A) 191
B) 192
C) 193
D) 194
Correct Answer: D
Solution :
(d): \[\left( x+\frac{1}{x} \right)=4\] Squaring we get; \[\left( {{x}^{2}}+\frac{1}{{{x}^{2}}}+2 \right)=16={{x}^{2}}+\frac{1}{{{x}^{2}}}=14\] Squaring again \[{{x}^{4}}+\frac{1}{{{x}^{4}}}+2\left[ {{x}^{2}}+\frac{1}{{{x}^{2}}}+{{x}^{2}}.2+\frac{1}{{{x}^{2}}}.2 \right]+4=256\]\[\Rightarrow {{x}^{4}}+\frac{1}{{{x}^{4}}}+2\left[ 1+2\left\{ {{x}^{2}}+\frac{1}{{{x}^{2}}} \right\} \right]=256\] \[\Rightarrow {{x}^{4}}+\frac{1}{{{x}^{4}}}+2\left[ 1+28 \right]=256\,\,\,\Rightarrow {{x}^{4}}+\frac{1}{{{x}^{4}}}=194\]You need to login to perform this action.
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