Let $[t]$ denote the greatest integer $\leq t$ and $\lim_{x \to 0} x[\frac{4}{x}] = A$. Then the function $f(x) = [x^2] \sin(\pi x)$ is discontinuous when $x$ is equal to

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 a function $f$ is defined by $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4 x}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4} \end{cases}$ and is continuous at $x = \frac{\pi}{4}$,then $k = $

The function $f(x) = |x| + \frac{|x|}{x}$ is

If $f(x)$ is continuous and $f\left( \frac{9}{2} \right) = \frac{2}{9}$, then $\lim_{x \to 0} f \left( \frac{1 - \cos 3x}{x^2} \right)$ is equal to:

Let $f, g: R \rightarrow R$ be functions defined by $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = x f(x)$. Consider the following statements: $(i)$ $f(x)$ is continuous at $x = 0$ but not differentiable at $x = 0$. $(ii)$ $g(x)$ is differentiable at $x = 0$, but $g'(x)$ is not continuous at $x = 0$. Then, which one of the following is true?