If $f$ is defined by $f(x) = \begin{cases} \frac{1-\cos ax}{x \sin x}, & x \neq 0 \\ \frac{1}{2}, & x = 0 \end{cases}$ and $f$ is continuous at $x=0$,then $a^{2} =$ . . . . . . .

The number of points in the interval $[2, 4]$, at which the function $f(x) = [x^2 - x - 1/2]$, where $[·]$ denotes the greatest integer function, is discontinuous, is ————

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

The function $f(x) = [x] \cdot \cos \left( \frac{2x - 1}{2} \right) \pi$,where $[\cdot]$ denotes the greatest integer function,is discontinuous at

If $f(x) = \begin{cases} [x] + [-x], & x \ne 2 \\ \lambda, & x = 2 \end{cases},$ then $f$ is continuous at $x = 2,$ provided $\lambda$ is (where $[.]$ is the Greatest Integer Function).

Let $k$ be a non-zero real number. If $f(x) = \begin{cases} \frac{(e^x - 1)^2}{\sin (x/k) \log (1 + x/4)}, & x \neq 0 \\ 12, & x = 0 \end{cases}$ is a continuous function,then the value of $k$ is