Question:
Let \(f: A \to B\) be a function defined as \(f(x) = \dfrac{x-1}{x-2}\), where \(A = \mathbf{R} - \{2\}\) and \(B = \mathbf{R} - \{1\}\). Then f is :
Let \(f: A \to B\) be a function defined as \(f(x) = \dfrac{x-1}{x-2}\), where \(A = \mathbf{R} - \{2\}\) and \(B = \mathbf{R} - \{1\}\). Then f is :
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Check one-one and onto, then solve y = (x-1)/(x-2) for x.
Updated On: Nov 11, 2026
- invertible and \(f^{-1}(y) = \dfrac{3y-1}{y-1}\)
- invertible and \(f^{-1}(y) = \dfrac{2y-1}{y-1}\)
- invertible and \(f^{-1}(y) = \dfrac{2y+1}{y-1}\)
- not invertible
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The Correct Option is B
Solution and Explanation
Let \(y = \dfrac{x-1}{x-2}\). Then \(y(x-2) = x-1 \Rightarrow x(y-1) = 2y-1 \Rightarrow x = \dfrac{2y-1}{y-1}\), defined for \(y \neq 1\).
One-one: \(f(x_1)=f(x_2)\) gives \((x_1-1)(x_2-2) = (x_2-1)(x_1-2)\), which simplifies to \(x_1 = x_2\).
Onto: for each \(y \neq 1\), \(x = \frac{2y-1}{y-1}\) exists and \(x \neq 2\) (else \(2y-1 = 2y-2\), impossible).
So f is bijective, invertible, with \(f^{-1}(y) = \dfrac{2y-1}{y-1}\).
Answer: (2) invertible and \(f^{-1}(y) = \dfrac{2y-1}{y-1}\)
One-one: \(f(x_1)=f(x_2)\) gives \((x_1-1)(x_2-2) = (x_2-1)(x_1-2)\), which simplifies to \(x_1 = x_2\).
Onto: for each \(y \neq 1\), \(x = \frac{2y-1}{y-1}\) exists and \(x \neq 2\) (else \(2y-1 = 2y-2\), impossible).
So f is bijective, invertible, with \(f^{-1}(y) = \dfrac{2y-1}{y-1}\).
Answer: (2) invertible and \(f^{-1}(y) = \dfrac{2y-1}{y-1}\)
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