课时规范训练(37)
2022届高三数学(理)一轮总复习课时规范训练:第五章 数列 5-4 Word版含答案

课时规范训练[A 级 基础演练]1.若数列{a n }的通项公式是a n =(-1)n·(3n -2),则a 1+a 2+…+a 10=( ) A .15 B .12 C .-12D .-15解析:选A.记b n =3n -2,则数列{b n }是以1为首项,3为公差的等差数列,所以a 1+a 2+…+a 9+a 10=(-b 1)+b 2+…+(-b 9)+b 10=(b 2-b 1)+(b 4-b 3)+…+(b 10-b 9)=5×3=15.故选A.2.(2021·河北承德模拟)等差数列{a n }的前n 项和为S n (n =1,2,3,…),当首项a 1和公差d 变化时,若a 5+a 8+a 11是一个定值,则下列各数中为定值的是( )A .S 17B .S 18C .S 15D .S 16解析:选C.由等差数列的性质得a 5+a 11=2a 8,所以a 5+a 8+a 11为定值,即a 8为定值.又由于S 15=15(a 1+a 15)2=15×2a 82=15a 8,所以S 15为定值.故选C.3.已知数列{a n }的通项公式是a n =n 2sin ⎝ ⎛⎭⎪⎫2n +12π,则a 1+a 2+a 3+…+a 2 016=( )A.2 015×2 0162B .2 016×2 0172C.2 015×2 0152D .2 016×2 0162解析:选B.a n =n 2sin ⎝ ⎛⎭⎪⎫2n +12π=⎩⎪⎨⎪⎧-n 2n2(n 为奇数),(n 为偶数),∴a 1+a 2+a 3+…+a 2 016=-12+22-32+42-…-2 0152+2 0162=(22-12)+(42-32)+…+(2 0162-2 0152)=1+2+3+4+…+2 016=2 016×2 0172.4.设S n 为等差数列{a n }的前n 项和,S 8=4a 3,a 7=-2,则a 9=( ) A .-6 B .-4 C .-2D .2解析:选A.由等差数列性质及前n 项和公式,得 S 8=8(a 1+a 8)2=4(a 3+a 6)=4a 3,所以a 6=0.又a 7=-2,所以公差d =-2,所以a 9=a 7+2d =-6.5.数列{a n }满足a n +1+(-1)na n =2n -1,则{a n }的前60项和为( )A .3 690B .3 660C .1 845D .1 830解析:选D.当n =2k 时,a 2k +1+a 2k =4k -1, 当n =2k -1时,a 2k -a 2k -1=4k -3, ∴a 2k +1+a 2k -1=2,∴a 2k +1+a 2k +3=2, ∴a 2k -1=a 2k +3, ∴a 1=a 5=…=a 61.∴a 1+a 2+a 3+…+a 60=(a 2+a 3)+(a 4+a 5)+…+(a 60+a 61)=3+7+11+…+(4×30-1)=30×(3+119)2=30×61=1 830.6.已知数列{a n }中,a 1=1,a n +1=(-1)n(a n +1),记S n 为{a n }的前n 项和,则S 2 017= . 解析:由a 1=1,a n +1=(-1)n(a n +1)可得该数列是周期为4的数列,且a 1=1,a 2=-2,a 3=-1,a 4=0,a 5=1,所以S 2 017=504(a 1+a 2+a 3+a 4)+a 2 017=504×(-2)+1=-1 007.答案:-1 0077.(2021·江西八所中学联考)在数列{a n }中,已知a 1=1,a n +1+(-1)na n =cos(n +1)π,记S n 为数列{a n }的前n 项和,则S 2 017= .解析:∵a n +1+(-1)na n =cos(n +1)π=(-1)n +1,∴当n =2k 时,a 2k +1+a 2k =-1,k ∈N *,∴S 2 017=a 1+(a 2+a 3)+…+(a 2 016+a 2 017)=1+(-1)×1 008=-1 007.答案:-1 0078.等差数列{}a n 的前n 项和为S n ,已知a 1=10,a 2为整数,且S n ≤S 4. (1)求{}a n 的通项公式; (2)设b n =1a n a n +1,求数列{}b n 的前n 项和T n .解:(1)由a 1=10,a 2为整数,知等差数列{}a n 的公差d 为整数. 又S n ≤S 4,故a 4≥0,a 5≤0,于是10+3d ≥0,10+4d ≤0. 解得-103≤d ≤-52.因此d =-3.数列{}a n 的通项公式为a n =13-3n .(2)b n =1(13-3n )(10-3n )=13⎝ ⎛⎭⎪⎫110-3n -113-3n .于是T n =b 1+b 2+…+b n=13⎣⎢⎡⎦⎥⎤⎝ ⎛⎭⎪⎫17-110+⎝ ⎛⎭⎪⎫14-17+…+⎝ ⎛⎭⎪⎫110-3n -113-3n =13⎝ ⎛⎭⎪⎫110-3n -110=n10(10-3n ).9.(2021·辽宁五校联考)已知等差数列{}a n ,公差d >0,前n 项和为S n ,S 3=6且满足a 3-a 1,2a 2,a 8成等比数列.(1)求{}a n 的通项公式;(2)设b n =1a n ·a n +2,求数列{}b n 的前n 项和T n .解:(1)由S 3=6,得a 2=2. ∵a 3-a 1,2a 2,a 8成等比数列,∴2d ·(2+6d )=42,解得,d =1或d =-43.∵d >0,∴d =1,∴数列{}a n 的通项公式为a n =n . (2)∵b n =1a n ·a n +2=1n (n +2),∴T n =11×3+12×4+13×5+…+1n (n +2)=12⎣⎢⎡⎦⎥⎤⎝ ⎛⎭⎪⎫1-13+⎝ ⎛⎭⎪⎫12-14+⎝ ⎛⎭⎪⎫13-15+…+⎝ ⎛⎭⎪⎫1n -1n +2 =12⎝ ⎛⎭⎪⎫32-1n +1-1n +2=3n 2+5n 4(n +1)(n +2). [B 级 力量突破]1.中国古代数学著作《算法统宗》中有这样一个问题:“三百七十八里关,初行健步不犯难,次日脚痛减一半,六朝才得到其关,要见次日行里数,请公认真算相还.”其意思为:有一个人走378里路,第一天健步行走,从其次天起脚痛每天走的路程为前一天的一半,走了6天后到达目的地,请问其次天走了( )A .192里B .96里C .48里D .24里解析:选B.由题意,知每天所走路程形成以a 1为首项,公比为12的等比数列,则a 1⎝ ⎛⎭⎪⎫1-1261-12=378,解得a 1=192,则a 2=96,即其次天走了96里.故选B.2.已知数列5,6,1,-5,…,该数列的特点是从其次项起,每一项都等于它的前后两项之和,则这个数列的前16项之和S 16等于( )A .5B .6C .7D .16解析:选C.依据题意这个数列的前7项分别为5,6,1,-5,-6,-1,5,6,发觉从第7项起,数字重复消灭,所以此数列为周期数列,且周期为6,前6项和为5+6+1+(-5)+(-6)+(-1)=0.又由于16=2×6+4,所以这个数列的前16项之和S 16=2×0+7=7.故选C. 3.数列{a n }的通项为a n =(-1)n(2n +1)sin n π2+1,前n 项和为S n ,则S 100= .解析:由a n =(-1)n(2n +1)sinn π2+1可得全部的偶数项为1,奇数项有以下规律:⎩⎪⎨⎪⎧a 1=-2,a 5=-10,a 9=-18,…⎩⎪⎨⎪⎧a 3=8,a 7=16,a 11=24,…所以a 1+a 5+…+a 97=25×(-2)+25×242×(-8)=-2 450,a 3+a 7+…+a 99=25×8+25×242×8=2 600,a 2+a 4+…+a 100=50×1=50 所以S 100=-2 450+2 600+50=200. 答案:2004.(2021·昆明调研)已知等差数列{}a n 中,a 2=4,a 4是a 2与a 8的等比中项. (1)求数列{}a n 的通项公式; (2)若a n +1≠a n ,求数列{}2n -1·a n 的前n 项和.解:(1)由a 2=4,且a 4是a 2,a 8的等比中项可得a 1+d =4,a 24=a 2a 8,即(4+2d )2=4(4+6d ),化简得d 2-2d =0, 则d =0或d =2,由于a 2=4,当d =0时,a n =4; 当d =2时,a 1=2,则a n =2n . (2)∵a n +1≠a n ,∴a n =2n ,则2n -1a n =2n -1·2n =2n ·n ,∵S n =21+2×22+3×23+…+(n -1)·2n -1+n ·2n,(*1)(*1)×2得,2S n =22+2×23+3×24+…+(n -1)·2n+n ·2n +1,(*2)(*1)-(*2)得,-S n =21+22+23+…+2n -n ·2n +1=2(1-2n)1-2-n ·2n +1,∴S n =(n -1)·2n +1+2.5.在等比数列{}a n 中,a 1>0,n ∈N *,且a 3-a 2=8,又a 1、a 5的等比中项为16. (1)求数列{}a n 的通项公式;(2)设b n =log 4a n ,数列{}b n 的前n 项和为S n ,是否存在正整数k ,使得1S 1+1S 2+1S 3+…+1S n<k 对任意n ∈N*恒成立?若存在,求出正整数k 的最小值;不存在,请说明理由.解:(1)设数列{}a n 的公比为q ,由题意可得a 3=16,∵a 3-a 2=8,则a 2=8,∴q =2.∴a n =2n +1.(2)∵b n =log 42n +1=n +12,∴S n =b 1+b 2+…+b n =n (n +3)4.∵1S n=4n (n +3)=43⎝ ⎛⎭⎪⎫1n -1n +3,∴1S 1+1S 2+1S 3+…+1S n=43⎝ ⎛⎭⎪⎫11-14+12-15+13-16+…+1n -1n +3=43⎝⎛⎭⎪⎫1+12+13-1n +1-1n +2-1n +3<43⎝ ⎛⎭⎪⎫1+12+13=229, ∴存在正整数k ,其最小值为3.。
高中数学试题含答案-课时规范练37 空间向量及其运算

课时规范练37 空间向量及其运算基础巩固组1.(2020江西南昌八一中学质检)已知向量a =(-2,x ,2),b =(2,1,2),c =(4,-2,1).若a ⊥(b -c ),则x 的值为( )A.-2B.2C.3D.-32.在下列条件中,使M 与A ,B ,C 一定共面的是( ) A.OM ⃗⃗⃗⃗⃗⃗ =OA ⃗⃗⃗⃗⃗ −OB ⃗⃗⃗⃗⃗ −OC⃗⃗⃗⃗⃗ B.OM ⃗⃗⃗⃗⃗⃗ =15OA ⃗⃗⃗⃗⃗ +13OB ⃗⃗⃗⃗⃗ +12OC ⃗⃗⃗⃗⃗ C.MA ⃗⃗⃗⃗⃗⃗ +MB ⃗⃗⃗⃗⃗⃗ +MC ⃗⃗⃗⃗⃗⃗ =0 D.OM ⃗⃗⃗⃗⃗⃗ +OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ +OC⃗⃗⃗⃗⃗ =0 3.(多选)给出下列命题,其中正确命题有( ) A.空间任意三个不共面的向量都可以作为一个基底B.已知向量a ∥b ,则a ,b 与任何向量都能构成空间的一个基底C.A ,B ,M ,N 是空间四点,若BA ⃗⃗⃗⃗⃗ ,BM ⃗⃗⃗⃗⃗⃗ ,BN ⃗⃗⃗⃗⃗⃗ 不能构成空间的一个基底,那么A ,B ,M ,N 共面D.已知向量{a ,b ,c }是空间的一个基底,若m =a +c ,则{a ,b ,m }也是空间的一个基底 4.下列向量与向量a =(1,-√2,1)共线的单位向量为 ( )A.(-12,-√22,-12)B.(-12,-√22,12)C.(-12,√22,-12) D.(12,√22,12) 5.(多选)已知点P 是△ABC 所在的平面外一点,若AB ⃗⃗⃗⃗⃗ =(-2,1,4),AP ⃗⃗⃗⃗⃗ =(1,-2,1),AC ⃗⃗⃗⃗⃗ =(4,2,0),则( ) A.AP ⊥AB B.AP ⊥BP C.BC=√53 D.AP ∥BC6.(2020四川三台中学实验学校高三月考)如图,设OA ⃗⃗⃗⃗⃗ =a ,OB ⃗⃗⃗⃗⃗ =b ,OC ⃗⃗⃗⃗⃗ =c ,若AN ⃗⃗⃗⃗⃗⃗ =NB ⃗⃗⃗⃗⃗⃗ ,BM ⃗⃗⃗⃗⃗⃗ =2MC ⃗⃗⃗⃗⃗⃗ ,则MN ⃗⃗⃗⃗⃗⃗⃗ =( ) A.12a +16b -23c B.-12a -16b +23c C.12a -16b -13cD.-12a +16b +13c7.若a =(2,-3,5),b =(-3,1,2),则|a -2b |=( ) A.7√2B.5√2C.3√10D.6√38.(多选)已知向量a =(1,-1,m ),b =(-2,m-1,2),则下列结论中正确的是( ) A.若|a |=2,则m=±√2 B.若a ⊥b ,则m=-1 C.不存在实数λ,使得a =λb D.若a ·b =-1,则a +b =(-1,-2,-2)9.已知a =(3,2λ-1,1),b =(μ+1,0,2μ).若a ⊥b ,则μ= ;若a ∥b ,则λ+μ= . 10.(2020上海七宝中学期末)在正方体ABCD-A 1B 1C 1D 1中,给出下面四个命题:①(A 1A ⃗⃗⃗⃗⃗⃗⃗ +A 1D 1⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ +A 1B 1⃗⃗⃗⃗⃗⃗⃗⃗⃗ )2=3(A 1A ⃗⃗⃗⃗⃗⃗⃗ )2;②AD 1⃗⃗⃗⃗⃗⃗⃗ 与A 1B ⃗⃗⃗⃗⃗⃗⃗ 夹角为120°;③A 1C ⃗⃗⃗⃗⃗⃗⃗ ·C 1D ⃗⃗⃗⃗⃗⃗⃗ =0;④正方体的体积是|AB ⃗⃗⃗⃗⃗ ·BC ⃗⃗⃗⃗⃗ ·CC 1⃗⃗⃗⃗⃗⃗⃗ |,则所有正确的命题的序号是 .11.如图,在长方体ABCD-A 1B 1C 1D 1中,O 为AC 的中点. (1)化简:A 1O ⃗⃗⃗⃗⃗⃗⃗⃗ −12AB ⃗⃗⃗⃗⃗ −12AD ⃗⃗⃗⃗⃗ ; (2)设E 是棱DD 1上的点,且DE ⃗⃗⃗⃗⃗ =23DD 1⃗⃗⃗⃗⃗⃗⃗⃗ ,若EO ⃗⃗⃗⃗⃗ =x AB ⃗⃗⃗⃗⃗ +y AD ⃗⃗⃗⃗⃗ +z AA 1⃗⃗⃗⃗⃗⃗⃗ ,试求实数x ,y ,z 的值.综合提升组12.已知向量{a ,b ,c }是空间向量的一个基底,向量{a +b ,a -b ,c }是空间向量的另外一个基底,若一向量p 在基底{a ,b ,c }下的坐标为(1,2,3),则向量p 在基底{a +b ,a -b ,c }下的坐标为( ) A.(12,32,3) B.(32,-12,3) C.(3,-12,32)D.(-12,32,3)13.已知空间直角坐标系O-xyz 中,OA ⃗⃗⃗⃗⃗ =(1,2,3),OB⃗⃗⃗⃗⃗ =(2,1,2),OP ⃗⃗⃗⃗⃗ =(1,1,2),点Q 在直线OP 上运动,则当QA ⃗⃗⃗⃗⃗ ·QB ⃗⃗⃗⃗⃗ 取得最小值时,点Q 的坐标为( ) A.(12,34,13)B.(12,32,34)C.(43,43,83)D.(43,43,73)14.(2020山东烟台高三期末)如图所示的平行六面体ABCD-A 1B 1C 1D 1中,已知AB=AA 1=AD ,∠BAD=∠DAA 1=60°,∠BAA 1=30°,N 为A 1D 1上一点,且A 1N=λA 1D 1.若BD ⊥AN ,则λ的值为 ;若M 为棱DD 1的中点,BM ∥平面AB 1N ,则λ的值为 .创新应用组15.如图,在四棱锥P-ABCD 中,PA ⊥底面ABCD ,AD ⊥DC ,AB ∥DC ,AD=DC=AP=2,AB=1,点E 为棱PC 的中点.(1)证明:BE ⊥PD ;(2)若F 为棱PC 上一点,满足BF ⊥AC ,求线段PF 的长.参考答案课时规范练37 空间向量及其运算1.A ∵b -c =(-2,3,1),∴a ·(b -c )=4+3x+2=0,解得x=-2.故选A .2.C M 与A ,B ,C 一定共面的充要条件是OM ⃗⃗⃗⃗⃗⃗ =x OA ⃗⃗⃗⃗⃗ +y OB ⃗⃗⃗⃗⃗ +z OC⃗⃗⃗⃗⃗ ,x+y+z=1, 对于A 选项,由于1-1-1=-1≠1,所以不能得出M ,A ,B ,C 共面; 对于B 选项,由于15+13+12≠1,所以不能得出M ,A ,B ,C 共面;对于C 选项,由于MA ⃗⃗⃗⃗⃗⃗ =-MB ⃗⃗⃗⃗⃗⃗ −MC ⃗⃗⃗⃗⃗⃗ ,则MA ⃗⃗⃗⃗⃗⃗ ,MB ⃗⃗⃗⃗⃗⃗ ,MC⃗⃗⃗⃗⃗⃗ 为共面向量,所以M ,A ,B ,C 共面; 对于D 选项,由OM ⃗⃗⃗⃗⃗⃗ +OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ +OC ⃗⃗⃗⃗⃗ =0,得OM ⃗⃗⃗⃗⃗⃗ =-OA ⃗⃗⃗⃗⃗ −OB ⃗⃗⃗⃗⃗ −OC ⃗⃗⃗⃗⃗ ,而-1-1-1=-3≠1,所以不能得出M ,A ,B ,C 共面.故选C .3.ACD 选项A,根据空间基底的概念,可得任意三个不共面的向量都可以作为一个空间基底,所以A 正确;选项B,根据空间基底的概念,可得B 不正确;选项C,由BA ⃗⃗⃗⃗⃗ ,BM ⃗⃗⃗⃗⃗⃗ ,BN ⃗⃗⃗⃗⃗⃗ 不能构成空间的一个基底,可得BA ⃗⃗⃗⃗⃗ ,BM ⃗⃗⃗⃗⃗⃗ ,BN ⃗⃗⃗⃗⃗⃗ 共面, 又由BA ⃗⃗⃗⃗⃗ ,BM ⃗⃗⃗⃗⃗⃗ ,BN⃗⃗⃗⃗⃗⃗ 过相同点B ,可得A ,B ,M ,N 四点共面,所以C 正确; 选项D,由{a ,b ,c }是空间的一个基底,则基向量a ,b 与向量m =a +c 一定不共面,所以可以构成空间另一个基底,所以D 正确.故选ACD . 4.C 由|a |=√1+2+1=2,∴与向量a 共线的单位向量为(12,-√22,12)或(-12,√22,-12).故选C .5.AC 因为AP⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ =0,故A 正确;BP ⃗⃗⃗⃗⃗ =AP ⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ =(3,-3,-3),AP ⃗⃗⃗⃗⃗ ·BP ⃗⃗⃗⃗⃗ =3+6-3=6≠0,故B 不正确;BC ⃗⃗⃗⃗⃗ =AC ⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ =(6,1,-4),|BC ⃗⃗⃗⃗⃗ |=√62+12+(-4)2=√53,故C 正确;AP ⃗⃗⃗⃗⃗ =(1,-2,1),BC ⃗⃗⃗⃗⃗ =(6,1,-4),则AP ⃗⃗⃗⃗⃗ 与BC ⃗⃗⃗⃗⃗ 不平行,故D 不正确.故选AC .6.A 由题可知,MN ⃗⃗⃗⃗⃗⃗⃗ =MB ⃗⃗⃗⃗⃗⃗ −NB ⃗⃗⃗⃗⃗⃗ =23CB ⃗⃗⃗⃗⃗ −12AB ⃗⃗⃗⃗⃗ =23(OB ⃗⃗⃗⃗⃗ −OC ⃗⃗⃗⃗⃗ )-12(OB ⃗⃗⃗⃗⃗ −OA ⃗⃗⃗⃗⃗ )=12OA ⃗⃗⃗⃗⃗ +16OB ⃗⃗⃗⃗⃗ −23OC ⃗⃗⃗⃗⃗ =12a +16b -23c ,故选A . 7.C ∵a =(2,-3,5),b =(-3,1,2),∴a -2b =(8,-5,1),∴|a -2b |=√82+(-5)2+12=3√10.故选C . 8.AC 对于A,由|a |=2,可得√12+(-1)2+m 2=2,解得m=±√2,故A 正确;对于B,由a ⊥b ,可得-2-m+1+2m=0,解得m=1,故B 错误;对于C,若存在实数λ,使得a =λb ,则{1=-2λ,-1=λ(m -1),m =2λ,显然λ无解,即不存在实数λ,使得a =λb ,故C 正确;对于D,若a ·b =-1,则-2-m+1+2m=-1,解得m=0,于是a +b =(-1,-2,2),故D 错误.故选AC . 9.-35710因为a ⊥b ,则a ·b =3(μ+1)+0+2μ=0,解得μ=-35.若a ∥b ,则a =m b ,即(3,2λ-1,1)=m (μ+1,0,2μ),故{3=m (μ+1),2λ-1=0,1=2mμ,解得{λ=12,μ=15.故λ+μ=710. 10.①②③设正方体的棱长为1.建立空间直角坐标系,如图,A 1A ⃗⃗⃗⃗⃗⃗⃗ =(0,0,1),A 1D 1⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ =(1,0,0),A 1B 1⃗⃗⃗⃗⃗⃗⃗⃗⃗ =(0,1,0),则A 1A ⃗⃗⃗⃗⃗⃗⃗ +A 1D 1⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ +A 1B 1⃗⃗⃗⃗⃗⃗⃗⃗⃗ =(1,1,1),故(A 1A ⃗⃗⃗⃗⃗⃗⃗ +A 1D 1⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ +A 1B 1⃗⃗⃗⃗⃗⃗⃗⃗⃗ )2=|A 1A ⃗⃗⃗⃗⃗⃗⃗ +A 1D 1⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ +A 1B 1⃗⃗⃗⃗⃗⃗⃗⃗⃗ |2=3,3(A 1A ⃗⃗⃗⃗⃗⃗⃗ )2=3|A 1A ⃗⃗⃗⃗⃗⃗⃗ |2=3.故①正确;AD 1⃗⃗⃗⃗⃗⃗⃗ =(1,0,-1),A 1B ⃗⃗⃗⃗⃗⃗⃗ =(0,1,1),设AD 1⃗⃗⃗⃗⃗⃗⃗ 与A 1B ⃗⃗⃗⃗⃗⃗⃗ 夹角为θ,所以cos θ=AD 1⃗⃗⃗⃗⃗⃗⃗⃗⃗ ·A 1B ⃗⃗⃗⃗⃗⃗⃗⃗⃗ |AD 1⃗⃗⃗⃗⃗⃗⃗⃗⃗ ||A 1B ⃗⃗⃗⃗⃗⃗⃗⃗⃗ |=√2×√2=-12, 因为0°≤θ≤180°,所以AD 1⃗⃗⃗⃗⃗⃗⃗ 与A 1B ⃗⃗⃗⃗⃗⃗⃗ 夹角为120°,故②正确; A 1C ⃗⃗⃗⃗⃗⃗⃗ =(1,1,1),C 1D ⃗⃗⃗⃗⃗⃗⃗ =(0,-1,1),A 1C ⃗⃗⃗⃗⃗⃗⃗ ·C 1D ⃗⃗⃗⃗⃗⃗⃗ =0-1+1=0,故③正确;正方体ABCD-A 1B 1C 1D 1的体积为|AB ⃗⃗⃗⃗⃗ ||A 1A ⃗⃗⃗⃗⃗⃗⃗ ||AD ⃗⃗⃗⃗⃗ |,但是|AB ⃗⃗⃗⃗⃗ ·BC ⃗⃗⃗⃗⃗ ·CC 1⃗⃗⃗⃗⃗⃗⃗ |=0,故④错误.11.解(1)∵AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ =AC ⃗⃗⃗⃗⃗ ,∴A 1O ⃗⃗⃗⃗⃗⃗⃗⃗ −12AB ⃗⃗⃗⃗⃗ −12AD ⃗⃗⃗⃗⃗ =A 1O ⃗⃗⃗⃗⃗⃗⃗⃗ −12(AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ )=A 1O ⃗⃗⃗⃗⃗⃗⃗⃗ −12AC ⃗⃗⃗⃗⃗ =A 1O⃗⃗⃗⃗⃗⃗⃗⃗ −AO ⃗⃗⃗⃗⃗ =A 1A ⃗⃗⃗⃗⃗⃗⃗ .(2)∵EO ⃗⃗⃗⃗⃗ =ED ⃗⃗⃗⃗⃗ +DO ⃗⃗⃗⃗⃗⃗ =23D 1D ⃗⃗⃗⃗⃗⃗⃗⃗ +12DB ⃗⃗⃗⃗⃗⃗ =23D 1D ⃗⃗⃗⃗⃗⃗⃗⃗ +12(DA ⃗⃗⃗⃗⃗ +AB ⃗⃗⃗⃗⃗ )=23A 1A ⃗⃗⃗⃗⃗⃗⃗ +12DA ⃗⃗⃗⃗⃗ +12AB⃗⃗⃗⃗⃗ =12AB ⃗⃗⃗⃗⃗ −12AD ⃗⃗⃗⃗⃗ −23AA 1⃗⃗⃗⃗⃗⃗⃗ , ∴x=12,y=-12,z=-23.12.B 设向量p 在基底{a +b ,a -b ,c }下的坐标为(x ,y ,z ),则p =a +2b +3c =x (a +b )+y (a -b )+z c =(x+y )a +(x-y )b +z c ,所以{x +y =1,x -y =2,z =3,解得{ x =32,y =-12,z =3,故p 在基底{a +b ,a -b ,c }下的坐标为(32,-12,3).故选B .13.C 设Q (x ,y ,z ),由点Q 在直线OP 上,可得存在实数λ使得OQ⃗⃗⃗⃗⃗⃗ =λOP ⃗⃗⃗⃗⃗ , 即(x ,y ,z )=λ(1,1,2),可得Q (λ,λ,2λ),所以QA ⃗⃗⃗⃗⃗ =(1-λ,2-λ,3-2λ),QB ⃗⃗⃗⃗⃗ =(2-λ,1-λ,2-2λ),则QA ⃗⃗⃗⃗⃗ ·QB ⃗⃗⃗⃗⃗ =(1-λ)(2-λ)+(2-λ)(1-λ)+(3-2λ)(2-2λ)=2(3λ2-8λ+5),根据二次函数的性质,可得当λ=43时,取得最小值-23,此时Q (43,43,83).故选C .14.√3-1 23(1)取空间中一个基底:AB ⃗⃗⃗⃗⃗ =a ,AD ⃗⃗⃗⃗⃗ =b ,AA 1⃗⃗⃗⃗⃗⃗⃗ =c ,设AB=AD=AA 1=1,因为BD ⊥AN ,所以BD ⃗⃗⃗⃗⃗⃗ ·AN ⃗⃗⃗⃗⃗⃗ =0,因为B D ⃗⃗⃗⃗⃗⃗ =AD ⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ =b -a ,AN ⃗⃗⃗⃗⃗⃗ =AA 1⃗⃗⃗⃗⃗⃗⃗ +A 1N ⃗⃗⃗⃗⃗⃗⃗⃗ =c +λb ,所以(b -a )·(c +λb )=0,所以12+λ-√32−λ2=0,所以λ=√3-1. (2)在AD 上取一点M 1使得A 1N=AM 1,连接M 1N ,M 1M ,M 1B ,因为A 1N ∥AM 1,且A 1N=AM 1,所以四边形AA 1NM 1是平行四边形,所以AA 1∥NM 1,AA 1=NM 1,又AA 1∥BB 1,AA 1=BB 1,所以BB 1∥NM 1,BB 1=NM 1,所以四边形BB 1NM 1是平行四边形,所以NB 1∥M 1B ,NB 1=M 1B ,又因为M 1B ⊄平面AB 1N ,NB 1⊂平面AB 1N ,所以M 1B ∥平面AB 1N ,又因为BM ∥平面AB 1N ,且BM ∩M 1B=B ,所以平面M 1MB ∥平面AB 1N ,所以MM 1∥平面AB 1N ,又因为平面AA 1D 1D ∩平面AB 1N=AN ,且MM 1⊂平面AA 1D 1D ,所以M 1M ∥AN ,所以△AA 1N ∽△MDM 1,所以A 1N DM 1=AA 1MD =λA 1D 1(1-λ)A 1D 1=2,所以λ=23. 15.(1)证明 ∵PA ⊥底面ABCD ,AD ⊥AB ,∴以A 为原点,AB 为x 轴,AD 为y 轴,AP 为z 轴,建立空间直角坐标系, 由题意B (1,0,0),P (0,0,2),C (2,2,0),E (1,1,1),D (0,2,0),BE ⃗⃗⃗⃗⃗ =(0,1,1),PD ⃗⃗⃗⃗⃗ =(0,2,-2), 则BE ⃗⃗⃗⃗⃗ ·PD ⃗⃗⃗⃗⃗ =0+2-2=0,∴BE ⃗⃗⃗⃗⃗ ⊥PD ⃗⃗⃗⃗⃗ ,即BE ⊥PD. (2)解 ∵BC⃗⃗⃗⃗⃗ =(1,2,0),CP ⃗⃗⃗⃗⃗ =(-2,-2,2),AC ⃗⃗⃗⃗⃗ =(2,2,0), 由点F 在棱PC 上,设CF⃗⃗⃗⃗⃗ =λCP ⃗⃗⃗⃗⃗ =(-2λ,-2λ,2λ),0≤λ≤1, ∴BF⃗⃗⃗⃗⃗ =BC ⃗⃗⃗⃗⃗ +CF ⃗⃗⃗⃗⃗ =(1-2λ,2-2λ,2λ), ∵BF ⊥AC ,∴BF ⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ =2(1-2λ)+2(2-2λ)=0,解得λ=34, ∴|PF ⃗⃗⃗⃗⃗ |=1-34|PC⃗⃗⃗⃗⃗ |=14×√4+4+4=√32,即线段PF 的长为√32.。
