Discussion
Let \(f : [0, 1] \to \mathbf{R}\) be continuous and \(\int_{0}^{x} f = \int_{x}^{1} f\) for all \(x \in [0, 1]\). Then :
Answer: B
\(f\) is constant (in fact identically zero).
Write \(F(x) = \int_{0}^{x} f\) and \(I = \int_{0}^{1} f\). The condition says \(F(x) = I - F(x)\), so \(F(x) = I/2\) for every \(x\). Putting \(x = 0\) gives \(I/2 = F(0) = 0\), so \(F \equiv 0\).
Since \(f\) is continuous, \(F' = f\) by the fundamental theorem of calculus, so \(f \equiv 0\), a constant function.
\(f(t) = t\) fails: \(\int_{0}^{x} t\,dt = x^{2}/2\) but \(\int_{x}^{1} t\,dt = (1 - x^{2})/2\). \(t^{2}\) fails likewise, and a strictly monotonic function cannot be identically zero.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development