Discussion
Define a function \(f : [0, 1] \to \mathbf{R}\) as \(f(x) = x^{2} + e^{x} - 2\). Then :
Answer: C
The correct statement is \(f\) is increasing.
\[ f'(x) = 2x + e^{x} \]
On [0, 1] both terms are non-negative and \(e^{x} \ge 1\), so \(f'(x) \gt 0\) throughout and \(f\) is strictly increasing.
That rules out the others. \(f'\) never vanishes on [0, 1], and a strictly increasing function is one-one. Also \(f(0) = -1 \lt 0\) and \(f(1) = e - 1 \gt 0\), so by the intermediate value theorem \(f(x) = 0\) does have a root in [0, 1].
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