绵阳市高中2020级第二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．DDCAABCDBACA二、填空题：本大题共4小题，每小题5分，共20分．13．214．115．316．[1，3)三、解答题：本大题共6小题，共70分．17．解：（1）由3(bacosC)a2sinC，及正弦定理可得，3sinB3sinAcosCasinAsinC，················································2分∵3sinBACACAC3sin()3sincos3cossin·······································4分∴3cosAsinCasinAsinC，································································6分即asinA3cosA，且A，可得a3；·············································8分311（2）由BAACcbcos(A)cb，可得cb1，·······················10分22由余弦定理b2c2a22bccosA4.···················································12分．（）由题意知，2，①分18解：12Sn=an+an···············································1当时，2，则；分n=12a1=a1+a1a11···························································2当≥时，2，②分n22Sn1=an1+an1·····················································3①②相减可得，22，分2an=an−an1+an−an1················································4∴22，则-，an+an1=an−an1anan1=1∴数列是以为首项，为公差的等差数列，分ana111··································5所以，an=n(n∈N).········································································6分∗2（2）abn()n，············································································7分nn3设，则2n2n13n2n1，分cnanbnccn()(n1)()()·························8nn13333∴当时，，所以，分n3cncn10cncn1············································9当时，，所以，分n3cncn10cncn1·············································10文科数学参考答案第1页（共6页）当时，，所以，分n3cncn10cncn1··············································11则c1c2c3c4c5，∴存在或，使得对任意的，≤恒成立．分m23nNanbnambm·····················1252219．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由1.90.12111nˆ，得nˆ0.5690.57，即yˆ0.12u0.57，···········································································11分∴所求非线性回归方程为：yˆ0.12x20.57．········································12分．（）设，，，，20解：1B()x1y1C()x2y21直线BC的方程为：xmy4，其中m=，········································1分kxmy4联立x2y2，消x整理得：(3m24)y224my360，····················2分14324m36所以：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)36文科数学参考答案第2页（共6页）36213m436m2144m24363m243m241所以：kk为定值．········································································5分124y1（2）直线AB的方程为：y(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4144|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）解：(1)a=2时，f(x)lnxx23x2，2x23x1(2x1)(x1)f()x，·······················································2分xx11由f(x)0解得：x>1或0x；由f(x)0解得：x1．·················3分2211故f(x)在区间(1，)，(0，)上单调递增，在区间(，1)上单调递减．···········4分2213所以f(x)的极大值是f()ln2，极小值是f(1)=0；·······························5分24文科数学参考答案第3页（共6页）ax2(a1)x1(ax1)(x1)（2）f()x，且x1≥0，································6分xx(ax1)(x1)①当a≥1时，ax1≥0，f()x≥0，x故f(x)在区间[1，2]上单调递增，所以f(x)minh(a)0，·····························7分1(ax1)(x1)②当0a≤时，ax10，f()x≤0，2x故f(x)在区间[1，2]上单调递减，a1所以f(x)h(a)f(2)ln21≥0，显然h()a在区间(0，]上单调递增，min2213故h(a)≤h()ln2<0．······································································9分24111③当a1时，由f(x)0解得：x≤2；由f(x)0解得：1≤x．2aa11故f(x)在区间(，2]上单调递增，在区间[1，)上单调递减．aa1a1111(a1)2此时f(x)minf()h(a)lna，则h()a≥0，a22a2a2a22a21故h()a在区间(，1)上单调递增，故h(a)