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2023~2024学年度第一学期期中学业水平诊断高三数学参考答案一、选择题：1.B2.A3.D4.A5.D6.A7.C8.C二、选择题9.ACD10.AB11.BC12.BCD三、填空题13.2614.815.

7416.(5,4)−−四、解答题17.解：（1）由题知，π22T=，所以，2ππT==，所以，2=.·················2分所以，π()2sin(2)4fxx=+.···························

································3分所以，πππ2π22242kxk−++，即3ππππ88kxk−+，·················4分故()fx的单调递

增区间为3ππ[π,π]()88kkk−+Z.································5分（2）将函数()fx图像上所有点横坐标伸长到原来的2倍（纵坐标不变），得π2sin()4yx=+，再向右平移π4个单位长度，得()2singxx=.···············6分

所以()2sin(sincos)hxxxx=+2π22sin2sincossin(2)42xxxx=+=−+，···································8分因为π02x，ππ3π2

444x−−，所以ππ242x−=，3π8x=时，()hx取得最大值为212+.·························································································10分18.解：（1）当1n

=时，2112aa=，则10a=或12a=，因为11a，所以12a=；································································2分当2n时，22112122nnnnSanSan−−=+−=+−

，两式相减得，22121nnnaaa−=−+，即221(1)nnaa−=−，因为1na，所以11nnaa−=−，即11nnaa−−=，·····4分故数列{}na是以2为首项，1为公差的等差数列.···································5分（2）由（

1）知，2(1)11nann=+−=+，所以12,1,(2)nnnbnnn+=+为奇数为偶数，··················································

·········7分21232nnTbbbb=++++1321242()()nnbbbbbb−=+++++++242111(222)()24462(22)nnn=++++++++4(14)1111111[()()()]1422446222nnn−

=+−+−++−−+·······················10分所以，1244344nnnTn+−=++.··············································

········12分19.解：（1）由题知，每年的追加投入是以80为首项，14155−=为公比的等比数列，所以，41()4580400400()4515nnna−==−−；································

··············3分同理，每年牧草收入是以60为首项，15144+=为公比的等比数列，所以，51()5460240()2405414nnnb−==−−.······························

··················6分（2）设至少经过n年，牧草总收入超过追加总投入，即0nnba−，即5454240()240(400400())240()400()64004545nnnn−−−=+−，·······

·····8分令4()(01)5ntt=，则上式化为2404006400tt+−，即25830tt−+，·······························································

················9分解得305t，即43()55n，所以，43lglg55n，即3lglg3lg5lg3lg2152.24lg4lg53lg21lg5n−+−==−−，所以3n.·························11分所以，至少经过3年，牧草总收

入超过追加总投入.····································12分20.解：若选①：（1）由正弦定理得，3sinsin3sincosBCAC=+,···············1分因为sinsin

()BAC=+,所以3sin()sin3sincosACCAC+=+，即3cossinsinACC=，又因为(0,π)C，sin0C，·······················3分所以1cos3A=.····················

··························································4分（2）在ABC中，1cos3A=，则22sin3A=，sinsin()sincoscos

sin221sinsinsin3tan3bBACACACcCCCC++====+.···········6分因为ABC是锐角三角形，所以π02π02BC，即π2π02ACC+,即ππ022CA−，所以πsin()πcos22tantan()π

2sin4cos()2AACAAA−−===−，所以1022tanC，···································································7分所以1(,3)3bc

.··············································································8分设btc=，则221122222bcbctbccbt+=+=+，令1

22tyt=+，1(,3)3t，则222111222tytt−=−=，令0y=，则1t=，则y在1(,1)3上单调递减，在(1,3)上单调递增，··································10分所以1

151223tt+，即222bcbc+的取值范围为5[1,)3.··························12分若选②：（1）因为2222()Sabc=−−，所以22()220bcaS−−+=，所以22222sin0bcabcbcA+−−+=，···········

································1分所以2cos22sin0bcAbcbcA−+=，所以sin22cosAA=−.··················································

···········3分又22sincos1AA+=，解得1cos3A=或cos1A=（舍），所以1cos3A=.·····································································

·········4分（2）在ABC中，1cos3A=，则22sin3A=，sinsin()sincoscossin221sinsinsin3tan3bBACACACcCCCC++====+，··········6分因为ABC是锐角三角形，所以π02π02BC

，即π2π02ACC+，即ππ022CA−，所以πsin()πcos22tantan()π2sin4cos()2AACAAA−−===−，所以1022tanC，·············

······················································7分所以1(,3)3bc.··············································································

8分设btc=，则221122222bcbctbccbt+=+=+，令122tyt=+，1(,3)3t，则222111222tytt−=−=，令0y=，则1t=，则y在1(,1)3上单调递减，在(1,3)上单调递增，···········

