$\quad$ Let $f(x)=|| x-1|-1|$ if $x<1$ and $f(x)=[x]$ if $x \geq 1,$ where, for any real number $x,[x]$ denotes the largest integer $\leq x$ and $|y|$ denotes absolute value of $y .$ Then, the set of discontinuity-points of the function f consists of

Let us first check at $x=1$
$\lim f(x)$ as $x\to1-=\lim || x-1|-1|$ as $x\to 1-=1$
$\lim f(x)$ as $x\to1+=\lim [x]$ as $x\to1+=1$
So, continuous at $x=1$
Let us now check at $x=0$
$\lim f(x)$ as $x\to 0-=\lim || x-1|-1|$ as $x\to 0-=0$
$\lim f(x)$ as $x\to 0+=\lim || x-1|-1|$ as $x\to 0+=0$
So continuous at $x=0$

Can you now finish the problem ..........

Let us now check at $x=2$
$\lim f(x)$ as $x\to 2-=\lim [x]$ as $x\to 2-=1$
$\lim f(x)$ as $x\to 2+=\lim [x]$ as $x \rightarrow 2+=2$
Discontinuous at $x=2$
Option (c) is correct.

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