函数与导数---双变量与极值点偏移问题(一)专练 1.已知定义在[0 ,) 上的函数21( )cos2f xxaxx. (1)若( )f x 为定义域上的增函数,求实数a 的取值范围; (2)若1a ,12()()0f xf x,12xx,0()f x为( )f x 的极小值,求证:1202xxx. 解:(1)由21( )cos2f xxaxx,得( )sinfxxax, ( )f x 为[0 ,) 上的增函数, 0x,( )sin0fxxax,sinaxx, 设( )sin(0)g xxx x,( )cos1 0g xx , ( )g x为减函数,( )(0)0g xg, 0a时( )f x 为定义域上的增函数, 故实数a 的取值范围是[0 ,) ; (2)证明:1a ,21( )cos2f xxxx,( )1sinfxxx , 设( )1sinh xxx ,( )1cos0h xx ,( )fx 为增函数, (0)01sin 010f ,( )1sin0f , 0(0,)x,0()0fx,当0(0,)xx时,( )0fx,( )f x 递减, 当0(xx,) 时,( )0fx,( )f x 递增,0()f x为( )f x 的极小值, 设12xx,(0)10f,2( )102f ,1020xxx, 设00( )()()F xf xxf xx,(0)x, 00( )222sincosF xxxx,0(( ))2sinsinF xxx , 0sin0x ,( )F x为增函数, 0000000(0)222sincos0222sin2(1sin)2()0Fxxxxxxfx , ( )(0)0F xF,( )F x 为增函数, 00( )(0)(0)(0)0F xFf xf x, 00()()f xxf xx,(0)x, 1202002002()()[()][()](2)f xf xf xxxf xxxfxx, 又()1sin202222f ,022xx, 02002xxx,1022xxx,即1202xxx. 2.已知函数( )sinxf xxe. (Ⅰ)求函数( )f x 在3[,2 ]2 的最大值; (Ⅱ)证明:函数1( )2( )2xg xxef x在(0,2 ) 有两个极值点1x ,2x ,并判断12xx与2 的大小关系. (Ⅰ)解:函数( )sinxf xxe, 所以( )cosxfxxe,则( )sinxfxxe , 所以当3[,2 ]2x时,sin0x,故( )0fx, 所以函数( )fx在3[,2 ]2 上单调递增, 又3()02f...