2016陕西铁路工程职业技术学院单招数学模拟试题
第I卷(选择题,共60分)
一、选择题(本大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中,只有一个是符合题目要求的):
1.下列函数中,周期为π,且为偶函数的是 ( )
A.415290769594460e2e485922904f345d.png = | sin9dd4e461268c8034f5c8564e155c67a6.png |B.415290769594460e2e485922904f345d.png = 2sin9dd4e461268c8034f5c8564e155c67a6.png·cos9dd4e461268c8034f5c8564e155c67a6.pngC.415290769594460e2e485922904f345d.png = cos D.415290769594460e2e485922904f345d.png=cos41e4eeeeffc63436aa240249fc78b9a1.png
2.已知全集U = Z ,A={1,3,5},B={ 9dd4e461268c8034f5c8564e155c67a6.png | 9dd4e461268c8034f5c8564e155c67a6.png3 - 29dd4e461268c8034f5c8564e155c67a6.png2 - 39dd4e461268c8034f5c8564e155c67a6.png = 0},则B∩CuA等于( )
A.{1,3}B.{0,-1} C.{1,5} D.{0,1}
3.双曲线中心在原点,实轴长为2,它的一个焦点为抛物线415290769594460e2e485922904f345d.png2 = 89dd4e461268c8034f5c8564e155c67a6.png的焦点,则此
双曲线方程为 ( )
A.04e3ef10ae6fc472a2177ffc66c4e9f4.png-y2 = 1B.ee07fd80c208685629635db91a455413.png-x2 = 1 C.y2 -04e3ef10ae6fc472a2177ffc66c4e9f4.png = 1 D.x2 -ee07fd80c208685629635db91a455413.png = 1
4.设a.b为两条直线,ab410a966ac148e9b78c65c6cdf301fd.png.β为两个平面,则下列命题正确的是 ( )
A.a.b与ab410a966ac148e9b78c65c6cdf301fd.png成等角,则a//b;B.若a∥ab410a966ac148e9b78c65c6cdf301fd.png,b∥β,ab410a966ac148e9b78c65c6cdf301fd.png∥β则0cc175b9c0f1b6a831c399e269772661.png∥b;
C.a 47c79240e185e467039a4197e19b0361.pngab410a966ac148e9b78c65c6cdf301fd.png,bc51a88011fa20bbb93b65d2a915137b5.pngβ,a∥b则ab410a966ac148e9b78c65c6cdf301fd.png∥β;D.ab6c28e2395ad4321cb259c6a31f06238.pngab410a966ac148e9b78c65c6cdf301fd.png,bb6c28e2395ad4321cb259c6a31f06238.pngβ,ab410a966ac148e9b78c65c6cdf301fd.png∥β则0cc175b9c0f1b6a831c399e269772661.png∥b.
5.设a1 = 2,数列{1+2an}是以3为公比的等比数列,则a4的值为 ( )
A.67 B.77 C.22 D.202
6.已知向量89e9c902fed032a267e19bcbbe710b75.png= (-1,2),7f711ac1314a21599dc5a9c5bb757eff.png= (2,1),则89e9c902fed032a267e19bcbbe710b75.png与7f711ac1314a21599dc5a9c5bb757eff.png的位置关系是 ( )
A.平行且同向 B.不垂直也不平行 C.垂直 D.平行且反向
7.在5b59e3c12b2cc4500e102dd7d0ad9d01.png的展开式中,常数项为15项,则n的值为 ( )
A.6 B.5 C.4 D.3
8.若8fa14cdd754f91cc6554c9e71929cce7.png(9dd4e461268c8034f5c8564e155c67a6.png)= 32151399be389f2b65beb0b274b69ed29.png的反函数为g(9dd4e461268c8034f5c8564e155c67a6.png),且g(a)+g(b)=2,则07ea9eb1f4232484e23c7ec7420df172.png+09d396adb41a8af8f97997736a8e073c.png的最小值为 ( )
