Question

Find the number of integers x which satisfies the inequality $\frac{3}{1+\sqrt{3}}$ < x < $\frac{3}{\sqrt{5}-\sqrt{3}}$

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is C

4

Rationalising both the terms we get,

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

$\frac{3}{\sqrt{5}-\sqrt{3}}$ = $\frac{3(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$

Thus the inequality becomes,

$\frac{3(\sqrt{3}-1)}{2}$ < x < $\frac{3(\sqrt{5}+\sqrt{3})}{2}$

Upon simplification this becomes

$\sqrt{3}-1$ < $\frac{2x}{3}$ < $\sqrt{5}+\sqrt{3}$

Substitute x to be 1, 2, 5, 6 you will find that 1< x< 6.

Hence, the answer is (C) 4.