Let f : A $ \to $ B be a function defined as f(x) = ${{x - 1} \over {x - 2}},$ Where A = R $-$ {2} and B = R $-$ {1}. Then f is :
invertible and ${f^{ - 1}}(y) = $ ${{3y - 1} \over {y - 1}}$
invertible and ${f^{ - 1}}\left( y \right) = {{2y - 1} \over {y - 1}}$
invertible and ${f^{ - 1}}\left( y \right) = {{2y + 1} \over {y - 1}}$
not invertible
