A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
\[x+\frac{1}{x}=2\] \[\Rightarrow \]\[{{x}^{2}}+1=2x\Rightarrow {{x}^{2}}-2x+1=0\] \[\Rightarrow \]\[{{(x-1)}^{2}}=0\Rightarrow x=1\] \[\therefore \]\[{{x}^{2}}+\frac{1}{{{x}^{3}}}=1+1=2\]You need to login to perform this action.
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