Let \( A=R-\{2\} \) and \( B=R-\{1\} \) and a function \( f:A\to B \) defined by \( f(x)=\displaystyle\frac{x-1}{x-2}\ \forall x\in A \), then the function \( f(x) \) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Both injective and surjective
Injective: suppose \( f(x_1)=f(x_2) \). Then
\( (x_1-1)(x_2-2)=(x_2-1)(x_1-2) \)
\( -2x_1-x_2=-2x_2-x_1 \)
\( x_1=x_2 \)
Surjective: take any y in B (so y ≠ 1) and solve y = f(x):
\( y(x-2)=x-1 \)
\( \displaystyle x=\frac{2y-1}{y-1} \)
This x is defined because y ≠ 1, and it is never 2 (2y − 1 = 2y − 2 is impossible), so x lies in A. Every y in B has a pre-image.
Hence f is both injective and surjective.