Question 53

If $$x + \frac{1}{x} = 5, x \neq 0$$ then the value of $$\frac{x^4 + \frac{1}{x^2}}{x^2 - 3x + 1}$$ is equal to:

Solution

Given, $$x + \frac{1}{x} = 5$$

$$\frac{x^4+\frac{1}{x^2}}{x^2-3x+1}=\frac{x\left(x^3+\frac{1}{x^3}\right)}{x\left(x-3+\frac{1}{x}\right)}$$

$$=\frac{x^3+\frac{1}{x^3}}{x+\frac{1}{x}-3}$$

$$=\frac{\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)}{5-3}$$

$$=\frac{\left(5\right)^3-3\left(5\right)}{2}$$

$$=\frac{125-15}{2}$$

$$=\frac{110}{2}$$

$$=55$$

Hence, the correct answer is Option D

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