Question : If $\left (a+\frac{1}{b} \right)=1$ and $\left (b+\frac{1}{c} \right)=1$, the value of $\left (c+\frac{1}{a} \right)$ is:
Option 1: 0
Option 2: 1
Option 3: – 1
Option 4: 2
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Correct Answer: 1
Solution :
Given:
$a+\frac{1}{b}=1$ and $b+\frac{1}{c}=1$
⇒ $a =1-\frac{1}{b}=\frac{(b-1)}{b}$
⇒ $\frac{1}{a}=\frac{b}{(b-1)}=\frac{-b}{(1-b)}$
Also, $b+\frac{1}{c}=1$
⇒ $\frac{1}{c} =1-b$
⇒ $c = \frac{1}{(1-b)}$
Now, $c + \frac{1}{a}$
$=\frac{1}{(1-b)}-\frac{b}{(1-b)}$
$=\frac{(1-b)}{(1-b)}$
$= 1$
Hence, the correct answer is 1.
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