If $f(x) = \frac{x^2 - 10x + 25}{x^2 - 7x + 10}$ for $x \neq 5$ and $f$ is continuous at $x = 5$,then $f(5) = $

The function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$ is not defined at $x = 0$. The value which should be assigned to $f$ at $x = 0$ so that it is continuous at $x = 0$ is:

If $f(x) = \begin{cases} ax^2 - b, & 0 \le x < 1 \\ 2, & x = 1 \\ x + 1, & 1 < x \le 2 \end{cases}$ is continuous at $x = 1$,then the most suitable values of $a$ and $b$ are:

$f(x) = \begin{cases} \frac{e^{\alpha x} - e^{x} - x}{x^{2}}, & x \neq 0 \\ \frac{3}{2}, & x = 0 \end{cases}$ Find the value of $\alpha$ for which the function $f$ is continuous.

Consider the function $f(x) = [x] + |1 - x|$ for $-1 \le x \le 3$,where $[x]$ is the greatest integer function.
Statement $1$: $f$ is not continuous at $x = 0, 1, 2$ and $3$.
Statement $2$: $f(x) = \begin{cases} -1 - x, & -1 \le x < 0 \\ 1 - x, & 0 \le x < 1 \\ 1 - x, & 1 \le x < 2 \\ 2 + x - 2, & 2 \le x < 3 \\ 3, & x = 3 \end{cases}$ (Note: The provided Statement $2$ in the prompt is incorrect).

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: