If $f(x) = \begin{cases} 1 + \cos x, & x \le 0 \\ a - x, & 0 < x < 2 \\ (x - b)^2, & x \ge 2 \end{cases}$ is continuous at $x=0$ and $x=2$,then find the value of $a^2+b^2$.

If the function $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right), & x < 0 \\ k, & x = 0 \\ \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2}+1}-1}, & x > 0 \end{cases}$ is continuous at $x = 0$,then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to:

Let $f(x) = \begin{cases} |x|, & -\infty < x < 2 \\ |2x-4|, & 2 \leq x \leq 20 \end{cases}$. If $x=a$ is a point where $f(x)$ is continuous but not differentiable and $x=b$ is a point where $f(x)$ is not differentiable $(a \neq b)$,then $a+b=$

If the function $f(x) = \frac{1 - \cos 4x}{8x^2}$ for $x \ne 0$ and $f(x) = k$ for $x = 0$ is a continuous function at $x = 0$,then the value of $k$ is:

For what value of $k$ is the function $f(x) = \begin{cases} \frac{\text{log}(1+2x) \sin x^{\circ}}{x^2}, & x \neq 0 \\ k, & x = 0 \end{cases}$ continuous at $x = 0$?

If a function $f(x)$ defined by $f(x) = \begin{cases} ax^2 + bx + c, & x \leq -1 \\ 2x^2 + 4x + 1, & -1 < x < 1 \\ cx^2 + bx + a, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$,and $\lim_{x \rightarrow \frac{3}{2}} f(x) = 14$,then find $\lim_{x \rightarrow -2} f(x)$.