推荐-重庆市南开中学2018—2018学年度第一学期高三年级月考数学理科试题 精品 - 下载本文

p2pp22 (2)|DE|?x?2?2xcos60??x?2?p

xxx22p2p2p2?p,由(t?)??1?2知, 令x?t,t?[1,4],|DE|?t?ttx22p2 t?在(0,p)单减,(p,??)单增

t当4?p时,|DE|?2pp?p?4,此时t?4即x?2,y? 42故D点与B点重合,E为AC中点; 当1?p?4时,|DE|2?p,此时t?p,即x?故D,E两点均在距离A点

p,y?p,

p米处

当0?p?1时,|DE|2?p2?p?1此时t?1,即x?1,y?p 故D点为AB中点,E点与C点重合

a?4x?a?2a?2x?a?2,得f(x)?. 20.解:(1)由f(2x)?xx4?b2?b?f(x)是R上的奇函数,?f(0)?2a?2?0,得a?1. 1?b2x?11?x,得f?1(x)?log2. 又f(?1)?f(1) ?b?1 ?f(x)?x1?x2?1由此得2?x1?y?0,??1?y?1. 故反函数f1?y?1(x) 揎义域为(-1,1)

(2)当x?[,]时,f12?1(x)?g(x)恒成立,

231?x1?x1?x1?x2?log2?log2,即?()

1?xk1?xk1?x12?0,x?[,],?1?x?0,1?x?0,且k?0,?k2?1?x2,令h(x)?1?x2 由k23则h(x)min?h()?23555,?k2?,故0?k?. 993?2b2x2?1?a?2??y2?1 21.解(1)由条件得?a,所以方程??4?b?1?2b?a? (2)易知直线l斜率存在,令l:y?k(x?1),A(x1,y1),B(x2,y2),E(?4,y0)

?y?k(x?1)?2222由?x2?(1?4k)x?8kx?4k?4?02??y?1?48k24k2?4x1?x2??,x1x2? 221?4k1?4k由AQ??QB?(?1?x1,?y1)??(x2?1,y2)即?由

??48k2?16?0

??(x1?1)??(x2?1)(1)

?y1???y2??(x1?4)??(x1?4)(2)AE??EB?(?4?x1,y0?y1)??(x2?4,y2?y0)即?

y?y??(y?y)120?0由(1)???x1?1x?4 ,由(2)???1x2?1x2?4??????(x1?1)(x2?4)?(x1?4)(x2?1)2xx?5(x1?x2)?8 ??12(x2?1)(x2?4)(x2?1)(x2?4)8k24k2?4,x1x2?将x1?x2??代入有 221?4k1?4k8k2?840k28k2?8?40k2?8?32k2??8221?4k1?4k2??????1?4k???0

(x2?1)(x2?4)(x2?1)(x2?4)22.解:(1)f[f(x)]?f(x)?c?(x?c)?c

222f(x2?1)?(x2?1)2?c?(x2?c)2?c (2)易得A点为(0,1)

?c?1?f(x)?x2?1

f(ai)?1ai2f(ai?1)?f(ai)ai?1?ai2?pi???ai?id,qi???(2i?1)d

aiaiai?1?aid?bn?qn?pn?(n?1)d,bn?bn?1?d ?{bn}也为等差数列

(3)当n?m时,设n?m?k(k?N)

2cn?1bn?1a2bn?1n?22dn?2m?1m?k?2m?1m?2m?1 ??a???????1 2bncnn?1n?1m?2m?k?1m?2m?1m?2bna

?cn从第m项开始递减

当n?m时,设n?m?k(k?N,k?1)

cn?1bn?1a2bn?1 ?? 2bncnbna

n?22dn?2mm?k?2mm?k?2?(k?1)ma???????1 n?1n?1m?1m?k?1m?1m?k?1?(k?1)m?1 ?cn从1到m项递增, ?cn有唯一最大项cm

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