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If $f(x)=[x^2]-[x]^2$ where [.] is the largest integer function then

$\begin{array}{1 1}(a)\;\text{f(x) is discontinuous for all integral value of x}\\(b)\;\text{f(x) is discontinuous only at x=0,1}\\(c)\;\text{f(x) is continuous only at x=1}\\(d)\;\text{None of these}\end{array}$

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$f(x)=[x^2]-[x]^2$
$f(1^+)=\lim\limits_{h\to 0}([(1+h)^2]-[1+h]^2)$
$\Rightarrow 1-1=0$
$f(1^-)=\lim\limits_{h\to 0}([(1-h)^2]-[1-h]^2)$
$\Rightarrow 0-0=0$
$\therefore f(x)$ is continuous at $x=1$
Hence (c) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 

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