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 $a + b + c =$

$f(x) = \begin{cases} \frac{\log x}{x-1}, & \text{if } x \neq 1 \\ k, & \text{if } x=1 \end{cases}$ is continuous at $x=1$,then the value of $k$ is

If the function $f(x) = \begin{cases} x - \frac{|x|}{x}, & x < 0 \\ x + \frac{|x|}{x}, & x > 0 \\ 1, & x = 0 \end{cases}$,then which of the following is true?

If the function $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right), & x < 0 \\ k, & x = 0 \\ \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2}+1}-1}, & x > 0 \end{cases}$ is continuous at $x = 0$,then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to:

Let $f(x) = \begin{cases} (x - 1)^{\frac{1}{2 - x}}, & x > 1, x \neq 2 \\ k, & x = 2 \end{cases}$. The value of $k$ for which $f$ is continuous at $x = 2$ is

If $f(x) = \begin{cases} \frac{x - 1}{2}, & 0 \leqslant x < 1 \\ 1/2, & 1 \leqslant x < 2 \end{cases}$ and $g(x) = (2x + 1)(x - k) + 3$ for $0 \leqslant x < \infty$,then $g(f(x))$ will be continuous at $x = 1$ if $k$ is equal to: