If \( A=R-\{2\} \) and \( B=R-\{1\} \) and function \( f:A\to B \) defined by \( f(x)=\displaystyle\frac{x-1}{x-2} \), then \( f \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → one-one and onto
Writing f in a simpler form,
\( \displaystyle f(x)=\frac{x-1}{x-2} \)
\( \displaystyle =1+\frac{1}{x-2} \)
One-one: if \( f(x_{1})=f(x_{2}) \), then \( \frac{1}{x_{1}-2}=\frac{1}{x_{2}-2} \), so \( x_{1}=x_{2} \).
Onto: for \( y\in B \), solving \( y=f(x) \) gives
\( \displaystyle x=\frac{2y-1}{y-1} \)
This is defined for every \( y\ne1 \), and it is never 2, since \( \frac{2y-1}{y-1}=2 \) would mean \( -1=-2 \). So every element of B has a pre-image in A.
So f is one-one and onto.