Let $f(x) = \frac{1 - x(1 + |1 - x|)}{|1 - x|} \cos \left(\frac{1}{1 - x}\right)$ for $x \neq 1$. Then

The function $f(x) = [x]$,where $[x]$ denotes the greatest integer function,is continuous at:

Let $f(x) = \begin{cases} -2 \sin x, & \text{if } x \leq -\frac{\pi}{2} \\ A \sin x + B, & \text{if } -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & \text{if } x \geq \frac{\pi}{2} \end{cases}$. For what values of $A$ and $B$ is $f$ continuous?

The function $f(x) = |x| + \frac{|x|}{x}$ is

If $f(x) = \begin{cases} x^2 + \alpha, & x \ge 0 \\ 2\sqrt{x^2 + 1} + \beta, & x < 0 \end{cases}$ is continuous at $x = 0$ and $f(\frac{1}{2}) = 2$,then $\alpha^2 + \beta^2$ is

If $f(x) = \begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & \text{for } -1 \leq x < 0 \\ 2x^2+3x-2, & \text{for } 0 \leq x \leq 1 \end{cases}$ is continuous at $x=0$,then $k$ is equal to