A) \[2\]
B) \[3\]
C) \[2\sqrt{2}\]
D) \[2\sqrt{3}\]
Correct Answer: D
Solution :
| [d] Given,\[x=\sqrt{3}+\sqrt{2}\] \[\therefore \] \[\frac{1}{x}=\frac{1}{\sqrt{3}+\sqrt{2}}\] \[\frac{1}{x}=\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\] (on rationalizing) \[=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}\] \[\therefore \] \[x+\frac{1}{x}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\] |
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