【5月山东潍坊高三二模数学】2020年山东省潍坊市高考模拟(二模)数学试题含答案
高考数学模拟试题-第16讲 变化率与导数、导数的计算(原卷版)
第16讲 变化率与导数、导数的计算学校:___________姓名：___________班级：___________考号：___________【基础巩固】1．（2022·全国·高三专题练习）若函数()()e ln 1axf x x =++，()04f '=，则a =( )A ．0B ．1C ．2D ．32．（2022·全国·高三专题练习）已知函数()()221sin 1x xf x x ++=+，其导函数记为()f x '，则()()()()389389389389f f f f ''++---=（ ） A ．2B ．2-C ．3D ．3-3．（2022·全国·高三专题练习）下列函数求导运算正确的个数为（ ）①)(333log xxe '=；①)(21log ln 2x x '=；①)(x xe e '=；①1ln x x '⎛⎫=⎪ ⎭⎝；①)(1x x xe e '=+． A ．1 B ．2 C ．3 D ．44．（2022·湖南·长沙县第一中学模拟预测）函数()2ln 1sin y x x =++的图象在0x =处的切线对应的倾斜角为α，则sin2α=（ ） A ．310B ．±310 C ．35D ．±355．（2022·湖北·黄冈中学模拟预测）已知a ，b 为正实数，直线y x a =-与曲线ln()y x b =+相切，则14a b+的最小值为（ ） A ．8B ．9C ．10D ．136．（2022·湖北·襄阳五中模拟预测）过点()1,2P 作曲线C ：4y x=的两条切线，切点分别为A ，B ，则直线AB 的方程为（ ） A ．280x y +-= B ．240x y +-=C ．240x y +-=D ．240x y +--=7．（2022·辽宁·沈阳二中模拟预测）函数()y f x =的图像如图所示，下列不等关系正确的是（ ）A ．()()()()02332f f f f ''<<<-B ．()()()()02323f f f f ''<<-<C ．()()()()03322f f f f ''<<-<D ．()()()()03232f f f f ''<-<<8．（2022·重庆一中高三阶段练习）已知偶函数()f x ，当0x >时，()()212f x x f x '=-+，则()f x 的图象在点()()2,2f --处的切线的斜率为（ ） A ．3-B ．3C ．5-D ．59．（2022·江苏·南京外国语学校模拟预测）若两曲线y =x 2-1与y =a ln x -1存在公切线，则正实数a 的取值范围为（ ） A ．(]0,2eB ．(]0,eC ．[)2,e +∞D ．(],2e e10．（多选）（2022·江苏·高三专题练习）下列求导数运算正确的有（ ） A ．(sin )cos x x '= B ．211()x x'=C ．31(log )3ln x x'=D ．1(ln )x x'=11．（多选）（2022·湖南·长郡中学高三阶段练习）下列曲线在x =0处的切线的倾斜角为钝角的是（ ）A ．曲线2sin y x x =-B ．曲线2sin y x x =-C ．曲线()2e xy x =-D ．曲线11e x y x -=+12．（2022·福建省福州格致中学模拟预测）已知函数()()()()20e 01x f x f x f x '=+--，则函数()f x =___________.13．（2022·广东·模拟预测）已知2ln ()1xf x x=+，则曲线在(1,1)处的切线方程为________． 14．（2022·北京市第一六一中学模拟预测）写出一个同时具有下列性质①①①的函数f (x )=___________： ①1212()()()f x x f x f x =：①当()0,x ∞∈+时，()0f x '>； ①()f x '是偶函数．15．（2022·全国·高考真题）曲线ln ||y x =过坐标原点的两条切线的方程为____________，____________． 16．（2022·全国·高考真题）若曲线()e x y x a =+有两条过坐标原点的切线，则a 的取值范围是________________．17．（2022·山东威海·三模）已知曲线212:e ,:2(0)x C y x C y x x a a =+=-++>，若有且只有一条直线同时与1C ，2C 都相切，则=a ________．18．（2022·浙江·高三专题练习）已知函数ln y x x =. （1）求这个函数的导数；（2）求这个函数的图象在点1x =处的切线方程.19．（2022·全国·高考真题（文））已知函数32(),()f x x x g x x a =-=+，曲线()y f x =在点()()11,x f x 处的切线也是曲线()y g x =的切线． (1)若11x =-，求a ； (2)求a 的取值范围．【素养提升】1．（2022·湖北·模拟预测）若过点()(),0m n m <可作曲线3y x =-三条切线，则（ ） A ．30n m <<-B ．3n m >-C ．0n <D ．30n m <=-2．（2022·山东潍坊·三模）过点()()1,P m m ∈R 有n 条直线与函数()e x f x x =的图像相切，当n 取最大值时，m 的取值范围为（ ） A ．25e e m -<< B ．250e m -<< C ．10em -<<D ．e m <3．（多选）（2022·湖南·长沙市南雅中学高三阶段练习）已知函数()e xx f x =(e为自然对数的底数)，过点(,)a b 作曲线()f x 的切线.下列说法正确的是（ ）A ．当0a =时，若只能作两条切线，则24e b = B ．当0a =，24e b >时，则可作三条切线 C ．当02a <<时，可作三条切线，则24e e a a a b -<< D ．当2a =，0b >时，有且只有两条切线4．（2022·安徽·合肥市第八中学模拟预测（理））若曲线31:C y x =与曲线2:e (0)xC y a a =>存在2条公共切线，则a 的值是_________．5．（2022·河北邯郸·二模）已知点P 为曲线ln exy =上的动点，O 为坐标原点．当OP 最小时，直线OP 恰好与曲线ln y a x =相切，则实数a ＝___．。
高考数学复习历年考点题型专题讲解38--- 数列中的通项公式(解析版)
高考数学复习历年考点题型专题讲解38数列中的通项公式一、题型精讲 解题方法与技巧 题型一、由S a n n 与的关系求通项公式例1、（2020届山东省烟台市高三上期末）已知数列{}n a 的前n 项和n S 满足()()21n n S n a n N *=+∈，且12a =.求数列{}n a 的通项公式；【解析】因为2(1)n n S n a =+,n *∈N , 所以112(2)n n S n a ++=+,n *∈N ,两式相减得112(2)(1)n n n a n a n a ++=+-+, 整理得1(1)n n na n a +=+,即11n n a a n n +=+,n *∈N ,所以n a n ⎧⎫⎨⎬⎩⎭为常数列, 所以121n a a n ==,所以2n a n =例2、（2020届山东省枣庄、滕州市高三上期末）已知等比数列{}n a 满足1,a 2,a 31a a -成等差数列，且134a a a =；等差数列{}n b 的前n 项和2(1)log 2nn n a S +=.求：（1）,n a n b ；【解析】设{}n a 的公比为q. 因为1,a 2,a 31a a -成等差数列， 所以()21312a a a a =+-，即232a a =.因为20a ≠，所以322a q a ==. 因为134a a a =，所以4132a a q a ===. 因此112n n n a a q-==.由题意，2(1)log 2n n n a S +=(1)2n n+=.所以111b S ==，1223b b S +==，从而22b =.所以{}n b 的公差21211d b b =-=-=.所以1(1)1(1)1n b b n d n n =+-=+-⋅=.例3、（2020届山东省德州市高三上期末）已知数列{}n a 的前n 项和为n S ，且0n a >，242n n n S a a =+.求数列{}n a 的通项公式；【解析】当1n =时，211142a a a =+，整理得2112a a =，10a >，解得12a =；当2n ≥时，242n n n S a a =+①，可得211142n n n S a a ---=+②，①－②得2211422n n n n n a a a a a --=-+-，即()()221120n n n n a a a a ----+=，化简得()()1120n n n n a a a a --+--=，因为0n a >，10n n a a -∴+>，所以12n n a a --=，从而{}n a 是以2为首项，公差为2的等差数列，所以()2212n a n n =+-=； 题型二、由a a n n 与1+的递推关系求通项公式例3、【2019年高考全国II 卷理数】已知数列{a n }和{b n }满足a 1=1，b 1=0，1434n n n a a b +-=+，1434n n n b b a +-=-.（1）证明：{a n +b n }是等比数列，{a n –b n }是等差数列； （2）求{a n }和{b n }的通项公式.【解析】（1）由题设得114()2()n n n n a b a b +++=+，即111()2n n n n a b a b +++=+． 又因为a 1+b 1=l ，所以{}n n a b +是首项为1，公比为12的等比数列． 由题设得114()4()8n n n n a b a b ++-=-+，即112n n n n a b a b ++-=-+．又因为a 1–b 1=l ，所以{}n n a b -是首项为1，公差为2的等差数列． （2）由（1）知，112n n n a b -+=，21nn a b n -=-． 所以111[()()]222n n n n n na ab a b n =++-=+-， 111[()()]222n n n n n n b a b a b n =+--=-+．例4、（2020届山东省德州市高三上期末）对于数列{}n a ，规定{}n a ∆为数列{}n a 的一阶差分数列，其中()*1n n n a a a n +∆=-∈N ，对自然数()2k k ≥，规定{}kn a ∆为数列{}n a 的k 阶差分数列，其中111k k k n n n a a a --+∆=∆-∆.若11a =，且()2*12n n n n a a a n +∆-∆+=-∈N ，则数列{}n a 的通项公式为（）A ．212n n a n -=⨯ B ．12n n a n -=⨯C ．()212n n a n -=+⨯D ．()1212n n a n -=-⨯【答案】B【解析】根据题中定义可得()()2*1112n n n n n n n n a a a a a a n a +++∆-∆+=∆-∆-∆+=-∈N ，即()1122nn n n n n n n a a a a a a a ++-∆=--=-=-，即122nn n a a +=+，等式两边同时除以12n +，得111222n n n n a a ++=+，111222n n n n a a ++∴-=且1122a =， 所以，数列2n n a ⎧⎫⎨⎬⎩⎭是以12为首项，以12为公差的等差数列，()1112222n n a n n ∴=+-=， 因此，12n n a n -=⋅.故选：B.例5、【2019年高考天津卷理数】设{}n a 是等差数列，{}n b 是等比数列．已知1122334,622,24a b b a b a ===-=+，．（Ⅰ）求{}n a 和{}n b 的通项公式；（Ⅱ）设数列{}n c 满足111,22,2,1,,k k n kk c n c b n +=⎧<<=⎨=⎩其中*k ∈N ． （i ）求数列(){}221nna c -的通项公式；【解析】（1）设等差数列{}n a 的公差为d ，等比数列{}n b 的公比为q ．依题意得2662,6124,q d q d =+⎧⎨=+⎩解得3,2,d q =⎧⎨=⎩故14(1)331,6232n n n n a n n b -=+-⨯=+=⨯=⨯． 所以，{}n a 的通项公式为{}31,n n a n b =+的通项公式为32n n b =⨯． （2）（i ）()()()()22211321321941nnnn n n n a c a b -=-=⨯+⨯-=⨯-．所以，数列(){}221nna c -的通项公式为()221941nnn a c -=⨯-．题型三、新定义题型中通项公式的求法例6、【2020年高考江苏】已知数列{}()n a n ∈*N 的首项a 1=1，前n 项和为S n ．设λ与k 是常数，若对一切正整数n ，均有11111kk k n nn S S a λ++-=成立，则称此数列为“λ～k ”数列．（1）若等差数列{}n a 是“λ～1”数列，求λ的值； （2）若数列{}n a”数列，且0n a >，求数列{}n a 的通项公式； 【解析】（1）因为等差数列{}n a 是“λ～1”数列，则11n n n S S a λ++-=，即11n n a a λ++=，也即1(1)0n a λ+-=，此式对一切正整数n 均成立．若1λ≠，则10n a +=恒成立，故320a a -=，而211a a -=-，这与{}n a 是等差数列矛盾．所以1λ=．（此时，任意首项为1的等差数列都是“1～1”数列）（2）因为数列*{}()n a n ∈N是“”数列，==．因为0n a >，所以10n n S S +>>1-=．n b，则1n b -=221(1)(1)(1)3n n n b b b -=->． 解得2n b =，即2=，也即14n nS S +=， 所以数列{}n S 是公比为4的等比数列．因为111S a ==，所以14n n S -=．则21(1),34(2).n n n a n -=⎧=⎨⨯≥⎩例7、【2019年高考北京卷理数】已知数列{a n }，从中选取第i 1项、第i 2项、…、第i m 项（i 1<i 2<…<i m ），若12mi i i a a a <<⋅⋅⋅<，则称新数列12mi i i a a a ⋅⋅⋅，，，为{a n }的长度为m 的递增子列．规定：数列{a n }的任意一项都是{a n }的长度为1的递增子列．（1）写出数列1，8，3，7，5，6，9的一个长度为4的递增子列； （2）已知数列{a n }的长度为p 的递增子列的末项的最小值为0m a ，长度为q的递增子列的末项的最小值为0n a ．若p <q ，求证：0m a <0n a ；（3）设无穷数列{a n }的各项均为正整数，且任意两项均不相等．若{a n }的长度为s 的递增子列末项的最小值为2s –1，且长度为s 末项为2s –1的递增子列恰有2s -1个（s =1，2，…），求数列{a n }的通项公式．【解析】（1）1，3，5，6.（答案不唯一）（2）设长度为q 末项为0n a 的一个递增子列为1210,,,,q r r r n a a a a -.由p <q ，得1pq r r n a a a -≤<.因为{}n a 的长度为p 的递增子列末项的最小值为0m a ，又12,,,pr r r a a a 是{}n a 的长度为p 的递增子列，所以0pm r a a ≤.所以0m n a a <·（3）由题设知，所有正奇数都是{}n a 中的项.先证明：若2m 是{}n a 中的项，则2m 必排在2m −1之前（m 为正整数）.假设2m 排在2m −1之后.设121,,,,21m p p p a a a m --是数列{}n a 的长度为m 末项为2m −1的递增子列，则121,,,,21,2m p p p a a a m m --是数列{}n a 的长度为m +1末项为2m 的递增子列.与已知矛盾.再证明：所有正偶数都是{}n a 中的项.假设存在正偶数不是{}n a 中的项，设不在{}n a 中的最小的正偶数为2m . 因为2k 排在2k −1之前（k =1，2，…，m −1），所以2k 和21k -不可能在{}n a 的同一个递增子列中.又{}n a 中不超过2m +1的数为1，2，…，2m −2，2m −1，2m +1，所以{}n a 的长度为m +1且末项为2m +1的递增子列个数至多为1(1)22221122m m m --⨯⨯⨯⨯⨯⨯=<个.与已知矛盾.最后证明：2m 排在2m −3之后（m ≥2为整数）.假设存在2m （m ≥2），使得2m 排在2m −3之前，则{}n a 的长度为m +1且末项为2m +l 的递增子列的个数小于2m .与已知矛盾.综上，数列{}n a 只可能为2，1，4，3，…，2m −3，2m ，2m −1，…. 经验证，数列2，1，4，3，…，2m −3，2m ，2m −1，…符合条件.所以1,1,n n n a n n +⎧=⎨-⎩为奇数，为偶数.二、达标训练1、（2020届浙江省温州市高三4月二模）已知数列{}n a 满足：12125 1,6n n n a a a a n -≤⎧=⎨-⎩()*n N ∈)若正整数()5k k ≥使得2221212k k a a a a a a ++⋯+=⋯成立，则k =（）A ．16B ．17C ．18D ．19【答案】B【解析】当6n ≥时，()1211111n n n n n a a a a a a a +--==+-，即211n n n a a a +=-+，且631a =.故()()()222677687116......55n n n n a a a a a a a a a n a a n +++++=-+-++-+-=-+-，2221211...161k k k a a a a k a +++++=+-=+，故17k =.故选：B .2、（2020届山东省潍坊市高三上学期统考）设数列{}n a 的前n 项和为n S ，且21n S n n =-+，在正项等比数列{}n b 中22b a =，45b a =．求{}n a 和{}n b 的通项公式；【解析】当1n =时，111a S ==， 当2n ≥时，1n n n a S S -=- =22(1)[(1)(1)1]n n n n -+----+=22n -，所以1(1)22(2)n n a n n =⎧=⎨-≥⎩．所以22b =，48b =于是2424b q b ==，解得2q 或2q =-（舍）所以22n n b b q-=⋅=12n -．3、（2020届山东省日照市高三上期末联考）已知数列{}{},n n a b 满足：1112,,2n n n n a a n b a n b ++=+-==.（1）证明数列{}n b 是等比数列，并求数列{}n b 的通项； 【解析】证明：因为n n b a n -=，所以n n b a n =+．因为121n n a a n +=+- 所以()()112n n a n a n +++=+ 所以12n n b b +=．又12b =，所以{}n b 是首项为12b =，公比为2的等比数列，所以1222n n n b -=⨯=．4、（2020·山东省淄博实验中学高三上期末）已知数列{}n a 的各项均为正数，对任意*n ∈N ，它的前n 项和n S 满足()()1126n n n S a a =++，并且2a ，4a ，9a 成等比数列.求数列{}n a 的通项公式；【解析】对任意*n ∈N ，有()()1126n n n S a a =++，①∴当1a =时，有()()11111126S a a a ==++，解得11a =或2. 当2n ≥时，有()()1111126n n n S a a ---=++.② ①-②并整理得()()1130n n n n a a a a --+--=. 而数列{}n a 的各项均为正数，13n n a a -∴-=. 当11a =时，()13132n a n n =+-=-，此时2429a a a =成立；当12a =时，()23131n a n n =+-=-，此时2429a a a =，不成立，舍去.32n a n ∴=-，*n ∈N .5、（2020届山东师范大学附中高三月考）设等差数列{}n a 前n 项和为n S ，满足424S S =，917a =.（1）求数列{}n a 的通项公式；（2）设数列{}n b 满足1212112n n n b b b a a a +++=-…，求数列{}n b 的通项公式 【解析】（1）设等差数列{}n a 首项为1a ，公差为d ．由已知得11914684817a d a d a a d +=+⎧⎨=+=⎩，解得112a d =⎧⎨=⎩．于是12(1)21n a n n =+-=-．（2）当1n =时，1111122b a =-=． 当2n ≥时，1111(1)(1)222n n n n nb a -=---=， 当1n =时上式也成立．于是12n n nb a =． 故12122n n n n n b a -==． 6、（2020·浙江温州中学3月高考模拟）已知各项均为正数的数列{}n a 的前n 项和为n S ，且11a =，n a =*n N ∈，且2n ≥）求数列{}n a 的通项公式；【解析】由n a =1n n S S --=+1(2)n =≥，所以数列1==为首项，以1为公差的等差数列，1(1)1n n =+-⨯=，即2n S n =，当2n ≥时，121n n n a S S n -=-=-，当1n =时，111a S ==，也满足上式，所以21n a n =-；7、【2019年高考浙江卷】设等差数列{}n a 的前n 项和为n S ，34a =，43a S =，数列{}n b 满足：对每个12,,,n n n n n n n S b S b S b *++∈+++N 成等比数列．（1）求数列{},{}n n a b 的通项公式；【解析】（1）设数列{}n a 的公差为d ，由题意得11124,333a d a d a d +=+=+，解得10,2a d ==． 从而*22,n a n n =-∈N ． 所以2*n S n n n =-∈N ，，由12,,n n n n n n S b S b S b +++++成等比数列得()()()212n n n n n n S b S b S b +++=++．解得()2121n n n n b S S S d++=-． 所以2*,n b n n n =+∈N ．8、【2019年高考江苏卷】定义首项为1且公比为正数的等比数列为“M－数列”.（1）已知等比数列{a n }()n *∈N 满足：245132,440a a a a a a =-+=，求证:数列{a n }为“M－数列”；（2）已知数列{b n }()n *∈N 满足:111221,n n n b S b b +==-，其中S n 为数列{b n }的前n项和．①求数列{b n }的通项公式；【解析】解：（1）设等比数列{a n }的公比为q ，所以a 1≠0，q ≠0.由245321440a a a a a a =⎧⎨-+=⎩，得244112111440a q a q a q a q a ⎧=⎨-+=⎩，解得112a q =⎧⎨=⎩．因此数列{}n a 为“M—数列”.（2）①因为1122n n n S b b +=-，所以0n b ≠．由1111,b S b ==，得212211b =-，则22b =. 由1122n n n S b b +=-，得112()n n n n nb b S b b ++=-，当2n ≥时，由1n n n b S S -=-，得()()111122n n n nn n n n n b b b b b b b b b +-+-=---，整理得112n n n b b b +-+=．所以数列{b n }是首项和公差均为1的等差数列. 因此，数列{b n }的通项公式为b n =n ()*n ∈N .。
