Consider $f(x) = \begin{cases} \frac{x^2}{|x|}, & x \ne 0 \\ 0, & x = 0 \end{cases}$

Let $[t]$ denote the greatest integer $\leq t$. The number of points where the function $f(x)=[x]|x^{2}-1|+\sin \left(\frac{\pi}{[x]+3}\right)-[x+1]$ for $x \in(-2,2)$ is not continuous is:

The value of $k$,for which the function $f(x) = \begin{cases} (\frac{4}{5})^{\frac{\tan 4x}{\tan 5x}}, & 0 < x < \frac{\pi}{2} \\ k + \frac{2}{5}, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,is:

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{x + 2}{x^2 + 3 x + 2}, & x \in R - \{-1, -2\} \\ -1, & x = -2 \\ 0, & x = -1 \end{cases}$ then $f$ is continuous on the set

Given $f(x) = b ([x]^2 + [x]) + 1$ for $x \geq -1$ and $f(x) = \sin(\pi(x+a))$ for $x < -1$,where $[x]$ denotes the greatest integer function,for what values of $a$ and $b$ is the function continuous at $x = -1$?

Let $f(x) = \begin{cases} \frac{x - 4}{|x - 4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x - 4}{|x - 4|} + b, & x > 4 \end{cases}$. Then $f(x)$ is continuous at $x = 4$ when