If $f(x) = \begin{cases} \frac{\sqrt{\pi} - \sqrt{\cos^{-1} x}}{\sqrt{x+1}}, & x \neq -1 \\ \frac{1}{\sqrt{\lambda \pi}}, & x = -1 \end{cases}$ is right continuous at $x = -1$,then $\lambda = $

If $f(x) = \begin{cases} \frac{1 - [x]}{1 + x}, & x \ne -1 \\ 1, & x = -1 \end{cases}$,then the value of $f(|2k|)$ will be (where $[.]$ denotes the greatest integer function). Which of the following statements is true?

The function $f(x) = \frac{\log(1+ax) - \log(1-bx)}{x}$ is not defined at $x=0$. The value which should be assigned to $f$ at $x=0$ so that it is continuous at $x=0$ is

If $f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ \frac{1}{2}, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then the value of $k$ is . . . . . . .

If the function $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{a(1-\cos 2x)}{x^2}, & x < 0 \\ b, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}, & x > 0 \end{cases}$ is continuous at $x = 0$, then $a+b=$

If the function $f(x) = \begin{cases} \frac{k\cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $k = $