Question : If $\left(3 y+\frac{3}{y}\right)=8$, then find the value of $\left(y^2+\frac{1}{y^2}\right)$.
Option 1: $5\frac{1}{9}$
Option 2: $4\frac{5}{6}$
Option 3: $7\frac{1}{9}$
Option 4: $9\frac{1}{9}$
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Correct Answer: $5\frac{1}{9}$
Solution : Given: $\left(3 y+\frac{3}{y}\right)=8$ $⇒(y+\frac{1}{y})=\frac{8}{3}$ Squaring both sides, $⇒(y^2+\frac{1}{y^2}+2)=\frac{64}{9}$ $⇒(y^2+\frac{1}{y^2})=\frac{64}{9} - 2$ $⇒(y^2+\frac{1}{y^2})=\frac{64-18}{9} $ $⇒(y^2+\frac{1}{y^2})=\frac{46}{9}=5\frac{1}{9}$ Hence, the correct answer is $5\frac{1}{9}$.
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Question : If $\left(4y-\frac{4}{y}\right)=11$, find the value of $\left(y^2+\frac{1}{y^2}\right)$.
Option 1: $7 \frac{9}{16}$
Option 2: $5 \frac{9}{16}$
Option 3: $9 \frac{11}{16}$
Option 4: $9 \frac{9}{16}$
Question : If $\left(y^2+\frac{1}{y^2}\right)=74$ and $y>1$, then find the value of $\left(y-\frac{1}{y}\right)$.
Option 1: $6 \sqrt{2}$
Option 2: $-2 \sqrt{19}$
Option 3: $2 \sqrt{19}$
Option 4: $-6 \sqrt{2}$
Question : If $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:
Option 1: $\frac{2 \sqrt{6}-6}{5}$
Option 2: $\frac{4 \sqrt{6}-6}{5}$
Option 3: $\frac{4 \sqrt{6}-6}{3}$
Option 4: $\frac{4 \sqrt{3}-6}{5}$
Question : Find the value of the given expression. $\left(2 \frac{1}{2}÷ 1 \frac{7}{8}\right) ÷\left(9 \frac{3}{8}÷ 11 \frac{2}{3} \text { of } \frac{1}{8}\right)$
Option 1: $\frac{33}{135}$
Option 2: $\frac{11}{135}$
Option 3: $\frac{28}{135}$
Option 4: $\frac{57}{135}$
Question : The value of $\left(\frac{2+7 \times 7 \div 9 \text { of } 9+6 \div 6 \times 2}{4 \div 4 \text { of } 5+7 \times 7 \div 7-6+4}\right)$ is:
Option 1: $\frac{1734}{2407}$
Option 2: $\frac{1865}{2106}$
Option 3: $\frac{1857}{2507}$
Option 4: $\frac{1664}{2305}$
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