A) \[\frac{\sqrt{261}}{5}\]
B) \[\sqrt{\frac{50}{211}}\]
C) 4
D) \[\sqrt{\frac{211}{50}}\]
Correct Answer: A
Solution :
\[x+\frac{1}{x}=3\frac{4}{5}=\frac{19}{5}\] Squaring both sides we get \[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2\times x\times \frac{1}{x}={{\left( \frac{19}{5} \right)}^{2}}\] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2=\frac{361}{25}\] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=\frac{361}{25}-2\] \[=\frac{361-50}{25}=\frac{311}{25}\] Subtracting 2 from both sides \[{{x}^{2}}+\frac{1}{{{x}^{2}}}-2=\frac{311}{25}-2\] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}-2\times x\times \frac{1}{x}=\frac{311-50}{25}\] \[=\,\,\frac{261}{25}\,\,\,\therefore \,\,\,{{\left( x-\frac{1}{x} \right)}^{2}}=\frac{261}{25}\] \[x-\frac{1}{x}=\underline{\frac{\sqrt{\mathbf{261}}}{\mathbf{5}}}\]
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