If the function $f(x)$,defined below,is continuous everywhere,then $k$ equals: $f(x)=\begin{cases} \frac{2^x-1}{\sqrt{1+x}-1}, & x \neq 0 \\ k, & x=0 \end{cases}$

For every integer $n$,let $a_n$ and $b_n$ be real numbers. Let function $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x) = \begin{cases} a_n + \sin \pi x, & \text{for } x \in [2n, 2n+1] \\ b_n + \cos \pi x, & \text{for } x \in (2n-1, 2n) \end{cases}$,for all integers $n$. If $f$ is continuous,then which of the following hold$(s)$ for all $n$?

For what value of $\lambda$ is the function defined by $f(x) = \begin{cases} \lambda(x^2 - 2x), & \text{if } x \le 0 \\ 4x + 1, & \text{if } x > 0 \end{cases}$ continuous at $x=0$? What about continuity at $x=1$?

Given $f(x) = \begin{cases} cx + 1, & x \leq 3 \\ dx + 3, & x > 3 \end{cases}$. If $f$ is continuous at $x = 3$,then $d - c =$ . . . . . . .

Let $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\ 2x \cot x+b, & \frac{\pi}{4} \leq x < \frac{\pi}{2} \\ a \cos 2x-b \sin x, & \frac{\pi}{2} \leq x \leq \pi \end{cases}$. If $f(x)$ is continuous for $0 \leq x \leq \pi$,then:

Let $f : [-1,3] \to R$ be defined as $f(x) = \begin{cases} |x| + [x], & -1 \leq x < 1 \\ x + |x|, & 1 \leq x < 2 \\ x + |x|, & 2 \leq x \leq 3 \end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. Then,$f$ is discontinuous at: