Official solution
Given, f(x) = 2x - 1; f : R \[\to\] R
\[g(x) = {{x - 1/2} \over {x - 1}};g:R - \{ 1) \to R\]
\[f[g(x)] = 2g(x) - 1\]
\[ = 2 \times \left( {{{x - {1 \over 2}} \over {x - 1}}} \right) - 1 = 2 \times \left( {{{2x - 1} \over {2(x - 1)}}} \right) - 1\]
\[ = {{2x - 1} \over {x - 1}} - 1 = {{2x - 1 - x + 1} \over {x - 1}} = {x \over {x - 1}}\]
\[\therefore\] \[f[g(x)] = 1 + {1 \over {x - 1}}\]
Now, draw the graph of \[1 + {1 \over {x - 1'}}\]
\[\because\] Any horizontal line does not cut the graph at more than one points, so it is one-one and here, co-domain and range are not equal, so it is into.
Hence, the required function is one-one into.