北京2013届朝阳高三二模数学理科试题及答案

北京2013届朝阳高三二模数学理科及答案 北京市朝阳区高三第二次综合练习 数学学科测试(理工类) 2013.5 一、选择题:本大题共8小题，每小题5分，共40分(在每小题给出的四个选项中，选出符合题目要求的一项. MN:1、已知集合，集合，则= M,0,1,3NxxaaM,,,3,,,,, A. B. C. D. 00,31,3,90,1,3,9,,,,,,,, 12m2、若，则实数的值为 ()d0xmxx，,,0 12 A(, B(, C(,1 D(,2 33 163、执行如图所示的程序框图(若输出的结果是，则判断框内的条件是 n,6n,7n,8n,9A. ? B. ? C. ? D. ? 开始 S = 0 1 n = 1 S=S+n 1 1 侧视图 正视图 n=n+2 否 是 俯视图 输出S (第3题图) (第5题图) 结束 (第3题图) 22xy2,,,,1(0,0)ab4、若双曲线的渐近线与抛物线有公共点，则此双曲线的离心率的取值范yx,，222ab 围是 A([3,)，, B((3,)，, C((1,3] D((1,3) 5、某三棱锥的三视图如图所示，则该三棱锥的体积为 1111A( B( C( D( 632 6、某岗位安排3名职工从周一到周五值班，每天只安排一名职工值班，每人至少安排一天，至多 pipe into the box, lateral lock, installed at the inner side protection. (9) dark distribution pipe should be laid along the lines of recent, and should reduce bending, buried within the walls of the pipe from the surface-distance should not be less than 15mm. (10) under normal circumstances, it tube bending radius shall meet the following requirements: 1) specify timing, there is generally no less than 6 times the pipe diameter, such as when only one bend, not less than 6 times the pipe diameter. 2) dark time, not less than ... (E) wire into the switch box and Cabinet, Board, should have a long widening of margins; (F) when the wires into the electrical equipment should remain at the Centre of 0.3M margin; (G) the floor tube to the power box, tube 1M of allowances; (H) the wires connected to the power supply and into and out of time, tube 1.5M of allowances. (I) wore metal pipe with wire retainer should be well set, retainer must not cut place, match the specification and pipe. (J) after the end of the wire, you should check the insulation resistance, the resistance shall be not less than 0.5M. (K) the dark construction of the works at the time of completion and acceptance of the delivery, and laid wiring changes in the construction should be part of the actual position (including junction boxes and junction boxes and pipeline specifications), fixed in the as-built drawings and indicate 安排两天，且这两天必须相邻，那么不同的安排方法有 101836A(种 B(种 C(种 D(种 12 fxx(),0,,,x7、已知函数，定义函数 给出下列命题: Fx(),fxaa()21(0),,，,,,,fxx(),0., a,0mn,0mn，,0?; ?函数是奇函数;?当时，若，，总有Fxfx()(),Fx()FmFn()()0，, 成立，其中所有正确命题的序号是 (? B(?? C(? D(?? A ,,,,,,,,, 8、点是棱长为1的正方体的底面上一点，则的取值范围是 PABCDABCD,ABCDPAPC 111111111 1111A([1,],, B( C( D( [,],,[,0],[1,0],4242 第二部分(非选择题 共110分) 二、填空题:本大题共6小题，每小题5分，共30分.把答案填在答题卡上. 3i，,9、i为虚数单位，计算 ( 1i， x,2cos,,,l,10、若直线AB与圆(为参数)相交于，两点， C:,y,,，12sin,, lAB且弦的中点坐标是，则直线的倾斜角为 ( (1,2), PCOCOPAB11、如图，切圆于点，割线经过圆心，， PCPB,,4,8 tan，,COPOBC则 ，?的面积是 ( 60032x12、某公司一年购买某种货物吨，每次都购买吨，运费为万元/次，一年的总存储费用为 x 万元，若要使一年的总运费与总存储费用之和最小，则每次需购买 吨( 3419,xy，,, ,xy,x,1,13、将一个质点随机投放在关于的不等式组所构成的三角形区域内，则该质点到此三角形, ,y,1, 1的三个顶点的距离均不小于的概率是 ( n,nn14、数列A的前项组成集合，从集合中任取nAn,,,{1,3,7,,21}()？N{21},1,3,7,,21？,nn kkT个数，其所有可能的个数的乘积的和为(若只取一个数，规定乘积为此数本身)，(1,2,3,,)kn,？k n,1n,2STTT,，，，？A,{1}T,1S,1A,{1,3}T,，13记(例如当时，，，;当时，，，nn1211112 n,3T,，13S,，，，,13137S,S,，.则当时， ;试写出 ( 223n 三、解答题:本大题共6小题，共80分.解答应写出文字说明，演算步骤或证明过程. 15、(本小题满分13分) AAAA22ABCfA()2cossin()sin,,,，,cosABC,,abc,,在?中， 所对的边分别为，且. 2222 fA()(?)求函数的最大值; ,,fACa()0,,6,,,(?)