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2021年“山东学情”高三10月联合考试数学试题(B)答案1.答案:C解析:由0562xx可知0)5(1xx)(,从而得到A=51,.而B=543,,,两者取交集,选C.2.答案:C解析:设2)1(,2)(,6)(23fd
dfddfV3.答案:B解析:由条件可知:ABCCBsinsinsincossin,即)(CBBCCBsinsinsincossinCBBCsincossinsin0sinCBBcossin1tanBB为ABC
内角4B22,2445sin4sin2RBbR4.答案:B解析:令.0,2ttx9282xx即为,98tt得到0892tt得108tt或.解出.03xxx或的范围是从而选B.5.答案:A解析:函数()sin(2)
6fxx的图像向右平移0个单位后得函数=sin2+6gxxsin226x,因为gx为奇函数,所以00g,即sin206,解得=+122k,kZ.6.答案:
A解析:,72272)(744717aaaaS66111111122112)(11bbbbT3373111711764117baTS7.答案:C解析:函数)0()1(222mmnxmnmxmxy是开口向上且关于直线1x对称的二次函
数;函数)0(11peepyxx关于直线1x对称,且在)1,(上单调递增,在),1(上单调递减,方程xxeepnmxmx1122的解的情况可能是无解,一解,两解,且解关于1x对称.8.答案:B解析:由题意得21()efxa-=有8个不同的实数解
,又()fx是定义在,,00上的偶函数,由()fx得图像得22211142eeea---<<,解得24a<<9.答案:ABD解析:A.函数yfx定义域为R,关于原点对称.2200xxfxxx<,由图象可知,函数yfx既是奇函数又是增函数;B.函
数ygx定义域为R,关于原点对称.122xxgx,显然,函数ygx既是奇函数又是增函数;C.函数yx定义域为,00∪,+,关于原点对称.由图象可知,函数不关于原点中心对称,函数yx不满足既是奇函数又是增函数;D.函数yhx定义域为R,关于原
点对称.因为22111222log1log1log10hxhxxxxx,yhx为奇函数,又1221log1hxxx在0,+单调递增,
00h且函数在定义域上图象连续,函数yhx在R上单调递增,函数yhx既是奇函数又是增函数.10.答案:AD解析:由条件可知:BAABcossin5sinsin20sinABBcos5sin2又1cossin22BB32cos,35s
inBB,在ABC中,BBCABBCABcos2AC222得6BC.A.666222362442ABBACcos222ACABBCAC;B.52356221sin21BBCABSABC;C.由角平分线性质可知:31BC
ABECAE26AE.234cos2BE222AAEABAEAB215)66(2622230BE.D.在ABD中,94cos2AD222BBDABBDAB5323225AD.11.答案
:AB解析:A.由M、N关于C对称可得0,3-C,则2)65(32T,T,选项A正确;B.由A知,2,图象过0,3-C,可得3,)3-2sin(xAyZ3,33342,则图象关于点)(0,3
4成中心对称,选项B正确;C.函数单调递减,Zkkxk2323222,即Zkkxk1211125,选项C错误;D.函数)(xf在64x上,则03265x,则值域为0,A,选项D错误.12.答案:ABC解
析:2111111)(.Annnnnnnnnnnnnaaaaaaaaaaaaa,故A正确;2022202120212020202220214354324321322122021232221.Baaaaaaaaaa
aaaaaaaaaaaaa故B正确;123234211112.Caaaaaaaaaaaaaaannnnnnnnn,,,迭加得112024202222maSSaann故C正确
;22543222642.Dnnaaaaaaaaaa11212112212nnnnaSaSaa故D错误.13.答案:38解析:由条件可知:01-cos-cos22CC,可得21-cosC或1cosC(舍)C为ABC内角
32C32,cos2222abCabbac38sin21CabS14.答案:21解析:因为(1)(2)fxfx-=+,所以()fx的对称轴为32x=,所以5541()()()3332fff-=-=-=-15.答案:231011
解析:当2n时113nnnaa,,311nnaa又当1n时213aa23a{}na的奇数项是以11a为首项,以3q为公比的等比数列,偶数项是以23a为首项,以3q为公比的等比数列
,2021132021242020++Saaaaaa……10111010133(13)13131011101131332210113216.答案:),3(解析:当xxfexln1)(,0
,切点,1,1)(),ln1,11'11xkxxfxx(切线,)(1ln1111xxxxy即11ln21xxxy1ln2OxP当1ln)(,2xxfexe,切点,1)(),1ln,'22xxfxx(,122xk切线,)(11l
n222xxxxy即22ln21xxxy22ln22lnOQxx切线互相垂直,11112121xxxx,2221ln2ln2ln2ln2OQOPxxxx,令2,1,l
n2txt2,1,24124)2(22)(tttttttf)(tf在2,1上单调递增,),(),,(3OQOP3)(tf17.解析:(1)数列}{nb为等比数列412bbq
142nnb………………2122log42loglog122122nbannnn………………5(2))121121(241nnaacnnn……………71222)]1211
21()7151()5131()311[(2nnnSn124nn……………1018.解析:(1)………………3结合正弦函数的图象与性质,可得当222242kxk即388kxk时,函数单调递增,……………………4
所以函数yfx的单调递增区间为388kkkZ,……………………………5(2)①令24tx,当53,248x时,,6t,111sin-,242yt(如图
).……………………7要使yfxm在区间53,24821sin2sincos84421cos2214=sin2222212221cos2sin2cos22442222sin2cos2441sin224fxxxxxxxxxxxx
xy0π2π�1412�π�上恰有两个零点,m的取值范围为102m<..……………………9②设1t,2t是函数1sin2ytm的两个零点(即1124tx
,2224tx),由正弦函数图象性质可知12tt,即122244xx.124xx,…………………………………………11122sin2xx..…………………………………………1219.解析:(1)定义域),0(xxxxxxf)1)
(12(112)(....................................................1令0)(xf得21x列表如下:..............................
......................3所以,)(xf的极小值点是21,无极大值点.....................................................4(2
)xmexxxxgx221ln)(222))(12()(xmexxxxgx.......................5)(xg在),1[上单调递减0)(xg在),1[上恒
成立022xmexx在),1[上恒成立....................6),1[,)(max22xexxmx.......................7令),1[,)(22
xexxxhx.......................8021)(22xexxh在),1[上恒成立)(xh在),1[上单调递减.......................102max2)1()(ehxh
.......................11实数m的取值范围是22em.......................1220.解析:(1)12nnSna当1n时122aa22a...............................
......................1x)21,0(21),21()(xf0)(xf递减极小值递增当2n时12nnSna①12nnSna②.....................
................................2①②得12(1)nnnanana11nnanan(2n)又2121aa满足上式11nnanan..................................................
...3当2n时213212321nnaaaaanan…将这1n个式子相乘得.............................................
........41nananan当1n时11a满足上式nan.....................................................6(2)(1)2nnbn2312232422(1)2nn
nTnn…①234122232422(1)2nnnTnn…②......................................7①②得1322)
1(22222nnnnT......................................82112(12)4(1)212nnn11442(1)2nnn
12nn......................................1012nnTn......................................1221.解析
:(1)在ABC中,由余弦定理得:13120cos31291cos2222BBCABBCABAC60120,0coscosDBDB………………2则在ACD
中,DCDADCDADACcos2222即CDADCDADCDADCDAD260cos21322故13CDAD当且仅当CDAD时,等号成立………………4431343sin21
CDADDCDADSACD故ACD面积的最大值为4313…………………………………………5(2)取BCDBAD,则在ABD中,cos45cos2222ADABADABBD在BCD
中,cos1213cos2222CDCBCDCBBD即cos1213cos452coscos3①………………7取四边形ABCD的面积为S,则有sin21sin21CDCBADABS即Ssin3sin②……………
……………………………8①2+②2得:222coscos3sin3sin4S)cos(610即)cos(662S则当时,3212max2maxSS,……………………………………10
此时,则2cos4coscos3coscos321cos故77cos12132BDBD即对角线BD长为7…………………………………………1222、解析:(1)令)0()(1xxexgx..
..........................................................................11)(1xexg令0)(xg,得1x列表如下:x)1,0(1),1()(
xg0)(xg递减极小值递增01)1()(0minegxg......................................................................................2xex1,当且仅当1x时取等号...
.......................................................................3再令)0(sin)(xxxxh...............
........................................................................40cos1)(xxh在),0(上恒成立,)(xh在),
0(上单调递增0)0()(hxh................................................................................
.......................5)0(sinxxx....................................................................................................6
当0x时,xxexsin1当0x时,0)(xf..................................................................................
...........7(2)由(1)知,当0x时,xex1,即)0(1lnxxx,当且仅当1x时取等号.......8Nn,且2n,1ln22nn............................................
................................9211lnnnn..........................................................................................
..............114)1321(21)1ln(22nnniini.........................................................
.......12当Nn,且2n时,4)1ln(22nniini.........................................................12