In order that the function $f(x) = (x + 1)^{1/x}$ is continuous at $x = 0$,$f(0)$ must be defined as

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

Let $f(x) = \begin{cases} x^3 - x^2 + 10x - 5, & x \le 1 \\ -2x + \log_2(b^2 - 2), & x > 1 \end{cases}$. The set of values of $b$ for which $f(x)$ has the greatest value at $x = 1$ is given by:

Let $f: R \rightarrow R$ be the function defined by $f(x) = \begin{cases} 5, & \text{if } x \leq 1 \\ a+bx, & \text{if } 1 < x < 3 \\ b+5x, & \text{if } 3 \leq x < 5 \\ 30, & \text{if } x \geq 5 \end{cases}$. Then $f$ is:

If the function $f(x) = \begin{cases} \frac{\log_{e}(1-x+x^{2}) + \log_{e}(1+x+x^{2})}{\sec x - \cos x}, & x \in (-\frac{\pi}{2}, \frac{\pi}{2}) - \{0\} \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k$ is equal to.

The value of $a$ for which the function $f(x) = \begin{cases} \frac{1-\cos 4 x}{x^2}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0 \end{cases}$ is continuous at $x=0$, is