若，求b的值( 12 pipe into the box, lateral lock, installed at the inner side protection. (9) dark distribution pipe should be laid along the lines of recent, and should reduce bending, buried within the walls of the pipe from the surface-distance should not be less than 15mm. (10) under normal circumstances, it tube bending radius shall meet the following requirements: 1) specify timing, there is generally no less than 6 times the pipe diameter, such as when only one bend, not less than 6 times the pipe diameter. 2) dark time, not less than ... (E) wire into the switch box and Cabinet, Board, should have a long widening of margins; (F) when the wires into the electrical equipment should remain at the Centre of 0.3M margin; (G) the floor tube to the power box, tube 1M of allowances; (H) the wires connected to the power supply and into and out of time, tube 1.5M of allowances. (I) wore metal pipe with wire retainer should be well set, retainer must not cut place, match the specification and pipe. (J) after the end of the wire, you should check the insulation resistance, the resistance shall be not less than 0.5M. (K) the dark construction of the works at the time of completion and acceptance of the delivery, and laid wiring changes in the construction should be part of the actual position (including junction boxes and junction boxes and pipeline specifications), fixed in the as-built drawings and indicate 16、(本小题满分14分) ABCDABCDG如图，四边形是正方形，平面，，，，， EA,PDADPDEA,,,22FHEA PC分别为，，的中点( PBEBP FG(?)求证:平面; PED FGHPBC(?)求平面与平面所成锐二面角的大小; H PC(?)在线段上是否存在一点,使直线与直线 MFM F ,所成的角为,若存在，求出线段的长;若 PAPM60E D C 不存在，请说明理由. 17、(本小题满分13分) G 为提高学生学习数学的兴趣，某地区举办了生“数 B 独比赛”.比赛成绩共有90分，70分，60分，40分，30分A 五种，按本次比赛成绩共分五个等级(从参加比赛的学生中 随机抽取了30名学生，并把他们的比赛成绩按这五个等级进行了统计，得到如下数据表: 成绩等级 A B C D E 成绩(分) 90 70 60 40 30 人数(名) 4 6 10 7 3 A(?)根据上面的统计数据，试估计从本地区参加“数独比赛”的小学生中任意抽取一人，其成绩等级为“ B或”的概率; X(?)根据(?)的结论，若从该地区参加“数独比赛”的小学生(参赛人数很多)中任选3人，记表示抽 ABEXX到成绩等级为“或”的学生人数，求的分布列及其数学期望; 20(?)从这30名学生中，随机选取2人，求“这两个人的成绩之差大于分”的概率( 18、(本小题满分13分) mx2axm,0fx(),，1已知函数()，. gxxa()e(),,R2x，1 (?)求函数的单调区间; fx() m,0xx,[0,2],fxgx()(),(?)当时，若对任意，恒成立，求的取值范围. a1212 19、(本小题满分14分) 22,,,,,,,,,xyFC:1，,BB,已知椭圆的右焦点为，短轴的端点分别为,且. (0)ab,,(1,0)FBFBa,,,121222ab C(?)求椭圆的方程; klMNFDx(?)过点且斜率为(0)k,的直线交椭圆于MN,两点，弦的垂直平分线与轴相交于点.设弦 DPMNP的中点为，试求的取值范围( MN pipe into the box, lateral lock, installed at the inner side protection. (9) dark distribution pipe should be laid along the li2013朝阳高三二模数学理科 第 3 nes of recent, and should reduce bending, buried within the walls of the pipe from the surface-distance should not be less than页 共 10 页 15mm. (10) under normal circumstances, it tube bending radius shall meet the following requirements: 1) specify timing, there is generally no less than 6 times the pipe diameter, such as when only one bend, not less than 6 times the pipe diameter. 2) dark time, not less than ... (E) wire into the switch box and Cabinet, Board, should have a long widening of margins; (F) when the wires into the electrical equipment should remain at the Centre of 0.3M margin; (G) the floor tube to the power box, tube 1M of allowances; (H) the wires connected to the power supply and into and out of time, tube 1.5M of allowances. (I) wore metal pipe with wire retainer should be well set, retainer must not cut place, match the specification and pipe. (J) after the end of the wire, you should check the insulation resistance, the resistance shall be not less than 0.5M. (K) the dark construction of the works at the time of completion and acceptance of the delivery, and laid wiring changes in the construction should be part of the actual position (including junction boxes and junction boxes and pipeline specifications), fixed in the as-built drawings and indicate 20、(本小题满分13分) n,2已知实数()满足，记( xxx,,,？||1(1,2,3,,)xin,,？Sxxxxx(,,,)？,,12ni12nij,,,1ijn 2(?)求及的值; S(1,1,),,S(1,1,1,1),,3 n,3(?)当时，求的最小值; Sxxx(,,)123 (?)求的最小值( Sxxx(,,,)？12n 注:表示中任意两个数,()的乘积之和. 1,,,ijnxxxx,,,？xxx,ijj12ni1,,,ijn 北京市朝阳区高三年级第二次综合练习 数学学科测试答案(理工类) 2013., 一、选择题: 1 2 3 4 5 6 7 8 题号 答案 D B C A A C D D 二、填空题: 题9 10 11 12 13 14 号 418,,nn(1)，答1, ， 30 2i, 221,63， 案 43512(注:两空的填空，第一空3分，第二空2分) 三、解答题: 15、(本小题满分13分) AAAA,22fA()2cossinsincos,，,,,,,sincos2sin()AAA解:(?)因为. 22224 ,,,,0,,,A,,,,AA因为为三角形的内角，所以，所以. 444 ,,3,A,,2A,所以当，即时，取得最大值，且最大值为. „„„6分 fA()424 ,,fAA()2sin()0,,,sin()0A,,(?)由题意知，所以( 44 ,,,,,,,,,,AA,A,,0又因为，所以，所以( 44444 ,,,C,B,又因为，所以( 123 ,6sin,abaBsin3,b,,,3由正弦定理得，( „„„„13分 ,sinsinABsinAsin4 16、(本小题满分14分) pipe into the box, lateral lock, installed at the inner side protection. (9) dark distribution pipe should be laid along the lines of recent, and should reduce bending, buried within the walls of the pipe from the surface-distance should not be less than 15mm. (10) under normal circumstances, it tube bending radius shall meet the following requirements: 1) specify timing, there is generally no less than 6 times the pipe diameter, such as when only one bend, not less than 6 times the pipe diameter. 2) dark time, not less than ... (E) wire into the switch box and Cabinet, Board, should have a long widening of margins; (F) when the wires into the electrical equipment should remain at the Centre of 0.3M margin; (G) the floor tube to the power box, tube 1M of allowances; (H) the wires connected to the power supply and into and out of time, tube 1.5M of allowances. (I) wore metal pipe with wire retainer should be well set, retainer must not cut place, match the specification and pipe. (J) after the end of the wire, you should check the insulation resistance, the resistance shall be not less than 0.5M. (K) the dark construction of the works at the time of completion and acceptance of the delivery, and laid wiring changes in the construction should be part of the actual position (including junction boxes and junction boxes and pipeline specifications), fixed in the as-built drawings and indicate (?)证明:因为GFG,分别为，的中点，所以. FPBBEPE FG又平面，平面， PEDPEPED,, FG所以平面. „„„„4分 PED ABCD(?)因为平面，， EA,PDEA ABCD所以平面， PD, z PDCD,所以，. PDAD, P ABCD又因为四边形是正方形， ADCD,所以. H F 如图，建立空间直角坐标系， D E 因为ADPDEA,,,22, C y 所以D，P，A， 0,0,00,0,22,0,0，，，，，，G A CB0,2,0，2,2,0，( B E(2,0,1)x ，，，， „„„„5分 GPCFPBEBH因为，， 分别为，，的中点， ,,,,,,,,111GFH所以1,1,1，，. 所以，. GF,,(1,0,)GH,,(2,0,)(2,1,)(0,1,1)，，222 1,,,,,,，,xz011,,n,,GF0,,21FGH设为平面的一个法向量，则,即， n,(,,)xyz,,,,,,11111n,,GH0,,1,,，,20xz11,,2 ,,,,,,,, 再令y,1，得n,(0,1,0).,. PB,,(2,2,2)PC,,(0,2,2)11 ,,,,,n,,PB0,2PBCn,(,,)xyz设为平面的一个法向量,则, ,,,,,2222n,,PC0,2, 2220xyz，,,nn,,222212cos,nnz,1n,(0,1,1)即，令，得.