A) 2 |
B) 3 |
C) \( \Large 2\sqrt{2} \) |
D) \( \Large 2\sqrt{3} \) |
D) \( \Large 2\sqrt{3} \) |
Given, \( \Large x = \sqrt{3} + \sqrt{2} \)
Therefore, \( \Large \frac{1}{x} = \frac{1}{ \left(\sqrt{3}+\sqrt{2}\right) } \times \frac{ \left(\sqrt{3} - \sqrt{2}\right) }{ \left(\sqrt{3} - \sqrt{2}\right) } \) [rationalizing]
\( \Large = \frac{1}{x} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} \)
Therefore, \( \Large x+\frac{1}{x} = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} = 2\sqrt{3} \)