黑龙江省大庆铁人中学2015届高三高考模拟题(四)_理科数学_Word版含答案

黑龙江省大庆铁人中学2015届高三10月月考数学（文）试题（解析版）【试卷综评】突出考查数学主干知识 ,侧重于中学数学学科的基础知识和基本技能的考查； 侧重于知识交汇点的考查。

全面考查了考试说明中要求的内容，如复数、简易逻辑试卷都有 所考查。

一、选择题（每小题5分，共60分）【题文】1．设集合A ＝{(x ，y )|x 24－y 216＝1}，B ＝{(x ，y ) |y ＝x)23(}，则A ∩B 的子集的个数是( )A ．8B ．4C ．2D ．1 【知识点】子集与真子集；交集及其运算.A1【答案解析】A 解析：结合双曲线x 24－y 216＝1的图形及指数函数y ＝x)23(的图象可知，有3个交点，故A ∩B 子集的个数为8．故选A ．【思路点拨】结合双曲线=1的图形及指数函数y=的图象可知，有3个交点．对于有限集合，我们有以下结论：若一个集合中有n 个元素，则它有2n个子集． 【题文】2．在等比数列}{n a 中，4231,4a a a a ⋅==，则=6a （ ）A ．81或—8 B ．81或81- C ．81-或8 D ．41或161【知识点】等比数列的性质.D3【答案解析】B 解析：由已知23423a a a a =⋅=，所以41,11323===a a q a ，所以81336±=⋅=q a a ，故选B ．【思路点拨】利用等比数列的性质，先求出a 3，再求出公比，即可求出a 6．【题文】3．已知中心在原点，焦点在y 轴上的双曲线的离心率为5，则它的渐近线方程为( )A ．y ＝±2xB ．y ＝±52x C ．y ＝±12x D ．y ＝±6x 【知识点】双曲线的简单性质．H6 【答案解析】C 解析：设双曲线的方程为y 2a 2－x 2b 2＝1(a >0，b >0)，∵e ＝c a＝5，c ＝a 2＋b 2，∴a 2＋b 2a 2＝1＋b a2＝5，∴b a ＝2，∴双曲线的渐近线方程为y ＝±12x ，故选C.【思路点拨】可设方程为：y 2a 2－x 2b2＝1(a >0，b >0)，由离心率和abc 的关系可得b 2=2a 2，而渐近线方程为y ＝±12x ．【题文】4．已知圆C 的方程为x 2＋y 2＋2x －2y ＋1＝0，当圆心C 到直线kx ＋y ＋4＝0的距离最大时，k 的值为( )A.13B.15 C ．－13 D ．－15【知识点】点到直线的距离公式；直线与圆的位置关系．H2 H4【答案解析】D 解析：圆C 的方程可化为(x ＋1)2＋(y －1)2＝1，所以圆心C 的坐标为(－1,1)，又直线kx ＋y ＋4＝0恒过点A (0，－4)，所以当圆心C 到直线kx ＋y ＋4＝0的距离最大时，直线CA 应垂直于直线kx ＋y ＋4＝0，直线CA 的斜率为－5，所以－k ＝15，k ＝－15.【思路点拨】圆心为C （﹣1，1）半径r=1，直线恒过定点B （0，﹣4），当直线与BC 垂直时，圆心C 到直线kx+y+4=0的距离最大，由斜率公式易得BC 的斜率，再由垂直关系可得． 【题文】5．函数f (x )＝2cos 2x －3sin2x (x ∈R )的最小正周期和最小值分别为 ( )A ．2π，3B ．2π，－1C ．π，3D ．π，－1【知识点】三角函数中的恒等变换应用；三角函数的周期性及其求法．C3 C7【答案解析】D 解析：由题可知，f (x )＝2cos 2x －3sin2x ＝cos2x －3sin2x ＋1＝2sin(π6－2x )＋1，所以函数f (x )的最小正周期为T ＝π，最小值为－1，故选D.【思路点拨】首先，结合已有的知识，得到f （x ）=2sin(π6－2x )＋1，然后，结合正弦函数的性质，得到相应的结果．【题文】6．已知函数f (x )是定义在R 上的奇函数，且是以2为周期的周期函数．若当x ∈[0,1)时，f (x )＝2x－1，则)6(log 21f 的值为( )A ．－52B ．－5C ．－12D ．－6【知识点】函数的周期性；函数奇偶性的性质；函数的值．B4【答案解析】C 解析：∵)(x f 为奇函数，6log 6log 221-=，且)(x f 周期为2∴21)12()23(log )26(log )6(log )6(log 23log 222212-=--=-=--=-=f f f f故选 C ．【思路点拨】由题意可得：6log 6log 221-=，结合函数的周期性可得：f （log 26）=f （log 2），再根据题中的条件代入函数解析式可得答案．【题文】7．若函数f (x )＝ln x －12ax 2－2x 存在单调递减区间，则实数a 的取值范围是（ ）A ．)1,(-∞B ．]1,(-∞C ．),1(+∞-D ．),1[+∞- 【知识点】利用导数研究函数的单调性．B12【答案解析】C 解析：f ′(x )＝1x －ax －2＝1－ax 2－2xx，由题意可知f ′(x )<0在(0，＋∞)内有实数解．即1－ax 2－2x <0在(0，＋∞)内有实数解． 即a >1x 2－2x 在(0，＋∞)内有实数解．∵x ∈(0，＋∞)时，1x 2－2x ＝(1x－1)2－1≥－1，∴a >－1.故选C ．【思路点拨】f ′(x )＝1x －ax －2＝1－ax 2－2x x ，化为a >1x 2－2x在（0，+∞）内有实数解，求1x2－2x的值域. 【题文】8．已知数列{a n }的前n 项和S n ＝n 2－6n ，数列{|a n |}的前n 项和T n ，则nT n的最小值是( )A ．626-B ．513 C ．25D ．3 【知识点】等差数列的前n 项和；数列的求和．D2 D4 【答案解析】C 解析：由已知 <<<<<-=43210,72a a a a n a n⎪⎩⎪⎨⎧≥+-=-≤+-=-=)4(1862)3(6232n n n S S n n n S T n n n ，⎪⎩⎪⎨⎧≥-+≤+-=)4(618)3(6n n n n n n T n 当4=n 时，有最小值25，故选：C. 【思路点拨】由题意可得a n =2n ﹣7，进而可得⎪⎩⎪⎨⎧≥-+≤+-=)4(618)3(6n n n n n n T n ，由函数的性质可得最值．【题文】9．若满足条件AB ＝3，C ＝π3的三角形ABC 有两个，则边长BC 的取值范围是( )A ．(1，2)B ．(2，3)C ．(3，2)D ．(2，2) 【知识点】解三角形．C8【答案解析】C 解析：若满足条件的三角形有两个，则32＝sin C <sin A <1，又因为BC sin A ＝ABsin C＝2，故BC ＝2sin A ，所以3<BC <2，故选C.【思路点拨】由已知条件C 的度数，AB 及BC 的值，根据正弦定理用a 表示出sinA ，由C 的度数及正弦函数的图象可知满足题意△ABC 有两个A 的范围，然后根据A 的范围，利用特殊角的三角函数值即可求出sinA 的范围，进而求出BC 的取值范围．【题文】10．已知x ，y 满足不等式组⎩⎪⎨⎪⎧x ＋y ≤2，y －x ≥0，x ≥0.目标函数z ＝ax ＋y 只在点(1,1)处取最小值，则有( )A ．a >1B ．a >－1C ．a <1D ．a <－1 【知识点】简单线性规划．E5【答案解析】D 解析： 作出可行域如图阴影部分所示．由z ＝ax ＋y ，得y ＝－ax ＋z .只在点(1,1)处z 取得最小值，则斜率－a >1， 故a <－1，故选D.【思路点拨】画出约束条件表示的可行域，利用目标函数取得的最小值，确定直线的斜率的范围，得到结果．【题文】11．函数f (x )的定义域是R ，f (0)＝2，对任意x ∈R ，f (x )＋f ′(x )>1，则不等式e x·f (x )>e x＋1的解集为( )A ．{x |x >0}B ．{x |x <0}C ．{x |x <－1，或x >1}D ．{x |x <－1，或0<x <1} 【知识点】函数单调性的性质；导数的运算．B11 B12【答案解析】A 解析：构造函数g (x )＝e x·f (x )－e x，因为g ′(x )＝e x·f (x )＋e x·f ′(x )－e x ＝e x [f (x )＋f ′(x )]－e x >e x －e x ＝0，所以g (x )＝e x ·f (x )－e x 为R 上的增函数．又g (0)＝e 0·f (0)－e 0＝1，所以原不等式转化为g (x )>g (0)，解得x >0.故选A【思路点拨】构造函数g （x ）=e x•f（x ）﹣e x，结合已知可分析出函数g （x ）的单调性，结合g （0）=1，可得不等式e x•f（x ）＞e x+1的解集．【题文】12．已知点P 是椭圆x 216＋y 28＝1(x ≠0，y ≠0)上的动点，F 1、F 2分别为椭圆的左、右焦点，O 是坐标原点，若M 是∠F 1PF 2的平分线上一点，且01=⋅F ，则||的取值范围是( )A ．[0,3)B ．(0,22)C ．[22，3)D ．(0,4] 【知识点】椭圆的简单性质；椭圆的定义．H5【答案解析】B 解析：延长F 1M 交PF 2或其延长线于点G ，∵01=⋅F ，∴01=⊥F 又MP 为∠F 1PF 2的平分线，∴|PF 1|＝|PG |且M 为F 1G 的中点，∵O 为F 1F 2的中点， ∴OM //F 2G .，且|OM|=12|F 2G|. ∵|F 2G |＝||PF 2|－|PG ||＝||PF 2|－|PF 1||，∴||OM ＝12|2a －2|PF 2||＝|4－|PF 2||.∵4－22<|PF 2|<4或4<|PF 2|<4＋22，∴||OM ∈(0,22)． 故选B ．【思路点拨】结合椭圆x 216＋y 28＝1的图象，当点P 在椭圆与y 轴交点处时，点M 与原点O 重合，此时|OM|取最小值0．当点P 在椭圆与x 轴交点处时，点M 与焦点F 1重合，此时|OM|取最大值．由此能够得到|OM|的取值范围．第II 卷（非选择题）二、填空题（每小题5分，共20分）【题文】13．若关于x 的不等式2－x 2=|x －a |至少有一个负数解，则实数a 的取值范围是________．【知识点】根的存在性及根的个数判断．B9 【答案解析】9,24⎡⎫-⎪⎢⎣⎭解析：y ＝2－x 2是开口向下的抛物线，y ＝|x －a |是与x 轴交于(a,0)点的“V 字形”折线，显然当a ＝2时，y ＝2－x 2(x <0)的图象都在折线下方，由2－x 2＝x －a 得x 2＋x －a －2＝0，由Δ＝1＋4a ＋8＝0得a ＝－94，此时y ＝x －a 与y ＝2－x 2(x <0)相切，故－94≤a <2.故答案为：[﹣，2）．【思路点拨】关于x 的不等式2－x 2=|x －a |至少有一个负数解化为y ＝2－x 2与y ＝|x －a |至少有一个横坐标为负数的交点，从而解得．【题文】14．已知21,e e 是互相垂直的两个单位向量，若向量21e e t a +⋅=与向量21e t e b ⋅+=的夹角是钝角，则实数t 的取值范围是【知识点】平面向量数量积的运算．F3【答案解析】()(),11,0-∞-- 解析：∵向量a 与向量b 的夹角是钝角，∴0<⋅b a ，且π>≠<,由0)()(2121<⋅+⋅+⋅e t e e e t ，且0,1||||2121=⋅==e e e e ，得0<t令0),(2121<⋅+=+⋅λλe t e e e t ，则⎩⎨⎧⋅==t t λλ1，于是1-=t ,故，0<t ，且1-=t故答案为：（﹣∞，﹣1）∪（﹣1，0）【思路点拨】利用向量21e e t +⋅=与向量21e t e ⋅+=是的夹角是钝角得到它们的数量积小于0，并且注意当向量的夹角为π时数量积也小于0要排除． 【题文】15．已知1,0,0=+>>b a b a ，则)1)(1(bb a a ++的最小值是 【知识点】基本不等式．E6 【答案解析】254解析：由已知ab b a 21≥+=，∴410≤<ab2112)(1)1)(1(2222222-+=+-++=+++=++abab ab ab b a b a ab b a b a b b a a 当且仅当41=ab 时，取最小值425 故答案为：．【思路点拨】利用基本不等式的性质、函数的单调性即可得出． 【题文】16．下列结论：①已知直线l 1：ax ＋3y －1＝0，l 2：x ＋by ＋1＝0，则l 1⊥l 2的充要条件是a b＝－3； ②命题“设a ，b ∈R ，若a ＋b ≠6，则a ≠3或b ≠3”是一个假命题； ③函数f (x )＝lg(x ＋1＋x 2)是奇函数；④在△ABC 中，若sin A cos B ＝sin C ，则△ABC 是直角三角形；⑤“m >n >0”是“方程mx 2＋ny 2＝1表示焦点在y 轴上的椭圆”的充要条件；⑥已知a 、b 为平面上两个不共线的向量，p ：|a ＋2b |＝|a －2b |；q ：a ⊥b ，则p 是q 的必要不充分条件．其中正确结论的序号为________． 【知识点】命题的真假判断与应用．A2【答案解析】③④⑤ 解析：当b ＝a ＝0时，有l 1⊥l 2，故①不正确；②的逆否命题为“设a ，b ∈R ，若a ＝3且b ＝3，则a ＋b ＝6”，此命题为真命题，所以原命题也是真命题，②错误；f (－x )＝lg(－x ＋1＋x 2)＝lg(1x ＋1＋x 2)＝－f (x )，所以③正确；由sin A cos B ＝sin C 得sin A cos B ＝sin(A ＋B )＝sin A cos B ＋cos A sin B ，所以cos A sin B ＝0，所以cos A ＝0，即A ＝π2，所以△ABC 是直角三角形，所以④正确；∵m >n >0，∴0<1m <1n ，方程mx 2＋ny 2＝1化为x 21m ＋y 21n＝1，故表示焦点在y 轴上的椭圆，反之亦成立．∴⑤是真命题；由于|a ＋2b |＝|a －2b |⇔(a＋2b )2＝(a －2b )2⇔a ·b ＝0⇔a ⊥b ，因此p 是q 的充要条件，∴⑥是假命题．故答案为：③④⑤【思路点拨】当b=a=0时，有l 1⊥l 2，即可判断①；考虑②的逆否命题的真假，即可判断②；由奇偶性的定义，即可判断③；运用三角恒等变换公式，化简即可判断三角形ABC 的形状，即可判断④；由条件，化简曲线方程为标准方程，即可判断⑤；运用向量的数量积的性质，两边平方，化简即可得到，再由充分必要条件定义即可判断⑥． 三、解答题（共70分） 【题文】17．（本小题满分10分）若函数f (x )＝－x 3＋6x 2－9x ＋m 在区间[0,4]上的最小值为2，求它在该区间上的最大值． 【知识点】函数的最值及其几何意义．B3【答案解析】6解析：f ′(x )＝－3x 2＋12x －9＝－3(x －1)(x －3)，----------------------------------2分由f ′(x )=0得， x =1或x =3，f (x )的值随x 的变化情况如下表：x0 (0,1) 1 (1,3) 3 (3,4) 4 f ′(x )－ 0＋ 0－f (x )m 递减m －4 递增m 递减m －4-------------6分 由已知f (x )的最小值为f (1)＝f (4)＝m －4＝2，∴m ＝6 ------------8分 ∴f (x )在[0,4]上的最大值为f (0)＝f (3)＝m ＝6 -------------10分【思路点拨】先求导数f′（x ）=﹣3x 2+12x ﹣9=﹣3（x ﹣1）（x ﹣3），由f′（x ）=0得，x=1或x=3；x=1与x=3把区间[0，4]分成（0，1）、（1，3）、（3，4），在每个区间上研究函数的单调性． 【题文】18．（本小题满分12分）已知函数f (x )的定义域是(0，＋∞)，当x >1时，f (x )>0，且f (x ·y )＝f (x )＋f (y )． (1)证明：f (x )在定义域上是增函数；(2)如果f (13)＝－1，求满足不等式f (x )－f (1x －2)≥2的x 的取值范围．【知识点】抽象函数及其应用；函数单调性的判断与证明．B3【答案解析】(1) 见解析；(2) [1＋10，＋∞)．解析：(1)令x ＝y ＝1，得f (1)＝2f (1)，故f (1)＝0. -------------2分 令y ＝1x ，得f (1)＝f (x )＋f (1x )＝0，故f (1x)＝－f (x )． -------------4分任取x 1、x 2∈(0，＋∞)，且x 1<x 2，则f (x 2)－f (x 1)＝f (x 2)＋f (1x 1)＝f (x 2x 1)，由于x 2x 1>1，则f (x 2x 1)>0，从而f (x 2)>f (x 1)，∴f (x )在(0，＋∞)上是增函数．------6分 (2)由于f (13)＝－1，而f (13)＝－f (3)，故f (3)＝1，在f (x ·y )＝f (x )＋f (y )中，令x ＝y ＝3，得f (9)＝f (3)＋f (3)＝2， --------8分 又由(1)知－f (1x －2)＝f (x －2)， 故所给不等式可化为f (x )＋f (x －2)≥f (9)，即f [x (x －2)]≥f (9)， ------------10分∴⎩⎪⎨⎪⎧x >0，x －2>0，x x －2≥9，解得x ≥1＋10，∴x 的取值范围是[1＋10，＋∞)． ------------12分【思路点拨】(1) 设x 1，x 2∈（0，+∞）且x 1＜x 2，则x 2x 1>1，，故f (x 2x 1)>0，由此导出f (x 2)－f (x 1)＝f (x 2)＋f (1x 1)＝f (x 2x 1)＜0，从而能够证明f （x ）在（0，+∞）上是增函数． (2) 令x ＝y ＝3，得f (9)＝f (3)＋f (3)＝2，故所给不等式可化为f (x )＋f (x －2)≥f (9)，即f [x (x －2)]≥f (9)，由此能求出x 的范围．【题文】19．（本小题满分12分）在△ABC 中，a 、b 、c 分别是角A 、B 、C 的对边，向量)2,(c a b m -=，)cos ,(cos C B n =，且n m //(1)求角B 的大小；(2)设f (x )＝cos ⎝ ⎛⎭⎪⎫ωx －B 2＋sin ωx (0<ω)，且f (x )的最小正周期为π，求f (x )的单调区间．【知识点】正弦定理；平面向量共线（平行）的坐标表示．C8 F2【答案解析】(1) B ＝π3 ；(2) 单调递增区间是Z k k k ∈++],65,3[ππππ； 单调递减区间是Z k k k ∈+-],3,6[ππππ解析：(1)由m ∥n 得，b cos C ＝(2a －c )cos B ， ∴b cos C ＋c cos B ＝2a cos B .由正弦定理得，sin B cos C ＋sin C cos B ＝2sin A cos B ， 即sin(B ＋C )＝2sin A cos B .又B ＋C ＝π－A ，∴sin A ＝2sin A cos B . ------------2分 又sin A ≠0，∴cos B ＝12，而B ∈(0，π)，∴B ＝π3. ------------4分(2)由题知f (x )＝cos(ωx －π6)＋sin ωx ＝32cos ωx ＋32sin ωx ＝3sin(ωx ＋π6)， -----6分 由已知得πωπ=||2，∵0<ω，∴2-=ω，f (x )＝－3sin(2x －π6)，------------8分由226222πππππ+≤-≤-k x k ，得Z k k x k ∈+≤≤-,36ππππ由2326222πππππ+≤-≤+k x k ，得Z k k x k ∈+≤≤+,653ππππ 故，函数f (x )的单调递增区间是Z k k k ∈++],65,3[ππππ； 单调递减区间是Z k k k ∈+-],3,6[ππππ ------------12分【思路点拨】（1）通过向量平行，推出关系式，利用正弦定理以及两角和的正弦函数，化简通过三角形内角，即可求出B 的大小．（2）利用（1）B 的值，以及两角和的正弦函数，化简函数的表达式，通过函数的周期求出ω，通过正弦函数的单调区间求解即可． 【题文】20．（本小题满分12分） 已知二次函数y ＝f (x )的图象经过坐标原点，其导函数f ′(x )＝2x ＋2，数列{a n }的前n 项和为S n ，点(n ，S n )(n ∈N *)均在函数y ＝f (x )的图象上． (1)求数列{a n }的通项公式；(2)设b n ＝2n·a n ，T n 是数列{b n }的前n 项和，求T n . 【知识点】数列的求和；导数的运算．B11 D4【答案解析】(1) a n ＝2n ＋1； (2) T n ＝(2n －1)·2n ＋1＋2.解析：(1)设f (x )＝ax 2＋bx ，f ′(x )＝2ax ＋b ＝2x ＋2，∴a ＝1，b ＝2，f (x )＝x 2＋2x ， ------------2分 ∴S n ＝n 2＋2n ，∴当n ≥2时，a n ＝S n －S n －1＝(n 2＋2n )－[(n －1)2＋2(n －1)]＝2n ＋1， 又a 1＝S 1＝3，适合上式，∴a n ＝2n ＋1. ------------6分 (2)b n ＝(2n ＋1)·2n，∴T n ＝3·21＋5·22＋7·23＋…＋(2n ＋1)·2n， ∴2T n ＝3·22＋5·23＋7·24＋…＋(2n ＋1)·2n ＋1， ------------8分相减得－T n ＝3·21＋2·(22＋23＋…＋2n )－(2n ＋1)·2n ＋1＝6＋2·4·1－2n －11－2－(2n ＋1)·2n ＋1＝(1－2n )·2n ＋1－2，∴T n ＝(2n －1)·2n ＋1＋2. ------------12分【思路点拨】(1) 由已知设函数f （x ），结合导函数可求函数解析式，进而可得s n ，然后利用当n≥2时，a n =s n ﹣s n ﹣1，a 1=S 1，可求通项；(2) 可求b n ，然后利用错位相减可求数列的和。

