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?

解析:∵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.