2016年5月7日福州高三质检文科数学含答案
2k
，
1 4k2
4k2 k
,所以圆心M坐标为 ，
22 1 4k1 4k
所以y1 y2 k x1 x2 2
PQ 1 x2 ···························· 9分 2
4 1 k2 1 3k2 4k2 k
y 所以 M的方程为 x ． ················ 10分 2 2 221 4k 1 4k 1 4k
2
222222
4k2 k 4 1 k 1 3k 4k k 1 0 0， ·因为 2 ············· 11分 2 2 22221 4k1 4k 1 4k 1 4k
所以点A在 M外． ················································································ 12分 （21）本小题主要考查导数的几何意义、函数的单调性、导数及其应用、不等式等基础知识，考查推理论证能力、运算求解能力等，考查函数与方程思想、化归与转化思想、分类与整合思想、数形结合思想等．满分12分．
解：（Ⅰ）f x ex a，依题意，设切点为(x0,0)， ······································ 1分
x
f(x0) 0, e0 a(x0 1) 0,
则 即 x
0 f(x) 0,0 e a 0,
x 0,解得 0 ······························································································ 3分
a 1,
所以f x ex 1，
所以，当x 0时，f x 0；当x 0时，f x 0．
所以，f x 的单调递减区间为( ,0)，单调递增区间为 0, ． ······················· 5分 （Ⅱ）令g(x) f(x) mx2， 则g (x) ex 2mx 1，
令h(x) g (x)，则h (x) ex 2m， ······························································· 7分
1
（ⅰ）若m ，
2
因为当x 0时，ex 1，所以h (x) 0， 所以h(x)即g (x)在[0, )上单调递增．
又因为g (0) 0，所以当x 0时，g (x) g 0 0， 从而g(x)在[0, )上单调递增，
而g(0) 0，所以g(x) g 0 0，即f(x) mx2成立． ······································ 9分
1
， 2
令h (x) 0，解得x ln(2m) 0，
当x (0,ln(2m))，h (x) 0，所以h(x)即g (x)在[0,ln(2m))上单调递减， 又因为g (0) 0，所以当x (0,ln(2m))时，g (x) 0， （ⅱ）若m
