MCQ
The function $f(x) = {[x]^2} - [{x^2}]$, (where $[y]$ is the greatest integer less than or equal to $y$),is discontinuous at
  • A
    All integers
  • B
    All integers except $0$ and $1$
  • C
    All integers except $0$
  • All integers except $1$

Answer

Correct option: D.
All integers except $1$
d
(d) Given $f(x) = {[x]^2} - [{x^2}]$

$ - 1 < x < 0,\,\,f(x) = {( - 1)^2} - 0 = 1$

$x = 0,\,\,f(x) = {0^2} - 0 = 0$

$0 < x < 1,\,\,f(x) = {0^2} - 0 = 0$

$x = 1,\,\,f(x) = {1^2} - 1 = 0$

$1 < x < \sqrt 2 ,\,\,f(x) = {1^2} - 1 = 0$

$x = \sqrt 2 ,\,\,f(x) = {1^2} - 2 = - 1$

$\sqrt 2 < x < \sqrt 3 ,\,\,f(x) = {1^2} - 2 = - 1$

$x = \sqrt 3 ,\,\,f(x) = {1^2} - 3 = - 2$

$\sqrt 3 < x < 2,\,\,f(x) = {1^2} - 3 = - 2$

$x = 2,\,\,f(x) = 4 - 4 = 0$; 

$2 < x < \sqrt 5 ,\,\,f(x) = 4 - 4 = 0$

$x = \sqrt 5 ,\,\,f(x) = 4 - 5 = - 1$

Hence function is discontinuous at all integers except $1$.

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