Question

已知定义域为R的函数f(x)满足f(x)+f(x+2)=2x²-4x+2,f(x+1)-f(x-1)=4(x-2).若f(t-1),-,f(t)成等差数列,则t的值为_______________.

Answer 1

2或3

解:∵f(t-1),-,f(t)成等差数列,

∴f(t-1)+f(t)=-1.

f(x+1)-f(x-1)=4(x-2)中,令x-1=m,则x=m+1.

得f(m+2)-f(m)=4(m-1),即f(x+2)-f(x)=4x-4.                      ①

又f(x+2)+f(x)=2x²-4x+2.                                     ②

②-①得

f(x)=(2x²-8x+6)=x²-4x+3.

∴f(t-1)+f(t)=t²-2t+1-4t+4+3+t²-4t+3=2t²-10t+11=-1,

即t²-5t+6=0,解得t=2或t=3.