B
C D B A C C A 第二部分（非选择题共110分）
二、填空题共6小题，每小题5分，共30分． （9）1（10）10
（11）1；23（12）
7 3（13）答案不唯一，a?0或a?4的任意实数（14）
25 5
三、解答题共6小题，共80分．解答应写出文字说明，演算步骤或证明过程． （15）（本小题13分） 解：（Ⅰ）A?2，??2，????······································································· 7分 ． 3?（Ⅱ）由（Ⅰ）得，f(x)?2sin(2x?)．
3?1········································· 8分 因为f(?)?1，所以sin(2??)?． ·
325?2???································· 9分 ,)，所以2???(,?)． ·因为??(12332?5·································································· 11分 所以2????， ·
367········································································ 12分 所以2???， ·
6所以cos2??cos???763····················································· 13分 ． 2
16. （本小题共13分）
解：（Ⅰ）这10名学生的考核成绩（单位：分）分别为：
93，89.5，89，88，90，88.5，91.5，91，90.5，91．
· 1分 其中大于等于90分的有1号、5号、7号、8号、9号、10号，共6人. ·
所以样本中学生考核成绩大于等于90分的频率为：
6?0.6， ································································ 3分 10从该校高二年级随机选取一名学生，估计这名学生考核成绩大于等于90分的概率为0.6. ································································································································ 4分
（Ⅱ）设事件A：从上述考核成绩大于等于90分的学生中再随机抽取两名同学，这两
····················································· 5分 名同学两轮测试成绩均大于等于90分. ·
由（Ⅰ）知，上述考核成绩大于等于90分的学生共6人，其中两轮测试成绩均大于等
········································· 6分 于90分的学生有1号，8号，10号，共3人. ·
C3231?. ························································· 9分 所以，P(A)?2?C615522·································································· 13分 （Ⅲ）x1?x2，s1?s2. ·
17.（本小题共14分） 解：（Ⅰ）因为AB1⊥平面ABC，AC?平面ABC，
··············································································· 1分 所以AB1?AC． ·
因为AC1?AC，AB1AC1?A，AB1，AC1?平面AB1C1，
····························································· 3分 所以AC?平面AB1C1． ·因为B1C1?平面AB1C1，
······································································ 4分 所以AC?B1C1． ·
（Ⅱ）法一：取A1B1的中点M，连接MA、ME． 因为E、M分别是B1C1、A1B1的中点， 所以ME∥A1C1，且ME?1A1C1． ························································ 5分 2C A1C1，且AD?在三棱柱ABC?A1B1C1中，AD1AC11， 21 E B 1 A 1 M 所以ME∥AD，且ME=AD，
············· 6分 所以四边形ADEM是平行四边形， ·
······································ 7分 所以DE∥AM． ·
又AM?平面AA1B1B，DE?平面AA1B1B， ························· 9分 所以DE//平面AA1BB． ·
注：与此法类似，还可取AB的中点M，连接MD、MB1． 法二：取AB的中点M，连接MD、MB1． 因为D、M分别是AC、AB的中点，
C B D A C 1 A1
B1 E 1BC． ·················· 5分 21在三棱柱ABC?A1B1C1中，B1EBC，且B1E?BC，
2C 所以MD∥B1E，且MD=B1E，
··········· 6分 所以四边形B1E DM是平行四边形， ·
····································· 7分 所以DE∥MB1． ·
又MB1?平面AA1B1B，DE?平面AA1B1B，
所以MD∥BC，且MD?························· 9分 所以DE//平面AA1BB． ·
法三：取BC的中点M，连接MD、ME． 因为D、M分别是CA、CB的中点，
D B M A ··································································· 5分 所以，DM//AB． ·
在三棱柱ABC?A1B1C1中，BC//B1C1，BC?B1C1，
C1 因为E、M分别是C1B1和CB的中点，
所以，MB//EB1，MB?EB1，
·········· 6分 所以，四边形MBB1E是平行四边形， ·
·································· 7分 所以，ME//BB1． ·又因为MEA1
E B1 MD?M，BB1AB?B，
C M D B A ME,MD?平面MDE，BB1,AB?平面AA1B1B，
············· 8分 所以，平面MDE//平面AA1B1B． ·因为，DE?平面MDE，
······················ 9分 所以，DE//平面AA1BB． ·
（Ⅲ）在三棱柱ABC?A1B1C1中，BC//B1C1， 因为AC?B1C1，所以AC?BC．
在平面ACB1内，过点C作Cz//AB1， 因为，AB1?平面ABC，
·························· 10分 所以，Cz?平面ABC． ·
建立空间直角坐标系C-xyz，如图．则
zC1EB1A1CDyABxC(0,0,0),B(2,0,0),B1(0,2,2),C1(?2,2,2),D(0,1,0),E(?1,2,2).
··························· 11分 DE?(?1,1,2)，CB?(2,0,0)，CB1?(0,2,2)． ·设平面BB1C1C的法向量为n?(x,y,z)，则
??2x?0?n?CB?0，即， ???2y?2z?0??n?CB1?0···························· 12分 得x?0，令y?1，得z??1，故n?(0,1,?1)． ·设直线DE与平面BB1C1C所成的角为θ， 则sinθ＝cos?DE,n??DE?n|DE|?|n|?3， 63. ·························· 14分 6所以直线DE与平面BB1C1C所成角的正弦值为
18. （本小题共14分）
x2解：（Ⅰ）在椭圆C：?y2?1中，a?2，b?1，
4···························································· 2分 所以c?a2?b2?3， ·
····················································· 3分 故椭圆C的焦距为2c?23， ·
c3·································································· 5分 ． ·?a2（Ⅱ）法一：设P(x0,y0)（x0?0，y0?0），
离心率e?xx22T················· 6分 ?y0?1，故y0?1?． ·
P443222222所以|TP|?|OP|?|OT|?x0?y0?1?x0，
4OF3································· 8分 所以|TP|?x0， ·
213·········· 9分 S?OTP?|OT|?|TP|?x0． ·
2413··················· 10分 又O(0,0)，F(3,0)，故S?OFP?OF?y0?y0． ·
223x0······························· 11分 因此S四边形OFPT?S?OFP?S?OTP??(?y0) ·
22则
23x032???x0y0?y0??1?x0y0． 24222x0x022由?y0?1，得2?y0?1，即x0?y0?1，
4436········································· 13分 所以S四边形OFPT?， ·?1?x0y0?222x0122················· 14分 当且仅当时等号成立. ?y0?，即x0?2，y0?4222020yx（Ⅱ）法二：设P(2cos?,sin?)（0????2········································ 6分 ），
222222则|TP|?|OP|?|OT|?4cos??sin??1?3cos?，
所以|TP|?······························································· 8分 3cos?， ·
13········································ 9分 |OT|?|TP|?cos?． ·
2213··············· 10分 又O(0,0)，F(3,0)，故S?OFP?OF?y0?sin?． ·
223························ 11分 因此S四边形OFPT?S?OFP?S?OTP??(cos??sin?) ·
26?6········································· 13分 ， ·?sin(??)?242?2···················· 14分 当且仅当??时，即x0?2，y0?时等号成立．
42S?OTP?
19.（本小题共13分）
axax················· 1分 解：（Ⅰ）法一：f'(x)?a?e?a?a?(e?1)(a?0,x?R)， ·
····························································· 2分 令f'(x)?0，得x?0． ·
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