If the function $f(x) = \begin{cases} \frac{x^2-(A+2)x+A}{x-2} & \text{for } x \neq 2 \\ 2 & \text{for } x=2 \end{cases}$ is continuous at $x=2$,then:

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 $f(x) = \begin{cases} \frac{(e^{3x}-1) \sin x^{\circ}}{x^2} & x \neq 0 \\ \frac{\pi}{60} & x = 0 \end{cases}$,then:

Let $[x]$ denote the greatest integer less than or equal to $x$. If $f(x) = [x \sin \pi x]$,then $f(x)$ is

The function $f(x) = |x - 24|$ is

Let $f: R \rightarrow R$ be defined as:
$f(x) = \begin{cases} \frac{\lambda|x^{2}-5x+6|}{\mu(5x-x^{2}-6)}, & x < 2 \\ \mu, & x = 2 \\ e^{\frac{\tan(x-2)}{x-[x]}}, & x > 2 \end{cases}$
Where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x = 2$,then $\lambda + \mu$ is equal to: