Question : If $x^2+y^2=29$ and $xy=10$, where $x>0,y>0$ and $x>y$. Then the value of $\frac{x+y}{x-y}$ is:

Option 1: $- \frac{7}{3}$

Option 2: $\frac{7}{3}$

Option 3: $\frac{3}{7}$

Option 4: $-\frac{3}{7}$

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: $\frac{5}{3}$

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 $x=\frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}+\sqrt{4}}$ and $y=\frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}}$ then the value of $\frac{x^2-x y+y^2}{x^2+x y+y^2}=$?

Option 1: $\frac{361}{363}$

Option 2: $\frac{341}{343}$

Option 3: $\frac{384}{387}$

Option 4: $\frac{321}{323}$

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: $\sqrt{127}$