黑龙江省大庆铁人中学2015届高三高考模拟试题（四）生物一、选择题（本题共20小题）1、用两个纯合的圆形块根萝卜作亲本进行杂交，F l全为扁形块根，F l自交后代F2中扁形块根、圆形块根、长形块根的比例为9：6：l，则F2圆形块根中纯合子所占比例为（）A．2/3 B．1/3 C．8/9 D．1/42、某种昆虫长翅(A)对残翅(a)、直翅(B)对弯翅(b)、有刺刚毛(D)对无刺刚毛(d)为显性，控制这三对性状的基因位于常染色体上。

右下图表示某一个体的基因组成，以下判断正确的是（）A．控制长翅和残翅、直翅和弯翅的基因遗传时遵循自由组合定律B．该个体的一个初级精母细胞所产生的精细胞基因型有四种C．该个体细胞有丝分裂后期，移向细胞同一极的基因为AbD或abdD．该个体与隐性个体测交，后代基因型比例为l：1：1：13、一位患有单基因遗传病的男子与正常女子结婚，如生女孩，则l00％患病；如生男孩，则100％正常。

这种遗传病属于（）A．常染色体显性遗传B．常染色体隐性遗传C．伴X染色体显性遗传D．伴X染色体隐性遗传4、某研究小组从蛙的卵巢中提取了一些细胞(无突变发生)，测定细胞中染色体数目，结果如右图所示。

下列分析正确的是（）A．甲组细胞叫做次级卵母细胞B．乙组细胞叫做卵原细胞或初级卵母细胞C．乙组细胞中可能有一部分正在进行DNA复制D．丙组细胞可能正在发生基因重组5、下列有关基因分离定律和自由组合定律的叙述正确的是（）A．可以解释一切生物的遗传现象B．体现在杂合子形成雌、雄配子的过程中C．研究的是所有两对等位基因的遗传行为D．两个定律间不存在着必然的联系6、与有丝分裂相比，减数分裂过程中染色体最显著的变化之一是（）A.染色体移向细胞两极B.同源染色体联会C.有纺锤体形成D.着丝点分开7、人体内某一细胞正在进行减数分裂，其中有44条常染色体和2条X染色体，则此细胞最可能是（）A．初级精母细胞或次级精母细胞B．初级精母细胞或初级卵母细胞C．次级精母细胞或初级卵母细胞D．次级精母细胞或卵细胞8、狗毛颜色褐色由b基因控制,黑色由B基因控制。

大庆铁人中学高三年级第二轮复习套题（四）命题人：陈永辉审题人: 于淑杰第Ⅰ卷第一部分：阅读理解阅读下列短文，从每篇短文后所给各题的四个选项(A、B、C和D)中，选出最佳选项，并在答题卡上将该项涂黑。

AWhile wearing T-shirts, jeans and a baseball cap are classic American stylechoices; the preppy(学院风格) look is what sophisticated(高雅时髦的) Britishyouths wear. But when it comes to the best-dressed nation, neither the US nor theUK tops CNN’s list. So which country has the best fashion sense? Take a look.ItalyItaly is a nation that values outward appearance. Elegance(高雅) is in Italian blood and most Italians would give up comfort in order to look graceful(优雅的). They dress well even for simple trips to the grocery store.Both men and women have a good understanding of style, dressing themselves up with well-groomed(讲究的) hair and a smart pair of shoes. It is no wonder that so many of the best fashion designers are Italian!SwedenRetail(零售) giant H & M continues to spread affordable Swedish style around the world. What ha s been called the “Swedish fashion miracle” actually comes from the “less is more” philosophy(信条).The Swedes tend to adopt a practical approach to what they wear, but they don’t lack creativity.FranceThe French have a chic(别致的) air about them. They appear casual and elegant without being over-dressed or under-dressed, and always seem to give off confidence in their style choices. “The laissez-faire(自由放任的) approach to fashion is something many people love —and the looks they wear are something we all want to emulate on a daily basis,” said The Huffington Post.21. Which of the following statements is TRUE according to the text?A. Wearing jeans is American style.B. Wearing T—shirts is American style.C. British young people like the preppy style.D. The UK is the best-dressed country in theworld.22. What can we learn from Swedish dressing style?A. It is practical and creative.B. It is unaffordable.C. It lacks confidence.D. It lacks creativity.23. What’s t he character of French dressing style?A. Over—dressed.B. Free and easy; elegant and confident.C. Under-dressed.D. Lovely.24. What’s the meaning of the underlined word “emulate” in the last paragrap h?A. Imitate(模仿)B. Thrill.C. Like.D. Hate.BOne day, when I was working as psychologist in England, an adolescent boy showed up in my office. It was David. He kept walking up and down restlessly, his face pale, and his hands shaking slightl y. His head teacher had referred him to me. “This boy has lost his family,” he wrote. “He is understandably very sad and refuses to talk to others, and I’m very worried about him. Can you help?”I looked at David and showed him to a chair. How could I help him? There are problems psychology doesn’t have the answer to, and which no words can describe. Sometimes the best thing one can do is to listen openly and sympathetically.The first two times we met, David didn’t say a word. He sat there, only looking up to look at the children’s drawings on the wall behind me. I suggested we play a game of chess. He nodded. After that he played chess with me every Wednesday afternoon—in complete silence and without looking at me. It’s not easy to cheat in chess, but I ad mit I made sure David won once or twice.Usually, he arrived earlier than agreed, took the chess board and pieces from the shelf and began setting them up before I even got a chance to sit down. It seemed as if he enjoyed my company. But why did he never look at me?“Perhaps he simply needs someone to share his pain with,” I thought. “Perhaps he senses that I respect his suffering.” Some months later, when we were playing chess, he looked up at me suddenly.“It’s your turn,” he said.After that day, David started talking. He got friends in school and Joined a bicycle club. He wrote to me a few times, about his biking with some friends, and about his plan to get into university. Now he had really started to live his own life.Maybe I gave David something. But I also learned that one—without any word—can reach out to another person. All it takes is a hug, a shoulder to cry on, a friendly touch, and an ear that listens.25. When he first met the author, David .A. felt a little excitedB. walked energeticallyC. looked a little nervousD. showed up with his teacher26. As a psychologist, the author .A. was ready to listen to DavidB. was skeptical about psychologyC. was able to descri be David’s problemD. was sure of handling David’s problem27. What can be inferred about David?A. He recovered after months of treatment.B. He liked biking before he lost his family.C. He went into university soon after starting to talk.D. He got friends in school before he met the author.CIf you think that Internet shopping on a Sunday gets you the best deals, then thinkagain. Tuesday is actually the best day of the week if you want to grab a bargain, according to an exhaustive analysis of on—line shopping. Thursday is the second best day followed by Friday—but Sunday is when you are least likely to make a saving.The study also looked at months and found that the pre—Christmas rush in November is when there are most bargains around on the web—and not the January sales. US retailer(零售商) reporting service Sum All examined 3,000 companies and half a billion transactions(交易) as part of its report. It worked out the average discount(折扣) offered to consumers on each day of the week and over different times of the year.The best day of the week was Tuesday where savings were 4.81 percent, followed by Thursday on 4.80 percent. In third place was Friday at 4.65 percent, then Wednesday at 4.13percent and Monday at 4.11 percent. Only the foolish would shop on a Saturday where the discounts were typically 3.84 percent—and Sunday was the worst with 3.37 percent.The analysis by month is even more surprising as most people think that waiting until the post—Christmas sales would yield the biggest discounts. In fact, November is when you will get the largest savings at 5.99 percent on average. January came in second at 4.95 percent, followed by May at 4.50 percent.The worst month to buy online was March where you could expect to save just 2.76 percent. Consumers and experts have been speculating for years about the best time to shop online.Booking a flight has almost become a science with some claiming that buying six weeks before you travel guarantees you the best price. However, a recent analysis by the US Airlines Reporting Commission found that was not actually the case.Not only did the six—week window fail to pile up, in some cases the best fares were on sale as long as 24 weeks before departure. The ARC said that booking well in advance is best advised for busy routes or long haul flights(长途航班), as the price only tends to go up.28. If a buyer wants to buy cheap goods online, he’d better go to the shop on .A. TuesdayB. ThursdayC. FridayD. Sunday29. Most web sellers often provide bargain on the web .A. after New Year’s DayB. during Valentine’s DayC. on Mother’s DayD. before Christmas30. Which of the following statements is True according to the third paragraph?A. The best day of the week was Tuesday where savings were 4.8 percent.B. In third place was Thursday where savings were 4.65 percent.C. Sunday where the discounts were 3.37 percent was the worst.D. Customers can get the best sayings on Saturday.31. What is the meaning of the underlin ed word “speculating” in the fifth paragraph?A. Quarrelling.B. Objecting.C. Searching.D. Considering.DIt is widely known that any English conversation begins with The Weather. Such a fixation with the weather finds expressio n in Dr. Johnson’s famous comment that “when two English meet, their first talk is of weather.” Though Johnson’s observation is as accurate now as it was over two hundred years ago, most commentators fail to come up with a convincing explanation for thisEnglish weather—speak.Bill Bryson, for example, concludes that, as the English weather is not at all exciting, the obsession with it can hardly be understood. He argues that “To an outsider, the most striking thing about the English weather is that there i s not very much of it.” Simply, the reason is that the unusual and unpredictable weather is almost unknown in the British Isles.Jeremy Paxman, however, disagrees with Bryson, arguing that the English weather is by nature attractive. Bryson is wrong, he says, because the English preference for the weather has nothing to do with the natural phenomena. “The interest is less in the phenomena themselves, but in uncertainty.” According to him. the weather in England is very changeable and uncertain and it attracts the English as well as the outsider.Bryson and Paxman stand for common misunderstanding about the weather—speak among the English. Both commentators, somehow, are missing the point. The English weather conversation is not really about the weather at all. English weather—speak is a system of signs, which is developed to help the speakers overcome the natural reserve(内向) and actually talk to each other. Everyone knows conversations starting with weather—speak are not requests for weather data. Rather, they are routine greetings, conversation starters or the blank “fillers”. In other words, English weather—speak is a means of social bonding.32. The author mentions Dr. Johnson’s comment to show that .A. most commentators agree with Dr. JohnsonB. Dr. Johnson is famous for his weather observationC. the comment was accurate two hundred years agoD. English conversations usually start with the weather33. What does the underlined word “obsession” most probably refer to?A. A social trend.B. An strong interest.C. A historical concept.D. An unknown phenomenon.34. According to the passage, Jeremy Paxman believes that .A. Bill Bryson has little knowledge of the weatherB. there is nothing special about the English weatherC. the English weather attracts people to the British IslesD. English people talk about the weather for its uncertainty35. What is the author’s main purpose of writing the passage?A. To explain what English weather—speak is about.B. To analyse misconceptions about the English weather.C. To find fault with both Bill Bryson and Jeremy Paxman.D. To convince people that the English weather is changeable.第二部分：七选五(每小题2分, 满分10分) 根据短文内容, 从短文后的选项中选出能填入空白处的最佳选项, 并在答题卡上写出答案。

大庆铁人中学高三学年考前模拟训练数学试题（理科)考试时间：120分钟 总分：150分 一、选择题（本大题共12小题，每题5分，共计60分，在每题给出的四个选项中，只有一个是正确的） 1．已知集合}|{2x y y M ==，}2|{22=+=y x y N ，则N M =（ ）.A )}1,1(),1,1{(- .B }1{ .C ]1,0[ .D ]2,0[2．复数ii321+-在复平面内对应的点位于（ ）.A 第一象限 .B 第二象限 .C 第三象限 .D 第四象限3．已知p ：a ＞3，q ：∃x ∈R ，使x 2＋ax ＋1＜0是真命题，则p 是q 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4．设,,l m n 表示不同的直线，αβγ，，表示不同的平面，给出下列四个命题： ①若m ∥l ，且.m α⊥则l α⊥； ②若m ∥l ，且m ∥α.则l ∥α；③若,,l m n αββγγα===，则l ∥m ∥n ； ④若,,,m l n αββγγα===且n ∥β,则l ∥m . 其中正确命题的个数是( ).A 1 .B 2 .C 3 .D 45．已知数列}{n a 中，11=a ，n a a n n +=+1，若利用如图所示的程序框图计算该数列的第10项，则判断框内的条件是（ ）.A ?8≤n .B ?9≤n.C ?10≤n .D ?11≤n6．已知向量(1,2),(4,)a x b y =-=，若a b ⊥，则93x y +的最小值为（ ）.A.B 12 .C 6 .D7．已知函数()()sin 2f x x ϕ=+，其中02ϕπ<<，若()6f x f π⎛⎫≤∈ ⎪⎝⎭对x R 恒成立，且()2f f ππ⎛⎫> ⎪⎝⎭，则ϕ等于 （ ）.A 6π .B 56π.C 76π .D 116π 8．已知三棱锥的三视图如图所示，则它的外接球表面积为（ ）.A 16π .B 4π .C 8π .D 2π9．从0到9这10个数字中任取3个数字组成一个没有重复数字的三位数,则这个数能被 3整除的概率为 （ ）.A 5419 .B 5438 .C 5435.D 604110．函数)(x f y =为定义在R 上的减函数，函数)1(-=x f y 的图像关于点（1,0）对称，y x ,满足不等式)2(2x x f -0)2(2≤-+y y f ，)2,1(M ，),(y x N ， O 为坐标原点，则当41≤≤x 时， ON OM ⋅的取值范围为 （ ）.A [)+∞,12 .B []3,0 .C []12,3 .D []12,011．已知双曲线1322=-x y 与抛物线ay x =2有相同的焦点F ，O 为原点，点P 是抛物线准线上一动点，点A 在抛物线上，且4=AF ，则PO PA +的最小值为（ ） .A 132 .B 24 .C 133 .D 6412．已知定义在R 上的可导函数)(x f 满足：0)()('<+x f x f ，则122)(+--m m em m f 与)1(f（e 是自然对数的底数）的大小关系是（ ）.A122)(+--m m em m f >)1(f .B122)(+--m m em m f <)1(f.C122)(+--m m em m f ≥)1(f .D 不确定二、填空题（本大题共4小题，每小题5分，共20分，把答案填在题中横线上）13．在边长为1的等边ABC ∆中，D 为BC 边上一动点，则AB AD ⋅的取值范围是 ．14．（x x +2）6)1(x -的展开式中x 的系数是 15．抛物线342-+-=x x y 及其在点)0,1(A 和点)0,3(B 处的切线所围成图形的面积为16．函数)(x f 的定义域为D ，若存在闭区间],[b a D ⊆，使得函数)(x f 满足：（1）)(x f 在],[b a 内是单调函数；（2）)(x f 在],[b a 上的值域为]2,2[b a ，则称区间],[b a 为函数)(x f y =的“和谐区间”。

资料概述与简介 大庆铁人中学2015年高三语文综合训练试题（四） 第Ⅰ卷阅读题 甲必考题 一、现代文阅读（每小题3分，共9分） 阅读下面的文字，完成1～ 2.下列表述不符合原文意思的一项是（3分）( )A．当务之急不是国学这个概念是不用还是“缓行”的问题，而是在约定俗成的意义上争取对其内涵和外延达成基本的共识。

B．“通”是中国传统学术分类的前提，这种分类到《隋书?经籍志》时形成了经、史、子、集四部这样一套完整体现中国传统学问知识系统的分类法。

C．狭义的国学概念既是一个相当笼统，同时内涵又十分丰富复杂的概念，是近代分科之学形成之前的中国传统学术文化之总和、总称。

D．以成人之道为中心的中国古代的学问其实就是“通人之学”，包括天地人贯通、文史哲贯通、儒释道贯通、真善美贯通、道学政贯通等等。

3.根据原文信息，下列推断正确的一项是（3分）( )A.《四库全书总目》包括经、史、子、集四部，每部又分为若干类，每类都内涵丰富，譬如“子部”的“医家”，收录历代中医古籍一万三千余部，可归入“集”部。

B.国学是一个不同于现代学科体系的传统学科体系，将国学与文学、史学、哲学等并列设立为一级学科的做法是不正确的，因为国学的地位更高，影响力更大。

C.现代学科体系与国学存在大量交叉重合的现象，这是现代学科相互渗透的结果，说明现代学科具有很强的交融性。

D.对于中国的国学，我们唯一正确的态度就是既要对它心存敬意，又要对它进行分析、开放、前瞻、批判和创新，从而创建适应我们时代需要的新国学。

二、古诗文阅读（共19分） （一）阅读下面文言文，完成4～．对下列句子中加点词的解释，不正确的一项是（）（3分） A．将朝廷?讽：劝说B．方觉吾徒之不也? 德：道德高尚 C．士谦竭家资?罄：尽D．葬者万余人 ?会：恰好 ．对一项是（）（3分） A．有兄弟分财不均至相阋讼士谦闻而出财补其少者令与多者相埒兄弟愧惧更相推让卒为善士。

B．有兄弟分财不均至相阋讼士谦闻而出财补其少者令与多者相埒兄弟愧惧更相推让卒为善士。

大庆铁人中学高三年级第二轮复习套题（四）命题人：陈永辉审题人: 于淑杰第Ⅰ卷第一部分：阅读理解阅读下列短文，从每篇短文后所给各题的四个选项(A、B、C和D)中，选出最佳选项，并在答题卡上将该项涂黑。

AWhile wearing T-shirts, jeans and a baseball cap are classic American stylechoices; the preppy(学院风格) look is what sophisticated(高雅时髦的) Britishyouths wear. But when it comes to the best-dressed nation, neither the US nor theUK tops CNN’s list. So which country has the best fashion sense? Take a look.ItalyItaly is a nation that values outward appearance. Elegance(高雅) is in Italian blood and most Italians would give up comfort in order to look graceful(优雅的). They dress well even for simple trips to the grocery store.Both men and women have a good understanding of style, dressing themselves up with well-groomed(讲究的) hair and a smart pair of shoes. It is no wonder that so many of the best fashion designers are Italian!SwedenRetail(零售) giant H & M continues to spread affordable Swedish style around the world. What ha s been called the “Swedish fashion miracle” actually comes from the “less is more” philosophy(信条).The Swedes tend to adopt a practical approach to what they wear, but they don’t lack creativity.FranceThe French have a chic(别致的) air about them. They appear casual and elegant without being over-dressed or under-dressed, and always seem to give off confidence in their style choices. “The laissez-faire(自由放任的) approach to fashion is something many people love —and the looks they wear are something we all want to emulate on a daily basis,” said The Huffington Post.21. Which of the following statements is TRUE according to the text?A. Wearing jeans is American style.B. Wearing T—shirts is American style.C. British young people like the preppy style.D. The UK is the best-dressed country in theworld.22. What can we learn from Swedish dressing style?A. It is practical and creative.B. It is unaffordable.C. It lacks confidence.D. It lacks creativity.23. What’s t he character of French dressing style?A. Over—dressed.B. Free and easy; elegant and confident.C. Under-dressed.D. Lovely.24. What’s the meaning of the underlined word “emulate” in the last paragrap h?A. Imitate(模仿)B. Thrill.C. Like.D. Hate.BOne day, when I was working as psychologist in England, an adolescent boy showed up in my office. It was David. He kept walking up and down restlessly, his face pale, and his hands shaking slightly. His head teacher had referred him to me. “This boy has lost his family,” he wrote. “He is understandably very sad and refuses to talk to others, and I’m very worried about him. Can you help?”I looked at David and showed him to a chair. How could I help him? There are problems psychology doesn’t have the answer to, and which no words can describe. Sometimes the best thing one can do is to listen openly and sympathetically.The first two times we met, David didn’t say a word. He sat there, only looking up to look at the children’s drawings on the wall behind me. I suggested we play a game of chess. He nodded. After that he played chess with me every Wednesday afternoon—in complete silence and without looking at me. It’s not easy to cheat i n chess, but I admit I made sure David won once or twice.Usually, he arrived earlier than agreed, took the chess board and pieces from the shelf and began setting them up before I even got a chance to sit down. It seemed as if he enjoyed my company. But why did he never look at me?“Perhaps he simply needs someone to share his pain with,” I thought. “Perhaps he senses that I respect his suffering.” Some months later, when we were playing chess, he looked up at me suddenly.“It’s your turn,” he said.After that day, David started talking. He got friends in school and Joined a bicycle club. He wrote to me a few times, about his biking with some friends, and about his plan to get into university. Now he had really started to live his own life.Maybe I gave David something. But I also learned that one—without any word—can reach out to another person. All it takes is a hug, a shoulder to cry on, a friendly touch, and an ear that listens.25. When he first met the author, David .A. felt a little excitedB. walked energeticallyC. looked a little nervousD. showed up with his teacher26. As a psychologist, the author .A. was ready to listen to DavidB. was skeptical about psychologyC. w as able to describe David’s problemD. was sure of handling David’s problem27. What can be inferred about David?A. He recovered after months of treatment.B. He liked biking before he lost his family.C. He went into university soon after starting to talk.D. He got friends in school before he met the author.CIf you think that Internet shopping on a Sunday gets you the best deals, then think again. Tuesday is actually the best day of the week if you want to grab a bargain, according to an exhaustive analysis of on—line shopping. Thursday is the second best day followed by Friday—but Sunday is when you are least likely to make a saving.The study also looked at months and found that the pre—Christmas rush in November is when there are most bargains around on the web—and not the January sales. US retailer(零售商) reporting service Sum All examined 3,000 companies and half a billion transactions(交易) as part of its report. It worked out the average discount(折扣) offered to consumers on each day of the week and over different times of the year.The best day of the week was Tuesday where savings were 4.81 percent, followed by Thursday on 4.80 percent. In third place was Friday at 4.65 percent, then Wednesday at 4.13percent and Monday at 4.11 percent. Only the foolish would shop on a Saturday where the discounts were typically 3.84 percent—and Sunday was the worst with 3.37 percent.The analysis by month is even more surprising as most people think that waiting until the post—Christmas sales would yield the biggest discounts. In fact, November is when you will get the largest savings at 5.99 percent on average. January came in second at 4.95 percent, followed by May at 4.50 percent.The worst month to buy online was March where you could expect to save just 2.76 percent. Consumers and experts have been speculating for years about the best time to shop online.Booking a flight has almost become a science with some claiming that buying six weeks before you travel guarantees you the best price. However, a recent analysis by the US Airlines Reporting Commission found that was not actually the case.Not only did the six—week window fail to pile up, in some cases the best fares were on sale as long as 24 weeks before departure. The ARC said that booking well in advance is best advised for busy routes or long haul flights(长途航班), as the price only tends to go up.28. If a buyer wants to buy cheap goods online, he’d better go to the shop on .A. TuesdayB. ThursdayC. FridayD. Sunday29. Most web sellers often provide bargain on the web .A. after New Year’s DayB. during Valentine’s DayC. on Mother’s DayD. before Christmas30. Which of the following statements is True according to the third paragraph?A. The best day of the week was Tuesday where savings were 4.8 percent.B. In third place was Thursday where savings were 4.65 percent.C. Sunday where the discounts were 3.37 percent was the worst.D. Customers can get the best sayings on Saturday.31. What is the meaning of the underlined word “speculating” in the fifth paragraph?A. Quarrelling.B. Objecting.C. Searching.D. Considering.DIt is widely known that any English conversation begins with The Weather. Such a fixation with the weather finds expression in Dr. Johnson’s famous comment that “when two English meet, their first talk is of weather.” Though Johnson’s observation is as accurate now as it was over two hundred years ago, most commentators fail to come up with a convincing explanation for thisEnglish weather—speak.Bill Bryson, for example, concludes that, as the English weather is not at all exciting, the obsession with it can hardly be understood. He argues that “To an outsi der, the most striking thing about the English weather is that there is not very much of it.” Simply, the reason is that the unusual and unpredictable weather is almost unknown in the British Isles.Jeremy Paxman, however, disagrees with Bryson, arguing that the English weather is by nature attractive. Bryson is wrong, he says, because the English preference for the weather has nothing to do with the natural phenomena. “The interest is less in the phenomena themselves, but in uncertainty.” According to him. the weather in England is very changeable and uncertain and it attracts the English as well as the outsider.Bryson and Paxman stand for common misunderstanding about the weather—speak among the English. Both commentators, somehow, are missing the point. The English weather conversation is not really about the weather at all. English weather—speak is a system of signs, which is developed to help the speakers overcome the natural reserve(内向) and actually talk to each other. Everyone knows conversations starting with weather—speak are not requests for weather data. Rather, they are routine greetings, conversation starters or the blank “fillers”. In other words, English weather—speak is a means of social bonding.32. The author mentions Dr. Johnson’s comm ent to show that .A. most commentators agree with Dr. JohnsonB. Dr. Johnson is famous for his weather observationC. the comment was accurate two hundred years agoD. English conversations usually start with the weather33. What does the unde rlined word “obsession” most probably refer to?A. A social trend.B. An strong interest.C. A historical concept.D. An unknown phenomenon.34. According to the passage, Jeremy Paxman believes that .A. Bill Bryson has little knowledge of the weatherB. there is nothing special about the English weatherC. the English weather attracts people to the British IslesD. English people talk about the weather for its uncertainty35. What is the author’s main purpose of w riting the passage?A. To explain what English weather—speak is about.B. To analyse misconceptions about the English weather.C. To find fault with both Bill Bryson and Jeremy Paxman.D. To convince people that the English weather is changeable.第二部分：七选五(每小题2分, 满分10分) 根据短文内容, 从短文后的选项中选出能填入空白处的最佳选项, 并在答题卡上写出答案。

黑龙江省大庆铁人中学2015届高三考前模拟试题（理）一、选择题（本大题共12小题，每题5分，共计60分，在每题给出的四个选项中，只有一个是正确的）1．已知集合}|{2x y y M ==，}2|{22=+=y x y N ，则N M =（ ）.A )}1,1(),1,1{(- .B }1{ .C ]1,0[ .D ]2,0[【知识点】数集与点集的区别；交集.【答案解析】 D 解析 ：由于集合M 、N 都是数集，所以{}{|0,|M y y N y y =≥=≤≤，则N M = {|0y y ≤≤，故选D.【思路点拨】先确定集合M 、N 都是数集，避免出现解方程组的错误，然后再求交集. 2．复数ii321+-在复平面内对应的点位于（ ） .A 第一象限 .B 第二象限 .C 第三象限 .D 第四象限【知识点】复数；复数的实部与虚部复平面.【答案解析】C 解析：解：可化为()()()22123151513131323i i i ii ----==---，所以在第三象限. 故选C.【思路点拨】可依据题意先把复数化简为实部加虚部的形式，对应坐标可知结果. 3．已知p ：a ＞3，q ：∃x ∈R ，使x 2＋ax ＋1＜0是真命题，则p 是q 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【知识点】必要条件、充分条件与充要条件的判断． 【答案解析】 A 解析 ：【思路点拨】根据二次函数的图象和性质，可得命题q ：∃x ∈R ，使x 2+ax+1＜0是真命题，表示对应函数的最小值小于0，即对应方程有两个实根，进而构造不等式求出a 的范围，再根据充要条件的定义可得答案．【典型总结】本题考查的知识点是充要条件，存在性问题，其中根据存在性问题与极值问题的关系，求出命题q 为真时a 的范围，是解答的关键．4．设,,l m n 表示不同的直线，αβγ，，表示不同的平面，给出下列四个命题：①若m ∥l ，且.m α⊥则l α⊥；②若m ∥l ，且m ∥α.则l ∥α； ③若,,l m n αββγγα===，则l ∥m ∥n ；④若,,,m l n αββγγα===且n ∥β,则l ∥m .其中正确命题的个数是( ).A 1 .B 2.C 3 .D 4【知识点】线线平行、线面平行、线面垂直等定理.【答案解析】B 解析：解：①平行线中的一条垂直于一个平面则另一条也垂直于这个平面m ⊥α则l ⊥α正确.②l 可能属于α，所以不正确.③l,m,n 可能交于一点，所以不正确. ④n ∥β∴n ∥l ∴l ∥α∴l ∥m ∴正确.【思路点拨】可由直线与平面平行的判定定理和性质定理推出各种说法的正误. 5．已知数列}{n a 中，11=a ，n a a n n +=+1，若利用如图所示的程序框图计算该数列的第10项，则判断框内的条件是（ ）.A ?8≤n .B ?9≤n .C ?10≤n .D ?11≤n【知识点】当型循环结构，程序框图 【答案解析】B 解析 ：【思路点拨】n=1，满足条件，执行循环体，S=2，依此类推，当n=10，不满足条件，退出循环体，从而得到循环满足的条件．【典型总结】本题主要考查了当型循环结构，算法和程序框图是新课标新增的内容，在近两年的新课标地区高考都考查到了，这启示我们要给予高度重视，属于基础题． 6．已知向量(1,2),(4,)a x b y =-=，若a b ⊥，则93x y +的最小值为（ ）.A .B 12 .C 6 .D【知识点】向量的运算；基本不等式. 【答案解析】C 解析：解：()142022a b x y x y ⊥∴-⋅+=⇒+=又293336x x y y +=+≥=，233x y =时等号成立即2x y =时等号成立.【思路点拨】本题可由向量的基本运算求出x 和y 的关系，利用基本不等式即可.7．已知函数()()sin 2f x x ϕ=+，其中02ϕπ<<，若()6f x f π⎛⎫≤∈⎪⎝⎭对x R 恒成立，且()2f f ππ⎛⎫>⎪⎝⎭，则ϕ等于 （ ） .A 6π.B 56π.C 76π.D 116π【知识点】y=Asin （ωx+φ）的部分图象确定其解析式． 【答案解析】 C 解析 ：若()|()|6f x f π≤对x ∈R 恒成立，则f （6π）等于函数的最大值或最小值,即2×6π+φ=kπ+2π，k ∈Z则φ=kπ+6π，k ∈Z,又f (2π)＞f (π)，即sinφ＜0，0＜φ＜2π当k=1时，此时φ=76π，满足条件故选C ． 【思路点拨】由()|()|6f x f π≤对x ∈R 恒成立，结合函数最值的定义，求得f （6π）等于函数的最大值或最小值，由此可以确定满足条件的初相角φ的值，结合f （2π）＞f （π），易求出满足条件的具体的φ值．8．已知三棱锥的三视图如图所示，则它的外接球表面积为（ ）.A 16π .B 4π .C 8π.D 2π【知识点】三视图 ；球的表面积公式.【答案解析】B 解析：解：由三棱锥的三视图可知中点位置即为球心，因为斜边长为，斜边上的中线等于斜边的一半，三棱锥的高为1，所以三棱锥的外接球的半径为1，所以根据球的表面积公式可得2S=4R =4ππ【思路点拨】由几何体的三视图可求出底面三角形为直角三角形，斜边长为2，高为1，所以可得球的半径为1，代入公式可求.9．从0到9这10个数字中任取3个数字组成一个没有重复数字的三位数,则这个数能被 3整除的概率为 （ ）.A 5419 .B 5438 .C5435.D 6041【知识点】古典概型及其概率计算公式．【答案解析】 A 解析 ： 0到9这10个数字中，任取3个数字组成一个没有重复数字的三位数，所有的三位数的个数为32109A A -=648个．将10个数字分成三组，即被3除余1的有{1，4，7}、被3除余2的有{2，5，8}，被3整除的有{3，6，9，0}．若要求所得的三位数被3整除，则可以分类讨论：①三个数字均取第一组，或均取第二组，有2 33A =12个；②若三个数字均取自第三组，则要考虑取出的数字中有无数字0，共有3243A A -=18个； ③若三组各取一个数字，第三组中不取0，有11133333•••C C C A =162个， ④若三组各取一个数字，第三组中取0，有112332••2C C A =36个，这样能被3整除的数共有228个．故这个三位数能被3整除的概率是2281964854=，故选A ． 【思路点拨】由题意可得所有的三位数有32109A A -=648个，然后根据题意将10个数字分成三组：即被3除余1的有1，4，7；被3除余2的有2，5，8；被3整除的有3，6，9，0，若要求所得的三位数被3整除，则可以分类讨论：每组自己全排列，每组各选一个，再利用排列与组合的知识求出个数，进而求出答案．10．函数)(x f y =为定义在R 上的减函数，函数)1(-=x f y 的图像关于点（1,0）对称，y x ,满足不等式)2(2x x f -0)2(2≤-+y y f ，)2,1(M ，),(y x N ， O 为坐标原点，则当41≤≤x 时， ON OM ⋅的取值范围为 （ ）.A [)+∞,12 .B []3,0 .C []12,3 .D []12,0【知识点】函数的奇偶性；线性规划；向量.【答案解析】D 解析：解：函数y=f （x-1）的图象关于点（1，0）对称，所以f （x ）为 奇函数．22220f x x f y y ∴-≤-+≤（）（），2222x x y y ∴-≥-+，222214x x y y x ⎧-≥-⎨≤≤⎩即()()2014x y x y x -+-≥⎧⎪⎨≤≤⎪⎩由图可得可行域为2OM ON x y ⋅=+可取的范围是[]0,12.故选D ．【思路点拨】本题是考查函数性质的一综合题，多个知识点交汇的典型题型，利用函数的性质把两个变量的关系转化成可行域，利用用线性规划的方法可解.11．已知双曲线1322=-x y 与抛物线ay x =2有相同的焦点F ，O 为原点，点P 是抛物线准线上一动点，点A 在抛物线上，且4=AF ，则PO PA +的最小值为（ ）.A 132 .B 24 .C 133 .D 64【知识点】双曲线与抛物线的几何性质；抛物线的几何意义；【答案解析】 A 解析 ：由已知得双曲线1322=-x y 的焦点坐标为（0，2）， （0，-2），与抛物线ay x =2有相同的焦点F ，所以a=8或-8.不妨令a=8，即28x y =， ∵|AF|=4，由抛物线的定义得，∴A 到准线的距离为4，即A 点的纵坐标为2，又点A 在抛物线上，∴从而点A 的坐标A （4，2）；坐标原点关于准线的对称点的坐标为B （0，-4）则|PA|+|PO|的最小值为：=故选A ．【思路点拨】利用抛物线的定义由|AF|=4得到A 到准线的距离为4，即可求出点A 的坐标，根据：“|PA|+|PO|”相当于在准线上找一点，使得它到两个定点的距离之和最小，最后利用平面几何的方法即可求出距离之和的最小值．【典型总结】此题考查学生灵活运用抛物线的简单性质解决最小值问题，灵活运用点到点的距离、对称性化简求值.12．已知定义在R 上的可导函数)(x f 满足：0)()('<+x f x f ，则122)(+--m m em m f 与)1(f（e 是自然对数的底数）的大小关系是（ ）.A122)(+--m m em m f >)1(f .B122)(+--m m em m f <)1(f.C122)(+--m m em m f ≥)1(f .D 不确定【知识点】利用导数判断单调性；构造新函数，不等式.【答案解析】A 解析 ：解：''()(),()[()()]x x g x e f x g x e f x f x =∴=+令, 因为0)()('<+x f x f ，所以'()g x <0, ()g x 在R 上是减函数，()22213m m 11024m m m ⎛⎫--=--+=---< ⎪⎝⎭，2m m 1∴-<，则2(m m )(1)g g ->，即2m m 2e(m m )(1)f ef -->，转化为122)(+--m m em m f >)1(f . 故选A ．【思路点拨】设出()(),x g x e f x =是本题的关键，然后利用函数()g x 的单调性即可. 二、填空题（本大题共4小题，每小题5分，共20分，把答案填在题中横线上） 13．在边长为1的等边ABC ∆中，D 为BC 边上一动点，则AB AD ⋅的取值范围是 ．【知识点】平面向量数量积的运算．两个向量的加减法的法则，以及其几何意义，两个向量的数量积的定义，【答案解析】1[,1]2解析 ：由题意可得AB 与BD 的夹角等于120°，∴AB AD ⋅=AB ⋅( AB BD +)=2AB AB BD +⋅=1+1×||BD cos120°=1-12•||BD ．由于D 为BC 边上一动点，故 0≤|BD|≤1，∴12≤1-12•||BD ≤1，即AB AD ⋅的取值范围是1[,1]2，故答案为1[,1]2【思路点拨】由题意可得AB 与BD 的夹角等于120°，利用两个向量的数量积的定义计算AB AD ⋅=1-12•||BD ，结合 0≤|BD|≤1 求得AB AD ⋅ 的取值范围．14．（x x+2）6)1(x -的展开式中x 的系数是 【知识点】二项式定理；二项式展开式的系数. 【答案解析】31解析：解：x 的系数为2x与6)1(x -展开式中2x 项的乘积的系数加上x 与6)1(x -展开式中常数项的乘积的系数，6)1(x -的展开式中常数项为1，2x 的项为4426115C x x ⎛⎫-= ⎪⎝⎭所以221530x x x ⨯=，13031x x x ⨯+=，所以x 的系数为31. 【思路点拨】可依据展开式中一次项得到的过程进行分析，是由那些项合并得到，就可以分开求出一次项，最后合并，找出系数即可.15．抛物线342-+-=x x y 及其在点)0,1(A 和点)0,3(B 处的切线所围成图形的面积为【知识点】直线与圆锥曲线的关系． 【答案解析】23解析 ：对y=x 2-4x+3求导可得，y′=2x -4∴抛物线y=x 2-4x+3及其在点A （1，0）和B （3，0）处的两条切线的斜率分别为-2，2 从而可得抛物线y=x 2-4x+3及其在点A （1，0）和B （3，0）处的两条切线方程分别为 l 1：2x+y-2=0，l 2：2x-y-6=0 （2）由2226y x y x =-+⎧⎨=-⎩可得交点P （2，-2）S=221432[2]x x x dx -+--+⎰（）（）+23243[26]x x x dx -+--⎰（）（）=22121[x x dx -+⎰（） +23269[x x dx -+⎰（）=3213x x x -+（）|21+321393x x x -+（）|32=23，故选A 【思路点拨】欲求切线的方程，只须求出其斜率的值即可，故先利用导数求出在切点处的导函数值，再结合A （1，0），B （3，0）都在抛物线上，即可求出切线的方程，然后可得直线与抛物线的交点的坐标和两切线与x 轴交点的坐标，最后根据定积分在求面积中的应用公式即可求得所围成的面积S 即可．16．函数)(x f 的定义域为D ，若存在闭区间],[b a D ⊆，使得函数)(x f 满足：（1）)(x f 在],[b a 内是单调函数；（2）)(x f 在],[b a 上的值域为]2,2[b a ，则称区间],[b a 为函数)(x f y =的“和谐区间”。

黑龙江省大庆市铁人中学2015届高三12月月考（期中）数学（理）试题第I 卷（选择题）一.选择题（每小题5分，共60分）1.设集合A ＝{x |y ＝3x －x 2}，B ＝{y |y ＝2x ，x >1}，则A ∩B 为( )A ．[0,3]B ．(2,3]C ．[3，＋∞)D ．[1,3]2.命题“∃x ∈R,2x ＋x 2≤1”的否定是( )A ．∀x ∈R,2x ＋x 2>1，假命题B ．∀x ∈R,2x ＋x 2>1，真命题C ．∃x ∈R,2x ＋x 2>1，假命题D ．∃x ∈R,2x ＋x 2>1，真命题3. 已知△ABC 中，tanA ＝－512，则cosA ＝( ) A.1213 B.513 C ．－513 D ．－12134. 若奇函数f (x )(x ∈R)满足f (3)＝1，f (x ＋3)＝f (x )＋f (3)，则f ⎝⎛⎭⎫32等于( )A ．0B ．1 C.12 D ．－125. 已知函数f (x )＝sin(2x －π4)，若存在α∈(0，π)使得f (x ＋α)＝f (x ＋3α)恒成立，则α等于( ) A.π6 B.π3 C.π4 D.π26.已知抛物线y 2＝2px (p >0)的准线与曲线x 2＋y 2－6x －7＝0相切，则p 的值为( )A ．2B ．1 C.12 D.147.圆心在直线y ＝x 上，经过原点，且在x 轴上截得弦长为2的圆的方程为( )A ．(x －1)2＋(y －1)2＝2B ．(x －1)2＋(y ＋1)2＝2C ．(x －1)2＋(y －1)2＝2或(x ＋1)2＋(y ＋1)2＝2D ．(x －1)2＋(y ＋1)2＝或(x ＋1)2＋(y －1)2＝28. 设O 为坐标原点，F 为抛物线y 2＝4x 的焦点，A 为抛物线上一点,若＝－4，则点A 的坐标为( )A ．(2，±22)B ．(1，±2)C ．(1,2)D ．(2,22)9.函数y ＝log a x ＋1(a >0且a ≠1)的图象恒过定点A ，若点A 在直线x m ＋y n－4＝0(m >0，n >0)上，则m ＋n 的最小值为( )A ．2＋ 2B ．2C ．1D ．410．若满足条件AB ＝3，C ＝π3的三角形ABC 有两个，则边长BC 的取值范围是( ) A ．(1，2) B ．(2，3) C ．(3，2) D ．(2，2)11．函数f (x )的定义域是R,f (0)＝2,对任意x ∈R ，f (x )＋f ′(x )>1,则不等式e x ·f (x )>e x ＋1的解集为( )A ．{x |x >0}B ．{x |x <0}C ．{x |x <－1，或x >1}D ．{x |x <－1，或0<x <1}12．已知点P 是椭圆x 216＋y 28＝1(x ≠0，y ≠0)上的动点，F 1、F 2分别为椭圆的左、右焦点，O 是坐标原点，若M 是∠F 1PF 2的平分线上一点，且，则的取值范围是( )A ．[0,3)B ．(0,22)C ．[22，3)D ．(0,4]第II 卷（非选择题）二.填空题（每小题5分，共20分）13. 设不等式组⎩⎪⎨⎪⎧ 0≤x ≤2，0≤y ≤3，x ＋2y －2≥0，所表示的平面区域为S ，若A 、B 为区域S 内的两个动点，则|AB |的最大值为________．14.已知|a |＝|b |＝2，(a ＋2b )·(a －b )＝－2，则a 与b 的夹角为________．15.两个等差数列的前n 项和之比为5n ＋102n －1，则它们的第7项之比为________． 16.函数f (x )＝x e x －a 有两个零点，则实数a 的取值范围是________．三、解答题（共70分）17. （10分）已知函数f (x )＝2sin x cos(x ＋π6)－cos2x ＋m . (1)求函数f (x )的最小正周期；(2)当x ∈[－π4，π4]时，函数f (x )的最小值为－3，求实数m 的值18. （12分）已知a 、b 、c 分别为△ABC 三个内角A 、B 、C 的对边，c ＝3a sin C －c cos A .(1)求A ；(2)若a ＝2，△ABC 的面积为3，求b 、c .19．（12分）各项都为正数的数列{a n }，满足a 1＝1，a 2n ＋1－a 2n ＝2.(1)求数列{a n }的通项公式；(2)求数列{a 2n 2n }的前n 项和S n .20. （12分）椭圆C 的中心在原点，一个焦点F (－2,0)，且短轴长与长轴长的比是(1)求椭圆C 的方程；(2)设点M (m,0)在椭圆C 的长轴上，点P 是椭圆上任意一点．当最小时，点P 恰好落在椭圆的右顶点，求实数m 的取值范围．21. （12分）点F 为(1,0)，M 点在x 轴上，P 点在y 轴上，且，(1)当点P 在y 轴上运动时，求N 点的轨迹C 的方程；(2)设A(x 1，y 1)、B(x 2，y 2)、D(x 3，y 3)是曲线C 上的三点，且、|、成等差数列，当AD 的垂直平分线与x 轴交于E(3,0)时，求B 点的坐标．22.（12分）已知.（1）若，函数在其定义域内是增函数，求的取值范围；（2）当，时，证明函数只有一个零点；（3）的图象与轴交于, ()两点，中点为，求证：．参考答案三.解答题（共70分）17[解析] (1)∵f (x )＝2sin x cos(x ＋π6)－cos2x ＋m ＝2sin x (32cos x －12sin x )－cos2x ＋m ＝3sin x cos x －sin 2x －cos2x ＋m ＝32sin2x －1－cos2x 2－cos2x ＋m ＝32sin2x －12cos2x －12＋m ＝sin(2x －π6)－12＋m .∴f (x )的最小值为－1－12＋m . 由已知，有－1－12＋m ＝－3，∴ m ＝－32. 18. [解析] (1)由c ＝3a sin C －c cos A 及正弦定理得，3sin A sin C －cos A sin C －sin C ＝0.由于sin C ≠0，所以sin(A －π6)＝12. 又0<A <π，故A ＝π3. (2)△ABC 的面积S ＝12bc sin A ＝3，故bc ＝4. 而a 2＝b 2＋c 2－2bc cos A ，故b 2＋c 2＝8.解得b ＝c ＝2.19．[解析] (1)因为a 2n ＋1－a 2n ＝2，a 21＝1，所以数列{a 2n }是首项为1，公差为2的等差数列．所以a 2n ＝1＋(n －1)×2＝2n －1，因为a n >0，所以a n ＝2n －1(n ∈N *)． (2)由(1)知，a n ＝2n －1，所以a 2n 2n ＝2n －12n ， 于是S n ＝12＋322＋523＋…＋2n －32n －1＋2n －12n ，① 12S n ＝122＋323＋524＋…＋2n －32n ＋2n －12n ＋1，② ①－②得，12S n ＝12＋222＋223＋224＋…＋22n －2n －12n ＋1＝12＋2(122＋123＋124＋…＋12n )－2n －12n ＋1＝12＋2×14－12n －11－12－2n －12n ＋1＝32－2n ＋32n ＋1， 所以S n ＝3－2n ＋32n . 20.[解析] (1)设椭圆C 的方程为x 2a 2＋y 2b2＝1(a >b >0)， 由题意⎩⎪⎨⎪⎧ a 2＝b 2＋c 2，a b ＝23，c ＝2.解得a 2＝16，b 2＝12.所以椭圆C 的方程为x 216＋y 212＝1. (2)设P (x ，y )为椭圆上的动点，由于椭圆方程为x 216＋y 212＝1，故－4≤x ≤4. 因为MP →＝(x －m ，y )，所以|MP →|2＝(x －m )2＋y 2＝(x －m )2＋12×⎝⎛⎭⎫1－x 216.＝14x 2－2mx ＋m 2＋12＝14(x －4m )2＋12－3m 2. 因为当|MP →|最小时，点P 恰好落在椭圆的右顶点，也即当x ＝4时，|MP →|2取得最小值．而x ∈[－4,4]，故有4m ≥4，解得m ≥1.又点M 在椭圆的长轴上，即－4≤m ≤4.故实数m 的取值范围是m ∈[1,4]．21.解(1)∵MN →＝2MP →，故P 为MN 中点．又∵PM →⊥PF →，P 在y 轴上，F 为(1,0)，故M 在x 轴的负半轴上，设N(x ，y)，则M(－x,0)，P ⎝⎛⎭⎫0，y 2，(x>0)， ∴PM →＝⎝⎛⎭⎫－x ，－y 2，PF →＝⎝⎛⎭⎫1，－y 2，又∵PM →⊥PF →，∴PM →·PF →＝－x ＋y24＝0， ∴y2＝4x(x>0)是轨迹C 的方程．(2)抛物线C 的准线方程是x ＝－1，由抛物线定义知|AF →|＝x1＋1，|BF →|＝x2＋1，|DF →|＝x3＋1，∵|AF →|、|BF →|、|DF →|成等差数列，∴x1＋1＋x3＋1＝2(x2＋1)，∴x1＋x3＝2x2又y21＝4x1，y22＝4x2，y23＝4x3，故y21－y23＝(y1＋y3)(y1－y3)＝4(x1－x3)，∴kAD ＝y1－y3x1－x3＝4y1＋y3，∴AD 的中垂线为y ＝－y1＋y34(x －3) AD 的中点⎝⎛⎭⎫x1＋x32，y1＋y32在其中垂线上， ∴y1＋y32＝－y1＋y34⎝⎛⎭⎫x1＋x32－3.∴x2＝x1＋x32＝1. 由y22＝4x2.∴y2＝±2.∴B 点坐标为(1,2)或(1，－2)．22. 解：（1）依题意：f (x )＝ln x ＋x 2－bx ．∵f (x )在(0，＋∞)上递增，∴1()20f x x b x'=+-≥对x ∈(0，＋∞)恒成立，……1分 即对x ∈(0，＋∞)恒成立，只需． …………2分∵x ＞0，∴，当且仅当时取“＝”，∴，∴b 的取值范围为． ………………4分（2）当a ＝-1，b ＝－1时，f (x )＝ln x +x 2＋x ，其定义域是(0，＋∞)，分）上单调递增，又，在（分602)1(,0111)1(0)(5012121)(22----->=<++-=∞+-------------->++=++='f ee ef x f x x x x x x f ． ∴函数f (x )只有一个零点．……7分（3）由已知得221111111222222111()ln 0ln ()ln 0ln f x x ax bx x ax bx f x x ax bx x ax bx ⎧⎧=--==+⎪⎪⇒⎨⎨=--==+⎪⎪⎩⎩， 两式相减，得11212122ln ()()()x a x x x x b x x x =+-+-112122ln ()[()]x x x a x x b x ⇒=-++．…………9分 由及2x 0＝x 1＋x 2，得10012012121221221()2[()]ln xf x ax b a x x b x x x x x x x x '=+-=-++=-++-11212111212212222(1)2()11[ln ][ln ](1)x x x x x x x x x x x x x x x x --=-=--+-+。

