Question : If $(4 y-\frac{4}{y})=13$, find the value of $(y^2+\frac{1}{y^2})$.
Option 1: $12 \frac{11}{16}$
Option 2: $10 \frac{9}{16}$
Option 3: $12 \frac{9}{16}$
Option 4: $8 \frac{9}{16}$
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Correct Answer: $12 \frac{9}{16}$
Solution :
Given: $4(y -\frac{1}{y}) = 13$
Squaring on both sides of the above expression, we get,
$⇒4^2(y -\frac{1}{y})^2 = 13^2$
$⇒16(y^2+\frac{1}{y^2}-2) = 169$
$⇒(y^2+\frac{1}{y^2}) = \frac{169}{16}+2$
$⇒(y^2+\frac{1}{y^2}) = \frac{169+32}{16}$
$⇒(y^2+\frac{1}{y^2}) = \frac{201}{16} = 12 \frac{9}{16}$
Hence, the correct answer is $12 \frac{9}{16}$.
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