Question

已知f(x)是定义在R上的不恒为零的函数,且对于任意的a,b属于R都满足f(ab)=af(b)+bf(a),(1)求f(

Answer 1

①

∵ f(ab)=af(b)+bf(a)

∴ f(0) = f(0*0)=0f(0)+0f(0) = 0

f(1) = f(1*1)=1f(1)+1f(1)= 2f(1)

∴ f(1)=0

②

0=f(1)=f(-1*-1)=-1*f(-1)-1*f(-1)=-2f(-1)

∴ f(-1)=0

f(-a)=f(-1*a)= -1f(a)+af(-1) = -f(a)

∴ f(x)为奇函数.