上海市奉贤区2016届高三上学期期中考试(暨区六校联考)数学试卷 Word版含答案
“六校联盟”2016届高三上学期第三次联考数学试卷(理科) 含解析
2015—2016学年广东省“六校联盟"高三(上)第三次联考数学试卷(理科)一、选择题:本大题共12小题,每小题5分,共60分。
在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知集合M={x∈R|},N={x∈R|y=ln(x﹣1)},则M∩N()A.∅B.{x|x≥1}C.{x|x>1}D.{x|x≥1或x<0}2.已知两条直线m,n,两个平面α,β,给出下面四个命题:①m∥n,m⊥α⇒n⊥α②α∥β,m⊂α,n⊂β⇒m∥n③m∥n,m∥α⇒n∥α④α∥β,m∥n,m⊥α⇒n⊥β其中正确命题的序号是()A.①③B.②④C.①④D.②③3.下列四个条件中,p是q的必要不充分件的是()A.p:a>b,q:a2>b2B.p:a>b,q:2a>2bC.p:非零向量与夹角为锐角,q:D.p:ax2+bx+c>0,q:﹣+a>04.设函数f(x)=x﹣lnx﹣,则函数y=f(x)()A.在区间(),(1,e)内均有零点B.在区间(),(1,e)内均无零点C.在区间()内有零点,在区间(1,e)内无零点D.在区间()内无零点,在区间(1,e)内有零点5.要得到函数y=cosx的图象,需将函数y=sin(2x+)的图象上所有的点的变化正确的是()A.横坐标缩短到原来的倍(纵坐标不变),再向左平行移动个单位长度B.横坐标缩短到原来的倍(纵坐标不变),再向右平行移动个单位长度C.横坐标伸长到原来的2倍(纵坐标不变),再向左平行移动个单位长度D.横坐标伸长到原来的2倍(纵坐标不变),再向右平行移动个单位长度6.已知{a n}是等比数列,a2=,a5=4,则a1a2+a2a3+…+a n a n=()+1A.(2n﹣1)B.(2n+4) C.(4n﹣1)D.(4n﹣2) 7.如果点P在平面区域,点Q在曲线x2+(y+2)2=1上,那么|PQ|的最小值为()A.﹣1 B.2﹣1 C.2 D.﹣18.已知函数y=f(x)是定义在R上的奇函数,当x≤0时f(x)=2﹣x+m﹣1(m∈R),a=f (log45),b=(log23),c=f(m),则a,b,c的大小关系为()A.a<b<c B.b<a<c C.c<a<b D.c<b<a9.在△ABC中,己知D是AB边上一点,若=λ,=+μ(λ,μ∈R),则λ=() A.﹣2 B.﹣1 C.1 D.210.已知函数f(x)=f′(1)x2+2x f(x)dx+1在区间(a,1﹣2a)上单调递增,则实数a的取值范围是()A.(,)B.[,) C.(﹣∞,)D.[,+∞)11.一个正三棱锥的四个顶点都在直径为2的球面上,其中底面的三个顶点在该球的一个大圆上,则该正三棱锥的体积是()A.2B.C.D.12.已知定义在(0,+∞)上的连续函数y=f(x)满足:xf′(x)﹣f(x)=xe x且f(1)=﹣3,f(2)=0.则函数y=f(x)()A.有极小值,无极大值B.有极大值,无极小值C.既有极小值又有极大值D.既无极小值又无极大值二、填空题:本大题共4小题,每小题5分,满分20分13.在△ABC中,内角A、B、C的对边长分别为a、b、c,已知a2﹣c2=3b,且sinAcosC=2cosAsinC,则b=.14.已知数列{a n}的前n项和为S n,且满足S n=2a n﹣1(n∈N*),则数列{na n}项和T n.15.已知某个几何体的三视图如图,根据图中标出的尺寸,可得这个几何体的体积等于,全面积为.16.若不等式(﹣1)n a<n+对任意n∈N*恒成立,则实数a的取值范围是.三、解答题:包括必做题和选做题,第17题到第21题为必做题,解答应写出文字说明,证明过程或演算步骤.17.已知函数f(x)=cos(2x﹣)+2sin(x﹣)sin(x+).(1)求函数f(x)的最小正周期和图象的对称轴方程;(2)求函数f(x)在区间[﹣,]上的最值.18.等差数列{a n}各项均为正数,其前n项和为S n,a2S3=75且a1,a4,a13成等比数列.(1)求数列{a n}的通项公式a n;(2)若数列{a n}为递增数列,求证:≤.19.如图,在三棱锥P﹣ABC中,PA⊥平面ABC,底面ABC是直角三角形,PA=AB=BC=4,O是棱AC的中点,G是△AOB的重心,D是PA的中点.(1)求证:BC⊥平面PAB;(2)求证:DG∥平面PBC;(3)求二面角A﹣PC﹣B的大小.20.已知点P是圆O:x2+y2=1上任意一点,过点P作PQ⊥y轴于点Q,延长QP到点M,使.(1)求点M的轨迹的方程;(2)过点C(m,0)作圆O的切线l,交(1)中曲线E于A,B两点,求△AOB面积的最大值.21.已知函数f(x)=ln(x+1)+ax2﹣x(a∈R).(1)若a=,求曲线y=f(x)在点(0,f(0))处的切线方程;(2)讨论函数y=f(x)的单调性;(3)若存在x0∈[0,+∞),使f(x)<0成立,求实数a的取值范围.请考生在第(22),(23),(24)三题中任选一题作答.注意:只能做所选定的题目.如果多做,则按所做的第一个题目计分,作答时请用2B铅笔在答题卡上将所选题号后的方框涂黑.选修4—1:几何证明选讲22.如图,已知圆O是△ABC的外接圆,AB=BC,AD是BC边上的高,AE是圆O的直径.过点C作圆O的切线交BA的延长线于点F.(Ⅰ)求证:ACBC=ADAE;(Ⅱ)若AF=2,CF=2,求AE的长.选修4-4:坐标系与参数方程.23.在平面直角坐标系xOy中,直线l的参数方程为(t为参数).在以原点O为极点,x轴正半轴为极轴的极坐标中,圆C的方程为ρ=2sinθ.(1)写出直线l的普通方程和圆C的直角坐标方程;(2)设点P(3,),直线l与圆C相交于A、B两点,求+的值.选修4-5:不等式选讲。
上海市奉贤区2016届高三上学期期末调研数学试卷
2015学年奉贤区高三数学一模调研测试卷(考试时间:120分钟,满分150分)一.填空题(本大题满分56分)本大题共有14题,考生应在答题纸相应编号的空格内直接写结果,1-14题每个空格填对得4分)1、复数()1i i +(i 是虚数单位)的虚部是__________.2、已知点()1,5A -和向量()2,3a =,若a AB 3=,则点B 的坐标为__________.3、方程9360xx+-=的实数解为__________.4、已知集合{}2230M x x x =--≤,{}lg N x y x ==,则M N ⋂=__________.5、若81x x ⎛⎫+ ⎪⎝⎭展开式中含2x 的项的系数是__________.6、若圆x y x y 22++2-4=0被直线x y a 3++=0平分,则a 的值为__________. 7、若抛物线22(0)y px p =>的准线经过双曲线221x y -=的一个焦点,则p =_________. 8、数列}{n a 是等差数列,2a 和2014a 是方程01652=+-x x 的两根,则数列}{n a 的前2015项的和为__________.9、函数sin y x x =+,,3x ππ⎡⎤∈-⎢⎥⎣⎦的值域是__________. 10、已知b a ,是常数,0ab ≠,若函数3()arcsin 3f x ax b x =++的最大值为10,则)(x f 的最小值为__________. 11、函数()sin 4f x x πω⎛⎫=+ ⎪⎝⎭在,2ππ⎛⎫⎪⎝⎭上单调递减,则正实数ω的取值范围是_________.12、设αβ、都是锐角,1cos ,cos()7ααβ=+=,请问cos β是否可以求解,若能求解,求出答案,若不能求解简述理由_________________________________________________________________________________________________________________________.13、不等式()()21430x x x +-+>有多种解法,其中有一种方法如下,在同一直角坐标系中作出11y x =+和2243y x x =-+的图像然后进行求解,请类比求解以下问题:设,a b Z ∈,若对任意0x ≤,都有2(2)(2)0ax x b ++≤,则a b +=__________.14、线段AB 的长度为2,点A 、B 分别在x 非负半轴和y 非负半轴上滑动,以线段AB 为一边,在第一象限内作矩形ABCD (顺时针排序),1BC =,设O 为坐标原点,则OD OC ⋅的取值范围是__________.二.选择题(本大题满分20分)本大题共有4题,每题有且只有一个正确答案,考生应在答题纸的相应编号上,将代表答案的小方格涂黑,选对得5分,否则一律得零分.15、下面四个条件中,使a >b 成立的必要而不充分的条件是…………( ). 22.1.22..lg lg a b A a b B C a b D a b +>>>>16、已知数列sin2n n a n π=⋅,则123100a a a a ++++= …………( ). .A 48-; .B 50-; .C 52-; .D 49-17、已知直角三角形的三边长都是整数且其面积与周长在数值上相等,那么这样的直角三角形有…( )..A 0; .B 1; .C 2; .D 318、设函数2()min{1,1,1}f x x x x =-+-+,其中min{,,}x y z 表示,,x y z 中的最小者.若(2)()f a f a +>,则实数a 的取值范围为…………( )..A ()1,0-; .B []2,0-; .C ()(),21,0-∞-- ; .D [)2,-+∞三.解答题(本大题满分74分)本大题共有5题,解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤. 19、如图,已知四边形ABCD 是矩形,1AB =,2BC =,PD ⊥平面ABCD ,且3PD =, PB 的中点E ,求异面直线AE 与PC 所成角的大小.(用反三角表示)P A B C DE20、设ABC ∆的内角A 、B 、C 所对的边分别为,,a b c ,且满足cos 2A =,3=⋅AC AB (1)、求ABC ∆的面积;(2)、求a 的最小值.212(),x y 对应点的曲线方程是C .(1)、求C 的标准方程;(2)、直线1:0l x y m -+=与曲线C 相交于不同两点,M N ,且满足MON ∠为钝角,其中O 为直角坐标原点,求出m 的取值范围.22、已知函数()x f y =是单调递增函数,其反函数是()1y f x -=.(1)、若⎪⎭⎫ ⎝⎛>-=2112x x y ,求()1y f x -=并写出定义域M ; (2)、对于(1)的()1y f x -=和M ,设任意2121,,x x M x M x ≠∈∈,求证:()()212111x x x f x f-<---;(3)、若()x f y =和()1y f x -=有交点,那么交点一定在x y =上.23、数列{}n a 的前n 项和记为n S 若对任意的正整数n ,总存在正整数m ,使得n m S a =, 则称{}n a 是“H 数列”.(1)、若数列{}n a 的通项公式2nn a =,判断{}n a 是否为“H 数列”;(2)、等差数列{}n a ,公差0d ≠,12a d =,求证:{}n a 是“H 数列”; (3)、设点()1,n n S a +在直线()1q x y r -+=上,其中120a t =>,0≠q .若{}n a 是“H 数列”,求,q r 满足的条件.2016年奉贤区高三数学一模参考答案一、填空题(每题4分,56分)1、1;2、()5,14B ;3、3log 24、(]0,3;5、56;6、1a =;7、; 8、1209;9、2⎡⎤⎣⎦; 10、4-;11、15,24⎡⎤⎢⎥⎣⎦12、(),0,,,cos y x ααβπααβ+∈<+= 在()0,π上递减,而()cos cos αβα+>,所以条件错误,不可解13、1- 14、[]1,3二、选择题(每题5分,20分)15、A ; 16、B ; 17、C ; 18、C ; 三、解答题(12+14+14+16+18=74分)19、取BC 的中点F ,连接,EF AF 、AEE 、F 是中点,EF ∴是PBD ∆的中位线 EF ∴∥PBAEF ∴∠(或者其补角)为异面直线AE 与PC 所成角 3分 在Rt PAB ∆中,PB ==5分PC EF == 6分 AF = ,AE =,52AE = 7分 由余弦定理可知222cos2AE EF AF AEFAE EF+-∠=⋅==10分AEF ∴∠=11分 异面直线AE 与PC 所成角的大小. 12分20、解:(1)因为cos2A =,所以23cos 2cos125A A =-=, 2分 4sin 5A =3分 又因为3AB AC ⋅=,得cos 3bc A = 4分PABCDEFcos 35bc A bc =⇒= 5分 1sin 22ABC S bc A ∆⇒== 7分(2)2222235,2cos 255bc a b c bc A b c =∴=+-=+-⨯⨯10分 2226a b c ∴=+- 11分 222222min 662102a b c b c a bc a ∴=+-⇒+=+≥=∴= 12分当且仅当b c==时a 最小值是2 14分21、(1)4= 1分所以点(),P x y 对应的曲线方程C 是椭圆 2分24,2a a =∴= . 3分 1c = 4分2,1,a c b ∴=== 5分22143x y += 6分 (2)、联立方程组220143x y m x y -+=⎧⎪⎨+=⎪⎩消去y ,得22784120x mx m ++-= 7分()2226428412336480m m m ∆=--=-> 8分27m ∴< 9分设1122(,),(,)M x y N x y得2124127m x x -= 10分方法一可计算2123127m y y -= 11分由MON ∠为钝角,则0OM ON ⋅<,12120x x y y +<22412312077m m --+< 12分所以2247m <13分m <<14分 方法二或者()()()21212121212122x x y y x x x m x m x x m x x m +=+++=+++ 11分()222241287240777m m m m m--=-+=< 12分所以2247m <13分m << 14分22、解:(1)、(),11+=-x x f ⎪⎭⎫⎝⎛+∞-=,43M 3+2=5分(2)、()()11112121212111+++-=+-+=---x x x x x x x f x f7分131,42x >-> ,211,4322>+∴->x x 9分 11121>+++∴x x ,1111021<+++<∴x x 10分 21212111x x x x x x -<+++-∴()()212111x x x f x f-<-∴-- 11分 (3)、设()b a ,是()x f y =和()1y f x -=有交点 即()()⎩⎨⎧==-a f b a f b 1,()()a f b b f a ==∴, 12分 当b a =,显然在x y =上 13分 当b a >,函数()x f y =是单调递增函数,()a b b f a f >∴>∴,)(矛盾 15分 当b a <,函数()x f y =是单调递增函数,()a b b f a f <∴<∴,)(矛盾 16分 因此,若()x f y =和()1y f x -=的交点一定在x y =上 16分23、解析:(1)111,2n a S ===当2n ≥时,122112nn n S -==-- 1分21n ∴-是奇数,2m 是偶数 2分 212nm∴-≠ 3分 ∴{}n a 不是“H 数列” 4分(2)1(1)(1)222n n n n n S na d dn d --=+=+ 6分对任意n *∈N ,存在m *∈N 使n m S a =,即11(1)(1)2n n na d a m d -+=+-(1)212n n m n -=-+8分 ,1n n -是一奇一偶,m ∴一定是自然数 10分 (3)2n ≥时()11n n q S a r +-+=,()11n n q S a r --+=()110n n n q a a a +-+-=1n n a qa +∴= 12分 ()212q t a r -⨯+=222a r qt t p =+-= 13分()()2212n n t n a p q n -⎧=⎪∴=⎨⋅≥⎪⎩ 14分 1q =时,()()212n t n a r n ⎧=⎪=⎨≥⎪⎩ ()21n S t n r r =+-=不恒成立 显然{}n a 不是“H 数列” 15分1q ≠时()11122111n n n p q p pq S t t qq q---=+=+---- 16分 111,n S a =={}n a 是“H 数列”,所以对任意2n ≥时,存在*m N ∈成立12211n m n p pq S t pq q q--∴=+-=-- 2q ∴=,2p t =,422,0r t t t r ∴+-== 2,0,0q r t ∴==>的正实数 18分。
2016届高三六校第一次联考
