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四川省南充高中2020-2021年度高三上期第四次月考(理科答案)一、选择题ABBDBCDACDBA二、填空题13.0x且0y14.2,015.4016.362三、解答题17.(每小题各6分)解:(1)()()062sin22cos2s
in3−=−=xxxxf则21222===T(2)由(1)知()−=6sin2xxf,则()−=62sin2xxg当125,6x时,()2,11,2162sin32,
662−−xgxx18.(第(1)小题4分,第(2)小题8分)解:(1)证明:AB是半圆O的直径,则BCAC⊥又5=AB,3=BC,则4=AC在ABC中,ACPAP
CACPA⊥=+222又ABPA⊥,AACAB=,故⊥PA平面ABC,从而BCPA⊥又ACBC⊥,APAAC=,故⊥BC平面PAC(2)建立如图所示的空间直角坐标系xyzC−,则()0,0,3B,()0,4,0A,
()3,4,0P()0,4,0=CA,()3,4,0=CP,()0,0,3=CB,点E是线段PB上靠近B点的三等分点则()=−+=+=1,34,21,34,10,0,331BPCBCE设(
)zyxm,,=是平面ACE的一个法向量,则=++===034204zyxCEmyCAm,令1=x得()2,0,1−=m设直线PC与平面ACE所成的角为,则2556556,cossin===mCP19.(每小题各4分)解:(1)由已知0na,()2
1+=nnnaaS,令1=n得:()12111111=+==aaaaS令2=n得:()221122222=+=+=aaaaS,令3=n得:()2212133333=+=++=aaaaS由此猜想nan=(2)由(1)知1=n时,na
n=成立假设()*,1Nkkkn=有kak=则1+=kn时,22222121212111kkaaaaaaSSakkkkkkkkk+−+=+−+=−=++++++即()()()10101111121+==+−+=+−−+++++kakakakkaakkkkk则1+=kn时nan
=成立,故nan=(3)471121=a成立当2n时,()()+−−=+−=−==12112121212414444112222nnnnnnnan473511312112112171515131
2111112232221+−+=+−−++−+−+++++nnnaaaan故4711112232221++++naaaa对任意*Nn恒成立.20.(第(1)小题4分),第(2)小题8分)解:(1)
由已知242==aa,又121===cace,则3222=−=cab故椭圆E的标准方程为13422=+yx(2)设()11,yxA,()22,yxB,直线mkxyl+=:,带入椭圆E的方程消去y得()0124843222=−+++mkmxxk则221438kkmxx+
−=+,222143124kmxx+−=,且()()2222224303431664kmmkmk+−+−=若OQBOQA=,则0444422112211=−++−+=−+−=+xmkxxmkxxyxykkQBQA即()()()()0441221=−++−+xmkxx
mkx()()()()08434843124208422222121=−+−−+−=−+−+mkkmkmkmkmxxkmxkx,整理得km−=满足2243km+则直线()1:−=−=xkkkxyl恒过定点()0,121.(每小题各4分)解:(1)当1=a时,()()
1ln1−+=xxxf,则()()2'''11lnxxxfxxxf−=+=当()1,0x时,()0''xf,()+,1x时,()0''xf()xf'在()1,0上单调递减,在()+,1上单调递增,故()()011''=fxf则函数()xf的单调递增区间是()+,0,无单调递
减区间.(2)当0=a时,()xxfln=在()+,0上单调递增,满足题意当0a时,()xf在()+,0上单调递增,则()0'xf(不连续等于0)恒成立()()2'''11lnxaxxfxxaxf−=+=当0a时,()0''xf,则()xf'在()+,0上单调递减而0111
1'+−=−−aaeef,当+−,1aex时,()0'xf,不合题意当0a时,()xf'在a1,0上单调递减,在+,1a上单调递增要使()0'xf(不连续等于0)恒成立,只需()0ln11ln1
'−=+=aaaaaaf,则ea0综上,实数a的取值范围是e,0(3)由(2)知()xxaxg1ln+=,()()011ln1ln1ln1212221121=−++=+=xxxxaxxaxxaxgxg又121=+xx,则0ln0ln21211212122112=−+
=+−++xxxxxxaxxxxxxxxa令112=xxt,即方程01ln=−+ttta在()+,1上有解.解法一:令()()++−=,1,1lnttttath,则()tttatattth
+−=−+−=1122',()21,1++ttt当2a时,()()thth0'在()+,1上单调递减,又()01=h,不合题意当2a时,当−+24,12aat时,()0'th;当+−+
,242aat时,()0'th;则−+24,12aat时,()()01=hth,而()()21122−+−+=aeaeeaehaaaa令()()212−+=xexxx,则()()022'''−=−=xxexexx,()x'在()+,2单调
递减,()()0422''−=ex()x在()+,2单调递减,则()()0522−=ex,即()0aeh故存在−+aeaat,2420,使得()00=th,故2a满足题意.综上,a的取值范围是()+,
2解法二:(分离变量洛必达法则)tttaln1−=,令()()+−=,1,ln1tttttm,则()()()2222222222'ln11ln1ln1ln1tttttttttttttm+−++=−++=令()()++−+=,1,11ln22ttt
ttM,则()()()()tMttttM+−=0112222'在()+,1上单调递增,()()01=MtM则()()tmtm0'在()+,1上单调递增,由洛必达法则有()2111limlim211=+=++→→tttmtt,故a的取值范围是()+,22
2.(每小题各5分)解:(1)由已知设抛物线C的标准方程为()02=aaxy,根据抛物线过点()1,2有2112==aa故抛物线C的直角坐标方程为xy212=由直线l的极坐标方程得010101cos22sin222
=−−=+−=+−yxxy即直线l的直角坐标方程为01=−−yx(2)点21,23P在直线l上,设直线l的参数方程为+=+=tytx22212223(t为参数),代入xy212=得0222
2=−+tt,设A,B两点所对应的参数分别为1t,2t,则2221−=+tt,121−=tt则()2234111121221212121=−+=−=+=+ttttttttttPBPA23.(每小题各5分)解:(1
)()()()133322====+−−+−−=mmmmxmxmxmxxf(2)根据基本不等式abba222+,bccb222+,caac222+有cabcabcba++++222则()()22223cbacba++
++(当且仅当cba==时等号成立)又1=++zyx,故()()()()()3111311122222azayxzayx+=−+−+−−+−+−即()231312−+aa或0a