If x - \(\frac{1}{x}\) = 1, find \(\left(\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x^2+1}-\frac{1}{x^2-1}\right)\)
Answer & explanation
Correct answer: option 2
\(\left(\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x^2+1}-\frac{1}{x^2-1}\right)\) = \(\frac{x+1-x+1}{(x+1)(x-1)}+\frac{1}{x^2-1}-\frac{1}{x^2-1}\)
⇒ \(\frac{2}{x^2-1}+\frac{1}{x^2+1}-\frac{1}{x^2-1}\)
⇒ \(\frac{1}{x^2-1}+\frac{1}{x^2+1}\)
⇒ \(\frac{x^2+1+x^2-1}{x^4-1}\) =\(\frac{2x^2}{x^4-1}\)
⇒ \(\frac{2x^2}{x^2(x^2-\frac{1}{x^2})}\) =\(\frac{2}{x^4-\frac{1}{x^2}}\) .....(1)
Now x - \(\frac{1}{x}\) = 1
x + \(\frac{1}{x}\) = \(\sqrt {1+4} = \sqrt {5}\)
Put in (1)
\(\frac{2}{(1)(\sqrt {5})}\)
±\(\frac{2}{\sqrt {5}}\)