福建省福州市2012年3月高三质量检查数学试题（理科）(word)

所以PB?(23?x)2?22?x2?2(x?3)2?10，

当x?3时，PBmin?10. 此时PO?3，OH?3.········································ 7分 由（Ⅰ）知，PO所以V四棱锥P?BFED??????平面BFED,

?S梯形BFED?PO?13?(34?4?21334?2)?23?3. ······················· 8分

（ⅱ）设点Q的坐标为?a,0,c?，

由（i）知，OP?3，则A(33,0,0)，B(3,2,0)，D(3,?2,0)，P(0,0,3).

????所以AQ?a?33,0,c????，QP??a,0,3?c????， ·················································· 9分

∵AQ=?QP，

?33a?,???a?33???a,???1∴?． ??3????c?3???cc????1?????????

∴Q(33??1,0,3???1,0,)， 3?)∴OQ?(????33??1??1． ·················10分

?设平面PBD的法向量为n?(x,y,z)，则n?PB?0,n?BD?0．

????∵PB????????????????3,2,?3，BD??0,?4,0??，∴??3x?2y?????4y?03z?0, ，

取x?1，解得：y?0,z?1，

?所以n?(1,0,1).

················································ 11分

设直线OQ与平面PBD所成的角?，

?????OQ?n?????sin??cos?OQ,n??∴??????OQ?n33??12?(2?3???12?3?)23??2?9??2

33??1?1)?(??1?129?6???9??221?6?9??2． ····························· 12分

又∵??0∴sin??∵??[0,]，∴??222．·················································································· 13分 ．

?4??4因此直线OQ与平面PBD所成的角大于20.（本小题满分13分）

，即结论成立． ································ 14分

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解：（Ⅰ）由已知可得△ABC为等边三角形. 因为CD?AD，所以水下电缆的最短线路为CD.

过D作DE?AB于E，可知地下电缆的最短线路为DE、AB. ·························· 3分 又CD?1,DE?32,AB?2，

故该方案的总费用为

1?4?32?2?2?0.5

?5?3（万元） ????6分

（Ⅱ）因为?DCE所以CE则y??EB?1cos?1?????0???3???, ?3?tan?cos?1cos?,ED?tan?,AE?. ··················································· 7分

?23?4??2????3?tan??2?2?2?3?sin?cos?， ······························ 9分

， ················· 10分

令g????3?sin?cos?,则g????cos???3?sin?cos?2???sin???3sin??1cos?2因为0???记sin?0?13?3，所以0?sin???3),

32，

,?0?(0,13当0?sin?当

13?，即0≤???0时，g?????0，

32?sin??，即?0

3?13?3时， g?????0，

所以g???min?g(?0)?223?22，从而y?42?23， ···································· 12分

此时ED?tan?0?24，

?24因此施工总费用的最小值为（42?23）万元，其中ED21．（本小题满分7分） 选修4-2，矩阵与变换 解：方程组可写为??3?4．····················· 13分

1??x??2?···························································· 2分 ??????， ·

2??y??3?系数行列式为3?2?4?1?2，方程组有唯一解.

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利用矩阵求逆公式得??3?41??2??1??1????2????1?2?···················································· ?， ·

3??2?5分

?1?x??因此原方程组的解为?????y???2??1?1?1??x?,????2??2?22 ??????，即?3??3??1?1?y?.????2??2??2

···························· 7分

（2）（本小题满分7分） 选修4-4：坐标系与参数方程 解：∵直线l的极坐标方程为?(cos??sin?)?1，

∴直线l的直角坐标方程为x?y?1?0， ························································ 2分

又圆C的普通方程为(x?1)2?(y?1)?r22，

所以圆心为(1,1)，半径为r. ······································································· 4分 因为圆心到直线l的距离d?1?1?12??2222， ······················································· 6分

又因为直线l与圆C相切，所以r. ·························································· 7分

（3）（本小题满分7分）选修4-5：不等式选讲

（法一）解：∵ a，b，c?R，a2?b2?c2?1,

∴ (a?b?c)2?(a?1?b?1?c?1)2?(a2?b2?c2)(12?12?12)?3. ························· 5分 当且仅当a?b?c?33时，a?b?c取得最大值3．········································· 7分

?c?2ac2（法二）解：∵a2?b2?2ab，b2?c2?2bc，a2∴ (a?b?c)2?a2?b2?c2?2ab?2bc?2ac

222222222

················································· 3分 ?a?b?c?(a?b)?(b?c)?(a??c) ·∵ a2?b2?c2?1, ∴(a?b?c)2?3，当且仅当a?b?c?33时等号成立， ······································ 6分

∴a?b?c的最大值为3． ·····················································································

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