2016届高三六校第一次联考理科数学试题参考答案及评分标准一. 选择题:1、B2、A3、D4、B5、A6、C7、A8、C9、B 10、D 11、C 12、B 11、如图,易知BCD ∆的面积最大12、 解:令21()()2g x f x x =-,2211()()()()022g x g x f x x f x x -+=--+-= ∴函数()g x 为奇函数 ∵(0,)x ∈+∞时,//()()0g x f x x =-<,函数()g x 在(0,)x ∈+∞为减函数又由题可知,(0)0,(0)0f g ==,所以函数()g x 在R 上为减函数2211(6)()186(6)(6)()186022f m f m mg m m g m m m ---+=-+----+≥即(6)()0g m g m --≥∴(6)()g m g m -≥,∴6,3m m m -≤∴≥二、填空题:本大题共4小题,每小题5分,共20分13、2 14、 5 15、 73 16、2016 ∵(2016)(2013)3(2010)6(0)20162016f f f f ≤+≤+≤≤+= (2016)(2014)2(2012)4(0)20162016f f f f ≥+≥+≥≥+=(2016)2016f ∴=三、解答题(17—21为必做题)CDBA17、解:(1)由题意易知122n n n a a a --=+,---1分 即1231112n n n a q a q a q ---=+,--2分2210q q ∴--= 解得1q =或12q =- -------- 3分(2)解:①当1q =时,1n a =,n b n = n S =2)1(+n n ----------5分②当12q =-时,11()2n n a -=-11()2n n b n -=⋅- ---------------7分n S =012111111()2()3()()2222n n -⋅-+⋅-+⋅-++⋅--21n S = 12111111()2()(1)()()2222n n n n -⋅-+⋅-++-⋅-+⋅- 相减得21311111()()()()22222n n n S n -⎡⎤=-⋅-+-+-++-⎢⎥⎣⎦-------- 10分整理得 n S =94-(94+32n )·1()2n ------------------------12分18、解:设甲、乙、丙各自击中目标分别为事件A 、B 、C(Ⅰ)由题设可知0ξ=时,甲、乙、丙三人均未击中目标,即(0)()P P A B C ξ== ∴()()()21011515P m n ξ==--=,化简得()56mn m n -+=- ① ……2分同理, ()3113553P m n mn ξ==⨯⨯=⇒= ②……4分 联立①②可得23m =,12n = ……6分(Ⅱ)由题设及(Ⅰ)的解答结果得:(1)()P P A B C A B C A B C ξ==++()3311221211153253253210a P ξ∴===⨯⨯+⨯⨯+⨯⨯=……8分()3131111510530b ∴=-++= ……10分31353110123151030530E ξ∴=⨯+⨯+⨯+⨯= ……12分19.解法一:(1)如图:,,AC AC BD O =连设1.AP B G OG 1与面BDD 交于点,连 ……1分1111//,,PC BDD B BDD B APC OG =因为面面面故//OG PC .所以122m OG PC ==.又111,,AO DB AO BB AO BDD B ⊥⊥⊥所以面 ……3分 故11AGO AP BDD B ∠即为与面所成的角。
2016年奉贤区高三数学一模参考答案.doc
2016年奉贤区高三数学一模参考答案一、填空题(每题4分,56分)1、1;2、()5,14B ;3、3log 24、(]0,3;5、56;6、1a =; 7、 8、1209; 9、2⎡⎤⎣⎦; 10、4-;11、15,24⎡⎤⎢⎥⎣⎦12、(),0,,,cos y x ααβπααβ+∈<+=在()0,π上递减,而()cos cos αβα+>,所以条件错误,不可解13、1- 14、[]1,3二、选择题(每题5分,20分)15、A ; 16、B ; 17、C ; 18、C ; 三、解答题(12+14+14+16+18=74分)19、取BC 的中点F ,连接,EF AF 、AEE 、F 是中点,EF ∴是PBD ∆的中位线 EF ∴∥PBAEF ∴∠(或者其补角)为异面直线AE 与PC 所成角 3分 在Rt PAB ∆中,2PB ==5分PC EF == 6分AF =,2AE =,52AE = 7分由余弦定理可知222cos 2AE EF AF AEF AE EF+-∠=⋅222+-== 10分AEF ∴∠= 11分异面直线AE 与PC所成角的大小. 12分PA BCDEF20、解:(1)因为cos 25A =,所以23cos 2cos 125A A =-=, 2分 4sin 5A =3分 又因为3AB AC ⋅=,得cos 3bc A = 4分cos 35bc A bc =⇒= 5分1sin 22ABC S bc A ∆⇒== 7分(2)2222235,2cos 255bc a b c bc A b c =∴=+-=+-⨯⨯ 10分2226a b c ∴=+- 11分222222min 662102a b c b c a bc a ∴=+-⇒+=+≥=∴= 12分当且仅当b c==a 最小值是2 14分21、(1)4= 1分 所以点(),P x y 对应的曲线方程C 是椭圆 2分24,2a a =∴= . 3分 1c = 4分2,1,a c b ∴=== 5分22143x y += 6分 (2)、联立方程组220143x y m x y -+=⎧⎪⎨+=⎪⎩消去y ,得22784120x mx m ++-= 7分()2226428412336480m m m ∆=--=-> 8分27m ∴< 9分设1122(,),(,)M x y N x y得2124127m x x -= 10分方法一可计算2123127m y y -= 11分由MON ∠为钝角,则0OM ON ⋅<,12120x x y y +<22412312077m m --+< 12分 所以2247m <13分77m ∴-<<14分 方法二或者()()()21212121212122x x y y x x x m x m x x m x x m +=+++=+++ 11分()222241287240777m m m m m--=-+=< 12分所以2247m <13分m << 14分22、解:(1)、(),11+=-x x f⎪⎭⎫⎝⎛+∞-=,43M 3+2=5分(2)、()()11112121212111+++-=+-+=---x x x x x x x f x f 7分131,42x >->,211,4322>+∴->x x 9分11121>+++∴x x ,1111021<+++<∴x x 10分 21212111x x x x x x -<+++-∴()()212111x x x f x f-<-∴-- 11分(3)、设()b a ,是()x f y =和()1y f x -=有交点 即()()⎩⎨⎧==-a f b a f b 1,()()a f b b f a ==∴, 12分 当b a =,显然在x y =上 13分 当b a >,函数()x f y =是单调递增函数,()a b b f a f >∴>∴,)(矛盾 15分 当b a <,函数()x f y =是单调递增函数,()a b b f a f <∴<∴,)(矛盾 16分 因此,若()x f y =和()1y fx -=的交点一定在x y =上 16分23、解析:(1)111,2n a S ===当2n ≥时,122112nn n S -==-- 1分 21n ∴-是奇数,2m是偶数 2分212n m∴-≠ 3分∴{}n a 不是“H 数列” 4分(2)1(1)(1)222n n n n n S na d dn d --=+=+ 6分对任意n *∈N ,存在m *∈N 使n m S a =,即11(1)(1)2n n na d a m d -+=+-(1)212n n m n -=-+8分 ,1n n -是一奇一偶,m ∴一定是自然数 10分 (3)2n ≥时()11n n q S a r +-+=,()11n n q S a r --+=()110n n n q a a a +-+-=1n n a qa +∴= 12分 ()212q t a r -⨯+=222a r qt t p =+-= 13分()()2212n n t n a p q n -⎧=⎪∴=⎨⋅≥⎪⎩ 14分 1q =时,()()212n t n a r n ⎧=⎪=⎨≥⎪⎩ ()21n S t n r r =+-=不恒成立 显然{}n a 不是“H 数列” 15分1q ≠时 ()11122111n n n p q p pq S t t qq q---=+=+---- 16分 111,n S a =={}n a 是“H 数列”,所以对任意2n ≥时,存在*m N ∈成立12211n m n p pq S t pq q q--∴=+-=--2q ∴=,2p t =,422,0r t t t r ∴+-== 2,0,0q r t ∴==>的正实数 18分。
2016年奉贤区高三(上)六校期中联考试卷(附答案及分析)
2016年奉贤区⾼三(上)六校期中联考试卷(附答案及分析)2015年上海市⾼三年级六校联考英语试卷试卷难度中上,整体来说⽐较容易做出来,但是基本每个模块都有⼏道难题。
语法填空35空填连词although学⽣较难想出来;⼗⼀选⼗给的词如E、H不常见,学⽣基本不认识;完型选项相对较长,但是属于常见词;阅读⽐较中规中矩,但是需要学⽣细⼼;翻译第3句倒装句容易出错。
第Ⅰ卷(共103分)I.Listening ComprehensionPart A Short ConversationsDirections: In Part A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. At an airport. B. At a railroad station.C. At a library. .D. At a bank.2. A. He can‘t finish his work now. B. He can‘t give the woman his computer.C. He has run out of his paper.D. He can‘t get his computer repaired.3. A. Being hungry. B. Work. C. The weather. D. Cooking4. A. The woman should ask others.B. He wonder if the woman needs help.C. They can work together the next day.D. He didn‘t hear her well.5. A. Reading the paper. B. Taking a final exam.C. Helping the librarian.D. Studying.6. A. Get another car.B. Ask someone else to help her.C. Buy something less expensive.D. Go to another repair shop.7. A. His team won the other night.B. He didn‘t go to the game.C. His team always loses.D. His team player for the first time.8. A. He wants to be invited to a card game.B. He told them what his favorite games are.C. He doesn‘t really enjoy playing cards.D. He doesn‘t know they‘re playing without him.9. A. Doctor and patient.B. Interviewer and interviewee.C. Teacher and student.D. Shopper and customer.10. A. She forgot to send a gift to Janet.B. They aren‘t obliged to buy a gift.C. She prefers to go shopping in a store.D. They should select an inexpensive gift.Part B PassagesDirections: In Part B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. To review material for a test. B. To introduce a new professor.C. To explain changes in the schedule.D. To describe the contents of a paper.12. A. At the beginning. B. In the middle.C. One week before the end.D. At the end.13. A. A regular class will be given. B. An optional review class will be given.C. An exam will be given resort.D. Class will be canceled.Questions 14 through 16 are based on the following passage.14. A. People who listen to the radio also buy newspaper.B. Radio is a substitute for newspapers in people‘s homes.C. Newspapers discourage people from listening to the radio.D. Many newspaper reporters also work in the radio industry.15. A. Movie attendance increased due to advertising on television.B. Old motion pictures were often broadcast on television.C. Television had no effect on movie attendance.D. Motion picture popularity declined.16. A. To illustrate another effect of television.B. To demonstrate the importance of televised sports.C. To explain why television replaced radio broadcasting.D. To provide an example of something motion pictures can‘t present.Part C Longer ConversationsDirections: In Part C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. Write your answers on your answer sheet.Complete the form. Write NO MORE THAN THREE WORDS for each answer. Ⅱ.Grammar and VocabularySection ADirections:After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.(A)It doesn‘t sound pleasant, but a powder ma de from the waste of gut (消化道) bacteria seems to stop people (25)______ (gain) weight.When we digest fiber, the bacteria in our guts give off a fatty acid, (26)______ makes our body release a certain material. As a result, we feel full.Gary Frost of Imperial College London and his colleagues have made a more (27) ______ (concentrate) form of that material.―That the powder gives you eight times the amount of someone (28)______ (follow) a typical Western diet,‖ Frost told New Scientist.(29)______ (stabili ze) a person‘s weight could have a big health impact, said Frost.As people on Western diets grow older, they gain an average of 0.3 to 0.8 kilograms per year. The powder, when (30)______ (drink) daily mixed in with fruit juice or a milkshake, helped dozens of overweight volunteers stick (31)______ their weight over six months, while others on a normal diet continued to put on weight.It sounds easy and quick. But are you sure you want to drink this instead of simply exercising more (32)______ (lose) weight?(B)Every year dozens of films are released, yet (33)______ ______ are forgotten after six months? Movies come and go, as throwaway as popcorn bags left on the floor of a cinema. But of those few films that do stay in people‘s minds, there is one that is truly ―evergreen‖.(34)______ you‘re young or old, or where ver you are in the world, the 1939 classic Gone with the Wind never seems to become unpopular. December 2015 the film celebrated its 75th birthday.The movie is based on a best-selling book of the same name by US author Margaret Mitchell. Hollywood was soon interested in turning the novel into a movie.The story is set in the periods before, during and after the American Civil War (1861-65), (35)______ the war is more of a backdrop (背景)to the story than an important part of it. The story is about relations between members of high-class southernfamilies.At the heart of the film is Scarlett O‘Hara. Beautiful and strong-willed, Scarlett is in love with a man, Ashley Wilkes,(36)______ heart belongs to another, Melanie Hamilton. Still, she tries to win Ashley‘s heart.One man, Rhett Butler, is especially interested in Scarlett. Rhett is as wild in his own way as she is. But although she flirts(调情)with Rhett, and despite the fact that she eventually marries him, she never really loves Rhett. It‘s only when she finally realizes that she can‘t have Ashley (37)______ she turns back to her husband.But, (38)______ anyone who has seen the movie will know, by that point Rhett doesn‘t want her back and Scarl ett is left with nobody (39)______ (love).This Civil War period piece repaid the time and effort of the filmmakers who worked on it. Over two decades, it held the record for making the most money of any film ever (40)______ (make). It‘s the kind of movie that every studio dreams of making.Section BDirections:Complete the following passage by using the words in the box. EachInstitute of Age Research gave a very exciting answer: 500 years.Think about it. If that had been the case many years ago, a person could have lived from the Ming Dynasty until now.But that idea was 41 by scientists at a recent medical conference in the UK, The Times reported.Speaking at the conference, British neurobiologist Sir Colin Blakemore, 70, claimed there is a limit on how long humans can live, and how much the body can age. Instead of 500 years, 120 years might be a real 42 limit to human lifespan.His estimate was based on current 43 and a look at medical effects.The number of people living past 100 in the world has 44 by 71 percent in the past decade, the Daily Mail reported. Blakemore agreed that this figure would continue to rise. But people living for longer than 120 years is ―so rarely 45 ‖ that, even with medical and technological 46 , it is unlikely that we can expect more than that, he said.His claims 47 those made previously by researchers at the Buck Institute of Age Research in California.In 2013, they changed two genetic (基因的) pathways in a tiny worm, and the creature lived a life five times as long as a normal one.If that technology could one day be used on human beings, it could 48 human lives to 500 years, the Daily Mail reported. But no matter which number is the ceiling for our lifespans, 120 or 500 years, it would take years for scientists to make that true for most people.We don‘t have to completely rely on 49 breakthroughs to live a longer life,however.On Oct 21, the Gerontological Society of China published a report on centenarians (百岁⽼⼈)in China. There are 58,789 people above the age of 100 now in our country, with the oldest being 128 years old.And by looking at these people‘s lifestyles, the society worked out a 50 for a long life: outdoor activities, more communication with others, and a healthy diet containing lots of fruit, vegetables, and low-fat food.Ⅲ. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.While the full picture has yet to emerge, it appears that the reason for the loss of AirAsia flight QZ8501 is different from thelosses of MH370 and MH17 earlier this year. MH17, shot down in Ukraine, was clearly a man-made disaster;QZ8501‘s disappearance appears to be weather-related. MH370‘s loss remains a mystery.Even though cloud-computing(云计算) could help advance the status(状态)of the black box in terms of the investigation after such incidents, as we are only too aware, nature – in the form of bad weather – often plays a significant role in airline disasters. Is there anything we can do, on the ground, to lower the 51 of these weather-related incidents?Recently,Microsoft Research illustrated that by combining real-time data from nearby flights, it is possible to 52 the wind speed to an accuracy 10 times better than the weather simulations(模拟)by supercomputers. Augmenting the data collected from real-time sensors from the cloud, one can create a better simulation, 53 the advances in the internet of flying things.The internet of flying things refers to the technology that is ready for adoption by agencies on the ground that want to get a bird‘s-eye view of weather conditions. The basic technology is already 54 for less than £500: equipping a drone(⽆⼈机) or unmanned airborne vehicle with a GoPro quality camera, enthusiastic fans can already survey the neighborhood from the air.If we look at the air crash incidents caused by bad weather conditions, can the killer technology of cloud computing and augmented reality be used as a life saver?W hen you fly into such wind speeds, is it not difficult to change one‘s actions 55 ? Isn‘t the flight simply doomed? Not necessarily. In this case, had nearby flying objects logged(记录)the abnormal wind speeds earlier, they would have been able to inform air traffic control in time to 56 a warning to flight QZ8501. In these situations, often timely interventions(⼲预) can save lives.But before this idea can 57 be realized, there are at least three obstacles to overcome if we want to use the power of the internet of flying objects.The first thing to note is that these flying objects shouldn‘t be 58 to aircraft. We could be talking about weather balloons, drones – anything in the air, in short, but these objects need to be identifiable. Only though 59 can messages from these flying objects be recognized and trusted by authorities such as FAA. So, for example,the drones that – it is imagined – will be carrying goods to households must be 60 and their call signs logged by the authorities before they can be of any practical help. At present they are not. In other words, the autonomous(⾮政府的)flying objects are required to collaborate with air traffic controllers if we want to build a picture that will deliver a secure and trustworthy solution.61 , these regulated and registered flying objects need to be effectively networked, so that – through the 62 of real-time data –the crowd-sourced(基于⼤众资源的)information delivers as accurate a picture as possible. Resolving any conflicts arising from information coming in from multiple sources requires a good computational model that can assign 63 weight to the various sources.And this collected data needs to depict a physical truth to decision makers –whether they are in front of the desk in the air traffic control center, on the flight deck of a nearby aircraft or in command of the rescue team. The task of