Question

Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be functions satisfying
$f(x+y)=f(x)+f(y)+f(x) f(y)$ and $f(x)=x g(x)$
for all $x, y \in R .$ If $\displaystyle\lim _{x \rightarrow 0} g(x)=1$, then which of the following statements is/are TRUE?

• A. $f$ is differentiable at every $x \in R$
• B. If $g(0)=1$, then $g$ is differentiable at every $x \in R$
• C. The derivative $f^{\prime}(1)$ is equal to $1$
• D. The derivative $f^{\prime}(0)$ is equal to $1$

Answer 1

Answer: (A), (B), (D)

$f(x)=x \cdot g(x) \forall x \in R$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} f(x)=\left(\displaystyle\lim _{x \rightarrow 0} x\right)\left(\displaystyle\lim _{x \rightarrow 0} g(x)\right)$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} f(x)=0$
Also $\displaystyle\lim _{x \rightarrow 0}(f(x+y))=\displaystyle\lim _{x \rightarrow 0}(f(x)+f(y)+f(x) \cdot f(y))$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} f(x+y)=f(y)$
$\Rightarrow f(x)$ is continuous $\forall x \in R$
$\Rightarrow f(0)=0$
$f^{\prime}(x)=\displaystyle\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} $
$\Rightarrow f^{\prime}(x)=\displaystyle\lim _{h \rightarrow 0} \frac{f(h)+f(x) \cdot f(h)}{h} $
$\Rightarrow f^{\prime}(x)=\displaystyle\lim _{h \rightarrow 0} \frac{f(h)}{h} \cdot(1+f(x)) $
$\Rightarrow \int \frac{f^{\prime}(x) d x}{1+f(x)}=\int 1 d x$
$\Rightarrow \ln |1+f(x)|=x+c$
$ f(0)=0 \Rightarrow c=0$
$\Rightarrow f(x)=e^{x}-1$
$\Rightarrow f^{\prime}(x)=e^{x}$
$\Rightarrow f(x)$ is differentiable and $f^{\prime}(0)=1$
Also $g(x)=\frac{f(x)}{x}, x \neq 0$
Now if $g(0)=1$
$\Rightarrow g(x)=\begin{cases}\frac{e^{x}-1}{x}, & x \neq 0 \\ 1, & x=0\end{cases} $
$\Rightarrow g(x)$ is continuous at $x=0$
$\Rightarrow g^{\prime}(0)=\displaystyle\lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{\frac{e^{h}-1}{h}-1}{h}=\frac{1}{2}$
$\Rightarrow g ( x )$ is diff. $\forall x \in R$ if $g (0)=1$