Question

Let $f,g,g^{\prime }$ be continuous and differentiable functions such that $f\left(x\right)=e^{x}g\left(x\right),$ $f(x+y)=f\left(x\right)+f\left(y\right),$ $g\left(0\right)=0,$ $g^{\prime }\left(0\right)=4.$ 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 $x$

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