Question

Let $f\left(x\right)=[x{]}^2+\sqrt{\left\{x\right\}}$, where [x] is greatest integer function and {x} is the fractional part function, then

the function f(x) is discontinuous at.

• A. All integer points
• B. x = 0
• C. x = 1
• D. All integer points except x = 1

Answer 1

The correct option is D

All integer points except x = 1

We are dealing with greatest integer function and fractional part function. So seeing the bahavior of f(x)

on and around integer points is very important.

Let try to do this by taking a 'p' on an integer I and on the left and right side of I i.e., LHL and RHL.

If $PϵI$

${lim}_{x\to p}+f\left(x\right)={lim}_{h\to 0}\left[\right[p+h{]}^2+\sqrt{\{p+h\}}]$

$=p^2+0$

$=p^2$

${lim}_{x\to p}-f\left(x\right)={lim}_{h\to 0}\left[\right[p-h{]}^2+\sqrt{\{p-h\}}]$

$=(p-1{)}^2+1$

For continuity at an integer p

${lim}_{x\to p}-f\left(x\right)={lim}_{x\to p}+f\left(x\right)=f\left(p\right)$

$i.e.,p^2=(p-1{)}^2+1$

$i.e.,p^2=p^2-2p+1+1$

$p=1$

$\therefore f\left(x\right)$ is continuous only at x=1. And it's discontinous at any other integer points.