Question : If $a+\frac{1}{a}=2$ and $b+\frac{1}{b}=-2$, then what is the value of $a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2} ?$
Option 1: 0
Option 2: 2
Option 3: 8
Option 4: 4
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Correct Answer: 4
Solution : $a+\frac{1}{a}=2$ ⇒ $a^2+1 = 2a$ ⇒ $a^2-2a+1 = 0$ ⇒ $(a-1)^2 = 0$ ⇒ $a = 1$ Similarly, $b+\frac{1}{b}=-2$ ⇒ $b^2+1 = -2b$ ⇒ $b^2+2b+1 = 0$ ⇒ $(b+1)^2 = 0$ ⇒ $b = -1$ $a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2} ?$ $\therefore 1^2+\frac{1}{1^2}+(-1)^2+\frac{1}{(-1)^2}$ = $1+1+1+1 = 4$ Hence, the correct answer is 4.
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