高考生物模拟试题及解析-课时规范练37 传统发酵技术和发酵工程

课时规范练37传统发酵技术和发酵工程一、基础练1.下列有关泡菜制作的实验叙述中,正确的是()A.泡菜腌制时间长短会影响亚硝酸盐含量,但温度和食盐的用量不影响其含量B.泡菜发酵过程中,会产生多种酸,其中主要是亚硝酸,还有少量的乳酸C.发酵初期,水槽内产生的气泡是酵母菌等利用氧气进行呼吸作用产生的CO2D.泡菜中乳酸的质量百分比为4%~8%时口味、品质最佳2.(2020江西都昌一中期中)传统发酵技术在实践生产生活中有广泛的应用。
下列有关叙述正确的是()A.家庭制作果酒、果醋和腐乳通常都不是纯种发酵B.果醋制作过程中发酵液pH逐渐降低,果酒制作过程中情况相反C.利用乳酸菌制作酸奶的过程中,先通气培养,后密封发酵D.参与腐乳发酵的微生物主要是毛霉,是一种能分泌蛋白酶的原核生物3.泡菜发酵利用的微生物主要是乳酸菌,而在发酵初期,水槽内经常有气泡产生,这些气泡产生的原因及成分分别是()A.乳酸菌是兼性厌氧型微生物,初期进行有氧呼吸产生CO2;气体为CO2B.发酵初期活动强烈的是酵母菌,其利用氧产生CO2;气体为CO2C.因腌制过程中的盐进入蔬菜使蔬菜体积缩小,气体被排出;气体为空气D.乳酸菌在发酵过程中产生了热量,使坛内温度升高,空气受热膨胀排出;气体为空气4.下图是草莓酒制作的步骤,下列叙述错误的是()A.步骤⑤为加入酵母菌液B.步骤②是对草莓进行严格的灭菌处理C.在草莓酒发酵后期放气间隔时间可延长D.发酵过程中发酵液密度和pH会逐渐减小5.下列评价果酒和果醋制作是否成功的方法,不合理的是()A.通过观察相关微生物的数量变化进行鉴定B.通过向果酒发酵液中加入酸性重铬酸钾试剂进行鉴定C.通过检测果醋发酵前后发酵液的pH变化进行鉴定D.通过检测果酒发酵后发酵液的温度变化进行鉴定6.(2020河北深州中学期末)我们的生活离不开微生物。
醋酸菌可酿醋,毛霉菌可制作腐乳,酵母菌可做馒头。
下列有关三种微生物的叙述,正确的是()A.三者细胞中都没有成形的细胞核B.三者都有线粒体,可进行有氧呼吸C.醋酸菌的遗传物质是裸露的环状,后两者的遗传物质与蛋白质结合成染色体D.三者合成蛋白质时,mRNA都通过核孔进入细胞质中发挥作用7.传统发酵技术和现代发酵工程都是利用各种微生物的代谢来实现相关产品的生产的,下列微生物在发酵生产产品过程中需要提供空气或氧气的是()①酵母菌②醋酸菌③乳酸菌④毛霉A.④B.①②④C.①②③D.②④8.下列关于发酵工程的说法,不正确的是()A.在菌种选育过程中,可以利用人工诱变、细胞工程和基因工程等手段B.青霉素是抗生素,青霉素的生产过程不需要严格灭菌C.谷氨酸棒状杆菌发酵可以得到谷氨酸D.发酵工程的产品可以采用过滤、沉淀、蒸馏、萃取等方法提取9.杨梅果实风味独特,酸甜适中,具有很高的营养和保健价值。
高一语文必修下课时规范训练(十七)

一、基础知识1.对下列各句中加点词的解释,不正确的一项是( )A .窃.以为与君实游处相好之日久 窃:私下 B .冀.君实或见恕也 冀:希望 C .重念蒙君实视遇..厚 视遇:看待 D .非特.朝廷士大夫而已 特:特别 解析:D 项,特:仅,只是。
答案:D2.下列各组句子中,加点词语的意义和用法相同的一组是( )A .⎩⎨⎧恩所加则思无因.喜以谬赏罚所及则思无因.怒而滥刑 B .⎩⎨⎧亏无为之大道哉.岂人主之子孙则必不善哉.C .⎩⎨⎧恩所.加则思无因喜以谬赏山峦为晴雪所.洗 D .⎩⎨⎧将有作则.思知止以安人此则.岳阳楼之大观也 解析:A 项,“因”都是“因为”的意思。
B 项,表疑问;表反问。
C 项,“所”字结构;与“为”连用,被。
D 项,连词,那么;动词,是。
答案:A3.下列各句中,加点词语的意义与现代汉语不同的一项是( )A .非特朝廷士大夫而已..B .人习于苟且..非一日 C .辟邪说..,难壬人,不为拒谏 D .重念蒙君实视遇厚,于反覆..不宜卤莽 解析:A 项,罢了;B 项,得过且过,没有长远打算;C 项,不正确的言论;D项,古义指书信往返,今义是重复的意思。
答案:D4.下列各句中,加点词语的用法与其他三项不同的一项是()A.将有作则思知止以安.人B.想谗邪则思正.身以黜恶C.乐.盘游则思三驱以为度D.总此十思,弘.兹九德解析:C项,意动用法,其余三项为使动用法。
答案:C二、语言文字运用阅读下面的文字,完成5~7题。
隋唐一统,特别是唐太宗平东突厥、平高昌,促进了丝绸之路贸易发展。
唐太宗对来自昭武九姓的使者说:“西突厥已降,商旅可行矣。
”于是,“诸胡大悦”(《新唐书·西域传下》)。
唐人文献和小说笔记里,商胡或胡商,是出现____________甚高的词语。
吐鲁番出土文书中,对于贸易物品的规格和价格管理,____________,就是为____________和外商进行边境贸易的需要而定,当地居民不可能有如此巨大的需求。
2019-2020学年数学人教A版选修1-1课时规范训练:3.2.3基本初等函数的导数公式及导数的运算法则(二)

91 9 3 =2x2-1+2x-2.
3 (2)y=x2+x-2+x-2cos x,
35 y′=2x-2x-2+(x-2)′cos x+x-2(cos x)′
35 =2x-2x-2-2x-3cos x-x-2sin x.
能力提升
9.已知函数 f(x)的导函数为 f′(x),满足关系式 f(x)=x2+3xf′(2)+ln x,则 f′(2)的值为( )
A.1
B.-1
2 C. 2
2 D.- 2
【答案】A
4.若过点(2,0)的直线与曲线 y=x3 和 y=ax2+7x-4 都相切,则 a 的值为( )
A.2 49
C.2 或-16
5 B.16
5 D.3 或16
【答案】C
5.(2017 年天津)已知 a∈R,设函数 f(x)=ax-ln x 的图象在点(1,f(1))处的切线为 l,则 l 在 y 轴上
第三章 3.2 3.2.3
基础练习
1.函数 y=(x-2)2 在 x=1 处的导数等于( )
A.-1
B.-2
C.-3
D.-4
【答案】B
2.设 f(x)=xln x,若 f′(x0)=2,则 x0 的值为( )
A.e2
B.e
ln 2 C. 2
D.ln 2
【答案】B
( ) ( ) π
π
3.(2019 年广东深圳期末)已知函数 f(x)=f′ 4 cos x+sin x,则 f 4 的值为( )
解:f′(x)=aex+(ax+b)ex=(ax+a+b)ex,
∴Error!∴a=b=1.g′(x)=-2x+c,
∴Error!∴c=2,d=1.
高中政治必修一课时规范训练综合练(3) Word版含解析

第三课综合练一、选择题1.下面四幅图中,在其他条件不变时,相对比较正确地反映了家庭消费与居民收入之间的关系的是( )解析:选C。
收入是消费的前提和基础,所以是正函数,但收入并不是影响消费的唯一因素,影响消费的因素除了收入之外还有物价、消费心理等,因此变化应该是平滑式的抛物线,答案为C项。
2.一般说来,当人均收入达到一定水平时,这个地方就会出现拆旧房盖新房的高潮。
改革开放以来,我国许多农民把草房改建成瓦房,又从瓦房搬进了楼房。
这一现象证明( )A.家庭消费是个人消费行为,与社会无关B.改变消费结构就能提高消费水平C.收入水平的提高将导致消费水平的提高D.消费为生产创造动力,生产是消费的目的答案:C3.近年来,我国高档商品的消费以年均两位数的速度增长。
这表明( )①我国居民发展资料和享受资料的消费逐渐增加②消费结构会随着经济的发展、收入的增加而不断变化③居民当前的可支配收入增多④居民消费已从发展资料消费向享受资料消费转变A.①②③B.②③④C.①②④D.①③④解析:选A。
本题可用排除法解答。
根据近年来高档商品的消费以年均两位数的速度增长就说居民消费已从发展资料消费向享受资料消费转变太断章取义了。
高档商品的增长是事实,但我国多数居民的消费还达不到转向享受资料消费的水平,故应排除④,选A 项。
4.张女士由于工作流动性较大,每到一地都是租房、租车,非常省心。
可每到一处都需要家具,而购置一套简单的家具至少也需要花去她3个月的工资,令其烦恼,于是她决定也租家具。
对此,下列说法正确的是( )A.张女士通过租赁消费提高了商品的使用价值B.张女士通过租赁消费提高了商品的价值量C.张女士对租到的物品暂时拥有其使用权D.张女士对租到的物品暂时拥有其所有权解析:选C。
租赁消费不会提高商品的使用价值和价值量,故A、B两项不选。
租赁消费只是暂时获得该商品的使用权,而不是所有权,故选C项,排除D项。
5.消费有不同的类型,我们最常见的有钱货两清的消费。
课时规范训练(八)

课时规范训练A基础巩固练1.(多选题)下列命题是全称量词命题的是()A.任意一个自然数都是正整数B.有的菱形是正方形C.梯形有两边平行D.∃x∈R,x2+1=0解析:AC选项A中的命题含有全称量词“任意”,是全称量词命题,选项C中,“梯形有两边平行”是全称量词命题.2.以下四个命题既是存在量词命题又是真命题的是()A.锐角三角形的内角是锐角或钝角B.至少有一个实数x,使x2≤0C.两个无理数的和必是无理数D.存在一个负数x,使1 x>2解析:B A中,锐角三角形的内角是锐角或钝角是全称量词命题;B中,当x=0时,x2=0,所以B既是存在量词命题又是真命题;C中,因为3+(-3)=0,所以C是假命题;D中对于任一个负数x,都有1x<0,所以D是假命题.3.下列命题中是全称量词命题并且是真命题的是()A.每个二次函数的图象都开口向上B.存在一条直线与已知直线不平行C.对任意实数a,b,若a-b≤0,则a≤bD.存在一个实数x,使x2-2x+1=0成立解析:C B,D是存在量词命题,故应排除;对于A,二次函数y=ax2+bx +c(a<0)的图象开口向下,也应排除,故应选C.4.下列命题中形式不同于其他三个的是()A.∀x∈Z,x2-9<x2B.∃x∈R,x2-2x+1≠0C.每一个正数的倒数都大于0D.∀x<2,x-3<0解析:B ACD均为全称量词命题,B为存在量词命题.5.(多选题)下列命题中是真命题的是()A.∀x∈R,2x2-3x+4≠0B.∀x∈{1,-1,0},2x+1>0C.∃x∈N,使x≤xD.不存在x∈N*,使x为29的约数解析:AC因为Δ=(-3)2-4×2×4<0,故A正确;若x=-1,则2x+1=-1<0,故B错误;取x=4∈N,有4≤4成立,故C正确;1,29都是29的约数,故D错误.故选AC.6.已知命题p:∃x∈R,x2+4x+a=0,若命题p是假命题,则实数a的取值范围是()A.0<a<4B.a>4C.a<0 D.a≥4解析:B因为p是假命题,所以方程x2+4x+a=0没有实数根,即Δ=16-4a<0,即a>4.7.写出命题“正方形是菱形”的另一种表述方法________.答案:所有正方形都是菱形(答案不唯一).8.给出下列命题(1)∀x∈R,x2>0;(2)∃x∈R,x+1≤0;(3)∃a∈∁R Q,b∈∁R Q,使得a+b∈Q.其中真命题的个数为________.解析:(1)当x=0时,x2=0,是假命题;(2)存在x=-2,使得x+1≤0,是真命题;(3)当a=2-2,b=3+2时,a+b=5,是真命题.答案:29.已知命题“∃-3≤x≤2,3a+x-2=0”为真命题,求实数a的取值范围.解:由3a+x-2=0,得3a-2=-x,∵-3≤x≤2,∴-2≤-x≤3,∴-2≤3a -2≤3,即0≤a ≤53, 故实数a 的取值范围是⎩⎨⎧⎭⎬⎫a ⎪⎪⎪0≤a ≤53. B 能力进阶练10.下列命题中真命题的个数是( )①∃x ∈R ,x ≤0; ②至少有一个整数,它既不是合数也不是质数;③∃x ∈{x |x 是无理数},x +5是无理数.A .0B .1C .2D .3解析:D ①∃x ∈R ,x ≤0,是真命题;②至少有一个整数,它既不是合数也不是质数,是真命题;③∃x ∈{x |x 是无理数},x +5是无理数,是真命题,例如:x =π.综上可得①②③都是真命题.11.(多选题)命题p :存在实数x ∈R ,使得数据1,2,3,x ,6的中位数为3.若命题p 为真命题,则实数x 的取值集合可以为( )A .{3,4,5}B .{x |x >3}C .{x |x ≥3}D .{x |2≤x ≤6}解析:ABC 根据中位数定义可知,只需x ≥3,则1,2,3,x ,6的中位数必为3,A ,B ,C 中的取值集合满足x ≥3,D 不满足.故选ABC .12.已知不等式x +3≥0的解集是A ,则使命题“∀a ∈M ,a ∉A ”为真命题的集合M 是( )A .{a |a ≥-3}B .{a |a >-3}C .{a |a ≤-3}D .{a |a <-3}解析:D 因为x +3≥0,所以A ={x |x ≥-3}.又因为对∀a ∈M ,都有a ∉A ,所以a <-3.故选D .13.若对任意x >3,x >a 恒成立,则a 的取值范围是________.解析:对于任意x >3,x >a 恒成立,即大于3的数恒大于a ,所以a ≤3.答案:{a |a ≤3}C 探索创新练14.设命题p :∃x ∈R ,x 2-2x +m -3=0,命题q :∀x ∈R ,x 2-2(m -5)x +m 2+19≠0.若p ,q 都为真命题,则实数m 的取值范围为________.解析:若命题p :∃x ∈R ,x 2-2x +m -3=0为真命题,则Δ=4-4(m -3)≥0,解得m ≤4;若命题q :∀x ∈R ,x 2-2(m -5)x +m 2+19≠0为真命题,即方程x 2-2(m -5)x +m 2+19=0无实数根.因此,Δ=4(m -5)2-4(m 2+19)<0,解得m >35.又p ,q 都为真命题,所以实数m的取值范围是{m |m ≤4}∩⎩⎨⎧⎭⎬⎫m ⎪⎪⎪m >35=⎩⎨⎧⎭⎬⎫m ⎪⎪⎪35<m ≤4. 答案:⎩⎨⎧⎭⎬⎫m ⎪⎪⎪35<m ≤4。
课时规范训练(一)

课时规范训练A基础巩固练1.(多选题)下列各组对象能构成集合的有()A.接近于1的所有正整数B.小于0的实数C.(2023,1)与(1,2023)D.某校高一(1)班的聪明学生解析:BC A中,接近于1的所有正整数标准不明确,故不能构成集合;B 中小于0是一个明确的标准,能构成集合;C中(2023,1)与(1,2023)是两个不同的点,是确定的,能构成集合;D中“某校高一(1)班的聪明学生”中“聪明”的标准不确定,因而不能构成一个集合.2.给出下列关系:①13∈R;②5∈Q;③-3∉Z;④-3∉N,其中正确的个数为()A.1B.2C.3D.4解析:B 13是实数,①正确;5是无理数,②错误;-3是整数,③错误;-3是无理数,④正确.故选B.3.下列说法中正确的是()A.某学校高一(8)班比较漂亮的女生能确定一个集合B.由1,32,64,⎪⎪⎪⎪⎪⎪-12,0.5组成的集合有5个元素C.将小于100的自然数按从小到大的顺序排列和按从大到小的顺序排列分别得到两个不同的集合D.方程x2+1=2x的解集中只有一个元素解析:D A是错误的,因为“漂亮”是个模糊的概念,因此不满足集合中元素的确定性;B是错误的,32=64,⎪⎪⎪⎪⎪⎪-12=0.5,根据互异性,得由1,32,64,⎪⎪⎪⎪⎪⎪-12,0.5组成的集合只有3个元素:1,32,0.5;C是错误的,根据集合中元素的无序性可知,小于100的自然数无论按什么顺序排列,构成的集合都是同一集合;D是正确的,方程x2+1=2x有两个相等的解,即x1=x2=1,其解集中只有一个元素,故D正确.4.下列各组中,集合P与Q表示同一个集合的是()A.P是由元素1,3,π构成的集合,Q是由元素π,1,|-3|构成的集合B.P是由π构成的集合,Q是由3.141 59构成的集合C.P是由2,3构成的集合,Q是由有序数对(2,3)构成的集合D.P是满足不等式-1≤x≤1的自然数构成的集合,Q是方程x2=1的解集解析:A由于A中P,Q的元素完全相同,所以P与Q表示同一个集合,而B,C,D中P,Q的元素不相同,所以P与Q不表示同一个集合.5.已知集合M是方程x2-x+m=0的解组成的集合,若2∈M,则下列判断正确的是()A.1∈M B.0∈MC.-1∈M D.-2∈M解析:C由2∈M知2为方程x2-x+m=0的一个解,所以22-2+m=0,解得m=-2.所以方程为x2-x-2=0,解得x1=-1,x2=2.故方程的另一根为-1.6.(多选题)集合A中含有三个元素2,4,6,若a∈A,且6-a∈A,那么a 可以为()A.2 B.-2C.4 D.6解析:AC若a=2,则6-2=4∈A;若a=4,则6-4=2∈A;若a=6,则6-6=0∉A.7.已知集合P中元素x满足:x∈N,且2<x<a,又集合P中恰有三个元素,则整数a=________.解析:因为x∈N,2<x<a,且集合P中恰有三个元素,易知a=6.答案:68.若由a ,b a ,1组成的集合A 与由a 2,a +b ,0组成的集合B 相等,则a 2023+b 2023的值为________.解析:由已知可得a ≠0,因为两集合相等,又1≠0,所以b a =0,所以b =0,所以a 2=1,即a =±1,又当a =1时,集合A 不满足集合中元素的互异性,舍去,所以a =-1. 所以a 2023+b 2023=-1.答案:-19.已知集合A 含有两个元素a -3和2a -1,a ∈R .(1)求实数a 的取值范围;(2)若a ∈A ,求实数a 的值.解:(1)如果a -3=2a -1,则a =-2.由于a -3,2a -1是集合A 含有的两个元素,所以实数a ≠-2.(2)因为a ∈A ,所以a =a -3或a =2a -1.当a =a -3时,0=-3,不成立;当a =2a -1时,a =1,此时A 中有两个元素-2,1,符合题意,综上所述,实数a 的值为1.B 能力进阶练10.若集合A 的元素y 满足y =x 2+1,集合B 的元素(x ,y )满足y =x 2+1(A ,B 中x ∈R ,y ∈R ),则下列选项中元素与集合的关系都正确的是( )A .2∈A 且2∈BB .(1,2)∈A 且(1,2)∈BC .2∈A 且(3,10)∈BD .(3,10)∈A 且2∈B解析:C 集合A 中的元素为y ,是数集,又y =x 2+1≥1,故2∈A ;集合B 中的元素为点(x ,y )且满足y =x 2+1,经验证,(3,10)∈B ,故选C .11.(多选题)由a 2,2-a ,4组成一个集合A ,且集合A 中含有3个元素,则实数a 的取值不可能是( )A .1B .-2C .-1D .2解析:ABD 由题意知a 2≠4,2-a ≠4,a 2≠2-a ,解得a ≠±2,且a ≠1,即a 的取值不可能是1,±2.12.已知集合A 含有两个元素1和2,集合B 表示方程x 2+ax +b =0的解组成的集合,且集合A 与集合B 相等,则a =________,b =________.解析:因为集合A 与集合B 相等,且1∈A ,2∈A ,所以1∈B ,2∈B ,即1,2是方程x 2+ax +b =0的两个实数根.所以⎩⎨⎧1+2=-a ,1×2=b ,所以⎩⎨⎧a =-3,b =2.答案:-3 213.已知集合A 的元素是a ,b ,2,集合B 的元素是2,b 2,2a ,若A =B ,求实数a ,b 的值.解:由已知A =B 得⎩⎨⎧a =2a ,b =b 2,① 或⎩⎨⎧a =b 2,b =2a ,② 解①得⎩⎨⎧a =0,b =0或⎩⎨⎧a =0,b =1.解②得⎩⎨⎧a =0,b =0或⎩⎪⎨⎪⎧a =14,b =12.又由集合元素的互异性,得⎩⎨⎧a =0,b =1,或⎩⎪⎨⎪⎧a =14,b =12.C 探索创新练14.已知a ∈A 且4-a ∈A ,a ∈N 且4-a ∈N ,则:(1)若A 中只有1个元素,则a =________;(2)若A有且只有2个元素,则集合A的个数是________.解析:因为a∈A且4-a∈A,a∈N且4-a∈N,若a=0,则4-a=4,此时A满足要求;若a=1,则4-a=3,此时A满足要求;若a=2,则4-a=2.此时A含1个元素.答案:(1)2(2)2。