2024届山东省潍坊市高三下学期二模英语模拟试题(含答案)
2024届山东省潍坊市高三下学期二模英语模拟试题潍坊市高考模拟考试英语2024.4注意事项:1.答题前，考生务必将自己的姓名、座号、考号填写在答题卡和试卷指定位置上。
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第一部分阅读(共两节,满分50分)第一节(共15小题;每小题2.5分，满分37.5分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项。
AThe backpack you take can make or break your trip when you go traveling. Here are the four best travel backpacks on the market.Amazon Basics 70LIt's much cheaper than many travel bags on the market and does not sacrifice any of the practical uses or ce that comes with more expensive bags. The bag may not be as luxury as some of the more high-end bags, but its simple style lets you focus on the main thing you need to focus on when traveling: the moment.Eurohike Nepal 65LThe Eurohike Backpack is a great choice because of how adaptable it is. Besides having a great amount of storage, it comes with an internal security pocket. It weighs just 1.38kg as opposed to other backpacks, which can weigh up to nearly 2kg. If you're going to go hiking when you travel, then it is perfect.Mountain Warehouse Tor 65LFirst , its brand is one of the most trusted in the industry ,so quality is guaranteed. Second, thebackpack's adjustable back allows you to change how the bag fits according to your needs. Available in both blue and green, this is a great choice if you want a bag that you can depend on.Osprey Europe Farpoint 70LOsprey is one of the best brands for backpacks. Its frame(框架)suspension, which can be adjusted to different needs, allows you to travel more comfortably. Whether you're visiting Switzerland in a thick, wool coat or the south of France in shorts and a T-shirt, the bag will match your look. This bag does say it is marketed for men, but, of course , it can be unisex.1. What is the selling point of the Amazon Basics?A. Its luxury style.B.Its fashionable design.C. Its huge ce for use.D. Its good value for money.2. What do Mountain Warehouse Tor 65L and Osprey Europe Farpoint 70L have in common?A. They are rich in color.B. They have the same capacity.C. They can be adjusted as needed.D. They are targeted for male customers.3. Which will you choose if anti-theft function is a concern?A. Amazon Basics 70L. B .Eurohike Nepal 65L.C. Mountain Warehouse Tor 65L,D. Osprey Europe Farpoint 70L.BAt just seven years old, Angelina Tsuboi discovered her passion for innovation. It all began with a simple game she programmed in her Los Angeles public school's Grade 2class. Today,at18,the Grade 12 student's initial curiosity has evolved into a deep-seated desire to use technology to decode(解码)real-world problems.In 2021, she co-developed Megaphone, one of her first apps, to tackle unanswered post- class questions and poor communication about events and announcements. Her problem-solving ability kept building from there.When she took online CPR classes at the start of the pandemic, she figured it couldn't be just her who was struggling with the steps. So she created an app called CPR Buddy―a winner in the2022 Apple Swift Challenge―which guides users through CPR using vibrations(震动) to regulate breath. After winning the award, Angelina presented her work to Apple CEO Tim Cook, a highlight in her young career, but one she didn't lose her cool over. “There's no point putting people on a pedestal (神坛),”she says.The next year, Angelina built an app called Lilac, designed to assist nonEnglish-speaking single parents with resources for housing, job opportunities and translation support. She was inspired by her own experiences as a child of a single mother who immigrated to the US.When Angelina decided to pursue pilot training at the age of 16, she was struck by how difficult it was to find financial support, which encouraged her to create yet another app, Pilot Fast Track, which helps those longing to be pilots find scholarships for flight training.Looking to the future, besides applying to colleges with great labs, Angelina is exploring the field of aeroce cybersecurity and mechatronics―combining computer science, electrical engineering and mechanical engineering.“There's not enough optimism in the world," she says. “I have also been in situations in my life where I've lost a lot of hope. But in the end, it is a mindset, and there are ways in any situation you're in to make it somewhat better."4.What is Angelina's pursuit?A. To design games for kids.B. To stimulate teen's curiosity.C. To address problems through technology.D. To find innovative approaches to digital challenges.5.What can we learn about Angelina from Paragraph 3?A. She couldn't breathe regularly.B. She was inspired by celebrities.C. She replaced CPR with an app.D. She was humble about her success.6.What was the primary goal of developing Pilot Fast Track?A. To direct pilots' career paths.B. To help to-be pilots find funds.C. To pair future pilots with airlines.D. To evaluate pilot training schools.7. What might be the best title?A. Breaking the codeB. Bearing growing painsC. Facing life as it isD. Following role modelsCSome people today might be early risers because of DNA they take after Neanderthals tens of thousands of years ago, suggests new research.When early humans migrated from Africa to Eurasia roughly 70,000 years ago, some of them mated with Neanderthals, who had already adapted to the colder, darker climates of the north. The ripple(涟漪) effects of that intermating still exist today: Modern humans of non- African ancestry(血统)have between 1 and 4 percent Neanderthal DNA. Some of that DNA relates to sleep more specifically, the internal body clock known as the circadian rhythm.For the new study, researchers compared DNA from today's humans and DNA from Neanderthal fossils(化石).In both groups, they found some of the same genetic variants involved with the circadian rhythm. And they found that modern humans who carry these variants also reported being early risers.For Neanderthals, being “morning people” might not have been the real benefit of carrying these genes. Instead, scientists suggest, Neanderthals’ DNA gave them faster, more flexible internal body clocks, which allowed them to adjust more easily to annual changes in daylight. This connection makes sense in the context of human history. When early humans moved north out of Africa, they would have experienced variable daylight hours--shorter days in the winter and longer days in the summer-for the first time. The Neanderthals' circadian rhythm genes likely helped early humans' offspring(后代)adapt to this new environment.Notably ,the findings do not prove that NU.K.erthal genes are responsible for the sleep habits of all early risers. Lots of different factors beyond genetics can contribute , including social and environmental influences. The study also only included DNA from a database called the U.K. Biobank-so the findings may not necessarily apply to all modern humans. Next, the research team hopes to study other genetic databases to see if the same link holds true for people of other ancestries. If the findings do apply more broadly , they may one day be useful for improving sleep in the modern world, where circadian rhythms are disturbed by night shifts and glowing smartphones.8.What does the new research focus on?A. DNA's dramatic changes.B. Genes’ influence on early risers.C. Neanderthals’ sleeping patterns.D. Ancestors’ environmental adaptability.9.What is paragraph 2 intended to show concerning the new research?A. Historical context.B. Additional proof.C. Sample analysis.D. Studying process.10. What is the real benefit of carrying Neanderthal's DNA for modern humans?A .Getting up earlier. B. Having healthier daily routines.C. Being more flexible in their work.D. Possessing a better circadian rhythm.11. What can be inferred about the findings from the last paragraph?A. They get proof from other studies.B. They are confirmed by early risers.C. They suggest potential applications.D. They reveal factors in sleeping disorders.DI had to say something after reading The Anxious Generation. It is going to sell well , because Jonathan Haidt is telling a scary story about children's development many parents are led to believe. However, the book's repeated suggestion that digital technologies are rewiring our children's brains and causing the epidemic (流行病)of mental illness is unsupported by science. Worse , the rude proposal that social media is to blame might distract (分心)us from effectively responding to the real causes of the current mental-health crisis in young people.Researchers have searched for the effects suggested by Haidt. Our efforts have produced a mix of no, small and mixed associations. Most data are correlative. When associations over time are found, they suggest not that social-media use predicts or causes depression, but that young people who already have mental-health problems use such platforms more often or in different ways from their healthy peers.We are not alone here. Several analyses and systematic reviews centralize on the same message. An analysis done in 72 countries shows no consistent or measurable associations between well-being and social media globally. Moreover, studies from some authorities finds no evidence of intense changes associated with digital-technology use.As a psychologist studying children's and adolescents’ mental health, I appreciate parents’frustration(沮丧)and desire for simple answers. As a parent of adolescents, I would also like to identify a simple source for the pain this generation is reporting. There are, however, no simple answers. The beginning and development of mental disorders are driven by a complex set of genetic and environmental factors.More young people are talking openly about their mental-health struggles than ever before. But insufficient services are available to address their needs. In the United States, there is, on average, one school psychologist for every 1,119 students. We have a generation in crisis and in desperate need of the best of what science and evidence-based solutions can offer. Unfortunately, our time is being spent telling stories that are unsupported by research and that do little to support young people who need, and deserve, more.12.What is presented in The Anxious Generation?A. Scary stories affect children's brains.B. Parents are responsible for children's health.C. Teen's mental illness results from screen time.D. The epidemic of mental illness is unavoidable.13.What does “the same message ”underlined in paragraph 3 refer to?A. Many countries do research in mental health.B. Well-being and social media are closely related.C. The young are trapped in the mental-health crisis,D. Social media don't necessarily cause mental illness.14. What is implied in the last paragraph?A. Effective actions need to be taken.B. Positive stories should be shared.C. Financial support needs to be provided.D. Broader research should be done.15.What is the author's purpose in writing the text?A. To suggest ways to help those in need.B. To encourage parents to brave the crisis.C. To recommend a newly-published book.D. To give a voice to children's mental issues.第二节(共5小题;每小题2.5分,满分12.5分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
山东省2014届高三文科数学备考之2013届名校解析试题精选分类汇编5：数列 Word版含答案
山东省2014届高三文科数学一轮复习之2013届名校解析试题精选分类汇编5：数列一、选择题1 ．（【解析】山东省青岛一中2013届高三1月调研考试文科数学）已知数列{n a }满足*331log 1log ()n n a a n ++=∈N ,且2469a a a ++=,则15793log ()a a a ++的值是 （ ）A ．15-B ．5-C ．5D ．15【答案】B 【解析】由*331log 1log ()n n a a n ++=∈N ,得313log log 1n n a a +-=,即13log 1n na a +=,解得13n n a a +=,所以数列{}n a 是公比为3的等比数列.因为3579246()a a a a a a q ++=++,所以35579933a a a ++=⨯=.所以5515791333log ()log 3log 35a a a ++==-=-,选 B ．2 ．（【解析】山东省德州市2013届高三3月模拟检测文科数学）若正项数列{}n a 满足1111n n ga ga +=+,且a 2001+a 2002+a 2003+a 2010=2013,则a 2011+a 2012+a 2013+a 2020的值为（ ）A ．2013·1010B ．2013·1011C ．2014·1010D ．2014·1011【答案】A 由条件知1111lg1n n n n a ga ga a ++-==,即110n naa +=为公比是10的等比数列.因为102001201020112020()a a q a a ++=++ ,所以1020112020201310a a ++=⋅ ,选A ．3 ．（【解析】山东省实验中学2013届高三第一次诊断性测试数学（文）试题）在各项均为正数的等比数列{}n a 中,31,1,s a a ==则2326372a a a a a ++=（ ）A ．4B ．6C ．8D．8-【答案】C 【解析】在等比数列中,23752635,a a a a a a a ==,所以22232637335522a a a a a a a a a ++=++22235()11)8a a =+=+==,选C ．4 ．（【解析】山东省济宁市2013届高三1月份期末测试（数学文）解析）已知函数()()2cos f n n n π=,且()()1,n a f n f n =++则123100a a a a +++⋅⋅⋅+=（ ）A ．100-B ．0C ．100D ．10200【答案】A 解:若n 为偶数,则()()221=(1)(21)na f n f n n n n =++-+=-+,为首项为25a =-,公差为4-的等差数列;若n 为奇数,则()()221=(1)21n a f n f n n n n =++-++=+,为首项为13a =,公差为4的等差数列.所以123100139924100()()a a a a a a a a a a +++⋅⋅⋅+=+++++++ 50495049503450(5)410022⨯⨯=⨯+⨯+⨯--⨯=-,选A ． 5 ．（【解析】山东省济南市2013届高三3月高考模拟文科数学）等差数列}{n a 中,482=+a a ,则它的前9项和=9S （ ）A ．9B ．18C ．36D ．72【答案】B 在等差数列中,28194a a a a +=+=,所以1999()941822a a S +⨯===,选 B ．6 ．（【解析】山东省实验中学2013届高三第三次诊断性测试文科数学）已知各项为正的等比数列{}n a 中,4a 与14a 的等比数列中项为22,则1172a a +的最小值 （ ）A ．16B ．8C ．22D ．4【答案】B 【解析】由题意知224149a a a ==,即9a =.所以设公比为(0)q q >,所以22971192228a a a a q q +=+=+≥=,2=,即42q =,所以q =,所以最小值为8,选B ．7 ．（【解析】山东省德州市2013届高三上学期期末校际联考数学（文））在各项均为正数的数列{a n }中,对任意m 、*n N Î都有m n m a a +=·n a 若636,a =则9a 等于 （ ）A ．216B ．510C ．512D ．l024【答案】A 解:由题意可知26336a a ==,所以36a =,所以93636636216a a a a +===⨯= ,选A ．8 ．（【解析】山东省潍坊市2013届高三上学期期末考试数学文（a ））如果等差数列{}n a 中,15765=++a a a ,那么943...a a a +++等于 （ ）A ．21B ．30C ．35D ．40【答案】C 【解析】在等差数列中,由15765=++a a a 得663155a a ==，.所以3496...=77535a a a a +++=⨯=,选C ．9 ．（山东省淄博市2013届高三复习阶段性检测（二模）数学（文）试题）已知等差数列{}n a 的前n 项和为n S ,满足1313113a S a ===，则 （ ）A ．14-B ．13-C ．12-D ．11-【答案】D 在等差数列中,1131313()132a a S +==,所以1132a a +=,即113221311a a =-=-=-,选 D ．10．（【解析】山东省枣庄市2013届高三3月模拟考试 数学（文）试题）两旅客坐火车外出旅游,希望座位连在一起,且仅有一个靠窗,已知火车上的座位的排法如表格所示,则下列座位号码符合要求的是（ ）A ．48,49B ．62,63C ．84,85D ．75,76【答案】C 根据座位排法可知,做在右窗口的座位号码应为5的倍数,所以C 符合要求.选 C ．11．（山东省威海市2013届高三上学期期末考试文科数学）{}n a 为等差数列,n S 为其前n 项和,已知77521a S ==，，则10S =（ ）A ．40B ．35C ．30D ．28【答案】【答案】A 设公差为d ,则由77521a S ==，得1777()2a a S +=,即17(5)212a +=,解得11a =,所以716a a d =+,所以23d =.所以1011091092101040223S a d ⨯⨯=+=+⨯=,选 （ ）A ．12．（【解析】山东省济宁市2013届高三1月份期末测试（数学文）解析）已知在等比数列{}n a 中,1346510,4a a a a +=+=,则该等比数列的公比为 （ ）A ．14B ．12C ．2D ．8【答案】B 解:因为31346()a a q a a +=+,所以34613514108a a q a a +===+,即12q =,选B ．13．（【解析】山东省实验中学2013届高三第三次诊断性测试文科数学）已知等差数列{}n a 的公差为d 不为0,等比数列{}n b 的公比q 是小于1的正有理数,若211,d b d a ==,且321232221b b b a a a ++++是正整数,则q 的值可以是 （ ）A ．71 B ．-71 C ．21 D ．21-【答案】C 【解析】由题意知21312,23a a d d a a d d =+==+=,22222131,b b q d q b b q d q ====,所以2222221232222212349141a a a d d d b b b d d q d q q q ++++==++++++,因为321232221b b b a a a ++++是正整数,所以令2141t q q=++,t 为正整数.所以2114t q q ++=,即21014t q q ++-=,解得q ===,因为t 为正整数,所以当8t =时,12122q -+===.符合题意,选C ．14．（【解析】山东省滨州市2013届高三第一次（3月）模拟考试数学（文）试题）已知数列{}n a 为等差数例,其前n 项的和为n S ,若336,12a S ==,则公差d = （ ）A ．1B ．2C ．3D ．53【答案】B 在等差数列中,13133()3(6)1222a a a S ++===,解得12a =所以解得2d =,选 B ． 15．（【解析】山东省济南市2013届高三上学期期末考试文科数学）已知数列{}n a 的前n 项和为n S ,且122-=n S n , 则=3a（ ）A ．-10B ．6C ．10D ．14【答案】C 解:22332231(221)10a S S =-=⨯--⨯-=,选 C ．16．（【解析】山东省临沂市2013届高三3月教学质量检测考试（一模）数学（文）试题）已知等差数列{n a }中,74a π=,则tan(678a a a ++)等于（ ）A ．B ．C ．-1D ．1【答案】C 在等差数列中6787334a a a a π++==,所以6784tan()tan14a a a π++==-,选 C ． 17．（【解析】山东省烟台市2013届高三5月适应性练习（一）文科数学）已知等比数列{a n }的公比q=2,前n硕和为S n .若S 3=72,则S 6等于 （ ）A ．312B ．632C ．63D ．1272【答案】B 【解析】3131(12)77122a S a -===-,所以112a =.所以6161(12)6363122a S a -===-,选 B ．二、填空题18．（【解析】山东省青岛市2013届高三第一次模拟考试文科数学）设n S 是等差数列{}n a 的前n 项和,1532,3a a a ==,则9S =_____________ ;【答案】54- 由1532,3a a a ==得1143(2)a d a d +=+,即12d a =-=-,所以919899298542S a d ⨯=+=⨯-⨯=-. 19．（山东省青岛即墨市2013届高三上学期期末考试 数学（文）试题）等比数列}{n a ,2=q ,前n 项和为=24a S S n ，则____________. 【答案】215解:在等比数列中,4141(12)1512a S a -==-,所以4121151522S a a a ==.20．（【解析】山东省实验中学2013届高三第一次诊断性测试数学（文）试题）数列{}n a 满足113,1,n n n n a a a a A +=-=表示{}n a 前n 项之积,则2013A =_____________.【答案】1-【解析】由113,1,n n n a a a a +=-=得11n n na a a +-=,所以231233a -==,312a =-,43a =,所以{}n a 是以3为周期的周期数列,且1231a a a =-,又20133671=⨯,所以6712013(1)1A =-=-.21．（山东省淄博市2013届高三复习阶段性检测（二模）数学（文）试题）在如图所示的数阵中,第9行的第2个数为___________.【答案】66 每行的第二个数构成一个数列{}n a ,由题意知23453,6,11,18a a a a ====,所以3243543,5,7,a a a a a a -=-=-=12(1)123n n a a n n --=--=-,等式两边同时相加得22[233](2)22n n n a a n n -+⨯--==-,所以()222223,2n a n n a n n n =-+=-+≥,所以29929366a =-⨯+=.22．（【解析】山东省泰安市2013届高三第一轮复习质量检测数学（文）试题）正项数列{}n a 满足:()222*121171,2,2,2,n n n a a a a a n N n a +-===+∈≥=则______.【答案】因为()222*112,2n n n a a a n N n +-=+∈≥,所以数列2{}n a 是以211a =为首项,以2221413d a a =-=-=为公差的等差数列,所以213(1)32n a n n =+-=-,所以1n a n =≥,所以7a ==23．（【解析】山东省潍坊市2013届高三第一次模拟考试文科数学）现有一根n 节的竹竿,自上而下每节的长度依次构成等差数列,最上面一节长为10cm,最下面的三节长度之和为114cm,第6节的长度是首节与末节长度的等比中项,则n=_____.【答案】16 设对应的数列为{}n a ,公差为,(0)d d >.由题意知110a =,12114n n n a a a --++=,261n a a a =.由12114n n n a a a --++=得13114n a -=,解得138n a -=,即2111(5)()n a d a a d -+=+,即2(105)10(38)d d +=+,解得2d =,所以11(2)38n a a n d -=+-=,即102(2)38n +-=,解得16n =.24．（【解析】山东省济宁市2013届高三第一次模拟考试文科数学 ）已知等差数列{n a }中,35a a +=32,73a a -=8,则此数列的前10项和10S =____.【答案】190【解析】由7348a a d -==,解得2d =,由3532a a +=,解得110a =.所以101109101902S a d ⨯=+=. 25．（【解析】山东省潍坊市2013届高三第二次模拟考试文科数学）已知等差数列{}n a 的前n 项和为n S ,若2,4,3a 成等比数列,则5S =_________.【答案】40因为2,4,3a 成等比数列,所以232416a ==,所以38a =.又153535()525584022a a a S a +⨯====⨯=. 26．（【解析】山东省烟台市2013届高三上学期期末考试数学（文）试题）已知等比数列{a n }中,6710111,16a a a a ==g g ,则89a a g 等于_______【答案】4【解析】在等比数列中2676()10a a a q ==>g ,所以0q >,所以289670a a a a q =>g .所以67101116a a a a =,即289()16a a =g ,所以894a a =g .27．（【解析】山东省泰安市2013届高三上学期期末考试数学文）下面图形由小正方形组成,请观察图1至图4的规律,并依此规律,写出第n 个图形中小正方形的个数是___________.【答案】(1)2n n +【解析】12341,3,6,10a a a a ====,所以2132432,3,4a a a a a a -=-=-=, 1n n a a n --=,等式两边同时累加得123n a a n -=+++ ,即(1)122n n n a n +=+++=,所以第n 个图形中小正方形的个数是(1)2n n + 三、解答题28．（【解析】山东省烟台市2013届高三上学期期末考试数学（文）试题）已知数列{a n }的前n 项和为S n ,且22n n S a =-.(1)求数列{a n }的通项公式;(2)记1213(21)n n S a a n a =+++-g g L g ,求S n【答案】29．（【解析】山东省潍坊市2013届高三上学期期末考试数学文（a ））设数列{}n a 为等差数列,且9,553==a a ;数列{}n b 的前n 项和为n S ,且2=+n n b S . (I)求数列{}n a ,{}n b 的通项公式; (II)若()+∈=N n b a c nnn ,n T 为数列{}n c 的前n 项和,求n T . 【答案】30．（【解析】山东省滨州市2013届高三第一次（3月）模拟考试数学（文）试题）已知数列{}n a 的前n 项和是n S ,且11()2n n S a n *+=∈N (Ⅰ)求数列{}n a 的通项公式;(Ⅱ)设113log (1)()n n b S n *+=-∈N ,令122311n T b b b b =++11n n b b ++,求n T . 【答案】31．（【解析】山东省临沂市2013届高三5月高考模拟文科数学）已知点(1,2)是函数()(01)x f x a a a =≠＞且的图象上一点,数列{}n a 的前n 项和()1n S f n =-. (Ⅰ)求数列{}n a 的通项公式;(Ⅱ)将数列{}n a 前2013项中的第3项,第6项,,第3k 项删去,求数列{}n a 前2013项中剩余项的和.【答案】解:(Ⅰ)把点(1,2)代入函数()x f x a =,得2a =.()121,n n S f n ∴=-=-当1n =时,111211;a S ==-= 当2n ≥时,1n n n a S S -=-1(21)(21)n n -=---12n -=经验证可知1n =时,也适合上式,12n n a -∴=.(Ⅱ)由(Ⅰ)知数列{}n a 为等比数列,公比为2,故其第3项,第6项,,第2013项也为等比数列,首项31324,a -==公比32012201328,2a ==为其第671项∴此数列的和为67120134(18)4(21)187--=- 又数列{}n a 的前2013项和为2013201320131(12)21,12S ⨯-==--∴所求剩余项的和为2013201320134(21)3(21)(21)77----=32．（【解析】山东省实验中学2013届高三第三次诊断性测试文科数学）已知数列}{n a 的前n 项和为n S ,且)(14*∈+=N n a S n n . (Ⅰ)求21,a a ;(Ⅱ)设||log 3n n a b =,求数列{}n b 的通项公式.【答案】解:(1)由已知1411+=a S ,即31,14111=∴+=a a a ,又1422+=a S ,即91,1)42221-=∴+=+a a a a （;(2)当1>n 时,)1(41)1(4111+-+=-=--n n n n n a a S S a ,即13--=n n a a ,易知数列各项不为零(注:可不证不说),311-=∴-n n a a 对2≥n 恒成立, {}n a ∴是首项为31,公比为-31的等比数列,n n n n a ----=-=∴3)1()31(3111,n a n n -==∴-3log ||log 33,即n b n -=33．（【解析】山东省泰安市2013届高三上学期期末考试数学文）在等差数列{}n a 中,13a =,其前n 项和为n S ,等比数列{}n b 的各项均为正数,11b =,公比为q ,且222212,,n n S b S q a b b +==求与; 【答案】34．（【解析】山东省济宁市2013届高三1月份期末测试（数学文）解析）设数列{}n a 的前n 项和为n S ,若对于任意的正整数n 都有23n n S a n =-.(I)设3n n b a =+,求证:数列{}n b 是等比数列,并求出{}n a 的通项公式; (II)求数列{}n nb 的前n 项和T n .【答案】35．（【解析】山东省德州市2013届高三3月模拟检测文科数学）数列{}n a 是公差不小0的等差数列a 1、a 3,是函数2()1(66)f x n x x =-+的零点,数列{}n b 的前n 项和为n T ,且*12()n n T b n N =-∈ (1)求数列{}n a ,{}n b 的通项公式;(2)记n n n c a b =,求数列{}n c 的前n 项和S n .【答案】36．（【解析】山东省德州市2013届高三上学期期末校际联考数学（文））已知数列{a n }的公差为2的等差数列,它的前n 项和为n S ,且1321,1,1a a a +++成等比数列. (I)求{a n }的通项公式; (2)13{},.4n n n n T T S <记数列的前项求证: 【答案】37．（【解析】山东省济南市2013届高三上学期期末考试文科数学）已知等差数列{}n a 的前n 项和为n S ,且满足24a =,3417a a +=. (1)求{}n a 的通项公式; (2)设22n a n b +=,证明数列{}n b 是等比数列并求其前n 项和n T .【答案】解:(1)设等差数列{}n a 的公差为d .由题意知3411212317,4,a a a d a d a a d +=+++=⎧⎨=+=⎩解得,11a =,3d =, ∴32n a n =-(n N *∈) (2)由题意知, 2322n a n n b +==(n N *∈),3(1)33122n n n b ---==(,2n N n *∈≥)∴333312282n n n n b b --===(,2n N n *∈≥),又18b = ∴{}n b 是以18b =,公比为8的等比数列()()818881187n nn T -==-- 38．（山东省烟台市2013届高三3月诊断性测试数学文）设{a n }是正数组成的数列,a 1=3.