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2018届高三唐山二模理科数学答案

唐山市 2017—2018 学年度高三年级第二次模拟考试 理科数学参考答案 一.选择题: A 卷:BACDB CDBDC AB B 卷:BACDC CDBDB AB 二.填空题: 1 (13)15 (14) (15)1 2 (16)[2,+∞) 三.解答题: 17.解: (1)由题意可知,AD=1. …1 分 在△ABD 中,∠DAB=150° ,AB=2 3,AD=1,由余弦定理可知, 3 BD2=(2 3)2+12-2×2 3×1× - =19, 2 BD= 19. …6 分 (2)由题意可知,AD=2cos θ,∠ABD=60° -θ, 在△ABD 中,由正弦定理可知, AD AB = , sin∠ABD sin∠ADB 2cos θ 即 =4 3, …10 分 sin(60° -θ) 2 3 整理得 tan θ= . …12 分 3 ( ) 18.解: (1)应该选择模型①. (2)剔除异常数据,即组号为 4 的数据, 1 剩下数据的平均数- x = (18×6-18)=18; 5 1 - y = (12.25×6-13.5)=12. 5 i=1 5 i=1 …3 分 …5 分 ∑ xiyi=1283.01-18×13.5=1040.01; 2 ∑ x2 i =1964.34-18 =1640.34. n 5 …7 分 ∑ xiyi-n·xy 1040.01-5×18×12 i=1 ? b= n = ≈-1.97, 2 2 1640.34-5×182 ∑ x i -nx i=1 …10 分 ?a ? =y -b x =12+1.97×18≈47.5, 所以 y 关于 x 的线性回归方程为:y ? =-2.0x+47.5. …12 分 高三理科数学参考答案第 1 页 19.解: (1)因为平面 AA1C1C⊥平面 ABC,交线为 AC,又 BC⊥AC, 所以 BC⊥平面 AA1C1C, 因为 C1C?平面 AA1C1C, 从而有 BC⊥C1C. …2 分 因为∠A1CC1=90° ,所以 A1C⊥C1C, 又因为 BC∩A1C=C, B1 所以 C1C⊥平面 A1BC, z C1 A1 A1B?平面 A1BC, 所以 CC1⊥A1B. …5 分 (2)如图,以 C 为坐标原点,分别以 C x B A y → CB ,→ CA 的方向为 x 轴,y 轴的正方 向建立空间直角坐标系 C-xyz. 由∠A1CC1=90° ,AC= 2AA1 得 A1C=AA1. 不妨设 BC=AC= 2AA1=2, 则 B(2,0,0),C1(0,-1,1),A(0,2,0),A1(0,1,1), → → → 所以A 1C1=(0,-2,0), BC1 =(-2,-1,1), AB =(2,-2,0), 设平面 A1BC1 的一个法向量为 m, → → 由A 1C1·m=0, BC1 ·m=0,可取 m=(1,0,2). 设平面 ABC1 的一个法向量为 n, 由→ BC1 ·n=0,→ AB ·n=0,可取 n=(1,1,3). m·n 7 55 cos?m,n?= = , 55 |m||n| 又因为二面角 A1-BC1-A 为锐二面角, 7 55 所以二面角 A1-BC1-A 的余弦值为 . 55 20.解: (1)设直线 l 的方程为 x=my+1,A(x1,y1),B(x2,y2), ?y2=4x, 由? 得 y2-4my-4=0, ?x=my+1, y1+y2=4m,y1y2=-4. 4 4 4(y1+y2) 所以 kOA+kOB= + = =-4m=4. y1 y2 y1y2 所以 m=-1, 所以 l 的方程为 x+y-1=0. 1 (2)由(1)可知,m≠0,C 0,- ,D(2m2+1,2m). m 则直线 MN 的方程为 y-2m=-m(x-2m2-1),则 M (2m2+3,0),N (0,2m3+3m),F (1,0), …10 分 …11 分 …8 分 …12 分 …3 分 …6 分 ( ) …8 分 高三理科数学参考答案第 2 页 (m2+1)(2m2+1)2 1 1 1 S△NDC= ·|NC|·|xD|= ·|2m3+3m+ |·(2m2+1)= , 2 2 m 2|m| 1 1 S△FDM= ·|FM|·|yD|= ·(2m2+2)·2|m|=2|m| (m2+1), …10 分 2 2 2 2 S△NDC (2m +1) 1 则 = =m2+ 2+1≥2, 4m2 4m S△FDM 1 1 当且仅当 m2= 2,即 m2= 时取等号. 4m 2 S△NDC 所以, 的最小值为 2. …12 分 S△FDM 其它解法参考答案给分. 21.解: 1 1- -ln x x (1)f ?(x)= . (x-1)2 1 1 1 1-x 令 h (x)=1- -ln x,则 h ?(x)= 2- = 2 ,x>0, x x x x 所以 0<x<1 时,h ?(x)>0,h (x)单调递增, 又 h (1)=0,所以 h (x)<0, 即 f ?(x)<0,所以 f (x)单调递减. - - - (2)g ?(x)=axln a+axa 1=a(ax 1ln a+xa 1), 1 - - - - 当 0<a≤ 时,ln a≤-1,所以 ax 1ln a+xa 1≤xa 1-ax 1. e ln x ln a - - 由(Ⅰ)得 < ,所以(a-1)ln x<(x-1)ln a,即 xa 1<ax 1, x-1 a-1 所以 g ?(x)<0,g (x)在(a,1)上单调递减, 即 g (x)>g (1)=a+1>1. 1 当 <a<1 时,-1<ln a<0. e 令 t (x)=ax-xln a-1,0<a<x<1,则 t ?(x)=axln a-ln a=(ax-1)ln a>0, 所以 t (x)在(0,1)上单调递增,即 t

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