所以==. ,1222220yz,,2nn,22,12 ,FGHPBC所以平面与平面所成锐二面角的大小为. „„„„9分 4 ,PCMFMPA60(?)假设在线段上存在一点,使直线与直线所成角为. ,,,,,,,,,,,,,,,,,, 01,,,PMPC,,依题意可设,其中.由,则. PC,,(0,2,2)PM,,(0,2,2),,,,,,,,,,,,,,,,,,,,,,,,, FMFPPM,，又因为,，所以. FP,,,(1,1,1)FM,,,,(1,21,12),, ,,,,,FMPA60因为直线与直线所成角为，， PA,,(2,0,2) pipe into the box, lateral lock, installed at the inner side protection. (9) dark distribution pipe should be laid along the li2013朝阳高三二模数学理科 第 5 nes of recent, and should reduce bending, buried within the walls of the pipe from the surface-distance should not be less than页 共 10 页 15mm. (10) under normal circumstances, it tube bending radius shall meet the following requirements: 1) specify timing, there is generally no less than 6 times the pipe diameter, such as when only one bend, not less than 6 times the pipe diameter. 2) dark time, not less than ... (E) wire into the switch box and Cabinet, Board, should have a long widening of margins; (F) when the wires into the electrical equipment should remain at the Centre of 0.3M margin; (G) the floor tube to the power box, tube 1M of allowances; (H) the wires connected to the power supply and into and out of time, tube 1.5M of allowances. (I) wore metal pipe with wire retainer should be well set, retainer must not cut place, match the specification and pipe. (J) after the end of the wire, you should check the insulation resistance, the resistance shall be not less than 0.5M. (K) the dark construction of the works at the time of completion and acceptance of the delivery, and laid wiring changes in the construction should be part of the actual position (including junction boxes and junction boxes and pipeline specifications), fixed in the as-built drawings and indicate ,,,,,,,,,,,，224,115所以cos,FMPA=，即，解得. ,,,22282212(21),，,, ,,,,,,,,,,5552所以，. PM,,(0,,)PM,444 52,PC所以在线段上存在一点,使直线与直线所成角为，此时.„„„14分 MFMPA60PM,4 17、(本小题满分13分) 46101解:(?)根据统计数据可知，从这30名学生中任选一人，分数等级为“A或B”的频率为( ，,,3030303 1 从本地区小学生中任意抽取一人，其“数独比赛”分数等级为“ 或”的概率约为(„„„„3分 AB3 (?)由已知得，随机变量X的可能取值为0，1，2，3( 12812124003112 所以;; PXC(0)()(),,,,PXC(1)()(),,,,,33332733279 1262121221330;( PXC(2)()(),,,,,PXC(3)()(),,,,33332793327 X随机变量的分布列为 0 1 2 3 X 8421 P 279927 81261 所以( „„„„„9分 EX,，，，，，，，,0123127272727 20(?)设事件M:从这30名学生中，随机选取2人，这两个人的成绩之差大于分( 设从这30名学生中，随机选取2人，记其比赛成绩分别为mn,( 2显然基本事件的总数为( C30 不妨设， mn, 1111m,90n,604030当时，或或，其基本事件数为; CCCC,，，()41073 111m,704030当时，n,或，其基本事件数为; CCC,，()673 11m,60n,30当时，，其基本事件数为; CC,103 111111111CCCCCCCCC,，，，,，，,()()3441073673103所以( PM(),,2C8730 3420所以从这30名学生中，随机选取2人，这两个人的成绩之差大于分的概率为(„„„„„13分 87 18、(本小题满分1 3分) 2mxmxx()()()111,,，,R解:(?)函数fx()的定义域为，.„„„„1分 fx(),,2222()()xx，，11 ,m,0xfx()fx()?当时，当变化时，，的变化情况如下表: pipe into the box, lateral lock, installed at the inner side protection. (9) dark distribution pipe should be laid along the lines of recent, and should reduce bending, buried within the walls of the pipe from the surface-distance should not be less than 15mm. (10) under normal circumstances, it tube bending radius shall meet the following requirements: 1) specify timing, there is generally no less than 6 times the pipe diameter, such as when only one bend, not less than 6 times the pipe diameter. 