A.7964c6a339acf2ddea25a5ef0552b97e.png B.6ca8c824c79dbb80005f071431350618.png C.9df743fb4a026d67e85ab08111c4aedd.png D.1
9.定义运算8ca53808da3df53a5df1c941c53e995a.png若| m– 2 | 3eb02fff4997ea78e9c47152ebbc24c3.pngm = | m-2|,则m的取值范围是 ( )
A.(-b63148f54c8e3d3cfd0603baae351208.png,1)B.[1,+b63148f54c8e3d3cfd0603baae351208.png] C.(0,+b63148f54c8e3d3cfd0603baae351208.png) D.(-b63148f54c8e3d3cfd0603baae351208.png,0)
10.在△ABC中,三边为a,b,c且a=2b·sinA,则B的大小为 ( )
A.81f3b33c4c6a63cea9158f20c9e0b24b.png或c87d41c12d441153b97f3593f330c121.png B.c87d41c12d441153b97f3593f330c121.png或cd3ba7dbe6650fbf0b092d3d3c833d5d.png C.c87d41c12d441153b97f3593f330c121.png或15a638c09d3e9bb4597ce8bf69141c36.png D.81f3b33c4c6a63cea9158f20c9e0b24b.png或d5d141e4e881e3e513453cfa0f09abd4.png
11.不等式log3( | 9dd4e461268c8034f5c8564e155c67a6.png– 5 | + | 9dd4e461268c8034f5c8564e155c67a6.png + 4 | ) >a对于9dd4e461268c8034f5c8564e155c67a6.png3659f7cbbee83f93790ff7f8c7480168.pngR恒成立,则a的取值范围是( )
A.(-b63148f54c8e3d3cfd0603baae351208.png,9) B.(-b63148f54c8e3d3cfd0603baae351208.png,2) C.(2,9) D.[1,+b63148f54c8e3d3cfd0603baae351208.png]
12.有n支球队参加单循环赛,其中两个队各赛了三场就退出了比赛,且此两队之间未进行比赛,这样到比赛结束时共赛了34场,那么n等于 ( )
A.12 B.11 C.10 D.9
第II卷(非选择题,共90分)
二、填空题(本大题共4小题,每小题4分,共16分)把答案填在横线上
13.某工厂生产A.B.C三种不同型号的产品,产品数量之比依次为3:4:7现用
分层抽样方法取出一个容量为n的样本,样本中B型号产品有28件,那么此样本
的容量n=.
14.设实数9dd4e461268c8034f5c8564e155c67a6.png.415290769594460e2e485922904f345d.png满足6eaf4ce81103b4eb01917410647ed092.png 则aae37404f1219a49dafcaf015d5965f3.png的最大值为.
15.定义运算6ce6681a8c902876e865075444cde2a2.png4951ffe137a99f697d71eb9421599f00.png= ad–bc,则满足条件08e2339d1f1d7b654b86605f47c22282.png186af28884ea9e8c4877e0a9e19277c4.png = 0的点p的轨迹方程为.
16.点P在正方形ABCD所在的平面外,PDb6c28e2395ad4321cb259c6a31f06238.png平面ABCD,且PD=AD,则PA与BD所成角的大小为.
3、解答题(本大题6个小题,共74分.解答应写出文字说明.证明过程或演算步骤)17.(12分)某地一天从6时到14时的温度变化曲线如图示,它近似满足函数
415290769594460e2e485922904f345d.png=Asin(45bf03a575f6e81359314e906fb2bff3.png9dd4e461268c8034f5c8564e155c67a6.png+6c4dbec1c9102e1076a6a6ca04576cf0.png)+b.
(1)求这段时间的最大温差;
(2)试求这段曲线的函数解析式.
18.(12分)袋中有大小相同的5个白球和3个 黑球,现从中任意摸出4个,求下列事件发生的概率:
(1)摸出2个或3个白球;
(2)至少摸出一个黑球.
19.(12分)如图,在三棱锥P - ABC中,△ABC是边长为2的等边三角形,且∠PCA=∠PCB
(1)求证:PCb6c28e2395ad4321cb259c6a31f06238.pngAB;
(2)若O为△ABC的中心,G为△PAB的重心,求证:GO∥平面PAC;
20.(12分)已知函数8fa14cdd754f91cc6554c9e71929cce7.png(9dd4e461268c8034f5c8564e155c67a6.png)=a9dd4e461268c8034f5c8564e155c67a6.png3+b9dd4e461268c8034f5c8564e155c67a6.png2+c(a,b,c∈R,a≠0) 的图像过点P(-1,2),且在点P处的切线与直线9dd4e461268c8034f5c8564e155c67a6.png-3415290769594460e2e485922904f345d.png=0垂直.
(1)若c=0试求函数8fa14cdd754f91cc6554c9e71929cce7.png(9dd4e461268c8034f5c8564e155c67a6.png) 的单调区间;
(2)若a>0,b>0且 (-b63148f54c8e3d3cfd0603baae351208.png,m),(n,+b63148f54c8e3d3cfd0603baae351208.png)是8fa14cdd754f91cc6554c9e71929cce7.png(9dd4e461268c8034f5c8564e155c67a6.png) 的单调递增区间,试求n-m的范围.
21.(12分)设椭圆a94e0d187523e019b19a0694e6e2551a.png+25e310c5f6d7daf582a112c19744a2db.png = 1( a>b> 0 )的左焦点为F,上顶点为A.过A做直线2db95e8e1a9267b7a1188556b2013b33.pngb6c28e2395ad4321cb259c6a31f06238.pngAF,l分别交椭圆和9dd4e461268c8034f5c8564e155c67a6.png轴正半轴于P、Q两点,若P分AQ所成的比为8∶5.
(1)求椭圆的离心率;
(2)若过A、Q、F三点的圆恰好与直线9dd4e461268c8034f5c8564e155c67a6.png +91a24814efa2661939c57367281c819c.png415290769594460e2e485922904f345d.png + 3 = 0相切,求椭圆方程.
22.(14分)已知Pn( an ,bn)( n∈N* )都在直线2db95e8e1a9267b7a1188556b2013b33.png∶y = 29dd4e461268c8034f5c8564e155c67a6.png + 2上,P1为d41d8cd98f00b204e9800998ecf8427e.png直线2db95e8e1a9267b7a1188556b2013b33.png与9dd4e461268c8034f5c8564e155c67a6.png轴的交点,数列{an}为等差数列,公差为1.
(1)求数列{an}、{bn}的通项公式;
(2)若8fa14cdd754f91cc6554c9e71929cce7.png(n) = a25d40b4e073860f2c7e67baa6eb404f.png是否存在8ce4b16b22b58894aa86c421e8759df3.png∈N*,使得8fa14cdd754f91cc6554c9e71929cce7.png(8ce4b16b22b58894aa86c421e8759df3.png+5)=28fa14cdd754f91cc6554c9e71929cce7.png(8ce4b16b22b58894aa86c421e8759df3.png)-2成立?
若存在,求出8ce4b16b22b58894aa86c421e8759df3.png值;若不存在,说明理由;
(3)求证:cb589aa9146ff13d73e2b0ec7a9760f2.png+ 76d3fcd01cb0bc58a5d607c145571a65.png + … + 0ee4cfb4fc2bb6d20056b990b2600b2c.png<add2b5c8b974155f65e931df2054a985.png,(n ≥ 2,n ∈ N3740008fd58c07137202b5975d74378d.png)
参考答案
一、 选择题(本大题共12小题,每小题5分,共60分)
13.98 14.bd8eacd6ef8c460fea72f998c06d4e7e.png 15.(理)-2±d21848cdd835abcb491be1f151e9b6c6.png865c0c0b4ab0e063e5caa3387c1a8741.pngd41d8cd98f00b204e9800998ecf8427e.png(文)(9dd4e461268c8034f5c8564e155c67a6.png-1)2 + 4415290769594460e2e485922904f345d.png2 = 1 16.c87d41c12d441153b97f3593f330c121.png
三、解答题
17.解:(1)由图示,这段时间的最大温差是30-10=20(998c43702c76b9d02152bcb64bab8f12.png)…………………………4′
(2)图中从6时到14时的图像是函数415290769594460e2e485922904f345d.png=Asin(45bf03a575f6e81359314e906fb2bff3.png9dd4e461268c8034f5c8564e155c67a6.png+6c4dbec1c9102e1076a6a6ca04576cf0.png)+b的半个周期的图像.
