If $f$ is a real-valued differentiable function satisfying $|f(x) - f(y)| \le (x - y)^2$ for all $x, y \in R$ and $f(0) = 0$,then $f(1)$ is equal to:

Let $g: R \rightarrow R$ be a differentiable function with $g(0)=0, g^{\prime}(0)=0$ and $g^{\prime}(1) \neq 0$. Let $f(x)=\begin{cases} \frac{x}{|x|} g(x), & x \neq 0 \\ 0, & x=0 \end{cases}$ and $h(x)=e^{|x|}$ for all $x \in R$. Let $(f \circ h)(x)$ denote $f(h(x))$ and $(h \circ f)(x)$ denote $h(f(x))$. Then which of the following is (are) true?
$(A)$ $f$ is differentiable at $x=0$
$(B)$ $h$ is differentiable at $x=0$
$(C)$ $f \circ h$ is differentiable at $x=0$
$(D)$ $h \circ f$ is differentiable at $x=0$

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 $1$

If $f(x) = \begin{cases} \frac{x^2 \ln \cos x}{\ln(1 + x^2)}, & x \neq 0 \\ 0, & x = 0 \end{cases}$,then $f(x)$ is

If $f(x) = |x|,$ then $f'(0) = $

Let $f(x) = \begin{cases} -1, & -2 \le x < 0 \\ x^2 - 1, & 0 \le x \le 2 \end{cases}$ and $g(x) = |f(x)| + f(|x|)$. Then,in the interval $(-2, 2)$,$g$ is