Let $f(x) = \begin{cases} \frac{1 + \cos 2\pi x}{1 - \sin \pi x}, & x < \frac{1}{2} \\ p, & x = \frac{1}{2} \\ \frac{\sqrt{2x - 1}}{\sqrt{4 + \sqrt{2x - 1}} - 2}, & x > \frac{1}{2} \end{cases}$. If $f(x)$ is discontinuous at $x = \frac{1}{2}$,then:

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{\cos 3x - \cos x}{x^2}, & \text{for } x \neq 0 \\ \lambda, & \text{for } x = 0 \end{cases}$ and if $f$ is continuous at $x = 0$,then $\lambda$ is equal to

If the function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$,$x \neq 0$ is continuous at $x = 0$,then $f(0) = $ . . . . . .

Let $f(x) = \begin{cases} (1 + |\sin x|)^{a/|\sin x|}, & -\pi/6 < x < 0 \\ b, & x = 0 \\ e^{\tan 2x/\tan 3x}, & 0 < x < \pi/6 \end{cases}$. If $f$ is continuous at $x = 0$,then the values of $a$ and $b$ are respectively:

$A$ function $f(x)$ is defined as $f(x) = \begin{cases} x^m \sin \frac{1}{x} & x \neq 0, m \in N \\ 0 & x = 0 \end{cases}$. The least value of $m$ for which $f'(x)$ is continuous at $x = 0$ is

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).