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资中二中高2022届第五学期9月月考理科数学参考答案题号123456789101112答案AADDCDBACDDA13.814.2−15.116.]22ln1,1[+e17(1)32+e;(2)23.18(1)()2log10x−,得11x−,解得2x,
)2,A=+---------------------------------------------2分对于函数xxg2)(=,该函数为增函数,因为10x,则221x,即()12gx,1,2B
=,----------------------------------------------4分因此,2AB=;-----------------------------------------------------------
----------------------------------------6分(2)CBB=,CB.-------------------------------------------------------
---------------------------------7分当21aa−时,即当1a时,C=,满足条件;------------------------------------------------
-----------9分当21aa−时,即1a时,要使CB,则1212aa−,解得312a--------------------------------11分综上所述,实数a的取值范围
为3,2−.---------------------------------------------------------------------12分19解:(1)因为()2fxxbxc=++为二次函数,且𝑓(2−𝑥)=𝑓(2+𝑥
),所以()fx的图象的对称轴方程为𝑥=2,------------------------------------------------------------------------2分又()fx的图象过点(1,0),故=++=−0122cbb,解得=−=
34cb,----------------------------------------------5分所以34)(2+−=xxxf--------------------------------------
-----------------------------------------------------6分(2)由题意可得)(xf在]4,1[的值域是)(xg在]4,1[的值域的子集-------------------------------------7分)(xf在]4,
1[的值域为]3,1[−;-------------------------------------------------------------------------------------8分①:当04=−a时,即4−=a;3)(−=xg,
不符合题意.------------------------------------------------9分②:当04−a时,即4−a;)(xg在]4,1[上单调递增;)(xg的值域为]134,1[++aa,有+−+313411aa,解得]2,25[−−
a;--------------------------------10分③当04−a时,即4−a;)(xg在]4,1[上单调递减;)(xg的值域为]1,134[++aa;有−++113431aa,得a无解;---------
---------------------------------11分综上a的取值范围为]2,25[−−.----------------------------------------------------------------------------
----------2分20(1)由题知,()()()()()()1111gxfxfxfxfxgx−=−−+=−+−−=−,------------------------4分且满足1010xx−
+,即11x−,故函数()gx为奇函数.------------------------------------------------------6分(2)函数()3logfxx=单调递增,题干中不等式等价于420xxtt−,对任意1,2x恒成立,即2141
222xxxxxtt−=++,对任意1,2x恒成立,--------------------------------------------------------------7分2xy=在1,2上递增,所以122t=.----------------
-----------------------------------------------------------8分令()()1,01xxhxexe=+,1201xx,()()12121211xxxxhxhxeee
e−=+−−()1212121212121xxxxxxxxxxxxeeeeeeeeeeee−−=−−=−,由于xye=在0,1上递增,12120,10xxxxeeee−−,所以()()120hxhx−,故()hx在1,2上
递增,值域为112222,22−−++,所以2211111,222222xx−−−+++,所以1112225t−=+.-----------------------------------------------------11分综上所述,2,2
5t.-----------------------------------------------------------------------------------------------12分21(1)因为函数)(xfy=在)+,0(上单调递减,所
以022ln2)('++=axxxf在)+,0(上恒成立;------------------------------------------------------1分等价于xxxa1ln−−在)+,0(上恒成立;令xxxxh1ln)(−−=,即)(xha
在)+,0(的最小值,.---------------------------------------------------2分222ln1ln1)('xxxxxxh−=+−−=,------------------
---------------------------------------------------------------3分0)('xh,得10x,)(xh单调递减;0)('xh,得1x,)(xh单调递增;)(xh的最小值为1)1(−=h,所以1−a-----
-------------------------------------------------------------------5分(2)令)()()(xgxfxF−=,由0)(xF,得1a
.-------------------------------------------------------6分当1a时,1ln)2(2)(2−+−xxxxF.令1ln)2(2)(2−+−=xxxxm,)12
(ln22)2(ln2)('+−+=+−+=xxxxxxxxm-----------------------------------------------------------8分令)()('xmxt=,0)211(2)(2'++=xxxt,所以)('xm
在)+,0(上单调递增.--------------------------------------------------------------------------------.9分因为0)1('=m,所以1x时,0)('xm,)(xm单调递减;1x时,0)('xm,)(xm单调
递增.故0)1()(=mxm,满足条件;---------------------------------------------------------------------------------11分综上所诉:1a-----------
------------------------------------------------------------------------------------------12分22.(1)22:1,:1043xyClxy+=
−−=;-----------------------------------------------------------------------5分(2)222:(212xtltyt=+=+
为参数),将其代入椭圆方程,有27102402tt++=,,MN对应的参数分别为12,tt,有12122028,77tttt+=−=,--------------------------------------------.8分所以12202|||||
|7PMPNtt+=+=.-------------------------------------------------------------------------10分23.(1)52,2()1,2325,3xxfxxxx−
=−,()5fx等价于5252xx−或2553xx−,解得0x或5x,所以不等式解集为{|0xx或5}x--------------------------------------
-----------5分(2).[1,2]x,()|4|fxx−等价于2||4xxax−+−−,等价于[1,2]x,||2xa−,即[1,2]x,2ax+或2ax−,从而1a−或4a.-----------
-------------------------------------------------------------------------------------10分