The function $f(x) = [x]$,where $[x]$ denotes the greatest integer not greater than $x$,is

Let $[x]$ represent the greatest integer not more than $x$. The discontinuous points of the function $f(x) = \frac{5+[x]}{\sqrt{11+[x]-6 \sqrt{2+[x]}}}$ lie in the interval

If $f(x) = \begin{cases} \frac{x^2 - 4x + 3}{x^2 - 1}, & x \ne 1 \\ 2, & x = 1 \end{cases}$,then:

If the function $f(x)$,defined below,is continuous on the interval $[0, 8]$,then
$f(x) = \begin{cases} x^{2} + ax + b, & 0 \le x < 2 \\ 3x + 2, & 2 \le x \le 4 \\ 2ax + 5b, & 4 < x \le 8 \end{cases}$

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} [e^x], & x < 0 \\ a e^x + [x - 1], & 0 \leq x < 1 \\ b + [\sin(\pi x)], & 1 \leq x < 2 \\ [e^{-x}] - c, & x \geq 2 \end{cases}$ where $a, b, c \in R$ and $[t]$ denotes the greatest integer less than or equal to $t$. Then,which of the following statements is true?

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: