\[{{x}^{2}}=y+z,\]\[{{y}^{2}}=z+x\]and \[{{z}^{2}}=x+y,\] then the value of \[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\]is [SSC (CPO) 2013] |
A) \[-\,\,1\]
B) \[1\]
C) \[2\]
D) \[4\]
Correct Answer: B
Solution :
\[{{x}^{2}}=y+z\] |
On adding x both sides, we get |
\[{{x}^{2}}+x=x+y+z\]\[\Rightarrow \]\[x\,\,(x+1)=x+y+z\] |
\[\frac{1}{x+1}=\frac{x}{x+y+z}\] (i) |
Similarly, \[\frac{1}{y+1}=\frac{y}{x+y+z}\] (ii) |
and \[\frac{1}{z+1}=\frac{z}{x+y+z}\] ... (iii) |
On adding Eqs. (i), (ii) and (iii), we get |
\[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=\frac{x}{x+y+z}+\frac{y}{x+y+z}\] |
\[+\,\frac{z}{x+y+z}\] |
\[=\frac{x+y+z}{x+y+z}=1\] |
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