If $$x + \frac{1}{x} = 1{\text{,}}$$   then the value of $$\frac{{{x^2} + 3x + 1}}{{{x^2} + 7x + 1}}$$   is?

A. $$\frac{1}{2}$$

B. $$\frac{3}{7}$$

C. 2

D. 3

Answer: Option A

Solution(By Examveda Team)

$$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = 1 \cr & {\text{Find }}\frac{{{x^2} + 3x + 1}}{{{x^2} + 7x + 1}} \cr & {\text{From equation (i)}} \cr & \Rightarrow x + \frac{1}{x} = 1 \cr & \Rightarrow {x^2} + 1 = x \cr & \Rightarrow \frac{{\left( {{x^2} + 1} \right) + 3x}}{{\left( {{x^2} + 1} \right) + 7x}} \cr & \Rightarrow \frac{{x + 3x}}{{x + 7x}} \cr & \Rightarrow \frac{{4x}}{{8x}} \cr & \Rightarrow \frac{1}{2} \cr} $$