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课时规范训练(三十七)Ⅰ.阅读理解A(2016·高考全国乙卷)I am Peter Hodes,a volunteer stem cell courier.Since March 2012,I've done 89 trips—of those,51 have been abroad.I have 42 hours to carry stem cells (干细胞) in my little box because I've got two ice packs and that's how long they last.In all,from the time the stem cells are harvested from a donor (捐献者) to the time they can be implanted in the patient,we've got 72 hours at most.So I am always conscious of time.I had one trip last year where I was caught by a hurricane in America.I picked up the stem cells in Providence,Rhode Island,and was meant to fly to Washington then back to London.But when I arrived at the check-in desk at Providence,the lady on the desk said:“Well,I'm really sorry,I've got some bad news for you—there are no flights from Washington.” So I took my box and put it on the desk and I said:“In this box are some stem cells that are urgently needed for a patient—please,please,you've got to get me back to the United Kingdom.”She just dropped everything.She arranged for a flight on a small plane to be held for me,re-routed(改道) me through Newark and got me back to the UK even earlier than originally scheduled.For this courier job, you're consciously aware that in that box you've got something that is potentially going to save somebody's life.1.Which of the following can replace the underlined word “courier” in Paragraph 1?A.provider B.delivery man C.collector D.medical doctor 2.Why does Peter have to complete his trip within 42 hours?A.He cannot stay away from his job too long.B.The donor can only wait for that long.C.The operation needs that much time.D.The ice won't last any longer.3.Which flight did the woman put Peter on first?A.To London. B.To Newark. C.To Providence. D.To Washington.BFor foreigners, entering a public toilet in China can be a horrifying experience. In the 1990s, a third of all complaints to tourism officials in Beijing concerned the design, and the bad smell of public toilets.China has battled smelly toilets for at least 2,000 years. In the Kingdom of Wei (220—266 A.D.), visitors to the palace bathrooms would find boxes to cover their nose and ward_off unpleasant smells. Both flushing (冲水) toilets and toilet paper were invented in China, however, only for the use of the emperor.Today Chinese scientists have claimed victory in their battle to improve public restrooms, introducing a bacterial spray that can almost get rid of all the bad smell. First, a set of six types of bacteria work to break down the smelly compounds and then a perfume made from orange skin lightly scents (使芳香) the air. The “smell-free toilet” study from the Chinese Academy of Science was declared the “ultimate” cure to an_“urgent”_national_issue.“Five scientists have worked on this from 2011 to the beginning of this year,”said Dr. Yan Zhiying, a bacteriologist with the academy's Chengdu Institute of Biology, adding that they had spent &140,000 on the project.“Some local government officials here visited a sewage(污水) plant and saw that the treatment technology had come from Japan. They wanted a home-grown solution so they asked us to work on it,” he added.“We obtained bacteria from all types of waste from humans, pigs, chicken and ducks, and we tested our compounds one by one,” he said.“The smells coming out of public toilets, or trash cans, are made up of more than 160 different compounds,”he explained, adding that their bacteria spray can change and absorb many of them.He boasted that the Chinese formula (配方), which costs only around £5 per liter, has no side effects and can be used to remove any bad smell.4.The underlined words “ward off” in Paragraph 2 can be replaced by ________.A.keep out B.take out C.give ont D.come out 5.The underlined phrase “an ‘urgent' national issue” refers to ________.A.the treatment of sewage B.a horrifying experienceC.the bad smell of public toilets D.the 160 smelly compounds 6.Which has not been mentioned as an advantage of the bacteria spray?A.Low price. B.No side effects.C.Removing bad smells. D.Beautiful design.7.Which can we infer after the spray is used in toilets in China?A.People needn't use flushing toilets.B.A perfume is given off in the air.C.Tourists' complaints will be reduced.D.Bacteria break down smelly compounds.CEveryone looks forward to progress, whether in one's personal life or in the general society. Progress indicates a person's ability to change the way he is living at the moment. Progress must lead to a better life and a better way of doing things. All these, however,remain true only in so far as people want to accept technology and move forward by finding new and more efficient ways of doing things.However, at the back of the minds of many people,especially those who miss the “good old days”,efficiency comes with a price. When communication becomes more efficient, people are able to contact one another no matter where they are and at whatever time they wish to. The click of a button allows people miles apart to talk or see each other without even leaving their homes.With the communication gadgets,_such as mobile phones and iPads, people often do not take the effort to visit one another personally. A personal visit carries with it the additional feature of having to be in the person's presence for as long as the visit lasts. We cannot unnecessarily excuse ourselves or turn the other person off.With efficiency also comes mass production. Such is the nature of factories and the success of industrialization today. Factories have improved efficiency. Unskillful tasks are left to machines and products are better made and produced with greater accuracy than any human hand could ever have done. However, with the improvements in efficiency also comes the loss of the personal touch when making these products. For example, many handicrafts (手工艺品) are now produced in a factory.Although this means that supply is better able to