江蘇省蘇州市高三期中調(diào)研數(shù)學(xué)試卷Word版含答案

1、高考數(shù)學(xué)精品復(fù)習(xí)資料 2019.5蘇州市20xx20xx學(xué)年第一學(xué)期高三期中調(diào)研試卷 數(shù) 學(xué) 20xx.11注意事項(xiàng)：1本試卷共4頁(yè)滿分160分，考試時(shí)間120分鐘2請(qǐng)將填空題的答案和解答題的解題過程寫在答題卷上，在本試卷上答題無效3答題前，務(wù)必將自己的姓名、學(xué)校、準(zhǔn)考證號(hào)寫在答題紙的密封線內(nèi)一、填空題(本大題共14小題，每小題5分，共70分，請(qǐng)把答案直接填寫在答卷紙相應(yīng)的位置)1已知集合，則 2函數(shù)的定義域?yàn)?3設(shè)命題；命題，那么p是q的 條件（選填“充分不必要”、“必要不充分”、“充要”、“既不充分也不必要”）4已知冪函數(shù)在是增函數(shù)，則實(shí)數(shù)m的值是 5已知曲線在處的切線的斜率為2，則實(shí)數(shù)a

2、的值是 6已知等比數(shù)列中，則 7函數(shù)圖象的一條對(duì)稱軸是，則的值是 8已知奇函數(shù)在上單調(diào)遞減，且，則不等式的解集為 9已知，則的值是 10若函數(shù)的值域?yàn)?，則實(shí)數(shù)a的取值范圍是 11已知數(shù)列滿足，則 12設(shè)的內(nèi)角的對(duì)邊分別是，d為的中點(diǎn)，若且，則面積的最大值是 13已知函數(shù)，若對(duì)任意的實(shí)數(shù)，都存在唯一的實(shí)數(shù)，使，則實(shí)數(shù)的最小值是 14已知函數(shù)，若直線與交于三個(gè)不同的點(diǎn)（其中），則的取值范圍是 二、解答題(本大題共6個(gè)小題，共90分，請(qǐng)?jiān)诖痤}卷區(qū)域內(nèi)作答，解答時(shí)應(yīng)寫出文字說明、證明過程或演算步驟)15(本題滿分14分)已知函數(shù)的圖象與x軸相切，且圖象上相鄰兩個(gè)最高點(diǎn)之間的距離為（1）求的值；（2）求

3、在上的最大值和最小值16(本題滿分14分)在中，角a,b,c所對(duì)的邊分別是a,b,c，已知，且（1）當(dāng)時(shí)，求的值；（2）若角a為銳角，求m的取值范圍17(本題滿分15分) 已知數(shù)列的前n項(xiàng)和是，且滿足，（1）求數(shù)列的通項(xiàng)公式；（2）在數(shù)列中，若不等式對(duì)有解，求實(shí)數(shù)的取值范圍18(本題滿分15分)如圖所示的自動(dòng)通風(fēng)設(shè)施該設(shè)施的下部abcd是等腰梯形，其中為2米，梯形的高為1米，為3米，上部是個(gè)半圓，固定點(diǎn)e為cd的中點(diǎn)mn是由電腦控制可以上下滑動(dòng)的伸縮橫桿（橫桿面積可忽略不計(jì)），且滑動(dòng)過程中始終保持和cd平行當(dāng)mn位于cd下方和上方時(shí)，通風(fēng)窗的形狀均為矩形mngh（陰影部分均不通風(fēng)）（1）設(shè)mn

4、與ab之間的距離為且米，試將通風(fēng)窗的通風(fēng)面積s（平方米）表示成關(guān)于x的函數(shù)；（2）當(dāng)mn與ab之間的距離為多少米時(shí)，通風(fēng)窗的通風(fēng)面積取得最大值？ 19(本題滿分16分)已知函數(shù)（1）求過點(diǎn)的的切線方程；（2）當(dāng)時(shí)，求函數(shù)在的最大值；（3）證明：當(dāng)時(shí)，不等式對(duì)任意均成立（其中為自然對(duì)數(shù)的底數(shù),）20(本題滿分16分)已知數(shù)列各項(xiàng)均為正數(shù)，,，且對(duì)任意恒成立，記的前n項(xiàng)和為（1）若，求的值；（2）證明：對(duì)任意正實(shí)數(shù)p，成等比數(shù)列；（3）是否存在正實(shí)數(shù)t，使得數(shù)列為等比數(shù)列若存在，求出此時(shí)和的表達(dá)式；若不存在，說明理由20xx20xx學(xué)年第一學(xué)期高三期中調(diào)研試卷數(shù) 學(xué) (附加) 20xx.11注意事

5、項(xiàng)：1本試卷共2頁(yè)滿分40分，考試時(shí)間30分鐘2請(qǐng)?jiān)诖痤}卡上的指定位置作答，在本試卷上作答無效3答題前，請(qǐng)務(wù)必將自己的姓名、學(xué)校、考試證號(hào)填寫在答題卡的規(guī)定位置21【選做題】本題包括a、b、c、d四小題，請(qǐng)選定其中兩題，并在相應(yīng)的答題區(qū)域內(nèi)作答若多做，則按作答的前兩題評(píng)分解答時(shí)應(yīng)寫出文字說明、證明過程或演算步驟a(幾何證明選講) （本小題滿分10分）如圖，ab為圓o的直徑，c在圓o上，于f，點(diǎn)d為線段cf上任意一點(diǎn)，延長(zhǎng)ad交圓o于e，（1）求證：；（2）若，求的值b(矩陣與變換)（本小題滿分10分）已知矩陣，求的值c(極坐標(biāo)與參數(shù)方程) （本小題滿分10分）在平面直角坐標(biāo)系中，直線的參數(shù)方程

6、為(為參數(shù))，以原點(diǎn)為極點(diǎn)，軸正半軸為極軸建立極坐標(biāo)系，圓的極坐標(biāo)方程為（1）求直線和圓的直角坐標(biāo)方程；（2）若圓c任意一條直徑的兩個(gè)端點(diǎn)到直線l的距離之和為，求a的值d(不等式選講)（本小題滿分10分）設(shè)均為正數(shù)，且，求證：【必做題】第22、23題，每小題10分，共計(jì)20分請(qǐng)?jiān)诖痤}卡指定區(qū)域內(nèi)作答，解答時(shí)應(yīng)寫出文字說明、證明過程或演算步驟22(本小題滿分10分)在小明的婚禮上，為了活躍氣氛，主持人邀請(qǐng)10位客人做一個(gè)游戲第一輪游戲中，主持人將標(biāo)有數(shù)字1,2,10的十張相同的卡片放入一個(gè)不透明箱子中，讓客人依次去摸，摸到數(shù)字6,7,10的客人留下，其余的淘汰，第二輪放入1,2,5五張卡片，讓留

7、下的客人依次去摸，摸到數(shù)字3,4,5的客人留下，第三輪放入1,2,3三張卡片，讓留下的客人依次去摸，摸到數(shù)字2,3的客人留下，同樣第四輪淘汰一位，最后留下的客人獲得小明準(zhǔn)備的禮物已知客人甲參加了該游戲（1）求甲拿到禮物的概率；（2）設(shè)表示甲參加游戲的輪數(shù)，求的概率分布和數(shù)學(xué)期望23(本小題滿分10分)（1）若不等式對(duì)任意恒成立，求實(shí)數(shù)a的取值范圍；（2）設(shè)，試比較與的大小，并證明你的結(jié)論20xx20xx學(xué)年第一學(xué)期高三期中調(diào)研試卷數(shù) 學(xué) 參 考 答 案一、填空題(本大題共14小題，每小題5分，共70分)1 2 3充分不必要 41 564 7 8 9 10 11 12 13 14二、解答題(本大

8、題共6個(gè)小題，共90分)15(本題滿分14分)解：（1）圖象上相鄰兩個(gè)最高點(diǎn)之間的距離為，的周期為，·········································

9、3;····························2分，·····················&

10、#183;·················································&

11、#183;··········································4分此時(shí)，又的圖象與x軸相切，····

12、3;·················································

13、3;6分；·················································&

14、#183;·················································&

15、#183;······8分（2）由（1）可得，當(dāng)，即時(shí)，有最大值為；·······································&#

16、183;·········11分當(dāng)，即時(shí)，有最小值為0······································

17、··················14分16(本題滿分14分)解：由題意得，····························

18、83;·················································

19、83;2分（1）當(dāng)時(shí)，解得或；···············································

20、83;················································6分（2），

21、83;···························8分a為銳角，·····················&#

22、183;······························11分又由可得，··················

23、··················································

24、·····················13分····························

25、83;·················································

26、83;······················14分17(本題滿分15分)解：（1），························&

27、#183;·················································&

28、#183;··············2分又當(dāng)時(shí)，由得符合，······························3分?jǐn)?shù)列是以1為首項(xiàng)，3為公比的等比數(shù)列

29、，通項(xiàng)公式為；·····················5分（2），是以3為首項(xiàng)，3為公差的等差數(shù)列，····················7分，···

30、83;·················································

31、83;·······························9分，即，即對(duì)有解，·················

32、;·················10分設(shè)，當(dāng)時(shí)，當(dāng)時(shí)，·······························

33、··················································

34、··········14分·······································

35、83;·················································

36、83;···················15分18(本題滿分15分)解：（1）當(dāng)時(shí)，過作于（如上圖），則，由，得，；························&

37、#183;······································4分當(dāng)時(shí)，過作于，連結(jié)（如下圖），則，········

38、··················································

39、············8分綜上：；·····································

40、····························9分（2）當(dāng)時(shí)，在上遞減，；···················

41、3;·················································

42、3;··························11分當(dāng)時(shí)， 當(dāng)且僅當(dāng)，即時(shí)取“”，此時(shí)，的最大值為，···················&

43、#183;························14分答：當(dāng)mn與ab之間的距離為米時(shí)，通風(fēng)窗的通風(fēng)面積取得最大值···················

44、3;15分19(本題滿分16分)解：（1）設(shè)切點(diǎn)坐標(biāo)為，則切線方程為，將代入上式，得，切線方程為；··········································

45、;··················································

46、;···2分（2）當(dāng)時(shí)，·············································

47、83;······························3分當(dāng)時(shí)，當(dāng)時(shí)，在遞增，在遞減，·················

48、;··················································

49、;··········5分當(dāng)時(shí)，的最大值為；當(dāng)時(shí)，的最大值為；····································

50、83;···································7分（3）可化為，設(shè)，要證時(shí)對(duì)任意均成立，只要證，下證此結(jié)論成立，當(dāng)時(shí)，········

51、83;··············································8分設(shè)，則，在遞增，又在區(qū)間上的圖象是一條

52、不間斷的曲線，且，使得，即，···············································

53、3;····11分當(dāng)時(shí)，；當(dāng)時(shí)，；函數(shù)在遞增，在遞減，····························14分在遞增，即，當(dāng)時(shí)，不等式對(duì)任意均成立··········

54、3;···············16分20(本題滿分16分)解：（1），又，；·······························

55、;········2分（2）由，兩式相乘得，從而的奇數(shù)項(xiàng)和偶數(shù)項(xiàng)均構(gòu)成等比數(shù)列，····································

56、3;······························4分設(shè)公比分別為，則，··················

57、;····················5分又，即，·····························

58、;······························6分設(shè)，則，且恒成立，數(shù)列是首項(xiàng)為，公比為的等比數(shù)列，問題得證；··············

59、83;·····················8分（3）法一：在（2）中令，則數(shù)列是首項(xiàng)為，公比為的等比數(shù)列，·······················&

60、#183;·············································10分且，數(shù)列為等比數(shù)列，即，即解得（舍去），&#

61、183;·················································&#

62、183;······································13分，從而對(duì)任意有，此時(shí)，為常數(shù)，滿足成等比數(shù)列，當(dāng)時(shí)，又，綜上，存在使數(shù)列為等比數(shù)列，此時(shí)···

63、83;··················16分法二：由（2）知，則，且，數(shù)列為等比數(shù)列，即，即解得（舍去），··························

64、;··················································

65、;···········11分，從而對(duì)任意有，····································13分，此時(shí)

66、，為常數(shù)，滿足成等比數(shù)列，綜上，存在使數(shù)列為等比數(shù)列，此時(shí)······················16分21【選做題】本題包括a、b、c、d四小題，請(qǐng)選定其中兩題，并在相應(yīng)的答題區(qū)域內(nèi)作答若多做，則按作答的前兩題評(píng)分解答時(shí)應(yīng)寫出文字說明、證明過程或演算步驟a(幾何證明選講，本小題滿分10分)解：（1）證明 ：連接，又，為等邊三角形，為中邊上的中線，；··

67、··················································

68、··················5分（2）解：連接be，是等邊三角形，可求得，為圓o的直徑，又，即··························

69、83;·················································

70、83;·····10分b(矩陣與變換，本小題滿分10分)解：矩陣a的特征多項(xiàng)式為，令，解得矩陣a的特征值，·····································

71、·······················2分當(dāng)時(shí)特征向量為，當(dāng)時(shí)特征向量為，························

72、·············6分又，····································&#

73、183;·················································&#

74、183;·······8分··········································

75、·································10分c(極坐標(biāo)與參數(shù)方程，本小題滿分10分)解：（1）直線的普通方程為；···········

76、;··················································

77、;·············3分圓c的直角坐標(biāo)方程為；··································

78、3;····························6分（2）圓c任意一條直徑的兩個(gè)端點(diǎn)到直線l的距離之和為，圓心c到直線l的距離為，即，··············

79、83;········································8分解得或········

80、3;·················································

81、3;····································10分d(不等式選講，本小題滿分10分)證：，··········

82、;··················································

83、;························10分22(本題滿分10分)解：（1）甲拿到禮物的事件為，在每一輪游戲中，甲留下的概率和他摸卡片的順序無關(guān)，則，答：甲拿到禮物的概率為；··············

84、83;·················································

85、83;······················3分（2）隨機(jī)變量的所有可能取值是1,2,3,4·······················

86、83;·············································4分，隨機(jī)變量的概率分布列為：123·

87、83;···········································8分4p所以·····

88、83;·················································

89、83;············10分23(本題滿分10分)解：（1）原問題等價(jià)于對(duì)任意恒成立，令，則，當(dāng)時(shí)，恒成立，即在上單調(diào)遞增，恒成立；當(dāng)時(shí)，令，則，在上單調(diào)遞減，在上單調(diào)遞增，即存在使得，不合題意；綜上所述，a的取值范圍是······································