Functional Equation and Differentiability

Detailed Solution

We are given that $f:\mathbb R\to\mathbb R$ is differentiable and satisfies $$ f(x+y)=f(x)+f(y)+f'(0)xy. $$ Our objective is to determine the general form of $f$. Step 1: Find the value of $f(0)$. Putting $y=0$ in the given functional equation, we get $$ f(x)=f(x)+f(0)+f'(0)\cdot x\cdot0. $$ Therefore, $$ \boxed{f(0)=0}. $$ Step 2: Differentiate the functional equation. We have $$ f(x+y)=f(x)+f(y)+f'(0)xy. $$ Differentiating both sides with respect to $y$, while treating $x$ as a constant, gives $$ f'(x+y)=f'(y)+f'(0)x. $$ Now put $y=0$. Then $$ f'(x)=f'(0)+f'(0)x. $$ Hence, $$ \boxed{f'(x)=f'(0)(1+x)}. $$ Step 3: Integrate to obtain the general form. Integrating with respect to $x$, $$ f(x)=f'(0)\left(x+\frac{x^2}{2}\right)+C, $$ where $C$ is an arbitrary constant. Using $f(0)=0$, we obtain $$ C=0. $$ Consequently, $$ \boxed{f(x)=f'(0)x+\frac{f'(0)}{2}x^2}. $$ If we put $$ a=f'(0),\qquad b=\frac{f'(0)}{2}, $$ then $$ \boxed{f(x)=ax+bx^2}. $$ Thus, option (A) is necessarily true. Step 4: Verification of the options. (A) TRUE. We have proved that $$ f(x)=ax+bx^2 $$ where $a,b\in\mathbb R$. (B) FALSE. The function need not be linear because the quadratic term may be nonzero. For example, take $f(x)=x+\frac{x^2}{2}$. (C) FALSE. The function need not be identically zero. The same example, $$ f(x)=x+\frac{x^2}{2}, $$ satisfies the given functional equation and is nonzero. (D) FALSE. The general solution contains a linear term as well as a quadratic term: $$ f(x)=f'(0)x+\frac{f'(0)}{2}x^2. $$ Therefore, it is not necessarily of the form $ax^2$. Final Answer: $$ \boxed{\text{Option (A)}} $$