A) \[a=-3\] and \[b=1\]
B) \[a=1\] and \[b=-\frac{1}{3}\]
C) \[a=\frac{1}{6}\] and \[b=\frac{1}{2}\]
D) None of these
Correct Answer: B
Solution :
[b] The given relation is possible when \[a-\frac{{{a}^{2}}}{3}+\frac{{{a}^{3}}}{9}+...=1+b+{{b}^{2}}+...\] Also \[-1\le a-\frac{{{a}^{2}}}{3}+\frac{{{a}^{3}}}{9}+...\le 1\] and \[-1\le 1+b+{{b}^{2}}+...\le 1\] \[\Rightarrow \left| b \right|<1\Rightarrow \left| a \right|<3\] and \[\frac{a}{1+\frac{a}{3}}=\frac{1}{1-b}\] \[\Rightarrow \frac{3a}{a+3}=\frac{1}{1-b}.\] There are infinitely many solutions. But in the given options, it is satisfied only when \[a=1\] and \[b=-\frac{1}{3}.\]
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