2023届绵阳二诊 理科数学答案

绵阳市高中2020级第二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．DACBDABCADCA二、填空题：本大题共4小题，每小题5分，共20分．113．914．15．516．[1，3)3三、解答题：本大题共6小题，共70分．17．解：（1）由3acosCa2sinC3b，可得3sinAcosCasinAsinC3sinB，··················································2分又3sinBACACAC3sin()3(sincoscossin)，····································4分∴asinA3cosA，5分且A，可得a3；·······································································6分311（2）由BAACcbcos(A)cb，可得cb1，······················8分22由余弦定理c2b2a22bccosA4.····················································9分1∵AT()ABAC，21平方可得(AT)2[(AB)2(AC)22ABAC]，··································10分4155即(AT)2(c2b22bccosA)，所以AT．·······························12分44252218．解：（1）因为0.92<0.99，根据统计学相关知识，R越大，意味着残差平方和()yiyˆi1越小，那么拟合效果越好，因此选择非线性回归方程②yˆmxˆ2nˆ进行拟合更加符合问题实际．·························································································4分（）令2，则先求出线性回归方程：，分2uixiyˆmuˆnˆ·································514916250.81.11.52.43.7∵u=11，y1.9，······················7分55n222222，分(uiu)(111)(411)(911)(1611)(2511)=374············9i1n(uu)(yy)ii45.1∴i1，分mˆn0.121···················································102374()uiui1理科数学参考答案第1页共6页由1.90.12111nˆ，得nˆ0.5690.57，·············································11分即yˆ0.12u0.57，∴所求非线性回归方程为：yˆ0.12x20.57．········································12分8219．解：设{}b的公比为q，则bbq2，所以q2，n312732所以q，·····················································································2分32因为{}b的各项都为正，所以取q，···················································3分n32所以b()n．···················································································4分n3若选①：由，得≥，分2anSn1(n1)2an1Sn11(n2)·····························5两式相减得：，整理得≥，分2an2an1an0an2an1(n2)························6因为，所以是公比为，首项为的等比数列，分a110{}an21·····················7∴n1，分an2·····················································································814∴ab()n，··············································································9分nn2314∵y()x在R上为增函数，···························································10分23∴数列单调递增，没有最大值，分{}anbn··················································11∴不存在，使得对任意的，≤恒成立．分mNnNanbnambm···················12若选择②：因为2，且，，an1an1an(n2)a10a20∴为等比数列，分{}an···········································································6a1公比q2，················································································7分a141∴a()n1，···················································································8分n4122n112所以ab()n1()n44()n≤4．·································10分nn433n6632当且仅当n1时取得最大值，··························································11分3∴存在，使得对任意的，≤恒成立．分m1nNanbnambm··························12若选择③：因为，所以≥，分an1an1anan11(n2)·····························5∴是以为公差的等差数列，又，分{}an1a11············································62∴an，所以abn()n，····························································8分nnn3理科数学参考答案第2页共6页223n2设cab，则ccn()n(n1)()n1()n1，····················9分nnnnn13333∴当n3时，cncn10，所以cncn1，当时，，所以，n3cncn10cncn1当时，，所以，n3cncn10cncn1则，分c1c2c3c4c5································································11∴存在，使得对任意的，≤恒成立．分m2或3nNanbnambm······················12，，20．解：（1）设B()x1y1，C()x2y2，1直线BC的方程为：xmy4(m=)，···················································1分kxmy422联立x2y2，消x整理得：(3m4)y24my360，·····················2分14324m36∴yy，yy······················································3分123m24123m24y1y2y1y2从而：k1k2x12x22(my16)(my26)yy122my1y26m(y1y2)3636213m436m2144m24363m243m241∴kk为定值.················································································5分124y（2）直线AB的方程为：y1(x2)，················································6分x126y16y1令x4，得到yM，······················································7分x12my166y2同理：yN．··········································································8分my266y16y2从而|MN||yMNy|||my16my2636|yy|122······················································9分|my1y26m(y1y2)36|12m24又|yy|(yy)24yy，1212123m24理科数学参考答案第3页共6页144|m2yy6m(yy)36|，·····················································10分12123m24所以|MN|3m24，·······································································11分1因为：m[3，4]，所以|MN|[35，63]，k即线段MN长度的取值范围为[35，63]．·············································12分21．解：（1）由a=1时，f(x)(x1)(ex1)，·············································1分由f(x)0解得：x>0或x<−1；由f(x)0解得：−1