Let \(A = R - \{-1\}\) and \(B = R - \{2\}\), where R is the set of real numbers. Let \(f : A \to B\) be defined by
\( \displaystyle f(x) = \frac{2x + 1}{x + 1} \)
Then f is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → one-one and onto
One-one: let \(f(a) = f(b)\). Cross-multiplying,
\((2a + 1)(b + 1) = (2b + 1)(a + 1)\)
\(2ab + 2a + b + 1 = 2ab + 2b + a + 1\)
\(a = b\)
Onto: take any y in B, so \(y \ne 2\), and solve \(f(x) = y\):
\(2x + 1 = y(x + 1)\)
\(x(2 - y) = y - 1\)
\( \displaystyle x = \frac{y - 1}{2 - y} \)
This x is a real number because \(y \ne 2\), and \(x \ne -1\), since \(x = -1\) would need \(y - 1 = y - 2\), which is impossible. So x lies in A and \(f(x) = y\).
Hence f is one-one and onto.