If the function $f(x) = \begin{cases} \frac{\cos ax - \cos 9x}{x^2}, & x \neq 0 \\ 16, & x = 0 \end{cases}$ is continuous at $x = 0$, then $a =$

Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx^2, & \text{if } x \le 2 \\ 3, & \text{if } x > 2 \end{cases}$ at $x=2$.

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 $k =$

Let the function $f(x)$ be defined as: $f(x) = \begin{cases} [\tan(\frac{\pi}{4} + x)]^{\frac{1}{x}}, & x \neq 0 \\ k, & x = 0 \end{cases}$. If $f(x)$ is continuous at $x = 0$, then the value of $k$ is...

Let $m$ and $n$ be the number of points at which the function $f(x) = \max \{x, x^3, x^5, \dots, x^{21}\}$,$x \in R$,is not differentiable and not continuous,respectively. Then $m + n$ 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