The function $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ is not defined at $x = \pi$. The value of $f(\pi)$,so that $f(x)$ is continuous at $x = \pi$,is

If the function $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right), & x < 0 \\ k, & x = 0 \\ \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2}+1}-1}, & x > 0 \end{cases}$ is continuous at $x = 0$,then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to:

Given the function $f(x) = 2x \sqrt{x^3 - 1} + 5 \sqrt{x} \sqrt{1 - x^4} + 7x^2 \sqrt{x - 1} + 3x + 2$,then:

If $f(x)$ is continuous at point $x=0$ where $f(x) = \begin{cases} \frac{3 \sin x + 5 \tan x}{a^x - 1} & , x < 0 \\ \frac{2}{\log 2} & , x = 0 \\ \frac{8x + 2x \cos x}{b^x - 1} & , x > 0 \end{cases}$ then the values of $a$ and $b$,respectively,are

$f(x) = \begin{cases} [x^2] - [-x^2], & x \neq 3 \\ k, & x = 3 \end{cases}$ is continuous at $x = 3$,then $k = $ where $[\cdot]$ is the greatest integer function.

If $f(x) = \operatorname{sgn}((x^2 - kx + 6)(\sin x - 1/2))$ (where $k > 0$) has exactly $4$ points of discontinuity in $(0, 6)$,then the maximum integral value of $k$ is: