| If \[x+\frac{1}{x}=3,\]then \[{{x}^{5}}+\frac{1}{{{x}^{5}}}\]is equal to [SSC (CPO) 2013] |
A) 123
B) 83
C) 92
D) 112
Correct Answer: A
Solution :
| \[x+\frac{1}{x}=3\] ... (i) |
| On squaring both sides, we get |
| \[{{\left( x+\frac{1}{x} \right)}^{2}}={{(3)}^{2}}\]\[\Rightarrow \]\[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2=9\] |
| \[\Rightarrow \] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=7\] ... (ii) |
| Again, squaring both sides, we get |
| \[{{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{2}}={{(7)}^{2}}\]\[\Rightarrow \]\[{{x}^{4}}+\frac{1}{{{x}^{4}}}+2=49\] |
| \[\Rightarrow \] \[{{x}^{4}}+\frac{1}{{{x}^{4}}}=47\] ... (iii) |
| On cubing both sides, we get |
| \[{{\left( x+\frac{1}{x} \right)}^{3}}={{(3)}^{3}}\] |
| \[\Rightarrow \]\[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\left( x+\frac{1}{x} \right)=27\] |
| \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+9=27\] \[\left[ \because \left( x+\frac{1}{x} \right)=3 \right]\] |
| \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}=18\] (iv) |
| On multiplying Eqs. (i) and (iii), we get |
| \[\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)\left( x+\frac{1}{x} \right)=47\times 3\] |
| \[\Rightarrow \] \[{{x}^{5}}+\frac{1}{{{x}^{5}}}+{{x}^{3}}+\frac{1}{{{x}^{3}}}=141\] |
| \[\Rightarrow \] \[{{x}^{5}}+\frac{1}{{{x}^{5}}}+18=141\] [from Eq. (iv)] |
| \[\Rightarrow \] \[{{x}^{5}}+\frac{1}{{{x}^{5}}}=123\] |
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