confirming available 64 against any possible internal flaws or external tampering(⼲扰) would require that network security levels are brought to another level.These three 65 are basic, but if they can be overcome, they might offer us a better opportunity to use today‘s technology to provide safer air transport in the future.51. A. risk B. disturbance C. uncertainty D. threat52. A. check B. follow C. predict D. monitor53. A. thanks to B. in spite of C. with reference to D. other than54. A. feasible B. possible C. profitable D. available55. A. secretly B. accordingly C. ambitiously D. unexpectedly56. A. ignore B. withdraw C. issue D. reserve57. A. practically B. financially C. academically D. purposely58. A. committed B. entitled C. limited D. presented59. A. administration B. identification C. communication D. indication60. A. sponsored B. selected C. eliminated D. regulated61. A. By contrast B. In addition C. As a result D. In general62. A. preparation B. protection C. exchange D. description63. A. appropriate B. similar C. extra D. tremendous64. A. instruction B. outlook C. measure D. evidence65. A. arrangements B. requirements C. achievements D. arguments Section BDirections:Read the following four passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C, and D. Choose the one that fits best according to the information given in the passage you have just read.(A)HE ―is having the time of his life as ?America‘s favoritenerd‘,‖ said US magazine Adweek. Sheldon Cooper is a cleverand confident scientist who understands the laws of quantumphysics in the hit US sitcom The Big Bang Theory. The show will come back for its eighth season on Sept 22, 2014.Jim Parsons, 41, who has brought this popular character to life, collected his fourth Emmy Award for Lead Actor in a Comedy Series in last month.Unlik e Sheldon, Parsons doesn‘t read comic books or worship the space captains of Star Trek, but he says that he does have a few uncool interests – mainly things like crossword puzzles and poetry.However, there are some other similarities between the two: they‘re both smart and they both use hard work to do well in their chosen jobs.Parsons earned a master‘s degree in classical theater from the University of San Diego. But he was a struggling actor for years before he found fame, searching constantly for a role in either television, film or theater. After he moved to New York in the early 2000s, he would audition(试镜)for as many as 15 TV shows per season.Then, in 2007, Parsons was given a script for The Big Bang Theory. After reading it, he felt immediately t hat the role of Sheldon would be very good for him. ―There was something in his inability to understand sarcasm, his inability to read emotions … in a general sense, that I understood,‖ he told The New York Times.To play Sheldon, Parsons was required to understand the complicated rhythm of Sheldon‘s speech and use his clumsy physical actions.Some of the show‘s difficult language, especially its technical terms from science, reminded Parsons of reading Shakespeare while at university. He memorized his dialogue carefully beforehand, writing out his lines on white cards. He created exact physical worlds for Sheldon, even down to the spots where he would place his hat or coat.―The way I was forced to approach Shakespeare is very similar in my mind to the way I‘m forced to approach some aspects of Sheldon,‖ he told US magazine Playbill.Parsons‘ hard-working and intellectual approach has won him loyal fans. ―Every time we run a story on him, our hits go way up,‖ Clifford Pugh, editor-in-chief of US website Cult ureMap told CNN. ―He‘s up there with Beyonce and Lady Gaga.‖66. Jim Parsons is ______.A. an America‘s favorite nerdB. a clever and confident scientistC. an actor with four Emmy AwardsD. a popular character in The Big Bang67. We does the underlined sentence ―…who has brought this popular character tolife‖ mean ______.A. Sheldon Cooper is a clever but lifeless nerdB. Parson‘s acting as the character is true to lifeC. Sheldon Cooper has the appearance of being aliveD. Parson saved Sheldon Co oper‘s life68. Before Parson‘s success in The Big Bang, ______.A. He was long unrecognized to the publicB. He enjoyed his fame as a struggling actorC. His master‘s degree little helped him get rolesD. He auditioned for 15 TV shows in New York69. What can we know about the way Parson played Sheldon?A. He couldn‘t understand the complicated rhythm of Sheldon‘s speech.B. He learned how to understand technical terms by reading Shakespeare again.C. He created clumsy physical actions for Sheldon.D. He referred to his college experience of studying Shakespeare.(B)A NEW study from the Programme for International Student Assessment (PISA) will land on the desks of policymakers around the world next month. It will make sobering reading for political leaders in many countries. In Sweden Jan Bjorklund, the education minister, is prepared for poor marks too.The triennial study by the OECD, a think-tank, measures the reading, maths and science proficiency of 15-year-olds. In the first study, in 2000, Swedish pupils performed a lot better than those in most other countries. But even as the country‘s schools inspired imitators elsewhere, their results have deteriorated. In 2009 Sweden‘s overall score fell below the OECD average. Other rankings show a similar trend.―I assume the results will continue falling. It will take several more years before the positive effects of our policy begin to show in global ratings,‖ says Mr Bjorklund, referring to an overhaul of Sweden‘s education system. Since coming to p ower in 2006, the centre-right coalition government has introduced reforms such as a new national curriculum. Mr Bjorklund, who heads the Liberal party, is convinced he can reverse the decline. But will voters have the patience to wait? With universities complaining that students arrive unprepared and companies worrying that Sweden will lose out to other countries, a sense ofurgency is in the air. Education will be important in next year‘s election.What went wrong? Money is not the problem. Free education from primary school to university has long been a pillar of Sweden‘s welfare system, and public spending on education is among the world‘s highest, according to the OECD. Immigration is high, though this according to Skolverket—the National Agency for Education—had only a marginal effect on overall results.Mr Bjorklund blames the poor results on the period when the Social Democrats were in charge. Others say poorly paid teachers are at fault. The profession, once highly regarded, has seen salaries fall far behind other jobs requiring a higher-education degree. The student demand for teaching programs is so low that almost anyone applying will be accepted. As many teachers approach retirement, unions warn of a teacher crisis ahead. In hopes of making the job more attractive, a career programme with better pay was launched this year.A growing gap between schools is another reason, says Skolverket. Sweden is now one of the few countries to show both worse results and more inequality. Free school choice is a contributing factor. The system, introduced 20 years ago, allows parents to choose between municipal schools and independent schools, all financed by tax money. The aim was to increase quality by competition, but it has also led to the best students flocking to the same schools.70. The PISA‘s new study indicates that ______.A. Many other countries imitate Sweden‘s free education patternB. Sweden‘s rankings declined compared to many years agoC. Swedish pupils fall behind those in other countries for yearsD. Sweden ranked low in social science, history and English71. Which of the following is not true according to the passage?A. Universities are unsatisfied with new students‘ academic proficiency.B. V oters long for the immediate result of the education reform.C. It won‘t be long before Sweden revives in the global rating.D. Education will have influence on the following year‘s election.72. What does the author say about the education in Sweden?A. No students apply for teaching programs because teachers are poorly paid.B. The Social Democrats weren‘t to blame for the poor results of pupils.C. Immigration is a big barrier to overcome in Swedish education system.D. The highest expenditure on education distinguishes Sweden from others.73. The free school choice widens the gap between schools because ______.A. the best students prefer the same schoolB. it was aimed to increase competitionC. it allows parents to choose between different schoolsD. municipal and independent schools are all tax-funded(C)BACK in the rosy mid-2000s, immigrants were tolerated and sometimes even welcomed in Britain. Times were good, and the hard-grafting folk who rushed in after Poland joined the EU in 2004 fared well. But five years of economic slump have hardened attitudes. Polls suggest the locals are turning frosty just as a new wave of immigrants, from Bulgaria and Romania, rumbles towards the nation‘s shores. Politicians reflect those fears. Yet the evidence suggests more immigration would be a boost, not a drag.Britain‘s native population is remarkably stable. Since the early 1990s births have tallied so closely with deaths and emigration that the head count has been flat. Almost all of the country‘s population growth is down to immigration. Thenumber of non-natives living in Britain rose from 4.8m in 1995 to 13.4m in 2011. Immigrants might make up a quarter of the country‘s population by 2015.The mixed background of Britain‘s immigrants helps explain their effect on the labour market. The Migration Advisory Committee, an official commission, looked across a range of statistical studies and rubbished the notion that aliens push down wages and lift unemployment. There is little evidence of a relationship between immigration and average wages (indeed, some studies find that wages rise). If there is a deleterious effect on employment, it seems to come from a specific type of immigrant—those from outside the EU—and to affect low-skilled natives.Immigration does, however, seem to accentuate income inequality in Britain.New arrivals tend to have hugely variable skills, from hotshot programmers to manual labourers. They add disproportionately to the top and bottom of the wage scale, making it seem more polarised. And the perception of stiffening competition for low-paid work may be driving public opinion. In 2005 YouGov, a pollster, found that 56% of Britons supported the freedom of EU citizens to live and work where they chose. In the latest poll just 38% did.This is not, however, a good reason to block immigration more strenuously, as both the Conservative and Labour parties pledge to do. Other factors are more important in pushing down wages in basic jobs. New technology has reduced the need for unskilled work. Income inequality is rising across the world, not just in Britain.But immigration can be part of the solution. Immigrants are, on average, better educated than natives, according to a forthcoming paper by Christian Dustmann of University College London and Tommaso Frattini of the University of Milan. This means Britain gains from their skills without having to invest in schools. And they help balance the books. The researchers carefully allocated fiscal costs and revenues to natives and immigrants. They found some similarities: newcomers tend to be about as likely to live in social housing as natives. But immigrants are much less likely to receive any kind of state benefits. Immigrants from Europe and those who arrived after 2000, whatever their origin, are especially cheap.74. Which of the following is not true according to the first paragraph?A. Britain used to tolerate and welcome immigrants in the mid-2000s.B. The evidence suggests more immigrants will slow down the economy.C. Politicians fear that there will be a huge number of new immigrants.D. Local people in Britain hold cold attitudes towards new immigrants now.75. In the second paragraph, the author doesn‘t indicate ______.A. Three out of four people in Britain will possibly be non-natives by 2015B. Immigration is the contributing factor to the population growth in BritainC. There is no obvious change in emigration out to other countriesD. Birth and death rates have remained at almost equal levels for a long time76. What can be inferred from the passage?A. The Conservative and Labour parties don‘t want to block immigration.B. The notion that aliens push down wages and lift unemployment is acceptable.C. New technology is more important in lower wages and income inequality.D. The harmful effect of immigration on low-paid work changed public opinion.77. What does the author imply in paragraph 6 by saying ―But immigration can bepart of the solution‖?A. Immigrants is a solution to unemployment.B. Immigrants can‘t receive any state benefits.C. Immigrants are the cheap labour for Britain.D. Immigrants have to invest in school education.Section CDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.SPACE travel has officially entered a new era. We still can‘t fly through space at the speed of light, but two new and exciting spacecrafts could be carrying astronauts into space as early as 2016.The US space agency NASA has awarded Boeing and SpaceX the task of building the next generation of space taxis for its astronauts.Boeing‘s CST-100 is just over 4.5 meters wide, slightly wider than SpaceX‘s Dragon V2 (at 3.7 meters). While they might not look large enough to be comfortable for even the smallest astronaut, believe it or not, both spacecrafts are designed to carry a maximum of seven people.Both spacecrafts have similar overall designs. NASA‘s iconic shuttles were designed like a space plane. The CST-100, however, is based on capsule designs. It is similar to the one that George Clooney flew in the movie Gravity.It is SpaceX that is the cooler company and may have made the cooler spacecraft. Its founder, Elon Musk, is the CEO of the electric car company Tesla Motors.According to reports, the Dragon V2 will be able to land with ―the accuracy of a helicopter‖ by using eight thruster engines to control its landing. Inside the spacecraft, the pilot‘s pull-down control screen is futuristic, yet simple and beautiful. It looks like it could have been designed by Apple.Boeing, on the other hand, has an impressive safety record with its airplanes, and the CST-100 is no exception. Unlike the Dragon V2, it needs parachutes and airbags for its landing. Both spacecrafts are reusable a number of times, which is set to ―revolutionize‖ the industry, according to experts.Of course, no capsule or space plane is useful unless it has a rocket to launch it into space. SpaceX has already sent a rocket into space and will use its existing model, while Boeing is currently developing a rocket to send its capsule into space.With NASA‘s support for private space companies marking a new era, space tourism is closer than ever before. So, in the words of American singer Frank Sinatra, ―fly me to the moon!‖(Note: Answer the questions or complete the statements in NO MORE THAN 10 WORDS.)78. Boeing and SpaceX are designated by NASA to manufacture ________________.79. In addition to nearly the same size, what do CST-100 and Dragon V2 have incommon?80. Different from Boeing‘s CST-100, the Dragon V2 will use ________________for its accurately safe landing.81. What is the advantage of SpaceX over Boeing?第Ⅱ卷(共47分)I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1. 街对⾯曾经有⼀家⾯包店。
上海市2016届高三上学期期中考试数学理试题Word版含答案
南模中学2021学年第一学期高三期中考试数学学科(理)试卷一、填空题:(每题4分)1.函数23()(0)f x x x -=<的反函数是1()f x -=___________.2. 已知1sin cos 2αα=+,且0,2πα⎛⎫∈ ⎪⎝⎭,那么cos 2sin()4απα-的值为3.函数3()sin())22f x x x ππ=-++,方程()0f x k -=在[0,]x π∈上有两个不等的实根,那么实数k 的取值范围为 .4.关于函数()sin 2cos 2f x x x =-有以下命题:①函数()y f x =的最小正周期为π; ②直线4x π=是()y f x =的一条对称轴; ③点(,0)8π是()y f x =的图象的一个对称中心;④将()y f x =的图象向左平移4π个单位,可取得2y x =的图象. 其中真命题的序号是 .5. 某船在A 处看灯塔S 在北偏东30°方向,它以每小时18海里的速度向正北方向航行,通过40分钟航行到B 处,看灯塔S 在北偏东75°方向,那么现在该船到灯塔S 的距离约为 海里.6. 设A 是自然数集的一个非空子集,关于k A ∈,若是2k A ∉,A 那么k 是A 的一个“酷元”,给定集合{}2lg(36),S x y x x N ==-∈,设集合M 由集合S 中的两个元素组成,且集合M 中的两个元素都是“酷元”,那么如此的集合M 有 个.7.某企业有甲、乙两个研发小组,他们研发新产品成功的概率别离是32和53. 现安排甲组研发新产品A ,乙组研发新产品B. 设甲、乙两组的研发彼此独立. 假设新产品A 研发成功,预计企业可获利润120万元;假设新产品B 研发成功,估量企业可取得利润100万元. 那么该企业可获利润的数学期望为 万元.8. .设函数()sin f x x x π=+,则1240264027()()()()2014201420142014f f f f ++++= . 9.关于函数()f x ,假设在概念域内存在实数x ,知足()()f x f x -=-,那么称()f x 为“局部奇函数”.若 ()2x f x m =+是概念在[1,1]-上的“局部奇函数”,则实数m 的取值范围是 .10.假设不等式23log 0a x x -<在1(0,)3x ∈内恒成立,则a 的取值范围是 .11.设180,0,102y x y x x y>>+++=,那么2x y +的最大值为 . 12. 已知偶函数()f x 知足对任意的x R ∈均有(1)(3)f x f x +=-,且2(1)[0,1]()1(1,2]m x x f x x x ⎧-∈=⎨-∈⎩,假设方程3()f x x =恰有5个实数解,那么实数m 的取值范围是 .13.已知函数2()22(4)1f x mx m x =--+, ()g x mx =,假设关于任一实数x()f x 与()g x 至少有一个为正数,那么实数m 的取值范围是 .14.已知函数()(2)f x x a x =+,且关于x 的不等式()()f x a f x +<的解集为A ,假设11[,]22A -⊆,那么实数a 的取值范围是 .二、选择题: (每题5分)15. 若是关于任意实数x ,[]x 表示不超过x 的最大整数. 例如 []3.273=,[]0.60=.那么“[][]x y =”是“1x y -<”的( )A .充分而没必要要条件B .必要而不充分条件C .充要条件D .既不充分也没必要要条件16.以下函数中,既是偶函数,又在区间()1,2内是增函数的为 ( ) A. 22log 2x y x -=+ B. cos 2y x = C. 222x xy --= D. 2log y x = 17. 在△ABC 中,内角A ,B ,C 所对的边别离a ,b ,c ,给出以下命题:①A>B >C ,那么sinA >sinB >sinC ;②必存在A ,B ,C ,使tanAtanBtanC <tanA+tanB+tanC 成立;③若tanAtanB >1,那么△ABC 必然是钝角三角形;④若a=40,b=20,B=25°,△ABC 必有两解.其中真命题个数为( )A .0B .1C .2D .3 18.已知函数2212(1),,1,12()111,0,.362x x x x f x x x ⎧⎛⎤-+-∈ ⎪⎥+⎪⎝⎦=⎨⎡⎤⎪-+∈⎢⎥⎪⎣⎦⎩函数π()sin()22(0)6g x a x a a =-+>,假设存在[]12,0,1x x ∈, 使得12()()f x g x =成立,那么实数a 的取值范围是( )A .14,23⎡⎤⎢⎥⎣⎦ B .10,2⎛⎤ ⎥⎝⎦ C .24,33⎡⎤⎢⎥⎣⎦ D .1,12⎡⎤⎢⎥⎣⎦三、解答题:19. (此题12分) 函数22()log 1x f x x -=-的概念域为集合A,关于x 的不等式2212()()2ax a x a R +<∈的解集为B,求使A B B ⋃=的实数a 的取值范围.20、(此题14分)已知函数21()2cos 22f x x x =--, (1)求函数()f x 在[0,]2π的最大值和最小值,并给出取得最值时的x 值;(2)设ABC ∆的内角A 、B 、C 的对边别离为a ,b ,c ,且c =,()0f C =,假设sin 2sin B A =,求a ,b 的值.21、(此题14分)已知函数22()()()6x x f x e a e a -=-+-- , x R ∈(1)求()f x 的最小值; (2)假设函数()f x 在R 上存在零点,求实数a 的取值范围.22. (此题16分)已知函数()22f x x a x x =-+,a R ∈(1)若0a =,判定函数()y f x =的奇偶性,并说明理由;(2)假设函数()f x 在R 上是增函数,求实数的取值a 范围;(3)假设存在实数[2,2]a ∈-,使得关于x 的方程()(2)0f x tf a -=有三个不相等的实数根,求实数t 的取值范围.23.(此题18分)已知函数()y f x =,x D ∈,若是关于概念域D 内的任意实数x ,关于给定的非零常数m ,总存在非零常数T ,恒有()()f x T mf x +>成立,那么称函数()y f x =是D 上的m 级类增周期函数,周期为T .假设恒有()()f x T mf x +=成立,那么称函数()f x 是D 上的m 级类周期函数,周期为T .(1)已知函数2()f x x ax =-+是[3,)+∞上的周期为1的2级类增周期函数,求实数a 的取值范围;(2)已知T=1,()y f x =是[0,)+∞上m 级类周期函数,且()y f x =是 [0,)+∞上的单调递增函数,当[0,1)x ∈时,()2x f x =,求实数m 的取值范围;(3)是不是存在实数k ,使函数()cos f x kx =是R 上的周期为T 的T 级类周期函数,假设存在,求出实数k 和T 的值,假设不存在,说明理由.南模中学2021学年第一学期高三期中考试数学学科(理)试卷参考答案1. -32(0)x x-> 2.142- 3.3[,3)24. ①③5. 626. 57. 1408. 40279.5[,1]4-- 10.1[,1)2711. 1812.837415415837 (,)(,) 6666++++--⋃13. (0,8) 14. (-1,0)15.A 16.D 17.C 18. A 19.20.22.考点函数奇偶性的判断;函数单调性的性质.专题函数的性质及应用.分析(1)若a=0,根据函数奇偶性的定义即可判断函数y=f(x)的奇偶性;(2)根据函数单调性的定义和性质,利用二次函数的性质即可求实数a的取值范围;(3)根据方程有三个不同的实数根,建立条件关系即可得到结论.解答解:(1)函数y=f(x)为奇函数.当a=0时,f(x)=x|x|+2x,∴f(﹣x)=﹣x|x|﹣2x=﹣f(x),∴函数y=f(x)为奇函数;(2)f(x)=,当x≥2a时,f(x)的对称轴为:x=a﹣1;当x<2a时,y=f(x)的对称轴为:x=a+1;∴当a﹣1≤2a≤a+1时,f(x)在R上是增函数,即﹣1≤a≤1时,函数f(x)在R上是增函数;(3)方程f(x)﹣tf(2a)=0的解即为方程f(x)=tf(2a)的解.①当﹣1≤a≤1时,函数f(x)在R上是增函数,∴关于x的方程f(x)=tf(2a)不可能有三个不相等的实数根;…(9分)②当a>1时,即2a>a+1>a﹣1,∴f(x)在(﹣∞,a+1)上单调增,在(a+1,2a)上单调减,在(2a,+∞)上单调增,∴当f(2a)<tf(2a)<f(a+1)时,关于x的方程f(x)=tf (2a)有三个不相等的实数根;即4a<t﹣4a<(a+1)2,∵a>1,∴.设,∵存在a∈[﹣2,2],使得关于x的方程f(x)=tf(2a)有三个不相等的实数根,∴1<t<h(a)max,又可证在(1,2]上单调增∴<h(a)max=,∴1<t<③当a<﹣1时,即2a<a﹣1<a+1,∴f(x)在(﹣∞,2a)上单调增,在(2a,a﹣1)上单调减,在(a﹣1,+∞)上单调增,∴当f(a﹣1)<tf(2a)<f(2a)时,关于x的方程f(x)=tf (2a)有三个不相等的实数根;即﹣(a﹣1)2<t﹣4a<4a,∵a<﹣1,∴,设,∵存在a∈[﹣2,2],使得关于x的方程f(x)=tf(2a)有三个不相等的实数根,∴1<t<g(a)max,又可证在[﹣2,﹣1)上单调减,∴g(a)max=,∴1<t<;综上:1<t<.23 .。
2016届高三第一次六校联考数学试题-Word版含答案
2016届高三六校第一次联考文科数学试题命题学校:珠海一中 2015,9,7本试题共4页,第1至21题为必做题,从第22、23、24三个小题中选做一题, 满分150分,考试用时120分钟。
一、选择题:本大题共12小题,每小题5分,满分60分,在每小题给出的四个选项中,只有一项是符合题目要求的。
1.已知全集U ={1,2,3,4,5},集合A ={2,3,4},B ={1,4},则(?U A )∪B 为( ) A .{1} B .{1,5} C .{1,4} D .{1,4,5} 2.若z 是z 的共轭复数,且满足i i z 24)1(2+=-⋅,则=z ( ) A .i 21+- B .i 21-- C .i 21+ D .i 21-3.已知命题q p ,,则“q p ∧是真命题”是“p ⌝为假命题”的( ) A .充分而不必要条件 B .必要而不充分条件 C .充要条件 D .既不充分也不必要条件4.设等比数列}{n a 的公比21=q ,前n 项和为n S ,则=33a S ( ) A .5 B .7 C .8 D .155.下列四个函数中,既是偶函数又在(0,+∞)上为增函数的是( ) A .x x y 22-= B .3x y = C .21ln x y -= D .1||+=x y6.已知双曲线的渐近线方程为x y 2±=,焦点坐标为)(0,6),0,6(-,则双曲线方程为( )A .18222=-y xB .12822=-y xC .14222=-y xD .12422=-y x7.函数)0)(3sin()(>+=ωπωx x f 相邻两个对称中心的距离为2π,以下哪个区间是函数)(x f 的单调减区间( )A .]0,3[π- B .]3,0[πC .]2,12[ππ D .]65,2[ππ8.曲线x x y 2ln -=在点)2,1(-处的切线与坐标轴所围成的三角形的面积是( )A .21B .43 C .1 D .29.在边长为2的正方体内部随机取一个点,则该点到正方体8个顶点的距离都不小于1的概率为( )A .61 B .65 C .6π D .6-1π10.一个空间几何体的三视图如下图,其中正视图是边长为2的正三角形,俯视图是边长分别为1,2的矩形,则该几何体的侧面积为( )A .43+B .63+C .432+D .632+11.执行如右图所示的程序框图,若输出的n =9,则输入的整数p 的最小值是( ) A .50 B .77 C .78 D .30612.已知抛物线x y =2上一定点B(1,1)和两个动点P 、Q ,当P 在抛物线上运动时,BP ⊥PQ ,则Q 点的纵坐标的取值范围是( ) A .),,(∞+⋃∞-2[]2- B .),,(∞+⋃∞-3[]1- C .),,(∞+⋃∞-3[]0 D .),,(∞+⋃∞-4[]1 二、填空题:本大题共4小题,每小题5分,满分20分。
2016-2017学年上海市浦东新区高三(上)期中数学试卷
2016-2017学年上海市浦东新区高三(上)期中数学试卷学校:___________姓名:___________班级:___________考号:___________一、填空题(本大题共12小题,共60.0分)1.设全集U=R,集合A={x|x<2},B={y|y=x2+1},则A∪∁U B= ______ .【答案】(-∞,2)【解析】解:∵集U=R,集合A={x|x<2}=(-∞,2),B={y|y=x2+1}=[1,+∞),∴∁U B=(-∞,1),∴A∪(∁U B)=(-∞,2),故答案为:(-∞,2).求出关于A、B的范围,得到B的补集,从而求出A∪(∁U B)即可.本题考查了函数的值域的解法,考查集合的运算,是一道基础题.2.函数f(x)=x2-1(x≥0)的反函数为f-1(x),则f-1(2)= ______ .【答案】【解析】解:根据函数与它的反函数的定义域和值域互换,令函数f(x)=x2-1=2,其中x≥0,解得x=;所以f-1(2)=.故答案为:.根据函数与它的反函数的定义域和值域互换,令f(x)=2,且x≥0,求出x的值即可.本题考查了函数与它的反函数的性质与应用问题,是基础题目.3.x>1,则函数y=x+的值域是______ .【答案】[3,+∞)【解析】解:∵x>1,则,x-1>0,>;那么:函数y=x+=x-1++1≥=3,当且仅当x=2时取等号.所以函数y的值域是[3,+∞).利用不等式法求值域即可.本题考查了函数值域的求法.高中函数值域求法有:1、观察法,2、配方法,3、反函数法,4、判别式法;5、换元法,6、数形结合法,7、不等式法,8、分离常数法,9、单调性法,10、利用导数求函数的值域,11、最值法,12、构造法,13、比例法.要根据题意选择.4.已知集合A={x||x|≤2,x∈R},B={x|≤4,x∈Z},则A∩B= ______ .【答案】{0,1,2}.【解析】解:∵集合A={x||x|≤2,x∈R}={x|-2≤x≤2},B={x|≤4,x∈Z}={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16},∴A∩B={0,1,2}.故答案为:{0,1,2}.先分别求出集合A和B,由此能求出A∩B.本题考查交集的求法,是基础题,解题时要认真审题,注意交集性质的合理运用.5.如图,正三棱柱ABC-A1B1C1中,有AB=AA1,则AC1与平面BB1C1C所成的角的正弦值为______ .【答案】【解析】解:取BC的中点E,连接C1E,AE则AE⊥BC,正三棱柱ABC-A1B1C1中,∴面ABC⊥面BB1C1C,面ABC∩面BB1C1C=BC,∴AE⊥面BB1C1C,∴∠AC1E就是AC1与平面BB1C1C所成的角,在R t△AC1E中,∵AB=AA1,sin∠AC1E=.故答案为:.根据题,过取BC的中点E,连接C1E,AE,证明AE⊥面BB1C1C,故∴∠AC1E就是AC1与平面BB1C1C所成的角,解直角三角形AC1E即可.考查直线和平面所成的角,求直线和平面所成的角关键是找到斜线在平面内的射影,把空间角转化为平面角求解,属基础题6.已知一组数据7、8、9、x、y的平均数是8,则这组数据的中位数是______ .【答案】8【解析】解:由题意知:(7+8+9+x+y)÷5=8,化简可得又因为该组数据为5个,则中位数对应位置(5+1)÷2=3.①当x=y时,得x=y=8.显然,改组数据中位数为8.②当x≠y时,不妨设x<y,又因为x+y=16,可以得到x<8<y,此时中位数也为8.发散思维,考查学生对于定义的明确掌握.先求出关于未知数的式子,再分情况讨论问题.学生要学会具体情况具体分析,将题目分情况讨论,便于解题过程的明晰.7.若不等式|3x-b|<4的解集中的整数有且仅有1,2,3,则b的取值范围______ .【答案】5<b<7【解析】解:因为<<<<<,又由已知解集中的整数有且仅有1,2,3,故有<<<<<<.故答案为5<b<7.首先分析题目已知不等式|3x-b|<4的解集中的整数有且仅有1,2,3,求b的取值范围,考虑到先根据绝对值不等式的解法解出|3x-b|<4含有参数b的解,使得解中只有整数1,2,3,即限定左边大于0小于1,右边大于3小于4.即可得到答案.此题主要考查绝对值不等式的解法问题,题目涵盖知识点少,计算量小,属于基础题型.