The value$(s)$ of $x$ for which the function $f(x) = \begin{cases} 1-x, & x < 1 \\ (1-x)(2-x), & 1 \leq x \leq 2 \\ 3-x, & x > 2 \end{cases}$ fails to be continuous is(are):

If $f(x)$ defined as given below is continuous on $R$,then the value of $a+b$ is equal to: $f(x) = \begin{cases} \sin x, & x \leq 0 \\ x^2+a, & 0 < x < 1 \\ b x+3, & 1 \leq x \leq 3 \\ -3, & x > 3 \end{cases}$

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?

The function $f: R - \{0\} \to R$,given by $f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1}$ can be made continuous at $x = 0$ by defining $f(0)$ as:

If $f(x) = \begin{cases} \frac{a|x|+x^2-2(\sin |x|)(\cos |x|)}{x} & , x \neq 0 \\ b & , x=0 \end{cases}$ is continuous at $x=0$,then $a+b$ is equal to:

Is the function $f$ defined by $f(x) = \begin{cases} x, & \text{if } x \le 1 \\ 5, & \text{if } x > 1 \end{cases}$ continuous at $x=0$? At $x=1$? At $x=2$?