Question : If $x^2-5 x+1=0$, then the value of $\left(x^4+\frac{1}{x^2}\right) \div\left(x^2+1\right)$ is:
Option 1: 21
Option 2: 22
Option 3: 25
Option 4: 24
Correct Answer: 22
Solution :
$x^2 - 5x + 1 = 0$
$⇒x(x - 5 + \frac{1}{x}) = 0$
$⇒(x - 5 + \frac{1}{x}) = 0$
$⇒(x + \frac{1}{x}) = 5$
Cubing both sides, we get,
$⇒(x + \frac{1}{x})^3 = 5^3$
$\because$ $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$⇒x^3 + \frac{1}{x^3} + 3.x.\frac{1}{x} (x + \frac{1}{x}) = 125$
$⇒x^3 + \frac{1}{x^3} + 3(5) = 125$
$⇒x^3 + \frac{1}{x^3} + 15 = 125$
$⇒x^3 + \frac{1}{x^3} = 125 - 15=110$
Now, $\frac{x^4 + \frac{1}{x^2}}{x^2 + 1}$
Dividing the numerator and denominator by $x$,
$=\frac{\frac{1}{x}(x^4 + \frac{1}{x^2})}{(x^2 + 1)\frac{1}{x}} $
$= \frac{(x^3 + \frac{1}{x^3})}{(x + \frac{1}{x})}$
$=\frac{110}{5}$
$=22$
Hence, the correct answer is 22.
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