Aspire Faculty ID #16425 · Topic: NIMCET 2009 · Just now
NIMCET 2009

Let $f(x) = \lfloor x^2 - 3 \rfloor$ where $\lfloor \cdot \rfloor$ is the greatest integer function. Number of points in $(1,2)$ where $f$ is discontinuous:

Solution

Discontinuity occurs when $x^2 - 3$ is an integer 
Let $x^2 - 3 = k \Rightarrow x = \sqrt{k+3}$ 
We need $1 < x < 2$ → square both sides: $1 < \sqrt{k+3} < 2$ 
$\Rightarrow 1 < k+3 < 4$ 
$\Rightarrow -2 < k < 1$ 
 Possible integer values: $k = -1, 0$ 
 So number of discontinuities = $2$

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