Question : If $x^4+\frac{16}{x^4}=15617, x>0$, then find the value of $x+\frac{2}{x}$.
Option 1: $\sqrt{121}$
Option 2: $\sqrt{129}$
Option 3: $\sqrt{123}$
Option 4: $\sqrt{127}$
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Correct Answer: $\sqrt{129}$
Solution : Given: $x^4+\frac{16}{x^4}=15617$ ⇒ $(x^2)^2 + (\frac{4}{x^2})^2 + 2\times x^2 \times \frac{4}{x^2} = 15617 + 2\times x^2 \times \frac{4}{x^2}$ ⇒ $(x^2+\frac{4}{x^2})^2 = 15625$ ⇒ $x^2 + \frac{4}{x^2} = 125$ ⇒ $x^2 + (\frac{2}{x})^2 + 2\times x \times \frac{2}{x} = 125 + 2\times x\times \frac{2}{x}$ ⇒ $(x+\frac{2}{x})^2 = 129$ ⇒ $x+\frac{2}{x} = \sqrt{129}$ Hence, the correct answer is $\sqrt{129}$.
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Question : If $x=\frac{1}{x-3},(x>0)$, then the value of $x+\frac{1}{x}$ is:
Option 1: $\sqrt{11}$
Option 2: $\sqrt{17}$
Option 3: $\sqrt{15}$
Option 4: $\sqrt{13}$
Question : If $2x=\sqrt{a}+\frac{1}{\sqrt{a}}, a>0$, then the value of $\frac{\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}$ is:
Option 1: $a+1$
Option 2: $\frac{1}{2}(a+1)$
Option 3: $\frac{1}{2}(a–1)$
Option 4: $a–1$
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 $(x+\frac{1}{x})=6$ and $x>1$, find the value of $(x^2–\frac{1}{x^2})$.
Option 1: $18 \sqrt{2}$
Option 2: $30 \sqrt{2}$
Option 3: $24 \sqrt{2}$
Option 4: $12 \sqrt{10}$
Question : If $x>0$ and $x^4+\frac{1}{x^4}=254$, what is the value of $x^5+\frac{1}{x^5}?$
Option 1: $717 \sqrt{2}$
Option 2: $723 \sqrt{2}$
Option 3: $720 \sqrt{2}$
Option 4: $726 \sqrt{2}$
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