A) 0
B) \[\frac{1}{2}\]
C) 1
D) 2
Correct Answer: C
Solution :
\[a+\frac{1}{b}=1\]\[\Rightarrow \]\[a=1-\frac{1}{b}=\frac{b-1}{b}\] |
\[\Rightarrow \]\[\frac{1}{a}=\frac{b}{b-1}\]and\[b+\frac{1}{c}=1\]\[\Rightarrow \]\[\frac{1}{c}=1-b\] |
\[\Rightarrow \]\[c=\frac{1}{1-b}\] |
\[\therefore \]\[c+\frac{1}{a}=\frac{1}{1-b}+\frac{b}{b-1}=\frac{1}{1-b}-\frac{b}{1-b}=\frac{1-b}{1-b}=1\] |
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