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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