎝⎭]=π3sin x 35⎛⎫-=- ⎪⎝⎭故选C .【点睛】本题主要考查了三角函数诱导公式的应用,属于基础题. 6.若0.5a e =,ln 2b =,2log 0.2c =,则有( ) A .a b c >> B .b a c >>C .c a b >>D .b c a >>【答案】A【解析】利用指数函数和对数函数的单调性比较a 、b 、c 三个数与0、1的大小关系,从而可得出这三个数的大小关系.【详解】指数函数x y e =为增函数,则0.501a e e =>=; 对数函数ln y x =为增函数,则ln1ln 2ln e <<,即01b <<; 对数函数2log y x =为增函数,则22log 0.2log 10c =<=. 因此,a b c >>. 故选:A.【点睛】本题考查指数式与对数式的大小比较,一般利用指数函数和对数函数的单调性得出各数与中间值0、1的大小关系,考查推理能力,属于基础题.7.已知偶函数f (x )在区间[)0+,∞ 单调递增,则满足1(21)()3f x f -<的 x 取值范围是( ) A .12(,)33B .12[,)33C .12(,)23D .12[,)23【答案】A【分析】由偶函数性质得函数在(,0]-∞上的单调性,然后由单调性解不等式. 【详解】因为偶函数()f x 在区间[)0,∞+上单调递增,所以()f x 在区间(,0)-∞上单调递减,故x 越靠近y 轴,函数值越小, 因为()121(3f x f -<),所以1213x -<,解得:1233x <<.故选:A .8.已知函数()1log 4(0x a f x x a -=->且1)a ≠在10,2⎛⎤ ⎥⎝⎦上无零点,在1,12⎛⎫ ⎪⎝⎭上有零点,则实数a 的取值范围为( ) A .10,4⎛⎫⎪⎝⎭B .()1,11,4⎛⎫⋃+∞ ⎪⎝⎭C .10,4⎛⎤ ⎥⎝⎦D .1,14⎛⎫ ⎪⎝⎭【答案】D【分析】将问题转化成研究方程1log 4x a x -=在10,2⎛⎤ ⎥⎝⎦上无实数根,在1,12⎛⎫⎪⎝⎭上有实数根,即考查函数()()1log ,4x a g x x h x -==的交点情况,作出函数图像数形结合即可得到答案.【详解】函数()f x 在10,2⎛⎤ ⎥⎝⎦上无零点,在1,12⎛⎫⎪⎝⎭上有零点,即方程()0f x =在10,2⎛⎤ ⎥⎝⎦上无实数根,在1,12⎛⎫⎪⎝⎭上有实数根,即1log 4x a x -=在10,2⎛⎤ ⎥⎝⎦上无实数根,在1,12⎛⎫ ⎪⎝⎭上有实数根,设()()1log ,4x a g x x h x -==,函数()h x 在R 上单调递增,且()()1110,,11422h h h ⎛⎫=== ⎪⎝⎭,()140x h x -=>恒成立,若1a >,则在()0,1x ∈时,()log 0a g x x =<,故不满足条件.由于()g x 与()h x 的图象在10,2⎛⎤ ⎥⎝⎦上无交点,在1,12⎛⎫⎪⎝⎭上有交点,根据函数的图像可知011122a g h <<⎧⎪⎨⎛⎫⎛⎫> ⎪ ⎪⎪⎝⎭⎝⎭⎩,解得114a <<故选:D.二、多选题9.已知幂函数()()2mf x m x =-,则( )A .3m =B .定义域为[)0,∞+C .( 1.5)( 1.4)m m -<- D2【答案】AC【分析】根据()f x 为幂函数得m 可判断A ;根据幂函数的解析式可判断B ;利用单调性可判断C ;D.【详解】()f x 为幂函数,21m ∴-=,得()33,=∴=m f x x ,A 对;函数()f x 的定义域为R ,B 错误;由于()f x 在R 上为增函数,331.5 1.4,( 1.5)( 1.4)-<-∴-<-,C 对;()3228f ==,=D 错误,故选:AC.10.下列选项中,正确的是( )A .函数()12x f x a -=-(0a >且1a ≠)的图象恒过定点1,2B .若不等式230ax bx ++>的解集为{}13x x -<<,则1a b +=C .若p :N n ∃∈,22n n >,则p ⌝:N n ∀∈,22n n ≤D .函数()ln 2f x x x =+-恰有1个零点 【答案】BCD【分析】对于A ,根据指数函数的图像与性质即可求解;对于B ,根据一元二次不等式的解法即可判断;对于C ,由特称命题的否定为全称命题即可得出结论;对于D ,由函数零点存在定理和函数单调性即可求解.【详解】对A :函数()12x f x a -=-(0a >且1a ≠)的图象恒过定点1,1,即A 错误;对B: 若不等式230ax bx ++>的解集为{}13x x -<<, 则a<0,且1-和3是方程230ax bx ++=的两个实数根,所以13313b a a ⎧-+=-⎪⎪⎨⎪-⨯=⎪⎩,解得1,2a b =-=,所以1a b +=,即B 正确;对于C :若p :N n ∃∈,22n n >,则p ⌝:N n ∀∈,22n n ≤,故C 选项正确;对于D :易知函数()ln 2f x x x =+-在,()0x ∈+∞上为单调递增, 又()1ln11210f +-=-<=,()2ln 222ln 20f +-=>=;所以由函数零点存在定理可得存在唯一0(1,2)x ∈,使得()00f x =; 所以,D 正确. 故选:BCD.11.下列既是奇函数,又在[2,2]-上是增函数的是( ) A .()13f x x = B .()12x f x -=C .()11x x e f x e -=+D .()()2ln1f x x x =+-【答案】AC【分析】对于ABC ,根据函数的定义域、奇偶性及单调性等性质即可判断. 对于D ,根据奇偶函数的定义和复合函数单调性即可得出D 错误.【详解】对于A ,()f x 的定义域为R ,()()()1133=f x x x f x -=--=-,则()f x 为奇函数,由幂函数的性质知: ()13f x x =在R 上单增,所以A 正确.对于B ,()f x 的定义域为R ,()()12x f x f x ---=≠-,所以()f x 不是奇函数,所以B 错误. 对于C ,()f x 的定义域为R ,()()1111x xxxe ef x f x e e -----===-++,则()f x 为奇函数,又因为()1+122=1111x x x x x e e f x e e e --==-+++,1x e +为增函数,21x e +为减函数,21x e -+为增函数, 211x e -+为增函数,所以C 正确. 对于D ,()()2ln 1f x x x =+-有210x x +->解得:R x ∈.()()()()2221ln 1ln ln11f x x x x x f x x x ⎛⎫⎛⎫-=-++==-+-=-⎪ ⎪⎝⎭+-⎝⎭,则 ()()2ln1f x x x =+-是奇函数,令22111t x x x x=+-=++在区间[2,2]-上单调递减,而ln y t =为增函数,故()()2ln 1f x x x =+-在[2,2]-上是减函数,所以D 错误.故选:AC. 12.已知函数函数()y f x a =-有四个不同的零点1x ,2x ,3x ,4x ,且1234x x x x <<<,则( )A .a 的取值范围是()0,1B .21x x -的取值范围是()0,1C .344x x +=D .1234222x x x x +=+ 【答案】AC【分析】结合()f x 的图象,由图可知01a <<,10x <,201x <<,由二次函数的对称性,可得344x x +=,可得答案.【详解】()y f x a =-有四个不同的零点1x ,2x ,3x ,4x ,即方程()f x a =有四个不同的解. ()f x 的图象如图所示,由图可知01a <<,10x <,201x <<,所以210x x ->, 即21x x -的取值范围是()0,+∞,由二次函数的对称性,可得344x x +=.因为121221x x -=-,所以12222x x +=,故12342212x x x x +=+. 故选:AC.三、填空题 13.计算:7πsin 4=______. 【答案】22-【分析】利用诱导公式即得. 【详解】7π7ππ2sinsin 2πsin 444⎛⎫=-=-= ⎪⎝⎭. 故答案为:22-. 14.声强级L (单位:dB )由公式1210lg 10-⎛⎫= ⎪⎝⎭I L 给出,其中I 为声强(单位:W/m 2). 声强级为60dB 的声强是声强级为30dB 的声强的______倍.【答案】1000【分析】根据已知公式,应用指对数的关系及运算性质求60dB 、30dB 对应的声强,即可得结果.【详解】由题设,601210lg()6010I -=,可得66010I -=, 301210lg()3010I -=,可得93010I -=, ∴声强级为60dB 的声强是声强级为30dB 的声强的60301000I I =倍. 故答案为:1000.15.已知定义在R 上的减函数()f x 满足()()()0,1,2f x f x P +-=-是其图象上一点,那么()12f x +<的解集为__________. 【答案】(2,0)-【分析】由题意可得()()()111f f x f <+<-,结合条件,利用奇偶性和单调性可解出不等式,得到答案.【详解】由()()0f x f x +-=知()f x 为奇函数, 由()12,f x +<即()212f x -<+<又()()()()()12,12,111f f f f x f -=∴=-∴<+<-, 又知函数()f x 在R 上为减函数,可得111x -<+<, 解得20,x -<<∴解集为()2,0-. 故答案为:()2,0-16.设a 为实数,若关于x 的方程129310x x a a ++⋅++=有实数解,则a 的取值范围是___________.【答案】,⎛-∞ ⎝⎦【分析】令3x t =()0t >,将原问题转化为方程22310t at a +++=有正根,利用判别式及韦达定理列出不等式组求解即可得答案.【详解】解:方程129310x x a a ++⋅++=可化为()2233310x x a a +⋅++=,令3x t =,则0t >, 所以原问题转化为方程22310t at a +++=有正根,设两根分别为12,t t ,则()2221212Δ94101030a a t t a t t a ⎧=-+⎪⎪⋅=+>⎨⎪+=->⎪⎩,解得a ≤ 所以a的取值范围是,⎛-∞ ⎝⎦,故答案为:,⎛-∞ ⎝⎦.四、解答题 17.(1)计算:1213lg15lg 42-⎛⎫+- ⎪⎝⎭;(2)已知4cos sin 13sin 2cos 4αααα-=+,求tan α的值.【答案】(1)1 (2)2【分析】(1)利用指数、对数的运算及其运算性质计算求解. (2)4cos sin 13sin 2cos 4αααα-=+分子分母同时除以cos α,把弦化切进行求解.【详解】(1)原式=()121233122lg 1523-⨯⨯⎛⎫⎛⎫+-+⨯ ⎪ ⎪⎝⎭⎝⎭=()1112lg102-⎛⎫+-+ ⎪⎝⎭=221-+ =1 (2)因为4cos sin 13sin 2cos 4αααα-=+,且cos 0α≠,所以分子分母同除以cos α有:4cos sin 4tan 13sin 2cos 3tan 24αααααα--==++,即3tan 2164tan αα+=-,7tan 14α=解得 tan 2α=.18.已知集合{}{}225120,31(0)xA x x xB y y x =--≥==+>.(1)求集合A B ⋂,()R C A B ⋃;(2)若集合{}22C x m x m =-≤≤且()R C A C C =,求m 的取值范围.【答案】(1){}|4x x ≥,32x x ⎧⎫-⎨⎬⎩⎭;(2)()1,2,22m ⎛⎫∈-∞-⋃ ⎪⎝⎭.【详解】试题分析:(1)利用一元二次不等式的解法化简集合A ,利用指数函数的性质化简集合B ,从而可求出R C A ,再利用集合交集与并集的定义求解即可;(2)()R C A C C ⋂=等价于()R C C A ⊆,结合(1)的结论,利用集合的包含关系,分两种情况讨论,分别列不等式组求解即可求得m 的取值范围.试题解析:(1)()()2325120234042x x x x x x --≥⇒+-≥⇒≥≤-或,∴342A x x x ⎧⎫=≥≤-⎨⎬⎩⎭或,{}2B y y =>,∴{}()34,2R A B x x C A B x x ⎧⎫⋂=≥⋃=>-⎨⎬⎩⎭.(2)∵()R C A C C ⋂=,()R C C A ⊆,342R C A x x ⎧⎫=-<<⎨⎬⎩⎭,①当C =∅时,22,2m m m -><-即时满足()R C C A ⊆∴2m <-; ②当C ≠∅时,要使()R C C A ⊆,则22231122222242m m m m m m m m -≤≥-⎧⎧⎪⎪⎪⎪->-⇒>⇒<<⎨⎨⎪⎪<<⎪⎪⎩⎩综上所述,()1,2,22m ⎛⎫∈-∞-⋃ ⎪⎝⎭.19.已知函数()f x 是定义在R 上的偶函数,当0x 时,()()2log 33f x x x =---. (1)求函数()f x 的解析式; (2)解不等式()()312f x f x -<+.【答案】(1)()()()22log 33,0,log 33,0.x x x f x x x x ⎧---⎪=⎨++->⎪⎩ (2)1342xx ⎧⎫-<<⎨⎬⎩⎭∣【分析】(1)根据函数的奇偶性即可求得函数在0x >时的解析式;(2)根据(1)中的函数解析式判断函数的单调性,利用函数的奇偶性解不等式即可.【详解】(1)设0x >,则0x -<,()f x 是定义在R 上的偶函数,()()()2log 33f x f x x x ∴=-=++-,22log (3)30()log (3)30x x x f x x x x ---≤⎧∴=⎨++->⎩,,; (2)由(1)知,0x >时,()()2log 33f x x x =++-,()2log 3y x =+与3y x =-在()0,∞+上都是增函数,f x 在()0,∞+上为增函数,f x 在(),0∞-上为减函数,2(31)(2)|31||2|81030f x f x x x x x ∴-<+⇒-<+⇒--<, 解得1342x -<<. ∴该不等式的解集为1342x x ⎧⎫-<<⎨⎬⎩⎭∣. 20.已知函数()2121x x f x -=+是定义在R 上的奇函数. (1)用定义法证明()f x 为增函数;(2)对任意()1,x ∈+∞,都有22130k f x f kx x x ⎛⎫⎛⎫+++-> ⎪ ⎪⎝⎭⎝⎭恒成立,求实数k 的取值范围. 【答案】(1)证明见解析(2)()-+∞【分析】(1)根据函数单调性的定义及指数函数的单调性与值域即可证明;(2)由已知条件,利用函数()f x 的奇偶性和单调性,可得2213k x k x x++>-对1x >恒成立,然后分离参数,利用基本不等式求出最值即可得答案.【详解】(1)证明:设12x x <,则()()()()()1212121212222212121212121x x x x x x x x f x f x ----=-=++++, 由12x x <,可得1222x x <,即12220x x -<,又1210x +>,2210x +>, 所以()()()121222202121x x x x -<++,即()()12f x f x <,则()f x 在R 上为增函数;(2)解:因为任意(1,)x ∈+∞,都有22130k f x f kx x x ⎛⎫⎛⎫+++-> ⎪ ⎪⎝⎭⎝⎭恒成立,且函数()f x 是定义在R 上的奇函数, 所以2213k k f x f kx f kx x x x ⎛⎫⎛⎫⎛⎫++>--=- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭对1x >恒成立, 又由(1)知函数()f x 在R 上为增函数,所以2213k x k x x++>-对1x >恒成立, 由11,01x x><<,有10x x -<, 所以22131x x k x x++>-对1x >恒成立, 设1,1t x x x=->,由1t x x =-在(1,)+∞递减,可得0t <, 所以22213551x t x t t t x x +++==+≤-=--t= 所以k >-k 的取值范围是()-+∞.21.为保护环境,污水进入河流前都要进行净化处理.我市工业园区某工厂的污水先排入净化池,然后加入净化剂进行净化处理.根据实验得出,在一定范围内,每放入1个单位的净化剂,在污水中释放的浓度y (单位:毫克/立方米)随着时间x (单位:小时)变化的函数关系式近似为321,0318,321x x x y x -⎧+≤≤⎪=⎨>⎪+⎩.若多次加进净化剂,则某一时刻净化剂在污水中释放的浓度为每次投放的净化剂在相应时刻所释放的浓度之和.由实验知,当净化剂在污水中释放的浓度不低于4(毫克/立方米)时,它才能起到净化污水的作用.(1)若投放1个单位的净化剂4小时后,求净化剂在污水中释放的浓度;(2)若一次投放4个单位的净化剂并起到净化污水的作用,则净化时间约达几小时?(结果精确到0.1,参考数据:lg 20.3≈,lg17 1.23≈)(3)若第一次投放1个单位的净化剂,3小时后再投放2个单位的净化剂,设第二次投放t 小时后污水中净化剂浓度为()g t (毫克/立方米),其中03t <≤,求()g t 的表达式和浓度()g t 的最小值.【答案】(1)6毫克/立方米(2)7.1(3)()()1822121t t g t =+++,03t <≤; ()g t 的最小值为12毫克/立方米【分析】(1)由函数解析式,将4x =代入即可得解;(2)分03x ≤≤和3x >两种情况讨论,根据题意列出不等式,从而可得出答案;(3)根据题意写出函数()g t 的解析式,再根据基本不等式即可求得最小值.【详解】(1)解:由321,0318,321x x x y x -⎧+≤≤⎪=⎨>⎪+⎩, 当4x =时,4318621y -==+, 所以若投放1个单位的净化剂4小时后,净化剂在污水中释放的浓度为6毫克/立方米;(2)解:因为净化剂在污水中释放的浓度不低于4(毫克/立方米)时,它才能起到净化污水的作用,当03x ≤≤时,令()4214x +≥,得20x ≥恒成立,所以当03x ≤≤时,起到净化污水的作用,当3x >时,令3184421x -⋅≥+,得32118x -+≤,则2lg17 1.233log 17 4.1lg 20.3x -≤=≈=, 所以37.1x <≤,综上所述当07.1x ≤≤时,起到净化污水的作用,所以若一次投放4个单位的净化剂并起到净化污水的作用,则净化时间约达7.1小时;(3)解:因为第一次投入1个单位的净化剂,3小时后再投入2个单位净化剂,要计算的是第二次投放t 小时后污水中净化剂浓度为()g t ,所以()()()()3318182212212121t t t x g t +-=++=++++,03t <≤, 因为210t +>, 所以()182211221t t ++≥=+, 当且仅当()1822121t t =++,即1t =时取等号, 所以()()1822121t t g t =+++,03t <≤, 当1t =时,()g t 取最小值12毫克/立方米.22.对于函数()f x ,若在其定义域内存在..实数x ,满足()()f x f x -=-,则称()f x 为“局部奇函数”. (1)若()2x f x m =-是定义在区间[1,1]-上的“局部奇函数”,求实数m 的取值范围.(2)若12()423x x f x n n +=-⋅+-为定义域R 上的“局部奇函数”,求实数n 的取值范围.【答案】(1)5[1,]4(2)1⎡⎣【分析】(1)根据()f x 是定义在区间[1,1]-上的“局部奇函数”可知()()f x f x -=-,由此求出m 的解析式,利用换元法,构造函数()h t ,求出函数()h t 的值域,即可求得m 的取值范围.(2)根据()f x 是定义在区间R 上的“局部奇函数”可知()()f x f x -=-,由此求出n 的解析式,利用换元法,构造函数()g t ,求出函数()g t 的值域,即可求得n 的取值范围.【详解】(1)解:因为()2x f x m =-是定义在区间[1,1]-上的“局部奇函数”所以()()f x f x -=-,即2(2)x x m m --=--所以2220x x m -+-=于是问题转换成方程2220x x m -+-=在[1,1]-有解令2x t =,[1,1]x ∈-,则1[,2]2t ∈ 所以12m t t=+在1[,2]2t ∈上有解 根据1()h t t t =+在1[,1]2t ∈为减函数,在[1,2]t ∈为增函数,且min ()(1)2h t h ==,max 15()()(2)22h t h h === 所以5()[2,]2h t ∈,即5222m ≤≤,解得514m ≤≤ 所以实数m 的取值范围为5[1,]4; (2)因为12()423x x f x n n +=-⋅+-是定义在区间R 上的“局部奇函数”所以()()f x f x -=-,即1212423(423)x x x x n n n n --++-⋅+-=--⋅+-整理可得:2442(22602)x x x x n n --++-⋅+-=令22x x t -=+,则[2,)t ∈+∞又222)2244(2x x x x t --+=+-=-所以方程222280t nt n -+-=在[2,)t ∈+∞上有解设22()228g t t nt n =-+-,[2,)t ∈+∞当(2)0g ≤时,222280t nt n -+-=在[2,)t ∈+∞上有解,此时22440n n --≤,解得:11n ≤当(2)0g >时,222280t nt n -+-=在[2,)t ∈+∞上有解,此时()22(2)0(2)22Δ44280g n n n ⎧>⎪-⎪->⎨⎪⎪=--≥⎩,解得:1n +≤综上所述:实数n的取值范围1⎡⎣.。

