Let $f(x) = \begin{cases} \int_{0}^{x} |1-t| dt, & x > 1 \\ x - \frac{1}{2}, & x \leq 1 \end{cases}$. Then:

Let $[x]$ denote the greatest integer function,and let $m$ and $n$ respectively be the numbers of the points,where the function $f(x) = [x] + |x - 2|$,$-2 < x < 3$,is not continuous and not differentiable. Then $m + n$ is equal to:

$A$ function $f(x) = \begin{cases} 1 + x, & x \le 2 \\ 5 - x, & x > 2 \end{cases}$ is

If $f(x) = 4x^3 - x^2 - 2x + 1$ and $g(x) = \begin{cases} \min \{f(t) : 0 \le t \le x\} & ; 0 \le x \le 1 \\ 3 - x & ; 1 < x \le 2 \end{cases}$,then the value of $g\left( \frac{1}{4} \right) + g\left( \frac{3}{4} \right) + g\left( \frac{5}{4} \right)$ is:

Let $f(x)=x^2+a x+b$,where $a, b \in R$. If $f(x)=0$ has all its roots imaginary,then the roots of $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0$ are

Match the items of List-$I$ with those of List-$II$.

List-$I$	List-$II$
$A$. If $y = |x| + |x - 2|$, then at $x = 2$, $\frac{dy}{dx} =$	$I$. $2$
$B$. If $f(x) = |\cos 2x|$, then $f'(\frac{\pi}{4} +) =$	$II$. $0$
$C$. If $f(x) = \sin(\pi[x])$, where $[x]$ is the greatest integer function, then $f'(1-) =$	$III$. $-2$
$D$. If $f(x) = \log|x - 1|$, $x \neq 1$, then $f'(\frac{1}{2}) =$	$IV$. does not exist