(优辅资源)江苏省扬州中学高二上学期期末考试数学试卷

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2??ax1?lnx1?x1?0a(x1?x2)?(lnx1?lnx2)?(x12?x22)?0 ∴? 两式相减得：2??ax2?lnx2?x2?0∴a?lnx1?lnx2?(x1?x2) ……………………9分

x1?x2∴h'(x0)?h'(?x1??x2)?a?1?2(?x1??x2) ?x1??x2?lnx1?lnx2lnx1?lnx211?(x1?x2)??2(?x1??x2)??(2??1)(x1?x2)? x1?x2?x1??x2x1?x2?x1??x2∵

????0且

????1 ∴2??1?0 ∵

0?x1?x2 ∴

(2??1)(x1?x2)?0………………11分

研究：lnx1?lnx2xx?x1?的符号，即判断ln1?12的符号．

x1?x2?x1??x2x2?x1??x2令t?x1xx?xt?1t?1,t?(0,1)，ln1?12?lnt?,t?(0,1) ，设H(t)?lnt?x2x2?x1??x2?t???t??1(?t??)??(t?1)11?2t2?(2???1)t??2???∴H'(t)?? tt(?t??)2(?t??)2t(?t??)2222t?方法（一）设F(t)??t?(2???1)t??，其对称轴为：1?2??1?2?(1??)1?2???1??1 2?22?22?2222∴F(t)在(0,1)上单调减，则F(t)?F(1)???2???1???(???)?1?0，即H'(t)?0在(0,1)上恒成立 ∴H(t)在(0,1)上单调增 ∴H(t)?H(1)?0，即xx?x21ln1?? 0……………14分 x2?x??x12∵x1?x2?0 ∴lnx1?lnx21??0 x1?x2?x1??x2优质文档

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∴lnx1?lnx21?(2??1)(x1?x2)??0，即h'(x0)?0 x1?x2?x1??x2∴在点M(x0,h(x0))处的切线斜率为正． ……………………16分

?2t2?(2???1)t??2(t?1)(?2t??2)?方法（二）H'(t)? t(?t??)2t(?t??)222∵????0，0?t?1 ∴t?1?0,?t???0 ∴H'(t)?0在(0,1)上恒成立

∴H(t)在(0,1)上单调增 ∴H(t)?H(1)?0，即lnx1x?x?12?0 ……………14分 x2?x1??x2∵x1?x2?0 ∴lnx1?lnx21??0 x1?x2?x1??x2∴lnx1?lnx21?(2??1)(x1?x2)??0，即h'(x0)?0 x1?x2?x1??x2∴在点M(x0,h(x0))处的切线斜率为正． ……………………16分

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