Suppose $f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases}$ and if $\lim_{x \to 1} f(x) = f(1)$,what are the possible values of $a$ and $b$?

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 the function $f(x)$,defined below,is continuous in the interval $[0, \pi]$,then find the values of $a$ and $b$.
$f(x) = \begin{cases} x + a\sqrt{2}(\sin x), & 0 \le x < \frac{\pi}{4} \\ 2x(\cot x) + b, & \frac{\pi}{4} \le x \le \frac{\pi}{2} \\ a(\cos 2x) - b(\sin x), & \frac{\pi}{2} < x \le \pi \end{cases}$

The function $f(x) = \frac{x^2 - 1}{x^3 - 1}$ is not defined at $x = 1$. If the function is continuous at $x = 1$,then the value of $f(1)$ will be:

If a real valued function $f(x) = \begin{cases} \log(1+[x]), & x \geq 0 \\ \sin^{-1}[x], & -1 \leq x < 0 \\ k([x]+|x|), & x < -1 \end{cases}$ is continuous at $x = -1$,then $k =$

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, $\lim _{x \rightarrow a} \frac{x^2-9}{x^3-5 x^2+9 x-9} =$