A continuous function `f(x)onRvec` satisfies the relation `f(x)+f(2x+y)+5x y=f(3x-y)+2x^2+1forAAx ,y in RdotT h e nfin df(x)dot`

Explanation

We have,
`f(x) +f(2x+y)+5xy=f(3x-y)+2x^(2)+1 AA x,y in R` …(1)
To get function we should eliminate `f(2x+y)` and `f(3x-y).`
For that let `2x+y=3x-y`
`implies y=(x)/(2)`
So, putting `y=(x)/(2),` we get
`f(x)+f((5x)/(2))+(5x^(2))/(2)=f((5x)/(2))+2x^(2)+1`
`implies f(x)=1-(x^(2))/(2)`