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.
Related Questions
Know More about
Staff Selection Commission Sub Inspector ...
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Get Updates Brochure
Your Staff Selection Commission Sub Inspector Exam brochure has been successfully mailed to your registered email id “”.
