If $f(x) = \begin{cases} 3 + x; & x \geqslant 0 \\ 2 - 3x; & x < 0 \end{cases}$,then $\lim_{x \to 0} f(f(x))$ is equal to -

Assertion $(A)$: $f(x)=|x-a|+|x-b|$ is continuous on $R$. Reason $(R)$: $\frac{|x-\alpha|}{x-\alpha}$ is continuous at $x \in R-\{\alpha\}$. The correct option among the following is:

Let $[t]$ denote the greatest integer less than or equal to $t$. If the function $f(x) = \begin{cases} b^2 \sin \left(\frac{\pi}{2} \left[\frac{\pi}{2}(\cos x + \sin x) \cos x\right]\right), & x < 0 \\ \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases}$ is continuous at $x = 0$,then $a^2 + b^2$ is equal to

Examine the following function for continuity: $f(x) = \frac{x^{2} - 25}{x + 5}, x \neq -5$.

If the function $f$ defined by $f(x) = \begin{cases} \cos x, & \text{if } x \leq 0 \\ 3x + \alpha, & \text{if } 0 < x < 2 \\ \beta x + 3, & \text{if } 2 \leq x \leq 4 \\ 11, & \text{if } x > 4 \end{cases}$ where $\alpha$ and $\beta$ are real constants,is continuous on $R$,then $\alpha^2 + \beta^2 =$

Discuss the continuity of the function $f$ given by $f(x) = x^{3} + x^{2} - 1$.