If $f(x) = \begin{cases} \frac{x - 1}{2}, & 0 \leqslant x < 1 \\ 1/2, & 1 \leqslant x < 2 \end{cases}$ and $g(x) = (2x + 1)(x - k) + 3$ for $0 \leqslant x < \infty$,then $g(f(x))$ will be continuous at $x = 1$ if $k$ is equal to:

Let $f(x) = [2x^2 + 1]$ and $g(x) = \begin{cases} 2x - 3, & x < 0 \\ 2x + 3, & x \geq 0 \end{cases}$,where $[t]$ denotes the greatest integer function $\leq t$. Then,in the open interval $(-1, 1)$,the number of points where $f(g(x))$ is discontinuous is equal to:

If $f(x) = \begin{cases} x^2 + \alpha, & x \ge 0 \\ 2\sqrt{x^2 + 1} + \beta, & x < 0 \end{cases}$ is continuous at $x = 0$ and $f(\frac{1}{2}) = 2$,then $\alpha^2 + \beta^2$ is

If $f(x) = \frac{e^{x^2} - \cos x}{x^2}$ for $x \neq 0$ is continuous at $x = 0$,then $f(0) = $.

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 $\lim_{x \rightarrow 0} f(x)g(x)$

Let $f(x) = \begin{cases} \frac{\sin \pi x}{5x}, & x \ne 0 \\ k, & x = 0 \end{cases}$. If $f(x)$ is continuous at $x = 0$,then $k =$