大庆铁人中学高三地理综合测试试题4一、单项选择题（每题4分，共11小题，共44分）读某地区人口年龄结构示意图，图中a、b、c示意该地区1935年、1975年和2015年的人口年龄结构状况。

并回答1—2题。

1．图中表示1935年、1975年和2015年人口年龄结构的曲线分别为A．a、b、c B．b、a、c C．a、c、b D．c、b、a2．该地区人口变化趋势的突出特征是A．劳动力过剩 B．自然增长率上升C．人口出生率较高 D．人口老龄化现象显著下面三幅图中，图甲为南疆公路旁的草方格沙障，图乙为行走在沙丘上的驼队，图丙为兰州郊区种植白兰瓜的砂田(含砂砾层)。

据图完成3～4题。

3.据图文信息可以推断( )A.图甲地区中河流径流量年际变化较小B.图乙中风力侵蚀作用明显C.图丙中降水丰富D.三地降水均小于200毫米4.下列关于图示景观，叙述错误的是( )A.图甲中的沙障能增加地表粗糙度，削减风力，从而保护公路不被风沙侵袭B.图乙中的驼队沿等高线行走，虽线路较长，但省力且视野开阔，安全性好C.图丙中的砂砾层能减少土壤水分蒸发，增强水的渗透能力和增大昼夜温差D.三地农业生产过程中不合理的人类活动容易导致土地荒漠化和次生盐碱化左图为“某区域示意图”，右图为A地河谷及地质剖面示意图，读图回答5-7题。

5.关于A地所在河流位置及流向说法正确的是A.北半球、由南向北流B.北半球、由北向南流C.南半球、由北向南流D.南半球、由南向北流6.A地河谷的地质构造为A.断层B.背斜C.地垒D.地堑7.图中A地河流水面最高时，下列说法正确的是A. B. 华北地区正值小麦收割季节C. D. 我国北方寒冷干燥图4为中国西南某区域不同地理要素图层。

读图，回答第8--9题。

8. 该区域A.高级住宅区集中在②地B.③地适宜种植小麦C.①地以常绿阔叶林为主D.钢铁厂布局在④地9. 据图推断A.居民点向西、南方向扩展B.森林面积扩大，酸雨危害减弱C.GPS可应用于道路规划D.湿地面积减少，地下水源增加2014年APEC会议于11月5日至11日在北京召开。