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The function $f(x)=[x]^2-[x^2]$ where $[x]$ is the greatest integer less than or equal to x, is discontinuous at

$\begin{array}{1 1}(a)\;\text{all integers}\\(b)\;\text{all integers except 0 and 1}\\(c)\;\text{all integers except 0}\\(d)\;\text{all integers except 1}\end{array}$

1 Answer

We have $f(x)=[x]^2-[x^2]$
At $x=0$
LHL=$\lim\limits_{h\to 0}f(-h)=\lim\limits_{h\to 0}[-h^2]$
$\Rightarrow \lim\limits_{h\to 0}f(-1)^2-[h^2]=\lim\limits_{h\to 0}1-0=1$
RHL=$\lim\limits_{h\to 0}f(h)=\lim\limits_{h\to 0}[h]^2-[h^2]$
$\Rightarrow \lim\limits_{h\to 0}0-0=0$
LHL $\neq$ RHL
$f(x)$ is not continuous at $x=0$
At $x=1$
LHL=$\lim\limits_{h\to 0}f(1-h)=\lim\limits_{h\to 0}[1-h]^2-[(1-h)^2]$
$\Rightarrow \lim\limits_{h\to 0}1-1=0$
$f(1)=[1]^2-[1^2]=1-1=0$
LHL=RHL=f(1)
$f(x)$ is continuous at $x=1$
Clearly at other integral points f(x) is not continuous.
Hence (d) is the correct answer.
answered Jan 3, 2014 by sreemathi.v
 

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