海淀区高三年级第一学期期末练习数学(文)答案及评分参考2011.1第Ⅰ卷(选择题共40分)一、选择题(本大题共8小题,每小题5分,共40分)题号12345678答案CAACBDBD第II卷(非选择题共110分)二、填空题(本大题共6小题,每小题5分,共30分.有两空的题目,第一空3分,第二空2分)9.240xy10.1911.(3,0)212yx12.2513.214.43三、解答题(本大题共6小题,共80分)15.(共13分)解:(I)xxxfcos23sin21)()3sin(x,...............................3分)(xf的周期为2(或答:0,,2kZkk).................................4分因为xR,所以3xR,所以)(xf值域为]1,1[................................5分(II)由(I)可知,)3sin()(AAf,...............................6分23)3sin(A,...............................7分A0,3433A,..................................8分2,33A得到3A................................9分,23ba且BbAasinsin,....................................10分32sin32bbB,1sinB,....................................11分B0,2B.....................................12分6BAC.....................................13分16.(共13分)解:(I)围棋社共有60人,...................................1分由150301260可知三个社团一共有150人....................................3分(II)设初中的两名同学为21,aa,高中的3名同学为321,,bbb,...................................5分随机选出2人参加书法展示所有可能的结果:1211121321{,},{,},{,},{,},{,},aaabababab2223121323{,},{,},{,},{,},{,}ababbbbbbb,共10个基本事件...................................8分设事件A表示“书法展示的同学中初、高中学生都有”,..................................9分则事件A共有111213212223{,},{,},{,},{,},{,},{,}abababababab6个基本事件....................................11分53106)(AP.故参加书法展示的2人中初、高中学生都有的概率为35.................................13分17.(共13分)解:(I)四边形ABCD为菱形且ACBDO,O是BD的中点....................................2分又点F为1DC的中点,在1DBC中,1//BCOF,...................................4分OF平面11BCCB,1BC平面11BCCB,//OF平面11BCCB....................................6分(II)四边形ABCD为菱形,ACBD,...................................8分又BD1AA,1,AAACA且1,AAAC平面11ACCA,.................................10分BD平面11ACCA,................................11分BD平面1DBC,平面1DBC平面11ACCA.................................13分18.(共13分)解:3332222()()2axafxxxx,0x..........................................2分(I)由题意可得3(1)2(1)0fa,解得1a,........................................3分此时(1)4f,在点(1,(1))f处的切线为4y,与直线1y平行.故所求a值为1.........................................4分(II)由()0fx可得xa,0a,........................................5分①当01a时,()0fx在(1,2]上恒成立,所以()yfx在[1,2]上递增,.....................................6分所以()fx在[1,2]上的最小值为3(1)22fa.........................................7分②当12a时,x(1,)aa(,2)a()fx-0+()fx极小由上表可得()yfx在[1,2]上的最小值为2()31faa.......................................11分③当2a时,()0fx在[1,2)上恒成立,所以()yfx在[1,2]上递减.......................................12分所以()fx在[1,2]上的最小值为3(2)5fa......................................13分综上讨论,可知:当01a时,()yfx在[1,2]上的最小值为3(1)22fa;当12a时,()yfx在[1,2...