对于此类基础考点在高考中属于得分内容,同学们一定要掌握.8.(1+x)7的展开式中x2的系数是______ .【答案】21【解析】解:由题意,二项式(1+x)7的展开式通项是T r+1=x r故展开式中x2的系数是=21故答案为:21.由题设,二项式(1+x)7,根据二项式定理知,x2项是展开式的第三项,由此得展开式中x2的系数.本题考查二项式定理的通项,熟练掌握二项式的性质是解题的关键.9.从总体中抽取一个样本:3、7、4、6、5,则总体标准差的点估计值为______ .【答案】【解析】解:样本数据:3、7、4、6、5的平均数为:=×(3+7+4+6+5)=5,方差为s2=×[(3-5)2+(7-5)2+(4-5)2+(6-5)2+(5-5)2]=2,所以标准差为s=.故答案为:.根据平均数与方差、标准差的计算公式,即可求出结论.本题考查了平均数,方差与标准差的计算问题,是基础题目.10.已知y=f(x)+x2是奇函数,且f(1)=1,若g(x)=f(x)+2,则g(-1)= ______ .【答案】-1【解析】解:由题意,y=f(x)+x2是奇函数,且f(1)=1,所以f(1)+1+f(-1)+(-1)2=0解得f(-1)=-3所以g(-1)=f(-1)+2=-3+2=-1故答案为:-1.由题意,可先由函数是奇函数求出f(-1)=-3,再将其代入g(-1)求值即可得到答案本题考查函数奇偶性的性质,利用函数奇偶性求值,解题的关键是根据函数的奇偶性建立所要求函数值的方程,基本题型.11.已知f(x)=log a(x+1),g(x)=log a(1-x),a>0且a≠1,则使f(x)-g(x)>0成立的x的集合是______ .【答案】当0<a<1时,原不等式的解集为{x|-1<x<0};当a>1时,原不等式的解集为{x|0<x<1}【解析】解:f(x)-g(x)>0,即log a(x+1)-log a(1-x)>0,log a(x+1)>log a(1-x).当0<a<1时,上述不等式等价于>><,解得-1<x<0;当a>1时,原不等式等价于>>>,解得0<x<1.综上所述,当0<a<1时,原不等式的解集为{x|-1<x<0};当a>1时,原不等式的解集为{x|0<x<1}.故答案为:当0<a<1时,原不等式的解集为{x|-1<x<0};a>1时,原不等式的解集为{x|0<x<1}.利用函数的奇偶性整理不等式为log a(x+1)>log a(1-x),对底数a分类讨论得出x的范围.本题考查不等式的解法,对底数a的分类讨论是关键.12.在R t△ABC中,两直角边分别为a、b,设h为斜边上的高,则=+,由此类比:三棱锥S-ABC中的三条侧棱SA、SB、SC两两垂直,且长度分别为a、b、c,设棱锥底面ABC上的高为h,则______ .【答案】+【解析】解:∵PA、PB、PC两两互相垂直,∴PA⊥平面PBC.设PD在平面PBC内部,且PD⊥BC,由已知有:PD=,h=PO=,∴,即.故答案为:.立体几何中的类比推理主要是基本元素之间的类比:平面⇔空间,点⇔点或直线,直线⇔直线或平面,平面图形⇔平面图形或立体图形,故本题由平面上的直角三角形中的边与高的关系式类比立体中两两垂直的棱的三棱锥中边与高的关系即可.类比推理是指依据两类数学对象的相似性,将已知的一类数学对象的性质类比迁移到另一类数学对象上去.其思维过程大致是:观察、比较联想、类推猜测新的结论.三、解答题(本大题共5小题,共70.0分)19.用一个半径为10cm的半圆纸片卷成一个最大的无底圆锥,放在水平桌面上,被一阵风吹倒,如图所示,求它的最高点到桌面的距离.【答案】解:如图所示,设PAB为轴截面,过点A作AD⊥PB,π•AB=10π,解得AB=10,∴△PAB是等边三角形,∴AD=AB•sin60°=10×=5.∴它的最高点到桌面的距离为5cm.【解析】如图所示,设PAB为轴截面,过点A作AD⊥PB,利用圆的周长公式π•AB=10π,解得AB=10,可得△PAB是等边三角形,即可得出.本题考查了圆锥的轴截面的性质、圆的弧长与周长计算公式、等边三角形的性质,考查了空间想象能力、推理能力与计算能力,属于中档题.20.已知全集U=R,集合A={x|4x-9•2x+8<0},B={x|},C={x||x-2|<4},求A∪B,C U A∩C.【答案】解:由1<2x<8,得A=(0,3).(2分)由,得B=(-2,3].(4分)由|x-2|<4-2<x<6,得C=(-2,6).(6分)所以A∪B=(-2,3],(8分)C U A∩C=(-2,0]∪[3,6).(12分)【解析】由1<2x<8,得A=(0,3).由,得B=(-2,3].由|x-2|<4-2<x<6,得C=(-2,6).由此能求出A∪B,C u A∩C.本题考查交、并、补集的混合运算,是基础题.解题时要认真审题,仔细解答.21.如图:三棱锥P-ABC中,PA⊥底面ABC,若底面ABC是边长为2的正三角形,且PB与底面ABC所成的角为.若M是BC的中点,求:(1)三棱锥P-ABC的体积;(2)异面直线PM与AC所成角的大小(结果用反三角函数值表示).【答案】解:(1)因为PA⊥底面ABC,PB与底面ABC所成的角为所以∠因为AB=2,所以(2)连接PM,取AB的中点,记为N,连接MN,则MN∥AC所以∠PMN为异面直线PM与AC所成的角计算可得:,MN=1,∠异面直线PM与AC所成的角为【解析】(1)欲求三棱锥P-ABC的体积,只需求出底面积和高即可,因为底面ABC是边长为2的正三角形,所以底面积可用来计算,其中a是正三角形的边长,又因为PA⊥底面ABC,所以三棱锥的高就是PA长,再代入三棱锥的体积公式即可.(2)欲求异面直线所成角,只需平移两条异面直线中的一条,是它们成为相交直线即可,由M为BC中点,可借助三角形的中位线平行于第三边的性质,做出△ABC的中位线,就可平移BC,把异面直线所成角转化为平面角,再放入△PMN中,求出角即可.本题主要考查了在几何体中求异面直线角的能力.解题关键再与找平行线,本题主要通过三角形的中位线找平行线,如果试题的已知中涉及到多个中点,则找中点是出现平行线的关键技巧.22.甲厂以x千克/小时的速度运输生产某种产品(生产条件要求1≤x≤10),每小时可获得利润是100(5x+1-)元.(1)写出生产该产品t(t≥0)小时可获得利润的表达式;(2)要使生产该产品2 小时获得的利润不低于3000元,求x的取值范围.【答案】解:(1)设生产该产品t (t ≥0)小时可获得利润为f (t ),则f (t )=100t (5x +1-)元,t ≥0,1≤x ≤10.(2)由题意可得:100×2×(5x +1-)≥3000,化为:5x 2-14x -3≥0,1≤x ≤10. 解得3≤x ≤10.∴x 的取值范围是[3,5]. 【解析】(1)设生产该产品t (t ≥0)小时可获得利润为f (t ),可得f (t )=100t (5x +1-)元. (2)由题意可得:100×2×(5x +1-)≥3000,解出即可得出.本题考查了函数的应用、二次函数的单调性,考查了推理能力与计算能力,属于中档题.23.已知函数f (x )=|x +|-|x -|;(1)作出函数f (x )的图象;(2)根据(1)所得图象,填写下面的表格:(3)关于的方程2()+|()|+=0(,∈R )恰有6个不同的实数解,求的取值范围.【答案】解:函数f (x )=|x + |-|x -|=, ,< < , <,< ,作出函数f (x )的图象如图:(2)由函数的图象得函数的定义域为{x|x≠0},函数的值域为(0,2],在(-∞,-1]和(0,1)上单调递增,在[1,+∞)和(-1,0),单调递减,函数关于y轴对称,是偶函数,函数与x轴没有交点,无零点.(3)∵0<f(x)≤2,且函数f(x)为偶函数,∴令t=f(x),则方程等价为t2+mt+n=0,则由图象可知,当0<t<2时,方程t=f(x)有4个不同的根,当t=2时,方程t=f(x)有2个不同的根,当t≤0或t>2时,方程t=f(x)有0个不同的根,若方程f2(x)+m|f(x)|+n=0(m,n∈R)恰有6个不同的实数解,等价为方程f2(x)+mf(x)+n=0(m,n∈R)恰有6个不同的实数解,即t2+mt+n=0有两个不同的根,其中t1=2,0<t2<2,则n=t1t2∈(0,4).【解析】(1)利用分段函数求出f(x)的表达式,然后作出函数f(x)的图象,(2)结合函数的图象判断相应的性质,(3)根据图象利用换元法将条件进行转化,利用数形结合即可得到结论.本题主要考查函数零点的应用,利用条件求出函数f(x)的表达式,利用数形结合是解决本题的关键,综合性较强,难度较大.二、选择题(本大题共6小题,共30.0分)13.“a>1”是“f(x)=(a-1)•a x在定义域内为增函数”的()条件.A.充分不必要B.必要不充分C.充要D.既不充分也不必要【答案】A【解析】解:当a>1时,a-1>0,a x在定义域内为增函数,则f(x)=(a-1)•a x在定义域内为增函数”成立,即充分性成立,若0<a<1,a-1<0,a x在定义域内为减函数,满足f(x)=(a-1)•a x在定义域内为增函数”,此时a>1不成立,即必要性不成立,故“a>1”是“f(x)=(a-1)•a x在定义域内为增函数”的充分不必要条件,故选:A根据函数单调性的性质结合充分条件和必要条件的定义进行判断即可.本题主要考查充分条件和必要条件的判断,根据函数单调性的性质以及充分条件和必要条件的定义是解决本题的关键.14.如图,直线a、b相交于点O且a、b成60°角,过点O与a、b都成60°角的直线有()A.1条B.2条C.3条D.4条【答案】C【解析】解:在a、b所确定的平面内有一条如图,平面外有两条.如图故选C本题宜用作出相应的图象来辅助说明过点O与a、b都成60°角的直线存在情况,一图一析,本题考查异面直线所成的角,确定与直线所成角为多少度的线的个数,本题用语言不易说明,比较抽象,故采取了用图象辅助的方法,图象语言,直观易懂,要学会熟练使用图象语言来表述问题.15.有5本不同的书,其中语文书2本,数学书2本,物理书1本,若将其随机地并排放到书架的同一层上,则同一科目的书都相邻的概率为()A. B. C. D.【答案】A【解析】解:有5本不同的书,其中语文书2本,数学书2本,物理书1本,将其随机地并排放到书架的同一层上,基本事件总数n==120,同一科目的书都相邻包含的基本事件个数m==24,∴同一科目的书都相邻的概率为p===.故选:A.先求出基本事件总数n=,再求出同一科目的书都相邻包含的基本事件个数m=,由此能求出同一科目的书都相邻的概率.本题考查概率的求法,是中档题,解题时要认真审题,注意排列知识的合理运用.16.已知三个球的半径R1、R2、R3满足R1+2R2=3R3,则它们的表面积S1、S2、S3满足的等量关系是()A.S1+2S2=3S3B.+=C.+2=3D.+4=9【答案】C【解析】解:因为S1=4πR12,所以=2,同理:=2,=2,由R1+2R2=3R3,得+2=3.故选:C.表示出三个球的表面积,求出三个半径,利用R1+2R2=3R3,推出结果.本题考查球的表面积,考查计算能力,是基础题.17.已知函数,则不等式f(x)≥x2的解集是()A.[-1,1]B.[-2,2]C.[-2,1]D.[-1,2]【答案】A【解析】解:①当x≤0时;f(x)=x+2,∵f(x)≥x2,∴x+2≥x2,x2-x-2≤0,解得,-1≤x≤2,∴-1≤x≤0;②当x>0时;f(x)=-x+2,∴-x+2≥x2,解得,-2≤x≤1,∴0<x≤1,综上①②知不等式f(x)≥x2的解集是:-1≤x≤1,故选A.已知分段函数f(x)求不等式f(x)≥x2的解集,要分类讨论:①当x≤0时;②当x>0时,分别代入不等式f(x)≥x2,从而求出其解集.此题主要考查一元二次不等式的解法,在解答的过程中运用的分类讨论的思想,是一道比较基础的题目.18.我们定义渐近线:已知曲线C,如果存在一条直线,当曲线C上任意一点M沿曲线运动时,M可无限趋近于该直线但永远达不到,那么这条直线称为这条曲线的渐近线:下列函数:①y=x;②y=2x-1;③y=lg(x-1);④y=;其中有渐近线的函数的个数为()A.1B.2C.3D.4【答案】C【解析】解:对于:①y=x,根据渐近线的定义,不存在渐近线;对于②y=2x+1是由y=2x的图象向上平移1个单位得到,其渐近线方程为y=1;对于③y=log2(x-1)是由y=log2x向右平移一个单位得到,其渐近线方程为x=1;对于④y==(1-),其渐近线方程为x=,y=;综上,有渐近线的个数为3个故选:C.利用图象的变换规律,结合初等函数的图象特点,即可得到结论.本题考查新定义,考查学生分析解决问题的能力,属于中档题.。
2016届高三六校第一次联考
2016届高三六校第一次联考理科数学试题参考答案及评分标准一. 选择题:1、B2、A3、D4、B5、A6、C7、A8、C9、B 10、D 11、C 12、B 11、如图,易知BCD ∆的面积最大12、 解:令21()()2g x f x x =-,2211()()()()022g x g x f x x f x x -+=--+-= ∴函数()g x 为奇函数 ∵(0,)x ∈+∞时,//()()0g x f x x =-<,函数()g x 在(0,)x ∈+∞为减函数又由题可知,(0)0,(0)0f g ==,所以函数()g x 在R 上为减函数2211(6)()186(6)(6)()186022f m f m mg m m g m m m ---+=-+----+≥即(6)()0g m g m --≥∴(6)()g m g m -≥,∴6,3m m m -≤∴≥二、填空题:本大题共4小题,每小题5分,共20分13、2 14、 5 15、 73 16、2016 ∵(2016)(2013)3(2010)6(0)20162016f f f f ≤+≤+≤≤+= (2016)(2014)2(2012)4(0)20162016f f f f ≥+≥+≥≥+=(2016)2016f ∴=三、解答题(17—21为必做题)CDBA17、解:(1)由题意易知122n n n a a a --=+,---1分 即1231112n n n a q a q a q ---=+,--2分2210q q ∴--= 解得1q =或12q =- -------- 3分(2)解:①当1q =时,1n a =,n b n = n S =2)1(+n n ----------5分②当12q =-时,11()2n n a -=-11()2n n b n -=⋅- ---------------7分n S =012111111()2()3()()2222n n -⋅-+⋅-+⋅-++⋅--21n S = 12111111()2()(1)()()2222n n n n -⋅-+⋅-++-⋅-+⋅- 相减得21311111()()()()22222n n n S n -⎡⎤=-⋅-+-+-++-⎢⎥⎣⎦-------- 10分整理得 n S =94-(94+32n )·1()2n ------------------------12分18、解:设甲、乙、丙各自击中目标分别为事件A 、B 、C(Ⅰ)由题设可知0ξ=时,甲、乙、丙三人均未击中目标,即(0)()P P A B C ξ== ∴()()()21011515P m n ξ==--=,化简得()56mn m n -+=- ① ……2分同理, ()3113553P m n mn ξ==⨯⨯=⇒= ②……4分 联立①②可得23m =,12n = ……6分(Ⅱ)由题设及(Ⅰ)的解答结果得:(1)()P P A B C A B C A B C ξ==++()3311221211153253253210a P ξ∴===⨯⨯+⨯⨯+⨯⨯=……8分()3131111510530b ∴=-++= ……10分31353110123151030530E ξ∴=⨯+⨯+⨯+⨯= ……12分19.解法一:(1)如图:,,AC AC BD O =连设1.AP B G OG 1与面BDD 交于点,连 ……1分1111//,,PC BDD B BDD B APC OG =因为面面面故//OG PC .所以122m OG PC ==.又111,,AO DB AO BB AO BDD B ⊥⊥⊥所以面 ……3分 故11AGO AP BDD B ∠即为与面所成的角。
上海市奉贤区2016届高三上学期期末调研化学试卷.pdf
解速率的影响:在相同条件下,向一支试管中加入2mL5%H2O2和1mLH2O,向另一支试管中加入2mL5%H2O2和1mLFeCl3溶液
,观察并比较实验现象
16、某同学将光亮的镁条放入盛有NH4Cl溶液的试管中,有大量气泡产生。为探究该反应原理,该同学做了以下试
验并观察到相关现象,由此得出的结论不合理的是:
途径a:CCO+H2CO2+H2O
途径b:CCO2
C.食物中可加入适量的食品添加剂,如香肠中可以加少量的亚硝酸钠以保持肉质新鲜
D.生石灰、铁粉、硅胶是食品包装中常用的干燥剂
5、对于复分解反应:XY(Z+W,下列叙述正确的是
A.若Z是强酸,则X和Y必有一种是强酸
B.若X是强酸,Y是盐,反应后可能有强酸或弱酸生成
七、(本题共13分)
38、AC 2分
39.
40. Cr2O72-+ H2O2CrO42- + 2H+。2分
41.。
42.K2Cr2O72Cr2O7小(或四种物质中K2Cr2O7的溶解度最小)。
43. 。
八、(本题共8分)
44.
45.2HCHO+2H2O 2分
46.
47.
2分
九、(本题共14分)
48.羟基、醛基;酯化(或取代)
49. CH2=CHCHO 、C(CH2OH)4
2分
51.3
52. CH3-CH=CH-CH2CHO(或CH3-CH2-CH=CH-CHO) 2分
53.
