A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{5}\]
Correct Answer: D
Solution :
\[x+\frac{1}{x}=5\] Then \[\frac{2x}{3{{x}^{2}}-5x+3}=\frac{\frac{2x}{3}}{\frac{3{{x}^{2}}}{x}-\frac{5x}{x}+\frac{3}{x}}\] \[=\frac{2}{3x+\frac{3}{x}-5}\] \[=\frac{2}{3\left( x+\frac{1}{x} \right)-5}=\frac{2}{3\times 5-5}=\frac{2}{10}=\frac{1}{5}\]
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