∴93b05c90d14a117ba52da1d743a43ab1.png·11ce3ac0eaeb890ee6179a8e9beeecc0.png= 14-6,解得45bf03a575f6e81359314e906fb2bff3.png= 8d8020fa29ca34d91ece3432f7dfc29f.png…………………………………………………………6′
由图示A = 93b05c90d14a117ba52da1d743a43ab1.png(30 - 10)= 10,b = 93b05c90d14a117ba52da1d743a43ab1.png(30+10) = 20,这时415290769594460e2e485922904f345d.png=10sin(8d8020fa29ca34d91ece3432f7dfc29f.png9dd4e461268c8034f5c8564e155c67a6.png+ 45bf03a575f6e81359314e906fb2bff3.png)+ 20…
………………………………………………………………………………………………8′
将9dd4e461268c8034f5c8564e155c67a6.png= 6,415290769594460e2e485922904f345d.png = 10代入上式可取6c4dbec1c9102e1076a6a6ca04576cf0.png=9df743fb4a026d67e85ab08111c4aedd.png31bf0b12546409e15021243132fc7574.png,…………………………………………… 10′
综上所求的解析式为415290769594460e2e485922904f345d.png=10sin(8d8020fa29ca34d91ece3432f7dfc29f.png9dd4e461268c8034f5c8564e155c67a6.png+ 9df743fb4a026d67e85ab08111c4aedd.png31bf0b12546409e15021243132fc7574.png)+ 20,9dd4e461268c8034f5c8564e155c67a6.png∈[6,14]. ………………………12′
18.解:(1)设摸出的4个球中有2个白球、3个白球分别为事件A、B,则P(A)= 42675ba6546315b1004815a9619f0c0c.png = d41d8cd98f00b204e9800998ecf8427e.png,P(B)= fdf93807a06bb1e4815e11b99d5e46ef.png = a784243d8211e519a1071acd55f1f3b0.png.………………………………………………………4′
∵A、B为两个互斥时间,∴P(A+B)= P(A)+P(B)= 37f005b930d41345b05b2e01fe53c651.png.
即摸出的4个球中有2个或3个白球的概率为37f005b930d41345b05b2e01fe53c651.png………………………………………6′
(2)设摸出的4个球中全是白球为事件C,则P(C)= de7abc0f2ea7e168d8cbacaa37d20c03.png = 9ef51575e73ac02eeac7df8736c37717.png,……………10′
“至少摸出一个黑球”为事件C的对立事件,其概率为P = 1-9ef51575e73ac02eeac7df8736c37717.png = 1167b022ae94e65119ed32271cf2c3f4.png. ………12′
19.证明:(1)设H为AB中点,连PH、CH.……………………………………………2′
∠PCA=258aaf6d14d7de164d9cb9e59de99c0e.pngc747017dd963feee68eba362e4fd95e6.png△PCAa905f2cdb8d21b5d137cdf36e48ded62.png△PCBc747017dd963feee68eba362e4fd95e6.png
在等边三角形ABC中, e1e6b5009ed0fb1f58277e1f623b3ed0.png平面PCH927000895ecb082a3260b23419ec18ea.png……
…………………………………………………………………………………理8′(文12′)
(2)点G.O分别在PH.CH上,0943a9eb9e17351e9cb3213d52a2310f.png平面PAC
(理)(3)由(1)可知∠PHC=7943b5fdf911af3ffcf9d8f738478e8a.png为二面角P – AB – C的平面角,7943b5fdf911af3ffcf9d8f738478e8a.png为锐角,cos7943b5fdf911af3ffcf9d8f738478e8a.png> 0.
在等边三角形ABC中,CH=91a24814efa2661939c57367281c819c.png,PG=8ec254d216dc97b460b754f8de2403a9.pngc747017dd963feee68eba362e4fd95e6.pngPH = bd8eacd6ef8c460fea72f998c06d4e7e.pngPG=291a24814efa2661939c57367281c819c.png,
设PC =9dd4e461268c8034f5c8564e155c67a6.png,则9dd4e461268c8034f5c8564e155c67a6.png2 = 3 + 12 - 12 cos7943b5fdf911af3ffcf9d8f738478e8a.pngc747017dd963feee68eba362e4fd95e6.png cos7943b5fdf911af3ffcf9d8f738478e8a.png = bfaa773e6bfd641c127db3a696dcb41d.png> 0,
2aed61798d0f470ebbf899eaad7fa539.png 即8479c9dfb370ba79747b665a4edf774f.pngc747017dd963feee68eba362e4fd95e6.png91a24814efa2661939c57367281c819c.png<9dd4e461268c8034f5c8564e155c67a6.png<4a68ac0136e5c8446ca046abe9e88800.png.……………12′
20.解:(1)由8fa14cdd754f91cc6554c9e71929cce7.png(9dd4e461268c8034f5c8564e155c67a6.png)过点P得-a + b + c = 2,8fa14cdd754f91cc6554c9e71929cce7.pngˊ(9dd4e461268c8034f5c8564e155c67a6.png)=3a9dd4e461268c8034f5c8564e155c67a6.png2 + 2b9dd4e461268c8034f5c8564e155c67a6.png, ………………2′
因为8fa14cdd754f91cc6554c9e71929cce7.png(9dd4e461268c8034f5c8564e155c67a6.png)在P处的切线与9dd4e461268c8034f5c8564e155c67a6.png- 3415290769594460e2e485922904f345d.png = 0垂直,所以3a – 2b = -3.
又c = 0,解得a = 1,b = 3,所以8fa14cdd754f91cc6554c9e71929cce7.png′(9dd4e461268c8034f5c8564e155c67a6.png)=39dd4e461268c8034f5c8564e155c67a6.png2 + 69dd4e461268c8034f5c8564e155c67a6.png.………………………………4′
令8fa14cdd754f91cc6554c9e71929cce7.pngˊ(9dd4e461268c8034f5c8564e155c67a6.png) = 0得9dd4e461268c8034f5c8564e155c67a6.png1 = 0, 9dd4e461268c8034f5c8564e155c67a6.png2 = -2;
当x>0或9dd4e461268c8034f5c8564e155c67a6.png< -2,8fa14cdd754f91cc6554c9e71929cce7.pngˊ(9dd4e461268c8034f5c8564e155c67a6.png) > 0,当 –2 <9dd4e461268c8034f5c8564e155c67a6.png< 0 ,8fa14cdd754f91cc6554c9e71929cce7.pngˊ(9dd4e461268c8034f5c8564e155c67a6.png) < 0,
所以(-b63148f54c8e3d3cfd0603baae351208.png,-2),(0,+b63148f54c8e3d3cfd0603baae351208.png)是f(9dd4e461268c8034f5c8564e155c67a6.png)的单调递增区间,(-2,0)是8fa14cdd754f91cc6554c9e71929cce7.png(9dd4e461268c8034f5c8564e155c67a6.png)的单调递减区间.
…………………………………………………………………………………………… 6′
(2)由8fa14cdd754f91cc6554c9e71929cce7.png′(9dd4e461268c8034f5c8564e155c67a6.png) = 3a9dd4e461268c8034f5c8564e155c67a6.png2 + 2b9dd4e461268c8034f5c8564e155c67a6.png =0,得9dd4e461268c8034f5c8564e155c67a6.png1=0,9dd4e461268c8034f5c8564e155c67a6.png2 = -7199d5d0c8733c4e1309a1972cbb3eb8.png.……………………………… 8′
又因为a > 0,b > 0所以当9dd4e461268c8034f5c8564e155c67a6.png> 0,或χ<852f97d5dd12be0330196b8c7a273a2e.png,8fa14cdd754f91cc6554c9e71929cce7.pngˊ(9dd4e461268c8034f5c8564e155c67a6.png) > O,
因此(-b63148f54c8e3d3cfd0603baae351208.png,-7199d5d0c8733c4e1309a1972cbb3eb8.png),( 0,+b63148f54c8e3d3cfd0603baae351208.png)是8fa14cdd754f91cc6554c9e71929cce7.png(9dd4e461268c8034f5c8564e155c67a6.png)的单调递增区间,………………………………10′
于是有n – m = 0 -(-7199d5d0c8733c4e1309a1972cbb3eb8.png) = 7199d5d0c8733c4e1309a1972cbb3eb8.png.由(1)知-a + b + c = 2,且3a - 2b = -3,
所以a = 1 - 2c > 0,b = 3 - 3c > 0,从而得c <93b05c90d14a117ba52da1d743a43ab1.png.
