The function $f(x) = 2x - |x - x^2|$ is

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} [e^x], & x < 0 \\ a e^x + [x - 1], & 0 \leq x < 1 \\ b + [\sin(\pi x)], & 1 \leq x < 2 \\ [e^{-x}] - c, & x \geq 2 \end{cases}$ where $a, b, c \in R$ and $[t]$ denotes the greatest integer less than or equal to $t$. Then,which of the following statements is true?

Let $a, b \in R, (a \ne 0)$. If the function $f$ defined as
$f(x) = \begin{cases} \frac{2x^2}{a}, & 0 \le x < 1 \\ a, & 1 \le x < \sqrt{2} \\ \frac{2b^2 - 4b}{x^3}, & \sqrt{2} \le x < \infty \end{cases}$
is continuous in the interval $[0, \infty)$,then an ordered pair $(a, b)$ is

The function $f(x) = p[x + 1] + q[x - 1],$ where $[x]$ is the greatest integer function,is continuous at $x = 1$ if:

If $f(x)$ is continuous at point $x=0$ where $f(x) = \begin{cases} \frac{3 \sin x + 5 \tan x}{a^x - 1} & , x < 0 \\ \frac{2}{\log 2} & , x = 0 \\ \frac{8x + 2x \cos x}{b^x - 1} & , x > 0 \end{cases}$ then the values of $a$ and $b$,respectively,are

The number of point / points where the function $f(x) = \frac{1}{x^2 - 5|x| + 6}$ is discontinuous is......