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Let f(x)= casesx3+8; & x<0,\ x2-4; & x 0, cases and g(x)= cases(x-8)1/3; & x<0,\ (x+4)1/2; & x 0. cases Then the number of points, where the function g f is discontinuous, is .
JEE Main 2026 (Online) 6th April Evening Shift
Previous year questions with solutions · 2025 – 2026
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 13 | 19 | 0.68 | 2.74% ↑64.2% |
| 2025 | 5 | 12 | 0.42 | 1.67% |
Let f(x)= casesx3+8; & x<0,\ x2-4; & x 0, cases and g(x)= cases(x-8)1/3; & x<0,\ (x+4)1/2; & x 0. cases Then the number of points, where the function g f is discontinuous, is .
The number of points in the interval [2, 4], at which the function f(x)= [x2-x- 12 ], where [ ] denotes the greatest integer function, is discontinuous, is .
The number of points of discontinuity of the function f(x) = [ x22 ] - [ x], x [0, 4], where [ ] denotes the greatest integer function, is .
If the function f(x) = ( x) - ( x) x - x is continuous at x = 0, then f(0) is equal to .