Question : If $\left(x^2+\frac{1}{x^2}\right)=7$, and $0<x<1$, find the value of $x^2-\frac{1}{x^2}$.
Option 1: $3 \sqrt{5}$
Option 2: $4 \sqrt{5}$
Option 3: $-4\sqrt{3}$
Option 4: $-3\sqrt{5}$
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Correct Answer: $-3\sqrt{5}$
Solution :
$(x-\frac{1}{x})^{2}=(x^2+\frac{1}{x^2}-2)$
⇒ $(x-\frac{1}{x})^{2}=7-2$
⇒ $(x-\frac{1}{x})^{2}=5$
⇒ $(x-\frac{1}{x})=-\sqrt{5}$ since $0<x<1$
Again, $(x+\frac{1}{x})^{2}=(x^2+\frac{1}{x^2}+2)$
⇒ $(x+\frac{1}{x})^{2}=(7+2)$
⇒ $(x+\frac{1}{x})^{2}=9$
⇒ $(x+\frac{1}{x})=3$ since $0<x<1$
Now we know $(x-\frac{1}{x})(x+\frac{1}{x})= x^2-\frac{1}{x^2}$
⇒ $x^2-\frac{1}{x^2}=-\sqrt{5}\times 3$
⇒ $x^2-\frac{1}{x^2}=-3\sqrt{5}$.
Hence the correct answer is $-3\sqrt{5}$.
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