Let $a$ be a positive real number. If a real valued function $f(x) = \begin{cases} \frac{6^x-3^x-2^x+1}{1-\cos \left(\frac{x}{a}\right)} & \text{if } x \neq 0 \\ \log 3 \log 4 & \text{if } x=0 \end{cases}$ is continuous at $x=0$,then $a=$

If $f: R \rightarrow R$ is defined by $f(x) = x - [x]$, where $[x]$ is the greatest integer not exceeding $x$, then the set of points of discontinuity of $f$ is

Let $a, b \in \mathbb{R}$ $(a \neq 0)$. If the function $f$ is defined as $f(x) = \begin{cases} \frac{2x^2}{a}, & 0 \leq x < 1 \\ a, & 1 \leq x < \sqrt{2} \\ \frac{2b^2-4b}{x}, & \sqrt{2} \leq x < \infty \end{cases}$ is continuous in the interval $[0, \infty)$,then an ordered pair $(a, b)$ is

Consider the function $f(x) = \begin{cases} \frac{P(x)}{\sin(x-2)}, & x \neq 2 \\ 7, & x = 2 \end{cases}$ where $P(x)$ is a polynomial such that $P''(x)$ is always a constant and $P(3) = 9$. If $f(x)$ is continuous at $x = 2$,then $P(5)$ is equal to:

Let $f: R \to R$ be a function defined by $f(x) = [x] \cos \left( \frac{2x - 1}{2} \pi \right)$,where $[x]$ denotes the greatest integer function. Then $f$ is:

Let $f: R \rightarrow R$ be a function given by $f(x) = \begin{cases} \frac{1-\cos 2x}{x^2} & , x < 0 \\ \alpha & , x = 0 \\ \frac{\beta \sqrt{1-\cos x}}{x} & , x > 0 \end{cases}$. If $f$ is continuous at $x = 0$,then $\alpha^2 + \beta^2$ is equal to: