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2020高三摸底数学试题参考答案一、单选题1-4CBAC5--8CBBA二、选择题9.AD10.BCD11.AD12.BCD三、填空题13.43−;14.3;15.116.-12四、解答题17.解:(1)1tan,0cossin==−=⊥xxxn
mnm,……………(3分)由2,0x,所以4=x…………(4分)(2)=−==12cossincosxxnmnm)4sin(−x………………………(6分))22,22(cos,44420−−−xx………………………(8分)(
),0,由于余弦函数图象得43,4………………………(10分)18.(1)解:设等比数列na的公比为q选①;0)1)(2(,6)1(223213=−+=++=++=qqqqaaaS,()122,2,1−−=−=nnaqq()nnna211−=−…
………………(5分)选②;711)1)(1(1111,133333636−=+=−−+=−−−−=qqqqqqqqSSq,()122,2−−=−=nnaq,()nnna211−=−……………………(5分)选③;042,42,42341121523=+−+=+=qqqaqaqa
aaa,0)22)(2(2=+−+qqq,()122,2−−=−=nnaq()nnna211−=−………(5分)(2)()nnnnnnab211=−=−……………………………………(6分)132
32122)1(222122232221++−+++=++++=nnnnnnnTnT………………………(9分)两式相减得,22)1(22222222111132+−=−−=−++++=−++++nnnnnnnnTnnT………(12分)19.解:(
1)0sinsinsincoscos222=++−CBCBA,0sinsinsin)sin1()sin1(222=++−−−CBCBA0sinsinsinsinsin222=++−CBCAB,……………(2分
)0222=++−bccab,……………(4分)212cos222−=−+=bcacbA32=A……………………………………(6分)(2),3432sin22,2,32====RaA分)9(………………)3sin(34)cos23si
n21(34)sin21cos23(sin34))3sin((sin34)sin(sin2+=+=−+=−+=+=+BBBBBBBBCBRcb1)3sin(23,3233,30++BBB,3342+
cb33424+++cba,即ABC周长的取值范围为+33424,。………(12分)20.解:(1)由题意得,,7500%)501(50001tta−=−+=tataannn−=−+=+23%)
501(1.…….……(2分)当时即037500225001−=−ttat232323221=−−=−−+tatatatannnn…………(4分)tan2−是以tta3750021−=−为首项,23为公比的等比数列。…………
…(5分)当时即0225001=−=tattan2−不是等比数列……………(6分)(2)当1500=t时,由(1)知,12330003000−=−nna……………(8分)210003000)23(30001+=−mm
a,即6231−m,……………(10分)法一:易知xy=23单调递增,又632243)23(,61681)23(54==,51−m,6m,m的最小值为6。…………………………(12分)法
二:42.41761.07781.03010.04771.04771.03010.02lg3lg3lg2lg23lg6lg6log123=−+−+==−m,6m,m的最小值为6。…………………………(12
分)21.解:(1)由题意得2,12==pp,抛物线的方程为xy42=。………………(4分)(2)设直线MN方程为:),(),,(2211yxNyxMcmyx,+=,联立=+=xycmyx42得0442=−−cmyy,−==+
cyymyy4402121……………(6分)以MN为直径的圆过点P,1,2−==NPMPkkMPN………………(7分)2414212121111+=−−=−−=yyyxykMP,同理242+=ykNP……………………………(8分)1242421−=
++yy,即0164)(22121=++++yyyy,52,02084+==++−mcmc……………………………(10分)验证0]4)1[(16)52(16)(16222++=++=+=mmmcm5)2(52++=++=+=ymmmycmyx,直线MN经过定点()2,5−。
…………………………………………………(12分)22.解:(1)当2=时,xexfxln2)(2−=,2)1(ef=,22)1(,22)(2'2'−=−=efxexfx,切线方程为)1)(22(22−−=−xe
ey,即02)1(222=+−−−eyxe………………………………………(3分)(2)当1=时,xexfxexfxx2)(,ln2)('−=−=,易知)('xf在()+,0单调递增,且02)1(,04)21(''−=−=efef,)('
xf存在唯一零点1,210x,020xex=满足且当()0,0xx时,)(,0)('xfxf单调递减,当()+,0xx时,)(,0)('xfxf单调递增。对020xex=两边取对数,得:00ln2lnxx−=02ln242
ln22222ln222ln2)()(000000min0−=−−+=−==xxxxxexfxfx)(xf无零点。………………………………………(7分)(3)由题意得,xxxex−−2
ln2,即22lnxxxex++,即2lnln2xexexx++,易知函数xeyx+=单调递增,2lnxx,…(9分)xxln2,令xxxhln2)(=,则2'ln22)(xxxh−=,令0)('=xh得ex=,列表得,eeehxh22)()(max
==,.………………………………………(12分)x()e,0e()+,e)('xh+0−)(xh单调递增极大值单调递减