Discussion
For positive non-zero real variables \(x\) and \(y\), if
\[ \ln\left(\frac{x+y}{2}\right)=\frac{1}{2}[\ln(x)+\ln(y)] \]
then, the value of \(\frac{x}{y} + \frac{y}{x}\) is
Answer: C
The value is 2.
The right-hand side is \(\tfrac{1}{2}\ln(xy) = \ln\sqrt{xy}\). Since ln is one-to-one, the condition says
\[ \frac{x + y}{2} = \sqrt{xy} \]
The arithmetic mean of two positive numbers equals their geometric mean only when the numbers are equal, since \(\tfrac{x + y}{2} - \sqrt{xy} = \tfrac{1}{2}(\sqrt{x} - \sqrt{y})^{2}\). So \(x = y\), and
\[ \frac{x}{y} + \frac{y}{x} = 1 + 1 = 2 \]
In fact \(x/y + y/x \geq 2\) for all positive \(x, y\), so 1 and 1/2 are impossible, and 4 would need \(x/y = 2 \pm \sqrt{3}\), which breaks the condition.
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