·······················10分所以1151223tt+，即222bcbc+的取值范围为5[1,)3.··························12分若选③：（1）由正弦定理得，coscos()42cossinaAaBCbAC+−=，

····1分因为coscos()ABC=−+，所以cos()cos()42cossinaBCaBCbAC−++−=，所以2sinsin42cossinaBCbAC=，所以2sinsinsin42sincossinABCBAC=.······

····························3分又因为,(0,π)BC，sin0,sin0BC，所以sin22cos0AA=，又22sincos1AA+=，解得1cos3A=.····

··········································4分（2）在ABC中，1cos3A=，则22sin3A=，sinsin()sincoscossin221sinsinsin3tan3bBACACACcCCCC++==

==+，··········6分因为ABC是锐角三角形，所以π02π02BC，即π2π02ACC+，即ππ022CA−，所以πsin()πcos22tantan()π2si

n4cos()2AACAAA−−===−，所以1022tanC，···································································7分所以1(,3)3bc.·························

·····················································8分设btc=，则221122222bcbctbccbt+=+=+，令122tyt=+，1(,3)3t，则222111222tytt−=−=，令0y=，则1t=，则y在1(,1

)3上单调递减，在(1,3)上单调递增，··································10分所以1151223tt+，即222bcbc+的取值范围为5[1,)3.·······················

···12分21.解：（1）由题知，()(1)(e)xfxxa=+−，············································1分所以，当0a时，e0xa−恒成立，所以，令()0fx

=，解得1x=−.所以，当(,1)x−−时，()0fx，()fx在(,1)−−上单调递减；当(1,)x−+时，()0fx，()fx在(1,)−+上单调递增；·················3分当0a时，令()0fx=，解得1x

=−或lnxa=，所以，当ln1a−，即1ea时，(1,ln)xa−时，()0fx，()fx在(1,ln)a−上单调递减，当(,1)(ln,)xa−−+时，()0fx，()fx在(,1)−−和(ln,)a+上单调递增；·······

·················································································4分当ln1a−，即10ea时，(ln,1)xa−时，()0fx，()fx在(ln,1)a−上

单调递减，当(,ln)(1,)xa−−+时，()0fx，()fx在(,ln)a−和(1,)−+上单调递增；············································································

···············5分当ln1a=−时，()0fx在(,)−+上恒成立，所以，()fx在(,)−+上单调递增.·····················································6分（2）由（1）知，当1a时，()fx在(

1,ln)a−上单调递减，在(,1)−−和(ln,)a+上单调递增，且当x→−时，()fx→−，当x→+时，()fx→+，所以，若方程()fxb=始终有三个不相等的实根，则(ln)(1)fabf−，即21(

ln)22eaaab−−在(1,)a+上恒成立.········································8分当1a时，显然1112e2ea−−.···········

················································9分令2()(ln)2agaa=−，则21()(ln)ln2gaaa=−−，因为1a，所以，ln0a，所以，21()

(ln)ln02gaaa=−−恒成立，所以，()ga在(1,)+上单调递减，所以，()(1)0gag=.··············································································11分综上，若方程()fx

b=始终有三个不相等的实根，b的取值范围为1102eb−.····························································12分22.解：（1）由题得，2ln(),(0)xxafxxx−−=，··············

·····················1分令()ln,(0)hxxxax=−−，则函数()fx有两个极值点，即方程()0hx=有两个正实数根.·········································································

······················2分因为11()1xhxxx−=−=，所以当(0,1)x时，()0hx，()hx单调递减，当(1,)x+时，()0hx，()hx单调递增，所以，min()(1)1hxha==−，且

当0x→时，()hx→+，x→+时，()hx→+.······································································4分所以，方程()0hx=有两个正实数

根，只需(1)10ha=−，解得1a，···················································································5分即函

数()fx有两个极值点时，a的范围为(1,)+.································6分（2）若12xx且123xx，则令21(1,3]xtx=，由（1）知，12()()0hxhx==，即1122lnlnaxxxx=−=−，则2211ln

lnxxxtx−==，即11lntxxt−=，解得，1ln1txt=−，所以，21ln1ttxtxt==−.···················8分所以，1212(1)lnlnln21lnln11tttttttxxaxxt=++++=−=−+−，·················9分令(1)ln

,(1,1(3)]ttttt+−=，则()2211(ln)(1)(1)ln2ln(1)()1tttttttttttt++−−+−+−=−−=，·························10分令12ln()

tPttt−+−=，则22221(1)10,(1,)3(]tttttPt−=−++=所以函数()Pt在(1,3]上单调递增，且(1)0P=，所以，()0Pt，··········11分所以，当(1,3]t时，()0t，所以，()t在(1,3]上单调递增，所以，

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