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
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Correct Answer: 8
Solution : Given: $2x+\frac{1}{2x}=2$ $⇒4x^{2}-4x+1=0$ $⇒\left (2x-1 \right)^{2}=0$ $⇒x=\frac{1}{2}$ So, $\sqrt{2\left (\frac{1}{x}\right)^{4}+\left (\frac{1}{x}\right)^{5}}\;=\sqrt{2\left (2\right)^{4}+\left (2\right )^{5}}\;=\sqrt{32+32}\;=8$ Hence, the correct answer is 8.
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Question : If $\left(x+\frac{1}{x}\right)=\sqrt{6}$ and $x>1$, what is the value of $\left(x^8-\frac{1}{x^8}\right)$?
Question : If $\left(x+\frac{1}{x}\right)=5$, and $x>1$, what is the value of $\left(x^8-\frac{1}{x^8}\right)?$
Question : If $\left(x^2 - \frac{1}{x^2}\right) = 4 \sqrt{6}$ and $x>1$, what is the value of $\left(x^3 - \frac{1}{x^3}\right)?$
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