Suppose that f is a differentiable function with the property that f(x + y) = f(x) + f(y) + xy and $\underset{h→0}{\lim}\frac{1}{h}f(h)=3$, then
Answer & explanation
Correct answer: option 3
$f'(x)=\underset{h→0}{\lim}\frac{f(x+h)-f(x)}{h}=\underset{h→0}{\lim}\frac{f(x)+f(h)+xh-f(x)}{h}$
$=\underset{h→0}{\lim}\frac{1}{h}f(h)+x=3+x$
Hence $f(x)=3x+\frac{x^2}{2}+c$
Putting x = y = 0 in the given equation, we have
$f(0)=f(0+0)=f(0)+f(0)+0⇒f(0)=0$
Thus c = 0 and $f(x)=3x+\frac{x^2}{2}$