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

Let $f(x) = \begin{cases} \frac{ax^{2}+2ax+3}{4x^{2}+4x-3}, & x \neq -\frac{3}{2}, \frac{1}{2} \\ b, & x = -\frac{3}{2}, \frac{1}{2} \end{cases}$ be continuous at $x=-\frac{3}{2}$. If $f(f(x)) = \frac{7}{5}$, then $x$ is equal to:

If the function $f(x)=\begin{cases} \frac{\tan a(x-1)}{x-1}, & \text{if } 0 < x < 1 \\ \frac{x^3-125}{x^2-25}, & \text{if } 1 \leq x \leq 4 \\ \frac{b^x-1}{x}, & \text{if } x > 4 \end{cases}$ is continuous in its domain, then $6a + 9b^4 = $

Given $f(x) = b ([x]^2 + [x]) + 1$ for $x \geq -1$ and $f(x) = \sin(\pi(x+a))$ for $x < -1$,where $[x]$ denotes the greatest integer function,for what values of $a$ and $b$ is the function continuous at $x = -1$?

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:

$A$ point in the domain of a function where the discontinuity cannot be removed by redefining the function at that point is called: