The function \(f : R - \{-1\} \to R - \{1\}\) defined by \(f(x) = \displaystyle\frac{x}{x+1}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Both 1-1 and onto
One-one: let \(f(x_1) = f(x_2)\).
\( \displaystyle \frac{x_1}{x_1+1} = \frac{x_2}{x_2+1} \)
\( x_1(x_2 + 1) = x_2(x_1 + 1) \)
\( x_1x_2 + x_1 = x_1x_2 + x_2 \)
\( x_1 = x_2 \)
So f is one-one.
Onto: take any y in \(R - \{1\}\) and solve \(y = \frac{x}{x+1}\) for x:
\( xy + y = x \)
\( x(1 - y) = y \)
\( \displaystyle x = \frac{y}{1 - y} \)
This x exists because \(y \ne 1\), and \(x \ne -1\), since \(\frac{y}{1-y} = -1\) would give \(y = y - 1\), which is impossible. So every element of the codomain \(R - \{1\}\) has a pre-image, and f is onto.