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

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 $1$

The total number of points of non-differentiability of $f(x) = \min \{ |\sin x|, |\cos x|, \frac{1}{4} \}$ in $(0, 2\pi)$ is

$A$ function $f$ is defined as follows:
$f(x) = \begin{cases} \sin x & \text{if } x \le c \\ ax + b & \text{if } x > c \end{cases}$
where $c$ is a known quantity. If $f$ is derivable at $x = c$,then the values of $a$ and $b$ are . . . . . . and . . . . . . respectively.

The function $f(x) = (x^2 - 1)|x^2 - 3x + 2| + \cos(|x|)$ is not differentiable at

Let $g: [-2, 2] \rightarrow R$ and $f: [-2, 2] \rightarrow R$ be two functions defined as $g(x) = \begin{cases} -1, & \text{if } -2 \le x < 0 \\ x^2 - 1, & \text{if } 0 \le x \le 2 \end{cases}$ and $f(x) = |g(x)| + g(|x|) + 2$. In the interval $(-2, 2)$,$f$ is not differentiable at $x = $