A) 123
B) 83
C) 92
D) 112
Correct Answer: A
$ \Rightarrow $ $ x^{2}+\frac{1}{x^{2}}=7 $ … (ii)
Again, squaring both sides, we get
$ {{( x^{2}+\frac{1}{x^{2}} )}^{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 $ {{( x+\frac{1}{x} )}^{3}}={{(3)}^{3}} $
$ \Rightarrow $ $ x^{3}+\frac{1}{x^{3}}+3( x+\frac{1}{x} )=27 $
$ \Rightarrow $ $ x^{3}+\frac{1}{x^{3}}+9=27 $ $ [ \because ( x+\frac{1}{x} )=3 ] $
$ \Rightarrow $ $ x^{3}+\frac{1}{x^{3}}=18 $ (iv) On multiplying Eqs. (i) and (iii), we get $ ( x^{4}+\frac{1}{x^{4}} )( x+\frac{1}{x} )=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 $
BETA
