$f(x) = \begin{cases} [x^2] - [-x^2], & x \neq 3 \\ k, & x = 3 \end{cases}$ is continuous at $x = 3$,then $k = $ where $[\cdot]$ is the greatest integer function.

Examine the following function for continuity: $f(x) = |x - 5|$.

If $f(x) = \begin{cases} \frac{\sin [x]}{[x] + 1}, & \text{for } x > 0 \\ \frac{\cos \frac{\pi }{2}[x]}{[x]}, & \text{for } x < 0 \\ k, & \text{at } x = 0 \end{cases}$; where $[x]$ denotes the greatest integer less than or equal to $x$,then in order that $f$ be continuous at $x = 0$,the value of $k$ is

Define $f: R \to R$ by $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ Then the value of $a$ so that $f$ is continuous at $x = 0$ is:

If $f(x) = [x] - [\frac{x}{4}]$,$x \in R$,where $[x]$ denotes the greatest integer function,then:

If $f(x) = \begin{cases} \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x} & \text{for } -1 \le x < 0 \\ 2x^2 + 3x - 2 & \text{for } 0 \le x \le 1 \end{cases}$ is continuous at $x = 0$,then $k = $