若点()2*11,2()n n n a aa n N ++-∈在函数321()23f x x x =+-的导函数()y f x '=图像上. (1)求数列{a n }的通项公式; (2)设12n n nb a a +=⋅,是否存在最小的正数M,使得对任意n *N ∈都有b 1+b 2++b n <M 成立?请说明理由.【答案】39．（【解析】山东省济宁市2013届高三第一次模拟考试文科数学 ）(本小题满分l2分)设数列{n a }满足:a 1=5,a n+1+4a n =5,(n ∈N*)(I)是否存在实数t ,使{a n +t }是等比数列?(Ⅱ)设数列b n =|a n |,求{b n }的前2013项和S 2013.【答案】解:(I)由+1+4=5n n a a 得+1=4+5n n a a -令()+1+=4+n n a t a t -,得+1=45n n a a t -- 则5=5t -,=1t - 从而()+11=41n n a a --- .又11=4a -, {}1n a ∴-是首项为4,公比为4-的等比数列,∴存在这样的实数=1t -,使{}+n a t 是等比数列(II)由(I)得()11=44n n a --⋅- ()=14nn a ∴--{1+4, 41==n n n n n n b a -∴为奇数，为偶数()()()()()123420132013122013=++=1+4+41+1+4+41++1+4S b b b ∴--1232013=4+4+4++4+1 201420144441=+1=143--- 40．（【解析】山东省枣庄市2013届高三3月模拟考试 数学（文）试题）已知等比数列13212{}1,6,,8n a q a a a a a >=-的公比且成等差数列.(1)求数列{a n }的通项公式;(2)设(1),: 1.n n nn n b b a +=≤求证 【答案】41．（【解析】山东省青岛市2013届高三第一次模拟考试文科数学）已知N n *∈,数列{}n d 满足2)1(3nn d -+=,数列{}n a 满足1232n n a d d d d =+++⋅⋅⋅+;数列{}n b 为公比大于1的等比数列,且42,b b 为方程064202=+-x x 的两个不相等的实根.(Ⅰ)求数列{}n a 和数列{}n b 的通项公式;(Ⅱ)将数列{}n b 中的第．1a 项,第．2a 项,第．3a 项,,第．n a 项,删去后剩余的项按从小到大的顺序排成新数列{}n c ,求数列{}n c 的前2013项和.【答案】解:(Ⅰ)2)1(3n n d -+= ,∴1232n n a d d d d =+++⋅⋅⋅+3232nn ⨯== 因为42,b b 为方程064202=+-x x 的两个不相等的实数根. 所以2042=+b b ,6442=⋅b b 解得:42=b ,164=b ,所以:n n b 2=(Ⅱ)由题知将数列{}n b 中的第3项、第6项、第9项删去后构成的新数列{}n c 中的奇数列与偶数列仍成等比数列,首项分别是12b =,24b =公比均是,8201313520132462012()()T c c c c c c c c =+++⋅⋅⋅+++++⋅⋅⋅+ 1007100610062(18)4(18)208618187⨯-⨯-⨯-=+=-- 42．（【解析】山东省潍坊市2013届高三第一次模拟考试文科数学）已知数列{}n a 的各项排成如图所示的三角形数阵,数阵中每一行的第一个数1247,,,,a a a a ⋅⋅⋅构成等差数列{}n b ,n S 是{}n b 的前n 项和,且1151,15b a S ===( I )若数阵中从第三行开始每行中的数按从左到右的顺序均构成公比为正数的等比数列,且公比相等,已知916a =,求50a 的值; (Ⅱ)设122111n n n nT S S S ++=++⋅⋅⋅+,求n T.【答案】解:(Ⅰ){}n b 为等差数列,设公差为155,1,15,51015,1d b S S d d ==∴=+== 1(1)1.n b n n ∴=+-⨯=设从第3行起,每行的公比都是q ,且0q >,2294,416,2,a b q q q === 1．+2+3++9=45,故50a 是数阵中第10行第5个数, 而445010102160.a b q ==⨯= (Ⅱ)12n S =++ (1),2n n n ++=1211n n n T S S ++∴=++21nS +22(1)(2)(2)(3)n n n n =++++++22(21)n n ++11112(1223n n n n =-+-+++++11)221n n +-+ 1122().121(1)(21)n n n n n =-=++++43．（山东省青岛即墨市2013届高三上学期期末考试 数学（文）试题）等差数列}{n a 中,9,155432==++a a a a . (Ⅰ)求数列}{n a 的通项公式;(Ⅱ)设213+=n a n b ,求数列},21{n n b a +的前n 项和n S 【答案】解:(Ⅰ)设数列{}由题意得首项的公差为,1a d a n且⎩⎨⎧=+=+⎩⎨⎧==++941563915115432d a d a a a a a 即 解得⎩⎨⎧==211d a所以数列{}12-=n a a n n 的通项公式为 (Ⅱ)由(Ⅰ)可得n n n a b 3231==+ 所以n n n n b a 3..21=+ 所以+++=323.33.23.11n S 13.+n n两式相减得++++-=433333(22n S 13.)3+++n n n 10 分43).12(323..1233.31313111+++-+=-+=+---=n n n n n n S n n n 即）（）（44．（【解析】山东省潍坊市2013届高三第二次模拟考试文科数学）某工厂为扩大生产规模,今年年初新购置了一条高性能的生产线,该生产线在使用过程中的维护费用会逐年增加,第一年的维护费用是4万元,从第二年到第七年,每年的维护费用均比上年增加2万元,从第八年开始,每年的维护费用比上年增加25%(I)设第n 年该生产线的维护费用为n a ,求n a 的表达式; (Ⅱ)设该生产线前n 年维护费为n S ,求n S .【答案】45．（山东省威海市2013届高三上学期期末考试文科数学）已知数列{}n a ,15a =-,22a =-,记()A n =12n a a a +++ ,23()B n a a =+1n a +++ ,()C n =342+n a a a +++ (*N n ∈),若对于任意*N n ∈,()A n ,()B n ,()C n 成等差数列.(Ⅰ)求数列{}n a 的通项公式; (Ⅱ) 求数列{}||n a 的前n 项和.【答案】解:(Ⅰ)根据题意()A n ,()B n ,()C n 成等差数列∴()+()2()A n C n B n =整理得2121253n n a a a a ++-=-=-+= ∴数列{}n a 是首项为5-,公差为3的等差数列 ∴53(1)38n a n n =-+-=- (Ⅱ)38,2||38,3n n n a n n -+≤⎧=⎨-≥⎩记数列{}||n a 的前n 项和为n S .当2n ≤时,2(583)313222n n n n S n +-==-+ 当3n ≥时,2(2)(138)313714222n n n n S n -+-=+=-+综上,2231322231314322n n n n S n n n ⎧-+≤⎪⎪=⎨⎪-+≥⎪⎩ 46．（【解析】山东省实验中学2013届高三第一次诊断性测试数学（文）试题）已知{}n a 是公比大于1的等经数列,13,a a 是函数9()10f x x x=+-的两个零点(1)求数列{}n a 的通项公式;(2)若数列{}n a 满足312312,80n n b og n b b b b =+++++≥ 且,求n 的最小值.【答案】47．（【解析】山东省济南市2013届高三3月高考模拟文科数学）正项等比数列}{n a 的前n 项和为n S ,164=a ,且32,a a 的等差中项为2S . (1)求数列}{n a 的通项公式; (2)设12-=n n a n b ,求数列}{n b 的前n 项和n T .【答案】解:(1)设等比数列}{n a 的公比为)0(>q q ,由题意,得⎪⎩⎪⎨⎧+=+=)(2161121131q a a q a q a q a ,解得⎩⎨⎧==221q a所以n n a 2= (2)因为12122--==n n n n a n b ,所以12753224232221-+++++=n n nT , 121275322123222141+-+-++++=n n n nn T , 所以12127532212121212143+--+++++=n n n n T122411)411(21+---=n n n 12233432+⋅+-=n n故2181612992n n nT ++=-⋅ 48．（山东省淄博市2013届高三复习阶段性检测（二模）数学（文）试题）等比数列．．．．{}n c 满足(){}1*1104,n n n n c c n N a -++=⋅∈数列的前n 项和为n S ,且2log .n n a c =(I)求,n n a S ;(II)数列{}{}1,41n n n n n b b T b S =-满足为数列的前n 项和,是否存在正整数m,()1m >,使得16,,m m T T T 成等比数列?若存在,求出所有m 的值;若不存在,请说明理由.【答案】解: (Ⅰ)40,103221=+=+c c c c ,所以公比4=q10411=+c c 得21=c121242--=⋅=n n n c所以212log 221n n a n -==-21()[1(21)]22n n n a a n n S n ++-=== (Ⅱ)由(Ⅰ)知211114122121n b n n n ⎛⎫==- ⎪--+⎝⎭于是11111112335212121n n T n n n ⎡⎤⎛⎫⎛⎫⎛⎫=-+-++-= ⎪ ⎪ ⎪⎢⎥-++⎝⎭⎝⎭⎝⎭⎣⎦假设存在正整数()1m m >,使得16,,m m T T T 成等比数列,则216213121m m m m ⎛⎫=⨯ ⎪++⎝⎭, 整理得24720m m --=, 解得14m =-或 2m = 由,1m N m *∈>,得2m =, 因此,存在正整数2m =,使得16,,m m T T T 成等比数列49．（【解析】山东省临沂市2013届高三3月教学质量检测考试（一模）数学（文）试题）已知等比数列{n a }的首项为l,公比q≠1,n S 为其前n 项和,a l ,a 2,a 3分别为某等差数列的第一、第二、第四项.(I)求n a 和n S ;(Ⅱ)设21n n b log a +=,数列{21n n b b +}的前n 项和为T n ,求证:34n T <.【答案】50．（【解析】山东省烟台市2013届高三5月适应性练习（一）文科数学）在等差数列{}n a 中,a 1 =3,其前n项和为S n ,等比数列{b n }的各项均为正数,b 1 =1,公比为q,且b 2 +S 2 =12, q=22S b . (1)求a n 与b n ; (2)设数列{C n }满足c n =1nS ,求{n c }的前n 项和T n . 【答案】51．（【解析】山东省青岛一中2013届高三1月调研考试文科数学）已知等差数列{}n a 的首项1a =1,公差d>0,且第2项、第5项、第14项分别为等比数列{}n b 的第2项、第3项、第4项. (1)求数列{}n a 与{}n b 的通项公式; (2)设数列{n c }对n ∈N +均有11c b +22c b ++nnc b =1n a +成立,求1c +2c 3c ++2012c . 【答案】.解答:(1)由已知得2a =1+d, 5a =1+4d, 14a =1+13d,∴2(14)d +=(1+d)(1+13d), ∴d=2, n a =2n-1又2b =2a =3,3b = 5a =9 ∴数列{n b }的公比为3,n b =3⋅23n -=13n -(2)由11c b +22c b ++nnc b =1n a + (1) 当n=1时,11c b =2a =3, ∴1c =3当n>1时,11c b +22c b ++11n n c b --= n a (2) (1)-(2)得nnc b =1n a +-n a =2 ∴n c =2n b =2⋅13n - 对1c 不适用∴n c =131232n n n -=⎧⎨∙≥⎩∴123c c c +++2012c =3+2⋅3+2⋅23++2⋅20113=1+2⋅1+2⋅3+2⋅23++2⋅20113=1+2⋅20121313--=2012352．（【解析】山东省泰安市2013届高三第一轮复习质量检测数学（文）试题）设等比数列{}n a 的前n 项和为,415349,,,n S a a a a a =-成等差数列.(I)求数列{}n a 的通项公式;(II)证明:对任意21,,,k k k R N S S S +++∈成等差数列.【答案】。
2024山东省潍坊市高三下学期二模英语试题及答案
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第一部分阅读(共两节,满分50分)第一节(共15小题;每小题2.5分，满分37.5分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项。
AThe backpack you take can make or break your trip when you go traveling.Here are the four best travel backpacks on the market.Amazon Basics70LIt's much cheaper than many travel bags on the market and does not sacrifice any of the practical uses or space that comes with more expensive bags.The bag may not be as luxury as some of the more high-end bags,but its simple style lets you focus on the main thing you need to focus on when traveling:the moment.Eurohike Nepal65LThe Eurohike Backpack is a great choice because of how adaptable it is.Besides having a great amount of storage,it comes with an internal security pocket.It weighs just1.38kg as opposed to other backpacks,which can weigh up to nearly2kg.If you're going to go hiking when you travel,then it is perfect.Mountain Warehouse Tor65LFirst,its brand is one of the most trusted in the industry,so quality is guaranteed.Second,the backpack's adjustable back allows you to change how the bag fits according to your needs.Available inboth blue and green,this is a great choice if you want a bag that you can depend on.Osprey Europe Farpoint70LOsprey is one of the best brands for backpacks.Its frame(框架)suspension,which can be adjusted to different needs,allows you to travel more comfortably.Whether you're visiting Switzerland in a thick,wool coat or the south of France in shorts and a T-shirt,the bag will match your look.This bag does say it is marketed for men,but,of course,it can be unisex.1.What is the selling point of the Amazon Basics?A.Its luxury style.B.Its fashionable design.C.Its huge space for use.D.Its good value for money.2.What do Mountain Warehouse Tor65L and Osprey Europe Farpoint70L have in common?A.They are rich in color.B.They have the same capacity.C.They can be adjusted as needed.D.They are targeted for male customers.3.Which will you choose if anti-theft function is a concern?A.Amazon Basics70L.B.Eurohike Nepal65L.C.Mountain Warehouse Tor65L,D.Osprey Europe Farpoint70L.BAt just seven years old,Angelina Tsuboi discovered her passion for innovation.It all began with a simple game she programmed in her Los Angeles public school's Grade2class.Today,at18,the Grade12 student's initial curiosity has evolved into a deep-seated desire to use technology to decode(解码)real-world problems.In2021,she co-developed Megaphone,one of her first apps,to tackle unanswered post-class questions and poor communication about events and announcements.Her problem-solving ability kept building from there.When she took online CPR classes at the start of the pandemic,she figured it couldn't be just her who was struggling with the steps.So she created an app called CPR Buddy―a winner in the2022Apple Swift Challenge―which guides users through CPR using vibrations(震动)to regulate breath.After winning theaward,Angelina presented her work to Apple CEO Tim Cook,a highlight in her young career,but one she didn't lose her cool over.“There's no point putting people on a pedestal(神坛),”she says.The next year,Angelina built an app called Lilac,designed to assist nonEnglish-speaking single parents with resources for housing,job opportunities and translation support.She was inspired by her own experiences as a child of a single mother who immigrated to the US.When Angelina decided to pursue pilot training at the age of16,she was struck by how difficult it was to find financial support,which encouraged her to create yet another app,Pilot Fast Track,which helps those longing to be pilots find scholarships for flight training.Looking to the future,besides applying to colleges with great labs,Angelina is exploring the field of aerospace cybersecurity and mechatronics―combining computer science,electrical engineering and mechanical engineering.“There's not enough optimism in the world,"she says.“I have also been in situations in my life where I've lost a lot of hope.But in the end,it is a mindset,and there are ways in any situation you're in to make it somewhat better."4.What is Angelina's pursuit?A.To design games for kids.B.To stimulate teen's curiosity.C.To address problems through technology.D.To find innovative approaches to digital challenges.5.What can we learn about Angelina from Paragraph3?A.She couldn't breathe regularly.B.She was inspired by celebrities.C.She replaced CPR with an app.D.She was humble about her success.6.What was the primary goal of developing Pilot Fast Track?A.To direct pilots'career paths.B.To help to-be pilots find funds.C.To pair future pilots with airlines.D.To evaluate pilot training schools.7.What might be the best title?A.Breaking the codeB.Bearing growing painsC.Facing life as it isD.Following role modelsCSome people today might be early risers because of DNA they take after Neanderthals tens of thousands of years ago,suggests new research.When early humans migrated from Africa to Eurasia roughly70,000years ago,some of them mated with Neanderthals,who had already adapted to the colder,darker climates of the north.The ripple(涟漪) effects of that intermating still exist today:Modern humans of non-African ancestry(血统)have between1 and4percent Neanderthal DNA.Some of that DNA relates to sleep more specifically,the internal body clock known as the circadian rhythm.For the new study,researchers compared DNA from today's humans and DNA from Neanderthal fossils(化石).In both groups,they found some of the same genetic variants involved with the circadian rhythm.And they found that modern humans who carry these variants also reported being early risers.For Neanderthals,being“morning people”might not have been the real benefit of carrying these genes.Instead,scientists suggest,Neanderthals’DNA gave them faster,more flexible internal body clocks, which allowed them to adjust more easily to annual changes in daylight.This connection makes sense in the context of human history.When early humans moved north out of Africa,they would have experienced variable daylight hours--shorter days in the winter and longer days in the summer-for the first time.The Neanderthals'circadian rhythm genes likely helped early humans'offspring(后代)adapt to this new environment.Notably,the findings do not prove that Neanderthal genes are responsible for the sleep habits of all early risers.Lots of different factors beyond genetics can contribute,including social and environmental influences.The study also only included DNA from a database called the U.K.Biobank-so the findings may not necessarily apply to all modern humans.Next,the research team hopes to study other genetic databases to see if the same link holds true for people of other ancestries.If the findings do apply more broadly,they may one day be useful for improving sleep in the modern world,where circadian rhythms are disturbed by night shifts and glowing smartphones.8.What does the new research focus on?A.DNA's dramatic changes.B.Genes’influence on early risers.C.Neanderthals’sleeping patterns.D.Ancestors’environmental adaptability.9.What is paragraph2intended to show concerning the new research?A.Historical context.B.Additional proof.C.Sample analysis.D.Studying process.10.What is the real benefit of carrying Neanderthal's DNA for modern humans?A.Getting up earlier.B.Having healthier daily routines.C.Being more flexible in their work.D.Possessing a better circadian rhythm.11.What can be inferred about the findings from the last paragraph?A.They get proof from other studies.B.They are confirmed by early risers.C.They suggest potential applications.D.They reveal factors in sleeping disorders.DI had to say something after reading The Anxious Generation.It is going to sell well,because Jonathan Haidt is telling a scary story about children's development many parents are led to believe. However,the book's repeated suggestion that digital technologies are rewiring our children's brains and causing the epidemic(流行病)of mental illness is unsupported by science.Worse,the rude proposal that social media is to blame might distract(分心)us from effectively responding to the real causes of the current mental-health crisis in young people.Researchers have searched for the effects suggested by Haidt.Our efforts have produced a mix of no, small and mixed associations.Most data are correlative.When associations over time are found,they suggest not that social-media use predicts or causes depression,but that young people who already have mental-health problems use such platforms more often or in different ways from their healthy peers.We are not alone here.Several analyses and systematic reviews centralize on the same message.An analysis done in72countries shows no consistent or measurable associations between well-being and social media globally.Moreover,studies from some authorities finds no evidence of intense changes associated with digital-technology use.As a psychologist studying children's and adolescents’mental health,I appreciate parents’frustration(沮丧)and desire for simple answers.As a parent of adolescents,I would also like to identify a simple source for the pain this generation is reporting.There are,however,no simple answers.The beginning and development of mental disorders are driven by a complex set of genetic and environmental factors.More young people are talking openly about their mental-health struggles than ever before.But insufficient services are available to address their needs.In the United States,there is,on average,one school psychologist for every1,119students.We have a generation in crisis and in desperate need of the best of what science and evidence-based solutions can offer.Unfortunately,our time is being spent telling stories that are unsupported by research and that do little to support young people who need,and deserve, more.12.What is presented in The Anxious Generation?A.Scary stories affect children's brains.B.Parents are responsible for children's health.C.Teen's mental illness results from screen time.D.The epidemic of mental illness is unavoidable.13.What does“the same message”underlined in paragraph3refer to?A.Many countries do research in mental health.B.Well-being and social media are closely related.C.The young are trapped in the mental-health crisis,D.Social media don't necessarily cause mental illness.14.What is implied in the last paragraph?A.Effective actions need to be taken.B.Positive stories should be shared.C.Financial support needs to be provided.D.Broader research should be done.15.What is the author's purpose in writing the text?A.To suggest ways to help those in need.B.To encourage parents to brave the crisis.C.To recommend a newly-published book.D.To give a voice to children's mental issues.第二节(共5小题;每小题2.5分,满分12.5分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
2020年山东省潍坊市高考化学二模试卷(含答案解析)
2020年山东省潍坊市高考化学二模试卷一、单选题（本大题共13小题，共32.0分）1.高分子材料在疫情防控和治疗中起到了重要的作用。
下列说法正确的是A. 聚乙烯是生产隔离衣的主要材料，能使酸性高锰酸钾溶液褪色B. 聚丙烯酸树脂是3D打印护目镜镜框材料的成分之一，可以与NaOH溶液反应C. 天然橡胶是制作医用无菌橡胶手套的原料，它是异戊二烯发生缩聚反应的产物D. 聚乙二醇可用于制备治疗新冠病毒的药物，聚乙二醇的结构简式为2.曲酸和脱氧曲酸是非常有潜力的食品添加剂，具有抗菌抗癌作用，其结构如图所示。
下列叙述错误的是A. 两物质都能使溴的四氯化碳溶液褪色B. 曲酸经消去反应可得脱氧曲酸C. 1mol 脱氧曲酸最多能与发生加成反应D. 曲酸与脱氧曲酸中含有相同种类的官能团3.用化学沉淀法去除粗盐中的杂质离子，包括粗盐溶解、加沉淀剂、过滤、调节pH、蒸发结晶等步骤。
下列说法错误的是A. 沉淀剂的添加顺序可以是NaOH溶液、溶液、溶液B. 向滤液中滴加盐酸，调节pH至滤液呈中性C. 蒸发结晶时，当蒸发皿中出现较多固体时，停止加热，利用余热将滤液蒸干D. 溶解、过滤、蒸发结晶等过程都使用玻璃棒搅拌溶液4.最新发现是金星大气的成分之一，化学性质与CO相似。
分子中不含环状结构且每个原子均满足8电子稳定结构。
下列叙述错误的是A. 元素的第一电离能：B. 3p轨道上有1对成对电子的基态X原子与基态O原子的性质相似C. 中C原子的杂化方式为D. 分子中键和键的个数比为1：15.芘经氧化后可用于染料合成。
芘的一种转化路线如图，下列叙述正确的是A. 芘的一氯代物有4种B. 甲分子中所有碳原子一定都在同一平面上C. 1mol乙与足量NaOH溶液反应，最多消耗D. 甲催化氧化后，再发生酯化反应也能得到乙6.短周期主族元素X、Y、Z、W的原子序数依次增大。
Z原子2p轨道上有3个未成对电子。
甲、乙、丙、丁、戊是这四种元素之间形成的化合物，的甲溶液常用于消毒，戊是Z和X组成的10电子分子，常温下己溶液显中性，它们有如下转化关系，则下列说法中错误的是A. 四种元素原子半径由大到小的顺序为：B. W的氢化物中的某一种具有漂白性C. 丙的电子式为D. W的氢化物沸点一定比Y的氢化物沸点高7.是合或“中国蓝”的重要原料之一。
山东省潍坊市2015届高三数学二模试卷(理科)
2015年山东省潍坊市高考数学二模试卷（理科）一、选择题：本大题共10小题．每小题5分，共50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．设全集U=R，集合A={x||x|≤1}，B={x|log2x≤1}，则∁U A∩B等于（）A．（0，1] B． C．（1，2] D．（﹣∞，﹣1）∪2．设i是虚数单位，若复数a﹣（a∈R）是纯虚数，则a的值为（）A．﹣3 B．﹣1 C．1 D．33．已知命题p：∀x＞0，x+≥4：命题q：∃x0∈R+，2x0=，则下列判断正确的是（）A．p是假命题B．q是真命题C．p∧（￢q）是真命题D．（￢p）∧q是真命题4．已知m、n是两条不同的直线，α、β是两个不同的平面，则下列命题中正确的是（）A．若m⊥α，n⊥β，且m⊥n，则α⊥βB．若m∥α，n∥β，且m∥n，则α∥βC．若m⊥α，n∥β，且m⊥n，则α⊥βD．若m⊥α，n∥β，且m∥n，则α∥β5．若，且，则tanα=（）A．B．C．D．6．已知定义在R上的函数y=f（x）满足f（x+2）=2f（x），当x∈时，，则函数y=f（x）在上的大致图象是（）A．B．C． D．7．已知三棱锥S﹣ABC的所有顶点都在球O的球面上，△ABC是边长为1的正三角形，SC为球O的直径，且SC=2，则此棱锥的体积为（）A．B．C．D．8．某公司新招聘5名员工，分给下属的甲、乙两个部门，其中两名英语翻译人员不能分给同一部门；另三名电脑编程人员不能都分给同一个部门，则不同的分配方案种数是（）A．6 B．12 C．24 D．369．已知圆C：（x﹣3）2+（y﹣4）2=1和两点A（﹣m，0），B（m，0）（m＞0），若圆C上存在点P，使得∠APB=90°，则m的最大值为（）A．7 B．6 C．5 D．410．已知函数，若函数f（x）的零点都在（a＜b，a，b∈Z）内，则b﹣a的最小值是（）A．1 B．2 C．3 D．4二、填空题：本大题共5小题，每小题5分，共25分．11．某校对高三年级1600名男女学生的视力状况进行调查，现用分层抽样的方法抽取一个容量是200的样本，已知样本中女生比男生少10人，则该校高三年级的女生人数是．12．当输入的实数x∈时，执行如图所示的程序框图，则输出的x不小于103的概率是．13．已知G为△ABC的重心，令，，过点G的直线分别交AB、AC于P、Q两点，且，，则= ．14．抛物线C：y2=2px（p＞0）的焦点为F，点O是坐标原点，过点O，F的圆与抛物线C的准线相切，且该圆的面积为36π，则抛物线的方程为．15．定义在（0，+∞）上的函数f（x）满足：对∀x∈（0，+∞），都有f（2x）=2f（x）；当x∈（1，2]时，f（x）=2﹣x，给出如下结论：①对∀m∈Z，有f（2m）=0；②函数f（x）的值域为其中所有正确结论的序号是：．（请将所有正确命题的序号填上）三、解答题：本大题共6小题，共75分，解答时应写出必要的文字说明、证明过程或演算步骤.16．已知向量，把函数f（x）=化简为f（x）=Asin（tx+ϕ）+B的形式后，利用“五点法”画y=f（x）在某一个周期内的图象时，列表并填入的部分数据如表所示：x ①tx+ϕ 0 2πf（x） 0 1 0 ﹣1 0（Ⅰ）请直接写出①处应填的值，并求ω的值及函数y=f（x）在区间上的值域；（Ⅱ）设△ABC的内角A，B，C所对的边分别为a，b，c，已知，c=2，a=，求．17．如图，边长为的正方形ADEF与梯形ABCD所在的平面互相垂直，其中AB∥CD，AB⊥BC，DC=BC=AB=1，点M在线段EC上．（Ⅰ）证明：平面BDM⊥平面ADEF；（Ⅱ）判断点M的位置，使得平面BDM与平面ABF所成锐二面角为．18．已知等比数列数列{a n}的前n项和为S n，公比q＞0，S2=2a2﹣2，S3=a4﹣2．（Ⅰ）求数列{a n}的通项公式；（Ⅱ）令，T n为数列{c n}的前n项和，求T2n．19．某公司采用招考的方式引进人才，规定考生必须在B、C、D三个测试点中任意选取两个进行测试，若在这两个测试点都测试合格，则可参加面试，否则不被录用．已知考生在每个测试点的测试结果只有合格与不合格两种，且在每个测试点的测试结果互不影响．若考生小李和小王一起前来参加招考，小李在测试点B、C、D测试合格的概率分别为，，，小王在上述三个测试点测试合格的概率都是．（Ⅰ）问小李选择哪两个测试点测试才能使得可以参加面试的可能性最大？请说明理由；（Ⅱ）假设小李选择测试点B、C进行测试，小王选择测试点B、D进行测试，记ξ为两人在各测试点测试合格的测试点个数之和，求随机变量ξ的分布列及数学期望Eξ．20．已知椭圆E的中心在坐标原点O，其焦点与双曲线C：的焦点重合，且椭圆E 的短轴的两个端点与其一个焦点构成正三角形．（Ⅰ）求椭圆E的方程；（Ⅱ）过双曲线C的右顶点A作直线l与椭圆E交于不同的两点P、Q．①设M（m，0），当为定值时，求m的值；②设点N是椭圆E上的一点，满足ON∥PQ，记△NAP的面积为S1，△OAQ的面积为S2，求S1+S2的取值范围．21．设f（x）=alnx+bx﹣b，g（x）=，其中a，b∈R．（Ⅰ）求g（x）的极大值；（Ⅱ）设b=1，a＞0，若|f（x2）﹣f（x1）|＜||对任意的x1，x2∈（x1≠x2）恒成立，求a的最大值；（Ⅲ）设a=﹣2，若对任意给定的x0∈（0，e]，在区间（0，e]上总存在s，t（s≠t），使f （s）=f（t）=g（x0）成立，求b的取值范围．2015年山东省潍坊市高考数学二模试卷（理科）参考答案与试题解析一、选择题：本大题共10小题．每小题5分，共50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．设全集U=R，集合A={x||x|≤1}，B={x|log2x≤1}，则∁U A∩B等于（）A．（0，1] B． C．（1，2] D．（﹣∞，﹣1）∪考点：交、并、补集的混合运算．专题：集合．分析：求出A与B中不等式的解集确定出A与B，找出A补集与B的交集即可．解答：解：由A中不等式解得：﹣1≤x≤1，即A=，由B中不等式变形得：log2x≤1=log22，解得：0＜x≤2，即B=（0，2]，∴∁U A=（﹣∞，﹣1）∪（1，+∞），则（∁U A）∩B=（1，2]，故选：C．点评：此题考查了交、并、补集的混合运算，熟练掌握各自的定义是解本题的关键．2．设i是虚数单位，若复数a﹣（a∈R）是纯虚数，则a的值为（）A．﹣3 B．﹣1 C．1 D．3考点：复数的基本概念．专题：计算题．分析：利用复数的运算法则把a﹣（a∈R）可以化为（a﹣3）﹣i，再利用纯虚数的定义即可得到a．解答：解：∵=（a﹣3）﹣i是纯虚数，∴a﹣3=0，解得a=3．故选D．点评：熟练掌握复数的运算法则和纯虚数的定义是解题的关键．3．已知命题p：∀x＞0，x+≥4：命题q：∃x0∈R+，2x0=，则下列判断正确的是（）A．p是假命题B．q是真命题C．p∧（￢q）是真命题D．（￢p）∧q是真命题考点：命题的真假判断与应用．专题：简易逻辑．分析：利用基本不等式求最值判断命题p的真假，由指数函数的值域判断命题q的真假，然后结合复合命题的真值表加以判断．解答：解：当x＞0，x+≥，当且仅当x=2时等号成立，∴命题p为真命题，￢P为假命题；当x＞0时，2x＞1，∴命题q：∃x0∈R+，2x0=为假命题，则￢q为真命题．∴p∧（￢q）是真命题，（￢p）∧q是假命题．故选：C．点评：本题考查了命题的真假判断与应用，考查了复合命题的真假判断，考查了利用基本不等式求最值，是中档题．4．已知m、n是两条不同的直线，α、β是两个不同的平面，则下列命题中正确的是（）A．若m⊥α，n⊥β，且m⊥n，则α⊥βB．若m∥α，n∥β，且m∥n，则α∥βC．若m⊥α，n∥β，且m⊥n，则α⊥βD．若m⊥α，n∥β，且m∥n，则α∥β考点：空间中直线与平面之间的位置关系．专题：空间位置关系与距离．分析：利用线面垂直的性质，面面垂直的判定以及面面平行的判定定理分别分析选择．解答：解：若m⊥α，n⊥β，且m⊥n，则α⊥β，故A正确若m∥α，n∥β，且m∥n，则α与β平行或相交，故B错误若m⊥α，n∥β，且m⊥n，则α与β平行或相交，所以C错误．若m⊥α，m∥n，则n⊥α，又由n∥β，则α⊥β，故D错误；故选：A点评：本题考查直线与直线的位置关系及直线与平面的位置关系的判断、性质．解决此类问题的关键是熟练掌握空间中线面、面面得位置关系，以及与其有关的判定定理与性质定理．5．若，且，则tanα=（）A．B．C．D．考点：同角三角函数基本关系的运用．专题：三角函数的求值．分析：由条件利用诱导公式、二倍角公式，同角三角函数的基本关系求得3tan2α+20tanα﹣7=0，解方程求得tanα的值．解答：解：若，且，则cos2α﹣sin2α=（cos2α+sin2α），∴cos2α﹣sin2α﹣2sinαcosα=0，即 3tan2α+20tanα﹣7=0．求得tanα=，或 tanα=﹣7（舍去），故选：B．点评：本题主要考查同角三角函数的基本关系，诱导公式、二倍角公式的应用，以及三角函数在各个象限中的符号，属于基础题．6．已知定义在R上的函数y=f（x）满足f（x+2）=2f（x），当x∈时，，则函数y=f（x）在上的大致图象是（）A．B．C．D．考点：函数的图象．专题：函数的性质及应用．分析：由题意求出函数f（x）在上的解析式，问题得以解决．解答：解：∵f（x+2）=2f（x），∴f（x）=2f（x﹣2），设x∈，则x﹣2∈，∴f（x）=，当x∈，f（x）=﹣2x2+12x﹣16，图象过点（3，2），（4，0）的抛物线的一部分，故选：A点评：本题考查了函数的解析式的求法和函数的图象的识别，属于基础题，7．已知三棱锥S﹣ABC的所有顶点都在球O的球面上，△ABC是边长为1的正三角形，SC为球O的直径，且SC=2，则此棱锥的体积为（）A．B．C．D．考点：球内接多面体；棱柱、棱锥、棱台的体积．专题：压轴题．分析：先确定点S到面ABC的距离，再求棱锥的体积即可．解答：解：∵△ABC是边长为1的正三角形，∴△ABC的外接圆的半径∵点O到面ABC的距离，SC为球O的直径∴点S到面ABC的距离为∴棱锥的体积为故选A．点评：本题考查棱锥的体积，考查球内角多面体，解题的关键是确定点S到面ABC的距离．8．某公司新招聘5名员工，分给下属的甲、乙两个部门，其中两名英语翻译人员不能分给同一部门；另三名电脑编程人员不能都分给同一个部门，则不同的分配方案种数是（）A．6 B．12 C．24 D．36考点：计数原理的应用．专题：排列组合．分析：分类讨论：①甲部门要2个电脑编程人员和一个英语翻译人员；②甲部门要1个电脑编程人员和一个英语翻译人员，分别求得这2个方案的方法数，再利用分类计数原理，可得结论解答：解：由题意可得，有2种分配方案：①甲部门要2个电脑编程人员，则有3种情况；两名英语翻译人员的分配有2种可能；根据分步计数原理，共有3×2=6种分配方案．②甲部门要1个电脑编程人员，则有3种情况电脑特长学生，则方法有3种；两名英语翻译人员的分配方法有2种；共3×2=6种分配方案．由分类计数原理，可得不同的分配方案共有6+6=12种，故选：B．点评：本题考查计数原理的运用，根据题意分步或分类计算每一个事件的方法数，然后用乘法原理和加法原理计算，是解题的常用方法9．已知圆C：（x﹣3）2+（y﹣4）2=1和两点A（﹣m，0），B（m，0）（m＞0），若圆C上存在点P，使得∠APB=90°，则m的最大值为（）A．7 B．6 C．5 D．4考点：直线与圆的位置关系．专题：直线与圆．分析：根据圆心C到O（0，0）的距离为5，可得圆C上的点到点O的距离的最大值为6．再由∠APB=90°，可得PO=AB=m，可得m≤6，从而得到答案．解答：解：圆C：（x﹣3）2+（y﹣4）2=1的圆心C（3，4），半径为1，∵圆心C到O（0，0）的距离为5，∴圆C上的点到点O的距离的最大值为6．再由∠APB=90°可得，以AB为直径的圆和圆C有交点，可得PO=AB=m，故有m≤6，故选：B．点评：本题主要直线和圆的位置关系，求得圆C上的点到点O的距离的最大值为6，是解题的关键，属于中档题．10．已知函数，若函数f（x）的零点都在（a＜b，a，b∈Z）内，则b﹣a的最小值是（）A．1 B．2 C．3 D．4考点：函数的单调性与导数的关系；函数零点的判定定理．专题：计算题；函数的性质及应用；导数的综合应用．分析：首先可判断f（0）=1＞0，f（﹣1）=1﹣1﹣﹣﹣…﹣＜0；再判断f（x）在（0，+∞）上单调递增，在（﹣∞，﹣1）上单调递增，从而说明没有零点，从而解得．解答：解：∵，∴f（0）=1＞0，f（﹣1）=1﹣1﹣﹣﹣…﹣＜0；故在上有零点；f′（x）=1﹣x+x2﹣x3+ (x2014)易知f′（1）=1，当x＞0且x≠1时，f′（x）=1﹣x+x2﹣x3+…+x2014==＞0，故f（x）在（0，+∞）上单调递增，且f（0）＞0；故f（x）在（0，+∞）上没有零点，当x＜﹣1时，f′（x）=1﹣x+x2﹣x3+…+x2014==＞0，故f（x）在（﹣∞，﹣1）上单调递增，且f（﹣1）＜0，故f（x）在（﹣∞，﹣1）上没有零点；综上所述，函数的零点都在区间上，故选A．点评：本题考查了导数的综合应用及函数的零点的判断，属于基础题．二、填空题：本大题共5小题，每小题5分，共25分．11．某校对高三年级1600名男女学生的视力状况进行调查，现用分层抽样的方法抽取一个容量是200的样本，已知样本中女生比男生少10人，则该校高三年级的女生人数是760 ．考点：分层抽样方法．专题：应用题；概率与统计．分析：先计算出样本中高三年级的女学生人数，再根据分层抽样的性质计算出该校高三年级的女生的人数．解答：解：根据题意，设样本中高三年级的女生人数为x，则（x+10）+x=200，解得x=95，所以该校高三年级的女生人数是1600×200=760．故答案为：760．点评：本题考查分层抽样，先计算中样本中高三年级的男女学生的人数是解决本题的关键，属基础题．12．当输入的实数x∈时，执行如图所示的程序框图，则输出的x不小于103的概率是．考点：程序框图．专题：图表型；算法和程序框图．分析：由程序框图的流程，写出前三项循环得到的结果，得到输出的值与输入的值的关系，令输出值大于等于103得到输入值的范围，利用几何概型的概率公式求出输出的x不小于103的概率．解答：解：设实数x∈，经过第一次循环得到x=2x+1，n=2，经过第二循环得到x=2（2x+1）+1，n=3，此时输出x，输出的值为4x+3，令4x+3≥103得x≥25，由几何概型得到输出的x不小于103的概率为P==．故答案为：．点评：解决程序框图中的循环结构时，一般采用先根据框图的流程写出前几次循环的结果，根据结果找规律，属于基础题．13．已知G为△ABC的重心，令，，过点G的直线分别交AB、AC于P、Q两点，且，，则= 3 ．考点：平面向量的基本定理及其意义．专题：平面向量及应用．分析：显然，根据G点为重心，从而可以用表示，而和共线，从而，而已知，从而会最后得到关于的式子：，从而得到，两式联立消去x即可求出答案．解答：解：如图，=；∴；G为△ABC的重心；∴，；∴；整理得，；∴；消去x得，；∴．故答案为：3．点评：考查向量加法、减法的几何意义，共线向量基本定理，重心的性质：重心到顶点距离是它到对边中点距离的2倍，以及向量加法的平行四边形法则，向量的加法、减法运算，平面向量基本定理．14．抛物线C：y2=2px（p＞0）的焦点为F，点O是坐标原点，过点O，F的圆与抛物线C的准线相切，且该圆的面积为36π，则抛物线的方程为y2=16x ．考点：抛物线的简单性质．专题：圆锥曲线的定义、性质与方程．分析：由题意画出图形，结合三角形的面积求出半径，再由M的坐标相等求得p，则抛物线方程可求．解答：解：如图，由题意可知，圆的圆心M在抛物线上，又圆的面积为36π，∴半径|OM|=6，则|MF|=，即，又，∴，解得：p=8．∴抛物线方程为：y2=16x．故答案为：y2=16x．点评：本题考查了抛物线的几何性质，考查了数学结合的解题思想方法，训练了抛物线焦半径公式的应用，是中档题．15．定义在（0，+∞）上的函数f（x）满足：对∀x∈（0，+∞），都有f（2x）=2f（x）；当x∈（1，2]时，f（x）=2﹣x，给出如下结论：①对∀m∈Z，有f（2m）=0；②函数f（x）的值域为时，f（x）=2﹣x，∴∀x∈（1，2]，f（x）≥f（2）=0，又∵∀x∈（0，+∞），f（2x）=2f（x），∴∀x∈（0，+∞），f（x）≥f（2）=0，∴②正确；对于③，∵f（2n+1）=2n+1﹣2n﹣1，假设存在n使f（2n+1）=9，即存在x1，x2，﹣=10，又2x变化如下：2，4，8，16，32，显然不存在满足条件的值，∴③错误；对于④，根据②知，当x⊆（2k，2k+1）时，f（x）=2k+1﹣x为减函数，∴函数f（x）在区间（a，b）上单调递减的充分条件是“存在k∈Z，使得（a，b）⊆（2k，2k+1），④正确．综上，正确的命题是①②④．故答案为：①②④．点评：本题考查了抽象函数及其应用问题，考查了利用赋值法证明等式的问题，此类题的特征是根据题中所给的相关性质灵活赋值以达到求值或者证明命题的目的，是综合性题目．三、解答题：本大题共6小题，共75分，解答时应写出必要的文字说明、证明过程或演算步骤.16．已知向量，把函数f（x）=化简为f（x）=Asin（tx+ϕ）+B的形式后，利用“五点法”画y=f（x）在某一个周期内的图象时，列表并填入的部分数据如表所示：x ①tx+ϕ 0 2πf（x） 0 1 0 ﹣1 0（Ⅰ）请直接写出①处应填的值，并求ω的值及函数y=f（x）在区间上的值域；（Ⅱ）设△ABC的内角A，B，C所对的边分别为a，b，c，已知，c=2，a=，求．考点：由y=Asin（ωx+φ）的部分图象确定其解析式；平面向量数量积的运算．专题：三角函数的求值；三角函数的图像与性质；平面向量及应用．分析：（Ⅰ）由三角函数恒等变换化简解析式可得f（x）=sin（2），由T=2（）=π，可求ω，由x∈，可求2x﹣的范围，即可求得f（x）的值域．（Ⅱ）由f（）=sin（A+）=1，根据A+的范围，可解得A，由余弦定理解得b，cosB，利用平面向量数量积的运算即可得解．解答：解：（Ⅰ）①处应填…1分f（x）=m•n+=sinωxcosωx﹣cos2ωx+=sin2ωx﹣+=sin2ωx﹣cos2ωx=sin（2）…3分因为T=2（）=π，所以由，ω=1．∴f（x）=sin（2x﹣）．因为x∈，所以﹣≤2x﹣≤，所以﹣1≤sin（2x﹣）≤，∴f（x）的值域为…6分（Ⅱ）因为f（）=sin（A+）=1，因为0＜A＜π，所以＜A+＜，所以A+=，A=，由余弦定理a2=b2+c2﹣2bccosA，得（）2=b2+22﹣2×，即b2﹣2b﹣3=0，解得b=3或b=﹣1（舍去），∴cosB==．所以=||||cosB=2×=1…12分点评：本题主要考查了由y=Asin（ωx+φ）的部分图象确定其解析式，平面向量数量积的运算，考查了余弦定理的应用，属于中档题．17．如图，边长为的正方形ADEF与梯形ABCD所在的平面互相垂直，其中AB∥CD，AB⊥BC，DC=BC=AB=1，点M在线段EC上．（Ⅰ）证明：平面BDM⊥平面ADEF；（Ⅱ）判断点M的位置，使得平面BDM与平面ABF所成锐二面角为．考点：二面角的平面角及求法；平面与平面垂直的判定．专题：空间角．分析：（Ⅰ）由已知三角形的半径关系得到AD⊥BD，再由面面垂直的性质得到ED⊥面ABCD，进一步得到BD⊥ED，利用线面垂直的判定得到BD⊥面ADEF，由BD⊂面BDM，利用面面垂直的判定得到平面BDM⊥平面ADEF；（Ⅱ）在面DAB内过D作DN⊥AB，垂足为N，则可证得DN⊥CD，以D为坐标原点，DN所在直线为x轴，DC所在直线为y轴，DE所在直线为z轴，建立空间直角坐标系，求出所用点的坐标，结合E，M，C三点共线得到，把M的坐标用含有λ的代数式表示，求出平面BDM的法向量，再由平面ABF的法向量为，由平面BDM 与平面ABF所成锐二面角为求得．则点M的坐标可求，位置确定．解答：（Ⅰ）证明：如图，∵DC=BC=1，DC⊥BC，∴BD=，又∵AD=，AB=2，∴AD2+BD2=AB2，则∠ADB=90°，∴AD⊥BD．又∵面ADEF⊥面ABCD，ED⊥AD，面ADEF∩面ABCD=AD，∴ED⊥面ABCD，则BD⊥ED，又∵AD∩DE=D，∴BD⊥面ADEF，又BD⊂面BDM，∴平面BDM⊥平面ADEF；（Ⅱ）在面DAB内过D作DN⊥AB，垂足为N，∵AB∥CD，∴DN⊥CD，又∵ED⊥面ABCD，∴DN⊥ED，∴以D为坐标原点，DN所在直线为x轴，DC所在直线为y轴，DE所在直线为z轴，建立空间直角坐标系，∴B（1，1，0），C（0，1，0），E（0，0，），N（1，0，0），设M（x0，y0，z0），由，得，∴x 0=0，，则M（0，λ，），设平面BDM的法向量，则，∴，令x=1，得．∵平面ABF的法向量，∴，解得：．∴M（0，），∴点M的位置在线段CE的三等分点且靠近C处．点评：本题主要考查直线与平面之间的平行、垂直等位置关系，二面角的概念、求法等知识，以及空间想象能力和逻辑推理能力，训练了利用空间向量求二面角的平面角，是中档题．18．已知等比数列数列{a n}的前n项和为S n，公比q＞0，S2=2a2﹣2，S3=a4﹣2．（Ⅰ）求数列{a n}的通项公式；（Ⅱ）令，T n为数列{c n}的前n项和，求T2n．考点：数列的求和；数列递推式．专题：等差数列与等比数列．分析：（I）利用等比数列的通项公式即可得出．（II）由（I）可得：c n=．可得T2n=（c1+c3+…+c2n﹣1）+（c2+c4+…+c2n），对奇数项与偶数项分别利用“裂项求和”、“错位相减法”即可得出．解答：解：（I）∵S2=2a2﹣2，S3=a4﹣2．∴S3﹣S2=a4﹣2a2=a3，∴，a2≠0，化为q2﹣q﹣2=0，q＞0，解得q=2，又a1+a2=2a2﹣2，∴a2﹣a1﹣2=0，∴2a1﹣a1﹣2=0，解得a1=2，∴．（II）由（I）可得：c n=．∴T2n=（c1+c3+…+c2n﹣1）+（c2+c4+…+c2n），记M=（c2+c4+…+c2n）=+…+=+…+，则=+…+，∴=+…+﹣=﹣=，∴M=﹣．∴T2n=+M=+M=+﹣．点评：本题考查了“错位相减法”、等比数列的通项公式及其前n项和公式、“裂项求和”，考查了推理能力与计算能力，属于中档题．19．某公司采用招考的方式引进人才，规定考生必须在B、C、D三个测试点中任意选取两个进行测试，若在这两个测试点都测试合格，则可参加面试，否则不被录用．已知考生在每个测试点的测试结果只有合格与不合格两种，且在每个测试点的测试结果互不影响．若考生小李和小王一起前来参加招考，小李在测试点B、C、D测试合格的概率分别为，，，小王在上述三个测试点测试合格的概率都是．（Ⅰ）问小李选择哪两个测试点测试才能使得可以参加面试的可能性最大？请说明理由；（Ⅱ）假设小李选择测试点B、C进行测试，小王选择测试点B、D进行测试，记ξ为两人在各测试点测试合格的测试点个数之和，求随机变量ξ的分布列及数学期望Eξ．考点：离散型随机变量的期望与方差；相互独立事件的概率乘法公式；离散型随机变量及其分布列．专题：概率与统计．分析：（Ⅰ）设考生小李在B，C，D各测试点测试合格记为事件B、C、D，且各事件相互独立，已知．求出小李在（B、C），（B、D），（C、D）测试点测试参加面试的概率，由概率的大小得答案；（Ⅱ）记小李在测试点B、C合格为事件B、C，小王在测试点B、D合格为事件B1、D1，由题意得到，求出ξ的所有取值，然后利用相互独立事件和定理重复试验求得概率，列出分布列，然后由期望公式求期望．解答：解：（Ⅰ）设考生小李在B，C，D各测试点测试合格记为事件B、C、D，且各事件相互独立，由题意，．若选择在B、C测试点测试，则参加面试的概率，若选择在B、D测试点测试，则参加面试的概率，若选择在C、D测试点测试，则参加面试的概率．∵P2＞P1＞P3，∴小李在B、D测试点测试，参加面试的可能性大．（Ⅱ）记小李在测试点B、C合格为事件B、C，小王在测试点B、D合格为事件B1、D1，则，且ξ的所有取值为0，1，2，3，4．P（ξ=0）=，P（ξ=1）==，P（ξ=2）==，P（ξ=3）==，P（ξ=4）=．ξ的分布列为：ξ 0 1 2 3 4P∴数学期望Eξ=．点评：本题考查了离散型随机变量的期望的应用，离散型随机变量的期望表征了随机变量取值的平均值，考查了相互独立事件和独立重复试验，是中档题．20．已知椭圆E的中心在坐标原点O，其焦点与双曲线C：的焦点重合，且椭圆E 的短轴的两个端点与其一个焦点构成正三角形．（Ⅰ）求椭圆E的方程；（Ⅱ）过双曲线C的右顶点A作直线l与椭圆E交于不同的两点P、Q．①设M（m，0），当为定值时，求m的值；②设点N是椭圆E上的一点，满足ON∥PQ，记△NAP的面积为S1，△OAQ的面积为S2，求S1+S2的取值范围．考点：直线与圆锥曲线的综合问题．专题：综合题；圆锥曲线的定义、性质与方程．分析：（Ⅰ）设方程为，确定c，利用椭圆E的短轴的两个端点与其一个焦点构成正三角形，可得a=2b，利用a2=b2+c2，求出a，b，即可求椭圆E的方程；（Ⅱ）①分类讨论，设l的方程为y=k（x﹣1），代入椭圆方程，利用韦达定理，结合向量的数量积公式，可得结论；②确定S1+S2=S△OPQ，求出|PQ|，可得面积，换元确定面积的范围即可求S1+S2的取值范围．解答：解：（Ⅰ）由题意椭圆的焦点在x轴上，设方程为，其左右焦点为F1（﹣，0），F2（，0），∴c=，∵椭圆E的短轴的两个端点与其一个焦点构成正三角形，∴a=2b，∵a2=b2+c2，∴a=2，b=1，∴椭圆E的方程为；（Ⅱ）①双曲线C右顶点为A（1，0），当直线l的斜率存在时，设l的方程为y=k（x﹣1），代入椭圆方程得（4k2+1）x2﹣8k2x+4k2﹣4=0，设直线l与椭圆E交点P（x1，y1），Q（x2，y2），则x1+x2=，x1x2=，∴•=m2﹣m（x1+x2）+x1x2+y1y2==（4m2﹣8m+1）+，当2m﹣=0，即m=时，•=．当直线l的斜率不存在时，直线l的方程为x=1，代入椭圆方程可得x=1，y=±．不妨设P（1，），Q（1，﹣），由M（，0）可得=（，﹣），=（，），∴•=，综上所述，m=时，•为定值；②∵ON∥PQ，∴S△NAP=S△OAP，∴S1+S2=S△OPQ，∵|PQ|=4•，∵原点O到直线PQ的距离为d=（k≠0），∴S△OPQ==令t=4k2+1，则k2=（t＞1），∴S==，∵t＞1，∴0＜＜1，∴0＜﹣+4＜3，∴0＜S＜．当直线l的斜率不存在时，S△OPQ==，综上所述，S1+S2的取值范围是（0，]．点评：本题考查椭圆方程，考查直线与椭圆的位置关系，考查向量知识的运用，考查韦达定理，有难度．21．设f（x）=alnx+bx﹣b，g（x）=，其中a，b∈R．（Ⅰ）求g（x）的极大值；（Ⅱ）设b=1，a＞0，若|f（x2）﹣f（x1）|＜||对任意的x1，x2∈（x1≠x2）恒成立，求a的最大值；（Ⅲ）设a=﹣2，若对任意给定的x0∈（0，e]，在区间（0，e]上总存在s，t（s≠t），使f （s）=f（t）=g（x0）成立，求b的取值范围．考点：利用导数研究函数的极值；利用导数研究函数的单调性；利用导数求闭区间上函数的最值．专题：函数的性质及应用；导数的综合应用；不等式的解法及应用．分析：（Ⅰ）求出g（x）的导数，令导数大于0，得增区间，令导数小于0，得减区间，进而求得g（x）的极大值；（Ⅱ）当b=1，a＞0时，求出f（x）的导数，以及h（x）=的导数，判断单调性，去掉绝对值可得f（x2）﹣f（x1）＜h（x2）﹣h（x1），即f（x2）﹣h（x2）＜f（x1）﹣h（x1），F（x）=f（x）﹣h（x），F（x）在递减，求得F（x）的导数，通过分离参数，求出右边的最小值，即可得到a的范围；（Ⅲ）求出g（x）的导数，通过单调区间可得函数g（x）在（0，e]上的值域为（0，1]．由题意，当f（x）取（0，1]的每一个值时，在区间（0，e]上存在t1，t2（t1≠t2）与该值对应．a=﹣2时，f（x）=b（x﹣1）﹣2lnx，求出f（x）的导数，由题意，f（x）在区间（0，e]上不单调，所以，0＜＜e，再由导数求得f（x）的最小值，即可得到所求范围．解答：解：（Ⅰ）g′（x）==，当x＞1时，g′（x）＜0，g（x）在（1，+∞）递增；当x＜1时，g′（x）＞0，g（x）在（﹣∞，1）递减．则有g（x）的极大值为g（1）=1；（Ⅱ）当b=1，a＞0时，f（x）=alnx+x﹣1，x＞0，f′（x）=+1=＞0在恒成立，f（x）在递增；由h（x）==，h′（x）=＞0在恒成立，h（x）在递增．设x1＜x2，原不等式等价为f（x2）﹣f（x1）＜h（x2）﹣h（x1），即f（x2）﹣h（x2）＜f（x1）﹣h（x1），F（x）=f（x）﹣h（x），F（x）在递减，又F（x）=alnx+x﹣1﹣，F′（x）=+1﹣≤0在恒成立，故h（x）在递增，a≤•﹣x，令G（x）=•﹣x，3≤x≤4，G′（x）=•﹣1=e x﹣1（﹣+1）﹣1=e x﹣1﹣1＞e2﹣1＞0，G（x）在递增，即有a≤e2﹣3，即a max=e2﹣3；（Ⅲ）g′（x）=e1﹣x﹣xe1﹣x=（1﹣x）e1﹣x，当x∈（0，1）时，g′（x）＞0，函数g（x）单调递增；当x∈（1，e]时，g′（x）＜0，函数g（x）单调递减．又因为g（0）=0，g（1）=1，g（e）=e2﹣e＞0，所以，函数g（x）在（0，e]上的值域为（0，1]．由题意，当f（x）取（0，1]的每一个值时，在区间（0，e]上存在t1，t2（t1≠t2）与该值对应．a=﹣2时，f（x）=b（x﹣1）﹣2lnx，f′（x）=b﹣=，当b=0时，f′（x）=﹣＜0，f（x）单调递减，不合题意，当b≠0时，x=时，f′（x）=0，由题意，f（x）在区间（0，e]上不单调，所以，0＜＜e，当x∈（0，]时，f'（x）＜0，当（，+∞）时，f'（x）＞0所以，当x∈（0，e]时，f（x）min=f（）=2﹣a﹣2ln，由题意，只需满足以下三个条件：①f（x）min=f（）=2﹣b﹣2ln＜0，②f（e）=b（e﹣1）﹣2≥1，③∃x0∈（0，）使f（x0）＞1．∵f（）≤f（1）=0，所以①成立．由②f（x）=b（x﹣1）﹣2lnx→+∞，所以③满足，所以当b满足即b≥时，符合题意，故b的取值范围为[，+∞）．点评：本题考查导数的运用：求单调区间和极值，主要考查不等式恒成立和存在性问题，注意运用参数分离和构造函数通过导数判断单调性，求出最值，属于难题．。
