If the function $f(x)$ defined by $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k = . . . . . .$

Let $f:(0,1) \rightarrow \mathbb{R}$ be the function defined as $f(x)=[4x](x-\frac{1}{4})^2(x-\frac{1}{2})$,where $[x]$ denotes the greatest integer less than or equal to $x$. Then which of the following statements is(are) true?
$(A)$ The function $f$ is discontinuous exactly at one point in $(0,1)$
$(B)$ There is exactly one point in $(0,1)$ at which the function $f$ is continuous but $NOT$ differentiable
$(C)$ The function $f$ is $NOT$ differentiable at more than three points in $(0,1)$
$(D)$ The minimum value of the function $f$ is $-\frac{1}{512}$

If $f(x)= \begin{cases} \frac{x-[x]}{x-2}, & x>2 \\ b, & x=2 \\ \frac{|x^2-x-2|}{a(2+x-x^2)}, & -1 < x \leq 2 \\ 2a-b, & x \leq -1 \end{cases}$ is continuous on $R$,then $\lim _{x \rightarrow 0} \frac{\sin ^2 ax+x \tan bx}{x^2}=$

If the function defined by $f(x) = \frac{\log (1+x)^{1+x}}{x^2} - \frac{1}{x}, x \neq 0$ is continuous at $x=0$,then $6 f(0)$ is equal to

Find all points of discontinuity of the function $f$ defined by $f(x) = |x| - |x + 1|$.

Is the function defined by $f(x) = x^{2} - \sin x + 5$ continuous at $x = \pi$?