If the function $f$ defined as $f(x) = \frac{1}{x} - \frac{k - 1}{e^{2x} - 1}$,$x \neq 0$,is continuous at $x = 0$,then the ordered pair $(k, f(0))$ is equal to?

The function defined by $f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x-4}{|x-4|} + b, & x > 4 \end{cases}$ is continuous at $x = 4$,then:

If $f(x) = \max(\sin x, \sin^{-1}(\cos x))$,then

If $f(x) = \begin{cases} x, & 0 \le x \le 1 \\ 2x - 1, & x > 1 \end{cases}$,then

If $f(x) = \begin{cases} \frac{8^x - 4^x - 2^x + 1}{x^2} & , \text{if } x > 0 \\ e^x \sin x + kx + \lambda \log 4 & , \text{if } x \le 0 \end{cases}$ is continuous at $x = 0$,then the value of $500 e^\lambda$ is

Let $f : [a, b] \rightarrow [1, \infty)$ be a continuous function and let $g : \mathbb{R} \rightarrow \mathbb{R}$ be defined as $g(x) = \begin{cases} 0 & \text{if } x < a \\ \int_a^x f(t) dt & \text{if } a \leq x \leq b \\ \int_a^b f(t) dt & \text{if } x > b \end{cases}$. Then: