Question : If $y+\frac{1}{y}=11$, then the value of $y^3-\frac{1}{y^3}$ is:
Option 1: $345 \sqrt{13}$
Option 2: $360 \sqrt{13}$
Option 3: $352 \sqrt{13}$
Option 4: $368 \sqrt{13}$
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Correct Answer: $360 \sqrt{13}$
Solution : Given, $y+\frac{1}{y}=11$ Squaring both sides, we get: $⇒(y+\frac{1}{y})^2=11^2$ $⇒(y-\frac{1}{y})^2+4(y)(\frac{1}y)=121$ $⇒(y-\frac{1}{y})^2+4=121$ $⇒(y-\frac{1}{y})^2=121-4$ $⇒(y-\frac{1}{y})=3\sqrt{17}$ Cubing both sides, we get: $⇒(y-\frac{1}{y})^3=(3\sqrt{17})^3$ $⇒y^3-\frac{1}{y^3}-3(y)(\frac{1}{y})(y-\frac{1}{y})=351\sqrt{13}$ $⇒y^3-\frac{1}{y^3}-3(3\sqrt{17})=351\sqrt{13}$ $⇒y^3-\frac{1}{y^3}=351\sqrt{13}+9\sqrt{13}$ $\therefore y^3-\frac{1}{y^3}=360\sqrt{13}$ Hence, the correct answer is $360\sqrt{13}$.
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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:
Option 1: $2-2 \sqrt{2}$
Option 2: $4+2 \sqrt{2}$
Option 3: $4-2 \sqrt{2}$
Option 4: $2+2 \sqrt{2}$
Question : If $(x-\frac{1}{3})^2+(y-4)^2=0$, then what is the value of $\frac{y+x}{y-x}$?
Option 1: $\frac{11}{13}$
Option 2: $\frac{13}{11}$
Option 3: $\frac{16}{9}$
Option 4: $\frac{9}{16}$
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 : Let $x=\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}$ and $y=\frac{1}{x}$, then the value of $3x^{2}-5xy+3y^{2}$ is:
Option 1: 1717
Option 2: 1177
Option 3: 1771
Option 4: 1171
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: $\frac{321}{323}$
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