南京市2016屆高三年級(jí)第三次模擬考試數(shù)學(xué)參考答案

1、南京市2016屆高三年級(jí)第三次模擬考試數(shù)學(xué)參考答案及評(píng)分標(biāo)準(zhǔn)說明：1本解答給出的解法供參考如果考生的解法與本解答不同，可根據(jù)試題的主要考查內(nèi)容比照評(píng)分標(biāo)準(zhǔn)制訂相應(yīng)的評(píng)分細(xì)則2對(duì)計(jì)算題，當(dāng)考生的解答在某一步出現(xiàn)錯(cuò)誤時(shí)，如果后續(xù)部分的解答未改變?cè)擃}的內(nèi)容和難度，可視影響的程度決定給分，但不得超過該部分正確解答應(yīng)得分?jǐn)?shù)的一半；如果后續(xù)部分的解答有較嚴(yán)重的錯(cuò)誤，就不再給分3解答右端所注分?jǐn)?shù)，表示考生正確做到這一步應(yīng)得的累加分?jǐn)?shù)4只給整數(shù)分?jǐn)?shù)，填空題不給中間分?jǐn)?shù)一、填空題（本大題共14小題，每小題5分，計(jì)70分. 不需寫出解答過程，請(qǐng)把答案寫在答題紙的指定位置上）15 23i 30.02 4 58 67

2、4 8 94 101，3 11 123 13(1，2) 14 二、解答題（本大題共6小題，計(jì)90分.解答應(yīng)寫出必要的文字說明，證明過程或演算步驟，請(qǐng)把答案寫在答題紙的指定區(qū)域內(nèi)）15(本小題滿分14分)解：（1）因?yàn)閙·n3bcosB，所以acosCccosA3bcosB由正弦定理，得sinAcosCsinCcosA3sinBcosB，·····················

3、······································3分所以sin(AC)3sinBcosB，所以sinB3sinBcosB因?yàn)锽是ABC的內(nèi)角，所以sinB0，所以cosB··

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

5、;7分（2）因?yàn)閍，b，c成等比數(shù)列，所以b2ac由正弦定理，得sin2BsinA·sinC ·········································&#

6、183;·····································9分因?yàn)閏osB，B是ABC的內(nèi)角，所以sinB········&#

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

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

9、3;···········14分16(本小題滿分14分)證明：（1）因?yàn)锳BAC，點(diǎn)D為BC中點(diǎn)，所以ADBC ·······························

10、83;·················2分 因?yàn)锳BCA1B1C1 是直三棱柱，所以BB1平面ABC 因?yàn)锳DÌ平面ABC，所以BB1AD ·······················&

11、#183;···························4分 因?yàn)锽CBB1B，BCÌ平面BCC1B1，BB1Ì平面BCC1B1， 所以AD平面BCC1B1 因?yàn)锳DÌ平面ADC1，所以平面ADC1平面BCC1B1 ······&#

12、183;······································6分(2)連結(jié)A1C，交AC1于O，連結(jié)OD，所以O(shè)為AC1中點(diǎn) ······

13、;·······································8分因?yàn)锳1B平面ADC1，A1BÌ平面A1BC，平面ADC1平面A1BCOD，所以A1BOD ··

14、················································12分因?yàn)镺為AC1中點(diǎn)

15、，所以D為BC中點(diǎn)，所以1 ···············································

16、3;··················14分17(本小題滿分14分)解：（1）由題意，得，1，解得a26，b23所以橢圓的方程為1 ························

17、;··········································2分（2）解法一 橢圓C的右焦點(diǎn)F(，0)設(shè)切線方程為yk(x)，即kxyk0，所以，解得k&

18、#177;，所以切線方程為y±(x)······························4分由方程組解得或 所以點(diǎn)P，Q的坐標(biāo)分別為(，)，(，)，所以PQ ··········&#

19、183;······················6分因?yàn)镺到直線PQ的距離為，所以O(shè)PQ的面積為 因?yàn)闄E圓的對(duì)稱性，當(dāng)切線方程為y(x)時(shí)，OPQ的面積也為綜上所述，OPQ的面積為 ················&

20、#183;················8分解法二 橢圓C的右焦點(diǎn)F(，0)設(shè)切線方程為yk(x)，即kxyk0，所以，解得k±，所以切線方程為y±(x)······················

21、·········4分把切線方程 y(x)代入橢圓C的方程，消去y得5x28x60設(shè)P(x1，y1) ，Q(x2，y2)，則有x1x2 由橢圓定義可得，PQPFFQ2ae( x1x2)2××·····················6分因?yàn)镺到直線PQ的距離為，所以O(shè)PQ的面積為

22、 因?yàn)闄E圓的對(duì)稱性，當(dāng)切線方程為y(x)時(shí)，所以O(shè)PQ的面積為綜上所述，OPQ的面積為 ·································8分解法一：(i)若直線PQ的斜率不存在，則直線PQ的方程為x或x當(dāng)x時(shí)，P (，)，Q(，)因?yàn)?#183;0，所

23、以O(shè)POQ當(dāng)x時(shí)，同理可得OPOQ ·································10分(ii) 若直線PQ的斜率存在，設(shè)直線PQ的方程為ykxm，即kxym0因?yàn)橹本€與圓相切，所以，即m22k22將直線PQ方程代入橢圓方程，得(12k2) x24k

24、mx2m260.設(shè)P(x1，y1) ，Q(x2，y2)，則有x1x2，x1x2·································12分因?yàn)?#183;x1x2y1y2x1x2(kx1m)(kx2m)(1k2)x1x2km(x1x2)m2(1k2)

25、5;km×()m2將m22k22代入上式可得·0，所以O(shè)POQ綜上所述，OPOQ ·····································14分解法二：設(shè)切點(diǎn)T(x0，y0)，則其切線方程為x

26、0xy0y20，且xy2 (i)當(dāng)y00時(shí)，則直線PQ的直線方程為x或x當(dāng)x時(shí)，P (，)，Q(，)因?yàn)?#183;0，所以O(shè)POQ當(dāng)x時(shí)，同理可得OPOQ ··································10分(ii) 當(dāng)y00時(shí)，由方程

27、組消去y得(2xy)x28x0x86y0設(shè)P(x1，y1) ，Q(x2，y2)，則有x1x2，x1x2 ······························12分所以·x1x2y1y2x1x2因?yàn)閤y2，代入上式可得·0，所以O(shè)POQ綜上所述，OPOQ ·

28、83;···································14分18(本小題滿分16分)解：(1)由題意，可得AD12千米 由題可知|， ········

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

30、;···································4分(2) 解法一：經(jīng)過t小時(shí)，甲、乙之間的距離的平方為f(t)由于先乙到達(dá)D地，故2，即v8 ·······

31、;·········································6分當(dāng)0vt5，即0t時(shí)，f(t)(6t)2(vt)22×6t×vt×cos

32、DAB(v2v36) t2因?yàn)関2v360，所以當(dāng)t時(shí)，f(t)取最大值，所以(v2v36)×()225，解得v ·······································

33、3;·9分當(dāng)5vt13，即t時(shí)，f(t)(vt16t)29(v6) 2 (t)29因?yàn)関8，所以，(v6) 20，所以當(dāng)t時(shí)，f(t)取最大值，所以(v6) 2 ()2925，解得v ··································

34、;······13分當(dāng)13vt16， t時(shí)，f(t)(126t)2(16vt)2，因?yàn)?26t0，16vt0，所以當(dāng)f(t)在(，)遞減，所以當(dāng)t時(shí)，f(t)取最大值，(126×)2(16v×)225，解得v 因?yàn)関8，所以 8v ························

35、·····················16分解法二：設(shè)經(jīng)過t小時(shí)，甲、乙之間的距離的平方為f(t)由于先乙到達(dá)D地，故2，即v8 ·····················&#

36、183;···························6分以A點(diǎn)為原點(diǎn)，AD為x軸建立直角坐標(biāo)系， 當(dāng)0vt5時(shí)，f(t)(vt6t)2(vt)2由于(vt6t)2(vt)225，所以(v6)2(v)2對(duì)任意0t都成立，所以(v6)2(v)2v2，解得v ·····

37、83;·········································9分當(dāng)5vt13時(shí)，f(t)(vt16t)232由于(vt16t)23225，所以4vt16t4對(duì)任意

38、t都成立，即對(duì)任意t都成立，所以解得v ··············································

39、3;13分當(dāng)13vt16即t，此時(shí)f (t)(126t)2(16vt)2由及知：8v，于是0126t12124，又因?yàn)?16vt3，所以f (t)(126t)2(16vt)2423225恒成立綜上可知8v ·································

40、;············16分19(本小題滿分16分)解：（1）當(dāng)m1時(shí)，f(x)x3x21f (x)3x22xx(3x2)由f (x)0，解得x0或x所以函數(shù)f(x)的減區(qū)間是(，0)和(，) ·······················

41、83;··············2分（2）依題意m0因?yàn)閒(x)x3mx2m，所以f (x)3x22mxx(3x2m)由f (x)0，得x或x0 當(dāng)0x時(shí)，f (x)0，所以f(x)在(0，)上為增函數(shù)；當(dāng)xm時(shí)，f (x)0，所以f(x)在(，m)上為減函數(shù)；所以，f(x)極大值f()m3m ·············

42、83;···································4分當(dāng)m3mm，即m，ymaxm3m···········

43、83;···································6分當(dāng)m3mm，即0m時(shí)，ymaxm綜上，ymax ··········&

44、#183;·······································8分（3）設(shè)兩切點(diǎn)的橫坐標(biāo)分別是x1，x2則函數(shù)f(x)在這兩點(diǎn)的切線的方程分別為y(x13mx12m)(3x122mx1

