Try this beautiful problem based on Discontinuity, useful for ISI B.Stat Entrance.

## Discontinuity Problem | ISI B.Stat TOMATO 734

$\quad$ 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

- all integers $\geq 0$
- all integers $\geq 1$
- all integers $> 1$
- the integer 1

**Key Concepts**

Limit

Calculas

Continuous

## Check the Answer

But try the problem first…

Answer: \(C\)

TOMATO, Problem 734

Challenges and Thrills in Pre College Mathematics

## Try with Hints

First hint

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 ……….

Second Hint

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.

## Other useful links

- https://www.cheenta.com/algebraic-equation-amc-10a-2001-problem-10/
- https://www.youtube.com/watch?v=fRj9NuPGrLU&t=331s

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