A) \[2\]
B) \[1\frac{1}{5}\]
C) \[\frac{1}{5}\]
D) \[\frac{3}{5}\]
Correct Answer: C
Solution :
| [c] Given, \[x+\frac{1}{x}=5\] Now, \[\frac{2x}{3{{x}^{2}}-5x+3}\] On dividing numerator and denominator by x \[\frac{2x}{3{{x}^{2}}-5x+3}=\frac{\frac{2x}{x}}{\frac{3{{x}^{2}}}{x}-\frac{5x}{x}+\frac{3}{x}}\] \[=\frac{2}{3x-5+\frac{3}{x}}=\frac{2}{3\left( x+\frac{1}{x} \right)-5}\] \[=\frac{2}{3(5)-5}=\frac{2}{15-5}=\frac{2}{10}=\frac{1}{5}\] |
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