Discussion
Using differentials, \( \sqrt{16.1} \) is approximately
Answer: C
Take \( y = \sqrt{x} \) at \( x = 16 \), where \( y = 4 \) and \( dy/dx = 1/(2\sqrt{x}) = 1/8 \). With \( dx = 0.1 \),
\[ \sqrt{16.1} \;\approx\; 4 + \frac{1}{8}(0.1) \;=\; 4 + 0.0125 \;=\; 4.0125. \]
The true value is 4.01248, so the linear estimate is high by two parts in a million
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