3分
十、(本题共14分)
54.14.44 2分
A.在离子化合物中不可能存在非极性共价键
B.由电子定向移动而导电的物质是金属晶体
C.有键能很大的共价键存在的物质熔沸点一定很高D.只含有共价键的物质不一定是共价化合物
上海市奉贤区2016届高三第一学期期中考试(区六校联考)语文试卷(无答案)
2015学年第一学期期中考试(暨区六校联考)高三语文试卷命题人:万涛学校:曙光中学 2015年11月一、阅读(80分)(一)阅读下文,完成第1-6题。
(17分)网络文学:什么才是人们该关注的①关于网络文学,本人一向不敢发声,之所以如此胆怯,是因为自己对其作品的阅读太少,说来惭愧:就连《诛仙》、《鬼吹灯》、《盗墓笔记》这样的明星产品都未读过,如此缺乏阅读积累,那说啥心里还不虚得很?不过不敢发声不等于无视或漠视,那么一个“庞然大物”客观上存在着,你的无视与漠视除了暴露自己的孤陋寡闻外,别无其他。
坦率地说,为了弥补这种无知,我倒是很想从他人的评述中获取些许相关的滋养..。
结果,一番搜寻阅读下来,从社会整体有关网络文学的关注中我获取的最大量信息概括起来或可用如下两字而蔽之。
②第一个字谓之为“量”。
据不完全统计,中国目前网络文学签约作者数量超过百万,每年上传作品的数量也接近一百万部,如起点中文网、红袖小说网等多家在线中文写作平台,每日更新的字数就多达一个亿。
同时,网络文学单个作品动辄三四部五六部者甚多。
点击量高指的是某部网络文学作品获得数以百千万的点击量都并非天方夜谭。
至于吸金量强则是说网络作家的“造富能力”好生了得,据“2013第八届中国作家富豪榜·网络作家富豪榜”发布:起点中文网白金级作家唐家三少以2650万元版税蝉联状元宝座,天蚕土豆、血红、我吃西红柿、梦入神机等“网络大神”则紧随其后。
③第二个字谓之为“体”。
具体地说就是各家都在力图为网络文学定性,试图回答网络文学的本体是什么?概括起来大体上亦不外乎如下四说:一曰传统文学之类型说,即网络文学无非就是传统文学中的类型文学,且多为玄幻、武侠、言情和戏说历史等四类;二曰非文学说,即网络文学多芜杂,压根儿还谈不上是文学;三曰互渗说,即网络文学与传统文学并非完全对立而是相互渗透;四曰另起炉灶说,即传统文学的那套话语已无力诠释网络文学,因而需要新建另一套话语体系方能言说。
2016届高三数学上学期期中考试理科试卷(带答案)
2016届高三数学上学期期中考试理科试卷(带答案)汕头市金山中学2015-2016学年度第一学期期中考试高三理科数学试题卷本试题分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,满分150分,时间120分钟. 第Ⅰ卷(选择题共60分)一、选择题:(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的) 1.已知为虚数单位,若复数是纯虚数,则实数的值为() A. B. C. D. 2.“ ”是“ ”的() A.充分不必要条件 B.必要不充分条件 C.充要条件 D.既不充分也不必要条件 3.已知数列为等比数列,,则() A. B.或C. D. 4.如图,正方体中,为棱的中点,用过点的平面截去该正方体的上半部分,则剩余几何体的左视图为() D 5.设双曲线的渐近线方程为,则该双曲线的离心率为() A. B.2 C. D. 6.已知平面向量,,,要得到的图像,只需将的图像() A.向左平移个单位长度 B.向右平移个单位长度 C.向左平移个单位长度D.向右平移个单位长度 7.设满足约束条件,若目标函数的最大值为8,则的最小值为() A. B. C. D. 8.定义平面向量的正弦积为,(其中为、的夹角),已知中,,则此三角形一定是() A.等腰三角形 B.直角三角形 C.锐角三角形 D.钝角三角形 9.运行如图所示的流程图,则输出的结果是()A. B. C. D. 10.如图,矩形内的阴影部分是由曲线及直线与轴围成,向矩形内随机投掷一点,若落在阴影部分的概率为,则的值是() A. B. C. D. 11.已知向量的夹角为在时取得最小值,当时,夹角的取值范围是()A. B. C. D. 12.设定义在上函数.若曲线上存在点使得,则实数的取值范围是() A. B. C. D.第Ⅱ卷(非选择题共90分)二、填空题:(本大题共4小题,每小题5分,共20分) 13.已知函数,则。
14.已知,则的值是. 15.已知函数,在中,分别是角的对边,若,则的最大值为。
2015-2016学年上海市六校高三(上)第一次联考数学试卷(理科)(解析版)
2015-2016学年上海市六校高三(上)第一次联考数学试卷(理科)一、填空题:本大题满分56分。
本大题共有14题,只要求直接填写结果,每题填对得4分,否则一律得零分。
1.函数f(x)=x2﹣1(x≥1)的反函数是f﹣1(x)=.2.已知||=2,||=1,的夹角为,则=.3.幂函数y=f(x)的图象经过点(4,),则=.4.方程log2(x﹣3)=log4(5﹣x)的解为.5.若直线l1的一个法向量=(1,1),若直线l2的一个方向向量=(1,﹣2),则l1与l2的夹角θ=.(用反三角函数表示)6.直线l:x+交圆x2+y2=2于A、B两点,则|AB|=.7.已知α∈(0,π),且tan()=,则cosα=.8.无穷等比数列{a n}的前n项和为S n,若S3=6,S6=3,则=.9.已知f(x)=kx﹣|x﹣1|有两个不同的零点,则实数k的取值范围是.10.已知a、b、c是△ABC中∠A、∠B、∠C的对边,若a=7,A=60°,△ABC的面积为10,则△ABC的周长为.11.奇函数f(x)的定义域为R,若f(x+2)为偶函数,且f(1)=1,则f(100)+f(101)=.12.已知等比数列{a n}的前n项和为S n,若S4、S2、S3成等差数列,且a2+a3+a4=﹣18,若S n≥2016,则n的取值范围为.13.设m∈R,过定点A的动直线x+my=0和过定点B的动直线mx﹣y﹣m+3=0交于点P(x,y).则|PA|•|PB|的最大值是.14.不等式(2﹣|x|)(2+x)>0的解集为.15.设[x]表示不超过x的最大整数,若[π]=3,[﹣1.2]=﹣2.给出下列命题:①对任意的实数x,都有x﹣1<[x]≤x.②对任意的实数x、y,都有[x+y]≥[x]+[y].③[lg1]+[lg2]+[lg3]+…+[lg2014]+[lg2015]=4940.④若函数f(x)=[x[x]],当x∈[0,n)(n∈N*)时,令f(x)的值域为A,记集合A中元素个数为a n,则的最小值为,其中所有真命题的序号为.二、选择题:本大题满分20分。
奉贤区高三数学期中试卷
一、选择题(每题5分,共50分)1. 已知函数$f(x) = \sqrt{1-x^2}$,其定义域为()。
A. $[-1, 1]$B. $[-1, 0) \cup (0, 1]$C. $(-1, 1)$D. $(-\infty, -1] \cup [1, +\infty)$2. 在三角形ABC中,角A、B、C的对边分别为a、b、c,若$cosA + cosB + cosC = \frac{3}{2}$,则三角形ABC为()。
A. 直角三角形B. 钝角三角形C. 等腰三角形D. 等边三角形3. 下列函数中,在区间(0,+∞)上单调递增的是()。
A. $y = e^{-x}$B. $y = \ln x$C. $y = x^2$D. $y =\sqrt{x}$4. 已知数列$\{a_n\}$的通项公式为$a_n = 3^n - 2^n$,则数列$\{a_n\}$的前n项和$S_n$的值等于()。
A. $3^n - 2^n$B. $3^n - 2^{n-1}$C. $3^n - 2^{n+1}$D. $3^n - 2^n + 1$5. 若向量$\vec{a} = (1, 2)$,向量$\vec{b} = (2, -1)$,则向量$\vec{a}\cdot \vec{b}$的值为()。
A. 5B. -5C. 3D. -36. 已知函数$f(x) = x^3 - 3x^2 + 4x - 1$,若$f(x) = 0$的三个根分别为$a$、$b$、$c$,则$a+b+c$的值为()。
A. 3B. 2C. 1D. 07. 在等差数列$\{a_n\}$中,若$a_1 = 2$,公差$d = 3$,则数列$\{a_n\}$的第10项$a_{10}$的值为()。
A. 25B. 30C. 35D. 408. 若复数$z = 1 + i$,则$|z|^2$的值为()。
A. 2B. 3C. 4D. 59. 在直角坐标系中,点P的坐标为$(2, 3)$,点Q在直线$y = 2x + 1$上,且$\overrightarrow{OP} \cdot \overrightarrow{OQ} = 7$,则点Q的坐标为()。
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2015学年第一学期期中考试(暨区六校联考)
高三数学试卷
命题人:丁欢锋 学校 曙光中学 2015年11月
一、填空题(本大题满分56分,每小题4分)
1.函数sin 2cos2y x x =⋅的最小正周期是 .
2.已知全集{}2,1,0=U ,{}+0A x x m ==,如果U C A ={}1,0,则=m .
3.已知数列{}n a 中,11a =,*13,(2,)n n a a n n N -=+≥∈,则通项公式n a = .
4.幂函数)(x f y =的图像经过点)21,4(,则1(
)16f 的值为 . 5. 函数)1(log )(2-=x x f (2x >)的反函数=-)(1x f . 6.已知函数5()2x f x x m
-=+的图像关于直线y x =对称,则m = . 7.已知4cos 5α=,则cos()2sin()22tan()cot()2
παπαππαα-+-+++= . 8.函数()arcsin 1arccos y x x =-+的值域是 .
9.已知{}n a 是各项均为正数的等比数列,且1a 与5a 的等比中项为3,则42a a +的最小值等于 .
10.已知)(x f y =是定义在R 上的奇函数,且当0≥x 时,x x x f 2141)(+-
=,则此函数的值域为 .
11.在等差数列{}n a
中,1a =,公差不等于零,且942a a a 、、恰好是某一个等比数列的前三项,那么该等比数列的公比的值等于 .
12.函数x x f πsin 2)(=
与函数()g x 的图像所有交点的橫坐标之和为 .
13.已知函数()2x f x =,对于实数m 、n 、p 有)()()(n f m f n m f +=+,
)()()()(p f n f m f p n m f ++=++,则p 的最大值等于 .
14.已知不等式23344
a x x
b ≤-+≤的解集为[,]a b ,则a b +的值为 .
二、选择题(本大题满分20分,每小题5分)
15.设,,a b R a b ∈>,则下列不等式一定成立的是 ( )
A .22a b >
B . 11a b
< C .2a ab > D . 22a b > 16.已知命题α:如果2x <,那么5x <;命题β:如果2x ≥,那么5x ≥;命题γ:如果5x ≥,那么2x ≥.关于这三个命题之间的关系,下列三种说法正确的是 ( )
① 命题α是命题β的否命题,且命题γ是命题β的逆命题.
② 命题α是命题β的逆命题,且命题γ是命题β的否命题.
③ 命题β是命题α的否命题,且命题γ是命题α的逆否命题.
A .①③;
B .②;
C .②③
D .①②③
17.函数2x y =的定义域为[,]a b ,值域为[1,16],a 变动时,方程()b g a =表示的图形可以是 ( )
A .
B .
C .
D .
18.已知集合()(){},M x y y f x ==,若对于任意()11,x y M ∈,存在()22
,x y M ∈,使得12120x x y y +=成立,则称集合M 是“垂直对点集”.给出下列四个集合: ①
()1,M x y y x ⎧⎫==⎨⎬⎩⎭; ②(){},sin 1M x y y x ==+; ③(
){}2,log M x
y y x ==;
④(){}
,2x M x y
y e ==-.其中是“垂直对点集”的序号是 ( ) A .①② B .②③ C .①④ D .②④
三、解答题(本大题满分74分)
19.(本题满分12分,第一小题满分5分,第二小题满分7分)
已知集合21|
1,1x A x x R x -⎧⎫=≤∈⎨⎬+⎩⎭,集合{}2,B x x a x R =-≤∈. (1)求集合A ; (2)若R B C A B = ,求实数a 的取值范围.
20.(本题满分14分,第一小题满分4分,第二小题满分10分)
已知)sin ,cos (ααA .)sin ,cos (ββB ,其中α、β为锐角,且5
10=AB . (1)求)cos(βα-的值;
(2)若2
12tan =α
,求αcos 及βcos 的值.
21.(本题满分14分,第一小题满分6分,第二小题满分8分)
钓鱼岛及其附属岛屿是中国固有领土,如图:点A 、B 、C 分别表示钓鱼岛、南小岛、黄尾屿,点C 在点A 的北偏东47°方向,点B 在点C 的南偏西36°方向,点B 在点A 的南偏东79°方向,且A 、B 两点的距离约为3海里。
(1)求A 、C 两点间的距离;(精确到0.01)
(2)某一时刻,我国一渔船在A 点处因故障抛锚发出求救信号。
一艘R 国
舰艇正从点C 正东10海里的点P 处以18海里/小时的速度接近渔船,其
航线为P →C →A (直线行进),而我东海某渔政船正位于点A 南偏西60°
方向20海里的点Q 处,收到信号后赶往救助,其航线为先向正北航行8
海里至点M 处,再折向点A 直线航行,航速为22海里/小时。
渔政船能否
先于R 国舰艇赶到进行救助?说明理由.
22.(本题满分16分,第1小题满分4分,第2小题满分6分,第3小题满分6分) 已知函数2)(++=x
m x x f (m 为实常数). (1)若函数)(x f y =图像上动点P 到定点)2,0(Q 的距离的最小值为2,求实数m 的值;
P
(2)若函数)(x f y =在区间),2[∞+上是增函数,试用函数单调性的定义求实数m 的取值范围;
(3)设0<m ,若不等式kx x f ≤)(在⎥⎦⎤⎢
⎣⎡∈1,21x 有解,求k 的取值范围.
23. (本题满分18分,第1小题满分5分,第2小题满分6分 ,第3小题满分7分)
已知数列{}n a 中,13a =,132n n n a a ++=⋅,*
n N ∈. (1)证明数列{}
2n n a -是等比数列,并求数列{}n a 的通项公式; (2)在数列{}n a 中,是否存在连续三项成等差数列?若存在,求出所有符合条件的项;若不存在,请说明理由;
(3)若1r
s <<且r ,*s N ∈,求证:使得1a ,r a ,s a 成等差数列的点列(),r s 在某一直
线上.。