湖南省长沙市长郡中学2019届高三下学期第一次适应性考试(一模)数学(文)试题(图片版)

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湖南省长沙市长郡中学2019届高三下学期第一次适应性考试(一模)数学(文)试题(图片版)

高三一模数学(文科)参考答案 第 1 页 2019 数学(文科)参考答案 一、选择题(本大题共 12 小题,每小题 5 分,共 60 分) 题号 1 2 3 4 5 6 7 8 9 10 11 12 答案 D B C A B B D A D B D C 二、填空题(本大题共 4 小题,每小题 5 分,共 20 分) 13.- 3 2 a+ 3 1 b 14.1+ 5 15. 25 72 16.a3+b3-c3<0 三、解答题(共 70 分。解答应写出文字说明、证明过程或演算步骤) 17.解:(Ⅰ)∵ nS2 是 an 与 an+1 的等比中项, ∴   nnnnn aaaaS  212 , 当 n=1 时, 1 2 112 aaa  ,∴ a1=1.·····································(2 分) 当 n≥2 时, 1 2 1 2 1222   nnnnnnn aaaaSSa , 整理得   0111   nnnn aaaa .······································(3 分) 又 an>0,∴  211   naa nn , 即数列 na 是首项为 1,公差为 1 的等差数列. ∴     nndnaan  1111 .·····································(6 分) (Ⅱ)               1 1111 121 11 nnnn nb nn n ,··························(9 分) ∴T2n=b1+b2+b3+…+b2n=(1+ 2 1 )-( 2 1 + 3 1 )+( 3 1 + 4 1 )-…+( 12 1 n + n2 1 )-( n2 1 + 12 1 n )=1- 12 1 n <1.··································(12 分) 18.解:(Ⅰ)调整前 y 关于 x 的表达式为 y=          80005000,1.0500045 50003500,03.03500 3500,0 xx xx x , ········································(2 分) 调整后 y 关于 x 的表达式为 y=       80005000,03.05000 5000,0 xx x .··········(4 分) (Ⅱ)由频数分布表可知从[3000,5000)及[5000,7000)的人群中按分层抽样抽取 7 人,其中[3000,5000)中占 3 人,分别记为 A,B,C,[ 5000,7000)中占 4 人,分别 记为 1,2,3,4,再从这 7 人中选 2 人的所有组合有:AB,AC,A1,A2,A3,A4,BC, 高三一模数学(文科)参考答案 第 2 页 B1,B2,B3,B4,C1,C2,C3,C4,12,13,14,23,24,34,共 21 种情况, 其中不在同一收入人群的有:A1,A2,A3,A4,B1,B2,B3,B4,C1,C2,C3,C4, 共 12 种, 所以所求概率为 P= 21 12 = 7 4 .···········································(8 分) (Ⅲ)由于小红的工资、薪金等税前收入为 7500 元, 按调整起征点前应纳个税为 1500×3%+2500×10%=295 元;···············(10 分) 按调整起征点后应纳个税为 2500×3%=75 元, 由此可知,调整起征点后应纳个税少交 220 元, 即个人的实际收入增加了 220 元, 所以小红的实际收入增加了 220 元.······································(12 分) 19.解:(Ⅰ)过点 P 作 PO⊥AD,垂足为 O.··································(1 分) 由于点 P 在平面 ABCD 内的射影恰好在 AD 上, ∴PO⊥平面 ABCD. ∴PO⊥AB.···························································(2 分) ∵四边形 ABCD 为矩形,∴AB⊥AD. 又 AD  PO=O,∴AB⊥平面 PAD, ∴AB⊥PD.···························································(3 分) 又由 AB=3,PB=3 2 ,可得 PA=3, 同理 PD=3.··························································(4 分) 又 AD=3 2 , ∴PA 2+PD2=AD2, ∴PA⊥PD,且 PA  AB=A, ∴PD⊥平面 PAB.······················································(5 分) (Ⅱ)设点 E 到底面 QBC 的距离为 h, A B D C P Q E P A B D C O 高三一模数学(文科)参考答案 第 3 页 则 hSVV QBCQBCEEBCQ   3 1 .········································(7分) 由 PBPE 3 1 ,可知 3 2BP BE , ∴ 22 23 3 2 3 2  hPO h .·······································(8 分) 又 2 293232 1 2 1  ABBCS QBC ,···························(10 分) ∴ 322 29 3 1 3 1   hSV QBCEBCQ .·····························(12 分) 20.( Ⅰ)设 xMF 2 ,则 MFF 21 内, 由余弦定理得 2 22 5 14120cos222      xx , 化简得 05 6 5 16            xx ,解得 5 6x , 故 42 21  MFMFa , ∴ 2a ,得 3222  cab , 所以椭圆 C 的标准方程为 134 22  yx .···································(4 分) (Ⅱ)已知 A(-2,0), B(2,0), 设 T(x,y), P(x1,y1), Q(x2,y2), 由 22 1 1  x y x ykk PATA ,① 22 2 2  x y x ykk QBTB ,② 两式相除得 2 2 1 1 2 22 2 y x x y x x   .·······································(6 分) 又   4 3 4 44 3 422 2 1 2 1 2 1 2 1 1 1 1 1    x x x y x y x y , 故 1 1 1 1 2 4 3 2 y x x y  , 高三一模数学(文科)参考答案 第 4 页 故    21 21 2 2 1 1 22 4 32 22 2 yy xx y x x y x x   ,③ ························(8 分) 设 PQ 的方程为 1 myx ,代入 134 22  yx 整理, 得   09643 22  myym ,  >0 恒成立.························································(10 分) 把        43 9 43 6 221 221 myy m myy 代入③, 得      3 1 43 9 143 6 43 9 4 3 1 4 311 4 3 2 2 2 22 2 21 2121 2 21 21              m m mmmm yy yymyym yy mymy x x , 得到 x=4,故点 T 在定直线 x=4 上.····································(12 分) 21.解:(Ⅰ)因为 f(x)为奇函数,其图象关于原点对称,所以只需考虑 x∈(0,+∞) 上的极值点个数,       ,0,1ln9 1 23 xxxxxxf 时,   2 22 2 2 1 1113 1 1 113 1 x xx x xxf         .·······················(1 分) 令   1113 1 22       xxxh ,   22 3 1 3 3 3 3 1 3 1 x xxx x xx xh                   , ∴当       3 3,0x 时,    xhxh ,0 单调递减,当        ,3 3x 时,    xhxh ,0 单 调递增, ∴   003 3       hh .··················································(3 分) 取   017161163 16,6       hx , 高三一模数学(文科)参考答案 第 5 页 ∴在区间       ,3 3 上存在唯一的 x0 使   00 xh .··························(4 分) ∴f(x)在区间 0,0 x 上单调递减,在区间 ,0x 上单调递增.··············(5 分) 又 f(x)为奇函数, ∴f(x)在区间 0, x 上单调递增,在区间 00, xx 上单调递减,在区间 ,0x 上 单调递增, ∴f(x)的极值点共 2 个.···············································(6 分) (Ⅱ)由(Ⅰ)可知 f(x)在区间       3 3,0 内单调递减,且 f(x)<0 恒成立.··(7 分) ∴       3 3,0x 时,   01ln9 1 23  xxxx , 即得   xxxx  23 1ln9 1 .··········································(8 分) 又令           3 3,04 1 n x , 得 nnnn na                           4 1 4 114 1ln4 1 9 1 23 .························(10 分) ∴ 3 1 4 113 1 4 11 4 114 1 4 1…4 1 4 1 4 1… 32 321                                        n n n naaaa .··(12 分) 22.解:(Ⅰ)      32 12:1 ty txC 消去 t,得 4 yx .又        sin cos y x ,代入 4 yx 得: 04sincos   . ∴ 0224sin: 044sin204cossin 1               C .················(2 分) C2:p=2a cos 化为:(x-a)2+y2=a2(a>0),又 C2 关于 C1:x-y=4 对称, ∴(a,0)∈C1,∴a=4,∴C2:( x-4)2+y2=16.························(4 分) 高三一模数学(文科)参考答案 第 6 页 (Ⅱ)C2 向左平移 4 个单位长度得:x2+y2=16,按      yy xx 2 3 变换后得:x2+ 2 3 2      y =16 16 2x + 12 2y =1.······················································(6 分) ∴C3: + =1,∴令 A(4,0), B(0,2 3 ),∴ AB =2 7 . 易得:lAB: 3 x+2y-4 3 =0,设 P(4cos ,2 sin )到 lAB 的距离为 d. 则 d= 7 34cos34sin34   = 7 14sin234      ≤   7 1234  . ······································(8 分) 当 sin( + 4  )=-1  + 4  =  2 3 = 4 5 时,d 有最大值   7 1234  . ∴( S△ABP)max= AB2 1 d= 2 1 ×2 7 ×   7 1234  =4 3 +4 6 .········(10 分) 23.解:(Ⅰ)               2 1,13 2 10,1 0,13 12 xx xx xx xxxf , 由   4xf 解得 1x 或 3 5x .········································(5 分) (Ⅱ)∵ 38 3628 36 8 3168 333 2222  abababababbaabba . 当 a=b=2 时等号成立,即知   312  xxxf . 解方程,分情况讨论:①当 x≤0 时,-3x+1<3,故 3 2 <x≤0;②当 x≥ 2 1 时,3x -1<3,故 2 1 ≤x< 3 4 ;③当 0<x< 2 1 ,满足 1-x<3. ∴x 的取值集合为 M=    3 4 3 2 xx .·································(10 分)
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