The number of linear functions f satisfying f(x+f(x)) =x+f(x)∀x∈R is
0
1
2
3
Let f(x)=ax+b------1
⇒f(ax+b+x)=x+ax+b⇒f((a+1)x+b)=(a+1)x+b Replace (a+1)x+b by y, we have ⇒f(y)=(a+1)y−ba+1+b
or f(x)=(a+1)x−ba+1+b-----2
Now compare (1) and (2), we get a=±1,b=0
There are two linear functions. Hence, the least value is 2.