If $f(x) = \begin{cases} x^2, & \text{when } x \le 1 \\ x + 5, & \text{when } x > 1 \end{cases}$,then

If function $f$ is continuous at point $x = \pi$ and $f(x) = \begin{cases} kx+1; & x \leq \pi \\ \cos x; & x > \pi \end{cases}$ then the value of $k$ is $\dots \dots \dots$

If the function $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{\sin(a + 1)x + \sin x}{x}, & x < 0 \\ b, & x = 0 \\ \frac{\sqrt{x + x^2} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}$ is continuous on $R$,then $a + b =$

$f(x) = \begin{cases} \frac{1-\cos kx}{x^2}, & x \le 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is:

Let $f(x) = \frac{1 - \tan x}{4x - \pi }, x \ne \frac{\pi }{4}, x \in [0, \frac{\pi }{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi }{2}]$,then $f(\frac{\pi }{4})$ is

If a function $f$ is defined by $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4 x}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4} \end{cases}$ and is continuous at $x = \frac{\pi}{4}$,then $k = $