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

46、)代入兩條切線方程，得t(x13mx12m)(3x122mx1)(2x1)，t(x23mx22m)(3x222mx2)(2x2)因?yàn)楹瘮?shù)f(x)圖象上有且僅有兩個(gè)不同的切點(diǎn)，所以方程t(x3mx2m)(3x22mx)(2x)有且僅有不相等的兩個(gè)實(shí)根···········12分整理得t2x3(6m)x24mxm設(shè)h(x)2x3(6m)x24mxm，h (x)6x22(6m)x4m2(3xm)(x2)當(dāng)m6時(shí)，h (x)6(x2)20，所以h(x)單調(diào)遞增，顯然不成立當(dāng)m6時(shí)， h (x)

47、0，解得x2或x列表可判斷單調(diào)性，可得當(dāng)x2或x，h(x)取得極值分別為h(2)3m8，或h()m3m2m 要使得關(guān)于x的方程t2x3(6m)x24mxm有且僅有兩個(gè)不相等的實(shí)根，則t3m8，或tm3m2m ·······························14分因?yàn)閠0，所以

48、3m80，（*），或m3m2m0（*）解（*），得m，解（*），得m93或m93因?yàn)閙0，所以m的范圍為(0，93，) ··································16分20(本小題滿分16分)解：（1）因?yàn)?b1，2b2，b3成等差數(shù)

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數(shù)n，使得an1bnan2 ·················&#

52、183;·······················10分（2）因?yàn)?，所?設(shè)f(n)，n2，nN*則f(n1)f(n)，因?yàn)閝2，n2，所以(q1)n22(q2)n3n2310，所以f(n1)f(n)0，即f(n1)f(n)，即f(n)單調(diào)遞增··········&

53、#183;·······················12分所以當(dāng)r2時(shí)，tr2，則f(t)f(r)，即，這與互相矛盾所以r1，即 ···················

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

55、3;·················16分南京市2016屆高三年級(jí)第三次模擬考試 數(shù)學(xué)附加題參考答案及評(píng)分標(biāo)準(zhǔn) 2016.05 說明：1本解答給出的解法供參考如果考生的解法與本解答不同，可根據(jù)試題的主要考查內(nèi)容比照評(píng)分標(biāo)準(zhǔn)制訂相應(yīng)的評(píng)分細(xì)則2對(duì)計(jì)算題，當(dāng)考生的解答在某一步出現(xiàn)錯(cuò)誤時(shí)，如果后續(xù)部分的解答未改變?cè)擃}的內(nèi)容和難度，可視影響的程度決定給分，但不得超過該部分正確解答應(yīng)得分?jǐn)?shù)的一半；如果后續(xù)部分的解答有較嚴(yán)重的錯(cuò)誤，就不

56、再給分3解答右端所注分?jǐn)?shù)，表示考生正確做到這一步應(yīng)得的累加分?jǐn)?shù)4只給整數(shù)分?jǐn)?shù)，填空題不給中間分?jǐn)?shù)21【選做題】在A、B、C、D四小題中只能選做2題，每小題10分，共計(jì)20分請(qǐng)?jiān)诖鹁砜ㄖ付▍^(qū)域內(nèi)作答解答應(yīng)寫出文字說明、證明過程或演算步驟A選修41：幾何證明選講證明：(1)連接AB因?yàn)镻A是半圓O的切線，所以PACABC因?yàn)锽C是圓O的直徑，所以ABAC又因?yàn)锳HBC，所以CAHABC，所以PACCAH，所以AC是PAH的平分線 ··············&

57、#183;····························5分(2)因?yàn)镠是OC中點(diǎn)，半圓O的半徑為2，所以BH3，CH1又因?yàn)锳HBC，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是半圓O的切線，

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

59、83;10分B選修42：矩陣與變換解：設(shè)曲線C上的任意一點(diǎn)P(x，y)，P在矩陣A對(duì)應(yīng)的變換下得到點(diǎn)Q(x，y)則 ， 即x2yx，xy，所以xy，y ·····································&#

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

61、183;············10分C選修44：坐標(biāo)系與參數(shù)方程解：M的極坐標(biāo)為(1，)，故直角坐標(biāo)為M(0，1)，且P(2cos，sin)，所以PM，sin1，1 ·················5分當(dāng)sin時(shí)，PMmax，此時(shí)cos±所以，PM的最大值是，此時(shí)點(diǎn)P的坐標(biāo)是(±，)&

62、#183;······························10分D選修45：不等式選講 解：函數(shù)定義域?yàn)?，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 當(dāng)且僅當(dāng)5，即x時(shí)，取等號(hào)所以，函數(shù)f(x)5的最大值為6 ··································10分【必做題】第22題、第23題，每題10分，共計(jì)20分 22（本小題滿分10分）解：（1）記“X是奇數(shù)”為事件A，能組成的三位數(shù)的個(gè)數(shù)是48 ········&#