If $f(x) = \begin{cases} \frac{\log (1 + 2ax) - \log (1 - bx)}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is

The value of $f(0)$,so that the function $f(x) = \frac{(27 - 2x)^{1/3} - 3}{9 - 3(243 + 5x)^{1/5}}, (x \ne 0)$ is continuous,is given by

If $f: [-2, 2] \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{\sqrt{1 + cx} - \sqrt{1 - cx}}{x}, & -2 \leq x < 0 \\ \frac{x + 3}{x + 1}, & 0 \leq x \leq 2 \end{cases}$ is continuous on $[-2, 2]$,then $c$ is equal to

Let $f(x) = \begin{cases} (3 - \sin(1/x))|x|, & x \ne 0 \\ 0, & x = 0 \end{cases}$. Then at $x = 0$,$f$ has a

If $f(x) = \operatorname{sgn}((x^2 - kx + 6)(\sin x - 1/2))$ (where $k > 0$) has exactly $4$ points of discontinuity in $(0, 6)$,then the maximum integral value of $k$ is:

If $f(x) = \begin{cases} \frac{9^x - 2 \cdot 3^x + 1}{\log(1 + 3x) \cdot \tan 2x} & , x \neq 0 \\ a(\log b)^c & , x = 0 \end{cases}$ is continuous at $x = 0$,then $a + b + c =$