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: $9 \frac{9}{16}$
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Correct Answer: $9 \frac{9}{16}$
Solution : Given: $(4 y-\frac{4}{y})=11$ ⇒ $(y-\frac{1}{y})=\frac{11}{4}$ Squaring both sides, we get: $(y-\frac{1}{y})^2=\frac{121}{16}$ ⇒ $y^2+\frac{1}{y^2}-2×y×\frac{1}{y}=\frac{121}{16}$ ⇒ $y^2+\frac{1}{y^2} = \frac{121}{16} +2$ $= \frac{121+32}{16}$ $= \frac{153}{16}$ $= 9\frac{9}{16}$ Hence, the correct answer is $9\frac{9}{16}$.
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