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):

The function $f(x) = |x| + \frac{|x|}{x}$ is

If $f(x) = \frac{\log_e(1 + x^2 \tan x)}{\sin x^3}, x \neq 0$ is to be continuous at $x = 0$,then $f(0)$ must be equal to

$f(x) = \begin{cases} \frac{\log x}{x-1}, & \text{if } x \neq 1 \\ k, & \text{if } x=1 \end{cases}$ is continuous at $x=1$,then the value of $k$ is

The function $f : R \rightarrow R$ defined by $f(x) = \lim_{n \rightarrow \infty} \frac{\cos(2 \pi x) - x^{2n} \sin(x-1)}{1 + x^{2n+1} - x^{2n}}$ is continuous for all $x$ in.

The values of $a$ and $b$,so that the function $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b, & \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous for $0 \leq x \leq \pi$,are respectively given by