The number of points of discontinuity of the function f(x)= [frac x 2 2 ]-[√(x)], x in[0,4] , where [•] denotes the greatest integer function is \ \ \ \ .

Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 7 April, Morning Shift — Question 46

The number of points of discontinuity of the function $f(x)=\left[\frac{x^{2}}{2}\right]-[\sqrt{x}], x \in[0,4]$ , where [•] denotes

the greatest integer function is $\_\_\_\_$ .

Numerical answer — enter this value.

Probable values of $x$ where $\left[\frac{x^{2}}{2}\right]$ may be discontinuous on $x \in[0,4]$ are $=1,2,3,4,5,6,7,8$

$x=\sqrt{2}, 2, \sqrt{6}, 2 \sqrt{2}, \sqrt{10}, 2 \sqrt{3}, \sqrt{14}, 4$

And for $[\sqrt{x}]$ corresponding values are

$x=1,2$

On checking for continuity at these points we get the $f(x)$ is discontinuous at

$x=1, \sqrt{2}, 2, \sqrt{6}, 2 \sqrt{2}, \sqrt{10}, 2 \sqrt{3}, \sqrt{14}$

Hence, $f(x)$ is discontinuous for 8 values of $x \in[0,4]$

Answer key and solution verified before publishing.

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