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: $\pm\frac{\sqrt{5}}{2}$
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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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Question : If $x+\left [\frac{1}{(x+7)}\right]=0$, what is the value of $x-\left [\frac{1}{(x+7)}\right]$?
Option 1: $3\sqrt{5}$
Option 2: $3\sqrt{5}-7$
Option 3: $3\sqrt{5}+7$
Option 4: $8$
Question : If $\left(x^2 - \frac{1}{x^2}\right) = 4 \sqrt{6}$ and $x>1$, what is the value of $\left(x^3 - \frac{1}{x^3}\right)?$
Option 1: $20 \sqrt{2}$
Option 2: $24 \sqrt{2}$
Option 3: $18 \sqrt{2}$
Option 4: $22 \sqrt{2}$
Question : If $\left(x^2+\frac{1}{x^2}\right)=7$, and $0<x<1$, find the value of $x^2-\frac{1}{x^2}$.
Option 1: $3 \sqrt{5}$
Option 2: $4 \sqrt{5}$
Option 3: $-4\sqrt{3}$
Option 4: $-3\sqrt{5}$
Question : If $2x+\frac{1}{2x}=2,$ what is the value of $\sqrt{2\left (\frac{1}{x}\right)^{4}+\left (\frac{1}{x}\right)^{5}}\; ?$
Option 1: 1
Option 2: 2
Option 3: 4
Option 4: 8
Question : If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$ is:
Option 1: $\frac{3}{4}$
Option 2: $\frac{4}{3}$
Option 3: $\frac{3}{5}$
Option 4: $\frac{5}{3}$
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