Question

Suppose that $f$ is a differentiable function with the property that $f(x+y)=f\left(x\right)+f\left(y\right)+xy$ and $\underset{h\to 0}{lim}\frac{1}{h}f\left(h\right)=3$ then

• A. $f$ is a linear function
• B. $f\left(x\right)=3x+x^2$
• C. $f\left(x\right)=3x+\frac{x^2}{2}$
• D. None of these

Answer 1

Answer: (C)

$f\left(x\right)=3x+\frac{x^2}{2}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)+xh-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{1}{h}f\left(h\right)+x=3+x$
Hence $f\left(x\right)=3x+\frac{x^2}{2}+c.$
Substituting $x=y=0$ in the given equation,
We have $f\left(0\right)=f(0+0)=f\left(0\right)+f\left(0\right)+0\Rightarrow f\left(0\right)=0$
Thus $c=0$ and $f\left(x\right)=3x+\frac{x^2}{2}$