Question : If $x+\frac{1}{x}=3$, then the value of $\frac{3x^{2}-4x+3}{x^{2}-x+1}$ is:
Option 1: $\frac{4}{3}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{5}{3}$
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Correct Answer: $\frac{5}{2}$
Solution :
Given: $x+\frac{1}{x}=3$
To find: $\frac{3x^{2}-4x+3}{x^{2}-x+1}$
Dividing with $x$ on both numerator and denominator,
= $\frac{3x-4+\frac{3}{x}}{x-1+\frac{1}{x}}$
= $\frac{3(x+\frac{1}{x})-4}{(x+\frac{1}{x})-1}$
= $\frac{(3×3)-4}{3-1}$
= $\frac{5}{2}$
Hence, the correct answer is $\frac{5}{2}$.
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