2023届绵阳二诊文科数学答案_2.2025数学总复习_数学高考模拟题_2023年模拟题_老高考_四川省绵阳市高2023届高三第二次诊断性考试绵阳二诊数学

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． DDCAA BCDBA CA 二、填空题：本大题共4小题，每小题5分，共20分． 13． 2 14．1 15．3 16．[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 ∵3sinB3sin(AC)3sinAcosC3cosAsinC ·······································4分 ∴3cosAsinC asinAsinC ，································································6分  即asinA3cosA，且A ，可得a 3；·············································8分 3 1 1 （2）由BAAC cbcos(A) cb ，可得cb1，·······················10分 2 2 由余弦定理b2 c2 a2 2bccosA4.···················································12分 18．解：（1）由题意知，2S =a2+a ，①···············································1分 n n n 当n=1时，2a =a2+a ，则a 1；···························································2分 1 1 1 1 当n≥2时，2S =a2 +a ，②····················································· 3分 n1 n1 n1 ①②相减可得，2a =a2−a2 +a −a ，················································4分 n n n1 n n1 ∴a +a = a2−a2 ，则a -a =1， n n1 n n1 n n1 ∴数列a 是以a 1为首项，1为公差的等差数列，··································5分 n 1 所以，a =n(n∈N ).········································································ 6分 n ∗ 2 （2）a b n( )n，············································································ 7分 n n 3 2 2 3n 2 设c a b ，则 c c n( )n (n1)( )n1  ( )n1，·························8分 n n n n n1 3 3 3 3 ∴当n3时，c c 0，所以c c ，············································9分 n n1 n n1 当n3时，c c 0，所以c c ，·············································10分 n n1 n n1 文科数学参考答案 第1页（共6页）当n3时，c c 0，所以c c ，··············································11分 n n1 n n1 则c c c c c ， 1 2 3 4 5 ∴存在m2或3，使得对任意的nN，a b ≤a b 恒成立．·····················12分 n n m m 5 19．解：（1）因为0.92<0.99，根据统计学相关知识， R2越大，意味着残差平方和(y  yˆ)2 i i1 越小，那么拟合效果越好，因此选择非线性回归方程②yˆ mˆx2 nˆ 进行拟合更加符合问题实际．································································4分 （2）令u x2，则先求出线性回归方程：yˆ mˆunˆ，·································5分 i i 1491625 0.81.11.52.43.7 ∵u =11，y 1.9，······················7分 5 5 n (u u)2 (111)2 (411)2 (911)2 (1611)2 (2511)2 =374，············9分 i i1 n (u u)(y y) i i 45.1 ∴mˆ  i1   0.121，·················································10分 n 374 (u u)2 i i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 20．解：（1）设B(x，y )，C(x，y )， 1 1 2 2 1 直线BC的方程为：xmy4，其中m= ， ········································ 1分 k xmy4  联立x2 y2 ，消x整理得：(3m2 4)y2 24my360，····················2分   1  4 3 24m 36 所以：y  y  ，y y  ，·············································3分 1 2 3m2 4 1 2 3m2 4 y y y y 从而k k  1  2  1 2 1 2 x 2 x 2 (my 6)(my 6) 1 2 1 2 y y  1 2 m2y y 6m(y y )36 1 2 1 2 文科数学参考答案 第2页（共6页）36 3m2 4 1   36m2 144m2 4  36 3m2 4 3m2 4 1 所以：k k 为定值 ．········································································5分 1 2 4 y （2）直线AB的方程为： y  1 (x2)，··············································6分 x 2 1 6y 6y 令x4，得到 y  1  1 ，··················································7分 M x 2 my 6 1 1 6y 同理：y  2 ．··········································································8分 N my 6 2 6y 6y 从而|MN|| y y || 1  2 | M N my 6 my 6 1 2 36| y y |  1 2 ······················································9分 |m2y y 6m(y y )36| 1 2 1 2 12 m2 4 又| y y | (y  y )24y y  ， 1 2 1 2 1 2 3m2 4 144 |m2y y 6m(y y )36| ，····················································10分 1 2 1 2 3m2 4 所以|MN|3 m2 4 ，·······································································11分 1 因为：m [3，4]，所以|MN|[3 5，6 3]， k 即线段MN长度的取值范围为[3 5，6 3]．·············································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 x x 1 1 由 f(x)0解得：x>1或0 x ；由 f(x)0解得：  x1．·················3分 2 2 1 1 故f(x)在区间(1，)，(0， )上单调递增，在区间( ，1)上单调递减．···········4分 2 2 1 3 所以f(x)的极大值是 f( ) ln2，极小值是f(1)=0；·······························5分 2 4 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 x x (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) h(a)0， ····························· 7分 min 1 (ax1)(x1) ②当0a≤ 时，ax10， f(x) ≤0， 2 x 故f(x)在区间[1，2]上单调递减， a 1 所以 f(x) h(a) f(2) ln21≥0 ，显然h(a)在区间(0，]上单调递增， min 2 2 1 3 故h(a)≤h( )ln2 <0．······································································9分 2 4 1 1 1 ③当 a1时，由 f(x)0解得：  x≤2；由 f(x)0解得：1≤x ． 2 a a 1 1 故f(x)在区间( ，2]上单调递增，在区间[1，)上单调递减． a a 1 a 1 1 1 1 (a1)2 此时 f(x)  f( )h(a) lna ，则h(a)    ≥0 ， min a 2 2a 2 a 2a2 2a2 1 故h(a)在区间( ，1)上单调递增，故h(a)