2) dark time, not less than ... (E) wire into the switch box and Cabinet, Board, should have a long widening of margins; (F) when the wires into the electrical equipment should remain at the Centre of 0.3M margin; (G) the floor tube to the power box, tube 1M of allowances; (H) the wires connected to the power supply and into and out of time, tube 1.5M of allowances. (I) wore metal pipe with wire retainer should be well set, retainer must not cut place, match the specification and pipe. (J) after the end of the wire, you should check the insulation resistance, the resistance shall be not less than 0.5M. (K) the dark construction of the works at the time of completion and acceptance of the delivery, and laid wiring changes in the construction should be part of the actual position (including junction boxes and junction boxes and pipeline specifications), fixed in the as-built drawings and indicate x(,),,,1(,),11(,)1，, , ,,fx() ， fx() 所以，函数的单调递增区间是，单调递减区间是，.„„„„3分 fx()(,),11(,),,,1(,)1，, ,m,0?当时，当变化时，，的变化情况如下表: xfx()fx() x(,),,,1(,),11(,)1，, ,, fx() ，， fx() 所以，函数的单调递增区间是，，单调递减区间是. fx()(,),,,1(,)1，,(,),11 „„„„„5分 m,0m,0(?)依题意，“当时，对于任意，恒成立”等价于 “当 时，对于xx,[0,2],fxgx()(),1212 任意， 成立”. fxgx()(),x,[0,2]minmax m,0当时，由(?)知，函数在上单调递增，在上单调递减， fx()[0,1][1,2] 2m因为f(2)11,，,，，所以函数的最小值为. f(0)1,fx()f(0)1,5 所以应满足. „„„„„„„„„„„„„„„„„„„„„„„6分 gx()1,max 2ax2ax,因为，所以. „„„„„7分 gxx()e,gxaxx()(+2)e, 2a,0?当gxg()(2)4,,时，函数，，， ,,x[0,2]gxx(),max a,0gx()1,显然不满足，故不成立. „„„„„8分 max 2,a,0x,,x,0?当时，令得，，. gx()0,21a 2,,,,10a,,2(?)当，即时，在上，所以函数在上单调递增， [0,2]gx()0,gx()[0,2]a 2a 所以函数gxg()(2)4e,,. max 2aa,,ln2,,,,1ln2a4e1, 由得，，所以. „„„„„10分 2a,,102,,,(?)当，即时, a 22,,(,2],[0,),gx()0,gx()0,在上，在上， aa pipe into the box, lateral lock, installed at the inner side protection. (9) dark distribution pipe should be laid along the li2013朝阳高三二模数学理科 第 7 nes of recent, and should reduce bending, buried within the walls of the pipe from the surface-distance should not be less than页 共 10 页 15mm. (10) under normal circumstances, it tube bending radius shall meet the following requirements: 1) specify timing, there is generally no less than 6 times the pipe diameter, such as when only one bend, not less than 6 times the pipe diameter. 2) dark time, not less than ... (E) wire into the switch box and Cabinet, Board, should have a long widening of margins; (F) when the wires into the electrical equipment should remain at the Centre of 0.3M margin; (G) the floor tube to the power box, tube 1M of allowances; (H) the wires connected to the power supply and into and out of time, tube 1.5M of allowances. (I) wore metal pipe with wire retainer should be well set, retainer must not cut place, match the specification and pipe. (J) after the end of the wire, you should check the insulation resistance, the resistance shall be not less than 0.5M. (K) the dark construction of the works at the time of completion and acceptance of the delivery, and laid wiring changes in the construction should be part of the actual position (including junction boxes and junction boxes and pipeline specifications), fixed in the as-built drawings and indicate 22所以函数在上单调递增，在上单调递减， (,2],[0,),gx()aa 24所以. gxg()(),,,max22aae 42a,,1由得，，所以. „„„„„11分 ,1a,,22aee 2,a,0(?)当，即时,显然在上， ,,0[0,2]gx()0,a 2a函数在上单调递增，且. gx()[0,2]gxg()(2)4e,,max 2aa,0 显然不成立，故不成立. „„„„„12分 gx()4e1,,max 综上所述，的取值范围是. „„„„„13分 a(,ln2],,, 19、(本小题满分14分) ,,,,,,,,, 解:(?)依题意不妨设，，则，. Bb(0,),Bb(0,)FBb,,,(1,)FBb,,(1,)2112 ,,,,,,,,,222由，得1,,,ba.又因为ab,,1， FBFBa,,,12 解得. ab,,2,3 22xyC，,1所以椭圆的方程为. „„„„„4分 43 l(?)依题直线的方程为. ykx,,(1) ykx,,(1),,,222222由得. (34)84120，,，,,kxkxk,xy，,1,43, 228k412k,xx，,xx,设Mxy(,)，Nxy(,)，则，. „„„„6分 112212122234，k34，k 243kk,MNP(,)所以弦的中点为. „„„„„7分 223434，，kk 2222MNxxyykxxxx,,，,,，，,()()(1)[()4]所以 12121212 42644(412)kk,2,，,(1)[]k 222(34)34，，kk 212(1)k，,. „„„„„9分 243k， 2314kkPDyx，,,,()直线的方程为， 224343kkk，， pipe into the box, lateral lock, installed at the inner side protection. (9) dark distribution pipe should be laid along the lines of recent, and should reduce bending, buried within the walls of the pipe from the surface-distance should not be less than 15mm. (10) under normal circumstances, it tube bending radius shall meet the following requirements: 1) specify timing, there is generally no less than 6 times the pipe diameter, such as when only one bend, not less than 6 times the pipe diameter. 2) dark time, not less than ... (E) wire into the switch box and Cabinet, Board, should have a long widening of margins; (F) when the wires into the electrical equipment should remain at the Centre of 0.3M margin; (G) the floor tube to the power box, tube 1M of allowances; (H) the wires connected to the power supply and into and out of time, tube 1.5M of allowances. (I) wore metal pipe with wire retainer should be well set, retainer must not cut place, match the specification and pipe. (J) after the end of the wire, you should check the insulation resistance, the resistance shall be not less than 0.5M. (K) the dark construction of the works at the time of completion and acceptance of the delivery, and laid wiring changes in the construction should be part of the actual position (including junction boxes and junction boxes and pipeline specifications), fixed in the as-built drawings and indicate 22kk由，得，则， x,D(,0)y,02243k，43k， 223(1)kk，所以. „„„„11分 DP,243k， 223(1)kk，22DP1k1143k，,,所以. „„„„„12分 ,,122212(1)k，MNk41，41k， 243k， 12又因为，所以. k，,1101,,2k，1 111所以01,,,. 2414k， DP1所以的取值范围是. „„„„„„„„„„„„„„„14分 (0,)4MN 20、(本小题满分13分) 222解:(?)由已知得( S(1,1,)11,,,,，,,,333 ( „„„„„3分 S(1,1,1,1)1111112,,,,,,,，,, (?)设( SSxxx,(,,)123 n,3当时，( SSxxxxxxxxxxx,,,，，(,,),123121323ij,,,13ij若固定xx,，仅让x变动，此时Sxxxxxxxxxxx,，，,，，()， 12132323123231 因此SSxxSxx,,min{(1,,),(1,,)}( 2323 同理SxxSxSx(1,,)min{(1,1,),(1,1,)},,( 2333 SxxSxSx(1,,)min{(1,1,),(1,1,)},,,,,( 2333 S,1xxx,,以此类推，我们可以看出，的最小值必定可在某一组取值的所达到， 123 SSxxx,min{(,,)}于是( 123x,,1kk,1,2,3 12222Sxxxxxx,，，,，，[()()]x,,1当(k,1,2,3)时， 123123k2 132,，，,()xxx( 12322 13S,,1S,,,,1||1xxx，，,xx,,1x,,1因为，所以，且当，时，( 12312322 S,,1因此( „„„„„8分 min pipe into the box, lateral lock, installed at the inner side protection. (9) dark distribution pipe should be laid along the li2013朝阳高三二模数学理科 第 9 nes of recent, and should reduce bending, buried within the walls of the pipe from the surface-distance should not be less than页 共 10 页 15mm. (10) under normal circumstances, it tube bending radius shall meet the following requirements: 1) specify timing, there is generally no less than 6 times the pipe diameter, such as when only one bend, not less than 6 times the pipe diameter. 2) dark time, not less than ... (E) wire into the switch box and Cabinet, Board, should have a long widening of margins; (F) when the wires into the electrical equipment should remain at the Centre of 0.3M margin; (G) the floor tube to the power box, tube 1M of allowances; (H) the wires connected to the power supply and into and out of time, tube 1.5M of allowances. (I) wore metal pipe with wire retainer should be well set, retainer must not cut place, match the specification and pipe. (J) after the end of the wire, you should check the insulation resistance, the resistance shall be not less than 0.5M. (K) the dark construction of the works at the time of completion and acceptance of the delivery, and laid wiring changes in the construction should be part of the actual position (including junction boxes and junction boxes and pipeline specifications), fixed in the as-built drawings and indicate (?)设 SSxxxxx,,(,,,)？,12nij,,,1ijn . ,，，，，，，，，xxxxxxxxxxxx？？？121312321nnnn,固定，仅让变动，此时 xxx,,,？x23n1 ， Sxxxxxxxxxx,，，，,，，，，，()()？？？2312321nnnn,因此( SSxxxSxxx,,min{(1,,,,),(1,,,,)}？？2323nn 同理( SxxxSxxSxx(1,,,,)min{(1,1,,,),(1,1,,,)}？？？,,2333nnn ( SxxxSxxSxx(1,,,,)min{(1,1,,,),(1,1,,,)},,,,,？？？2333nnn S以此类推，我们可以看出，的最小值必定可在某一组取值,1的所达到，于是xxx,,,？12n SSxxx,min{(,,,)}？( 12nx,,1k,？kn1,2,, 12222当Sxxxxxx,，，，,，，，？？()时， [()()]x,,1kn,1,2,,？nn1212k2 1n2( ,，，，,()xxx？n1222 nS,,?当为偶数时，， n2 nnS,,S,,若取，，则，所以( xxx,,,,？1xxx,,,,,？1min12nnnn，，1222222 1Sn,,,(1)?当为奇数时，因为，所以， ||1xxx，，，,？n12n2 1Sn,,,(1)若取xxx,,,,？1，xxx,,,,,？1，则， 121n,nnn,,11，，122222 1Sn,,,(1)所以( „„„„„„„„„„13分 min2 pipe into the box, lateral lock, installed at the inner side protection. (9) dark distribution pipe should be laid along the lines of recent, and should reduce bending, buried within the walls of the pipe from the surface-distance should not be less than 15mm. (10) under normal circumstances, it tube bending radius shall meet the following requirements: 1) specify timing, there is generally no less than 6 times the pipe diameter, such as when only one bend, not less than 6 times the pipe diameter. 2) dark time, not less than ... (E) wire into the switch box and Cabinet, Board, should have a long widening of margins; (F) when the wires into the electrical equipment should remain at the Centre of 0.3M margin; (G) the floor tube to the power box, tube 1M of allowances; (H) the wires connected to the power supply and into and out of time, tube 1.5M of allowances. (I) wore metal pipe with wire retainer should be well set, retainer must not cut place, match the specification and pipe. (J) after the end of the wire, you should check the insulation resistance, the resistance shall be not less than 0.5M. (K) the dark construction of the works at the time of completion and acceptance of the delivery, and laid wiring changes in the construction should be part of the actual position (including junction boxes and junction boxes and pipeline specifications), fixed in the as-built drawings and indicate