If the function $f(x)=\begin{cases} \frac{\tan a(x-1)}{x-1}, & \text{if } 0 < x < 1 \\ \frac{x^3-125}{x^2-25}, & \text{if } 1 \leq x \leq 4 \\ \frac{b^x-1}{x}, & \text{if } x > 4 \end{cases}$ is continuous in its domain,then $6a + 9b^4 = $

For real $x$ with $-10 \leq x \leq 10$,define $f(x) = \int_{-10}^x 2^{[t]} dt$,where for a real number $r$,we denote by $[r]$ the greatest integer less than or equal to $r$. The number of points of discontinuity of $f$ in the interval $(-10, 10)$ is

If the function $f(x) = \begin{cases} x + a \sqrt{2} \sin x & \text{if } 0 \leq x \leq \frac{\pi}{4} \\ 2x \cot x + b & \text{if } \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{if } \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous in $[0, \pi]$,then $a - b = $

If the function $f(x) = \begin{cases} \frac{2}{x} \{\sin(k_1+1)x + \sin(k_2-1)x\} & , x < 0 \\ 4 & , x = 0 \\ \frac{2}{x} \log_e \left(\frac{2+k_1x}{2+k_2x}\right) & , x > 0 \end{cases}$ is continuous at $x = 0$,then $k_1^2 + k_2^2$ is equal to

If $f(x) = \begin{cases} x, & x > 1 \\ x^2, & x < 1 \end{cases}$,then $\lim_{x \to 1} f(x) = $

Consider the function $f(x) = \begin{cases} \frac{x+5}{x-2}, & \text{if } x \neq 2 \\ 1, & \text{if } x=2 \end{cases}$. Then,$f(f(x))$ is discontinuous