Let $f:R\to R$ a continuous function and $F$ a primitive of $f$. Prove that there is no $a,b \in R^*$ so $f(ax+bF(x))=ax+b$, $\forall x\in R$.

Question

Assuming that a and b exists: $F(ax+bF(x))=\int f(ax+bF(x))(a+bf(x))dx+C=\int (ax+b)(a+bf(x))dx=a^2{x^2/2}+abx+b^2F(x)+ab\int xf(x) dx+C$ I know that I have to find a contradiction, or maybe a value of x for which this relationship is not true, but I am stuck at this point. I would be extremely grateful for any ideas. thank you.