A) \[\sqrt{2}\]
B) \[\sqrt{3}\]
C) 2
D) 1
Correct Answer: C
Solution :
| \[x=\frac{2\sqrt{2}}{1+\sqrt{2}}\] |
| \[\Rightarrow \] \[x+\sqrt{2}=+\frac{2\sqrt{2}}{1+\sqrt{2}}+\sqrt{2}\] |
| \[=\frac{2\sqrt{2}-(\sqrt{2}+2)}{1+\sqrt{2}}=\frac{3\sqrt{2}+2}{1+\sqrt{2}}\] |
| \[\Rightarrow \] \[x-\sqrt{2}=\frac{2\sqrt{2}}{1+\sqrt{2}}-\sqrt{2}\] |
| \[=\frac{2\sqrt{2}-(\sqrt{2}+2)}{1+\sqrt{2}}=\frac{\sqrt{2}-2}{1+\sqrt{2}}\] |
| \[\Rightarrow \] \[x+1=\frac{2\sqrt{2}}{1+\sqrt{2}}+1\] |
| \[=\frac{2\sqrt{2}+(1+\sqrt{2})}{1+\sqrt{2}}=\frac{1+3\sqrt{2}}{1+\sqrt{2}}\] |
| \[\Rightarrow \] \[x-1=\frac{2\sqrt{2}}{1+\sqrt{2}}-1\] |
| \[=\frac{2\sqrt{2}-(1+\sqrt{2})}{1+\sqrt{2}}=\frac{\sqrt{2}-1}{1+\sqrt{2}}\] |
| So, \[\frac{x+\sqrt{2}}{x-\sqrt{2}}+\frac{x+1}{x-1}=\frac{3\sqrt{2}+2}{\sqrt{2}-2}+1+\frac{3\sqrt{2}}{\sqrt{2}-1}\] |
| \[=\frac{(3\sqrt{2}+2)(\sqrt{2-1})+(1+3\sqrt{2})(\sqrt{2-2})}{(\sqrt{2}-2)(\sqrt{2}-1)}\] |
| \[=\frac{(6+2\sqrt{2}-3\sqrt{2}-2)+(\sqrt{2}+6-2-6\sqrt{2})}{2-2\sqrt{2}-\sqrt{2}+2}\] |
| \[=\frac{(4-\sqrt{2})+(4-5\sqrt{2})}{4-3\sqrt{2}}\] |
| \[=\frac{8-6\sqrt{2}}{4-3\sqrt{2}}=\frac{2\,\,(4-3\sqrt{2})}{4-3\sqrt{2}}=2\] |
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