a) Let $a,b\in R$, $a <b$. Prove that $x \in (a,b)$ if and only if there exists $\lambda \in (0,1)$ such that $x=\lambda a +(1-\lambda)b$. b) If the function $f: R \to R$ has the property: $$f (\lambda x+(1-\lambda) y) < \lambda f(x) + (1-\lambda)f(y), \forall x,y \in R, x\ne y, \forall \lambda \in (0,1), $$prove that one cannot find four points on the function’s graph that are the vertices of a parallelogram