A) 3
B) 5
C) 7
D) 9
Correct Answer: A
Solution :
(a): \[x+\frac{1}{x}=3\] On squaring both sides, \[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2=9\] \[\Rightarrow {{x}^{2}}+\frac{1}{{{x}^{2}}}=9-2=7\]???.(i) Expression \[=\frac{{{x}^{4}}+3{{x}^{3}}+5{{x}^{2}}+3x+1}{{{x}^{4}}+1}=\frac{{{x}^{4}}+1+3{{x}^{3}}+3x+5{{x}^{2}}}{{{x}^{4}}+1}\] \[=\frac{{{x}^{2}}\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)+3{{x}^{2}}\left( x+\frac{1}{x} \right)+5{{x}^{2}}}{{{x}^{2}}\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}=\frac{{{x}^{2}}\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)+3\left( x+\frac{1}{x} \right)}{{{x}^{2}}+\frac{1}{{{x}^{2}}}}\] \[=\frac{7+3\times 3+5}{7}=\frac{21}{7}=3\]
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