Question

Let f$f:R\to R$and $g:R\to R$ be function sastisfying $f(x\hspace{0.17em}+y)=f\left(x\right)+f\left(y\right)+f\left(x\right).f\left(y\right)$and $f\left(x\right)=xg\left(x\right)$ for all $x,y\in R.$ If $\underset{x\to 0}{\lim }g\left(x\right)=1$, then which of the following statements is/are True?

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

Answer 1

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

The derivative $\text{'}\left(0\right)$is equal to $1$

Step 1: Given information

$f(x\hspace{0.17em}+y)=f\left(x\right)+f\left(y\right)+f\left(x\right).f\left(y\right)$

$f\left(x\right)=xg\left(x\right)$

$\underset{x\to 0}{\lim }g\left(x\right)=1$

Step 2: Checking the truth of Option (A)

$f(x\hspace{0.17em}+y)=f\left(x\right)+f\left(y\right)+f\left(x\right).f\left(y\right)$

Let $x=0,y=0:$

$$\begin{gathered} f(0\hspace{0.17em}+0)=f\left(0\right)+f\left(0\right)+f\left(0\right).f\left(0\right) \\ \Rightarrow f\left(0\right)=f\left(0\right)+f\left(0\right)+f\left(0\right).f\left(0\right) \\ \Rightarrow f^2\left(0\right)+f\left(0\right)=0 \\ \Rightarrow f\left(0\right)(f(0)+1)=0 \\ \Rightarrow f\left(0\right)=0or-1 \end{gathered}$$

Now, differentiating the equation:

$$\begin{gathered} f(x\hspace{0.17em}+y)=f\left(x\right)+f\left(y\right)+f\left(x\right).f\left(y\right) \\ \Rightarrow f(x\hspace{0.17em}+y)-f\left(x\right)=f\left(y\right)+\hspace{0.17em}f\left(x\right).f\left(y\right) \\ \Rightarrow \frac{f(x\hspace{0.17em}+y)-f\left(x\right)}{y}=\frac{f\left(y\right)+\hspace{0.17em}f\left(x\right).f\left(y\right)}{y} \\ \Rightarrow \underset{y\to 0}{\lim }\frac{f(x\hspace{0.17em}+y)-f\left(x\right)}{y}=\underset{y\to 0}{\lim }\frac{f\left(y\right)+\hspace{0.17em}f\left(x\right).f\left(y\right)}{y} \\ \Rightarrow f\text{'}\left(x\right)=\underset{y\to 0}{\lim }\frac{f\left(y\right)+\hspace{0.17em}f\left(x\right).f\left(y\right)}{y} \\ \Rightarrow f\text{'}\left(x\right)=(1+f(x)\underset{y\to 0}{\lim }\frac{f\left(y\right)}{y}\left[\because \underset{y\to 0}{\lim }\frac{f\left(y\right)}{y}=1\right] \\ \Rightarrow f\text{'}(x)=(1+f\left(x\right) \end{gathered}$$

Therefore, f(x) is differentiable at every $x\in R$

Hence, Option (A) is true.

Step 3: Checking for Option (D):

Calculating $f\text{'}\left(0\right)$

$$\begin{gathered} f\text{'}\left(0\right)=1+f\left(x\right) \\ =1+f\left(0\right) \\ =1+0or1+(-1) \\ =1or0 \end{gathered}$$

Therefore, f'(0) equals to $1$ is possible.

Hence Option (D) is True.

Step 4: Checking for Option (C):

$f\text{'}\left(1\right)=1+f\left(1\right)$

Finding $f\left(1\right)$ :

$$\begin{gathered} f\text{'}\left(x\right)=1+f\left(x\right) \\ \Rightarrow \frac{f\text{'}\left(x\right)}{f\left(x\right)}=1 \\ \text{Integrating both sides:} \\ \int \frac{f\text{'}\left(x\right)}{f\left(x\right)}dx=\int dx \\ \Rightarrow \ln (1+f(x\left)\right)=x+C \\ Puttingx=0, \\ \ln (1+f(0\left)\right)=0+C \\ \Rightarrow C=0 \\ \therefore \ln (1+f(x\left)\right)=x \\ \Rightarrow 1+f\left(x\right)=e^x \\ \Rightarrow f\left(x\right)=e^x-1 \\ \therefore f\text{'}\left(x\right)=e^x \\ \therefore f\text{'}\left(1\right)=e \end{gathered}$$

Hence, Option (C) is false.

Step 5: Checking for option (B)

$$\begin{gathered} f\left(x\right)=xg\left(x\right) \\ \Rightarrow g\left(x\right)=\frac{f\left(x\right)}{x} \\ \Rightarrow g\left(x\right)=\frac{e^x-1}{x} \end{gathered}$$

Checking differentiable of $g\left(x\right):$

$$\begin{gathered} g\text{'}\left(0^{+}\right)=\underset{h\to 0}{\lim }\frac{g(x+h)-g\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{e^h-1}{h}}-1}{h} \\ =\underset{h\to 0}{\lim }\frac{e^h-1-h}{h^2} \\ =\frac{1}{2} \\ g\left(0^{-}\right)=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{e^{-h}-1}{-h}-1}}{-h} \\ =\underset{h\to 0}{\lim }\frac{e^{-h}-1+h}{h^2} \\ =\frac{1}{2} \\ =g\left(0^{+}\right) \end{gathered}$$

Hence, $g\left(x\right)$ is differentiable at every $x\in R$.

Therefore, Option (B) is True.

Hence, Option (A), (B), (D) are true.