n– m = 7199d5d0c8733c4e1309a1972cbb3eb8.png = 6ca8c824c79dbb80005f071431350618.png·d908a029bd844c2c98674e089d3560d0.png = 1 - 460c00453894c124ac6d5ee4c2570ee5.png> 1,故n – m >1.……………………12′
21.解:(1)由F(-c,0),A(0,b)知直线AP方程为415290769594460e2e485922904f345d.png– b = - 0428e36c75e29a8e87ce9c81e3f6e6e2.png9dd4e461268c8034f5c8564e155c67a6.png,令415290769594460e2e485922904f345d.png = 0得
Q(7413e631c0eade1965552dc8029279fa.png,0)………………………………………………………………………………2′
设P(9dd4e461268c8034f5c8564e155c67a6.png0,415290769594460e2e485922904f345d.png0),P分AQ所成的比为6af8e2f02f674b41b6ccf43debc252d2.png= b0e1326365729673e417839e572d1ced.png,
word/media/image122.gif781c52f73a7ae49a5c1d96c30e0982f3.png
409fdbd6005a46ce24cc36a3305d4a26.png
代入a94e0d187523e019b19a0694e6e2551a.png + 25e310c5f6d7daf582a112c19744a2db.png = 1 中得2b2 = 3ac,又b2 = a2-c2,解得离心率c =93b05c90d14a117ba52da1d743a43ab1.png.………………6′
(2)Rt△AOF中,| AF | = a,sin∠FAO = 093f8a967400f274ba09d084a57ef1af.png = 988f5adcded469a44a397e792c6692d4.png∠FAO = 81f3b33c4c6a63cea9158f20c9e0b24b.png,∠AQF = 81f3b33c4c6a63cea9158f20c9e0b24b.png,则
| FQ | = 2| AF |= 2a = 4c,故圆心B(c,0),
∴Rt△QAF的外接圆方程为(9dd4e461268c8034f5c8564e155c67a6.png– c )2 + 415290769594460e2e485922904f345d.png2 = a2,……………………………………10′
该圆与9dd4e461268c8034f5c8564e155c67a6.png+91a24814efa2661939c57367281c819c.png415290769594460e2e485922904f345d.png + 3 = 0相切,则d = 90967633e18e12bd3d36df332f018d5c.png = a .
即c + 3 = 2a = 2×2cc747017dd963feee68eba362e4fd95e6.pngc = 1,则a =2,b2 = 3.
∴所求椭圆方程为fdc64c74176032b356803f9ad87d0a74.png+ee07fd80c208685629635db91a455413.png = 1.……………………………………………………12′
22.解(1)(理)P1(a1,b1)为直线415290769594460e2e485922904f345d.png = 2χ+ 2与9dd4e461268c8034f5c8564e155c67a6.png轴交点,则a1 = -1,b1 = 0………2′
由已知9dd4e461268c8034f5c8564e155c67a6.png、415290769594460e2e485922904f345d.png∈(0,+b63148f54c8e3d3cfd0603baae351208.png),都有g(x·415290769594460e2e485922904f345d.png) = g(9dd4e461268c8034f5c8564e155c67a6.png) + g(415290769594460e2e485922904f345d.png)成立,又g(2) = 1,
得g(4) = =g(26392228661363e75c352077a2cfe66d7.png2) = g(2) + g(2) = 2,
因为n ≥ 2时,bn > 0,且g(Sn) = g(bn) + g(2+bn) - 2,( n∈N* )
所以2 + g( Sn ) = g( bn ) + g( 2+bn ),即g(4) +g( Sn ) = g( bn ) + g( 2+bn ).
