Question : If $(a^2 = b + c)$, $(b^2 = a + c)$, $(c^2 = b + a)$. Then, what will be the value of $(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})$?
Option 1: –1
Option 2: 2
Option 3: 1
Option 4: 0
Correct Answer: 1
Solution :
Given: $(a^2 = b + c)$, $(b^2 = a + c)$ and $(c^2 = b + a)$ ------------ (1)
The given expression is $(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})$
= $(\frac{1×a }{(a+1)×a}+\frac{1× b}{(b+1)×b }+\frac{1×c }{(c+1)×c })$
The first term of the expression is $\frac{1× a}{(a+1)×a } = \frac{a}{a^{2}+a}$
The second term of the expression is $\frac{1× b}{(b+1)× b} = \frac{b}{b^{2}+b}$
The third term of the expression is $\frac{1× c}{(c+1)× c} = \frac{c}{c^{2}+c}$
Substitute the given values from equation (1) in the expression, we get,
$\frac{a}{a^{2}+a}+\frac{b}{b^{2}+b}+\frac{c}{c^{2}+c}= \frac{a}{(b+c)+a}+\frac{b}{(a+c)+b}+\frac{c}{(b+a)+c}=\frac{a+b+c}{a+b+c}$ = 1
Hence, the correct answer is 1.
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