Discussion
The function \(f(x) = x^{2}(x - 2)\) is strictly decreasing in
Answer: E
\(f\) is strictly decreasing in \(\left(0, \frac{4}{3}\right)\).
\[ \begin{aligned} f(x) &= x^{3} - 2x^{2} \\ f'(x) &= 3x^{2} - 4x = x(3x - 4) \end{aligned} \]
\(f'(x) \lt 0\) between the roots 0 and \(\frac{4}{3}\), so \(f\) decreases there and increases outside. (1,2) runs past \(\frac{4}{3}\) into the increasing part; (-1,0) and (-1,1) include negative \(x\), where \(f' \gt 0\); \(\left(\frac{4}{3}, \infty\right)\) is where \(f\) increases.
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