【正文】 xxfxfffxxcbxxxf??????????????[例 2] 4 D. 3 C. 2 B. 1 A.) ( )]([,0)]([,)()()( 2 0 0 4 是不可能則有實數(shù)解方程且上的函數(shù)都是定義在實數(shù)集和若年浙江高考題xfgxgfxRxgxf??[例 3] 51 D. 51 C.51 B. 51 A.2222??????xxxxxx) ] ,([)]([).()]([) ] } ,([{)() ] ,([,0)]([ 000000xgfxfgxgxxfgxxgfgxgxgfxxxgfx????????有實數(shù)解即方程于是則有實數(shù)解設方程[解析 ] .51,51。,.,0]43)21[(4)1(40)1(230)(39。),21)(4()(,21:0)1(39。1)()( )2( )( )1( .2)(,)()( )( 20 05 2的取值范圍求實數(shù)上是增函數(shù)在若解不等式的解析式；求函數(shù)且的圖象關于原點對稱和已知函數(shù)年浙江高考題???????????xfxgxhxxfxgxgxxxfxgxf[例 2] ,2,2,)(),(.,0202),(),()( 1)( 2200000000xxyxxyxfyyxQyyxxyyxxyxPyxQxfy???????????????????????????即的圖象上在函數(shù)點即則關于原點的對稱點一點的圖象上任設函數(shù)?[解析 ] .2)( 2 xxxg ???故.2)( 2 xxxg ???故].21,1[,211,012,1。)()(39。0)(39。 )1( ).)(4()(, )( 2020 2的取值范圍求都是遞增的上和在若上的最大值和最小值；在求若；求導數(shù)為實數(shù)已知年浙江高考題axfxffxfaxxxfa???????????[例 3] .423)(39。1,]1,1[14)(,1 1 .1)1(2)1()( )3( 2?????????????????????????????????????????????????????綜上解得時當解得時當對稱軸的方程為時當上是增函數(shù)在時當xxxhxxxh.,)