2024届福州市5月质检文科数学试题及答案(2)

ss5uKu（Ⅰ）求圆C的方程；KKs55u

yCA（Ⅱ）过点M任作一条直线与椭圆?:xy??1相交于4822TA、B两点，连接AN、BN，求证：?ANM??BNM．

OMNx

22．（本小题满分14分）

Kss5uKuKs55u

B第21题图 已知函数f(x)?ln(x?1)?ax?a?R?． x?1（Ⅰ）当a?2时，求函数y?f?x?的图象在x?0处的切线方程； （Ⅱ）判断函数f(x)的单调性；

（Ⅲ）若函数f?x?在?a,a?1?上为增函数，求a的取值范围．

2012年福州市高中毕业班综合练习 文科数学试卷参考答案及评分参考

一、选择题(本大题共10小题,每小题5分,共50分)

1.B 2. B 3.A 4.A 5. C 6. C 7. D 8. B 9.C 10. D 11. C 12. B 二、填空题(本大题共5小题,每小题4分,共20分) 13. ?1,?2? 14. 3 15. ?5 16. （1）an?2?an?1?an，a1?1，a2?1；或直接列举4n?12出数列各项；（前2项不是主要的）（2）an?2?an?an?1???1?和

an?0.618（不唯an?1一，关键要反映“64=65”的一般关系和拼接后以假乱真的原因）

三、解答题(本大题共6小题，共80分) 17．（本小题满分12分）

解：（Ⅰ）估计该校高三学生质检数学成绩在125～140分之间的概率p1为

33?1?2?6?8?320， ··································································· 2分

又设样本容量为m，则

63?，解得，m?40． ································· 4分 m20（Ⅱ）样本中成绩在65～80分之间的学生有80～95分之间的学生

1?40=2人，记为x,y；成绩在202?40=4人，记为a,b,c,d， ···································· 5分20

从上述6人中任选2人的所有可能情形有：

?x,y?,?x,a?,?x,b?,?x,c?,?x,d?,?y,a?,?y,b?,?y,c?,?y,d?,

····························· 8分?a,b?,?a,c?,?a,d?,?b,c?,?b,d?,?c,d?，共15种， ·

至少有1人在65～80分之间的可能情形有

···· 11分?x,y?,?x,a?,?x,b?,?x,c?,?x,d?,?y,a?,?y,b?,?y,c?,?y,d?,共9种， ·

93?． ·因此，所求的概率p2?···················································· 12分 15518．（本小题满分12分）

解：（Ⅰ）∵ ?是锐角，sin??3， 54∴ cos??1?sin2??． ······························································ 2分

5根据三角函数的定义，得cos??又∵ ?是锐角， ∴ sin??1?cos2??5，13

12．····························································· 4分 134531216∴ cos??????cos?cos??sin?sin???????． ·················· 6分

51351365（Ⅱ）由题意可知，OA?(cos?，sin?)，OC?(23,?2)．

?∴ f(?)?OA?OC?23cos??2sin??4cos(??)， ····························· 8分

6?∵ 0???，

2??2?∴ ， ····································································· 9分 ????663∴ ?1?3?cos(a?)?，从而?2?f(?)?23， ···························· 11分 262∴ 函数f(?)的值域为(?2,23)． ·················································· 12分 19．（本小题满分12分）

解：（I）设甲公司第n年市场占有率为an，依题意，?an?是以a1?A为首项，以

d?A为公差的等差数列． ····································································· 2分 2∴ an?A?(n?1)?AAA······················································ 3分 ?n?． ·

222设乙公司第n年市场占有率为bn，根据图形可得：

bn?A?1111···················································· 5分 A?2A?3A?...?n?1A ·

22221????2?n?1?A． ··········································································· 6分

2??（II）依题意，2012年为第20年，则

a20?AA211······························· 9分 ?20??A?10A，b20?(2?19)A?2A， ·

2222b202A??20%，即b20?20%?a20， ············································ 11分 a2010A∴

∴ 2012年会出现乙公司被甲公司兼并的局面． ································· 12分 20．（本小题满分12分）

解：（Ⅰ）分别取PC、PD中点E、F，连结EF，则EF即为所求，下证之：1分 ∵ E、F分别为PC、PD中点，

∴ EF//CD． ····································· 2分 ∵ EF?平面ABCD，CD?平面ABCD,…3分 ∴ EF//平面ABCD．··························· 4分 （作法不唯一）

（Ⅱ）由三视图可知，PA?平面ABCD，

BC?2AD?2CD?2，四边形ABCD为直角梯形．

过点A作AG?BC于G，则AG?CD?1,GC?AD?1．

∴ AC?AD2?CD2?2,AB?AG2?BG2?12?(2?1)2?2，

∴ AC2?AB2?BC2，故AC?AB． ················································· 6分 ∵ PA?平面ABCD,AC?平面ABCD,

∴ PA?AC. ··············································································· 7分 ∵ PAAB?A，

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