2016年5月7日福州高三质检文科数学含答案
从而g(x)在[0,ln(2m))上单调递减，
而g(0) 0，所以当x (0,ln(2m))时，g(x) g 0 0，即f(x) mx2不成立．
1
综上所述，k的取值范围是( ,]． ··························································· 12分
2
请考生在第（22）,（23）,（24）题中任选一题作答，如果多做，则按所做的第一题计分，作答时请写清题号．
（22）选修4 1：几何证明选讲
本小题主要考查圆周角定理、相似三角形的判定与性质、切割线定理等基础知识，考查推理论证能力、运算求解能力等，考查化归与转化思想等．满分10分．
解：（Ⅰ）设 ABE外接圆的圆心为O ，连结BO 并延长交圆O 于G点，连结GE， 则 BEG 90 , BAE BGE．
=FC ，所以 FBE BAE， ·因为AF平分∠BAC，所以BF······················· 2分 所以 FBG FBE EBG BGE EBG 180 BEG 90 ，
所以O B BF，所以BF是 ABE外接圆的切线． ······································ 5分
（Ⅱ）连接DF，则DF BC，所以DF是圆O的直径，
因为BD2 BF2 DF2，DA2 AF2 DF2， 2222
所以BD DA AF BF． ······································· 7分 因为AF平分∠BAC，所以 ABF∽ AEC,
ABAF所以，所以AB AC AE AF (AF EF) AF, AEAC
因为 FBE BAE，所以 FBE∽ FAB，从而BF2 FE FA，
所以AB AC AF2 BF2，
所以BD2 DA2 AB AC 6． ····························································· 10分 （23）选修4 4；坐标系与参数方程
本小题考查极坐标方程和参数方程、伸缩变换等基础知识，考查运算求解能力，考查数形结合思想、化归与转化思想等．满分10分．
x 2 2cos ,
解：（Ⅰ）将 消去参数 ，化为普通方程为(x 2)2 y2 4，
y 2sin 即C1:x2 y2 4x 0， ··············································································· 2分 x cos ,将 代入C1:x2 y2 4x 0，得 2 4 cos ， ································· 4分
y sin
所以C1的极坐标方程为 4cos ． ······························································ 5分
x 2x ,
（Ⅱ）将 代入C2得x 2 y 2 1，
y y
所以C3的方程为x2 y2 1． ········································································ 7分 C3的极坐标方程为 1，所以|OB| 1．
π
又|OA| 4cos 2，
3
所以|AB| |OA| |OB| 1． ········································································ 10分 （24）选修4 5：不等式选讲
