南京市2016届高三年级第三次模拟考试数学参考答案

1、南京市2016届高三年级第三次模拟考试数学参考答案及评分标准说明：1本解答给出的解法供参考如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制订相应的评分细则2对计算题，当考生的解答在某一步出现错误时，如果后续部分的解答未改变该题的内容和难度，可视影响的程度决定给分，但不得超过该部分正确解答应得分数的一半；如果后续部分的解答有较严重的错误，就不再给分3解答右端所注分数，表示考生正确做到这一步应得的累加分数4只给整数分数，填空题不给中间分数一、填空题（本大题共14小题，每小题5分，计70分. 不需写出解答过程，请把答案写在答题纸的指定位置上）15 23i 30.02 4 58 67

2、4 8 94 101，3 11 123 13(1，2) 14 二、解答题（本大题共6小题，计90分.解答应写出必要的文字说明，证明过程或演算步骤，请把答案写在答题纸的指定区域内）15(本小题满分14分)解：（1）因为m·n3bcosB，所以acosCccosA3bcosB由正弦定理，得sinAcosCsinCcosA3sinBcosB，·····················

3、······································3分所以sin(AC)3sinBcosB，所以sinB3sinBcosB因为B是ABC的内角，所以sinB0，所以cosB··

4、;··················································

5、;7分（2）因为a，b，c成等比数列，所以b2ac由正弦定理，得sin2BsinA·sinC ·········································&#

6、183;·····································9分因为cosB，B是ABC的内角，所以sinB········&#

7、183;·············································11分又···

8、3;·················································

9、3;···········14分16(本小题满分14分)证明：（1）因为ABAC，点D为BC中点，所以ADBC ·······························

10、83;·················2分 因为ABCA1B1C1 是直三棱柱，所以BB1平面ABC 因为ADÌ平面ABC，所以BB1AD ·······················&

11、#183;···························4分 因为BCBB1B，BCÌ平面BCC1B1，BB1Ì平面BCC1B1， 所以AD平面BCC1B1 因为ADÌ平面ADC1，所以平面ADC1平面BCC1B1 ······&#

12、183;······································6分(2)连结A1C，交AC1于O，连结OD，所以O为AC1中点 ······

13、;·······································8分因为A1B平面ADC1，A1BÌ平面A1BC，平面ADC1平面A1BCOD，所以A1BOD ··

14、················································12分因为O为AC1中点

15、，所以D为BC中点，所以1 ···············································

16、3;··················14分17(本小题满分14分)解：（1）由题意，得，1，解得a26，b23所以椭圆的方程为1 ························

17、;··········································2分（2）解法一 椭圆C的右焦点F(，0)设切线方程为yk(x)，即kxyk0，所以，解得k&

18、#177;，所以切线方程为y±(x)······························4分由方程组解得或 所以点P，Q的坐标分别为(，)，(，)，所以PQ ··········&#

19、183;······················6分因为O到直线PQ的距离为，所以OPQ的面积为 因为椭圆的对称性，当切线方程为y(x)时，OPQ的面积也为综上所述，OPQ的面积为 ················&

20、#183;················8分解法二 椭圆C的右焦点F(，0)设切线方程为yk(x)，即kxyk0，所以，解得k±，所以切线方程为y±(x)······················

21、·········4分把切线方程 y(x)代入椭圆C的方程，消去y得5x28x60设P(x1，y1) ，Q(x2，y2)，则有x1x2 由椭圆定义可得，PQPFFQ2ae( x1x2)2××·····················6分因为O到直线PQ的距离为，所以OPQ的面积为

22、 因为椭圆的对称性，当切线方程为y(x)时，所以OPQ的面积为综上所述，OPQ的面积为 ·································8分解法一：(i)若直线PQ的斜率不存在，则直线PQ的方程为x或x当x时，P (，)，Q(，)因为·0，所

23、以OPOQ当x时，同理可得OPOQ ·································10分(ii) 若直线PQ的斜率存在，设直线PQ的方程为ykxm，即kxym0因为直线与圆相切，所以，即m22k22将直线PQ方程代入椭圆方程，得(12k2) x24k

24、mx2m260.设P(x1，y1) ，Q(x2，y2)，则有x1x2，x1x2·································12分因为·x1x2y1y2x1x2(kx1m)(kx2m)(1k2)x1x2km(x1x2)m2(1k2)

25、5;km×()m2将m22k22代入上式可得·0，所以OPOQ综上所述，OPOQ ·····································14分解法二：设切点T(x0，y0)，则其切线方程为x

26、0xy0y20，且xy2 (i)当y00时，则直线PQ的直线方程为x或x当x时，P (，)，Q(，)因为·0，所以OPOQ当x时，同理可得OPOQ ··································10分(ii) 当y00时，由方程

27、组消去y得(2xy)x28x0x86y0设P(x1，y1) ，Q(x2，y2)，则有x1x2，x1x2 ······························12分所以·x1x2y1y2x1x2因为xy2，代入上式可得·0，所以OPOQ综上所述，OPOQ ·

28、83;···································14分18(本小题满分16分)解：(1)由题意，可得AD12千米 由题可知|， ········

29、;······································2分解得v ···········

30、;···································4分(2) 解法一：经过t小时，甲、乙之间的距离的平方为f(t)由于先乙到达D地，故2，即v8 ·······

31、;·········································6分当0vt5，即0t时，f(t)(6t)2(vt)22×6t×vt×cos

32、DAB(v2v36) t2因为v2v360，所以当t时，f(t)取最大值，所以(v2v36)×()225，解得v ·······································

33、3;·9分当5vt13，即t时，f(t)(vt16t)29(v6) 2 (t)29因为v8，所以，(v6) 20，所以当t时，f(t)取最大值，所以(v6) 2 ()2925，解得v ··································

34、;······13分当13vt16， t时，f(t)(126t)2(16vt)2，因为126t0，16vt0，所以当f(t)在(，)递减，所以当t时，f(t)取最大值，(126×)2(16v×)225，解得v 因为v8，所以 8v ························

35、·····················16分解法二：设经过t小时，甲、乙之间的距离的平方为f(t)由于先乙到达D地，故2，即v8 ·····················&#

36、183;···························6分以A点为原点，AD为x轴建立直角坐标系， 当0vt5时，f(t)(vt6t)2(vt)2由于(vt6t)2(vt)225，所以(v6)2(v)2对任意0t都成立，所以(v6)2(v)2v2，解得v ·····

37、83;·········································9分当5vt13时，f(t)(vt16t)232由于(vt16t)23225，所以4vt16t4对任意

38、t都成立，即对任意t都成立，所以解得v ··············································

39、3;13分当13vt16即t，此时f (t)(126t)2(16vt)2由及知：8v，于是0126t12124，又因为016vt3，所以f (t)(126t)2(16vt)2423225恒成立综上可知8v ·································

40、;············16分19(本小题满分16分)解：（1）当m1时，f(x)x3x21f (x)3x22xx(3x2)由f (x)0，解得x0或x所以函数f(x)的减区间是(，0)和(，) ·······················

41、83;··············2分（2）依题意m0因为f(x)x3mx2m，所以f (x)3x22mxx(3x2m)由f (x)0，得x或x0 当0x时，f (x)0，所以f(x)在(0，)上为增函数；当xm时，f (x)0，所以f(x)在(，m)上为减函数；所以，f(x)极大值f()m3m ·············

42、83;···································4分当m3mm，即m，ymaxm3m···········

43、83;···································6分当m3mm，即0m时，ymaxm综上，ymax ··········&

44、#183;·······································8分（3）设两切点的横坐标分别是x1，x2则函数f(x)在这两点的切线的方程分别为y(x13mx12m)(3x122mx1

45、)(xx1)，y(x23mx22m)(3x222mx2)(xx2) ···········································10分将(2，t

46、)代入两条切线方程，得t(x13mx12m)(3x122mx1)(2x1)，t(x23mx22m)(3x222mx2)(2x2)因为函数f(x)图象上有且仅有两个不同的切点，所以方程t(x3mx2m)(3x22mx)(2x)有且仅有不相等的两个实根···········12分整理得t2x3(6m)x24mxm设h(x)2x3(6m)x24mxm，h (x)6x22(6m)x4m2(3xm)(x2)当m6时，h (x)6(x2)20，所以h(x)单调递增，显然不成立当m6时， h (x)

47、0，解得x2或x列表可判断单调性，可得当x2或x，h(x)取得极值分别为h(2)3m8，或h()m3m2m 要使得关于x的方程t2x3(6m)x24mxm有且仅有两个不相等的实根，则t3m8，或tm3m2m ·······························14分因为t0，所以

48、3m80，（*），或m3m2m0（*）解（*），得m，解（*），得m93或m93因为m0，所以m的范围为(0，93，) ··································16分20(本小题满分16分)解：（1）因为3b1，2b2，b3成等差数

49、列， 所以4b23b1b3，即4×3(2ad)， 解得， ····································4分 由an1bnan2，得anda(n1)d，整理得 ···

50、3;····································6分解得n， ············

51、83;···························8分由于1且0 因此存在唯一的正整数n，使得an1bnan2 ·················&#

52、183;·······················10分（2）因为，所以 设f(n)，n2，nN*则f(n1)f(n)，因为q2，n2，所以(q1)n22(q2)n3n2310，所以f(n1)f(n)0，即f(n1)f(n)，即f(n)单调递增··········&

53、#183;·······················12分所以当r2时，tr2，则f(t)f(r)，即，这与互相矛盾所以r1，即 ···················

54、3;···············14分若t3，则f(t)f(3) ·，即，与相矛盾于是t2，所以，即3q25q50又q2，所以q ·························

55、3;·················16分南京市2016届高三年级第三次模拟考试 数学附加题参考答案及评分标准 2016.05 说明：1本解答给出的解法供参考如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制订相应的评分细则2对计算题，当考生的解答在某一步出现错误时，如果后续部分的解答未改变该题的内容和难度，可视影响的程度决定给分，但不得超过该部分正确解答应得分数的一半；如果后续部分的解答有较严重的错误，就不

56、再给分3解答右端所注分数，表示考生正确做到这一步应得的累加分数4只给整数分数，填空题不给中间分数21【选做题】在A、B、C、D四小题中只能选做2题，每小题10分，共计20分请在答卷卡指定区域内作答解答应写出文字说明、证明过程或演算步骤A选修41：几何证明选讲证明：(1)连接AB因为PA是半圆O的切线，所以PACABC因为BC是圆O的直径，所以ABAC又因为AHBC，所以CAHABC，所以PACCAH，所以AC是PAH的平分线 ··············&

57、#183;····························5分(2)因为H是OC中点，半圆O的半径为2，所以BH3，CH1又因为AHBC，所以AH2BH·HC3，所以AH在RtAHC中，AH，CH1，所以CAH30°由(1)可得PAH2CAH60°，所以PA2由PA是半圆O的切线，

58、所以PA2PC·PB，所以PC·(PCBC)(2)212，所以PC2 ··········································

59、83;10分B选修42：矩阵与变换解：设曲线C上的任意一点P(x，y)，P在矩阵A对应的变换下得到点Q(x，y)则 ， 即x2yx，xy，所以xy，y ·····································&#

60、183;··········5分代入x22xy2y21，得y22y·2()21，即x2y22，所以曲线C1的方程为x2y22 ······························&#

61、183;············10分C选修44：坐标系与参数方程解：M的极坐标为(1，)，故直角坐标为M(0，1)，且P(2cos，sin)，所以PM，sin1，1 ·················5分当sin时，PMmax，此时cos±所以，PM的最大值是，此时点P的坐标是(±，)&

62、#183;······························10分D选修45：不等式选讲 解：函数定义域为0，4，且f(x)0 由柯西不等式得52()2()()(5··)2，········

63、··············5分 即27×4(5··)2，所以56 当且仅当5，即x时，取等号所以，函数f(x)5的最大值为6 ··································10分【必做题】第22题、第23题，每题10分，共计20分 22（本小题满分10分）解：（1）记“X是奇数”为事件A，能组成的三位数的个数是48 ········&#