Q38 / 100 · BITSAT 2024 (Memory Based Paper 1) · Mathematics
If arithmetic mean of two distinct positive real number $a$ and $b(a \gt b)$ be twice their geometric mean, then $a: b=$
- Correct answerA$(2+\sqrt{3}):(2-\sqrt{3})$
- B$(2+\sqrt{5}):(2-\sqrt{5})$
- C$(2+\sqrt{2}):(2-\sqrt{2})$
- DNone of these
Solution
By the given condition,
$\frac{a+b}{2}=2 \sqrt{a b} \Rightarrow a+b=4 \sqrt{a b}$
Now, $(a-b)^2=(a+b)^2-4 a b=16 a b-4 a b=$ $12 a b$
$\therefore a-b=\sqrt{12 a b}=2 \sqrt{3} \sqrt{a b}$.
(Taking + ve sign only as $a \gt b$)
$\therefore \frac{a+b}{a-b}=\frac{4 \sqrt{a b}}{2 \sqrt{3} \sqrt{a b}}=\frac{2}{\sqrt{3}}$
By componendo and dividendo,
$\frac{2 a}{2 b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$ or $\frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Hence, the correct answer is option: 'A'.