2023年10月四川绵阳高三一诊理数答案

绵阳市高中2021级第一次诊断性考试理科数学参考答案及评分意见一、选择题：本大题共12小题，每小题5分，共60分．BCDACADBBDCC二、填空题：本大题共4小题，每小题5分，共20分．13．714．2215．916．−1三、解答题：本大题共6小题，共70分．217．解：（1）由a1，a2，a4成等比数列，则a2=a1a4，································2分2∴(a1+2)=a1(a1+6)，可解得a1=2，·················································································3分n(n−1)2∴数列{an}的前n项和S=na+d=n+n；······························5分n12（）an2nn①，分2bn+bn+1=(2)=(2)=2···············································6当n=1时，b1+b2=2，可得b2=1,·······················································7分n+1可得bn+1+bn+2=2②，····································································8分n+1nn由②式－①式，得bn+2−bn=2−2=2，············································9分∴b2n=(b2n−b2n−2)+(b2n−2−b2n−4)+(b4−b2)+b22n−22n−42分=2+2+2+1·····································································114(1−4n−1)=+114−41n−=．························································································12分38318．解：（1）∵T==，则=，·····················································1分38又f()=tan(+)=1，||，·························································2分382∴=，·······················································································4分83∴f(x)=tan(x+)；······································································5分8833（2）由题意，g(x)=tan(x++)，··················································6分888∵−f(0)=−tan=tan(−)································································7分8833∴由g()=−f(0)，得tan(++)=tan(−)····································8分43288837∴+=−+k，kZ，·····························································9分8328118k∴=−+，kZ，又0，····················································10分1237∴的最小值为．········································································12分419．解：（1）∵fx()(2=xmxm2+)(−+2)=2x3−2(m−2)xmxmm2+−(−2)为奇函数，−2(m−2)=0∴，解得：m=2．··························································5分−mm(−2)=0（2）当m>0时，2x2+m>0，∴函数f(x)=(2x2+m)(x−m+2)不可能有两个零点．······························6分m当m<0时，由fx()=0，解得：x=−或m−2，·······························7分2m要使得f(x)仅有两个零点，则m−2=−−，·········································8分2即2mm2−7+8=0，此方程无解．故m=0，即f(x)=+2x324x，······························································9分令h(x)=f(x)−3=2x32+4x−3，则h(x)=6x2+8x=2x(3x+4)，44hx()0，解得：x0或x−，hx()0解得：−x0，3344故hx()在()−，−，(0，+)上递增，在(−，0)上递减，··························10分33417又h(−)=−0，327故函数y=−f(x)3仅有一个零点．·······················································12分20．解：（1）∵cos(C−B)sinA=cos(C−A)sinB∴(cosCcosB+sinCsinB)sinA=(cosCcosA+sinCsinA)sinB································2分∴cosCcosBsinA=cosCcosAsinB······························································3分又∵△ABC为斜三角形，则cosC≠0，∴cosBsinA=cosAsinB，·······································································5分∴sin(A−B)=0，又A，B为△ABC的内角，∴A=B；·························································································6分a（2）由△ABC的面积S=，211∴S=absinC=，则bsinC=1，即=sinC，········································7分2b1由S=acsinB=，则csinB=1，即=sinB，·········································8分c由（1）知A=B则a=b，11∴−=sin2B−sin2C，····································································9分ca22又sinC=sin(A+B)=sin2B，∴=sin2B−sin22B=sin2B−4cos2Bsin2B=sin2B−4(1−sin2B)sin2B···············10分令sin2B=t，令f(t)=t−4(1−t)t=4t2−3t，又因为0＜sin2B<1，即0＜t<1，39∴当t=时，f(t)取最小值，且f(t)min=−，··········································11分816119综上所述：−的最小值为−.······················································12分ca221621．解：（1）当a=2时，f(x)=(lnx−2x+2)lnx，1lnx−2x+2−2(x−1)(lnx+1)f(x)=(−2)lnx+=，··································2分xxx11令fx()0得：x1；令fx()0得：0x或x1，······················3分ee11∴fx()的单调递减区间为：(0，)和(1，+)；单调递增区间为：(1)，．····5分eeex（2）fxxaxa()≤−+−2等价于e(ln)(ln1)0xx−ln2−−+−−xxaxx≥（*）········6分xx−1令tg=x=x−x()ln，则gx()=，x∴gx()在(0，1)上递减，在上递增。∴的最小值为g(1)1=，即：t≥1，·················································8分（*）式化为：e(t1−)+t0a2−t≥，当t=1时，显然成立．te2−tte2−t当t1时，a≥，令h(tt)(1=)，则a≥ht()，····················9分t−1t−1max−(tt−2)(et−)ht()=，当时，易知et−t0，(t−1)2故易得：h(t)在(1， 2)上单调递增，在（2，+）上单调递减，······················10分2∴h(t)max=h(2)=4−e，····································································11分∴实数a的取值范围为：ae≥4−2．·····················································12分1x=t+①22．解：（1）曲线C的参数方程为C：t（t为参数），111y=t−②t2222由①−②得C1的普通方程为：x−y=4；····································2分x=2+2cos曲线C2的参数方程为C2：（为参数），y=2sin22所以C2的普通方程为：(x−2)+y=4；··········································4分2222k（2）曲线C1的极坐标方程为：cos−sin=4(+)，··············5分424∴2=，···············································································6分cos2=6由得：=22，4A2=cos2∴射线：=(0)与曲线C1交于A(22，)，···································7分662222曲线C2的极坐标方程为cos+sin−4cos=0=4cos，=，由6得：B=23，=4cos，∴射线：=(0)与曲线C2交于B(23，)，··································9分661则S△PAB=S△POB−S△POA=−||()sinOP=3−2．·····················10分26BA2x+8，x5，23．解：（1）f(x)=3x+3−x−5=4x−2，−1x5，····································1分−2x