福建省厦门市2020届高中毕业班第一次质量检查数学（文）试题

厦门市 2020 届高中毕业班第一次质量检查参考答案 文科数学 一、选择题： CADAC DCACB CB 二、填空题： 13． 2− 14．1 15． 6 3 16． 6 7 ，10 3 2 21− 三、解答题： 17．本题主要考查等差、等比数列的定义，考查分组求和法、等比数列的求和运算以及对数运算；考查运算求解 能力；考查化归与转化思想等.满分 12 分. 解：（1）因为1， na ， +1na 成等差数列，所以 12 1n na a += + ， ······················································1 分 当 1n = 时，有 1 22 1=6a a= + ，得 1 3a = ，···········································································2 分 所以 +1 1 2( 1)n na a− = − ，又 1 1 2a − = ，所以 1 1 2 1 n n a a + − = − ， 所以 1na − 是首项为 2，公比为 2的等比数列. ····································································5 分 （2）由（1）知 1na − 是首项为 2，公比为 2的等比数列， 所以 11 2 2 2n n n a −− =  = ，所以 2 1 n n a = + ． ········································································6 分 所以 1 2 3(2 1) (2 1) (2 1) (2 1)n n S = + + + + + + + + ··································································7 分 1 2 3(2 2 2 2 )n n= + + + + + 2(1 2 ) 1 2 n n − = + − 12 2n n+= + − ， ······················································9 分 所以 2log 10nS  即 1 102 2 2n n+ + −  ， ················································································10 分 因为 1(2 2) [2 ( 1) 2] 2 1 0n n nn n+ + − − + − − = +  ，所以数列 1{2 2}n n+ + − 为递增数列． 当 9n = 时， 10 102 9 2 2+ −  ，不满足，当 8n = 时， 9 102 8 2 2+ −  满足． 所以满足不等式 2log 10nS  的最大的正整数 n的值为8 ． ······················································12 分 18．本题考查直线的方程、抛物线的定义及轨迹方程、直线与圆锥曲线的关系等知识；考查运算求解能力、推 理论证能力等；考查数形结合思想、函数与方程思想、化归与转化思想等．满分 12 分. 解：（1）法一：依题意：平面内动点 E 到定点 F (0,1)和到定直线 1y = − 的距离相等， ····················1 分 根据抛物线的定义，曲线C 是以点 F 为焦点，直线 1y = − 为准线的抛物线， 其方程为 2 4x y= ． ········································································································3 分 法二：设点 E ( , )x y ，依题意有： 1EF y= + ， ·································································1 分 即 2 2( 1) 1x y y+ − = + ，化简得到C 的方程为 2 4x y= . ························································3 分 （2）法一：依题意可设直线 l 的方程为： 1y kx= + ， A 1 1( , )x y