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 :
JEE · Math · previous-year question
- A.invertible and $${f^{ - 1}}(y) = $$ $${{3y - 1} \over {y - 1}}$$
- B.invertible and $${f^{ - 1}}\left( y \right) = {{2y - 1} \over {y - 1}}$$correct
- C.invertible and $${f^{ - 1}}\left( y \right) = {{2y + 1} \over {y - 1}}$$
- D.not invertible
Answer
B. invertible and $${f^{ - 1}}\left( y \right) = {{2y - 1} \over {y - 1}}$$
Explanation
Assume, y = f(x) $$ \Rightarrow $$ y = $${{x - 1} \over {x - 2}}$$ $$ \Rightarrow $$ yx - 2y = x - 1 $$ \Rightarrow $$ (y - 1)x = 2y - 1 $$ \Rightarrow $$ x = $${{2y - 1} \over {y - 1}}$$ = f -1(y) As on the given domain the function is invertible and its inverse can be computed as shown above.
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