JEE Main 7 Apr 2025 Shift 1 Paper

Probable values of x where [‌

x2

2

] may be discontinuous on x∈[0,4] are
‌=1,2,3,4,5,6,7,8
‌x=√2,2,√6,2√2,√10,2√3,√14,4
And for [sqrt‌x] corresponding values are x=1,2
On checking for continuity at these points we get the f(x) is discontinuous at
x=1,√2,2,√6,2√2,√10,2√3,√14
Hence, f(x) is discontinuous for 8 values of x∈[0,4]