The function $f: R - \{0\} \to R$,given by $f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1}$ can be made continuous at $x = 0$ by defining $f(0)$ as:

Let $f: R \rightarrow R$ be such that $f(2x-1) = f(x)$ for all $x \in R$. If $f$ is continuous at $x = 1$ and $f(1) = 1$,then:

Let $R$ be the set of all real numbers and $\alpha \in R$ be positive. Define a function $f: R \rightarrow R$ by $f(0)=0$ and $f(x)=|x|^\alpha \sum \limits_{n=0}^{\infty}\left(1+x^2\right)^{-n}$,for $x \neq 0$. Then the set of real numbers $\alpha$ for which $f$ is continuous at $x = 0$ has

If $f(x) = \lim_{n \to \infty} \frac{[x^2] + [(2x)^2] + [(3x)^2] + \cdots + [(nx)^2]}{n^3}$,then $f(x)$ is (Where $[\cdot]$ is the Greatest Integer Function).

Let $a, b, c$ be three real numbers. If the function $f(x) = \begin{cases} \cos(2x + \pi) & \text{if } x \leq 0 \\ ax^2 + b & \text{if } 0 < x < 1 \\ cx + 4 & \text{if } 1 \leq x \leq 2 \\ 3a + 1 & \text{if } x \geq 2 \end{cases}$ is continuous everywhere,then $b^2 - bc + c^2 =$

Find all points of discontinuity of $f,$ where $f$ is defined by
$f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases}$