广东省实验中学深圳学校2023-2024学年高一上学期10月第一次段考英语试题

广东省实验中学深圳学校2023-2024学年高一上学期10月第一次段考英语试题

广东省实验中学深圳学校2023-2024学年高一上学期10月第一次段考英语试题学校:___________姓名:___________班级:___________考号:___________一、单项选择1.Which of the following words is created the same way as sunflower!A.homework B.winner C.untidy D.musician2.I wonder if she ________ us tomorrow, but I think if she ________ us, we will be able to complete the task ahead of time.A.helps; helps B.will help; has helpedC.helps; will help D.will help; helps3.Her ________ face suggested that she was ________ by some ________ things .A.frightening, frightened, frighteningB.frightened, frightened, frighteningC.frightening, frightening, frighteningD.frightening, frightened, frightened4.We ________ each other for about twenty years since we left Japan.A.didn’t see B.don’t meet C.haven’t seen D.hadn’t seen 5.The sports meeting is _____. We are doing what we can to prepare for it.A.in the way B.drawingC.this way D.approaching6.Between the two rows of trees ________ the teaching building.A.stands B.stand C.standing D.having stood 7.________for the NBA in 2009, Stephen joined the Golden State Warriors.A.Selected B.To selectC.Having selected D.Selecting8.The old man stood there with his eyes ______ on the picture.A.to focus B.focusing C.focused D.focus9.He was extremely ______ to survive in the terrible fire.A.fortunate B.comfortable C.confident D.scared 10.Next after coffee, Americans are ________ to drink cola drinks, such as Coke or Pepsi, or other soft drinks.A.possible B.likely C.probable D.as well 11.We cannot always ________ the traditional rules handed down to us from the past.A.build B.observe C.create D.forgive 12.I’ve just _______ an old friend in the park while taking a walk with my parents.A.come out B.come about C.come across D.come over 13.The reason ________ Tom failed in the exam was ______ he didn’t work hard.A.why; that B.that; why C.why; because D.for which;becauseA.to be breathed B.to breatheC.breathing D.being breathed15.I have lost my wallet. I ________ it somewhere just now.A.must drop B.must have been droppedC.must have dropped D.should have dropped16.The Universal Studio Resort, ________ seven theme lands, can offer its visitors ______ fun experience.A.consisted of; a variety of B.consisting of; varieties ofC.comprising; various D.comprised of; various of17.---Have you heard that Ben and Bill had a fight yesterday?---Yes. Now they don’t say a single word to each other. It seems as if they ______ each other before.A.didn’t know B.don’t know C.have never known D.had neverknown18.It can be hard for you to imagine what trouble I have had _______ the problems _______ in the meeting held last week.A.to deal with; refer to B.dealing with; referring toC.dealing with; referred to D.to deal with; referred to19.Not only ______, but also he kept it.A.has he made a promise B.he made a promiseC.did he make a promise D.he had made a promise20.________ the wealth of the country increasing, more waste will be produced.A.As B.With C.Because D.For二、阅读理解This Is the Way Learning CenterThe staff, parents and children enrolled (招收) at This Is the Way Learning Center are one big family. We take the health and safety of our family very seriously. This Is the Way Learning Center is committed to maintaining a safe environment for your children. We are following all safety rules. Should you have any questions regarding the processes in place please call our center. We will be happy to answer your questions.This Is the Way Learning Center cares for and educates children aged 6 weeks and up. We are enrolling for all programs (Preschool and Nursery School) in addition to basic childcare. There will be no enrollment fees and registration fees. We are devoted to the future of your child and creating a lifetime love of learning at every age and stage. What can be more precious or important than the early developmental and learning years of our children?This Is the Way Learning Center is located across the street from Henry Barnard School at 18 Shaker Road. It is conveniently located just minutes from 91 North and South, in the town of Enfield.Hours:• 6:30 am through 5:30 pm• Monday through Friday• Closed only on major holidays.Ages:• Children through kindergarten• School ageFeatures:• All-day, all-year daycare• Homelike setting• Age-appropriate play areas• Child-friendly curriculum• Computer education• Trained and experienced teachersYou may email us at ***************************or call us at 860-253-0010 to ask your questions and request a free brochure.21.What is This Is the Way Learning Center?A.It’s a big family.B.It’s a training center.C.It’s a safety center.D.It’s a local program.22.Who can teach in This is the Way Learning Center?.A.Everyone enrolled at it.B.Age-appropriate coaches.C.Computer education teachers.D.Trained and experienced teachers. 23.For whom is this text probably written?A.Parents.B.Visitors.C.Students.D.Professors.The waste management park at Water Beach in Cambridgeshire handles 400,000 tons of recycling every year, and is the biggest centre of its kind in the East of England.Fran Hawes is standing on the edge of a mountain of dry recycling. She says, “Some think it’s a discouraging thing, but for me personally, it is a huge responsibility.” Fran, who started to do the job at the age of 26, takes her job very seriously. “I am that annoying person at a party who will get everyone’s recycling and make sure everything goes in the right bin,” she admits. “That’s my job. That’s my mission. That’s what I need to do. It allows me to find my leadership style.”Bin loader Richard Hughes and his colleagues clock on (打卡上班) at the waste service garage in Peter borough at 6:30 and are out emptying bins within half an hour. Richard works four 12-hour days and admits the pay and the condition are not great — but there are three days off which he can spare to his partner and children. His colleague Crystal Teal has been working as a bin loader for two and a half months. “I didn’t feel my last job gave me self-worth and general happiness, so I decided to change and do something completely different,” she says. “I decided to be a binman. It doesn’t bother me that men are mainly doing it,” Crystal adds. “It can be smelly some days, but go home and have a shower, you’ll be all right.”Their biggest problem? When householders put the wrong material — like rotting food and nappies — in the recycling bins. “It makes me feel angry that they are just putting it out there. We’re doing our job to serve them. We are providing service to take their rubbish away and they can’t do their part to put the right things in,” said Richard. “Any problem has a solution — therefore I’m not afraid of a pile of waste,” Fran says. “To any normal person, this might be intimidating (吓人的). It’s a challenge. But this will be gone by Monday morning.”24.Why does Fran Hawes make others annoyed at a party?A.She still performs her task of garbage sorting.B.She hosts the party in a leadership way.C.She asks everyone to be responsible for their jobs.D.She might be smelly at parties. 25.Why did Crystal choose to be a bin loader?A.She needed a well-paid job.B.She preferred the three-day-off holiday.C.She liked to do something meaningful.D.She didn’t find loading bins dirty and smelly.26.What made Richard sometimes angry?A.People didn’t take out their rubbish.B.People threw away rubbish everywhere.C.People didn’t sort their rubbish out rightly.D.People picked out useful things from the bins.27.Which of the following can best describe Fran?A.Brave.B.Amazing.C.Anxious.D.Optimistic.As a young child, I was painfully shy. I’d watch other children play in the park, wishing I could join them, but I was too scared to approach. Eventually, my mother would come to the rescue. She’d ask the other kids if I could play, too. Today, I feel comfortable giving public lectures in large halls and having conversations in small groups, but I still tend to avoid situations in which I’m expected to spend time with a roomful of strangers.There could be many reasons. For one thing, I might be carrying some childhood fear of rejection. But beyond that possibility, one likely element is that I tend to underestimate how much people like me after I meet them, as most of us do.A new research paper reports that the common concern that new people may not like us, or that they may not enjoy our company, is largely unfounded.Erica Boothby of Cornell University and her colleagues conducted a series of studies to find out what our conversation partners really think of us. In doing so, they discovered a new cognitive illusion (认知错觉) they call “the liking gap”: our failure to realize how much strangers appreciate our company after a bit of conversation.The researchers observed the gap in a variety of situations: strangers getting acquainted in the research laboratory, first-year college students getting to know their dormitory mates over the course of many months, and community members meeting fellow participants inpersonal development workshops. In each circumstance, people consistently underestimated how much others liked them. For much of the academic year, as dormitory mates got to know each other and even started to develop enduring friendships, the liking gap persisted.The data also revealed some of the potential reasons for the illusion: we are often more severe with ourselves than with others, and our inner critic prevents us from appreciating how positively other people evaluate us. Not knowing what our conversation partners really think of us, we use our own thoughts as a proxy (代理人). This is a mistake, because our thoughts tend to be more negative than reality.28.Why does the author mention his childhood experience?A.To show how his character changed.B.To explain what he was like when he was young.C.To show an example of why people are shy of communication.D.To emphasize the important role of a mother in one’s childhood.29.What does the underlined word “unfounded” probably mean?A.Careless.B.Baseless.C.Selfless.D.Meaningless. 30.What do we know about the liking gap from the text?A.It indicates what strangers really think of us.B.It begins and ends quickly among strangers.C.It disappears when strangers get to know each other.D.It states our misunderstanding of how much others like us.31.Which of the following is the best title for this text?A.People Like You More than You KnowB.How to Get Along Well with StrangersC.The Way to Know What Others Think of UsD.Having Conversations with Strangers Benefits UsWhen it comes to making healthy lifestyle changes, which should come first — changing your diet or becoming more physically active?Previous studies suggested that providing people with too much information about nutrition and physical activity at once tends to be discouraging. That has led to the popularity of advising people to make changes gradually, and set smaller goals.So the scientists divided 200 inactive participants who were age 45 or older into fourgroups. One group was instructed about making diet and fitness changes at the same time, the second group was taught about diet changes first, then fitness changes four months later, the third group changed their exercise habits first and made changes in their eating habits four months later, and the final, control group was not instructed about either diet or fitness changes but about how to manage their stress.The researchers tracked the groups for a full year. Compared to the control group, the three intervention (干预) groups made healthy changes in their diet. Those who changed their fitness habits first significantly increased the amount of exercise they received daily compared to the other groups after four months. However, at the end of the year, the group that changed both diet and exercise at the same time was the only one that met the nationally recommended targets for both exercise and nutrition levels, while those who worked on improving their nutrition first were unable to meet the recommended levels of fitness after a year. The researchers suspect that changes to diet are easier to make than changes to physical activity.The findings show, however, that pairing diet and exercise changes may help to overcome some of the barriers people face in adding more physical activity into their lives. If folks change diet and exercise orderly, the scientists say, they may end up placing more importance on the first set of behavior changes and feel less pressured to address the second set.32.The advice mentioned in the second paragraph seemed .A.popular B.dangerous C.scientific D.unsatisfactory 33.Which group made progress earlier than other groups in increasing the amount of daily exercise?A.The first group.B.The second group.C.The third group.D.The control group.34.The reason why those improving their nutrition first failed to meet the levels of fitnessA.they might be very lazyB.they wouldn’t like to change their behaviorC.it is much harder to change physical activityD.they might put on weight before changing their physical activity35.Which of the following does the author probably approve of?A.Pairing diet and exercise changes.B.Becoming more physically active first.C.Changing diet and exercise following a fixed order.D.Changing diet first and placing more importance on it.Simple ways to improve your written EnglishMany people think it is really difficult to improve (提高) their writing in English. Don’t worry, though. Here are some simple steps that you can take to improve your written English.● Increase your vocabulary.To express yourself clearly, you need a good active vocabulary. That’s not just being able to know lots of words — it means actually being able to use them correctly. 36 Tip: When you learn a new word, try to learn all the forms of that word.● 37People often say that we learn to write best by reading. Reading in English is useful in many ways. It is a great way to get an idea of the different styles of writing and see how to use words properly.Tip: 38 Learning shouldn’t be boring. Read each text several times to make sure you understand how to use new words and expressions in the text.● Improve your grammar.Grammar is very important because it improves the quality (质量) of your writing.Tip:39 The first time, look for general mistakes and the second time look for mistakes with the grammar point you are studying at the moment.● Just do it!The best way to improve your writing is to get a pen and paper and write. Be prepared to write several versions (版本) of each text. 40A.Know your readers.B.Read widely and often.C.Always check your writing twice.D.Remember, practice makes perfect!E.Choose books or articles that interest you.F.Use simpler language and shorter sentences to show your ideas.G.Do this by learning new words with example sentences, not just word lists.How to Do Well in Science ClassDoing well in science depends on developing useful study skills and learning to participate (参与) in class. 41 If you’ve got good study skills that you’ve developed in other courses, you’ll be able to use many of these to do well in science.Read the required material. When your teacher requires reading from a textbook or a website, take time to read it before you come to class. If you don’t have time to read it well, it’s a good idea to look quickly at it so that you have an idea of what it’s about. 42 Take clear and organized notes. The notes you take in class will help you know what to study between classes. 43 Instead, pay careful attention to any information your teacher stresses, which is likely to be included on tests.44 Many science classes include laboratory experiments, which are practice of techniques you’ve learned about in your textbook or lectures. Your teacher will expect you to show up ready to start your experiment. Read the instructions for the laboratory in advance of class.Reread your notes after class. If there’s anything in your notes that confuses you, or which you feel might be wrong, check with your teacher or a classmate to ensure you have the right information. 45 This tests your understanding of the material in a way that ensures you’ve really grasped (理解) the meaning.A.Be prepared for the laboratory experiment.B.Work in a group with others in class.C.Doing a good job in laboratories is also a good way to improve your performance in science.D.Recording a lecture will allow you to hear it again.E.Rereading can help you to rewrite your notes in a better way.F.Don’t try to write down everything that the teacher says.G.It also helps you be prepared to answer questions and have discussions.三、完形填空on my way to college, I had great friends and a loving family. There was nothing missing. ButI spoke too soon.It was around 9:25 a.m. that I heard the news that would 47 my life forever. My brother Zach had been in a car accident. He fought for five days before he 48 . That day, I became an only child. I felt extremely 49After Zach’s death, I found 50 in food. I ate, then I slept, then I ate again. I couldn’t cry. I could hardly feel anything, and I was 51 . I stopped building relationships for fear that they would end just as suddenly as Zach’s life. Also, I became nervous about any potentially 52 situations — driving late at night — but I couldn’t express this fear of life because I wanted to be strong for my parents. I saw my parents’ 53 worse than mine because of the losing of their son. I didn’t want them to 54 me. I also experienced a lot of 55 , because I was angry about why the sadness had happened to me, and I never escape from this emotion.Now, it has been nearly five years since Zach’s death. I don’t 56 life anymore: I face it bravely. I 57 my friendships and began socializing more. I even share Zach’s story with people around me. Although my new friends never met him, they know about Zach.One lesson I learned from losing my brother was never to be 58 to say, “I love you.” I loved my brother, but it was too late to 59 it loudly. The last time I remember telling my brother I loved him was when he was dying. Don’t make this 60 like me.46.A.admit B.finish C.approach D.determine 47.A.regret B.change C.beautify D.ignore 48.A.took away B.gave away C.passed away D.flew away 49.A.lonely B.tired C.bored D.nervous 50.A.interest B.benefit C.improvement D.relief 51.A.funny B.hopeful C.patient D.senseless 52.A.difficult B.particular C.dangerous D.tense 53.A.pain B.discouragement C.willingness D.memory 54.A.think about B.dream about C.talk about D.care about 55.A.surprise B.anger C.disappointment D.doubt 56.A.damage B.choose C.fear D.leave 57.A.produced B.rebuilt C.communicated D.raised58.A.stubborn B.satisfied C.brave D.afraid 59.A.explore B.express C.spread D.pray 60.A.mistake B.opinion C.explanation D.difference四、语法填空阅读下面短文,在空白处填入1个适当的单词或括号内单词的正确形式。

2020-2021广东实验中学高三数学上期末试卷含答案

2020-2021广东实验中学高三数学上期末试卷含答案

2020-2021广东实验中学高三数学上期末试卷含答案一、选择题1.设,x y 满足约束条件 202300x y x y x y --≤⎧⎪-+≥⎨⎪+≤⎩,则46y x ++的取值范围是A .3[3,]7- B .[3,1]- C .[4,1]-D .(,3][1,)-∞-⋃+∞2.已知数列{}n a 的前n 项和2n S n =,()1nn n b a =-则数列{}n b 的前n 项和n T 满足( ) A .()1nn T n =-⨯ B .n T n = C .n T n =-D .,2,.n n n T n n ⎧=⎨-⎩为偶数,为奇数3.等比数列{}n a 的前n 项和为n S ,若36=2S =18S ,,则105S S 等于( ) A .-3B .5C .33D .-314.已知数列{}n a的首项110,1n n a a a +==+,则20a =( ) A .99 B .101C .399D .4015.若直线()10,0x ya b a b+=>>过点(1,1),则4a b +的最小值为( ) A .6B .8C .9D .106.在△ABC 中,若1tan 15013A C BC ︒===,,,则△ABC 的面积S 是( ) ABCD7.数列{}n a 中,对于任意,m n N *∈,恒有m n m n a a a +=+,若118a =,则7a 等于( ) A .712 B .714 C .74D .788.已知,,a b R +∈且115a b a b+++=,则+a b 的取值范围是( )A .[1,4]B .[)2,+∞C .(2,4)D .(4,)+∞9.数列{}n a 为等比数列,若11a =,748a a =,数列1n a ⎧⎫⎨⎬⎩⎭的前n 项和为n S ,则5(S = )A .3116B .158C .7D .3110.在ABC ∆中,内角,,A B C 所对的边分别为,,a b c ,且()cos 4cos a B c b A =-,则cos2A =( ) A .78B .18C .78-D .18-11.若变量x ,y 满足约束条件1358x y x x y ≥-⎧⎪≥⎨⎪+≤⎩,,,则2yz x =-的取值范围是( ) A .113⎡⎤-⎢⎥⎣⎦,B .11115⎡⎤--⎢⎥⎣⎦,C .111153⎡⎤-⎢⎥⎣⎦, D .3153⎡⎤-⎢⎥⎣⎦,12.在等差数列 {}n a 中, n S 表示 {}n a 的前 n 项和,若 363a a += ,则 8S 的值为( )A .3B .8C .12D .24二、填空题13.已知向量()()1,,,2a x b x y ==-r r ,其中0x >,若a r 与b r 共线,则yx的最小值为__________.14.已知0,0x y >>,1221x y +=+,则2x y +的最小值为 . 15.在ABC △中,角A ,B ,C 所对的边分别为a ,b ,c ,若三角形的面积2223)4S a b c =+-,则角C =__________. 16.已知x ,y 满足3010510x y x y x y +-≤⎧⎪-+≥⎨⎪-+≤⎩,则2z x y =+的最大值为______.17.已知数列{a n }的前n 项和S n =n 2+2n +1(n ∈N *),则a n =________. 18.设正项数列{}n a 的前n 项和是n S ,若{}n a 和{}nS 都是等差数列,且公差相等,则1a =_______.19.设(32()lg 1f x x x x =++,则对任意实数,a b ,“0a b +≥”是“()()0f a f b +≥”的_________条件.(填“充分不必要”.“必要不充分”.“充要”.“既不充分又不必要”之一)20.设x ,y 满足则220,220,20,x y x y x y --≤⎧⎪-+≥⎨⎪++≥⎩则3z x y =-的最小值是______.三、解答题21.已知在等比数列{}n a 中, 11a =,且2a 是1a 和31a -的等差中项. (1)求数列{}n a 的通项公式;(2)若数列{}n b 满足()*21n n b n a n N=-+∈,求{}nb 的前n 项和nS.22.在等差数列{}n a 中,36a =,且前7项和756T =. (1)求数列{}n a 的通项公式;(2)令3nn n b a =⋅,求数列{}n b 的前n 项和n S .23.已知数列{}n a 中,11a =,121n n a a n +=+-,n n b a n =+. (1)求证:数列{}n b 是等比数列; (2)求数列{}n a 的前n 项和n S .24.已知a ,b ,c 分别为△ABC 三个内角A ,B ,C 的对边,且acos C +3asin C -b -c =0.(1)求A ;(2)若AD 为BC 边上的中线,cos B =17,AD =1292,求△ABC 的面积. 25.已知各项均为正数的等比数列{}n a 的首项为12,且()3122123a a a -=+。

2020-2021学年广东省实验中学高一(上)期末数学试卷 (解析版)

2020-2021学年广东省实验中学高一(上)期末数学试卷 (解析版)

2020-2021学年广东省实验中学高一(上)期末数学试卷一、单项选择题(共8小题).1.设集合A={x|1≤x+1<5},B={x|x≤2},则A∩(∁R B)=()A.{x|0≤x<4}B.{x|0≤x≤2}C.{x|2<x<4}D.{x|x<4}2.下列四组函数中,表示同一函数的一组是()A.y=|x|,u=B.y=,s=()2C.D.3.已知a=log3,b=ln3,c=2﹣0.99,则a,b,c的大小关系为()A.b>a>c B.a>b>c C.c>a>b D.b>c>a4.在△ABC中,“”是“”的()A.充分必要条件B.充分而不必要条件C.必要不充分条件D.既不充分也不必要条件5.已知函数f(x+2)=2x+x﹣2,则f(x)=()A.2x﹣2+x﹣4B.2x﹣2+x﹣2C.2x+2+x D.2x+2+x﹣26.在同一直角坐标系中,函数y=,y=log a(x+)(a>0且a≠1)的图象可能是()A.B.C.D.7.函数f(x)=A sin(ωx+φ)(A>0,ω>0,0<φ<)的部分图象如图所示,将其向右平移个单位长度后得到的函数解析式为()A.y=sin2x B.y=sin(2x+)C.y=sin(2x﹣)D.y=sin(2x﹣)8.方程cos x=log8x的实数解的个数是()A.4B.3C.2D.1二、多项选择题(共4小题).9.下列各式中,值为的是()A.cos2﹣sin2B.C.2sin195°cos195°D.10.已知a,b为正实数,则下列判断中正确的是()A.B.若a+b=4,则log2a+log2b的最大值为2C.若a>b,则D.若a+b=1,则的最小值是811.已知函数f(x)=|cos x|+cos|2x|,下列说法正确的是()A.若x∈[﹣π,π],则f(x)有2个零点B.f(x)的最小值为C.f(x)在区间上单调递减D.π是f(x)的一个周期12.已知函数f(x)=a sin x+b cos x,其中a,b∈R,且ab≠0,若对一切x∈R恒成立,则()A.B.C.是偶函数D.是奇函数三、填空题(本大题共4小题,每小题5分,共20分)13.已知函数(ω>0)的最小正周期是π,则ω=,单调递增区间是.14.命题“所有三角形都有内切圆”的否定是.15.已知角θ的终边在直线y=﹣3x上,则=.16.已知函数,若a、b、c、d、e(a<b<c<d<e)满足f(a)=f(b)=f(c)=f(d)=f(e),则M=af(a)+bf(b)+cf(c)+df(d)+ef(e)的取值范围为.四、解答题(本题共6小题,共70分.解答时应写出必要的文字说明、证明过程或演算步骤)17.计算下列各式的值:(1);(2).18.已知幂函数f(x)=(m2+2m﹣2)x m+2,且在(0,+∞)上是减函数.(1)求f(x)的解析式;(2)若(3﹣a)m>(a﹣1)m,求a的取值范围.19.已知函数.(1)求函数f(x)的最小正周期、对称轴和对称中心;(2)若锐角α满足,且β满足,求cosβ的值.20.某生物研究者于元旦在湖中放入一些凤眼莲(其覆盖面积为k),这些凤眼莲在湖中的蔓延速度越来越快,二月底测得凤眼莲的覆盖面积为24m2,三月底测得凤眼莲的覆盖面积为36m2,凤眼莲的覆盖面积y(单位:m2)与月份x(单位:月)的关系有两个函数模型y=ka x(k>0,a>1)与可供选择.(1)试判断哪个函数模型更合适并求出该模型的解析式;(2)求凤眼莲的覆盖面积是元旦放入凤眼莲面积10倍以上的最小月份.(参考数据:lg2≈0.3010,lg3≈0.4711).21.已知定义域为R的函数,是奇函数.(1)求a,b的值;(2)判断f(x)单调性并证明;(3)若∀t∈[﹣1,4],不等式f(t2+2)+f(2t2﹣kt)<0恒成立,求k的取值范围.22.已知函数为f(x)的零点,为f(x)图象的对称轴.(1)若f(x)在[0,2π]内有且仅有6个零点,求f(x);(2)若f(x)在上单调,求ω的最大值.参考答案一、单项选择题(本大题共8小题,每小题5分,共40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.设集合A={x|1≤x+1<5},B={x|x≤2},则A∩(∁R B)=()A.{x|0≤x<4}B.{x|0≤x≤2}C.{x|2<x<4}D.{x|x<4}解:因为集合A={x|1≤x+1<5}={x|0≤x<4},B={x|x≤2},∴∁R B={x|x>2},∴A∩(∁R B)={x|2<x<4},故选:C.2.下列四组函数中,表示同一函数的一组是()A.y=|x|,u=B.y=,s=()2C.D.解:A.y=|x|和的定义域都是R,对应关系也相同,是同一函数;B.的定义域为R,的定义域为[0,+∞),定义域不同,不是同一函数;C.的定义域为{x|x≠1},m=n+1的定义域为R,定义域不同,不是同一函数;D.的定义域为{x|x≥1},的定义域为{x|x≤﹣1或x≥1},定义域不同,不是同一函数.故选:A.3.已知a=log3,b=ln3,c=2﹣0.99,则a,b,c的大小关系为()A.b>a>c B.a>b>c C.c>a>b D.b>c>a解:∵,∴a<0,∵ln3>lne=1,∴b>1,∵0<2﹣0.99<20=1,∴0<c<1,∴b>c>a,故选:D.4.在△ABC中,“”是“”的()A.充分必要条件B.充分而不必要条件C.必要不充分条件D.既不充分也不必要条件解:在△ABC中,A∈(0,π),考虑充分性,“”推不出“”,如当A=时,sin A=,所以“”不是“”的充分条件;再考虑必要性,“”⇒A∈()⇒“”,所以“”是“”的必要条件;故选:C.5.已知函数f(x+2)=2x+x﹣2,则f(x)=()A.2x﹣2+x﹣4B.2x﹣2+x﹣2C.2x+2+x D.2x+2+x﹣2解:设t=x+2,则x=t﹣2,∴f(t)=2t﹣2+t﹣2﹣2=2t﹣2+t﹣4,∴f(x)=2x﹣2+x﹣4.故选:A.6.在同一直角坐标系中,函数y=,y=log a(x+)(a>0且a≠1)的图象可能是()A.B.C.D.解:由函数y=,y=log a(x+),当a>1时,可得y=是递减函数,图象恒过(0,1)点,函数y=log a(x+),是递增函数,图象恒过(,0);当1>a>0时,可得y=是递增函数,图象恒过(0,1)点,函数y=log a(x+),是递减函数,图象恒过(,0);∴满足要求的图象为:D故选:D.7.函数f(x)=A sin(ωx+φ)(A>0,ω>0,0<φ<)的部分图象如图所示,将其向右平移个单位长度后得到的函数解析式为()A.y=sin2x B.y=sin(2x+)C.y=sin(2x﹣)D.y=sin(2x﹣)解:由函数图象知,A=,=﹣=,解得T=π,所以ω==2,所以函数f(x)=sin(2x+φ);因为f()=sin(+φ)=﹣sin(+φ)=﹣,所以+φ=+2kπ,k∈Z;解得φ=+2kπ,k∈Z;又0<φ<,所以φ=;所以f(x)=sin(2x+);将函数的图象向右平移个单位长度后,得y=sin[2(x﹣)+]的图象,即y=sin(2x﹣).故选:C.8.方程cos x=log8x的实数解的个数是()A.4B.3C.2D.1解:方程cos x=log8x的实数解的个数,即函数y=cos x的图象和函数y=log8x的图象交点的个数.数形结合可得函数y=cos x的图象和函数y=log8x的图象(图中红色曲线)交点的个数为3,故选:B.二、多项选择题(本大题共4小题,每小题5分,共20分.在每小题给出的四个选项中,有多个选项是符合题目要求的.全部选对的得5分,选对但不全的得3分,有选错的得0分)9.下列各式中,值为的是()A.cos2﹣sin2B.C.2sin195°cos195°D.解:对于A,cos2﹣sin2=cos=;对于B,=tan45°=;对于C,2sin195°cos195°=sin390°=sin30°=;对于D,==.故选:BC.10.已知a,b为正实数,则下列判断中正确的是()A.B.若a+b=4,则log2a+log2b的最大值为2C.若a>b,则D.若a+b=1,则的最小值是8解:已知a,b为正实数,(a+)(b+)=ab+++≥2+2=4,当且仅当a=b =1是取等号,故,所以A正确;因为正实数a,b满足a+b=4,∴4≥2,化为:ab≤4,当且仅当a=b=2时取等号,则log2a+log2b=log2(ab)≤log24=2,其最大值是2.则log2a+log2b的最大值为2,所以B正确;若a>b,a,b为正实数,有不等式性质有,所以C正确;若a+b=1,+=(+)•(a+b)=1+4++≥5+2=9,所以D不正确;故选:ABC.11.已知函数f(x)=|cos x|+cos|2x|,下列说法正确的是()A.若x∈[﹣π,π],则f(x)有2个零点B.f(x)的最小值为C.f(x)在区间上单调递减D.π是f(x)的一个周期解:根据函数f(x)=|cos x|+cos|2x|,整理得f(x)=2cos2x+|cos x|﹣1,对于A:若x∈[﹣π,π],当x=±π或时,满足函数f(x)=0,则f(x)有4个零点,故A错误;对于B:由于t∈[0,1],当t=0时,f(x)的最小值为﹣1,故B错误;对于C:利用函数的关系式转换为f(x)=g(x)+h(x),由于函数g(x)=|cos x|在(0,)上单调递减,函数h(x)=|cos2x|在(0,)上单调递减,故f(x)在区间上单调递减,故C正确;对于D:因为f(x+π)=f(x),所以f(x)的周期T=π,故D正确;故选:CD.12.已知函数f(x)=a sin x+b cos x,其中a,b∈R,且ab≠0,若对一切x∈R恒成立,则()A.B.C.是偶函数D.是奇函数解:由题意函数f(x)=a sin x+b cos x=sin(x+φ),其中a,b∈R,ab≠0.因为=1,对一切x∈R恒成立,可知f()=±1,所以+φ=kπ+,k∈Z,可得φ=kπ+,k∈Z,可得φ=,f()=sin(+),f()=sin(+),故f()>f(),或f()<f(),故A错误;因为f(x﹣)=sin(x﹣+)=sin x,所以f(x)为奇函数,故C错误;因为f(x+)=sin(x++)=sin(x+)=cos x,又因为cos x是偶函数,所以f(x)为偶函数,故D错误;f(﹣x)=sin(﹣x)=sin(x﹣),故B正确;故选:B.三、填空题(本大题共4小题,每小题5分,共20分)13.已知函数(ω>0)的最小正周期是π,则ω=2,单调递增区间是.解:由周期的求解方法可知;π=,可得ω=2;可得函数f(x)=2sin(2x+),令﹣+2kπ≤2x+≤+2kπ∴+kπ≤x≤+kπ,(k∈Z)即函数f(x)的递增区间为:[﹣+kπ,+kπ](k∈Z),故答案为2,[+kπ,+kπ](k∈Z)14.命题“所有三角形都有内切圆”的否定是“存在一个三角形没有内切圆”.解:全称命题“所有三角形都有内切圆”,它的否定是特称命题:“存在一个三角形没有内切圆”.故答案为:“存在一个三角形没有内切圆”.15.已知角θ的终边在直线y=﹣3x上,则=.解:∵角α的终边在直线y=3x上,∴tanα=3,∴====.故答案为:.16.已知函数,若a、b、c、d、e(a<b<c<d<e)满足f(a)=f(b)=f(c)=f(d)=f(e),则M=af(a)+bf(b)+cf(c)+df(d)+ef(e)的取值范围为(0,9).解:函数f(x)的图象如图所示:由图可得a+d=2,b+c=2,5<e<6,所以M=(a+b+c+d+e)f(e)=(4+e)(6﹣e)=﹣e2+2e+24=﹣(e﹣1)2+25,因为5<e<6,所以函数M在(5,6)上单调递减,又e=5时,M=9,e=6时,M=0,所以M的取值范围为(0,9),故答案为:(0,9).四、解答题(本题共6小题,共70分.解答时应写出必要的文字说明、证明过程或演算步骤)17.计算下列各式的值:(1);(2).解:(1)原式=;(2)原式=3+lg100+2=3+2+2=7.18.已知幂函数f(x)=(m2+2m﹣2)x m+2,且在(0,+∞)上是减函数.(1)求f(x)的解析式;(2)若(3﹣a)m>(a﹣1)m,求a的取值范围.解:(1)∵函数是幂函数,∴m2+2m﹣2=1,即m2+2m﹣3=0,解得m=1或m=﹣3,∵幂函数f(x)在(0,+∞)上是减函数,∴m+2<0,即m<﹣2,∴m=﹣3,∴f(x)=x﹣1,(2)令g(x)=x﹣3,因为g(x)的定义域为(﹣∞,0)∪(0,+∞),且在(﹣∞,0)和(0,+∞)上均为减函数,∵(3﹣a)﹣3>(a﹣1)﹣3,∴3﹣a<a﹣1<0或0<3﹣a<a﹣1或3﹣a>0>a﹣1,解得2<a<3或a<1,故a的取值范围为:{a|2<a<3或a<1}.19.已知函数.(1)求函数f(x)的最小正周期、对称轴和对称中心;(2)若锐角α满足,且β满足,求cosβ的值.解:f(x)=sin2x﹣×(1+cos2x)+=sin2x﹣cos2x=sin(2x﹣),则(1)f(x)的最小正周期T=,由2x﹣=kπ+得2x=kπ+,得x=+,k∈Z,即函数的对称轴为{x|x=+,k∈Z}.由2x﹣=kπ得2x=kπ+,得x=+,k∈Z,即函数的对称中心为(+,0),k∈Z.(2)若锐角α满足,且β满足,则sin[2(α+)﹣]=﹣,得sin(2α+)=cos2α=﹣,即2cos2α﹣1=﹣,得2cos2α=,即cos2α=,则cosα=,sinα=,∵,∴cos(α+β)=±,当cos(α+β)=时,cosβ=cos(α+β﹣α)=cos(α+β)cosα+sin(α+β)sinα=×+×=,当cos(α+β)=﹣时,cosβ=cos(α+β﹣α)=cos(α+β)cosα+sin(α+β)sinα=﹣×+×=.20.某生物研究者于元旦在湖中放入一些凤眼莲(其覆盖面积为k),这些凤眼莲在湖中的蔓延速度越来越快,二月底测得凤眼莲的覆盖面积为24m2,三月底测得凤眼莲的覆盖面积为36m2,凤眼莲的覆盖面积y(单位:m2)与月份x(单位:月)的关系有两个函数模型y=ka x(k>0,a>1)与可供选择.(1)试判断哪个函数模型更合适并求出该模型的解析式;(2)求凤眼莲的覆盖面积是元旦放入凤眼莲面积10倍以上的最小月份.(参考数据:lg2≈0.3010,lg3≈0.4711).解:(1)函数y=ka x(k>0,a>1)与在(0,+∞)上都是增函数,随着x的增加,函数y=ka x(k>0,a>1)的值增加的越来越快,而函数的值增加的越来越慢,由于凤眼莲在湖中的蔓延速度越来越快,因此选择模型y=ka x(k>0,a>1)符合要求.根据题意可知x=2时,y=24;x=3时,y=36,∴,解得.故该函数模型的解析式为,1≤x≤12,x∈N*;(2)当x=0时,,元旦放入凤眼莲的覆盖面积是m2,由>10•,得>10,∴x>=≈5.9,∵x∈N*,∴x≥6,即凤眼莲的覆盖面积是元旦放入凤眼莲面积10倍以上的最小月份是六月份.21.已知定义域为R的函数,是奇函数.(1)求a,b的值;(2)判断f(x)单调性并证明;(3)若∀t∈[﹣1,4],不等式f(t2+2)+f(2t2﹣kt)<0恒成立,求k的取值范围.解:(1)由于定义域为R的函数是奇函数,则即,解得,即有f(x)=,经检验成立;(2)f(x)在(﹣∞,+∞)上是减函数.证明:设任意x1<x2,f(x1)﹣f(x2)=﹣=,由于x1<x2,则2x1<2x2,即有>0,则有f(x1)>f(x2),故f(x)在(﹣∞,+∞)上是减函数;(3)不等式f(t2+2)+f(2t2﹣kt)<0,由奇函数f(x)得到f(﹣x)=﹣f(x),f(2t2﹣kt)<﹣f(2+t2)=f(﹣t2﹣2),再由f(x)在(﹣∞,+∞)上是减函数,则2t2﹣kt>﹣t2﹣2,即有3t2﹣kt+2>0对t∈[﹣1,4]恒成立,当t=0时,2>0,显然成立;当0<t≤4时,k<=3t+,3t+≥2,当且仅当t=时,取得等号,则k<2;当﹣1≤t<0时,k>=3t+,又3t+=﹣[(﹣3t)+]≤﹣2,当且仅当t=﹣∈[﹣1,0)时,取得等号,则k>﹣2;综上可得k的范围是(﹣2,2).22.已知函数为f(x)的零点,为f(x)图象的对称轴.(1)若f(x)在[0,2π]内有且仅有6个零点,求f(x);(2)若f(x)在上单调,求ω的最大值.解:(1)因为f(x)在[0,2π]内有且仅有6个零点,则6个零点间有周期,所以①,又8个零点间的一定比[0,2π]的区间长度大,即②,由①②可得,又为f(x)的零点,所以,k1∈Z③,为f(x)图象的对称轴,则,k2∈Z④,④﹣③可得,即ω=2(k2﹣k1)+1,因为k1∈Z,k2∈Z,所以ω为奇数,故ω=3,由③可得φ=,k1∈Z,又|φ|,所以φ=﹣,故;(2)由(1)可知,ω=2(k2﹣k1)+1,k1,k2∈Z,故ω为奇数,因为f(x)在上单调,则,解得ω≤12,所以ω的最大值可能为11,9,7,…,当ω=11时,φ=k1π,又|φ|,所以φ=﹣,故,此时函数f(x)在上不单调;当ω=9时,φ=k1π,又|φ|,所以φ=,故,此时函数f(x)在上单调递减,符合题意.综上可得,ω的最大值为9.。

【含高考模拟卷13套】广东省实验中学2020-2021学年高三第一次阶段考试数学试题含解析

【含高考模拟卷13套】广东省实验中学2020-2021学年高三第一次阶段考试数学试题含解析

广东省实验中学2020-2021学年高三第一次阶段考试数学试题注意事项:1. 答题前,考生先将自己的姓名、准考证号填写清楚,将条形码准确粘贴在考生信息条形码粘贴区。

2.选择题必须使用2B 铅笔填涂;非选择题必须使用0.5毫米黑色字迹的签字笔书写,字体工整、笔迹清楚。

3.请按照题号顺序在各题目的答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试题卷上答题无效。

4.保持卡面清洁,不要折叠,不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。

一、选择题:本题共12小题,每小题5分,共60分。

在每小题给出的四个选项中,只有一项是符合题目要求的。

1.已知集合{}2|320M x x x =-+≤,{|N x y ==若M N M ⋂=,则实数a 的取值范围为( ) A .(,1]-∞B .(,1)-∞C .(1,)+∞D .[1,)+∞2.若集合M ={1,3},N ={1,3,5},则满足M ∪X =N 的集合X 的个数为( ) A .1 B .2 C .3D .43.已知函数()cos (0)f x x x ωωω=->,()y f x =的图象与直线2y =的两个相邻交点的距离等于π,则()f x 的一条对称轴是( ) A .12x π=-B .12x π=C .3x π=-D .3x π=4.一辆邮车从A 地往B 地运送邮件,沿途共有n 地,依次记为1A ,2A ,…n A (1A 为A 地,n A 为B 地).从1A 地出发时,装上发往后面1n -地的邮件各1件,到达后面各地后卸下前面各地发往该地的邮件,同时装上该地发往后面各地的邮件各1件,记该邮车到达1A ,2A ,…n A 各地装卸完毕后剩余的邮件数记为(1,2,,)k a k n =….则k a 的表达式为( ).A .(1)k n k -+B .(1)k n k --C .()n n k -D .()k n k -5.已知函数()sin()(0,0)3f x x πωφωφ=+><<满足()(),()12f x f x f ππ+==1,则()12f π-等于( )A .-2B .2C .-12D .126.设a ,b ,c 分别是ABC ∆中A ∠,B ,C ∠所对边的边长,则直线sin 0A x ay c ⋅--=与sin sin 0bx B y C +⋅+=的位置关系是( )A .平行B .重合C .垂直D .相交但不垂直7.若函数()222y sin x ϕϕπ⎛⎫<⎪⎝+⎭=的图象经过点012π⎛⎫⎪⎝⎭,,则函数()()()22f x sin x cos x ϕϕ=-+-图象的一条对称轴的方程可以为( ) A .24x π=-B .3724x π=C .1724x π=D .1324x π=-8.已知双曲线2222:1(0,0)x y C a b a b-=>>的右焦点为F ,若双曲线C 的一条渐近线的倾斜角为3π,且点F到该渐近线的距离为3,则双曲线C 的实轴的长为 A .1 B .2 C .4D .8559.若集合{}(2)0A x x x =->,{}10B x x =->,则A B =A .{}10x x x ><或B .{}12x x << C .{|2}x x >D .{}1x x >10.如图,在ABC ∆中,23AN NC =,P 是BN 上一点,若13AP t AB AC =+,则实数t 的值为( )A .23B .25C .16D .3411.已知集合{}2230A x x x =--≤{}2B x x =<,则A B =( )A .()1,3B .(]1,3C .[)1,2-D .()1,2-12.已知双曲线2222:1(0,0)x y E a b a b-=>>满足以下条件:①双曲线E 的右焦点与抛物线24y x =的焦点F 重合;②双曲线E 与过点(4,2)P 的幂函数()f x x α=的图象交于点Q ,且该幂函数在点Q 处的切线过点F 关于原点的对称点.则双曲线的离心率是( ) A 31+ B 51+ C .32D 51二、填空题:本题共4小题,每小题5分,共20分。

专题4-2 三角函数图像与性质归类-(原卷版)

专题4-2 三角函数图像与性质归类-(原卷版)

专题4-2 三角函数图像与性质归类目录一、热点题型归纳【题型一】平移1:正弦←→余弦 (1)【题型二】平移2:识图平移 (3)【题型三】平移3:恒等变形平移 (4)【题型四】平移4:中心对称,轴对称,单调性等性质 (5)【题型五】平移5:最小平移 (6)【题型六】平移6:求w 最值 (7)【题型七】正余弦函数对称轴 (8)【题型八】正余弦对称中心 (9)【题型九】三角函数周期 (9)【题型十】单调性与最值 (11)【题型十一】正余弦“和”与“积”性质、最值 (11)【题型十二】三角函数零点 (12)【题型十三】图像与性质:x1与x2型 (13)【题型十四】三角函数最值 (14)【题型十五】万能代换与换元 (15)【题型十六】图像和性质综合 (15)二、真题再现 (16)三、模拟检测 (178)【题型一】平移1:正弦←→余弦【典例分析】(2022·安徽省太和中学高三阶段练习)已知函数()()πcos 0,02f x x ωϕωϕ⎛⎫=+><< ⎪⎝⎭,若()f x 的图象向右平移π12个单位后,得到函数()2πsin 23g x x ⎛⎫=+ ⎪⎝⎭的图象,则( )A .6π=ϕB .π4ϕ= C .π3ϕ= D .2π5ϕ=1(2023·全国·高三专题练习)已知直线8x π=是函数()2sin(2)||2πϕϕ⎛⎫=+< ⎪⎝⎭f x x 的图像的一条对称轴,为了得到函数()y f x =的图像,可把函数2cos 26y x π⎛⎫=- ⎪⎝⎭的图像( )A .向左平移24π个单位长度B .向右平移24π个单位长度C .向左平移12π个单位长度 D .向右平移12π个单位长度2.(2022·全国·高三专题练习)为得到函数cos 23y x π⎛⎫=- ⎪⎝⎭的图象,只需将函数sin 24y x π⎛⎫=-- ⎪⎝⎭图象上所有的点( )A .向左平移712π个单位长度B .向右平移712π个单位长度 C .向左平移724π个单位长度D .向右平移724π个单位长度3.(2023·全国·高三专题练习)为了得到函数πsin 24y x ⎛⎫=+ ⎪⎝⎭的图象,可以将函数πcos 23y x ⎛⎫=+ ⎪⎝⎭的图象( )A .向左平移5π24个单位 B .向右平移7π24个单位 C .向右平移5π24个单位D .向左平移7π24个单位【题型二】平移2:识图平移【典例分析】(2022·陕西·渭南市华州区咸林中学高三开学考试(理))如图,函数()()π2sin 0,||2f x x ωϕωϕ⎛⎫=+>< ⎪⎝⎭的图像过()π,0,2π,22⎛⎫⎪⎝⎭两点,为得到函数()()2cos g x x ωϕ=-的图像,应将()f x 的图像( )A .向右平移7π6个单位长度 B .向左平移7π6个单位长度 C .向右平移5π2个单位长度D .向左平移5π2个单位长度()++(0)0Asin x b A ,的步骤和方法:确定函数的最大值M 和最小值2M mA ,2M mb; :确定函数的周期T ,则可2T得=; :常用的方法有代入法和五点法. 把图象上的一个已知点代入(此时A b ,,已知)或代入图象与直线y b =的交点求解注意交点是在上升区间上还是在下降区间上).五点法”中的某一个点为突破口.【变式演练】1.(2022·河南·高三阶段练习(理))函数()()2sin f x x ωϕ=+(0>ω且0πϕ<<)在一个周期内的图象如图所示,将函数()y f x =图象上的点的横坐标伸长为原来的2倍,再向右平移π4个单位长度,得到函数()y g x =的图象,则π3g ⎛⎫= ⎪⎝⎭( )AB .1C .-1D .2.(2022·全国·长垣市第一中学高三开学考试(理))将函数sin y x =的图象上所有点的横坐标变为原来的(0)m m >倍,纵坐标不变,再将所得函数图象向左平移(0)ϕϕπ<<个单位长度,最后将所得函数图象上所有点的纵坐标变为原来的(0)n n >倍,横坐标不变,得到如图所示的函数()f x 的部分图象,则,,m n ϕ的值分别为( )A .22,2,3m n πϕ===B .12,2,23m n πϕ===C .2,2,3m n πϕ===D .1,2,23m n πϕ===3.(2022·四川省内江市第六中学模拟预测(文))已知函数()()cos 0,0,2f x A x A πωϕωϕ⎛⎫=+>>< ⎪⎝⎭,将函数()f x 的图象向左平移34π个单位长度,得到函数()g x 的部分图象如图所示,则3f π⎛⎫= ⎪⎝⎭( )A .12 B .12-C D .【题型三】平移3:恒等变形平移【典例分析】(2022·湖北·高三开学考试)要得到2()sin 43g x x π⎛⎫=+⎪⎝⎭的图象,只需要将22()cos 2sin 2f x x x =-的图象( ) A .向左平移24π个单位长度 B .向右平移24π个单位长度 C .向左平移12π个单位长度D .向右平移12π个单位长度【变式演练】1.(2023·全国·高三专题练习)已知函数()2sin cos f x x x =+的图象向左平移()0ϕϕ>个单位长度后得到函数()sin 2cos g x x x =+的图象,则()g ϕ=( )A .65B .115C .15 D .852.(2022·全国·高三专题练习)为了得到函数2cos2y x =的图象,只需把函数2cos 2y x x =+的图象( ) A .向左平移3π个单位长度 B .向右平移3π个单位长度C .向左平移6π个单位长度 D .向右平移6π个单位长度3.(【百强校】2015届浙江省宁波市镇海中学高三5月模拟考试理科数学)设()cos 22f x x x =,把()y f x =的图像向左平移(0)ϕϕ>个单位后,恰好得到函数()cos 22g x x x =-的图象,则ϕ的值可以为( ) A .6π B .3πC .23πD .56π【题型四】平移4:中心对称,轴对称,单调性等性质【典例分析】(2022·安徽·高三开学考试)将函数()sin(2)(0)f x x ϕϕπ=+<<的图象向右平移6π个单位长度得到()g x 的图象,若()g x 的图象关于直线3x π=对称,则6g π⎛⎫= ⎪⎝⎭( )A .B .12-C .0D .12)+)00((Asin x A ,两个点关于中心对称,则函数值互为相反数。

2020-2021学年广东省实验中学高三(上)期末英语试卷

2020-2021学年广东省实验中学高三(上)期末英语试卷

2020-2021学年广东省实验中学高三(上)期末英语试卷1.(问答题,1.5分)Where does this conversation probably take place?A.In a bookstore.B.In a classroom.C.In a library.2.(问答题,1.5分)At what time will the film begin?A.7:20.B.7:15.C.7:00.3.(问答题,1.5分)What are the two speakers mainly talking about?A.Their friend Jane.B.A weekend trip.C.A radio programme.4.(问答题,1.5分)What will the woman probably do?A. Catch a train.B. See the man off.C. Go shopping.5.(问答题,1.5分)Why did the woman apologize?A.She made a late delivery.B.She went to the wrong place.C.She couldn't take the cake back.6.(问答题,3分)(1)Whose CD is broken?A. Kathy's.B. Mum's.C. Jack's.(2)What does the boy promise to do for the girl?A. Buy her a new CD.B. Do some cleaning.C. Give her 10 dollars.7.(问答题,3分)(1)What did the man think of the meal?A.Just so-so.B.Quite satisfactory.C.A bit disappointing.(2)What was the 15% on the bill paid for?A.The food.B.The drinks.C.The service.8.(问答题,4.5分)(1)Why is the man at the shop?A.To order a camera for his wife.B.To have a camera repaired.C.To get a camera changed.(2)What color does the man want?A.Pink.B.Black.C.Orange.(3)What will the man do afterwards?A.Make a phone call.B.Wait until further notice.C.Come again the next day.9.(问答题,6分)(1)What would Joe probably do during the Thanksgiving holiday?A. Go to a play.B. Stay at home.C. Visit Kingston.(2)What is Ariel going to do in Toronto?A. Attend a party.B. Meet her aunt.C. See a car show.(3)Why is Ariel in a hurry to leave?A. To call up Betty.B. To buy some DVDs.C. To pick up Daniel.(4)What might be the relationship between the speakers?A. Classmates.B. Fellow workers.C. Guide and tourist.10.(问答题,6分)(1)Where does Thomas Manning work?A. In the Guinness Company.B. At a radio station.C. In a museum.(2)Where did the idea of a book of records come from?A. A bird-shooting trip.B. A visit to Europe.C. A television talk show.(3)When did Sir Hugh's first book of records?A. In 1875.B. In 1950.C. In 1955.(4)What are the two speakers going to talk about next?A.More records of unusual facts.B.The founder of the company.C.The oldest person in the world.11.(填空题,7.5分)Guardian Weekend magazine is launching a writing competition for UK women aged 16-21 on the theme of conversations.How to enterAll you have to do is submit a 700-word personal essay that shows off your talents--on thetheme of conversations.Did you have an unforgettable conversation with your grandmother about her youth that changed how you viewed her?Do you find having certain conversations really hard, and if so, why?Is there a conversation you regret,or one you regret you never had?We're keen to hear about your personal experiences.***************************************************************.The PrizesThere will be one winner and two runners-up.The three winners will each receive £250.The winners will be notified(通知)by email or telephone on or before 30 March 2021 and given details of how to claim their prizes.As part of the editing process, the three winners will participate in a video call with a Guardian Weekend editor to discuss and edit their essay for publication.The one overall winner will also receive a 1-1 work shop with a Guardian editor. RulesFollow all rules carefully to prevent disqualification.■Only one entry is permitted per person.Entries on behalf of another person will not be accepted and joint submissions are not allowed.■The Competition opens at 09:00 on 22 February 2021 and closes at 23:59 on 9 March 2021.Entries received outside this time period will not be considered.■Your entry must not be copied,and must not contain any third-party materials or content that you do not have permission to use.■You must include your name, age and contact details, inc luding your email address and phone number.(1)What's the theme of the writing competition?___A.Regrets.B.Conversations.C.Grandmother's youth.D.Personal experiences.(2)What extra prize will the overall winner receive?___A.An additional £250.B.A video of the competition.C.A prior notification of the win.D.A 1-1 workshop with an editor.(3)Which of the following will result in disqualification?___A.Co-authoring an entry.B.Including contact details.C.Mailing your entry on 1 March.ing others' content with permission.12.(填空题,10分)April Fools' Day is supposed to be a day to play jokes on others in hopes of getting a good laugh and making one feel like a fool.However,the April Fools' Day of 2019 was quite different for my mom and me.That day my friend Jimmy and I were playing a game.I had dropped down from a bar(横木)many times in the past without ever having a problem,but that day the simple act of dropping to the ground became a nightmare(噩梦).I broke my arm.Jimmy's dad heard my crying and rushed out to see what was going on.When he saw the problem,he quickly put me into his truck and went inside to telephone my mom and let her know he would take me to the hospital.As that day was April Fools' Day,Mom was not buying it and really thought all this was a big joke.Mom was finally convinced by Jimmy's mom.When she saw me,she broke down in tears because she felt so bad-she simply believed that was just a big trick.I guess one could compare this to the story The Boy Who Cried Wolf.Since I had played bad tricks before,it was no wonder that my mom didn't believe it.I as well as my mom was made to look like a fool that day.We both learned a valuable lesson.(1)What happened to the writer on the April Fools' Day of 2019?___A.He dreamed a terrible dream.B.He was hurt by Jimmy.C.He had an accident.D.He fooled his mom with his friend.(2)Why did Jimmy's dad phone the writer's mom?___A.Because he wanted to play a joke on her.B.Because he wanted her to go to the hospital to pay the money.C.Because he wanted to see how deeply she loved her son.D.Because he wanted to inform her of the accident.(3)What does the underlined sentence "Mom was not buying it" mean?___A.She didn't believe what Jimmy's dad said.B.She would not like to pay the money.C.She had no preparation for the bad news.D.She thought her son deserved (应受) it.(4)Why did the writer mention the story The Boy Who Cried Wolf?___A.Because he thought it was his mom's fault.B.Because he had a lesson like that.C.Because he was proud of his story.D.Because he felt he was luckier than that boy.13.(填空题,10分)A city in Netherlands is planning to construct a pretty bike path made of recycled wood.The first of its kind in the world, the path near the city of Emmen will be surfaced not with the usual asphalt(沥青) but with wood chips packed together with organic resin(有机树脂). The idea behind the path is to cut the use of conventional,less eco-friendly materials such as concrete,which is very difficult to recycle.And the creation of the wood chips will require no direct cutting down of trees;the company leading the experiment,Grontmij,plans to use waste products from sawmills(锯木厂).The idea of creating a permanent road from a material that is celebrated for its biodegradability(生物降解性)might seem ridiculous. But the engineers working on the path insist that the wood and resin surfacing will stay in good condition for a long time, with a working life at least as long as concrete or asphalt. Rudi van Hedel, project manager of bio-based economy at Grontmij, explained that the light weight of the material also makes it far easier to move.However,van Hedal said,"At present,the material costs of the path are higher than those of traditional paths made of asphalt or concrete,but the construction costs are comparable or perhaps slightly cheaper. We expect that as production ability increases, the costs will go down. In the future, we hope to use bio-fibres(生物纤维)that are cheaper than the wood fibreswe're currently experimenting with, and we expect that biodegradable materials will be able to compete with asphalt and concrete."(1)What can we learn about the path?___A.It uses some asphalt.B.It aims to protect trees.C.It is a pioneering project.D.It is being built by a sawmill.(2)What do the engineers stress about the path in Paragraph 4?___A.Its material is heavy in weight.B.It can stay in service for long.C.Its length may break a record.D.It is environmentally friendly.(3)According to van Hedal, what is a disadvantage of wood fibres?___A.Their production process is complex.B.They are not as hard as asphalt.C.They are not always available.D.Their costs are relatively high.(4)Where is the text most likely from?___A.A diary.B.A novel.C.A magazine.D.A guidebook.14.(填空题,10分)A new study published this week in the journal Nature Communications has concluded that a 100 percent change to organic (有机的) food production in England and Wales would actually lead to a great increase in greenhouse gas emissions (排放). In turn, this would contribute to further climate change.Although organic farming directly pours out fewer emissions than conventional farming-around 20 percent lower for crops and 4 percent for farm animals-it produces notably less food. As to this study's findings, total organic agriculture in England and Wales would produce 40 percent less food. With less food in the market, the countries would need to increase food imports, which would produce more global greenhouse gas emissions.Organic farming also increases the amount of absorbing carbon,a process where carbon dioxide(CO2) is "absorbed" out of the atmosphere and captured by plants and stored in the soil. However,even a total change to organic farming would only be equal to a tiny part of the higher emissions from overseas land use."We predict a drop in total food production of 40 percent under a fully organic farming process,compared to conventional farming, if we keep to the same national diet," Dr Adrian Williams,lead author and reader in Agri-Environmental Systems at Cranfield University,said in a statement. "This results from lower crop quantity, because output is limited by a lower supply of nitrogen, which is mainly from other crops or solid waste from cattle on the grassland."Nevertheless,it is important to note that organic farming still holds some useful benefits for the environment,such as reducing exposure to chemicals and improving the varieties of creatures. In conclusion,the study suggests that organic farming will continue to play a key role in resolving the world's environmental problems. However, it's just one part of a much wider solution.(1)What will total organic agriculture bring to England?___A.More main food.B.More species crop.C.More food imports.D.More fresh oxygen.(2)How does organic farming increase the amount of absorbing carbon?___A.By taking in CO2.B.By changing CO2.C.By giving off CO2.D.By producing CO2.(3)What is the last paragraph mainly about?___A.The ways to reduce organic farming.B.The results caused by organic farming.C.The solution to the environment problems.D.The advantages of organic farming.(4)Which of the following is the best title of the text?___anic farming, green foodanic farming, our hope in futureanic farming, a mistake we madeanic farming, a double-edged sword15.(填空题,12.5分)Can creativity be taught?That's a question without a simple yes or no answer.Creativity may not be able to be taught directly,but what you can get better at is frequently targeting at the circumstances of life which bring up the greatest chances for true creative expression.(1)___ ,but more like something which manifests(显现)itself inside those who learn to develop it and create the right conditions for it.Limit your selection of tools to only the most vital.(2)___ .You'll be sharper than someone who merely fights with a larger set of tools.Learn how to be resourceful.(3)___ .Creativity is not just about creating something new but making old things work better as well.Think of crazy possibilities as well as practical ones.You might find inspiration for a workable solution in one of your ideas.Don't listen to feedback(反馈)and keep following your own path.The problem of asking for feedback is that the feedback will be given according to the person's past experience.Others will unconsciously push you in a direction that they see as best.(4)___ .Just don't let criticism (even the constructing type)destroy your creativity during the creative process.(5)___ .Routines are positive if they strengthen a healthy creative consciousness and negative if they destroy that.The key is to discover a creative routine that puts you in a more creative mindset.A.Creativity is not like a lightning strikeB.Having a routine is actually not a bad ideaC.You can literally do anything you like with themD.Resourcefulness is about making the most of what you have to work withE.The more limited your set of tools is,the more creative the output will beF.This is done with good intentions,but it actually hurts your natural creativityG.While breaking your routine once in a while to force new ways of thinking is good16.(填空题,15分)There was once a wonderful old man who loved everythinganimals,spiders,insects and all sorts of living things.One day while walking through the woods,the old man found a cocoon(面).He decided to take the cocoon home to watch its magic process of(1)___ from a little cocoon to beautiful(2)___ .A few days later the cocoon stared to move.It moved frantically(狂乱地).A small(3)___ appeared.He sat and watched it struggle and struggle for several hours to(4)___ its body through the little hole.But it seemed not to make any progress.Then the old man felt(5)___ for the little butterfly inside,and rushed to its aid.With a pair of scissors and(6)___ ,he cut the little opening big enough in the cocoon for the butterfly to come out.And then the butterfly came out of its cocoon,but it had a(7)___ body and small,fragile wings.The man continued to watch the butterfly because he expected that,(8)___ the wings would expand and be able to(9)___ the body,but nothing happened!It never was able to fly.What the man,in his(10)___ and eagerness,did not understand was that the restricting cocoon and the struggle(11)___ for the butterfly to get through the tiny opening werenature's way of forcing fluid(液体) from the body of the butterfly into its wings(12)___ it would be ready for flight once it achieved its freedom from the cocoon.Sometimes(13)___ are exactly what we need in our life.If we were allowed to(14)___ our life without any obstacles and difficulties,it would weaken or even damage us over time both physically and mentally.We would not be as(15)___ as we could have been.In fact,it is necessary to live with some difficulties.And it is these obstacles and difficulties that make us strong and get prepared for the greater challenges in our life.(1)A.advancing B.growing C.transforming D.developing (2)A.dog B.butterfly C.cat D.snake(3)A.spot B.passage C.opening D.butterfly (4)A.expand B.push C.shrink D.force(5)A.sorry B.moved C.excited D.happy(6)A.cheerfully B.gently C.casually D.randomly (7)A.powerful B.awkward C.swollen D.flexible(8)A.in no way B.in no case C.by any chance D.at anymoment(9)A.support B.protect C.control D.lift(10)A.carefulness B.willingness C.kindness D.sadness (11)A.arranged B.required C.intended D.prepared (12)A.if only B.as if C.as though D.so that (13)A.struggles B.pains C.failures D.sufferings (14)A.adapt to B.care for C.go through D.reflect on (15)A.potential B.strong C.patient D.energetic 17.(填空题,15分)The game of Go(围棋)is an important board game with origins in China from more than 4,000 years ago.In China.Go (1)___ (recognize) as "hand conventions" as well,through which players communicate with each other.(2)___ other words,they" talk" through the placing of pieces on the board.The metaphor(比喻) first (3)___ (use)by Zhi Daolin,a Buddhist master of the Jin Dynasty,reflects the nature of thegame.While scholars of his time often lost (4)___ (they) in philosophical(哲学的)debates on life and universe,he preferred to play Go,(5)___ he believed was full of hows and whys of life.Maser Go players often feel as if they were playing a real life game:sometimes one can move forward,while at other times,one must slow down;sometimes one can(6)___ (direct) face the challenger,at other time,one must take an indirect approach.Appropriate placement of each tiny Go piece is similar to(7)___ one might solve a difficult problem of life.The purpose of(8)___ (play) Go is not just to win but also,mere importantly,to seek(9)___ (wise)throughthe process.The players unite with each other on the board.Instead of fighting as (10)___ (enemy),they cooperate to play a good game.One thoughtless move could ruin the enjoyment.18.(问答题,15分)假如你是李华.你的美国笔友即将随其父母来中国,并中国度过中秋节.他来信向你询问有关中秋节的习俗,请你告诉他有关中秋节的习俗.要点如下:1 历史悠久.2 中国人独有的传统节日.3 家庭团圆.4 共进晚餐.5 吃月饼.6 赏月.要求:1 词数100 左右.2 可适当添加细节.3 开头已给出,不计入总词数.Dear Mike,I am glad to know that you are coming to China with your parents and spend the Mid-AutumnDay here.___Yours,Li Hua.19.(问答题,25分)阅读下面材料,根据其内容和所给段落开头语续写两段,使之构成一篇完整的短文。

2022-2023学年广东省佛山市实验中学高二上学期第一次段考化学试题

2022-2023学年广东省佛山市实验中学高二上学期第一次段考化学试题

2022-2023学年广东省佛山市实验中学高二上学期第一次段考化学试题1.下列说法不正确的是A.焓变是一个与反应能否自发进行有关的因素,放热反应具有自发进行的倾向B.在同一条件下物质有不同的熵值,其体系的混乱程度越大,熵值越大C.自发反应是指不需要条件就能发生的反应D.—个反应能否自发进行与焓变和熵变的共同影响有关2.下列属于吸热反应的是A.金属钠与水反应B.铁丝在氧气中燃烧C.NaOH溶液与盐酸反应D.Ba(OH) 2 ·8H 2 O晶体与NH 4 Cl晶体反应3.已知可逆反应A2(g)+B2(g) 2AB(g)的能量变化如图所示,则下列叙述中不正确的是A.该反应的活化能为a kJ·mol -1B.该反应的逆反应的活化能为b kJ·mol -1C.该反应的ΔH=+(a-b) kJ·mol -1D.使用催化剂,a的数值减小,ΔH也减小4.依据下列热化学方程式得出的结论中,正确的是A.已知,则氢气的燃烧热为B.已知C(石墨,s)=C(金刚石,s) ,则金刚石比石墨稳定C.已知,则含的稀溶液与稀盐酸完全中和,放出的热量D.已知;;则5.在测定中和反应反应热的实验中,下列说法正确的是A.测量酸溶液的温度后,未冲洗温度计就测碱溶液的温度,会使求得的中和热的数值偏大B.0.1mol/L盐酸与某强碱稀溶液中和放出的热量随反应物的用量改变而改变,但中和反应反应热不变C.分多次把NaOH溶液倒入盛有稀盐酸的量热计的内筒中D.玻璃搅拌器沿内筒壁顺时针移动,使酸碱溶液混合6.恒温下,反应aX(g)bY(g)+cZ(s)达到平衡状态,把容器容积缩小到原来的一半,且达到新的平衡状态时,X的物质的量浓度从0.1 mol/L增大到0.19 mol/L。

下列判断中正确的是A.a>b B.a>b+c C.a=b+c D.a<b7.已知 2SO2(g)+O2(g) 2SO3(g) ΔH<0。

导数构造函数十二种题型归类(学生版)

导数构造函数十二种题型归类(学生版)

导数构造函数十二种题型归类内容速递一、知识梳理与二级结论二、热考题型归纳【题型一】 导数四则运算基础【题型二】 幂函数与f(x)积型【题型三】 幂函数与f(x)商型【题型四】 指数函数与f(x)积型【题型五】 指数函数与f(x)商型【题型六】 正弦函数与f(x)型【题型七】 余弦函数与f(x)型【题型八】 对数函数与f(x)型【题型九】 一元二次(一次)与f(x)线性【题型十】 指数型线性【题型十一】对数型线性【题型十二】综合构造三、高考真题对点练四、最新模考题组练知识梳理与二级结论一、导数的运算(1)基本初等函数的导数公式原函数导函数f(x)=c(c为常数)f′(x)=0 f(x)=xα(α∈Q,且α≠0)f′(x)=αxα-1 f(x)=sin x f′(x)=cos xf(x)=cos x f′(x)=-sin x f(x)=a x(a>0,且a≠1)f′(x)=a x ln a f(x)=ex f′(x)=e xf(x)=log a x(a>0,且a≠1)f′(x)=1 x ln af(x)=ln x f′(x)=1 x(2)导数的四则运算法则法则和差[f(x)±g(x)]′=f′(x)±g′(x)积[f(x)g(x)]′=f'x g x +f x g'x ,特别地,[cf(x)]′=cf′(x) 商f(x)g(x)′=f(x)g(x)-f(x)g (x)g(x)2(g(x)≠0)(3)简单复合函数的导数一般地,对于两个函数y=f(u)和u=g(x),如果通过中间变量u,y可以表示成x的函数,那么称这个函数为函数y=f(u)和u=g(x)的复合函数,记作y=f(g(x)). 它的导数与函数y=f(u),u=g(x)的导数间的关系y ′x =y ′u ·u ′x即y 对x 的导数等于y 对u 的导数与u 对x 的导数的乘积.二、导数构造规律(1)、关系式为“加”型,常构造为乘法①fx +f x ≥0,构造F x =e xf x ,Fx =e xf x +fx ,②xfx +f x ≥0,构造F x =xf x ,Fx =xfx +f x ,③xfx +nf x ≥0,构造F x =x nf x ,Fx =x n -1xfx +nf x ;(2)、关系式为“减”型,常构造为除法①fx -f x ≥0,构造F x =f x e x ,F x =f x -f x ex,②xf x -f x ≥0,构造F x =f x x ,Fx =xfx -f x x 2,③xf x -nf x ≥0,构造F x =f x x n ,Fx =xf x -nf x xn +1.热点考题归纳【题型一】导数四则运算基础【典例分析】1(2022春·北京·高三模拟)若f x =e x ln x ,则f x =()A.e xln x +e xxB.e x ln x -e xxC.e x xD.e x ln x 2(2023春·黑龙江伊春·高三模拟)函数y =e x sin2x 的导数为()A.y =2e x cos2xB.y =e x sin2x +2cos2xC.y =2e x sin2x +cos2xD.y =e x 2sin2x +cos2x【提分秘籍】基础求导公式:C=0;x α=αx α-1;a x=axln a ;log a x=1x ln a ;sin x=cos xcos x=sin x【变式演练】3(2022春·北京·高三清华附中校考)函数f x =sin xx的导数是()A.x sin x +cos xx 2B.x cos x +sin xx 2C.x sin x -cos x x 2D.x cos x -sin xx 24(2023春·四川资阳·高三联考)已知函数y =x ⋅tan x 的导函数为()A.y =sin x cos x +xcos 2x B.y =sin x cos x +x cos2xcos 2xC.y =sin x cos x +1cos 2xD.y =sin x cos x +cos2xcos 2x【题型二】幂函数与f (x )积型【典例分析】1设函数f x 是定义在0,+∞ 上的可导函数,其导函数为f x ,且有2f x +xf x >0,则不等式x -20212f x -2021 -f 1 >0的解集为()A.2020,+∞B.0,2022C.0,2020D.2022,+∞2(黑龙江省大庆实验中学2020-2021学年高三数学试题)函数f x 是定义在区间0,+∞ 上的可导函数,其导函数为f x ,且满足xf x +2f x >0,则不等式(x +2020)f (x +2020)3<3f (3)x +2020的解集为()A.x |x >-2017 B.x |x <-2017C.x |-2020<x <0D.x |-2020<x <-2017【提分秘籍】若已知对于xf(x )+kf (x )>0(<0),构造g (x )=x k∙f (x )分析问题;【变式演练】3(江西省赣州市八校协作体2020-2021学年高三联考数学(理)试题)已知定义在R 上的奇函数f (x ),其导函数为f (x ),当x ≥0时,恒有x3f (x )+f (x )>0.则不等式x 3f (x )-(1+2x )3f (1+2x )<0的解集为().A.{x |-3<x <-1} B.x -1<x <-13C.{x |x <-3或x >-1}D.{x |x <-1或x >-13}4(山西省忻州市岢岚县中学2020-2021学年高三4月数学(理)试题)设函数f x 是定义在(-∞,0)上的可导函数,其导函数为f 'x ,且有2f x +xf 'x >x 2则不等式x +2019 2f x +2019 -4f -2 <0的解集为()A.(-2019,-2017)B. (-2021,-2019)C.(-2019,-2018)D.(-2020,-2019)5(安徽省黄山市屯溪第一中学2020-2021学年高三数学试题)已知函数f (x )是定义在R 上的奇函数,其导函数为f x ,若对任意的正实数x ,都有x f x +2f (x )>0恒成立,且f 2 =1,则使x 2f (x )<2成立的实数x 的集合为()A.-∞,-2 ∪2,+∞B.-2,2C.-∞,2D.2,+∞【题型三】幂函数与f (x )商型【典例分析】1(2022届湖南省衡阳市高三上学期期末考试数学试卷)函数f x 在定义域0,+∞ 内恒满足:①f x >0,②2f x <xf x <3f x ,其中f x 为f x 的导函数,则() A.14<f 1 f 2<12 B.116<f 1 f 2<18 C.13<f 1 f 2<12 D.18<f 1 f 2<142(黑龙江省哈尔滨市第三中学2021-2022学年高三第一次阶段性测试数学试题)已知偶函数f x 的导函数为f x ,且满足f 2 =0,当x >0时,xf x >2f x ,使得f x >0的x 的取值范围为【提分秘籍】对于x ∙f (x )-kf (x )>0(<0),构造g (x )=f (x )x k【变式演练】3(河南省郑州市示范性高中2020-2021学年高三阶段性考试(三)数学(理)试题)已知函数f x 的导函数为f x ,若f x <x ,f x <2,f x -x 对x ∈0,+∞ 恒成立,则下列个等式中,一定成立的是()A.f 2 3+12<f 1 <f 2 2 B.f 2 4+12<f 1 <f 2 2C.3f 2 8<f 1 <f 2 3+12D.f 2 4+12<f 1 <3f 2 84(江西省上高二中2021届高三上学期第四次月考数学试题)已知定义在R 上的偶函数f x ,其导函数为f x ,若y ,f -2 =1,则不等式f x x 2<14的解集是()A.-2,2B.-∞,-2 ∪2,+∞C.-2,0 ∪0,2D.-∞,0 ∪0,25设f x 是偶函数f x x ≠0 的导函数,当x ∈0,+∞ 时,y ,则不等式4f x +2019 -x +2019 2f -2 <0的解集为()A.-∞,-2021B.-2021,-2019 ∪-2019,-2017C.-2021,-2017D.-∞,-2019 ∪-2019,-2017【题型四】指数函数与f (x )积型【典例分析】1(【全国百强校】广东省阳春市第一中学2022届高三第九次月考数学(理)试题)已知函数f (x )(x ∈R )的导函数为f (x ),若2f (x )+f (x )≥2,且f (0)=8,则不等式f (x )-7e -2x >1的解集为()A.(-∞,0)B.(0,+∞)C.(-∞,-1)∪(0,+∞)D.(1,+∞)2(广东省普宁市华美实验学校2020-2021学年高三第一次月考数学试题)已知f x 是R上可导的图象不间断的偶函数,导函数为f x ,且当x>0时,满足f x +2xf x >0,则不等式e1-2x f x-1> f-x的解集为()A.12,+∞B.-∞,12C.-∞,0D.0,+∞【提分秘籍】对于f (x)+kf(x)>0(<0),构造g(x)=e kx∙f(x)【变式演练】3(2020届河南省八市重点高中联盟领军考试高三11月数学(理)试题)已知定义在R上的函数f x 的导函数为f x ,若f1 =1,ln f x +f x +1>0,则不等式f x ≥e1-x的解集为()A.-∞,1B.-∞,eC.1,+∞D.e,+∞4已知函数f x 的导函数为f x ,且对任意的实数x都有f x =e-x2x+5 2-f x (e是自然对数的底数),且f0 =1,若关于x的不等式f x -m<0的解集中恰有唯一一个整数,则实数m的取值范围是()A.-e2,0B.-e2,0C.-3e4,0D.-3e4,92e【题型五】指数函数与f(x)商型【典例分析】1定义在(-2,2)上的函数f(x)的导函数为f x ,满足:f x +e4x f-x=0,f1 =e2,且当x>0时,f (x)>2f(x),则不等式e2x f(2-x)<e4的解集为()A.(1,4)B.(-2,1)C.(1,+∞)D.(0,1)2已知定义在R上的函数f(x)的导函数为f'(x),且满足f'(x)-f(x)>0,f(2021)=e2021,则不等式f1 e ln x<e x的解集为()A.e2021,+∞B.0,e2021C.e2021e,+∞D.0,e2021e【提分秘籍】对于f (x)-kf(x)>0(<0),构造g(x)=f(x) e kx【变式演练】3(天一大联考高三毕业班阶段性测试(四)理科数学)定义在R上的函数f x 的导函数为f x ,若f x <2f x ,则不等式e4f-x>e-8x f3x+2的解集是()A.-12,+∞B.-∞,12C.-12,1D.-1,124已知定义在R上的函数f(x)的导函数为f (x),且满足f (x)-f(x)>0,f(2021)=e2021,则不等式f1 3ln x<3x的解集为()A.(e6063,+∞)B.(0,e2021)C.(e2021,+∞)D.(0,e6063)5(贵州省凯里市第三中学2022届高三上学期第二次月考数学(理)试题)已知函数f(x)是定义域为R,f (x)是f(x)的导函数,满足f (x)<f(x),且f(1)=4,则关于不等式f(x)-4e x-1>0的解集为()A.(-∞,1)B.1e ,1C.1e,eD.1e,+∞【题型六】正弦函数与f(x)型【典例分析】1(【衡水金卷】2021年普通高等学校招生全国统一考试高三模拟研卷卷四数学试题)已知定义在区间0,π2上的函数f x ,f x 为其导函数,且f x sin x-f x cos x>0恒成立,则()A.fπ2>2fπ6 B.3fπ4 >2fπ3C.3fπ6<fπ3 D.f1 <2fπ6 sin12(【市级联考】广西玉林市2018-2019学年高三上学期考试数学试题)已知f'(x)为函数y=f(x)的导函数,当x x∈0,π2是斜率为k的直线的倾斜角时,若不等式f(x)-f'(x)⋅k<0恒成立,则()A.{x22-m ln x2-2mx2=0x22-ln x2-m=0B.f(1)sin1>2fπ6C.f(x)=x2+6x-10D.3fπ6-fπ3 >0【提分秘籍】对于sin x∙f (x)+cos x∙f(x)>0(<0),构造g(x)=f(x)∙sin x对于sin x∙f (x)-cos x∙f(x)>0(<0),构造g(x)=f(x) sin x【变式演练】3(贵州省遵义航天高级中学222届高三第五次模拟考试数学试题)已知定义在0,π2上的函数,f(x)为其导函数,且f(x)sin x<f (x)cos x恒成立,则()A.f π2 >2f π6B.3f π4>2f π3 C.3f π6 <f π3 D.f (1)<2f π6 sin14已知奇函数f x 的导函数为f x ,且f x 在0,π2上恒有f (x )cos x -f (x )sin x <0成立,则下列不等式成立的()A.2f π6>f π4 B.f -π3 <3f -π6 C.3f -π4 <2f -π3D.22f π3 <3f π4 5(广东省七校联合体2021届高三下学期第三次联考(5月)数学试题)设f x 是定义在-π2,0 ∪0,π2 上的奇函数,其导函数为f x ,当x ∈0,π2 时,f x -f x cos xsin x<0,则不等式f x <233f π3sin x 的解集为()A.-π3,0 ∪0,π3 B.-π3,0 ∪π3,π2C.-π2,-π3 ∪π3,π2D.-π2,-π3 ∪0,π3【题型七】余弦函数与f (x )型【典例分析】1(2023春·新疆克孜勒苏·高三模拟)已知函数y =f x 对于任意的x ∈-π2,π2满足f x cos x +f x sin x >0(其中fx 是函数f x 的导函数),则下列不等式成立的是()A.f 0 >2f π4 B.2f -π3 >f -π4 C.2f π3 >f π4D.f 0 >2f π3 2(2023·全国·高三专题练习)定义在0,π2上的函数f x ,已知f x 是它的导函数,且恒有cos x ⋅f x +sin x ⋅f x <0成立,则有()A.3x -y -1=0B.3f π6>f π3C.f π6>3f π3D.2f π6<3f π4【提分秘籍】对于cos x ∙f (x )-sin x ∙f (x )>0(<0),构造g (x )=f (x )∙cos x ,对于cos x ∙f (x )+sin x ∙f (x )>0(<0),构造g (x )=f (x )cos x【变式演练】3(四川省成都市第七中学2022-2023学年高三上学期10月阶段考试理科数学试题)已知偶函数f (x )是定义在[-1,1]上的可导函数,当x ∈[-1,0)时,f (x )cos x +f (x )sin x >0,若cos (a +1)f (a )≥f (a +1)cos a ,则实数a 的取值范围为()A.[-2,-1]B.-1,-12C.-12,0D.-12,+∞ 4(四川省南充高级中学2021-2022学年高三考试数学试题)已知偶函数f (x )的定义域为-π2,π2,其导函数为f '(x ),当0<x <π2时,有f (x )cos x +f (x )sin x <0成立,则关于x 的不等式f (x )<2f π3 cos x 的解集为()A.0,π3B.π3,π2C.-π3,0 ∪0,π3D.-π2,-π3 ∪π3,π2【题型八】对数与f (x )型【典例分析】1已知函数f ′(x )是奇函数f (x )(x ∈R )的导函数,且满足x >0时,ln xf (x )+1xf (x )<0,则(x -2019)f (x )>0的解集为()A.(-1,0)∪(1,2019)B.(-2019,-1)∪(1,2019)C.(0,2019)D.(-1,1)2(【全国百强校】重庆市巴蜀中学20-20学年高三下考试理科数学试题)定义在0,+∞ 上的函数f x 满足x ⋅f 'x ⋅ln x +f x >0(其中f 'x 为f x 的导函数),则下列各式成立的是()A.ef e>π-f 1π>1 B.ef e<π-f 1π<1 C.ef e>1>π-f 1πD.ef e<1<π-f 1π【提分秘籍】对于f (x )ln x +f (x )x>0(<0),构造g x =ln x ∙f (x )【变式演练】3(江西省新余市第四中学2023届高三上学期第一次段考数学试题)已知定义在[e ,+∞)上的函数f (x )满足f (x )+x ln xf ′(x )<0且f (2018)=0,其中f ′(x )是函数f x 的导函数,e 是自然对数的底数,则不等式f (x )>0的解集为()A.[e ,2018)B.[2018,+∞)C.(e ,+∞)D.[e ,e +1)4(山东省招远一中2019届高三上学期第二次月考数学试题)定义在(0,+∞)上的函数f (x )满足xf '(x )ln x +f (x )>0(其中f '(x )为f (x )的导函数),若a >1>b >0,则下列各式成立的是()A.af (a )>bf (b )>1 B.af (a )<bf (b )<1 C.af (a )<1<bf (b )D.af (a )>1>bf (b )5(2023重庆渝中·高三重庆巴蜀中学校考阶段练习)已知函数f x 是奇函数f x x ∈R 的导函数,且满足x >0时,ln x ⋅f x +1x f x <0,则不等式x -985 f x >0的解集为()A.985,+∞B.-985,985C.-985,0D.0,985【题型九】一元二次(一次)与f (x )线性【典例分析】1(2021届云南省昆明第一中学高中新课标高三第三次双基检测数学试题)函数y =f (x )的定义域为R ,其导函数为f (x ),∀x ∈R ,有f (x )+f (-x )-2x 2=0在(0,+∞)上f (x )>2x ,若f (4-t )-f (t )≥16-8t ,则实数t 的取值范围为()A.[-2,2]B.[2,+∞)C.[0,+∞)D.(-∞,2]2(2020届黑龙江省实验中学高三上学期期末考试数学(理)试题)设函数f x 在R 上存在导函数f x ,∀x ∈R ,有f x -f -x =x 3,在0,+∞ 上有2f x -3x 2>0,若f m -2 -f m ≥-3m 2+6m -4,则实数m 的取值范围为()A.-1,1B.-∞,1C.1,+∞D.-∞,-1 ∪1,+∞【提分秘籍】二次构造:f (x )×÷r (x )±g (x ),其中r (x )=x n,e nx,sin x ,cos x 等【变式演练】3(江苏省盐城中学2020-2021学年高三上学期第二次阶段性质量检测数学试题)已知定义在R 上的函数f (x )的导函数为f (x ),且对任意x ∈R 都有f (x )>2,f (1)=3,则不等式f (x )-2x -1>0的解集为()A.(-∞,1)B.(1,+∞)C.(0,+∞)D.(-∞,0)4(吉林省蛟河市第一中学校2021-2022学年高三下学期第三次测试数学试题)已知定义在R 上的可导函数f (x ),对于任意实数x 都有f (-x )=f (x )-2x 成立,且当x ∈(-∞,0]时,都有f '(x )<2x +1成立,若f (2m )<f (m -1)+3m (m +1),则实数m 的取值范围为()A.-1,13B.(-1,0)C.(-∞,-1)D.-13,+∞ 5(【市级联考】福建省龙岩市2021届高三第一学期期末教学质量检查数学试题)已知定义在R 上的可导函数f (x )、g (x )满足f (x )+f (-x )=6x 2+3,f (1)-g 1 =3,g (x )=f (x )-6x ,如果g (x )的最大值为M ,最小值为N ,则M +N =()A.-2B.2C.-3D.3【题型十】指数型线性【典例分析】1(安徽省阜阳市第三中学2021-2022学年高三上学期第二次调研考试数学试题)设函数f x 定义域为R ,其导函数为f x ,若f x +f x >1,f 0 =2,则不等式e x f x >e x +1的解集为()A.-∞,0 ∪0,+∞B.-∞,0C.2,+∞D.0,+∞2(黑龙江省哈尔滨市第六中学2020-2021学年高三3月阶段性测试数学试题)已知函数f x =e 2x -ax 2+bx -1,其中a ,b ∈R ,e 为自然对数底数,若(0,1],f x 是f x 的导函数,函数f x 在0,1 内有两个零点,则a 的取值范围是()A.2e 2-6,2e 2+2B.e 2,+∞C.-∞,2e 2+2D.e 2-3,e 2+1【提分秘籍】对于f (x )-f (x )>k (<0),构造g x =e x f x -k【变式演练】3(金科大联考2020-2021学年高三10月质量检测数学试题)设函数f (x )的定义域为R ,f (x )是其导函数,若f (x )+f (x )>-e -x f (x ),f 0 =1,则不等式f (x )>2e x +1的解集是()A.(0,+∞)B.(1,+∞)C.(-∞,0)D.(0,1)4(2023春·福建龙岩·高三联考)∀x ∈R ,f x -f x =-2x +1 e x ,f 0 =-3,则不等式f x >-5e x 的解集为()A.-2,1B.-2,-1C.-1,1D.-1,25(2023春·四川眉山·高三模拟)函数f x 的定义域是R ,f 1 =2,对任意x ∈R ,f x +f x >1,则不等式e x f (x )>e x +e 的解集为()A.x |x >1B.x |x <1C.{x |x <-1或0<x <1}D.{x |x <-1或x >1}【题型十一】对数型线性【典例分析】1(2023春·安徽合肥·高三合肥一中校考)已知函数f x 的定义域为0,+∞ ,其导函数为f x ,若xf x -1<0,f e =2,则关于x 的不等式f e x<x +1的解集为()A.0,1B.1,eC.1,+∞D.e ,+∞2(2022春·江西赣州·高三赣州市赣县第三中学校考阶段练习)定义在(0,+∞)的函数f (x )满足xf x -1<0,f 1 =0,则不等式f e x-x <0的解集为()A.(-∞,0)B.(-∞,1)C.(0,+∞)D.(1,+∞)【提分秘籍】y =ln (kx +b )与y =f (x )的加、减、乘、除各种结果逆向思维【变式演练】3(2023·全国·高三专题练习)若函数f x 满足:x -1 fx -f x =x +1x-2,f e =e -1,其中f x 为f x 的导函数,则函数y =f x 在区间1e,e的取值范围为()A.0,eB.0,1C.0,eD.0,1-1e4(2021年全国高中名校名师原创预测卷新高考数学(第八模拟))已知函数f (x )的定义域为R ,且f (x +2)是偶函数,f (x )>12x -1+ln (x -1)(f (x )为f (x )的导函数).若对任意的x ∈(0,+∞),不等式f -t 2+2t +1 ≥f 12 x-2 恒成立,则实数t 的取值范围是()A.[-2,4]B.(-∞,-2]∪[4,+∞)C.[-1,3]D.(-∞,-1]∪[3,+∞)【题型十二】综合构造【典例分析】1(河北省沧州市沧县中学2020-2021学年高三数学)已知定义在R 上的可导函数f (x )的导函数为f '(x ),对任意实数x 均有(1-x )f (x )+xf '(x )>0成立,且y =f (x +1)-e 是奇函数,不等式xf (x )-e x >0的解集是()A.1,+∞B.e ,+∞C.-∞,1D.-∞,e2(江西省吉安市重点高中2020-2021学年高三5月联考数学试题)已知函数f x 是定义域为0,+∞ ,fx 是函数f x 的导函数,若f 1 =e ,且xfx -1+x f x >0,则不等式f ln x <x ln x 的解集为()A.0,eB.e ,+∞C.1,eD.0,1【变式演练】3(2022·高三测试)已知定义在R 上的函数f (x )的导函数是f (x ),若f (x )+xf (x )-xf (x )>0对任意x ∈R 成立,f 1 =e .则不等式f (x )<e xx 的解集是()A.(1,+∞)B.(-1,0)∪(0,1)C.(-1,0)D.(0,1)4(2023·四川·校联考模拟预测)定义在0,+∞ 上的函数f x 的导函数为f x ,且x 2+1 f x <x -1x f x ,若θ∈0,π4 ,a =tan θ,b =sin θ+cos θ,则下列不等式一定成立的是()A.f 1 <f a B.f 1 >2bf b2+sin2θC.f 1 >f a sin2θD.f a 2+sin2θ <f b 1sin θ+1cos θ5(2023春·江西吉安·高三模拟)若定义在R 上的可导函数f (x )满足(x +3)f (x )+(x +2)f (x )<0,f (0)=1,则下列说法正确的是()A.f (-1)<2eB.f (1)<23eC.f (2)>12e 2D.f (3)>25e 3高考真题对点练一、单选题1(浙江·高考真题)设f x 是函数f x 的导函数,y =f x 的图象如图所示,则y =f x 的图象最有可能的是()A .B .C .D .2(江西·高考真题)已知函数y =xf (x )的图象如图所示(其中f (x )是函数f (x )的导函数),则下面四个图象中,y =f x 的图象大致是()A. B.C. D.3(陕西·高考真题)f x 是定义在(0,+∞)上的非负可导函数,且满足xf ′x +f x ≤0.对任意正数a ,b ,若a <b ,则必有()A.af b ≤bf aB.bf a ≤af bC.af a ≤f bD.bf b ≤f a4(湖南·高考真题)设f (x )、g (x )分别是定义在R 上的奇函数和偶函数,当x <0时,f (x )g (x )+f (x )g (x )>0.且g (-3)=0,则不等式f (x )g (x )<0的解集是()A.(-3,0)∪(3,+∞)B.(-3,0)∪(0,3)C.(-∞,-3)∪(3,+∞)D.(-∞,-3)∪(0,3)5(2015·福建·高考真题)若定义在R 上的函数f x 满足f 0 =-1,其导函数f x 满足f x >k >1,则下列结论中一定错误的是()A.f 1k<1kB.f 1k>1k -1C.f 1k -1<1k -1D.f 1k -1>kk -16(2013·辽宁·高考真题)设函数f x 满足x 2fx +2xf x =e x x ,f 2 =e 28,则x >0时,f x ()A.有极大值,无极小值B.有极小值,无极大值C.既有极大值又有极小值D.既无极大值也无极小值7(2015·全国·高考真题)设函数f '(x )是奇函数f (x )(x ∈R )的导函数,f (-1)=0,当x >0时,xf '(x )-f (x )<0,则使得f (x )>0成立的x 的取值范围是A.(-∞,-1)∪(0,1)B.(-1,0)∪(1,+∞)C.(-∞,-1)∪(-1,0)D.(0,1)∪(1,+∞)8(辽宁·高考真题)函数f x 的定义域为R ,f -1 =2,对任意x ∈R ,f x >2,则f x >2x +4的解集为()A.-1,1B.-1,+∞C.-∞,-1D.-∞,+∞最新模考真题一、单选题1(2023·西藏日喀则·统考一模)如图,已知函数f x 的图象在点P 2,f 2 处的切线为直线l ,则f 2 +f 2 =()A.-3B.-2C.2D.12(2023·陕西榆林·统考三模)定义在0,+∞ 上的函数f x ,g x 的导函数都存在,f x g x +f (x )g x =2x -1x ln x +x +1x2,则曲线y =f x g x -x 在x =1处的切线的斜率为()A.12 B.1 C.32D.23(2023·四川成都·统考模拟预测)已知定义在R 上的函数f x 的导函数为f x ,若f x <e x ,且f 2 =e 2+2,则不等式f ln x >x +2的解集是()A.0,2B.0,e 2C.e 2,+∞D.2,+∞4(2023·陕西咸阳·校考模拟预测)已知函数f x 是定义在R 上的可导函数,其导函数记为f x ,若对于任意实数x ,有f x >f x ,且f 0 =1,则不等式f x <e x 的解集为()A.-∞,0B.0,+∞C.-∞,e 4D.e 4,+∞5(2023·河南·校联考模拟预测)已知函数f x 的定义域为R ,f x 为函数f x 的导函数,当x ∈0,+∞ 时,sin2x -f x >0,且∀x ∈R ,f -x +f x -2sin 2x =0,则下列说法一定正确的是()A.f π3-f π6 >12 B.f π3-f π4 <14C.f π3 -f 3π4 <14 D.f π3 -f -3π4 >146(2023·黑龙江大庆·大庆实验中学校考模拟预测)已知函数f x 的定义域为0,+∞ ,f x 为函数f x 的导函数,若x 2f x +xf x =1,f 1 =0,则不等式f 2x -3 >0的解集为()A.0,2B.log 23,2C.log 23,+∞D.2,+∞7(2023·山东烟台·统考二模)已知函数f x 的定义域为R ,其导函数为f x ,且满足f x +f x =e -x ,f 0 =0,则不等式e 2x -1 f x <e -1e的解集为( ).A.-1,1eB.1e ,e C.-1,1 D.-1,e8(2023·安徽·校联考模拟预测)已知函数f x 、g x 是定义域为R 的可导函数,且∀x ∈R ,都有f x >0,g x >0,若f x 、g x 满足f x f x <g xg x ,则当x 1<x <x 2时下列选项一定成立的是()A.f x 2 g x 1 >f x 1 g x 2B.f x g x 1 >f x 1 g xC.f x 2 -g x 2 f x 1 -g x 1 <g x 2 g x 1 D.f x 2 g x 2 <f x 1 +f x 2g x 1 +g x 2二、多选题9(2022·重庆九龙坡·重庆市育才中学校考模拟预测)已知函数f (x )对于任意的x ∈0,π2都有f (x )cos x -f (x )sin x >0,则下列式子成立的是()A.3f π6>2f π4 B.2f π4<f π3 C.2f (0)<f π4 D.2f (0)>f π3 10(2020·山东泰安·校考模拟预测)定义在0,π2 上的函数f (x ),f x 是f (x )的导函数,且fx <-tan x ⋅f (x )恒成立,则() A.f π6>2f π4B.3f π6 >f π3C.f π6>3f π3D.2f π6>3f π411(2023·黑龙江·黑龙江实验中学校考三模)已知函数f x 在R 上可导,其导函数为f x ,若f x 满足:x -1 fx -f x >0,f 2-x =f x e 2-2x ,则下列判断不正确的是()A.f 1 <ef 0B.f 2 >e 2f 0C.f 3 >e 3f 0D.f 4 <e 4f 012(2023·辽宁锦州·校考一模)定义在R 上的函数f x 满足xf x -f x =1,则y =f x 的图象可能为()A. B.C. D.三、填空题13(2024·四川成都·石室中学校考模拟预测)已知函数f x 的定义域为-π2 ,π2,其导函数是f x .有f x cos x+f x sin x<0,则关于x的不等式f(x)>2fπ3cos x的解集为.14(2023·广东佛山·统考模拟预测)已知f x 是定义在R上的偶函数且f1 =2,若f x <f x ln2,则f x -2x+2>0的解集为.15(2023·广东广州·广州市从化区从化中学校考模拟预测)设函数y=f x 在R上存在导数y=f x ,对任意的x∈R,有f x -f-x=2sin x,且在0,+∞上f x >cos x.若fπ2-t-f t >cos t-sin t.则实数t的取值范围为.16(2023·山东·模拟预测)定义在0,π2上的可导函数f x 的值域为R,满足f x tan x≥2sin x-1f x ,若fπ6=1,则fπ3 的最小值为.。

2020-2021学年广东省深圳高级中学高一(上)期中数学试卷(附答案详解)

2020-2021学年广东省深圳高级中学高一(上)期中数学试卷(附答案详解)

2020-2021学年广东省深圳高级中学高一(上)期中数学试卷一、单选题(本大题共8小题,共40.0分)1. 已知集合A ={x ∈R|3x +2>0},B ={x ∈R|(x +1)(x −3)>0},则A ∩B =( )A. (−∞,−1)B. (−1,−23)C. ﹙−23,3﹚D. (3,+∞)2. 如果a <b <0,那么下列各式一定成立的是( )A. |a|<|b|B. a 2<b 2C. a 3<b 3D. 1a <1b3. 德国数学家秋利克在1837年时提出“如果对于x 的每一个值,y 总有一个完全确定的值与之对应,则y 是x 的函数,“这个定义较清楚地说明了函数的内涵,只要有一个法则,使得取值范围中的每一个值,有一个确定的y 和它对应就行了,不管这个对应的法则是公式、图象、表格还是其它形式.已知函数f(x)由如表给出,则f(f(2020))的值为( )A. 1B. 2C. 3D. 20184. 若命题“∃x 0∈R ,使得x 02+mx 0+2m −3<0”为假命题,则实数m 的取值范围是( )A. [2,6]B. [−6,−2]C. (2,6)D. (−6,−2)5. 设a =0.60.3,b =0.30.6,c =0.30.3,则a ,b ,c 的大小关系为( )A. b <a <cB. a <c <bC. b <c <aD. c <b <a6. 若实数a ,b 满足1a +4b =√ab ,则ab 的最小值为( )A. √2B. 2C. 2√2D. 47. 已知函数f(x)={2x ,x ≥2(x −1)2,x <2,若关于x 的方程f(x)=k 有三个不同的实根,则数k 的取值范围是( )A. (0,1)B. (1,2)C. (0,2)D. (1,3)8. 已知函数f(x)=2+x2+|x|,x ∈R ,则不等式f(x 2−2x)<f(2x −3)的解集为( )A. (1,2)B. (1,3)C. (0,2)D. (1,32]二、多选题(本大题共4小题,共20.0分)9.下列函数中,最小值是2的是()A. y=a2−2a+2a−1(a>1) B. y=√x2+2+1√x2+2C. y=x2+1x2D. y=x2+2x10.下列四个结论中正确的是()A. 命题“∃x0∈R,sinx0+cosx0<1”的否定是“∀x∈R,sinx+cosx≥1”B. 命题“至少有一个整数n,n2+1是4的倍数”是真命题C. “a>5且b>−5”是“a+b>0”的充要条件D. 当α<0时,幂函数y=xα在区间(0,+∞)上单调递减11.如图1是某条公共汽车线路收支差额y与乘客量x的图象(收支差额=车票收入−支出费用).由于目前本条线路亏损,公司有关人员将图1变为图2与图3,从而提出了扭亏为盈的两种建议.下面有4种说法中正确的是()A. 图2的建议是:减少支出,提高票价B. 图2的建议是:减少支出,票价不变C. 图3的建议是:减少支出,提高票价D. 图3的建议是:支出不变,提高票价12.对∀x∈R,[x]表示不超过x的最大整数.十八世纪,y=[x]被“数学王子”高斯采用,因此得名为高斯函数,人们更习惯称为“取整函数”,则下列命题中的真命题是()A. ∃x∈R,x≥[x]+1B. ∀x,y∈R,[x]+[y]≤[x+y]C. 函数y=x−[x](x∈R)的值域为[0,1)D. 若∃t∈R,使得[t3]=1,[t4]=2,[t5]=3…,[t n]=n−2同时成立,则正整数n的最大值是5三、单空题(本大题共4小题,共20.0分)13.已知函数f(x)=a x−2−4(a>0,a≠1)的图象恒过定点A,则A的坐标为.14.若函数f(x)=ax2+2ax+1在[1,2]上有最大值4,则a的值为.15.y=f(x)是定义域R上的单调递增函数,则y=f(3−x2)的单调递减区间为.16.对于函数f(x),若在定义域存在实数x,满足f(−x)=−f(x),则称f(x)为“局部奇函数”.若函数f(x)=4x−m⋅2x−3是定义在R上的“局部奇函数”,则实数m 的取值范围为.四、解答题(本大题共6小题,共70.0分)17.化简求值:(1)0.064−13−(−18)0+1634+0.2512(2)12lg25+lg2+(13)log32−log29×log32.18.设函数y=√−x2+7x−12的定义域为集合A,不等式1x−2≥1的解集为集合B.(1)求集合A∩B;(2)设p:x∈A,q:x>a,且p是q的充分不必要条件,求实数a的取值范围.19.已知函数f(x)=a x(a>0且a≠1)在区间[1,2]上的最大值与最小值的和为6.(1)求函数f(x)解析式;(2)求函数g(x)=f(2x)−8f(x)在[1,m](m>1)上的最小值.20.已知函数f(x)是R上的偶函数,当x≥0时,f(x)=x3.(1)求x<0时f(x)的解析式;(2)解关于x的不等式f(x+1)≥8f(x).21.为了研究某种药物,用小白鼠进行试验,发现药物在血液内的浓度与时间的关系因使用方式的不同而不同.若使用注射方式给药,则在注射后的3小时内,药物在白鼠血液内的浓度y1与时间t满足关系式:y1=4−at(0<a<43,a为常数),若使用口服方式给药,则药物在白鼠血液内的浓度y2与时间t满足关系式:y2={√t,0<t<13−2t,1≤t≤3,现对小白鼠同时进行注射和口服该种药物,且注射药物和口服药物的吸收与代谢互不干扰.(1)若a=1,求3小时内,该小白鼠何时血液中药物的浓度最高,并求出最大值?(2)若使小白鼠在用药后3小时内血液中的药物浓度不低于4,求正数a的取值范围.22. 定义在R 上的函数g(x)和二次函数ℎ(x)满足:g(x)+2g(−x)=e x +2e x −9,ℎ(−2)=ℎ(0)=1,ℎ(−3)=−2. (1)求g(x)和ℎ(x)的解析式;(2)若对于x 1,x 2∈[−1,1],均有ℎ(x 1)+ax 1+5≥g(x 2)+3−e 成立,求a 的取值范围;(3)设f(x)={g(x),x >0ℎ(x),x ≤0,在(2)的条件下,讨论方程f[f(x)]=a +5的解的个数.答案和解析1.【答案】D【解析】【分析】本题考查一元二次不等式的解法,交集及其运算,考查计算能力,属于基础题.先求出集合B和A,然后利用交集运算求解A∩B.【解答】解:因为B={x∈R|(x+1)(x−3)>0}={x|x<−1或x>3},},又集合A={x∈R|3x+2>0}={x|x>−23}∩{x|x<−1或x>3}={x|x>3},所以A∩B={x|x>−23故选:D.2.【答案】C【解析】【分析】本题考查了不等式的基本性质,属基础题.根据条件取特殊值a=−2,b=−1,即可排除ABD;由不等式的基本性质,即可判断C.【解答】解:由a<b<0,取a=−2,b=−1,则可排除ABD;由a<b<0,根据不等式的基本性质可知C成立.故选:C.3.【答案】C【解析】【分析】本题考查函数值的求法,是基础题,解题时要认真审题,注意函数性质的合理运用.先求出f(2020)=2018,从而f(f(2020))=f(2018),由此能求出结果.【解答】解:由题意知:f(2020)=2018,f(f(2020))=f(2018)=3.故选:C.4.【答案】A【解析】【分析】本题考查存在量词命题的真假,二次不等式恒成立,考查转化思想.先写出原命题的否定,再根据原命题为假,其否定一定为真,利用不等式对应的是二次函数,结合二次函数的图象与性质建立不等关系,即可求出实数m的取值范围.【解答】解:命题“∃x0∈R,使得x02+mx0+2m−3<0”的否定为:“∀x∈R,都有x2+mx+2m−3≥0”,由于命题“∃x0∈R,使得x02+mx0+2m−3<0”为假命题,则其否定为真命题,∴Δ=m2−4(2m−3)≤0,解得2≤m≤6.则实数m的取值范围是[2,6].故选:A.5.【答案】C【解析】【分析】本题主要考查了幂函数和指数函数的性质,是基础题.利用幂函数y=x0.3在(0,+∞)上单调递增,比较出a,c的大小,再利用指数函数y=0.3x 在R上单调递减,比较出b,c的大小,从而得到a,b,c的大小关系.【解答】解:∵幂函数y=x0.3在(0,+∞)上单调递增,且0.6>0.3,∴0.60.3>0.30.3,即a>c,∵指数函数y=0.3x在R上单调递减,且0.6>0.3,∴0.30.6<0.30.3,即b<c,∴b<c<a,故选:C.6.【答案】D【解析】【分析】本题考查了利用基本不等式求最值,属于基础题.由已知得a,b>0,利用√ab=1a +4b≥2√1a⋅4b即可得出ab≥4,验证等号成立的条件.【解答】解:实数a,b满足1a +4b=√ab,则a,b>0.∴√ab=1a +4b≥2√1a⋅4b,可得ab≥4,当且仅当1a =4b,a=1,b=4时取等号.则ab的最小值为4.故选:D.7.【答案】A【解析】【分析】本题考查函数零点与方程根的关系,考查数形结合思想,属于中档题.题目等价于函数y=f(x)的图象与直线y=k有3个交点,作出图象,数形结合即可【解答】解:作出函数f(x)的图象如图:若关于x 的方程f(x)=k 有三个不同的实根,即函数y =f(x)的图象与直线y =k 有三个交点,根据图象可知,k ∈(0,1). 故选:A .8.【答案】A【解析】 【分析】本题考查分段函数的性质以及应用,注意将函数解析式写出分段函数的形式,属于中档题.根据题意,将函数的解析式写出分段函数的形式,据此作出函数的大致图象,据此可得原不等式等价于{x 2−2x <0x 2−2x <2x −3,解可得x 的取值范围,即可得答案.【解答】解:根据题意,函数f(x)=2+x2+|x|={−4x−2−1,x <01,x ≥0,其图象大致为:若f(x 2−2x)<f(2x −3),则有{x 2−2x <0x 2−2x <2x −3,解可得:1<x <2,即不等式的解集为(1,2);故选:A.9.【答案】AC【解析】【分析】本题考查了基本不等式的应用,关键掌握应用基本不等式的基本条件,一正二定三相等,属于基础题.根据应用基本不等式的基本条件,分别判断即可求出.【解答】解:对于A:a−1>0,y=a2−2a+2a−1=(a−1)2+1a−1=(a−1)+1a+1≥2√(a−1)⋅1a−1=2,当且仅当a−1=1a−1,即a=2时取等号,故A正确;对于B:y=√x2+2√x2+2≥2,当且仅当√x2+2=√x2+2,即x2=−1时取等号,显然不成立,故B错误;对于C:y=x2+1x2≥2√x2⋅1x2=2,当且仅当x=±1时取等号,故C正确;对于D:当x<0时,无最小值,故D错误.故选:AC.10.【答案】AD【解析】【分析】本题考查命题的真假的判断,考查充要条件,命题的否定,幂函数的性质等知识的应用,是基本知识的考查.利用命题的否定判断A;令n=2k和n=2k+1,k∈Z分析n2+1是不是4的倍数判断B;根据充要条件判断C;由幂函数的性质判断D即可.【解答】解:命题“∃x0∈R,sinx0+cosx0<1”的否定是“∀x∈R,sinx+cosx≥1”,满足命题的否定形式,所以A正确;令n=2k,k∈Z,则n2+1=4k2+1不是4的倍数,令n=2k+1,k∈Z,则n2+1=4k2+4k+2不是4的倍数,所以“至少有一个整数n,n2+1是4的倍数”是假命题,所以B不正确;“a>5且b>−5”推出“a+b>0”成立,反之不成立,如a=5,b=−4,满足a+ b>0,但是不满足a>5且b>−5,所以“a>5且b>−5”是“a+b>0”的充要条件不成立,所以C不正确.当α<0时,幂函数y=xα在区间(0,+∞)上单调递减,满足幂函数的性质,所以D正确;故选:AD.11.【答案】BD【解析】【分析】本题考查了用函数图象说明两个量之间的变化情况,主要根据实际意义进行判断,考查了读图能力和数形结合思想.根据题意知图象反应了收支差额y与乘客量x的变化情况,即直线的斜率说明票价问题;当x=0的点说明公司的支出情况,再结合图象进行说明.【解答】解:根据题意和图(2)知,两直线平行即票价不变,直线向上平移说明当乘客量为0时,收入是0但是支出的变少了,即说明了此建议是减少支出而保持票价不变;由图(3)看出,当乘客量为0时,支出不变,但是直线的倾斜角变大,即相同的乘客量时收入变大,即票价提高了,即说明了此建议是提高票价而保持支出不变,故选:BD.12.【答案】BCD【解析】【分析】本题考查函数新定义,正确理解新定义是解题基础,由新定义把问题转化不等关系是解题关键.由新定义得[x]≤x <[x]+1,可得函数f(x)=x −[x]值域判断C ;根据题意,若n ≥6,则不存在t 同时满足1≤t <√23,√46≤t <√56,n ≤5时,存在t ∈[√35,√23)满足题意,判断D . 【解答】解:∀x ∈R ,x <[x]+1,故A 错误;由“取整函数”定义可得,∀x ,y ∈R ,[x]≤x ,[y]≤y ,由不等式的性质可得[x]+[y]≤x +y ,所以[x]+[y]≤[x +y],B 正确;由定义得[x]≤x <[x]+1,所以0≤x −[x]<1,所以函数f(x)=x −[x]的值域是[0,1),C 正确;若∃t ∈R ,使得[t 3]=1,[t 4]=2,[t 5]=3,…[t n ]=n −2同时成立,则1≤t <√23,√24≤t <√34,√35≤t <√45,√46≤t <√56,…√n −2n ≤t <√n −1n ,因为√46=√23,若n ≥6,则不存在t 同时满足1≤t <√23,√46≤t <√56,只有n ≤5时,存在t ∈[√35,√23)满足题意,故选:BCD .13.【答案】(2,−3)【解析】 【分析】本题主要考查指数函数的性质,利用a 0=1的性质是解决本题的关键.比较基础. 根据指数函数的性质,令指数为0进行求解即可求出定点坐标. 【解答】解:由x −2=0得x =2,此时f(2)=a 0−4=1−4=−3, 即函数f(x)的图象过定点A(2,−3), 故答案为:(2,−3)14.【答案】38【解析】 【分析】口向上和向下两种情况判定函数值在何时取最大值,并根据最大值为4,即可求出对应的实数a的值【解答】解:当a=0时,f(x)=1,不符合题意,舍去.当a≠0时,f(x)的对称轴方程为x=−1,(1)若a<0,则函数图象开口向下,函数在[1,2]递减,当x=1时,函数取得最大值4,即f(1)=a+2a+1=4,解得a=1(舍).(2)若a>0,函数图象开口向上,函数在[1,2]递增,当x=2时,函数取得最大值4,即f(2)=4a+4a+1=4,解得a=3,8,综上可知,a=38.故答案为:3815.【答案】[0,+∞)【解析】【分析】本题考查了复合函数的单调性问题,考查二次函数的性质,属于中档题.根据复合函数单调性“同增异减”的原则,问题转化为求y=3−x2的单调递减区间,求出即可.【解答】解:根据复合函数单调性“同增异减”的原则,因为y=f(x)是定义域R上的单调递增函数,要求y=f(3−x2)的单调递减区间,即求y=3−x2的单调递减区间,而函数y=3−x2在[0,+∞)单调递减,故y=f(3−x2)的单调递减区间是[0,+∞),故答案为:[0,+∞).16.【答案】[−2,+∞)【分析】本题考查函数与方程的关系,关键是理解“局部奇函数”的定义,属于拔高题.根据“局部奇函数“的定义便知,若函数f(x)是定义在R上的“局部奇函数”,只需方程(2x+2−x)2−m(2x+2−x)−8=0有解.可设2x+2−x=t(t≥2),从而得出需方程t2−mt−8=0在t≥2时有解,从而设g(t)=t2−mt−8,由二次函数的性质分析可得答案.【解答】解:根据题意,由“局部奇函数”的定义可知:若函数f(x)=4x−m⋅2x−3是定义在R上的“局部奇函数”,则方程f(−x)=−f(x)有解;即4−x−m⋅2−x−3=−(4x−m⋅2x−3)有解;变形可得4x+4−x−m(2x+2−x)−6=0,即(2x+2−x)2−m(2x+2−x)−8=0有解即可;设2x+2−x=t(t≥2),则方程等价为t2−mt−8=0在t≥2时有解;设g(t)=t2−mt−8=0,必有g(2)=4−2m−8=−2m−4≤0,解可得:m≥−2,即m的取值范围为[−2,+∞);故答案为:[−2,+∞).17.【答案】解:(1)0.064−13−(−18)0+1634+0.2512=0.43×(−13)−1+24×34+0.52×12=2.5−1+8+0.5=10;(2)12lg25+lg2+(13)log32−log29×log32=lg5+lg2+3−log32−2(log23×log32)=1+12−2=−12.【解析】本题考查了指数幂和对数的运算的性质,属于基础题.(1)根据指数幂的运算性质计算即可;(2)根据对数的运算性质计算即可.18.【答案】解:由题意得:−x2+7x−12≥0,解得:3≤x≤4,故A=[3,4],∵1x−2≥1,∴x−3x−2≤0,解得:2<x≤3,故B=(2,3],(1)A∩B={3};(2)设p:x∈A,q:x>a,且p是q的充分不必要条件,即[3,4]⫋(a,+∞),故a<3,故a的取值范围是(−∞,3).【解析】本题考查了一元二次不等式的求解,集合的交集运算,考查了充分必要条件,考查了推理能力与计算能力,属于基础题.(1)分别求出集合A,B,求出A∩B即可;(2)根据集合的包含关系求出a的范围即可.19.【答案】解:(1)函数f(x)=a x(a>0且a≠1)在区间[1,2]上的最大值与最小值之和为6,则a+a2=6,即a2+a−6=0,解得a=2或a=−3(舍),故a=2,∴f(x)=2x;(2)g(x)=f(2x)−8f(x)=22x−8⋅2x,令2x=t,则原函数化为ℎ(t)=t2−8t,t∈[2,2m],其对称轴方程为t=4,当2m≤4,即1<m≤2时,函数最小值为(2m)2−8⋅2m=4m−8⋅2m;当2m>4,即m>2时,函数的最小值为42−8×4=−16.∴g(x)=f(2x)−8f(x)在[1,m](m>1)上的最小值为g(x)min={4m−8⋅2m,1<m≤2−16,m>2.【解析】本题考查指数函数的解析式、单调性与最值,二次函数的性质,是中档题.(1)根据指数函数的性质建立方程a+a2=6,即可求a的值,进一步得到函数解析式;(2)求出函数g(x)=f(2x)−8f(x)的解析式,换元后对m分类,利用二次函数的性质求最值.20.【答案】解:(1)根据题意,设x <0,则−x >0,则f(−x)=(−x)3=−x 3,又由f(x)为偶函数,则f(x)=f(−x)=−x 3, 故x <0时f(x)的解析式为f(x)=−x 3; (2)根据题意,f(x)为偶函数,则f(x)=f(|x|), 所以8f(x)=8f(|x|)=8×|x|3=(2|x|)3=f(2|x|), 又由当x ≥0时,f(x)=x 3,在[0,+∞)上为增函数;则f(x +1)≥8f(x)⇔f(|x +1|)≥f(|2x|)⇒|x +1|≥|2x|, 变形可得:3x 2−2x −1≤0,解可得:−13≤x ≤1,即不等式的解集为[−13,1].【解析】本题考查函数的奇偶性的性质以及应用,涉及绝对值不等式的解法,属于中档题.(1)根据题意,设x <0,则−x >0,由函数的解析式可得f(−x)=(−x)3=−x 3,结合函数的奇偶性分析可得答案;(2)根据题意,由函数的奇偶性以及解析式分析可得原不等式等价于|x +1|≥|2x|,解可得x 的取值范围,即可得答案.21.【答案】解:(1)当a =1时,药物在白鼠血液内的浓度y 与时间t 的关系为:y =y 1+y 2={−t +√t +4,0<t <17−(t +2t),1≤t ≤3; ①当0<t <1时,y =−t +√t +4=−(√t −12)2+174,所以当t =14时,y max =174;②当1≤t ≤3时,∵t +2t ≥2√2,当且仅当t =√2时取等号, 所以y max =7−2√2(当且仅当t =√2时取到),因为174>7−2√2, 故当t =14时,y max =174.(2)由题意y ={−at +√t +4(0<t <1)7−(at +2t )(1≤t ≤3) ① −at +√t +4≥4 ⇒ −at +√t ≥0 ⇒ a ≤√t ,又0<t <1,得出a ≤1;令u =1t ,则a ≤−2u 2+3u,u ∈[13,1],可得(−2u 2+3u )min =79 所以a ≤79, 综上可得0<a ≤79, 故a 的取值范围为(0,79].【解析】本题考查学生的函数思想,考查学生分段函数的基本思路,用好分类讨论思想,注意二次函数最值问题,基本不等式在求解该题中作用.恒成立问题的处理方法.用好分离变量法.(1)建立血液中药物的浓度与时间t 的函数关系是解决本题的关键,要根据得出的函数关系式采取合适的办法解决该浓度的最值问题;二次函数要注意对称轴和区间的关系、还要注意基本不等式的运用;(2)分段求解关于实数a 的范围问题,注意分离变量法的应用.22.【答案】解:(1)∵g(x)+2g(−x)=e x +2e x −9,∴g(−x)+2g(x)=e −x +2e x −9, 由以上两式联立可解得,g(x)=e x −3; ∵ℎ(−2)=ℎ(0)=1,∴二次函数的对称轴为x =−1,故设二次函数ℎ(x)=a(x +1)2+k , 则{a +k =14a +k =−2,解得{a =−1k =2,∴ℎ(x)=−(x +1)2+2=−x 2−2x +1;(2)由(1)知,g(x)=e x −3,其在[−1,1]上为增函数,故g(x)max =g(1)=e −3,∴ℎ(x 1)+ax 1+5≥e −3+3−e =0对任意x 1∈[−1,1]都成立,即x 12+(2−a)x 1−6≤0对任意x ∈[−1,1]都成立,∴{1−(2−a)−6≤01+(2−a)−6≤0,解得−3≤a ≤7, 故实数的a 的取值范围为[−3,7];(3)f(x)={e x −3,x >0−x 2−2x +1,x ≤0,作函数f(x)的图象如下,令t=f(x),a∈[−3,7],则f(t)=a+5∈[2,12],①当a=−3时,f(t)=2,由图象可知,此时方程f(t)=2有两个解,设为t1=−1,t2=ln5∈(1,2),则f(x)=−1有2个解,f(x)=ln5有3个解,故共5个解;②当−3<a<e2−8时,f(t)=a+5∈(2,e2−3),由图象可知,此时方程f(t)=a+5有一个正实数解,设为t3=ln(a+8)∈(ln5,2),则f(x)=t3=ln(a+8)有3个解,故共3个解;③当a=e2−8时,f(t)=a+5=e2−3,由图象可知,此时方程f(t)=a+5有一个解t4=2,则f(x)=t4=2有2个解,故共2个解;④当e2−8<a≤7时,f(t)=a+5∈(e2−3,12],由图象可知,此时方程f(t)=a+5有一个解t5=ln(a+8)∈(2,ln15],则f(x)=t5有1个解,故共1个解.【解析】本题考查函数解析式的求法,考查不等式的恒成立问题及函数零点与方程解的关系,旨在考查数形结合及分类讨论思想,属于中档题.(1)运用构造方程组法可求g(x),运用待定系数法可求ℎ(x);(2)原问题等价于x12+(2−a)x1−6≤0对任意x1∈[−1,1]都成立,进而求得实数a的取值范围;(3)作出函数f(x)的图象,结合图象讨论即可.。

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2020-2021学年广东省实验中学高三(上)第一次段考数学试卷一、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选项中,只有一项是符合要求的.
1.(5分)已知A={y|y=log2x,x>1},,则A∩B=()A.[,+∞)B.(0,)
C.(0,+∞)D.(﹣∞,0)∪[,+∞)
2.(5分)“sin2α=”是“tanα=2”的()
A.充分不必要条件B.必要不充分条件
C.充要条件D.既不充分也不必要条件
3.(5分)为了规定工时定额,需要确定加工某种零件所需的时间,为此进行了5次试验,得到5组数据:(x1,y1),(x2,y2),(x3,y3),(x4,y4),(x5,y5),由最小二乘法求得回归直线方程为.若已知x1+x2+x3+x4+x5=250,则y1+y2+y3+y4+y5=()
A.75B.155.4C.375D.442
4.(5分)命题p:变量(x,y)满足约束条件,则z=的最小值为,命题q:直线x=2的倾斜角为,下列命题正确的是()
A.p∧q B.(¬p)∧(¬q)C.(¬p)∧q D.p∧(¬q)5.(5分)已知两个单位向量,若,则的夹角为()A.B.C.D.
6.(5分)已知α∈(0,),β∈(﹣,0),cos(α+)=,cos(﹣)=,则cos(α+)=()
A.B.﹣C.D.﹣
7.(5分)已知长方形的四个顶点:A(0,0),B(2,0),C(2,1),D(0,1).一质点从点A出发,沿与AB夹角为θ的方向射到BC上的点P1后,依次反射到CD、DA和AB 上的点P2、P3、P4(入射角等于反射角).设P4的坐标为(x4,0),若1<x4<2,则tanθ
的范围是()
A.B.C.D.
8.(5分)设n∈N*,函数,函数(x>0).若函数y=f(x)与函数y =g(x)的图象分别位于直线y=1的两侧,则n的取值集合为()
A.{1,2}B.{2,3}C.{1,3}D.{1,2,3}
二、选择题:本题共4小题,每小题5分,共20分,在每小题给出的四个选项中,只有多项符合题目要求.全选对的得5分,有选错得得0分,部分选对得得3分.
9.(5分)已知函数,x∈R,则()A.﹣2≤f(x)≤2
B.f(x)在区间(0,π)上只有1个零点
C.f(x)的最小正周期为π
D.x=为f(x)图象的一条对称轴
10.(5分)已知空间中不同直线m、n和不同平面α、β,下列命题中是真命题的是()A.若m、n互为异面直线,m∥α,n∥α,m∥β,n∥β,则α∥β
B.若m⊥n,m⊥α,n∥β,则α⊥β
C.若n⊥α,m∥α,则n⊥m
D.若α⊥β,m⊥α,n∥m,则n∥β
11.(5分)设公差不为0的等差数列{a n}的前n项和为S n,若S17=S18,则下列各式的值为0的是()
A.a17B.S35C.a17﹣a19D.S19﹣S16
12.(5分)1970年4月24日,我国发射了自己的第一颗人造地球卫星“东方红一号”,从此我国开始了人造卫星的新篇章.人造地球卫星绕地球运行遵循开普勒行星运动定律:卫星在以地球为焦点的椭圆轨道上绕地球运行时,其运行速度是变化的,速度的变化服从面积守恒规律,即卫星的向径(卫星与地球的连线)在相同的时间内扫过的面积相等.设椭圆的长轴长、焦距分别为2a,2c,下列结论正确的是()
A.卫星向径的取值范围是[a﹣c,a+c]
B.卫星在左半椭圆弧的运行时间大于其在右半椭圆弧的运行时间
C.卫星向径的最小值与最大值的比值越大,椭圆轨道越扁
D.卫星运行速度在近地点时最大,在远地点时最小
三、填空题:本题共4小题,每小题5分,共20分
13.(5分)已知等腰三角形的底边长为6,一腰长为12,则它的内切圆面积为.14.(5分)大学在高考录取时采取专业志愿优先的录取原则.一考生从某大学所给的8个专业中,选择3个作为自己的第一、二、三专业志愿,其中甲、乙两个专业不能同时兼报,则该考生有种不同的填报专业志愿的方法(用数字作答).
15.(5分)如图,在四棱锥S﹣ABCD中,SA⊥平面ABCD,底面ABCD是菱形,且∠DAB =60°,SA=AB=1,则异面直线SD与BC所成的角的余弦值为,点C到平面SAD的距离等于.
16.(5分)点P在双曲线(a>0,b>0)的右支上,其左、右焦点分别为F1、
F2,直线PF1与以坐标原点O为圆心、a为半径的圆相切于点A,线段PF1的垂直平分线恰好过点F2,则该双曲线的离心率为.
四、解答题:本题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤.17.(10分)设{a n}是公差大于零的等差数列,已知a1=3,a3=a22﹣27.(1)求{a n}的通项公式;
(2)设{b n}是以函数y=4sin2πx的最小正周期为首项,以2为公比的等比数列,求数列
{a n b n}的前n项和S n.
18.(12分)在①;②这两个条件中任选一个,补充到下面问题中,并进行作答.
在△ABC中,内角A,B,C的对边分别为a,b,c,,,______.(1)求角A,B,C的大小;
(2)求△ABC的周长和面积.
19.(12分)在多面体ABCC1A1B1中,四边形ABB1A1为菱形,∠B1BA=60°,平面ABB1A1⊥平面ABC,=,AC⊥BC,AB⊥B1C.
(1)若O是线段AB的中点,证明:平面ABC⊥平面B1OC;
(2)求二面角C1﹣AC﹣B的正弦值.
20.(12分)为了了解居民的家庭收入情况,某社区组织工作人员从该社区的居民中随机抽取了100户家庭进行问卷调查,经调查发现,这些家庭的月收入在3000元到10000元之间,根据统计数据作出:
(1)经统计发现,该社区居民的家庭月收入Z(单位:百元)近似地服从正态分布N(μ,196),其中μ近似为样本平均数.若Z落在区间(μ﹣2σ,μ+2σ)的左侧,则可认为该家庭属“收入较低家庭“,社区将联系该家庭,咨询收入过低的原因,并采取相应措施为该家庭提供创收途径.若该社区A家庭月收入为4100元,试判断A家庭是否属于“收入较低家庭”,并说明原因;
(2)将样本的频率视为总体的概率;
①从该社区所有家庭中随机抽取n户家庭,若这n户家庭月收入均低于8000元的概率不
小于50%,求n的最大值;
②在①的条件下,某生活超市赞助了该社区的这次调查活动,并为这次参与调在的家庭
制定了贈送购物卡的活动,贈送方式为:家庭月收入低于μ的获赠两次随机购物卡,家
庭月收入不低于μ的获赠一次随机购物卡;每次赠送的购物卡金额及对应的概概率分别为:
赠送购物卡金额(单位:元)100200300概率
则A家庭预期获得的购物卡金额为多少元?(结果保留整数)
21.(12分)已知曲线C上的动点M到y轴的距离比到点F(1,0)的距离小1,(Ⅰ)求曲线C的方程;
(Ⅱ)过F作弦PQ、RS,设PQ、RS的中点分别为A、B,若,求最小时,弦PQ、RS所在直线的方程;
(Ⅲ)是否存在一定点T,使得?若存在,求出P的坐标,若不存在,试说明理由.
22.(12分)已知函数f(x)=(2x2﹣4ax)lnx+x2.
(Ⅰ)当a=1时,求函数f(x)在[1,+∞)上的最小值;
(Ⅱ)若函数f(x)在[1,+∞)上的最小值为1,求实数a的取值范围;
(Ⅲ)若a>,讨论函数f(x)在[1,+∞)上的零点个数.。

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