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

Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined as $f(x) = \begin{cases} 5, & \text{if } x \le 1 \\ a + bx, & \text{if } 1 < x < 3 \\ b + 5x, & \text{if } 3 \le x < 5 \\ 30, & \text{if } x \ge 5 \end{cases}$. Then $f$ is

Let $f : [-1,3] \to R$ be defined as $f(x) = \begin{cases} |x| + [x], & -1 \leq x < 1 \\ x + |x|, & 1 \leq x < 2 \\ x + |x|, & 2 \leq x \leq 3 \end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. Then,$f$ is discontinuous at:

If the function $f(x) = \begin{cases} x + a \sqrt{2} \sin x & \text{if } 0 \leq x \leq \frac{\pi}{4} \\ 2x \cot x + b & \text{if } \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{if } \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous in $[0, \pi]$,then $a - b = $

Given $f(x) = \begin{cases} cx + 1, & x \leq 3 \\ dx + 3, & x > 3 \end{cases}$. If $f$ is continuous at $x = 3$,then $d - c =$ . . . . . . .

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