Let $f(x) = \begin{cases} x^p \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$. Then $f(x)$ is continuous but not differentiable at $x = 0$ if:

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} x-1, & \text{for } x \leq 1 \\ 2-x^2, & \text{for } 1 < x \leq 3 \\ x-10, & \text{for } 3 < x < 5 \\ 2x, & \text{for } x \geq 5 \end{cases}$,then the set of points of discontinuity of $f$ is

If $f(x) = \begin{cases} 6 \beta - 3 \alpha x, & \text{if } -4 \leq x < -2 \\ 4x + 1, & \text{if } -2 \leq x \leq 2 \end{cases}$ is continuous on $[-4, 2]$,then $\alpha + \beta = $

If $f(x) = \begin{cases} x, & \text{if } x \text{ is irrational} \\ 0, & \text{if } x \text{ is rational} \end{cases}$,then $f$ is

The function $f$ is defined by $f(x) = \begin{cases} 2x - 1, & \text{if } x > 2 \\ k, & \text{if } x = 2 \\ x^2 - 1, & \text{if } x < 2 \end{cases}$. If $f$ is continuous at $x = 2$,then the value of $k$ is equal to:

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: