$f(x) = \begin{cases} \frac{1-\cos kx}{x^2}, & x \le 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is:

Let $f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x - 2)^2}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases}$. If $f(x)$ is continuous for all $x$,then $k =$

If $f(x) = \begin{cases} \frac{1 - [x]}{1 + x}, & x \ne -1 \\ 1, & x = -1 \end{cases}$,then the value of $f(|2k|)$ will be (where $[.]$ denotes the greatest integer function). Which of the following statements is true?

Given $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & \text{if } x < 0 \\ a, & \text{if } x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text{if } x > 0 \end{cases}$
If $f(x)$ is continuous at $x=0$,then the value of $a$ is:

The function $f(x) = 2x - |x - x^2|$ is

Let $f(x) = x^2 + x \sin x - \cos x$. Then