A) \[42\]
B) \[43\]
C) \[45\]
D) \[47\]
Correct Answer: D
Solution :
\[\because \] \[\left( x+\frac{1}{x} \right)=3\] \[\therefore \] \[{{\left( x+\frac{1}{x} \right)}^{2}}={{(3)}^{2}}\] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2\times x\times \frac{1}{x}=9\] \[\Rightarrow \] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=9-2=7\] Squaring both sides again, \[\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)={{(7)}^{2}}\] \[\Rightarrow \] \[{{x}^{4}}+\frac{1}{{{x}^{4}}}+2\times {{x}^{2}}\times \frac{1}{{{x}^{2}}}=49\] \[\Rightarrow \] \[{{x}^{4}}+\frac{1}{{{x}^{4}}}+2=49\] \[\Rightarrow \] \[{{x}^{4}}+\frac{1}{{{x}^{4}}}=49-2=47\]
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