At $x=0$, the function $f(x) = \begin{cases} \frac{x}{|x|+2x^2}, & x \neq 0 \\ k, & x=0 \end{cases}$ is:

Let $f(x) = \begin{cases} 0, & \text{if } -1 \leq x < 0 \\ 1, & \text{if } x = 0 \\ 2, & \text{if } 0 < x \leq 1 \end{cases}$ and let $F(x) = \int_{-1}^{x} f(t) \, dt, -1 \leq x \leq 1$. Then:

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 $f(x) = \begin{cases} x \sin x, & 0 < x \le \frac{\pi}{2} \\ \frac{\pi}{2} \sin(\pi + x), & \frac{\pi}{2} < x < \pi \end{cases}$,then

If $[x]$ denotes the greatest integer not exceeding $x$ and if the function $f$ defined by $f(x)= \begin{cases} \frac{a+2 \cos x}{x^2} & , x < 0 \\ b \tan \frac{\pi}{[x+4]} & , x \geq 0 \end{cases}$ is continuous at $x=0$, then the ordered pair $(a, b)$ is equal to

If $f(x) = \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}$ is continuous at $x = 0$,then $f(0) = $