所以4Sn=bn(2+bn)c747017dd963feee68eba362e4fd95e6.pngb2 = 2, b2– b1 = 2;
由4Sn = bn (2+bn)及4Sn+1 = bn+1(2 +bn+1)c747017dd963feee68eba362e4fd95e6.pngbn+1 - bn = 2
所以{bn}是以0为首项,2为公差的等差数列,∴bn = 2n-2 ……………………4′
因为Pn( an,bn)( n ∈ N3740008fd58c07137202b5975d74378d.png)在直线y = 29dd4e461268c8034f5c8564e155c67a6.png + 2上,
则bn = 2an + 2,∴an = n - 2.……………………………………………………………6′
(1)(文)解:P1=(a1,b1)为直线415290769594460e2e485922904f345d.png = 29dd4e461268c8034f5c8564e155c67a6.png + 2与9dd4e461268c8034f5c8564e155c67a6.png轴交点,则a1 = -1,b1 = 0 ……2′
∴an = -1 + ( n – 1 ) = n – 2,d89572eaa5cda62e7545156c0fd62375.png(n∈N*)在直线415290769594460e2e485922904f345d.png = 29dd4e461268c8034f5c8564e155c67a6.png + 2上,
则bn=2an+ 2,∴bn= 2n - 2.……………………………………………………………4′
(2)8ce4b16b22b58894aa86c421e8759df3.png为偶数时,8fa14cdd754f91cc6554c9e71929cce7.png(8ce4b16b22b58894aa86c421e8759df3.png + 5) = ak+ 5 =8ce4b16b22b58894aa86c421e8759df3.png+ 3,28fa14cdd754f91cc6554c9e71929cce7.png(8ce4b16b22b58894aa86c421e8759df3.png) – 2 = 2( 28ce4b16b22b58894aa86c421e8759df3.png– 2 ) – 2 = 48ce4b16b22b58894aa86c421e8759df3.png- 6
由8ce4b16b22b58894aa86c421e8759df3.png+ 3 = 48ce4b16b22b58894aa86c421e8759df3.png- 6c747017dd963feee68eba362e4fd95e6.png8ce4b16b22b58894aa86c421e8759df3.png= 3 ,与8ce4b16b22b58894aa86c421e8759df3.png为偶数矛盾,
8ce4b16b22b58894aa86c421e8759df3.png 为奇数时,8fa14cdd754f91cc6554c9e71929cce7.png (8ce4b16b22b58894aa86c421e8759df3.png+5) = bk+5 = 28ce4b16b22b58894aa86c421e8759df3.png+ 8,2 ƒ (8ce4b16b22b58894aa86c421e8759df3.png) – 2 = 28ce4b16b22b58894aa86c421e8759df3.png- 6
由28ce4b16b22b58894aa86c421e8759df3.png+ 8 = 28ce4b16b22b58894aa86c421e8759df3.png- 6得8ce4b16b22b58894aa86c421e8759df3.png不存在.故满足条件的8ce4b16b22b58894aa86c421e8759df3.png不存在.…………………理10′(文9′)
(3)| P1Pn|2 =( n – 1 )2 + ( 2n – 2 )2 = 5( n – 1 )2,n ≥ 2,
005f129ded6a76e7c80aefffbe8ec768.png + 588e8dc8dd8434d79a4c2e3ef6884ebc.png + … + 2728cc8dfe3fe093a67e09b569a3860c.png = 22417f146ced89939510e270d4201b28.png[05b2a665353dfe3236decd3ffce0f4dd.png+e46b11c8c9444c2ee03fe6ca3151e324.png + … + 2a0c5a864fe23692a9e96c5301b54c13.png]
≤22417f146ced89939510e270d4201b28.png[05b2a665353dfe3236decd3ffce0f4dd.png + 92ba4449e73e31e451961e53f5630ba5.png… + b8071096a662704695204d2ddcc82948.png]
= 6767ee6dfa4a45a15db0c0a61e7c9a96.png
∴7649665e656d409d7fec5f10b72b99f8.png… + ea0c1a34dc55f195fffbfc001bb67721.png………………………14′
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