Let $f: R \rightarrow R$ be defined as $f(x)=\begin{cases} \frac{a-b \cos 2 x}{x^2} & ; x<0 \\ x^2+c x+2 & ; 0 \leq x \leq 1 \\ 2 x+1 & ; x>1 \end{cases}$. If $f$ is continuous everywhere in $R$ and $m$ is the number of points where $f$ is $NOT$ differentiable,then $m+a+b+c$ equals:

Let $f:(-2,2) \rightarrow \mathbb{R}$ be defined by $f(x) = \begin{cases} x[x] & , -2 < x < 0 \\ (x-1)[x] & , 0 \leq x < 2 \end{cases}$ where $[x]$ denotes the greatest integer function. If $m$ and $n$ respectively are the number of points in $(-2,2)$ at which $y = |f(x)|$ is not continuous and not differentiable,then $m + n$ is equal to $...........$.

Let $f: R \rightarrow R$ be given by
$f(x) = \begin{cases} x^5+5x^4+10x^3+10x^2+3x+1, & x < 0 \\ x^2-x+1, & 0 \leq x < 1 \\ \frac{2}{3}x^3-4x^2+7x-\frac{8}{3}, & 1 \leq x < 3 \\ (x-2)\log_e(x-2)-x+\frac{10}{3}, & x \geq 3 \end{cases}$
Then which of the following options is/are correct?
$(1)$ $f^{\prime}$ has a local maximum at $x = 1$
$(2)$ $f$ is onto
$(3)$ $f$ is increasing on $(-\infty, 0)$
$(4)$ $f^{\prime}$ is $NOT$ differentiable at $x = 1$

If $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$ then $f(x)$ at $x = 0$ is

Let $f:\left[-\frac{1}{2}, 2\right] \rightarrow R$ and $g:\left[-\frac{1}{2}, 2\right] \rightarrow R$ be functions defined by $f(x)=\left[x^2-3\right]$ and $g(x)=|x| f(x)+|4 x-7| f(x)$,where $[y]$ denotes the greatest integer less than or equal to $y$ for $y \in R$. Then
$(A)$ $f$ is discontinuous exactly at three points in $\left[-\frac{1}{2}, 2\right]$
$(B)$ $f$ is discontinuous exactly at four points in $\left[-\frac{1}{2}, 2\right]$
$(C)$ $g$ is $NOT$ differentiable exactly at four points in $\left(-\frac{1}{2}, 2\right)$
$(D)$ $g$ is $NOT$ differentiable exactly at five points in $\left(-\frac{1}{2}, 2\right)$

Let $f(x)$ be a non-negative differentiable function on $[0, \infty)$ such that $f(0)=0$ and $f^{\prime}(x) \leq 2 f(x)$ for all $x>0$. Then,on $[0, \infty)$: