Let $f:[-1,2] \rightarrow \mathbb{R}$ be given by $f(x)=2x^2+x+[x^2]-[x]$,where $[t]$ denotes the greatest integer less than or equal to $t$. The number of points where $f$ is not continuous is:

If $f(x) = [x] - [\frac{x}{4}]$,$x \in R$,where $[x]$ denotes the greatest integer function,then:

If $f(x)= \begin{cases}-2 \sin x & , \quad x \leqslant-\frac{\pi}{2} \\ a \sin x+b & , \quad \frac{-\pi}{2} < x < \frac{\pi}{2} \\ \cos x & , \quad x \geqslant \frac{\pi}{2}\end{cases}$ is continuous at $x=-\frac{\pi}{2}$ and $x=\frac{\pi}{2}$,then the value of $2a+b$ is

Let $f(x) = \begin{cases} x^p \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$. Then $f(x)$ is continuous but not differentiable at $x = 0$ if:

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 $f(x) = \begin{cases} \frac{2^x + 2^{3-x} - 6}{\sqrt{2^{-x}} - 2^{1-x}} & \text{if } x > 2 \\ \frac{x^2 - 4}{x - \sqrt{3x - 2}} & \text{if } x < 2 \end{cases}$. Determine the nature of the function at $x = 2$.