| If \[x=\frac{\sqrt{2}+1}{\sqrt{2}-1}\]and \[y=\frac{\sqrt{2}-1}{\sqrt{2}+1},\]then the value of \[({{x}^{2}}+{{y}^{2}})\]is |
A) 34
B) 36
C) 32
D) 38
Correct Answer: A
Solution :
| Since, \[x=\frac{\sqrt{2}+1}{\sqrt{2}-1}\] |
| and \[y=\frac{\sqrt{2}-1}{\sqrt{2}+1}\] |
| \[\therefore \] \[x=\frac{\sqrt{2}+1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{{{(\sqrt{2}+1)}^{2}}}{2-1}\] |
| \[=\frac{2+1+2\sqrt{2}}{1}=3+2\sqrt{2}\] |
| Similarly, \[y=\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}\] |
| \[=\frac{2+1-2\sqrt{2}}{2-1}=\frac{3-2\sqrt{2}}{1}\] |
| Now, \[({{x}^{2}}+{{y}^{2}})={{(3+2\sqrt{2})}^{2}}+{{(3-2\sqrt{2})}^{2}}\] |
| \[=9+8+2\times 3\times 2\sqrt{2}\] |
| \[+\,\,9+8-2\times 3\times 2\sqrt{2}\] |
| \[=18+16=34\] |
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