| If \[x+\frac{1}{x}=2,\]then what is the value of \[x-\frac{1}{x}\]? |
A) \[0\]
B) \[1\]
C) \[2\]
D) \[-\,\,2\]
Correct Answer: A
Solution :
| Given that, \[x+\frac{1}{x}=2\] (i) |
| On squaring both sides, we get |
| \[{{\left( x+\frac{1}{x} \right)}^{2}}=4\] |
| \[\Rightarrow \]\[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2=4\]\[\Rightarrow \]\[{{x}^{2}}+\frac{1}{{{x}^{2}}}=2\] |
| Now, \[{{\left( x-\frac{1}{x} \right)}^{2}}=\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)-2\] |
| \[\Rightarrow \]\[{{\left( x-\frac{1}{x} \right)}^{2}}=2-2=0\] [from Eq.(ii)] |
| \[\therefore \] \[x-\frac{1}{x}=0\] |
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