Question

Let $f:R\to R$ and $g:R\to R$ be functions satisfying
$f(x+y)=f(x)+f(y)+f(x)f(y)$ and $f(x)=xg(x)$
for all $x,y\in R.$ If $\underset{x\to 0}{\lim }g(x)=1$, 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)

$f(x)=x\cdot g(x)\forall x\in R$
$\Rightarrow \underset{x\to 0}{\lim }f(x)=\left(\underset{x\to 0}{\lim }x\right)\left(\underset{x\to 0}{\lim }g(x)\right)$
$\Rightarrow \underset{x\to 0}{\lim }f(x)=0$
Also $\underset{x\to 0}{\lim }(f(x+y))=\underset{x\to 0}{\lim }(f(x)+f(y)+f(x)\cdot f(y))$
$\Rightarrow \underset{x\to 0}{\lim }f(x+y)=f(y)$
$\Rightarrow f(x)$ is continuous $\forall x\in R$
$\Rightarrow f(0)=0$
$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$
$\Rightarrow f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(h)+f(x)\cdot f(h)}{h}$
$\Rightarrow f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(h)}{h}\cdot (1+f(x))$
$\Rightarrow \int \frac{f^{\prime }(x)dx}{1+f(x)}=\int 1dx$
$\Rightarrow \ln |1+f(x)|=x+c$
$f(0)=0\Rightarrow c=0$
$\Rightarrow f(x)=e^x-1$
$\Rightarrow f^{\prime }(x)=e^x$
$\Rightarrow f(x)$ is differentiable and $f^{\prime }(0)=1$
Also $g(x)=\frac{f(x)}{x},x\ne 0$
Now if $g(0)=1$
$\Rightarrow g(x)=\{\begin{array}{ll}\frac{e^x-1}{x},& x\ne 0\\ 1,& x=0\end{array}$
$\Rightarrow g(x)$ is continuous at $x=0$
$\Rightarrow g^{\prime }(0)=\underset{h\to 0}{\lim }\frac{g(0+h)-g(0)}{h}$
$=\underset{h\to 0}{\lim }\frac{\frac{e^h-1}{h}-1}{h}=\frac{1}{2}$
$\Rightarrow g(x)$ is diff. $\forall x\in R$ if $g(0)=1$