Let $f : R \to R$ be differentiable at $c \in R$ and $f(c) = 0$. If $g(x) = |f(x)|$,then at $x = c$,$g$ is

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 $\lim _{x \rightarrow 0} g(x)=1$,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 $1$

For the function $f(x) = e^{\sin |x|} - |x|$, $x \in R$, consider the following statements:
Statement $I$: $f$ is differentiable for all $x \in R$.
Statement $II$: $f$ is increasing in $(-\pi, -\frac{\pi}{2})$.
In the light of the above statements, choose the correct answer from the options given below:

Let $g(x) = x \cdot f(x)$,where $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$. Discuss the differentiability of $g$ at $x = 0$.

If $f(x) = \min \{1, x^2, x^3\}$,then

Define $f(x) = \begin{cases} x^2 + bx + c, & x < 1 \\ x, & x \geq 1 \end{cases}$. If $f(x)$ is differentiable at $x = 1$,then $(b - c)$ is equal to