A) \[\frac{4}{3}\]
B) \[\frac{3}{2}\]
C) \[\frac{5}{2}\]
D) \[\frac{5}{3}\]
Correct Answer: C
Solution :
[c] Given, \[\left( x+\frac{1}{x} \right)=3,\] |
Given function \[=\,\,\frac{3{{x}^{2}}-4x+3}{{{x}^{2}}-x+1}=\frac{x\left( 3x-4+\frac{3}{x} \right)}{x\left( x-1+\frac{1}{x} \right)}\] |
\[=\,\,\frac{3x+\frac{3}{x}-4}{x+\frac{1}{x}-1}=\,\,\frac{3\left( x+\frac{1}{x} \right)-4}{\left( x+\frac{1}{x} \right)-1}\] |
\[=\frac{3\times 3-4}{3-1}=\frac{9-4}{2}=\frac{5}{2}\] |
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