Question

If $a=\frac{1}{2-\sqrt{3}},b=\frac{1}{3-\sqrt{3}}andc=\frac{1}{\sqrt{5}+1}$ then the ascending order is

• A. a < c < b
• B. a < b < c
• C. c < b < a
• D. b < c < a

Answer 1

The correct option is C c < b < a

Given, $a=\frac{1}{2-\sqrt{3}},b=\frac{1}{3-\sqrt{3}}andc=\frac{1}{\sqrt{5}+1}$

$a=\frac{1}{2-\sqrt{3}}$

$\Rightarrow a=\frac{1}{(2-\sqrt{3})}\times \frac{(2+\sqrt{3})}{(2+\sqrt{3})}=\frac{(2+\sqrt{3})}{(4-3)}=2+\sqrt{3}\approx 3.73$

$b=\frac{1}{(3-\sqrt{3})}=\frac{1}{(3-\sqrt{3})}\times \frac{(3+\sqrt{3})}{(3+\sqrt{3})}=\frac{3+\sqrt{3}}{6}=\frac{4.73}{6}$

$c=\frac{1}{(\sqrt{5}+1)}=\frac{1}{(\sqrt{5}+1)}\times \frac{(\sqrt{5}-1)}{(\sqrt{5}-1)}=\frac{\sqrt{5}-1}{4}=\frac{1.23}{4}$

$Thus,a>1,\frac{1}{2}<b<1andc<\frac{1}{2}$
Therefore, the ascending order is c < b < a.
Hence, the correct answer is option c.