Question : If $x^{2}+1=2x$, then the value of $\frac{x^{4}+\frac{1}{x^{2}}}{x^{2}-3x+1}$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –2
Correct Answer: –2
Solution : $x^{2}+1=2x$ $⇒x+\frac{1}{x}= 2$----------(i) Now, $\frac{x^{4}+\frac{1}{x^{2}}}{x^{2}-3x+1}$ = $\frac{x(x^{3}+\frac{1}{x^{3}})}{x(x+\frac{1}{x}-3)}$ = $\frac{(x+\frac{1}{x})^3-3(x+\frac{1}{x})}{x+\frac{1}{x}-3}$ Substituting the value of equation (i), we get, = $\frac{2^3-3×2}{2-3}$ = $-2$ Hence, the correct answer is –2.
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