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: $\sqrt{13}$
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Correct Answer: $\sqrt{13}$
Solution : Given: $x=\frac{1}{x-3},(x>0)$ Or, $x^2-3x-1=0$ Or, $x=\frac{3\pm\sqrt{13}}{2}$ So, $x = \frac{3+\sqrt{13}}{2}$ Now, $x+\frac{1}x = \frac{3+\sqrt{13}}{2}+\frac{2}{3+\sqrt{13}}$ Rationalising, = $\frac{3+\sqrt{13}}{2}+\frac{2}{3+\sqrt{13}}\times \frac{3-\sqrt{13}}{3-\sqrt{13}}$ = $\frac{3+\sqrt{13}}{2}+2\times \frac{3-\sqrt{13}}{-4}$ = $\sqrt13$ Hence, the correct answer is $\sqrt{13}$.
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Question : The value of $\frac{1}{4-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+\frac{1}{\sqrt{14}-\sqrt{13}}-\frac{1}{\sqrt{13}-\sqrt{12}}+\frac{1}{\sqrt{12}-\sqrt{11}}-\frac{1}{\sqrt{11}-\sqrt{10}}+\frac{1}{\sqrt{10}-3}-\frac{1}{3-\sqrt{8}}$ is:
Question : If $(x-\frac{1}{3})^2+(y-4)^2=0$, then what is the value of $\frac{y+x}{y-x}$?
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