高中数学压轴题一
备战2013
高考
地理事物空间分布特征语文高考下定义高考日语答题卡模板高考688高频词汇高考文言文120个实词
数学――压轴题跟踪演练系列一
1((12分)已知抛物线、椭圆和双曲线都经过点,它们在轴上有共同焦点,椭圆和双曲线M1,2x,,
的对称轴是坐标轴,抛物线的顶点为坐标原点. (?)求这三条曲线的方程;
,(?)已知动直线过点,交抛物线于两点,是否存在垂直于轴的直线被以为直AB,APP3,0llx,,
,径的圆截得的弦长为定值,若存在,求出的方程;若不存在,说明理由. l
2p,2解:(?)设抛物线方程为,将代入方程得 ypxp,,20M1,2,,,,
2………………………………………………(1分) ?, 抛物线方程为: yx4
由题意知椭圆、双曲线的焦点为…………………(2分) FF,?1,0,1,0, c=1,,,,21
222对于椭圆, 2112114222aMFMF,,,,,,,,,,,,,,12
?,, a12
22?,,,, a12322,,
………………………………(4分) 222?,,,, bac222
22xy?,, 椭圆方程为: 1322222,,
,2222aMFMF,,,,对于双曲线, 12
,?,, a21
2,?,, a322
222………………………………(6分) ,,,?,,,, bca222
22xy?,, 双曲线方程为: 1322222,,
,,(?)设的中点为,的方程为:,以为直径的圆交于两点,中点为 APAPDE,DEHCllxa,
xy,3,,11令………………………………………………(7分) Axy,,, C?,,11,,22,,
1122?,,,, DCAPxy3,,1122 x,311 CHaxa,,,,,23,,122
21122222,,,,?,,,,,,,, DHDCCHxyxa323,,,,111,,,,442 ,,,axaa-23,,1
2…………(12分) 当时aDH,,,,,2462,为定值;?,, DEDH222为定值
,此时的方程为:lx ,2
2a,62((14分)已知正项数列中,,点在抛物线上;数列中,点abyx,,1Aaa,,,,,,,1nnnnn,1
在过点,以方向向量为的直线上. Bnb,0,11,2,,,,,,nn
(?)求数列的通项公式; ab,,,,,nn
,a, n为奇数,,n,(?)若fn,,问是否存在,使成立,若存在,求出fkfk,,274kN,k,,,,,,,b, n为偶数,,,n,
值;若不存在,说明理由;
nn,1aa(?)对任意正整数,不等式成立,求正数的取值范围. ,,0na,,,,,,,,2111nan111,,,,,,,,,bbb,,,,,,12n
2解:(?)将点代入中得 yx,,1Aaa,,,nnn,1
aaaad,,?,,,11 nnnn,,11
…………………………………………(4分) ?,,,,,, aann115,,n1
直线lyxbn:21,21,,?,, n
,n,5, n为奇数,,,(?)fn,………………………………(5分) ,,,21n,, n为偶数,,,,
当为偶数时,为奇数,kkfkfk,,,27274 ,,,,?,,,,?, kkk275421,4,,
当为奇数时,为偶数,kk,27……………………(8分)
35?,,,,?, 227145,kkk舍去,,,,,,2综上,存在唯一的符合条件。k,4
nn,1aa(?)由 ,,0,,,,,,,,2111nan111,,,,,,,,,bbb,,,,,,12n
,,,,,,1111即a,,,,111,,,,,,bbb23,n,,,,12n,,
,,,,,,1111记,,,,111fn,,,,,,,,bbb23,n12n,,,,,,
,,,,,,,,11111?,,,,,, fn11111,,,,,,,,,,bbbb25,n121nn,,,,,,,,,
fn,1,,,,231232424nnnn,,,,?,,,,,, 1,, 23,fnbn,,2525nn,,2523nn,,,n1,,,
241616nn,, ,,1241615nn,,
?,, fnfnfn1,即递增,,,,,,,
1445?,,,, fnf1,,,,,min3155
45?,, 0a15
………………………………(14分)
223.(本小题满分12分)将圆O: 上各点的纵坐标变为原来的一半 (横坐标不变), x,y,4得到曲线C.
(1) 求C的方程;
(2) 设O为坐标原点, 过点的直线l与C交于A、B两点, N为线段AB的中点, F(3, 0)
延长线段ON交C于点E.
|AB| ,3求证: 的充要条件是. OE,2ON
,x,x,,,,P(x, y)(x, y)解: (1)设点, 点M的坐标为,由题意可知………………(2分) ,,y,2y,,
2x22222,,x,4y,4,,y,1又?. x,y,4,4
2x2,y,1所以, 点M的轨迹C的方程为.………………(4分) 4
(2)设点, , 点N的坐标为, (x, y)A(x, y)B(x, y)001122
?当直线l与x轴重合时, 线段AB的中点N就是原点O,
不合题意,舍去; ………………(5分) ?设直线l: x,my,3,
,x,my,3,由消去x, ,22,x,4y,4,
22得………………? (m,4)y,23my,1,0
3my,,,?………………(6分) 02,m4
223m3m,4343x,my,3,,,,?, 00222m,4m,4m,4
433m(, ,)?点N的坐标为.………………(8分) 22,,m4m4
8323m(, ,)?若, 坐标为, 则点E的为, 由点E在曲线C上, OE,2ON22,,m4m4
24812m4222得, 即 ?舍去). m,4m,32,0,m,8 (m,,4,,12222(m,4)(m,4)
22212m,4m,164m,1|y,y|,,,1,由方程?得 1222m,4m,4
又 |x,x| , |my,my| , |m(y,y)|,121212
2|AB| , m,1|y,y| ,3?.………………(10分) 12
24(m,1)2|AB| ,3m,8. ,3,?若, 由?得? 2m,4
362(, ,)y,,x (x,0)?点N的坐标为, 射线ON方程为: , 362
,23,2x,,y,,x(x,0) 236,,3(, ,),由 解得 ?点E的坐标为 2,,33622,,x,4y,4y,,,,3,
?. OE,2ON
|AB| ,3综上, 的充要条件是.………………(12分) OE,2ON
1(x,R)f(x),4.(本小题满分14分)已知函数. x4,2
11f(x)(1) 试证函数的图象关于点(, )对称; 24
n(2) 若数列的通项公式为, 求数列的前m项和a,f() (m,N, n,1, 2, ?,m){a}{a}n,nnm
S;m
11112(3) 设数列满足: , . 设. bT,,,?,,b,b,b{b}n1n,1nnnb,1b,1b,1312n
若(2)中的满足对任意不小于2的正整数n, 恒成立, 试求m的最大值. S,TSnnn
11P(x, y)f(x)解: (1)设点是函数的图象上任意一点, 其关于点的对称点为. P(x, y)(, )00024x,x1,0x,1,x,,,0,,,22由 得 ,,1y,y1y,,y.00,,,2,,24,
1所以, 点P的坐标为P.………………(2分) (1,x, ,y)002
1f(x)由点在函数的图象上, 得y,. P(x, y)0000x04,2
xx00144? ,,,,f(1x),01,xxx000,,,,424242(42)
x011114f(x) ?点P在函数的图象上. (1,x, ,y),y,,,,000xx002224,2,2(42)
11f(x)?函数的图象关于点对称. ………………(4分) (, )24
1kk1f(x),f(1,x),(2)由(1)可知, , 所以, f(),f(1,), (1,k,m,1)2mm2km,k11即………………(6分) f(),f(), , ?a,a,,km,kmm22
由, ……………… ? S,a,a,a,?,a,am123m,1m
得 ………………? S,a,a,a,?,a,a,mm,1m,2m,31m
1m,11m1由?,?, 得 2S,(m,1),,2a,,2,,,,mm22626
1?………………(8分) S,(3m,1).m12
12(3) ?, ………………? b,,b,b,b,b(b,1)1n,1nnnn3
?对任意的. ………………? n,N, b,0,n
1111111,,,,,,由?、?, 得即. ,,b1bbbb(b1)bb1,n,1nnnnnnn,1
111111111T,(,),(,),?,(,),,,3,?.……………(10分) nbbbbbbbbb1223nn,11n,1n,1
2??数列是单调递增数列. b,b,b,0, ?b,b,{b}n,1nnn,1nn
?关于n递增. 当, 且时, . n,2Tn,NT,Tn,n2
11144452? b,,b,(,1),, b,(,1),,12333399981
175?………………(12分) T,T,3,,.n2b521
238417575?即? ?m的最大值为6. ……………(14分) S,,(3m,1),,m,,6,m5212523939
225((12分)E、F是椭圆的左、右焦点,是椭圆的右准线,点,过点E的直线交椭Pl,lxy,,24
圆于A、B两点.
AEAF,,AEF(1) 当时,求的面积;
y(2) 当AB,3时,求AFBF,的大小;
AP
(3) 求,EPF的最大值.
M
FExOmn,,4B,1解:(1) ,,,Smn2,,AEF22mn,,82,
,,,AEAF4,(2)因,,,,ABAFBF8, ,BEBF,,4,,
AFBF,,5.则
tanEPFtanEPMFPM,,,,,()(1) 设 Ptt(22,)(0),
32232222223,t,,,,,,,()(1), 221,tttttt,,663
3tanEPFEPF,,,,,30当时, t,63
22S1nan,26((14分)已知数列中,,当时,其前项和满足, a,Sna,,,n1nn321S,n
anlim(2) 求的表达式及的值; Sn2n,,Sn
a(3) 求数列的通项公式; ,,n
11nN,n,2(4) 设,求证:当且时,. b,,ab,nnn33nn,,(21)(21)
22S11n解:(1) ,,,,,,,,,,aSSSSSSn22(2),,,nnnnnnn111,21SSS,nnn1
,,11所以是等差数列.则. S,,,nSn,21n,,
a22n. ,,,,limlim22,,,,nn,,SSS212lim1nnn,,n
112,(2)当时,, n,2aSS,,,,,nnn,12nnn212141,,,
1,n,1,,,,3综上,a,. ,n2,n,2,,2,14,n,
1110,,,baab,,,(3)令,当时,有 (1) n,2
32121nn,,
1111法1:等价于求证. ,,,332121nn,,2121nn,,,,,,
111230,,,fxxxx,,,,,0,当时,令 n,2,,213n,3
33132,, fxxxxxxx,,,,,,,,,,232(1)2(1)2(1)0,,2223
1(0,]fx则在递增. ,,3
1110,,,又,
21213nn,,
11gg()(),,所以即. ab,nn332121nn,,
11112233法(2) abbaba,,,,,,,,,()()nn33nn,,2121nn,,(21)(21)
22 (2) ,,,,,,()()abababab
ababba22 (3) ,,,,,,,()[()()]abaabb,,,,,,,()[(1)(1)]abaabb2222
baaba333因,所以 aabb(1)(1)0,,,,,,ba,,,,,,,,,,,,11111022222223
由(1)(3)(4)知. ab,nn
1,a22,gbababab,,,,,法3:令,则 gbbab,,,,,,210,,,,2
22gbmaxggamaxaaaa,,,,0,,32所以 ,,,,,,,,,,
121422323()3()0aaaaa,,,,,,0,,,a因则, aaaa,,,,10,,3393
22所以 (5) gbababab,,,,,,0,,
由(1)(2)(5)知 ab,nn
7( (本小题满分14分)
22yx设双曲线=1( a > 0, b > 0 )的右顶点为A,,22ab
P是双曲线上异于顶点的一个动点,从A引双曲线的两条渐近线的平行线与直线OP分别交于Q和R两点.
,,,2OP(1) 证明:无论P点在什么位置,总有|| =
,,,,,OR?| ( O为坐标原点); |OQ
(2) 若以OP为边长的正方形面积等于双曲线实、虚轴围成的矩形面积,求双曲线离心率的取值范围;
第21题 b解:(1) 设OP:y = k x, 又条件可设AR: y = (x – a ), a
,,,,,,ab,kababkabOR 解得:= (,), 同理可得= (,), OQak,bak,bak,bak,b
222,,,,,,abab,kabkabab(1,k)OR?|?| =|+| =. 4分 OQ222ak,bak,bak,bak,b|ak,b|
,,,
OP 设 = ( m, n ) , 则由双曲线方程与OP方程联立解得:
22222abkab22m =, n = , 222222b,akb,ak
22222222,,,ab(1,k)kabab222OP? || = :m + n = + = , 222222222b,akb,akb,ak
222?点P在双曲线上,?b – ak > 0 .
,,,,,,,,2OPOR ?无论P点在什么位置,总有|| = |?| . 4分 OQ
222ab(1,k)(2)由条件得:= 4ab, 2分 222b,ak
24b,ab172即k = > 0 , ? 4b > a, 得e > 2分 24ab,4a