Find all the points of discontinuity of the function $f$ defined by
$f(x) = \begin{cases} x + 2, & \text{if } x < 1 \\ 0, & \text{if } x = 1 \\ x - 2, & \text{if } x > 1 \end{cases}$

If the function defined by $f(x) = \frac{\log (1+x)^{1+x}}{x^2} - \frac{1}{x}, x \neq 0$ is continuous at $x=0$,then $6 f(0)$ is equal to

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

The values of $a$ and $b$ for which the function $f(x) = \begin{cases} 1+|\sin x|^{a/|\sin x|}, & -\pi / 6 < x < 0 \\ b, & x = 0 \\ e^{\tan 2 x / \tan 3 x}, & 0 < x < \pi / 6 \end{cases}$ is continuous at $x = 0$ are

If the real valued function $f(x) = \begin{cases} \frac{\cos 3x - \cos x}{x \sin x} & \text{if } x < 0 \\ p & \text{if } x = 0 \\ \frac{\log(1 + q \sin x)}{x} & \text{if } x > 0 \end{cases}$ is continuous at $x = 0$, then $p + q =$

Show that the function defined by $g(x)=x-[x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$.