If x² - 5x + 1 = 0, then the value of $$\left( {{x^4} + \frac{1}{{{x^2}}}} \right) \div \left( {{x^2} + 1} \right)$$     is:

A. 21

B. 22

C. 25

D. 24

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & {x^2} - 5x + 1 = 0 \cr & \left( {{x^4} + \frac{1}{{{x^2}}}} \right) \div \left( {{x^2} + 1} \right) = ? \cr & {x^2} - 5x + 1 = 0 \cr & \Rightarrow x + \frac{1}{x} = 5 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 110 \cr & \frac{{x\left( {{x^3} + \frac{1}{{{x^3}}}} \right)}}{{x\left( {x + \frac{1}{x}} \right)}} = \frac{{110}}{5} = 22 \cr} $$