第二节 二项式定理
第二节二项式定理考试要求1．理解二项式定理，二项式系数的性质.2.会用二项式定理解决与二项展开式有关的简单问题．[知识排查·微点淘金]知识点1二项式定理(1)二项式定理：(a＋b)n＝C0n a n＋C1n a n－1b＋…＋C k n a n－k·b k＋…＋C n n b n(n∈N*)；上述公式叫做二项式定理．[微思考](a＋b)n与(b＋a)n的展开式有何区别与联系？提示：(a＋b)n的展开式与(b＋a)n的展开式的项完全相同，但对应的项不相同而且两个展开式的通项不同．(2)通项公式：T k＋1＝C k n a n－k b k叫做二项展开式的通项，它表示展开式的第k＋1项；(3)二项式系数：二项展开式中各项的系数C0n，C1n，…，C n n叫做二项式系数．知识点2二项式系数的性质[微提醒]易混淆二项式中的“项”“项的系数”“项的二项式系数”等概念，注意项的系数是指非字母因数所有部分，包含符号，二项式系数仅指C k n(k＝0，1，…，n).[小试牛刀·自我诊断]1．思考辨析(在括号内打“√”或“×”)(1)C k n a n－k b k是(a＋b)n的展开式中的第k项．(×)(2)二项展开式中，系数最大的项为中间一项或中间两项．(×)(3)(a ＋b )n 的展开式中某一项的二项式系数与a ，b 无关．(√)(4)通项公式T k ＋1＝C k n an －k b k中的a 和b 不能互换．(√) (5)(a ＋b )n 的展示式中某项的系数是该项中非字母因数部分，包括符号等，与该项的二项式系数不同．(√)2．(链接教材选修2－3 P 37A 组T 5)二项式⎝⎛⎭⎪⎫3x ＋12x 8的展开式的常数项是 ．答案：73．(链接教材选修2－3 P 37A 组T 8)在二项式⎝⎛⎭⎫x －1x n 的展开式中只有第5项的二项式系数最大，则展开式中含x 2项的系数是 ．答案：－564．(链接教材选修2－3 P 40A 组T 8)若⎝⎛⎭⎫x 3＋1x n的展开式的所有二项式系数的和为128，则n ＝ .答案：75．(混淆项的系数与二项式系数)在二项式⎝⎛⎭⎫x 2－2x n 的展开式中，所有二项式系数的和是32，则展开式中各项系数的和为 ．答案：－1一、基础探究点——求展开式中的特定项或特定项的系数(题组练透)1．(2020·北京卷)在(x －2)5的展开式中，x 2的系数为( ) A ．－5 B ．5 C ．－10D ．10解析：选C 由二项式定理得(x －2)5的展开式的通项T r ＋1＝C r 5(x )5－r (－2)r ＝C r 5(－2)rx5－r2，令5－r2＝2，得r ＝1，所以T 2＝C 15(－2)x 2＝－10x 2，所以x 2的系数为－10，故选C ． 2．(2020·全国卷Ⅰ)⎝⎛⎭⎫x ＋y2x (x ＋y )5的展开式中x 3y 3的系数为( ) A ．5 B ．10 C ．15D ．20解析：选C 解法一：∵⎝⎛⎭⎫x ＋y 2x (x ＋y )5＝⎝⎛⎭⎫x ＋y2x (x 5＋5x 4y ＋10x 3y 2＋10x 2y 3＋5xy 4＋y 5)，∴x 3y 3的系数为10＋5＝15.解法二：当x ＋y 2x 中取x 时，x 3y 3的系数为C 35， 当x ＋y 2x 中取y 2x时，x 3y 3的系数为C 15， ∴x 3y 3的系数为C 35＋C 15＝10＋5＝15.故选C ．3．(2021·北京卷)⎝⎛⎭⎫x 3－1x 4的展开式中常数项是 ． 解析：由二项式的展开式可得C 34·(x 3)1·⎝⎛⎭⎫－1x 3＝－4. 答案：－44．(2021·江西南昌模拟)已知(x －1)(ax ＋1)6的展开式中含x 2项的系数为0，则正实数a ＝ .解析：(ax ＋1)6的展开式中含x 2项的系数为C 46a 2，含x 项的系数为C 56a ，由(x －1)(ax ＋1)6的展开式中含x 2项的系数为0，可得－C 46a 2＋C 56a ＝0，因为a 为正实数，所以15a ＝6，所以a ＝25.答案：255. (x 2＋x ＋y )5的展开式中，x 5y 2项的系数为( ) A ．10 B ．20 C ．30D ．60解析：选C 解法一：(x 2＋x ＋y )5＝[(x 2＋x )＋y ]5，含y 2的项为T 3＝C 25(x 2＋x )3y 2.其中(x 2＋x )3中含x 5的项为C 13x 4·x ＝C 13x 5.所以x 5y 2的系数为C 25×C 13＝30. 解法二：(x 2＋x ＋y )5表示5个x 2＋x ＋y 之积，所以x 5y 2可从其中5个因式中，2个取因式中的x 2，剩余的3个因式中1个取x, 2个因式取y ，因此x 5y 2的系数为C 25C 13C 22＝30.1.求二项展开式中的特定项问题，实质是考查通项T k ＋1＝C k n an －k b k 的特点，一般需要先建立方 程求k ，再将k 的值代回通项求解，注意k 的取值范围(k ＝0,1,2，…，n )．2．求三项展开式中某些特定项的系数的方法：(1)通过变形先把三项式转化为二项式，再用二项式定理求解；(2)两次利用二项式定理的通项公式求解；(3)由二项式定理的推证方法知，可用排列、组合的基本原理去求，即把三项式看作几个因式之积，要得到特定项看有多少种方法从这几个因式中取因式中的量．二、综合探究点——二项式系数与各项系数和问题(思维拓展)[典例剖析][例](1)在二项式(1－2x)n的展开式中，偶数项的二项式系数之和为128，则展开式的中间项的系数为()A．－960B．960C．1120 D．1680解析：根据题意，奇数项的二项式系数之和也应为128，所以在(1－2x)n的展开式中，二项式系数之和为256，即2n＝256，解得n＝8，则(1－2x)8的展开式的中间项为第5项，且T5＝C48(－2)4x4＝1120x4，即展开式的中间项的系数为1120.故选C．答案：C(2)若(1－2x)8＝a0＋a1x＋a2x2＋…＋a8x8，则|a0|＋|a1|＋|a2|＋|a3|＋…＋|a8|＝()A．28－1 B．28C．38－1 D．38解析：由题可知，x的奇数次幂的系数均为负数，所以|a0|＋|a1|＋|a2|＋|a3|＋…＋|a8|＝a0－a1＋a2－a3＋…＋a8.因为(1－2x)8＝a0＋a1x＋a2x2＋…＋a8x8，令x＝－1得a0－a1＋a2－a3＋…＋a8＝38，则|a0|＋|a1|＋|a2|＋|a3|＋…＋|a8|＝38.故选D．答案：D(3)(2021·浙江卷)已知多项式(x－1)3＋(x＋1)4＝x4＋a1x3＋a2x2＋a3x＋a4，则a1＝，a2＋a3＋a4＝.解析：(x－1)3的展开式的通项为T r＋1＝C r3x3－r·(－1)r，(x＋1)4的展开式的通项为T r＋1＝C r4x4－r1r，则a1x3＝C03x3·(－1)0＋C14x311＝5x3，所以a1＝5.同理，a2x2＝C13x2(－1)1＋C24x212＝－3x2＋6x2＝3x2，a3x＝C23x1(－1)2＋C34x113＝3x＋4x＝7x，a4＝C33x0(－1)3＋C44x014＝0，所以a2＝3，a3＝7，a4＝0，所以a2＋a3＋a4＝10.答案：5101．赋值法的应用二项式定理给出的是一个恒等式，对于x，y的一切值都成立．因此，可将x，y设定为一些特殊的值．在使用赋值法时，令x ，y 等于多少，应视具体情况而定，一般取“1，－1或0”，有时也取其他值．如：(1)形如(ax ＋b )n ，(ax 2＋bx ＋c )m (a ，b ∈R )的式子，求其展开式的各项系数之和，只需令x ＝1即可．(2)形如(ax ＋by )n (a ，b ∈R )的式子，求其展开式各项系数之和，只需令x ＝y ＝1即可． 2．二项展开式系数最大项的求法如求(a ＋bx )n (a ，b ∈R )的展开式系数最大的项，一般是采用待定系数法，设展开式各项系数分别为A 1，A 2，…，A n ＋1，且第k 项系数最大，应用⎩⎪⎨⎪⎧A k ≥A k －1，A k ≥A k ＋1，求解出正整数k 即可．[学会用活]1．(2021·安徽宣城调研)若(2－x )7＝a 0＋a 1(1＋x )＋a 2(1＋x )2＋…＋a 7(1＋x )7，则a 0＋a 1＋a 2＋…＋a 6的值为( )A ．1B ．2C ．129D ．2188解析：选C 令x ＝0得a 0＋a 1＋a 2＋…＋a 7＝27＝128，又(2－x )7＝[3－(x ＋1)]7，则a 7(1＋x )7＝C 77·30·[－(x ＋1)]7，解得a 7＝－1.故a 0＋a 1＋a 2＋…＋a 6＝128－a 7＝128＋1＝129. 2．(2021·广西高三5月联考)若(a ＋x 2)(1＋x )n 的展开式中各项系数之和为192，且常数项为2，则该展开式中x 4的系数为( )A ．30B ．45C ．60D ．81解析：选B 令x ＝0，得a ＝2，所以(a ＋x 2)(1＋x )n ＝(2＋x 2)(1＋x )n .令x ＝1，得3×2n＝192，所以n ＝6.故该展开式中x 4的系数为2C 46＋C 26＝45.故选B ．3．已知m 为正整数，(x ＋y )2m 展开式的二项式系数的最大值为a ，(x ＋y )2m＋1展开式的二项式系数的最大值为b .若13a ＝7b ，则m 等于( )A ．5B ．6C ．7D ．8解析：选B 由题意可知，a ＝C m 2m ，b ＝C m2m ＋1，∵13a ＝7b ，∴13·2m ！m ！m ！＝7·2m ＋1！m ！m ＋1！，即137＝2m ＋1m ＋1，解得m ＝6.限时规范训练 基础夯实练1．(2021·河北唐山二模)在⎝⎛⎭⎫x －2x 6的展开式中，常数项为( ) A ．20 B ．－20 C ．160D ．－160解析：选D ⎝⎛⎭⎫x －2x 6展开式的通项T k ＋1＝C k 6x 6－k ⎝⎛⎭⎫－2x k ＝(－1)k 2k C k 6x 6－2k ，令6－2k ＝0，得k ＝3，所常数项T 3＋1＝(－1)323C 36＝－160，故选D ．2．(2021·北京东城区二模)已知(2x ＋a )5的展开式中x 2的系数为－40，那么a ＝( ) A ．－2 B ．－1 C ．1D ．2解析：选B (2x ＋a )5的展开式通项为T r ＋1＝C r 5·(2x )5－r ·a r ＝C r 5·25－r a r x 5－r ，令5－r ＝2，可得r ＝3，所以，C 35·22a 3＝40a 3＝－40，解得a ＝－1.故选B ． 3．(2021·四川乐至中学月考)(1＋2x )5的展开式中，各项二项式系数的和是( ) A ．1 B ．－1 C ．25D ．35解析：选C 由题得各项二项式系数和为C 05＋C 15＋C 25＋C 35＋C 45＋C 55＝25.故选C ．4．(2021·陕西西安模拟)若(2－x )10展开式中二项式系数和为A ，所有项系数和为B ，一次项系数为C ，则A ＋B ＋C ＝( )A ．4095B ．4097C ．－4095D ．－4097解析：选C 由(2－x )10展开式的通项公式为T r ＋1＝C r 10·210－r ·(－x )r ＝(－1)r ·210－r C r 10·x r ，所以一次项系数C ＝(－1)1·29·C 110＝－5120，二项式系数和A ＝210＝1024，令x ＝1，则所有项的系数和B ＝(2－1)10＝1，所以A ＋B ＋C ＝－4095.故选C ．5.⎝⎛⎭⎫x －x2y (x ＋2y )5的展开式中x 2y 4的系数为( )A ．24B ．36C ．48D ．72解析：选C 因为⎝⎛⎭⎫x －x 2y (x ＋2y )5＝x (x ＋2y )5－x2y(x ＋2y )5，可得(x ＋2y )5的展开式通项为T r ＋1＝C r 5x 5－r (2y )r ＝2r C r 5x5－r y r， 令r ＝4可得x 2y 4的系数为24C 45＝80，令r ＝5，可得x 2y 4的系数为－25C 55＝－32，故展开式中x 2y 4的系数为80－32＝48.故选C ．6．(2021·福建福州二模)在(x ＋y ＋z )6的展开式中，xyz 4的系数是( ) A ．15 B ．30 C ．36D ．60解析：选B 因为(x ＋y ＋z )6＝[(x ＋y )＋z ]6，所以[(x ＋y )＋z ]6的通项公式为C r 6·(x ＋y )6－r·z r ，令r ＝4，所以C 46·(x ＋y )2·z 4＝15(x 2＋2xy ＋y 2)z 4，因此xyz 4的系数是15×2＝30，故选B ． 7．(2021·广东韶关一模)已知(1＋x )10＝a 0＋a 1(2＋x )＋a 2(2＋x )2＋…＋a 10(2＋x )10，则a 9＝( )A ．－10B ．10C ．－45D ．45解析：选A (1＋x )10＝[1－(2＋x )]10＝a 0＋a 1(2＋x )＋a 2(2＋x )2＋…＋a 10(2＋x )10，T r ＋1＝C r 10[－(2＋x )]r ，a 9＝C 910(－1)9＝－10.故选A ．8．(2021·山东潍坊二模)已知正整数n ≥7，若⎝⎛⎭⎫x －1x (1－x )n 的展开式中不含x 5的项，则n 的值为( )A ．7B ．8C ．9D ．10解析：选D (1－x )n 的二项展开式中第k ＋1项为T k ＋1＝C k n (－1)k x k，又因为⎝⎛⎭⎫x －1x (1－x )n ＝x (1－x )n －1x (1－x )n 的展开式不含x 5的项，所以x C 4n (－1)4x 4－1x C 6n(－1)6x 6＝0，C 4n x 5－C 6n x 5＝0，即C 4n ＝C 6n，所以n ＝10，故选D ． 9．(2021·湖南岳阳二模)若(1＋x )(1－2x )7＝a 0＋a 1x ＋a 2x 2＋…＋a 8x 8，则a 1＋a 2＋…＋a 7＋a 8的值为 ．解析：令x ＝1，得a 0＋a 1＋a 2＋…＋a 7＋a 8＝－2，令x ＝0，得a 0＝1，则a 1＋a 2＋…＋a 7＋a 8＝－2－1＝－3.答案：－3综合提升练10．“杨辉三角”是我国古代重要的数学成就，它比西方的“帕斯卡三角形”早了300多年，如图是一个三角形数阵，记a n 为图中第n 行各数之和，则a 5＋a 11的值为( )1 1 1 12 1 13 3 1 14 6 4 1 15 10 10 5 1……A ．528B ．1020C ．1038D ．1040解析：选D a 5＝C 04＋C 14＋C 24＋C 34＋C 44＝24＝16，a 11＝C 010＋C 110＋C 210＋…＋C 1010＝210＝1024，所以a 5＋a 11＝1040.故选D ．11．(2021·河北饶阳中学模拟)(x ＋x ＋1)⎝⎛⎭⎫x －2x 6的展开式中x 2的系数为( )A ．72B ．60C ．48D ．36解析：选C ⎝⎛⎭⎫x －2x 6的展开式的通项公式为T r ＋1＝C r 6(x )6－r ·⎝⎛⎭⎫－2x r ＝(－2)r ·C r 6·x 3－r (r ＝0,1,2,3,4,5,6)．令3－r ＝1，得r ＝2；令3－r ＝32，得r ＝32∉Z ，舍去；令3－r ＝2，得r ＝1.故(x ＋x ＋1)·⎝⎛⎭⎫x －2x 6的展开式中x 2的系数为(－2)2·C 26＋(－2)1·C 16＝60－12＝48.故选C ．12．1－90C 110＋902C 210－903C 310＋…＋(－1)k 90k C k 10＋…＋9010C 1010除以88的余数是( )A ．－1B ．1C ．－87D ．87解析：选B 1－90C 110＋902C 210－903C 310＋…＋(－1)k 90k C k 10＋…＋9010C 1010＝(1－90)10＝8910＝(88＋1)10＝8810＋C 110889＋…＋C 91088＋1，∵前10项均能被88整除，∴余数是1.13．(2021·广东梅州模拟)记(1－x )6＝a 0＋a 1(1＋x )＋a 2(1＋x )2＋a 3(1＋x )3＋a 4(1＋x )4＋a 5(1＋x )5＋a 6(1＋x )6，则a 4＝ .解析：(1－x )6＝(－1＋x )6＝[－2＋(1＋x )]6，展开式的通项公式为T r ＋1＝C r 6(－2)6－r(1＋x )r ，令r ＝4 即可，a 4＝C 46(－2)2＝4C 26＝60.答案：6014．(2021·黑龙江哈尔滨三模)在⎝⎛⎭⎫x ＋ax n 的展开式中，只有第六项的二项式系数最大，且所有项的系数和为0，则含x 6项的系数为 ．解析：∵⎝⎛⎭⎫x ＋ax n 的展开式中，只有第六项的二项式系数C 5n 最大，∴n ＝10，再令x ＝1，可得所有项的系数和为(1＋a )10＝0，∴a ＝－1.故二项展开式的通项公式为T r ＋1＝C r 10·(－1)r ·x 10－2r ，令10－2r ＝6，求得r ＝2，可得含x 6项的系数为C 210＝45.答案：4515．(2021·浙江绍兴模拟)二项展开式(2x ＋4)5＝a 0＋a 1x ＋a 2x 2＋a 3x 3＋a 4x 4＋a 5x 5，则a 1＝ ；a 0＋a 2＋a 4＝ (可采用指数的形式或数字的方式作答).解析：因为(2x ＋4)5的展开式的通项为C r 5(2x )5－r 4r ＝C r 5·25－r ·4r ·x 5－r ， 令r ＝4，则a 1＝C 45×21×44＝2560，令r ＝5，则a 0＝C 55×20×45＝1024，令r ＝3，则a 2＝C 35×22×43＝2560，令r ＝1，则a 4＝C 15×24×41＝320，故a 0＋a 2＋a 4＝1024＋2560＋320＝3904.答案：2560 390416．已知⎝⎛⎭⎫mx 2－4＋x 25的展开式中所有项的系数和为1，则x 4的系数为 ． 解析：令x ＝1，则(m －3)5＝1，解得m ＝4，∴⎝⎛⎭⎫m x 2－4＋x 25＝⎝⎛⎭⎫4x 2－4＋x 25，⎝⎛⎭⎫4x 2－4＋x 25展开式的通项公式为C r 5⎝⎛⎭⎫4x 2－45－r (x 2)r ；∵⎝⎛⎭⎫4x 2－45－r 展开式通项公式为C k 5－r ⎝⎛⎭⎫4x 25－r －k (－4)k ，∴当k ＝1，r ＝3时，展开式中的项为 －320x 4；当k ＝3，r ＝2时，展开式中的项为－640x 4；∴x 4的系数为－320－640＝－960.答案：－960创新应用练17．(2021·湖北黄冈月考)若(x ＋2)8＝a 0＋a 1x ＋a 2x 2＋a 3x 3＋a 4x 4＋a 5x 5＋a 6x 6＋a 7x 7＋a 8x 8，则a 1－2a 2－4a 4＋5a 5－6a 6＋7a 7－8a 8＝ (用数字作答)．解析：∵(x ＋2)8＝a 0＋a 1x ＋a 2x 2＋a 3x 3＋a 4x 4＋a 5x 5＋a 6x 6＋a 7x 7＋a 8x 8，∴等式两边求导得8(x＋2)7＝a1＋2a2x＋3a3x2＋4a4x3＋5a5x4＋6a6x5＋7a7x6＋8a8x7.令x＝－1，有8×(－1＋2)7＝a1－2a2＋3a3－4a4＋5a5－6a6＋7a7－8a8，即a1－2a2＋3a3－4a4＋5a5－6a6＋7a7－8a8＝8.又a3＝C5825＝1792，故所求值为8－1792×3＝－5368.答案：－5368。
2024届山东潍坊高三二模化学试题+答案
试卷类型：A潍坊市高考模拟考试化 学2024.4注意事项：1．答题前，考生先将自己的学校、班级、姓名、考生号、座号填写在相应位置。
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可能用到的相对原子质量：H 1 B 11 C 12 N 14 O 16 F 19 Na 23 S 32 Fe 56 Cu 64 As 75 Sr 88一、选择题：本题共10小题，每小题2分，共20分。
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1．化学与科技密切相关。
下列说法正确的是（ ）A ．超分子粘合剂具有很高的粘合强度，超分子以共价键连接其他分子B ．催化电解工业废水中的硝酸盐制取氨气，实现了氮的固定C ．弹道二维23In Se 晶体管有优异的光学性能，其成分是一种合金材料D ．全氟磺酸质子膜广泛用于水电解制氢，磺酸基具有很好的亲水性 2．下列物质的性质与用途对应关系错误的是（ ）A ．晶体硅的导电性介于导体和绝缘体之间，常用于制造晶体管、集成电路等B ．蛋白质遇酶能发生水解，用加酶洗衣粉洗涤真丝织品可快速去污C ．血液中233H CO NaHCO −平衡体系能缓解外来酸碱的影响，可以平衡人体的pHD ．金属电子从激发态跃迁回基态时，释放能量而呈现各种颜色，常制作烟花 3．以下实验装置正确，且能达到实验目的的是（ ）A ．图1装置观察钠在空气中燃烧现象B ．图2装置测定碘的浓度C ．图3装置探究温度对反应速率的影响D ．图4装置制备乙酸乙酯4．下列分子的空间结构相同的是（ ） A ．3XeO 和3NHB ．2COCl 和2SOClC ．3ClF 和3BFD ．2CO 和2SCl5．4SiCl 的水解反应可用于军事工业中烟雾剂的制造，其部分反应机理如图。
山东省潍坊市2024届高三下学期4月高考模拟考试(二模)英语试题(含答案)
潍坊市高考模拟考试英语2024.4注意事项:1.答题前，考生务必将自己的姓名、座号、考号填写在答题卡和试卷指定位置上。
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第一部分阅读(共两节,满分50分)第一节(共15小题;每小题2.5分，满分37.5分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项。
