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