A) \[2\frac{1}{3}\]
B) \[2\frac{2}{3}\]
C) \[2\frac{4}{9}\]
D) \[2\frac{5}{9}\]
Correct Answer: C
Solution :
| [c] Given,\[x\left( 3-\frac{2}{x} \right)=\frac{3}{x}(x\ne 0)\] \[\Rightarrow \] \[3x-\frac{3}{x}=2\] \[\Rightarrow \] \[3\left( x-\frac{1}{x} \right)=2\] \[\Rightarrow \] \[x-\frac{1}{x}=\frac{2}{3}\] On squaring both sides, we get \[{{x}^{2}}+\frac{1}{{{x}^{2}}}-2=\frac{4}{9}\] \[\Rightarrow \] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=\frac{4}{9}+2=\frac{22}{9}\] \[\therefore \] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=2\frac{4}{9}\] |
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