| If \[{{x}^{2}}+{{y}^{2}}+\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}=4,\]then the value of is [SSC (CPO) 2013] |
A) 2
B) 4
C) 8
D) 16
Correct Answer: A
Solution :
| \[\because \]\[{{x}^{2}}+{{y}^{2}}+\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}-4=0\] |
| \[\Rightarrow \]\[{{x}^{2}}+\frac{1}{{{x}^{2}}}-2+{{y}^{2}}+\frac{1}{{{y}^{2}}}-2=0\] |
| \[\Rightarrow \]\[{{\left( x-\frac{1}{x} \right)}^{2}}+{{\left( y-\frac{1}{y} \right)}^{2}}=0\] |
| If \[x-\frac{1}{x}=0\]\[\Rightarrow \]\[{{x}^{2}}-1=0\] |
| \[\therefore \] \[x=1\] |
| Similarly \[y=1\] |
| \[\therefore \] \[{{x}^{2}}+{{y}^{2}}=1+1=2\] |
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