Question : If $x=\sqrt{a}+\frac{1}{\sqrt{a}}$ and $y=\sqrt{a}-\frac{1}{\sqrt{a}}, (a>0),$ then the value of $(x^{4}+y^{4}-2x^{2}y^{2})$ is:
Option 1: 16
Option 2: 20
Option 3: 10
Option 4: 5
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Correct Answer: 16
Solution : $x=\sqrt{a}+\frac{1}{\sqrt{a}}$ Taking square on both sides, $x^2 = a+\frac{1}{a}+2$---------------(i) $y=\sqrt{a}-\frac{1}{\sqrt{a}}$ Taking square on both sides, $y^2 = a+\frac{1}{a}-2$---------------(ii) Subtracting (ii) from (i), $x^2-y^2 = (a+\frac{1}{a}+2)-(a+\frac{1}{a}-2)$ ⇒ $x^2-y^2 = 4$ Taking square on both sides, ⇒ $(x^2-y^2)^2 = 4^2$ ⇒ $x^4-2x^2y^2+y^4 = 16$ Hence, the correct answer is 16.
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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:
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:
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:
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 $\frac{x^2-x y+y^2}{x^2+x y+y^2}=$?
Question : If $x^4+\frac{16}{x^4}=15617, x>0$, then find the value of $x+\frac{2}{x}$.
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