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If \( \Large a^{2}= \left(b+c\right),\ b^{2}= \left(c+a\right),\ c^{2}= \left(a+b\right) \) then the value of \( \Large \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} \) is equal to
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A) 1 |
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B) -1 |
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C) 0 |
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D) \( \Large - \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \) |
Correct Answer:
| D) \( \Large - \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \) |
Description for Correct answer:
Given : \( \Large a^{2}= \left(b+c \right)\, b^{2}= \left(c+a\right)\ and\ c^{2}= \left(a+b\right) \)
Subtracting first two relations
\( \Large a^{2}-b^{2}=b-a=- \left(a-b\right) \)
=> \( \Large \left(a+b\right) \left(a-b\right)=- \left(a-b\right) \)
\( \Large \therefore a+b = -1 \)
=> \( \Large a+1 = -b \)
Similarly, \( \Large b+1=-c,\ and\ c+1=-a \)
\( \Large \therefore \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=-\frac{1}{b}-\frac{1}{c}-\frac{1}{a} \)
\( \Large - \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \)
Part of solved Polynomials questions and answers : >> Elementary Mathematics >> Polynomials
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