Question : If $x-\frac{1}{x}=1$, then what is the value of $\left (\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x^{2}+1}-\frac{1}{x^{2}-1} \right)\;$?
Option 1: $\pm \sqrt{5}$
Option 2: $\frac{2}{5}$
Option 3: $\pm\frac{2}{\sqrt{5}}$
Option 4: $\pm\frac{\sqrt{5}}{2}$
Correct Answer: $\pm\frac{2}{\sqrt{5}}$
Solution :
Given: $x-\frac{1}{x}=1$
$⇒x^{2}-1=x$
$⇒x^{2}+1=x+2$
$\left (\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x^{2}+1}-\frac{1}{x^{2}-1} \right)$
$=(\frac{2}{x^{2}-1}+\frac{1}{x^{2}+1}-\frac{1}{x^{2}-1})$
$=(\frac{1}{x^{2}-1}+\frac{1}{x^{2}+1})$
$=(\frac{2x^{2}}{(x^{2}-1)(x^{2}+1)})$
Using $x^{2}-1=x$ and $x^{2}+1=x+2$, we get,
$=(\frac{2x^{2}}{(x)(x+2)})$
$=\frac{2x}{x+2}$
Solving $x^{2}-x-1=0$, we get $x=\frac{1\pm\sqrt{5}}{2}$
So, $x+2=\frac{5\pm\sqrt{5}}{2}$
When $x=\frac{1+\sqrt{5}}{2}$, $\frac{2x}{x+2}=\frac{2}{\sqrt{5}}$
When $x=\frac{1-\sqrt{5}}{2}$, $\frac{2x}{x+2}=-\frac{2}{\sqrt{5}}$
Hence, the correct answer is $\pm\frac{2}{\sqrt{5}}$.
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