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Question
Let \[f\left( x \right) = \frac{1}{1 - x} . \text{Then}, \left\{ f o \left( fof \right) \right\} \left( x \right)\]
Options
\[\text{x for all x} \in R\]
\[\text{x for all x} \in R - \left\{ 1 \right\}\]
\[\text{x for all x} \in R - \left\{ 0, 1 \right\}\]
none of these
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Solution
\[\text{Domain of f}:\]
\[1 - x \neq 0\]
\[ \Rightarrow x \neq 1\]
\[\text{Domain of f} = R - \left\{ 1 \right\}\]
\[\text{Range of f}: \]
\[y = \frac{1}{1 - x}\]
\[ \Rightarrow 1 - x = \frac{1}{y}\]
\[ \Rightarrow x = 1 - \frac{1}{y}\]
\[ \Rightarrow x = \frac{y - 1}{y}\]
\[ \Rightarrow y \neq 0\]
\[\text{Range of f} = R - \left\{ 0 \right\}\]
\[So,f: R - \left\{ 1 \right\} \to R - \left\{ 0 \right\} andf: R - \left\{ 1 \right\} \to R - \left\{ 0 \right\} \]
\[\text{Range of f is not a subset of the domain of f}.\] \[\text{Domain}\left( fof \right)=\left\{ x: x\text{in domain of f and f}\left( x \right) \ \text{in domain of f} \right\}\]
\[\text{Domain}\left( fof \right)=\left\{ x: x \ in R - \left\{ 1 \right\}\text{and}\frac{1}{1 - x} \in R - \left\{ 1 \right\} \right\}\]
\[ \text{Domain}\left( fof \right)=\left\{ x: x \neq 1 \text{and}\frac{1}{1 - x} \neq 1 \right\}\]
\[\text{Domain}\left( fof \right)=\left\{ x: x \neq 1 \text{and}1 - x \neq 1 \right\}\]
\[\text{Domain}\left( fof \right)=\left\{ x: x \neq 1 \text{and}x \neq 0 \right\}\]
\[\text{Domain}\left( fof \right)=R - \left\{ 0, 1 \right\}\]
\[\left( \text{f of} \right)\left( x \right) = f\left( f\left( x \right) \right) = f\left( \frac{1}{1 - x} \right) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{1 - x}{1 - x - 1} = \frac{1 - x}{- x} = \frac{x - 1}{x}\]
\[\text{For range of f of}, x \neq 0\]
\[\text{Now,f of} : R - \left\{ 0, 1 \right\} \to R - \left\{ 0 \right\} \text{and}f: R - \left\{ 1 \right\} \to R - \left\{ 0 \right\}\]
\[\text{Range of f of is not a subset of domain of f}.\]
\[\text{Domain}\left( f o\left( fof \right) \right)=\left\{ x: x \text{in domain of f of and}\left( fof \right)\left( x \right) \text{in domain of f} \right\}\]
\[\text{Domain}\left( f o\left( \text{f of} \right) \right)=\left\{ x: x \in R - \left\{ 0, 1 \right\}\text{and}\frac{x - 1}{x} \in R - \left\{ 1 \right\} \right\}\]
\[ \text{Domain}\left( f o\left( fof \right) \right)=\left\{ x: x \neq 0, 1 \text{ and }\frac{x - 1}{x} \neq 1 \right\}\]
\[\text{Domain}\left( f o\left( fof \right) \right)=\left\{ x: x \neq 0, 1 \text{ and }x-1 \neq x \right\}\]
\[\text{Domain}\left( f o\left( fof \right) \right)=\left\{ x: x \neq 0, 1 \text{ and }x \in R \right\}\]
\[\text{Domain}\left( f o\left( f of \right) \right)=R - \left\{ 0, 1 \right\}\]
\[\left( fo\left( fof \right) \right)\left( x \right) = f\left( \left( fof \right)\left( x \right) \right)\]
\[ = f\left( \frac{x - 1}{x} \right)\]
\[ = \frac{1}{1 - \frac{x - 1}{x}}\]
\[ = \frac{x}{x - x + 1}\]
\[ = x\]
\[\text{So},\left( fo\left( fof \right) \right)\left( x \right) = x, \text{where}x \neq 0, 1\]
So, the answer is (c).
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