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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