$f(x) = \begin{cases} [x] + [-x], & \text{when } x \neq 2 \\ \lambda, & \text{when } x = 2 \end{cases}$
If $f(x)$ is continuous at $x = 2$, the value of $\lambda$ will be

If the function $f(x) = \frac{1 - \cos 4x}{8x^2}$ for $x \ne 0$ and $f(x) = k$ for $x = 0$ is a continuous function at $x = 0$,then the value of $k$ is:

If $f(x) = \begin{cases} x e^{-\left( \frac{1}{|x|} + \frac{1}{x} \right)}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then $f(x)$ is

If $f(x) = \begin{cases} \frac{x^2 - (a+2)x + a}{x-2} & x \neq 2 \\ 2 & x = 2 \end{cases}$ is continuous at $x = 2$,then the value of $a$ 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:

If the function $f(x) = \begin{cases} k_{1}(x-\pi)^{2}-1, & x \leq \pi \\ k_{2} \cos x, & x>\pi \end{cases}$ is twice differentiable,then the ordered pair $(k_{1}, k_{2})$ is equal to