increase demand, now that the supply is quick and efficient, the demand might fall because mass production lowers the quality of the handicraft and it is difficult to find unique designs on each item.Nevertheless, we must not commit the mistake of analyzing progress only from one point of view. In fact,progress has allowed tradition to keep up. It is only with progress and the invention of new technology that many old products can be brought back to their old state. New technology is required for old products to stay old.It is people's attitude towards progress that causes the type of influence that technology has on society.Technology is flexible. There is no fixed way of making use of it. Everything depends on people's attitude. The worst effects of progress will fall on those who are unable to rethink their attitudes and views of society. When we accept progress and adapt it to suit our needs, a new “past” is creat ed.8.The underlined word “gadgets” in Para. 2 is closest in meaning to “_______”.A.tools B.messages C.barriers D.skills 9.Compared with home-made handicrafts, machine-made products _______.A.lack great accuracy B.lack the personal touchC.are of high value D.are quite welcome10.What can be learned about technology from Paragraph 4?A.It can destroy old traditions.B.It can lead to social progress.C.It can be used to correct mistakes.D.It can be used to preserve old products.11.What can be concluded from the last paragraph?A.Progress can suit the needs of daily life.B.People review the past with great regret.C.Technology should be introduced in a fixed way.D.People's attitude decides the use of technology.D(2017·河北衡水中学模拟)I have two kids, a boy and a girl. I don't worry about my girl;I'm sure she'll be well trained in mixed martial arts. I worry about my son. I'm pretty sure he's going to be feminine(女性化的). Sorry to say that, but let's face it. No dad wants his son to go feminine.At the rate we're moving in a couple of years you won't be able to tell the difference between boys and girls. Sound extreme? Think about this. In every moviewhere advanced time-traveling beings come to our planet there's one constant:You can't tell the male aliens from the female aliens!But that's the future. Kids today are soft and fat. People ask why. Is it junk food? No. Junk food has been around for fifty years. Is it video games? No. Video games have been around for thirty years.None of the kids playing them back in the day were terribly obese.We're all scratching our heads trying to figure out what we've introduced to society to ruin our kids. But it's not anything we've added that has ruined our kids. It's stuff we've gotten rid of.Take the gym rope for example. Remember that thing that stretched from the floor to the ceiling in your gym class that you could never climb?Most of the kids couldn't make it to the top. But that wasn't the point; the point was you had to try while some middle-aged guy who couldn't make it up a flight of stairs shouted at you. We should have put our son on that rope, and given him a head start. But we didn't want to shame the boy, so we took it down.Taking down the rope would be a good idea if there were no ropes in life. But they're everywhere. You just can't see them.They're in every goal unrealized and expectation not met. The point everyone missed about the rope is you weren't supposed to make it to the top. It was there to create a fire that burned in the oversized belly of every kid.12.What is the example of aliens used to show?A.Girls are growing like boys.B.Boys are going feminine.C.There are really time-traveling beings.D.Alien movies are popular with children.13.What causes kids today to be soft and fat?A.Junk food. B.Video games.C.Stuff introduced to society. D.Things removed from society. 14.How does the author find teenagers today?A.They have too much dignity. B.They are overprotected.C.They are less independent. D.They are thought too highly of. 15.What does the rope probably symbolize?A.Goals. B.Expectations. C.Competition. D.Assistance.Ⅱ.七选五Being energy and resource efficient and conscious of improving the health and well-being of the work environment will likewise reduce the costs of running one business.___1___Conserve your computer's energy.For many people with desk jobs the computer is necessary to preserve things done.___2___It is estimated that people waste over $1 billion on electricity every year just in computer use! To conserve energy from your computer use, turning off your computer when it's not in use may be a good choice.3____It is already bad enough that you have to worry about air pollution every time you walk outside, but it is likewise a big priority when you work inside. So, open your windows to increase airflow. If you can't open windows, be sure to take outdoor breaks throughout the day. And don't smoke in or near the office.Green your desk, cubicle (小房间), office ...4____Or, even better, buy plants for all of your neighbors. They will not only see this as one friendly gesture, but they'll likewise have cleaner air to breathe! Plants absorb indoor air pollution and increase the flow of oxygen, therefore winning a green decoration to complement your desk!Recycle.There are many items in your office that you can recycle.___5___You are capable of acquiring a few bins and posting recycling guidelines on them. Some of the items to be collected may include paper products,cardboard boxes, plastic bags, etc.A.Maintain healthy airflow.B.Yet this “efficiency” comes at a cost.C.It's time to “go green” at work, and here is how.D.Literally earn a plant and place it on or near your desk.E.It's natural to accept similar measures at work to save energy.F.If you do not have one recycling station at work, start one on your own!G.Setting the computer to the sleep mode when you're away temporarily makes good sense.。