2023年10月四川绵阳高三一诊文数答案

绵阳市高中2021级第一次诊断性考试文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．BBCADBACBCBC二、填空题：本大题共4小题，每小题5分，共20分．13．714．515．[1−)+，16．1三、解答题：本大题共6小题，共70分．17．解：（1）由S1，2S2，3S3成等差数列，则4S2=S1+3S3，得3a3=a2，···············3分1∴数列{an}的公比q=，···································································4分3n−14−n由a1=27，数列{an}的通项公式an=a1q=3；································6分4−n（2）令bn=log3an，则bn=log33=4−n，·······································8分∴当n4时，bn0，·······································································9分∴当n=3或4时，Tn取得最大值：T3=T4=3+2+1=6．··························12分18．解：（1）∵f()=tan(+)=1，38∴+=+k，而||，···························································2分8423∴=，即f(x)=tan(x+)，························································3分8888∴fx()的最小正周期为：T==；·················································4分333（2）由题意，g(x)=tan(x++)，··················································5分888∵−f(0)=−tan=tan(−)，8833∴由g()=−f(0)，得tan(++)=tan(−)，·································7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）∵fxxmxmxmxmxm()(2)(2)=22(2)(2)=+−+−−+−−232m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fxxmxm()(2)(2)=+−+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（2）在△ABC中，由（1）知，a=b，bc1sinC由正弦定理=，则=，··············································7分sinsBCinbcBsin1又=sinB，即cBsin1=，c11∴===+=sinsin()sinCABB2，ab11∴−==sin2B−sin22B，·······························································9分c2a211∴−=sin2B−sin22B=sin2B−4cos2Bsin2B=sin2B−4(1−sin2B)sin2B，···········10分c2a2令sin2B=t，令f(t)=t−4(1−t)t=4t2−3t，·····················································11分又因为0＜sin2B<1，即0＜t<1，39∴当t=时，f(t)取最小值，且f(t)min=−，816119综上所述：−的最小值为−．··················································12分c2a21621．解：（1）方法一：f(x)=ex−1−x2+ax−a，·························································1分因为fx()在(1，+)上单调递增，∴fx()≥0恒成立，x21−ex−故：当x1时，a≥恒成立．····················································3分x−1x21−ex−设g(x)=(x1)，则a≥g()x，x−1max−(xx−2)(ex−1−)则gx()=，(x−1)2易知exx+1，所以ex−1x，故令g(x)0得到：1x2；令g(x)0得到：x2．∴gx()在(2，+)上递减；在(1，2)上递增．···········································5分故：当x1时，gmax(x)=g(2)=4−e．∴实数a的取值范围：a≥4e−．·························································6分方法二：f(x)=ex−12−x+ax−a，因为fx()在(1，+)上单调递增，所以fx()≥0恒成立，xa2x−+a等价于：−10≤在[1，)+上恒成立，·······································2分ex−1xaxa2−+设gxx()1(1)=−，则gx()0≤，ex−1max−(x−a)(x−2)gx()=，ex−1当a=2时，gx()0，∴gx()在上递减，gxmaxg()(1)==0，符合题意．··························3分当a2时，易知在(1，2)上递减，在(2，)a上递增，在（2,+）上递减，因为g(1)0=，a故只需满足ga()=−10（由exx+1易得），符合题意．··················4分ea−1当1a2时，易知在(1，a)上递减，在(a，2)上递增，在上递减，4−a因为g(1)=0，故只需满足g(2)=−10，即4−ea2，e当a1时，易知在(1，2)上递增，在（2，+）上递减，·······················5分4−ag(x)=g(2)=−10，不符合题意．maxe综上：实数a的取值范围：a≥4e−．···················································6分（2）的极值点个数等价于fx()的变号零点个数，x2−+axa令gx()=−1，则等价于gx()的变号零点个数，··························7分ex−1当x→−时，gx()→+；当x→+时，g(x)→−1，−(x−a)(x−2)由（1）可知gx()=，g(1)=0，ex−1当a=2时，易知在（−，+）上递减，故有唯一变号零点1；·····8分当a2时，易知在（−，2）上递减，在（2，a）上递增，在上递减，a因为gg(2)=(1)0，ga()=−10，故有唯一变号零点1；ea−1当a2且a1时，易知在()−，a上递减，在(a，2)上递增，在（2，+）上递减，···········································································································9分a4−ag(a)=−10，g(2)1=−，ea−1e若g(2)≤0，即4e−2≤a时，有唯一变号零点1；··································10分若g(2)0，即ae−4且a1时，gx()有三个变号零点1，x2，x3，且12xx23。当a=1时，易知gx()在(1)−，上递减，在(1，2)上递增，在(2，+)上递减，····················································11分3由于g(1)0=，g(2)1=0−，有唯一变号零点x，且x2.e00综上：当a4−e且a1时，fx()有三个极值点；当或a≥4e−时，有唯一极值点．············································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