Question

Let $f$ be a differential function satisfying $f\left(x+y\right)=f\left(x\right)+f\left(y\right)+\left(e^x-1\right)\left(e^y-1\right)\forall x,y\in R$ and $f^{\prime }\left(0\right)=2$. Identify the correct statement(s)?

• A. $\underset{x\to 0}{lim}\frac{f\left(f\left(x\right)\right)}{f\left(x\right)-x}=4$
• B. $\underset{x\to 0}{lim}{\left(f\left(x\right)+\cos x\right)}^{\frac{1}{e^x-1}}=e^2$
• C. Number of solutions of the equation $f\left(x\right)=0$ is $2$.
• D. Range of the function $y=f\left(x\right)$ is $\left(-\infty ,\infty \right)$.

Answer 1

Answer: (A), (B)

The correct options are
A $\underset{x\to 0}{lim}\frac{f\left(f\left(x\right)\right)}{f\left(x\right)-x}=4$
B $\underset{x\to 0}{lim}{\left(f\left(x\right)+\cos x\right)}^{\frac{1}{e^x-1}}=e^2$

$f\left(x+y\right)=f\left(x\right)+f\left(y\right)+\left(e^x-1\right)\left(e^y-1\right)$
from this, we can say
$\begin{array}{c}f\left(x\right)=e^{2x}-1\\ \underset{x\to 0}{lim}\frac{f\left(f\left(x\right)\right)}{f\left(x\right)-x}=\underset{x\to 0}{lim}\frac{e^{2\left(e^{2x}-1\right)}-1}{e^{2x}-1-x}\\ =\underset{x\to 0}{lim}\frac{2e^{2\left(e^{2x}-1\right)}\left(2e^{2x}\right)-0}{2e^{2x}-1}\\ =\frac{2\left(1\right)2\left(1\right)}{(2-1)}=4\\ \\ \underset{x\to 0}{lim}{\left(f\left(x\right)+\cos x\right)}^{\frac{1}{e^x-1}}\end{array}$
It is in the form $1^{\infty }$
$\begin{array}{c}{e}^{\underset{x\to 0}{lim}{\left(f\left(x\right)+\cos x\right)}^{\frac{1}{e^x-1}}}=e^{\underset{x\to 0}{lim}\left(\frac{e^{2x}+\cos x-2}{e^x-1}\right)}\\ =e^{\underset{x\to 0}{lim}\left(\frac{2e^{2x}-\sin x}{e^x}\right)}=e^{\left(\frac{2\left(1\right)-0}{1}\right)}=e^2\end{array}$