Question:
The AM between two positive numbers a and b (a > b) is twice their GM. Then, a : b = ?
दो धनात्मक संख्याओं a और b (a > b) के बीच समांतर माध्य (AM) उनके गुणोत्तर माध्य (GM) का दोगुना है। तो a : b = ?
Options:
- \((2 + \sqrt{3}) : (2 – \sqrt{3})\)
- 2 : 3
- \((3 + \sqrt{2}) : (3 – \sqrt{2})\)
- 3 : 2
- \((2 + \sqrt{3}) : (2 – \sqrt{3})\)
- 2 : 3
- \((3 + \sqrt{2}) : (3 – \sqrt{2})\)
- 3 : 2
Correct Answer: a)
Explanation:
Given \(\text{AM} = 2\text{GM} \implies \dfrac{a+b}{2} = 2\sqrt{ab} \implies \dfrac{a+b}{2\sqrt{ab}} = \dfrac{2}{1}\).
Applying componendo and dividendo:
\(\dfrac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \dfrac{2+1}{2-1} = 3 \implies \left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right)^2 = 3 \implies \dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \dfrac{\sqrt{3}}{1}\).
Applying componendo and dividendo again:
\(\dfrac{\sqrt{a}}{\sqrt{b}} = \dfrac{\sqrt{3}+1}{\sqrt{3}-1}\).
Squaring both sides:
\(\dfrac{a}{b} = \dfrac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)^2} = \dfrac{4+2\sqrt{3}}{4-2\sqrt{3}} = \dfrac{2+\sqrt{3}}{2-\sqrt{3}}\).
Hence, \(a : b = (2+\sqrt{3}) : (2-\sqrt{3})\).
दिया गया है: \(\text{AM} = 2\text{GM} \implies \dfrac{a+b}{2} = 2\sqrt{ab} \implies \dfrac{a+b}{2\sqrt{ab}} = \dfrac{2}{1}\)।
योगान्तरानुपात (Componendo & Dividendo) प्रमेय से:
\(\dfrac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \dfrac{3}{1} \implies \left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right)^2 = 3 \implies \dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \sqrt{3}\)।
पुनः योगान्तरानुपात लगाने पर:
\(\dfrac{\sqrt{a}}{\sqrt{b}} = \dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)।
दोनों पक्षों का वर्ग करने पर:
\(\dfrac{a}{b} = \dfrac{4+2\sqrt{3}}{4-2\sqrt{3}} = \dfrac{2+\sqrt{3}}{2-\sqrt{3}}\)।
अतः \(a : b = (2+\sqrt{3}) : (2-\sqrt{3})\)।