Question : If $\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$, then find the value of $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$.
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: 3
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Correct Answer: 2
Solution :
Given: $\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$
⇒ $a^2=b+c$, $b^2=c+a$, and $c^2=a+b$
Now, $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Multiplying the numerator and denominator of the first, second, and third term by $a$, $b$, and $c$ respectively.
$=\frac{2a}{a+a^2}+\frac{2b}{b+b^2}+\frac{2c}{c+c^2}$
$=\frac{2a}{a+b+c}+\frac{2b}{b+a+c}+\frac{2c}{c+a+b}$
$=\frac{2a+2b+2c}{a+b+c}$
$=\frac{2(a+b+c)}{a+b+c}$
$=2$
Hence, the correct answer is 2.
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