If x = 2 + \(\sqrt {3}\)
Find \(\frac{x^2 - x + 1}{x^2 + 1 + x}\)
Answer & explanation
Correct answer: option 4
Formula → If x = a + b and a2 - b2 = 1, then \(\frac{1}{x}\) = a - b always
⇒ x = 2 + \(\sqrt {3}\)
⇒ \(\frac{1}{x}\) = 2 - \(\sqrt {3}\)
Now put the values and find,
\(\frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1}\) = \(\frac{2 + \sqrt {3} + 2 - \sqrt {3} - 1}{2 + \sqrt {3} + 2 - \sqrt {3} + 1}\) = \(\frac{3}{5}\)