Question

Let $f:R\to R$ be defined as $f\left(x\right)=(|x-1|+|4x-11|)[x^2-2x-2],$ where $[.]$ denotes the greatest integer function. Then the number of points of discontinuity of $f\left(x\right)$ in $(\frac{1}{2},\frac{5}{2})$ is

Answer 1

$f\left(x\right)=(|x-1|+|4x-11|)[x^2-2x-2]$
or, $f\left(x\right)=(|x-1|+|4x-11|)[{(x-1)}^2-3]$
or, $f\left(x\right)=\underset{g\left(x\right)}{\underset{}{(|x-1|+|4x-11|)}}\underset{h\left(x\right)}{\underset{}{(\left[{(x-1)}^2\right]-3)}}$


$g\left(x\right)$ is continuous for all real values of $x.$
Possible points of discontinuity of $h\left(x\right)$ are the points where $(x-1{)}^2$ takes integral values.
At $x=1,$
$h\left(1^{-}\right)=h\left(1^{+}\right)=h\left(0\right)=-3$
So, $f\left(x\right)$ is continuous at $x=1$

At $x=2,$
$h\left(2^{-}\right)=0-3=-3$
$h\left(2^{+}\right)=1-3=-2$
$h$ is discontinuous at $x=2,$ so $f\left(x\right)$ is discontinuous at $x=2$

At $x=\sqrt{2}+1,$
$h\left(\right(\sqrt{2}+1{)}^{-})=1-3=-2$
$h\left(\right(\sqrt{2}+1{)}^{+})=2-3=-1$
$h$ is discontinuous at $x=\sqrt{2}+1,$ so $f\left(x\right)$ is discontinuous at $x=\sqrt{2}+1$

$\therefore$ In $(\frac{1}{2},\frac{5}{2}),$ there are two points of discontinuity.