?定义域为(-?,0)?(0,+?) 2(?x)2?12x2?1且f(?x)?????f(x)?xx?f(x)为奇函数...........6'..........8'2x12?12x22?1(3)任取1?x1?x2,则f(x1)?f(x2)? ?x1x212(x?x)(xx?)1212x2?x12?2(x1?x2)??x1x2x1x2..........10'
12(x1?x2)(x1x2?)2?0?x1?x2?0,x1?1,x2?1?x1x2?1?x1x2
?f(x1)?f(x2)?f(x)在[1,??)上是增函数.19.解(1) 当x?(0,10]时,fmax(x)?f(10)?110,
..........12'当x?(10,35]时,f(x)?110,当x?(35,60]时,f(x)?110故开讲后10分钟,学生能力最强,能维持25分钟...........3' ..........4'(2)由f(5)?105,f(38)?107故学生在开讲后38分钟比开讲后5分钟接受能力强一些.
..........6'..........8'(3)当x?(35,60]时,?x?145?105,故x?40,故每节课最长为40分钟比较合理20.解 ?2x2?3x?10...........12...........1'
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1?()x?24?2x?3x?10?2?2x?42?x2?3x?10??2x?4??3?x??2...........2'...........3'
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(1)?f(x)?(2?x)2?6?2?x?1而x?[?3,?2],令t=2?x?[4,8]，则y?t2?6t?1...........4'...........6'...........8'42?x...........10' ...........12'...........13'?y?[?7,17]即f(x)?[?7,17](2)?(2?x)2?a?2?x?4?0?a?2?x?4令t?2?x?[4,8]则y?t?在[4,8]递增t4?ymin?4??3?a?(??,3]421.解
(1)?f(x)为定义在R上的奇函数?x2?2x,x?0,??f(x)??2???x?2x,x?0.(2)f(x)在R上单调递增.证明如下:?f(0)?0...........1'当x<0时,?x?0?f(x)??f(?x)??[(?x)2?2(?x)]?f(x)??x2?2x...........3'...........4'当x?[0,??)时,任取x1,x2?[0,??)且x1 ...........9'...........12' 7 而此不等式组对于k?[0,1]恒成立,?(x?1)?0?(x2?1)?0?2?(x?1)?1?(x?1)?0??2?2x?0?(x?2)?0?2x?1?(x2?2)?0??x?R?x?R?即???2?x?2???1?3?x??1?3 ?x?(?2,?1?3)8 ...........14'