AThe backpack you take can make or break your trip when you go traveling. Here are the four best travel backpacks on the market.Amazon Basics 70LIt's much cheaper than many travel bags on the market and does not sacrifice any of the practical uses or space that comes with more expensive bags. The bag may not be as luxury as some of the more high-end bags, but its simple style lets you focus on the main thing you need to focus on when traveling: the moment.Eurohike Nepal 65LThe Eurohike Backpack is a great choice because of how adaptable it is. Besides having a great amount of storage, it comes with an internal security pocket. It weighs just 1.38kg as opposed to other backpacks, which can weigh up to nearly 2kg. If you're going to go hiking when you travel, then it is perfect.Mountain Warehouse Tor 65LFirst , its brand is one of the most trusted in the industry ,so quality is guaranteed. Second, the backpack's adjustable back allows you to change how the bag fits according to your needs. Available inboth blue and green, this is a great choice if you want a bag that you can depend on.Osprey Europe Farpoint 70LOsprey is one of the best brands for backpacks. Its frame(框架)suspension, which can be adjusted to different needs, allows you to travel more comfortably. Whether you're visiting Switzerland in a thick, wool coat or the south of France in shorts and a T-shirt, the bag will match your look. This bag does say it is marketed for men, but, of course , it can be unisex.1. What is the selling point of the Amazon Basics?A. Its luxury style.B.Its fashionable design.C. Its huge space for use.D. Its good value for money.2. What do Mountain Warehouse Tor 65L and Osprey Europe Farpoint 70L have in common?A. They are rich in color.B. They have the same capacity.C. They can be adjusted as needed.D. They are targeted for male customers.3. Which will you choose if anti-theft function is a concern?A. Amazon Basics 70L. B .Eurohike Nepal 65L.C. Mountain Warehouse Tor 65L,D. Osprey Europe Farpoint 70L.BAt just seven years old, Angelina Tsuboi discovered her passion for innovation. It all began with a simple game she programmed in her Los Angeles public school's Grade 2class. Today ,at18,the Grade 12 student's initial curiosity has evolved into a deep-seated desire to use technology to decode(解码)real-world problems.In 2021, she co-developed Megaphone, one of her first apps, to tackle unanswered post- class questions and poor communication about events and announcements. Her problem-solving ability kept building from there.When she took online CPR classes at the start of the pandemic, she figured it couldn't be just her who was struggling with the steps. So she created an app called CPR Buddy―a winner in the 2022 Apple Swift Challenge―which guides users through CPR using vibrations(震动) to regulate breath. After winning theaward, Angelina presented her work to Apple CEO Tim Cook, a highlight in her young career, but one she didn't lose her cool over. “There's no point putting people on a pedestal (神坛),”she says.The next year, Angelina built an app called Lilac, designed to assist nonEnglish-speaking single parents with resources for housing, job opportunities and translation support. She was inspired by her own experiences as a child of a single mother who immigrated to the US.When Angelina decided to pursue pilot training at the age of 16, she was struck by how difficult it was to find financial support, which encouraged her to create yet another app, Pilot Fast Track, which helps those longing to be pilots find scholarships for flight training.Looking to the future, besides applying to colleges with great labs, Angelina is exploring the field of aerospace cybersecurity and mechatronics―combining computer science, electrical engineering and mechanical engineering.“There's not enough optimism in the world," she says. “I have also been in situations in my life where I've lost a lot of hope. But in the end, it is a mindset, and there are ways in any situation you're in to make it somewhat better."4.What is Angelina's pursuit?A. To design games for kids.B. To stimulate teen's curiosity.C. To address problems through technology.D. To find innovative approaches to digital challenges.5.What can we learn about Angelina from Paragraph 3?A. She couldn't breathe regularly.B. She was inspired by celebrities.C. She replaced CPR with an app.D. She was humble about her success.6.What was the primary goal of developing Pilot Fast Track?A. To direct pilots' career paths.B. To help to-be pilots find funds.C. To pair future pilots with airlines.D. To evaluate pilot training schools.7. What might be the best title?A. Breaking the codeB. Bearing growing painsC. Facing life as it isD. Following role modelsCSome people today might be early risers because of DNA they take after Neanderthals tens of thousands of years ago, suggests new research.When early humans migrated from Africa to Eurasia roughly 70,000 years ago, some of them mated with Neanderthals, who had already adapted to the colder, darker climates of the north. The ripple(涟漪) effects of that intermating still exist today: Modern humans of non- African ancestry(血统)have between 1 and 4 percent Neanderthal DNA. Some of that DNA relates to sleep more specifically, the internal body clock known as the circadian rhythm.For the new study, researchers compared DNA from today's humans and DNA from Neanderthal fossils(化石).In both groups, they found some of the same genetic variants involved with the circadian rhythm. And they found that modern humans who carry these variants also reported being early risers.For Neanderthals, being “morning people” might not have been the real benefit of carrying these genes. Instead, scientists suggest, Neanderthals’ DNA gave them faster, more flexible internal body clocks, which allowed them to adjust more easily to annual changes in daylight. This connection makes sense in the context of human history. When early humans moved north out of Africa, they would have experienced variable daylight hours--shorter days in the winter and longer days in the summer-for the first time. The Neanderthals' circadian rhythm genes likely helped early humans' offspring(后代)adapt to this new environment.Notably ,the findings do not prove that Neanderthal genes are responsible for the sleep habits of all early risers. Lots of different factors beyond genetics can contribute , including social and environmental influences. The study also only included DNA from a database called the U.K. Biobank-so the findings may not necessarily apply to all modern humans. Next, the research team hopes to study other genetic databases to see if the same link holds true for people of other ancestries. If the findings do apply more broadly , they may one day be useful for improving sleep in the modern world, where circadian rhythms are disturbed by night shifts and glowing smartphones.8.What does the new research focus on?A. DNA's dramatic changes.B. Genes’ influence on early risers.C. Neanderthals’ sleeping patterns.D. Ancestors’ environmental adaptability.9.What is paragraph 2 intended to show concerning the new research?A. Historical context.B. Additional proof.C. Sample analysis.D. Studying process.10. What is the real benefit of carrying Neanderthal's DNA for modern humans?A .Getting up earlier. B. Having healthier daily routines.C. Being more flexible in their work.D. Possessing a better circadian rhythm.11. What can be inferred about the findings from the last paragraph?A. They get proof from other studies.B. They are confirmed by early risers.C. They suggest potential applications.D. They reveal factors in sleeping disorders.DI had to say something after reading The Anxious Generation. It is going to sell well , because Jonathan Haidt is telling a scary story about children's development many parents are led to believe. However, the book's repeated suggestion that digital technologies are rewiring our children's brains and causing the epidemic (流行病)of mental illness is unsupported by science. Worse , the rude proposal that social media is to blame might distract (分心)us from effectively responding to the real causes of the current mental-health crisis in young people.Researchers have searched for the effects suggested by Haidt. Our efforts have produced a mix of no, small and mixed associations. Most data are correlative. When associations over time are found, they suggest not that social-media use predicts or causes depression, but that young people who already have mental-health problems use such platforms more often or in different ways from their healthy peers.We are not alone here. Several analyses and systematic reviews centralize on the same message. An analysis done in 72 countries shows no consistent or measurable associations between well-being and social media globally. Moreover, studies from some authorities finds no evidence of intense changes associated with digital-technology use.As a psychologist studying children's and adolescents’ mental health, I appreciate parents’frustration(沮丧)and desire for simple answers. As a parent of adolescents, I would also like to identify a simple source for the pain this generation is reporting. There are, however, no simple answers. The beginning and development of mental disorders are driven by a complex set of genetic and environmental factors.More young people are talking openly about their mental-health struggles than ever before. But insufficient services are available to address their needs. In the United States, there is, on average, one school psychologist for every 1,119 students. We have a generation in crisis and in desperate need of the best of what science and evidence-based solutions can offer. Unfortunately, our time is being spent telling stories that are unsupported by research and that do little to support young people who need, and deserve, more.12.What is presented in The Anxious Generation?A. Scary stories affect children's brains.B. Parents are responsible for children's health.C. Teen's mental illness results from screen time.D. The epidemic of mental illness is unavoidable.13.What does “the same message ”underlined in paragraph 3 refer to?A. Many countries do research in mental health.B. Well-being and social media are closely related.C. The young are trapped in the mental-health crisis,D. Social media don't necessarily cause mental illness.14. What is implied in the last paragraph?A. Effective actions need to be taken.B. Positive stories should be shared.C. Financial support needs to be provided.D. Broader research should be done.15.What is the author's purpose in writing the text?A. To suggest ways to help those in need.B. To encourage parents to brave the crisis.C. To recommend a newly-published book.D. To give a voice to children's mental issues.第二节(共5小题;每小题2.5分,满分12.5分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。