Question

Let $f$ and $g$ be functions satisfying $f\left(x\right)=e^{x}g\left(x\right)$, $f(x+y)=f\left(x\right)+f\left(y\right),g\left(0\right)=0$, $g^{\prime }\left(0\right)=4,g$ and $g^{\prime }$ are continuous at $0$
Then

• A. $f\left(x\right)=0$ for all $x$
• B. $f\left(x\right)=x$ for all $x$
• C. $f\left(x\right)=x+4$ for all $x$
• D. $f\left(x\right)=4x$ for all $x$

Answer 1

The correct option is D $f\left(x\right)=4x$ for all $x$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{e^{h}g\left(h\right)}{h},(,\frac{0}{0})$
form
Thus applying L-Hospital's rule,
$f^{\prime }\left(x\right)={lim}_{h\to 0}{e}^h{g}^{\prime }\left(h\right)+e^{h}g\left(h\right)$
$=g^{\prime }\left(0\right)+g\left(0\right)=4$
Hence $f\left(x\right)=4x+c$, but $f\left(0\right)=0=c$
So $f\left(x\right)=4x\forall x$