已知函数f(x)在R上满足y=f(x)=2f(2-x)+ex-1+x2,则曲线y=f(x)在点(1,f(1))处的切线方程是()A.2x-y-1=0B.x-y-3=0C.3x-y-2=0D.2x+y-

题目简介

已知函数f(x)在R上满足y=f(x)=2f(2-x)+ex-1+x2,则曲线y=f(x)在点(1,f(1))处的切线方程是()A.2x-y-1=0B.x-y-3=0C.3x-y-2=0D.2x+y-

题目详情

已知函数f(x)在R上满足y=f(x)=2f(2-x)+ex-1+x2,则曲线y=f(x)在点(1,f(1))处的切线方程是(  )
A.2x-y-1=0B.x-y-3=0C.3x-y-2=0D.2x+y-3=0
题型:单选题难度:偏易来源:不详

答案

f(2-x)=2f(x)+e1-x+(2-x)2     ①
f(x)=2f(2-x)+ex-1+x2,②
联立①②解得:f(x)=-class="stub"1
3
(2e1-x+ex-1+3x2-8x+8 )
f(1)=-2,f'(1)=1
∴曲线y=f(x)在点(1,f(1))处的切线方程是x-y-3=0
故选B.

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