If $f(x) = \begin{cases} ax^2 - b, & 0 \le x < 1 \\ 2, & x = 1 \\ x + 1, & 1 < x \le 2 \end{cases}$ is continuous at $x = 1$,then the most suitable values of $a$ and $b$ are:

If $f(x) = \operatorname{sgn} \left( 3\cos x - \frac{a}{3} \right)$ is continuous for all $x$,then the least positive integral value of $a$ is - (where $\operatorname{sgn}(x)$ denotes the signum function of $x$)

If $f(x) = \frac{\ln(e^{x^2} + 2\sqrt{x})}{\sqrt{x}}$ is continuous at $x = 0$,then $f(0)$ must be equal to:

Let $f:(0,1) \rightarrow R$ be a function defined as $f(x) = \sqrt{n}$ if $x \in \left[\frac{1}{n+1}, \frac{1}{n}\right)$ where $n \in N$. Let $g:(0,1) \rightarrow R$ be a function such that $\int_{x^2}^x \sqrt{\frac{1-t}{t}} dt < g(x) < 2\sqrt{x}$ for all $x \in (0,1)$. Then $\lim_{x \rightarrow 0} f(x)g(x)$

If $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ for $x \neq \pi$ is continuous at $x = \pi$,then $f(\pi) =$

Let $f: R \rightarrow R$ be defined by $f(x) = \begin{cases} a - \frac{\sin [x-1]}{x-1} & \text{if } x > 1 \\ 1 & \text{if } x = 1 \\ b - \left[ \frac{\sin [x-1] - [x-1]}{([x-1])^3} \right] & \text{if } x < 1 \end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. If $f$ is continuous at $x = 1$,then $a + b =$