Q66 / 125 · BITSAT 2011 · Mathematics

If \(f(x)=\frac{1-x}{1+x}\) the domain of \(f^{-1}(x)\) is

  • A
    \(\mathrm{R}\)
  • Correct answer
    B
    \(\mathrm{R}-\{-1\}\)
  • C
    \((-\infty,-1)\)
  • D
    \((-1, \infty)\)

Solution

Let \(f(x)=y\). Then \(\frac{1-x}{1+x}=y\) \(\Rightarrow x=\frac{1-y}{1+y} \Rightarrow f^{-1}(y)=\frac{1-y}{1+y}\) Thus, \(f^{-1}(x)=\frac{1-x}{1+x}\) Clearly, \(\mathrm{f}^{-1}(\mathrm{x})\) is defined for \(1+\mathrm{x} \neq 0\). Hence, domain of \(\mathrm{f}^{-1}(\mathrm{x})\) is \(\mathrm{R}-\{-1\}\)

Hence, the correct answer is option: 'B'.

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