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Question: Answered & Verified by Expert
Let $f: \mathrm{A} \rightarrow$ B be a function defined as $f(x)=$ $\frac{x-1}{x-2}$, where $A=R-\{2\}$ and $B=R-\{1\}$. Then $f$ is
MathematicsFunctionsJEE MainJEE Main 2018 (15 Apr Shift 2 Online)
Options:
  • A
    invertible and $f^{-1}(y)=\frac{2 y+1}{y-1}$
  • B
    invertible and $f^{-1}(y)=\frac{3 y-1}{y-1}$
  • C
    no invertible
  • D
    invertible and $f^{-1}(y)=\frac{2 y-1}{y-1}$
Solution:
1967 Upvotes Verified Answer
The correct answer is:
invertible and $f^{-1}(y)=\frac{2 y-1}{y-1}$
Suppose $y=f(x)$
$$
\begin{aligned}
&\Rightarrow y=\frac{x-1}{x-2} \\
&\Rightarrow y x-2 y=x-1 \\
&\Rightarrow(y-1) x=2 y-1 \\
&\Rightarrow x=f^{-1}(y)=\frac{2 y-1}{y-1}
\end{aligned}
$$
As the function is invertible on the given domain and its inverse can be obtained as above.

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