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:

Let $f(x) = \begin{cases} x^2 + k, & \text{when } x \ge 0 \\ -x^2 - k, & \text{when } x < 0 \end{cases}$. If the function $f(x)$ is continuous at $x = 0$,then $k =$

Let $f(x) = x \cdot \left[ \frac{x}{2} \right]$ for $-10 < x < 10$,where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of $f$ is equal to

If $f(x) = \begin{cases} [x] + [-x], & x \ne 2 \\ \lambda, & x = 2 \end{cases},$ then $f$ is continuous at $x = 2,$ provided $\lambda$ is (where $[.]$ is the Greatest Integer Function).

If $a$ is the point of discontinuity of the function $f(x) = \begin{cases} \cos 2 x, & \text{for } -\infty < x < 0 \\ e^{3 x}, & \text{for } 0 \leq x < 3 \\ x^2-4 x+3, & \text{for } 3 \leq x \leq 6 \\ \frac{\log (15 x-89)}{x-6}, & \text{for } x>6 \end{cases}$ Then, $\lim _{x \rightarrow a} \frac{x^2-9}{x^3-5 x^2+9 x-9} =$

Let $f(x) = \begin{cases} 0, & x=0 \\ 2-x, & 0 < x < 1 \\ 2, & x=1 \\ \frac{1}{2}-x, & 1 < x < 2 \\ \frac{-3}{2}, & x \geq 2 \end{cases}$ then which of the following is true?