问题 选择题
已知函数f(x)满足f(2x-1)=
1
2
f(x)+x2-x+2
,则函数f(x)在(1,f(1))处的切线是(  )
A.2x+3y+12=0B.2x-3y+10=0C.2x-y+2=0D.2x-y-2=0
答案

f(2x-1)=

1
2
f(x)+x2-x+2,

再边对x求导,∴2f'(2x-1)=

1
2
f'(x)+2x-1.令x=1,

∴2f'(1)=

1
2
f'(1)+1.

∴f'(1)=

2
3

∴y=f(x)在(1,f(1))处的切线斜率为k=

2
3

又在f(2x-1)=

1
2
f(x)+x2-x+2中令x=1,得f(1)=4

∴函数y=f(x)在(1,f(1))处的切线方程为y-4=

2
3
(x-1),

即2x-3y+10=0.

故选B.

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