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If f(x+f(y))=f(x)+yAAx ,y in R a n df(0...

If `f(x+f(y))=f(x)+yAAx ,y in R a n df(0)=1,` then find the value of `f(7)dot`

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The correct Answer is:
1
`f(x+f(y))=f(x)+y,f(0)=1`
Putting `y=0,` we get
`f(x+f(0))=f(x)+0`
or ` f(x+1)=f(x) AA x in R`
Thus, `f(x)` is periodic with 1 as one of its period. Hence,
`f(7)=f(6)=f(5)= …=f(1)=(0)=1`
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