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Let $D=R-\{0,1\}$ and $f: D \rightarrow D, g: D \rightarrow D$ and $h: D \rightarrow D$ be three functions defined by $f(x)=\frac{1}{x} ; g(x)=1-x$ and $h(x)=\frac{1}{1-x}$. If $j: D \rightarrow D$ is such that (gojof) $(x)=f(x)$ for all $x \in D$, then which one of the following is $j(x)$ ?
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Verified Answer
The correct answer is:
$g(x)$
We have,
$\begin{aligned}
& (\text { goj of })(x)=f(x) \text { for all } x \in D \\
& \Rightarrow \quad g((j o f)(x))=\frac{1}{x} \\
& \Rightarrow \quad 1-(j o f)(x)=\frac{1}{x} \\
& \Rightarrow \quad 1-j(f(x))=\frac{1}{x} \\
& \Rightarrow \quad 1-j\left(\frac{1}{x}\right)=\frac{1}{x} \\
& \Rightarrow \quad j\left(\frac{1}{x}\right)=1-\frac{1}{x} \\
& \Rightarrow \quad j(x)=1-x=g(x) \\
\end{aligned}$
$\begin{aligned}
& (\text { goj of })(x)=f(x) \text { for all } x \in D \\
& \Rightarrow \quad g((j o f)(x))=\frac{1}{x} \\
& \Rightarrow \quad 1-(j o f)(x)=\frac{1}{x} \\
& \Rightarrow \quad 1-j(f(x))=\frac{1}{x} \\
& \Rightarrow \quad 1-j\left(\frac{1}{x}\right)=\frac{1}{x} \\
& \Rightarrow \quad j\left(\frac{1}{x}\right)=1-\frac{1}{x} \\
& \Rightarrow \quad j(x)=1-x=g(x) \\
\end{aligned}$
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